Exercises • In Real· Analysis with Solutions' YU TrQng Tuan

..

Exercises • In

Real· Analysis with

Solutions' BY A GROUP OF GRADUATE STUDENTS,UNIVERSITY OF SAIGON

EDITED BY YU TrQng Tuan

. I t6nghQP� _________________ � __ �

FOREWORDS

Most of the exercises contained herein are

taken ·from W .Rudin' s treatise "Real and Complex

Analysis H (Mac Graw Hill, N.Y. 1966 & 1974). In the preparation of this monograph, we

have drawn from several sources; among these, Hewitt & Stromberg's treatise "Real and Abstract Analysis "

(Springer Verlag N.Y. 1965) must be singled out.

Saigon, September 1974. V.T.T.

TMoughou.t .th.i..J. eha.p.te.It, 1I-.i..J. a. P04.i..uVe. mU4wtt ,all a.

mta..4ulta.bl.e 4Pa.C.e. (X,QlZ )

EXERCI5:E ],.- L e t janl and I bn ! b. Bequ.nce" in [-00, 00 1 • Pl"OVQ th. fottowin(lB

(a) l im BUP (-an ) - - lim inf an n+oo n�oo

(c) If an " bn fO l" a,tt n, t h.n lim inf an" l im inf bn Show by an e=ample tha t B trict inequal i ty can hold in (b):

SOLUTION

Cal' , (b) , Cc) are trivi a l . See for example Ca). Firs t , note that sup I - x : XEE I - -inf E, E c R. By

repeate.d appl i ca ti ons of thi s identity., we have

l im sup (-a ) � inf sup ( - a ) - inf (-inf a ) n k n;> l< n l< n »ok n

--sup inf an l< n;>1e

We can take the fo l l owing example

- - l im,inf a • n Ie !Q.1e

let an-C- l ) n,b�-(- l) n�l

Then l im sup an - l im sup bn - I, but

li � Can + bn} - 0 < 1 + � - l i m sup an + l i m sup bn Therefore , strict inequal i ty can hold in Cb).

EXERCHE 2 . - Put fn - 'X.E if n i" odd, fn. - 1 - I'e if n i$ BIJlm. What iB the l",leva"c, of thi" e=ample to Patou'. lBmma ?

SOLUTION

It i s clear that lim inf fn - 0, and that

l i m inf r f dfl. J1 n . This shows that in Fatou's l emma we may have s trict

ine'qual i ty.

5

EXERCISE 3.- Suppose fn : X ,-+,[0,,001 is measurabZ e,

'1 �f2> . . . . . . >0 and

fn (z)�/(z) as n-....oo , z,eX. Suppose f1 e L1 ( II- ) , prove that then

l im conaZusion does not ho Zd if "f1 e

SOLUTIOII

- Ix fdll- •

£1 ( 11-)" i s Show that this

omi tted.

Let f 1 be in L 1 , and let fn (x) � f (x) . x e X. ' The conclu

sion follows fr'om Lebesgue's dominated convergence theorem, since Ifnlis majorized by fl'

Now suppose the condition " f 1 E Ll" is ami t ted. Cons ider the following example:

1 fn- ii' l.{O,oo{ Then f decreases uniformly to 0, but ,n

r JR/n(X)dx '.00 • n • 1 .Z .......

EXERCISE If, - Prove that ths ss t of pain ts at whiah a sequenae of measurab Ze rea Z funations aonverges is a mea surab Z e set .

SO LUTI 011

Let (fn) be a sequence of real measurable functions. Then. ,

g .. lim sup fn and h � lim, inf fn are measurable.

Recall that a sequence'of real numbers (an) converges if, and only if lim sup an • lim inf an • hence lim fn(x) exists (in {--,""']) iff g(x) • hex). Since g and h are measurable, the set ! x : fn �x) converges I ;;; I x : g ex) • hex) I is measurable.

This completes the proof.

EXER(I�E S.- Let X be an un aountabZe set, and'Zet

C'IIl- ! A ex : either A or A C is at most aountab Z e l Define !'-<A) - 0 if A i s at most coun tab Z e,

�A) - 1 if 'Ao ,is at /fIos t countabZe. trovs that CRl i. a 0- atgebra, «nd II- i.e a ",easure on C1R.

7

SOLUTION

L e t (1. I A e X : A i s a t m o s t coun tab l- e 1 , t;J. I A e X : AC i s a t Dl o s t coun t a b l e I.

Sin ce X i s un coun tabl e . (p an d fiJ ar e d i sj o int. hence

I (p'.� I form a p art i ti on of GIl . By defi n i t ion of (P. we hav e � e (P ; A E (P and Be (P im p�r

that the s e t s A U B . A n B and A - B a l l b e l ong t o (P •

S ince fiJ c ons i s t s ex act ly of a l l A wi th ACE (P. we hav e X E fiJ ; AE fiJ' and BE fiJ imp l y that the set s A n B. A U B and B-A al l b e l ong to.� .

( 1 )

N ow suppos e A E (P an d B e fiJ • T hen t he fol l owing s hold An B c A. hence A n.B.E<;1 (A U B)c • AC n BC c BC • thu s ( A U B) C E (p. henc e A U B'E fiJ

T herefor e we have (aeGll· xeClll· (Aecm and � eGll) � (A U B . A n B • . and A - B b e long to.QlJ. } Oa

N ow l e t A eC1ll, n . 1 . Z • • • • • • • and A · U A . n n n=l If A i s at m o s t c oun t a b l e for a l l l!. then A i s n· at m o s t

cou n table. hence AE <;1 . 00

I f A E;; fiJ for s om e n •• t hen AC . n AC c AC , with no n =l n no A� E (J , henc e AC e (J • henc e A E t;J o

W e c onc lu de that 00

(2 ) t\,EC1ll n:::::l 1, 2, . .... t im p l ie s U An b e l ong s to.C1ll n=l

( 1) a nd ( 2 ) s how that C1ll i s a 0- alg eb ra on X.

By defi n i ti on of II. • I/-(A) . 0 i f Ae (J and IL ( B) � 0 i f BEt;J.

In p articu l ar 1/-(0) . 0 and IL(X) . 1 . L e t us p rove that I/- i s comple t e l y add i t iv e. F i r s t note that

if A e (J and B e (p then A U B i s a proper sub s e t of X. Hen ce i f ;.. e t;J and B i s d isj o int fr om A , t h e n B e c;J. L e t (An) be a p ar t i t i on of A, with AneC1ll; n- l, 2, ... I f A E (J, then A n e (p for a lJ. n. H ence

o : IIo (A) ·r I/-(An) s ince I/-{A)· O,n • 1, 2, ... n=l

If A E fiJ· , t hen ther e i s s ome n. with Ano

E � ( for o therwise A E q.). By the prec eding rem ark, f or n � n. we hav e AnE c;J •

8 Hence

,,(A) • 1 •

Thus �. is completely additive, hence � is a measure on the measurable space (x.C1R).

Q.E.D.

EXERCISE b.- Let C1/l.be a a - al.gebra on a //tit X. S�ppOIltl CiIli.s a t most "o�ntabl.�. prove' that� has Z·n e Lements for some i.nteger n .

SOLUTION

Suppose C1Tl is at most countable. Define an equivalence relation as follows :

(x'",y ) iff ( XEM implies ye� , for Me em) Then the equivalence class x of x is the intersecti.on of

all measurable sets �ontaining �. Since there are at most countably many measurable sets in X, th�s implies that x is measurable.

Let F be the quotient space XI", • I x : x e X I : Fe: C1Tl, hence F is at most countable. Define �:C1Tl -+- qJ(F) by .(M)· I x : x eM ! •

Then the followings hold : F is at'most countable , hence • is surjective. x -E Ml - M2 implies. r e: Ml - M2, for Ml and M2 in C11l, so

that • is injective. Thus • is bijective. If C1fl is fini te, then F is fini te and has n elements ,

which implies tliat qJ(F) has 2" elements, and hence that C1Tlhas 2n

elements. Suppose Clfl is countable. By the foregoing result , F is

countable , hence F is equipotent to N; but�) is uncountable (See Remark below), and . : C11l- (j)(F) is a bijection , a contradiction. Thus F is finite, and 'the preceding case occurs. Q. E. D. REMAR�.- 'or every set r ,E is not equipotent to qJ(E) .

Indeed , if fI E _ (j)(E) is a mapping, let I JI -! r E El r Ii! f(1t) I .

Then A Ii! fIE). This sho .. s that CardIEl < a.rd((j)(E ) ).

In t h e case B _ N, not e t h a t (j)(N ) is equi potent to the s e t

10,11 N o� all' mappi ngs s I II -IO,l!, and the la t t er i s equi potent

to the tri adi c can tor s e t . rina.ll'l, w e h ave card((j)(II ) ) " - Card(.RJ .

Se .. BrlJr ci se [IT] (IIJ and (b), ch. VIII.

EXERCISE 7.- Suppou /I-(X)<oo , (fn ) a uquene. of bounded eomple:z: meaeul'ab le funetione on I, and fn _ f unifol'm ly on X. Pl'ove tha.t

and ehow that t he hypotheeis W /I-(X)<� � eannot be omi tted.

SOLUTION

Suppose /I-(X) is finite.

9

Since (fn) converges uniformly to f . there is an n. such that

hence I f (x) - f '(x) l < 1 for xeX and n;;..n • • n n. I f (x)l<l + I fn (x) 1 for X.EX and n;;"n • • n 0

Since f n o is bounded and /I-(X) finite, 1 + I f I is integrable on X. . no By Lebesgue's dominated convergence 'theore m • the result

now follows , 1 "

Next. let X • rOt 00.[' and fo • ii lrO .n]' Then (fn ) con-verges uniformly to 0 on X. but

j[;:dm. 1 for n - 1.Z • • • • • • •

o

This shows that the hypothesis " ",.(X)<oo .. cannot be ami tted.

Q.E. D.

EXERCISE !.- Let EI( I T< - 1 . 2 . . . . . . be meaeul'able eets . L e t

E - l :z: ex : :z: be longs t o infinite ly many

A ..

00 00 n u

Pl'ove that E - A .

Bence, if

in at moet finite l y many of the e e te E/(.

SOLUTION

� U Elc • Then

k=n Let x be in E. Then for

x e An for all n, hence x e A.

Al :, A2 :l . . . . . .. and A - n An'" n=l

all n. x e Elc for some k > n. Thus

� . •

10

Conversely, if x e A then x E An for all n, he nce XEEk(n)

for some k�n) > n. This implies x belongsto infinitely many Ek•

Thus E = A Q.E.D.

EXERCISE '1.- LBt fELl ( )L). PraVB that to eaah c> 0 there eri s t8

all> a suoh that r J

_lfld)L < t fJhenBvsr )LIE) < b.

E

SOLUTIOW

There is a sequence (s�) of simple measurable functions

such that

o� sn(x) t If(x)1 as n- .... x e x. By Lebesgue's monotone convergence theorem. I

Hence to each c > 0, there exists an n. such that

o � JUf l - s )d)L no E

Since

such'·that

is simple and integrable, there is an M < ....

s (x) � M a. e. no Put b . � •• Then if )L{E) < b • we have

f Ifld)L �, I(lfl - sn ) d)L + ISnd)L E E E

� 1: (If I - sn ) d)L

+ LM d)L

. X � & + & • t ! !

Q.E.D.

EXE R C I SE 10.- Suppose fI. ie a positive meaaure on x, f : X -+ [0,001

is measurable , '1 fdlL = r:, ,where 0 < r: < 00 , and ", is a r:onstant

Prove tJiat

oOif 0 < '" < 1

c if ex -

o if 1 < '" < 00

Bint If "'�1, t he integrands are dominated by ",f.

If '" < 1, Fatou' s lemma can be app lie d.

SO LUTI ON

Let <f(t) • Log (1 + t), t > -1. Then 'f(0) • 0, ,,' (t) •

an d .,' (0) • lim I ' Il'f(h) • 1. This gives

u lim A Log (1 + X) • u A_cO (u > 0)

Take A • n, this gives

(1) 1 im n Log (1 + �) ;, u' n n ... oo Take A • noe, '" > 0, then A ____ ..0 as n ____ 00 . Thus

lim n"'-l n Log () +

nU

: ) •

,u oe n.oo

If "' > I, n",-l_oc:> as n_oo; if 0 <'" <1, n"'-l ____ O

Thus we have

(2 ) lim n Log (1 + (*r) • 0 1 < • < 00 n ... oo

1.

l+t

(3 ) lim n Log (1 + (*rJ • 00 (0 < 01 < 1 , O<u <oo). n ... 00

Put gn (x) • n Log [1 + (7-fJ ' n • 1 ,2, • . •

Case 1 � 01< 00

Since x : f (x) .oo ! has measure 0, (1) and (2) gives

(4 ) For almost all x lim gn (x) • f(x). if "' . 1 n ... oo

Now for 0 � t < 00 , we have

- 0 if 1 <",

If '" �1, 1 + t'" '�(1 + t)"" , hence Log (1 + t"') � oeLog ( 1 + t).

12 Bu t

Log II + t ) � t I t f oll ows that

Log ( 1 + t'"') � OIt ( 1 � .. <oo. O� t<oo ) h enc e

(S) hex) � .. f (x) ( 1 � .. < 00) .

(4 ) an d (5) sh ow tha t w e c an a pply Lebes gu e' s dom ina t e d conv e rg en ce theor em . W e ob ta in

CaSB 0 < '" < 1.

By (3 ), E · \ x /

· 0 if 1< .. .

f (x) > 0 I ha s nonzero measure, s inc e jf d'r- > O.

(7 ) 1 im gn (x ) • 00 X E E (0 < .. < 1 ) n"oo

By F at ou's

I t £ ol 1 ow s Thu s

gn (x) • 0 x � E.

l emma, l im i nf ign d'r- � foo dll- ·00 ('r- (E ) > 0). X E .

that l im in£!gnd ll- .00, henc e l i mfgn dll- ·00

it it ..

(0<01< 1 ) .

Q.E. D.

14

EXE R C I SE ],.- Given 0 < � � .1, construct an opsn sst E c [0.1]

which is densB in [O,lJ such that m(EJ-£(To say that A is dsnBB

in B means that every open sst of B contains a point of A )

SOLUTI ON

L e t (An) be the s equ enc e of open s e t s def ined induc t i v el y as f oll ow s:

( i ) Al is t he open interv al of c enter i and of l ength � ( i i ) The c om pl ement of (AI U • • • • • U'An ) in [O,lJ is the

u n i on of 2n d isj oi nt cl osed int e rv al s of eq ual l eng th ;>- £ , and

Z2n of c en ters an , l • an,2 , • . • . . an,2n

( iii ) An+l has m e asu re ( and is th e u n ion of Zn o p en "2n+ l i nt erv al s of c en t er a • • • . • a e ac h of l en gth n,l n,2n'

One c a� see the f oll owings ( a ) ( ii ) hol ds w i t h n • 1

(b) £';;;1, henc e if s tatem en t (ii i ) holds, then s t a t em en t (ii) hol ds wit h ( n + 1 ) i ns t ead of n.

� (

(c ) By ( i i i), t he An' s are mu tu all y disj oint. L e t E • Al U A2 U • • • • • • GO

By (E) and (c ) , m(E) - 'r m (An ) , henc e b y (ii i ) , m(E) - � . t.i By (E),

.!. )m[O,l] 2n

an y in t erv al w hic h does n o t int ersec t E ha s l en g t h henc e e v ery open s e t c ontains a p o i n t of E.

Zn' 1

Q .E. D.

I!EMAR�S (a) Let I be a segment of l eng th l .By a d i l ation of

15 coeffi c i en t l, the above construction 9ives an open set E

in I, dense in I, a n d has measure El. (b ) Ever� open subset of E i n

(c ) I - E i s perfect (havin9 no isol a ted poi n t ) and

cont�lns no nonemptg open segment, hence I - � has no connec ted

component consistin9 of more than one poine.

(d) m(I-E) - (l-£) l •

EX ERC ISE 2.- Given 0 � t' < 1 • cons truc t a tota l ly di sconns ctsd Slit X in [0.1) su ch that mIX) - �'. (X is to havOI no connolctOld subss t consisting of morOl than on� p oint ) .

Suppose a < �'.If v is lowOIr ssmi con tinous ( l.s : c ) and v � l:1(.8how that v�O.Bsncs tk cannot bOl approzimatOld from

beZow by l.s.c functions. in ths ssnss of thOl Vi tali-CarathBodory thOlorsm.

sO LunON

L e t E. - 1 - £' , and con s tru ct t he correspond ing open s e t, which i s d en se in (0, 1) , as in t he ' s o lu t ion tom.

Let K -[0, 1] - E, t hen m(X) - [', and X i s com p act. B y Rema rk (c) toill. X has th e req u i red propert i es. N ow suppose v :R - R i s l. s . c . an d v � Xl< T he in eq ual i ty vex) >0 im pl ie s ll«x ) >0 , hence Ik(x) • 1

�') t hat XE K. S in ce v i s lo s. c,

,�x: v (x) > 0 't i s an op'en sub se t of K.

But X i s tota l ly d i s con nected, hence X conta ins no open segm en t] or • '[ with .. < ',". T hus

�X: vex ) > o ! i s e m p ty,

hen ce vex) � a for all x.

Su ppos e m (X ) - £. > 0'. T hen

H en ce

� l\ -v) dm > £/ Xk - v)\im

1 .fi. Xl< -v ) dm > f' Xkdm - E;' > 0 R 0

T herefore, ll< canno t b e app rox im ated fr om b e l ow by l . s. c . function s .

Q. E.D.

16 1

[X[P.CI�E: 3.- L.t 0< 0 � DO. ConB tru.c t a Bor.Z BQt E ,c R Buch thcrt

O<m ( E n I)< m (I)

for IIvllrl/ nonQmp toy .lIgmQn t I, and m (E) - c.

SOLUTION

We observe that the problem is to construct in each [n,n+l[ a Borel set E(nl of measure cn ; with

o < c < 1 n o <IIICE n 1)<111(1)

DO L cn • c, and

n:...ao

for every· open seglllent I c [n,n+l[! n · 0, �l • �2, •••••

Therefore , it su�fices to consider the case n • 0, and (A transla don aives tAe general case) put .c • • b

Firit. choose a sequence ( em) with b < em < 1.

, � b as III -+ DO m For III • 1. 2 ••••••• Em is an open set constructed inducti--'

vely as follows:

(i) E1

is the open dense subset of [O.�] Qf lIIeasure '1

' constructed as in the solution to CD.

(ii) Each'Em is the union of coun�ably many disjoint open intervals I L of length t L k · l .2 •••••• with 11,11. m, 1I. ft . r;

le=l m ,le m

(iii) Em+l c E. is such that Em+1 n 1m ,le is open and dense

in Im,le ' has lIIeasure em+1 t L '

em m ,"

constructed as in Remark (a) to CD. Let E • E

1 n E2 n

By eiii) , Em+1 has lIIeas�re (byei».

Hence (ii) is valid with m+l ins�ead of m.

By (iii) and Remark (b) in [!), every interval in 1m ,le has

length � 2-m• Moreover,

m(I - E ) . (1 - �+l )t is>O. m,le m+1 em m,le

Now let I • }a-6 , a+6 [ be an interval in [0,1].

Let no be sucn that 2-no < ! .It is cl�ar that E is dense 4 no in [0, 1] •

so that there .exi s t s a point be E loIi th lb-al < 2-no no

fof s ome k - 1 .2 •• • • : but

so that l e I . no .k:: By (i ii) , m ( I kn En)-no,

m (l k n. E ) -no '

� t , n � no. hence 2: n,k no c

_0 t > 0 (1) t n.k no

(1) implie s (1 m(l k n EC) _ - �) no' �o By (1), mCI n E ) > 0 •

By (2) , r.I (I n EC) > O. hence

m (I n E ) <: m (l ) .

W e have there fore prov e d t h a t O < m (l n E) <mel).

t k> 0 (2) n,

Q.E.D.

EXERCISE 4.- ShOLJ that th4lrfl ars unoountabt. utll E c: [0,1]

LJith Jl-rE) - �' , 0"';; 2:' < 1.

SOLUTIOII

Let K be the comp act s e t in [0,11. constructed in" 0.

By Remark (c) of ill . K has no isolated point. Hence by Baire's t heorem. K is uncoun t able.

EXERCISE 5.- If f i8 a L4Ib4l1lgU8 m8allwrabt41 ocmpl41z fU"otion on R� pro�. that th8r. ill a Borst fwnotion g on Rl 8wah that

SOLUTIOII

f - g a •••

Since f - R.f + i Imf, �t suffices to consider the case of real f .

17

18 For r e Q, pu t A

r � ! x : f(x) >r I,

f is Lebesgue measurable, so Ar is Lebesgue measurable,V r eQ.

No te that Ar CAr' if r � r'.

Let ! r1, r2 ' • • • • l be an enumeration ofQ.

We choose , by indu ction on n , a Borel measurable

s�t A such tha t r n and su ch tha t

Br c .Br, if r>r' , r and r' e Q.

I t is �lear t ha t this choice is pos sible .

Pu t g(x) - sup ! reQ : x E§: Br I for x e g(x ) - 0, otherwis e,

then g is Borel measurable.

U B r

By con s t ruct ion, f"(x ) - g (x) if x·e; U ( A A Br· ) _ N , reQ r

and N is a s e t of z eta measu re. Hence f - g a,e,

AN OTnER SO LUTI ON

Q.E. D.

For re al f, f - f+ � f- , f+ and f are Lebesgue me as urab l e Thus, i t suffice s to consider the.c a s e f � O.

There is a s equence (sn) of Lebes gue-simp le functions wch that

o � So (x ) t f (x) as n � 00

For e ach n, 3 a Bo rel - simple function S such that '" n "n

and Sn differ only on a s e t Bn of measure O. Pu t g(x) - �i�inf Sn (x) . I t is clear that g is Dore l

measu rable, and if 00

xf/! 1'1 Bn ' then

so that g (x) - f (x) a, e

Q;E . D .

(X(RCIS( b.� Construct a sequenc, of continuous functions (fn

) on

[ a. 1] su c h t ha t o � fn � 1 and

Zim ff (x)dx - a n .. oo O . n

but su"h that the sequsn"s I fn (x) j "onv8rgss for no % e [o,l]

SOLUTION

Far k • 1,2 , • • • • • , d e f ine gk by

gk (t) • 1 far I t I � ir • 0 far I t I > ire

I

1 1 2-4k l tl 'for A � It l� *- 2k 4k

I , I ,

/,' , I '-

1 1 if( 2k

Now define gk,m (t) • gk (t -�) for t E [ O,l ] , m · 1 , 2 , • • • , 2k .

Ths graph of gk

Cons ider the s equence \ inl as

It is clear, by cons truction , that 1

(i ) � gk.� (t) dt � j(,fk ( t ) dt < t, hence

. 1

l im J f dIn • O . o n

(ii) For t E [0,1], and for k • 1 , 2, • • • , there is an m1 such that

gk m Cx) • 1 , , 1

and there i s an m2 wi th

so that g k m (x) • 0,

• 2

l im inf fn (t). O·

%1+00 hence fn Ct) does not�onverge .

Q.E.D.

EXERCISE 7.- If � is an arbitrary positivlI measure and if

f E L� �), proVe that

lias a -fini te measur ••

I %: f (x) tl 0 \

19

20

.SOLUTIOII

It is clear that

I x: f(x) � 0 I . I x: I f ex) I > 0 I . Al U A2 U • • • with

A . I x: If ex) I > �! n • 1 • 2 ..... n Since I f (x) I > .!. for x E n A n .we have

Hence ",{An) < - lin. It fo11011's that the set I x : f(�) � 0 I is the union of countabIy many sets of finite measure, that is it has G.- finite ·measure.

Q.E.D.

EXE:RCISE: II�- Lilt (E,<C) a topoZogiaaZ.

spaall, (P, d ) a "'lItria

IIpaa. , and f:E�P a mapping .If VeE, dsfin •

SOLUTIOII

• (V) - SUP �d(f(U),f(lI)):� � � Por':t:EE, t.t <t>(:t:) - inf I .(V): V is a I'Il1ighborhoad of:t: l Ca) Prove that <t> : (E,�)- [0, 00] is upper s.",i-continuous

(14. S. c. )

t (b) ProvlI that f is continuous at a point :t: iff �(z) - o.

(a) Sho", that the •• t of points of aontinuity of f is a Ga

.ea) ·Suppose 4>ex.) < 11<, 0 < '" � 00 .By definition of <t>

there exists an open set Y containg Xo with .ey) < II< • So

x e Y implies

Therefore. if .�(xo) < 11<. then 3 open set Y, x. E Y and

�(x) < .. for x E Y.

This shows that I x: .ex) < 01 is a neighborhood of every points of itself, hence

I x: ·ex) < 011 is open for OI.E ] - 00 , .oj It follows that f is u.s.c.

21

(b) Suppose f is continuous at x., that is

Vr - I x: d(f(x) , f(x.) ) < r l is a neighborhood of x., r>O

It is clear that ,(Vr) �Zr. and x. e V

r ' so that <!r(x.) - O.

Conversely, suppose <!r(x.) � O.Hence by definition of <!r, for r> 0 there exists a neighborhood V of x. with .(V) < r;

Therefore YC Yr, hence Yr is a neighborhood of x • .

Thus f is continuous at x ••

(c) Let Aor. - x: <!rex) < 00 I • > 0 Thus A or. is open , 00 > O. and Aoo C A� if or � , •

It is clear that I x: <!rex) - 0 is the intersection

of the I 1,2 hence Al/ns, n - , . .. . , by (b) , the set of po'ints of

continuity of f is 00 n n=l

Q.E.D.

EXERCISE �.- Let Ifni bs a ssquenaB of rBa Z nonn egative funo�ions

on R 1 . and aons ide r the I (a) If f1 an d fZ (b) If f1 and fz (0) If each fn i s

fo Z io lJing four arB u. B. c . thBn are z. s. a. then

u. s . c • thBn .." L fn i s 1

t . s . c. t h�n 00

L2 f is 1 n

s tatemBn t s: f1 +fZ is u . 49 . o.

f1+fZ i s t. s . c .

u. s . c .

t . s . c.

SholJs that (a). (b). (d) =B truB. and (c ) is fa Zu SholJs that (d) i. fats. if wnonn.gativ.w is om itt.d. Is t he tru t h 9f t h. s ta temen t. affected if R1 is rep Z aced

.by a genera Z t op o Zogi caZ spacor ?

SOLUTION

(a) Suppose f1 and f2 are u.s.c. renl functions (not necessar ily nonnegative) . 1 .

Let 00 e ] -00 , 00 ] and x. e R with f1 (x.) +f2 (x.) < •.

(.

22 There exi s t s.!: > 0. with flexo) + f2 ex.) + 2 £ < IX. and

an opf'n s e t V containing Xo with fiex ) < flex.) + IX for x e V

and i • 1. 2 . Thi s shows that fl ( x) + f2 (x) < IX for x E V • so that

W. I x: f 1 (x) + f 2 (x) < IX

is a neig hborhood of . eve ry Xo E W. hence W i s open.

Thus f1 + f� i. U . S . c .

(b) Recal l that by definition we have the fol lowing.

f is u . s . c . i ff ( -f) i s l . s . c . , f • real funct ion . This shows that ( b ) i s a direct consequence 'of (a)

( c ) (c) i s false. a s shown t h e following e:<ample:

"Le t n -+ rnbe an enume rati on of E • Q n] a ,b( • a < b

n • I • 2 •• • . ••• and

Since f is the characteristic function of a closed n set , i t i s u . s . c . , n • 1. 2 •• • • • • But we have

f n=l fn • IE whi ch i s not u . s . c . ,for,E i s not closed in ] a ,b(.

(d) To prove (d) , note that f 1+·····+fN i s 00

Then ,ince f i s pos i t ive , this imp l ie s L f i s n 1 n

los . c . by (b)

sup � f N 4" n

Recall that the supremum of any collect ion of l . s . c . real functions i s l . s . c . s o we see imme d iate ly

Ii .f: f • sup L f i s los . c . 1 n N 1 n

Now . let gn • -fn with fn in the examp l e of (c) . Then

s i n ce

g . - " f i s not los . c. n � ·n

fn i s not u . s . c .

Therefore the word "nonneg ative" may not be omitted in the hypothe s i s of (d) .

Nex t . we obs erve that the proof of (a) ,(c) and (d) may be •

app lied to i general top olog i cal space X in p lace of R1 , with " open set in I" instead of " open set in Rl" .

. Q.E.D.

EXERCISE �a.- L8t � b8 a regu lar Bore l m8asur8 on a comp act Hausdorff space X, as sume �(X) - 1.Prove th a t t h e r8 is a compact s e t K C X (The carrier or support of �) s u ch that � (X) - 1 but �(H) < 1 for every p roper compact s u b s e t B of K ( comp are E:r:ercis8 11).

so L UTI ON

Let g(be the collection of all compact sets )(. C X wi th

�(K) • 1. g( is nonempty since X e �. � is wri tten as

Now for K1 and 1C2 in g{, we have

1I-[(K1 n 1C2)C] • �(IC� U IC� ) • 0 hence �(K1 n 1C2) • 1 , so that K1 n K20e 17(.

Thus this family C;Xof closed sets in the compact space X has the "finite inter&ection property ",so that the intersection K of all members of r;}(., n X ... ,is nonempty ,and compact. _eA

If V is an open set containing K. then the collection

consisting of V and X:;', 01 e A , form an opel\ covering of X. X is

compact, therefore we can extract a finite subcovering

V U ICc U U ICc . X OIl 01 n This implies

IC OIU .. ... . . . . U IC 01 c V, 1 n

so that �(V) . l, since

lCo n . . . ... . . . n K belon� to 17( 01 01 1 n Thus we have proved that �(V) . 1 whenever V is an open

set containing IC . But � is regular, so that we have

�(V): V open :l K I - 1

The preceding proves the following assertions

H is compact in X and H e g( implies IC C H

H compact and H is a propero subset of IC implies H fI! �.

23

In the second case • H fI! �. hence �(H) f 1 .,hence �(H) < 1

Q.E.D.

24 kXERCISE 11.- (a) Show that every compact subset of Rl is the .upport of a Bore l mearsure.

(b) If X is a c losed ,et in a me tric sp.cs X •

show that K is the support of a continuous funct ion iff X is the clo.urs of an opsn set.

( c ) Prov� the fo l lowing gsnera�i.ation of (b): If X is a norma l topo logica l space • then a c lossd se t X in X �B the support of a oontinuous funotion f:X � R iff there u:i.ts a set V .nd

( i ) V is open , ( ii ) V is a Pa sst , ( i ii ) X i. the closure of V.

SOLUTlOIJ

with

with

set ,

(a) For every n - 1,2,:.we may flnd x nl , • . . ,x in K Ok

)( c: 'u B (x ,i-) j=l �j Therefor'e we find a sequence I x l'x 2' • • • ! • A in K

A. - K

We define fI. b'y the fOI'lllula fI.(1 x n I) - ; o, and for E Borel II-(E) - � 2 -n " the summation being taken over all n such

1 that Xo e E. I t is clear that II- is a Borel for V open in . Rl, we have .,.(V) > a iff is dense in )(.

measure on R .Moreover , V n )( is nonempty , for A

This shows that )( is the support of fI. in the fol lowing sense: )( is the smalle st cl osed set in Rl with the property that .,.()(o)_O.

and that

(b) Let F be a closed set in the aetric space (X, pl. Define Pr: X -R by Pr(x) - inf ! (x,y): x e F! . I t is known (exercise� ) that Pr is uniformly continuous ,

P (x) - a iff x E F r Now let )( be a closed set in X. If )( is the support of

the continuous function f:x-R , then by definition , IC is the closure of the set

v - I x: f (x)" a ! which is open ,s ince f i s continuous.

Conversely , SUppo se )( - V for V open. Let F _Vc , and, we see tha� Pr:X- R is continuous , and PF(x). 0 iff x e F, hence PrCx) > a iff x eV'

Therefore K · V is the closure of the set

x:Pr(x) ; 0 that is , K is the support of Pi'

(c) Suppose X is a normal space, and let K be a closed set in X

First, if K is the closure of V • I x: f(x) ; 0\ f: X--+-Ris continuous,then

V ., 'J I x: I f (x) I � � \ • lJ F n ;

�here

each Pn is closed since f is continuous. Therefore we have proved the sufficient coridition.

Conversel�,suppose K is the closure of an open set V, and V · F1'U'P

2 U � • • • • ,with P

II closed,n· 1,2,;.�.

By Urysohn's lemmma , there exists for e very n • 1,2, • • •

a continuous function f : X-[O,2-1I] ,with n

Put f,· � f � n •

fl· 0 II VC

fl. 2-n II i n

Then it is clear that f is the uniform limit of the sequence (iN) , with iN • f1 + f2+ .... + fN, so that f:X --+- [0 , 1] is a continuous function.

If x E V , tb.en there exists an n such that xe Fn, hence

If Therefore hence K • V

x � V ,then fn (x) • 0 I

Ix: f(x); 0 \ . V, is the support of f.

, n • 1,2, . • . . hence f (x) • 0

Q.E. D.

r.XERCISE :'2.- Le t X be a me tri a spaae • lJith me tri a p. FoZ" any non�mp ty Eel . defi?1l

"PE (z) • inf I p(z,1I): 11 E E l · SltOIJ tltat PE is a u.niformly continuous function on X

and tltat PEr::) - 0 iff % belongs to tits closurs E of E.

If A and F are disjoint �onsmp ty a l o�sd subsets of X. ,zamins tit .. re lllvancQ of tile func tion

PA (:: ) fez) - PA (z) + Palz)

to Url/Bohn' s l emma.

26

SOLUTIO" Let x,x' E X and y E E • ;..-e have

p(x,y) ,,:;; p(x,x') - p(x' ,y) (tri ang le inequality)

henc e f1: (x) ,,:;; p(x,x') - p(x' ,y)

Take the infimum of the right sid!! of tlds inequality as y ranges over E, we obtain

Pt (x) ''':;; p(x, x') - PE (x ' ) , 0 r

Interchanging the roles of x and x', we have

PE (x ' ) I PE (x' )

Therefo'tc, f1: is

- PE(x)

- P E (x) I ,,:;; ,,:;;

uniiormly

p(x,x'). Thus

p(x ,x' ).

continuous

Moreover, it is clear tha\ the following statements

are equivalent

(i) x belongs'

to the closure � of E.

(ii)

(i i i)

lienee

For every n • 1 ,2 such that

p(xn ,x) < � PE (x) • 0

iff

, . . . . ,

x e r- .

there exists a x E E n

Now let A ' and B be disj oint ·nonempty closed subsets

of X • We observe that x E A implies x fi B , so that

(interchanging the roles of A and B if necessary)

PA(X) .. ps(x) >.0

for all x in X •

so that

Therefore[ ' 1 ;] is a continuous function of x, PA (x) .. Ps (X�

f(x) r 'J is continuous P A (x) .. Ps (x)

It is clear that f(x) e [0,1] ,and

on X •

iff x e A) i.e. f(x) • 1

X E B iff PB(x). 0 iff xe B) i.e. f(x)· 0 (. Therefore f is the function whose existence is asserted_in

the Urysobn lemma. Q.E.D.

EX ERCISE 13.- Examine the proof of Ri e sa' t he orem (Theore m

2.14 illand pro�e t he foLLowings:

(al If El C V1

' E2

C V2

IJheI'e V1

and V2

are disjo i n t open Bet s, then

",( E J U E 2

I - ",( E 1) + .)1.( E 2

I :

sve n if El

and 82 aI'S n o t ill C1ll. Ibl If EE � then E - N U Xl

U K2 U • • • • ,where

I xd is a countabLe coLLection of di sjo in t aompaat s e ts

and /1-( N I a 0 •

SO/.UTION

(a) Suppose Et and Vi as ass erted i - I .2.

I t i s cl ear tha t <

fI-(E U· E ) � "(E ) + " (E2 1 • 1 2 ,.. 1 ,..

C onv e rs ely . let V be an open s e t containing E· �l U E2

Note that Vl and V2 are d· isjoint • T he. refn r� we have

fI-(V) �",(V n Vl ) + '",CV nV2)�",(Ell + fI-(E2)

S ince w e hav e

",(E) - inf ! ",(V): V open, V�' E ! .

this g ives the rev e rs ed inequal i ty

",(E U E ) > '" (E 1 + ",(E) 1 2 . 1 2

This c omple tes the p roof of (a ) .

(b) R e c all the defi n i t ion o f C'IIlp• w e have th e f ollowing

(EE�l iff ( ",(E) i s equal to sup I ",(K):K comp a c t . KCE ! and is finit e

T hus w e c an choose a compact Kl C E wi t h

fI-(E) � ",(K 1) + 1 and for each n • we choose induct iv ely a K n+l w i th

K cE- K UK U n+l 1 2

fI-(E ) � ",(K ) + 1 that i s n n n'

U'J( n) - En+l and

1 ",(E� - Kill � n'

L e t N - E - (K l U K2U • • • • ) , hence

N · (E-K l) n(E-1zl n • • • • • N · E l n E 2n • • • • • ;

but E n+l • E n - )(0 has m easu re ��. this impl i e s

",(N) • 11111 ",(E 0) • 0

1. Rererenee BueN •• - rheorem 2.14- mean s Theorem 2.l41n·Re.l

and CoZIJpl"r An.ll/sis", W. Rudin, lie Gr .• v Hi ll 1966}.

2.7

28 .EXERCISE l�,- Let X be the p Lane with th� foLLowing topoLogy A Be t i. �pen if and onZy if its in t.�section with svs�y v.�tiaal Lins i s an opsn subse t of that Line. �ith �s spsat to tl,. usua l topoLOSlY of RI,

Sho� that this X is a locaL Ly oompact Bausdo�ff spaae. If·f eCo(X); let'::!" "'::n be those vaZue s of :: fo � whiah

f(:: . y ) tl 0 for at lsast one y (the�" ars· o'll.y fin i tely many suah ::1) and define

Let '" be the msasu�e aseooiated .with t hie A by the Rieu thso�em,·

If 8 is ths ::-o::i8, .�o� that ",(8) - 00 a l t hough ",(X) - 0 for svsry 'compact xc: E.

SOLUTION

L et�l be the discrete topology on Rl (that is. everY singleton I x I is an open set in �l)' Then every point x e Rl : has the compact set I x I as a neighborhood. so tha t (�, �l j is a �ocally compact Hausdorff space. It is clear �y the s�atement in the exercise that the topological space (X.�) is exactly the topOlogical product of �s:l\) and

' �'�

2)'

�2

being the usual topology of R:. Therefore i� is clear that (X.�) is locally' compact and Hausdorff. ,

If X is compact in X. the first projection pr1(X) is compact in � .(�> hence is a fin! te set. Therefore K is a finite union

Xi is a compact set in Rl for i'. 1. 2 . .. . . n.

If f : X --+ C has compact support. the'n the support

of f is contained in jxl ... .. xnl x Rl'.as show by the preceding remark.

Thus for feC (X) (f continuous with compact support) . c . .

h£ . i:[f(Xj,Y)dY j=l

is a linear functio�al on Cc(X) . and hf � 0 if f. � D.

29

By the proof of Rie s z ' s t heorem,the measure � defined by ( .. ) an d ( .. ) I

",(V) z sup z sup

(*) I ",(E ) '" i:nf Xc V, X comp act !

� (V) E c V , V open I i s a repre sent ing me asure for A • , We observe "t hat i f K • \ x! x X ' , where X' i s compact i n RI, t hen by (*) for JJ.. and (*) I for m ins t e ad of .. ,,,, (X) i s exac t ly m ( X'} ( m stands for Lebesgue me asure on RI).

Thu s .. i s characterized by the ident i ty ",Ox\ x la , b[) • b- a , x e R •

Th i s i denti ty now g ives .. (R x \ 0\)·00 as fol lows Le t V be an open s e t con t a in ing R x 10j. Then for

xeR, (x,O)e V, so there ex i s ts an £x>O wi th

I x j xl- Ex' I:x [ C V. Thi s implie s that there mus t be an n with uncountab ly

many £ � !. x n (If this i s not the cas e , t hen Ex i s > a for a t mo s t countably many x, whi ch contradicts the fact that � i s uncount able) .

Ex Ex Let Xx • I x I x ] - L""� L"" I, for I: � !.. For x n X • X LJ • • • U X , ", (X) i s � � , x

l xm

hence ", (V) � sup ! i : m�l ,2 , . . . 1 c 00.

Thu s we have proved that if V i s an open set containing R x \0\, then ", (V) '" 00; by (to) ' , this imp l ies

",(Rxlo\) • 00.

Now if X i s a compa ct sub set of R x \ 0 j (X compact in the space X), t hen

hence ev iden tly .. (X) os o. Therefore, for E • "RI X 10 \, we find that

",( E ) • . 00 and sup I fL{X): X compact , K c E I . 0

that i s � i s not ( inner) regu lar .

Q . E . D .

30

EXERCISE �5.- Let E be a well-ordered uncountable set with first element a. For u,v in E, the set l .r E E : u�.r�v I is denoted by (u,v] (and similarly for other kinds of segment).

Let loll a min l .r E E : [a,.rl] is uncountable l .

It follows that X � [a, loll] is well-ordered, uncoun

tablli, and for .r E [a. wI [ �hen [a • .r] is at most countable.

For ", e x, let p .. -'[a, or[ and S'" a lor, loll]' Consider

the topology on X generated by tho P", 's and the S", 's. co E X (the order topology); Prove the following statements

�. __ .:-.. ··(a) X is a compact Hau�dor!rspace. (b) X- \ 1o,J1 I is not fJ-compact. (c) Let X be a 'ca{pact sutset of X. Then X is uncoun-'

tablll iff loll is a l imi t poin t of x. (In 0 ther words. every S", contai.ns a � of X, .. . <.J < w1J •

.... .........." (d) If (Xn) 71 -· 1.2 • • • • is a sequence of uncountable

compact subsets of X. then n Xn2 XI n X2 n .... is uncountablll.

(.) irlliu'y Bliquence (an) .. i'f! '(a. loll [ contains a subee-

(a ) which conver�es to a point in [a; 1011[' nk .. �",,!,'1.) If f e C(X). tJ.ue corresponds an or < loll such that f is constant on Soc

Wof define em and � as fonows

� E E C1!l if either E U ! 1011 1 contains an uncountable compact set-In this aaS8, \.(E ) - 1 - or

EC U \wil contains an uncoutabl.ll compact set-In this case. ).(E) • o.

(g)C1!lis a a-algebra which contai.ns all Borel sets in X.

tI. i.s a mllasure on em "'hieh is not regular (every neighborhood of

for eV8ry f e crx). !(wi) - Ix fd)'

OQ.e�i.b. thli reguLar measure which the Rillsz.representation thlio�em associates with �his linear f��etiDnal.

SOLUTION

Suppose I Ei I i E I i s an open covering of X. Let A � IXEX : (x, w J i s covered .by finitely many of the E"s l 1 ' 1 �

W1E A, hence A i& nonempty . Let c-minA. We shall prove that c • a (the first el�ment of X).

31

Suppose a < c . F i rs t , there ex i s t s an i. with ce E. 10

By def i n i t ion o f the topology on X (and a<c) , there exis ts an ex , wi th ]ex , c] c: the open s e t E . • Bu t", belong s to some 19 Eex hence [ex , wI] i s cbvered by fIn i tely many Ei 'so There -1 , fore '" E A wIth .. < c, a con t rad i c t i on . Thi s contrad i c tion shows that a � c , hence [a, wI] is covered by a f in i te number of the Ei's� Thu s X is compact .

Now le t '" < � be in X . Let Y .' m in S", , t hen ", < Y • hence ex e Py •

If Y • " , we take U · Py , V • So< I f Y < � , take U · Py , V " S", Then U and V are d i sjo int open s e t s with .. E U, , E V. There fore X i s Hausdorff . The proof of (a) is comp leted . Now l e t K be a compact s e t in [a,w1[ • The collection

! [ a ,x [: x < w1\ form an open covering of K from which we can ex trac t a f i n i t e subcovering [a ,xl [ ' • • • • [a,�n[ . Therefore K i s con tained in a [a, '" [ , .. < IIll •

By the preceding resu l t , K is at mos t countable , if K i s any compact sub set of [a , w1[ . S ince [a , w1[ i s uncountable , i t may not be a coun table union of such compact s e t s . The refore [a , w1[ is no t a- compact, hence (b) i s p roved.

Nex t le t K be 'a compac t s e t in X , and consider the statemen t s ( i ) to (iv):

(i) K i s uncountable . ( i i ) K is not contained in any [a,x], x < wI ( i i i ) K ,n S.. contains a po int � < "1 ' for '" < w1 ( iv ) Eve ry ne ighborhood of wI contains a point

, E K. , � different to wI' It is clear that ( i i ) , ( iii ) and ( iv) are equ ivalent. But by the result in the p roof of ( b ) , if w1 in an

i s o lated po int of K, then (K- I w1\ is compact l K i s coun table, Hence it f o l l ows eas i ly that (i ) and (i i ) are equ ivalent.

The equ ivalence of ( i ) and (iv) is the as sertion o f (c) . Hence (c ) is proved.

Le t (�) be an increas ing sequence in X. X i9 compact , hence the re' is a subsequence (an ) o f (an) ' which conve rges k to be X. (an) i s incre.1s ing, therefore ant b as n_ <>0. This imp lies t hat ! b, a1,a2" • • ! i s countab le and compac t . Therefore a1< a2< " ,<w1 imp l ie s b · sup an ex i s t s , and b<w1•

le t Le t K1 , K2 be

ex <wI ' By (c) ,

n uncoun table compact subsets of X, and wI is a l im i t po int of both Kl and K2,

32 'l"herefore w e can f ind ( a

n) c K

l• ( bn ) c K2.n • 1 . 2 • • • • • such

tha t

"By the p r e c eding re sul t� • sup an

• sup bn

ex i s ts.

and � < "'1 • But � • sup an

' hence � e Kl• and s im ilarly , e K

2• T�eref ore K

l n K

2"c on tains a , w i th ex < ,<00,11'

T h i s s how s t ha t Kl n K2

i s uncountabl e.

L e t (Kn

) be as sequence of uncoun tabl e comp ac t s e ts

in X. By the p r ec eding resul t . X' • Kl n . • n K i s uncountabl e . n n � � �

Moreov e r n K n_l n n K'

n=l n Therefore. to cons ider n K

n= 1 n

i t suffi ces to con s ider K' • and then w e c an suppose n Kl � K2

�" "

L e t '" < "'1 . Each K i s un coun table. hence we r; find' a s equ ence (a

n) • a e K . K

l n n K

n' w i th n n

< "'1

Therefore � 2 l im an

e x i st s. � < 00,11

• and

can

� e K n n c 1,2, ... 00

T h i s shows tha t 00,11 i s a l i m i t p o in t of K a n K n= 1 n'

hence K i s un coun tabl e. T hu s ( d ) is proved,

No� l et us p rove (e). Suppose (an) i5 a sequence in

X. an

" < 00,11 for n • 1.Z • • • • •

L e t b - min l a l . a 2 • • • • ! . n 1 nT n+ H e n ce b

n < 00,11' an d bl

< b2

< • " .

By the re sul t in the p roof of ( d) . bn t b < 00,11' I t i s clear tha t (b

n) contains a su bsequ ence (ank) of (an)' hence

(e) is prov ed .

REMARK.- Note that by (c) and Ie ) , the sta temen t

• E 1s compact 1ff every sequence 1n E conta1ns a

subsequence whlch con verges ln E •

vhlch holds for me t ric spaces, is not valid for the topOlogical

space E - [a,wl

[.

L e t f b e a coml e x valued con t inuous fu n c t i on on X, \n d y f(w

l) . We denote by On the open set I z : Iz-yl< lln i inC.

f is con tinuous and ° i s open . hence f - 1 (D ) is an open s e t in n n X which contains , "'1

I t follows that there ex i s t CII: , n :. . 1 , Z , • • • • • , n . 1 '"I < .. 2 < · · · · <011 , w i th 5 .. e C (On ) n

Le t .. • sMP "n ' Then .. < OIl by a preceding remark , and -1 5 .. e 5 .. e f (Dn ) . Hence f (S .. ) e On Y n .

Therefor� f (Sca) e (l On - I y \ • that is f (x ) · f ( "' l) wheneve'r X E S .. . Thus (f ) i s prov e d .

L e t ClTl a n d ). ,be defined as in the exercise • Define

33

<5 - I E e x : E U l wi l cont a in s an uncoun t ab l e compact s e t ! � . l E e X : EC E' 6

Then X E 6 , jl E �, an d I 6 , 0" l i s a partit ion of C1Tl.We le ave the reader to ve rify that C)U is an algebra.

Note that by (d) , 6 is closed under the operation of countab le inte rsect ion , and then , we ver ify eas ily tha t CNl is a a-algebr a .

Moreo�er , for .. <w1 , 5 .. contains ( � , w1J e 6 , w i t h .. < �<"':! hence s .. E ClTl .5 im i larly , for a < .. , P� • [ .. , 011] , hence Poe E C1fl. .

(X ,cm )

Thus Qll contains a l l open sets o n X , hence contains a l l

We le ave t he 'reader , that 1 1011 1 e � , and

By the preceding result

to v erify that :leA) - I iff A

:leA) - 0 iff A

, for .. < 011 '

Bore l sets on X

� is a me asure on E 6 E 0"

:l(S .. ) . 1 Thus . eve ry neighborhood of 011 has measure I , but :l( "'1) - O .

Now if f e C (X ) , theri by ( f ) • there corresponds an .. < "'1

such that f is constant on 5" But ),(S� ) w 0 (5 .. E 6 ) , hence

Finally . we verify eas i ly tha t . if � i s the regu lar

measure associated to the l inear fun ct ional

f -- f ( WI ) - Lfd)'

the sense of the Riesz repr� sentat ion the orem , t hen � ( E ) - I i f f WI E E , E i s a Borel set Q . E . n .

34

EXERCISE I b . - Co n s tru o t a s e q u e n oe of p o s i t i v e oon tinuo u s fu n o� i o n 8 f on R

l , s u o h t h a t f (:z: ) _00 i f a'nd on Ly if :z: i s ra t i ona L . n n

SO L U T I O f.l For n • 1 .2 • • • • • l e t Dn be the s e t

Then we

L e t L n

\ � . I n ! ' m E z ! '" (�!) Z

have

1 (n + 2) !

\ 00

Q . U D . n = l n and U • n • I x+y : X E Dn · ly l ,. D + ] - 1.

n · Zn [

n

Thu s 1) n U n+k • D + ] - L n+k • Ln+k [ • hence n n 00 II Up . D and n 00 (tt Up)

00

.U U D - Q n = l n = l n

< L n I

Now for n - 1 . 2 • • • • 'd :hoos e fn R - [ O . n] s a t i sfy ing

( i ) cont inuous . f n (il) fn (x ) - n if x E Dn '

( i i i ) fn (x ) • 0 if x i! Un ' Th is. can be done . by c ons truct ing (u s e the Urys ohn lemma)

fn l C O , l] - gn . "then def ine fn (x + k ) � gn (X ) . k E Z, x e [ 0 . 1 ] .

We h ave

imp l i e s fN+k (x ) • N + k . k 1 . 2 . ...

if )II' @ Q oofn (x ) _ oo as n -;;: 00 . = a n u . then x i£ n U . for a l l n . Th e re fore

Thu s

I f x ft D n = l: p =n, p p =n p . : "this shows that x � Up ., ' Up • • • • wi t h P I < P 2 < . . . . . • " n d k < Pk

I 2 There f ore f (x j" & O . hence l im inf fn (x ) = O . t h a t i s

P k f

n(x ) . . doe s: - not conve rge , .to 00 as n _ oo .

· Q . E . D .

C EXE RCISE 1 7 . -Co n s tru o t a s e quenoe (fn ) of.nonn ega tive oon tinucus

fu n o tion s on RI

such that

if an d on Ly if :z: i s i rrat ionaL

S O L U T I OIJ

Le t Dn and Un as in U}] . and u s e the Urysohn lemma to construct a sequen"ce gn : R ->- ( 0 . 1 ] • g n cont inuous . su ch tha t gn (x ) � a i f x E Dn and

gn (x ) a 1 if x E Un Let fn ex) g l ex ) + g 2 ex ) + • • • • 0 + gn (x )

Dn c: Dn+ 1c: • • • • , . hence x E Dn imp l i es gn"+ k ex ) • a for k � 0 . 1 .2 • • • • • • • hence fn+k (x) � n ,

Thu s x E Q i mpi ies that (f (x ) ) is bounded . n Since

i mpl ies x E inf in i t e ly many Uk • For the se k • gk (x ) • 1 •

Therefore fn (x ) t 00 as n ->- 00

Thus fn (x ) -+- 00 if and only i f x E Q •

Q . E . D .

3 5

E XERCISE :l c'l.- A pos i t i ve me aeure o n a se t X i s ea Z Z e d ., - fi n i te i f X i s a eoun tab Ze uni o n o f s e ts X . wi t h fL{X . } < 00 . Prove t h a t fL i s 1. 1.1 ., - fi n i t e i f an d on ly i f the re e :r:i s ts f E L {fL} s u eh t h a t f{:J:} > 0 for e v e ry :J: E X.

S O L UTI OIJ 00 ( i ) Supp ose fL i s ., · finite . Hen ce X • U X

n = l n '

Let E a X n n di s j oint and has finite me asure.

De fine f by f(x) . 1

n 2)1- ( E ) n

• 1

v n .

XE f (x) > a for every x E X , and f • L n2 fL�En)

the sum i s taken

ove r a l l n such t h at )I- (En ) > O . Lebes gue ' s monotone �onveTgence theorem g ives

36

't' x e X. Thus i f � i s a - finite , we can find f e Ll ( �) with f ( x) > 0

(ii) By E xercise ITJ. if f e Ll ( �) . then

I x : I f (x ) I > 0 I has a -finite measure.

The - converse i mplication of the exercise follows immedi ately from this .

* * " Q . E . D.

CHAPTER

EX ERCISE 1.- Pro v s that thB 8upremum of any co Z Z e c t i o n of con v ex fun �t i on s on l a, b[ i8 c on VBX on l a , b [ and that poi n twi s e and upper Z im i t s o f sequsnce of convex fun c t i on s are convex.

Give an example of the Zower Z im i t of a sequence of con v ex fun c t i o n s w h i c h i s n o t convex.

SO L U T I O N

Let a < x ,,;; y < b , an d % • X X + ( 1 - X ) Y wit h 0 ,,;; x ,,;; l .

Let ( fi ) l e r be a co l lection of convex functions , and f (x ) • syP f i ex)

If either f (x ) or f ey ) is +"" , then evident ly

f e z ) ,,;; x f (x ) + ( l - x ) f (y) Now let f (x) , f ey ) be finite . Then

f i ( % ) ,,;; X f i (x) + ( l - A.j fi (y) ,,;; x f (x ) + ( 1 - It. ) f (y)

This implies f e z ) ,,;; xf (x ) + ( 1 - It. ) f (y )

Thu s sup f . i s convex on ] a , b [ . i �

i e I

Now let f � lim fn be the pointwise limit of a sequence of convex fun ctions . Since for each n

fn ( z ) ,,;; Xfn (x) + ( 1 - A. ) fn (y ) , we have by passing to the limit

f e z ) ,,;; x f (x ) + ( 1 - x ) f (y ) Hence 1 im. f is convex n I f f • lim sup fn ' then f • lim g , with g = sup fk n n k � n Hence if f is convex for a l l n then g i� convex by the preceding n

resu l t s .

39 To g ive an ex amp l e whi c h s at i s fi es all requ irements of

the exe rci s e , note that hex) - - Ix l is not convex on R1 , and h ex ) - inf I x , -x l , x e R

Therefore we con s ider

Then

f 2 p (x ) f 2 p +1 (x )

l im inf fn (x)

• x ,. -x , p - 1 , Z , . . . .

h ex ) , and fn i s convex for a l l n .

Q . E . D .

E X E RC I SE 2. - If of' i s "ontlez o n J a, b e and i f t i s "ontlez and 7'1 0 7'1 -de i!r e as ing on the range of ., ,protle t h a t t o ., i s "ontlez on J a, b [ . For Of > 0 , s hOLl t h a t the . "ontlezi ty of Z og ., imp Lies the "ontlezi ty of • , bu t n o t tli"e versa.

SOLUTION

Let a < x y < b , Z - 1..X + (l-:\.) y , :l. E [ O , l j . Then . ( z) � :I. 'f(x) + (l -).) <, (y) ( ., convex )

t(.,(z) ) � .(). . (x) + ( l-).) t(y) ) ( + non de cre asing) . (:\.'f (x)+ ( 1-:1.) .,(y) ) � :I. .�. (x ) ) + ( l - :l.) .( .,(y) ) ( 4) convex )

Thus t o . is convex .

Now Log ., i s · def ined for . > 0 , and ·t (x) - expx i s convex and nonde cre a s i ng • By the preceding resu lt , . .. • oLog . i s convex i f Log . i s convex •

The convers e i s not true is not a convex funct ion .

. (x ) c x i s convex , bu t Logx

Q . E . D .

EXERCISE 3 . - A S SUM e that ., i s a "on t in u o u s re a Z fun " t ion on J a. b [ su"h t ha t

. (:..:r' ) � t . (z ) + t . (y ) for a Z Z % and y e Ja, b ( • Pro tl e t h a t ., i s "ontlez .

SOLUTION

Con s ider the set of dyadic numbers 1 m : n e N , m e Z ! !I D zn, which is dense in [ 0 , 1] .

40

By repeated app l i cat ion5 of the hypothe s i s , we p r o v e

eas i ly by induction on n that

( * ) ., (rx + (1 - r) y ) � r ., (x ) + ( 1 - r ) • (y) , r - � Zn

Def ine f , g on [ 0 , 1 ] by f ( x ) - . ( X x + ( 1 - X ) 7' ) g ( x ) - x " (x ) + (1 - � ) ., (y) .

Then f i s l . s . c . and g i s con tinuous . •

Let U - · ! X e [ O , l ) : f ( X ) � g p . ) I U is c l osed , s ipce U i s the �omp lement of the set

I X e [ 0 , 1 ) : f ( :\.) > g ( � ) and s ince f-g i s l . s . c .

But by ( * ) , U contain s D, hence U - [ 0 , 1 ) .

Therefore for 0 � � � 1 , f ( X x + ( 1 - ;\. ) y ) � x f (x) + ( 1 - ). ) f (y) .

hence f is conve x .

REM ARKS

(a ) In the proof , we s uppose on l r; t h a t f i s l o ... e r s e micon ti:n uo us •

(b) rhe con cl usion of t he ex erci se does not fo l l o ... i f con ti n u i t r; o f f i s omi t te d · from t h e h r;po t h es i � , a s sho ... n the

fo l l olo'ini exampl e .

EXAM P L E : {Su [!J c.h. V I I I J C h o o s e a ( Ham el ) b a s i s of

b a s e b e ( e i ) i e I ' an d d e f i n e

an d extend ., t o a Q-l i n e ar m a p o f

I f ., i s c onvex , t hen ., i s

t h e Q-ve ct o r s p a c'e R . L e t t h i s

., ( e i o) = I , ., ( e i ) = a if i # i o

R i n t o Q

boun de d ab ove on ] a , b [

hen ce ., is bounde d i n a n e ig h bo r h o o d ' of 0 , h e n c e ., i s con t i n u o u s

T h e c o n t i n u i t y of ., now im pl i e s t h a t ., ( R ) cont a ins a n i n t e rval ,

c ontradi cts t h e f a ct ., (R) = Q ( No t e t h a t ., i s not c o n s t a n t on

] -l ,l[).

T h e r e fore ., i s not convex , but

EX ERCISE 4 . - Suppose f is a comp lez meaeurab l e function on X, � is a poei tive me asure on X • and

(0 < p < "",. ) .

41

( a ) If r < p < s • r e E. and s , e 0, prove that p e 0 (b) Prove tha t L og ¥ is conve% in the i n t s rior of E and

that " i s con tinuous on E.

(e) By (a) , E is eonn s o t e d. I s E n e oe s s ari Zy open ? C Zo s e d ?

Can E oon s i s t of a s i ng Z e p o i n t ? Can E be any conne o t e d subse t of

] O; oo t ?

SO LUTION

g ives

Hence

p e E .

(d) If r < p < s , p r o v " t h a t I f l ,,;; ",a% ( If I , 'If I ) . Hence p r s L r ( fA. ) n L s ( fA. ) c LP ( /l. ) •

( e ) A s su",,, t h a t I f l r < ao for s o"''' r < ao and prove that

I f lp If I .., as p -)0 00

(a) For \ � 0 , 0 "';;; r < p < s imp l ies t.P";; lI!ax (t1" , ts ) ,,;; t1" +ts . Hence I f I P';;; l f l 1" + l f I S , so that " (P ) "';;; ,, (r ) + 'I ( s ) Thu s r < p < s and r e E s e E imply p e E . (b) Let 0 < :\0 < 1 and p " :\o r + ( 1 - :\o ) s ,for some a < � < 1 . Pu t oo � 1 and II . 1 so 1 + 1 . 1 and 1 < ca , . < 00 ; � "{'I"='D , 00 II" P

The Holder inequal i ty app ly ing to U . 1 f(�1" and v .1 fl ( 1 - � )s

. (p ) • Ix uvd ... ";; l u l oo I v lll • ( � (r) J� : ( _,, (s ) )l- :I.

Log , ( � r + ( 1 - },, ) s ) ,;;; A Log ,, (r ) + (1 - ), ) Log ,, (s ) Thu s Log ., is convex in the interior of E . Now sUJl'pose (Pn ) is a sequen ce in E which converges to

We have (see (a) ) ,

I f lP n ,;;; I f lm i n ( P n ,J» + I f l max ( p , p ) - 1 Z n n " . " By the dominated convergence theorem , , (� ) converges to

, (p ) a� n -)0 ao , hence " is con tinuous at p (c ) The answer to the l ast que s t ion of (c) is affirmative

hence so are -the other ones To prove this , i t suff ices to cons truct X as the disj o int un ion of two among the following s

1 - A • (0 , 1 ] , f lex ) • x -l/a E 2 ] 0 , a [ Z - B • [ 1 , 00 [ , f (x )

2 - x -lib J E • ] b , 00 [

3 - C • [ a ,}] , f' 3 (X ) 2 -l i e [x (L ogx ) ) J E • ] 0 , c]

42

2 - l id 4 - D • [ e , ..., [ , f4 (x ) • [x (Logx ) ] ; E • [ d , ..., [

(a , b , c , d are nonegative , and I'- i s m , Lebesgue measure on the correspond ing space)

The verification of the assert i ons concerning E in I_ to 4 _

are immediate . Note that for X · [ 0 , 1] and f • 1 , then E � ] 0 , "" [ .

since

This completes the answer to (c ) .

(d ) By (b ) , LoU

Log I f I � � L og l f l + - p p r

L og ,, (p ) • pLog IfD p Now � + �s

• 1 hence p p ,

Log I f Ip � lI)ax (Log I f Ir , L og I f l s ) s o that I f l � max ( I f I , I f I ) s ince exp i s nondecre as ing .

p r s

(e ) Suppose 1 f I.,., > 0 , and I f ir < 00

Thus for ° < a [h ' f I Pdl'-]l/P � a [1'-( E)r/P

Since lim [)L(E)] lip . 1 , it fo l l ows that p+ oo -

lim inf I f In � a n .. ""

(*) lim inf I f ln � I f U"", . n-+-...,

(n - 1 , 2 , • • •

Now by Lebesgue ' s mono tone the orem

On the othe r III ; ' ) �I'-E

hand ,

Ir ('! ' ) P d l'-t ° as p ->- 00

>£ ill dl'- > 0 i f p > 1 a

hence we have

l im sup[i( ill)n dl'-]l/n • lim sup [ r (ill) n dl'-fin n ..... oo X a n + OO JE a

But

so thi s - give s

hence

C # )

l i m sup Wn .::: Woo a � a

l im sup I f l � A r loo n n+oo By C*) and C I ) , we obtain the i dent i ty

l im I f in • H I 00 n+oo an d now Cd ) give s the des ired i dentity

Q . E . D.

EXERCISE 5 . - A s aums '" i n addi t i on to the h ypo t hes i s of �:re1"cise 0 " that I'-(X) • 1

(a) P1"ove that I f l1" � l f I8 if 0 < " !' "< s � 00.

(b) Unde1" what condi tions do e s it happen t ha t 0 < 1" < s � 00

and I f l1" · I f ls < oo ?

( c ) P1"ove that L1""(I'- ) ::l L8 ( I'- ) i f 0 < 1" < s • Und81" what

con di t ions do t h e s e two spaces contain t he s ame fun ct i o n s ?

(d)" A s sume that I f l1"< 00 f0 1" s om e 1" > 0 , and p1"ove tha t

l im I f 1 • e:rp ( J�LOf1 1 f I d l'- ) p + o F

if e:rp ( - oo ) i s defi n e d to b8 o .

SO lunON 5 Ca) J,.e t 0 < r < 5 < 00 , and let p - r • p > I , hence

. Cx ) • xP i s convex on [ 0 , 00 [ .

Applying the Jensen inequal i ty to g _ l f l T , with . Cx ) - xP

43

44

we get Hence

( O < r < s < oo ) . The inqua l i ty i s triv ial for s - 00 . Thu s (a) i s completed

(b) ' In the complement to this exercise • we s ha l l prove the following theorem THE O R E M .

SUppOSIl • is s trictty co"vex , ,that i s

' ( A X + ( 1 - A ) y ) < A'(x ) + ( l - ). ) , (V ) for 0 < ), < The" ' squa t i ty ho tds i" Je"sll" ' s i"llqu a t i t y _ tha t is

. ( hg d ll- ) - Ix of og d ll- ( Il-(X) - l ) .

if a"d O " ty if g i s co"sta"t a . 1I ( Il- ] '

Now if equal i ty holds in (a) wi th 0 < r < s < 00 . then s ince ,ex ) xP i s s tr ictly convex o n R+ . , the preced ing theorem now

imp l'ies ( g J - I f : r is cons tant a . e ( Il- ] , hence f f \ i s cons tant a . e [ 'Il- ] . We conclude that the s i tuat ion in (b ) occurs if 'and only if

f i s ' constant a . e ( flo ) ,

(c) Let 0 < r < s < 'oo • I f f E L 5 ( 1l- ) . then by (a)

Hen ce I f lr � I f Is

< ,00 , Lr C Il- ) :::> L 5 ( 1l- ) ,

Now we claim that the fol lowing s tatements are equ ivalent

ex i s t s an � > 0 • such that flo CA) > 0 impl ies '" � II-CA)

In fact if (i i ) holds • then we c an par t i t i on X into finitely many atom s . each of measure � . • [An atom A in X i s such that ,

, i

B c A impl ie5 e ither flo{B) or \l-CA-B ) , is 0 ] This, has been d one then LP ( LL ) .

C n f C Q' rJ ·j, �"' �' wi " .... : c.{;...., A � = k"-. t:< ... � .(,, � .... 1 S or some n . ' � ;1- ' . ' ' , v t..... A . Therefore Lr ( flo ) - LS C flo ) and Cii) impl i e s Ci ) . 1.� 1":" "1�] -i.; ;;:-�

., c A :;i "<f4lkl'f'l: Now suppose )I. does not sat isfy (i i ) • that i s ' (;. ....... �r'-.... -.f; For n - 1 . 2 • • • • • • there exi s t s A c X such t hat

n

Then 'one can eas i ly reduce the s i tuation t o the case in which the An' s A�� : ()<'HA� )< }(, are . pa i'Twi se d is j o int ,

C , r(A.() < �(A:) > t , . . , o<y0�, ) <. t15{J

�

. �)�

45

L e t I'-(An ) · an ' an � 0 • Hence we m a,y choose an infin i t e subsequence lan

k \ of .\ an \ su ch that

we have

( * )

N ow define f l A n k Jc. 1 , 2 , . . . ,

f • 0 o therwi s e , I t i s cl ear t hat f II! L 9 ( 1'- ) , and b y ( * ) Thus � f ( U ) does not hold , t hen L r ( flo ) ; L 5 ( 1'- )

Therefore ( i ) imp l i e s ( U ) , Th i s comp letes (c) .

(d) App ly Jensen ' s inequa l i ty to I f l P , w i th ,(x ) · ·L ogx ,

By (a) ( 1 ) L o g I f l � fxL Og l f l d l'- ' P

, lI f ll decre ases as p --+- O , S i n ce L og is increas ing .

on 10, ao [ , we P Ilill �� ., .t;. .... 11ft '" conclude that . f r-. � p .( ..... n t il • Lc,u i� woM./

It ... Do %, �'� 1/ tiL Log l im O f D • l im L og O f I

, p.�Q P p+O P (pu t Log 0 • • 00 ).l L: .... (r �" ) ,, ;:,� (r(r App ly the inequa l i ty Logx � x • 1 wi th x .. .. (p ) , we obta in_ I' ' 1 '// , itlJ) (, r � - ,�J '1,, 1::/1 '

H -:-,!-. IIH iLl 1 pI X ( l f I P ' l ) du. " " "' '; r ,,!: ,

" T L og I f lp � p( <f(p) . l ) · r ...-'"

Hence L og u n �l l f ' P • Idl'-P X P O < p < ao )

I t i s Jcnown that • .1. , I",.: iJ..... l'l � tX: f� 0 (t P • 1 ) t Log t as p � 0 , p

t he m ono tone convergence theorem now imp l ie s

(2) l i m L ogo R i l � iLOg I f I d ", P+o P

By ( 1) and (2) , we obtain the de s ired resu lt

lim If I • exp rx

L O g l f l d ", p+O P J�

COM P L EM ENT. Proof of t he Suppose )I.(X) • 1

c onstant a , e . Therefore , 3c e R

then L e t

Q . E . D .

46

By J e n s e n ' s i n e qu a l i ty

., (x ) .:;; ;: L "o g d )l-

'P(y ) .:;; 1

fAc ., ogd\L r-:i He n c e

• ( jl d )l- ) • ...( A. X + ( 1 _ A. ) y ) �< A. ., (x ) + ( 1 - A. ) ,, ( y )

.:;; Ix � g d l'-

Therefore if g i s n o t c on s t a n t a . e • t h e n

Q . E . D .

EX ERCISE b . - Let m be Lebesg ue meaSUl"e on [ 0,1 ) • FiMd an !,,,, otions on [ 0, 1 ) Bu ch t hat the reZation

1 ( l im M rl ) - f ( � o tJ dm

p � O p a ho Z ds fo r eva l"Y boudad , m �adur� , p o s i tive ! on [ 0, 1 )

S O L UTION

Suppo s e s a t i sf i e s the requ i r e d prop e r ty . By · ( d ) of IT] �e conclude t h a t sa t i s f i e s the i n e qu a l i ty

( * ) 41 ( ex p Jo\Og l f1 dm ) • i1 ( 41 o f ) dm For 0 < c < 1 • de f i n e f ( t ) • x for 0 � t .:;; c • a nd

f ( t ) • 1 for c < t .:;; 1

By ( * ) • we have for any po s i t i v e x and 0 < c < 1

 (x c ) � C. (x ) + ( l - c ) <1> ( 1 )

hence i f .,(x ) · (x ) - • ( 1 ) • t h e n " (X C ) . c ' (x ) •

Therefore ., (x ) • 01 Logx , s o that we have

. ( x ) • 01 L ogx + Il>

Q . E . D .

EX ERCISE 7 . - L�t � bQ a p os i t ive measure on a sp acs cx,� } . Sho� the foZ t o�ing as s e r t i ons

( a) ( t7. ) ( l' < s im p Zi e s L r ( 1'- ) C L s ( 1'- ) ) . i s e qui va Z e 'I t to ( , ) There is 17'1 1: , 0 < .f: < 00 such that if E e C1Il

wi t h 0 < !L(E) < 00 the'l I'-(E) > -c ( b ) ( Y' ) l' < S imp Z i e s '[,1' ( 1'- ) :J [, S ( I'- ) h o Z ds if flnd

SO L U T I O N

on Z y i f the fo Z Z owi'l� ho Z ds :

( a ) There i s an H , 0 < H < 00 such t ha t if E e C1Tl i s a ., -fi n i te s e t , than I'-(E) � H .

( a ) Supp ose ( , ) does not hold , i . e , there i s a s equence

47

of measurable se t s (En ) with 0 < I'-(En ) < }n (s ee Exe rc i s e ill Ccl l We may suppose that the En ! s are p a i rw i se d l s j o in t , that I'-(En l . " an conv e rge to 0 as n -+ 00 , and tha t

< CoO

for an in f in i t e subs equence (a ) . ,, 1 nk De f ine f • L... a-I l: , t hen f e L r ( 1'- ) - L 5 C I'- l .

k nk En� Thu s ( � ) does not ho ld •

Therefore ( 0< ) imp l i e s ( , ) Conversely , suppose ( , ) h o l d s , an'd suppose f e .Lr C l'- l •

The support of f , u . ! x : .£ (x ) f 0 I

i s a a - f inite s e t .

LEMM A .

Supp ose ( , ) ho 'ds and U i s a a -fi n i t e s e t . Then t here ex i s t s a part i t ion (En ) of U ,Buch that

(i ) I'-(En » . C for aZ Z 'I (i i ) En i s an atom for I'- • ( E i s an atom fo r I'- means, if

B .c E , B me asurab Z e � then , e i th e r !L ( B ) o r !L ( E-B } is e quaZ to 0 }

PROOF OF THE LEMM A .

We s k e t c h the proof , the det a i l s are l e f t t o the readers It suff i ce s t o prove the lemm a wi th I'-( E ) f i n i t e , and f i n i te

part i t i o n i n s t e ad of U a - f i n i t e and coun t a b l e par t i t i on (En ) ' I.� t 1'- ( \ 1 ) � n c , then the p ro o f i s by i nuuct ion on n

Q . E . D .

48 Now by the l emma , the s uppo rt U of f has a part i t i on ( E n )

(fin i t e or c ou t ab l e ) , e,ach En ' b e i ng a )I.- atom a n d /L ( E n ) 01> E •

By ( i i ) o f the lemma , f I E i s constant a . e . The re fore n (each )I.(En ) � & > 0 henc e ) f i s bounded a . e . by a constant M ,

Thu s

s o that jl f l r d ",< - imp l i e s jr f l s d )l. < 00 ,

We have proved that ( � ) imp l i e s ( "' ) , ' The proof of ( a )

is comp l e t e d (b) Suppose ", d o e s not s at i s fy ( 11 ) , t h a t i s , there are

me'asurab le s e t s ' En ' e ac h E being a - f ini te , and )I.(E ) t ..., n n By def ini t i �n of a - f in i tene s s , we may supp ose each

)I. (En ) < ..., ,Moreover , we may supp o s e the En I S are , p a i rwi s e disj o int , ' ,

Let an /L (En) an t oa • We can f ind a subs equence (a ) of ( an ) ' with nk � 1-!.

and then let .of" ank S < 00

f - Lk

a ; X En nk k I t i s clear that f e LS ( )I. ) - Lr ( ", ) , hence ( V ) does n o t

hold . Therefore ( y ) imp l i e s ( b ) •

The c onv erse part of (b ) is s im i lar to Ii] ( a) • Supp o s � U - I x : ( f (x ) l :;:. o ! is a - f i n i te , hence b y ( 1I )

U has measure � M , By m (a ) ,

REMUK

, �! I • /L (U) •

We have prov t' d that ( /I ) imp l i e s ( y ) , hence ( b ) i s p rove d Q . E , D .

Thl! sea el!mei H s ( � ) /lnd ( '/I ) o f t h l! exerc:l. se ari se from

the 'cases of ( [ O , l J , Ll!besgue m l! a sure ) and (N , count:l.ng m e a sure ) .

EXE�CISE e . -If g 1. e a poei tive function on ] O, I [ euch that g (z ) __ � ae z -.. 0 , then there ie a co" v ez fun ction h on ] 0, 1 [ such that h � g and h (z ) _ 00 as z __ 0 • Tru e or fa lse ? Is the prob l em changed if ) O, 1 [ i e rep Zaced by ] 1 , "" [ , and z -.. o i s rep Zaclld by z __ ..., ?

s O L U T I O'"' By hypothe s i s on g , there ex i s t s a s equence ( !n ) in

] O , l [ such t ha t ( i ) &1 > &2 > · · . . • • •

( i i ) g (x ) .,. n if a < x < &n ( i i i ) En � a as n ---+ oa Define

Then h n hn � g , hence

hn and h by

hn (x ) • n ( 1 . - r ) J. ] 0 & fX) n , n

h • sup hn n i s COl)vex (draw the graph of hn ) , hence h is convex h � g , and by ( i ) to '( i i i ) , ' hn (X ) - DO as x - a .

Now cons ider ' the fol lowin g p rop o s i t i on

Let h · : ] 0 , 00 [ ---+ R . Then

49

there ex i s ts a convex g ] 0 , co l ---+ R wi:th l im g (x ) · 00 and g � h) i f and only if Ln"h (x ) l im inf � • or > a X"" 00

X -DO

This propo s i t ion s o lves the remaining of the exercise , and is not d iffi�u l t to prove .

Therefo�e , in general , there does not ex ists an h as requ i r e d in the exe rcise (We may cons ider the examp le g (x ) • Logx ) .

Q .E . D . E X E R C I S E � . - Suppose f : ] 0, 1 [-'[ 0, 00 [ i s measurable and no t essen tia l ly b(,'unded . 8y W (e ) , II f H - oa as p ---+ ..,

P Can you find f such that D flp -- DO arbi trari ly s low ly More precise ly , is i t true that to every p o s i tive function on ] O , ool: such that (p )-+oo as p ---+ 00 , one can find an f such that n np-� DO as p _ <>0 , but D f U � (p ) for a l l s uffi cien t ly . p large p

A VSW E R . The answer to .the que s t ion i s ' affirmative , and we can

constru c t such an f : X :""'" [ 0 , 00 [ if X cont ain s sets of a rb i t rari ly sm all me asure .

( fn )

Now suppose X contains sets E C n ) w i th a < I'-(E C n ) ) � � We sha l l show that there is an a E ' 0 , <>0 [ , and a sequence

of me asurab l e function s I fn i such that the fo l l owing hold : n ( 1 ) £:n - 2:: k l E • E are mutua l ly dis j o i n t , k k = l k

50 ! m (E1< ) " &1< > 0 with &1< < 1< = 1

(2) I f l < �(p ) 1 V p � a - p - Z-n

Note that - by ( 1 ) , the sequence (fo ) i s increas ing , hence (f�) is increas ing . Hence wi th f • l �m flf ' then If Uoo " "" , and the mono tone converg ence theorem g ives u s

n f I - m f I a s n - . o p p , p g iven � 1

Hence by (2) I f I � 4>(p) , p � a

p Now , we proceed the constru c t ion Since 4>(x) - 00 a� x - 00 , there is a number a > 0 such

that

p -

4>(x ) > 1 if Pu t f • I E

00 , by m ( e ) : Q f 1a x

x � a , with m (E 1 ) • &1 > 0 • Then U f 1 n p -� The number E1 i s chosen such that 1

& Ix < 4>(x) - .!. if x � a • _ 1 2

as

Sup�o se f 1 , . . . . . f h,ave been constructed choose \ � 0 0 as k _ "" , 4\1< > 0 , n

L E . + 4\ < .1 k . 1 , 2 , . ; . . , n � k r i= 1

Let g 1< " i l E i + (n+ l ) X 1 = 1 E�1<

m (E 4\J . .\ and E4\ n E . " 0 for i . 1 , Z t • , • , n 1< 1 k Pu t Glo: (x )

" D g 1< " x . ( & + 2x E + • • • • + nX & + (n+ l ) x 4\ )l

1 2 0 k x

( 3 ) By Exerc i s e 8] (e) , Gk (x ) -� (n + 1 ) as x _ ""

The sequence (Gk (X )� k converge s po intwise to n f nllx '

1 . 2 • • • • • • i s decreas ing . and

F ix a k • by ( 1 ) t here i s a number M ;> a su ch that ( 4 ) Gk (x) < n+2 if x � M

S ince 4>(x ) _ ..;, a s x _ coo . there is a number b � M su ch that (5) 4>(x ) - 2- (n+1 » n + 2 if x � b On t he compac t interval [ a . b) • the sequence of funct i on

(Gk ) of the variab le x • k • 1 . 2 . . . . . . . decreases and converges to Fn ' 'hence (Gk (x ) ) converg es uniformly t o Fn (x ) as k --� 00

by Dint ' s theorem

Therefore , there

( 6 ) Gx (x ) 0

Le t E . c1 n +l X o ,

Then by ( 4 ) , (5) ,

ex i st s aO ko ... k - F

n (x ) <

E n + l

f 0 n + l and

• E c} X o n+ l . 2:> i = l

( 6 ) we

1 zn+l , ko

"t E • 1

have

such that

x e [ a , b)

in ( 6 ) Pu t

I fn+ ll x < �(x ) + Z - ( n +l l (x ;.. a)

Therefore , we have con stru ,: ted by indu ct ion on n the

des ired sequence (fn

) .

Q . E . IJ .

EX E R CISE 1 0 . - Su pposs fn e LP ( fI. ) n = 1 , 2 , . . . . , and I fn

- f np- o and fn- g a . R as 11 - 00 . What r e Z a t i on e:r:i s t s be tween f end g ?

ANS W E R

Und er the hypothe s i s' of the exerci se , t he re ex i s ts a subsequence (f ) of (f ) which converges po i ntw i se to f a . e nk n (Theor em ) . l Z ) . Therefore f • g a . e .·

Q . E . D .

EXE RCI�E 11 . - Suppose fI. ( 0 ) 3 1 and suppose f and g p o s i t i ve mea surab Ze fu n r! tions on 0 sUr!h t h a t fg � 1 . Prov .. t h a t L fd � • Lg d� �

SO L U T I ON

° fg ;.. 1 , hence

1 � !af' gt d� hence

1 �

1 1 f7 gr� l . By �older ' s inequal i ty

� [ fo(f�t f l dfl.� [ L(g' ) 2 d� ] t

Q . E . D .

E X E R CI�E 1 2 . - Sup p o s e fI. ( Q ) - 1 a n d h : Jl - 0[0, 00 ] i s measura b Z e

If A - Lhd fl. , p r o v ll t h a t

5 1

52

..;;-:-;z � L� dll- � 1 + A

PrO V IJ that _qu a l i ty ho lds ' 7:n t ile first i nequa l i ty iff

h - aon s t an t a , IJ , . �nd ho lds in the s e aond one iff h - 0 a . e ,

S O LUTI0� � The func t i on , (x ) - ( 1 + x2) i s convex . and s tr i c tly

convex . Hence Jensen ' s inequal i ty g ives us the firs t inequ ality

hence

Since (1 + h2 ) � � 1 + h • the s econd inequal i,ty fo l l ows . By the theorem 1n IT} (b)

"";1 + A2 - fa � d ll- iff h is constant a , e

Furt:hermore (1 + h2 ) t - 1 + h iff h • 0 a , e

L� dll- • 1 + A iff h • 0 a . e •

Q . E . D .

EXERCISE 103 . - Un der lJhat aondi t i ons on I and g dOB S equa l i t y ho lds i n

( 1 )

( 2 )

A�SQlER

I fS l 1 � Inp. I g lq (Hi:ilde r ' s inequa. l i t y )

I f + g lp � B f l

p + I s l

p (Mi n /(OlJsT d ' s inequ a l i t y )

In the case 1 � P < _ • we refer to Theorem 3 , 5 to conclude that equal i ty ho lds in (1 ) iff

.( 1 ' ) There ex i s t s con'stant oc and � .with .. . � � 0 • and

oc f? • � ( g'� ) a . e .

Therefore equ a l i ty ho lds in ( 1 ) iff ( 1 ' ) holds .

Now cons ide? the case p . 00 in ( 2 ) • Put

a • 1 £ 1 _ , b • h 100 and c • a + b •

We will prove that the fol lowings are equivalent

(a ) H£ + g loo · D f l..: h loo (� + b • c )

(b ) F o r every neighborhood V o f 1 i n C • there i s an CIt with I "d · 1 • such that II- (C1 ( Ol aV) n g -l ( oc bV� ) >. 0

(c ) There ex i s t s an CIt e C • l oc i · 1 • such that tor every ne ighborhood V o f .. in C • we have

53

P1\O O F

( i ) (a ) imp l i es ( b ) L e t V be any neighb orhood of 1 • and let V ' be a neighbo r

hood of 1 s u c h t h a t I f I Z l l < a and 1 � 2 1 < b w i t h z l + z 2 e "' c V' , then

zl E or. aV· . z 2 E '" bV ( i t is c l ear that such a V' ex i s ts ) The coll e c t i Oil ! or. CV ' '" e" T I form an open cove r i ng

of CT . ! z : I Z I • c ! • a c omp a c t .s e t . Hence there ex i s ts a fini te cover ing

"' l CV' U or.2 CV ' U . . . . U "'n cV ' "::l" cT THis union con t a ins a set A of the form

he n ce I z l e ] c - & . c] \ (f + g ) (x) E A I has non zero me asure .

There fore • t here i s an '" • "'i such that

I x : ( f + g ) (x ) e ",: cV I I has non z e ro m e a sure • that i s

fI- ( f - l ( ", aV) () g - l ( or. bV) ) > 0

Hence ( a ) imp l i e s (b ) ( i i ) ( b ) imp l i es ( c ) L e t Vn • D ( l ; *) . ! Z E C I Z - 1 1 < k l By (b) for n · 1 , 2 . . . . . . . there i s a� "' n,d th I "' n l a 1

su ch that o

-The 'S equence ( "' n) in the c omp act s e t -l z subs equenee ( orn ) converg ing to '" with I '" I · 1 .

k

I z I " 1 I has a

For every ne ighborhood V o f 1 in C • there i s an nk such that '" n V n C or V Hence k k

II'(C 1 ( '" aV) () g- l ( ", bV) ) > Therefore (b) imp l i e s (el , ( i i i ) (c ) imp l i e s (a)

y ,& .> 0 • there i s a neighborhood V of 1 such that l x: I f+g r - 1 (] c - e • ..0 [ ) ! ? (Cl (aV) () g- l ( bV) )

54

hence

hence

H f + d ",, :> c - 1: . t: :> a n f + g g 00. :> I f I 00 + n g g 00 � I f + g I 00

Thu s ( c ) � ( a ) Q . E . D

We conclude :ha t (a ) i s equ i v a l ent to ( c ) • that i s •

ftf + g l oo 2 l n oo + h 1 00 i ff ( c ) There i s :on Ql s e i8 • a � 8 < 2 lt • su c h

t h a t f o r eve ry n e ighborhood Y of � • w e have

lL(r1 (aY) n g - 1 (bY)) :> a This comp l e tes t he answer t o the qu e s t i o n .

EXERCISE 1 4 . - Suppose 1 < p < 00 , f e LP • LP ( J 0, 00 [ )

re l a t i v e to Lebesgue measure . an d

to

a . e .

x

Hf (x ) - F (x ) - � J "' ( t ) d t x O ' ( O < x < oo )

(a) Pro v e Ha rdy ' 5 inequal-i ty

w hi ch s hows t ha t H carri e s LP i n to LP

Sugg � s t i on : as&ume . fi rs t that f > 0 and f e Cc ( ] O, oo [ ) In te g rat i n g by p arts g i v e s .

N o t e t h a t :::F ' = f - F , a n d app ly Ho lde r ' s inequ a l i ty

Then de r i v e -the g e neral case .

Prove t ha t equa l i t y ho lds i n ( a ) on ly if f = 0

( c) Pro v e that the cons tan t q = -L can n o t be rep la ::-e.i. p-1

by a sma l l e r one . (Con s i de r f ix ) x--1. in ] O, 1[ , P < l'

SO Lun ON

1 - - 1 (d) I f f :> 0 a n d f e L , pro v e that F E L •

(a) SPE C I A L CASE : f � 0 i s con t i nuous with ,compact support (f e � ) .

We may app ly the Leibni z formu l a to differi ate F :

hence

x x f l ex ) · · h jf ( t ) dt + } h ( f f ( t) dt )

a a 1:. F (x) + 1:. f (x ) ( 0 < x < 00 ) x x

( 1 ) xF ' · f • F Intergrating by parts g ives

fooFPch. • XFPJoo. r oo pF ' FP-1xdx a 0 Jo

Rec a l l that f has support in [ a , b) c ) 0 , � [ so that xfP (x ) - a as x - 0 ·

Furthermo re xFP • (xF ) pxl-p , and

Hence l im . xFP (x) • a

Hen ce

then

)(+00 I t fol lows that

(2 ) l<>OFP dx .

a By ( 1 ) and ( 2 ) , we have £OOFPdx c - p £00 fFP-1dx + p IaooFPdX , hence

a ( 3 ) foO FPdx • ---E- IfFP- 1dx

0 p - l a Us ing Hol der ' s inequal i ty , with q • -E- one has� p - l '

( 4 )

foo I fFP-l l dx :0:;;; I f I IFP - 1 1 • I f I IFP I �l . P q P o

I FP li-� :o:;;; q lf lp , or

I F I :0:;;; q lf l P P

Therefore if f > a i s cont inuous with compact support

I Hf I :0:;;; q lf l P P

55

56

GENERAL CASE . By defin i tion . one has I H (f ) 1 � H ( I F I ) .

Therefore i t suff ices to prove the Hardy inequality

in the case f ;> a

By Lu s in ' s theorem there exi s ts a sequence ( fn ) in

Cc (l O , ao [ ) wh ich conv erges to f in LP norm .

(fo ) i s a Cauchy sequence in LP , and the inequal i ty

( 4 ) appl i ted to I fm - fo l instead of f shows that (H (fo ) )

is a Cauchy sequence in LP • . Thu s there i s a subs equence

(H (f ) ) whi ch converge s a . e . to a funct ion g E LP satisfy ing nle h lp � q 1 1m sup I £o l

p - ql £ lp r

Now l a ,x [ has finite measure (Recall ffi ) hence the

. inequal i ty

Ix L 1 · I xHf (x) - xHf (x) l � I f - f l dt � xP I f - £ I n o n P

o shows that Hfn (x) converges to Hf (x) • x E J 0 , ao [

We conc lude that g - Hf a . e r and

( 4 ' ) I Hf I � ql £ l P p

Thi s completes the p ro o f o f ( a ) •

f ;> a

. ( b ) Suppose f e LP . and I H£ I • ql £ l , l a •

By ( a ) H i s cont inuous . We cons ider the proof of C a ) .

The formu l a (3 ) ext ends from the case f > 0 , f -E Cc to the

case f ;> a , f E LP ( It suffices to I1 rove that the r i g h t

hand s id e o f (3 ) i s a con t inuous function of f > a in L P

Therefore • f o r f ;> n in LP ,

Hence the equal ity I Hf I • p

� ql f . P

ql £ l , f � a P that equ a l i ty holds in Hol de r ' s inequal i ty

imp lies

J�FP-ldx o

Thus there exi s ts .. � a w i th "'fP • (FP-l ) q. FP a . e .

But F is a cont inuous funct ion • hence f i s continuous

so that · we may app ly equal i ty (1) :

"'lx f ' • (1 - "'l ) f

hence

c • a

then (ft)

f ' �1" ! . ! . hence f (x) • cx� •

f exl x x Th". function ·cx � be longs to LP , 1 < P < DO,only i f

Therefore inequ a l i ty .holds

(c ) . Le t r b e in ) P . DO [ , q -1

I f f (x) • x r o < x < 1 . 0 1 � x < _ 1· f\-� f ' f ' Pdx dx

0 0 Now F • Hf

x i s ... a • and

in (a) only if f • o a . e •

. � . s • r .r:T .

• _r_ r-p

F (x ) · � J o

1 _ .1. u- Y du · . sx r

Therefore

I DOFPdx � r l FP dx • s p_r_ ,(by ( ft ) ) •

L r-p o 0

Hence I F I > si t I . P P

This shows that for every s < q • there exi s ts f E LP

with I F I � sl £ I . P P

Therefore , the cons tan t q • � cannot be rep laced

by a smaller one

(d) Suppose f > 0 , f E L l , and I f Ul > 0 •

The re i s some x . < 00 and x . f · f ( t ) dt · a > 0 . •

• 0 100 Thus F (x ) .... !. i f x > x . , hence F (x ) dx .. 00 x x Therefore F E Ll •

Q . E . D .

57

58

tXER CISE 1 5 . - Supp ose I an I i s a sequ ence of posi t iv e . n um b e r s .

Prove that � N ( • ) £... (L L a )p DO (..f...)p L aP if 1 < p < 00

N-1 N n,.l n p-1 n=l n

SOLUT ION

Suppose an � a • n - l . 2 • • • • iiith L a�· < oo.Thert.fore an -)0 0 as n ....... DO . so that .we m ay rearrange the sequence to

" obtain a sequence a a( n ) , a a( n ) � o as n -+ 00 . If an is rep laced b y a a( n ) .' t he left s ide of the

inequ a l i ty (* ) decreases .while the right one i s left invari ant •

and

Thus it suff i ce s to c ons ider the case a1 � a2 � • • • •

00 • Define f (x ) • an+ 1

Then F (x ) � � t, n=1

• n < x � n+1 • n • 1 . 2

an N -l < x � N

j7Pdx • L a� o

� . . . .

• and

Thu s we may app l y the Hardy inequal i ty to f • and

obta in the des i red result Q . E . D .

E X E R CI S E l b . - Prove Egoroff '� theorem : If · �X ) < 00 , if

(fn ) is a sequence �f comp lez me asurab le function8 which

converges pointOJis e at any every point of x, and i.f t > 0

there is a measurab Ze s e t E e x , with /I- ( X - E) < t , such tha t I fn l converge s unifo rm Zy on E • .

(Note . that pointwise .oon vergence on X may b e rep Zaced by convergence a lmo s t e v s rywhere on X . )

Does the resu Zt eztend to a -iinih spaces ?

SO L UTI ON

Le t � eX) < 00 and suppose (fn ) as in the hypothe s i s

Define S .... • n ix : I f r ex ) - f j (x ) I < } I n . " r , j > ,\ I f k i s fixed • and x e X g iven • then 3 n such that

X E S n , k • e fn ex ) conv e rg e s ) .

Therefore and U Sn . k n = 1

x

and hence ... ( 5" .k ) t ... (X) as n - 00 (Theorem 1 . 1 9 ( d) ) .

5 ince )L(X) < 00 , Y k · 1 , Z , . . . . , there i s an nk

such that

Let E 2 n 5 k = l nk , k E e Sn k ' we have :

k '

)I- (X - 5 ) < � nk , k . . Z ...

, then . )L (X - E) � c and s ince

,. X E E

Thu s (fn ) converg e s uniformly on E . Now con s i der the following examp le

We observe that if E e N s a t i s fying

is empty , hence E • N and it is clear that (In ) does not

converge uniformlY ( t o 0 ) on N

Therfore Egoroff ' s theorem cannot ex tend t o a - f in i t e

spaces Q . E . D .

59

E XE R CISE :107 . - (a) If 0 < p < 00 , pu t Yp 2 ma;:; ( l , 2P-1 ) , and shoLl that

( ' ) I .. - � IP � Yp ( I " IP + I � I P ) for arbitrary comp le:z: n umb ers .r an d � .

(b ) Supp ose ... is a posi t i ve meas ure on X , 0 < p < 00 1 E LP (/L) , f E LP (/L) , f (:z:) - f(:z:) a. e. and l r l - I f l as n -oo n n n p p ShOLl th at thlln lim I f - f l - 0 , by comp leteting the tLiO proofs that n p are ske tched be low :

(i. ) By Egoroff ' s theorem, X � A U B in such a Llay that

Ji.flP < £ , 1B) < 00 , and fn - f un if;rm lY on B. Fatou ' s lemma, app lie d t o

j! f ! p-, Ze ads t o B n

(ii ) Put lemma.

li m sup �f (Pdl'- < E . J � n

h - Y ( l f l P + I f I P ,) - I f -n p n

60

S O L UT I ON

hence

(a) If 0 «'(x) .. 0 ,

(x + y) p � Z p- l (xp + yp ) (x � O ,y � O ) . I t fo llows that fo r 0 < p < 00 , and Y • max ( l , Z p-l ) , then

p

hen ce ( 0 )

1", - p I P .� ( lor! + 1 � I ) p � Y C I .. I P+ I � I P ) P

(b ) ( i ) . Define Ale · j x : f < l f (x) I P < le ! le • l , Z , • • •

I f I P E LI , s o each '\ h as finite me asure . Le t �> O . Since l ' f , Pd • lim r l f l Pdfl. , the re is a le such

X le+ oo JAle th at l,, ' f I P d fl. < f. Egoroff ' $ theo rem app l i ed to tlie s e t .Ale Ale

£ ()L(Ale) < 00 ) gives B C '\ , )L-(� - B) < IT and fn -- f uniform ly on B .

Le t A • X - B • � U (Ale - B) . Then

+ le. !. 2Jc

• £ .

Therefore X " A U B , fI.(B) < ""' , }l f I Pd fl. < £ ; and f conve rge uni forrrt-A n ly t o f on B .

By Fatou ' s lemma , w i t h I f l P 2 l im I fn l P • l im i n f I fn l P

l' f , Pd )L- �' lim inf [l fn I P d)L- . lim inf(!x - JJ hence (see the first line of (i i ) )

Since I f I - I f I as n -- 00 , thi s gives · n P P fl f l Pd)L- � lim sup II f I Pd)L. A . A n

The inequalitY , C * J in C al g lve s

hence.

It i s cle ar' that lim fHn- f I Pdli- · · O (The convergence is unin+oo JB

fomly on B and fL (B) < 00 ) , hen ce finally

(1 ) lim sup Il f - ' f ! Pdll- �2 Y £ for every E > 0 n ... 00 X n P

that i s lim I f - f l • O . n .. oo n p

(ii ) Firs t no te that i f lim an • a i s finite , then we h ave lim inf (anbn ) • a - lim sup bn •

Consider h ., Y ( I f I P + I f I P ) - ( fn - f ( P, _ By (*) of ( a) n P n hn > 0 for every n , hence we may app ly FatQu ' s lellUDa. Sin ce lim h • 2 Y / f l P a . e . , we obtain n p

2 \ fl f l P dJL � lim inf [/Yp C l fn I P + I f l P ) dfL - fifn- f I PdfL ] • Hp· / i f/ P dJL - lim sup Il fn - f l PdfL '

by the preceding remark. I t follows that

l im sup II fn � fl P d'!'- � 0 !

hen ce this lim sup is O . thence lim I f - £I • 0 n+ 00 n p Q . E. D.

EXER CI�E 16 . - Le t JL bs a positivs m. asu�s on r . A sequenee !fn \ of oomp le: measur ab l e fun ctions is said to converge in me asure �o tl .. meaaurab l.B fun at ion f i.f to ev ery e > 0 t her" eo.zor".p ond. an

N suah that 1'- ( 1 : : I fn {: ) - f {z ) l > 1: I ) < E

.for a l � n > N [This no tion ie of impo�tan ee in probabi l i ty theory) A S llums }L{X) < 00 and prove ths fo l l owini sta tements (a ) If fn {z ) � f{z) a • • • • thsn . fn - I in measurll , ( b ) If fn E LP { IL ) and I I -/ 1 - 0 • then f - I n p n

in m.aswrej hs�. 1 � P � 00 '

(0) If In - I i.n msasure • then (fn ) hall a eub .sq, .. "",e

whioh eOmJergss: to f a , e ,

61

62

, In v e B ti�ate the aon v e r B B 8 of (a) and (b) • Nhat happe n 8 ,

t o (a ) , ( b ) and ( 0 ) i t ", (Z) - 00 , ta l' i n s tanaa i t '" i. L e ba 8iue m�a8wre on Rl !

SO LUT ION

(a) Suppo se fn (x) -,.. f (x) a . e and let & > 0 •

Eor AN · I x : I fn (x) - f (x) l � & Y,n >- N \ then

Al c A2 c and 'J An • X - U . )Io-(U) · 0 s ince fn (x) -+ f (x) a . e ,

Thus now imp l i e s

"'(�) t ",(X), as N -+ oo , the hypothe s i s

3N ",(A� ) < , I t i s clear that

I x : I fn (x) - f (x) I >,' \ c � if n >- N

Hence (a) is proved . \

", (X) < ....

(b ) Replacing f by f - f , we may suppose f • 0 and . n n II f f -- o n p as n -+ DO • Hence

3N ' flf I P < ,pH hence

Y £> 0 , n

The'refore Y £ > O

n • N+l • N+2 • • • • • • , _

3N : n >- N imp l i e s

p. ( l x : I fn (x) l > £ \ ) < r; , hence (fn ) converges in measure to 0 as des ired .

(c) Suppose (fn ) converges in measure to f • L et

An ,lt • , 1 x e X : l'in'(x) - f (x) 1 > f f Hence by defini tion of the' �onvergence in measure ,

Y Ic , 3nk such that with "k • \Ik , ltwe have )Io-(AIt) < f' Since ",(X) < 00 anJ AI ,It c' A2 ,Ie C • • • • nYI An •1t • X - N • ",(N) • 0

. If (nit) b chos en. such that n1 <: n:z < . . .. . . then (fnk) is • subsequence of '(fn ) • which clear ly converge to f' on · X - N

Thus , (e) is comp leted •

Now let us con s i der the conver ses of (a) and (b) .

(a ' ) There is · a sequence I fQ � in LP' ( [ O , l] ) such that If 1 -" 0 as n _ "':' , but f (x ) converges for no- x in [ 0 , 1] , n p n

(See Exerc ise ill . ch. I I ) • By (b) , fn --+ a in measure • Thu s the converse of (a) does not hold .

(b ' ) I f fn • n� l[ O , l /n ] then f n - 0 in measure

6J

but � f It · 1 n · 1 ; 2 , . . • Therefore the converse of (b) is f a l s e . n p Nex t we investigate (a) , (b i and (c) in the case tI-(X) . DO

(alt ) Cons ider fn V x .E R . But for 0 .;: £ < _ , we have

m ( I x : I f n (x ) ( >:- E b . !II ( l n ,n] ) • n

Hence (fn ) does not · converge to f . 0 in mp',,"sure

Thus (a) is false i f /Io(X) . 00 •

(bit ) The proof · of (b) i s unaffect .. d even if jI.(�) � >0 •

(cit ) f (x ) � o or

hence E has

Let E be the set of all x E X

f n .(x ) � o for some n . Then E

a parti tion ! En I ' each I'- (En ) at wh·ich either

i s a G -finite set , < 00 .

By (c) , there i s a subsequence (f ) o f (f ) which con-nl. n verges a . e . to f on El . By induction on n , (f ) is a sub-

nk1'l sequence of (f ) which converges t o f a. e , on Ek1'l • nk Therefore ( f nn n • 1 , Z , • • • is a subsequence of

which converg e s a . e : to f on E , hence oti X

Q . E . D •

. E:XE:R CISE: 1'. - Define tits .... n H a l rangs of f E Loo( 1'- ) ths u t Hf �onsis Hng of a l l w E e su clt that

1'- ( I z : ( f (z) "- w ( < E I ) > 0 for svsr� & > 0 • PrOVB that Hf is comp a�t • and

R f c· \ .. : ( .. ( � I f I DC ! "1 r Lst Af bs ths s s t of a � l avs r agss -;(iT JE fd �

E e C1Il and . I'- (Ei > 0 • What rs lations ezist bstl<issn Hf and Af? Ars thsrs ms asurss I'- suc lt that Af is �onvsz for s vsr� f E L 00 ( fI. ) ? A re t hsrs ",easurs s I'- such. that Af fa i l s to h fl �onvez for �O"'fl

f e LoO( 1'- ) BOl<i ars tltsu ruu Zts affe cted if L GO ( p. ) i s r.pf.a�sd b�

L l( ", ) . for ins t'lncs�

SO . u n O N

Suppose f e L 00 ( � ) , and ( III 0 ) i s a sequence in Rf which - w Hence '1 -1: > 0 , 3 �' > 0 and some n :

\ x : I f (x) - w 1 < £ 1 � j x : If ( x l - III 1 < £' n . Tl:erefore 'f E > O . the set \ x : I f (x) - 111 1 < E

p o d t i ve measure , hence w e Rf He)'lce Rf i s . closed , has

Now if h i > I f l oo • it is clear that for E - Hl z I �l f I ..;,) thr set \ x : I f (x) - z I < � 1 has zero measure . Hence z . Ii Rf •

Thi s shows that· Rf c: D' (O ; l f l oo) , therefore Rf is c omp

ac' • Now we shall prove that Af is contai?ed in the convex hul l Supp o s e 0 < p.(I;) < DO , and let K be the compact se t Rf

.1 E For every £ > 0 , ·there are w.1 ' • • • • 1 111 0 e K, and a

par t i t ion \ U1 , • • • , Un l of J( such that I z - "'i I<E for z e Ui If � - _1_ p. (f-1 (U ) n E ) , then ;\'i � O and i p.(E ) i

). + . . .. . . . . .. +). - I

X n For III ._ 1 fdp. e Af it is clear that ,

I'- (E ) n

1&1 ), . 111 1 - w l < E 1.

This shows that Af is contained in the convex hull

n � �"'i wi : n is s ome intege r , 0 .,,;; ;\.i '< 1 and L),i-I , "' I. e Rf '� of Rf ' as as serted •

. In the examp l e f (t) - t , . t e [ 0 , 1] ( I'- being Lebesgue mea,lUTe on [ 0 , 1 ] )

I f X - l a. 1 Af - ] O ,l [ i s not closed.

II- ( la I ) - 1 , then R - Af • I f (a) I f. A n.mtrivial eXaJIple i s the fol lowing I

EXA -I P L E . :x - '[ 0 , 1] , p. . III . Then every Af is' convex .

i s convex

If now X • \ a , b I , and ' II- i s the counting measure on X , the'.l Af cons is ts of 3 po in ts ,

Af - I f Ca) , f Cb) , �(f (a) + f eb ) )

..

Therefore Af is not convex if f : ! a , b I -+ C i s not constan t . · Now let us answer the last quest ion of the exerci se . If f e Ll t.hen Rf i s c losed in C , and

( * ) sup ! z : z e Rf - 1 £ I 00 Thus Rf is not compact if f 'E L 00 Note that ( * ) holds a l so in the case f e Loo• Simi larly , for f e Ll ,

- I f I "", sup t z : z e Af

(Cons ide r the average of f on E E - r l (U) and U c: the annulus . . '

j a < I z l < b l in C - .with .suitable a , b and U ) and Af a lways;'lies in �he convex hul l of Rf

. 1 E X E RC I S E 2 0 . - SUppOBfJ • i.s a r . a l furt"ti.on on R

1 . . . ' " 1 ' . . ( J f (Z ):U) � f r 'POfJdz o 0

for every real bounded me asurab l . f .

SO L UTION

Prove that • i.s t hen "onvez •

Let x ,y 'e Define We have

Rl and 0 < � < 1

f - x X['d ,,i) + Y l [ �,d ' . 1 1

Q . E . D .

su,,� . that

. ( �x + ( 1 - l. )y) - . ( '1 f (t) dt ) � fa '(f (t) ) dt

Thus

o � L' 9(x ) dt

o

. ( >. x + ( l - � )y ) � � .(x) + (l - � ) .(y)

for 0 < l. < 1 • Hence 9 is convex .

i '

Q. E . D.

65

66

EXE ItCISE i!l.. � S"pp os. ,.. i. a p ositive m. as "r. on X, �(X) < "",

f e L (�) . I f loo > o. and

DC - r [ f l "d� n Jx· (n - 1 . 3. . . . )

Pro",. that

- 1 , 1 ""

S O L UT I ON

Thus

This

Note that - n - 0 imp lies 'I f l ll • 0 a. e . , hence I f l",,· O •

• 11 ; O · 't n . We h ave

gives . (1)

. ·n+1 l im sup -- � I f I "". · 11

Let c · be a pos i tive numb e r , 0 < c � I f loo • Then

E • I x : c � I f (x) ! � 1 f.1 "" I hu nonzero me asure . On EC " I� f l < I , and I� fr � O � Since �(EC) < ,..(X) < "" , Lebes gue t s domin ated conve rgence theorem gives

lim r ilin d� • 0

n ... _ J&c cll On the other hand we h ave

It ' fo Hows that

.£ Ifrd� � � .(1�1I+.1

lim inf ---

Llfr

1dll- � II-(E) > O .

l ill

I f l�n. l I f I I f I n I f i ll But . c • c c � c on E. I t fo l l ows t hat

�Ifr+l �·[�fr. there fore the It. inf above is ;> 1 . This gives

lill inf .. c- (1I+1 ) n+l . ;> 1 -II • C II .

67

hen ce "' n + 1 lim inf -- � c. This holds for eve ry c < I f l oo, so that OI n

or (Z ) lim inf ;+ ! � I f l 00' n ( 1 ) and ( 2 ) give

lim OI n + ! • I £ I"", n-� 00 Ql n

Q.E. D .

£ XE �CIS£ 22 . - Ca l l a ms t � a spacs 1 a comp letion o f a ms tri c space

X if X is d�" s s ilt 1 and 1 comp ls te . I" Se a. :S. l S refsrs"cs was mad.

to "ths " comp lstiorl of a me tric spaas • . State and prove a lDIique"s"s

ths orems Llhi ch justifiss tht.s ts rmi"o toGI/.

SOLUTIO�

Oe. 6bl.i.,u01l. - Ls t (X, d) bs a .Il onsmptl/ me tri a spacs , The pair ( (I, a) , ')

is ca Z Zs d a comp leti em of :r if

(i ) (1, a) i s a comp Zets me tri a sp ace

( i i ) .: X - I is an isome trl/

(ii i ) , (X) is dellBS ill 1

Th e. OIle.m 1 . - Le t ( Il ' �l ) and (12, '2 ) bi. comp �etiorls of X. Theil thsre

s%i s ts a u"ique'

(is oms tri a) homeomorp hi sm

f : I1 - I2

such that

Le.mma.. - E is a comp t.t. metri a s pace

(a) If Z is dSIl .. i" 1,. and f1 : Z ->- E is lDIiform Zy aon ti

/lUOUS, t hs" f1 s% tsll ds ulliqus Zy to a lDIiform ZI/ cOll ti"uous g : 1 - E (b ) If (Y, .) is a comp Zstiem of X, alt d h : :r -E "i s lDIi

formZy co" ti"uous , ths" thsrs i s a ull iqus f1 : 1 --- E s u ch t h at

g • • - h.

PIl 0 0 6 0 6 the. temma..

(a) I f (xn ) i� a C�uchy s equence in Z , then (g (xn ) ) is a Cauchy sequence in Y . since g is uniformly continuous.

Thus if (xn ) i s a sequence in Z whi.ch converges to x e Y . g (x) • lim g (xn ) e x i s t s . s ince E is comp lete . I f x E E . the continuity of g shows that g (x) • g (x) . hence' g is an extension of i . I t b clear that g is .... e l l-defined • . "(proof : I f l im xn • x and lim x�.)t "examine the sequence ! x1 . x r .x2 .x2 , , , . ! whi ch converges to x) and that g is uni formly cont inuo�s .

I f &1 and &2 are extens i ons of . g and x E Y ,. there are (Xn) C 7 , lilll Xn • x . The continu.ity· of &1 and &2 give s

gl (x) • lim gl (xn ) • lim g (xn ) • l im g2 (xn) • �2 (x) Thus gl • &2 and & is therefore unique . (b) Put Z • ,, (X) in (a) .

Plt00 6 0 6 .t:he Olt em 1 . Jus t use . (b) of the lemma. Theoltem 2 . - Ls t U bs t it s ss1: of a n Cauchy ssqu8nces i n X. an J ds

fins IV on rJtx U by

(zn ) '" (I/n ) iff Hm d (zn ' Yn ) - 0 "+ 00

'flt8n IV i.. an "quival"n ,," rs lat i on

If z . y E u l '" ;0 I . dsfi.ns

d (z. y ) - lim d (zn' Yn ) n+ oo

(zn ) E z. (Yn ) E y .

(i ) d i s ws l l-ds finlld o n :t . an d i s a .ms t ri c o n I. (ii ) If 'f(z) i s thll "lass of tit. ss qusn cs ! z • ..: . . . . z

·. z . . . . !

tlt"n 'f : X - I i s an i s oms t ry . an d ,(X) i . dsn s s in I .

(Hi ) (I. d) is comp l s t s .

Pllo o 6 . I t is cle.:lr that IV is reflexive and symmetric. If (xn ) '" (Yn ) , (Yn) '" (%n ) ' then the triangle inequality

d (xn ' %n) � d (xn ,yn ) + d (Yn ' %n ) shows that lim d (xn ' %n ) • O . Therefore ", is transitive .

( i ) (xn ) '" (x�) and (Yn ) '" (Y�) . Then s ince d (xn ,yn ) � d (xn 'x� ) + d (x� ,y� ) + d (Yn 'y�)

we con clude that lim d (xn 'Yn ) � lim d (x� ,y� ) hence l im d (xn ,yn ) •

lim d (x� ,y� ) . Thus d is wel l-defined. I t is clear that d (x ,y) • 0 implies x • y . The triangle inequality is verified eas i ly .

Thus ( i ) is proved. ( i l ) It is cle ar that " is an is ometry. But if (xn ) is a

Cauchy sequence in X and t > 0 , there is n. such that d (xn ,xm) � t whenever m � n. and n � n • •

Thus lim d (x ,x ) � t , hence d (x , (x.) ) � t , X being the U+ CIIO Do m

cl ass o f (xn ) . Thus ,,(X) is dense in Y. ( i i i ) Suppose (x (n) ) is a Cauchy sequence in Y .

1 Choose xn such that d (x (n) , 'f(xn ) ) < n as in (ii ) , and let x be the clas s Of the sequence · ! x1 'x2 . . . . j . Then d (x .x (n ) ) .s;; n ' hence lim x (n) • x.

This c omp letes the proof of Theorem 2 .

6Q

Theorems 1 and Z state the exis tence and unicity of "the" complet ion of a met ri c sp ace X.

Q . E . D.

EXE RCISE 23.- For re al numbe rs % an d y. write % tv Y iff % - Y e Q . I t i. o l e ar t h a t tv i s an equivalence re Zati on on R . 'Lst E b s a s s t

in ]0.1 [ whi ch con tains s%act ly one p oint in every' squivaZen ce a lass Thsn E is n ot Leb esgue me asurab le (Confs r E%amP Zs 2 . 2 2 ) .

(a) ShoJ.J that · the I"e is an & > 0 s u ch that i f A c [0. 2 ] an d m ( [0. 2 J \ A) < E i mp liu that A n eE + l /n ) ' " � for a Z Z in tegers

n - 1 . 2 • • • • (Ba re Z \ A - 1 % e X : % Ii A • an d A + � - y-%� :: e A ! (b) Dali ne 9 : R :"" E by 'f(t) - Y i ff t rv y . t e ll. y e E.

ShoJ.J that 9 i S J.Je Z Z-delins d. and that .( [n .1I+1 [) - E tor every

(0) By (b ) . loIe "an de fine t : ', (0.00( .--+ [0 . 2 J by

t ( t ) - 9(t) + � . n-l � t < 11 . n - 1 � 2 • • � . ' D.fin. il fami ly 1ft ! t > 0 of fun otion s on ' (Oi2 J by '.

ft (%) - 1 if % - t( t ) ; ft (%) - 0 otherJ.Jiee Show t h a t ftaoh ft is Leb ugue me asurab ls . an d that lim It (%) - 0 for "1;78r1l % E [ 0 6 2 J. t+oo .

( d) Use (a) to (0) to shOJ.J that Egoroff ' s t he orem doea not e%tend to the "ase in J.Jhi oh If I 1 2

are rep iaoed by I f I n l n- . . . . t \ t > O an � tit. a .. ump tion is that ft (%) - f (% 1 as 't , --+ "". for e1l6I"y % e x. (Thus in t�e s o lution to lW . ms as urabi Zi ty of Sn . k is qf importan ce )

SO LUn ON

(a) Cons ider following s tatement ( I ) For eve ry n • l , Z , • • , there is a set Bn , BI1 is Lebesgue

me asurable , m (Bn)< l, and ij c B • n n The truth o f ( I ) shows that U c B • B 1 n B2 n • • • But m"cs) - 0

hence comp letene s s o f m shows that U i s Leb es gu� measurab le . Our set E i s n ot Lebes gue measurable (See 2 . Z 2 ) , hence (I) is not val i d i f U is rep l aced by E. I t fol l ows that

(1) The re i s an & > 0 , and m (B) < ' imp lies E \ B � 0. Suppose A c [ O , 2 ] , i s me asurab le and B . [ O , 2 ] ,\ A has, me asure < c Then m (B - 1) • m (B) < E. ( 1 ) shows that E contains a point of

1 n I l R \ (B - il) ' hence E + il contains a point of. R\B. Since E + il i s a subset of [ 0 ,2 ] , this implies

A n (E + l) � 0 n

70

We conclude that (Z ) There i s an & > 0 . such that if A c: [ 0 .Z J ' an d

IIIUO .Z] \ A ) < E then A n CE + .!.) ' � Ql • for eve ry n . · l . Z • • • • n . Cb) Fix t e It. Then E contains exactly one point y such that t � y. Thus .(t) i s we l l - de fined. This is true for all t . hence .(t) i s define d for a l l t e R.

I f Y e rt. then y e [ O . II . hence t • y + n e [n ,n+ l [ . n is an in teler . hence t I\J y. Thus y • f(t) .

I t follows that .({n .n+ l [ ) • E.

(c) ft is the characteri st i c function of the s ingleton ! t(t) l , hence f i s Lebe s gue lIIe asurable .

Le t us prove that l illl ft (x) • 0 for every x e [ O , Z ] . t+-

F i rs t note that y e E and r e Q illlp l i e s y + r I\J y , hence y + r e E if r � O. Thi s l ives

(3) (E . + .!.) n E . Ql n - I .Z • • • • n Next note that

tern- l .n £) - .( [n- l ,n [ ) + � - E + � by (b) Hence t ( [ O . -[) -0 (E + �) • F. . . n= l

Le t x e [O , Z ] . I f x e F , then ft (x) • 0 'It , hence l illl ft (x) • 0

t+-.I f x e F, then x e s alle E + �. Hence x e !! + t if k � n ,

hence x � t(t) for t e [k- l .k I . I t follows that ft (x) • a i f t > n , hen ce lill ft (x) • O.

t+ oo (d) Let X • [O , Z ] an d "' . III i s Lebes gue Vle asure on X. Let ! ft l

be as in (c) , and E as in (il) . By (a) if III (X\ A) < E,' then A con-

tains poi�ts y • x . + ! x e E n . n n ' n ·

Le't tn • xn + (n - 1) , hence t(tn ) • Yn ' thence ft (Yn) • 1 . I . follows that n

( 4) For eve ry M > e , 3t > M and y e A such that ft (y) • 1 . Th i s shows that ! ft I doe s not converge uni fomly on A to 0 .' The preceding gives the examp l e of a family I ft I t > 0 of

measurable fun ctions on X whtch converges pointwis e (to 0 ) on X, but does not conve rge un i fomly on the comp lelllerit of any subset of X of measure � £ •

* * *

Q. E . D.

72

' " . . :�

E X E R C ISE 1 . - If M i s a Zinear subspace of H , p ro v e that (H l) l. M SO L UTI O N

Le t E denotes a nonemp ty sub s e t of H , and suppose /oj i s a l inear subsp ace � f H .

Reca l l that E l i s determi ned by

E l . \ x e H : (x ,y ) • 0 for al l y e E ! The fol lowings are e i ther evi dent or d i re c t from thi s

de f i n i t i on : ( 1 ) E l is a c l osed subspace of H

'(2 ) A c IJ imp li es 5 1 C Al

(3 ) rl . E l

( 4 ) E e r e (E l l (by defini t i on , E c the c losed s e t (E ll ) Therefore Now M is a closed subspace of H

that i s x e H h as a unique representation as x • y + z , y e g , z e lol l

I f x e (Ml ) l , then z • x - y e (101 1 ) 1 , for y e M c (M l ) l

Therefore z e (M l ) l n �I l • 1 0 1 , hence z • O .

Thus x . ' y e M .This g ives (1-1 1)1 e M .

In conc lus ion , we have M (�1 1 ) 1

Q . E . D .

EXERCISE 2. - .For n - 1 , 2 , • • • • , l il t I !)n 1 b e a n i n dllp llnden t s e t of !) e ctors i n � • D e !) e Zop a cc ns t ru c t i v e p r o cll88 w h i c h g e n e ra t e 8 an orthonorma l 8 e t \ u n l 8 u c h that u n is a linear comb i na tioN of

73

U1 • V.2 • • • • • • • un

' iYo ttl t h a t t h i s l e ads to a proof of tho e:z: i s "tence

of a m a:z:ima Z orthonorna L. s e t i n s ep arab L e Hi L b e r t spaces Ll h i c h

m a k e s no app e a l t o the Ha� s Jo rff m a:z: im a l i ty p r i n ci p L e . ( A space � i s s eparab l e if i t ccn t�ins a coun tab l e de n s e s�bse t ) .

S O LU T I ON

Define inductive ly u1 , • • • ,un , • • • as f o l l ows

( i ) 1011 - v1 (i i ) wn · un • I W"'I ' n n ( i i i ) wn+ 1 � vn+ l - L (vnT1 , ui )ui n = l By l i i i ) wi th n ins te ad . of (n+ l ) 101 i s a l inear combi -n

nat i on of vn and u1 , • • • ,un_ 1 .Thus un i s a l inear comb inations of v1 • • • • • vn • wn+ 1 i s evi dently i O . s ince vn + l may not be a l i near combi nation of v 1 • • • • , vn (See ( i i i ) ) .

� By (i i ) I un I

- 1 for a l l n . . By (i i i ) (101 n+ l ' ui ) . a for . 1 , Z , . • • ; n .

Hence (un+ l ' ui ) - 0 for i . 1 , 2 , • • • , n .

�ie conclude tha t ! un I i s an orthonormal s e t i n H and

u n is a line al' comb ina t i on of v1 • . • • • vn • (As a conconsequence ,

u1 ' • • • , un I and I v1 ' • • • , vn ! gene rate the same subspace , s ince

u1 ' • • . , un l i s i ndependent ) .

Suppose H i s s eparab le J f H i s f ini te dimens iona l , then

the preceding cons tructi on g ives an orthonormal bas is of H.

Therefore i t suffi ces to cons i der the case in whi ch H i s

i nfini te dimens i onal .

Now le t l a� l be a dens e sub s e t of H • Define inductive ly

a subs equence (an ) of (a ) as f o l l ows . Ie n

(a) n 1 i s the smal l e s t integer such tha t a "I 0 • n 1 (b ) is the sma l l e s t integ e r such that I an , • • • , an 1 . 1 k+l I

i s independent • nk + 1 ex i s ts s ince H i s fini te dimens ional .

Moreover

a , • • • • , a ! . n 1 nk+ 1

a 2 , • • • , a l i e in the l i ne ar span of n k+1

· Th·e orthonormal i zati on d i s cus sed in e fi rs t part app ly-

a

ing to (vlc l , wi-th vic • a , gives an orthonormal s e t l u ! s u ch that "Ie . n

74

oy

• • • • • . un ! and Therefore I

! VI • . • • • , vn I genera te the same sub·space •

a1 • • • • • · a I lies in tne subspace generated nk . • hence the subspace generated by I Un l is dens e

in H • Thus H cont ains a maximal orthonormal set I Un l •

Q . E . D .

REMAR( . Th e process des cri be d i n the fi r s t p a r t o f the s o l u ti on

is " the GrAm - Schmi t t o r t h on orma l i z a ti o n " .

EXERCISE 3 .- ShoLl that LP (T) l.e BBparab Z" if 2 � P < 00 • bu t. that Loo (T ) ie no t 8 "para], Z" .

SO L U T I ON

. By Lus in ' s theorem • C(T) is dense in LP (T) if 1 � P < .... But the space P(T ) of all trigonometric polynomials is

dense in C(T ) . (See Theorems 3 . l 4 and 4 . 2 5 ) . Hence P is dense in LP (T) · . 1 � P < _ . But for

0 � t < 2 lt

I un I is a basis for T!lerefore the

P (n - .O . !: l . ! Z • • • • • ) . •

set of all trigonome tric polynomi als N

p - I cnun . n = - N form a countable den se subse t of P hence of LP (T) in the LP norm .

Therefore LP (T) i.5 separab le • 1 � P < 00

Now let J( • ! X [ o . x ] : 0 � x < 2 lt I C Loo(T) (We identi fy L-(T) to Loo( [ O , 2 % C ) ) . I f x' < x then

I X ( 0 ,x ] - X( 0 , x , ] I - I X ( O ,x 1 J 1 - ' 1

Moreover J( is uncountab le . Let S be a dense sub set of Loo(T) • Then for every x "' x ,

in [ 0 . 2 % [ . there .are s . s ' e S such that 1 X [ O , x ' ] - s ' I < 1 X ( O , x ] - s 100 0 , hence s",s ' • Thus we can define an inj ection from K to S (by the axiom of choice ) h.ence S is uncountab le . Thus L .... (T) i s nO.t separab le .

Q . E . D .

E X E R CISE � . - $how that H is separab Le if and on Ly if H contains a mazima L orthonormaL sys tem whi �h is at mos t cour. tab L6 .

S O L U T I O �

The necess ary condi t ion fol l ows from Exercis e 0 . Convers e ly . suppo s e l un l n - 1 . 2 • • • • be a max imal ortho

normal system in H whi ch is finite o r countab l e . Let !P

S - \ '" r U ' e Q ( e Q + l', Q ) '\' l &1 n n • r n or r n

then S is countab l e and dense in H • hence H i s separab l e . Q . E . D .

75

E X E R CISE 5 . - If M - ! z : Lz - 0 I whszo.t L is a continuous Li near ��notionaL on H • prous that /ofl is a vector space of dimension 1 (Un1see /of - H. i . B L - O J .

S O L U T I O �

L is continuous hence 101 - L-1 U O \ ) . is c losed . L ;. O . hence 101 ;. H • ' Thus 3 xo ,. 101 with L (x . ) ;. O. x . -y. + z . .... i th y. e lol l and z . e M . hence L (y. ) - L (x. ) . ;' o .

I f y e lol l , 3 � e C wi th Ly - >" Ly. ; hence L (y- >..y. ) -O j

hence y- � y. e 101 n loll � 0 • hence y - � y • •

Thus \ y. 1 i s a basis of lol l Q . E . D .

EXERCISE Ia . - rst I u I n - 1 . 2 • • • • bs a n orthonorma L s e t in H , n Sh'ow that this gives an sZaMp Le of a c Losed and bounded

s s t whi ch is no t compact . Lst Q bs ths sst of a l L Z e H of the form

Prove that Q is

More gBnBraZZIl.

ar.d Zst S b/l the se t ,of

Pro ve that S is

Prove that 'H i s

00

• whsr/l z - "' c u � n n n = l compact (Q ';'II ca l led the Hi Lbsrt cube )

Z s t

a H

Z -

l c\n I bs

Z e H of 00

a IIsquence

ths form

w;,ers L c u n n 1 compact if and onLy if

no t local ly COMpact.

of pOllitiv8 numberll

I c I � c\ • 00 n n L '

2 6n < 00 1

76

SOLU T i oN

We use the fol lowing theorem : A metric space K is compact iff every s equence in K has a

conve rgent subsequence Since I u - u n m • V2 (Pythagoras' theor�m) if n "f m and

is clearly closed and • bounded , but contains no conve rgent subsequence , hence i s not

compact . Now B' (0 , 1 ) contains the sequence I un l which has no convergent subsequence . This shows that B ' CO , l ) is not compact , hence a has no compact" neighborhood . Thus H i s not locally compact

No� let S be as in the exerci s e . Suppose L b� < 00 , and let (Xk

sequence in S .

We define by induction on m the integers N 1 , N 2 , • • • • • and ti'e s equences (xk , l ) ' (xk , 2 ) ' • • • as fo llows

00 ( i ) L b� < 1.2 , N 1 < N2 < . . . ( L b� < 00 ) •

n=Nm m (ii ) (xk , l ) is a subsequence of (xk ) , and (xk ,m1-1 ) a

subsequen ce of (x ) , m • 1 , 2 , • • • , �uch that , N k ,m

theorem

m (i i i ) � 1 c - c 1 2 k.. n , (k , lI ) n , ( k ' ,m ) n= l

< 1 -Z m

for all k , k '

By ( i ) and ( i i i ) , I x - x 1 2 k ,m k ' ," < Z

-2 for al l k , k ' m

Hence if jr • x� • then (y ) is a convergent s equence . m . ,. ... m m By (ii) , (Ym ) is a sub sequence of (xk ) . I f y __ Y • � c u as BI -- 00 , then m � n n (Ym 'un ) • Ym , n -- Yo • (Y ,un ) , hence I Ym , n l � cn Thus y e S . This show; that S i s compact . Now suppose � b� . 00 • Thus there are integers

�� • I < N2 < . . . . such that Nm+ 1" L b� > 1

n = H ., + l • c 1u1+ c 2u 2 + · . · · · + C H.,UHm ' then by Pythagoras' is an orthonol'1llal sys tem ) , for m < m' we have .

I x - X 1 1 2 111 m

77

Nm I 1I� '" i n =Nm+l This show.s that ex ' ) contains no converg ent m subsequence

I t i s clear that x m E S , hence S is not We conc lude that S is compact iff

compact . 00

L >� < -. 1

Q . E . D .

EX E R C I :i: E 7 . - Supposs I a I is ' a s equsnce o f posi. ti.vs nwnbu's n 2 such that " a b < 00 whensver b � 0 and " b < 00 • L. n n n L. n

Provs tha t I a� < 00 •

More genera l ly , suppose I an I i s a s equence of comp lez

numbers such that " a b converges whenlflvlflr " I b I P < _ L. n n L. n

SO L U T I O '"

Pr07Je tha't " I a I q < 00 • Bere 1 < p < 00, q - -Ll L. n p-

For further appl ications later , we s tate the fol lowing

PROPOS ITION . If l a1 , a2 , , , . numbers , then

is a sequence of comp lex

II • sup l l � anbn l : b l . " , bN N

E ct and there exis t s a sequ:nce (bn ) n= l , 2.;," . ' , .� I bn I � • 1 such that

( J:1 I an l q )t • �lanbn ( 1 < p < 00 , q. �) ,

As a consequence of the case p • q • Z , one has : In a Hi lbert space , I x l · sup j l (x ,y ) l : I y l · I ! .

PROOF OF THE PROPOSITION . The right s ide of the identity i s denoted b y A .

lIy the HClldt'!r A.

Obs erve that Therefore it

aN+1 • aN + 2 • • • • • • • • • o .

78

Put C • q q q . l al l + l a2 ! + . • • • . . + l aN ' • ��d suppose C > o �

( * ) b • a i f an • D . bn • C , l an l q- 2 a if a n N n n Since p • q\ • i t is clear that I lL I P • I •

1 n

A direct computat i on shows that I an bn • C*

Hence N q L l ao l � A . 1

This shows the first assertion of the lemma •

Note that i f

.,. O·

then we can apply ( * ) t o find a sequence (bn) such tnat

Let us prove the second as sertion of the exe rcise and the remaining part of the · pro,')sit i on .

Suppose (an) is a sequence of comp lex numbers such that

We shall find a s equence (bn ) such that Vn • and Ianbo dIverges •

Since such that

(' ) Ck

Nk + 1 � n � Nk+ 1 • k • 1 . 2 • • • • •

We verify eas i ly that "k+ 1 " Ib I P

n"�+ l 0

Thus I l bn l P n: 1

k • 1 . 2 • • _

a if an • O . and .,. O. and

diverges as desired. Th is completes the proofs of the propos i tion and .the exercise .

Q . E . D .

79 E X ERCISE 6 . - If H1 a�d 82 arB WO Hi lbert spac88. prOlJB tha t one o f thBm i s isomorphic t o a subspaca of tha o thBr.

SOLUTION

Let I u .. I .. E A and I v � I � E B be maximal orthonormal sets in HI and H2 respectively (See The orem 4. 18 ) .

Recal l that if A and B are nonempty sets , then there exi s ts either an inj ective map 9 : A - B or an inj ective map . : B --+ A .

(Zermel o ' s theorem , a consequence of the Hausdorff max imal i ty theorem)

Suppose the first' case occurs , define � by (I ) � (x) · ' L (x ,u .. ) v9 ( .. ) mE A I t is clear that � is l inear ,

by Bessel ' s inequal i ty . Therefore � is an 'is ome try of HI onto a subspace of H2

HI i s complete , hence � (HI) is closed . This shows that � is an isome tric isomorph ism which maps HI onto a closed subspace of H2 •

Q . E . D .

E X E R CISE , . - If A is a ma�surab �a subs B t of [ O. � � ] • prOlJB that

tim L cosn:r:d:t: " "' 00 A

- tim LSinn: d:t: - a n+oo A

SOLUTION

Le t Un (t ) • e i n t Q � t :;;; 27: maximal orthonormal bas is 'of L 2 (T) '.

f I cn l2 2 . I X A 12 , Therefore

n= -00 i cosnx dx 1 • !(cn + c_n ) , Thus £. cosnx dx and f", sinnx dx

, then

I cn l -)o Q

i s innx _ a as

-! u ! i s a n n=-oo

as l n l- lOG . But

dx • rt-Cc � c ) 1 n -n n - - .

Q . E . D .

80

EXER C I S E 10.- L s t n1 < n2 < . . . . bs posi tivs intsgers. and Zet E be the SB.t of a l l : e [0. 2X-] at. whioh I sin�k: I "onVergBB

Prdve tha ! m (E) - O .

SO LUTION

Define g (x) -lcti!, s innl? ' x e E . By [I] and the dominated convergence theorem , we hav� for every A C ·E,

g (x) dx - liin i s in n1ldx - 0 lc+oo . A

This shows that I g i - 0 a . e·. o n E . Since cos 2� - 1 - 2sin 2� , for x e E

cos 2nlcx _ 1 - 2g 2 (x ) as k _ 00 •

Similarly to the preceding part , we verify eas i ly that . . 1 1 - 2g 2 (x ) - 0 a . e . on E , hence I g l -\fZ a . e . on E

The identities g - 0 a.e and g . _1_ . On E soow that m(E) - 0 12

Q . E . D .

EXERCISE 11 . - Suppose E is a ve "tor spa"e o n C and (::. y ) ->- (::. y ) is an inner produot in E . Prove the identity

(In 4 (::. y ) - I: + y l2 _ !":: - y l2 + i I:: + iy I2_i 1::_iy l 2

If E ' is a "omp le:: ve "tor spa"e with inner produ"t ana

• : E - E ' is linear • provfJ that

( 1 . (:: ) 1 :: I:: 1 • '" e E ) is equi�a Zent to ( ( . (: ) • • (y ) )- (::. y ) ::. y e E )

Thus if H is a Hi lbert spa"e . (u",) � e A �s a ma::ima Z , orthonormaZ 8 � t in H • then the fo Z lowi�Gs ar� equiva Zent:

SOLUTION

1= 1= - )"' 1': ( 01 ) 1 2 (BuBe Z 's inequa Zity ) • ..... Ol g A

(:/:. y ) .. . L � ( .. )gr;:r Ol eA

Recal l for g iven y ,x -.:.. (x ,y) is linear, and that (x , y) - (y ,xT

Now we have

I x + y 1 2 - I x - Y 1 2 - (x + y ,x + y) - (x - y ,x - y)

- 2 (y , x ) + 2 (x , y ) - 4Rs (X ,y) 2 2

I x + iy l - I x � iy l - (x+ iy ,x + i y ) - (x - iy ,x - iy)

- 2 ( iy ,x ) + 2 (x , iy) - 2 i (y , x ) - 2 i (x , y )

- - Z i [ (x ,y ) - (x ,y ) ) - - Z i . Zilm (x �y )

- ·4Im (x ,y )

Thus : the right s ide of (I ) i s

4 (Rs (x ,y) + ilm (x , y ) ) - 4 (x . :')

By (I ) , I . (x ) l - I x l imp l i es ( _ (x) , _ (y ) - (x ,y )

The converse is evi dent: . 2 2

( _ (x ) , _ (x ) ) - (x ,x) hence 1 - (x ) 1 - I x l

81

To prove the las t assertion , l e t H' be the space cons i s ting

of all mappings s : A -+ C such that

s ( 00: ) � 0 for at most countably many 00: E A, and

L l s ( oo: ) 1 2 < 00 -EA

Us ing the preceding part with _ (x) (_) - s (or) - x (oo:) ,

we obtain iMmedi ately the las t assert ion .

tXERCISE 12 . - Define ak by

� . r ( �+ aost )kdt k - 1 . 2 • . . • • ak 0 2

Q . E . D .

ShoLl that . (* ) aO'nvezog s s a s k _. OCI . Hozoe pzoefJise Z !I .

SOLUT I ON

IJhow that

We have

O < • . k-1/2 &o1.m ck k ... OCI

l+cost -r-' -

2 cos u

< 00

u - f . Hence

i� cos 2kudu - Z Ik o

Integrating by parts , we cbtain

82

I _ 2k+ 2 . I Ie 2k+ l 1e+1

WI! have

(»

h··nce

Th.us •

w; th !of

�+ 1 -- -'Ie

Hence , for a

l ie �+1 -c

• Js.... Vk

Zk+Z k+1 -- - � Zk+ l k+1

Thus ale decreases as 11; -->' 00 , hence ale converges . By (*) we have , !:ince v'T+'i � 1 + ! . a > a

i . [ 1 + 1 ] � < 1 + 1 < 1 + 1 \+1 4 (� +k) 8 (k2 +k) 8)(2

1 < ale < 1 + 1 ale+1 8k2

s ince Log (l+a ) � a • a � 0

(I ) a < Logale - Loga k+! < 1 8k2

We deduce from (' ) that

o < Logal - Logale+! < HI + 1 + • • • • • • • + ;2 ) < 22

00 < 00' s ince the series L 1

• ;2 converg es . 1

Thus Log\+1 � 1.0g\ - LogM • LOg (-i-) • hence

a 1e+ 1 > � zr for k - 1 , 2 , • • •

This shows that

Q . E . D .

LogM

E X E R CISE 1 3 . - Suppoa. I ia a oontinuou8 funotion on Rl . �i th p� riod i. Prou. that

( o J 1.im � � I fn.) _ Sol f ( t ) dt

H +- oo 2.. n = 1

83

for . u . ry irra tionaL number � . Hint Do it fir�t for

f (t ) � .=p (2 � i k t )

k - O. t l . t2

SO LUTIO/ol

First cons ider the case f (t) • exp (Zlt ikt )

k • 0 , t l , t 2 , • • • • • •

(*) holds evidently if k - 0 ( in that case f - 1 ) .

If k � 0 , s ince 01 is irrational , exp (Zlt ik cr ) is � 0 . , So Ii II , � £;1 exp (Zlt ikn .. ) - � 6;J exp (ZX ik:. )]n

He!lce II

I N! " I �1 exp (Zlt ikn'cr ) en : ikllo. _ 1

e2lt ik cr _ 1

1 l e2lt ikll CII - 1 1 Z 2 1t ' k ..;; 2 ' k N I e '� cr _ 1 1 N I e 1' 1 ,or - 1 1

Thi5 implies

l im 11+ 00 II � L,exp (Zltikn .. ) - 0 -

, n = 1 It follows easily that

H

1 f e2T. ikt dt

o (k-O , t1 , • • • )

i f P (t) " - L �eXl) (ZXikt) , then (*) holds with P instead of f . k --H

Now if f is continuous with period 1 , then (Theorem 4 . Z5) H

Y t > 0 , there is a P L ckexp (Zlt ikt ) such that I f - p l,...,< t - II

Thus , with g I II

I! Lf (n cr ) -

N 1

- f - P , we have

[ If (t)�t l ..;; I � � g (n cr )

, II

+ I .� � P (n .. ) 1 - fa gdt l 1 - i pdx l

o The firs t term at the right s ide of the inequal ity is ";; Z t

since I g l oo ";; t Therefore

1 1 II 1 l im sup - L f (n cr ) - f. f (t ) dt l :s; Z&

hence 11 + 00 N 1 01 II i f (t)dt l l im ' I! L f en cr ) - 0 11+ 00 N 1 0 Q .E.D.

84

EXERCISE l � . - Supp�.a E i. a (eemp Z az ) v.e.er apae. with inn a r

produ a t ("" �) ' and"

u1, : . . , u

n I i. an orthon01'7fla L •• t in E. If

II i. th. Zinaar BubBpae. of E ganaratlld ·by . 1 u l' u2' . . · , un I ' and

� Z E � � .

Y - f.:l (z, uJ

) uJ

prOtlB that, (z- y ) e III an d

and . how that

SO Lurr OIi I t i s clear that (7 ,u j l • (x ,U j ) (U j ,U j ) • (y ,U j ) ' henct'

hence (x - y,u

j) • 0 j . l , • • • ,n

(x - y, z · ) • 0

. If z e M, then

z , E M

Bu t x - z • (x - y) + ( y - z ) • (x - y) . , z ,

(x - y , z ' ) . 0 , s o b y Py�agoras ' theorem

I x _ : 12

• I x _ y l2

+ I y _ z l2

"> Ix _ y l2

This proves the first as s er t ion of the exerc i s e .

Now suppose U E M l (that i s (z ,u) • 0 f o r z e M) . We h��e

(y ,u) • 0 (since y E M) , hence

Thus

But fO"l

I (x - y ,u ) l · l (x , u) 1 � Ix - y l , u l

I x - y I ....

U •

(by S chwarz '

sup l l (x . u ) l : l u i'

· 1 , u Ii ", 1

ex .Y) (x e M) , we have I x -y l

I x - y l • (x -y , u ) • (x ,u )

inequal i ty)

This proves the reve rs e inequal i ty in the case x e M .

EXERCISE 1 5 . -

an d find

Q . E . D .

"'=

S" � g (:r: )d;!; . 5 -1 -1 51 i g (:r: ) \ 2a:z: . 1 -1

SOLUTION

1 f :t?g (:r: )d;!; . 0; -1

Let E - C[ -1 , 1 ] , and define on E the inner product 1

(f ,g) - S f (x )iTxTdx - 1

Let . M be the subspace of E spanned by the functions ! 1 , x , x2 \

The Gram-Schmitt 0·rthon01"lllali zation described in 0 nOli .applying to I I , x • x2 \ give s us

ul (x) - (+) ' ; u2 (x) - (+) ' X ; u3 (x) - iv1(x2- }) . (To s impl ify the computations , Iwn ' is found by using

85

Pythagoras ' theorem . See [TIl and the notations 1

in ill . If we observe that f (x3 _bx) (a+cx2 ) dx • 0

- 1

M can be replaced by the l inear span of x • and we need not the expressions of ul ,u3 )

hence

Let " (x) • x3 • With the same notations as in II] we have 1

• Z { �) 2 V6 (" 'ul ) • (x ,u3 ) • 0 ; (" 'u2 ) n .. • � Thus

., • �2 ' Ix _ >' 1 2 • I x l2 _ 1 .... 2

2 Z 6 8 I" - .... . ..,. - IT • I'7'r ( Pythagoras ' theorem)

By� we have 1 min f \x3 - (a+bx+cx2 ) \ 2dx • rh

a , b . c -1 and s imultaneously

max

where the restrict ions of the ex e rc i s e are expres sed as g e Ml ; I g I - 1 Q . E . D .

86

EXERCISE 11. . - So t lJ e [II] wHit

dz l'upeatilJ e t y .

lOO an d e -:: d% instead

o of f l and

-1

SOLUTION

Le t E be the set of all polynomials P : [ O , oo [ - C , and

(f ,g ) • Loo f (x)nxTe-xdx .

o

We pratice the Gram-Schmitt process to ! l ,x ,x2 I and then

the method in � , whi ch ", (x) • x3

• Rec a l l tha t

(M h

hence

f";n e-xdx • n l • 1 . 2 • • • (n -l ).n o

Now v1 (x ) • 1 I v 12 1 • 1 u1 • v1 v

2 '(x )

I w 1 2 2

• x I v f ? • 2 (U1 , v 2

)

• Iv 12 -2

(u v ) 2 • l ' 2

1 ;

• 1 ; w2 (x)

hence u2 (x )

V3 (x ) • x2 ; I V3 12 • 24 ; (v3 ,u1) · 2 ;

• X -

• X -

.. 6-2 '. 4 ; w3 (x) • x2 -2 -4 (x -I ) • x 2 -4x +2 ;

- IV3 1 2 - (V3 ,u1) 2 - (v3 , u2)2 • 4 ;

w x2 u3

.� : u3 (x) • r- - 2x + 1 ;

1

1

.Now for ,, (x) - x3

, we have

("" U3) • 1 8 and 1 � 12 • 6 1 • 20 . 62

tl1e linear subspace of. E consisting of all polynom ials 'tI f degree � 2

2 2 2 • ,,2 - [(X ,U1)2 + (x , U2) 2 + (x , U3) 2 ] I" - lI" l • 1 ,. 1 - I ll" I

2 I" - ... � • 36

Q . E . D .

EXERCISE 1 7 . - If :: 0' e B and M i s a " l o eed t ineal'. subspaae o f B.

pl'OlJe tltat

"'in ; I::� o l : :: e .M 1 - maz � ( (:: o. I( J I : I( e Il l ; 1 1( 1 - 1 t

SO L u r r O Ii

If x . e 101 , then the equal ity holds immediately with sides equal to O .

Suppose Xo ri M. Then X . - y. + z . , y. e 101 , z. e eM is closed) . Thus for x e .M ,

x. - x - Cy. - x) + z. , nence I xo - x l2 - Iy . - X l2 + h. 1 2 ;> h. 12 and Ixo - x l 2 taxes its minimum value at x - y • •

Thus (1) JII in ! l x - x . I : X e M ! - l zo l

Now for y e lolL , one has Cx . ,y) - Cy.+z . ,y) - Czo ,y)

Thus I Cx . ,y) 1 - I Cz. ,y) 1 � I Zo l by Schwarz ' inequal ity. But if y -J7;1 ' then (x ,y) - I zo.1 Thus

(2) I Zo l · max � l cx . ,Y) 1 : y e lol L and I y l - I t Cl) toge ther with (2 ) give the desired equal ity

Q . E .D.

87

the

lol L

EXERCISE l a . -· Suppo s e H is an infin i t e aimsns ional Hi l bert sp a�.

Sho� th�t there is a on.-to-on. mapping Y of[ O, l ) into H s u ah that

Y (b )- Y (a ) is orthogonal to Y(d)- Y ( a ) �hsnBl1sr O � a � b � a � d � l .

( Y mall be �a Z Z e d a "aurll./J �ith orthogon al inareatu n t " ) . Hin t : 2 Take H ' - L ([ 0, 1 ) ) , and �on/Jid/Jr r (t ) - % [O, t)

S O L ur r O Ii

Let H ' - L2 C [ O , I ] ) , then H ' is separable , hence has a countable maximal orthonormal set . If

then r et ) - X [ O , t ) O � t � l

C* ) C r Cd) - r cc) , rCb ) - r Ca» - t X [ c , dPX[!l ,b) - 0 whenever 0 �

'a � b � ·c � d � 1 .

Now suppose H i s infinite dimens ional . Then H has a maximal orthonormal set Cv � ) , e B. isometric isomorphism

, B infinite. Thus there exists an

Since . : H ' - Ii

C .Cf) , .(j» • Cf , g ) CSee [!] ) . Define Y Ct) • • : C r Ct) )

f , I E L2 ( [ O , I ] ) it is clear f�om C* ) that Y has the requ ired property .

Q •. E . D .

sa

EXE RCISE 1 ' . - Le t E bs an unitary space , an d H thB comp Ze t i on of E. (E:ee rci u rz2l ah. J ) . If (% ) and (II ) are s e q uBnces in E which con ve rgE � n n to % an d II rs.ps ctivs ly, dil fine

choi �s

maks. H

( 0 ) (:e, y) - lim (:en , lIn ) n"'OO

(a) Show that t hi . lim i t B:ei s t s , (%, y ) is inde p e n ds n t of th e

01 (:e , II ) ant: that ( 0 ) ds fine s an inne r p ro du c t in H, which n n 2 i n t o a Hi lbe rt .p ace ( . h ow that d (%, O ) - (%, %) , :e E H ) .

(b) If (ei) " e I i. a ma:cima l o rthonorma l s e t in E, show

that this i. a l. o a ma:cima l ortho.norma l B6t in H.

(a) Show that the Gram-Schmidt o rthonorma UlIi on can be ap

p He d t o E ( i f E is sep arab le, of aourss ) .

SO LUTI ON

(a) (x ,y) - (x ' ,y' ) • (x-x ' ,y) + (x ' ,y-y ' )

By S chwar� ' inequality

I (x ,y) - (x ' ,y ' ),1 � I x-x ' i l y l + l x ' l I y-y ' I (x ,X' ,y ,y ' e E) .

I t foll ows that if (xn ) , (Yn ) are Cauchy sequences in E

then ( s in ce (hn ll • (d(xn ,O ) ) are b ounde d } if an • (xn 'Yn ) ' ( an ) i s

a C auchy s e quence in C, hence converges to a limit .

I f lim xn • x • lim x� , lim Yn • l im y� • y , then

11m x� • x and lim y� • y where

(x�) · l x1 ,xi ,x2 ,x2 , . . · ! , (Y�) . ! Yl 'yi 'Y2 ' Y2 ' ' ' ' ! The pre ceding p art shows that

lim (xn 'Yn ) • lim (x� ,y�) s in ce both are lim (x� ,y�) . Thus (x ,y) is wel l-de fine d on H , and i t s re s triction o n E is the g iven inner product (take xn • x, Yn • y (n • 1 , 2 , . . . ) )

In p articu l a r , if I I is the re sulting "n om" , then

(X ,X) • l im (xn 'xn ) • lim [ d (xn , O ) ] 2 ". [ d (x , O ) ] 2

s o that (x ,x) • 0 only if x '" O . The remaining requi rements o f an" inner product for ( , ) fol low "without diffi culti e s .

Note that i n the pre ce ding p roof, w e have

d (x ,y) • lim (x -y x -y ) 1 /2 • (x-y x_y) 1/2 • I x-y l o n ' n · o '

89 hence the me tri c of H is exact ly that given by the inner product ( * ) By de finition o f a comp let i on space , H is complete . Thus H is a Hi lbe rt space .

(b) E is dense in H , hence f • H , hence E l • 1 0 1

I f M i s dense in E , then M i s also dens e in H (proof ( In ) C; E , lim In • x E H; ch oose xn E M , I Xn - zn I � �. Then

lim xn • x) Here we s ay that B • (e i) i E li s a maximal orthonormal set

in E i ff the sub space M sp'anned by B is dense in E . This defin i t i on is compatib le with The orem 4 . 1 8 in the case o f Hi lbert space s . The pre ceding part and Theorem 4 . 1 8 s how that if B is a maximal orthonormal set in E , then B i s als o a maximal orthonormal set in H .

(c) W e could ex�ine the Gram-Schmidt orthonormalizati on in 0 , and we see that step (b) invo lves on ly inner product.

Q . E. D.

E X E R CI S E 2 0 . - Define 14 (tJ - ei s t for a U s E Rl

, t e Rl

. Le t X b e s

t he eomp Zez lIe ctol' sp aee eon s i s ting of a Z Z fini te tine ·zr eombi n ati ons

of the s e fun c t i ons ws ' If f E X and g E X, s h ow'that

A (f, g ) - tim J f ( t J g (tJ dt

A"'ao -.4

ezi s ts . Show that this inn e r p ro duot mak e s X i n t o a wni tary ( ')e p aee

WhOllf: eomp Ze ti on i.s a n on eep arab Z4 Ri Zbert space H. Show a l s o that ! Ws : s E R I i s'

a ma.:ima Z orthon ormaZ s e t in H.

SOLUTIO/ol

I t follows that

• 2A

cos A(s 1 - s2 )

5 1 - s2

( � Unl ear� sp a ce - inner p r o d u ct sp a ce - p �ehi lbert sp ace , See D e E 4 . 1 .

90

It fo l l ows e as i ly that the limit defining ( f , g ) ex i s ts , and that thi s inner p rodu ct makes X into an unit ary space .

By Exe rcise � , X i s dense in a Hi lbe rt space H , and ! U s : 5 E R I . be ing a ma:"(imal orthonormal set in X, is also a m ax i

m a l orthonormal set in H. H con t ains an uncountable max imal orthonormal set , hence H is not separ3hle (Exe rcise 111 ) .

* * *

Q. E. D.

92

EXE RCI SE 1. . - If 1 < p < 00 , pralle that the unit ba l l of LP ( 11- ) i s

s t rictly conve:; t h i s means t h a t i f

I f a - I g a - I • f ,J g • h - L2 ( f +g ) . p . p

then I h O < 1 (G .. oms tri ca l ly, the su rface of the ba l l con tains n o

s t raighl Z i n s s /, Sho!.) that this faHs i n e very L1 ( 11- ) . in e ve ry

L oo ( ", ) , and in e v e ry C(X) (Ignore tri v i a l i t i e s , s u ch as spaces

con si &ting o f only one p o in t ) .

SO L U T I O N

Suppose 1 < p < 00 , I f Ip - Ig Ip - U }(f+g ) l p - 1 .

Hence I f + fl - If. + Itl .. ,By [TI} of chap ter I I I ,

f - " g , " � O . I f I - I g l imp l i e s that f - g . This proves the first assertion

S ince C (or R1) i s s tr i c tly convex , in the sequel we suppose

the spaces L1 (",) , LoO(�) and C(X) are a t least two dimen s ional .

Consider L1 (1I-) . We can find d i s j o in t measurab l e sets A an d B

wi th �(A) an d II-(B) f in i te and > 0 . I f

I" l B 1 f - , g . - and h - -r(f + g ) , "' (A) ", (B)

it is evident that I � 11 - Ig 11 - I h 11 - 1 and f ; g This shows that L1 (�) is not strtctly convex.

Nex t , con s i der Loo(�) . There is a measurabl e set E such that

",(E) > 0 , and ", (Ec') > o . Therefore , if 1 , g - 1 , h - -r(f+g )

then I f 000 • n g 0 oo·� 1 and n h n • 1 , f ., g •

Final l y , suppose X i s comp a ct , f E C(X) i s non const an t .

Therefore I f 1 �akes a value b < D f n Let

A 2 \ x . : I f (x ) I � inf m + } b

B • I x : I f (x ) I � } Hf I + } b

00

Then A and B are nonemp ty , closed , d i s j o int sets in X . By Urysohn ' s l emma , there i s a con t inu �

<fI : X - [ 0 , 1 ] <fI IA E I ,

I f g • <fI f , then .g ., f , and

U fa · h i · D t( f+g ) ' Thus C(X) i s not s trictly convex

Q • . E .D . - � .

93

EXERCISE 2 . - L e t C be the space of a � L aon t inuous fun ctions on [ O, 1 1

wi t h the sup n o rm . Le t M con s i s t of a n f E. C for whi ch

L 1 f2 f { t J dt _ r f { t J dt � 1

a JL 2 Prove t ha t M is a c L o s e d c o n v e% subse t of C which con ta in s

no e Leme n t of m i n i m a L norm .

S O L UT I O N

I t is c l e a r t h a t M i s c l o s e d a n d convex . Con s ider the l as t assertion of the

Let gn (t )

If f n

exercise " 1 / ( 2 n + 1 ) . (1 - 2t)

g �l , i t i s c l e ar

1

, n • 1 , 2 • . • • , t E

that f E M , s ince n

hnl 1 - fo�g (t ) dt - l\ (t ) dt . 1 .

Now s ince I gn 1 00 - 1 and

Therefore

"2 hn 11 t I , we have

I f I � 1 as n --+ 00 • n 00

[ 0 , 1 ]

Suppose f E M with l E i 00 � 1 (This wi l l lead to a contradi c t io n ) . Thu s Re f e M , ..I Re f A co � I, so w e m ay suppose f i s real-valued

Since I E U l � 1 0 co ' the inequ a l i ty 1 l 1 . i If l dt � J.2f - L f - 1

a a 2 Ulpl ies U 100 - l f l1 - I, hence I f (t ) 1 - I fC1I' t e [ O , l J . f is

rea l . valued , hence f ( t) • ·1 o r - 1 . The cont inu ity of f imp l i e s

ei ther f . 1 or f ,; - I , hence f It M . T h i s contrad ict ion shows that

If f E M , then I f l co > 1

Thus m in I f I - 1 , and there i s no el e.ent i� M wb i ck f E H DO

ha s .in i.al norm .

REP1A R K . - R e ". l l tll .. t e ve ry "l oc e d "on v. " c . t i ll • Jli lbe r t cpa �e II .. " • IIn i q \le e l e IWen t o f .i n i .... l n o r • • 0 .n d 0 c ll o w t ll a t c·hi " f .. i l "

i n t h . ". " . of " OMe B .. n a ch c p a ". c .

E X E R CI SE 3 . - Le t /It b e the ,e t 0 1 a L L 1 E L1 ( [ 0, 1 ] ) , re l a ti v e to

Lebesgue ,.ea,ure, s u e h that

1 f f ( t ) d t - 1

o ih�w that /It i s a c L o ,e d �on�e� , e t Of L1 ( [ 0, 1 ) ) which con tain.

inlin i te L¥ ,.an¥ e L ement. of ,.ini,. a L norm.

SOLu r r o �

It is cl ear that f E M i.pl i�s 1 ;>. f f (t ) d t - I, and any f ;>. 0 in M has li O n! 1 .

o

Hence M contains infin i te l y .any elements of .ini.a l · lIorm .

(Note that in W and m , M i s the s e t I x : Lx - 1 \ , where L i s a

su itable continuous l inear funct ional , hence M i s closed and convex ) Q . E . D .

EXE � C I IE � . - Le t f be a bounded L i n .ar lu" c tion a L on a sub space of a Hi L b e rt space H. Prove that f has a uniqu. norm-preserving lz t.n

, iC79t tg Ii bounde d L in . a r funct ionaL on H • and that this ex te!l s ion v an i s / .. . on /Il l.

S O L U T I O N . (We do not use the Hahn-Banach theoreM ) . Suppose f : 104 - C is bounded . Iience ·f . is uniformly conti

nuous , hence f extends uniquely by continuity to Mi. Thus · we can suppose that 104 is closed , se that 104 is a Hilbert space itsel f .

Therefore , there exists a unique x e 104 such that f ey) • (y ,x) , y E M ; and I f l · I x l .

Define F by F (y) • (y ,x ) y e H

then F I M · f , F I M .L · 0 , and I F I • I x l • I f l .

Now. suppose Fl is a norm-pre serving extens ion of f •

Then Fl (y) • (� ,xl ) y e Ii Since Fl l M • f, we have

Y E 104

hence x - xl e .104 1 TJ:.�s?- by Pythagoras ' theorem ,

Therefore ,

iaply x - xl ·. 0 , hence '1 - F •

Q . E . D.

EXERCISE 5 . - Const�ct a boundsd linsar funotionaL on some subspace of soms L1 (/I-) which has two (huc. infinitaZy many ) distint .nOl'lll-pl'e

· • • rui�i Z inllar sz tsnsions tO L� (/I-) . S O L U T I O N

Let X - I a , b ! and. define /I-( \ a ! ) - /l-C \\\ ) • I , /1-(0) • a ili'd . .

", eX) - Z . Then Ll (}L) _ C2 , and I (x , y ) l l - I x I + I y t •

If f : Cx l o l -'C is defined by f (x , O ) - x , then f has "wo nOTlll-pre serving extensions ( Recal l that Ll (/I-) * - L CoO (/I-) )

fl (x , y) • x

Q. E . D .

tXERCISE b ' - L il t X be a norMlld � in ll ar space (MLS) , and L s t X ' b e ; t s du aL space, that i s ths u t of aU boun dlld Hnear funotiona Z " on I

95

9 6 with the norm U t i - sup I l f (z ) 1 : Iz U � ] I

(a) ProlJe that X" is a Banach spacs

(b) ProlJe that the mapping f -+ f (z ) . Ez (f} i s, for each

x E X, a boundsd Linear fun ctionaL on X ', of norm 11 .. 11 . (Tld'e givee the naturaZ embedding x -+ Ex' of X in ita " secon d dua Z "

X " , the dua L space of X ' )

(a) Prove that l l'::n l I i s bounded iff I f (zn ) \ is boundsd for

every f E X '

SO LunON

(a) Suppo s e ! fn I i s a Cauchy s e qu en ce in X· , that is

( * ) V t> O , 3 N : m and n > N imply U fn - fm I < t

Thi s s hows that for x E X , ! fn (x ) I is a Cauchy s e quen ce .

Thu s fn (x ) converg e s to f (x ) as n - 00 .

I t i s easily verified from ( * ) that f i s l inear.Mo reover

( I ) I fn (x) - f (xn • l im I fn (x ) - fm (x) l <t l x !if n > N -m-+- oo

Thu s B f - f i ->- 0 as n -+ 00 (Take I x l � 1 i n (* ) ) , hen ce n I fn I �onverg e s to f in X*.

REMARK In t he s a m e ma nner as { a } , we can prove tha t L (X / Y ) _

the spa ce of a l l bounded l i nea r op e r a t o r f. x .- y .d t h n o rm

I f I - s u p t I f (z ) 1 , l x l � l I 1s • Ba n a ch space i f Y i s SOlD.e Ban a ch spa ce. (The conv ers e is a ls o

t r u e , t h a t i s , i f X � 0 i s soie NLS a n d L (X / Y ) is comp l e te , then

r 1s compl e te } .

that

hen ce

(b ) It i s clear from the defin i t ion of " f tl , for f E X* , IE (f) I · I f (x ) I � l f l lx n x .

I E D �l x l x (Eviden t l y , E x i s l inear , this j us tifies the notat ion I E xH ) By t he Hahn Banach theorem , if x E X then there ex i s ts

an .f e X* su ch tha t I f I • 1 , an d f (x ) - I x l . Thu s I Ex" �lx l , hen ce

IE I - I x , x Now x _ E i s clearly a l inear map , hence this is an isom-x

me t ry o f X into X**

(c ) If 1 xII I i s a bounded sequence , then 1 £ (xlI ) I is bounded in C , s ince

1 f (x� ) 1 � 1 f I lxn I (f e X* )

9 7

Convers e l y , supp ose I X n l i s a s equen ce i n X su ch that ! f (xn ) 1 i s bounded in C, for every f e X* .

Therefore I Ex (f) 1 is boun ded for ev ery f e X* n Since X* i s a Banach space (by ( a) ) i t fo llows from the

Banach-Ste inhau s s theorem tha t I I Ex I I is bounded. By (b) n

l Ex I • I x I , hence ! I Xnl I is boun ded.

Thi s compl e te s the proof of ( c ) . Q. E . D.

EXERCISE 7 . - L4It 0 0 , Zl , and Z "" b41 thll Banaoh sp acBs oonsis ting of a Z Z oomp ZBz sBqllllnolls Z - l � I ' i - 1 , 2, • • • , d4lfin4ld as fo Z Zows

• Z e Zl if an d onZy if 1.: 11 -2: I li l < 00 z e Zoo if and onZy if I z i - sllp l l . I < ""

00 -i t-Co is thll subsp aoB of ZOO oon s i s ting of a 1 .1. Z e Z"" for

lJhioh li -0 as i - 00 . PrOVIl that C! : _ '/.1 an d that (Z l ) A Z oo , using, the

no tati on of 4Izll roiSB [!) bu t thcH ( Z "" ) 0 � Zl . PrOVB t ha t c! o and Z l are sBparabZB b u t that Z OO i s not . (ThB s tatement " ,c � - Zl

If m e an s , m·o re ezp Z i ci'tZy, t hat to Bvery bo unded Zinear fun ctionaZ A on 0 0 t hBrB �o rre sponds a uniquB s e quBnce I l. . I such t hat '10

"1 :\. . 1 -I A I an d A z �" l . X . for a 1 .1. ::: E C! o £... '10 £... '10 '10 ' 1 and con v e rs e Z y . That (Z oo ) O � Z depends on the fa ct t hat there

is a nontriviaZ boun ded linear fun ctionaZ ·on Z "" which v an i shes on a1.1. of c o) .

SO LUTYON

L e t en be the s equence (l � ) with

In • 1 , and ' m • 0 if m f n

Z l I t is clear th;;,

in c. o r

(*) l im I x - 2.,; � l<el< I • 0 , if X • 1/ ... 00 1 k = l (a) c·: • L • L e t A E c: , and X II •

( l.i) ), (e n) . By (* )

(1). •

I f xII e 1;. is the sequen ce (t k ) defined by

t k ),k if 1 � k � N and ), k I- 0 . �

l . o otherwise k then . I XlI l � l N • 1 , Z , . . . . . , and

N - l .. Z , • • • •

! A I

I t is clear that I 2}1l tu I � L l'Ao I SWP l to· 1 , hence

00 0

(3 ) I J\ I �L I �o l 0 = 1 The re fo re , II. ...... (�o ) . (�eo ) i s an isome t ry of cf

into 1.1 • and this map i s eviqently l in e a r . But i f (�o ) e 1. 1

the re corre s p on ds an A e c:t defined by (1 ) .

Thus II. _ (;\'0 ) i s an i some tric i somo rp hi sm of c: onto

l l . Thi s i somorphism does n o t depend on any cho ice (that 1'5 , depe

'n den t only on the defin i t i on of Co an d Z l) . 50 that we can wri t e

c ' . Z l

(b Y ( z1) * • 1.00

There c orre sponds to e a ch II. e ( t 1 ) * a sequen ce (;\'0 ) of

1.."'" define d by

then

(I ' ) • x

Let II. be a fixed e lemen t of ( 1.1) * , an d le t

Xli . [ � N J. ;\. lI II if� I- 0

XII . o o the rwise

I x II 11 � l eN

I • 1 , N • 1 ,2 , . • • , an d

N • 1 , Z , • • •. •

Thus l;\. lI l � 1 11. 1 N · 1 , Z , • • • • • , hence (Z ' ) sup l ;\.lI l � I II. I

II

I t i s c l e ar that I L A n �n I .:(; sup I;\.n

I L I �n I , hence

n (3 I ) I A I � sup

n l >-n I Thus , by (2) I arid (3 ) I , A · .......... (An) i s · an iso.e t ry

9 9

of

Z 1 * into Z OO<l . I t i s cl e ar that thi s .ap is l inear and surj e ct ive

and s imi l arly t · the . prece dini p art , we can wri te Z l*= Z OO

( c ) c. i s a prop er closed subsp ace o f Z OO • Thus the

Hahn - Banach the e , • • l iv e s Ci.eO ). � cr , that is

· the re exists a

(nontriv i a l ) l in e ar fUl\ ctional A on Z oo , It. � 0 , and A x • 0 for all x IE c • •

In terms e f the i dentity c , • t l , we cal\ wr ite

(t - ) * � Z 1

(d) I t i s . clear from (* ) that c. and Zl are sep arable .

N ow c on s ider � .., • It is knOlm that lP(N) i s un coul\tab l e

( s e e [?J chapter I ) . I f A i s a sub s e t o f N , define xA

t o be the

s equ ence ( l i ) wi th li . 0 i f i tt A li . I i f i E A

(that is , xA

is the 'characteri s t i c fun ction of A ) .

I t i s clear that

if A � A I , then I x A

- x A I 100" I

This s hows that if (Yi ) i E I i s a den s e sub s e t of Z ";' ,

then we can define an in j e ctive lII ap . : lP(N) -+ I su ch that

(by u s in, the axiom o f cho i ce ) .

hence I i s un countable , Therefore ZOO i s n o t separab l e .

�t"A�K . I t i • • imple to verif� t h e f o llo wing asser t i o ns :

(-) If O'n �n } E Z I for everl) ( l ) E c . ' t h e n (l..n ) E Z I

1 n 1 Z-. ( , ) If ().n �n

) E Z for everl) ( In

) E Z , then {An } e:

[XUCI!t I - If f i. a comp Ze", fUlI ctio1t defill _ d 011 tire comp act in � _ rv a i I - [ al b) c: Rl , d�fill e

1 00

� l /fJ - sup l iN: ) - 1 (11 ) I : I: - y I � b , : e I, 11 e I ! and lor 0 < 01 � 1 , deline

H. (I) - sup I Cilb (l)a - . : b > 0 ! and t he corresponding Lip schitll spaoe Lip oc .co n si s ts of a l l f fo r whi c h H .. ,ff) is fin i te . Define .

Zip '" - I f e Lip 0< : Lim a - OC Ol a tf) - 0 I a.O

Prove that Lip • is a Ban ach space , if I f D - I ! (a ) I + Hoc (f) , aLso if I f I - I f l .... + Moc(f.} . Prove that Zip '" i s a oZo sed su bBpace of Lip • • (The membe rs of Lip oc are said to satisfy a Lip s chitz condition of

o rds r 01 ) .

S O L UT IO�

hence

Let f e L ip • • Then

(* ) I f ex) - f ey) I Ix - y l '"

This shows that

w�(f) � -.. - � M (f)

a

.. 8 f 1 .... � l f (a) 1 • (b- a ) Moc (f)

, x e I , y e a · I x - y I > 0

1 f ( a ) 1 • M ( f) an d B f D"", . M ( f) are e·quivalent norms on Lip ...

Now suppose ! fn I is a Cau chy sequence in Lip .. . We have

( I ) Y & > 0 , there is an N : m and n > N impl ie s I fn - fm I "" < t . an d M (fm - fn ) < t •

Thu s we find an f e C ( I ) su ch that

I fm - f I 00 < t i f m > N

Le t · gmn • fm - fn ' gm • fm - f , an d write ( * ) w i th gmn in stead of f , we have

and l e t t in g n --+ .... , we obtain

I gm (x ) - gm (y ) I� t(b - a) " · if m > N

the refore M ( gm ) --+ O as m --+ "",. Thu s

1 01

I fm - £ I . D fm - Hoo + M (gm) .-0 as m .- 00

We con.clude tha t Lip -0< is a Banach space if 1 £ • • I f 100 + M .. ef ) •

hence a l so if I f I • I f (a ) 1 + M (f ) .

N ow suppose (fn ) is a s e quen ce in l ip .. which converg e s

in Lip co to f . Then f o r a g iven I: > 0 .

Since fH e l ip ....

3r > 0 : 0 < a < r impl ies �a(fH ) < £ a ":

Thus • i f 0 < a < r • w e see that

� a(f) � wa(fH - f) + wa(fN ) � 2 £ a'" - .. This shows that w a(f ) l! - 0 as a _ 0 . hence f e l ip "' ,

hence 1 ip .. i s closed in L ip .. .

Q . E . D.

REM A R K . Suppo s e U i � � b o u n d e d s e t in a m e tri c spa ce , and F 1 s

a Ban a ch sp a ce . If f e CBF

( U ) ( f bounde d , con tinuous U .- P) define

an d then definfl t b fl . correspon din g I.i p .. F ( U } , l i p <:6F (U} . It is e ,! s !l

t o e x t en d t h e proof of the exerci s e , t o g e t t h e comp l e teness o f

Lip an d l i p in thi s c a s e .

E X E RCISE , . - Le t X be a triang Z e in the p Zane, Ze t H be the se t

con s i st ing of t he v e r t i ce s of X, an d Z e t A be t he' se t of aU re a Z

fun c t ions f o n X. o f the form

f (:e. 1I } - o< :e + � y + '! (o<, � an d Y re a Z )

Show that to e a ch (:e • • y . ) e X there corresp on�s a unique

measure ". on H such tha t

f (:e • • y . ) - ;; fd ".

Rep Z ace X by a square , Z e t H again be the s e t of i t s v e r t i ce s '"

and Z e t A be as abo v e . Show tha t to each p o i n t of X the re sti H

oo rre sponds a me asure on H. w i th the abo v e p roperty. bu t that

un iquen e s s is now Z o s t .

Can you e:e trapo Zate t o a mo re g e n e raZ theo rem ?

<;)

1 02

SO L UTION I t i s easy t o prove t ha t each p o in t of K i s a convex

comb ination of t he .vertices o f K,

one has

(' ) M - >-1 A + �2 B + �3 C �i > 0 ; �l + �2 + �3 - I

If f (M) • -.x + �y + y , .... i th M- (x ,y) , t hen

f (M) - A(1)I) + Y . A is a l inear fun ct i on a l on R2 Thus

f (1)1) - �l f (A) +' �2 f (B) + �3 f (C) , hen ce

(* ) f fM) · L fd� if fL{A) - �l ' II-(B) - �2 ' I'-(C) - � 3

Conversely i f . (*) holds . then for every linear fun ct i on a l A .

A ( M - ( �A + A2 B + A3 C) ) - 0

This sho .... s that M · �lA + �2B '+ A3C , hence I'- i s un ique'

s in ce the represen tation ( I ) is ' un ique . (Note that here I'- i s pos itive)

For the case of a square each point of K has infin itely.

repre sent at10n as a l in e ar comb inat ion of the vertices of K , hen ce

there corre sponds infin i t e l y many repre sent ing me asures a s s o ciated

to each p o in t M in K ( an d among them , t here are at least t .... o p o s i -

ti .. e measure s ) .

Here .... e assert s ome gene ral i z at ions of this resuf t .

DEFINITION ; K i s said to be a

' polyhedra of vertices Al , • • • , An + l if

n + l n+ l K - I M . 5' AiAi : � i > O , t: Ai- 1 1 .-+-m. � 1.=1

I f moreove r I A1A! • • • • A1An+1 1 i s indepedent , then K is

said to be a n - s implex . ( In that case the represen tation

M

ASSERT ION .

• �\.. A . £... l. l. i s un ique ) "

(a) I f K i s a po lyhe dra and H is t he set of a l l yertices

of K then to every M e K t here corresponds a me asure carri ed by

H such that

f (M) · Lfdll'

wheneve r f (M) - A (M) + Y , A i s a l inear funct iona.l ,and ye R

1 03 (b) The represen t in g measure � i s un ique i f and only if K

is a n - s impl ex . for some n .

E X E R CI S E 1 0 . - Le t X b e a l o ca Z ly comp act o r a . comp l e te me tri c space (X " 0) . an d le t ! fn l b. a � equen ce of con tin uoue comp le", fun ctions

on X. s u ch th at f ("') - Zilll fn .("') e",i s ts (as a cOlllp le", n umber) for e ve 11l '" E X •

. (a) ProvfI th at there is an op en s e t V " 0 an d a n umber H < <>o s u ch that If ("') I < H for 'a n '" E V and fo r . n - 1.2 • • • • n

(b ) If � > o . p rove that ' there i s an open . se t V " 0 an d an i n t e g e r R such that 1 [ (",) - f ("') I ";; Eif '" E V an d n � N •

. 1'

SOL UT I ON

(a) Fo!"" ):: . n · 1 . 2 , . . . . . put

Ak ,n · I x E X : ! f ill (x) l ,,;; k i f m � it ! . The con tiunui ty of f . f 1 • • • • shows that Ak is the intersection n n+ - , n of closed sets . hence Ak i s closed. For x E X the conv� rgence o f ,n (fn (x) ) shows that (fn (x) ) is a b ounded set . hence x lies in some Ak ,n ' It f o l l ows that

<>0 00

X - U U A n=l k = l k ,n

i s a un i on of c ount ab ly many closed sets . Baire ' s theorem shows that s ome A . h as. n onempty inte ri.or.

· ko , no I f X is l oc a l ly comp act . l e t V be a nonempty open s e t in

Ak with V comp act. l f1 1 • • • • • ! f I i s b oun de d by A on V. hence by o ,no n o de fin i t i on of � ,n

X E V . n • 1 . 2 . . . .

I f X i s a comp le te me tri c space . an d - for fixed n .

X · 0 "I' x E X : I f (x) I ,,;; k ! • U Bk k=l n hence every open s e t w in X meets . ( the int eri or of) s ome Bk• This app lies to Ak an d f . . . . . f • and sh ows th at. the re is a ·n on-a ,n o 1 11 0 empty open set V in � such that f • • • • • f i s bounded on V. . . " ): o ,no 1 no Thus (fn ) i s bounde d on V.

1 04 (b) For N - 1 ,2 �3 , • • • , put

All - I x : I fm (x) - fn

(x) I � & i f m � N an d n � N ! N ?te that the continuity o f fm - fn s hows that '" x : I fm (x) - f

n (x) I � & I i s c l os e d , hen ce All

i s closed.

The conve rgen ce of I f (x) \ shows th at th i s is a C au chy s e II quen ce . hence x E s ome All ' Th i s gives

00 X : U -\.

11 = 1 By Baire ' s the orem . s ome A N contain s a nonempty open s e t

o

Q. E . D.

R E M A R� . - N o t e th a t (a ) 1s a coro l l a r y of (b ) .

E XE R CISE 11 . - ShOlJ t h at ths comp t l/ t i on of a norme d Une ar spacs

(sse [IE ch ap te r III). is a Ban a ch space.

S O L UT I ON

Le t X be a name d l in e ar sp ace (abbr. N LS) , and Y the com

p le t i on of X. Since Y is comp lete , to prove that Y is a B an a ch sp a ce

i t - suffi ces to show th at Y is a NLS.

Thu s it remains to show that Y i s a vector s p ace , and

d (). x . O ) - I x l d (x . O ) .

In fact . X i s den se in Y . an d i f lim xn - x . lim y

n - y

with (xn ) , (Yn ) in X � put

x + y - l im (xn+ Yn )

x x - 1 im (x xn ) •

Thi s makes y int o a l ine ar sp ace , a�d the s e con d l ine give s .

since d (). xn , O ) - I x l d (xll , O ) , that d (u , O ) • I x l d (x . D) .

Q . E . D.

EXE R CISE 1� . - Le t C be C([ O, l J ) wi th t hs supremum norm . Le t Xn bl/ t hs eKosp ace of C con sis ting of tho s e f for whi ch there ezi s t s a t E [ O, l J == I such that I f (t }

"- ! fs } I � n I s -t ! for a U s E - I

Piz n an d p rolls that each opsn Be t in C con tains an open s e t whi ch do es not in terse ct Xn • Show t hat this imp l ies t he ezis tence of a denss Ga in C which consi s ts entire l y of nowhere di fferen tiab l e ,,",nction s •

1 05 S O L U T I O '"

To prove the first assertion , i t suffice s to show that for each f in C , and for every E > 0 (n is g iven) ,

( i ) There exists an g E e , wi th I g-£ I 00 < '2 & , and (i i) There is an �, > 0 such that h E C and I h-g 1 < E'

impl ies h i!. Xn • In fact , f i s continuou s , hence uniformly con tinuous on

[ O , l J , so that there is a set p - I a. - . 0 < a1 < a2 < . . . < ak - 1 I

such that X · E J ai , aiT1 J implies ( f (x ) - f (ai ) I < E

mo reover l aiT1 - a t l < � for i - O , l , • • • , k- l

, and

. g i s defined by g (x ) - inf (f (ai ) + 2n (x - a i ) , f (ai T1 ) - 2n (a iT1. - X ) )

if x e [ ai , aiT1 J . I t i s clear that · g satisfies ( I ) , and moreover , the one

s ided derivatives o f g a t every point x E ( O , l J exis t s , with I g� (x ) I - ( gi (x) ( D 2n , thu s g i! Xn

By the con s truct ion of g we f ind a set P ' - \ 0 . bo < b1 < • • . < bm - 1 i

such that , g (b i T 1 ) - g (b i ) I

b > n

• 2n i - 0 ,1 , 2 • • ,m-l (b iTl - b i) ,

If II - m in ! (b1 - bo ) , (b2 - b1) , • • • , (bm - bm_1 ) ! , and

E' nil - 2

then I h - g l oo < " imp l ie s I h (b iT1 ) - h (b i ) ' > n (biTl - bi )

Thus if x e' J b i , b i T1 [ , then e i ther

I h (x ) - h (b i ) I Ih (x ) - h (b iTl) I or

This s hows that h � Xn • Thu s ( i i ) i s proved. Now by the preceding part XC i s open and den se in C , n

n - 1 , 2 , • • • therefore

X - n XC i s a dense se t in C n= l n

1 06

If f E X and x E [ 0 . 1 1 . then

� i s unbounded.

'� f (X) - f ey) : 'Y E [ 0 . 1 1 . y '; x � x - y

But i f f' (x) e xi s t s . it is clear that thi s set i s bounded. Hence f is not di fferentiab le at x.

Thus X i s a den se G4 in C . which cons i s t s en tirely of nowhere differen t iable functions.

Q . E . D .

EXERCISE 13 . - Le t , A - faij ) bQ an infin i te ma tri: wi th "omp Ze: .ntrie s, where i , j - 1 , 2 , • • • , A asso cia tes ' wi t h e ach .equ.nce (Sj ) cl se.qu ence ( ai ) de fin . d by

( - ) a . -'I.

provide d t h a t these seri e s converge.

i - 1 .. 2 , • • •

(a) Show t h a t 'if A lIIap S C o into ' c o. t h8n A E L (c.; c o )

an d t h at (i ) L illi a . . - o for e a cll j

1. -+-00 'l-J

(i i ) sup L l a . . l' < 00 i j 'I.J

(b) Show that L (c o ; c o ) can bQ i den tifi Q d wi t h t he s e t of a n A s a t i sfying r i ) an d (H ) • wi t h t he no l.""

I A I - sup 2:, a . . r • • 'I.J 'I. J

(c) L e t E be til. lIubspace of' Z OO con s i s t ing of a L L converge n t sequ8nces . De fin.

tllen

1\0" - Z im s , j_.oo J

if " -

Ito i s a boun de d, Hnotar functionaZ on E • If A opera te s sllow t ha� Ago A - 11.0 iff A s a tisfivs (i ) . ( i i � an d

(i i i ) H m f a • • - 1 i+ooj-1 'I.J

SIIo w t ?a t E i ll i somorplli c to C o, and find L fE;E)

as in

� (d) The proce ss �f palllling from (Sj ) to (�i ) is c a Z Z e d a lIummabi Z i ty me thod. Two .:amp Z . s are

1 �f 1 "- j "- i aiJ - i ' .... .... - o if i < j

and a • • - (1 - r . ) r�-l , O < r . < l , r . _ 1 as i -+- GO . 1.J 1. 1. 1. 1.

1 07

Prova t�lat aach of t hB BII a Z s o tran sforms lIom B di v B rglmt stl quBnctI ( tl v tln BomB un boun d. d en tl ll ) to - c07lv B rgtl7l t lItlqutlnctls (tJi ) '

S O LU TIO'"

The notation " denot e s a l<equence - (S j ) j - 1 . 2 . . . . . and

the corre spondini sequence ( tJi ) in (*) is deno ted by � . Thu s A is i.den·t ified with the corre sp ond ing l inear mapping x -+- Ax

( a) I f \ . A2 . . . . . - AnE Cf . an d

" x • (A1" ' � x • • • • • Ao')( ' O . O . . . . . ) . we ve

'rify e as i ly that 4> e L(c o ; c o ) and that

( 1 ) I . I - max ( l'\I . . A21 • • • • • IAol ) Suppose that Ax

(2 ) .A.x �

then it is c l e ar

E Co whenever x e co . O,efine

• tJi -� aij Sj • " - . (Sj ) e Co . that Ai E c: • and 1 Ai;" I . � I Ax icoo

I f Aox - (A1;. • • • • • \". 'O . 0 . . . . . . ) . then Ao e L (co ; c'o )

for n - · 1 , 2 , . ... . and 00

(3 ) lAo I . Ilax �j� l ai j l

But l Ao x I � IAx l . so that

i • 1 . 2 • • • • • n t !Aox lo:l . 2 • • • • • is bounded for

eve ry x e C o • Hence by the B anach-Steinhau s s theoreJII ,

(4 ) sup I Ao I - sup�! I a i j I : i - I . Z , • • • I < <>0

o ) = 1 \ It fo l l ows frolll the defin i t i on of norm in Co that

hence f inal ly we have

I A I � sup �� .I a i j l

hence A E L (c. ; co ) .

n • 1 , 2 • • • • l . x E Co

i - 1 , Z . . . . < 00

( i i ) i s n othing but (4 ) , so i t relllains only to prove ( i )

I f e 1 is the sequence (S j ) with

Sj - 0 if ' j � i , s 1 - 1

then e i E co . and Aej • (&1 j ) 1:1 . 2 , • • • •

!!

This shows ( i ) • hence the proof of (a) is comp l e t�d .

( b ) It i s s imp l e t o s e e that in (a) w e have • i f A map s c

1 08

into c • • then I A I '> I Anl . Hen ce in addition

( 5 ) I A I - sup �r I ai j I : i - 1 . 2 • • • • • • J

Le t X ' be the s e t o f a l l matrix (a i j ) i , j - 1 . 2 • • • • with t he norm

(5 ) . The p re ce d ing re su l t wil l be st ated precise ly as follows :

then

L e t L : X --+ L (co ; co ) defined by

L (A ) e L (co ; co ) • L (A) - <1> . an d i f " - (5 j ) , then

<1> " - ( ai ) • with ai - L a i j 5 j i - 1 . 2 . . . . j = l

by ( 5 ) • we have I LA ) - l A B . A e· X •

I f <l> e L (co ; co ) . then for .... E C • • <1> " - , (Al,, _ �" ' ' ' ' ) with Ai e c! i - 1 . 2 . . . . S in ce c: - 1, 1 . we c an find for each i a

sequen ce (a ') in 1, 1 • ' with i j ,j = 1 . 2 • • • •

- A , " • ai-! a . j s j l. j =1 l. This s hows that L : X � L ( co ; c o ) is onto . Therefore L is

an i some tric i s om0rphism o f X on to L (co ; c o ) . Sin ce L depends only

on the de f in i t i on o f co . we can then iden tify L (co ; co ) with X. an d

the corre sponding n o rm o f L (co ; co ) i s t herefo re the norm ( 5 ) o f X •

( c ) I t i s c l e a r that A o is l in e ar • and that

l Ao '" - I , l im s j l � sup I s j l - I x D . " - (Sj ) e E

hence l Ao I � 1 . J .... .." j

Now if � o A � Ao • then A map s C o into Co • hen c e A 5 a t i -j sfies ( i ) and ( i i ) . Moreove r . let

eo - (1 , 1 , • • . , 1 , 1 J • • • • )

then e o e E . A o e o - 1 . hence if Ae o - (oi ) ' then

but

l im a - 1

0i -&1 a1 j i �;;:u : A satisfies ( i i i ) .

Conve r s e l y . supp ose A satisfies ( i ) to ( i ii ) . Then A maps

c. into Co " and map s E in to E. By ( i i i ) . we see that

Ao (Ae o ) - Ao e o - 1

But e a ch " - ( S j ) E E spl i t s in to ,, - 'r' .. :z •

1 09

z ,;. lio-x e o , y . ( 5 1 - A o " , s 2 - Ao " " " ) E c .

y E c o , hen ce A.A., • Ao Y • 0 , so finally,

that is Ao oA • A o ,

The refore we have p rove d that A. oA - A. ho l d s if an d

on ly if A s at i sfies ( i l t o ( i i i )

that

Def ine T

,. -

E -4- c . and T' : Co -+ E by

(S j ) E E , T" - ( � i ) E Co is the sequence such

1< • ( �l ) e Co , T' " • ( S j ) e E i s the sequ ence

l 2 ' l3 ' • • • ) + II e o ,

It i s clearly v e rified that T ' - T-1 , an d I T ' n • I T I - 2 ,

Thu s T i s an i somorp hism o f E onto Co . [ I f we ob serve t hat " . ( ii ) e c . then

.� 00

" In en , an d ,, ' . ( s j ) e E i s (A"'J e o + � (s j -A", e j , n = l ) = 1

t hen T and T ' can be defined s imp ly as

Ten - en + 1 ' n3 O . 1 . . . . , T ' en +1 - 'n , n · O , l , � . . . ] T and T ' arise an i somo rp h i sm U of L (c . ; co ) on to L ( E ; E )

n ame ly U (A) • T ' . A 0 T .

Here we find anothe r form o f L (E ; E ) : L (E ; E) i s the set o f

al l pairs ( c .A) whe re c E C , an d A E L ( c . ; c . ) . and

If " - (Ao ,, ) e o + Y • and B - ( c .A) , then

Bx • (A. �) ce o + A�

we find e a sily that if x • ( � j ) E E ,

B" - ( lli ) , w i th "1i-hen ce II B I I sup I Ai I i - 1 , 2 . , . .

(We do not compu te D B I , for B

\

I •. A . " • then

c t + �ai j ( l ' ) = J

> I A B

E L ( E ; E ) )

- Z l - lI. , x 1.

( d ) The ex amp l e s g iv en in the ex e r c i s e ev i den tly s a t i s f y

( i ) , ( i i ) and ( i i i ) , hence t h e y t r an sform converg en t sequen ces

1 1 0

(5j ) to sequence,

( Cli ) whi ch converge to the s ame l im i t .

Now con s ider t he case in wh ich

if l � j � l

a i f i < j

and t he sequence

S2p _l . •

Then we see that

Pt , s • �p� p . 1 2 2p • . . . .

o CI • k! • 2k-l (2 k - 1 )

hence ( Clf) E C o • Thus t he prece din, A transforms an un bounded

(S j ) in to a sequen ce (-<I i ) in c.

S imil arly , con s ider t he case in which

l im ri - 1 1 ... 00

Take any c su ch that r1 < .!. ,;:: 1 j -l C -..;:

and if 5 j • (- c)

hence ( oi ) e c.

t hen

o • 1

i • 1 , 2 . . . .

This shows that A tran sforms some dive rgent sequen ces

to sequence s whi ch conv e rg e to 0 •

Q . E . D .

E X E R CISE l � . - SUppO Sll X and r are Ban ac:h spac:es • an d suppose II. is a l inear mapp ing of X in to r . wi th the fo l lowing property: Fo r e v ery sequence I zn I in X fo r whi c:h Z - l im zn an d y - l im lI. ;cn azi s t. i t i s tl'UI2 that y' . lI. z. Pl'OVe that II. is con tinuous

Thi s is the s&J -ca H a d "cl osed gl'aph theorem ". Ob'serve

tllat t hel's I2zi st (non l inear) mappings ( of gl onto .r . fo l' instancl2 ) who se gl'aph i s c loB.d a l though they al'e n o t con tinuou s . e . g .

S O L UT I ON

! (z) . ! , �f z � 0

f rO ) • 0

Suppo se X . Y are Ban ach space s , and �uppose II. is as in

the theorem .

We observe first that I (x , y ) i a I x lx + g y ly i s a norm on

Z • X x Y which make s Z in to a Banach space , and t he l ineari ty

of A now shows that

G " \ (x , J\ x)

is a l i�ear subspace of Z x E X I c z

Note that �o p rove the compl e tene s s o f Z , we need the

fo l l owing fact : (xn ' y n ) (x , y ) in Z H and only i f xn -+- x in X ,

and y n -+- y in Y , as n - 00 •

Now the hypothe s i s on A is th at

if (xn 'Yn ) E G , " and (xn ' Yn ) - (x ,y) as n - 00 ,

then (x ,y) E G

that i s G i s closed in Z • Sin ce Z i s c omp l e te , this sh ows that

G i s a Banach space .

S�nce for pr 1 (x , y) • x then I pr 1 I � 1 , the mapping

, : G --' X ., (x , A x ) • x

1 1 1

i s a bij e ctive hounded line ar mapp ing of the B"anach space G ont o th: B anach space X . B y the Banach the orem (Theorem 5 . l0 ) ,·l i s also

con tinuous . - 1 For pr 2 (x , y)o . y , I p r 2 I � 1 and we have " II. . pr 2 0 ,

Thi s shows the con tinu ity of A Q . E . D .

R E M A R I: . If f E L2 ( [ 0 ,1] ) , then I f 11 � I f l2

NOtt l e t X . ( L2 , 1 II ) lin d y . ( L2 , I 12 ) , lind l e t"

A f - f , A , x -+- Y. By the pre ce din g ,

' lI.f n - f I i -+- 0 J.mpl 1 es I f ZJ - f 11 - 0 •

hen ce t he vraph of II. i . cl o s e d .

BU!" ro r f n

.0 J\ J. . n o t bo u n d e d

Thi s s hotts tha t compl e t en e s s ( o f X I can n o t b e om"i t t e d

in the h yp o t h e s i . of the cl osed g r aph t h e o r e m .

1 1 2

EXERCI SE 15 . - SUppO S8 (An ) i s a sequence o f bounded linear tran cfozomati.?ns from a NLS X in to a Banach space Y. suppoae I I A nR �M < 00

for a l l r. • and suppose there i s a dense subs e t E of X such that I An:!: ! converges for each :!: e X

Prove t ha t ! An:!: ! opnv erg es for each :!: E X .

SO LUTION Let x e x , an d £ > 0 be g iven . Then

(1) There i s an Y e E su ch that I x - y m < E (E dense in X)

Since j AnY \ converg e , there is an N such t hat

(2) m , n > N imp li e s < £ •

By (1 ) , ana s ince II An " � M , n · 1 , 2 , . . . , we have

(3 ) Anx - And � M £

By (2 ) and (3 ) , i f m ,n > N then

Anx - Amx D � 1 Anx - AnY I + I AnY - AmY D + I AmY - Amx l

� (2M + 1 ) £

Thus ! Anx \ i s a Cauchy sequence in the Banach space y , . hence \ AnX \ converg e s

Q. E . D . RE M A RK .

In the pre ceding e x ercise , i f .,e s u ppose · An '" -"0 a s

n ->- 00 , '" e E " i n s t e a d of " ! An x \ con ver'l/e s , x e E . , t hen .,e ha ve An lt - O for every It E X .

EXERCISE 1b . - If II- i s a po s itive measu�e. eaoh f E Loo(fL} defines

a mU ltip l i oation operator Mf on L 2 ( fL} into L2 (fL} . suoh that

Mf (g ) - fg

Prov's that IMf l � I f loO • For whi"h measure fL is i t true

tha t aMf l - I f aoo

for a l Z f E Loo(fI.} ? For whi oh f E Loo(fL} does

Mf map L2 (fI.} into L 2 (fI.) ?

SO L UT I O N .

1 1 3

I t i s c l e ar that ! fg ! :O;; O f Boo! g ! , hence D fg B� :O;; U f lloo � d2 ' Th i s s hows that 1I�lf l :O;; U f U .,d

Now l e t u s def ine a k ind of general i z ed 0 - fini tene s s : D E F IN I nON.

� is said to be n weak ly 0 _ fin ite " if

(flo (A ) - DO) implies ( 3 B c A su ch that O < �(B) < "", ) .

We s h a l l prove the fo l l ow ing : ( 1 ) IMf a • I f a..., for a l l f e L DO i f f � i s weakly 0 - f i n i te F i r s t suppose I" i s we akly 0 - f in i te . I f ft H ..., < DO , then Y c < H U ..., , A · I x : ! f (x ) 1 > c \

has non z ero m e asure . Hence there i s a s e t B c A su ch that a < Ji,(B) < 00

Thu s i f g . XB , then I Mfg R2 � c l g ll2 Th i s sliows that O Mf H � c , hence n Mf K � l f a oO I f II f H ..., • 00 , then in t h e s ame manner we can cons ide r

I x n :o;; l f (x ) ! < n + l and w e c an prove that n �If n = DO •

Next supp o s e . � i s not weakly 0- f i n i t e . Then there ex i s ts a measurab l e set A such that /Io(A) = DO . and such that B c A

impl ies e i ther floCB ) � a or �(B ) • DO • Thu s i f g e L 2 ( "') . then g

i s a a . e . on A . I f we pu t f = IA then fg • a for all g E L 2 (�)

hence R Mf II = a • l">u t I I f O 00 < .;., Therefore our con c lus i on ( 1 ) i s prove d . T o d i s cu s s t h e l as t qu e s t ion , w e s t r i c t our at tent i o n

tG the case i n which � i s ... eakly 0 - f in i te . I t 1 ,,; seen t h at M f i s sur j e �tiv:e if and only if ( t) i s in L 00 (flo) . The verif icat i on o f � h i s as s e r t i on i s l�it t o t h e readers .

Q . E . D .

EXERCISE 17 . - If 8n i 8 the n-th p artiat 814m o f t he Fourier se�i8s of a function f e C rT ) . prove that

..2.- _ 0 Logn

.uniforomty , as n __ DO , for sach f E C rT) . :hat is, prove that:

....

1 1 4

Z• g S7l 1""

l.m --- • 0 71-+- "" Log71

071 t he o ther han d. if � -+ 0 • prove that there ezi a t a an Log71

! E CrT) su�h that the sequence 8 ([. 0 ) I " \ i s unbounded.

X n

,sO L U T I O " Re�a1 1 that the n-th Fourier coeffi cient of f i s

f (h) • _1_ f f (t ) e-in :t dt . n • O . ±l . ±2 . . . . . . . 2 11: T and that

Hence i f

n sn ( f .'t) •. 2: f (k) e- ik:t

1< = -n n . 2: e i1<:t • k =-n

s in (n+}) t t s in !

"then s · "f * D n n (a) Let u s now est imate U Dn l1

t t In { O . lt] . s in i s convex . hence s in ! >- i ; Ob serve that

1 J I Dn l dt o

In the above the f i rs t teTlll (j • 0) i s � � ( 2n+ 1 ) • 11: . and (2n+ 1)

the remaining te rms are maj ori zed re spe ct ive ly by ctive1y by

Hence

faj + l 1 Jl. I s in (n + ... ) t I d t a . \i. J a . j

1t K • k • 1 . 2 • • • • • 2n .

(1 ) 1 1 1 U Dn 1 1 .;;;; 2 X(1+ 1+r:r " . . . . + rn-) � 7Logn if n i s l arge enoug�.

( l im (l+}- . . . . . . +t - Logk) - the "Eu ler constant _ ex is ts and � O . 6) k ..... oo .

I t fol l ows

(2 )

I Dn h that- ';;;; M < 00 . n Logn "

sn (f) A f • ---, then , n . Logn

• 1 . 2 • • • • • hence i f

n • 1 ,"2 , • • • • •

1 1 5

s ince

I f f is a trigonome tric po lynom ial of degree N then

Snf • f if n � N, hence Anf -+ 0 uniformly .

Since the s e t � of al l trigonometric polynom i a l s i n

dense in C CT) , b y Exerci s e lill we have

proved

,\t ->, 0 uniformly as n - 00 , f e C(T ) .

Therefore the first conc lus ion of the ex ercise is

(b ) Nex t , we have , s ince I s inx l � 1 )( 1 ,

x xh I Dn ll . r I Dn 1 dt 4 r I S in � 2n+ l ) t l d t J_ Jo S lnt -x 1 xl';. ( n +"2)J:

� 4 f I S in (2n+ l ) t l dt - 4 J I s int l dt o t o t

f k = l

JeT. I I s in t l dt k X ( k - l )X

n 8 .L: ! > ! Logn . It k = l k It

Thus we have (3 ) I Dn 11

> � Logn

Bu t if g e - LI CT ) and . f - (f*g ) ( O ) , then

I I . U - li h U 1 CTh-i ..- i s simp l e to ver i fy ) . We can app ly this result to prove the

seco�d as sertion of the exercise as fo l lows

I f � Vgn

then

Henc -�

-+ 0 a s n --+ 00 , and

II . U -n

CC(T) , n I ) i s comp l e t e , hence the Banach-Ste inhau s s 00

theorem g ives the ex i s tence of an f E C(T) such that l�n ( f ) 1 i s

unbounded . Q . I; . D .

1 1 6

E XE RCISE 1 11 . - L e t E be a s ep arab le m e t ri c sp a ce , ctn d A a s ub s e t o f E. r E E i s s ai d t o b e a con de n s a t i on p oi n t of A i f f e ve ry n e i ghb orh ood o f ;:: con tai n s an un coun t ab le i n fi n i e s ub s e t of A. All den o t e s ths s e,t of con de n s at i on p o i n ts o f A .

( a ) Sh ow t h at . A - A # i s a t mos t coun t ab l e . an d A ll i s c l o u d. (b ) Sh ow t h a t (A Ii) 1I - A ll. De du ce that e ve ry c l ose d s e t in X

is the un i on of a p e rfe ct s e t an d a fin i te o r cO W1 t ab l e�se t , n ame ly A � A ' U (A - A ll) .

( c ) Supposs t h at E i s comp l e t e an d e v e ry b a l l B (r;�) in E con tains at le as t, 2 p o i n t s (E h as no i s o lat e d p o in t ) . Sh ow t h a t e ve ry B (r; r ) i s an un coun t ab le i n fi n i t e s e t i n E . SholJ t'h a t E # ;. E.

S O L UT r ON (a ) Supp ose x E A - All . The re i s an r > 0 , and B (x ; r) n A i s

a t mos t coun t ab l e . I f y E B (x ; r) and r ' • I' - d (x ,y ) , then B (y , r ' ) e B (x ; r) , hen ce B (y ; r ' ) n A i s a t mos t countab le , hence y 6!! A I.

Le t D be a coun tab l� subs e t o f E , whi ch i s dense in E . There is s ome t E D , and k , s u ch that x E B ( t ; l/ k ) C B (x ; r) . I t fol l ows that B ( t ; l/k) e E - All , and t h at t

A - 'AII C U U B ( t ; 1/11) n A z k

the uni on i s t aken ove r a sub s e t of D , and for s u i t ab l e k ·· l , Z , • • •

Thi s shows that A - All i s the coun t ab le uni.on of fin i te or coun t ab le sets , hen ce i t i s at mos t coun t able .

I f (x ) i s a s equence in A ll conve rg ing t o .x E E , every ne ighn borhood of x i s a ne ighborh.ood of s ome xn ' hence x i s in All .

1

(b) I t i s clear that a c on den s at i on point i s a l imi t point . AN is a c losed s e t , hence (AIi) 1I e All.

Conve rse ly , i f. x E All , then every ne i ghborhood V of x contains an un count able subset of A . Sin ce A - Ali is at mos t coun t ab le , V n AN i s also un coun t ab l e . I t fol lows that x E CAII)II .

Th-e i dentity Ailll 2 All shows th a.t in p arti cul ar All has n o i solated poin t , i . e . All i s p e rfe ct . I f A i s c losed , then A il e A , and A · All U tA - All) , wi th All pe rfe ct an d A - A ll finite or countab l e .

( c) First n ote that B ( a ; r) - \ b \ ·O < r ' � f ( Eve ry b a l l B (a ; r) i s an B' (x ; r) · I y : d ( x , y ) � r l .

contains B ' (c ; r ' ) , in fin i te s e t ) in w h i ch '

Let B ( a ; r) , I' > 0 , be a b a l l in E , and \ xn I be a sequence in B (a ; r) , Then ( i ) B (a ; r) - \ x l \ :J B ' ( a l ; rl ) , 0 < r l � f , and ( i i ) B (an ; rn ) - . l xn: n ! :;) B' ( an + l , rn + l) , 0 < rn+ l � rnlZ .

1 1 7

I t i s clear that lim an • y l ies in a l l B ' ( an ; rn) , hence y ; xn for all n',

Thi s prove s that B ( a ; r) i s not a coun table s e t . I f x E E , t he p re ce ding shows th at x i s a con dens ati on poin t of E , an d hen ce E " 3 E .

* * *

Q . E . D.

) ) 9 CHAPT!==R 6

COMPLEX JfEAS URE

E X E R CI SE ], . - I/ I" i s a comp l ez measure "n a a -algebra cnl and if

'E EClll define � (E J - sup 2: I I" (Ei J I

the supremum being taken over a l l finite partitions l Ei ! of E.

Pr.sve that � � I I" I • the to tal v ari tion of I" •

SOLUTI OI'

Recal l that

where the supremum i s t aken over all .f ini teO o r coun t a b l e par t i tions

I E1 \ o f E (E E cnl) . Hence

( l ) ),, (E ) � 1 11-1 (E ) , E E em

Now l e t ! E1 I be a countab l e p a r t i tion of E , and (. > a

The series I 1 I"(E 1 ) 1 i s c onverg ent , hence there i s an N such that

he nce

I I" (EII ) 1 + I I" (EII+1 ) 1 + . . . . . . < (.

De f ine the fin i t e par t i t i on (Fi) 6f E by

i " 1 , 2 , . . . , N- l

FII • Ell U EII+1 U • • • • •

I t f o l l ows e a s i ly that II

1 1" \ (E) � �(E ) + £ hence

( 2 ) 1 I"I (E) � �(E) , E E CIll.

I t f o l l ows from ( 1 ) and ( 2 ) that A · 1 1" 1

Q. E . D

1 20

EXERCISE 2 . -L e t m be the Lebesgue meas�re and � the eoun ting measure on ] O. l ( •

Prove tha t m « ).(i . s . A(E) = 0 imptiBB m (E) - 0 ) b u t there i s n o f e Ll (�) with dm � fd � • Henee a-fini teness eanno t be omi tted in the Radon-Nikodym theorem.

SO LUT I ON I t i s clear that m « � ( � (E ) - 0 imp l i es E - " ) .

L e t f e L l ( �) , then

S ,; I x : f (x ) � 0

i s at most countable , s ince � i s the count ing me asure

Therefore if E - ] O . l [ - 5 then m (E ) - I , bu t �fd� � 0

This shows that there does not exist any f e Ll C).) such

that m ( E ) - � fd ). E Bore 1 set

that is elm - fd� .

Thu s " a - fini te" c anno t be om i t ted in the hyp o thes i s of

the Radon-Nikodym theo rem . Q . E . D .

EXERCISE 3 . - Prove tha t the vs ctor space M (X) of a L L eomp Lsz regu Lar BoreL measures on a Loca LLy compact Hausdorff space X is a Banach spaee if I � 1= I �( (X) .

SO LUTION The Rie s z representat ion theo rem (Theo rem 6 . 19 ) de t e rmine�

the natural isomorphism of M (X ) onto Ce (X) * which as sociates to

each )l. E M (X ) a l inear func t i onal A on Co(X) ' " and

" . . . ( � . - A is onto and I � a - I � I CX) - u A n ) \""f�" .�� Recall that E * is compl e te , if E i s a NLs c'i�l chap t e r V) .

Thu s, M (X) wi th the no rm a � a· I )Io 1 (X) , being isome trically isomorphic

to Cc,(X ) * is a Banach sp ace .

,., E . D .

1 21

EXER CISE 4 , - Suppo s e 1 � P -;;; 00 , and q is the ezponen t �onjugate to F ; Suppose fI. i s a (J -fini t e measuroe and g is a measuroabLe fun�tion su�h that fg E Ll ( fl.) foro e ve roy f E LP (fI.} ,

Proove that then g e Lq (II-) .

SO L U T I O N

Ca) Firs t cas e : 1 � P < 00 ,

fI. i s (J - f in i te , hence X - Xl \J X 2 U . . . . , . Xl C X 2 C • •

.

•. , .

and fLCXn ) <00 n - 1 . 2 , • • • Let

A n \ g (x ) I � n I I t fo1101o/s that A1C

,A2 c . . . . fI. (An ) < oo n · 1 . 2 , • • • and

A l U A 2 U • • • • - X , and moreover Ix \ g \

n I qd fL • .h�g l qd fL � nq fl.(Xn ) <

hence if 00 •

A f n then

Thu,s by the Banach-S t e inhauss theorem I IAn ! ! is bounde d ,

s o that r \ g \ q d fl. � M < 00 n-1 . 2 . . . .

JAn Recall that Al U A2 U . . . . . X • and Al C A2 C . . . . so by

the monotone convergence theorem .

h l g l q dfl. - l im I I g l q dfl. � M < 00 hence g E L q (fl.) •

n_oo An (b ) Case p . 00. q - 1 .

Define f by f (x ) _.i0.. i f g (x ) � 0

g (x ) • 0 if g (x ) - 0

Then If l � 1 , 'and hence fg . I g I e 1.1 . oa

Q . E . D .

REMAR(. The pr e ce di n g r e su l t ca. b e er tende d t o the case i n which

1 22

flo i s weak l " a - fi n i te (See the d e fi n i ti on of th e weak a - fi n i"t e n e s s " in � chap t er V ) .. Le t us s t a te s ome u s e f u l conse que n ce s of "e llk

,, -fi n i t en e s ... . (i ) fIo {A' ) - sup j flo(E) , E a -f i n i t e , E c: A ! A me as u rab l E

, PR O O F O F ( .i l . I t suffices to cons ider the case flo (A ) a 00 •

L e t c be the r ight s ide of ( i ) , hence there are a-finite sets E

nc: A such that , fIo (En ) j c as n _ 00 . If

E - El U E2 U then E is a - finite , E c: A , "hence I"(E ) - c •

Thus i f c < 00 , then fIo (A - E ) - 00 , hence one can f ind

a set B in A - E with 0 < /I-(B) < 00 • The inequality

",(E) < ",(E U B )

would contradi ct the de fini t ion of c , therefore c - 00 , hence fIo(E) - 00 • Thi s shews ( i ) •

(i,i ) M g i s me asurabl e , then t h e re i s a G -fin.1 t e s e t E s u ch tha t I g i - D g X D (1 .;;; q ';;; oo) . q E q

PROOF OF (.ii I . I f q - 00 , then ( i i ) , is an easy consequence o f ( i ) . Hence suppose 1 .;;; q ,< 00 , and consider

� .;;; I g (x ) I .;;; n ! n - 1 , 2 , • • •

If each An i s a -f inite , n - 1 , 2 , • • • , then (E) is eV Ident . 1 on \i' ,

o h B -q

00 •

contains a a-finite s e t E such that ",(E) - 00

hence

( i .1 i ) tha t fg ,e Ll (/I-)

,If U g H - 00 , then t h e r e , i s some f E LP (u-l s u ch , q r-

PROOF O F (U.i I . By (E) . i t suffices to restritt ' ou r atten tion to the case in whi ch X is 'o -f in i t e .

',(a ' ) Case 1 < p < 00 • T h i s i s a n equ ivalent form of (a) .

Nbte that one can 'construct withou t d i fficulty such an f , this

g ives a proof of (a) whi ch does not involve the Banach-Ste inhauss

theorem . For an idea .of this proof , see m chapter IV

(b ' ) If P - '00 and I g 11 _ 00 , define

f (x) - � if g (x) � , O and - 0 i f : (x) - o . I g (x ) I

Then 1 £ 1 00 - l and fg - I g l � LI ( II-)

There fore we have prov e d the f o l l owing :

_l + _l 'a I f JI. is we akly O' -,f in i te " g is m easurab l e , 1 p q

1 23

and 1 � p � DO , then g E L·q C",) i f f fg e LI C ",) whenev er f E'LP C",) .

Thi s is the gene ral izat i on . of the ex e r c i s e that we hop� For more informa t i on� abo u t we ak 0'- f in i ten e s s , see the

Remark of II] below.

EXERCISE S . - Suppose Z aon si�ts of two poin ts a and b ; defin e

",r i a l ) - l ' - I'-( I b l J - )L (Z) - 00, and ",(0) - o. Is i t true fol' this I'- that L

oo( ",) i s the dua Z of Ll ( ",) 7

S O L u n o�

I t i s c l e ar that i f f E L I { I'-) imp l i e s f eb ) - 0 , and

1 £ 11 - · l f Ca ) 1 • Thus A- E' L I C ",) * imp l ie s

Ai - :\. f Ca) for a un i que >.. e C , and I AI - I >.. I

that i s LI ()L) * is i s omorphic t9 C

On the other s ide , L OO (I'-) can be i den t i f ied w i th

C2 - ! (>..1 ' >"2 ) : >"i e C , i - l , 2 I ' I ( XI , >"2 ) U - max C I >"1 1 , 1 >"2 1 )

Thus LI C",) * i s no t i s-omorphic to Loo( )L) .

�EMARI:: . It is known th. t LI I ",) * is i somorphi c to LooI ",) if I'is we . k l y 0'- fi n i te I • con.geqUe<l ce of Q] ) .

Now suppose ", i s n o t w .. .. k l l/ a-fi n i t e , th .. t i s t h e r e i s ' .. ..... . ur.bl .. s .. t B in' ]( s u ch t h " l; I'- ( B) - 00 , .. n d A C B impl i e s tha t ", IA) i s . i t h .. r 0 or 00 . Therelore i f f e Ll I ",) th .. n

,... I Jt e B , � < 1 f (:r ) I I - 0 , n - 1 , :1 , • • • ,

hence f iB - O . Thus i f 41 , L

oo(",) _ LI 11'-) . i s defin .. d b y

�('1) f - ffgdl'-

th .. n 4I l lB ) - 0 a l though I lB

I � I • Therefore Lao

(",) c .. n n o t be

i d .n tifi e d wi t h LI (",) * (1.n tbe n . t u r a l f •• h i en ) . Hen ce the fo l l owing hol ds

LI I)L) * _ L""'I",) if .. nd on l ", if I" is w ... k l ", 0' -fi n i t a .

1 24

E X E R CISE b . - Supp o se 1 < p < 00 an d p l'o v e t h a t Lq {fI-} i s the du a L space o f L P (IL) e ven i f )L i s n o t a -fi n i t e {A s u su a L , q - ....£....1 oj .

p-

S O L U T I ON

(Re call that by theo rem 6 . 16 we h ave

if )L is a -fini tel

We associ ate to each g e Lq ( )L) an A • T (g ) in LP (II-) *

defined by ( 1 )

By Holde r ' s

I f

Ai - Ix fgdfl- • f e LP ( u.) .

inequal i ty a A a � h 1'1

f (x ) • ilx) ! g (x ) , '1-2 if g (x) of 0 and f (x ) · 0 if g (x) • 0

'1 / p then � f lp • (h 1'1) • A f · ( I g n'1 ) '1

; hen ce

1 11. 1 > 1 & 11 Thus D Tg ! - h l'1 • hence T i s a l inear i some try of Lq (l'-)

into LP ( )L) * ' The refore to comp lete the proof of the i denti ty LP *-Lq

it remains to show that T is on to , that is e a ch A e LP ( I'-) * i s of

the form (1 ) for some g e Lq ( I'-) '

Suppose A e LP ()L) * . We shal l find a 0 - finite s e t A such

that (2 ) A f • A(f1.A) , f E �P ()L) .

Since LP (A) * L'1 (A) , if (2 ) holds then there ex i s t s a

g e L q (A) such that r ( 3 ) A (fXA) - JAfgd p. . f e LP ( �) .

I f g i s extended to a fun ction in . L q (X) by pu tting g / Ac - 0

then (2 ) and ( 3 ) g ive (1 ) • Therefore i t rema ins only to p rove

the exi stence of A •

Now if E i s a o -fini t e s e t in X • define

AEf • A (fl1; ) , f e LP (IL) •

Thus i f ! f R • 1 then I AE f k l AR I fXE I � I Aa , that i s P I AE I � I AI .

There exists a uni que gf: E L q (1'-) whi c h v an i shes on E C such that

IAE I . hE 'q

and

1 25

A Ef · Ix fgE d ", . f E LP ( II- ) ( L P CII-. E ) * • L q ( II- . E ) ) .•

Thu s i f E and F are disj o irt t • we have gE U F� gE + g r • imd hence

that i s

hen ce

and let

hence

on A ,

A and B

that i s

whi'ch i s

E . F a -f ini te . disj o in t

Now let f1 • f2 • • • • be i n LP ( II-) ,

I A I - � < Af n • � I A I

Q f l ' • 1 and n P n a 1 . 2 , • • •

D A) � I AEn I n · 1 . Z • • • • • hen ce Q AA I � a A D

ff A,.; 1 II A I I f now 1\ of \ , that i s

R f n p� 1 and Af > a ,

B • I x f (x ) of are disj o int and n � n

ft A I q · U A l q + A U B A

3 f E L P (II-) such that f vanishes

then for

a ! ' � M > a . Hence

U AA U B" > D A I absurd. There fore A .. AA • hen ce ( Z ) hol d s .

Q . E . D.

E X E R CI SE 7 . - Suppose II- is a comp Ze:c mBaSUI'B on [ O. 2 lt L , and define

AssumB that � (n ) - a as n -� oo and prolle that then

f (n ) -+ O as n -+ - 00

s O L un cm ( I f II- is real . then � (-n) · � ( n ) and the re su l t fol l ows

immediate ly)

By The orem 6 . 12 �e have d ", � hd l � 1 where h is a Bore l ftIn ction , . t h ( t ) I • 1 for all t E [ 0 , 2 lt[ .

1 26

"We have . s i n ce I II- l i s re a l .

hence �( -n ) . ( feint h ( t ) d l 1'-1 ( t ) r· je-in tFi(t ) d i l'- l ( t )

( 1 ) �( -n ) . j-in t (h(t ) ] 2 d JLct ) . (n -0 n . % 2 • • • • )

By Lu sin ' s t he o rem . C (T) i s dense in L1 (1'-) ' The s e t Cj) o f al l 2X- t r i g on ome t r i c po lynomial s i s uniformly den �e in C(T) . hen ce

Cj) i s den s e in L1 C 1 1'-D . L e t X be Ll ( l lI- l ) with t he norm

I f l • ./i f l d l JL I and define

A n f - jr(t) e-int dJLCt )

I "'n f l .� fl f l d l lL l We have

hence ( 2 )

I f g e t ) • DAn I � 1 • n • 1 . 2 , • • •

( f X . n - l , 2 • • • )

then An (g ) · ;Cn-Ic) • so that A n ( g ) - O a s n - ""' . Th i s s hows that

( 3 ) Anf -- 0 a s n __ oo for al l f E Cj) •

By (2 ) , (3 ) "and the Remarlc t o [!] ch.V t

An g __ 0 a s n --"",for al l g E X

hence in p a r t i cu l ar this ho l ds for g • � • s ince � i s a bounded Bo rel fun c t ion . By (1 ) . ; (-n ) · An? ) - • hence

Q . E . D .

E X E R C I SE a . - In the te:rmino logy of 0 . find a l l " JL su ch that ;;' i s periodic of pariod k [ This ' msan s t hat �(n+k ) - ;(n ) for a l l in ts(lers 'n : of cour.s. It i s also assumsd t o b e an in tege r).

SO LUT I ON

if dX{t)

� A r -in t - i kt I'-(n+lc) • JL(n) • J e (e -

• ( e -ikt• l ) d l'- ( t ) .

l ) d JL (t ) • � (n)

Thus ; (n+ Ic) . � (n ) i ff x(n) - 0 fo r all n. Bu t � . O i f f x van i she s Cj) . ht:n ce i ff x - 0

The re fo re � is k-period i c if and only i f

Support � c l · t : e lkt • 1 ! Q . E . D.

E X E R CISE � . - L. t � be a fin ite posi tive measure on is said to havs uniformLy absoLute Ly con tinuous integra L s if t o B aoh £ > 0 there corresponds a 6 > 0 such t hat

( 0 ) "' (E ) < 6 imp L i e s I f fd .,.. 1 < t'. • f e d1 • JE Prove the fo L Lo�ing theorem of Vi taLi : If j fn l has

u . a. c. i . and if f (Z ) _ f (z ) a . e . , then f e Ll (",) and 'I

S O L UT I O N

Lim �f - f l d � - O . Ji n

Suppose f i s real • f - f+ - f • and

E+ - \ x e E : f (x) > 0 I . E - - I x e E : f (x) < 0 I I f · ( * ) ho lds w i th f then

",(E) < 6 imp1 ies

There fore i t is easy to verify that

c:A has u . a . c . i • • iff \ I f l : f e c:A I has u . a . c . i .

( i!- ) Now suppose I fn i has u . a . c . i • • hence I l fn l \ has u . a . c . i .

Suppose fn (x ) - f (x ) a. e . Then by Ego roff ' s theorem , to

every 6 > 0 , there i s a s e t E such that jI-(E ) < 6 and fn - f

unifo rm l y on X - E .

If now 6 i s chosen such tha t f l f I dp. < 1 whenever JE n

�(E) < 6 • for a 1 1 n • 1 . 2 • • • • • then we have

' ( 1 ) h 1 f l dp. < 1 ( by Fatou ' s l emma ) , and

( 2 ) r I f l d.,.. · JX-E

if E i s c ho sen such that fn - f un i fo rll11y on X - E .

Thu s by ( 1 ) and ( 2 ) f e L l (p.) .

(b ) For the remain ing p ar t , s ince f e L1 (.,..) i t suffices t o consider the case f � O .

( ' ) Eq u i v a l en t l y , d1 i s un i form l y i n t e grab l e .

1 27

1 28

Suppose ! fn ! has u . a . c . i . and fn (x ) -- 0 a . e . We have to prove

that If n ll -- O. Since ! I fn l ! al so has u . a . c . i . , we can rep l ace

in by I f n I , hence c an suppo se fn � O.

By (3 )

Let & > 0 . There i s a ll , 0 < ll < £ , such that

( 3 ) JI.(E) < 1I imp l i e s 1: fn d Ji. < & n · 1 , 2 , . . . .

By· � (c) chap ter I I I , fn -- 0 in measure , hen ce

3N n > N imp l i e s 1'- 1 x : ' fn (x» l1 1 < 11

Thu s i f E I x fn (x ) > & ! ' .then n

· E c: I x : f n (x ) > ,) ! ' and n

f � & on E� ) n

n > N

Therefore

To e ach & > 0 there corre spon ds an N > 0 such that

Thus

Q . E . D.

EXERCI SE 1 0. - (a ) Show that Vi ta li 's theorem imp lie6 Lebesgue ' s dominated convergen ce theorem, for fin i te me asure space s. Con struct an IIzamp l e in which Vi ta l i 's t he orem app l ies al though the hyp o thesis of Lebe sgue 's theorem doe s n o t ho l d.

(b) Cons truct a seqltlln ce Ifn ! say an [ 0, 1 ] so t hat 'n ez ) -- 0 a . II . , fIn -- 0 bu t I fn I doe s no t havil u . a. c. i .

(c ) Supp oss JI. i s a po s i tive meas ure on X . I fn \ is a sequen , in L1 (JI.) , I E L1 (fL } , and

fo r e very me as urab le set E of finite me as ure . Pro ve that \ fn \ h as u. a. c. i . '

1 29 SOLUTI ON

(a ) I f g E L l ( fI.) . then I f : I f I <::;; I � I \ has u. a. c:i . (See [g] ch ap t e r I ) . Hence I fn l � g . g E L l ( fI.) imp l ies I fn ! h as u. a. c. i .

Thus i f fI. (X) < 00 • , then the hypothesis o f Lebes gue ' s dominated con- , ve rgence theorem imp li es the hyp othes i s of Vi t a li ' s t he orem • Hence Vi t a l i ' s the orem imp l ies Lebesgue ' s dominat,ed conve rgence theo rem in the case fI.(X) < 00 .

In the fo l l owing examp le . Vi t a li ' s theorem is app 1 i ci ab1e but L�b�sgue ' s the orem , i s not .

EXAM P L E f

Then

fn : [ 0 . 1 ) - R

fn (x) - 0 eve rywhere . 1 [ I fn l dx

o n + 1 • s o t h at u. a. c. i . 00

But i f fn <::;; g n · 1 . 2 • • • • • t hen t;.l

fn <::;;

th at the convergence of I fn ! i s ,not dominated.

g , hen ce g E! L 1 . s 0

but

� o that

2 (b) Le t fn (x) . nsinnx i f 0 <::;; x <::;; ii x .

fn (x) • 0 othe rw i s e ( n • 7 , 8 . 9 • • • •

Then fn (x) -+- 0 Y x E [ O . l J and

1 2 x/ n 1 fndx " f fn dX 2 0 0 0 jT-/n It

f dx � s in t dt • 2 n 0

I fn I do{'s n ot have u. a. c. i .

( c ) Let C3ll ' � l l E : E E Gl1l and fI.(E) < 00 \ c L 1 (1" ) '

I f f is the p oin twise limit a. e . of a s equence (l.AJ . then f ( x) is 0 or 1 for almost a l l x , .hence f " XA a . e . ( in whi ch A . A

1•

A2 , . . . are me asurab le sets ) I t fo l lows f rom Theore'm ( 3 . 12 ) that QT{ ' i s a c l os e d subset o f L 1 ( 1") .

The refore em" is a comp lete met ri c space ., an d the met ri c p on C3/{ : i s g i ven by

( 1 ) P(lA , tB ) · � lA - :t Bll l Z fI. (A l> B ) .

in whi ch A l> 8 (A - 8 ) U (8 - A) .

Every g E Ll (l") der"ines a m apping G on <.-o/{ ' ,

1 30

G U"E ) -.{ gdlL

Exe rcise W o f chap ter I shows tha t G i s a (uni formly) con tinuous map on the me t ri c sp ace (C�ll' , p) .

N ow supp ose I fn I i s a sequence in" L 1 C IL) , and

( 2 ) lim rfndlL - 0 , E E em and IL(E) < 00 . n + oo JE

Define ( 3 ) F (X ) - lfnd IL- XE E c�Ii.'

n E E Ea ch Fn i s a cont inuous funct i on on (C�U' , p) , and (2 ) gives

C 4 ) lim F (u) - 0 n ..... oo n for every u E C1l1 ' .

Exe rcise [!EJ Cb) ch apt er V shows that

(5 ) To eve ry & >0 , there correspond u E C1/l' . b > 0 . and n o such that v E C1ll ' and p(v , u) 0 , there correspond an Eo E " C1/l . b > O , and

no ;;.. 1 ; ILCE 6 E o ) < b imp l i e s

for every n � n o .

Take & , b , and no as in ( 6 ) • I f IL(A) < � , put B • E o · A , C - E . U A. Then IL(Eo 6 B ) and },-(E . 6 C) are � IL(A) < " "b , hence ( 6 )

g i ve s lIs fn dlL l � E , Ii fndlL l � & for n � n. , hen ce

11 fn d lLl � 2 E n � no

Exe rcise ill ch apte r I app lying to fl . . . . . fn o shows that 3 b' > 0 , and IL(A) < b' imp l i e s

I I fn dIL-l � 2 E n · 1 , 2 , • • , n.

The refore the fo l l owing holds

(7) ' & > 0 , 3 "1 > 0 n • 1 , 2 , • • •

• (7 ) shows that I fn I has u . a. c. i . , an d this g ives (c) in the case f • O. Sin ce f E LI ( }'-) : (c) fo l l ows � as i ly .

Q . E . D.

131 RE M ARK S TO (c ) •

(i ) Suppose in addi ti on that fI.(X } < 00 , an d fn ---+ f in meas ure as n ---+ 00 . Then [I!J ( b ) shows that

11m Iff n - d 1 - 0 n .... 00

(ii ) The fol l owin " theorem hol ds

The orem. - Suppose fI. is a fini te posi ti ve me asure on X. and (fn ) i s a s squence in L l (fl.) whi ch conv/l r�es in measure to f E L l ( fl.) .

Then the fo t .zOtJin�s a re equivalent : (a) lim II f - f ill - 0 (f ---+ f �s trong ly � in Ll (fl. ) } . n+c.:;) " n (b ) 'I g E Loo( tJ. } . l im r gf dfl. - r gfd.".

n .... ooJx n Jx . 1 (fn ---+ f "we ak.ly � in L ()L) ) •

P R O O F . I t i s cle ar that (a) impl ies (b ) even i f .". (X) • 00. That (b ) imp lies (a) i s Remark ( i ) .

N ote that i f X • N , and tJ. i s the coun ting me as ure on X , t hen L1 ( tJ.) � Z I , an d ( a) and (b) are e quivalent in this case , although .". i s not finite ( Se e W. Rudin , "Functi on al An alys is " , Mc Graw Hill 1 9 7 3 p . 82 Exe rcise m ( c) , ch . I I I ) .

( iii ) The fol lowing can verse of Vi taJi ' s the orem hal ds ,

Suppose fI.(X} < 00 , I f I i s a sequen ce i n ' If (fl.) whi ch con ve r-1 n

a . e . t o f e L (IL) . Then Ifn I h ItS u . a . c. i .

EXERCI SE 1 1 . - ( a ) Show that ill fai Is i f !L (X} - 00 . even i f j Ifnll ! i s boun ded.

( b ) (b ) Show th at poin twise convergen ce a. e . can be re-p Zace d by con ve rgen oe in meas ure in 0 .

S O L uT I ON

( a ) Cons ider X • [ 0 ' ,00 [ , fI. i s Lebe s gue me asure on [ O , oo [ , an d

f . 1. :1: f . O n n [ 0 , n ] ' . •

(b ) I f fn ---+ f in measure , Exercise [!1) ( c ) ch I I I g ives a subsequen ce (fnk ) wh i ch conve rge s pointwise , a. e . to f. The refore

p art W ( a) shows that f e L I ( fI.) . But in [) (b) , we need only that

fn ---+ f in me asure hen ce this can apply to our case.

Q. E . D.

1 3 2 E XE R CISE 102 . - Supp ose fL i s a fin i te positive meas ure on ( X. qr[) an d

p > 1. If I < c I h as ' u. a. c. i . p

(b ) If ! fn l is a s equen ce in dI whi ch con ve rge s in me as ure t o f show t hat Z im Ifn - fl 1 - o .

SO L UT r OIJ

hen.ce

hen ce

f e dI

n+oo .

Ca) By Exe rci se m (b) of ch. I I I , i f fL(E) > 0 then

l l f l [I I f I " ]l /P - d fL � - dft-

E fL (E) . "'" E �(E ) .

I f f e ·dI , �hen the pre ceding gives

l l f l d/L � ft-(E) YC

r � JE

l f l d/L < & whenever.

1ft-(El l P < r. / C , for e'very

Replace A by X, the hypothesis /Lex) < 00 shows (a) .

( b ) Th i s f o l l ows from the hyp othe s is fLeX) < 00 , Ca) . and Exe rcise GlJ .

� * *

Q . E . D .

C H A PTER 7

EXERCISE 1 . - Find an =amp l e of a ",.;notone alas s CJIl in a se t X su ah that 0 E CNl. , X E CJIl , "'.Ju t G"11l i s not a a -a lgebl'a

S O L UTION Suppo'se X i s a s e t hav ing more than one p o in t , a i s an

el ement o f X . Le t C11l con s i st s of '" , ! a I , an d X . CIIl i s a mono tone

c l as s , bu t l a l E CIll and X - l a ! Ii CIIl , hence Clll i s not a a-a l g e br a .

Q . E . D.

E X E R CI SE 2 . - Suppose f i s a Lebesgue measul'ab l e nonegative l'e a l fun ation o n W and A (f) i . the .01'dinate se t of f • This i � the se t of an points (::: , y ) E HZ fol' whi ah 0 < y < f(::: ) .

la ) Show that A (f) � Lebe sgue measul'able , in the two dimen siona l s� n 8 e .

( b ) Pl'ove that the integral o f f ovel' HZ is e qu a l t o t he ",e as'ure of A (f ) .

SO LUT I O N

( a ) Show that the i/l'aph of f i s a 'measul'ab l e subse t o f RZ (d) show t hat the meaSUl'e of the i/l'ap h is equ a l to O .

( a ) Supp o se g � 0 i s a s imp le me asurable fun c t i on . Thu s

g • c1X E + • • • • • • + ckXE

1 k where E 1 • • • • • E l< are p a i rwi se d i s j o in.t Borel sets and c i ;a. 0

i • 1 . 2 • • • • Hen ce

i s measurable in the two dimens i on a l sense .

Now i f f i s a Lebe sgue measura bl e nonegat ive real fun c t ion ,

t here i s a s e quen ce (fn ) of Le be sgue measura bl e noneg a t ive s imple

fon ct ions such th�t 0 � fn (x) t f (x ) as n - 00

1 3 S I t i s c l e at l y veri f i e d that

A ( f ) - A (f l ) U A (f 2 ) U

A ( f ) i s a coun table un ion of me asurable set s . hence i t i s me asura -bl e .

have ( b ) Le t Q = A ( f ) . U s ing the nota t i on s of De f in i t i on 7 . 7 . we

Qx - J y : (x . y ) E Q ! - [ O . f (x ) [ .

I t fo l lows that m2 (Q ) - 11m1 (Qx ) dx - J;�l f (X ) dx .

( c ) I t fa 1 10ws f rom (a ) tha t A (f + �) - ! (x .y ) : O < y < f (x) + � ,. A n

i s me asurab l e for n - 1 . 2 • • • • Thus 00

A ' -n0/n I (x .y ) : 0 < y � f (x ) ! i s me asurab l e . The graph of f i s t he un ion of A ' - A and

I (x . 0) : f (x ) - 0

whi ch i s measurabl e . Thi s shows that g raph f i s measurabl e .

( d) Sim I l arly t o (b ) . if Q - g r aph f . then Qx • I (x • f (x ) !

con s i s t s of only one poin t . fo r x E � . T�us

m2 (graphf ) • �Rlml (Qx )dx • O . Q . E . D. ,

E X E R CI �E 3 . - Find an B%amp Z e af a pa si t ive oan tinuaus fun o tian f in t hll apen un i t square in R2 • wha Be in tegrat (-re tative ta L e be sgue me asure ) is fin i te bu t s u oh tha t

J1 .(% ) ,. f (%. y ) dy

o i s infinite far same % E ] 0. 1 [ •

S O LUT I O" Put f (x . y ) • yU (

x ) • u (x ) • Vl x - } I - 1 , 0 � x < � • O < y < l

f i s a con t inuous fun ct ion on j O , 1 [x j O . 1 [ . an d f � O . We have

1 3 6 ,, ( x ) • 1

U+T

=- + 00 • Now

1 .f ,,(x ) dx • o

hen (;e f h a s t he requ i re d p rop e rty .

1 t -2 dt � 4

Q . E . D .

EXE R C I SE 4 . - (a ) Fo r an y fun ction f 07'1 Rk an d e v e ry y E Rk, t e t

fy

(r ) • f (r-y } , r E R k

(ly i s t he t ran s t a te of f) . If f E LP (R< ) , 1 ,,;: p < 00 , p ro v e t h a t

y __ fy

i s a � n i fo rm Z y con t i n u o u 8 m app i n g o f Rk i n t o LP (Rk ) . (Th i s i s T h e orem 9 : 5 )

( b ) If f an d g are m e a 8 u ra b t e fun c t i o n 8 07'1 Rk , de fin e

f -g (r ) ,. IRk f (x -y } g (y ) dy .

p ro v i de d t ha t t h i 8 in t e g r a t ex i s t s .

Suppo 8 e p and q are oon jug a t e exp 07'1 e n t s , 1 � P < 00 , a n d

eho w t h a t : .g i 8 a con ti�uo u 8 fun c t i o n if f E r,P IRk } an d q k k g E L cR } , anc t h a t f _g E C. IR } if 1 < p < oo •

(o ) Supp o e e A an d B c R� ha'Je 7'1 07'1 % 8 1' 0 Ms aeu rll 3 . L e t E b e t he 8 � b s e = o f Rk con s i 8 tin g o f el i , r f.:Jr ",' h i " h

(=-A ) n B iii ! b = x-a : a E A , b e E ! has 7'107'1 . g p: me asure . Show t h a t E i s a� c p 8 n se t and t h a t E i 8 n onemp t y . Eh�� t h a t

A + B _ ) r ,. a + b a E A a n d I: E E l con t ai n s a n on e -; ty " p e n ss t.

so L U T Z ON ( a) Su" p o , e 1 � P < 00 an d f E L'.' (Rk ) . Le t £ > O. Ill' Lu s i n I �

t heo rem , t h e re i s a fun c t ion g in Cc (R" ) su ch t hn t I f -& ! < c. l' N o w g van i s h e s ou t s i de some [ -A , Al K , A > 0 , and g : s u :) i f o rm l ,'

c o n t i n u ou s � n [ - 2A , 2 :1.1 � , hen c e 3 1» 0 , II < A , s u -: h t h a t

D t M < ll im p l i e s I g (x - t ) - g ( y l l l <

The r e f 0 re i f I s - t � < � ,

JRk i g (x - s ) - g (x - t ) I P dx

..

hence I & s - g t l p < £ (No te t h a t fh (X ) dX � J h (x -u ) dx ) I t fo l l ows t h a t .

1 37

I f s - ft l < I f s - g s l + I & s - g t l + I g t - f t A '" 2 1 £ - g ! + N g s -g t l he n c e

I f s - f t I � 3& wheneve r I s - t I < " . T h i s s hows ( a ) .

( b ) Supp o s e g E L q ( Rk ) . For h E LP , define A h · fh C -y ) g (y � dy

Then A i s a b oun ded l in e a r fun ct i on a l on LP CRk ) , an d II A U - II g ll<; . Bu t i t i s c l e a r t h a t

f* g ( x ) � A ( f _ x ) T hu s i t f o l l ows from ( a ) tha t x -+ f o g (x ) i s a cont inuou s

fun ct i o n . T h i s shows the f i r s t a s s e r t ion o f (b ) . N o t e that we have by t he p r e ce ding p roof

I f og 100 � I f I H g i P � Now i f f1 an d g l van i s h e s ou t s i de [ -A , A] k , then f 1 og 1

v an i s he s ou t s i de some [ - 2 A , 2A]k

• I f 1 < p < 00 , then 1 < q < 00

By Lu s in ' s theorem , there are f1 an d g in C CRk ) , whi c h van i s he s 1 c

ou t s i de some [ _A , A] k , s u c h t h a t

" '"

We have

, fg 1 - g B < &' , wi th "

m in ( I , ------�&------1 + II np + f g IIq

1 fog (x) - f l o g 1 (x ) 1 � I f U K g - g j, II + I f - f 1 n ! gft < t: •

The re fore i f x lit [ -A , A] Ie t he n I f og ( x ) 1 < & • Th i s shows that f og e C . (Rk ) .

( c ) F i r s t c on s i de r the c a s e 0 < m (A) < 00 • L e t f = XA an d g '" � B ' then f E L l an d g e Loo , hen ce f O g is a con t inuou s fun c t i on , Bu t s in c e

f og (x ) � JxA (X -y ) X B (y) dy '" m [ (x - A ) n B ]

i t f o l l ows t h a t E - I x : ( XA o .1:S ) (x » O !

an d the con t inu i ty o f XA * XB shows t h a t E i s open . By Fub ini ' s theorem

1 38

i s > 0 , hen - E i s non emp ty •

. In the general c as e , the regu l ari ty o f Lebe sgue measure

g iv e s t ha t m ( (x - A ) n B) > 0 i ff 3comp a c t s A i C A and B 1 C B su c h

t h a t X -A i 2 B l has p o s i t iv e m e a su re . I t f o l l ows eas i l y from t h e

prece d i n g c a s e t h a t E i s c p e n •

F i n a l l y , i f x e A . l\ then ( x - A ) n II i s emp t y , so t h a t

x • E . T h i s s hows t h a t A + R c on t a i n s t he n on em p t y o p e n se t E

Q . l. D .

ExERCI S E 5 . -1 k F � SUppO S8 1 ;S;; P ;S;; 00 • f E L rR ) an d " E L ' iR ) .

( a J Sho w t ha t the i n t e ,, �a � defi ning f .g ,

f '-; (::: ) � IRk f (::: -y ) g (y ) dy

8zi s t a fo r a Z mc s t a � l z , t h a t f ." E L" IRk ) , an d t h a t

I f '" Ip ;S;; ff ll I g I p •

( b J Shew t ha t e q"a l i t y can ho l d in ( J. J if P - 1 an d if p� oo

( e ) Ass"",e 1 0, an .! s ho .' t h a t the re e.:: i £ '- $

SO L U T I ON

( a ) The c a s e p � 00 h a s b e en d i �cu s se d in [IJ ( h ) . R e c a l l t h a t

b y t he pro o f o f Th e o re m 7 . 1 4 , f (x - y ) & ( y ) i s a m e a s u r a b l e fun c t i o n

o f (x , y ) i f f and g a re ( Bo r e l ) m e a su r a b l l' fun c t i on s . ( O n e c an

e x t end t h i s to thl' 'ca � c o r Rk ) . N o " c o " s i Je r t he r. a s e I < P < "'" .

By Ho l de r ' s i n e ql l a l i t y I' .1 -

.,(x ) � Jl f (X - Y J l l g (y ) l d \ '. - j l f l <;' l f lt l g l dY

r �. l J � .l � ( J ( l f l :; } , ] q ( I f f x - \, ) l l g ( y ) l r dy j P

= II f llt r Jl f (X-Y 1 1 1 , ( :· 1 ! ; -lY l f .. 'I' (x )

(x , y ) , -" I f (x - y l l l g (y J I P i s m e a su rab l e . Hence b y Fub i n i ' s t h e o rem ,

S.(X ) l' dx - I f l� IS ! f (x-y) l I g (y ) I P dx dy

- I f l � l f ll l g l P < 00 . 1 P

1 39

Thi s shows that + i s fin i t e a . e . , hen ce ", (x ) is fin i t e fo r almo st al l x , hen ce jf (x - y) g (Y) dY 'ex i s t s for almo st a l l x "

Let An - I x : � < l g (x ) 1 < n j , and gn

· g XA r.' g E LF'

hen ce gn E Ll n LP n Loo for all n • 2 , 3 , . "

By 0 (b) , f *gn i s con t inuou s , hen ce i s B ore l me asur ab le I f ", (x ) < "'" we can app ly t he domi nated conve rgence t heorem

to see that l im f *g

n( x ) • f .g (x )

n ... oo There fore the sequen ce o f con t inuous fun ct ion s £ *gn conve rge s a , e , to f*g , hence f*g i s a measurable func t i on .

I t i s c l e ar tha t I f *g l :0;:;; • --.:. t , hen ce

� + 1 • (p - 1 ) + 1 • p , thu s we have q 1I £ *g " :0;:;; II f lll ll d

p p F ina l ly i f p � 1 , the p re c e d ing proof can be appl i ed wi r.h

some s l ight simrl ifications ( s ince Q . 00) to ob ta in the same re s u l t .

f 1 and

(b) Case p • 1 .

( i ) I f f � 0 an d g � 0 , then by Fub in i I s t heo rem ,

I f *& l 1 � I f i l idl ( i i ) I f f (x ) • f1 (x ) e iO<x and g (x ) • g l (x) e iOtx wi t h

g � > 0 , then K f *g ' , .. 1 £ : *g l l · I f l lh l l

Case p Q 00

( i ) I f f E L 1 , f � 0 and g " 1 , then l £ *g IOO = I f "1 . 1 £ 1 1 1d oo:; '

( i i ) I f f 2 ;t[ 0 , 1 ] and g is a bounde d ( con t inuou s ) fun c -t i on on Rl such tha t 1 i m g (x ) • f g I ..., . then i f *g B "" h ft "" ' x+ oo

1 4 0 ( c ) Re c a l l t h a t, i f 1 < p < 00 and q - h ' I f g l1 = l i ft 1 9 B ( f ;"' O an d g � O ) i f an d o n l y i f OI f' " �g q p q

fa r some 01 ;... 0 an d � � 0 ( p . 63 No� suppo se � f n 1 = , g E LP an d

I f *g lp . I f l1 1 dp I t fo l l ows from t he p ro o f o f ( a ) t h a t 'I' a Ijo a. e "

so that OI l f (x -y ) l · � l f (x -y ) l I g (y ) I P a , e ,

More p re c i se l y , t h e r e i s a se t N su ch t ha t mk (N ) " 0 and f o r every

x E R" - � , I g ( y ) I i s con s t a n t a . e , re l a t ive to t he me asu re

! f (x -y ) l dy

L e t E - I y : f ( -y ) � 0 l ' E h a s p o s i t ive me asure ( s ince I f U1 1 ) hen c e by [IJ ( e )

3 " > 0 : I x - x ' I 0

:\0 .. ! � ( ," J I = ex for almo s t a l l y E x + E , an d f o r I x -x I I < h t he s e t (x + E ) n ;:x ' · :': ) h a s non zero me asure , hen c e Cx = Cx '

Tht'rc f o l'e f in a l l y we s e e tha t \ g l i s c on s t an t a , e , The hyp

o t h e s i s g E L ; ir.1jJl i e s \ g l · 0 a , e . , s i n ce m (Rk ) 2 00 . .

Thu s we have prov e d t h i s e qu iv a l e n t form of ( c )

! f o1 > 0 a n d I f wg i p - H f l1 h Dp ( 1 O . L e t t > 0 , an d

A 2 I x : ( 1 - r) l d "", o ! By de f i n i t ion of the e s sen t i al no rm , A h a s nonzero me a su re , I f we

put z _1_ JJ& f (x ) f o r x E A

m eA ) I g (x ) I . 0 for x i A

t hen H i 1 1 a 1 , an d

I f *g I ..... ( 1 - r) D & I "", I n t h e c ase l � p < """ we can u s e t he me t hod o f

x im a t e ide n t i t y " ( s e e Exe r c i s e � f (x ) h ( ) 1 ),. • ),. x • it ;\.2 + x 2

ch I X and t ake

( 0 < \ < 00 ) .

an "appro":

I f g E LP , then for � enoug h sma l l , we have

• h� * g - g Ip ';;; � I g Ip and the conclusion o f Cd) fo l l ows imme diately

REM A R K .

1 41

(i ) The i n " q u a l i t ll I f .g l -< I f il l g ' can be p rove d by u s i n g p p

J en s en ' s i n eq u a l i t y a s fo l l o w s (1 ';;; p < 00 ) :

It s u ffi ces to con s i de r t h e c a s e f � O , g � 0 , I f 11 • 1 an d p ro ve tha t l f .g l .;;; h i p p

If d ",, ( lI } • f ( x - !l } d y , ". i s p o s i t i ve a n d ",, ( R ic } - l .

te t 'fi t } - tP, hen ce 'f i s con vex ( s t ri c t l ll con vex i f l < p < oo ) .

T h u s by J"n s �n ' s i n e q u a l i t y

In t e gra t e thi s i n e q u a l i t y wi t h rep e c t t o x , on e h a s

t h e refore

I f o g S .;;; 1 9 B p p

(i i ) The p roof I i } of ( a ) s ho ws t h a t i f -I f 11 - l an d

I f .g l - S g l , Ole h a v e t h e con cl u s i on i n ( c ) (1 < p < oo ) : p p Fo r a l mo s t a l l x E R ic , I g l i s con s t an t a . e • • d t h r e s p e c t: t o

t h e m e a s u re I f ( x - Y l l d y .

Thi s i s an o t he r p ro o f o f ( c ) u s i n g t h e t he o rem in 0 ( b )

chap t e r III.

( i i i ) No t e that t h e con cl u s i on a t t h e e n d o f p . 6 3

con c e rn i n g t h e B� l de r ' s i n e q u a l i t y can be p r o v e d b y u s i n g t he

e q u a l i t y case of Jensen ' s i n e q u a l i t y , III W ( b ) , s i n ce 'I t } - exp t

i s s t r i ct l y con vex .

(i v ) I t can be sho wn ( u s i n g Hol de r ' s i n e q u a l i t y for 3 f u n c-

t i on s ) t h a t

i f :!. + 1 p q r . Se e " Ab s t r a c t Harmon i c "' a l !l s i s · ,

E . Be "i t t a.c d KJl o s !I , theorem 2 0 . l 8 , p 2 9 6

1 42

E X E R C I SE b . - L e t M be t he B an a c h space of a Z Z comp Ze", m e a s u re s

on Rk . Th e n orm in M i s a ",I a l "' l iRk ) . A s s o c i ate to , e a c h Ba rd se t

E c Rk t he se t

E (2 ) � ! ("' > lJ ) : '" + y E E \ c R\

Rk

If 1'-' an d A E M, de fin e t h e i r con v o l u tion "' . X to be t he s e t fun c t i o n g i tl lln by

(ll- 'A) ( E ) - ( ll- X ), ) (E (2 ) ) fo r II v e rll Bo re l lIe t E c Rk

; ",x A i e t he p r o du c t measure . ( a ) Pro v e t ha t ll- ' X E /of an d t h a t I ", A.x l ';;; I ll- Ih a ( b ) Prove t he formu Z a

(fl.' X ) (E) 2 jfl.(E-t ) d X( t )

for e ve ry fI. and X E /of an d e ve rll Bo re l Sil t E. Bere E - t 2 ! ", - t : "' E <: \

r c ) Pro ve t hat jfd v =

e v e r y f E C. !Rl ) . fo r

", A A i s the unique v E /of fff ("' +Y ) d",(", ) d >" ( y )

s u c h t h a t

( d ) Pro v e t h a t con v o l u t ion in /of i e commu t a t i v e , a s s o c i a tive and d i s t ri bu t i v e with re spe ct to addi t i on.

(e ) Define '" to be d i s crll te i f ll- i s concen tra t e d on a coun tab l e se t ; de fin e ll- to be con tinuous i f IL ( I ", I ) a 0 fo r e v e r!' p o in t

k k '" e R ; l e t r.t be ths Lll bs sg u e ," s a su re on R (n o te t ha t m Ii M ) . Pt'o v s t ha t ", A X i s dis cre t s i f bo t h fJ. an d :\. are di s c re te , t ha t ll- A :>.. i s con tinuou s if ll- i s aon tinu o u s and :>.. E M, and t h a t IL A :>" « m

if '" «. m.

(f) If d ", - fdm, d :>.. a g dm , ! E [, 1 IRk

) and g E L1 IR

k)

"

provs that d (ll- AA) - f Ag dm . '(g ) Sho w t ha t /of h a s a un i t , i . e . , lI·h C'uJ t ha t t he re e xi s t s a

a E M su"h t ha t a A IL = IL for al l ll- E M.

SO LUT I ON

F irst let u s p rove t h a t t he t o t a l v a r i a t i on o f IL x :>" i s ex act -

1 y 1 ", l x I x l .

The pol ar represen t at ion t he o r em g i v e s

d fl.(x ) - u (x ) d l ", l (x ) , d :>"(y) • v ( y ) d l >.. l l Y ) .

w i t h u :m ·1 v m e a s u ra b l e , and 1 1I 1 • h' l > 1 f , .. r" whe r .'

No t e that if d ' - fd. ( . be ing p r s i l ! n' "'. r. ' SU l e an a

f E L1 (d.J ) , t hen d l ' l - I f l d .

Here , we have

CII--x :>..) (E) - fIr d ",(x ) d :>.. (y) • �k [ £/"'(X )] d >.. (y)

1 43

hence d (��) (x , y) that

. - Ifr u (x ) d l ll- l (x ) v (y )d l � l (y)

u (x )v (y ) d l !l> l (x ) d l � l (y) . As luv l - l , it fo l l ows

d l!l>xA. l (x , y) - d l !l> l (x ) d l � l (y ) .

( a ) The addit ion + : RkxRk __ Rk i s con t inuou s , an d E ( 2 ) i s the set (+ ) -1 ( E ) . Thu s E ( 2 ) is a Bo rel se t if E i s Bo rel .• Now i f

I " ' ' ' ;; ; ' ,: ::) ;,; .:'�::; ,::, � ; :� ,:; ,:::;! ,; �l:e::; ;E: :"

hence !I> * A. i s comp l ete ly acd i t ive . Thu s !I> * � E. M . (I'-x� i s a complex measure , s o t he serie s in ( 1 ) converg e s ab s o lu tely ) .

I t fol lows from ( 1 ) that

L I !I>* :I. (En ) I � LI (!I>)(,\) (E� 2 » ) 1 � l!I>n l (E )

the re f� re we have I 1'-* A. I (E ) � I !I>XA. I (E) - I !I> I x I A. I ( E )

The refo re it fo l l ows (from the Fub ini ' s t he orem ) that

1!I> * A.I CRk ) � C 1 l'- l x l '\ l l CRkXRk ) 2 1 11- 1 1 :-. 1

that i s 11I- * A. I � I II- I I � a ( b ) Recal l that we have (De fin i t i cn 7 . 7 )

()Lx �) (Q) - J Lfoyd)L (X ) Jd A.(y) /Where QY - I x : (x , y) E Q !

I f E C Rk , t hen for )" E R "' , we have (E U » ) y _ ) z : (z , y ) E [ ( 2 ) \ - l z : z + y E E !

T hu '; ( E ( 2 » ) y � \ x -y : x E E i � [ - � .

. u. . �) ( I ' a j' " I E - t '" ,\ ( t ) ' r , - J Rk� "" ( e ) By de f i n t t i on o f E -y , i t fo l l ows that x+y E E i f a Id

,n l y I f x E E - ), . He � c e

XE (x + y ) - \E_Y ) (x ) , )( and Y E R ' No_ by (b) , we have

144

(JL*A.) (E) - rfx (x) d ll-(x ) d .\(y) JJ�(r._y ) hence

(2 ) (II-�") (E) - jJ lr. CX+Y) dIl- CX ) d " CY) CE is a Bore l set in Rk

It fo l l ows from ( Z ) that if s is a s impl e Bore l fun ct ion , then we have

(Zbi • ''''A t • if f

- ffS ( , > y)dfL(X )h(Y) . e C . (R)" ) " then 1 f t ':;; M , f i s Bore l measurable ,

hence' � c!l� fin'd ,J. sequen ce (so ) o f s impl e Borel function s such

that !or. Gx ) -... f (x) as n--+-oo , and I so (x ) I .:;; I f (x ) I .:;; M (A coroll ary

of The orem 1 . 1 7 ) lf the left s ide of ( Zb ) is transformed into in tegration with re spect to i ll- I * I � I • and the right one to in t�grat ion with re spect to I lI- lx l A. 1 • then we can app ly Lebe sgue ' s dom ina-ted convergence theorem to obt ain

(3 ) JRkfd (",*A.) - fff C:<+Y) dfL(X) d" CY) • f e C . (Rk )

Therefore the Rie s z represen ta t ion, theorem (for complex me asures ) shows that fL * A. i s the un ique v e M s at i s fying

and

C3b) �/ (�) d U (X) - fff (X+Y) dll-(X ) d"(Y)

(d) It is clear from (3 ) that (01 11-) * A. - "'* (0<" ) - 0< (II-*A.) . , . ( or e C) "' * ("1+"2 ) - 11- *"1 + 11- *"2 • ()l-l+)l-2 ) * A. -fL1 * " +fL2 * ;\.

thus * i s a b i l inear mapping of MxM into M.

Al so by (3 ) . s in ce x + Y - Y + x we have (By Fu bin i ' s t he o rem)

ffd (II-* ") • fff (X+Y) dfLCX ) dA.(Y)

-ff f (y+x) d fLCX ) d"Cy) • ffd (>"* II-)

hence fL * A. - A. * fL • L e • • convolut ion i s commutative .

Final ly it i s easily shown that both 11-1 * (11-2 *11-3 ) and (II-L*1I-2 ) *1I-3 satisfy JfdV - ffJf (x1+x2 +x3 ) d ll-1 (x1 ) dIl-2 (x2 ) dIl-3 (x3 ) ,

by (3 ) and Fubin i ' s theorem . <

Thus convo lution i s associative . (e) Suppose II- i s concen trated on ' A and A. i s concen trated on

B. ·'Thus' lI-x ).. is concentrated on AxB (by De finition 7 . 7 ) hence

(flo""') ( E ) z 0 i f E ( 2 ) n Ax B » 0 , that i s E n (A+ 8 ) » 0 • I t fo l lows that I fIo o . 1 ( E ) = 0 ,;henever E n (A+ 8 ) - 0 , hence (flo" >') i s con cen tr a t e d on A+ B

I f flo (Y) i s d i sre te , flo i s con cen trated on a countable set

1 4 5

A ( B ) . Bu t A+ 8 = + (Ax B ) i s a coun t a b l e se t . Thu s II- �.'" i s con ce n t ra · ted on a coun table se t . The r e f ore flo " >' i s d i sre te i f bo th II- and >. are di srete measure s .

Now suppo se that II- i a con t inuous me asu re and >. e M. By (b) , we see that for x e Rk

(flo ·>') C l x \ ) = f II- ( \ x \ - t ) d>. (t ) . jflo ( \X -t \ ) d >.(t )

hen ce (fIo " :\.) ( \ x j ) 0 s in ce the i n t e grand fIo C l x - t \ ) i s iden t i c a l l y o • Thu s I'- * :\. i s a con t inuou s me a su re .

Now i f f E: L 1 (dm ) an d >. e: �j , d e f in e

f . >. c kk f (X - t ) d >. ( t )

I f dlL • fdm , "then b y (b )

Since

hen ce

so tha t

(flo· >') (E) = f[ fE_t f (); ) dx}X (t )

Lebe sgue me a sure i s t r a s l a t i on invar i an t ,

JE_t f (x ) dx • fE f (x - t ) dx

ff (X -t ) dx d ).. ( t ) • f[J f (X - t ) d>. (t )] dX

(4 ) d (IL">') (x ) » (f o >.) (x ) dm (x ) I n part i cu l a r , th i s s hows t ha t i f 11- « m and >. e M , t hen !'- . :\. « m

(f ) Now i f d >. » g dm , g e L1 (dm ) , ( 4 ) g ive s

d ( fIo*>') (X ) • ( f .>.) (x ) • jf (X -t ) d >.(t ) " j f (X -y)g (Y) dY

t hu s d (1'- .>.) • ( f .g ) dm . (g ) I f II i s

& ( E ) - 1 • 0

the m e a su re if a e E , i f 0 e E

d e f ined by

( � is the Di rac measu re , that is t he measure of m a s s 1 con cen t ra t e d a t 0 ) . t hen i t fol lows f rom ( b ) t h a t

1 46 r jJ{E - t ) d b (t ) 2 r = fL ( E l

.JR , J \ o l fL . Thu s a i s a un i t for convo lu t i on i n M .

REMARK :;; •

(i ) H i s a v e c t o r sp" .: e . By ( d ) , • i s a " m u l t i p l i ca t i o n " i. n H I hnn ce M i s a n a l g e b r a . Mo reo v e r , M i s c o m m u t a t i v e .

( i i ) The i n e qu al i t y I fL . � B ,,; ( fL l l x I i n ( d ) s h o w s t h a t

(H , + , . , .) i s a n o rme d a l g e bra . Si n ce H i s a Ban a ch sp a ce , • m a ke s H i n to a comm u t a t i ve Ban a ch a l g e b ra .

( i i i ) H h a s a m u l t i pl i c a t i v e u n i t a s s ho ",n b y ( g ) . T h i s u n i t b a l so s a t i sfi e s I b l - b ( 1 0 I ) - 1 •

( i v ) Le t H d den o t e s t h e s e t o f a l l d i s c re te m e a s u r e s on R " I t i _� cl e a r t h a t H d i s a v e c t or s u b sp a c e o f M . ( e ) s ho ws t h a c • i s a m u l t i pl i c a t i on i n H d , hen ce Hd i s a s U b a l g e bra o f H . M o r e o ve r , i c can be s h o wn t h a t i f fL.7. f: Hd ' fL n i s con cen t r a t e d on a co un t a b l e se t Sn ar, d I fLn - fLl ---+- o , then fI. i s con ce n t r a t e d on U Sn - S

r. The r e fo re If d i s a c l o s e d s u b a l ge b r a o f H ( We se e t h a t b E H d , hen c,· Hd h a s .. un i t ) •

(v) Le t H be t h e se t of a l l con t in u o u s co m p l e x m e a s u r e s o n c Rk . By (f) , Hc i s an i de al in H. I t can be s h o wn t h a t H c i s cl c se d i n H , an d t h a t H - Hc e Hd '

(vi ) Le t Ha be t h e ss t of a l l comp l e x m .... s u re s w h i ch i s

a b s o l u t sl y con t i n u o u s wi t h r e spe c t t o Le be s g u e m e a s u re o n R " The m appin g f -> fdx !! d ll- i s a l i n e a r i so m e t r y o f LI (R " ) o n t o H Se. a Ha i s a cl o se d s u b a l g e bra of H .

No t e t h a t Ii � H ( i f k ;;" 1 ) , an d i t c a n b-e s ho ",n t h a t Ll e R-" ) a do e s n o t h a v e a m u l t i pl i ca t i ve un i t .

( vi i ) In chap t e r 8 , we s h a l l con s t r u c t a t h a t iL' 1 m . T h u s H a i s a p ro p e r s u b s e t of H c ( S e e

fL E M (R k ) s u ch c [Z] (b ) ch . V I I l ) _

E X E RCISE 7 . - (j] ( : O dsJ't o t e s t h e a -a Z g e : n. of a ! , Bord s e t s en a

t op o l o g i ca l 8 F � ce X. Le t X a n d Y b e t : p � l o g i ca Z s p a c e s , e a ch h a v ! n g a ooun t a b Z e � a s e fo r i t s top o l o gy . F ! ' � ,' e t Ii a t (13 ( x )( Y i -(j3 ( X ) x Vj ( y )

. '(13 IR"'+n ) In p a rt � �� Z a r \ (.B( R"' ) X<.B( R" ) . SO L U T I O,�

F i r s t �e c a l l the de f ini tion o f the p rodu c t topol ogy on ;(XV. T h i s i s t he co l l e c t ion of al l un i on s of se t s of the forr:1 \-x h'

",here V i s open in X and W i s open in Y .

J 47

Hence the p ro j e c t ions p r 1 an d p r 2 p r1 (x , y ) • X E X , p r2 (x , y ) • y E Y , (x , y ) E XxY •

are con t inuou s mapp i n g s o f Xx Y on t o X an d Y r e s p e c t iv e l y . Thu s i f A E <B(X) ( B E <B (y) ) • Ax Y (Xx B) i s a Bo r e l s e t

in XXY , hen ce Ax B G <B(Xx y ) .

<B (X ) x <B(Y) i s by d e f in i t i on the a - a lgebra generated by : Ax B : A E (R(X) • B E <B(Y) l hen ce the preceding s hows t h a t <B (X ) x <B(Y) c <B (Xx Y) .

Now suppose ) Vn l n = 1 . 2 . . . the topo l ogy of X ( re sp . Y )

: w", ! m = 1 , 2 . . . . ! i s a b a s e for

I f .. i s an open set in X x Y t he n 101 i s a su i t a b l e un ion o f s e t s V

n x Wm ' t h a t i s a coun t a b l e un i on of open s e t s . I n p a r t i cu l ar .. E (R(X ) x (B ( y ) . S in ce (B ( Xx Y ) i s gene r a t e d by the open s e t s on XX Y . thi s shows that <B (XxY) c <B (X ) x <B (Y) .

The refore we have (B ( X ) x <B (Y ) • <B ( X ) x <B(Y) i f e a ch of X and Y has coun tab le b a s e for i t s topo l ogy .

Q. E . D . R£ MARK .

The p r e ce d i n g p roo� s h o ws t ha t , i f X an d Y h a v e coun � a b l e ba s e s for the i r topo l o g i e s , t he n <B (Xx Y } i s gene r a t e d b y

! VxW , V open in X , W open in Y !

E X E R CHE a . - (Po t al' coo rd"� " a t e B i n RK ) . L e t Sk_1 bl! tl", u " i*t sphe I'e in Rk , i . e . the se t of a i l u E R k w h o s e d i s tan ce from o rigin 0 i s 1 . ShoLl t h a t e v e I'!! :r E R'

• e:r ce p t fo I', :r - 0, h a s a u n i q u e I'epI'e s e n t a t i o n of the fO I'm :r - 1'U Llhe I'e I' i s a po s i t i v e I'e a l n umbe r an d u E Sk _ 1 ' Sh� w t ha t c h i s g i v e s a homeomo rp h i sm o f Rk - ! 0 I on to J 0, 00 [ x Sk _ 1 •

t he

L e t m k be L e b e s g u e We a e U I'e on R k, an d de fine a me a S U I'e 0k_1 'V on 5k _1 as !o 1. l o w s : If A C S� _ 1 an d A i s a BO I'e l se t, l e t A b s the se t o f a l l p e in t s 1'U , whe I'e 0 < I' < 1 a n d U E A , a n d de fin e

<\J 0 k _1 (A ) � k m k (A ) .

PI'o v e t h � t the fO I'mu l a J. fdm k � f";k - 1 dI'i f ( I' u ) dttk _ 1 ( u ) .

Rk 0 5k _ l

. 148

is v a t i d fo r e v e ry n onnega t i v e Bo re : fun cti on f on Rk . Che ck that t h i s co i n ci de s wi th fam i l i ar re su t t s when k - 2 a n d k � J •

SO L UT I ON '

Suppose x E Rk - 1 0 1 • L e t r ex ) • I x ! (the Eu clidian norm of x ) , (thu s r e ] D , oc [ ) an d let u (x ) . x

I x n We have x • ru .

(hence lu I . 1 ).

Conversely if � • ru where r i s po s i t ive and l u i · 1 , then

Ix l · - ra u l · r

and hence u • � • Therefore the repre sentation . I x l

x • ru i s un ique . De f ine '9 : Rk _ l o l -.. ] O , .,., [x S;_l by

.,(x ) - (I x l , a � I )

I t i s e as l l y shown that . is continuous . S ince 1'-l (r ,u ) • ru i s cont inuou s , w e conclude that , i� a homeomorp hi sm .

A s a con sequence , � (Rk- I O I ) i s the a- algebra generated by � ,-1 (] r1 , r2 [xv) : 0 < r1 < r2 < 00 , V open in Sk _1 t (see remark

I of [2] ) s ince a homeomorphi sm maps Bo re l s e t s onto Bo re l sets and s o does it inverse mapping •

We denote by rE the set I rx ': x E E ! . The equal ity mk (rE ) • rkmk (E ) , r � a

holds if E i s a k-ce l l , hence ho l d s fdr every Lebe sgue me asurab l e set E .

N o w supp ose A is a Bore l set in 5k_1 an d 0 < r1 < r 2 < 00 , an d .1 et A' • .,-l (] O , l] XA) . I f E. - 'f-1 () r1 , r2 ] xA) , E i s t he set

t\J C'\J I'\J C'\J r2A ' - r1A I , r1A ' C r2A' , hence '" '" k k "', n:k (E) - mk (r2A ' ) - mk (r1A' ) a (r2 . - r1 ) mk (A ) •

I t fol l ows from the regul arity of Lebes gue me asure that

mk �A) • mk (A' ) ( A • .,-1 (] 0 , 1 [x A) ) and that if

E • 'f-1(] r1 , r2 [xA) , k �

k k '" r2 - I 1 then mk (E) • (r 2 - r 1 )mk (A) • -- - ak_1 (A) There fore r

I, ( 1 ) mk ( E ) a f 2

rk -1dr f dC\ _l r1 A k -1 d Le t A be the product measure r dr ak _1 on ] 0 , 00 [x Sk _1 ' and consider

1 49

( 2 ) (E ) = >..( � ( E ) ) E i s a Bo re 1 s e t in R k - 1 0 : ' fl i s a Bo rel m e a s u re on RJ.: _ \ o l ' ( 1 ) an d Fub in i ' s t he o rer:\ s ho w

t h a t mk ( E ) · fl e E ) i f E i s s o m e 1f'-1 ( ] r 1 , T 2 [ '< A ) whe re A i s a Bo re l

s e t in 5k _ l , I t fo l lows e a s i l y t h a t i f E • � - I (Vx W ) _ V open in

) 0 , 00 [ , W o pen in_

5k _1 - then m k ( E ) z fL (E ) , I t can be shown t h a t

the c o l l e c t ion o f a l l ( Bo re l ) s e t s. a t wh i ch t wo po s i t ive Bo r e l

m e a sure s are equal form a a - a l g e b r a , Thus mk an d fL co i n c i d e on

t he a - a l g e hr a g e n e r a t e d by t h e s e t s ,- 1 (VxW) (that is , by the

col l e c t i on o f all open s e t s in Rk _ ! O ! ) , hence m� ( E ) • fl e E )

whe n e v e r E i s a Borel se t in Rk _ I O \

(3 )

Thu s by ( 2 ) an d Fu b i n i ' s t h e o rem ,

(3 ) mk ( E ) a fL (E)

i s v al i d al so i f 0

(4 ) JRk s dmk

i f s i s any s imp l e po s i t iv e Bo re l fun c t ion ,

I f f i s a p o s i t iv e Bo re l me asu r a b l e fun c t ion , there i s a

sequence I Sn \ o f s imp l e Bo re l fun ct i on s su ch that 0 .,,:;; s 1 .;; s2 "':;;

and sn (x ) --+ f (x ) a s n - 00 . T hu s by Leb e s gue ' s monotone conve rgence

theorem, we o b t ain

( 5 ) f. f (x ) dx =

Rk J�k -1 dr i f (ru ) dok _ 1 (u )

o 5k _1 In t he case k z 2 , ( 5 ) ha s t he form

r1 roo 2 ';:

L R/ (X , Y ) dx dY • J/dr l f (rco s R , rs i n 9 ) d 9

In t he c a s e k z 3 , the e l emen tary c a l cu l u s course g ives

ffff (x , y , z) dxdydzz I;; dr[islti n a d 9

,{ ��rs in a c o s lf' , r5 ir. ij s i n If , r c o s � ) d, ] ltJ2lt

a , ( !>. ) = 1 X A ( e ,'f ) s i n 9 c.l 6 dlf C (1

T h e r e fo r e ou r f o r�u l a ( 5 ) i s a g e n e r a l i z a t i on of t h e C D s e s k 2 and k 3

Q . L . D .

1 50

E X E R C I SE '1 . - Suppo S Q (X. S. fL ) an d ( Y. J. >..J a r e (J -fi n i t02 me a su re sp a c e s , and suppo s e * i s a m e a su r e � ,, : S , J; s u c h t h a t

* (A x B ) - fL (A ) X ( B ) whe n e ve r A E S an d B E J. P,'o v e t h a t � h e n <II (E ) e v e ry F: E Sx J .

S O L U T I O N

Le t. CITl. 3 j E e SxJ : ojI (E) • ( fLx�) ( E ) I .

( fLXX) ( F: ) je r

� i s a mon o t o n e c l a s s . S in ce for A E S and B E J , we have

( fLx�) (Ax B) • fL (A ) x ( B )

t he hypo t he s i s of the ex e r c i se shows that CHl con t a i n s a l l me asu r -

a b l e rect an g l e s Ax B .

Bu t SxJ i s a l so the mon o tone c l a s s g e n e rated by t he m e asu r

ab l e re c tan g l e s , 5 0 SxJ c � , hence C!/l a SxJ .

I t fol lows that

+(E ) • ( fL X A) (E ) for eve ry E E SxJ .

Q . E . D .

E X E R C I SE 1. 0 . - (a ) Su p p o fl e E i.. a den s e se t in R1 and f a r ea l fun et i o n on R 2 flu c h t ha t f i s Bo re l measu rab l e fo r e ach % E E an d fll

% i s con tinUOU 8 fo r a l Z II E R1 . Prove that f i s Bo re l me asurab l e o n

R2.

(b ) Pro v e ( a ) L1i t h "Le b e sgue m e a surab l e " i n s t e ad of " Bore l me a su rab l e " • Then p ro v e thi s un de r t he w ea k e r hypo t h e s i s t h a t fll i ; co n ti n u o u s fo r a lm o s t a Z Z !I E R1 .

( e ) Suppo s e f as in t he first part of ( b ) • Supp o se g :R1_. R 1

i s co n ti n u o u s an d h (y ) - f (g (!I ) . y ) . Pro v e tha t h i s L e b e sg u e m e a su rab l e o n R1 . T h e n u s e L:.< s in ' s t h e C' re", t ::! o b t a i n t h e S al'! e res" Z t i f con t i n u i t y o f g L S re p l ac e d by m e a s u rab l i ty.

(d) Supp o s e g is a re a l fun ction on Rk L1hich is con t i n u c u s in e a a h v a r i a b Z e s e p a r a t e Z y . Mo re eZF L i a i t Z y . fo r e v e r!! c ho i a e o f ;r 2 . · . . . ;r �. t he mapping % 1 '-+ g (%1 . % 2 . . . . . %k ) i 6 con t i n u o u s . e t a .

FrOlle t ha t g i s a Bo re l fun ction .

SC L U T T O N

(3) Choo se

is dense · in RI .

00

d i s c r E' t e sets D: C D2 C . . . C E s u c h t h a t U Dn n = l D � < a _ I _ < a 0 < a 1 < . . . . I . and n { , o l , n , n �

D e f i n e fn (n � 1 , 2 , • • • l a s f o l l ows .

x E D • Ot h e r .... i se i f x i. D n n t here i s some i s u c !) t h a t 8 1 _1 , n · or n < X < � n • a i • n I e t

fn (x , y ) • � f ( ex , y ) +� f ( � , y ) � - '" � - or

Then .... e

� = �n ' "' : "'n )

1 5 1

( t h a t i s i f x · .\ or " (l -X ) � , l e t fn (x , y ) • x f ( "' , y) " ( l - X) f ( � , y ) )

In [ a . 1 , a . ] x R , (� , 7' ) f-,-..... f (ex . y ) = fex (Y ) i s Bo r e l m e a p l - , n l , n u r a b l e . Thu s

(x , y ) , - > �r ( ex , y ) . be i n g t h e p r odu c t o f a con t inuo u s � - or

fun ct i o n an d a Borel fUll c t i on , i s i t s e l f a Bo r e l fun c t i o n . S im i l a r l y

(x , y ) >-> �f ( � , y ) i s a Bo re l fun ct i on in [ a . 1 , a . ] x R � - or l - . n l , n

I t fo 1 1 0 .... s e a s i l y t h a t f i s Bo r e l m e a s u r a b l e , n . 1 , 2 , • • • n e U Un

, I f x , t h a t i s x E D f o r n "" N , t h en f n (x , y ) � f (x , y ) n

f o r n .... N , h�nce f n (x , y ) > -> f (x , y ) . 1 £ x (i U nn , s ince U Dn i s d e n s e i n R , the or r. ' 5 and �a ' 5 c onverge t o x a s n __ 00 •

The r e f o re t he con t inu i ty o f f Y imp l i e s t h a t fn ( x ,y ) - f (x , y ) a s

n - oo , s i nce fn (x , y ) • >' n f ("'n ' Y ) + ( l - Xn ) f ( �n ' Y ) '

[ We c ou l d choos e [n (x , y ) c f ( .. , y ) t o g e t a more s imp l e p ro o f ] .

Thu s u n d e r the hyp o t he s i s o f ( a ) , .... e c an f in d a sequence

( fn ) o f Bo re l fun c t i on s su ch that f (x , y ) - f (x ,y ) a s n _ 00 • n

fo r a l l (x ,y ) E R- . Th i s s ho .... s t h a t f i s Bo r e l m e a su ra b l e ( I n t he

t .... o d imen s i ona1 s e n s e ) .

( b ) In ( a ) i f we rep l a c e " Bo r e l m e .1 su r a b l e " by " Le b e s g u e

m e a su r abl e " , t h e n t h e

W e con c lu ce t h a t i f g - rab l e 'of x E E ( i i )gY su ra b l e .

r e s u l t in g fn i s Le h e s gue m e a su rabl e .

: R2 _ R i s s u c h t h ;; t ( i ) g i s Le b e s g u e me a s u x i s c on t i nuoll s 'l' y E R , t hen g i s Lebesgue m e a -

� o \o/ SUP Fo st' f i s a s i n t he hypo t he s i s o f ( b ) , t ha t i s

( i )fx i s L e b e s g u e m e a su T a b l e Yx E E

( i i ) ' T h e r e i s a s� t N an d m 1 (N )

nuou s fo r eve ry y E R -N .

U s u c h t h a t fY i s con t i -

De f ine g (x , y ) � f ( x , , ) i f y (i N , an d >: (x , y ) � 0 i f y E N .

1 52

We see t h a t l ic • gx an d f" d i ff e r on l y p o s s i b l y . on N . he n ce g

s a t i s f i e s ( i ) . Mo reove r g Y i s 0 i f y E N . g Y = fY i f y e N . hen c e

g s at i s f i e s ( i i ) . I t f o l l o w s t h a t & i s L e h e s g u e m e a s u r ab l e .

The s e t RxN h a s L e b e s gue m e a s u r e 0 ( s i n c e ml (� ) • 0 ) . and

f • g in the comp leme n t o f t h i s s e t . The r e f o re f i s a l so L e b e s g u e

me asurabl e .

( a ) .

( c ) Let Dn . ! . . . < ex < '" < '" < - l .n O .n l , n . . . . ! be a s i n

I f g (y ) E [ "'i . n ' '''i + l , n [ ' d e f i n e hn (Y ) t o be f ( "'i ; n 'y )

Thu s hn

i : Lebe s gu e me a su rabl e .

Now i f n _ oo . "' . ( y ) t g (y ) . and the con t iflu i ty o f f Y s h u \, s l . n 'that hn ( y ) - > f (g (y ) ; y ) . h e y ) . Thu s h i s t he po in twi s e l im i t o f

t he sequen ce ( hn ) o f m e asurable fun c t i o n s , he n c e h i s me a � u r a b l e .

Sup p o s e g i s a Le be sgue m e asur a b l e fun c t io n an d l e t us p ro v e

t h a t h ey) • f (g (r ) . y ) i s Lebe s g u e me asu r abl e .

By Lu s i n ' s t he o rem . the re i s a se t N o f m e as u re O . an d a

s e qu e n ce (!:n ) o f c o n t i. nu o u s fun (: t i on : on R s u c h t ha t g,

( y ) --> g (y )

as n - 00 , '1 y e N .

Let h ( n ) (Y) · f ( gn ( Y ) ' Y ) ' The p r e ced i n g re s u l t i s t h a t h ( n ) i s L e b.; sgue me asurabl e . I f y e R-N . t he n

h ey ) '" f Y ( g ( " ) ) • fY ( l im &n ( Y ) ) = l im fY ( gr ( Y ) '" l i.m h { n ) (Y ) n + oo n ..... oo n ·� oo

Thu s h i s t h e p a iL w i s e l im i t a . e . o f ( h ( n ' ) ' h e n c e h i s L e b e s g u e

me asurab l e .

(No t e t ha t w e n e e d t h e co n t in u i t y o f f Y f o r a l l r e x c e p t on

a s e t of me a s u re 0 ) .

(d) We w i l l p rove ( d ) by i n du c t io n on k . The r e sul t i s e d de n t i f k = 1 • ); o w sup p o s e ( d ) i s p rove d f o r k = n � 1 , an d we shall p ro v e t h i s i n t he case k '" n + 1 •

. Le t x • X l ' Y . • (x 2 .x3 ' • • • ,xl< + l ) • a n d

g,, (y ) 2 g Y ex ) '" g (x 1 . x 2 . . . . . x·J.:+ l ) .

By hyp o t he s i s gY i s c o n t inuou s fo r a l l y e Rn By t he indu c t io n hyp ot he s i s . - g" i s Bo re l m e asu rahl e f o r e ach x E R .

we d e f i n e . as fo l l o w s n I f .. , .. i ,c;. x oo , i t fo l l o.'s t h a t l , n gn (x , y ) --> g (x , y ) as n _ oo . Thu s g i s t he p o i n t w i s e l im i t o f a

sequen ce 'J f Bo re l ' fun c t i on s , h e n c e g i s i t s e l f B o r e l m e a s u r a b l e .

Q. E . D . E X E R CI ::> E 11 . - Use Fubi ll i ' s t h e o !'E:- all ': t h e !'e Z a ti oll

1.. x

, 00 f e - rt dt

o

(x > ? )

t o p !'otJe t h a t

S O LUTI ON

1A , 8:"X d.x

o 2

e - X�

s i n x has a p r i m i t i v e e - X Io. ( a c o s x + b s in x ) , an d t h i s i s

- x X e ( - c o s x - X s inx ) 1 + 10. 2

N ow by Fub i n i ' s t h e o re m , 00 l'� s i n x 1A 1 x t ---

x-- dx s s inx dx e- d t

0 0 0

00 x = A m _-_e ___ ( c: o s x +

f [ - x t

1 + t 2 t s inx ) ] dt

x= o o

C * ) rs�nx dx o

e - A t ] - --- ( c o s A + t s i nA) d t 1 + t

2

The r i g h t s i de o f C * ) t e n d s t o

I t f a l l ows t h a t A I , 1 s i n x d 1m --- x

A -,.. 00 0 x - !

2

fOO

d t ! � s o

, as A ---. 00 4

Q . E . D .

1 55

"

"

[XERCT�e: 1 . - Th" slImme tri a de riv a tive 'of a aomp l8z Bo re l m easure on Rk is defin ed to bs

(D !1- ) (z ) _ Zim !1-(B (zjr ) ) " s ym , l'+O m (B (z ; r ) )

101" ; r9 B (z; r) is t he open ban in Rk wi th , oen tar at z and radius r, Prove that The o ram 8. � imp H s s th. ana Zoio u s th.ore", fo r D 811""

SO LUTION

Le t Q be the co llection of all balls B (x ; r) where x r<:ngt' S over Ric an d a < r < 00 . Q is a substantial family i n the s e n s e o f

Defin i t i on , 8 . 2 , a n d t h e =orre sponding di fferen t i at i on i s Dsy�' Thus Theorem 8 . 6 g ives the fol lowing

Theo rem , If !1- i8 a "omp le:r Bo rf1 l me as u ra on Ric , t hen

whe r.

,fa ) (b )

(D 8':1ml'-) (:r ) ''''''fate a. e . I ", J D E Ll (Rk ) syror!1-

(c) Fo r e v e ,.y Ba r.' 1, set E,

1'- (& ) - 1'- . (E ) + r(D I'- ) (:r ) d:r /The L f1 b " Si ua df'<'C''''f, ,, ,. i . ... 0 11 Q .IE . BY "' ; , 0; !1- rR t a t i v e to ", i

E X E R CISE 2 . - SUPPO BS l !1-n I i. s a 8B qu�n cs of f'0·s i. t � v" Bo ,..Z me asurss on Rk an d

1 56

)l. ( E.' ) � I )l. (E J n=1 n

A s sume fL(R �: J < 00 • Show t h a t )l. i s a Bo re l m e a s u re . Wha t i s t h e rs l at i on b e t w e e n t h e L e b e s g u e de aomp o s i t i on s o f t he )l. " an � � h a t

o f )l. ? P ro v e t h a t 00 ([')l.) {,z ) � I ( Dfl-n J (: ) a . e .

SO L U T I ON Sup p o s e E 1 , E 2 , . . . i s a s equen ce o f p a i rw i s e d i s j o i n t Bo-

rel se t s i n R\ and E • El U E 2 U Then 00 OO OC 00 00::"1

)l. ( E ) : &/n(E ) • &J�tn (E m ) ) � �1 (�tn ( EM ) ) N o t e t h a t the l as t e qu a l i t y h o l d s s i n c e t he doub l e s e r i e s h a s p o s i tive t e rm s o n l y . T h i s s hows th a t

00 )l. (E ) = IlL([� )

r:I = 1 .. -

hence )l. i s c o m p l e t e l y a d d i t ive •

Thu s )l. i s a p o s i t i ve Bo re l m e a s u re on Rk . Sup p o s e t h e L e b e s g u e de comp o s i t i o n o f )l. n i s

(n I , 2 , . • • ) where " n i s con ce n t rat e d on a Bo re l s e t En st: ch t h a t

m (En ) Z O .

L e t :\. = I\ n . E a ch \ n i s a p o s i t i ve Bo re l me asu re , h e n c e by t he p re ce,: i n g p a rt J\. i s a p o s i t i\'e Bo re l m e a su re . Bu t ), i s con cen t r a -t e d 0n the Bo re l se t E = E 1 U E 2 U an d m ( E ) • 0 , he n ce 'I. .l m . Now

x , and

I II fn ll1 � I )l. n (Rk ) 2 fL (Rk ) < Th i s s hows t ha t th" se r i e s \' fn (x ) conve rg e s fo r a l m o s t al l

00

f (x ) : I f n ex ) n = l

" 1 � )l.(Rk ) (Thc o rer:l 1 . 3 8 )

N e x t i f E i s a Bo re l s e t i n RX , then

fLn ( E ) 2 r f elm + "� ( E ) J E n hen ce by t h e m on o t o ne conve rge n ce t h e o rem ( fn > D ) , we o b t a i n

)l. ( E ) .. J>dm' J\. ( E )

an d have t h e i den t i ty

1 5 7 II-

a - l>n , a II-s - III-n , s whe re t he s U D c r i p t s a and s den o t e the abs o lu t e ly c on tinuous and t he s in g u l ar p a rt ( re l a ti ve t o Le b e s gu e m e asu re ) re sp e c t iv e l y .

Fin a l l y l e t 0 b e a s ub s t an t i a l f am i l y in Rk , :m d l e t l: d,, no t e t he c o r re sp on d in g d i ff e r en t i a t i on .

By The o re m 8 . 6 , t o e a c h n the re co rre spon ds a s e t Ac

such t hat m eAn ) - 0 an d

S im i l arl y , the re i s a s e t A o f m e a su re 0 such t ha t

(DII-) (x ) • f (x )

I t fo l l ows that f o r B - A U ' \ U A2 U ; . . , t hen

D II- (x ) • I DII-n (x ) , X E Rk - B

he n c e D II-(x ) - I DII-n (X ) a. e .

Q . E. D .

EX E R C I SE 3 . - Supp o s e e a�h fn i s a p o.i t i ve � o n de �re aBi�w fun � tion 0'. R l . an d

fo r a t t z.

SO L U T I O "

Prove t hat 00 f ' (z ) - I f ' (X )

7'1-1 " a. e ..

L e t u b e a n o n d e c re a s ing fun c t i o n on RI . Let

u 1 (x ) - sup u (y ) y < x

u2 (x ) - i n f u ( y ) x < y

- l im u (y ) y � x y < x

• l im u (y ) y � x x < y

( a U (x - ) )

u i s nonde cre as ing , hen ce ul an d u2 d i f f e r on l y o n a f in i te o r coun -t a b l e s e t D , so t hat if

v (x ) - U I (X )

t hen v i s l e ft con t inuou s , v � u , an d

u (x ) - v ex ) + w (x )

whe re w i s p o s i t ive , w (x ) • 0 i f x � D.

1 58

Th i s s ho w s t hat i f u i s nonde cre a s i n g and bounded t hen

( . ) u (x ) • c + v (x, + w (x )

c · x�..i!' u (x ) , v E NBV , W E BV i s pos i t ive an d i s d i f fe rent from 0 on ly o n a fin ite o r coun t ab l e se t D ( Fo r we ha\ e

L w (x ) � the t o ta l vari ation o f u (3 l im (x ) · l im u (x ) ) x E D n x�oo x ... - oo tl

The l emm a t o B . 1 9 s hows that h ' (x ) � 0 a . e • Inve s t i gate di -re ctly t he quo t ien ts

f ey ) - f (x ) y - x

we o b ta in u ' (x ) � V t (x ) a . e .

N o w supp o s e fn an d f a s i n t he exe rc i s e . I f our at ten t ion i s re s t r i ct e d t o ] - oo , AJ (A < ool owe can supp o s e t hat f , f I ' f2 , • •

are boun de d. By the preceding o bs erv a t ion ,

( 1 ) fn (x ) . en + gn (x ) + hn (x ) whe re gn E N BV , cn � 0 , hn ( x ) • 0 i f x i! a coun table set Dn , and

Since (2 ) f� (x ) . g� (x ) a. e .

f (x ) • I Cn + I gn (x ) + I hn (x ) � c + and thI S i s a l s o the ( 0 ) decompos i t ion d f f , i t f o l l ows

each � n

(3 ) f ' (x ) • g ' (x ) a. e .

There fore i t rema i n s t o s how t hat

( ' ) I f

and

then

Le t u s

(4 )

in E NBV

g ·

g ' ( x ) ·

are nonde creas ing

L)n i s bounded

L >� ( x ) a . e .

now prove C * ) • By t heorem

fun c t ions

8 . 1 4 , in (x ) • II. n (] - 00 ,x [ ) n • 1 , 2 , • • . . .

n •

i s a fin i t e po s i tive Borel Measu re n • 1 , 2 , • • •

( 5 )

( 6 ) g (x ) • � ( ] - 00 ,x I ) . Theorem 8 . 1 8 and 8 . 1 show that

( 7 ) g� (x ) - DII-n (x )

i ' (x ) • DfI-(x )

a. e ,

a . e .

N o w ( 5 ) . ( 7 ) and exe r c i se 0 sho .... that ( I ) hol d s . Thi s comp l e t e s the proof of t he exe r c i s e .

g (x ) + h ex )

that

1 J 2 , • • • • •

I f

Q . E . D.

1 59 NOTES

(a) It can be s ho wn tha t e v e r y r e a l f E BV i s the di fferen ce

of t ... o n on de cre a s i n g fun c t i on s i n BV, n alll e l y f - " - "' , an d

"(r ) - Tf

(r )· + l i . f (r )

]( -+- - 00 ThUS· t he de composi ti on ( *) g i v e s u s

(.*) f (r ) - c + g (r ) + h � )

wi th c - l i m f (r ) , g e NBV , an d h i s 0 e r cept on a r+-oo

co un tabl e a e t IJ

(b) It CIUI be sho wn t h a t i f f spl i t s as in ( 0 *) t hen

( *) ' f ' (r ) - g ' (r) a . e .

(c) N o t e t h a t ( * *) an d ( 0 ) ' can b e appl i e d t o cO lllp l e r va l u e d

Theorem 8 . 1 3 (b) .

( d ) In ( * ) , u a n d v have the sallie t o t a l v a r i a t i Q n , n allle l y ,

l i m u (r ) - l illl a (_ ; % ... - 00

EX E R CISE Lf . - (a) Supposs (ai ) i s a seql':.II : i: of pos itille re aL numbers

f a . " 00 71-1 1.

and

Le t a >0 , oons truct a Beque·n ea (::: . ) in [ 0, 00 [ su ch that (i ) Eaeh 1. .

intarllal in [ 0, 00 [ whioh doe s not intersect ll r . 1 ' .. 1 2 ha s l eng tr 00 t i t - . , . .

< b . (ii ) [ O, OO [ c !JI:::C ai ' :::i+ ai r (i i i ) ::: 1<:1' , <' . . . . .

(b ) Suppose I �:l is a sequen oe of comp le::: numbers such that I: °i - "" . Show that there are segmen ts Ji - ] :::i - I ci l '::: i + l ei ' [

euch tha t :::i " :::j i f i " j , l :::i ! is dense in R l , and ellery '" e R l l i e s in infinitely many Ji ,namely (Ji ) , such t hat

k

SO L U T I ON

l im ::: . - ::: k .... 00 10k

Pu t f (::: i ) .. ci ( i - 1 , 2 , . . . ) , f (", ) - 0 fo r a Z Z other ::: . Prolle that f is nowhere diffaren tiab l e .

( a ) Let n • • 1 . Since L an - 00 , w e can -iind n1<n2 < . . . .

su ch that

a + • • • • • + a _ I ... 11

n k n k + l k • O , l , � • •

1 60

Le t an � k � . k • 1 . 2 • • • • Fo r e ach k • 0 . 1 . . . . . we can p l a-ce xn L + 1 • • • • • x in such a manner t l)at [x . x ] is

� n k+1 -1 n k n k+ l .

con tained in J U J l U • • • U J _l . whe re Ji - ] x . - a . ,x . + a ·i [ , n k nk+ n k+ 1 1 l. l. and s imul taneously X l < X 2 < • • •

This comp le tes t he construct ion of (a) .

(b) F ir st suppo se ! b 1 , b2 , . . . \ ·i s a sequen ce in } a , oo [ su ch that f b . i= 1 l.

• 000 .

' Then we can find I nk I su ch that n 1 z 1 , n 1 < n2< . . , and b +b + • • • + b n k n k + J. n k+ 1 _1 ;;" 1 k a 1 , 2 , • • •

Thus i f D � ! 1 , Z , . . . ,n2 - 1

then

Suppose

L b . - 00 i E D 1

and L b .. • OQ i E D , l

II c. 1 - OQ • 1= 1 ..

Repe ated appli cat ions of t he prece d-

ing show! that each fixed n ,

! 1 , 2 , • • • j ha s a part i t ion ! D1 , D; , . . . \ su ch that fo r

2: I c . 1 • 00 i E D 1 n We o b serve t hat in t a ) J 0 , <>0 [ can be repl aced by Rl 1£ ( i i i )

is om itte d , and t h a t the sequen ce (x i ) in ( a ) can be choosen to be out of a g ive.l count able s e t .

Thus we can d� fine by induct ion o n n the sequen ces (x i ) i E D such t h a t :

n ( i ) ' Every interv R I w h i ch do,", s not con tain any X i (i E Dn )

has l eng th < Z-n .

( i i ) ' R ci'e'

nJ i w i t h J i a ) x i - l ci l , xi + l c1 1 [ . n

( i.H ) ' J f .,. k , i t= Dn an d k E Dn ' them X i r xk • ( i v ) ' x i '" x j i f i E Dr. • J E D m an d m < n •

\..,JThUS hy ( i ) ' . ) x1 , x2 • • _ . ! i s dense in R.

By (i i ) ' , e ach x E R l ie s in i nfin i t e l y many o f t he .Tn ' and the l ast cond i t ion i s a consequence o f the con struct ion in ( (1 ) .

By (i i i ) ' and (iv ) ' , x . .,. x . i f i '.,. j • Th is g ives the 1 J con st ru ct ion as requ ired in ( h ) .

1 61

Now de fine f by f (x i ) � ci ; f (x ) - 0 for al l other x .

Let E - j x . : c . � 0 �\ . T I f x , E E , to every 1» 0 the re cor-, 1 1 . "" re sponds an x E R - E such that I x - x i 1 < 1> . Thu s ! f (X� - f (X i ) 1

� hl x i �

hen ce the quotient I f (xx

) - f (x t ) I i s not bounded in every neigh -

- x . � borhood of xi ' hen ce f i s n o t diff eren t i able a t x i ' If x (l E , to eve ry 6 > 0 , the re correspond an x , E R - E , x ' 1- x , and an xk E E such t h a t l x ' - x l < b , I X k - x l < 1\ and X ' E J k , H fo l lows that

I f (x ' ) - f (x ) I . 0 ; I f (x k ) - f (x ) ! i ck r ;;.. 1 · x ' - x x - x I �· I-S:

• k Thi s shows that

l im I f (y ) - f (x ) I .Y+-X Y - X

doe s not exist , hence f is not different i ahle at x .

There fore f i� nowhere di ffe rent i ab l e .

Q . E . D .

E X E R Cl � E 5 . - Supp o se P; c [ a , b ] , m (E) a O. Co n s t " u c t an anso L " t o! Ly

con t inuous mon o t o n i c fun ction f on [ a , b] s u ch t ll <l t J" ' Ix ) ., DO ta l' B T) ll l'Y ::: E E.

so LUTTON

then

such

Observe that if V is an open se t o f f in i te m � n s u r e a n d

h ex ) - f.�(t ) dt 00

h ' (x ) - I for al l x e V . Suppo se m (E) - O . We can find the open s e t s V1 -J V2 :::l . . ;:l E

that m (Vn ) <. 2-n , since Lebe sgue m e a � U T C i s ou t e r regt' l a r.

Let i -L X v ' Thus h ll � 1 . Le t n x

f (x ) - f g (t ) d t

a f i s absolutely con tinuous (Th e o r em 8 . 1 7 ) an d mono ton i c .

Suppose x e Vu . 3 II > 0 such that ) x - lI , x+ b l C Yr. ' hen ce i f

x ' e ] x - lI , x "'''' [ , w e have

1 6 2

I f (x ) - f (x ' ) 1 - lf Xg ( t ) dt l � !f X c/.. v + " , +X V ) dt l - n i x - x ' l .

x ' x ' 1 n

I t fo l l ows t hat f ' (x ) ex i s t s an d i s + 00 i f x E

f ' (x ) - 00 for eve ry x E E

n V , hen ce n

Q . E . D . \

E X E � C I SE b . - Show t h a t t h e p ro du c t of two a b s o l u t e l y con t i n u o u s fun ction s on ( a . b ] i s abso l u te l y con t in u o u s . USB � h i s to de r i v e a theorem a bo u t i n t eg ra t i o n by p a rt s .

SO L u n ON F i r s t o b s e rve t ha t by Theo rem 8 . 1 6 , F : [ a , b ) -- C is ab s o

lute l y c ont inuous i f and on ly i f the re i s an f E L 1 ( [ a , b ) , dm ) su c h that x

(1 ) F (x ) · F (a ) + J f (t ) dt a

(a :{ x � b )

Now supp ose f and g are integrab l e funct i on s on [ a , b ) . Sup p o se a ,;( a' � b ' ,.;; b . By Fub in i ' s theo rem we have

b ' y b ' x ( 2 ) f [f f (X ) dX]g (Y ) dY + I [ r g (Y ) dy]r (X ) dx

a ' a ' a ' . a t

If f (x ) g (y ) dx dy D U D '

b ' b '

t. I .•...•... ,---..,._....,. x � y

r / 0'

2 [� , f (x ) dx ] [i , g (y ) dy ] y � x " ........ !'-.---..,

a I b t I t fo l l ow s e a s i ly that i f F i s a s in ( l ) an d

x G (x ) " G ( a ) '" f g (t ) d t , then we have F ' 2 f a . e . an d

a

G ' - g a . e . , and by ( 2 ) b ' b '

(3 ) F (b ' ) G (b ' ) - F (a ) G ( 3 ) • f F (x ) g (x) c.L, + j G (x ) f (x ) dx a a

Since i Fg I "" .l1 l g I • Fg E Ll on [ a , b ] . S im i l arly G f E LI . Thu s FG i s absolutely con t inuou s .

The refo re the pro duct of two ab so l u t e l y cont inuou s fun c t i on s o n [ a ,b] i s a b s o l u t e l y con t inuou s .

Mo reov e r , (3 ) g iv e s the formu l a fo r integ rat ion b y part s

( 4 ) b J FG ' dx

a

b - FGJ� - � F ' Gdx

a in which F and G are absolu te ly con tinuous fun c t i ons on [ a ,b ] .

Q . E . D .

E X E R C I S E 7 . r If f i s a re a l fun c t io n o n [ O . l J an d

Y( t ) m t + i f ( t. J

1 63

t he l eng t h of t h e g rap h of f i s . b y de fin i t i o � , the t o t a l vari a t i on o f y on [ 0 . 1 ] . Sho w that t h i s l e n g t h i s fi n i te if an d o'fl l y if

f has b ounded vari a t i o n .

Su p p o s e f ( O ) = 0, f i s con t i nuo u s a n d n o n d e c re a si ng. a n d fa i s t;he s i n g u l ar part of f ( s e e t h e o rem 8. 1 8 ) . Prove t h a t t he l e n g t h o f t h e g rap h o f f i s 1

fs ( 1 ) + f VI + [ f ' ( t ) ) 2 dt o

How doe s t h i s form u l a change i f f E BV bu t f i s n o t n e c e s -s a r i l y m on o to n i c ? How l on g i s t he g rap h o f t h e fun c t i o n t ru. ct e d i n examp l e 8 . 2 0 (b ) ?

SO L UT I O"

cons·

It is seen that u , v e BV imp l i e s ll + V E BV. S in ce If (t ) • t i s of bounded variation on { O , l ] , f E BV iff Y E BV (on [ O , l ] ) - if y et ) • t + i f ( t ) . Therefore the len g th o f g rap h (f) i s f in i te i f

and o n l y if f E BV o n [ 0 , 1 ] . Now suppose f (O ) • 0 , f i s con t inuous and nonde c reas ing .

Thu s f e N BV . By Theorem 8 . 1 4 (b ) and ( c ) ( 1 ) f (x ) • fI- ( [ O ,x [ ) 0 � x � 1

and fI- i s a p os i t ive cont inuous measu re ( fI- (\ x I ) • ° . x ) I f the Lebe sgue decompo s i t ion of fI- i s d fl- · d fl- s + hdx , then

h i s a pos i t ive (con t inuou s) measure , h ex ) • f ' ( x ) a . e . (Theo rem 8 . 1 8) , and fs i s defined by x

(2 ) f (x ) · fs (x ) + f f ' (t ) dt , hen ce fs ( l ) . l iL s 1 ( ] O , 1 [ ) o

This shows that i f � i s associated to Y as in 8 . 1 4 (3 ) Y (x ) - x + H (x ) - � (] O ,x [ )

then ( 4 ) d �(x ) - idfl-s (x )

h�nce by Theo rem 8 . 1 4 (b ) , (l + i f ' (t ) ) dt

1 64 x

( 5 ) Ty (x ) . l ifLs 1 (] O ,X [ ) + 1 1 1 + f l ( t J l dt o

Sin ce the length o f graph (f) i s the t;ll tal v ar i a t ion a t Y on [ 0 , 1 ) , and l fl-s I Cl O , l [ ) is fs (l ) , thi s g ives

1 L · L (g raph (f) ) s fs (l ) + IVI + [ f ! ( t ) )

2dt

o Note that in the p rece ding p roof , we have used the iden t i t y

I v + (J I - I " I + I (J I if v.L (J in the imp l i cation (4 ) -.. ( 5 ) .

Suppo s e f i s cont inuous , f E BY , and f i s no t merely mono -ton i c . ( 5 ) s h o w s t h a t 1

L - l fl- s I C l O , l [ ) + iV'! + f ! ( t ) 2 dt o

Bu t now f i s no t in creas ing , hen ce fl-s needs not be p o s i t ive , s o that fs Cl ) m ay b e d ifferent t o I fLs l (] o , l [ ) .

F inal l y , i f f i s as in 8 , 2 0 (b) , then f • fs (s ince f ! 2 0

a. e . ) an d so L (grap h (f ) ) • 1 .

EXE R CISE 11 . - If E i s a L e b e sgue measurab l e s e t 'in R k, t he upp e r

an d l o w e r l im i t ! o f t h e q u o t i e n t m (E n 8 (::' j /) ) )

m ( 8 (z ; a ) ) ar. ca Z Z . d � h. upp e r an d l ow e r den s i t i e s o f E a t z , fiE (z ) an d £c: (z ) If t h , U Q are e q u a l , t he i r common v a l u e DE (z ) i s t h e den s i t y (som etime s ca l l e d t h e m e t ri c de ns i t y ) of E at z. If DE (z ) - 1 , z i s a poin t o f de n s i ty of E.

P ro v e t h a t DE (Z ) - 1 at a l mo ! t a l l z E E a n d DE (z ) - a fo r a lmo st a l l z e E .

Co n s t ru o t a s e t E c R1 s u c h t ha t £E ( O ) � DE ( O ) , Ca� i t hap p e n t h a t £c: ( 0 ) % a an d DE ( O ) - 1 ?

SO L U T I O N

There i s a Bo re l set E ' such that m (E t:. E ' ) 2 O . I f fl- i s define d by

fl- (A ) s m ( E ' n A)

t hen II- i s a Bo re l m e a sure on R k , A s imp le app l i c at ion o f [] (First re duce to the case m ( E ) < 00 ) and Theorem 8 , 6 g ives

DE (x ) • n fl-ex ) • 1 fo r almost a l l x E E , and

Dt (x ) • D fl-Cx ) - 0 for almost al l x e E .

Now l e t A be the open set 1 1 1 1 1 1

l rf 'n [ U ] :r! 'r, [ U • • • • U 1 ZiiT ' (2p -I) ! [ U . . . . . and E • A U ! -x : x E A ! .

I t can be verfie d that DE ( 0 ) • a an d i\ ( 0 ) C 1 .

1 65

Q . E . D.

EXERCISE ., . - Show (wi t h t he aid of t he Hau s do rff ' s 1I . .1::: im 12 : -· ,' " t iw '� rem ) t ha t there exi s t re a l di s co n t i n u o u s fun c t i o n v r o� R" . � �h t h a t

(1 ) f (x + y ) 2 f (x ) + f ry ) fo r a l l x and y E Rl .

Show t h a t i f (1 ) h o l ds and f i 6 L e b e s g u e me ao� ra L Z e . then f i 8 con tin u o u s .

Sho w t h a t i f ( 1 ) ho l ds an d the g rap h o f f i s n o t den se i n t he p l an e . t he n f i s con tinu o u s .

Fin d a 1.l con tinuous fun c ti o n s w h i ch s a t i sfy (1 ) .

S O L U T I ON

By the Hausdo rff l!Iaximal i t y t heorem , every (non ze ro ) ve c t or space V over a f ie l d X has a �-bas i s ; in part icu l ar , R ha s a Q- b as is j ed i E I •

I f f (e i o ) • 1 and f (e i ) • 0 , i E I , i tl i . , then we can extend f to a R - l ine ar map o f R in to Q. Thu s

f (rx + r ' x ' ) • rf (x ) + r ' f (x ' ) r , r ' E Q; X , X ' E R In p art icu lar (r • r ' • 1 ) we have

( 1 ) f (x ... y ) . f (x ) '" r'(y ) for a l l x , y E R I f f i s con t inuous t hen t he graph o f f con t ains the inte rval [ 0 , 1 ] which con t radi cts the f act that f (R) • Q.

This shows that f is d i scont inuou s . Suppose f s at isf iis ( 1 ) . A s imp l e con sequen ce of ( 1 ) is t hat ( 2 ) f (rx ) • rf (x ) r E Q

I f f is not cont inuous at x o , then by ( 1 ) , f is d i s con t inuous a t 0 hen ce there i s a sequence (xn ) with l im xn 0 , and £ >0 such that

I f (xn ) I > £ n • 1 , 2 , • • •

We can choose (nk ) su ch that l im kx • O . Bu t t hi s impl ies t ha t k .... ao n k

I f (kx ) I > k£ , n k and hence f i s no t bouded on any 1 a , b [ .

1 66 There fo re t h i s shows that i f f s a t i s f i e s (1 ) and f i s bou n d

e d o n some ] a , b [ , then f i s con t inuou s . I f x E R , x � l im rn .... here (rn ) i s a sequen ce o f rational

n ... 00 numbe rs . N ow by ( 2 ) f (x ) - l i m f e rn ) - l i m ( rn f ( l ) ) - xf ( 1 )

Thus f ( x ) - x f ( l ; Suppose f i s Le besgue me asu rab l e . Thu s

E - I x : - ). < f (x ) < ). I has po s i t ive me asu re for some ).. > 0 . Bu t then by exerci s e m ( c ) c h . VI I , E .. E :::l ] a , b [ fo r some inte rv a l ] a , b [ . By (I ) , I f (x ) 1 < 2 \ i f x e E + E , hen ce f i s boun ded on l a , b [ . Thu s f i s con t inuous by the pre ceding resu I t .

Suppose f s at i s fi e s ( 1 ) an d f ( 1 ) - l , Hence f (rx ) � rf (x ) , fo r reO . By the p re ce d in g , cont inu ity o f f imp l i e s tha t f (x ) a x . Hence i f f i s d i s con t inuous t hen 3 .. E R - 0 an d f (",, ) - l •

Note that I(r- r ' )"" + r ' : r E O I i s den se in Rl . Hen ce \( ( r - r ' ) ",+ r ' , r) : r , r ' E 0 1 is den se in R2 .

I t i s clear that t h i s s e t i s con t ained in the graph o f f. Hen ce the d iscon t inuity of f imp l i e s that graph (f ) i s dense in R2 .

The re fo re i f f s at i s fi e s ( I ) , and graph (f ) i s n o t den se in RZ then f (x ) - xf ( l ) , hen ce f is con t inuou s . Q . E . D .

EXE R C I S E ], [] . - Po r f E L co cW' ), defi n e f t (::: ) 8 f (::: - t ) ( t E Rk ) and assum e t hat

lim 1 f t - fl oo • 0 t+O

Thll n o rm i s the e s sen t i a l sup rem um . Pro ve that tm ,fe r t h e s e co n di

tion s t he re i s a u n i fo rm l y con tin u o u s fun c t i on g on Rk s u ch that

so LUTI ON

g (::: ) • 1 (::: ) a . e .

sincel f (x - t ) dt B ( O , r ) - r f (x + t ) d t � r f (t ) dt JB ( O , r l JB ( x , r )

t hen , i f f E Ll (Rk ) , .... e h ave : (see Remark 8 , 7 an d Exerc i se [] )

(1 ) 1 im r-O

l f f E LOO to g s ho .... s that

1 r f (x - t ) d t 2 f (x ) a . e , m (B ( O , r ) ) JB ( O , r )

l im r f (x - t ) dt - f (x ) r+O JB( o , r )

O < r < l

1 6 7 f o r almo s t a l l x E B (x • • l ) . It fo l l ows e a s i l y t hat i f f E L � the n (1 ) ho l ds fo r f .

Put h n --_,-_ X 1 m ( B (O ,�) ) B C C ' il) t hen

(f o hn ) (x ) " gn (x ) = I I I f (x - t ) d t m ( B (O ' n) ) B C O , l /n )

so that l i lll in (x ) • f (x ) a. e . n . oc

Now we have

gn (x - t ) - gn (x ) �k.k [ f (X - t -y ) - f (x -y) ] hn (y ) dy

. r ( f - f ) (x - Y ) hn (y ) dy J Rk t

By Exe r c i s e Ii] (b ) c hap ter VI I i t fo l l O'Pis t h a t

(3 ) I gn (x - t ) - gn (x ) I .:::;; 1 f t - f loclho l l • I f t - £ I ..,

Thus gn i s un i f o rm l y cont inuous an d \ gn I i s equi con tinuo u s . We have t he As c o l i ' s theo rem

THEOREM . Le t K be a eomp a et lO e t r- i c sp ace, an d A c C IK) . If

Ii ) Fo r- 8 v e r-y r E X 1 f Ir ) : f E A ! i 8 boun de d

Ii i ) A i s e q u i eon t inuou 8

T h e n . v . r-y Ife q u e n ee (gn

) in A eon tain s a s u b s e q u e n e e

eo n v e r-g B s un i fo rm l y I to a eO>1 t i n u ou s fun c t i on on X ) . ( g ) 4Ih i eh

n 1(

IIf X i 8 a lOe tr-i e sp a ee , A c CIX ) i s e a i d t o b e equi eon t i n u o u 8 i f

YE > O , 3 a > 0 : If lr ) - f ly ) 1 < E whe>1 e v e r d lr , y ) < II fo r a l l r, y E X a n d a l ! f E A )

We w i l l n o t prove this t he orem . A de t a i l e d proo f c an be fo un d in 1 7 . 2 3 of .. Prin c i p les of Mathem at i cal An alys i s " W . Ru d in Mc Graw Hi l l 1 9 6 4 •

By (3 ) (g n ) s a t i s f i e s ( i i ) o f t he the orem We have

I g n (x ) 1 � I ft,l hn i l • 1 f 1 00 (r. • 1 . 2 • • • • )

hen ce gn sat i s f i e s ( i ) . Thi s s ho ws that we can f i nd sequen ce s

(g 1 ) , (g 2 ) ' · • • • su c h t h at n , n , )

(4 ) (gn , k + l ) i s a suhs equen ce o f (gn , k J an d (gn , l ) i s a sub-

sequ e n ce o f (gn )

(5 ) (g n , k ) conve rges un i fo rm l y on t he c l o s e d bal l B' (O . k )

1 68

I f we of (g ) which

k n R . By ( 2 ) we

cons i de r ) g l , l ' g2 , 2 ' • • • ! ' thi s i s a subsequence converges pointwise to a con t inuous function g on have

(6) g (x ) • f (x ) a . e .

U gt - gft 00 · n f t - 0 00 The norm being the e s s ent ial supremum . Bu t g t - g i s con tinuous , and so

II gt - gfioo " sup ! I g (x - t ) - g (x ) 1 Thus l im II gt - g lloo · l im II ft - flloo • 0 shows that g i s uni form ly continuous .

Q . E . D .

EXERCISE 1 1 . - Sup p o s e G i s a s ubgro up of R (re l a t i v e t o addi t ion ) . G � R l an d G is Lebesgue me asurab l e . Pro v e t ha t m (G) - �

S O L U T I ON Suppose m eG ) > 0 and we wi l l prove �hat this wou ld l ead to

a contradi ct ion .By Ex. WCc) ch.VI I , G + G con tains an interval ] x - t,x+ t [ (hence x e G ) . Since G 1 5 a g roup , this g ives ] - t , t [ c G ,

hence ] -n t , n t [ c G by induction o n n ( ] - (n+ l ) t , Cn+l ) t [ - ] -n t , n t [+ ] - t , t [ ) , so that G • R. This con tradiction s hows that ·m (G ) • O .

Q . E . D.

EXER CISE 12 . - Ca l l t a period o! the fun ation f on Rl if

f (::: +- t ) - f (::: ) fo r a Z Z ::: e Rl . SUPP OSQ ! i s a real Lebesgue me asurab l e fun c t ion wi t h p e ri o ds s an d t such t hat sit i s i rra ti o n a l . Prove t h a t the re i s a cons t an t \ s u ch that

f (::: ) - :\, a • • •

but that ! n e e ds n o t be a·on s tan t.

S O LU T I ON

(i ) Suppo s e G i s an addit ive subgroup· of Rl . I f x l , x2 " • • •

are in G and l im xn • x, e G , then xn - x e G , an d l im xn-x - O . Hence G has n o l im i t point if and only i f 0 i s an i so l ated p o i n t o f G. Thus G i s not dense i n R iff 0 i s an i sola ted p o i n t of G .

If 0 i s i so l at e d i n G , and a · m in I x E G : x > 0 then i t is eas ily verified that G - aZ. - ! ·o ,� a , �2 a , • • • !

1 69

( i i ) Suppose s and t are two real number , and si t i s irrat ional . Le t

G . \ ms + n t : m ,� e Z ! G i s an add it ive subgroup o f R. By (i) , i f G is not dense in R then G • aZ, hen ce s • p a , t • qa , hen ce s i t · p/q is in Q. a contradiction . Thus ' G is den se in R .

( i i i ) Supp ose f a s in the exe r ci s e , s and t are two pe riods of f, and G as in ( i i ) . Hence every y e G is a period of f.

For .. e R . l e t Eo< · ) X � f (x ) > '" !

Ev�ry y e ' G i s a period of f . hen ce x + y e E", iff x e E� Thus for y e G . Ol e R we have E .. + Y • E ...

Suppose ", i s such that E � has a point o f den s i ty x • • We wil l prove ' that every x e R is a point of den s i ty of E ...

x . i s a poin t o f den s i ty of E ... hence DE (x . ) • 1 . hence ( see Exe rcise m ) , '"

( 1 ) To each 0 < £ 0 such that o < r < s imp l i e s

ZEr < m (B (x . ; r ) n E",) � Z r L e t f. and /) be a s in (1 ) . and fix x e R . n . < � , By ( i i ) .

G i s dense in R, hence x . + G i s dense in R. Hen ce for n • 1 , Z • • • 'we can find an Yn e G su ch that

I x - (x . + y n ) I < Z -n We have , sin ce Yn e G (hence E", + Yn 2 E ( Z ) B (x . + Yn ; r. ) n E o< · [ B (x . ; r. ) n E",l + Yn and (3 ) m (B (x ; r . ) A B (x . + Yn : r. ) ) < Z . Z -n

where A A B · (A - B) U ( B - A) . (1 ) . (Z ) and (3 ) g ive ( 4 ) m (B (x ; r . ) n E",) > Z u . - Z . Z -n

(3 ) ho lds fo r eve ry n " 1 . Z • • • • • hence ( 5 ) m (B (x ; r o ) n E", ) � Z£r • •

Therefore the fol lowing ho l ds ( 6 ) To each 0 < £ < 1 , there correspoJlds a lI > O such that

0 < r < 11 imp lies Z f.r � m (B (x : r )'n E",) " Zr I t fo l l ows that x i s a poin t of density of E ... (iv) (iii ) shows that i f E", has a po in t o f dens i ty x. then every ,

x E R i s a point o f den s i ty of EOl ' I n that case , w e s e e that DI (R - Eo) • 0 (Exercise 00 ) .

Otherwise t hen

1 70

Let J\. - sup ! .. E R : m (E� ) - ol . There is a sequence O<n t J\. with m CE� ) - O . (Note that there is at least an "'nwi th m (E�) - 0 ) .

I f 0< ", � , then Eo< c E, . I f now m (E;') - 0 , this implies that m CEJ) - 0 and m (E � -' EOt) - 0 , hence f (x ) ;;... .. for almost all x e E� , hence for almost al l x E R.

Let E - n E", , Ec • U E� , hence m (E c ) • O . I f x e E n = l n n = l n then f (x » o J\. , then m (Ec ) • 0, hence f (x ) :s;; J\. for almost al l x e R. We deduce from this that f (x ) :S;; J\. for almo st all x E R (by considering a sequence O<n � X ) . There fore finally we have f (x ) • J\. for almost all x e R.

Now let G - I ms + nt Define f by

m,n E Z ! , ' sIt being irrational . f (x) • 0 f (x) • 1

x e G x � G

It is clear that f has G as group of periods , and f is not ···n·' · tant • Therefore in the precedings f is constant a .e . but n c .: .; , not be constant .

Q . E . D .

E X E RCISE 1 3 . - Suppo.e f i. a aon tinuou. aomp Zez fv�ction o n [a . b ] �ith totaZ vari ati on V. Prove that t o 8ach � < V there corre sponds a a > 0 �i th the foZ Z OLling property :

If a - Z o < zl � " < zn - b and if IzC zi. _1 1 < II

fo r i - 1 • • • • • " . then L I f (z ... ) - f ez . 1 ) 1 > � . i-1 · � -

SOL UTl ON

I f P • I a • ·x. < �1 < • • • < xn • b ! we denote by L p the sum L I f (xi ) - f (x i _1 ) 1 i = l

I t is clear that P c P ' imp lies L p � Lp .

Q . I Let W < V • the to tal vari ation of f. By defini tion Ilf this total . variation , there is a parti tion

y. < Yl < . . . < Y� ! of [a , b] such that W < LQ • W + I: :s;; V

Cho06e a such that (1 ) � � �in ( I Y1- y. l , ly2 - y1 1 . . . . . l ylc - ylc -1 1 ) (2 ) I x - y l < lI implies I f (x) - f lY) l < �

If P • I a < Xl < • • • < xn • b I and (Xi - xi - 1) < b

1 71

(i • 1 , 2 , . . . , n ) then (1 ) imp l i e s that e a ch [x i _ 1 , x i [ contains at mo s t .an Yj • If p I i s the set cons isting o f b and all Xi such t ha t [xi _1 , x i [ contains s ome Yj , t hen p l . \ z . < % 1 < ' " < Zk l wi th

(3 ) I Z i - yi l W • Fin a l ly p I c: P , hen ce

Lp � Lp , > W There fore we have prove d the de s i re� re sul t .

Q. E . D.

E X E RC I�E 1 4 . - Supposs f i s a con tinuous r8a l function on [ a. b ] . Por Bach rs al y. ls t Mf (Y ) bs ths numbsr (fini ts o r infinits) of poin ts :r or. [ a, b ] '1 .- wlt�(!l1 f ez ) - Y (l1f /r"Y bE' c a l Z ed the mu Z tip l i. city function of n. Prous that Mf i s a Bo rs l fun ction and that !Mf (y ) dY i s ths tot aZ uari ation of f on [ a. b ] .

HINT : Th� �� u!t L� c1�a� 6 o � 6 un �on� who � e g �aph L� a 6LnLte

wUon 0 6 � t�tU.g h lLn e. Lnte.�va.!� . Ap,,�o r.Lmate 6 b!f a � uLtab! !f c.ho � e.n

� �q ue.nc.t 0 6 � ueh 6 un c.tion� ! TheL� m ultiplL �t !f 6 un c.tLon � ho ul d Ln

c.ua.H to that 0 6 6 1 . UH ITII SO L U T I ON

The result i s cle ar for function s, of the form

g (x) • � g (xi_1 ) + (1 - � ) g (xi ) if x · �xi_1 + (l -�) x i '

o ;:;; � ;:;; 1 . a • x . < Xl < • • • < xk • b . We w i l l de f ine s e t s P1 . P2 • • • • and

that if Pn · \ a · x . < x 1 < . . < xk • b I , then

where

whe re f

gn (x ) • � f (x i _ 1) + (1 - >. ) f (xi )

I t fo l l ows that

i f x • � x i_1 + (l - �)x i and 0 ;:;; � ;:;; 1 .

..Q,) . jMgn (y ) dy • the total variation o f gn • Lp �p i s as i n the pr e ce ding .exercis e . n

Nown

le t us de fine the s e t s p by induction on n as f o l l ows n (i ) Q1 · I a . b l . ( B ) I f � • I y . , . . . . yk ! t hen Pn is Qn join the po ints t akes i t s extremum in [x i_ � ,x i] , i • l, . . . . k.

I ! \ x. +.:J. I ( i i i ) If P • x . . . . . . x k ' le t Qn + l • l X . '- r . x l . . . ,x k •

172

(that is Q is P join the midn o ints of [xi l .x ! l . i • 1 . . . . . k) . n n - -By ( i i i ) . i t fo l lows that mesh (Pn ) de cre ases to O . hence the prece ding exe r c ise shows that I p t the to tal variation o f f Thus (1 ) s hows t.hat n

( 2 ) fM (y ) dy t V (-£) as n � -.:>. 'n

By the me an value theorem and the de finition of gn . i t fol lol,s that M � Mf • and that M � M � gn g 1 ' g2

Now suppose that y e R , and al • • • • • ak are in [ a . b [ with

There i s an n o , and Pn o • ! x • • x 1 . . . . . xt ! such that e ach [xi _l ,X i [ cont ains at mos t one point among al • • • • • ak •

I f aj e [x i _l ,x i [ .

x E ' [x 1_l , x 1 [ such that

the con struct ion of P 1 shows that n o + g l (x ) . f (aj ) • y . This shows that n o +

Mg (y) ... k . n o + l � R It follows from this t hat Mgn (y ) I Mf (y) as n -+- 00 , y e

the monotone convergence theorem g ives u s

(3 ) fMf (Y 1 dY . l im f�l (y) dy n+oO gn

Therefore ( 2 ) and

fMf (Y ) dY •

( 3 ) show t hat

the total vari ation of f.

Q . E . D. EXERCISE 1 5 . - (a) Le t A bs a aown tab le eubsst of R having the fo l l owing property :for every pair of s teme n ts %, y of A swah th�t % < y, there are e lements !.t, v ,w of A suah that u < % < v < II < w, Show that there i s a bijeation f of A on to the set Q of rational numbe rs , s u ah that % < Y imp li e s f(%) < fry ) .

(Prob lem 1 , se ation 2 . 2 , p . 21 of [ l J ) .

Ib ) Le t E be aompaat perfe at set (- aompaat and having no i so tats d poi n t ) of Rl whi ah is tota l ly dis aonns atsd. Uss (a) to show t hat the rs is a nonds areasing fun ation 'f : R -.. ( 0, 1 ] whi ah i ll aons tan t on evsry aonnsated aomponent of R - E, and .(E ) - [ O. l ]

U&8 t hi s to rsds aovsr the mapping of s%aMpte 8. 3 0 (b ) .

SO L U T I ON

(a ) Suppose A · ! al , a2 , . . . ! and Q · i rl ' r2 , . . . ! Cons ider the fo l l Owing con struct i on of b i ject ions fn' An - Bn :

[ 1 ] JPounda e.1 ons or Ifode rn A n al ysi s , J .Die udonne , /clldemi c Press 1 9 6 0

( i ) Al • ! a1 ! (i i ) Supp ose

A . ! x n 1

• Bl • I b1 \ \ a1 • • . . · an ! < • • • < xk \

fn (x Z ) • YZ , Z • 1 . 2 • • • • • k

C

•

f 1 ( a1 ) • b1 An ' \ b1 • • • • • bn ! C Bn • and . Bn • I Y 1 < • . . < Ylo: I '

1 7 3

Le t I . · J - oo . X 1 [ n A . I r • J X L ' x Z H [ n A . Ik • J xk • oo [ n A

J • • J - oo . y1 [ n Q JZ • J Y pYZ + l [ n o. J)c • ) )'k ' o<. [ n o

( r • 1 . 2 • • • • • k - l )

I f an+ 1 E An (bn + 1 E Bn ) . l e t Bn + l • Bn u j bn +1 1 • A u ja l l ) . n n + .

Othe rw i s e t hen an + 1 E some I Z (bn + 1 E JZ ) ' Z • a , 1 , • • k. Le t m (re sp . m ' ) be t he l e ast index i su ch that b i E J r (al E r r ) ' an d put Bn +1 • Bn U \ bnT1 • bm \ (AnT1 • An u ! an + 1 . am . p ·

We define fn + l as fo l l ows fO T1 IAn · fn ; fn + 1 (an + 1 ) • bm i f an +1 � An ' and fn + l (am ' ) • bm+ 1 i f bm +1 � Bn • I t i s c l e a r from

t he p re ce d ing con struct ion t ha t fn + 1 i s i n cre as ing an d b i j e c t ive .

We verifYoothat t h i s con s t ru c t�n i s . po s s i bl e . s in ce a1 • • • •

• • • an E An ' A • U An ' S im i larly Q" . U Bn ' I f f 1's de fined by

f (x ) • fn (x ) i f x E An t hen f i s we l l - de f i ne d . s in ce fn +1 1,\ • fn ' f (A ) • U Bn ' hence f i s surj e c t i v e . Now i f u < v are in A . then u . v E A . hen ce

f eu ) • fn (u ) < fn (v ) • f (v )

Thu s t he fun c t ion f h a s a l l requ i re d p rope rties .

(b ) No te that in ( a ) . 0 c an b e rep l aced by any coun tab l e -set having t he s ame p ropert i e s as (a) . I n the seque l . l e t D be e i t he r 0 n J O . l [ . o r D · ! m/ 2 n : n • 1 . 2 . . . . m • 1 . 2 . . . . 2n _ 1 ! . Thu s D i s dense in [ 0 . 1 ) . and the re i s a b i j e c t ion f : A - D such that u < v imp l ie s f (u ) < f (v ) .

Let a • min E . b • max E ; [ a . b ) - E i s the union o f coun tab l y m any open i n t e rval s )ar n ' �n [ � I n ' We t ake a po int an E I n ' for e ach n · 1 . 2 • • • • � an d l e t A · \ a1 . a2 . . . . ! . and we w i l l ve ri fy that A has the property as in (a) ,

Suppose an E I n ' am E lm ' an < am ; In · )" n ' � n [ ' . n E E •

• < �n • s i n ce E has no i so l at e d p o i n t . E is t o ta l ly d i s conn e c ted . hen ce [ a '�n ) con t a in s a conn e c ted componen t I r of [a . b ) - E , T hus

a Z\ < an Sim i l arl y . we c an find ap . aq in A . and

a t < an < ap < all < aq

1 74 Thus one can f ind f : A ---+ D . where 0 i s a den se coun table subset of J O . l [ . f b i j e c t ive and x < y imp l i e s f (x ) < f ey ) .

We de fine 'f o n [ a . b J as fo l l ows

( 1 ) If (x ) . f (an ) i f x E Jorn ' � n [ ( 2 ) , (x ) . inf ) ' (y ) y E E . Y > x 'f (b ) • 1

Since 0 i s den se in ] 0 . 1 [ . and f preserves the o rdering . (2 ) i s equ i valen t to

( 2 ) ' 'f (x ) • sup ! '(y) : y Ii E . y < x ! . (a ) • 0 I t i s c lear that ' i s nonde creas i ng . Sin ce If i s constant on each In i t remai n s t o show that ' i s con t inuous on E . But (Z ) ! hows that , i s r ight - con tinuou s . (2 ) ' shows that Of i s left -cont inuous (note that If i s n on de crea s ing ) . at a po int x E E. hen ce Of is a con t inuous fun ct ion . (N ote als o that cont inuity of If fol l ows from mon oton i ci ty o f If an d that If i s on to ) Now "' n ' n E E . " (or n ) • " ( �n ) • f ( an) • I t f o l l ows that 0 c .eE ) .

Since E i s comp act . 'f eE ) i s compact . hen ce closed. Thus 'f ( E ) :J !l' . [ O , l J . hence 'f eE ) • [ O . l J

Note that t h i s mapping 'f i s de fined o n [ a . b ] . an d c an extend to the who l e of R by put ting

'f ex ) • 0 x :;;;; a x ... b

F in al l y . l e t us re turn to EXaJllp le B . 2 0 (b ) . Re call that for. n ... O . En i s the un i on o f Zn 'closed interval s . e ach o f l ength Z -n r.n • De-le t ing a segment in the cen ter of e ach of the se 2

n interval s so

that e a ch o f t he remaining int ervals h a s l ength 2 -n - 16 l' F inally n t E .. n E • n

I t fo ll ows that En - En+ 1 • In • 1 U I n • 2 U . . . U In . 2n

and hence [ O . l J - E i s the un ion o f the p a i rwi se dis j o in t open segn men ts r k ' n • 1 . Z • • , • • and Ie • 1 . 2 • • • • • 2 • n .

The co rre sponding ! e t A has t he form

with A .. l a • • a1 . 1 . a 1 . 2 . a2 . 1 . · . . . a2 . ij , a3 . 1

. . . • t a1 • 1 < a . < a1 , 2

a2 • 1 < a1 1 < a2 2<a . < a2 3 < a 1 2 < a2 ij • • , . . , , . that i s . t he 2

n + 1_1 f i r s t poin ts dete Tllline 2n + 1 in terval s . each o f whi ch conta in s one o f the con se cutive 2n + 1 points fol lowing t he i r

?rder of enumerat ion . from l e f t t o rig h t .

T h e s ame enumerat ion c an b e app l i ed to the set o f dyadi c

numbe r s in ] 0 , 1 [ ,

D \ 1 • 17: 1 3

8 ' S ' 5 7 8 ' 8"

and t he re fo re o u r m app ing Of i s de f i n e d by

' ( 1 . ) 3 a \ ' ( II , I ) • I } I

1 7 5

, . . . . . !

Let f be t he mapp ing in examp l e 8 . 2 0 (b) . The con s t ru �t ion i s s o

symme t r i c t h a t f an d If a g ree on [ 0 , 1 ] - E • Thu s t h e con t inui ty o f

f a n d ' s h o w t h a t f • ., •

The re fo re we f in d an other way to define c on t inuous s i n gul ar

fun c t i on s . N ot e t ha t in t he p re ce din g , • is de fine d on E but m (E )

nl' C 0 S n o t be O .

Q . E . D.

E X E R C ISE � 6 . - l a ) Le t K b. t h e su b s e t of [ 0 , 1 ] co n s i s t in � Of a Z Z

" um /) e rs o f t h e fa ""

f �

� wi t h � • 0 o� � • 2 . "-1 3

n n De s � r i b e t h e �omp Zemen t of K in [ 0, 1 ] , an d s ho w t h a t K is t h e

"Can to r mi ddZ. t h i rds s e t W I t radi � Can tor se t ) .

(b ) r:;;fine f o n K a s fa Z l o ws :

i f z -L ..2. �n = 0 0 1' 2 fo r a Z Z n 2 1 3"

Show t h a t f i ll non de cr e a s i n g an d comp are f to t h e re s tri c t i on t o K

of t he !u" d i on in Ezamp l e 8 . 2 0 (bJ •

I c ) T he fo l l o w i n g i s a g6 n 6 ra U :& a t i on o f Ib ) . De fi ne

' : K - r. O , 1 ] k , , (x ) 2 1'1' 1 (z ) , . . . , 'k (z ) ) c 1 c1 + k c 1 +n k 1

'1 (z ) -4- + -8- + t --2- 2n +. c 2 +k

+ --8- + • • • + c 2+n k -· -2-

Ck +k ck + n k + --8- +. 0 • • + --2-

O f 3 - 1 3 - 2 = 0 0 1' 2 1. z · c1 + c2 + • • • • • , cn

1 2n + •

+ •

fo r a Z Z n .

Show t ha t ' i s :!on t i n u o u s , s u rj e c t i v e , an d tha t If mo s t 2 n_ to _ o n e I ' ex t en ds to a cont i n uou s B u rj e c t i ve map

[ 0 , 1 ] - [ O, 1 ] k , ca l L e d a "Pe an o CU rl"e " ) .

i s a t

1 76

so t u n ON

( a ) Le t u s f i rs t p rove t h ;1 t e a ch x i n [ 0 , 1 ] has a te rn a ry dev e l oppmen t 00

( 1 ) x · L C 3 -n c � 0 , 1 , 2 f o r al l n . n = 1 n n

The re are on l y t wo p o s s i b l i t i e s 1 1 2 2 * ] 0 , 1 ] 2 ] O ' r1 u l : vr! u 1 r , l J · I . U I 1 U I 1

I f X E I i ' l e t c1 = i ( i · 0 , 1 , 2 ) . Then d e v i d e I i in t o three i n t e rv a l s in t h e s ame m anne r ,

an d find c2 , • • • , and s o on . I t i s c a s y t o show t h ::t t t h i s g iV e s (1 ) .

that

1 1 2 2 * [ 0 , 1 [ · [ O ' r [ U [ 3 ' r [ U [ I , l l · J o U J 1 U .J 2 • If x E J i , l e t c i • i ( i 3 0 , 1 , 2 )

The s ame me t hod g iv e s �� , c� , _ . . , an J i t i s e asy t o see

i x a � x i oo . If x is not a k . 3 -n , the two repre sen t a t i on s ( l } a re the same ( s i n ce

both g i v e x 2 " c . 3 - i t x , an d O < x - x · < ·3 -n ) . I f x i s a n � 1 n k . 3 -� the re are e x a c t l y t wo rep resen t a t i o n s o f x as in ( l ) .

Therefore , t he p r e c e d i n g shows t h a t ; one rep re sen t a t ion ( 1 ) 0 f x su c h t hat c 1 -; 1 0 c cu r s i f an

'd 0 n 1 y i f , x ri 1 t ' i [ . Bu t

2 1 2 2 . 3' E K , an d f o r tun at e l y 3' Z 9' + IT + • • • E K , an d we hnd that 1 2 1 3" 3 [ i s a conne c t e d comp one n t o f [ 0 , 1 ] - K.

The s �e a rg ume n t g ive s 1 2 1 2 7 8 [ 0 , 1 ] - K · ] 3" 3' [ U ( ] g ,g [ U J ij" g [ ) U • • •

and we ve r i f y e a s i l y (hy i n du c t i on ) t ha t K i s t he t r i ad i c Can t o r set ( a s de s c r i be tl i n Ex amp l e 8 . 2 0 (h ) , w i t h ''n • (t) n ) .

( b ) Le t n be t he se t o f al l e x t r em i t i. e s o f conn e c t e d c omp o nen t s o f [ O , l J - K ;

x < y , and ck

s i nce

\ 1 2 1 7 B D � t 3' 3 ' 9 ' 9 ' 9 ' g, • • . \

N o t e t h e n

< dk I f x

l'n 3 - n , y • L dn 3 -n a s i n ( I ) , and t h a t i f x . \' .t:..... t h e r e i 5- s ome k su c h t h::t t c 1 d 1 , • • • , ck _ 1 • d k _ 1 '

E K an d y E K , t h i s imp l i e s Ck 2 0 an d dk • 2 . N ow 00

1 '\' - i -k - 1 2 2 k ( k + 1 ) 2 , £... c 1 2 � 2 (r + '4 + • • • ) z 2 - " dk 2 -1 = , t 1

thi s s hows t h a t f ( x ) �� f ey ) . �lo re . p re c i s e l y , 'f (x ) • 'f C y ) if an d on l y i f

1 7 7

th at . is x an d y are t he extrem i t i e s o f the same conn e c t e d componen t o f [ 0 , 1) - IC (he n ce in p articu l ar x , y E D , and ' (x ) 2 'f (y ) i s some k/Zn ) , and that if e ither x o r y is no t in D, then o« (x ) < o« (y ) (re call that we suppose x < y , x , y E K ) .

At the same t ime , t h i s s holi s t hat if y - x < 3 -k + 1 t hen

c1 - dI , c2 • d2 , . . ; , ck t - dk _ 1 , and ck � d k hen ce t hen

f ey ) - f (x ) < Z -�Z(. z - t -I I ) • Th i s s hows the con t inu i t y o f f . We verify e a s i l y that

f (�) f (Z ) 1 .) - r · l' 3 4 r ' "

and that every dyadi c numbe r k / Zo i s i n f e D) . Thu s f ( K ) i s den se

in [ 0 , 1 ) (f (O ) . 0 , f e l l - 1 , f n on d e cre as i n g )

I t is easy to veri fy that f i s undoub t e dly t he mapp ing o f Exalllp le 8 . Z 0 (b) , s in ce f c o i n c i d e wi th the l a t t e r on D .

(c ) The ve rifi cat i on of con t inu i ty an d surj e ct iv i ty o f 0( is the s ame as in ( b ) and i s l e f t t o the re ade r . To verify that '

, ' is at mos t Z - to - one is s im i lar to that f is at most 2 - to - o n e •

Note that f can exten d t o a mapp ing on [ 0 , 1 ] , b y d e f ining its v alue on ) Cl , � [ w i th Ol e IC , � E f ( f(Cl) . f ( � ) ! ) .

f ( h + ( l - � H ) • ).<f (Cl) + ( l - �) f( � ) e [ 0 , 1 ) ' e r O , l J ' c onve x ) This g ives t he e x amp le o f a con t inu ou s surj e ct ive map o f [ 0 , 1 ] on t o

[ 0 , 1 ) ' , called a "Pe ano curve " . Q . E . D.

E X E R CISE 1. 7. - Lil t X be a eomp a e t pe r-fe et se t in Rk . Sho �' t ha t t he re is a aomp aet p e rfe ot subse t E o f t hll tri adi e Can to r se t C , a n d a mapp i n g f : E - X, f i s surj e c t i v e an d i s a t me s t 2 k_ to _on e .

( b ) Use ( a ) a n d E%erci s e s [TI] , Q1] to eon s t ru " t a eon tin u o u s

Measure � on Rk � i t h supp o rt X. This sho�s t h a t e v e ry eomp ae t per fe et su bset of Rk i s the s u p p o r t of a

" con ti n u ou s m e a s u re .

SO LU T I O"

( a ) ( i ) F i r s t let u s de fine t he n o t i o n of .. con den sat ion point" { * l De f i n i t ion. L e t X be a s e p a rab 7. e me t ri e spaee ( X has a coun tab L e dlln s ll subse t ) , an d E e l. % E X i s s a i d t o b e a eon den s a t ion p o in t of E i f II V Q r� " ll i g hb o r ho o d of % eo n tai n s an u n eoun tab L e infin i te s u bu t 0/ E.

Let E# be t he set of a l l conden sati on po i n t s of E. X is s e parab le , hence E - E ' i s coun t ab l e , he n ce eve ry x E (EN )N i s i n E ll .

( . ) s .. c t i o n 3 . 9 , Probl em 4 , p . 4 0 , re fe ren ce I I I i n e x e r ci s e � . se e

a l s o Exerci s e Q!J ch a p t e r v .

1 78

In p a.r t i cu l a r , E N i s cl o s e d .

( i i ) Nex t , suppose X i s a c omp act p erfe ct m e t r i c sp ace . I f V i s a n onemp ty open s e t in X , then V i s a l o c a l ly comp act metric space wh i c h has no i s o l a ted p o i n t r Baire ' s theo rem (for l o cal ly comp act s p a ce ) shows that V i s un coun tab l e . Th i s s hows tha� every p o in t of X is a con de n s a t ion point of X.

( i i i } Now sup p o s e K is a . c omp a c t pe rfe ct set in Rk . ( i i ) s hows t hat K N • K (no ta t i on i n ( i ) ) . We m ay suppcse that K C [ O , l ] k

Exe r c i s eQJD ( c ) g iv e s a con t inuou s surj e c t ive map

C -+ [ 0 , 1 ] 1e

whe re C i s the t r i a d i c Cantor se t , and If i s a t most Z lc -t'o-one Le t A • 'f- 1 ( K ) , and E • A'. Thu s A - E i s at most coun ta ble . Thi s shows t hat Of ( E � • 'f (A)# . S i n ce (E ) " K , and K • K N by ( i i ) , th i s g iv e s 'P(E)K • K. Bu t E i s c l o s e d in C , hence comp a c t , and so '!' (E ) is c omp a c t . Thu s 'f eE ) • K.

Th i s comp l e t e s the proof o f (a) , s in c e AN is comp act ( c l o s e d in C ) , an d p e rfect ( having no i s o l at e d p o i n ts ) .

( b ) By Rem a rk ( i ) t o Exe r c i s e!m ' E i s t he support o f a cont inuous po s i t ive Bo re l me asu re � on R ( an d I � I • 1 ) .

(a) g iv e s a con tinuou s m ap 'f : E -+ Rk wh i c h i s f in i t e - to one , and 'fe E ) � K .

Pu t 1 k I'- (A) • q.- (Al l . . \ i s a Be ; � ; , 1 R . I t i s c l e ar that ", i s a p ' � i c ; ' a -e 01. RX .

If x E Rk , ., - l(x ) i s a fini te s.! t . Bu t � i s con t i nuou s , hen ce

�( .,- l (x ) ) • O. Thus I'- ( I x ! ) • 0 , he n ce fI. is a con tinuous measu r e •

., (Rk _ K) i s emp ty , hence fI. i s supp orted by K .

Fin al ly i f V i s a n open s e t wh i c h inte rse c t s K , then lf- 1 (V ) i s an open n onemp ty set on E . The con s tru c t ion of � shows t ha t eve ry n onemp ty o p e n set o� E h a s po s i t iv e me a su re ( support � • E ) . Thu s fI. (V ) > IT. The re fore support fI. . � K. This comp l e t e s (b ) .

E X E R CI S E ], ! . - L e t E an d f b e alt in Ezamp Z e 8 . 2 0 (b ) .

Q . E . D .

( a ) De fin e � ( [ O, z [ ) - f ez ) . Shot.} tha t ,. ezten ds to a con t in u o u 8 p o s i t i u , Bo re l m e a s u r e 0 71 R I wi t h s u p p o r t C .

Re aa Z Z t h a t ( 0, 1 ] - B i lt t he u n i on of aoun tab � y m an y di s

jo i n t i.n te rv a � s [ an , bn L an d ! (a.n ) - f (b n ) i s a n um b e r k/2n . Le t D = I 0. 1 l u I ,;z l ' b l ' a 2 ' b 2 • • • • !

1 79

D ' _ ) 0, 1 ! U 1 % _k Z-1I : 11 - 1 , 2 . . . , k - 1 , 2 , • • , 2" -1 ! ( b ) Show that f map s E - D hom e omo l'p hi ca l Z!f 011 to [ 0, 1 ] - D ' .

( c ) If A C E , . how that A i s a Bo re l B e t i f a1l d 01l l y i f

f ( A ) i s a B o rs l Be t .

( d ) Prope that thara iB a Ls be sgue m e asul'ab l e s u b s e t A o f

[ 0 , 1 ] , rsuch t h a t m eA ) - 0 ) w h i c h i. s 1I o t Bo re l m e a s u rab l e .

SO LUr r OIJ Ca) Define f (x ) • 0 if x � a , f (,x ) • 1 if x ;> 1 . The o rem

8 . 1 4 ( b ) g iv e s a fin ite Bore l me a sure ;>.. on R such that

;\ ( ] - ao , x [ ) • f (x ) hence

;\ ,( [ O , x [ ) • f (x ) . Theorem 8 . 1 s hows that ;\ ( I x \ J • a , x E R , hen ce ;>. i s con t inuou s f i s nonde cre a sing . henc� � is p o s i t ive .

:' i n a l l y . i f ]or . � [ n E "f " , then f (or ) < f U ) . Thu s E i s

containe d in the support of ;\ . Bu t f i s c on s t an t on e a ch l or . � [

whi ch does n o t inte rsect E . hen ce ;\ (E e ) • O . Thu s ). is sup p o r t e d

by E . The refore E • support ;>. •

( b ) I t can be shown e a s i ly t h a t f eD ) • D ' �� d that x E E - D imp l i e s f (x ) E [ 0 , 1 ] - D ' . Moreov e r , f ! E - D i s in j e c t iv e • .Le t ., be the re s trict ion o f f to E - D , con s i de r e d as a mapp ing E - D _ [ a , l ] - D ' Hen ce " i s con t inu ou s .

If . , � E D an d . < � then ' ( ] or , � [ - D) • ) ., ( .. ) , ., ( �) [ - D ' This shows that .,-1 is con t inuou s , hen ce " is a homeomo rphism .

( c ) Suppo se A C E . I f A i s Bo r e l me asurab l e , s o i s A - D (s ince D is coun t a b l e ) Thus ' (A-D) • [ ,- l ) - l (A_ D) i s Bo rel m e a s u r a b l e «,- 1 con t i -

nuo u s ) . Bu t f (A) • '(A- D) U ( a f in i te o r coun t a b l e se t ) . � n ce '!' ( A ) i s Borel measu rabl e .

" '

The converse is s im i l a r . f (A) Bo re l me asurabl e imp l ie s that A i s Bo re l measurah l e .

Th i s comp l � t e s the proof of ( c ) .

( d ) I t i s lm 010fTl (Ex amp l e 2 . 2 2 ) that [ 0 , 1 ] con t a i n s a s e t A ' whi c h i s n o t Lebesgu e me asu rab l e . He n c� A ' i s n o t Bo re l me asurabl e I f A · rl (A' ) n E , then f (A) • A ' , By l c ) , A i s n o t " Bo ni' l me asu'i� abl \: But m eE ) • a , hen ce 11\ (A) • a , hen ce A i s Leb e s gue -nleasurab l e .

, Q. E . D.

REM A RK . - (1. ) In (a ) to (c) , lie clln %'P l a ce , E by the s e t lin , e x e r c t _ � (b ) , an d v1. th D , D ' s u 1 t a bl !l ch o$en .

1 80

gue

�hu. 1n ( d ) , i t s u ffi ces to con si d e r t he ca se m {E ) - O . {i i } La t g - ., -1 , B • It- D . BI) (d ) , "' e s e e t h a t B 1 s Le b e s

m e a s u r ab l e , a l tho ugh g- l

{ B ) i s n o t , an d g i s a ho m e o m o rphi sm

(hence \I i . lA be . gue me a.u r a bl e ) .

e:Xe:R CISe: J.', - If f (z ) • 0 for a l l z .e [ 0 , 1 ] e z cep t fo r z • z l , z 2 ' " and l: I f (zn ) 1 < 00

n-1 then f ' {z ) - 0 a. s . by the L.,,,,,,,a t o The orsm 8. 1 9 . Can t h s re be an

un coun tab Z e • ., � E 8uch t h a t f i s n o t diffe rs n t i ab Z s a t any p o i n t of E ?

S O L UTT OIJ

The answe r i s affirm ative , as s hown the fol lowing ex amp l e .

Examp le . Re cal l this defin i t i on o f the tri adi c Can tor ·set :

Let 1II1 · w ] },fr . For n • 1 . Z , • • • , [ 0 , 1 ] - ("I U is the un i on of 2n d i s j oint closed i n te rv a l s of equ al

• • • U loin ) length

2 -n / . n (2 1 ) , s ay , I l , I 2 ' • • • • I 2n • De l e te an open in terval In , l<

nd n .

n , of length (1/3 ) 2 - (Z/3 ) n in t he cen te r o f In , k ' k • 1 . Z , • • • , Zn Take III n+ 1- In , l U • • • U J

n ,2 n 00 •

Then le t .. . . 1011 U 1012 U • • • • nl;Jl loin '. and. E • ( 0 . 1 ] - w· •

Now f i s defined as fol l ows :

f (O) • f (1 / 3 ) • f (Z / 3 ) • f el l • 1 / 3 .

I f In , k • l or , � ( , t ake f (ex) • f ( � ) • l ength I

n , k • ( 1 /3 ) Z -n ( Z / 3 ) n

n · l , 2 • • • • , k - l , Z " • • . , Z n f (x ) • 0 for a l l othe r x i n ( 0 , 1 ] . Le t D be the se t of al l x su ch t hat f (x ) of O . I f x e E - D , then for eve ry n t he re i s a e D , b e D .

a < x < b , an d b-a < ( 1 1 3 ) 2 -n ( 2 / 3 ) n . su ch t h a t

f (x ) - f (a) � - 1 , x - a

hence f i s no t d i ffe ren t i ab l e at x .

[(b ) - f (x)

b - x � 1

Th i s g ive s that f is not d i f f e r en t i ab l e at eve ry point o f E (and f ' (x ) • a fo r a l l X i! E ) .

Note that E i s comp act p e r fe c t , he n ce · hy Ba i re ' s the orem , E i s un coun tabl e .

1 81

EXE R CISE 2 0 . - Lil t I'- bIZ a re a Z Bore Z me as t.< re on Rk . Fer .. E Rk an d n - l , Z , • • • , Ze t t:.� lr) b ll thll infi mum of t he qt.<otien ts E;:.L , "'he rs

m l C ) c ra71ges ove r a l l ope n ct.<b es in Ii ", n os a II dglis are p a ra Z Z e Z to the co ordinats ars s an d havs llingth l ll . s t h an l in , s t.< ch that ,.. E c. De fin e

I �QI'-) Ir) - lim t:.� I:)

n+oO I t ha lowllr ct.<bi caZ de ri vativlI

of I'- at ,. ) .

Lilt J b s a compact intarval on t ha lins y - : in RZ , Zli t m l bll onll dime nsi on a l Lsbssgt.<e "'sas ure �", J, an d dsfine 1'-0:) • m 1 IE: n J ) for e Vll ry B.orll l s e t E: C RZ . Prove t h at I � QI'-) Ix) • a fo r eve ry t e R2 a Z t h ot.<gh I'- � m ICompars � i t h Ths o rs m 8 . 10 ) ,

s o LUT I ON , Supp ose J - · 1 (X , x ) a ,-;; x � b \ . I f x • (x ,y) i!! J and r • Iy - xl > 0 ,

v • j x- r , x+ r [ " ] y - r ,y+ r [ then V n J · 0

I t follows that ( �QI'-) (x) • O .

I f " • (X , x ) i!! J , L e . x i!! [ a , b ] , an d ] x- r , x+ r [ n [ a , b ] - .0 ,

in the s ame manner we find th at ( �QI'-) (x) • O .

a •

I f " • (x , x ) e J and l in be fixe d , Y • Y- j x - a , x+ b [ ] x -b , x+ a [

n - 3 , b • n -1 • a . Then p.(Y) •

I t fol lows that t:. � ()() ';;; _1_ , hen ce (� QfI.) (x) • o . fi n

Q . E . D.

EXE RCISE 21 . - Raca l l t hat a mli asurab Za re ct an g Ze is a carts sian p roduct: A >< B, in ", h i c h A E d and B e CO I Dllf. ? l ) .

ProvlI tnat t hs rll i s a a l osa d s a t E C RZ , ", i t h mZ IE) > 0 ,

"' h i an con t ain s n o ma ast.< rab Ze re c t a71 g Ze of p o s i t i vs measure.

S O L UT I O"

S olut i on to Exercise [I) ch . I I g ive s a comp act set K C [ 0 , 1 1 ,

m ( K ) • l, whi ch i s t o t a l ly d i s c onne c t e d . Thus K con t a i n s no non e mpt y open se gmen t .

1 8 2

De f i n e E • i l l. , y ) : x , y

Thu s E · (+ ) - 1 (K) i s

E · I y (x , y ) E : "

I t f o l l ows that E • x -x

E Rl, x+ y E K ; .

c l osed. If x E R,

E ( \ : x+ y E K I • ) y ) . ) 1 K, hence m (Ex) • m ( - K ) • m (K) • r'

By The oTem 7 . 6 , an d De f in i t i on 7 . 7 ,

m2 (E ) . fml (Ex) dX . f} dx > 0 .

Thus 1!12 (E ) > O . I f A x B e E , t hen A + B c K . I t f o l l ows t h a t A + B c on t a i n s

n o n onempt y open s e gmen t , hen ce m eA) � m ( B ) " 0 (Exe T c i se [D ( c ) o f

chapte T VI I ) .

N o t e t h at E n ( [ 0 , 2 ] x rO , 2 ] ) i s c omp a ct , having pos i t ive mea

sUTe , an d con t a i n s n o me asuTab l e Te c t ang l e of pos i t ive me aSUTe .

* * *

Q . E . D.

1 84

chdP �e�, we 4 h dll U4 e J(j�� ) dm r � ) r 6 . 9 ) (x l ·i� rx - t ) g (� l dm r t l

he ) .£7 r x J e -.i.. x� dm C d

• - 6 rd dx T j'OQ Iii 00

dl1d EXERCISE 1 . - Comp ut. t h. Fou r i e r t ran s fo rm of the ohara o te ri s t i o

funotion of an in te rva Z . Fo r n • 1 , 2 , 3 , • • , le t gn

b e t he oharao

teri s t i o fun o t i on o f [ -n , " ] , and oomp u te gn

' g1

exp l i oi t l y , ( The grap h i. 1J p ie oe!.li s e Z i n e ar ) , Sho!.l t ha t g

n' gl is t he Fo u rie r t ran s

form of a fun o t i on fn e [,1

, excep t fo r a mu l tip l i ca t ive oo n s tan t

f ( ) sinx s i nnx n

x · x

2

Sho !.l t hat I fn

11 - 00 an d oon cl ude that t he .,app ing

! - f m ap s L1 i n to a p r op e r sub se t o f Co .

SO L U r r OIJ

I f g •

Hence

"1. [ a , b ) ' "'� have

g (x ) • J' e - ix t dm et ) • a

__ l_ee - i x b - ix ,Zit

(1 ) g ex ) • i (e- ix b _ e - i X d ) x .rzr.

(xr O )

so that fo r g • g ;\. • X [ _;\. , ;\. ] ( ;\. � 0 ) , we have

(2 ) &:\ (x ) jf s in :\ x · - --- , x 11: X

"! 0 ,

How note that if f • XA and g • X s ' then

(f * g ) (x ) . !XA (X - t ) Xs (t ) dlll ( t ) · m ( (x - A) n B )

In our case A . [ -n ,n ] and B . [ - 1 , 1 ] , hence

(gn * g l ) (x ) • m( [x - n ,x + n ] n [ - 1 , 1 ] ) I t fol lows that for n � 1 ,

(gn * g l ) (x )

We have (gn *

• 0

· v1 1 · --Ii 11:

( n +

gJ' • gn ' &1

i f Ix 1 � n +

i f Ix I � n -

1 ) - I x l ) i f n - l

' hen ce by (2 )

� -Ix l � n+ l

1 8 5

( 3 ) (g g ) A(X ) • � s inx " s innx _ fn (x ) (f ( 0 ) • 2n ) n * 1 It x2 n 1::

Fo r I x l � I , I f (x ) 1 �-Z-2 • The con tinu i ty o f f at 0 n It X n

s hows that f E L1

(R) . Thu s the invers i on the o rem 9 . 1 1 g ives n

(g * g l ) (x ) ·l:ixt f (t )dm (t ) n oo n

Bu t fn ( t ) • fn ( - t ) as s hown by (3 ) , hence we obtain

(4 ) (g * g l ) (x ) ·l:- ix t f (t ) dm (t ) • f ( t ) n n n 00

Le t us s how t hat I fn ll - oo as n _ oo . Firs t of a l l ,

hence ( 5 )

s inx x

3 I f (x ) I � -2 n :It

� 3

2 11:

I �I I x I

11:

(6 ) II f I � l ( ! IS i nnX

\ n 1 1I: 2 J :It

i f Ix I � !! 3

i f I x I . � !! 3

o J:

dm (x ) • � 131�ldt -3" x

S i n ce joo! s i n t l d t ... 2:1 (n+ 1 ) 1: � dt

th"at 0 t nr. (n+l )1I:

11:2 n x t - 3

• � ",_1_ • 00 , ( 6 ) shows X Ln+l

1 86

( 7 ) l im I fnl l - oCI &5 de s i r e d . n ..... oo

Let A : L 1 ( R) - CO (R) defined by A f - f I t i s known that 1\ is one - t o -one ( t he inve r s ion t he o rem ) . If hel l ) a C. , The o rem 5 . 1 0

shows t hat there i s a ll > 0 such that

Bu t I t 1 00 t h i s con t ra d i c t s ( 7 ) , an d hence h 1 S n o t sur-n j e ct ive •

Con s i de r A f (1

o

the integral

(1 _ �) e - i t xlx

=,,

_

A - i t x = O i t A - e

t

We deduce f rom this that i f ( s e e Fi g . l )

f (x ) - m ax � O ,min (l + Bt , l - At )

1

then t he con j ug a t e Fourier tran sform o f _1. f is .,

-1 t il =-_--.:;;.e:::-__ +

At 2

- 0 (_1_ ) a s t 2

- e

B F i g 1 • T� graph o f f

i t B

/ \ F i g 2 • II funct i o n in

Thi s s hows that ( s ince ., e c.. ) . If e L 1 , hen ce by the inve r s ion t he o rem f i s t h e Fou r i e r transform o f • •

The t r an s l at i on inv ar i an ce of A(Ll ) (f e A imp l i e s ft E A) shows t ha t it con t a i n s all f un c t ions as in Fig 2 , s in ce this i s a comb inat ion of an f and a (gx * g l ) t ' I t fo l l ows e a s i l y that every p ie cewi s e l ine ar fun c t ion in C . (R) i s in h (L I ) , and we con clude that A ( Ll ) i s dense in C. (R) .

Q. E . D.

EXE R CI SE 2 . - Gi.ve u :amp Z e s of f e L2 - Ll , but the PLanchel'e Z tl'Cl1I S

fo l'11f I of f be l ong s to Ll . Undel' what ci.l'aum s t anclls can this happ en ?

ANSWE R then t he con j uga te F ourie r t ransfo nn o f g i s

l ex ) -l�lxt g (t ) dm (t ) coo

hen ce f - I e L2 , and the Pl ;m cherel t r an s form o f f i s in L 2 .

1 87

I f f � Ll , then f s at i s f i e s a l l requ i remen t s o f th� exerc i s e (see Theorem 9 . 1 4 ) .

We can take the fol lowini ex amp l e

i n w h i � h g n i s the characte r i s t i c fun ction o f [ -n , n ) .

Q . E . D .

E X E R CISE 3 , - If f E Ll an d fi t ! (t ) 1 dm ( t ) < 00 . prove t h a t f co inci des a . e . wi t h a differen t iJ b l e fun ction w ho se de rivat i v e i s

il�! ( t ) e i:r t dm (t ) . coo

S O L U T I O N

The con t inu i t y o f f in ( - 1 , 1 ) an d the inequ al i ty

I l ( t ) I � I d (t ) I I t I ;> 1 shows that { e Ll . I t f o llows f rom t he i nve r s ion t he o rem 9 . 1 1 that f (x ) coin c i de s a . e . with

g (x ) -l"""O; i txf (t ) dJn (t )

and Theorem 9 . 2 ( f ) (mo d i f i e d for the con jugate Fou r i e r tran s fo rm ) g ive s

1'00 g ' (x ) - i -00 tt (t ) e i tx dJn (t )

Q. E . D.

E X E R C I S E 4 . - Supp o s e f E Ll , f is diffe re n tiab l e almos t eve ryw here and f l E LI . Does i t fo l l ow t h a t t h e Fou r i e r t r an sfo 2'm of f' is

H f ( t ) ?

S O L U T I ON

The answer i s ne g �t ive : Cons i de r f - X( _ l ,l ) ' then f ' - a a . e .

and t i f ct ) - iA s in t , Bu t i t c an b � e as i ly shown t ha t i f f i s c on

t inuou s l y d i f f e ren t i ab l e , an d f ' E LI , t hen i' ( t ) - (i t ) f (t ) ( Integ rat ing by part s ) .

1 88

EXE RCISE 5 . - Le t S be th8 c l a s s of a l l fun c� i��s f on R1 which halJe th9 fo l l o win!1 P l'OP91'ty . : f i s infin i t e Z y diffe l'e n ti a b l e , an d the1'8. a1'.8 n umbe l'S Amn rf) < oo , fo l' m , n - 0 , 1 , 2 • • • • , such t hat

I :z:n �f (:z: ) I � A (fJ (:z: e R1 ) mn 88 1'8 D i s t h e o l'din al'Y diff81'en tiat ion opel'atol'. Pl'O lJ 8 that t he Po ul'iel' tl'an sfol'm map s S onto S •

Pin d e:z:amp l e s of m 9mbe l's of S.

S O L U T I O!; We shall p rove that the Fourier tran sform map s S in to S .

Thi s can al so appl y t o the conjugate Fourier t r an s form

and hence the inve rsion theorem g ives t hat 5 � s .

Recal l t he f o l l owing import an t re su l t s

- I f f e Ll , f has derivat ive a t eve ry po int , an d f l e Ll , then (f ' ) � ( t ) • ( it ) t (t ) . (Exe rcise @J ) .

- I f f e L l , an d g (x ) • ( - ix ) f (x ) i s in Ll • then (t) f ( t ) . g et ) (The o rem 9 . 2 ( f ) ) .

Fix f e S. S in ce

I xn nmf (x ) I � A 2x -2 (x ;. 0 ) m .n i' it fol lows that ( _ ix )n Omf (x ) i s i n Ll , and hen ce that

I tm on f (t ) I � B n , m

hence f e S . Thi s comp letes t he p roo f .

Ex a.pmle� 0 6 m e m b e t .l 0 6 S We cou l d con s ider

_x2 ( 1 ) f � P (x ) e • in whi ch P i s a polyn om i al in x , o r

( 2 ) f b e ing any Coo fun ction o f comp act support .

( See Exe rcse 12 )

1 8 9

LJ lle re " .... - L% i Yi �f " - (% 1 " " % k ) ' ,, - (Y1 " " ' Y k ) ' an d '" k is L e b fl s gu e m e asure on R , div i de d b y (2 � ) kI2 fo r aon v e n i e � ae .

Pro v e tha t 1 E c . fRk ) . PrOVfl t h e i n ve rs i o n the o rem an d t he

P Z an allf! re Z t he o r"," in t h i s aon te%t , a s LJ" Z Z as t he an a l o g u e o f The o r em 9 . 2 J .

S O L U T I OII The con t inu i t y o f f is a s imp l e con s e quence of Le b e s g u e ' s

domin a t e d convergence theorem :

s ince

l im I f ( ,. ) - f ( ") I � n l im r I f (,, ) I l e - i "n ' '' _ e - i " , " l d .. - O n�oo JRk f E L l .

Suppos e >I' - (yl . . . . ' Y k ) and Y I " 0 , l e t y' z (�l , 0 . . . . 0 ) .

We can im i t ate t h e p roof of Theo rem 9 . b to s e e t ha t

S i n c e ( h e re

't' t > 0 . 3 11 > 0 : 1 ,...' 1 < II illlp l i e s I f ", - f I < t ( S e .. ' � ( a ) c h . V J I ) .t h i s s hows t h a t

imp l i e s 1 2 £ ( ",) I < I:

R� p e at t h i s p roce s s fo r t he c a s e y 2 ' y 3 ' • • • • Yk 1 O . an d n o t e

t h a t a t l e a s t on e I y i I i s � 4J we o bt ain

VI: < 0 , 3A > 0 : 1 .. 1 > .-\ i mp l i e s I f ( y.) l < t

hen c e f v an i she s a t i n f in i ty .

Now pu t H ( ,, ) - e -I ,, ; . e - lxl l . . . -e - l xk l

( O < X < oo ) c hen

1 90

(Use Pub i n i I s theore. &l\ d the case k • 1 i n 11 . 7 ) hen cil

o < H (x) " 1 ,

The analolu e s o f 9 . 8 to 9 . 1 7 fo l lows e as i l y ,

obtain the inve r s i on tile o r •• for L l � ) hen ce we

If f e LI M,i< ) , _ d if f E Ll lfl<) g (>< ) • �I< j()I') I1 1. ..., cbtk ( ,,) ( !C e RI< )

titf1" g E C. a" d f ( ") · g (><) a • • •

and t he P l ache re l theo reM 9 . 1 2 hol d s , except in 9 . 1 2 ( d ) ,

Le t u s

. J f (><) e - 1" ' " dale

(x) [ -A , A ] �

J f(,,) e- l" . ,,� ( "J • ( _A , A ] l< now prove t he fo l l owing an al o gue o f 9 . 2 3 .

1 I< THEOREIo4 . - To ,t1,. I'W �"'p t.1I: 1It111!o",orp lli s," of L M. ) , u:�a p t fo r

Co - 0, t it" rl1 OOI'l'f1lfp o" d. tl "lt i ,? " • . " E �1 . " c: h that C1 l f ) - ft..; . (1 ) to ( 6 ) o f 9 . 2 2 fol l ows i_e d i a te I y , that i s the re i s II bO lnded cont inuous fun ction , : Rle - C su ch that

and e l l

I t reMa in s t o prove t h ll t � (>< l • e -l", . " for a su i t ab ' e " • \ '

Thi s i s , i_p I e : i f " • (xl

" "" xle l . Iln d I el . . . . . ek l i s the

st llndard bas i s of Rk . then " • X l e 1 .. . . . . .. xle.e

le • hen c e ( * ) s hows

that

, (x l · Hxlel ) · . . Hx k ek )

The p roo f o f 9 . B ( ( 7 ) to ( 1 1 ) of 9 . 2 2 ) g i v e s

( 1 - i x lY l , Xl e l • e . . . . .

hen c e fo r " • (yl

. . . . . Y k l • we i e t

H ") • e - 1" , "

This g i ve s �

as de s i re d . C1 ( f ) • J f ( x l e-

h· " dJak C .... ) • f ( ,,)

l{, E . .1.

. '\

1 91

EXERCISE: 7 . - If f E Ll M< ). A is a Z i" . ar operator of 1ft "" d '

g (11: ) - f (AI1: ) . how i . g · r. t a t. d t o f 1 If f ,i. s ill lIari aft t l<7\der r o t a ti o li ll . i . e . if f (l1: � dep e " ds O"tl( 0" th • • " " H di aK di s t a7l "e of 11: fro� t 1U. o ri g i" .

p rou. that th. B aM e is tF!<. fo r f. SOL UTI OIi

Suppose A : RIc _ Ric

is a linear

f • A ne eds not be in L1 �1c) . There fo re that A-1 exist s . Define Ae by

operato r . I f de t A • D , we con s ider only the case

(Ae i s 'the

Let

Thus

(A -1 .. ) . x . .. . Ae"

adj oint of A- 1) .

i - f • A . We Iiave

i (x) • fa f (A .. ) e-b'. x dm ( .. ) Ric , Ic .

i (,,) •

fa - 1 . •

Rlcl det A I . f ( .. ) e::- iA .... " dalc ( .. )

· I de t A I �1c

f ( ... ) � Ji .... A* '" dalc ( ... )

I det A I l (Ae,,) , hence (f • A}A • I de t A I ! • A e ,

Hote that A i s a ro tati on (about the o �i g in ) i n Ric if an d

only if Ax • A ... . " . ... , hence if and only i f A • A*.

Thu s if A i s a ro tation of Ric " then

(f • At .' f • A,

As a con sequen ce , if f is inv arian t under r o t a t ion s , �hen

the same h o l ds fo r f, Q . E . " ,

Lap ta�i "" of a fl<" ,,�io" f 07\ Rk i ll

k % 6f - L 4-

)-1 bJ provid8 d t 1l8 partiaZ d8ri v a tive B el1:i . t , What es t he r8 Z a t i O" i;" ,:ueen f IZPld g if g " , 6 f a?t d a Z Z ?t " " " s s arl( i n t "g rab i titl( con di ti o ?t " <l r' 1' sat e sfi e d 1 It e s " Z "ar t h a t t h " Lap Za"ia?t "om�u t " 8 w i t h tra?t c � t. ;on& Prov" t ha t e t a ho '"ommu t" B wi t " ro t a t i o" s . i . fI , . t"at

Mf 0 A) - (6f J 0 A "·"."" "" r f hall co?t ti?t u o"s s " "o" d d" l*i " a t i " " IJ and A i ii . a. r" tat" o'l .I: . , vj R ' ,

1 92

S O L U T I OII F ix an in dex 1 . 1 � j � Ie . Let " • ( yl .y j ) . w i t h

I f g •

,... 1 • ( • • • ·. Y.j - l ' Y j � l " " )

then we have

1 - 1 "' ''''' • e '

R.' - l

( /)f ) � The r e f o re -- f( x) • n IlX j then g ( >. ) . - .r x � f (") . I f

i = l J

( - ill j ) f (l() . I t fo l l ows that if g. t> f .

I x l denotes the eucl idi an d i s tance o f " from t he orIgin . then

Suppose f has con t inuous se cond de rivatives , and A i s a rotat ion ab ou t t h e o r i g in O .

I f f has comp act supp ort . then f e Ll (Rk ) . hen ce the Fou r i e r tr�n s f o rm o f f exi st s , an d ( 1 ) shows t hat

[ 6 (f • A) j "' (,, ) • _ 1 >< 12 ( f • A) " C" l • - I A x I2 (f • A) ( x) (f . A • f 0 A . cf ITh • (M ) " (,h) • [ (M ) • A( ( x)

The inj e ct i v ity of A shows t h � t

( 2 ) 6 ( f • A) • (M ) • A . Sin ce eve ry ro t a t i on i s the compos i t i on of a t r an s l a t i on an.d

a ro t a t ion about the o ri g in , ( 2 ) i s v al id i f f has cqllp ac! support an d A i s a ro t a t ion .

I n g en e ra l . con s ide r f . , • whe re , e Coc hav ing comp a c t sup port w i th ,, (x l · 1 for every I ,. I � R. ( , eC';;' . see � l . In B (O ; F. )

6 (f • • ) • t, f . and ( 2 ) g ives

c, (f • A ) .- (M ) • A •

Q . E . r;.

1 93

E X E R C I S E '1 . - Sh ow .that Q V B ry m e asuraL l e C!haraC! h r of Rk i s C!or. t i -

SO LU T I O" Suppose or : RK _ C i s Le bes gue me asurab le su ch t hat

( 1 ) '" (x + ". ) - "' Cx ) . .. C".)

(2 ) I .. C,, ) I - 1 for every ,.;

Define 01> : L 1 (Rk ) - C by

(3 ) oI>Cf ) .� jf cx) " CX ) d X

( 2 ) g iv e s ", e L 00 , hence <ZI i s a l in e ar fun ct ional o n L1 , o f nonn not exceeding 1 . (1 ) g ives

. <ZICf * g) • ffi Cx - r ) g C.".) d". cr{ ,, ) dx

- j(;;' Cx - ".� ex Cx - ". ) d x ) g C ". ) '" (". ) d ".

• oI>Cf ) oI>Cg ) hen ce <ZI i s a comp lex homomorp h i sm o f the a l gebra L 1 (Rk

) . The an al O l!: ue of Theorem 9 . 2 3 for R

K proved in Ex erci se @] s hows that there i s a � e Loo(Rk ) , an d ". e Rk

, su ch t ha t

� C X ) - exp C - i .. , "' ) ,

( 4 ) <ZI(,, ) � fi(xj "C".) dm C-) = i c".) .

Compa re (3 ) an d ( 4 ) , we get

", Cx ) - exp C - ix , ,,. ) for almo s t al l , e � . We wi l l p rove that "' .. � • that i s !X ( ' ) [ � ( ,, ) J - 1 - 1 for al l

" • To do thi s , i t suf f i ce s to s how that : i f 7. i s a measuTabl e cha racter o f R k , and "' C") - 1 for a lmo st a l l .... , then o< C ,, ) - 1 for eve ry

Le t

'" i s a characte r , hen ce G i s an add i t ive subg roup o f Rk . or i s Lebes gu e me asurab l e , hen ce G i s L ebe s gue measurab l e . I n add i t ion ,

(II) an d G

mK (RK - G ) - Q . Thu s G - R K by the fo l l owing ex t en s i on of Exe rcise [22J c h . V I I I Sup p o se G i s a Le besg u e measu rab l e a ddi t i v e s u b g roup of R

k.

k ,I R • Thlln mk (G) " O.

PJt o 0 6 06 ( � l . I f m C G ) > 0, then G - G + G con ta ins an open se t , s ay " + l - E , E [ k , ,.. e G . Th i s s ho ws t hat V - l -E , E [ k C G , hen ce

1 94

vD . y + y + . . . . + y (n t . n s ) c G n - l , 2 , • .-.

)Jt V • ] _n£ ,n£ [k

, this g ives G • R�. Thu s G ·; RK i.p l ie s � (G ) - O . D Th i s p rove s ( I ) .

aXE RCISE 10 0. - Sl4ppo .. f e Ll for a l l z: E Rk . Frou tltat

11(¥ ) I < 1( 0 ) for .... r¥ ¥ ,J O.

j().LUTI OI/ If I icy) I - £ ( 0 ) , that i s 3 }.. . R and

e 1}" jl (,,) e h .... d ,, · jf (,, )d'"

Q . E . D .

then by 1 .39 ( c ) , e i�e l � f (x) • f (x ) for al.o s t al l " , hen ce

e i}..e 1 ><.Y • 1 fot al.ost all " . The continui ty of the left s i de o f

thi s identity lives e n e l ........ . 1 .. x .

I t fo1 16ws • • s i ly that " . ... . 0 "''' , hence ... . O .

Q . E . D • .

P7'OV .. t lta t g E C o . What " an 1I0W tr ail abow t g i.f f • Loo ?

.sO LU T I ON Suppo se 1 � P � 00 and f E LP .

I f - 00 < a < b < 00 an d '" i s 1/ (b - a) tillle s Lebe s gue lIIe a s u re

on [ 3 , b ] , then II- C l a , b J ) · l . Hence Exercise III (c ) ch. I I I give s

!ben ce

jb I f l d l'- � [fb

l f I P d ",J / P

a a

fb - 1 [fb ] l / P ( 1 ) ' I f (t ) I d t � ( b - a ) ( l -p ) I f ( t ) I P d t

• a a ( (b - a) - 1 ) .• ·

Thu s the integral de fin ina g ex is ts for all x .

I f .-A � x ,x ' � A, the- dom in ate d convergence t heorem app l i e d

to " [ -A ,A] �ives (a - -A, b -A in (1 ) )

1 a I i (x ' ) - II (x ) I - 0 J: t .. x

benc. i i s con tinuou s .

1 96

1 ..;; p < 00 , aen celif ' Pdx - � I f l Pdx . Tke hypothg l i s f & LP 00

Ift+ 1

n.f-'-oo n -0

shows that l im In+,� , Pdx - 0 , and w& dec:b.u t hat I n I� oo n

x+ 1 I i. f \ f l P du - 0 . " Thi s and ( 1 ) i iv.

I x 1.-00 x

jX+1

l iB I f l d� - 0, hence II E C. (R) . I x 1.-00 x

Suppose f E Loo(R) f I f - 00 < x � x ' < coo • then

jX ' x ' + 1

I II (x ) - i Ct ' ) 1 � I f l dx 1 I f l dx � 2 (x ' - x l i i i.., . x x+1

Thu s II is unifonaly cont inuou s . Evidently , Ii 100 :s;; If 100

�n.ra li.ation . One can show e as i ly that if V i , a bounded .e a'UTable 58t in Rk , i (X ) - 1 f (x ) dx , md f Ii LP (WC ) , then

1t.+V (io) II I! C.�) if I � P < ..,

( i l ) II i s unifoI'1ll1y continuou s , and 1 9 1 .., :( I f 100 if p - 00

RE"A IttI:: . - If h- 1 (-l , O ) ' t hen g- f .h . The p re ce din g r •• 1I1 t. fol l o lt

• • • i l y from lO r . rei ... [D eh . VII .

E X E R CI SE 1 2 . - L: t �b. ths a las . of a l l infinite ly diffiren tiab l e

co",p l" !",. ction o n R1 . and let C� con s i s t of a l l g E Coo ",hO BB . w�port i. COMpact . Show that C;'doe8 not conei.t of 0 alone .

Let L� be thfi olau o! a l l ! loIhiah be long to L1 1 0ca l lll.

i . e. ! E Lio a

oCprotl id8d that ! i . measurab l e and fr l ! 1 < 00 for

every bo"rtdi d i n terval I.

I! ! E Li oo artd g E C;'. p rotle that ! . . g E Coo . PrOtli that the re arB ssqu,nos . ! gn ! in C;O s" ah that

I f . g n - f 11

- 0

a. n -+- oo . for svsr!! ! e L1 (Compars Ths..:>rsm 9 . 1 0 ) . Prove tha t

1 96

j g" I can a lllo bll cho Bll " t ha t (f • g,, ) (:r: ) - f (:r: ) a. e . , fo r e ve ry

f E L�o c ; i " fac t, .fo r . , d tab l ll l g" I t hll ",o" v e rg e " "'Q 0 "' CI<r8 a t

Q VQ r!l po i " t z a t r.ihi ch f i 8 t hll dQ r i v a t i v e of i t . i"dQ fi" i tQ i ,, !: e

:ira v . Prov ll t ha t (f . h ) (:r: ) - f (:r: ) a. Il . , if f E Ll aS A - O , a"d

that f • h E C�. a l thol<g h h A dOll 8 "ot hav e compa c t supp o r t ( h A is dQ fi" e d i " SIl C ti O" 9. ? ) (Tin:. i ll t h Q AbIl Z ' 8 /(11 !"" Q Z ) .

so LUTI ON If g (t ) · a for I t l :> 1 , and

(tlen g ( n ) (=.1 ) ex i s t , . and i s O. Thus doe s not con s i s t of 0 alone .

g (t ) • e xp (_. 2_1 __ ) for I t I < 1 , t - I

g E C� . Th i s shows tha t C;O

Nex t suppose f E Lio e ' g E C;o has support in [ -M , MJ . '"

F ix x E Rl . For 0 < I h l � 1 , we have x+M+ l 1$ * f l ex + h) - (g * f ) (X) .� g (x -y+ h) -g (x -Y) f (y ) dy

h x - (M+ I ) h

The integ rand has abs<? l u te v alue bounde d by 1 9 '1oo l f (y ) I (by the mean v a lue theorem) . S ince f i s in tegrable on [x - (M+ I ) ,x+ (M+ I ) ] we can appl y Lebe sgue ' s dom inated convergence t heorem , and . o btain

.l im (g * f)(x+h) - (g * flex) • � ' {X - y ) f (y )dy • � ' * f (x ) h- O h P

The re fore ( f * g ) ' • f * i ' . I t fo l l ows that f * g E C"': Now reca l l t he fun ct ion g g e t ) - 0 , I t I :> 1 ; g (t ) · exp r-2-1--) , I t I < 1 .

\ t - 1 Def ine u (x ) by I -x

u (x ) • f g (t ) dt ;' 0 -00

Then u (x ) • a for x � 2 ; u (x ) • c for x ,,:;; 0 ; and u � O . f coo We can use this u to show that t he re are un ct ion s gn in c

su ch that

• gn (x ) • n i f

- a i f

a ,,:;; gn (x ) � n

Ix l ,,:;; _1_ 2n

fx I > _1_ + 2 -n • an

2n

i :> nXB ( O ; 1/ 2n )

for every x . imp l ie s that

( 1 )

( 2 )

• ( B ( O ; 1/ 2n ) ) 1 f (x - t ) dt � (g * f ) ( x ) B (O , ; 1/ 2 n )

g � nx S ( o . .. ) wi t h an • n 2n

imp l i e s t h a t

� 2n sn � f (x - t ) d t m (B ( O ; an ) ) ' B ( O ; a

n)

(g * f ) (x ) • f ( x )

Since 2 n sn �

1 , (1 ) and ( 2 ) shows t h a t

1 97

(3 ) l im gn * f (x ) • f (x ) when e v e r l im _l_!: f (X - t ) dt • f ( x)

n- ..., r...o 2 r B ( a ; r )

(3 ) s hows that (g * f ) (x ) ->- f (x ) a . e . , by 0 ch VI I I . n . . Thi s case f � a imp l i e s t he gene r a l on e . F in a ll y , t h e i n t e

g r al defining f * g involve s o n l y t he re st r i c t ion o f f on a comp a ct i n te rv a l , and hen ce we o b t ain

(4 ) l im gn * f (x ) • f (x ) n _ 00

a . e . -if f E L l 1 0 c

N o t e that by the prece d ing p art , gn * f E C� .

That h • f _ f a. e . i s a con sequen ce o f t he fo l l ow ing theo r em _see The o r em (3 1 . 43 ) p 4 0 4 - 4 0 7 o f [ I ] •

Theo rem. If u i e a p o e i t iv/ con tin uou s fun ation on R su.ch t h a t

(i ) L:( t ) dm (t } - 1

and if t hsn

(ii ) u (- t ) - u (t )

l im u. >.. _ f ez ) - f (::: ) .I. - 0

fo r a l Z z in the Le b esgue se t of f ( t he se t a t whi ah df i 8 t he de ri vative of i t s i n defi n i te in t sgra l ) . In partiau lar

l im u .l. " f (::: ) - 1 (::: ) a. e .

PJto o 6 . Im itate t he pro of o f the the orem l a c. c i t . (In our case , h ). (t ) • !. h1 ( .I.t ) , whe re h1 (t ) 2 rr_I_2 ) ). V ;; l + t

Fin a l l y , to p ro ve the l a st a s s e r t ion

f • h). E COO , V f e Ll

we w i l l p rove t hat ( a ) II hi,n )11 00 < oQ

(b) I f f E Ll , g i s bounded w i t h !! g ' II "., < ..., t hen ( f .g ) ' - f *g ' . ( l J R e a l an d Ab s e r a c e :h a l ,{ s i s . E , llewi t t an d K , S t r ombe rg

Sp r i n ge r - V e rl a g , 1 9 6 5 ,

1 96

Thh has beeD dOJle , then _ c an apply (b ) w i t l! g . h �n ) .. snd

g e t (f * h �) ( n ... l ) • f * h�n "' l ) , heD ce f * h� Ii!: COO

P'I'o of of (b ) .

Ci .' f)(X+ h� - (g * f) C!) • jg(X -Y+h)h- g(x -y) f (y ) dy The int e g r an d is bounde d' in absolu te value by 1 9 ' 1 oo l f l and

-+- g ' (x -y ) f (y) a s h - O . Lebe sgue ' s do.ina t e d conve rgence the o reJII noli' g i ves us

I i. jg (X -y+h) - g (x -y) f (y) dy . jg l (X-Y ) f (Y) dY h -- O h

he n ce (g * f ) ' (x ) • (g ' * f ) (x ) .

Proof o f ( a J . I t su f f i ce s t o !lrove that for

,(x ) · p ex ) I de g P OS;; 2 (n - I )

[ 1 + x2J n t he n 1 ,' n OC) < OC)

Th i s give s (a ) w i thou t difficul ty . Q . ! . D.

' . 1 �" AlU: . - 1I � - H� . lIu t g (r ) - (2 r ) .� (r ) l e l . L , u d lI en ce fOh8 0 -

r • • 9 . 1 ( � ) g l ve e t h a t h�n ) 1 . a 'o u r l . r t r &ll e�or. , hence h/..n ). c .

fOhl . V1 pe . a. o th e r proof o � (4 ) .

EXE R CUE 13 . - Fi n d lim

fA � . i tzdt

II -+- 00 - A t

10111,, '1''' ;\ i. a p o s i t i lJ " "on . t an t .

S O L U T I OIJ

(-00 < :r < 00 )

F i rst o b s e rve that we have .(Exe r c i s e r::..£J ch . VI I ) A .

1m -- . -l ' j S 1 nu du It (the D i r i ch l e t i n t e g ral ) A ..... "'" 0 u 2

hen ce a change o f v a r i a b l e g iv e s A

l im J s i n Rudu A _ OC) 0 u

It ! I R ! 2

for R .; 0 , and is 0 fo r R • 0

Nex t we have A f s i n �t

- A t

A eixt dt . 2 f s inxt . co s tx dt

o t

A . f s i n (l.+x ) t

a t '

A dt +

f ,in (:\.-x ) t dt o t

I t fo 1 1 o,..s that A

l i_ f JiJl.).t .ixt dt is A-+ oo -A 1

o if Ix I > ). 1: r if I x I • >. x if Ix I < ).

199

Observe t hat if f • -Xr _). , :\.] ' t hen ; ( t ) .�' S i�lt (See ( 2 ) of Exerc ise [] ) • I f A 'A (x ) . J f (t ) eitxda ( t )

-A IlS in Theorem 9. 1 3 (d) . then

A J . s in).t e 1 t" d t • le +A (x )

-A t Hence the preceding resu l t sho,..s t hat

lim .A (x ), . t(f (x + ) + f (x - ) ) A_ oO

(,..he re f (x + ) • l illl f ey) • f (x - ) • l i_ f ey) ) Y-+ " Y ..... " ,, < y Y < "

This l im i t i s f (x ) i f x � �).. This i s relevant t o 9 . 13 (d ) .

Q . E . D.

E X E R CISE � 4 . - Fi n d aon di t io'n l! on f an d I or f w hi ch Q n s u re s t roQ oo rr. ct n s s s of thQ fo Z lo�ing fo �a l argumQn t : If

,(t ) - - f (z ) � liz 1 fO<:> -i tz 2 ,.. 00

F (z i - I f ez + UX) k- -oO .

t hen F is P Q ri o di c. wi th p e ri o d 2 lC . the n t h Fo u ri � r coeffi cien t of � inz F is . (n ) . hsncQ F (z ) - L , (n ) 1l '

,In p ar t i cu l ar

00 00 ( N ) L ' f (2 k x) - L ,(n )

k--CIIQ n-.. oo

go' r . gen e l'a Z lll 00 00 ( 0 ) L f ( k � ) •

k--oo ",L .,(n ",) " i f' ''' > O. � > �. "' � : 21t. n- ....,

200 What do e u ( - ) . a!l a bou t t hll Z im i t . a s or - 0 .. of the I' i g h t

han d s i de ( f o I' "n i oe " fun otion s . of CaUl'S " ) ? I s t hi s in ag I' lI am a " t

wi th t he i " v .. r s i on t h e o rem ?

so LUn 01< The fo l lowing theo rem i s due t o G . H. Hardy

The orem . - Le t f E Ll

(T) an d aeeumll

(1 / f in ) z 0 ( 1 /,, ) as n _ ao Thlln S

n (f. t ) and 0" (f. t ) o o" v e rg e fo I' t he same v a Zue s of t

an d to t he s am e Z im i t . A Z so , i f an

(f. t ) oon'1 6 rg e s u n i fo rm Z !I on 8 0 m e s .. t, s o doe s S

n ( f , t ) .

Co ro l l ary . - Le t f b e of bounded vari a t ion on T; t h e n the S,,

(f. t )

oon v e rg e to t (f ( t+ O ) + f i t - O J ) an d i n p a r ti ou Z ar t o f i t ) a t e v e r!l

p o i n t of oon t i n u i t !l . The con v e rge " oe is un i form on o Z o s e d in t e r v a Z s

of oon t i n u i t!l of f .

Thi s i s � 2 . 2 p . 5 2 - 5 3 in "An In troduction to Harmon i c Analys i s " , Y. Katzne lson , John Wi ley and Sons Inc , 1 96 8 . Here Sn i s the partia l sum o f the Fourier se r ie s o f f ,

Sn {f , t)

and a i s t he Ce s aro n

an (f , t )

n . L f (n ) e in t f en ) . lc = -n

means of the Sn f S , On · n

. L: ( 1 _ ill)f {n ) e in t lc = -n n+l

-l-f e - in tf (t ) dt • 2 11: 1 n+r(S. + S + · · · + S ) 1 . n

an (f , . ) i s the convolu t ion f * Kn ' Kn i s the Fe j e r ' s kernel on T ,

K (t ) • I ( 1 _ lli)eilct . 1 [Sin �] 2

n lc=-n n+l � sin i-In our case , we impo se the fo l l owing condit ions (2 ) F i s· con t inuous . has bounded variation s and I .(n ) I :::;;;...!:L I n I

fo r l arg e v a lue s of I n I •

I f f s a t i s f i e s (2 ) , then by the coro l l ary loc . ci t , t he Fourier series o f F converg e s uniformly to F , in part icul ar

(3 ) F (x ) . I, (n ) e inX

and (n i s ( 3 ) with x - o .

Note that (2 ) i s s a t i sf ied if f i s con tinuous and f E BV _

s in ce f( t ) - O (l / t ) . More over . i f

f (x) • Lg (t ) d t

201 f E Ll , t hen (2 ) i s s a t i s f i e d imme d i a t e l y .

N o w sup p o s e (Z ) , an d con s i d e r f l ( t ) . f eu ) p . > O ) . Then

"l( t ) • _1_ f f (x ) e - i tx dx • -l J f (il.x ) e - i t xdx 2 7C 1 2 r.:

. t 1 1 f - l );"

• - - f (x ) e dx )" 2 r.:

• !. If ( t f>.. ) iI.

( ' ) g iv e s u s 2'>1 (Zlt k ) . LIf1 (n ) , hen ce

f f (k. Z n) • 1:. f; ,(n/o:) k = -oo )"n = - oo

Pu t 0: . 1/ )" • J • Z7C/a< • Z lt )" , t h i s g iv e s

( * ) f f (Jc � ) · ", f lf(n or) k = -oo n = -oo

I f 0< _ 0 . t he r i g ht s i de of ( * ) t en ds t o t he R i e m ann i n t e g r al of If on R , an d t he l e f t on e t e n d s t o f ( 0 ) . T h i s i s in ag r e emen l w i t h t he inv e r s i on t he o rem :

f ro ) • jf ( t ) dm (t ) • z� jJ' f (x ) e - i t x dx d t .

Q . E . D .

E X E R CI S E 1 5 . - Tak e f (% )

t he i de n t i t lf

e - I% I i n t h e p 1'e cs dinJ € % s l' c i s € an d dS l' i v Q

S O L U T I O "

2 7C 0: e + ,,27C or _ 1

I t i s c l e a r t h a t t he s e r i e s If (x .. k � ) converg e s un i f orml ) on [ O , Z lC] . We have

L""'e- ' x ' e - i tx dx • zIa""'e -X co s ( t X ) rb:.

,\ p r i m i t iv e of e - X costx shou l d have t he f o m e - x ( aco s tx .. b s i n t x ) , an d i s

e -X ( - c osx t .. t s i nxt ) (1 .. t 2 )

I h i s g i v e s

f '"; -x c o s t x dx = 1 im ,;,e_-_x""C_-..;;,c_o,;;.,s..;;,tx_· _"_,:-t_s,",'""n,;;.,t;;.;.x�) ] f. o A-+ <>o ( 1 .. t ' ) 0

I t fo l l ows t h a t

1 �2

202

and hen ce <ret ) -. O ( I / t ) a s I t 1 -- 00

he n ce

Thu s the fonlU i a ( - ) In !xe r c i se G] g ive s

f f ( k �) Je - - MJ

GO 1 .. 2 e -� - 2 L 2 2

n z -ao r � .. 4x2n2 ] . U > 0 )

00 � . 2 L 2 � e ' - 1 n z - oo r � .. 4x2n2)

If � - 2 x or • t h i s g ive s

( or > 0 )

Q . E. D .

E X E � C I SE l b . - raka f (� j l'il 8U � ti'flg i da'fl ti t .. r

S O L U T I ON First let u s f i n d f • By The o re. 9 . 2 ( f ) , we have

( f ) l ( t ) -L-: - ib C-U ) f (X ) drI (X )

l°O-itx c . ) _ X 2 .l... ( ) t f- ( ) - e - lX e \.All x - - r t . -ao

Thi s s ho ws t hat f (t ) - f ( o ) e - t i I 4 • f l O ) i s found e a s i l y as f o l l o w s

Thu s le o ) •

_ ( !.) 2 f l O ) - f f (t ) dm (t ) - f ( O ) f e 2 dt - 2 f ( 0 ) e -t dt �

f 2

� 2 .. 1 2 [ f ( 0 ) J , hen c.e t ( 0 ) • rz

h t 2 t a t 00 2 I f -x - i t x dx 1 1 -I;' - e e - - - .ffX e -

2 x . 00 2 x rz

� can app ly the identity ( * ) of Exerci 5e [Al

f: exp (_k2 ,2 ) . 1! /i f' exp (_4lt2n2 ) !c - -DO ' ZxDr---oo 4 �2 hence

�( "AR� . - rbe c;'o.pa t.ti oJl of ; .110". tll",t i f 1

til.., � • g .

g (r ) - . r

- r

* * *

203

Exercises • In Real· Analysis with Solutions' YU TrQng Tuan

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