Solutions to Exercises

Chapter 2

2.1. U is convex: p((1− t)x + t y

)< (1− t) + t = 1 if x, y ∈ U and t ∈ (0, 1).

U is absorbing: p(x/2p(x)

) = 1/2 < 1 if x ∈ X and p(x) �= 0.U is balanced: p(kx) = |k|p(x) < 1 if x ∈ U and |k| ≤ 1.

2.2. p j (x) + p j (y) ≤ max{p1(x), . . . , pm(x)} + max{p1(y), . . . , pm(y)} if j =1, . . . ,m and x, y ∈ X . Hence p(x + y) ≤ p(x) + p(y) for x, y ∈ X .Let X := K

2, and let p1(x) := |x(1)| + 2|x(2)|, p2(x) = 2|x(1)| + |x(2)|for x := (x(1), x(2)) ∈ X . Then p1 and p2 are norms on X . But q is not aseminorm on X : If x := (1, 0) and y := (0, 1), then q(x + y) > q(x)+ q(y).

2.3. Let p ∈ {1, 2,∞}, x ∈ �p, xn := (x(1), . . . , x(n), 0, 0, . . .) ∈ c00, n ∈ N.If p := 1 or p := 2, then ‖xn − x‖p → 0.If p := ∞ and x ∈ c0, then ‖xn − x‖∞ → 0. Conversely, let (yn) be asequence in c0 such that yn → x in �∞, and let ε > 0. There is n0 ∈ N

such that ‖yn0 − x‖∞ < ε/2. Since yn0 ∈ c0, there is j0 ∈ N such that|yn0( j)| < ε/2 for all j ≥ j0. Then |x( j)| ≤ |x( j) − yn0( j)| + |yn0( j)| ≤‖x − yn0‖∞ + |yn0( j)| < ε/2 + ε/2 = ε for all j ≥ j0. Hence x ∈ c0.

2.4. For t ∈ (0, 1], let x(t) := 1/√t , and let x(0) := 0. Then x ∈ L1([0, 1]), but

x /∈ L2([0, 1]). Also, √x ∈ L2([0, 1]), but √x /∈ L∞([0, 1]).Let y(t) := x(t) if t ∈ [0, 1] and y(t) := 0 if either t < 0 or t > 1. Theny ∈ L1(R), but y /∈ L2(R)∪L∞(R). Let z(t) := √

x(t) if t ∈ [0, 1], z(t) := 0if t < 0, and z(t) := 1/t if t > 1. Then z ∈ L2(R), but z /∈ L1(R) ∪ L∞(R).Let u(t) := 1 if t ∈ R. Then u ∈ L∞(R), but u /∈ L1(R) ∪ L2(R).

2.5. For x ∈ C([0, 1]), ‖x‖1 ≤ ‖x‖2 ≤ ‖x‖∞. For n ∈ N, let xn(t) := 1 − ntif 0 ≤ t ≤ 1/n, and xn(t) := 0 if (1/n) < t ≤ 1. Then xn ∈ C([0, 1]) forn ∈ N, ‖xn‖∞ = 1, ‖xn‖1 = 1/2n and ‖xn‖2 = 1/

√3n.

2.6. Let x ∈ X . If ‖x‖′ = 0, then x is a constant function with x(a) = 0, that is,x = 0. For t ∈ (a, b], there is s ∈ (a, t) such that x(t) − x(a) = (t − a)x ′(s)by the mean value theorem, and so |x(t)| ≤ |x(a)| + (b − a)‖x ′‖∞. Hence

© Springer Science+Business Media Singapore 2016B.V. Limaye, Linear Functional Analysis for Scientists and Engineers,DOI 10.1007/978-981-10-0972-3

203

204 Solutions to Exercises

‖x‖1,∞ = max{‖x‖∞, ‖x ′‖∞} ≤ max{1, b − a}‖x‖′, ‖x‖′ ≤ ‖x‖1,∞ and‖x‖∞ ≤ ‖x‖1,∞ for all x ∈ X .Also, if for n ∈ N, we let xn(t) := (t − a)n/(b − a)n, t ∈ [a, b], thenxn ∈ X, ‖xn‖∞ = 1, and ‖x ′

n‖∞ = n/(b − a) → ∞.

2.7. For x ∈ X , let us write |||x + Z(F)||| := inf{‖x + z‖ : z ∈ Z(F)}. Let y ∈ Y .There is x ∈ X such that F(x) = y, and then |||x + Z(F)||| = inf{‖u‖ :u ∈ X and u − x ∈ Z(F)} = inf{‖u‖ : u ∈ X and F(u) = y} = q(y). Ifk ∈ K, then F(kx) = ky. Also, if y1, y2 ∈ Y and F(x1) = y1, F(x2) = y2,then F(x1 + x2) = y1 + y2. As in the proof of Proposition 2.5(i), q(ky) =|||kx + Z(F)||| = |k| |||x + Z(F)||| = |k|q(y), and q(y1 + y2) = |||x1 + x2 +Z(F)||| ≤ |||x1 + Z(F)||| + |||x2 + Z(F)||| = q(y1) + q(y2).Suppose Z(F) is a closed subset of X . Let y ∈ Y be such that q(y) = 0. Ifx ∈ X and F(x) = y, then |||x + Z(F)||| = q(y) = 0, and so x ∈ Z(F),that is, y = F(x) = 0. Hence q is a norm on Y . Conversely, suppose q is anorm on Y . Let (xn) be a sequence in Z(F) such that xn → x in X . Then|||x+Z(F)||| = |||x−xn+Z(F)||| ≤ ‖x−xn‖ → 0, and so |||x+Z(F)||| = 0.Let y := F(x). Then q(y) = |||x + Z(F)||| = 0, so that F(x) = y = 0, thatis, x ∈ Z(F). Thus Z(F) is a closed subset of X .

2.8. Suppose there are x0 ∈ X and r > 0 such that U (x0, r) ⊂ E . Since E iscompact, it is closed in X , and so U (x0, r) ⊂ E . But then the closed unit ballof X is a closed subset of the compact set s(E − x0), where s := 1/r .

2.9. The closed unit ball E of �2 is not compact since the normed space �2 isinfinite dimensional. Let (xn) be a sequence in the Hilbert cube C . There isa subsequence xn,n of (xn) such that for each j ∈ N, the sequence xn,n( j)converges in K to x( j), say. Let x := (x(1), x(2), . . .). Then x ∈ �2 since|x( j)| ≤ 1/j for each j ∈ N. Also,

‖xn,n − x‖22 ≤m∑

j=1

|xn,n( j) − x( j)|2 +∞∑

j=m+1

(2j

)2for every m ∈ N.

Let ε > 0. Choose m ∈ N such that the second term above is less than ε2/2,and then choose n0 ∈ N such that |xn,n( j) − x( j)|2 < ε2/2m for all n ≥ n0and j = 1, . . . ,m. It follows that ‖xn,n − x‖2 < ε for all n ≥ n0.

2.10. Suppose a norm ‖ · ‖ on a linear space X is induced by an inner product. Ifx, y ∈ X, ‖x‖ = 1 = ‖y‖ and x �= y, then

‖x + y‖2 = 2‖x‖2 + 2‖y‖2 − ‖x − y‖2 < 2‖x‖2 + 2‖y‖2 = 4.

Thus (X, ‖ · ‖) is strictly convex. The norm ‖ · ‖2 on �2, and the norm ‖ · ‖2 onL2([0, 1]) are induced by inner products.Hence they are strictly convex.On theother hand, let x := e1 and y := e2. Then ‖x‖1 = 1 = ‖y‖1 and ‖x+ y‖1 = 2,and ‖x + y‖∞ = 1 = ‖x − y‖∞ and ‖(x + y) + (x − y)‖∞ = 2. Hence �1

and �∞ are not strictly convex. Similarly, if x and y denote the characteristicfunctions of [0, 1/2] and (1/2, 1], then ‖x‖1 = 1/2 = ‖y‖1 and ‖x+y‖1 = 1,

Solutions to Exercises 205

and ‖x + y‖∞ = 1 = ‖x − y‖∞ and ‖(x + y) + (x − y)‖∞ = 2. HenceL1([0, 1) and L∞([0, 1]) are not strictly convex. Thus if p ∈ {1,∞}, then thenorms on �p and L p([0, 1]) are not induced by an inner product.

2.11. Suppose |〈x, y〉|2 = 〈x, x〉〈y, y〉, and define z := 〈y, y〉x − 〈x, y〉y. Pro-ceeding as in the proof of Proposition 2.13(i), we see that 〈z, z〉 = 0, and soz = 0, that is, 〈y, y〉x = 〈x, y〉y. Conversely, if 〈y, y〉x = 〈x, y〉y, then wecan readily check that |〈x, y〉| = ‖x‖‖y‖.Suppose ‖x+ y‖2 = (‖x‖+‖y‖)2. Then Re 〈x, y〉 = ‖x‖‖y‖. If either x = 0or y = 0, then clearly, ‖y‖x = ‖x‖y. Now let x �= 0 and y �= 0, and defineu := x/‖x‖, v := y/‖y‖. Then ‖u‖ = 1 = ‖v‖, and ‖u + v‖ = 2 since

‖u + v‖2 = ‖u‖2 + ‖v‖2 + 2Re 〈u, v〉 = 2 + 2Re 〈x, y〉‖x‖‖y‖ = 4.

By Exercise 2.10, (X, ‖ · ‖) is strictly convex, and so u = v, that is, ‖y‖x =‖x‖y. Conversely, if ‖y‖x = ‖x‖y, then we can readily check that ‖x + y‖ =‖x‖ + ‖y‖.

2.12. Let x, y, z ∈ X . By the parallelogram law,

‖x + y + z‖2 + ‖x + y − z‖2 + ‖x − y + z‖2 + ‖x − y − z‖2= 2‖x + y‖2 + 2‖z‖2 + 2‖x − y‖2 + 2‖z‖2= 4(‖x‖2 + ‖y‖2 + ‖z‖2).

2.13. Let x ∈ Z . If k ∈ K, then clearly kx ∈ Z . Let y ∈ X , and let ε > 0. Then

0 ≤ 〈x + εy, x + εy〉=〈x, x〉+ ε2〈y, y〉+2εRe 〈x, y〉=ε2〈y, y〉+2εRe 〈x, y〉,

and so 0 ≤ ε〈y, y〉 + 2Re 〈x, y〉. Let ε → 0 to obtain 0 ≤ Re 〈x, y〉. Replacex by −x to obtain Re 〈x, y〉 ≤ 0, and so Re 〈x, y〉 = 0. Replace x by i x toobtain Im 〈x, y〉 = 0, and so 〈x, y〉 = 0.In particular, if x, y ∈ Z , then 〈x + y, x + y〉 = 〈x, x〉+〈y, y〉+2 Re 〈x, y〉 =0 + 0 + 0 = 0, that is, (x + y) ∈ Z . Thus Z is a subspace of X .Suppose x + Z = x1 + Z and y + Z = y1 + Z . Then x1 − x, y1 − y ∈ Z ,and so 〈x1, y1〉 = 〈(x1 − x) + x, (y1 − y) + y〉, which is equal to

〈(x1 − x), (y1 − y)〉 + 〈(x1 − x), y〉 + 〈x, (y1 − y)〉 + 〈x, y〉 = 〈x, y〉.

Let 〈〈x + Z , y + Z〉〉 := 〈x, y〉 for x, y ∈ X . If 〈〈x + Z , x + Z〉〉 = 0, thenx ∈ Z , that is, x+Z = 0+Z . It follows that 〈〈·, ·〉〉 is an inner product on X/Z .Now |〈x, y〉|2 = |〈〈x + Z , y + Z〉〉|2 ≤ 〈〈x + Z , x + Z〉〉〈〈y + Z , y + Z〉〉 =〈x, x〉〈y, y〉 for all x, y ∈ X .

2.14. Let x, y ∈ X . Then 〈x + y, x + y〉−〈x − y, x − y〉 = 4Re 〈x, y〉. Replacingy by iy, 〈x + iy, x + iy〉 − 〈x − iy, x − iy〉 = 4Re 〈x, iy〉 = 4 Im 〈x, y〉.Hence the right side is equal to 4 Re 〈x, y〉 + i 4 Im 〈x, y〉 = 4〈x, y〉.

206 Solutions to Exercises

2.15. ‖x − y‖2 = ‖x‖2 + ‖y‖2 − 2Re 〈x, y〉 = ‖x‖2 + ‖y‖2 − 2 ‖x‖ ‖y‖ cos θx,y .

2.16. In Example 2.12(i), replace n by mn. Also, ‖In‖F = (1 + · · · + 1)1/2 = √n.

2.17. For x ∈ �2, ‖x‖2w = ∑∞j=1 w( j)|x( j)|2 ≤ ‖w‖∞‖x‖22. If v ∈ �∞, then

w( j) ≥ 1/‖v‖∞ for all j ∈ N, and so ‖x‖2w ≥ ‖x‖22/‖v‖∞. Conversely,suppose there is α > 0 such that ‖x‖2 ≤ α‖x‖w for all x ∈ �2. Then 1 ≤α w( j) by considering x := e j for j ∈ N. Hence ‖v‖∞ ≤ α.

2.18. The set {e1 + · · · + em, e1 − e2, . . . , e1 − em} is linearly independent. Next,y1 := 0, z1 := x1 = e1 + · · · + em, u1 := z1/‖z1‖2 = (e1 + · · · + em)/

√m

and y2 := 〈x2, u1〉u1 = 0, z2 := x2 − y2 = e1 − e2, u2 := z2/‖z2‖2 =(e1 − e2)/

√2, as desired. Let n ∈ {2, . . . ,m − 1} and assume that u j :=

(e1 + · · · + e j−1 − ( j − 1)e j )/√

( j − 1) j for j = 2, . . . , n. Then

yn+1 :=n∑

j=1

〈xn+1, u j 〉u j =n∑

j=2

〈xn+1, u j 〉u j =n∑

j=2

u j√( j − 1) j

,

and zn+1 := xn+1 − yn+1 = e1 − en+1 − ∑nj=2 u j/

√( j − 1) j . Now

e1 −n∑

j=2

u j√( j − 1) j

= e1 −n∑

j=2

e1 + · · · + e j−1 − ( j − 1)e j( j − 1) j

= 1

n

n∑

j=1

e j .

Hence zn+1 = (e1 + · · · + en)/n − en+1 = (e1 + · · · + en − nen+1)/n, andun+1 := zn+1/‖zn+1‖2 = (e1 + · · · + en − nen+1)/

√n(n + 1) as desired.

2.19. For j = 1, . . . , n, denote the j th column of A by x j ∈ Km . As in Theorem

2.17, obtain u1, . . . , un by theGram–Schmidt orthogonalization of x1, . . . , xn .Then x j = y j + ‖x j − y j‖u j , where y1 := 0, and

y j := 〈x j , u1〉u1 + · · · + 〈x j , u j−1〉u j−1 for j = 2, . . . , n.

Let Q denote the m × n matrix whose j th column is u j for j = 1, . . . , n, andlet R := [ri, j ] denote the n×n matrix, where ri, j := 〈x j , ui 〉 if 1 ≤ i ≤ j −1,r j, j := ‖x j − y j‖ and ri, j := 0 if j + 1 ≤ i ≤ n for j = 1, . . . , n. Thenfor j = 1, . . . ,m, x j = 〈x j , u1〉u1 + · · · + 〈x j , u j−1〉u j−1 + ‖x j − y j‖u j =r1, j u1 + · · · + r j−1u j−1 + r j u j , that is, A = [x1, . . . , xn] = [u1, . . . , un]R =QR.Uniqueness: Suppose A = Q′R′, where the columns u′

1, . . . , u′n of Q

′ form anorthonormal subset of K

m , and R′ := [r ′i, j ] is upper triangular with positive

diagonal entries. Then x1 = r ′1,1u

′1. Hence r ′

1,1 = ‖x1‖ = r1,1, and u′1 =

x1/r1,1 = u1. Next, let j ∈ {2, . . . , n}, and suppose we have shown u′1 =

u1, . . . , u′j−1 = u j−1. Then x j = r ′

1, j u′1 + · · · + r ′

j−1, j u′j−1 + r ′

j, j u′j =

r ′1, j u1 + · · · + r ′

j−1, j u j−1 + r ′j, j u

′j . Hence r ′

i, j = 〈x j , ui 〉 = ri, j for i =1, . . . , j − 1 and r ′

j, j = ‖x j − r1, j u1 − · · · − r j−1, j u j−1‖ = r j, j , so thatu′j = (x j − r1, j u1 −· · ·− r j−1, j u j−1)/r j, j = u j . Thus Q′ = Q and R′ = R.

Solutions to Exercises 207

If A is an infinite matrix whose columns form a linearly independent subsetof �2, then the above arguments hold.

2.20. Let m ∈ N. Then∑m

n=0 | ∫ 1−1 x(t)pn(t)dt |2 ≤ ∫ 1

−1 |x(t)|2dt by the Besselinequality. Also, equality does not hold here since x /∈ span {p0, p1, . . . , pm}.

2.21. If r := 0, then∫ T−T ur (t)dt = 2T , and if r �= 0, then

∫ T−T ur (t)dt =

(2 sin rT )/r for all T > 0. Hence limT→∞ 12T

∫ T−T ur (t)dt is equal to 1 if

r = 0, and it is equal to 0 if r �= 0. As a result, limT→∞ 12T

∫ T−T x(t)dt

exists for every x ∈ X . Let p, q ∈ X . Then p q ∈ X , and so 〈p, q〉 iswell-defined. Clearly, the function 〈· , ·〉 : X × X → K is linear in thefirst variable, it is conjugate symmetric, and 〈p, p〉 ≥ 0 for all p ∈ X .Let p := c1ur1 + · · · + cnurn , where r1, . . . , rn are distinct real numbers,and c1, . . . , cn ∈ C. Then 〈p, p〉 = |c1|2 + · · · + |cn|2. It follows thatp = 0 whenever 〈p, p〉 = 0. Thus 〈· , ·〉 is an inner product on X . Also,〈ur , us〉 = limT→∞ 1

2T

∫ T−T ur−s(t)dt . Hence {ur : r ∈ R} is an uncountable

orthonormal subset of X . (In particular, it is a linearly independent subset ofX .)

2.22. Since the map F : X → Y is linear, and since 〈· , ·〉Y is an inner producton Y , the function 〈· , ·〉X : X × X → K is linear in the first variable, it isconjugate-symmetric, and 〈x, x〉X ≥ 0 for all x ∈ X .(i) 〈· , ·〉X is an inner product on X if and only if x = 0 whenever 〈x, x〉X =〈F(x), F(x)〉Y = 0, that is, F(x) = 0.(ii) Since F is one-one, if F(uα) = F(uβ), then uα = uβ . Also, 〈uα, uβ〉X =〈F(uα), F(uβ)〉Y for all α,β.(iii) Since F is one-one, {F(uα)} is an orthonormal subset of Y . Also, since

F is onto, if {F(uα)} is a proper subset of an orthonormal subset E of Y , then{uα} is a proper subset of the orthonormal subset F−1(E) of X .

2.23. There are α > 0 and β > 0 such that β‖x‖ ≤ ‖x‖′ ≤ α‖x‖ for all x ∈ X .Hence (xn) is a Cauchy sequence in (X, ‖ · ‖′) if and only if (xn) is a Cauchysequence in (X, ‖ · ‖), and (xn) is a convergent sequence in (X, ‖ · ‖′) if andonly if (xn) is a convergent sequence in (X, ‖ · ‖).

2.24. c0 is the closure of c00 in the Banach space �∞ (Exercise 2.3).To show that c is a closed subspace of the Banach space �∞, let (xn) be asequence in c such that xn → x in �∞, and let ε > 0. There is n0 ∈ N suchthat ‖xn0 − x‖∞ < ε/3. Since xn0 is a Cauchy sequence in K, there is j0 ∈ N

such that |xn0(i) − xn0( j)| < ε/3 for all i, j ≥ j0. Then

|x(i)− x( j)| ≤ |(x − xn0)(i)− (x − xn0)( j)| + |xn0(i)− xn0( j)| <2ε

3+ ε

3= ε

for all i, j ≥ j0. Hence (x( j)) is Cauchy sequence in K, and so x ∈ c.To show that C0(T ) is a closed subspace of the Banach space C(T ), let (xn)

be a sequence in C0(T ) such that xn → x in C(T ), and let ε > 0. There

208 Solutions to Exercises

is n0 ∈ N such that ‖xn0 − x‖∞ < ε/2. Since xn0 ∈ C0(T ), there is acompact subset Tε of T such that |xn0(t)| < ε/2 for all t ∈ T \Tε. Then|x(t)| ≤ |x(t) − xn0(t)| + |xn0(t)| ≤ ‖x − xn0‖∞ + |xn0(t)| < ε/2 + ε/2 = εfor all t ∈ T \Tε, and so x ∈ C0(T ).

2.25. Since X has a denumerable basis, X is not a Banach space.

2.26. (i) Let Y be an m dimensional subspace of X . There is an orthonormal basisu1, . . . , um of Y . Let (xn) be a Cauchy sequence in Y . Then xn = 〈xn, u1〉u1+· · · + 〈xn, um〉um for n ∈ N. Fix j ∈ {1, . . . ,m}. For all n, p ∈ N, |〈xn, u j 〉 −〈xp, u j 〉| ≤ ‖xn − xp‖. Hence the Cauchy sequence (〈xn, u j 〉) converges inK to k j , say. Then xn → k1u1 + · · · + kmum in Y .(ii) Let X be infinite dimensional. By Theorem 2.17, there is a denumerableorthonormal subset {u1, u2, . . .} of X . The sequence (un) in the closed unitball of X does not have a convergent subsequence since ‖un − u p‖ = √

2 forall n �= p. Hence the closed unit ball of X is not compact.Conversely, suppose X is finite dimensional. By Theorem 2.23(iii), there is anisometry from X onto an Euclidean space whose closed unit ball is compactby the classical Heine–Borel theorem.(iii) Suppose X is complete, and it has a denumerable (Hamel) basis. ByTheorem 2.17, there is an orthonormal subset {u1, u2, . . .} of X which is a(Hamel) basis for X . Then x := ∑∞

n=1 un/n belongs to X by the Riesz–Fischer theorem, but it does not belong to span {u1, u2, . . .}.

2.27. Let x := (x(1), x(2), . . .) ∈ c0, and for m ∈ N, let sm(x) := ∑mn=1 x(n)en .

Then ‖sm(x)− x‖∞ = sup{|x(n)| : n = m + 1,m + 2, . . .} → 0 as m → ∞.Thus x = ∑∞

n=1 x(n)en . Also, if there are k1, k2, . . . in K such that x =∑∞n=1 knen , then k j = ∑∞

n=1 knen( j) = x( j) for each j ∈ N.Let x ∈ c, and x(n) → �x . Then y := x − �xe0 ∈ c0, and y = ∑∞

n=1 y(n)en .Hence x = �xe0 + ∑∞

n=1(x(n) − �x )en . Also, if x = ∑∞n=0 knen , where

k0, k1, k2, . . . are in K, then y0 := x − k0e0 = ∑∞n=1 knen belongs to the

closure of c00 in �∞, that is, y0 ∈ c0. It follows that k0 = �x and k j =∑∞n=1 knen( j) = x( j) − �xe0( j) = x( j) − �x for each j ∈ N.

Let x ∈ �1. Define kn := x(2n − 1) + x(2n) and �n := x(2n − 1) − x(2n)

for n ∈ N. For m ∈ N, let s2m(x) := k1u1 + �1v1 + · · · + kmum + �mvm ands2m−1(x) := k1u1+�1v1+· · ·+km−1um−1+�m−1vm−1+kmum . Then for allminN, s2m(x) = x(1)e1+· · ·+x(2m)e2m and s2m−1(x) = x(1)e1+· · ·+x(2m−2)e2m−2+kmum . Since km → 0,we obtain x = k1u1+�1v1+k2u2+�2v2+· · · .Also, if x = k ′

1u1 + �′1v1 + k ′

2u2 + �′2v2 + · · · , then x(2n − 1) = (k ′

n + �′n)/2

and x(2n) = (k ′n − �′

n)/2, that is, k′n = x(2n − 1) + x(2n) = kn and �′

n =x(2n − 1) − x(2n) = �n for n ∈ N.As �∞ and L∞([a, b]) are not separable, they do not have Schauder bases.

2.28. For j ∈ N and x( j) ∈ X j , define ‖x( j)‖ j := 〈x( j), x( j)〉1/2. The function〈· , ·〉 : X × X → K is well-defined since for x, y ∈ X ,

Solutions to Exercises 209

∞∑

j=1

|〈x( j), y( j)〉 j |≤∞∑

j=1

‖x( j)‖ j‖y( j)‖ j ≤( ∞∑

j=1

‖x( j)‖2j)1/2( ∞∑

j=1

‖y( j)‖2j)1/2

.

If x, y ∈ X , then x + y ∈ X since ‖x( j) + y( j)‖ j ≤ ‖x( j)‖ j + ‖y( j)‖ j and

( ∞∑

j=1

(‖x( j)‖ j + ‖y( j)‖ j)2

)1/2

≤( ∞∑

j=1

‖x( j)‖2j)1/2

+( ∞∑

j=1

‖y( j)‖2j)1/2

.

Also, it is easy to see that if x ∈ X and k ∈ K, then k x ∈ X . Further, it followsthat 〈· ,, ·〉 is an inner product on X . Suppose X is complete. Fix j ∈ N. If(xn( j)) is a Cauchy sequence in X j , define xn := (0, . . . , 0, xn( j), 0, 0, . . .),where xn( j) is in the j th entry, and note that (xn) is Cauchy sequence in X .If xn → x in X , then xn( j) → x( j) in X j . Thus X j is complete. Conversely,suppose X j is complete for each j ∈ N. Then the completeness of X followsexactly as in the proof of the completeness of �2 given in Example 2.24(ii).

2.29. The case k := 1 is treated in Examples 2.24(iii), (v) and 2.28(iv). We considerhere the case k := 2. The cases k > 2 are similar.(i) Let (xn) be a Cauchy sequence in (C2([a, b]), ‖·‖2,∞). Then (xn), (x ′

n) and(x ′′

n ) are Cauchy sequences in the Banach space (C([a, b]), ‖ · ‖∞). A well-known result in Real Analysis shows that there are x and y in C1([a, b]) suchthat ‖xn−x‖∞ → 0, ‖x ′

n−x ′‖∞ → 0, and ‖x ′n− y‖∞ → 0, ‖x ′′

n − y′‖∞ → 0.Then x ′ = y ∈ C1([a, b]), that is, x ∈ C2([a, b]), and ‖xn − x‖2,∞ =‖xn − x‖∞ + ‖x ′

n − x ′‖∞ + ‖x ′′n − x ′′‖∞ → 0.

(ii) Let (xn) be a Cauchy sequence in (W 2,1([a, b]), ‖ · ‖2,1). Then (xn), (x ′n)

and (x ′′n ) are Cauchy sequences in the Banach space (L1([a, b]), ‖ ·‖1). As we

have seen in Example 2.24 (v), there are absolutely continuous functions x andy on [a, b] such that ‖xn − x‖1 → 0, ‖x ′

n − x ′‖1 → 0, and ‖x ′n − y‖1 → 0,

‖x ′′n − y′‖1 → 0. Then x ′ = y is absolutely continuous on [a, b], that is,

x ∈ W 2,1([a, b]), and ‖xn−x‖2,1 = ‖xn−x‖1+‖x ′n−x ′‖1+‖x ′′

n −x ′′‖1 → 0.(iii) Let (xn) be a Cauchy sequence in (W 2,2([a, b]), ‖ · ‖2,2). Then (xn), (x ′

n)

and (x ′′n ) are Cauchy sequences in the Hilbert space (L2([a, b]), ‖ · ‖2). As

we have seen in Example 2.28(iv), there are absolutely continuous functions xand y on [a, b] such that ‖xn−x‖2 → 0, x ′ ∈ L2([a, b]), ‖x ′

n−x ′‖2 → 0, and‖x ′

n − y‖2 → 0, y′ ∈ L2([a, b]), ‖x ′′n − y′‖2 → 0. Then x ′ = y is absolutely

continuous on [a, b] and x ′′ = y′ ∈ L2([a, b]), that is, x ∈ W 2,2([a, b]), and‖xn − x‖2,2 = ‖xn − x‖2 + ‖x ′

n − x ′‖2 + ‖x ′′n − x ′′‖2 → 0.

2.30. (i)=⇒(ii): Let s := ∑∞n=1 xn . Then 〈s, xn〉 = 〈xn, xn〉 = ‖xn‖2 for n ∈ N.

(ii)=⇒(iii): Let un := 0 if xn = 0, and un := xn/‖xn‖ if xn �= 0. Then〈s, un〉 = ‖xn‖ for n ∈ N, and

∑∞n=1 ‖xn‖2 = ∑∞

n=1 |〈s, un〉|2 ≤ ‖s‖2.(iii)=⇒(i): Let sm := ∑m

n=1 xn, m ∈ N. Then for m > p, ‖sm − sp‖2 =‖xp+1+· · ·+xm‖2 = ‖xp+1‖2+· · ·+‖xm‖2. Thus (sm) is a Cauchy sequencein the Hilbert space H , and so it converges in H .

210 Solutions to Exercises

2.31. For each n ∈ N, En := {un, vn, wn} is an orthonormal subset of �2 andspan En = span {e3n−2, e3n−1, e3n}. (Let m := 3 in Exercise 2.18.) HenceE := {un : n ∈ N} ∪ {vn : n ∈ N} ∪ {wn : n ∈ N} is an orthonormal subset of�2 and span E = span {e j : j ∈ N}, which is dense in �2.

2.32. For each n ∈ N, En := {un, vn, wn} is an orthonormal subset of �2 andspan En = span {e3n−2, e3n−1, e3n}. (Compare Exercise 4.35.) Hence E :={un : n ∈ N} ∪ {vn : n ∈ N} ∪ {wn : n ∈ N} is an orthonormal subset of �2

and span E = span {e j : j ∈ N}, which is dense in �2.

2.33. It is easy to see that E := {u0, u1, v1, u2, v2, . . .} is an orthonormal subset ofL2([−π,π]). Let x ∈ E⊥. For k ∈ Z,

x(k) = 1

2π

∫ π

−π

x(t)e−ikt dm(t) = 1

2π

∫ π

−π

x(t)(cos kt − i sin kt)dm(t) = 0

since 〈x, u0〉 = 0 and 〈x, un〉 = 〈x, vn〉 = 0 for all n ∈ N. Hence x = 0 a.e.on [−π,π]. Thus E⊥ = {0}.Since 〈x, u0〉 = √

2π a0 and 〈x, un〉 = √π an, 〈x, vn〉 = √

π bn for all n ∈ N,the Fourier expansion x = 〈x, u0〉u0 + ∑∞

n=1

(〈x, un〉un + 〈x, vn〉vn)and the

Parseval formula ‖x‖22 = |〈x, u0〉|2 + ∑∞n=1

(|〈x, un〉|2 + |〈x, vn〉|2)yield the

desired results.

2.34. The subsets E := {u0, u1, u2, . . . , } and F := {v1, v2, . . .} of L2([0, 1]) areorthonormal. Given x ∈ L2([0, 1]), let y(t) := x(t/π) if t ∈ [0,π], andy(t) := x(−t/π) if t ∈ [−π, 0). Then y is an even function on [−π,π].Also, y ∈ L2([−π,π]). Further, ∫ π

−π y(t)dm(t) = 2π∫ 10 x(s)dm(s) = 2πa0.

For n ∈ N,∫ π

−π y(t) cos nt dm(t) = 2π∫ 10 x(s) cos nπs dm(s) = πan and∫ π

−π y(t) sin nt dm(t) = 0. By Exercise 2.33, y(t) = a0 + ∑∞n=1 an cos nt ,

the series converging in the mean square on [−π,π]. Hence x(s) = a0 +∑∞n=1 an cos nπs, the series converging in the mean square on [0, 1].

Similarly, given x ∈ L2([0, 1]), let z(t) := x(t/π) if t ∈ [0,π], and z(t) :=−x(−t/π) if t ∈ [−π, 0). Then z is an odd function on [−π,π]. As above, wemay use Exercise 2.33 to obtain z(t) = ∑∞

n=1 bn sin nt , the series convergingin the mean square on [−π,π], and so x(s) = ∑∞

n=1 bn sin nπs, the seriesconverging in the mean square on [0, 1].

2.35. The subspace G := span {v0, v1, . . .} of H consists of all polynomials in t3

defined on [−1, 1]. Let x be a continuous function on [−1, 1], and define y ∈ Xby y(t) := x(t1/3), t ∈ [−1, 1]. Let ε > 0. By the Weierstrass theorem, thereis a polynomial q defined on [−1, 1] such that ‖y − q‖∞ < ε. Define p(s) :=q(s3) for s ∈ [−1, 1]. Then p ∈ G and ‖x − p‖∞ = sup{|y(s3) − q(s3)| :s ∈ [−1, 1]} = ‖y−q‖∞ < ε. Further, ‖x− p‖2 ≤ √

2 ‖x− p‖∞ <√2 ε. By

Proposition 1.26(ii),G is dense in H . The calculations of v0, v1 v2 are routine.

2.36. If x ∈ H , then by Bessel’s inequality,∑

α |〈x, uα〉|2 ≤ ‖x‖2 < ∞. By theRiesz–Fischer theorem,

∑α〈x, uα〉vφ(α) ∈ G. Thus the map F : H → G is

well-defined. It is easy to see that F is linear. Let x j := ∑α〈x j , uα〉uα ∈ H

Solutions to Exercises 211

for j := 1, 2. By the orthonormality of the sets {uα} and {vφ(α)}, 〈x1, x2〉 =∑α〈x1, uα〉〈x2, uα〉 = 〈F(x1), F(x2)〉. In particular, F is continuous. To

show that F is onto, let y ∈ G. Then y = ∑β〈y, vβ〉vβ . By Bessel’s

inequality,∑

β |〈y, vβ〉|2 ≤ ‖y‖2 < ∞. By the Riesz–Fischer theorem,x := ∑

β〈y, vβ〉uφ−1(β) ∈ H . If α := φ−1(β), then F(uα) = vβ . HenceF(x) = ∑

β〈y, vβ〉F(uφ−1(β)) = ∑β〈y, vβ〉vβ = y by the continuity and the

linearity of F .

2.37. Let G denote the closure of span E . Since E⊥ = G⊥, we obtain E⊥⊥ =G⊥⊥ = G, as in the Projection Theorem.

2.38. Since y ∈ G, we obtain ‖x − y‖ ≥ d(x,G). Let w ∈ G. Then (y − w) ∈ G.Since (x − y) ∈ G⊥, we obtain (x − y) ⊥ (y − w), so that

‖x − w‖2 = ‖(x − y) + (y − w)‖2 = ‖x − y‖2 + ‖y − w‖2 ≥ ‖x − y‖2.

Thus ‖x − y‖ = d(x,G). If w ∈ G and ‖x −w‖ = d(x,G), then ‖x −w‖ =‖x − y‖, and so ‖y − w‖ = 0 by the above inequality, that is, w = y.

2.39. (i) Let x ∈ X\Y , and Z := span {x,Y }. Now Y �= Z , and since Y is finitedimensional, Y is a closed subspace of Z . By the lemma of Riesz, there iszn ∈ Z such that ‖zn‖ = 1 and (1 − 1/n) < d(zn,Y ) ≤ 1 for every n ∈ N.Now (zn) is a sequence in the closed unit ball of Z . By Theorem 2.10, thereis a convergent subsequence (znk ). Let znk → x1 in Z . Then ‖x1‖ = 1. Also,since (1 − 1/nk) ≤ |||znk + Y ||| ≤ 1, and |||znk + Y ||| → |||x1 + Y |||, we seethat d(x1,Y ) = |||x1 + Y ||| = 1.(ii) Let x ∈ H\G, and let y be the orthogonal projection of x on G. ByExercise 2.38, ‖x − y‖ = d(x,G). Let x1 := (x − y)/‖x − y‖.

2.40. The function 〈〈· , ·〉〉 : H/G×H/G → K is well-defined: Suppose x j +G =x ′j + G for j = 1, 2. Let x j = y j + z j and x ′

j = y′j + z′

j , where y j , y′j ∈ G

and z j , z′j ∈ G⊥ for j = 1, 2. Since z j − z′

j = (x j − x ′j ) + (y j − y′

j ), where(x j − x ′

j ) ∈ G and (y j − y′j ) ∈ G, we see that (z j − z′

j ) ∈ G for j = 1, 2.Hence (z1 − z′

1) ⊥ z2 and (z2 − z′2) ⊥ z′

1, and so

〈〈x1 + G, x2 + G〉〉 = 〈z1, z2〉 = 〈z′1, z2〉 = 〈z′

1, z′2〉 = 〈〈x ′

1 + G, x ′2 + G〉〉.

It is easy to see that 〈〈· , ·〉〉 is an inner product on H/G. Also, by Exercise2.38, 〈〈x1 + G, x1 + G〉〉 = 〈z1, z1〉 = ‖z1‖2 = ‖x1 − y1‖2 = (

d(x1,G))2 =

|||x1 + G|||2. Further, since H is complete, so is H/G.

Chapter 3

3.1. Suppose L := {x1, x2, . . .} is an infinite linearly independent subset of X , andlet B be a (Hamel) basis for X containing L . Define f (xn) := n‖xn‖ for n ∈ N,

212 Solutions to Exercises

and f (b) := 0 for b ∈ B\L . Let f : X → K denote the linear extension of thisfunction. Then f is not continuous. Similarly, define F(xn) := nxn for n ∈ N,and F(b) := b for b ∈ B\L . Let F : X → X denote the linear extension ofthis function. Then F is one-one and R(F) = X , but F is not continuous. LetY be a normed space, and y0 ∈ Y, y0 �= 0. If G(x) := f (x)y0, x ∈ X , thenG is linear but not continuous.

3.2. (i) If x := k1x1 + · · · + knxn , then ‖x‖ ≥ |||k j x j + X j ||| = |k j | |||x j + X j |||for j = 1, . . . , n, and so ‖F(x)‖ ≤ |k1|‖y1‖ + · · · + |kn|‖yn‖ ≤ α‖x‖.(ii) If y := �1y1 + · · · + �m ym , then ‖y‖ ≥ |||�i yi + Yi ||| = |�i | |||yi + Yi |||for i = 1, . . . ,m, and if x := �1x1 + · · · + �mxm , then F(x) = y and‖x‖ ≤ |�1|‖x1‖ + · · · + |�m |‖xm‖ ≤ γ‖y‖.

3.3. Suppose Y is a subspace of X such that Z( f ) ⊂ Y and Y �= Z( f ). Lety0 ∈ Y\Z( f ). Consider x ∈ X . Let k := f (x)/ f (y0) and y := x − k y0.Then x = y + k y0, where y ∈ Z( f ) ⊂ Y , and so x ∈ Y .Next, let Y denote the closure of Z( f ). Suppose f is continuous. Then Y =Z( f ). If Z( f ) is dense in X , then Z( f ) = Y = X . Conversely, suppose Z( f )is not dense in X . Then Z( f ) ⊂ Y and Y �= X , and so Y = Z( f ), that is,Z( f ) is a closed subspace of X . Hence f is continuous.

3.4. Consider the norm ‖ · ‖p on X := c00, where p ∈ {1, 2,∞}.p = 1: If r ≥ 0, then | fr (x)| ≤ ∑∞

j=1 |x( j)| = ‖x‖1 for all x ∈ X , andfr (e1) = 1, so that ‖ f ‖ = 1. Conversely, if r < 0, then fr (en) = n−r → ∞.p = 2: If r > 1/2, then α := ( ∑∞

j=1 j−2r)1/2

< ∞, and | fr (x)| ≤ α‖x‖2for all x ∈ X . Further, let xn := (1, 2−r , . . . , n−r , 0, 0 . . .) ∈ X and αn :=(∑n

j=1 j−2r)1/2

for n ∈ N. Then ‖xn‖2 = αn and fr (xn) = α2n for n ∈ N. If

r > 1/2, then fr (xn/αn) → α, and if r ≤ 1/2, then fr (xn/αn) → ∞.p = ∞: If r > 1, then β := ∑∞

j=1 j−r < ∞, and | fr (x)| ≤ β‖x‖∞ for allx ∈ X . Further, let xn := (1, . . . , 1, 0, 0, . . .) ∈ X and βn := ∑n

j=1 j−r forn ∈ N. Then ‖xn‖∞ = 1 and fr (xn) = βn for each n ∈ N. If r > 1, thenfr (xn) → β, and if r ≤ 1, then fr (xn) → ∞.

3.5. For x ∈ �1, ‖F(x)‖1 ≤ ∑∞i=1

( ∑∞j=i |x( j)|

)/ i2 ≤ ∑∞

i=1

( ∑∞j=1 |x( j)|)/ i2

= π2‖x‖1/6. Also, ‖F(en)‖1 = ‖(1, 1/22, . . . , 1/n2, 0, 0, . . .)‖1 → π2/6.Further, if x ∈ �1 and x(n) �= 0, then

∑∞j=n+1 |x( j)| <

∑∞j=1 |x( j)|, and so

‖F(x)‖1 < π2‖x‖1/6.3.6. (i) If P �= 0, then there is y ∈ R(P) with ‖y‖ = 1, and so ‖P‖ ≥ ‖P(y)‖ =

‖y‖ = 1. Suppose X is an inner product space. If P is an orthogonal projection,then ‖x‖2 = ‖P(x)‖2 + ‖x − P(x)‖2 ≥ ‖P(x)‖2, x ∈ X , and so ‖P‖ ≤ 1.Conversely, let ‖P‖ = 0 or ‖P‖ = 1. Then ‖P‖ ≤ 1. Let y ∈ R(P)

and z ∈ Z(P). If z := 0, then clearly 〈y, z〉 = 0. Let z �= 0, and assume‖z‖ = 1 without loss of generality. Define x := y − 〈y, z〉z. Then ‖x‖2 =‖y‖2 − |〈y, z〉|2 = ‖P(x)‖2 − |〈y, z〉|2 ≤ ‖x‖2 − |〈y, z〉|2, and so 〈y, z〉 = 0,that is y ⊥ z. Thus P is an orthogonal projection operator.(ii) |||Q(x)||| = |||x + Z ||| ≤ ‖x‖ for all x ∈ X , and so ‖Q‖ ≤ 1. If Z = X ,then X/Z = {0 + Z}, and so ‖Q‖ = 0. On the other hand, if Z �= X

Solutions to Exercises 213

and ε > 0, then by the Riesz lemma, there is x ∈ X such that ‖x‖ = 1 and|||Q(x)||| = d(x, Z) > 1−ε, and so ‖Q‖ > 1−ε, which shows that ‖Q‖ = 1.

3.7. Suppose M := [ki, j ] defines a map F from c00 to itself. Then the j th col-umn F(e j ) := (k1, j , k2, j , . . .) is in c00 for each j ∈ N. Conversely, sup-pose the j th column (k1, j , k2, j , . . .) of M is in c00 for each j ∈ N. Thenfor each j ∈ N, there is m j ∈ N such that ki, j = 0 for all i > m j .If x := (x(1), . . . , x(n), 0, 0, . . .) ∈ c00, then y(i) := ∑∞

j=1 ki, j x( j) =∑nj=1 ki, j x( j) ∈ K for all i ∈ N, and so y(i) = 0 if i > max{m1, . . . ,mn}.

Thus y := (y(1), y(2), . . .) ∈ c00.The result for the norm ‖ · ‖1 follows as in Example 3.14(i) since e j ∈ c00 forj ∈ N, and the result for the norm ‖ · ‖∞ follows as in Example 3.14(ii) byconsidering xi,m := (sgn ki,1, . . . , sgn ki,m, 0, 0, . . .) ∈ c00 for i,m ∈ N.

3.8. (i) Let x ∈ X . For i ∈ N,∑∞

j=1 |ki, j 〈x, u j 〉| ≤ ‖x‖β1(i), where β1(i) :=∑∞j=1 |ki, j |, and so let fi (x) := ∑∞

j=1 ki, j 〈x, u j 〉. Also, writing |ki, j 〈x, u j 〉| =|ki, j |1/2

(|ki, j |1/2|〈x, u j 〉|)for i, j ∈ N, Bessel’s inequality shows that

∞∑

i=1

| fi (x)|2 ≤∞∑

i=1

( ∞∑

j=1

|ki, j |)( ∞∑

j=1

|ki, j ||〈x, u j 〉|2)

≤ β1α1‖x‖2 < ∞.

(ii) Let x ∈ X . For i ∈ N,

∞∑

j=1

|ki, j 〈x, u j 〉| ≤( ∞∑

j=1

|ki, j |2)1/2( ∞∑

j=1

|〈x, u j 〉|2)1/2

≤ γ2,2‖x‖2,

and so let fi (x) := ∑∞j=1 ki, j 〈x, u j 〉. Also, Bessel’s inequality shows that

∞∑

i=1

| fi (x)|2 ≤∞∑

i=1

( ∞∑

j=1

|ki, j |2)( ∞∑

j=1

|〈x, u j 〉|2)

≤ γ22,2‖x‖2.

In both cases, y := ∑∞i=1 fi (x)vi belongs to Y by the Riesz–Fischer theorem

for the Hilbert space Y . Let F(x) := y. Then ‖F(x)‖2 ≤ ∑∞i=1 | fi (x)|2 again

by the Bessel inequality.Hence the matrix M defines F ∈ BL(X,Y ), and ‖F‖ ≤ √

α1β1 in case (i),while ‖F‖ ≤ γ2,2 in case (ii).

3.9. p = 1: For j ∈ N, α1( j) = ∑∞i=1 1/ i

2 j2 = π2/6 j2, and so ‖F‖ = α1 =π2/6.p = ∞: For i ∈ N, β1(i) = ∑∞

j=1 1/ i2 j2 = π2/6i2, and so ‖F‖ = β1 =

π2/6.p = 2: γ2

2,2 = ∑∞i=1

∑∞j=1 1/ i

4 j4 = (π4/90)2, and so ‖F‖ ≤ γ2,2 = π4/90.

Also, if we let x( j) := (−1) j/j2 for j ∈ N, then x ∈ �2, and ‖x‖2 = π2/√90,

whereas ‖F(x)‖2 = π6/(90)3/2. Hence ‖F‖ = π4/90.

214 Solutions to Exercises

3.10. Let p ∈ {1, 2,∞}, and (1/p) + (1/q) = 1. As in Example 3.13, fy is acontinuous linear map on (X, ‖ · ‖p) and ‖ fy‖ ≤ ‖y‖q .(i) p = 1: Let t0 ∈ (a, b), and for t ∈ [a, b], let xn(t) := n − n2|t − t0|if |t − t0| ≤ 1/n and xn(t) := 0 otherwise. Clearly, xn ∈ X, xn ≥ 0 and‖xn‖1 ≤ 1 for n ∈ N. Let An := {t ∈ [a, b] : t0 − (1/n) ≤ t ≤ t0} andBn := {t ∈ [a, b] : t0 ≤ t ≤ t0 + (1/n)} for n ∈ N. Then

fy(xn) =∫

An

(n + n2(t − t0)

)y(t)dt +

∫

Bn

(n − n2(t − t0)

)y(t)dt

= n∫

An∪Bn

y(t)dt + n2∫

An

(t − t0)y(t)dt − n2∫

Bn

(t − t0)y(t)dt.

Now n∫An

y(t)dt → y(t0),n∫Bn

y(t)dt → y(t0),n2∫An

(t − t0)y(t)dt →−y(t0)/2, and n2

∫Bn

(t − t0)y(t)dt → y(t0)/2 by the continuity of y at t0.Hence fy(xn) → 2y(t0) − y(t0)/2 − y(t0)/2 = y(t0). Thus ‖y‖∞ ≤ ‖ fy‖.(ii) p = ∞: Let xn := ny/(1 + n|y|), n ∈ N. Then xn ∈ X and ‖xn‖∞ ≤ 1.Since n|y|2/(1 + n|y|) → |y| pointwise and monotonically on [a, b],

fy(xn) =∫ b

a

n|y|21 + n|y|dm →

∫ b

a|y|dm = ‖y‖1.

Thus ‖y‖1 ≤ ‖ fy‖.(iii) p = 2: If x := y ∈ X , then ‖x‖2 = ‖y‖2 and fy(x) = ‖y‖22. Thus‖y‖2 ≤ ‖ fy‖.

3.11. If x ∈ X , then clearly F(x) ∈ X .(i) Let x ∈ X . Using the Fubini theorem, we obtain ‖F(x)‖1 ≤ α‖x‖1.Hence F ∈ BL(X) and ‖F‖ ≤ α1. On the other hand, let t0 ∈ (a, b), andconsider the sequence (xn) given in Exercise 3.10(i). For s ∈ [a, b], defineys(t) := k(s, t), t ∈ [a, b]. Then ys ∈ X , and

∫ ba xn(t)ys(t)dt → ys(t0) for

each s ∈ [a, b]. By the bounded convergence theorem,

‖F(xn)‖1 =∫ b

a

∣∣∣∣

∫ b

ak(s, t)xn(t)dt

∣∣∣∣ds →

∫ b

a|k(s, t0)|ds.

Thus∫ ba |k(s, t0)|dt ≤ ‖F‖ for every t0 ∈ (a, b). Also, note that the function

t �−→ ∫ ba |k(s, t)|ds is continuous on [a, b]. Hence α1 ≤ ‖F‖.

(ii) For s ∈ [a, b], define ys(t) := k(s, t), t ∈ [a, b]. Then ys ∈ X , and β1 =sup{‖ys‖1 : s ∈ [a, b]}. Consider the linear functional fys on X consideredin Exercise 3.10(ii). Then β1 = sup{‖ fys‖ : s ∈ [a, b]}. Further, F(x)(s) =fys (x) for all x ∈ X and s ∈ [a, b]. It follows that

‖F(x)‖∞ = sup{| fys (x)| : s ∈ [a, b]} ≤ β1‖x‖∞ for all x ∈ X.

Hence F ∈ BL(X) and ‖F‖ ≤ β1. On the other hand, for each s ∈ [a, b],

Solutions to Exercises 215

‖ fys‖ = sup{| fys (x)| : x ∈ X and ‖x‖∞ ≤ 1} ≤ ‖F‖

since | fys (x)| ≤ ‖F(x)‖∞ for all x ∈ X . Hence β1 ≤ ‖F‖.(iii) Let x ∈ X and s ∈ [a, b]. By the Schwarz inequality,

|F(x)(s)|2 ≤( ∫ b

a|k(s, t)|dt

)(∫ b

a|k(s, t)||x(t)|2dt

)

≤ β1

( ∫ b

a|k(s, t)||x(t)|2dt

).

By the Fubini theorem, ‖F(x)‖22 ≤ β1∫ ba

( ∫ ba |k(s, t)|ds)|x(t)|2dt ≤ β1α1

‖x‖22. Hence F ∈ BL(X) and ‖F‖ ≤ (α1β1)1/2.

3.12. (i) For i, j ∈ N, the series∑∞

n=1 k1(i, n)k2(n, j) converges in K since∑∞n=1 |k1(i, n)k2(n, j)| ≤ (

∑∞n=1 |k1(i, n)|2)1/2(∑∞

n=1 |k2(n, j)|2)1/2 < ∞.Consider x ∈ �2, and let y := F2(x) ∈ �2. Then for i ∈ N,

F(x)(i) = F1(y)(i) =∞∑

n=1

k1(i, n)y(n) =∞∑

n=1

k1(i, n)

∞∑

j=1

k2(n, j)x( j)

=∞∑

j=1

( ∞∑

n=1

k1(i, n)k2(n, j)

)x( j) =

∞∑

j=1

k(i, j)x( j).

Interchanging the order of summation is justified since

( ∞∑

n=1

∞∑

j=1

|k1(i, n)k2(n, j)x( j)|)2

≤( ∞∑

n=1

∞∑

j=1

|k1(i, n)x( j)|2)

γ22,2

=( ∞∑

n=1

|k1(i, n)|2)

‖x‖22γ22,2,

where γ22,2 := ∑∞

n=1

∑∞j=1 |k2(n, j)|2. (See, for example, [13, Proposition

7.21].) Hence the matrix M := [k(i, j)] defines the map F . Also,

‖F‖2 ≤∞∑

i=1

∞∑

j=1

|k(i, j)|2 ≤∞∑

i=1

∞∑

j=1

( ∞∑

n=1

|k1(i, n)k2(n, j)|)2

≤∞∑

i=1

∞∑

j=1

( ∞∑

n=1

|k1(i, n)|2)( ∞∑

n=1

|k2(n, j)|2)

=( ∞∑

i=1

∞∑

n=1

|k1(i, n)|2)( ∞∑

n=1

∞∑

j=1

|k2(n, j)|2)

.

216 Solutions to Exercises

(ii) Replace i, j, n ∈ N by s, t, u ∈ [a, b] respectively, and replace summationby Lebesgue integration in (i) above.

3.13. The seminorm p is discontinuous on X : Let xn(t) := tn/n for n ∈ N andt ∈ [0, 1]. Then xn ∈ X and ‖xn‖1,∞ = 1, but p(xn) = n − 1 → ∞.The seminorm p is countably subadditive on X : Let s := ∑∞

k=1 xk in X with∑∞k=1 p(xk) = ∑∞

k=1 ‖x ′′k ‖∞ < ∞. Let Y := C([0, 1]) with the sup norm.

Since Y is a Banach space, the absolutely summable series∑∞

k=1 x′′k of terms

in Y is summable in Y . Let y := ∑∞k=1 x

′′k ∈ Y . Define sn := ∑n

k=1 xkfor n ∈ N. Since sn → s in X , the sequence (s ′

n) in C1([0, 1]) converges(uniformly) to the function s ′ ∈ C1([0, 1]), and the derived sequence (s ′′

n ),where s ′′

n = ∑nk=1 x

′′k , n ∈ N, converges uniformly to the function y. By a

well-known theorem in Real Analysis, y = (s ′)′ = s ′′. Thus s ′′ = ∑∞k=1 x

′′k ,

and so p(s) = ‖s ′′‖∞ ≤ ∑∞k=1 ‖x ′′

k ‖∞ = ∑∞k=1 p(xk).

3.14. (i) If xn := kn,1y1 + · · · + kn,m ym → x := k1y1 + · · · + km ym in X , then

|p(xn)− p(x)| ≤ p(xn−x) ≤ |kn,1−k1|p(y1)+· · ·+|kn,m−km |p(ym) → 0.

(ii) Let p be a lower semicontinuous seminorm on a Banach space X . Withnotation as in Lemma 3.18, sn → s in X , and p(sn) ≤ ∑∞

k=1 p(xk) forn ∈ N. Hence p(s) ≤ limn→∞ inf{p(sm) : m ≥ n} ≤ ∑∞

k=1 p(xk). Thus p iscountably subadditive. By the Zabreiko theorem, p is continuous.

3.15. If F ∈ BL(X, �r ) and j ∈ N, then | f j (x)| = |F(x)( j)| ≤ ‖F(x)‖r ≤‖F‖‖x‖ for all x ∈ X , and so f j ∈ BL(X, K).Conversely, suppose f j ∈ BL(X, K) for all j ∈ N.r = 1: For n ∈ N, let pn(x) := | f1(x)| + · · · + | fn(x)|, x ∈ X . Then eachpn is a continuous seminorm on X , and for each x ∈ X , pn(x) ≤ ‖F(x)‖1for all n ∈ N. By Corollary 3.22, there is α > 0 such that pn(x) ≤ α‖x‖for all n ∈ N and x ∈ X , and so ‖F(x)‖1 ≤ α‖x‖ for all x ∈ X . HenceF ∈ BL(X, �r ). A similar argument holds if r ∈ {2,∞}:r = 2: For n ∈ N, let pn(x) := (| f1(x)|2 + · · · + | fn(x)|2

)1/2, x ∈ X .

r = ∞: For n ∈ N, let pn(x) := | fn(x)|, x ∈ X .Aliter: The map F : X → �r is closed: Let xn → 0 in X and F(xn) → y in �r .

Then F(xn)( j) → y( j), and also F(xn)( j) = f j (xn) → 0, so that y( j) = 0for each j ∈ N. By the closed graph theorem, F is continuous.

3.16. Let E be a totally bounded subset of X , and let ε > 0. Find x1, . . . , xm inE such that E ⊂ U (x1, ε) ∪ · · · ∪ U (xm, ε). Define F(x) := limn→∞ Fn(x)for x ∈ X . There is n0 such that ‖Fn(x j ) − F(x j )‖ < ε for all n ≥ n0 andj = 1, . . . ,m. Let x ∈ E , and choose x j such that ‖x − x j‖ < ε. By Theorem3.24, there is α > 0 such that ‖Fn‖ ≤ α for all n ∈ N, and ‖F‖ ≤ α. Hence

‖Fn(x) − F(x)‖ ≤ ‖Fn(x − x j )‖ + ‖Fn(x j ) − F(x j )‖ + ‖F(x j − x)‖≤ ‖Fn‖‖x − x j‖ + ‖Fn(x j ) − F(x j )‖ + ‖F‖‖x − x j‖≤ (2α + 1)ε

Solutions to Exercises 217

for all n ≥ n0. Thus (Fn(x)) converges to F(x) uniformly for x ∈ E .

3.17. Suppose ‖Fn‖ ≤ α for all n ∈ N, and there is E ⊂ X with span E densein X and (Fn(x)) is Cauchy in Y for each x ∈ E . Let X0 := span E . SinceY is a Banach space, (Fn(x)) converges in Y for each x ∈ E , and hence foreach x ∈ X0. Define F0(x) := limn→∞ Fn(x) for x ∈ X0. Then ‖F0(x)‖ ≤limn→∞ ‖Fn(x)‖ ≤ α‖x‖ for all x ∈ X0. Thus F0 ∈ BL(X0,Y ) and ‖F0‖ ≤α. By Proposition 3.17(ii), there is F ∈ BL(X,Y ) satisfying F(x0) = F0(x0)for all x0 ∈ X0 and ‖F‖ = ‖F0‖ ≤ α. Let x ∈ X and ε > 0. Since X0 is densein X , there is x0 ∈ X0 such that ‖x−x0‖ < ε. Also, Fn(x0) → F0(x0) = F(x0)in Y , and so there is n0 ∈ N such that ‖Fn(x0) − F(x0)‖ < ε for all n ≥ n0.Hence

‖Fn(x) − F(x)‖ ≤ ‖Fn(x − x0)‖ + ‖Fn(x0) − F(x0)‖ + ‖F(x0 − x)‖≤ ‖Fn‖ ‖x − x0‖ + ‖Fn(x0) − F(x0)‖ + ‖F‖‖x0 − x‖≤ (2α + 1) ε

for all n ≥ n0. Thus Fn(x) → F(x) in Y for all x ∈ X .Conversely, let F ∈ BL(X,Y ) be such that Fn(x) → F(x) in Y for each

x ∈ X . Then (‖Fn‖) is bounded by the Banach–Steinhaus theorem.

3.18. If n0 ∈ N and x ∈ R(Pn0), then R(Pn0) ⊂ R(Pn), and so Pn(x) = x forall n ≥ n0. Thus if E := ⋃∞

n=1 R(Pn), then Pn(x) → x for each x ∈ E . InExercise 3.17, let Y := X, Fn := Pn for n ∈ N, and F = I .

3.19. Define x0(t) := 1 and xn(t) := tn for n ∈ N and t ∈ [a, b]. In Polya’s theorem,let E := {x0, x1, x2, . . .}. Then span E is the linear space of all polynomialfunctions on [a, b], which is dense in C([a, b]) by the Weierstrass theorem.If all weights are nonnegative, then

∑mnj=1 |wn, j | = ∑mn

j=1 wn, j = Qn(x0) →Q(x0) = b − a.

3.20. (i) ‖x‖X ≤ ‖x‖F for all x ∈ X . Also, F is continuous if and only if there isα > 0 with ‖F(x)‖Y ≤ α‖x‖X , that is, ‖x‖F ≤ (1 + α)‖x‖X for all x ∈ X .(ii) By the closed graph theorem, F is continuous, and so ‖ · ‖F is equivalentto ‖ · ‖X by (i) above. Hence (X, ‖ · ‖F ) is a Banach space (Exercise 2.23).(iii) Let (xn) be a sequence in X such that ‖xn‖X → 0 and there is y ∈ Ywith ‖F(xn) − y‖Y → 0. Then (xn) is a Cauchy sequence in (X, ‖ · ‖F ) since(xn) is a Cauchy sequence in (X, ‖ · ‖X ) and (F(xn)) is a Cauchy sequence in(Y, ‖ · ‖Y ).Since (X, ‖ · ‖F ) is a Banach space, there is x ∈ X such that ‖xn − x‖F → 0,that is, ‖xn − x‖X → 0 and ‖F(xn) − F(x)‖Y = ‖F(xn − x)‖Y → 0. Hencex = 0 and F(x) = y, so that y = F(0) = 0.(iv) The comparable norms ‖ · ‖X and ‖ · ‖F are equivalent by the two-normtheorem. Hence F is continuous by (i) above.

3.21. F is linear: Let F(x1) := x2 and F(x1) := x2. Then F1(x1) = F2(x2) andF1(x1) = F2(x2). Since F1 and F2 are linear, F1(x1 + x1) = F2(x2 + x2), thatis, F(x1 + x1) = x2 + x2 = F(x1) + F(x1). Similarly, F(k x1) = kF(x1).

218 Solutions to Exercises

F is a closed map: Let x1,n → x1 in X1 and F(x1,n) → x2 in X2. Definex2,n := F(x1,n) for n ∈ N. Since F1 is continuous, F1(x1,n) → F1(x1), andsince F2 is continuous, F2(x2,n) → F2(x2). But F1(x1,n) = F2(x2,n) sinceF(x1,n) = x2,n for all n ∈ N. Hence F1(x1) = F2(x2), that is, F(x1) = x2.Since X1 and X2 are Banach spaces, F is continuous.

3.22. (i) If y ∈ �q , then ‖xy‖1 = ∑∞j=1 |x( j)y( j)| ≤ ‖x‖p‖y‖q , and so xy ∈ �1

for all x ∈ �p. Conversely, suppose xy ∈ �1 for all x ∈ �p. Let M := [ki, j ],where k1, j := y( j) for j ∈ N, and ki, j := 0 otherwise. Then M definesa map from �p to �1, and so its first row y = (y(1), y(2), . . .) is in �q byCorollary 3.26. Also, if we let fy(x) := ∑∞

j=1 x( j)y( j) for x ∈ �p, then‖F‖ = ‖ fy‖ = ‖y‖q .(ii) If y ∈ �r , then ‖xy‖r ≤ ‖x‖∞‖y‖r , and so xy ∈ �r for all x ∈ �∞, and‖F‖ ≤ ‖y‖r . Conversely, suppose xy ∈ �r for all x ∈ �∞. Let x := (1, 1, . . .).Since x ∈ �∞, y = xy ∈ �r and ‖y‖r = ‖xy‖r = ‖F(x)‖r ≤ ‖F‖.(iii) If y ∈ �∞ and p ≤ r , then ‖xy‖r ≤ ‖x‖r‖y‖∞ ≤ ‖x‖p‖y‖∞, and soxy ∈ �r for all x ∈ �p, and ‖F‖ ≤ ‖y‖∞. Conversely, suppose p ≤ r andxy ∈ �r for all x ∈ �p. Let M := diag (y(1), y(2), . . .). Then M defines thematrix transformation F from �p to �r , and so F is continuous by Proposition3.30. Hence |y( j)| = ‖F(e j )‖r ≤ ‖F‖ for j ∈ N, and so ‖y‖∞ ≤ ‖F‖.

3.23. (i) If y ∈ Lq , then ‖xy‖1=∫ 10 |x(t)y(t)|dm(t) ≤ ‖x‖p‖y‖q , and so xy ∈ L1

for all x ∈ L p. Conversely, suppose xy ∈ L1 for all x ∈ L p. For n ∈ N,let yn(t) := y(t) if |y(t)| ≤ n and yn(t) := 0 otherwise. Then yn ∈ L∞ ⊂Lq . For n ∈ N, define fn(x) := ∫ 1

0 x(t)yn(t)dm(t), x ∈ L p, and note that

‖ fn‖ = ‖yn‖q . If x ∈ L p, then fn(x) → ∫ 10 x(t)y(t)dm(t) by the dominated

convergence theorem, and so there is α > 0 such that ‖yn‖q = ‖ fn‖ ≤ α forall n ∈ N by Theorem 3.24. If p = 1, then the set

{t ∈ [0, 1] : |y(t)| > α} =∞⋃

n=1

{t ∈ [0, 1] : |yn(t)| > α}

is of measure zero, and so ‖y‖∞ = ess sup|y| ≤ α. If p = 2, then ‖yn‖22 →∫ 10 |y(t)|2dm(t) by the monotone convergence theorem, and so ‖y‖2 ≤ α. Ifp = ∞, then letting x(t) := 1 for t ∈ [a, b], we see that y = xy ∈ L1. Thusy ∈ Lq in all cases. Also, if we let fy(x) := ∫ 1

0 x(t)y(t)dm(t) for x ∈ L p,then ‖F‖ = ‖ fy‖ = ‖y‖q .(ii) If y ∈ Lr , then ‖xy‖r ≤ ‖x‖∞‖y‖r , and so xy ∈ Lr for all x ∈ L∞, and

‖F‖ ≤ ‖y‖r . Conversely, suppose xy ∈ Lr for all x ∈ L∞. Let x(t) := 1 fort ∈ [a, b]. Since x ∈ L∞, y = xy ∈ Lr and ‖y‖r = ‖xy‖r = ‖F(x)‖r ≤‖F‖.(iii) If y ∈ L∞, then ‖xy‖2 ≤ ‖x‖2‖y‖∞, and so xy ∈ L2 for all x ∈ L2,

and ‖F‖ ≤ ‖y‖∞. Conversely, suppose xy ∈ L2 for all x ∈ L2. We show thatF is a closed map. Let xn → x in L2 and F(xn) = xn y → z in L2. Since

Solutions to Exercises 219

‖xy− z‖1 ≤ ‖(x− xn)y‖1+‖xn y− z‖1 ≤ ‖x− xn‖2‖y‖2+‖xn y− z‖2 → 0,we see that z = xy = F(x). By the closed graph theorem, F is continuous.Let ε > 0, and let Eε := {t ∈ [0, 1] : |y(t)| > ‖F‖ + ε}. Assume for amoment that m(Eε) > 0. If x denotes the characteristic function of Eε, then|x |(‖F‖ + ε

) ≤ |xy| on [0, 1], and (‖F‖ + ε)‖x‖2 ≤ ‖xy‖2 = ‖F(x)‖2 ≤

‖F‖‖x‖2, where ‖x‖2 �= 0. This is impossible. Hence |y| ≤ ‖F‖ + ε a.e. on[0, 1]. Since this holds for every ε > 0, ‖y‖∞ = ess sup|y| ≤ ‖F‖.

3.24. Let X := C([a, b]). Then (X, ‖·‖∞) is a Banach space, and ‖xn−x‖∞ → 0 ifand only if (xn) converges to x uniformly on [a, b]. We show that the identitymap I from (X, ‖ · ‖∞) to (X, ‖ · ‖) is a closed map. Let ‖xn‖∞ → 0 and‖xn − y‖ = ‖I (xn) − y‖ → 0. Then y(t) = limn→∞ xn(t) = 0 for everyt ∈ [a, b], that is, y = 0. By the closed graph theorem, I is continuous, and sothere is α > 0 such that ‖x‖ ≤ α‖x‖∞ for all x ∈ X , that is, the norm ‖ · ‖∞is stronger than the norm ‖ · ‖. The norms ‖ · ‖∞ and ‖ · ‖ on X are completeand comparable, and so they are equivalent.

3.25. LetW := Y × Z , and ‖(y, z)‖′ := ‖y‖+‖z‖ for (y, z) ∈ W . Then (W, ‖ · ‖′)is a Banach space. Define F : W → X by F(y, z) := y + z for (y, z) ∈ W .Then F is linear, onto, and ‖F(y, z)‖ = ‖y+ z‖ ≤ ‖y‖+‖z‖ = ‖(y, z)‖′ for(y, z) ∈ W . By the open mapping theorem, F is an open map. Hence there isγ > 0 such that for every x ∈ X , there is (y, z) ∈ W with x = F(y, z) = y+zand ‖y‖ + ‖z‖ = ‖(y, z)‖′ ≤ γ‖x‖.

3.26. Since F is an open map, there is γ > 0 such that for each n ∈ N, there iszn ∈ X with F(zn) = yn − y and ‖zn‖ ≤ γ‖yn − y‖. Let xn := x + zn forn ∈ N. Then F(xn) = F(x) + F(zn) = yn and ‖xn − x‖ = ‖zn‖ → 0.

3.27. Note: In this exercise, the bounded inverse theorem and the open mapping the-orem are deduced directly from the Zabreiko theorem, and then closed graphtheorem follows.(i) Since F is continuous, one-one and onto, F and F−1 are closed maps.

In Remark 3.29, replace X, Y, F and p by Y, X, F−1 and q respectively. Ifboth X ad Y are Banach spaces, the continuity of q follows from the Zabreikotheorem.(ii) Let Z := Z(F). Then F ∈ BL(X/Z ,Y ) is one-one and onto. Also,

q(y) = ∣∣∣∣∣∣(F)−1(y)∣∣∣∣∣∣ for y ∈ Y . If X is a Banach space, then q is a count-

ably subadditive seminorm on Y , and if Y is also a Banach space, then q iscontinuous, as in (i) above, and so there is γ > 0 such that q(y) < γ‖y‖for every y ∈ Y . By the definition of an infimum, for every y ∈ Y , there isx ∈ X satisfying F(x) = y and ‖x‖ ≤ γ‖y‖. Hence F is an open map byProposition 3.41.(iii) � is continuous since ‖�(x, F(x))‖ = ‖x‖ ≤ ‖x‖ + ‖F(x)‖ for x ∈ X .

If X and Y are Banach spaces, and F is a closed map, then Gr(F) is a closedsubspace of the Banach space X × Y , and so �−1 ∈ BL(X,Gr(F)) by (i)above. Hence there is α > 0 such that ‖F(x)‖ ≤ ‖x‖ + ‖F(x)‖ ≤ α‖x‖ forall x ∈ X .

220 Solutions to Exercises

3.28. (i) Let p′ ≤ p and r ≤ r ′. Then ‖x‖p ≤ ‖x‖p′ and ‖F(x)‖r ′ ≤ ‖F(x)‖r forall x ∈ �p. Hence BL(�p, �r ) ⊂ BL(�p′

, �r′), and ‖F‖p′,r ′ = sup{‖F(x)‖r ′ :

x ∈ �p′and ‖x‖p′ ≤ 1} ≤ sup{‖F(x)‖r : x ∈ �p and ‖x‖p ≤ 1} = ‖F‖p,r

for all F ∈ BL(�p, �r ).Let F ∈ CL(�p, �r ), and let (xn) be a bounded sequence in �p′

. Then (xn)is a bounded sequence in �p, and so there is a subsequence (xnk ) such that(F(xnk )) converges in �r , and hence in �r

′. Thus F ∈ CL(�p′

, �r′).

(ii) Let p′ ≥ p and r ≥ r ′. Replace �p, �r , �p′, �r

′by L p, Lr , L p′

, Lr ′

respectively in (i) above.

3.29. Let F := k1F +· · ·+ knFn . Then F ∈ CL(X). Hence G = k0 I + F belongsto CL(X) if and only if k0 I belongs to CL(X), that is, k0 = 0.

3.30. Since P2 = P and P is a closed map, R(P) is a closed subspace of X . Thedesired result follows from Theorem 3.42.

3.31. (i) Suppose M defines a map F from �p to �r . Then F is continuous. Sinceαr ( j) = ‖F(e j )‖r for j ∈ N, αr = sup{‖F(e j )‖r : j ∈ N} ≤ ‖F‖. Next,let i ∈ N, and define fi (x) := ∑∞

j=1 ki, j x( j), x ∈ �p. Then | fi (x)| ≤‖F(x)‖r ≤ ‖F‖‖x‖p for all x ∈ �p, and so ‖ fi‖ ≤ ‖F‖. Since βq(i) = ‖ fi‖for i ∈ N, βq = sup{‖ fi‖ : i ∈ N} ≤ ‖F‖. (See Corollary 3.26.)(ii) Let p = 1, and r ∈ {1, 2,∞}.First suppose αr < ∞. We show that M defines a map from �1 to �r .r = 1: This is worked out in the text (Example 3.14(i)).r = 2: Let x ∈ �1 and i ∈ N. Then

∑∞j=1 |ki, j x( j)| ≤ α2‖x‖1. Let y(i) :=

∑∞j=1 ki, j x( j). Writing |ki, j x( j)| = (|ki, j ||x( j)|1/2

)|x( j)|1/2, we obtain( ∞∑

j=1

|ki, j x( j)|)2

≤( ∞∑

j=1

|ki, j |2|x( j)|) ∞∑

j=1

|x( j)|=( ∞∑

j=1

|ki, j |2|x( j)|)

‖x‖1.

Hence∑∞

i=1 |y(i)|2 ≤ ‖x‖1 ∑∞j=1

(∑∞i=1 |ki, j |2

)|x( j)| ≤ α22‖x‖21, and y :=

(y(1), y(2), . . .) ∈ �2. Thus M defines a map F from �1 to �2, and ‖F‖ ≤ α2.r = ∞: Let x ∈ �1 and i ∈ N. Then

∑∞j=1 |ki, j x( j)| ≤ α∞‖x‖1. Let

y(i) := ∑∞j=1 ki, j x( j). Since |y(i)| ≤ α∞‖x‖1, y := (y(1), y(2), . . .) ∈ �∞.

Thus M defines a map F from �1 to �∞, and ‖F‖ ≤ α∞.Conversely, if M defines a map F from �1 to �r , then αr ≤ ‖F‖ < ∞.

Lastly, let r ∈ {1, 2,∞}, and assume that αr ( j) → 0. For n ∈ N, let Mn

denote the infinite matrix whose first n columns are the same as those ofthe matrix M , and the remaining columns are zero. Then the matrices Mn

and M − Mn define maps Fn and F − Fn from �1 to �r respectively, and‖F − Fn‖ = sup{αr ( j) : j = n + 1, n + 2, . . .} for each n ∈ N. Since eachFn is of finite rank and ‖F − Fn‖ → 0, F ∈ CL(�1, �r ).(iii) Let r = ∞, and p ∈ {1, 2,∞}.First suppose βq < ∞. We show that M defines a map from �p to �∞.

Solutions to Exercises 221

p = ∞: This is worked out in the text (Example 3.14(ii)).

p = 2: Let x ∈ �2 and i ∈ N. Then∑∞

j=1 |ki, j x( j)| ≤ β2(i)‖x‖2. Let y(i) :=∑∞

j=1 ki, j x( j). Hence |y(i)| ≤ β2‖x‖2, and y := (y(1), y(2), . . .) ∈ �∞.Thus M defines a map F from �2 to �∞, and ‖F‖ ≤ β2.p = 1: Let x ∈ �1 and i ∈ N. Then

∑∞j=1 |ki, j x( j)| ≤ β∞(i)‖x‖1. Let y(i) :=

∑∞j=1 ki, j x( j). Hence |y(i)| ≤ β∞‖x‖1, and y := (y(1), y(2), . . .) ∈ �∞.

Thus M defines a map F from �1 to �∞, and ‖F‖ ≤ β∞.Conversely, if M defines a map F from �p to �∞, then βq ≤ ‖F‖ < ∞.

Lastly, let p ∈ {1, 2,∞}, and assume thatβq(i) → 0. Forn ∈ N, letMn denotethe infinite matrix whose first n rows are the same as those of the matrix M ,and the remaining rows are zero. Then the matrices Mn and M − Mn definemaps Fn and F − Fn from �p to �∞ respectively, and ‖F − Fn‖ = sup{βq(i) :i = n + 1, n + 2, . . .} for each n ∈ N. Since each Fn is of finite rank and‖F − Fn‖ → 0, F ∈ CL(�p, �∞).

3.32. We use Exercise 3.31.(i) For j ∈ N, α1( j) = j, α2( j) = √

j , and α∞( j) = 1, while for i ∈ N,β∞(i) = 1 and β2(i) = β1(i) = ∞. If M defines a map from �p to �r , thenαr < ∞ and βq < ∞, and so p = 1 and r = ∞. Conversely, supposep = 1 and r = ∞. Then M defines a map F ∈ BL(�1, �∞), and ‖F‖ =α∞ = β∞ = 1. But F /∈ CL(�1, �∞), since ‖e j‖1 = 1 for all j ∈ N, and‖F(e j ) − F(ek)‖∞ = ‖ek+1 + · · · + e j‖∞ = 1 for all j > k in N.

(ii) For j ∈ N, α1( j) = ∑ jk=1 1/k, α2( j) = (∑ j

k=1 1/k2)1/2

, and α∞( j) =1, while for i ∈ N, β∞(i) = 1/ i, β2(i) = β1(i) = ∞. If M defines a mapfrom �p to �r , then αr < ∞ and βq < ∞, and so p = 1 and r ∈ {2,∞}.Conversely, suppose p = 1 and r ∈ {2,∞}. Then M defines F in BL(�1, �r ),and ‖F‖ = α2 = π/

√6 if r = 2, and ‖F‖ = α∞ = β∞ = 1 if r = ∞. To see

that F ∈ CL(�1, �2), let Mn denote the infinite matrix whose first n rows arethe same as those of the matrix M , and the remaining rows are zero. Then thematricesMn andM−Mn definemaps Fn and F−Fn from �1 to �2 respectively,each Fn is of finite rank, and ‖F − Fn‖ = (∑∞

k=n+1 1/k2)1/2 → 0. Also,

F ∈ CL(�1, �∞) since β∞(i) → 0.

3.33. (i) Let p = 1. By Exercise 3.31(ii), the converse of Corollary 3.31 holds.Let p ∈ {2,∞}. Define M := [ki, j ], where k1, j := 1 for all j ∈ N, andki, j := 0 otherwise. Let r ∈ {1, 2,∞}. Then αr ( j) = 1 for all j ∈ N, and soαr = 1. If x( j) := 1/j for j ∈ N, then x ∈ �p, but the series

∑∞j=1 k1, j x( j)

does not converge in K. Hence M does not define a map from �p to �r .(ii) Let r = ∞. By Exercise 3.31(iii), the converse of Corollary 3.26 holds.

Let r ∈ {1, 2}. Define M := [ki, j ], where ki,1 := 1 for all i ∈ N, and ki, j := 0otherwise. Let p ∈ {1, 2,∞}. Then βq(i) = 1 for all i ∈ N, and so βq = 1.If x := e1, then x ∈ �p, and

∑∞j=1 ki, j x( j) = 1 for all i ∈ N. However,

(1, 1, . . .) /∈ �r . Hence M does not define a map from �p to �r .(iii) Let p ∈ {2,∞} and r ∈ {1, 2}. If p > r , then there is x ∈ �p\�r , and so

222 Solutions to Exercises

the identity matrix I does not define a map from �p to �r , although the r -normof each column of I and the q-norm of each row of I is equal to 1.If p = 2 = r , let M denote the matrix which has an n × n diagonal blockwith all entries equal to 1/

√n for each n = 1, 2, . . . in that order, and whose

all other entries are equal to 0. Then the 2-norm of each column as wellas each row of M is equal to 1. Assume for a moment that M defines anoperator F on �2. Then F ∈ BL(�2) by Proposition 3.30. For n ∈ N, letxn := (0, . . . , 0, 1/

√n, . . . , 1/

√n, 0, 0, . . .), where 1/

√n occurs only in the

n places numbered((n − 1)n/2

) + 1, . . . , n(n + 1)/2. Then ‖xn‖2 = 1, but‖F(xn)‖2 = √

n → ∞, which is impossible.

3.34. Let p ∈ {2,∞}, r ∈ {1, 2,∞}. SupposeM := [ki, j ] defines F ∈ CL(�p, �r ).Assume that ‖F(e j )‖r = αr ( j) �→ 0. Then there are j1 < j2 < · · · in N andthere is δ > 0 such that αr ( jk) ≥ δ for all k ∈ N. Since F ∈ CL(�p, �r ),there is a subsequence (em) of the sequence (e jk ), and there is y ∈ �r suchthat F(em) → y in �r . Fix i ∈ N. Now βq(i) ≤ βq ≤ ‖F‖ (Exercise3.31(i)), where q ∈ {1, 2}. Hence F(em)(i) = ki,m → 0, and so y(i) =limm→∞ F(em)(i) = 0. Thus y = 0. But ‖y‖r = limm→∞ ‖F(em)‖r ≥ δ.Hence αr ( j) → 0.Let p = 1. If M := [ki, j ], where k1, j := 1 for j ∈ N, and ki, j := 0 otherwise,

then M defines a map in CL(�1, �r ), but αr ( j) = 1 for all j ∈ N.

3.35. (i) Define a0 := c0 := 0. For j ∈ N, α1( j) = |a j | + |b j | + |c j−1| andα2( j)2 = |a j |2 +|b j |2 +|c j−1|2, while for i ∈ N, β1(i) = |ai−1|+ |bi |+ |ci |.Now M defines F ∈ BL(�1) if and only if α1 := sup{α1( j) : j ∈ N} < ∞,and M defines F ∈ BL(�∞) if and only β1 := sup{β1(i) : i ∈ N} < ∞. Also,if α1 < ∞ and β1 < ∞, then M defines F ∈ BL(�2), and conversely, if Mdefines F ∈ BL(�2), then α2 < ∞ (Exercise 3.31(i)). All these statementshold if and only if (a j ), (b j ), (c j ) are bounded sequences.Let a j → 0, b j → 0 and c j → 0. Then α1( j) → 0 and β1(i) → 0,

so that F ∈ CL(�p) for p ∈ {1, 2,∞}. To prove the converse, note thatfor j ∈ N, α∞( j) = max{|a j |, |b j |, |c j−1|}, while for i ∈ N, β∞(i) =max{|ai−1|, |bi |, |ci |}. If F ∈ CL(�1), then β∞(i) → 0 (Exercise 4.21), ifF ∈ CL(�2), then α2( j) → 0 (Exercise 3.34), and if F ∈ CL(�∞), thenα∞( j) → 0 (Exercise 3.34). In each case, a j → 0, b j → 0 and c j → 0.(ii) Let a j := c j := 0 and b j := k j for all j ∈ N in (i) above.

(iii) Let a j := w j and b j := c j := 0 for all j ∈ N in (i) above.

3.36. Let p ∈ {2,∞} and let r ∈ {1, 2}. For n ∈ N, let Mn denote the infinite matrixwhose first n rows are the same as those of the matrix M , and the remainingrows are zero.p = ∞ and r = 1: For x ∈ �∞,

∑∞i=1

∑∞j=1 |ki, j x( j)| ≤ γ1,1‖x‖∞. Hence M

defines F ∈ BL(�∞, �1), and‖F‖ ≤ γ1,1.Also,Mn defines Fn ∈ CL(�∞, �1),and ‖F − Fn‖ ≤ ∑∞

i=n+1

∑∞j=1 |ki, j | → 0. Hence F ∈ CL(�∞, �1).

p = ∞ and r = 2: For x ∈ �∞,∑∞

i=1

(∑∞j=1 |ki, j x( j)|

)2 ≤ γ21,2‖x‖2∞. Hence

Solutions to Exercises 223

M defines F ∈ BL(�∞, �2), and ‖F‖ ≤ γ1,2. Also, the matrix Mn definesFn ∈ CL(�∞, �2), and ‖F − Fn‖2 ≤ ∑∞

i=n+1

( ∑∞j=1 |ki, j |

)2 → 0. HenceF ∈ CL(�∞, �2).p = 2 and r = 1: For x ∈ �2,

∑∞i=1

∑∞j=1 |ki, j x( j)| ≤ γ2,1‖x‖2. Hence M

defines F ∈ BL(�2, �1), and ‖F‖ ≤ γ2,1. Also, Mn defines Fn ∈ CL(�2, �1),and ‖F − Fn‖ ≤ ∑∞

i=n+1 β2(i) → 0. Hence F ∈ CL(�2, �1).The case p = 2, r = 2 is treated in the text (Example 3.14(iii)).

Let γ1,1 < ∞. ThenM defines F ∈ CL(�∞, �1), andCL(�∞, �1) is contained

in CL(�p, �r ) for all p, r ∈ {1, 2,∞} (Exercise 3.28(i)).Note that γ2,2 ≤ γ1,2, γ2,1 ≤ γ1,1.

3.37. Let n ∈ N. Then kn(· , ·) ∈ C([0, 1]×[0, 1]), and for x ∈ X and s ∈ [0, 1],|Fn(x)(s)| ≤ ‖kn(· , ·)‖∞‖x‖1, so that ‖Fn(x)‖∞ ≤ ‖kn(· , ·)‖∞‖x‖1. ThusFn ∈ BL(X,Y ). For i = 0, 1, . . . , n, let yi (s) := si (1 − s)n−i , s ∈ [0, 1].Then yi ∈ Y for each i , and

Fn(x)(s)=n∑

i=0

ci yi (s), where ci :=(n

i

) n∑

j=0

k

(i

n,j

n

)(n

j

)∫ 1

0t j (1 − t)n− j dt

for all x ∈ X and s ∈ [0, 1]. Hence each Fn is of finite rank. Also, it followsthat ‖Fn − F‖ ≤ ‖kn(· , ·) − k(· , ·)‖∞ → 0. (See [6, p.10].) Since Y is aBanach space, F ∈ CL(X,Y ).

3.38. Let E denote the closure of F(U ).Suppose F ∈ CL(X,Y ), and let (yn) be a sequence in E . Then there isyn ∈ F(U ) such that ‖yn − yn‖ < 1/n, and there is xn ∈ U such thatF(xn) = yn for each n ∈ N. Let (ynk ) be a subsequence of (yn) such thatynk → y in Y . Then ynk → y, and y ∈ E . Hence E is a compact subset ofY . In particular, every sequence in F(U ) has a Cauchy subsequence, that is,F(U ) is totally bounded.Conversely, suppose E is a compact subset of Y . Let (xn) be a bounded

sequence in X . There is α > 0 such that xn/α belongs to U , and letyn := F(xn/α) for n ∈ N. Let (ynk ) be a subsequence of (yn) such thatynk → y in E . Then F(xnk ) → αy in Y . Hence we see that F ∈ CL(X,Y ).This conclusion also holds if we assume that F(U ) is totally bounded and Yis a Banach space, since a Cauchy subsequence in F(U ) converges in Y .

3.39. By Exercise 3.38, F(U ) is a totally bounded subset of Y , and by Exercise3.16, (Fn(y)) converges uniformly to F(y), y ∈ F(U ). Now ‖(Fn − F)F‖ =sup{‖(Fn − F)(F(x))‖ : x ∈ U } = sup{‖(Fn − F)(y)‖ : y ∈ F(U )} → 0.

3.40. By Exercise 4.31(iii),∑

k ‖A(uk)‖2 = ∑k ‖A∗(uk)‖2 = ∑

j ‖A(u j )‖2.(i) Let A ∈ BL(H,G) be a Hilbert–Schmidt map, and let {u1, u2, . . .} be acountable orthonormal basis for H such that

∑j ‖A(u j )‖2 < ∞. Consider

x = ∑j 〈x, u j 〉u j ∈ H . Then A(x) = ∑

j 〈x, u j 〉A(u j ). For n ∈ N, define

224 Solutions to Exercises

An(x) := ∑nj=1〈x, u j 〉A(u j ), x ∈ H . Since An ∈ BL(H,G) is of finite

rank, it is a compact linear map for each n ∈ N. Also, for all x ∈ X ,

‖A(x) − An(x)‖2 ≤( ∑

j>n

|〈x, u j 〉|‖A(u j )‖)2

≤( ∑

j>n

‖A(u j )‖2)

‖x‖2.

Hence ‖A − An‖2 ≤ (∑j>n ‖A(u j )‖2

) → 0, and so A is a compact map.(ii) Let A ∈ BL(�2). Suppose A is defined by a matrix M := [ki, j ]. SinceA(e j )(i) = ki, j for i, j ∈ N,

∑∞j=1 ‖A(e j )‖22 = ∑∞

j=1

( ∑∞i=1 |ki, j |2

) = γ22,2.

If γ2,2 < ∞, then clearly A is a Hilbert–Schmidt map. Conversely, supposeA is a Hilbert–Schmidt map on �2. Then there is a denumerable orthonormalbasis {e1, e2, . . .} for �2 such that

∑∞k=1 ‖A(ek)‖22 < ∞. Let ki, j := A(e j )(i)

for i, j ∈ N. Then γ22,2 = ∑∞

j=1 ‖A(e j )‖22 = ∑∞k=1 ‖A(ek)‖22 < ∞, and so A

is defined by the matrix M := [ki, j ] satisfying γ2,2 < ∞.(iii) Let A ∈ BL(L2). Let {u1, u2, . . .} be a denumerable orthonormal

basis for L2 consisting of continuous functions on [a, b]. For i, j ∈ N, letwi, j (s, t) := ui (s)u j (t). Then {wi, j : i, j ∈ N} is a denumerable orthonormalbasis for L2([a, b] × [a, b]).Suppose A is a Fredholm integral operator defined by a kernel k(· , ·) inL2([a, b] × [a, b]). Define

ci, j :=∫ b

a

∫ b

ak(s, t)wi, j (s, t)dm(s)dm(t) for i, j ∈ N.

Then 〈A(u j ), ui 〉 = ci, j for all i, j ∈ N. By Parseval’s formula,

∞∑

j=1

‖A(u j )‖22 =∞∑

j=1

( ∞∑

i=1

|〈A(u j ), ui 〉|2)

=∞∑

j=1

∞∑

i=1

|ci, j |2 = ‖k(· , ·)‖22.

Hence A is a Hilbert–Schmidt map. Conversely, suppose A is a Hilbert–Schmidt map. Define ci, j := 〈A(u j ), ui 〉 for i, j ∈ N. Arguing as in (ii)above, we obtain

∑∞i=1

∑∞j=1 |ci, j |2 = ∑∞

j=1 ‖A(u j )‖22 < ∞. The Riesz–Fischer theorem shows that the double series

∑∞i=1

∑∞j=1 ci, jwi, j converges

in L2([a, b]×[a, b]), to say, k(· , ·). Let B denote the Fredholm integral opera-tor on L2 defined by the kernel k(· , ·). Then 〈B(u j ), ui 〉 = ci, j = 〈A(u j ), ui 〉for all i, j ∈ N. Hence A = B.

Chapter 4

4.1. (i) Let a := (1, 0). Clearly, g is linear, continuous, and ‖g‖ = 1 = g(a). Afunction f : K

2 → K is a Hahn–Banach extension of g to K2 if and only if

Solutions to Exercises 225

f is linear on K2 and ‖ f ‖ = 1 = f (a), that is, there are k1, k2 ∈ K such that

f (x) = k1x(1)+ k2x(2) for all x := (x(1), x(2)) ∈ K2, ‖ f ‖ = |k1|+ |k2| =

1, and k1 = 1. Hence the only Hahn–Banach extension of g to K2 is given by

f (x) := x(1) for x := (x(1), x(2)) ∈ K2.

(ii) Let b := (1, 1). Clearly, h is linear, continuous, and ‖h‖ = 1 = h(b). A

function f : K2 → K is a Hahn–Banach extension of h to K

2 if and onlyif f is linear on K

2 and ‖ f ‖ = 1 = f (b), that is, there are k1, k2 ∈ K

such that f (x) = k1x(1) + k2x(2) for all x := (x(1), x(2)) ∈ K2, ‖ f ‖ =

|k1| + |k2| = 1, and k1 + k2 = 1. But for k1 ∈ K, |k1| + |1 − k1| = 1 if andonly if k1 ∈ [0, 1]. Hence the Hahn–Banach extensions of h to K

2 are givenby ft (x) := t x(1)+ (1− t)x(2) for x := (x(1), x(2)) ∈ K

2, where t ∈ [0, 1].4.2. Suppose there is α > 0 such that

∣∣∑s csks

∣∣ ≤ α∥∥ ∑

s cs xs∥∥ as stated. Let

Y := span {xs : s ∈ S}, and for y := ∑s cs xs ∈ Y , define g(y) := ∑

s csks .Then g ∈ Y ′ and ‖g‖ ≤ α, and so there is f ∈ X ′ such that ‖ f ‖ = ‖g‖ ≤ αand f (xs) = g(xs) = ks for all s ∈ S. The converse holds with α := ‖ f ‖.

4.3. By the Hahn–Banach extension theorem, E �= ∅.E is convex: Suppose f1, f2 ∈ E, t ∈ (0, 1), and f := (1− t) f1 + t f2. Then

f (y) = (1 − t) f1(y) + t f2(y) = (1 − t)g(y) + tg(y) = g(y) for all y ∈ Y ,and so ‖g‖ ≤ ‖ f ‖ ≤ (1 − t)‖ f1‖ + t‖ f2‖ = (1 − t)‖g‖ + t‖g‖ = ‖g‖.E is closed: Suppose ( fn) is in E , and f ∈ X ′ such that ‖ fn − f ‖ → 0.

Then f (y) = limn→∞ fn(y) = limn→∞ g(y) = g(y) for all y ∈ Y , and‖ f ‖ = limn→∞ ‖ fn‖ = limn→∞ ‖g‖ = ‖g‖.E is bounded and E contains no open ball: E ⊂ { f ∈ X ′ : ‖ f ‖ = ‖g‖}.E may not be compact: Let X := (C([0, 1]), ‖ · ‖∞), and let Y denote the

subspace of X consisting of all constant functions. Define g(y) := y(0) fory ∈ Y . Then g ∈ Y ′ and ‖g‖ = 1. Given t ∈ [0, 1], define ft (x) := x(t) forx ∈ X . Then each ft is a Hahn–Banach extension of g, and ‖ ft − fs‖ ≥ 1if t �= s, since there is x ∈ X such that ‖x‖∞ = 1, x(t) = 0 and x(s) = 1.Hence the sequence ( f1/n) in E does not have a convergent subsequence.

4.4. Suppose X ′ is strictly convex. Let Y be a subspace of X , g ∈ Y ′ with ‖g‖ = 1,and let f1 and f2 be Hahn–Banach extensions of g to X . Then ‖ f1‖ = ‖ f2‖ =‖g‖ = 1. Also, ( f1 + f2)/2 is a Hahn–Banach extension of g to X , and so‖( f1 + f2)/2‖ = ‖g‖ = 1. Hence f1 = f2.Conversely, suppose there are f1 �= f2 in X ′ such that ‖ f1‖ = 1 = ‖ f2‖ and‖ f1 + f2‖ = 2. Let Y := {x ∈ X : f1(x) = f2(x)}, and define g : Y → K byg(y) := f1(y) for y ∈ Y . Then ‖g‖ ≤ 1. It can also be shown that ‖g‖ ≥ 1.(See [11].) Hence f1 and f2 are distinct Hahn–Banach extensions of g.

4.5. If Y is a Banach space, then BL(X,Y ) is a Banach space (Proposition 3.17(i)).If BL(X,Y ) is a Banach space, then its closed subspaceCL(X,Y ) is a Banachspace. Now suppose CL(X,Y ) is a Banach space. Let a ∈ X be nonzero,and let f ∈ X ′ be such that f (a) = ‖a‖ and ‖ f ‖ = 1. Let (yn) be aCauchy sequence in Y , and for n ∈ N, define Fn : X → Y by Fn(x) :=f (x)yn, x ∈ X . Then Fn ∈ CL(X,Y ) and ‖Fn − Fm‖ = ‖yn − ym‖ for

226 Solutions to Exercises

all n,m ∈ N. Hence there is F ∈ CL(X,Y ) such that ‖Fn − F‖ → 0. Inparticular, ‖a‖yn = Fn(a) → F(a). Hence Y is a Banach space.

4.6. For j ∈ {1, . . . ,m}, let us define g j (y) := G(y)( j), y ∈ Y . Then G(y) =(g1(y), . . . , gm(y)) for y ∈ Y . By Lemma 2.8(ii), g j ∈ Y ′, and so there is aHahn–Banach extension f j ∈ X ′ of g j for j = 1, . . . ,m. Define F : X → K

m

by F(x) := ( f1(x), . . . , fm(x)) for x ∈ X . Then F(y) = G(y) for y ∈ Y , andby Lemma 2.8(ii), F ∈ BL(X, K

m). Note that ‖F‖ ≥ ‖G‖, and |F(x)( j)| =| f j (x)| ≤ ‖ f j‖‖x‖ = ‖g j‖‖x‖ for all x ∈ X and j = 1, . . . ,m.Consider the norm ‖ · ‖∞ on K

m . We show that ‖F‖ ≤ ‖G‖. For x ∈ X ,

‖F(x)‖∞ = max{|F(x)(1)|, . . . , |F(x)(m)|} ≤ max{‖g1‖, . . . , ‖gm‖}‖x‖,

while for y ∈ Y , |g j (y)| ≤ max{|g1(y)|, . . . , |gm(y)|} = ‖G(y)‖∞, and so‖g j‖ ≤ ‖G‖ for each j = 1, . . . ,m. Thus ‖F(x)‖∞ ≤ ‖G‖‖x‖ for all x ∈ X .Finally, suppose G ∈ BL(X, �∞). For j ∈ N, define f j ∈ X ′ as above, andlet F(x) := ( f1(x), f2(x), . . .) for x ∈ X . Since |F(x)( j)| = | f j (x)| ≤‖ f j‖‖x‖ = ‖g j‖‖x‖ ≤ ‖G‖‖x‖ for x ∈ X and j ∈ N, F(x) ∈ �∞. Clearly,F : X → �∞ is linear and F(y) = G(y) for all y ∈ Y . Also, on replacing‘max’ by ‘sup’ in the earlier argument, it follows that ‖F‖ = ‖G‖.

4.7. Suppose F : K3 → Y is linear, and F(y) = y for all y ∈ Y . Let F(e1) :=

(k1, k2, k3), where k1 + k2 + k3 = 0. Then F(e2) = F(e2 − e1) + F(e1) =e2−e1+(k1, k2, k3) = (k1−1, k2+1, k3), and F(e3) = F(e3−e1)+F(e1) =e3−e1+(k1, k2, k3) = (k1−1, k2, k3+1). Hence ‖F(e1)‖1 = |k1|+|k2|+|k3|,‖F(e2)‖1 = |k1−1|+|k2+1|+|k3| and ‖F(e3)‖1 = |k1−1|+|k2|+|k3+1|.We show that at least one of ‖F(e1)‖1, ‖F(e2)‖1, and ‖F(e3)‖1 is greater than1. Suppose ‖F(e3)‖1 ≤ 1. Then k3 �= 0, for otherwise |k1 − 1| + |k2| ≤ 0,that is, k1 = 1 and k2 = 0, and so k1 + k2 + k3 �= 0. If ‖F(e1)‖1 ≤ 1 also,then ‖F(e2)‖1 ≥ 1 − |k1| + 1 − |k2| + |k3| ≥ 1 + 2|k3| > 1. Thus ‖F‖ > 1.

4.8. Suppose the stated condition holds. Assume for a moment that a /∈ E . Thenthere is r > 0 such that U (a, r) ∩ E = ∅. Now U (a, r) and E are disjointconvex subsets of X , and U (a, r) is open. By the Hahn–Banach separationtheorem, there are f ∈ X ′ and t ∈ R such that Re f (a) < t ≤ Re f (x) for allx ∈ E . Let t �= 0, and g := f/t . If t > 0, then Re g(a) < 1 and Re g(x) ≥ 1for all x ∈ E , while if t < 0, then Re g(a) > 1 and Re g(x) ≤ 1 for allx ∈ E , contrary to the stated condition. If t = 0, there is s ∈ R such thatRe f (a) < s < t , and wemay consider g := f/s. Hence a ∈ E . The conversefollows by the continuity of f at a.

4.9. E ∩ Y = ∅ since E is open and E ∩ Y = ∅. Let Z := X/Y with the quotientnorm |||·|||, and let E := Q(E), where Q : X → Z is the quotient map. ThenE is an open convex subset of Z and 0 + Y /∈ E . Hence there is f ∈ Z ′ suchthat Re f (x + E) > 0 for all x + Y ∈ E . Let f := f ◦ Q.

4.10. Let r := inf{‖x1 − x2‖ : x1 ∈ E1 and x2 ∈ E2}. Assume for a moment thatr = 0. Then there is a convergent sequence (x1,n) in E1 and a sequence (x2,n)

Solutions to Exercises 227

in E2 such that ‖x1,n − x2,n‖ → 0. Also, if x1,n → x1, then x2,n → x1 aswell, and so x1 ∈ E1 ∩ E2, contrary to the hypothesis. Hence r > 0. DefineEr := E1 + U (0, r). Then Er = ⋃{x1 + U (0, r) : x1 ∈ E1} is an openconvex subset of X . If x1 ∈ E1, x ∈ U (0, r), and x2 := x1 + x ∈ E2, thenr ≤ ‖x1 − x2‖ = ‖x‖ < r , a contradiction. Hence Er ∩ E2 = ∅. By theHahn–Banach separation theorem, there are f ∈ X ′ and t2 ∈ R such thatRe f (x1) < t2 ≤ Re f (x2) for x1 ∈ Er and x2 ∈ E2. Since E1 is a compactsubset of X , Re f (E1) is a compact subset ofR, and so it is closed inR. Hencethere is t1 < t2 such that Re f (x1) ≤ t1 < t2 for all x1 ∈ E1.

4.11. (i) Replace the summation∑∞

j=1 in Example 4.18(ii), (i) and (iii) by thesummation

∑nj=1 to obtain an isometry � from (Kn, ‖ · ‖q) into the dual of

(Kn, ‖ · ‖p), where (1/p) + (1/q) = 1, and p = 1, 2,∞. This isometry isonto since both (Kn, ‖ · ‖q) and the dual of (Kn, ‖ · ‖p) have dimension n.(ii) Let y := (y(1), y(2), . . .) ∈ �1, and define fy(x) := ∑∞

j=1 x( j)y( j)

for x := (x(1), x(2), . . .) ∈ c0. Clearly, fy ∈ (c0)′, and ‖ fy‖ ≤ ‖y‖1.For n ∈ N, define xn := (sgn y(1), . . . , sgn y(n), 0, 0, . . .) ∈ c0. Now‖xn‖∞ ≤ 1, and ‖ fy‖ ≥ | fy(xn)| = ∑n

j=1 |y( j)| for all n ∈ N. Hence‖ fy‖ ≥ ‖y‖1. Define �(y) : �1 → (c0)′ by �(y) := fy, y ∈ �1.Then � is a linear isometry. To show that � is onto, let f ∈ (c0)′. Thenf (x) = f

(∑∞j=1 x( j)e j

) = ∑∞j=1 x( j) f (e j ) for x ∈ c0. Define y :=

( f (e1), f (e2), . . .). Then∑n

j=1 |y( j)| = ∑nj=1 xn( j)y( j) = f (xn) ≤ ‖ f ‖

for all n ∈ N. Thus y ∈ �1, and f = fy . (Compare the case p := ∞ ofExample 3.12, and also Example 4.18(iii).)

(iii) If p ∈ {1, 2}, then c00 is a dense subspace of �p, and if p := ∞, thenc00 is a dense subspace of c0 (Exercise 2.3). By Proposition 4.13(i), the dualspace of (c00, ‖ · ‖p) is linearly isometric to (�p)′, that is, to �q if p ∈ {1, 2},and to (c0)′, that is, to �1 if p := ∞.

4.12. (i) It is clear that � : c′ → (c0)′ is linear, and ‖�(x ′)‖ ≤ ‖x ′‖ for all x ′ ∈ c′.Since c0 is a closed subspace of c, and e0 := (1, 1, . . .) ∈ c\c0, there is x ′ ∈ c′such that x ′(y) = 0 for all y ∈ c0 and x ′(e0) = ‖e0‖∞ = 1. Then x ′ �= 0, but�(x ′) = 0. Hence � is not an isometry.(ii) Let y ∈ �1. Then | fy(x)| ≤ (|y(1)| + ∑∞

j=1 |y( j + 1)|)‖x‖∞ =‖y‖1‖x‖∞ for all x ∈ c, and so ‖ fy‖ ≤ ‖y‖1. On the other hand, definexn := (sgn y(2), . . . , sgn y(n), sgn y(1), sgn y(1), . . .) for n ∈ N. Thenxn ∈ c, ‖xn‖∞ ≤ 1, and

‖ fy‖ ≥ | fy(xn)| =∣∣∣∣|y(1)| +

n−1∑

j=1

|y( j + 1)| +∞∑

j=n

sgn y(1)y( j + 1)

∣∣∣∣

≥n∑

j=1

|y( j)| −∞∑

j=n+1

|sgn y(1)| |y( j)| for all n ∈ N.

228 Solutions to Exercises

Since∑∞

j=n+1 |y( j)| → 0, we see that ‖ fy‖ ≥ ‖y‖1. To show that � isonto, let f ∈ c′. For n ∈ N, let un := sgn f (e1)e1 + · · · + sgn f (en)en . Thenun ∈ c, ‖un‖∞ ≤ 1, and | f (e1)|+· · ·+| f (en)| = f (un) ≤ ‖ f ‖ for all n ∈ N.Hence the series

∑∞j=1 f (e j ) converges inK. Let s f ∈ K denote its sum. Con-

sider x ∈ c, and let x( j) → �x . Then x = �xe0 + ∑∞j=1(x( j) − �x )e j (Exer-

cise 2.27), and so f (x) = �x f (e0) + ∑∞j=1

(x( j) − �x

)f (e j ) = �x

(f (e0) −

s f) + ∑∞

j=1 x( j) f (e j ). Define y := ( f (e0) − s f , f (e1), f (e2), . . .). Then∑nj=1 |y( j)| = | f (e0) − s f | + ∑∞

j=1 | f (e j )| ≤ | f (e0) − s f | + ‖ f ‖ < ∞.Thus y ∈ �1, and f = fy . Hence � is a linear isometry from �1 onto c′.

4.13. Let H := W 1,2. Fix y ∈ H . Then fy(x) = 〈x, y〉1,2 for x ∈ H . (See Example2.28(iv).) As in the proof of Theorem4.14, fy ∈ H ′ and ‖ fy‖ = ‖y‖2 = ‖y‖2.Thus the map �(y) := fy, y ∈ H , gives a linear isometry from H to H ′.Next, if f ∈ H ′, then by Theorem 4.14, f = fy , where y := y f . Hence � isa linear isometry from H onto H ′.

4.14. Let p ∈ {1, 2}. Then (C([a, b]), ‖ · ‖p) is a dense subspace of L p([a, b]).Hence the dual space of (C([a, b]), ‖ · ‖p) is linearly isometric to the dualspace of L p([a, b]), that is, to Lq([a, b]), where (1/p) + (1/q) = 1.

4.15. Let X be a reflexive normed space, and letY := X ′. Since J is a linear isometryfrom X onto X ′′, and since X ′′ = BL(Y, K) is a Banach space, X is a Banachspace. Suppose X is separable as well. Then X ′′ = Y ′ is separable, and soY = X ′ is separable. Let X := �1. Then X is a separable Banach space. ButX is not reflexive. Otherwise X ′ would be separable, but X ′ is isometric to �∞which is not separable.

4.16. Suppose xnw→ x and xn

w→ x in X . Then x ′(x) = limn→∞ x ′(xn) = x ′(x) forall x ′ ∈ X ′. Hence x = x by Proposition 4.6(i).(i) Let E := {xn : n ∈ N}. Suppose xn

w→ x in X . Then x ′(E) is a boundedsubset of K for every x ′ ∈ X ′. Hence E is a bounded subset of X .Conversely, suppose there is α > 0 such that ‖xn‖ ≤ α for all n ∈ N, andthere is a subset D of X ′ whose span is dense in X ′ and x ′(xn) → x ′(x)for every x ′ ∈ D, that is, J (xn)(x ′) → J (x)(x ′) for every x ′ ∈ D. SinceJ (xn), J ∈ BL(X ′, K), and ‖J (xn)‖ = ‖xn‖ ≤ α for n ∈ N, Exercise 3.17shows that J (xn)(x ′) → J (x)(x ′) for every x ′ ∈ X ′, that is, xn

w→ x in X .(ii) If xn → x , then |x ′(xn) − x ′(x)| ≤ ‖x ′‖‖xn − x‖ → 0 for every x ′ ∈ X ′,

and so xnw→ x in X . Suppose X is an inner product space, xn

w→ x in X , and‖xn‖ → ‖x‖. Then ‖xn − x‖2 = ‖xn‖2 − 2Re 〈xn, x〉 + ‖x‖2 → 0.(iii) Let (xn) be a bounded sequence in X . Then the bounded sequence

(〈x1, xn〉) in K has a convergent subsequence (〈x1, x1,n〉) by the Bolzano–Weierstrass theorem for K. Next, the bounded sequence (〈x2, x1,n〉) has aconvergent subsequence (〈x2, x2,n〉), and so on. Define un := xn,n for n ∈ N.The diagonal sequence (un) is a subsequence of (xn). For each fixedm ∈ N, thesequence (〈xm, un〉) is convergent, and so the sequence (〈y, un〉) is convergentfor every y ∈ span {x1, x2, . . .}. Let Y denote the closure of span {x1, x2, . . .}.

Solutions to Exercises 229

It can be seen that for every y ∈ Y , (〈y, un〉) is a Cauchy sequence, and henceit is convergent in K. Further, if z ∈ Y⊥, then 〈z, un〉 = 0 for all n ∈ N. SinceX = Y ⊕ Y⊥, it follows that the sequence (〈x, un〉) is convergent for everyx ∈ X . Define f (x) := limn→∞〈x, un〉 for x ∈ X . Then f ∈ X ′, and so, thereis u ∈ X such that f (x) = 〈x, u〉 for all x ∈ X . Thus 〈x, un〉 → 〈x, u〉 forevery x ∈ X , that is, x ′(un) → x ′(u) for every x ′ ∈ X ′. Thus un

w→ u in X .Suppose (〈xn, x〉) is convergent in K for every x ∈ X . For n ∈ N, letfn(x) := 〈x, xn〉, x ∈ X . Then fn ∈ X ′ and ‖ fn‖ = ‖xn‖. Definef (x) := limn→∞ fn(x), x ∈ X . By the Banach–Steinhaus theorem, (‖ fn‖) isbounded, that is, (‖xn‖) is bounded, and so f ∈ X ′. As above, there is x ∈ Xsuch that xn

w→ x in X .(iv) Let xn

w→ 0 in X . If F ∈ BL(X,Y ), then y′◦F(xn) → 0 for every y′ ∈ Y ′,

and so F(xn)w→ 0 in Y . Now let F ∈ CL(X,Y ), and assume for a moment

that F(xn) �→ 0 in Y . By passing to a subsequence, if necessary, we mayassume that there is δ > 0 such that ‖F(xn)‖ ≥ δ for all n ∈ N. Since (xn)is a bounded sequence and F is a compact linear map, there is a subsequence(xnk ) such that F(xnk ) → y in Y . But since F(xnk )

w→ 0 in Y , we see thaty = 0. This is impossible since ‖F(xnk )‖ ≥ δ for all k ∈ N.(v) Let (un) be an orthonormal sequence in the Hilbert space X . As a conse-

quence of the Bessel inequality, 〈x, un〉 → 0 for every x ∈ X , that is, unw→ 0

in X . Let F ∈ CL(X,Y ). Then F(un) → 0 in Y by (iv) above.In particular, let X = Y := �2, and let an infinite matrix M define a mapF ∈ CL(�2). Then F(e j ) → 0, that is, α2( j) → 0. Also, the transpose Mt

of M defines a map Ft ∈ CL(�2), and so Ft (ei ) → 0, that is, β2(i) → 0.

4.17. (i) |x ′(x)| ≤ ‖x‖1 for all x ∈ �1, and so x ′ ∈ (�1)′. Clearly, x ′(en) = 1 �→ 0.(ii) The span of the set E := {e j : j ∈ N} is dense in �2, which is linearly

isometric to (�2)′. Hence the result follows from Exercise 4.16(i).(iii) Without loss of generality, we let x := 0. Assume for a moment that

xn �w→0 in X . Then there is x ′ ∈ X ′ such that x ′(xn) �→ 0, and so there isδ > 0 and there are n1 < n2 < · · · in N such that |x ′(xnk )| ≥ δ for allk ∈ N. Form ∈ N, define um := sgn x ′(xn1) xn1 +· · ·+ sgn x ′(xnm ) xnm . Since|um( j)| ≤ |xn1( j)| + · · · + |xnm ( j)| ≤ ∑∞

n=1 |xn( j)| ≤ α for all j ∈ N, wesee that ‖um‖∞ ≤ α for allm ∈ N, and som δ ≤ |x ′(xn1)|+ · · ·+ |x ′(xnm )| =|x ′(um)| ≤ α‖x ′‖ for all m ∈ N, which is impossible. Thus xn

w→ 0 in �∞. Inparticular, let xn := en for n ∈ N, and observe that

∑∞n=1 |en( j)| = 1 for all

j ∈ N. Hence enw→ 0 in �∞.

(iv) Let x ′ ∈ X ′. By Exercise 4.11(ii), there is y ∈ �1 such that x ′(x) =∑∞

j=1 x( j)y( j) for all x ∈ c0, and so x ′(xn) = ∑nj=1 y( j) → ∑∞

j=1 y( j).

Assume for a moment that xnw→ x in c0. Fix j ∈ N. Then xn( j) → x( j). But

xn( j) = 1 for all n ≥ j , and so x = (1, 1, . . .) which is not in c0. Thus thereis no x ∈ c0 such that xn

w→ x .(v) Let xn

w→ x in C([a, b]). Then (xn) is bounded in (C([a, b]), ‖ · ‖∞) by

230 Solutions to Exercises

Exercise 4.16(i). For t ∈ [a, b], define x ′t (x) := x(t), x ∈ X , so that x ′

t ∈ X ′.Hence xn(t) = x ′

t (xn) → x ′t (x) = x(t) for each t ∈ [a, b].

Conversely, suppose (xn) is uniformly bounded on [a, b], and xn(t) → x(t)for each t ∈ [a, b]. Let x ′ ∈ X ′. There is y ∈ BV ([a, b]) such that x ′(x) =∫ ba x dy for all x ∈ X . There are nondecreasing functions y1, y2, y3, y4 defined

on [a, b] such that y = y1− y2+ i(y3− y4), and further,∫ ba xn dyi → ∫ b

a x dyifor i = 1, . . . , 4 by the bounded convergence theorem for the Riemann–Stieltjes integration. Thus x ′(xn) = ∫ b

a xn dy → ∫ ba x dy = x ′(x).

4.18. (i)=⇒(ii)=⇒(iii) by the projection theorem.(iii)=⇒(iv): If Y⊥ = {0}, then Y = Y⊥⊥ = {0}⊥ = X .

(iv)=⇒(v): Let Y := Z( f ) in the proof of Theorem 4.14.

(v)=⇒(i): Let (yn) be a Cauchy sequence in X . For n ∈ N, define fn(x) :=〈x, yn〉, x ∈ X . Then fn ∈ X ′, and ‖ fn − fm‖ = ‖yn − ym‖ for all n,m ∈ N.Since X ′ is a Banach space, there is f ∈ X ′ such that ‖ fn − f ‖ → 0. Lety ∈ X be such that f (x) := 〈x, y〉, x ∈ X . Then ‖yn − y‖ = ‖ fn − f ‖ → 0.

4.19. Let H denote the completion of X .(i) Let span {uα} be dense in X . Since X is dense in H , we see that span {uα}is dense in H . By Theorem 2.31, x = ∑

n〈x, un〉unfor every x ∈ H , and in particular for every x ∈ X . The converse follows asin the proof of (iv)=⇒(i) of Theorem 2.31.(ii) Let G denote the closure of Y in H . Then G ∩ X = Y since Y is closed in

X . Let 〈〈· , ·〉〉 be the inner product on H/G which induces the quotient normon H/G (Exercise 2.40). Let 〈〈x1 + Y, x2 + Y 〉〉 := 〈〈x1 + G, x2 + G〉〉 forx1, x2 ∈ X . If x ∈ X and 〈〈x + Y, x + Y 〉〉 = 〈〈x + G, x + G〉〉 = 0, thenx ∈ G ∩ X = Y , that is x + Y = 0 + Y . It follows that 〈〈· , ·〉〉 is an innerproduct on X/Y , and 〈〈x + Y, x + Y 〉〉 = 〈〈x + G, x + G〉〉 = |||x + G|||2 =d(x,G)2 = d(x,Y )2 = |||x + Y |||2 for all x ∈ X .(iii) Let 〈· , ·〉′ denote the inner product on H which induces the norm on H ′,as in Corollary 4.16(i). Let f0, g0 ∈ X ′. By Proposition 3.17(ii), there areunique f, g ∈ H ′ such that f (x) = f0(x) and g(x) = g0(x) for all x ∈ X ,‖ f ‖ = ‖ f0‖ and ‖g‖ = ‖g0‖. Define 〈 f0, g0〉′ := 〈 f, g〉′. Then 〈· , ·〉′ is aninner product on X ′, and 〈 f0, f0〉′ = 〈 f, f 〉′ = ‖ f ‖2 = ‖ f0‖2.

4.20. If there are x ′1, . . . , x

′m in X ′ and y1, . . . , ym inY such that F(x) = ∑m

i=1 x′i (x)yi

for x ∈ X , then R(F) ⊂ span {y1, . . . , ym}, and so F is of finite rank.Conversely, suppose F is of finite rank. Let y1, . . . , ym be a basis forR(F). Then there are unique functions f1, . . . , fm from X to K such thatF(x) = f1(x)y1 + · · · + fm(x)ym for x ∈ X . Fix i ∈ {1, . . . ,m}. Clearly, fiis linear. Let Yi := span {y j : j = 1, . . . ,m and j �= i} and di := d(yi ,Yi ).Since di > 0 and | fi (x)|di ≤ ‖F(x)‖ ≤ ‖F‖‖x‖ for all x ∈ X , fi is contin-uous. Let x ′

i := fi ∈ X ′ for i = 1, . . . ,m.Suppose F(x) := ∑m

i=1 x′i (x)yi for x ∈ X as above, and let y′ ∈ Y ′. Then

F ′(y′)(x) = y′(F(x)) = ∑mi=1 x

′i (x)y

′(yi ) = ( ∑mi=1 y

′(yi )x ′i

)(x) for all

x ∈ X , that is, F ′(y′) = ∑mi=1 y

′(yi )x ′i .

Solutions to Exercises 231

4.21. Let p, r ∈ {1, 2}, and (1/p)+(1/q) = 1 and (1/r)+(1/s) = 1. The transposeMt of M defines a map Ft ∈ BL(�s, �q), which can be identified with F ′.Since F is compact, F ′ is compact, and so is Ft . Since s, q ∈ {2,∞}, thesequence of the columns of Mt , which is the sequence of the rows of M , tendsto 0 in �q by Exercise 3.34.Let p ∈ {1, 2,∞} and r := ∞. f M := [ki, j ], where ki,1 := 1 for i ∈ N, andki, j := 0 otherwise, then M defines F ∈ CL(�p, �∞), but βr (i) = 1, i ∈ N.

4.22. If p ∈ {1, 2,∞}, then ‖A(x)‖p ≤ ‖x‖p for all x ∈ L p, and so A ∈ BL(L p).Now let p ∈ {1, 2}, and (1/p) + (1/q) = 1. For y ∈ Lq , let

�(y)(x) :=∫ ∞

0x(t)y(t)dm(t), x ∈ L p.

Now � : Lq → (L p)′ is a linear isometry, and it is onto. (Compare Examples4.19(ii) and 4.24(ii).) Define At : Lq → Lq by At := (�)−1A′�. ThusA′ ∈ BL((L p)′) can be identified with At ∈ BL(Lq). Fix y ∈ Lq . Letz(s) := 0 if s ∈ [0, 1), and z(s) := y(s − 1) if s ∈ [1,∞). Then z ∈ Lq , and

�(At (y))(x) = A′(�(y))(x) = �(y)(A(x))

=∫ ∞

0x(t + 1)y(t)dm(t) =

∫ ∞

1x(s)y(s − 1)dm(s)

=∫ ∞

0x(s)z(s)dm(s) = �(z)(x).

for all x ∈ L p. Hence �(At (y)) = �(z), and in turn, At (y) = z, as desired.

4.23. Let n ∈ N. For y ∈ �q , let �(y)(x) := ∑∞j=1 x( j)y( j), x ∈ �p. Then � is

an isometry from �q onto (�p)′. Let Ptn := (�)−1P ′

n�. Thus P ′n ∈ BL((�p)′)

can be identified with Ptn ∈ BL(�q). Clearly, (Pt

n)2 = Pt

n . Fix y ∈ �q , and letyn := (y(1), . . . , y(n), 0, 0, . . .). Then yn ∈ �q , and

�(Ptn(y))(x) = P ′

n(�(y))(x) = �(y)(Pn(x)) =n∑

j=1

x( j)y( j) = �(yn)(x)

for x ∈ �p. Hence �(Ptn(y)) = �(yn), and in turn, Pt

n(y) = yn , as desired.

4.24. For x ′ ∈ X ′, let y′ := P ′(x ′). Then (P ′)2(x ′)(x) = P ′(y′)(x) = y′(P(x)) =x ′(P(P(x))) = x ′(P(x)) = P ′(x ′)(x) for all x ∈ X . Hence (P ′)2 = P ′.Also, R(P ′) = {x ′ ∈ X ′ : P ′(x ′) = x ′} = {x ′ ∈ X ′ : x ′(P(x) − x) =0 for x ∈ X} = Z0 and Z(P ′) = {x ′ ∈ X ′ : P ′(x ′)(x) = 0 for x ∈ X} ={x ′ ∈ X ′ : x ′(P(x)) = 0 for x ∈ X} = Y 0.

4.25. Let x ∈ X and y′ ∈ Y ′. Then

F ′′(JX (x))(y′) = JX (x)

(F ′(y′)

) = F ′(y′)(x) = y′(F(x)) = JY(F(x)

)(y′).

232 Solutions to Exercises

Hence F ′′(JX (x)) = JY

(F(x)

)for all x ∈ X , that is, F ′′ JX = JY F . Also,

‖F ′′‖ = ‖(F ′)′‖ = ‖F ′‖ = ‖F‖.Let Xc denote the closure of JX (X) in X ′′, and let Yc denote the closure ofJY (Y ) in Y ′′. Let x ′′ ∈ Xc. We show that F ′′(x ′′) ∈ Yc. Let (xn) be a sequencein X such that JX (xn) → x ′′ in X ′′. Then JY F(xn) = F ′′ JX (xn) → F ′′(x ′′)in Y ′′, where JY F(xn) ∈ JY (Y ). Thus F ′′(x ′′) ∈ Yc. Define Fc : Xc → Yc byFc(x ′′) = F ′′(x ′′), x ′′ ∈ Xc. Then Fc is linear, and Fc(JX (x)) = F ′′(JX (x)) =JY F(x) for all x ∈ X , that is, Fc JX = JY F . Also, ‖F‖ ≤ ‖Fc‖ ≤ ‖F ′′‖ =‖F‖. The uniqueness of Fc ∈ BL(Xc,Yc) follows by noting that JX (X) isdense in Xc.

4.26. Suppose F ′ ∈ CL(Y ′, X ′). Then F ′′ ∈ CL(X ′′,Y ′′). By Exercise 4.25,F ′′ JX = JY F , and so JY (F(U )) ⊂ {F ′′(x ′′) : x ′′ ∈ X ′′ and ‖x ′′‖ ≤ 1}.The latter set is a totally bounded subset of Y ′′ since F ′′ = (F ′)′ is compact(Exercise 3.38), and so JY (F(U )) is also a totally bounded subset of Y ′′. SinceJY is an isometry, F(U ) is a totally bounded subset of Y . Since Y is a Banachspace, F ∈ CL(X,Y ) (Exercise 3.38).

4.27. Let xn → x0 in H and A(xn) → y0 inG. Then for every y ∈ Y , 〈A(xn), y〉 →〈y0, y〉 on one hand, and on the other hand, 〈A(xn), y〉 = 〈xn, B(y)〉 →〈x0, B(y)〉 = 〈A(x0), y〉, and so 〈A(x0), y〉 = 〈y0, y〉. It follows that y0 =A(x0). Thus A is a closed map. By the closed graph theorem, A ∈ BL(H,G).Aliter: Let E := {A(x) : x ∈ H and ‖x‖ ≤ 1} ⊂ G. Consider y′ ∈ G ′. Thereis y0 ∈ G such that y′(y) = 〈y, y0〉 for all y ∈ G. Hence |y′(A(x)| =|〈A(x), y0〉| = |〈x, B(y0〉| ≤ ‖B(y0)‖ for all x ∈ H satisfying ‖x‖ ≤ 1. Bythe resonance theorem, E is a bounded subset of G, that is, A ∈ BL(H,G).Similarly, B ∈ BL(G, H). By the uniqueness of the adjoint, B = A∗.

4.28. R(A∗) ⊥ R(B) if and only if 〈A∗(y1), B(y2)〉 = 0 for all y1, y2 ∈ G. But〈A∗(y1), B(y2)〉 = 〈y1, AB(y2)〉 for y1, y2 ∈ G, and 〈y1, AB(y2)〉 = 0 for ally1, y2 ∈ G if and only if AB(y2) = 0 for all y2 ∈ G, that is, AB = 0.

4.29. A∗(G⊥) ⊂ G⊥ if and only if 〈A∗(x), y〉 = 0 for all x ∈ G⊥ and y ∈ G. But〈A∗(x), y〉 = 〈x, A(y)〉 for x ∈ G⊥ and y ∈ G = (G⊥)⊥, and 〈x, A(y)〉 = 0for all x ∈ G⊥ and y ∈ G⊥⊥ if and only if A(G) = A(G⊥⊥) ⊂ (G⊥)⊥ = G.

4.30. Suppose A ∈ BL(H,G) is one-one and onto. Replacing A by A∗ in Theorem4.27(i), we see that R(A∗) is dense in H . Also, by Theorem 4.27(ii), A∗ isbounded below. Let β > 0 be such that β‖y‖ ≤ ‖A∗(y)‖ for all y ∈ G.To show A∗ is onto, consider x ∈ H . Since R(A∗) is dense in H , there is asequence (yn) in G such that A∗(yn) → x in H . Then (A∗(yn)) is a Cauchysequence in H , and β‖yn−ym‖ ≤ ‖A∗(yn)−A∗(ym)‖ for all n,m ∈ N.Hence(yn) is a Cauchy sequence in G. Since G is complete, there is y ∈ Y such thatyn → y in G. Then A∗(yn) → A∗(y). Hence x = A∗(y) ∈ R(A∗). Thus A∗is onto. Since A∗ is also one-one, consider B := (A∗)−1 : H → G. Then B islinear. Also, if x ∈ H and x = A∗(y), then ‖B(x)‖ = ‖y‖ ≤ ‖A∗(y)‖/β =‖x‖/β, and so B ∈ BL(H,G). Hence A−1 = B∗ ∈ BL(G, H).Open mapping theorem: Suppose A ∈ BL(H,G) is onto. Let Z := Z(A),

Solutions to Exercises 233

and define A : H/Z → G by A(x + Z) := A(x) for x ∈ H . Then A isone-one and onto. As we saw above, A−1 ∈ BL(G, H/Z). Hence A is anopen map, and so is A = A ◦ Q since Q : H → H/Z is an open map.The closed graph theorem can be deduced from the open mapping theorem as

in Exercise 3.27(iii).

4.31. (i) Let A ∈ CL(H,G). Then A∗ ∈ BL(G, H), and so A∗A ∈ CL(H).Conversely, let A∗A ∈ CL(H). Consider a bounded sequence (xn) in H , andlet α > 0 be such that ‖xn‖ ≤ α for all n ∈ N. Since A∗A is compact, thereis a subsequence (xnk ) such that (A∗A(xnk )) converges in G. But

‖A(xnk ) − A(xn j )‖2 = |〈A∗A(xnk − xn j ), xnk − xn j 〉|≤ 2α‖A∗A(xnk ) − A∗A(xn j )‖ for all k, j ∈ N.

Now the Cauchy sequence (A(xnk )) converges in G. Thus A ∈ CL(H,G).(ii) Let A ∈ CL(H,G). Then (A∗)∗A∗ = AA∗ ∈ CL(G), and by (i) above,

A∗ ∈ CL(G, H). (Note: Theorem 4.21 of Schauder is not used.)(iii) Let A ∈ BL(H,G) be a Hilbert–Schmidt map. Let {u1, u2, . . .} be

a countable orthonormal basis for H such that∑

n ‖A(un)‖2 < ∞. Let{v1, v2, . . .} be an countable orthonormal basis for G. Then

∑

m

‖A∗(vm)‖2 =∑

m

∑

n

|〈A∗(vm), un〉|2 =∑

n

∑

m

|〈vm, A(un〉)|2

which is equal to∑

n ‖A(un)‖2. Hence A∗ is a Hilbert–Schmidt map.(This proof shows that if {u1, u2, . . .} is another countable orthonormal basisfor H , then

∑n ‖A(un)‖2 = ∑

m ‖A∗(vm)‖2 = ∑n ‖A(un)‖2.)

4.32. Let A ∈ BL(H). Define B := (A + A∗)/2 and C := (A − A∗)/2. Then Bis hermitian, C is skew–hermitian, and A = B + C . Suppose A = B1 + C1,where B1 is hermitian and C1 is skew–hermitian. Then A∗ = B1 −C1, and soB1 = (A + A∗)/2 = B and C1 = (A − A∗)/2 = C .Note that BC = (A2 − (A∗)2 − AA∗ + A∗A)/4 and CB = (A2 − (A∗)2 −A∗A + AA∗)/4. Hence BC = CB if and only if A∗A = AA∗, that is, Ais normal. Also, C = 0 if and only if A∗ = A, that is, A is hermitian, andB = 0 if and only if A∗ = −A, that is, A is skew–hermitian. Finally, note thatB2 −C2 = (AA∗ + A∗A)/2. Hence BC = CB and B2 −C2 = I if and onlyif A∗A = AA∗ and A∗A + AA∗ = 2I , that is, A∗A = I = AA∗.

4.33. Let A ∈ BL(H). Clearly, A∗A− AA∗ is self-adjoint. Hence A is hyponormalif and only if 〈A∗A(x), x〉 ≥ 〈AA∗(x), x〉, that is, ‖A(x)‖2 ≥ ‖A∗(x)‖2 forall x ∈ H . Since A is normal if and only if ‖A∗(x)‖ = ‖A(x)‖ for all x ∈ H ,and A∗ is hyponormal if and only if ‖A∗(x)‖ ≥ ‖A(x)‖ for all x ∈ H , A isnormal if and only if A and A∗ are both hyponormal.Let A denote the right shift operator on �2. Then A∗ is the left shift operator

234 Solutions to Exercises

on �2. Hence A∗A = I and AA∗(x) = (0, x(2), x(3), . . .) for all x ∈ �2. ThusA∗A ≥ AA∗, but A∗A �= AA∗.

4.34. For x in H, let B(x)( j) := x( j + 1) for all j ∈ Z. Then 〈x, B(y)〉 =∑∞j=−∞ x( j)y( j + 1) = ∑∞

j=−∞ x( j−1)y( j) = 〈A(x), y〉 for all x, y ∈ H .Hence A∗ = B, the left shift operator on H . Also, it is easy to see thatA∗A(x) = x = AA∗(x) for x ∈ H . Hence A is a unitary operator on H .

4.35. Note: ωnn = 1, but ω p

n �= 1 for p = 1, . . . , n − 1, |ωn| = 1 and ωn = ω−1n .

We show that the n columns of Mn form an orthonormal subset of Cn . Let

j, � ∈ {1, . . . , n}. Then the inner product of the j th and �th columns is equal to(∑np=1 ω

(p−1)( j−1)n ω

−(p−1)(�−1)n

)/n = ( ∑n

p=1 ω(p−1)( j−�)n

)/n, which is equal

to (1 + · · · + 1)/n = 1 if j = �, and which is equal to 0 if j �= � sinceω

j−�n �= 1 and (1 − ω

j−�n )(1 + ω

j−�n + · · · + ω

(n−1)( j−�)n ) = 1 − ω

( j−�)nn = 0.

Hence MtnMn = I = MnMt

n , and so the operator A is unitary. Also, Mtn = Mn

since k j,p := ω( j−1)(p−1)n /

√n = kp, j for p, j = 1, . . . , n.

4.36. (i) Let x ∈ H . For n ∈ N, let an(x) := 〈An(x), x〉. Then (an(x)) is a monoton-ically increasing sequence in R, and it is bounded above by α〈x, x〉. Henceit is a Cauchy sequence in R. For m ≥ n, define Bm,n := Am − An . Then0 ≤ Bm,n ≤ αI − A1 for all m ≥ n, and

‖Bm,n‖ = sup{〈Bm,n(x), x〉 : x ∈ H and ‖x‖ ≤ 1} ≤ |α| + ‖A1‖.

By the generalized Schwarz inequality, for all m ≥ n,

‖Bm,n(x)‖ = 〈Bm,n(x), x〉1/4〈B2m,n(x), Bm,n(x)〉1/4

≤ 〈Bm,n(x), x〉1/4‖Bm,n‖3/4‖x‖1/2≤ 〈Bm,n(x), x〉1/4(|α| + ‖A1‖)3/4‖x‖1/2.

It follows that (An(x)) is a Cauchy sequence in H . Let An(x) → y in H ,and define A(x) := y. Clearly, A : H → H is linear. Also, since A1 ≤An ≤ αI , we see that ‖An‖ ≤ |α| + ‖A1‖ for all n ∈ N, and so ‖A(x)‖ ≤(|α| + ‖A1‖)‖x‖ for all x ∈ H . Thus A ∈ BL(H). Since An(x) → A(x),we see that the monotonically increasing sequence (〈An(x), x〉) convergesto 〈A(x), x〉 for each x ∈ H . Thus A is self-adjoint, and An ≤ A for alln ∈ N. Finally, suppose A is self-adjoint, and An ≤ A for all n ∈ N. Then〈A(x), x〉 = limn→∞〈An(x), x〉 ≤ 〈 A(x), x〉 for all x ∈ H . Hence A ≤ A.The uniqueness of A is obvious.(ii) Let A−

n := −An , so that A−n ≤ A−

n+1 ≤ −β I for all n ∈ N. Then the

desired result follows from (i) above.

4.37. For x, y ∈ H , define 〈x, y〉A := 〈A(x), y〉. Then 〈· , ·〉A : H × H → K islinear in the first variable, is conjugate symmetric, and satisfies 〈x, x〉A ≥ 0for all x ∈ H since A is a positive operator. In Exercise 2.13, let X := H andreplace 〈· , ·〉 by 〈· , ·〉A. ThenG = {x ∈ H : 〈x, x〉A = 0}, andG is a subspace

Solutions to Exercises 235

of H . In fact, G is closed since A is continuous. For x + G, y + G ∈ H/G,let 〈〈x + G, y + G〉〉 := 〈x, y〉A. It follows that 〈〈· , ·〉〉 is an inner producton H/G. In particular, |〈x, y〉A|2 ≤ 〈x, x〉A〈y, y〉A, that is, |〈A(x), y〉|2 ≤〈A(x), x〉〈A(y), y〉 for all x, y ∈ H .

4.38. Let n,m ∈ N, m �= n. Since Pn Pm = 0, we see that R(Pm) ⊂ Z(Pn), andsince Pn is an orthogonal projection operator, Z(Pn) ⊥ R(Pn). Hence forall m �= n, R(Pn) ⊥ R(Pm). For m ∈ N, let Qm := P1 + · · · + Pm . ThenQm

∗ = P1∗ + · · · + Pm∗ = P1 + · · ·+ Pm = Qm , and so Qm is an orthogonalprojection operator. By Exercise 3.6(i), ‖Qm‖ = 0 or ‖Qm‖ = 1. Let x ∈ H .Then the Pythagoras theorem shows that for each m ∈ N,

‖P1(x)‖2 + · · · + ‖Pm(x)‖2 = ‖P1(x) + · · · + Pm(x)‖2 = ‖Qm(x)‖2 ≤ ‖x‖2.

By Exercise 2.30,∑

n Pn(x) is summable in H . Let P(x) := ∑∞n=1 Pn(x).

Clearly, P is linear, and ‖P(x)‖ ≤ ‖x‖ for all x ∈ H . Hence ‖P‖ ≤ 1.Let x ∈ H . Since Pn(P(x)) = Pn(x) for all n ∈ N, we obtain P(P(x)) =∑

n Pn(P(x)) = ∑n Pn(x) = P(x). Thus P is an orthogonal projection

operator on H (Exercise 3.6(i)).

Let G denote the closure of the linear span of⋃∞

n=1 R(Pn). Let x ∈ H . ThenPn(x) ∈ R(Pn) ⊂ G for all n ∈ N. Since G is a closed subspace of H , itfollows that P(x) ∈ G. Hence R(P) ⊂ G. Conversely, Pn(x) = P(Pn(x)) ∈R(P) for all n ∈ N. Hence R(Pn) ⊂ R(P). Since R(P) is a closed subspaceof H , it follows that G ⊂ R(P). Thus R(P) = G. If x ∈ H and Pn(x) = 0for all n ∈ N, then clearly P(x) = 0. Conversely, if x ∈ H , and P(x) = 0,then Pn(x) = Pn(P(x)) = Pn(0) = 0. Thus Z(P) = ⋂∞

n=1 Z(Pn).

4.39. Let n ∈ N, and let xn := (1, e−iθ, e−2iθ, . . . , e−(n−1)iθ, 0, 0, . . .)/√n, where

θ ∈ (−π,π]. Then A(xn) = (0, 1, e−iθ, e−2iθ, . . . , e−(n−1)iθ, 0, 0, . . .)/√n.

Clearly, xn ∈ X and ‖xn‖2 = 1. Hence 〈A(xn), xn〉 = (n − 1)eiθ/n ∈ ω(A).Letting n = 1, we see that 0 ∈ ω(A). Next, let k ∈ K satisfy 0 < |k| < 1.Then k = reiθ, where 0 < r < 1 and θ ∈ (−π,π]. There is n ∈ N such thatr < (n − 1)/n. Since 0 and (n − 1)eiθ/n belong to ω(A), and since ω(A) is aconvex subset of K, we see that k ∈ ω(A). Thus {k ∈ K : |k| < 1} ⊂ ω(A).Since ‖A‖ = 1, it follows that ω(A) ⊂ {k ∈ K : |k| ≤ 1}. We show that if

k ∈ K and |k| = 1, then k /∈ ω(A). Let x ∈ X . Then|〈A(x), x〉| ≤ ∑∞

j=1 |x( j)||x( j + 1)| ≤ 12

∑∞j=1(|x( j)|2 +|x( j + 1)|2). Thus

|〈A(x), x〉| ≤ ‖x‖2. Assume for a moment that ‖x‖2 = 1 = |〈A(x), x〉|.Since |x( j)||x( j + 1)| ≤ (|x( j)|2 + |x( j + 1)|)2/2 for every j ∈ N, |x( j)| =|x( j + 1)| for each j ∈ N. This is impossible since 0 <

∑∞j=1 |x( j)|2 < ∞.

Thus ω(A) = {k ∈ K : |k| < 1}.4.40. For x, y ∈ X , 〈A(x + y), x + y〉 − 〈A(x − y), x − y〉 = 2〈A(x), y〉 +

2〈A(y), x〉 and 〈A(x + iy), x + iy〉−〈A(x − iy), x − iy〉 = −2i〈A(x), y〉+2i〈A(y), x〉. Multiplying the second equality by i and adding it to the first, weobtain the generalized polarization identity.

236 Solutions to Exercises

Let H �= {0} be a Hilbert space over C, A ∈ BL(H), and ω(A) ⊂ R. Then

4〈A(y), x〉 = 〈A(y + x), y + x〉−〈A(y − x), y − x〉+i〈A(y + i x), y + i x〉−i〈A(y − i x), y − i x〉 for x, y ∈ H .Note that 〈A(z), z〉 ∈ R for every z ∈ H .Hence 4〈x, A(y)〉 = 4〈A(y), x〉 = 4〈A(x), y〉. Thus A is self-adjoint.

Chapter 5

5.1. Suppose k ∈ K, k �= 0. If k /∈ σ(A), then (A−1− k−1 I )((A − k I )−1A

) =−k−1(I − k A−1)(I − k A−1)−1 = −k−1 I and

((A− k I )−1A

)(A−1− k−1 I ) =

(I − k A−1)−1(−k−1)(I − k A−1) = −k−1 I , and so k−1 /∈ σ(A−1). ReplacingA by A−1, and k by k−1, we see that if k−1 /∈ σ(A−1), then k /∈ σ(A). Henceσ(A−1) = {λ−1 : λ ∈ σ(A)}.

5.2. Let p(t) := antn + an−1tn−1 + · · · + a1t + a0, where n ∈ N, an, . . . , a0 ∈ K,and an �= 0. Suppose λ ∈ σ(A). Assume for a moment that p(λ) /∈ σ(p(A)),that is, p(A) − p(λ)I is invertible. For m ∈ {1, . . . , n}, let qm(A) := Am−1 +λAm−2 + · · · + λm−2A + λm−1 I , so that Am − λm I = (A − λI )qm(A) =qm(A)(A − λI ). Define q(A) := anqn(A) + · · · + a1q1(A). It follows thatp(A)− p(λ)I = (A−λI )q(A) = q(A)(A−λI ). Hence A−λI is invertible.This contradiction shows that p(λ) ∈ σ(p(A)).Next, let K := C, and μ ∈ C. Then there are λ1, . . . ,λn ∈ C such that

p(t)−μ = an(t−λ1) · · · (t−λn). Suppose μ ∈ σ(p(A)). Then p(A)−μI =an(A − λ1 I ) · · · (A − λn I ). If A − λ j I is invertible for each j = 1, . . . , n,then so would be p(A) − μI . Hence there is λ j ∈ σ(A) such that μ = p(λ j ).

5.3. Let I − AB be invertible. Then (I − BA)(I + B(I − AB)−1A

) = I − BA+B

((I − AB)−1 − AB(I − AB)−1

)A = I − BA+ B(I − AB)(I − AB)−1A =

I − BA + BA = I . Similarly,(I + B(I − AB)−1A

)(I − BA) = I .

(This formula is conceived as follows: (I − BA)−1 = I + BA + BABA +BABABA+· · · = I +B

(I + AB+ ABAB+· · · )A = I +B(I − AB)−1A.)

Let k ∈ K be nonzero, and let A := A/k. Then AB − k I = −k(I − AB) isinvertible if and only if −k(I − B A) = BA − k I is invertible.

5.4. Let λ ∈ σ(A) = σe(A). Then there is nonzero x := (x(1), . . . , x(n)) ∈ Kn

such that A(x) − λx = 0. Suppose ‖x‖∞ = |x(i)|. We show that λ ∈ Di .Now ki,1x(1) + · · · + (ki,i − λ)x(i) + · · · + ki,nx(n) = 0, that is,

ki,i − λ = −ki,1x(1)

x(i)− · · · − ki,i−1

x(i − 1)

x(i)− ki,i+1

x(i + 1)

x(i)− · · · − ki,n

x(n)

x(i).

Since |x( j)|/|x(i)| ≤ 1 for all j ∈ {1, . . . , n}, we obtain |ki,i − λ| ≤ ri .

5.5. Suppose A(x) = λx and u = (g1(x), . . . , gn(x)). Then∑n

j=1 g j (x)x j = λx ,and hence

∑nj=1 g j (x)gi (x j ) = λgi (x), that is,

∑nj=1 u( j)gi (x j ) = λu(i) for

each i = 1, . . . , n. Thus Mut = λut . Also, x = (∑nj=1 u( j)x j

)/λ.

Solutions to Exercises 237

Conversely, suppose Mut = λut and x = ( ∑nj=1 u( j)x j

)/λ. Then for each

i = 1, . . . , n,∑n

j=1 gi (x j )u( j) = λu(i), and so

A(x) =n∑

i=1

gi (x)xi = 1

λ

n∑

i=1

( n∑

j=1

u( j)gi (x j )

)xi

= 1

λ

n∑

i=1

λu(i)xi =n∑

i=1

u(i)xi = λx .

Also, λu(i) = ∑nj=1 gi (x j )u( j) = gi

(∑nj=1 u( j)x j

) = gi (λx) = λgi (x),and so u(i) = gi (x) for all i = 1, . . . , n, that is, u = (g1(x), . . . , gn(x)).Also, x = 0 if and only if u = 0. Thus x is an eigenvector of A correspondingto λ if and only if ut is an eigenvector of M corresponding to λ.

5.6. Suppose λ ∈ σe(A). Then there is nonzero x ∈ X such that A(x) = λx ,that is,

(x0(t) − λ

)x(t) = 0 for all t ∈ [a, b]. Let t0 ∈ [a, b] be such that

x(t0) �= 0. Since x is continuous at t0, there is δ > 0 such that x(t) �= 0 forall t ∈ I := [a, b] ∩ (t0 − δ, t0 + δ). Hence x0(t) = λ for all t ∈ I .Conversely, supposeλ ∈ K, and x0(t) = λ for all t in a nontrivial subinterval I

of [a, b]. Then there is t0 ∈ [a, b], and there is δ > 0 such that (t0−δ, t0+δ) ⊂I . For t ∈ [a, b], define x(t) := 1 if |t − t0| ≤ δ/2, x(t) := 2(t − t0 + δ)/δif t0 − δ < t < t0 − δ/2, x(t) := 2(t0 − t + δ)/δ if t0 + δ/2 < t < t0 + δ,and x(t) := 0, if |t − t0| ≥ δ. Then x ∈ X and x �= 0. Since x0(t) − λ = 0for all t ∈ [a, b] satisfying |t − t0| < δ, and x(t) = 0 for all t ∈ [a, b]satisfying |t − t0| ≥ δ, we see that

(x0(t) − λ

)x(t) = 0 for all t ∈ [a, b], that

is, A(x) = λx .

5.7. Let E denote the essential range of x0. Suppose λ ∈ E . For n ∈ N, letSn := {t ∈ [a, b] : |x0(t) − λ| < 1/n}, and let xn denote the characteristicfunction of Sn . Then ‖xn‖22 = m(Sn) > 0, and

‖A(xn) − λxn‖22 =∫

Sn

|x0 − λ|2|xn|2dm ≤ m(Sn)

n2= ‖xn‖22

n2.

Hence A − λI is not bounded below, that is, λ ∈ σa(A). Thus E ⊂ σa(A).Conversely, suppose k /∈ E , that is, there is ε > 0 such that m({t ∈ [a, b] :|x0(t) − k| < ε}) = 0. Then |x0(t) − k| ≥ ε for almost all t ∈ [a, b], and thefunction 1/(x0−k) belongs to X . For y ∈ X, define B(y) := y/(x0−k).ThenB ∈ BL(X) and (A − k I )B = I = B(A − k I ). This shows that σ(A) ⊂ E .

Since σa(A) ⊂ σ(A), we obtain σa(A) = E = σ(A).

Next, suppose λ ∈ σe(A). Then there is nonzero x ∈ X such that A(x) = λx ,

that is, (x0(t) − λ)x(t) = 0 for almost all t ∈ [a, b]. Since x is nonzero,m({t ∈ [a, b] : x(t) �= 0}) > 0, and so m({t ∈ [a, b] : x0(t) = λ}) > 0.Conversely, let λ ∈ K, S := {t ∈ [a, b] : x0(t) = λ}, and suppose m(S) > 0.

238 Solutions to Exercises

Let x denote the characteristic function of the set S. Then x ∈ X and x �= 0.Since (x0(t) − λ)x(t) = 0 for all t ∈ [a, b], λ ∈ σe(A).

5.8. For n ∈ N, An−λn I = B(A−λI ), where B := An−1+λAn−2+· · ·+λn−2A+λn−1 I . Let λ ∈ σa(A). Then there is a sequence (xn) in X such that ‖xn‖ = 1for every n ∈ N, and A(xn) − λxn → 0. Consequently, An(xn) − λnxn =B(A(xn) − λxn) → 0, and so λn ∈ σa(An). Hence |λn| ≤ ‖An‖, and so|λ| ≤ inf

{‖An‖1/n : n ∈ N}.

5.9. (i) Let α > 0 be such that ‖A(x)‖ ≤ α‖x‖ for all x ∈ X . If k ∈ K and|k| > α, then ‖A(x)− kx‖ ≥ |k|‖x‖−‖A(x)‖ ≥ (|k|−α)‖x‖ for all x ∈ X ,and so k /∈ σa(A). Thus σa(A) ⊂ {k ∈ K : |k| ≤ α}.(ii) Let β > 0 be such that ‖A(x)‖ ≥ β‖x‖ for all x ∈ X . If k ∈ K and

|k| < β, then ‖A(x)− kx‖ ≥ ‖A(x)‖− |k|‖x‖ ≥ (β −|k|)‖x‖ for all x ∈ X ,and so k /∈ σa(A). Thus σa(A) ⊂ {k ∈ K : |k| ≥ β}.(iii) If A is an isometry, let α = β := 1 in (i) and (ii) above.

5.10. For x ∈ �1, ‖A(x)‖ ≤ 2‖x‖1, and ‖A(e2)‖1 = ‖2e3‖1 = 2. Hence ‖A‖ = 2.Further, A2(x) = (0, 0, 2x(1), 2x(2), . . .), and ‖A2(e1)‖1 = ‖2e3‖ = 2.Hence ‖A2‖ = 2. Since �1 is a Banach space, |λ| ≤ ‖A2‖1/2 = √

2 for everyλ ∈ σ(A).

5.11. Let α := sup{|λn| : n ∈ N}. Then ‖A(x)‖p ≤ α‖x‖p for all x ∈ X . Hence‖A‖ ≤ α. In fact, ‖A‖ = α since A(e j ) = λ j e j for each j ∈ N. This alsoshows that λ j ∈ σe(A) for each j ∈ N. Conversely, let λ ∈ σe(A), andA(x) = λx for a nonzero x ∈ X . Then there is j ∈ N such that x( j) �= 0.Since λ j x( j) = λx( j), we obtain λ = λ j . Thus σe(A) = {λ j : j ∈ N}.Let E denote the closure of {λ j : j ∈ N}. Since σa(A) is closed in K,

E ⊂ σa(A). Conversely, suppose k ∈ K\E , and let δ := d(k, E) > 0.Define B(y) := (y(1)/(k − λ1), y(2)/(k − λ2), . . .) for y ∈ X . For y ∈ X ,|B(y)( j)| = |y( j)|/|k − λ j | ≤ |y( j)|/δ for all j ∈ N, and so B(y) ∈ X , and‖B(y)‖p ≤ ‖y‖p/δ. Thus B ∈ BL(X). It is easy to check that (A − k I )B =I = B(A − k I ). Hence σ(A) ⊂ E . Thus σa(A) = σ(A) = E .

5.12. Since K is a separable metric space, so is its subset E . Let {λ j : j ∈ N}be a countable dense subset of E . For x := (x(1), x(2), . . .) in �2, defineA(x) := (λ1x(1),λ2x(2), . . .). Then σ(A) = E by Exercise 5.11.Suppose (λ j ) is a sequence in K such that λ j → 0, and define A as above.

Also, for n ∈ N, define An(x) := (λ1x(1), . . . ,λnx(n), 0, 0, . . .), x ∈ �2.Then An ∈ BL(�2) is of finite rank for each n ∈ N. Since ‖A − An‖ =sup{|λ j | : j = n + 1, n + 2, . . .} → 0, we see that A ∈ CL(�2). Further,σe(A) = {λ j : j ∈ N}, and σ(A) = E = {λn : n ∈ N} ∪ {0}.

5.13. Let k ∈ K with |k| > ‖A‖, and define A := A/k. Then ‖ A‖ < 1. HenceI − A is invertible, and (A − k I )−1 = −(I − A)−1/k = −( ∑∞

n=0 An)/k =

−∑∞n=0 A

n/kn+1. Also, ‖(A − k I )−1‖ ≤ 1/|k|(1 − ‖ A‖) = 1/(|k| − ‖A‖).Further, for n ∈ N, and x ∈ �p, An(x) = (0, . . . , 0, x(1), x(2), . . .), where the

first n entries are equal to 0. Hence (A−k I )−1(y)( j) = −∑∞n=0 A

n(y)( j)/kn+1 =−y( j)/k − · · · − y(1)/k j for y ∈ �p, and j ∈ N.

Solutions to Exercises 239

5.14. Clearly, ‖B‖ = 1. Hence σe(B) ⊂ σa(B) ⊂ σ(B) ⊂ {λ ∈ K : |λ| ≤ 1}. Letx ∈ X . For λ ∈ K, B(x) = λx if and only if x( j + 1) = λx( j) for all j ∈ N,that is, x := x(1)(1,λ,λ2, . . .). If X := �1, �2, or c0, then (1,λ,λ2, . . .) ∈ Xif and only if |λ| < 1; if X := �∞, then (1,λ,λ2, . . .) ∈ X if and only if|λ| ≤ 1, and if X := c, then (1,λ,λ2, . . .) ∈ X if and only if |λ| < 1 orλ = 1. In all cases, σa(B) = {λ ∈ K : |λ| ≤ 1} since σa(B) is a closed subsetof K. It follows that σ(B) = {λ ∈ K : |λ| ≤ 1}.

5.15. Clearly, ‖A‖ = 1. Hence σe(A) ⊂ σa(A) ⊂ σ(A) ⊂ {λ ∈ K : |λ| ≤ 1}. Letx ∈ L p. For λ ∈ K, A(x) = λx if and only if x(t + 1) = λx(t) for almost allt ∈ [0,∞), that is, x(t + j) = λ j x(t) for almost all t ∈ [0, 1) and all j ∈ N.Let p ∈ {1, 2}. Then ∫ j+1

j |x(t)|pdm(t) = |λ j |p ∫ 10 |x(s)|pdm(s) by the

translation invariance of the Lebesgue measure. It follows that ‖x‖pp =( ∫ 1

0 |x(s)|pdm(s)) ∑∞

j=0 |λp| j . The series∑∞

j=0 |λp| j is convergent if andonly if |λ| < 1. Hence σe(A) = {λ ∈ K : |λ| < 1}. Next, let p = ∞. Since{|λ| j : j ∈ N} is a bounded subset of K if and only if |λ| ≤ 1, we see thatσe(A) = {λ ∈ K : |λ| ≤ 1}. In all cases, σa(A) = {λ ∈ K : |λ| ≤ 1} sinceσa(A) is a closed subset of K. It follows that σ(A) = {λ ∈ K : |λ| ≤ 1}.

5.16. Let x ∈ X and x �= 0. Then 〈rA(x), x〉 = 〈A(x), x〉 − qA(x)〈x, x〉 = 0, thatis, rA(x) ⊥ x . It follows that for k ∈ K, ‖A(x)−k x‖2 = ‖rA(x)‖2+|qA(x)−k|2‖x‖2 ≥ ‖rA(x)‖2.

5.17. Since A ∈ CL(X), σ(A) = σa(A), which is a closed and bounded subset ofK by Proposition 5.5. Also, if λ ∈ σa(A), then |λ| ≤ inf{‖An‖1/n : n ∈ N}.(See Exercise 5.8.)

5.18. For x := (x(1), x(2), . . .) ∈ X , define D(x) := (w1x(1), w2x(2), . . .),R(x) := (0, x(1), x(2), . . .) and L(x) := (x(2), x(3), . . .). Then A = RDand B = LD. Since wn → 0, D is a compact operator (Exercise 3.35(ii)).Hence A and B are compact operators. Let x ∈ X and λ ∈ K.Now A(x) = λx if and only if 0 = λx(1) and w j x( j) = λx( j + 1) for all

j ∈ N. If λ �= 0 and A(x) = λx , then x = 0, and so λ /∈ σe(A). Since Ais compact, σ(A) = σe(A) ∪ {0} = {0}. Let E0 := {x ∈ X : A(x) = 0} ={x ∈ X : w j x( j) = 0 for all j ∈ N}. Hence E0 �= {0} if and only if thereis j ∈ N such that w j = 0. Also, e j ∈ E0 if and only if w j = 0. Thus if{ j ∈ N : w j = 0} is a finite set, then E0 = span {e j : j ∈ N and w j = 0},and otherwise dim E0 = ∞.Next, B(x) = λx if and only if w j+1x( j +1) = λx( j) for all j ∈ N. If λ �= 0

and B(x) = λx , then x = 0. For if there is j0 ∈ N such that x( j0) �= 0, thenw j �= 0 and x( j) �= 0 for all j ≥ j0, and in fact |x( j)| → ∞ as j → ∞.Hence σe(B) ⊂ {0}. Since B is compact, σ(B) = σe(B) ∪ {0} = {0}. LetG0 := {x ∈ X : B(x) = 0}. Then G0 ={x ∈ X : w j x( j) = 0 for all j ≥ 2}.Clearly, e1 ∈ G0, and so 0 ∈ σe(B). Also, for j ≥ 2, e j ∈ G0 if andonly if w j = 0. Thus if { j ∈ N : j ≥ 2 and w j = 0} is a finite set, thenG0 = {e1} ∪ span {e j : j ≥ 2 and w j = 0}, and otherwise dimG0 = ∞.

240 Solutions to Exercises

5.19. Let X := L2([a, b]). Since k(· , ·) ∈ L2([a, b]×[a, b]), we see that A is acompact operator on X . Let x ∈ X , and define

y(s) := A(x)(s) =∫ s

ax(t)dm(t), s ∈ [a, b].

Since x ∈ L1([a, b]), the fundamental theorem of calculus for Lebesgueintegration shows that y is absolutely continuous on [a, b], and y′ = x almosteverywhere on [a, b]. Also, y(a) = 0.Let A(x) = 0. Then y(s) = 0 for almost all s ∈ [a, b]. In fact, y(s) = 0 forall s ∈ [a, b] since y is continuous on [a, b]. Hence x(s) = y′(s) = 0 foralmost all s ∈ [a, b]. This shows that 0 /∈ σe(A). Next, let λ ∈ K, λ �= 0 andA(x) = λx . Then x = A(x)/λ = y/λ, and so x is absolutely continuous on[a, b]. By the fundamental theorem of calculus for Riemann integration, y is inC1([a, b]), and y′(s) = x(s) for all s ∈ [a, b]. Hence x = y/λ is inC1([a, b]),and λx ′(s) = y′(s) = x(s) for all s ∈ [a, b]. Also, x(a) = y(a)/λ = 0. Thusx satisfies Bernoulli’s differential equation λx ′ − x = 0, and also the initialcondition x(a) = 0. It follows that x = 0, and so λ /∈ σe(A). Thus σe(A) = ∅.Also, σa(A) = σ(A) = {0} since A is compact.

5.20. First, consider X := C([0, 1]). Then A ∈ CL(X) since k(· , ·) is continuous.Let x ∈ X , and define

y(s) := A(x)(s) =∫ s

0t x(t) dt + s

∫ 1

sx(t) dt, s ∈ [0, 1].

Then y(0) = 0. By the fundamental theorem of calculus for Riemann integra-tion, y ∈ C1([0, 1]), and

y′(s) = s x(s) − s x(s) +∫ 1

sx(t) dt =

∫ 1

sx(t) dt for all s ∈ [0, 1].

Then y′(1) = 0. Further, y′ ∈ C1([0, 1]), and y′′(s) = −x(s) for s ∈ [0, 1].Thus we see that if x ∈ X and y := A(x), then y ∈ C2([0, 1]), y′′ = −x andy(0) = 0 = y′(1). Conversely, suppose x ∈ X , and let y ∈ C2([0, 1]) satisfyy′′ = −x and y(0) = 0 = y′(1). Integrating by parts,

A(y′′)(s) =∫ s

0t y′′(t)dt + s

∫ 1

sy′′(t)dt

= s y′(s) − 0 y′(0) −∫ s

0y′(t)dt + s

(y′(1) − y′(s)

)

= −∫ s

0y′(t)dt = −y(s) + y(0) = −y(s)

Solutions to Exercises 241

for all s ∈ [0, 1]. Hence A(y′′) = −y, that is, A(x) = y.Let x ∈ X be such that A(x) = 0. Then 0′′ = −x , that is, x = 0. Hence

0 /∈ σe(A). Next, let λ ∈ K and λ �= 0. Let x ∈ X be such that A(x) = λx .Then it follows that λx ′′ = −x and λx(0) = 0 = λx ′(1), that is, λx ′′ +x = 0 and x(0) = 0 = x ′(1). Now the differential equation λx ′′ + x = 0has a nonzero solution satisfying x(0) = 0 = x ′(1) if and only if λ =4/(2n − 1)2π2, n ∈ N. In this case, the general solution is given by x(s) :=cn sin(2n − 1)πs/2, s is in [0, 1], where cn ∈ K. (If K := C, we must firstshow that λ ∈ R, as in the footnote in Example 5.23(ii).) Fix n ∈ N, let λn :=4/(2n − 1)2π2, xn(s) := sin(2n − 1)πs/2, s ∈ [0, 1], and let yn := λnxn .Then y′′

n = λnx ′′n = −xn and yn(0) = 0 = y′

n(1). Hence A(xn) = yn = λnxn .It follows that λn is an eigenvalue of A, and the corresponding eigenspaceof A is spanned by the function xn . There are no other eigenvalues of A. Itfollows that σe(A) = {1/(2n−1)2π2 : n ∈ N}. Since A is a compact operator,σa(A) = σ(A) = σe(A) ∪ {0}.Next, consider Y := L p([0, 1]), where p ∈ {1, 2,∞}. The arguments in this

case are exactly the same as the ones given in Example 5.23(ii).

5.21. Let k ∈ K. Suppose A + B − k I is invertible, and A − k I is one-one. SinceA − k I = A + B − k I − B = (A + B − k I )(I − (A + B − k I )−1B), wesee that (I − (A + B − k I )−1B) is one-one, that is, 1 is not an eigenvalue of(A + B − k I )−1B. Since B is compact, so is (A + B − k I )−1B. Hence 1 isnot a spectral value of (A + B − k I )−1B, that is, I − (A + B − k I )−1B isinvertible. It follows that A − k I is invertible.

5.22. Suppose A − k I is one-one, that is, k /∈ σe(A). Since A is compact, A − k Iis invertible. In particular, A − k I is onto.Conversely, suppose A − k I is onto. For n ∈ N, let Zn := Z((A − k I )n).Then Zn is a closed subspace of Zn+1 for all n ∈ N. Assume for a momentthat Zn � Zn+1 for all n ∈ N. Fix n ∈ N. By the Riesz lemma, there iszn+1 ∈ Zn+1 such that ‖zn+1‖ = 1 and d(zn+1, Zn) ≥ 1/2. It is easy to seethat (A − k I )(Zn+1) ⊂ Zn and A(Zn) ⊂ Zn . Hence for all z ∈ Zn ,

‖A(zn+1) − A(z)‖ = ‖kzn+1 + (A − k I )(zn+1) − A(z)‖ ≥ |k|/2 > 0.

In particular, ‖A(zn+1) − A(zm+1)‖ ≥ |k|/2 for all n,m ∈ N with n �= m.Now (zn+1) is a bounded sequence in X , but the sequence (A(zn+1)) has noconvergent subsequence. This contradicts the compactness of A. Hence thereis m ∈ N such that Zm+1 = Zm . Let Z0 := {0}. We show that Zm = Zm−1.

Let y ∈ Zm . Since A − k I is onto, there is x ∈ X such that y = (A − k I )(x).Now (A − k I )m+1(x) = (A − k I )m(y) = 0, that is, x ∈ Zm+1 ⊂ Zm .Thus (A − k I )m−1(y) = (A − k I )m(x) = 0, that is, y ∈ Zm−1. Similarly,Zm−1 = Zm−2, . . . , Z2 = Z1, and Z1 = Z0, that is, A − k I is one-one.(Compare the proof of Proposition 5.20.)

5.23. If k /∈ σa(A), then the result follows from Lemma 5.19. Now supposek is in σa(A). Since A is compact and k �= 0, we see that k ∈ σe(A),

242 Solutions to Exercises

and the corresponding eigenspace Ek is finite dimensional. Let {x1, . . . , xm}be a basis for Ek , and find x ′

1, . . . , x′m in X ′ such that x ′

j (xi ) = δi, j fori, j = 1, . . . ,m. Define Y := ⋂m

j=1 Z(x ′j ). Then Y is a closed subspace of X ,

and X = Y ⊕Z(A−k I ). Define B : Y → X by B(y) := (A−k I )(y), y ∈ Y .Then B is one-one. In fact, arguing as in the proof of Proposition 5.18, we seethat B is bounded below. Let β > 0 be such that β‖y‖ ≤ ‖(A − k I )(y)‖ forall y ∈ Y . We show that R(B) is a closed subspace of X . For n ∈ N, let yn ∈ Ybe such that (A(yn) − kyn) converges in X to, say, z. Let α > 0 be such that‖A(yn) − kyn‖ ≤ α for all n ∈ N. Since β‖yn‖ ≤ ‖A(yn) − kyn‖ ≤ α forall n ∈ N, (yn) is a bounded sequence in Y . By Lemma 5.17, (yn) has a con-vergent subsequence, and if it converges to y in Y , then A(y) − ky = z. Thusz ∈ R(B), and so R(A − k I ) = R(B) is a closed subspace of X . (Comparethe proof of Lemma 5.19.)

5.24. Suppose (i) holds. Then 1 /∈ σe(A). In fact, 1 /∈ σ(A) since A is compact.In this case, the inverse (I − A)−1 is continuous, that is, x := (I − A)−1(y)depends continuously on y ∈ X .Clearly, (ii) holds if and only if 1 ∈ σe(A). In this case, the eigenspace E1 :={x ∈ X : A(x) = x} of A corresponding to its nonzero eigenvalue 1 is finitedimensional, since A is compact.

5.25. (i) Since A is compact, 1 ∈ σe(A′) if and only if 1 ∈ σe(A).(ii) Let y ∈ X . Suppose there is x ∈ X such that x − A(x) = y. If

x ′ is in Z(A′ − I ), that is, if A′(x ′) = x ′, then x ′(y) = x ′(x) − x ′(A(x)) =x ′(x) − A′(x ′)(x) = 0. Conversely, let {x ′

1, . . . , x′m} be a basis for Z(I − A′),

which is finite dimensional since A′ is a compact operator, and suppose thatx ′j (y) = 0 for j = 1, . . . ,m. Then x ′(y) = 0 for all x ′ ∈ Z(I − A′).

Assume for a moment that there is no x ∈ X such that x − A(x) = y, thatis, y /∈ R(I − A). Clearly, y �= 0. Since R(I − A) is a closed subspace of X(Exercise 5.23), there is x ′ ∈ X ′ such that x ′(z) = 0 for all z ∈ R(I − A) andx ′(y) = ‖y‖. Then x ′(x) − A′(x ′)(x) = x ′(x − A(x)) = 0 for every x ∈ X ,that is, x ′ ∈ Z(I − A′), but x ′(y) = ‖y‖ �= 0. This is a contradiction.Next, suppose x0 ∈ X and x0−A(x0) = y. Then x ∈ X satisfies x−A(x) = y

if and only if x−x0−A(x−x0) = 0, that is, x−x0 ∈ Z(A− I ), which is finitedimensional since A is a compact operator. This is the same thing as sayingx := x0 + k1x1 + · · · + kmxm, where k1, . . . , km are in K, and {x1, . . . , xm} isa basis for Z(A − I ).

5.26. Let J denote the canonical embedding of X into X ′′. By Exercise 4.25, A′′ J =J A, and so (A′′ − k I )J = J (A − k I ) for k ∈ K. It follows that if A′′ − k Iis one-one, then A − k I is one-one, and if A′′ − k I is bounded below, thenA − k I is bounded below. Hence σe(A) ⊂ σe(A′′) and σa(A) ⊂ σa(A′′).Next, since X ′ is a Banach space, σ(A′′) = σ(A′). Also, σ(A′) ⊂ σ(A).

5.27. Let A be a normal operator on a Hilbert space H . By mathematical inductionon n ∈ N, we show that if λ ∈ K, x ∈ X and (A − λI )n(x) = 0, then(A−λI )(x) = 0. If n = 1, then this is obvious. Assume this holds form ∈ N.

Solutions to Exercises 243

Suppose (A−λI )m+1(x) = 0. Let y := (A−λI )(x). Then (A−λI )m(y) = 0.By the inductive assumption, (A − λI )(y) = 0. Now

‖(A − λI )(x)‖2 = 〈(A − λI )(x), (A − λI )(x)〉 = 〈(A∗ − λI )(A − λI )(x), x〉≤ ‖(A∗ − λI )(y)‖‖x‖ = ‖(A − λI )(y)‖‖x‖ = 0

since A is normal. Thus (A − λI )(x) = 0.

5.28. Assume for a moment that σe(A) is uncountable. For each λα ∈ σe(A), letuα be a corresponding eigenvector of A with ‖uα‖ = 1. Since A is normal,{uα} is an uncountable orthonormal subset of H . Since ‖uα − uβ‖ = √

2 forα �= β, no countable subset of H can be dense in H . This is a contradictionto the separability of H .

5.29. Let A ∈ BL(�2) be defined by an infinite matrix M := [ki, j ]. If M is diagonal,then it is clear that MtM = diag (|k1,1|2, |k2,2|2, . . .) = MMt . Hence A is anormal operator (Example 4.28(i)).Conversely, suppose A is a normal operator, and M is upper triangular. Note

that ki, j = 0 for all i > j . First, A(e1) = ∑∞i=1 ki,1ei = k1,1e1. Since A is

normal, e1 is an eigenvector of A∗ corresponding its eigenvalue k1,1, that is,A∗(e1) = k1,1e1. But since the matrix Mt defines the operator A∗, we seethat A∗(e1) = ∑∞

i=1 k1,i ei . Hence k1,2 = k1,3 = · · · = 0. Next, A(e2) =∑∞i=1 ki,2ei = k1,2e1 + k2,2e2 = k2,2e2, since k1,2 = 0 as we have just shown.

Again, since A is normal, A∗(e2) = k2,2e2. But A∗(e2) = ∑∞i=1 k2,i ei =∑∞

i=2 k2,i ei . Hence k2,3 = k2,4 = · · · = 0. In this manner, by mathematicalinduction, we obtain ki,i+1 = ki,i+2 = · · · = 0 for every i ∈ N. Thus M =diag (k1,1, k2,2, . . .).If A is a normal operator, and M is lower triangular, then the normal operatorA∗ is defined by the upper triangular matrix Mt . Hence Mt is a diagonalmatrix, that is, M is a diagonal matrix.

5.30. Let A ∈ BL(H) be unitary. Since A is normal, and A is an isometry, σ(A) =σa(A) ⊂ {k ∈ K : |k| = 1} by Exercise 5.9(iii). Let k ∈ K with |k| �= 1. Ify ∈ H and y = (A − k I )(x), x ∈ H , then

‖y‖ = ‖A(x) − kx‖ ≥ | |k| − 1| ‖x‖ = | |k| − 1| ‖(A − k I )−1(y)‖

since ‖A(x)‖ = ‖x‖. Hence ‖(A − k I )−1‖ ≤ 1/| |k| − 1|.5.31. Let k ∈ K\ω(A), and β := d

(k,ω(A)

). Since σ(A) ⊂ ω(A), A − k I is

invertible. If x ∈ H and ‖x‖ = 1, then ‖A(x) − kx‖ ≥ |〈A(x) − kx, x〉| =|〈A(x), x〉 − k| ≥ β. Let y ∈ H . If y = (A − k I )(x), where x ∈ H , then

‖y‖ = ‖A(x) − kx‖ ≥ β‖x‖ = β‖(A − k I )−1(y)‖.

Hence ‖(A − k I )−1‖ ≤ 1/β.Let A ∈ BL(H) be self-adjoint. Since (mA, MA) ⊂ ω(A) ⊂ [mA, MA],

244 Solutions to Exercises

we see that ω(A) = [mA, MA]. Let k ∈ K\[mA, MA]. Clearly, β = |Im k|if Re k ∈ [mA, MA], β = |k − mA| if Re k < mA, and β = |k − MA|if Re k > MA. Further, if K := R, and k ∈ R\σ(A), then (A − k I )−1

is self-adjoint, and so ‖(A − k I )−1‖ = sup{|μ| : μ ∈ σ((A − k I )−1)} =sup{|(λ − k)−1| : λ ∈ σ(A)} = 1/d, where d := d(k,σ(A)) (Exercise 5.1).(Note: If K := C, A is normal, and k ∈ C\σ(A), then (A − k I )−1 is normal,

and so the above proof works since ‖B‖ = sup{|μ| : μ ∈ σ(B)} for a normaloperator B on a Hilbert space H over C.)

5.32. Let A ∈ BL(H) be self-adjoint. Then ±i /∈ σ(A). Since A∗ = A, weobtain T (A)∗ = (A − i I )−1(A + i I ) = (A + i I )(A − i I )−1. It followsthat (T (A))∗T (A) = I = T (A)(T (A))∗, that is, T (A) is unitary. Also,1 /∈ σ(T (A)) since T (A)− I = (A−i I−A−i I )(A+i I )−1 = −2i(A+i I )−1

is invertible.Next, let B ∈ BL(H) be unitary, and suppose 1 /∈ σ(B). Since B∗ = B−1,

we obtain S(B)∗ = −i(I − B∗)−1(I + B∗) = −i(I − B−1)−1(I + B−1) =i(I + B)(I − B)−1 = S(B), that is, S(B) is self-adjoint.Further, it can be easily checked that S(T (A)) = i(I +T (A))(I −T (A))−1 =A and T (S(B)) = (S(B) − i I )(S(B) + i I )−1 = B.

5.33. If A ≥ 0, then σ(A) ⊂ ω(A) ⊂ [0,∞). Conversely, suppose σ(A) ⊂ [0,∞).Then inf ω(A) = mA ∈ σa(A) = σ(A) ⊂ [0,∞), and so ω(A) ⊂ [0,∞).If 0 ∈ σe(A), then 0 ∈ ω(A) sinceσe(A) ⊂ ω(A). Conversely, suppose A ≥ 0,and 0 ∈ ω(A), that is, there is x ∈ H such that ‖x‖ = 1 and 〈A(x), x〉 = 0. Bythe generalized Schwarz inequality, ‖A(x)‖ ≤ 〈A(x), x〉1/4〈A2(x), A(x)〉1/4,and so A(x) = 0. Hence 0 ∈ σe(A).

5.34. Define B(x) := (x(1) cos θ + x(2) sin θ,−x(1) sin θ + x(2) cos θ) for x :=(x(1), x(2)) ∈ R

2. It is easy to check that 〈A(x), y〉 = 〈x, B(y)〉 for allx, y ∈ R

2. Hence A∗ = B. Also, ‖A(x)‖22 = (x(1) cos θ − x(2) sin θ)2 +(x(1) sin θ + x(2) cos θ)2 = x(1)2 + x(2)2 = ‖x‖22 for all x ∈ R

2. HenceA : R

2 → R2 is a linear isometry. Since R

2 is finite dimensional, A is onto.Thus A is a unitary operator. Clearly, A is defined by the 2 × 2 matrix M :=[ki, j ], where k1,1 := cos θ, k1,2 := − sin θ, k2,1 := sin θ and k2,2 := cos θ.Now det(M − t I ) = t2 − 2t cos θ + 1, and it is equal to 0 if and only ift = cos θ ± (cos2 θ − 1)1/2 ∈ R. Hence σ(A) = σe(A) = {1} if θ := 0,σ(A) = σe(A) = {−1} if θ := π, and σ(A) = σe(A) = ∅ otherwise.

5.35. Since A is a Hilbert–Schmidt operator on H , it is a compact operator (Exercise3.40(i)). Henceσe(A) is countable.Also, the eigenspace corresponding to eachnonzero eigenvalue of A has a finite orthonormal basis. Further, since A is anormal operator, any two eigenspaces of A are mutually orthogonal. Let (λn)

be the sequence of nonzero eigenvalues of A, each eigenvalue being repeatedas many times as the dimension of the corresponding eigenspace. Then thereis a countable orthonormal subset {u1, u2, . . .} of H such that A(un) = λnunfor each n ∈ N. By Exercise 4.31(iii), A∗ is a Hilbert–Schmidt operator on H .Let {u1, u2, . . .} be an orthonormal basis for H such that

∑j ‖A∗(u j )‖2 < ∞.

Solutions to Exercises 245

Then

∑

n

|λn|2 =∑

n

‖A(un)‖2 =∑

n

∑

j

|〈A(un), u j 〉|2

=∑

j

∑

n

|〈un, A∗(u j )〉|2 ≤∑

j

‖A∗(u j )‖2

by the Parseval formula and the Bessel inequality.

5.36. Suppose A is normal, and let μ1, . . . ,μk be the distinct eigenvalues of A. Forj ∈ {1, . . . , k}, let E j := Z(A − μ j I ), and let Pj denote the orthogonalprojection operator on H with R(Pj ) = E j . If x ∈ X , then Pj (x) ∈ E j , andso APj (x) = μ j Pj (x) for j = 1, . . . , k. Also, if i �= j , then Ei ⊥ E j , and soR(Pj ) = E j ⊂ E⊥

i = Z(Pi ), that is, Pi Pj = 0.Let G := E1 + · · · + Ek . Then G⊥ = {0} as in the proof of Theorem

5.39, and so G = H . Consider x ∈ H . Then x = x1 + · · · + xk , wherex j ∈ E j = R(Pj ) for j = 1, . . . , k. Thus x = P1(x) + · · · + Pk(x)and A(x) = AP1(x) + · · · + APk(x) = μ1P1(x) + · · · + μk Pk(x), thatis, I = P1 + · · · + Pk and A = μ1P1 + · · · + μk Pk , as desired.The converse follows easily since every orthogonal projection operator is nor-mal, and a linear combination of normal operators is normal.

5.37. Suppose A is a nonzero compact self-adjoint operator on H . Let μ1,μ2, . . . bethe distinct nonzero eigenvalues of A. Since A is self-adjoint, each μ j is real,and since A is compact, either the set {μ1,μ2, . . .} is finite or μn → 0. Foreach j , let E j := Z(A − μ j I ), and let Pj denote the orthogonal projectionoperator on H with R(Pj ) = E j . Since A is compact, each Pj is of finite rank.If x ∈ X , then Pj (x) ∈ E j , and so APj (x) = μ j Pj (x) for each j . Also, ifi �= j , then Ei ⊥ E j , and so R(Pj ) = E j ⊂ E⊥

i = Z(Pi ), that is, Pi Pj = 0.Let G denote the closure of span (∪ j E j ), and let P denote the orthogonal pro-

jection operator on H with R(P) = G. By Exercise 4.38, P(x) = ∑j Pj (x)

for all x ∈ H . Also, H = Z(A) ⊕ G, as in the proof of Theorem 5.40.Let P0 denote the orthogonal projection operator on H with R(P0) = Z(A).Then for x ∈ H , x = P0(x) + P(x) = P0(x) + ∑

j Pj (x) and A(x) =AP0(x) + ∑

j APj (x) = ∑j μ j Pj (x).

In fact, A = ∑j μ j Pj . This is obvious if the set {μ1,μ2, . . .} is finite. Suppose

now that this set is infinite. First note that ‖x‖2 = ‖P0(x)‖2 + ‖P(x)‖2 =‖P0(x)‖2 + ∑∞

j=1 ‖Pj (x)‖2, and so∑n

j=1 ‖Pj (x)‖2 ≤ ‖x‖2 for all x ∈ Hand all n ∈ N. For n ∈ N, define An := ∑n

j=1 μ j Pj . Given ε > 0, find n0 ∈ N

such that |μ j | < ε for all n > n0. Then for all n ≥ n0,

246 Solutions to Exercises

‖A(x) − An(x)‖2 =∥∥∥∥

∞∑

j=n+1

μ j Pj (x)

∥∥∥∥

2

≤∞∑

j=n+1

|μ j |2‖Pj (x)‖2

< ε2∞∑

j=n+1

‖Pj (x)‖2 ≤ ε2‖x‖2, x ∈ H.

Thus ‖A − An‖ < ε for all n ≥ n0. Hence A = ∑∞j=1 μ j Pj in BL(H).

For the converse, note that each Pj is self-adjoint since it is an orthogonal

projection operator, each μ j ∈ R, and so A∗ = ∑j μ j Pj

∗ = ∑j μ j Pj = A.

Also, A is compact since An := ∑nj=1 μ j Pj is a bounded operator of finite

rank for each n and ‖An − A‖ → 0.

5.38. The kernel k(s, t) := min{1− s, 1− t}, s, t ∈ [0, 1], is a real-valued continu-ous function on [0, 1]×[0, 1], and k(t, s) = k(s, t) for all s, t ∈ [0, 1]. HenceA is a compact self-adjoint operator on H . Also, A �= 0. Let λ be a nonzeroeigenvalue of A, and let x ∈ H be a corresponding eigenvector of A. SinceA(x) ∈ C([0, 1]), x = A(x)/λ is continuous on [0, 1]. Now

y(s) := A(x)(s) = (1 − s)∫ s

0x(t) dt +

∫ 1

s(1 − t)x(t) dt, s ∈ [0, 1].

Clearly, y(1) = 0. By the fundamental theorem of calculus for Riemannintegration, y ∈ C1([0, 1]), and

y′(s) = (1 − s)x(s) −∫ s

0x(t) dt − (1 − s)x(s) = −

∫ s

0x(t) dt, s ∈ [0, 1].

Hence y′(0) = 0. Further, y′ ∈ C1([0, 1]), and y′′(s) = −x(s) for s ∈ [0, 1].Thus y ∈ C2([0, 1]), y′′ = −x and y′(0) = 0 = y(1). Since y = A(x) = λx ,we see that λx ′′ = −x and λx ′(0) = 0 = λx(1), that is, λx ′′ + x = 0 andx ′(0) = 0 = x(1). Now the differential equation λx ′′ + x = 0 has a nonzerosolution satisfying x ′(0) = 0 = x(1) if and only ifλ = 4/(2n−1)2π2, n ∈ N.In this case, the general solution is given by x(s) := cn cos(2n − 1)πs/2 fors ∈ [0, 1], where cn ∈ K. (If K := C, we must first show that λ ∈ R,as in the footnote in Example 5.23(ii).) Conversely, fix n ∈ N, let λn :=4/(2n − 1)2π2, xn(s) := cos(2n − 1)πs/2, s ∈ [0, 1], and let yn := λnxn .Clearly, y′′

n = λnx ′′n = −xn and y′

n(0) = 0 = yn(1). Integrating by parts,

A(y′′n )(s) = (1 − s)

∫ s

0y′′n (t)dt +

∫ 1

s(1 − t)y′′

n (t)dt

= (1 − s)(y′n(s) − y′

n(0)) − (1 − s)y′n(s) + yn(1) − yn(s)

= −yn(s) for s ∈ [0, 1].

Solutions to Exercises 247

Thus A(y′′n ) = −yn , that is, A(xn) = yn = λnxn . It follows that λn is in fact

an eigenvalue of A, and the corresponding eigenspace of A is spanned by thefunction xn . Hence A(x) = ∑∞

n=1 λn〈x, un〉un , where λn := 4/(2n − 1)2π2

and un(s) := √2 cos(2n − 1)πs/2, s ∈ [0, 1], for n ∈ N.

5.39. The kernel k(s, t) := min{s, t}, s, t ∈ [0, 1], is a real-valued continuousfunction on [0, 1] × [0, 1], and k(t, s) = k(s, t) for all s, t ∈ [0, 1]. Hence Ais a compact self-adjoint operator on L2([0, 1]). Also, A �= 0. By Exercise5.20, the nonzero eigenvalues of A are given by λn := 4/(2n − 1)2π2, n ∈ N,and the eigenspace of A corresponding to λn is span {xn}, where xn(s) :=sin(n−1/2)πs, s ∈ [0, 1]. Let un(s) := √

2 sin(n−1/2)πs, s ∈ [0, 1]. Then{un : n ∈ N} is an orthonormal basis for L2([0, 1]) consisting of eigenvectorsof A. For x ∈ L2([0, 1]), A(x) = ∑∞

n=1 λn〈x, un〉un , that is,∫ s

0t x(t) dm(t) + s

∫ 1

sx(t) dm(t) = 8

π2

∞∑

n=1

sn(x)

(2n − 1)2sin(2n − 1)

πs

2,

where sn(x) := ∫ 10 x(t) sin(n − 1/2)πt dm(t), and the series on the right side

converges in L2([0, 1]). We shall use Theorem 5.43.Let y ∈ L2([0, 1]) and μ ∈ K, μ �= 0. Consider the integral equation

x(s) − μ

(s∫ s

0t x(t) dm(t) + s

∫ 1

sx(t) dm(t)

)= y(s), s ∈ [0, 1].

Let μn := λ−1n = (2n − 1)2π2/4 for n ∈ N. If μ �= μn for any n ∈ N, then

there is a unique x ∈ L2([0, 1]) satisfying x − μA(x) = y. In fact, sinceμ/(μn − μ) = 4μ/

((2n − 1)2π2 − 4μ

)and 〈y, un〉 = √

2 sn(y) for n ∈ N,

x(s) = y(s) + 8μ∞∑

n=1

sn(y)

(2n − 1)2π2 − 4μsin(2n − 1)

πs

2, s ∈ [0, 1].

Further, ‖x‖ ≤ α‖y‖, where α := 1 + 4|μ|/minn∈N{|(2n − 1)2π2 − 4μ|}.Next, suppose μ := (2n1 − 1)2π2/4, where n1 ∈ N. There is x in L2([0, 1])satisfying x − μA(x) = y if and only if x ⊥ un1 , that is, sn1(y) = 0. In thiscase, since μ/(μn − μ) = (2n1 − 1)2/4(n − n1)(n + n1 − 1) for n �= n1,

x(s) = y(s) + (2n1 − 1)2

2

∑

n �=n1

sn(y)

(n − n1)(n + n1 − 1)sin(2n − 1)

πs

2

+ k1 sin(2n1 − 1)πs

2, s ∈ [0, 1], where k1 ∈ K.

5.40. Suppose A is a nonzero compact operator on H . Then A∗A is a compact self-adjoint operator on H . Also, since ‖A(x)‖2 = 〈A∗A(x), x〉 for all x in H , we

248 Solutions to Exercises

see that Z(A∗A) = Z(A). In particular, A∗A �= 0. Then H has an orthonormalbasis {uα} consisting of eigenvectors of A∗A. Let A∗A(uα) = λαuα for eachα. By Theorem 5.36(iii), the set S := {uα : λα �= 0} is countable. Let S :={u1, u2, . . .} and A∗A(un) = λnun for each n. If S is in fact denumerable, thenλn → 0. Also, λn ≥ 0 for each n since A∗A is positive. Let sn := √

λn > 0and vn := A(un)/sn for each n. Then snsm〈vn, vm〉 = 〈A(un), A(vm)〉 =〈A∗A(un), um〉 = λn〈un, um〉 for all n,m. It follows that {v1, v2, . . .} is anorthonormal subset of H . If {v1, v2, . . .} is denumerable, then so is S, andsn → 0. Let x ∈ H . Let sn ≤ s for all n. Then

∑n |sn|2|〈x, un〉|2 ≤ s‖x‖2, and

so∑

n sn〈x, un〉vn converges in H . Define B(x) := ∑n sn〈x, un〉vn for x ∈ H .

Then B(un) = snvn = A(un) for all n. Also, if uα ∈ Z(A), then uα ⊥ un forall n, and so B(uα) = 0 = A(uα). Hence A(x) = B(x) = ∑

n sn〈x, un〉vn forall x ∈ H .Conversely, let A(x) = ∑

n sn〈x, un〉vn for all x ∈ H , where {u1, u2, . . .} and{v1, v2, . . .} are countable orthonormal subsets of H , and s1, s2, . . . are positivenumbers such that sn → 0 if the set {v1, v2, . . .} is denumerable. First, suppose{v1, v2, . . .} is finite. Then there ism ∈ N such that A(x) = ∑m

n=1 sn〈x, un〉vnfor all x ∈ H , and so R(A) = span {v1, . . . , vm}. Since A is a boundedoperatorof finite rank, it is compact. Next, suppose {v1, v2, . . .} is denumerable. Letε > 0. There is m0 ∈ N such that |sn| < ε for all n > m0. For m ∈ N, letAm(x) := ∑m

n=1 λn〈x, un〉vn, x ∈ H . Then

‖A(x) − Am(x)‖2 =∥∥∥

∞∑

n=m+1

sn〈x, un〉vn∥∥∥2 =

∞∑

n=m+1

|sn|2|〈x, un〉|2

≤ ε2∞∑

n=m+1

|〈x, un〉|2 ≤ ε2‖x‖2 for x ∈ H and m ≥ m0.

Thus ‖A− Am‖ ≤ ε for all m ≥ m0. Since Am → A in BL(H), and each Am

is a bounded operator of finite rank, A is compact.In this case, 〈A∗(x), y〉 = 〈x, A(y)〉 = ∑

n sn〈x, vn〉〈un, y〉 = 〈C(x), y〉 forall x, y ∈ H , where C(x) := ∑

n sn〈x, vn〉un for x ∈ H . Hence A∗ = C .Let x ∈ H and y := A(x). Then 〈y, vn〉 = sn〈x, un〉 for all n, and soA∗A(x) = A∗(y) = C(y) = ∑

n sn〈y, vn〉un = ∑n s

2n 〈x, un〉un .

Also, ‖A(un)‖2 = 〈A∗A(un), un〉 = 〈λnun, un〉 = s2n for all n, and so∑

n s2n = ∑

n ‖A(un)‖2.Now suppose H is a separable Hilbert space. Then the orthonormal basis {uα}for H consisting of eigenvectors of A∗A is countable. Also, A∗A(uα) = 0 ifand only if A(uα) = 0. Hence

∑α ‖A(uα)‖2 = ∑

n ‖A(un)‖2. By Exercise4.31(iii), A is a Hilbert–Schmidt operator if and only if

∑n ‖A(un)‖2 < ∞,

that is,∑

n s2n < ∞.

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Index

AAbsolutely continuous function, 24Absolutely summable series, 53Absorbing set, 69, 126Adjoint, 141Almost everywhere, 20Angle, 71Antisymmetric relation, 2Approximate eigenspectrum, 160Approximate eigenvalue, 164Approximate solution, 104Arzelà theorem, 19Ascoli theorem, 19

BBaire theorem, 14Balanced set, 69Banach space, 53Banach–Steinhaus theorem, 93Basic inequality for an operator norm, 79Basis, 3Bernstein polynomial, 17, 118Bessel inequality, 50Best approximation, 74Bolzano–Weierstrass theorem, 13Bounded below, 77Bounded convergence theorem, 22Bounded inverse theorem, 101, 116, 157Bounded linear map, 76Bounded operator, 76Bounded sequence, 12Bounded set, 12Bounded variation, 25, 135

CCanonical embedding, 129Carathéodory condition, 20Cauchy sequence, 12Cayley transform, 201Characteristic function, 21Closed ball, 34Closed complement, 100Closed graph theorem, 97, 116, 157Closed map, 96Closed set, 10Closed unit ball, 34Closure, 10Compact linear map, 106Compact operator, 107Compact perturbation, 200Compact set, 14Compact support, 36Comparable norms, 37Complete metric space, 13Completion of a normed space, 129Componentwise convergence, 10Compression spectrum, 179Conjugate exponents, 81Conjugate-linear map, 44Conjugate-symmetric function, 44Conjugate-transpose of a matrix, 142Continuous function, 15Continuously differentiable function, 24Convergence in the mean, 28Convergent sequence, 10Convergent series, 53Convex set, 69, 125Coset, 4Countable set, 2Countably additive function, 20Countably subadditive seminorm, 89

© Springer Science+Business Media Singapore 2016B.V. Limaye, Linear Functional Analysis for Scientists and Engineers,DOI 10.1007/978-981-10-0972-3

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252 Index

DDense subset, 11Denumerable set, 2Diagonal operator, 117, 142, 188, 198Dimension, 4Direct sum, 68Dominated convergence theorem, 22Dot product, 44Dual basis, 128Dual space, 128

EEigenequation, 161Eigenspace, 160Eigenspectrum, 160Eigenvalue, 160Eigenvector, 160Energy principle, 153Equicontinuity of seminorms, 92Equicontinuous set of functions, 18Equivalence class, 2Equivalence relation, 2, 4Equivalent norms, 37Essential bound, 27Essentially bounded function, 27Essentially uniform convergence, 28Essential range, 198Essential supremum, 27

FFast Fourier transform, 157Féjer theorem, 30Finite dimensional linear space, 4Finite rank, 5Finite set, 2Fourier coefficient, 29, 30Fourier expansion, 64Fourier integral, 152Fourier matrix, 157Fourier-Plancherel transform, 153Fourier series, 29Fourier transform, 152Fredholm alternative, 200Fredholm integral equation, 196Fredholm integral map, 87Frobenius norm, 71, 86Fubini theorem, 23Function space, 36Fundamental theorem of calculus for Rie-

mann and Lebesgue integrations, 24

GGelfand–Mazur theorem, 170Generalized eigenvector, 200Generalized polarization identity, 158Generalized Schwarz inequality, 151General solution, 200Gershgorin disk, 198Gershgorin theorem, 198Gram–Schmidt orthonormalization, 48Graph of a function, 96Green’s function, 177

HHaar system, 61Hahn–Banach extension, 122Hahn–Banach extension theorem, 122, 133Hahn–Banach separation theorem, 127Hamel basis, 3Heine–Borel theorem, 15Helly theorem, 154Helmert basis, 71Hermite polynomials, 50Hermitian operator, 157Hilbert cube, 70Hilbert space, 62Hilbert space isomorphism, 68, 146Hilbert–Schmidt map, 99, 118, 157, 202Hilbert–Schmidt test, 113Homeomorphic spaces, 15Homeomorphism, 15Homogeneous equation, 200Hyperplane, 6Hyperspace, 6Hyponormal operator, 157

IInfimum, 8Infinite dimensional linear space, 4Inner product, 44Inner product space, 44Integrable function, 22Inverse Cayley transform, 201Inversion theorem, 31, 152Invertible operator, 160Isometric spaces, 15Isometry, 15

KKernel, 5, 87Kronecker symbol, 6

Index 253

LLaw of cosines, 71Lebesgue integral, 21Lebesgue measurable set, 20Lebesgue measure, 20Lebesgue outer measure, 20Left shift operator, 143, 156, 169, 170, 199Legendre polynomials, 49Light-like vector, 45Limit, 10Linear combination, 3Linear functional, 5Linearly dependent subset, 3Linearly independent subset, 3Linear map, 5Linear space, 3Lower triangular matrix, 201

MMatrix defining a linear map, 82Matrix transformation, 82Maximal element, 2Maximal orthonormal subset, 52Mean square convergence, 28Measurable function, 20, 22Metric, 7Metric space, 7Minimum residual property, 199Minkowski gauge, 126Minkowski inequality, 45Minkowski inequality for functions, 26Minkowski inequality for numbers, 8Minkowski space, 44Monotone convergence theorem, 22Multiplication operator, 165, 198Mutually orthogonal subspaces, 182

NNeumann expansion, 199Node, 95Norm, 33Normal operator, 146Normed space, 34Norm of the graph, 115Nullity, 5Null space, 5Numerical range, 149

OOblique projection operator, 100One-to-one correspondence, 2

Open ball, 11, 34Open map, 102Open mapping theorem, 103, 116, 157Open set, 10Open unit ball, 34Operator equation, 104Operator norm, 79Orthogonal complement, 69Orthogonal projection, 69Orthogonal projection operator, 100, 152Orthogonal set, 47Orthonormal basis, 65Orthonormal set, 48

PParallelepiped law, 70Parallelogram law, 47Parseval formula, 64Parseval identity, 68Partially ordered set, 2Partial order, 2Partial sum, 53Particular solution, 200Perturbation technique, 105Pointwise bounded set of functions, 18Pointwise convergence, 17Pointwise limit, 17Polarization identity, 70Polya theorem, 95Positive-definite function, 44Positive operator, 146Product norm, 39, 40Product space, 5Projection map, 40Projection operator, 99Projection theorem, 68Pythagoras theorem, 48

QQR factorization, 71Quadrature formula, 95Quotient map, 40Quotient norm, 38Quotient space, 4

RRange space, 5Rank, 5Rank-nullity theorem, 5Rayleigh quotient, 171, 199Real hyperplane, 127

254 Index

Reflexive normed space, 129Reflexive relation, 2Relation, 2Representer, 132Residual element, 199Resonance theorem, 130Riemann–Lebesgue lemma, 31Riesz-Fischer theorem, 63Riesz representation theorem, 132, 136Right shift operator, 142, 156, 157, 169, 170Ritz method, 184

SSaw-tooth function, 61Schauder basis, 60Schauder theorem, 140Schur result, 131Schur test, 113Schwarz inequality, 45Schwarz inequality for functions, 26Schwarz inequality for numbers, 8Second dual space, 129Self-adjoint operator, 146Seminorm, 33Separable metric space, 11Sequence, 10Sequence space, 35Signal analysis, 153Signum function, 81Simple function, 21Singular value, 202Skew-hermitian operator, 157Sobolev space, 58, 62, 73Space-like vector, 44Span, 3Spanning subset, 3Spectral radius formula, 170Spectral theorem, 189–191, 193Spectral value, 165Spectrum, 160Square-integrable function, 26Standard basis, 4Strictly convex normed space, 70, 154Sturm–Liouville problem, 177Sublinear functional, 120Subsequence, 10Subspace, 3Sum of a series, 53Summable series, 53Sup metric, 10Sup norm, 36Supremum, 8Symmetric relation, 2

TTaylor–Foguel theorem, 154Term of a sequence, 10Tietze extension theorem, 16Time-frequency equivalence, 153Time-like vector, 45Toeplitz-Hausdorff theorem, 150Tonelli theorem, 23Total energy, 153Totally bounded set, 12Totally ordered set, 2Total variation, 25, 135Transitive relation, 2Transpose, 136Transposed homogeneous equation, 200Triangle inequality, 7Tridiagonal operator, 117Trigonometric polynomials on R, 72Two-norm theorem, 101

UUncountable set, 2Uniform boundedness principle, 93Uniform convergence, 10Uniform limit, 17Uniformly bounded set of functions, 18Uniformly continuous function, 16Unique Hahn–Banach extension, 133Unit sphere, 34Unitary operator, 146Upper bound, 2Upper triangular matrix, 201Urysohn lemma, 16Usual inner product, 44

VVanishing at infinity, 36Volterra integration operator, 199

WWeak convergence, 155Weak limit, 155Weierstrass theorem, 17Weight, 95Weighted shift operator, 117, 199

ZZabreiko theorem, 91Zero space, 5Zorn lemma, 2