Integrable Distributions Multipliers and The space ofMultipliers with Segal Algebra

Adam M. AliKhartoum Academy for Aviation and Technology May 2014

P.O. Box 12475, Khartoum SudanAdamrahad54@gamil.com

Abstract : Here we obtain a class of distributions which are multiplierson for . We show that this class strictly contains

for and hence multiplier space is

larger than the multiplier space . Also, this classprovides a set of –multipliers which are not boundedmeasures.Terms: Multipliers, Segal Algebra, Banach Space.Introduction: Note that such distributions are also multipliers on Lp for 1¿p¿ ∞ and that the space is of such distributions is exactly themultiplier space of a certain Segal algebra. This gives anexample of a Segal algebra whose multipliers space can bedetermined and is larger than the space of bounded measures. We also prove that the multipliers of Wm,1 are integrabledistributions of order ≤1∨≤2accordingly as m is odd or even.

Proposition (1)[4]: Let and any non-negative integer. Then if is a

distribution in , then defines a multiplier from into . Proof:

If , it can be expressed in the form , where

. Consider any . Then

Since , for .Hence

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Hence as the above is true for anyrepresentation of .i.e.,

,on , the dense subspace of . Hence can be uniquelyextended to a continuous linear map from into ,satisfying

That is given as a convolution operator shows that islinear and commutes with translations. HenceT∈ (Wm,p,Lp ).Corollary (2) [3]: If Tis a distribution in (C0

∞)', then T defines a multiplier from Wm,p into Lp, when 1≤p≤∞.Proof:For any fj∈D we have

T∗∑j=1

∞fi=∑

|α|≤m∂αμα∗∑

j=1

∞fj, where μα∈M (Rn )

¿ ∑|α|≤m

∂α(μα∗∑j=1

∞fj)=∑

|α|≤mμα∗∑

j=1

∞∂αfj=∑

|α|≤m∑j=1

∞μα∗∂αfj

Since fj∈D⊂Wm,p and ∂αfj∈L

pfor |α|≤m, we have

‖T∗∑j=1

∞fj‖Lp

=‖∑j=1

∞T∗fj‖Lp

≤∑|α|≤m

∑j=1

∞

‖μα∗∂αfj‖Lp

≤∑|α|≤m

∑j=1

∞

‖μα‖‖∂αfj‖Lp

≤max {‖μα‖M}∑|α|≤m

∑j=1

∞

‖∂αfj‖Lp

≤‖T‖(C0∞ )' ∑|α|≤m∑j=1

∞

‖∂αfj‖Lp

≤‖T‖(C0∞ )'∑j=1

∞

‖∂αfj‖Lp

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Hence T is continuous linear map and ‖T‖≤‖T‖(C0∞ ) such thatT∈ (Wm,p,Lp )Proposition (3)[4]: Let and any non-negative integer if is a

distribution such that for , then defines amultiplier from into itself.Proof:From the above Proposition, for any such that , the

fact that shows that is a multiplier from into . Let us take . Then is in and is in if . We would like to show which would prove that . For any , clearly,

. Since we get

Hence we can extend this map, say, , continuously to thewhole of . Since

forand is defined on , by uniqueness of the extension wecan write as distributions. This is what wewished to prove.Notice that for

for each such that

Remark (4)[4]: In the context of Theorem (9) the above result shows that the

multiplier space for , contains distributions

satisfying , where , for every non-negativeinteger . Notice that by means of Proposition (4.2.7)(following), we show that the class of such distributions islarger than the space of bounded measures on .

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Put For any non-negative integer. Then we can make the comingobservation.Proposition (5)[4]:

(i) is contained in the space of pseudo-measures onRn.(ii) contains the measure algebra, M (Rn).(iii) is a Banach algebra under convolution, with a norm

given by

Proof:While (ii) is trivial and (iii) follows from forward

verification, (i) is a consequence of the fact that and the multipliers of are contained in the space of pseudomeasures. Now, when . For any , non-negative we have

and . We shall show that

is larger than by showing thatM (Rn)is strictlycontained in if . However, in the case of , we makethe following.Proposition (6)[4][1]:

(i) On the real line , the space coincides with M (Rn)for each .

(ii) OnR, the space of multipliers from into is

isomorphic to (C0m)′

for m≥1 .Proof:(i) isa consequence of Proposition (2). It can also be seen

directly from Schwartz(ii)Consider the function k (x)=e−1 x≥0=0 x<0

Clearly and k is a fundamental solution of thedifferential operator (d /dx+1) . Denote km the functionk∗k∗...∗k , convoluted m times. The map f→km∗f is an

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isomorphism of onto , the inverse being given bythe operator (d /dx+1)m .

Now, let T∈ (Wm,1,L' ) . We wish to show that T∈ (C0m)′

. ThatT is a distribution such that

T (g )=T∗g , g∈Wm,1 ,can be checked by standard arguments. If we take is in . Hence T∗km∗f belongs to . This show thatT∗km is actually a multiplier on . Note that km is also amultiplier on . This means that there exit suchthat

T∗km=μConsider

( ddx+1)m

(T∗km )=( ddx +1)mμ

But

( ddx+1)mμ∈ (C0m)

′

and

( ddx+1)m

(T∗km )=T∗( ddx +1)mkm=T∗δ=T

Hence T∈(C0m)′

That being already established, (ii) follows. In the case of higher dimensions (i) of the aboveProposition fails. We show in the following that strictly

contains if (see below ). This enable us to assertthat in , , the space of multipliers on is largerthan the multipliers on , unlike the case of for .Proposition (7) [4]:

The multiplier space strictly contains the space

of bounded measures, if , and .Proof:

As we know that is contained in and foreach , it is enough to show that is a space larger

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than for . This follows from the Proposition (6)above.Proposition (8)[4][1]:There exists a distribution in , which is not a boundedmeasure onRn, if . In fact may be chosen to have compactsupport.Proof:There exists a distribution such that is of order forevery , but is not a measure, if . For, one canconstruct an example of such a distribution on , usingthe one given onR2. Let us first show that is then a distribution of order

. If is the parametric of , which lies in , we canwrite

for some

As belongs to is of order . Sinceis in . Hence is of order .

Let be relatively compact open set such that therestriction of to is not a measure. Such exists,otherwise by using a partition of a unity we could conclude

that is a measure. Let be a cut-off function whichis on . Put . Since

and and are of order , we see that and are oforder . Moreover, since has compact support, and are actually Integrable of order . Hence , but is not

in . This completes the proof. It can be shown that every multiplier on is given byconvolution with a multiplier distribution. Here, we provethat a multiplier on is Integrable of order at most 2.This consequently gives the following Proposition on the class

.Proposition (9)[4]:

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Every distribution satisfying is Integrable of order if are multipliers on for , the above

proposition follows as an easy consequence of the result whichwe prove below.Proposition (10) [4]:

If then is Integrable of order or accordingly as is odd or even.Proof:For the differential operator , we can find a parametrixwhich lies in , for any positive integer . Now, suppose

and . Consider for some forsome .

Or

But as is a multiplier on . Hence

can be written as an element in , as . Also

, . Hence clearly . This showsthat is Integrable of order . Let for some . Nowtake and its parametrix which lies in . Thenclearly is in . Hence

Notice that can be written in the form

, where are in . Clearly .

Hence can be expressed as an element in whichimplies that is of order . The space is a dual space. We prove the result for amore general case. Let us first make the following definition.Definition (11)[4]: Let be a Banach space of distributions,continuously imbedded in . If we define

Then is a Banach space under the norm

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This is just a notion of Sobolev space carried over to a spaceof distributions. Let us recall the definition of for any Banachspace of distributions, given by

with the norm

Proposition (12) [4]: Let be a normal space of distributions which iscomplete under its own norm and assume that operates on

continuously by translations. Then the dual space of

is isomorphic to and the norm of and

are equivalent.Proof:

First, suppose that . Then since and

if , is also in . For , consider. We would like to show that is also in . As

, we have and so . For any ,

Since . Thus we obtain

This shows that defines a continuous linear functional on adense subspace of . We can extend, therefore, onto anelement of the dual space , satisfying

This being true for each , we get

(6)

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Conversely, let us take and show that this definesabounded linear functional on . If and has a

representation , then we define

We claim that is independent of the representation of .

If is a representation of zero isagain a representation of zero, if . Let us take an approximate identity for . Then

in for . Since are continuous on , we have

Now for every . By approximating by elements of , we can conclude

Hence

. Hence . This proves the claim. Clearly is linear. We shall now check continuity, if

Since this is true for any representation of , we get

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Hence

This together with (6) gives the equivalence of the norms

in and , and hence the isomorphism.Proposition (13)[4]:

(i) The space on is a ;dual space isomorphic to

.

(ii) The space is a dual

space isomorphic to Proof:Is immediate if we take and in Proposition (12)respectively to get (i) and (ii). Notice that

, is proved in thecourse of the proof of Theorem: [The space of multipliers

is isometrically isomorphic to the space .]

Remark (14)[4]: If for any Banach space which is a normal space ofdistributions on , we can define, for each non-negativeinteger ,

and

then is a Banach space. We can obtain for , resultsimilar to those involving .Now we characterize the multiplier space of the Sobolev space

as dual spaces. In the case when , by the resultof Translation invariant operators in , the multiplier space

is isometrically isomorphic to a dual space ,where

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is a Banach space under the norm

where the infimum is taken over all such representation of .As we have identified the multipliers on with those on if , we get immediately,

for , the equality sign stands for isomorphism between the spaces. We shall now consider the case and prove that the

multiplier space can be identified with a dualspace of a space of continuous functions. For simplicity, weprove the result in the case .Definition (15) [4]:

Let

where the infimum is taken over all representation of .Then is a Banach space.Theorem (16) [4]:The space of multipliers from into itself is isomorphic tothe dual space, with norms equivalent.Proof:

First, suppose . We shall show that induces abounded linear functional on . Define, for

(7)Since for any and we see that and

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The R.H.S of (7) is meaningful. We claim that is therequired continuous linear functional. Let us first verify that

is well defined. Take a representation of 0 in , say,

. If is an approximate identity for , with,

in , for each . Also

This shows that the convergence of the series is uniform with

respect to n. Hence . Notice thatsince is a multiplier,

This is zero as is zero. This shows that iszero.

is obviously linear. We shall prove the continuity. Take. Then

But this is valid for any representation of .Hence

Thus and satisfies

(8)

for every . Hence the inclusion map of

into is continuous.

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Conversely, take . We must show that induces amultiplier on . Fix . Define for any ,

Then

For every . This shows is isomorphic to. Hence note that is a map from into

satisfying

(9)for any and

for . As is dense in we get for

Linearity of implies the linearity of . Hence is amultiplier from into . For any we can write

for , using (9). As by the Segalalgebra property of . Hence if a continuously translatingelement of . But is the continuously translating

subspace of . Hence we get and .By (9), this proves that is actually a multiplier fromW1,1

into itself with

Hence .This together with (8) gives the isomorphismup to equivalent norm. From (8)

and from above

Implies that Remark (17) [4]:

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The proof of the above theorem can be carried out analogously

for to obtain as a dual space, say . In thedefinition of we use in the place of in ,where is given by

Remark (18) [4]: In the definition of , we can equivalently replace

by and by . We now show that is actually

contained in . Proposition (19) [4]:

is imbedded continuously in .Proof:

Let , be any element in .Consider . Since

for ,

is valid since in . Notice that for

Hence . Also

Since is any representation of ,

Since this is true for every , we get immediately

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Also, this is valid for every representation of . Hence

Note thatby the above proposition we obtain . Asseen already in Theorem (16) and Proposition (13), is

isomorphic to and is isomorphic to .Hence we arrive at the result of Proposition (3), for , and

, .Remark (20)[4]:We notice that the space of Definition (15) is actually themodule tensor product . Hence we have identified the

multiplier space as the dual space of tensor product.We shall be able to arrive at this result if we use the“Multipliers of Segal algebras, if we show that the Segalalgebra satisfies a certain property defined in the“Multipliers of Segal algebra”. By means of the result that

is the dual space of [see Proposition (13)] wecan show that does satisfy this property. We define a new Segal algebra on and show thatits multiplier space can be identified with the space of integrable distributions of order with also oforder , . This gives us an example of aSegalalgebra on whose multiplier space can be precisely

characterized and is larger than the space of boundedmeasures.As already observed (Proposition (5)), is a convolutionalgebra of distribution which can be thought of as a

generalization of . Hence is a “natural” spacemultipliers for a Segal algebra on .Definition (21)[4]:

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Let be the fundamental solution of the differentialoperator given by the Bessel Potential

for . It can be seen that belongs to . Further, lies in the space of rapidly decreasing distributions,defined in “Theory of Distribution by Schwartz. We define

Then is a Banach space under the norm

Where and the infimum is taken over all possiblerepresentations of in . Note that . In fact , then

Also the following proposition can be proved easily.Proposition (22) [4]:The space is a Segal algebra under the norm . Analogous to, we can define another space.

Definition (23) [4]:

Let . Then is aBanach space under the norm

where, as before, and the infimum is taken over allpossible representations of in . We now proceed to the multiplier results.Proposition (24) [4][2]:

The space of multipliers from into

is isometrically isomorphic to the space .Proof:

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Using the facts that is a dual space and is the

continuously translating subspace of , we can prove theproposition .Proposition (25) [4]:The space of multipliers from into the Segal algebra isisomorphic to the space .Proof: Observe that the differential operator is a

continuous bijection from onto which takes onto

. Hence we can identify with and with so that the proof is immediate, on account of Proposition (24)above.Proposition (26) [4]: The space of multipliers on the Segal algebra isisomorphic to the space .Proof:

Let, first, . As is identified with , foreach , we write

where . Define

As in the last part of the proof of Theorem (2) we see that is independent of the representation of and defines

a multiplier of into itself with for some . Conversely, let . Note that is given by apseudo measure , since is a Segal algebra, such that

Let and be the operators of convolution with and . Then and are multipliers from into whichare clearly given by convolution with and .

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Since , by Proposition (25) we have and

within : Applying the operator to theserelations and using we obtain that

so that . The equivalence of norms followsby the closed graph theorem.

References:[1] D. Ornstein, A non-inequality for differentialoperators in the -norm, Arch. Rational Mech.Anal. 11(1962),40-49.[2] E.M.Stein and G. Weiss, Introduction to FourierAnalysis on Euclidean Spaces, Princeton Univ.Press,Princeton,N.J.,197.[3] Shawgy Hussein and Adam M.Ali: Plancherel –type andOperator-Valued Fourier Multipliers of and Sobolev Spaces,Ph.D. Thesis, Sudan University of Science and Technology (2011).[4] S.Poornima, Multipliers of Sobolev spaces,J.Funct .Anal. 45(1982),1-28.semigroups. Proc. A. M. S.125 (197) No.3, 635-647.

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