The post Haar systems on real lines and adapted to an accretive function appeared first on Will Kwon.

]]>One of the fundamental problem is the solvability of the Dirichlet problem for the Poisson equation in a bounded domain(open and connected)

in :

Many authors have been studied the solvability of the problem in a various settings. One possible way to solve the problem is to use the method of layer potential. In this talk, we assume for simplicity. The fundamental solution for the Laplacian is defined by

where is the volume of the unit ball in . Suppose first that has -boundary. For , we define the *double layer potential* of by

where denotes the outward unit normal to the boundary . Then it is easy to check that is well-defined on and is harmonic in . Also, can be extended as a continuous function from to up to the boundary. Set for , . Since is of , one can easy to show that

(1)

for some constant .

For , we define

Due to (1), the above one is well-defined. We have the jump relation formula for the double layer potential:

for fixed , we have

Also, it can be shown that is compact, (and also , is compact). Hence by the Riesz-Schauder theory, the operator is invertible if and only if it is injective. Using some argument, one can show that it is injective

and so the Dirichlet problem has a solution. Uniqueness follows from the maximum principle. One can also obtain the solvability of the Dirichlet problem when the boundary data is in . We interpret the boundary condition via nontangential limits. We also note that the above argument holds when is a -domain.

The difficulty in extending these results to the case of and Lipschitz domains is that, in these cases, we have

So even the -boundedness, much less the compactness of is far from obvious. Another difficulty is that the operator we considered is of non-convolution type operators. So we cannot use the Fourier transform method to guarantee the -boundedness of the operator.

In Calder\’on’s remarkable paper, he proved that the Cauchy integral operator along a Lipschitz curve is bounded from to itself when the Lipschitz constant is small. This leads to the solvability of the Dirichlet problem for the Poisson equation in -domain due to Fabes-Rievere-Joedit. Later, Coifman-McIntosh-Meyer resolved the restrictive condition of Caderon’s result.

The first proof of theorem of Coifman-McIntosh-Meyer uses some inductive argument from Calder\’on’s result. One famous proof is to use the variant of -theorem, the -theorem due to David-Journ\’e-Semmes. But we do not pursuit in this direction. In 1989, Coifman-Jones-Semmes provide another proof using the theory of complex variables and some variant of the Littlewood-Paley theory. We also note that there is a geometric proof of theorem of Coifman-McIntosh-Meyer given by Melnikov and Verdera. Recently, Muscalu gives a new proof of Coifman-McIntosh-Meyer theorem whose methods can be extended to the case of multilinear operators.

In Section 1, we construct Haar system on the real line . In Section 2, we construct Haar system associated to an accretive function, which will be defined in later. This leads to the proof of Coifman-McIntosh-Meyer theorem. For those who are interested in, see the paper of Coifman-Jones-Semmes.

In this section, we construct the Haar system on the real lines. Recall that a dyadic interval in is an interval of the form

where and are integers. For we denote by the set of all dyadic intervals in whose side length is . We also denote by the set of all dyadic intervals in Then we have

Moreover, the -algebra of measurable subset of formed by countable unions and

complements of elements of is increasing as increases, i.e., is a filtration. For simplicity, we write . Observe that any two dyadic intervals of the same side length either coincide or their interior are disjoint. Moreover, either two given dyadic interval contains the other or their interiors are disjoint.

Given a locally integrable function on , we let

denote the* average* of over an interval . The* conditional expectation* of a locally integrable function on with respect to the increasing family of -algebras generated by is defined as

We also define the *dyadic martingale difference operator* as follows:

also for .

*Remark (Justification of the terminology). *First, we justify the terminology “conditional expectation”. We first show that for any , we have

Since , for any , either or . So

The identity holds for all . Hence the identity holds for all . This shows that

which justify the terminology “conditional expectation”.

For a dyadic interval and write , where and are the left and right halves of . The function

is called the *Haar function associated with the interval* . By construction, the Haar functions have norm equal to . Moreover, if , either or not. Clearly, if , then

(2)

If , we may assume . Then either is contained in the left or in the right half of , on either of which is contant. Hence (2) follows. We introduce the notation

for . Then

The following proposition shows that the dyadic difference operator is a projection to the space which is spanned by Haar functions.

Proposition 1.Let . For all , we have

and also

Proof. Observe that every interval in is either an or an for some unique . Thus, we write

Also,

So

which is easily checked to be equal to

This coplemtes the first part. Now from the orthogonality of , we get

This completes the proof of Proposition 1.

Using the dyadic difference operator, we can decompose a function in as follows:

Theorem 2.Every function can be written as(3)

where the series converges almost everywhere and in . We also have

(4)

Proof. It follows from the Lebesgue differentitation theorem that there exists a set of measure zero on such that for all , we have

whenever is a sequence of decreasing intervals such that . Given in , there exists a unique sequence of dyadic intervals such that . Then for all , we have

Hence a.e. as . Since

where denotes the dyadic maximal function, we have that . Hence due to the -boundedness of the dyadic maximal operator, it follows from the dominated convergence theorem that in . For a given and as before, we have

which tends to zero as , since the side length of each is . Since , by the dominated convergence theorem, we conclude that in as . Observe that

as and a.e. and in . By Proposition ??,

This proves identity (3). To prove (4), we rewrite

For , we have

Since the last integral is nonzero only when . If this is the case, then for some dyadic interval . Then the function is supported in the interval and the function is constant on any dyadic subinterval of . Then

since . Hence, whenever . This shows that

Now the identity (4) follows from Proposition 1. This completes the proof of Theorem 2.

The purpose of this section is to extend Proposition 1 and Theorem 2. This extension leads to a proof of the theorem of Coifman-McIntosh-Meyer. We introduce the following notion.

Definition.A bounded complex-valued function on is said to beaccretiveif there is a constant such that for almost all .

Let be an accretive function. For a measurable set in with , we define

Observe that

since is accretive. For each , we define

with a fixed choice of the square roots. If , then . For , we introduce a pseudo-inner product

on . By definition, each is supported on and is constant on and . Moreover,

The following theorem is a generalization of Theorem 2.

Theorem 3.Let . Then

where the sum converges in . Moreover,

for some constant .

Proof. We define

and

Then we first show that

Since is accretive, for any . Since , it follows from the Lebesgue differentiation theorem that

where is the unique dyadic interval such that for all and .

Since is accretive, we have

where with . Hence by the dominated convergence theorem, we get in as . Following the exactly same argument as in the proof of Theorem 2, we can also show that a.e. and in as . Also,

in and almost everywhere. Following the exact same argument in the proof of Proposition ??, we can show that

So the first conclusion of the lemma is verified.

Expanding out , we see that

since is accretive. So

Since , it follows from Theorem 1 that

It remains to show that

Recall that for some constant .

For , it follows from Theorem 1 that

Hence for any set , we have

(5)

(6)

Since

by (5),we have

Since

we get

This completes the first part of the proof. For the converse, let and set , .

Then

Since

Since is accretive, we get the desired result. This completes the proof.

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Lemma 1.Let and be two open sets satisfying

- for all , there is a ball such that and

- for all such that

we have .

Then for some constant depending only on the dimension.

*Proof.* Let . Then by (i), we have a ball such that and

Since is open, choose a maximal ball such that , and . If , then by (i).

If , there is a ball such that . By maximality of , . Then

from the contrapositive of second condition. Now choose a decreasing sequence of balls converging to . Then by continuity of measure, we have

So we construct a cover of the set so that for all ,

- ;
- ;
- .

Now we apply the Vitali covering lemma to get a countable subcollection of balls such that . Here denotes the cocentric ball with radious 5 times of the radius of ball . Since and , we have

Thus,

Hence,

and so

This completes the proof of Lemma 1.

The crawling ink spots lemma helps us to get -estimates. The application of crawling ink spots lemma can be found at e.g. Dong and Kim (arXiv:1806.02635v1).

Fix . Suppose that . For , define

and

Let and . Suppose that there exists a constant such that the following hold: and , if

then we have

Then by Lemma 1, we see that

Since

we have

So we obtain the following estimates:

for some constant .

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]]>

We denote by which is a collection of all open balls in . Let be an open subset of . By , we denote the homogeneous Sobolev space defined as the completion of the complex-valued functions in the seminorm . The dual space is .

The space `bounded mean oscillation’ is introduced by John and Nirenberg [2]. A locally integrable function is in if

One research direction on PDEs is to study the global well-posedness(GWP) for small data and local well-posedness(LWP) for large data . Note that the natural scaling of the Navier-Stokes equation is

(1)

So it is natural to find a function space which is scaling invariant in this scaling. As an example, and are scaling invariant in this scaling. Kato [3] proved GWP for small initial data and LWP for initial data of the Navier-Stokes equation in . Similar result holds for . This result is proved by Cannone [1] and Planchon [6]. In [4], Koch and Tataru considered this problem, which is the largest spaces for the well-posedness. For , note that

is scaling-invariant with respect to \eqref{eq:NS-scaling}. Motivated by this, to gurantee the well-posedness, we require satisfying

This is related to the Calerson measure characterization of the space .

See Stein [7].

Motivated by this, we define the norm by

The following theorem is proved in [4,Theorem 1].

Theorem 1.Let be a tempered distribution. Then if and only if there exist with .

In this note, we give an another characterization of which was proved by Maz’ya and Verbitsky [5]. In that paper, they proved the following the Helmholtz decomposition:

Theorem 2.Let . Suppose that there exists a constant such that

(2)

for all cube in . Then we have

where

The condition \eqref{eq:BMO-inverse-characterization} is motivated from the form boundedness of second-order elliptic operators. See [5] for more details. Now we state the main theorem of this note.

Theorem 3.Let be a tempered distribution in , . Then if and only if there exists a constant such that \eqref{eq:BMO-inverse-characterization} holds for all cube in .

Proof. By Theorem 2, we have

where and . Set . So and hence by Theorem 1.

Conversely, suppose that , where . Then for every supported on a cube , we have

for some constant which does not depend on and . Hence by – duality, we have

Here constants does not depend on . This completes the proof.

*Remark.* There is a Littlewood-Paley characterization of .

**References**

- Marco Cannone, A generalization of a theorem by Kato on Navier-Stokes

equations, Rev. Mat. Iberoamericana 13 (1997), no. 3, 515–541. MR 1617394 - F. John and L. Nirenberg, On functions of bounded mean oscillation, Comm. Pure Appl. Math. 14 (1961), 415–426. MR 0131498
- Tosio Kato, Strong L p -solutions of the Navier-Stokes equation in R m , with applications to weak solutions, Math. Z. 187 (1984), no. 4, 471–480. MR

760047 - Herbert Koch and Daniel Tataru, Well-posedness for the Navier-Stokes equations, Adv. Math. 157 (2001), no. 1, 22–35. MR 1808843
- V. G. Maz 0 ya and I. E. Verbitsky, Form boundedness of the general second-order differential operator, Comm. Pure Appl. Math. 59 (2006), no. 9, 1286–1329. MR 2237288
- F. Planchon, Global strong solutions in Sobolev or Lebesgue spaces to the

incompressible Navier-Stokes equations in R 3 , Ann. Inst. H. Poincar´ e Anal. Non Lin´eaire 13 (1996), no. 3, 319–336. MR 1395675 - Elias M. Stein, Harmonic analysis: real-variable methods, orthogonality, and oscillatory integrals, Princeton Mathematical Series, vol. 43, Princeton University Press, Princeton, NJ, 1993, With the assistance of Timothy S. Murphy, Monographs in Harmonic Analysis, III. MR 1232192

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Lemma 1.Let and be three Banach spaces with

Suppose is reflexive. Then for each , there is a constant such that

*Proof.* If the statement were not true, then there exists a number

and a sequence in such that

Note for each . So we may assume . By Banach-Alalogu theorem, there exists a subsequence of and we still denote it by the same indices. Since , we have in for some . Moreover, since

, it follows that in

. But

for all . Hence, letting , we obtain

which is a contradiction. This completes the proof.

Theorem (Aubin-Lions).Let and be three Banach spaces with

Suppose that are reflexive. Then for and , we have

*Proof.* Since and are reflexive, so are and . It suffices to show that if weakly both in and , then strongly in .

Observe that it suffices to show that strongly in . Indeed, if strongly in , then by Lemma 1, for each , we have

Since weakly in , is bounded in . Thus,

Since was arbitrary chosen, strongly in .

Suppose thus that weakly both in and . Since weakly in , is bounded in . Set

(1)

Since , is bounded in . Hence, to prove that

we will show that

Then the conclusion follows from the dominated convergence theorem.

Let be fixed. Since is continuous on , it is uniformly continuous and thus it is absolutely continuous. So for all ,

Integrating this over , we also have

and so integration by part gives

for all . By (1) and Hölder’s inequality, we obtain

Hence given , we can choose such that

For each , define

Then note that weakly in . Indeed, let . Then we have

Since weakly in and can be regarded as a function in , the right hand side converges to .

Thus, converges to 0 strongly in . Hence,

This completes the proof.

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]]>**Theorem 1** (Strichartz esitmates for Schrödinger)**.** Fix and and call a pair of exponents admissible if , and . Then for any admissible exponents and , we have the homogeneous Strichartz estimate

(1)

the dual homogeneous Strichartz estimate

(2)

and the inhomogeneous (or retarded) Strichartz estimate

(3)

We illustrate some application of Strichartz estimate. If is the solution to an inhomogeneous Schrödinger equation

where , is given by Duhamel’s formula

on some time interval containing . Apply to both sides and using Strichartz’s estimate, we have

for any admissible and .

Now we begin the proof of Theorem 1 when admissible pairs are non-endpoint.

*Proof of Theorem 1.* For , by Minkowski’s integral inequality, we have

For any Schwarz function in spacetime,

Then by \eqref{eq:dispersive-estimiate-Schrodinger}, we have

Since and satisfying we have .

Recall the Hardy-Littlewood-Sobolev inequality:

whenever

So by the Hardy-Littlewood-Sobolev inequalty, we have

Now by Hölder’s inequality, we have

On the left-hand side,

Thus,

This proves the second part (2).

Now we show the first part (1) by duality argument. Note

The last part comes from (2). Hence by duality, we get

For the last part, we use some abstract lemma: the Chirst-Kiselev lemma.

Lemma(Christ-Kiselev).Let be Banach spaces, let be a time interval, and let be a kernel taking values in the space of bounded operators from to . Suppose that satisfying

for all and some . Then we have

By (1)

Hence by Christ-Kiselev’s lemma,

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]]>**채점의 기준은 교과서입니다.**/ 증명이 안 좋은 것도 사실이고, 책의 증명이 틀린 부분도 있어서, 그 점은 제가 감안해서 반영합니다. 다만 제가 교수님이 뭘 세부적으로 수업을 했는지 알 수도 없는 상황이고, 그렇다고 제가 수업에 들어가서 그 수업을 듣는다는 것은 제 입장에서도 많은 시간낭비입니다.**문장으로 작성하십시오****. 각 문장 하나하나가 논리적 연결이 이루어지도록 답안을 작성하는 것을 원칙으로.****/**수학적 증명의 글 형식은 본질적으로 논설문이며, 여러분의 ‘답안지’가 저를 ‘설득’할 수 있어야 합니다. 해당 명제가 참 또는 거짓이라는 것을 ‘논리적’으로 입증해야 합니다. 자기만 알아들을 수 있는 랭귀지로 글을 쓰면 안됩니다. 속칭 화살표 증명을 할 경우, 제가 오해를 할 가능성이 매우 높다는 것을 알아주십시오.- “
**Obvious, Clear” is NOT FOR YOU**/ 왜 자명한지에 대한 간단한 설명조차 없는 경우 감점합니다. **반례를 제시할 때, 왜 반례가 되는 지 논증이 없으면 0점**/ 주장은 누구나 할 수 있습니다. 그에 대한 논증이 없으면 아무런 소용이 없습니다.- 고등학교 교육과정에서 배우지 않은 부등식을 사용할 경우(교과서에서, 참고서 아님) 반드시 증명을 제시할 것. 교과서 본문에 없는 정리를 사용할 경우 원칙적으로 점수를 부여하지 않음.
- 수식이나 과정이 길어지더라도, 절대로 ‘이러한 방법으로 구할 수 있다’ 라고 끝맺지 않습니다. 이러한 태도는 ‘나는 할 수 있는데 귀찮아서 안 했다’와 같으며, 이는 그를 못한 것에 대한 핑계에 불과합니다.
- 과제를 늦게 제출 할 경우,
**미리 메일로 제출전 저에게 공지를 해야 하며**, 1일 늦을 경우 원 점수의 90%, 2일 늦을 경우 70%, 3일 늦을 경우 25%, 4일 늦을 경우 0%를 획득합니다.**메일로 공지하지 않을 경우에도 100% 감점합니다.**

This is my grading policy. I will strictly follow this. If you don’t follow this rule, you may have a disadvantage for this.

**Grading is based on textbook**/ I know the proof is not good and there is some mistake in the proof in our textbook. I can not attend the class because it costs my time to take that class for me. I don’t have enough time to do that.**When you write a proof, write it in a sentence.**/ You must persuade me that your proof is right in a logical way. Do not write arrow and uncommon abbreviations.**‘Obvious, clear’ is not for you.**You must explain shortly why it is easy.**When you give a counterexample, you must justify it.****Otherwise, I give the zero point.****If you use some theorem which is not listed in our textbook, I cannot give any point if you don’t prove. Also, if you use some inequality which is not studied in calculus or high school, you must prove it.****You must complete the proof no matter the proof is long and complicated.****You must inform me via e-mail when you submit your homework late. Late homework makes you get a penalty.**You will get a ‘discounted’ score based on the following rules. 1 day lates (90%), 2 days lates (70%), 3 days lates (25%), 4 days lates (0%)**If there is no e-mail from you when you want to submit your homework, not on time, there will be no point at all.**

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where is a viscosity constant. We call this problem as (NS).

Assume . Now we consider a weak formulation of (NS).

A function is called a *weak solution* of (NS) if in and satisfies

for any .

The definition is well established. Take inner product to the first equation in (NS) with and integrate it by parts. For the first part,

(1)

For the second part,

For the thrid part, it vanishes since . Also, for , for , by the Sobolev embedding theorem and the boundedness of , . Hence \eqref{eq:weak-sol-st-NS} makes sense.

From now, we assume . In 1933, J. Leray proved the existence of weak solution of (NS). To present the theorem of Leray, we define some terminology and the fundamental fixed point theorem proved by Leray and Schauder.

Theorem(Leray-Schauder’s fixed point theorem).Let be a Banach space. Suppose that the compact operator satisfies the following: there exists a constant such that if is a solution of , and

then there exists in satisfying .

If is reflexive and is completely continuous, then is compact. So in this case, we can apply the Leray-Schauder principle for such operator. See this article for proof.

Theorem.Let be a bounded Lipschitz domain. Then there exists at least one weak solution of (NS).

*Proof.* Since is a bounded Lipschitz domain, due to Poincar\’e’s inequality, one can check that is an inner product on .

Also, for any , . Indeed, for any ,

So .

Now for each by the existence result on Stokes equation, there exists a unique weak solution satisfying

(2)

for all . By Lemma ??, the above identity holds for any .

Since , by the Riesz representation theorem, there exists a unique in such that

for all . By the uniqueness, the operator is well-defined. Similarly, there exists a unique such that

for all with . So \eqref{eq:perturb-Stokes} can be written as

i.e., in operator form. Note that the existence of weak solution of (NS) is equivalent to the existence of a fixed point of the above operator equation. If there exists such that , then . So becomes a weak solution of (NS).

To use the Leray-Schauder fixed point theorem, we need to show that the operator defined by is a compact operator and a priori estimate.

We show that is completely continuous on . Let in . Then there exists a constant such that for all . For simplicity, let . Since is bounded Lipschitz domain, the Rellich-Kondrashov theorem shows that is compactly embedded in . Thus, strongly in . Now

So as , in . Note

Put . Then we get

So

as . This shows that is completely continuous.

Now we left to show a priori estimate.

First, note that for any ,

The last identity holds because of divergence free condition. By density, for any .

Now suppose satisfies , . So

Here we used

(3)

Here the constant does not depend on . Therefore, by the Leray-Schauder fixed point theorem, has a fixed point . This is the desired weak solution of (NS).

Now we prove the uniqueness.

Theorem.Let be a bounded Lipschitz domain and suppose that

where is a constant in the inequality

(4)

for any . Then (NS) has a unique weak solution.

*Remark. * The inequality (4) holds since is bounded and the Sobolev embedding theorem.

*Proof. * Suppose and are two different solutions to (NS). Then

Apply (3) and (4)}. Then we get

Now by assumption, we get

which is a contradiction. Therefore, the solution of (NS) is unique.

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Theorem 1(Schauder fixed point theorem).Let be a compact convex set in a Banach space and let be a continuous mapping of into itself. Then has a fixed point, that is, for some .

*Proof.* Fix . Since is compact, has a finite subcover which covers . Write .

Let . Define by

The map is well-defined. Since is continuous, is continuous. Also,

Note that for such with , . Thus, for such , we have . Hence . Now is continuous, is compact and convex set. So is homeomorphic to . Hence by the Brouwer fixed point theorem, for some .

Now since and is compact, there exists a convergent subsequence which converges to in . We claim that is a fixed point. Note

as . Hence .

Using this theorem, we obtain another version of Schauder fixed point theorem.

Theorem 2.Let be a bounded and closed convex set in a Banach space and let be a compact mapping of into itself. Then has a fixed point.

*Proof.* Let be the closed convex hull of Then is convex and since the closed convex hull of a compact set is itself compact, is compact. Note that because and is closed. So . Since is convex and is the convex hull of , .

Thus, by the Schauder fixed point theorem, has a fixed point in . So we are done.

Now we prove the Leray-Schauder fixed point theorem. Here we consider the simplest form.

Theorem 3(Leray-Schauder).Let be a Banach space and let be a compact mapping of into itself. Also, suppose there exists a constant such that

for all and satisfying . Then has a fixed point.

*Proof.* We may assume . Define by

Let . Then maps to . We show that

is compact. First, is continuous. For those with , we

have

From the continuity of , is continuous on .

Next, let be a sequence in the ball . There are two cases:

- there exists a subsequence satisfying for all ;
- there exists a subsequence satisfying for all .

For the first case, since and is compact, there exists a subsequence such that converges strongly in . So converges strongly in .

For the second case, choose a subsequence so that and as by Bolzano-Weierstrass theorem on real numbers and compactness of . Hence

Thus, is compact.

Hence by Theorem 2, has a fixed point . We claim that is also a fixed point of . Suppose with . Then

and so , which contradicts the assumption. Hence and consequently .

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]]>We introduce some notations used in this article.

- As usual, stands the Euclidean space of points . For , multi-index , and a function we set
denote the space of smooth functions with compact support in .

- For , denote the space of all real-valued Lebesgue measurable functions so that

Let be a normed linear space and denote the dual space of . For and , we write instead of . A sequence in converges *weakly* to in if for any . We write in for simplicity.

Let us give some basic examples on this concept.

Example 1.When , note that where . Then a sequence in converges weakly to if for any . By the Riesz representation theorem on spaces, there exists such that

So weakly in if and only if

for any .

Proposition 2.If is compact Hausdorff space and is bounded in . Then for each

if and only if weakly in .

The proof of this fact uses the following Riesz representation theorem on . For the proof, see Rudin, RCA for example.

Theorem 3(Riesz representation theorem).Let be a compact Hausdorff space and let be the set of signed Radon measures on such that

is finite. Define a norm on by setting

Then the dual space of is . So

for any bounded linear functional on , there

exists a such that

*Proof of Proposition 2.* Suppose . Then for any bounded linear functional on , there exists a signed Radon measure with such that

for all . Then

Since is bounded in and , by the dominated convergence theorem, we conclude that

This shows converges weakly to in .

Conversely, suppose converges weakly to in . Then for each , one can easily check that is a Radon measure on . So

This shows for each

.

>We list one of the well-known properties of weak convergence.

Proposition 4.Let be a sequence in . Then

- If strongly in , then weakly in .
- If weakly in , then is bounded and .
- If weakly and if strongly in , then .

*Proof.* (i) Let be a bounded linear functional on . Then we have

Letting , we see that . So converges weakly to .

(ii) We prove this by using the Banach-Steinhaus theorem. For each , define by

and by

Then

So

Since is a Banach space, by the Banach-Steinhaus theorem,

Since

this shows that .

Note that there exists such that and . So from this, we have

Now taking liminf to above inequality to get

(iii) This follows from

From (ii), is bounded. So letting , we get the desired result.

Let be a sequence in . We say

converges to weakly-star (or weakly{*}) if

for any . One can easily deduce similar results like Proposition

??.

In arbitrary normed linear space, usually the unit ball is not compact in norm topology. Also, it may not be compact in weak topology. However, in weak star topology, the unit ball is always compact. This is due to Banach and Alalogu.

Theorem 5.Every bounded sequence in has a weakly-star convergent sequence.

*Proof. *For simplicity, we assume is separable. Then there exists a countable dense subset of . We may assume that satisfies for all . Now consider

Then for the first line, by the Bolzano-Weierstrass theorem, there exists a convergent subsequence of . For such , there exists a convergent subsequence of of . Continuing this process, then we see that

exists. Now define

Then exists for all . Since is bounded in , is a Cauchy sequence for each in . So from this, is Cauchy for all in . Hence,

exists. Then is linear and

and so

So is a bounded linear functional and . So weakly star in .

For the general case, one can find the proof, e.g., Rudin or Yosida. We omit the proof of general case.

We say is \emph{reflexive }if for every , there

is such that

for all . When is reflexive, we identify and

.

Theorem 6.If is reflexive, then every bounded sequence in has a weakly convergent sequence.

*Proof.* Let be a bounded sequence in . Since is reflexive, is a bounded linear functional on and for all . Hence by Theorem 5, there exists a convergent subsequence of in which converges to . So for any ,

i.e.,

This shows converges weakly to in .

Recall that a mapping is said to be *continuous* if in for any sequence in . A mapping is *weakly continuous* if weakly in whenever weakly in . A mapping is said to be *completely continuous* if strongly in whenever weakly in . Finally, a mapping is *compact* if is continuous and for every bounded sequence in , there exists a subsequence such that converges strongly in .

*Remark.* Let be a bounded set in and be linear. Suppose is compact in . Then is bounded. When is linear, to check the compactness, it suffices to check is compact in whenever is a bounded set in .

By definition and Proposition 4, one can check that completely continuity implies continuity. Also, it implies weak continuity.

Proposition 7.If is reflexive and is completely continuous, then is compact.

* Proof. * Let be a bounded sequence in . Then by Theorem ??, there exists a subsequence for which converges weakly to in . Since is completely continuous, converges strongly to . Hence is compact.

Example 8.Define by . Then is compact. Let be a bounded sequence in . Then is bounded. So by the Bolzano-Weierstrass theorem, there exists a subsequence

for which converges. So is compact. Define . Then forms an orthonormal basis for . Then from Parseval’s inequality, we see that for any . So converges weakly to . But

So is not weakly continuous. Hence is not completely continuous.

Example 9.The identity map

is completely continuous but not compact. Note that

is non-reflexive.

Theorem 10.Let be linear. Then

- if is continuous, then is weakly continuous.
- if is compact, then it is completely continuous.
- if are Banach spaces, then is weakly continuous if and only if is continuous.

Proof. (i) Let and let be a bounded linear functional on . Then is

linear and

So is a bounded linear functional on . Hence

i.e.,

So weakly in .

(ii) Let weakly in . Then by Proposition 4, is bounded in . Since is compact, there is a subsequence

for which converges strongly to in . Then by (i), and so weakly in . It remains to show .

Note that weakly in . So for any bounded linear functional on ,

So for any . Now by the consequence of Hahn-Banach theorem, there exists a bounded linear functional such that

So in . Hence, strongly in . Now we are left to show that strongly in Suppose not. Then there exists and a subsequence such that

(1)

Note that is a bounded sequence. So using the compactness of again, there exists a subsequence such that in . This contradicts (1). Hence, strongly in .

(iii) Suppose is weakly continuous. Consider

Then is weakly closed subspace of . Hence is strongly closed in . Thus, is continuous by the closed graph theorem.

A normed linear space is *continuously embedded* in if and there exists a constant such that for all . We write . We say is *compactly embedded* in if is continuously embedded in and if is bounded in , then there exists a subsequence which converges strongly in .

Example 11.By the Sobolev embedding theorem, the inclusion map , is linear and continuous. But the embedding is not compact. Indeed, take

Then for every and is bounded in .

Note

and similarly,

But

which cannot be bounded. So it cannot have a convergent subsequence in . Also, is linear and continuous but not compact. Hence it is not completely continuous.

Corollary 12.Let . If is compactly embedded in and converges weakly to in , then converges strongly to in . In addition, if is reflexive, then the converse holds.

*Proof. *Since is compactly embedded in , by (i) is completely continuous. Then the corollary is proved by the definition of completely continuous. If is reflexive, the inclusion map is compact by Proposition 7.

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]]>Lemma. Let . Then on ,

*Proof.* Set . Then we have

where , are certain constants, and

Take an and a such that in and outside and set

where is obtained in

Note that and are well-deifned since have compact supports and . Recalling some facts on Newtonian potential, we have

Note also that are infinitely differentiable and

Furthermore

So for any , we have

We estimate the integral oscillation of . Since in , we see that if , then

and

Note that

and for , , we have .

Therefore,

Note that

where is the surface measure on .

We write in polar coordinate. Then by using previous identity

By using mean value theorem, we obtain

Hence

So

Note that this estmiate allows shifting the origin. For this reason for any and any ball such that , we have

By taking the supremum of the left-hand side over all balls containing , we obtain the desired result.

The following result can be obtained by Calderon-Zygmund theory of singular integral by using the -boundedness of Riesz transform. However, by using Fefferman-Stein theorem, we can obtain it elegantly without using kernel method.

Theorem. Let , and . Then

We first prove the case . For the case , it can be obtained by using Fourier transform. For , by the Fefferman-Stein theorem and previous lemma and the -boundedness of Hardy-Littlewood maximal function, we obtain

For , we use duality argument. Assume that . Then integration by parts and Holder’s inequality give

Since is dense in ,

by density, we conclude that for any ,

we have

So by taking supremum to with , we obtain

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