# Monthly Archives: January 2018

## Strichartz estimates for Schrodinger equation

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

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,

한 학기동안 저의 절대적인 채점기준입니다. 이를 지키지 않을 경우 불이익은 학생에게 있습니다.

1. 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.
2. 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.
3. Obvious, Clear” is NOT FOR YOU / 왜 자명한지에 대한 간단한 설명조차 없는 경우 감점합니다.
4. 반례를 제시할 때, 왜 반례가 되는 지 논증이 없으면 0점  / 주장은 누구나 할 수 있습니다. 그에 대한 논증이 없으면 아무런 소용이 없습니다.
5. 고등학교 교육과정에서 배우지 않은 부등식을 사용할 경우(교과서에서, 참고서 아님) 반드시 증명을 제시할 것. 교과서 본문에 없는 정리를 사용할 경우 원칙적으로 점수를 부여하지 않음.
6. 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.
7. 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.

1. 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.
2. 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.
3. ‘Obvious, clear’ is not for you. You must explain shortly why it is easy.
4. When you give a counterexample, you must justify it. Otherwise, I give the zero point.
5. 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.
6. You must complete the proof no matter the proof is long and complicated.
7. 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.

## Existence and uniqueness of stationary Navier-Stokes equation

Let be a bounded domain in or . We consider the Dirichlet boundary value problem for the stationary Navier-Stokes equation:

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

So

(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.

## Leray-Schauder fixed point theorem

First, we derive the Schauder fixed point theorem.

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 .

## Continuous, completely continuous and compact mapping

### 1. Weak convergence and weakly compact

• 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

1. If strongly in , then weakly in .
2. If weakly in , then is bounded and .
3. 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 .

### 2. Compact operators and completely continuous

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

1.  if is continuous, then is weakly continuous.
2.  if is compact, then it is completely continuous.
3.  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.