Monthly Archives: January 2018

Strichartz estimates for Schrodinger equation

Theorem 1 (Strichartz esitmates for Schrödinger). Fix n\ge1 and \hbar=m=1 and call a pair \left(q,r\right) of exponents admissible if 2\le q,r\le\infty, \frac{2}{q}+\frac{n}{r}=\frac{n}{2} and \left(q,r,n\right)\neq\left(2,\infty,2\right). Then for any admissible exponents \left(q,r\right) and \left(\tilde{q},\tilde{r}\right), we have the homogeneous Strichartz estimate

(1)   \begin{equation*} \Norm{e^{it\triangle/2}u_{0}}_{\Leb q_{t}\Leb r_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)}\apprle_{n,q,r}\Norm{u_{0}}_{\Leb 2_{x}\left(\mathbb{R}^{n}\right)} \end{equation*}

the dual homogeneous Strichartz estimate

(2)   \begin{equation*} \Norm{\int_{\mathbb{R}}e^{-is\triangle/2}F\left(s\right)ds}_{\Leb 2_{x}\left(\mathbb{R}^{n}\right)}\apprle_{n,\tilde{q},\tilde{r}}\Norm F_{\Leb{\tilde{q}^{\prime}}_{t}\Leb{\tilde{r}^{\prime}}_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)} \end{equation*}

and the inhomogeneous (or retarded) Strichartz estimate

(3)   \begin{equation*} \Norm{\int_{t'<t}e^{i\left(t-t'\right)\triangle/2}F\left(t'\right)ds}_{\Leb q_{t}\Leb r_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)}\apprle_{n,q,r,\tilde{q},\tilde{r}}\Norm F_{\Leb{\tilde{q}^{\prime}}_{t}\Leb{\tilde{r}^{\prime}}_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)}. \end{equation*}

We illustrate some application of Strichartz estimate. If u:I\times\mathbb{R}^{n}\rightarrow\mathbb{C} is the solution to an inhomogeneous Schrödinger equation

    \begin{align*} i\partial_{t}u+\frac{1}{2}\triangle u & =F, \\ u\left(0,x\right) & =u_{0}\left(x\right), \end{align*}

where u_{0}\in\tSob s\left(\mathbb{R}^{n}\right), is given by Duhamel’s formula

    \[ u\left(t\right)=e^{\left(t-t_{0}\right)\frac{\triangle}{2}}u_{0}\left(t_{0}\right)+\int_{t_{0}}^{t}e^{i\left(t-s\right)\frac{\triangle}{2}}F\left(s\right)ds \]

on some time interval I containing 0. Apply \left\langle \nabla^{s}\right\rangle to both sides and using Strichartz’s estimate, we have

    \[ \Norm u_{\Leb q_{t}\Sob sr_{x}\left(I\times\mathbb{R}^{n}\right)}\apprle_{n,q,\tilde{q},\tilde{r},s}\Norm{u_{0}}_{\tSob s_{x}\left(\mathbb{R}^{n}\right)}+\Norm F_{\Leb{\tilde{q}^{\prime}}_{t}\Sob s{\tilde{r}^{\prime}}_{x}\left(I\times\mathbb{R}^{n}\right)} \]

for any admissible \left(q,r\right) and \left(\tilde{q},\tilde{r}\right).

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

Proof of Theorem 1. For G:\mathbb{R}\times\mathbb{R}\times\mathbb{R}^{n}\rightarrow\mathbb{C}, by Minkowski’s integral inequality, we have

    \begin{align*} \Norm{\int_{\mathbb{R}}G\left(t,s,x\right)ds}_{\Leb q_{t}\Leb r_{x}}^{q} & =\int_{\mathbb{R}}\left[\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}}G\left(t,s,x\right)ds\right)^{r}dx\right]^{\frac{q}{r}}dt\\ & \le\int_{\mathbb{R}}\left[\int_{\mathbb{R}}\Norm{G\left(t,s,\cdot\right)}_{\Leb r_{x}}ds\right]^{q}dt\\ & =\Norm{\int_{\mathbb{R}}\Norm{G\left(t,s,\cdot\right)}_{\Leb r_{x}}ds}_{\Leb q_{t}}^{q}. \end{align*}

For any Schwarz function F in spacetime,

    \begin{align*} \Norm{\int_{\mathbb{R}}e^{i\left(t-s\right)\triangle/2}F\left(s\right)ds}_{\Leb q_{t}\Leb r_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)} & \le\Norm{\int_{\mathbb{R}}\Norm{e^{i\left(t-s\right)\triangle/2}F\left(s\right)}_{\Leb r_{x}\left(\mathbb{R}^{n}\right)}ds}_{\Leb q_{t}\left(\mathbb{R}\right)} \end{align*}

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

    \begin{align*} & \apprle\Norm{\Norm F_{\Leb{r^{\prime}}_{x}\left(\mathbb{R}^{n}\right)}*\frac{1}{\left|t\right|^{n\left(\frac{1}{2}-\frac{1}{r}\right)}}}_{\Leb q_{t}\left(\mathbb{R}\right)}. \end{align*}

Since 2<r\le\infty and 2<q\le\infty satisfying \frac{2}{q}+\frac{n}{r}=\frac{n}{2}, we have n\left(\frac{1}{2}-\frac{1}{r}\right)=\frac{2}{q}.

Recall the Hardy-Littlewood-Sobolev inequality:

    \[ \Norm{F*\frac{1}{\left|t\right|^{\alpha}}}_{\Leb q_{t}}\le C\Norm F_{\Leb p_{t}} \]

whenever

    \[ \frac{1}{p}=\frac{1}{q}+\frac{n-\alpha}{n}. \]

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

    \[ \Norm{\Norm F_{\Leb{r^{\prime}}_{x}\left(\mathbb{R}^{n}\right)}*\frac{1}{\left|t\right|^{n\left(\frac{1}{2}-\frac{1}{r}\right)}}}_{\Leb q_{t}\left(\mathbb{R}\right)}\apprle\Norm{\Norm F_{\Leb{r^{\prime}}_{x}\left(\mathbb{R}^{n}\right)}}_{\Leb{q'}_{t}\left(\mathbb{R}\right)}=\Norm F_{\Leb{q'}_{t}\Leb{r^{\prime}}_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)}. \]

Now by Hölder’s inequality, we have

    \[ \left|\int_{\mathbb{R}}\int_{\mathbb{R}}\left\langle e^{i\left(t-s\right)\triangle/2}F\left(s\right),F\left(t\right)\right\rangle dsdt\right|\apprle_{n,q,r}\Norm F_{\Leb{q'}_{t}\Leb{r'}_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)}^{2}. \]

On the left-hand side,

    \[ \int_{\mathbb{R}}\int_{\mathbb{R}}\left\langle e^{i\left(t-s\right)\triangle/2}F\left(s\right),F\left(t\right)\right\rangle dsdt=\Norm{\int_{\mathbb{R}}e^{-is\triangle/2}F\left(s\right)ds}_{\Leb 2_{x}\left(\mathbb{R}^{n}\right)}^{2}. \]

Thus,

    \[ \Norm{\int_{\mathbb{R}}e^{-is\triangle/2}F\left(s\right)ds}_{\Leb 2_{x}\left(\mathbb{R}^{n}\right)}\apprle\Norm F_{\Leb{q^{\prime}}_{t}\Leb{r'}_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)}. \]

This proves the second part (2).

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

    \begin{align*} & \int_{\mathbb{R}}\int_{\mathbb{R}^{n}}e^{it\triangle/2}u_{0}\overline{F\left(t\right)}dxdt\\ & =\int_{\mathbb{R}^{n}}u_{0}\int_{\mathbb{R}}\overline{e^{-it\triangle/2}F\left(t\right)}dtdx\\ & \le\Norm{u_{0}}_{\Leb 2\left(\mathbb{R}^{n}\right)}\Norm{\int_{\mathbb{R}}e^{-it\triangle/2}F\left(t\right)dt}_{\Leb 2\left(\mathbb{R}^{n}\right)}\\ & \apprle\Norm{u_{0}}_{\Leb 2\left(\mathbb{R}^{n}\right)}\Norm F_{\Leb{q^{\prime}}_{t}\Leb{r^{\prime}}_{x}}. \end{align*}

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

    \[ \Norm{e^{it\triangle/2}u_{0}}_{\Leb q_{t}\Leb r_{x}}\apprle\Norm{u_{0}}_{\Leb 2\left(\mathbb{R}^{n}\right)}. \]

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

Lemma (Christ-Kiselev). Let X,Y be Banach spaces, let I be a time interval, and let K\in C^{0}\left(I\times I;B\left(X;Y\right)\right) be a kernel taking values in the space of bounded operators from X to Y. Suppose that 1\le p\le q\le\infty satisfying

    \[ \Norm{\int_{I}K\left(t,s\right)f\left(s\right)ds}_{\Leb q_{t}\left(I;Y\right)}\le A\Norm f_{\Leb p_{t}\left(I;X\right)} \]

for all f\in\Leb p_{t}\left(I;X\right) and some A>0. Then we have

    \[ \Norm{\int_{s\in I:s<t}K\left(t,s\right)f\left(s\right)ds}_{\Leb q_{t}\left(I;Y\right)}\apprle_{p,q}A\Norm f_{\Leb p_{t}\left(I;X\right)}. \]

By (1)

    \begin{align*} \Norm{\int_{\mathbb{R}}e^{i\left(t-s\right)\triangle/2}F\left(s\right)ds}_{\Leb q_{t}\Leb r_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)} & \Norm{e^{it\triangle/2}\int_{\mathbb{R}}e^{-is\triangle/2}F\left(s\right)ds}_{\Leb q_{t}\Leb r_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)}\\ & \apprle\Norm{\int_{\mathbb{R}}e^{-is\triangle/2}F\left(s\right)ds}_{\Leb 2_{x}}\\ & \apprle\Norm F_{\Leb{q'}_{t}\Leb{r'}_{x}}. \end{align*}

Hence by Christ-Kiselev’s lemma,

    \[ \Norm{\int_{s<t}e^{i\left(t-s\right)\triangle/2}F\left(s\right)ds}_{\Leb q_{t}\Leb r_{x}\left(\mathbb{R}\times\mathbb{R}^{n}\right)}\apprle\Norm F_{\Leb{q'}_{t}\Leb{r'}_{x}}.\qedhere \]

Grading policy

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.

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 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. 

Existence and uniqueness of stationary Navier-Stokes equation

Let \Omega be a bounded domain in \mathbb{R}^2 or \mathbb{R}^3. We consider the Dirichlet boundary value problem for the stationary Navier-Stokes equation:

    \begin{align*} -\nu\triangle u+\left(u\cdot\nabla\right)u+\nabla p & =f\qquad \text{in } \Omega \\ \Div u & =0\qquad \text{in } \Omega \\ u & =0 \qquad \text{on } \partial \Omega \end{align*}

where \nu>0 is a viscosity constant. We call this problem as (NS).

Assume f\in \Sob{-1}{2} (\Omega). Now we consider a weak formulation of (NS).

A function u\in \Sob{1}{2}_0(\Omega) is called a weak solution of (NS) if \Div u =0 in \Omega and u satisfies

    \[ \nu \int_\Omega \nabla u : \nabla v \intd{x} = \int_\Omega (u\otimes u) : \nabla v \intd{x} + \action{ f,v } \]

for any v\in C_{0,\sigma}^\infty(\Omega).

The definition is well established. Take inner product to the first equation in (NS) with v\in C_{0,\sigma}^\infty(\Omega) and integrate it by parts. For the first part,

(1)   \begin{equation*} -\int_\Omega \nu \triangle u v \intd{x} = \nu \int_\Omega \nabla u : \nabla v \intd{x}. \end{equation*}

For the second part,

    \begin{align*} \int_\Omega (u\cdot \nabla u) v \intd{x} &= \sum_{i,j=1}^3 \int_\Omega u^j \frac{\partial u^i}{\partial x_j} v^i \intd{x}\\ &= -\sum_{i,j=1}^3 \int_\Omega \frac{\partial (u^j v^i)}{\partial x_j} u^i \intd{x} \\ &=- \int_\Omega \left(\sum_{j=1}^3 \frac{\partial u^j}{\partial x_j}\right)\sum_{i=1}^3 v^i u^i \intd{x}-\sum_{i,j=1}^3 \int_\Omega u^i u^j \frac{\partial v^i}{\partial x_j} \intd{x} \\ & =- \int_\Omega (u\otimes u) : \nabla v \intd{x}. \end{align*}

For the thrid part, it vanishes since \Div v =0. Also, for u\in \Sob{1}{2}(\Omega), for n=2, 3, by the Sobolev embedding theorem and the boundedness of \Omega, u \in \Leb{4} (\Omega). Hence \eqref{eq:weak-sol-st-NS} makes sense.

From now, we assume \nu =1. 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 X be a Banach space. Suppose that the compact operator A:X\rightarrow X satisfies the following: there exists a constant M>0 such that if u\in X is a solution of u=\sigma Au, 0\leq \sigma < 1 and

    \[ \norm{u}{X}<M \]

then there exists u in X satisfying u=Au.

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

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

Proof. Since \Omega is a bounded Lipschitz domain, due to Poincar\’e’s inequality, one can check that [u,v] = \int_\Omega \nabla u : \nabla v \intd{x} is an inner product on \Sob{1}{2}_{0,\sigma} (\Omega).

Also, for any w\in \Sob{1}{2}_{0,\sigma} (\Omega), \Div w\otimes w \in \Sob{-1}{2}(\Omega). Indeed, for any v\in \Sob{1}{2}_{0,\sigma} (\Omega),

    \begin{align*} \action{\Div (w\otimes w) , v} = -\action{w\otimes w,\nabla v}&\le \norm{w\otimes w}{2;\Omega}\norm{\nabla v}{2;\Omega}\\ &\leq C\norm{w}{4;\Omega}\norm{w}{4;\Omega} \norm{\nabla v}{2;\Omega} \intertext{By the Sobolev embedding $\Sob{1}{2}(\Omega) \hookrightarrow \Leb{6}(\Omega)\hookrightarrow\Leb{4}(\Omega)$, we get} &\leq C\norm{w}{1,2;\Omega}^2 \norm{v}{1,2;\Omega}. \end{align*}

So \Div(w \otimes w)\in \Sob{-1}{2}(\Omega).

Now for each w\in \Sob{1}{2}_{0,\sigma}(\Omega) by the existence result on Stokes equation, there exists a unique weak solution u \in \Sob{1}{2}_{0,\sigma}(\Omega) satisfying

(2)   \begin{equation*} \int_{\Omega} \nabla u : \nabla v \intd{x} = \action{-\Div(w\otimes w),v} + \action{f,v} \end{equation*}

for all v\in C_{0,\sigma}^\infty (\Omega). By Lemma ??, the above identity holds for any v\in \Sob{1}{2}_{0,\sigma} (\Omega).

Since -\Div(w\otimes w) \in \Sob{-1}{2}(\Omega), by the Riesz representation theorem, there exists a unique A(w) in \Sob{1}{2}_{0,\sigma}(\Omega) such that

    \[ [A(w) , v] = -\action{\Div(w\otimes w) , v} \]

for all v\in \Sob{1}{2}_{0,\sigma} (\Omega). By the uniqueness, the operator A:\Sob{1}{2}_{0,\sigma}(\Omega) \rightarrow \Sob{1}{2}_{0,\sigma}(\Omega) is well-defined. Similarly, there exists a unique F\in \Sob{1}{2}_{0,\sigma}(\Omega) such that

    \[ [F,v] = \action{f,v} \]

for all v\in \Sob{1}{2}_{0,\sigma} (\Omega) with \norm{\nabla F}{2;\Omega}= \norm{f}{-1,2;\Omega}. So \eqref{eq:perturb-Stokes} can be written as

    \[ [u,v ] =[A(w),v] +[F,v],\]

i.e., u=A(w)+F 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 u such that u=A(u)+F, then [A(u),v]=\int_\Omega u\times u :\nabla v \intd{x}. So u becomes a weak solution of (NS).

To use the Leray-Schauder fixed point theorem, we need to show that the operator T:\Sob{1}{2}_{0,\sigma} (\Omega) \rightarrow \Sob{1}{2}_{0,\sigma} (\Omega) defined by T(w) =A(w)+F is a compact operator and a priori estimate.

We show that T is completely continuous on \Sob{1}{2}_{0,\sigma}(\Omega). Let u_n \rightharpoonup u in \Sob{1}{2}_{0,\sigma}(\Omega). Then there exists a constant C>0 such that \norm{w_k}{1,2;\Omega} \leq C for all k. For simplicity, let u_k = T(w_k). Since \Omega is bounded Lipschitz domain, the Rellich-Kondrashov theorem shows that \Sob{1}{2}_{0,\sigma}(\Omega) is compactly embedded in \Leb{4}(\Omega). Thus, w_k \rightarrow w strongly in \Leb{4}(\Omega). Now

    \begin{align*} &\relphantom{=} \norm{w_k \otimes w_k - w\otimes w}{\Leb{2}}\\ & = \norm{(w_k-w)\otimes w_k +w \otimes w_k -w\otimes w}{\Leb{2}}\\ & \leq C\norm{w_k-w}{\Leb{4}} \norm{w_k}{\Leb{4}} +\norm{w}{\Leb{4}} \norm{w_k-w}{\Leb{4}}. \end{align*}

So as k\rightarrow \infty, w_k \otimes w_k \rightarrow w\otimes w in \Leb{2}(\Omega). Note

    \[ [u_k-u_m,v] =\int_{\Omega} w_k\otimes w_k -w_m \otimes w_m : \nabla v \intd{x}\qquad \text{for all } v\in \Sob{1}{2}_{0,\sigma}(\Omega). \]

Put v=u_k -u_m. Then we get

    \[ \norm{\nabla u_k -\nabla u_m}{2;\Omega}^2 \leq \norm{\nabla u_k -\nabla u_m}{2;\Omega} \norm{w_k\otimes w_k -w_m\otimes w_m}{2;\Omega} \]

So

    \[ \norm{\nabla u_k -\nabla u_m}{2;\Omega} \leq \norm{w_k\otimes w_k -w_m\otimes w_m}{2;\Omega}\rightarrow 0 \]

as k,m\rightarrow \infty. This shows that T is completely continuous.

Now we left to show a priori estimate.

First, note that for any v\in C_{0,\sigma}^\infty(\Omega),

    \begin{align*} &\relphantom{=} \int_\Omega v \otimes v : \nabla v \intd{x} \\ &=\sum_{i,j=1}^3 \int_\Omega v^i v^j \frac{\partial v^i }{\partial x_j} \intd{x} \\ & =-\sum_{i,j=1}^3 \int_{\Omega} \frac{\partial v^j}{\partial x_j} v^i v^i \intd{x} -\sum_{i,j=1}^3 \int_\Omega v^i v^j \frac{\partial v^i }{\partial x_j} \intd{x} \\ & =-\sum_{i,j=1}^3 \int_\Omega v^i v^j \frac{\partial v^i }{\partial x_j} \intd{x}=-\int_\Omega v \otimes v : \nabla v \intd{x}. \end{align*}

The last identity holds because of divergence free condition. By density, \int_\Omega v \otimes v : \nabla v \intd{x} =0 for any v\in \Sob{1}{2}_{0,\sigma}(\Omega).

Now suppose u\in \Sob{1}{2}_{0,\sigma}(\Omega) satisfies u=\sigma (Au +F), 0\leq \sigma <1. So

    \begin{align*} [u,u] &= [\sigma (A(u)+F),u] = \sigma [A(u),u] + \sigma [F,u] \\ &=\sigma \int_\Omega (u\otimes u) :\nabla u \intd{x} + \sigma \action{f,u} \\ & \leq \norm{f}{-1,2;\Omega}\norm{\nabla u}{2;\Omega}. \end{align*}

Here we used

    \[ \int_\Omega v \otimes v : \nabla v \intd{x} =0 \qquad \text{for all } v\in \Sob{1}{2}_{0,\sigma} (\Omega). \]

So

(3)   \begin{equation*} \norm{\nabla u}{2;\Omega} \leq \norm{f}{-1,2;\Omega} <(\norm{f}{-1,2;\Omega} +1). \end{equation*}

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

Now we prove the uniqueness.

Theorem. Let \Omega be a bounded Lipschitz domain and suppose that

    \[ 2c_0^2(n,\Omega) \norm{f}{-1,2;\Omega} <1, \]

where c_0 (n,\Omega) is a constant in the inequality

(4)   \begin{equation*} \norm{v}{4;\Omega} \leq c_0 (n,\Omega) \norm{\nabla v}{2;\Omega} \end{equation*}

for any v\in \Sob{1}{2}_0 (\Omega). Then (NS) has a unique weak solution.

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

Proof. Suppose u_1 and u_2 are two different solutions to (NS). Then

    \begin{align*} \norm{\nabla u_1 -\nabla u_2}{2;\Omega}^2&= \int_{\Omega} (u_1\otimes u_1 - u_2\otimes u_2): \nabla (u_1-u_2) \intd{x} \\ & = \int_{\Omega} (u_1\otimes (u_1-u_2)): \nabla (u_1-u_2) \intd{x} \\ & \relphantom{=}+ \int_{\Omega} (u_1-u_2)\otimes u_2: \nabla (u_1-u_2) \intd{x}\\ & \leq (\norm{u_1}{4;\Omega} +\norm{u_2}{4;\Omega}) \norm{u_1-u_2}{4;\Omega} \norm{\nabla u_1 - \nabla u_2}{2;\Omega}. \end{align*}

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

    \begin{align*} & \leq 2c_0^2 \norm{f}{-1,2;\Omega} \norm{\nabla u_1- \nabla u_2}{2;\Omega}^2. \end{align*}

Now by assumption, we get

    \begin{align*} \norm{\nabla u_1 -\nabla u_2}{2;\Omega}^2& <\norm{\nabla u_1 -\nabla u_2}{2;\Omega}^2, \end{align*}

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 S be a compact convex set in a Banach space X and let T be a continuous mapping of S into itself. Then T has a fixed point, that is, Tx=x for some x\in S.

Proof. Fix k\in\mathbb{N}. Since S is compact, \left\{ B_{\frac{1}{k}}\left(x\right)\right\} _{x\in K} has a finite subcover which covers S. Write \left\{ B_{\frac{1}{k}}\left(x_{1}\right),\dots,B_{\frac{1}{k}}\left(x_{N}\right)\right\}.
Let S_{k}=\mathrm{conv}\left(x_{1},\dots,x_{N}\right). Define J_{k}:S\rightarrow S_{k} by

    \[ J_{k}x=\frac{\sum_{i}\mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)x_{i}}{\sum_{i}\mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)}. \]

The map is well-defined. Since \mathrm{dist } is continuous, J_{k}x is continuous. Also,

    \[ \norm{J_{k}x-x}{}\le\frac{\sum_{i}\mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)\norm{x_{i}-x}{}}{\sum_{i}\mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)}. \]

Note that for such x\in S with \mathrm{dist }\left(x,S\setminus B_{\frac{1}{k}}\left(x_{i}\right)\right)>0, x\in B_{\frac{1}{k}}\left(x_{i}\right). Thus, for such i, we have \norm{x_{i}-x}{}<\frac{1}{k}. Hence \norm{J_{k}x-x}{}<\frac{1}{k}. Now J_{k}\circ T:S_{k}\rightarrow S_{k} is continuous, S_{k} is compact and convex set. So S_{k} is homeomorphic to \mathbb{D}^{N}. Hence by the Brouwer fixed point theorem, J_{k}\circ T\left(x_{k}\right)=x_{k} for some x_{k}\in S_{k}.

Now since \left\{ x_{k}\right\} \subset S and S is compact, there exists a convergent subsequence \left\{ x_{n_{k}}\right\} which converges to x in S. We claim that x is a fixed point. Note

    \[ \norm{Tx_{n_{k}}-x_{n_{k}}}{}=\norm{J_{n_{k}}\circ T\left(x_{n_{k}}\right)-x_{n_{k}}}{}<\frac{1}{n_{k}}\rightarrow0 \]

as k\rightarrow\infty. Hence Tx=x.

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

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

Proof. Let A be the closed convex hull of \overline{TS}. Then A is convex and since the closed convex hull of a compact set is itself compact, A is compact. Note that A\subset S because TS\subset S and \mathcal{S} is closed. So \overline{TS}\subset S. Since S is convex and A is the convex hull of \overline{TS}, A\subset S.
Thus, by the Schauder fixed point theorem, T has a fixed point in A. So we are done.

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

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

    \[ \norm xX<M \]

for all x\in X and \sigma\in\left[0,1\right] satisfying x=\sigma Tx. Then T has a fixed point.

Proof. We may assume M=1. Define \tilde{T}:X\rightarrow X by

    \[ \tilde{T}\left(x\right)=\begin{cases} T\left(x\right) & \text{if }\norm{T\left(x\right)}{}<1\\ \frac{T\left(x\right)}{\norm{T\left(x\right)}{}} & \text{if }\norm{T\left(x\right)}{}\ge1. \end{cases} \]

Let \mathbb{D}=\left\{ x\in X:\norm x{}\le1\right\}. Then \tilde{T} maps \mathbb{D} to \mathbb{D}. We show that \tilde{T}:\mathbb{D}\rightarrow\mathbb{D}
is compact. First, \tilde{T}:\mathbb{D}\rightarrow\mathbb{D} is continuous. For those x with \norm{T\left(x\right)}{}=1, we
have

    \[ \tilde{T}\left(x\right)=\frac{T\left(x\right)}{\norm{T\left(x\right)}{}}=T\left(x\right). \]

From the continuity of T, \tilde{T} is continuous on \mathbb{D}.

Next, let \left\{ u_{n}\right\} be a sequence in the ball \mathbb{D}. There are two cases:

  •  there exists a subsequence \left\{ u_{n_{k}}\right\} satisfying \norm{\tilde{T}\left(u_{n_{k}}\right)}{}<1 for all k;
  •  there exists a subsequence \left\{ u_{n_{k}}\right\} satisfying \norm{\tilde{T}\left(u_{n_{k}}\right)}{}\ge1 for all k.

For the first case, since \tilde{T}\left(u_{n_{k}}\right)=T\left(u_{n_{k}}\right) and T is compact, there exists a subsequence \left\{ u_{\tilde{n}_{k}}\right\} such that \left\{ T\left(u_{\tilde{n}_{k}}\right)\right\} converges strongly in X. So \tilde{T}u_{\tilde{n}_{k}} converges strongly in X.

For the second case, choose a subsequence \left\{ u_{\tilde{n}_{k}}\right\} so that \frac{1}{\norm{T\left(u_{\tilde{n}_{k}}\right)}{}}\rightarrow\alpha and T\left(u_{\tilde{n}_{k}}\right)\rightarrow z as k\rightarrow\infty by Bolzano-Weierstrass theorem on real numbers and compactness of T. Hence

    \[ \tilde{T}\left(u_{\tilde{n}_{k}}\right)\rightarrow\alpha z. \]

Thus, \tilde{T} is compact.

Hence by Theorem 2, \tilde{T} has a fixed point x. We claim that x is also a fixed point of T. Suppose \norm{Tx}{}\ge1 with Tx=x. Then

    \[ x=\tilde{T}x=\sigma Tx\quad\text{with }\sigma=\frac{1}{\norm{Tx}{}} \]

and so \norm x{}=1, which contradicts the assumption. Hence \norm{Tx}{}<1 and consequently x=T^{*}x=Tx.

Continuous, completely continuous and compact mapping

1. Weak convergence and weakly compact

We introduce some notations used in this article.

  • As usual, \mathbb{R}^{n} stands the Euclidean space of points x=\left(x',x_{n}\right). For i=1,\dots,n, multi-index \alpha=\left(\alpha_{1},\dots,\alpha_{n}\right)\alpha_{i}\in\left\{ 0,1,2,\dots\right\}and a function u\left(x\right) we set

        \[ u_{x_{i}}=\frac{\partial u}{\partial x_{i}}=D_{i}u,\quad D^{\alpha}u-D_{1}^{\alpha_{1}}\cdots D_{n}^{\alpha_{n}}u,\quad\nabla u=\left(u_{x_{1}},\dots,u_{x_{n}}\right). \]

    C_{0}^{\infty}\left(\Omega\right) denote the space of smooth functions with compact support in \Omega.

  • For 1\le p<\infty, \Leb p\left(\Omega\right) denote the space of all real-valued Lebesgue measurable functions u so that

        \[ \norm u{\Leb p\left(\Omega\right)}:=\left(\int_{\Omega}\left|u\left(x\right)\right|^{p}dx\right)^{\frac{1}{p}}<\infty. \]

Let X be a normed linear space and X^{*} denote the dual space of X. For f\in X^{*} and x\in X, we write \left\langle f,x\right\rangle instead of f\left(x\right). A sequence \left\{ x_{n}\right\} in X converges weakly to x in X if \lim_{n\rightarrow\infty}\left\langle f,x_{n}\right\rangle =\left\langle f,x\right\rangle for any f\in X^{*}. We write f_{n}\rightharpoonup f in \Leb p for simplicity.

Let us give some basic examples on this concept.

Example 1. When 1<p<\infty, note that \left[\Leb p\left(\Omega\right)\right]^{\prime}=\Leb{p^{\prime}}\left(\Omega\right) where \frac{1}{p}+\frac{1}{p^{\prime}}=1. Then a sequence \left\{ f_{n}\right\} in \Leb p\left(\Omega\right) converges weakly to f if \left\langle l,f_{n}\right\rangle \rightarrow\left\langle l,f\right\rangle for any l\in\left(\Leb p\left(\Omega\right)\right)^{\prime}. By the Riesz representation theorem on \Leb p spaces, there exists g\in\Leb{p^{\prime}}\left(\Omega\right) such that

    \[ l\left(f\right)=\int_{\Omega}fgdx\quad\text{for all }f\in\Leb p\left(\Omega\right). \]

So f_{n}\rightharpoonup f weakly in \Leb p if and only if

    \[ \lim_{n\rightarrow\infty}\int_{\Omega}f_{n}gdx=\int_{\Omega}fgdx \]

for any g\in\Leb{p^{\prime}}\left(\Omega\right).

Proposition 2. If X is compact Hausdorff space and \left\{ f_{n}\right\} is bounded in C\left(X;\mathbb{R}\right). Then f_{n}\left(x\right)\rightarrow f\left(x\right) for each x
if and only if f_{n}\rightharpoonup f weakly in C\left(X;\mathbb{R}\right).

The proof of this fact uses the following Riesz representation theorem on C\left(X;\mathbb{R}\right). For the proof, see Rudin, RCA for example.

Theorem 3 (Riesz representation theorem). Let X be a compact Hausdorff space and let M\left(X\right) be the set of signed Radon measures \nu on X such that \left|\nu\right|
is finite. Define a norm on M\left(X\right) by setting

    \[ \norm{\nu}M=\left|\nu\right|\left(X\right). \]

Then the dual space of C\left(X\right) is M\left(X\right). So
for any bounded linear functional \ell on C\left(X\right), there
exists a \mu\in M\left(X\right) such that

    \[ \ell\left(f\right)=\int_{X}fd\mu. \]

Proof of Proposition 2. Suppose f_{n}\left(x\right)\rightarrow f\left(x\right). Then for any bounded linear functional \ell on C\left(X\right), there exists a signed Radon measure \mu with \left|\mu\right|\left(X\right)<\infty such that

    \[ \ell\left(f\right)=\int_{X}fd\mu \]

for all f\in C\left(X\right). Then

    \[ \left|\ell\left(f_{n}\right)-\ell\left(f\right)\right|=\left|\int_{X}\left(f_{n}-f\right)d\mu\right|\le\int_{X}\left|f_{n}-f\right|d\left|\mu\right|. \]

Since \left\{ f_{n}\right\} is bounded in C\left(X;\mathbb{R}\right) and \left|\mu\right|\left(X\right)<\infty, by the dominated convergence theorem, we conclude that

    \[ \lim_{n\rightarrow\infty}\ell\left(f_{n}\right)=\ell\left(f\right). \]

This shows f_{n} converges weakly to f in C\left(X;\mathbb{R}\right).

Conversely, suppose f_{n} converges weakly to f in C\left(X;\mathbb{R}\right). Then for each x\in\mathbb{R}, one can easily check that \delta_{x} is a Radon measure on X. So

    \[ \lim_{n\rightarrow\infty}f_{n}\left(x\right)=\lim_{n\rightarrow\infty}\int_{X}f_{n}d\delta_{x}=\int_{X}fd\delta_{x}=f\left(x\right). \]

This shows f_{n}\left(x\right)\rightarrow f\left(x\right) for each
x\in X.
>We list one of the well-known properties of weak convergence.

Proposition 4. Let \left\{ x_{n}\right\} be a sequence in X. Then

  1. If x_{n}\rightarrow x strongly in X, then x_{n}\rightharpoonup x weakly in X.
  2. If x_{n}\rightharpoonup x weakly in X, then \left\{ \norm{x_{n}}{}\right\} is bounded and \norm x{}\le\liminf_{n\rightarrow\infty}\norm{x_{n}}{}.
  3. If x_{n}\rightharpoonup x weakly and if f_{n}\rightarrow f strongly in X^{*}, then \left\langle f_{n},x_{n}\right\rangle \rightarrow\left\langle f,x\right\rangle.

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

    \[ \left|\left\langle f,x_{n}\right\rangle -\left\langle f,x\right\rangle \right|=\left|\left\langle f,x_{n}-x\right\rangle \right|\le\norm f{}\norm{x_{n}-x}{}. \]

Letting n\rightarrow\infty, we see that \lim_{n\rightarrow\infty}\left\langle f,x_{n}\right\rangle =\left\langle f,x\right\rangle. So x_{n} converges weakly to x.

(ii) We prove this by using the Banach-Steinhaus theorem. For each n\in\mathbb{N}, define T_{n}:X^{*}\rightarrow\mathbb{R} by

    \[ T_{n}\left(f\right)=\left\langle f,x_{n}\right\rangle \]

and T:X^{*}\rightarrow\mathbb{R} by

    \[ T\left(f\right)=\left\langle f,x\right\rangle . \]

Then

    \[ \lim_{n\rightarrow\infty}T_{n}\left(f\right)=\lim_{n\rightarrow\infty}\left\langle f,x_{n}\right\rangle =\left\langle f,x\right\rangle =T\left(f\right). \]

So

    \[ \sup_{n}\left|T_{n}\left(f\right)\right|<\infty\quad\text{for each }f\in X^{*}. \]

Since X^{*} is a Banach space, by the Banach-Steinhaus theorem,

    \[ \sup_{n}\norm{T_{n}}{}<\infty. \]

Since

    \[ \norm{T_{n}}{}=\sup_{f\in X^{*},\norm f{}\le1}\left|T_{n}\left(f\right)\right|=\norm{x_{n}}{}, \]

this shows that \sup_{n}\norm{x_{n}}{}<\infty.

Note that there exists f\in X^{*} such that \left\langle f,x\right\rangle =\norm x{} and \norm f{}=1. So from this, we have

    \[ \left\langle f,x_{n}\right\rangle \le\norm{x_{n}}{}. \]

Now taking liminf to above inequality to get

    \[ \norm x{}=\left\langle f,x\right\rangle =\liminf_{n\rightarrow\infty}\left\langle f,x_{n}\right\rangle \le\liminf_{n\rightarrow\infty}\norm{x_{n}}{}. \]

(iii) This follows from

    \begin{align*} \left|\left\langle f_{n},x_{n}\right\rangle -\left\langle f,x\right\rangle \right| & =\left|\left\langle f_{n},x_{n}\right\rangle -\left\langle f,x_{n}\right\rangle +\left\langle f,x_{n}\right\rangle -\left\langle f,x\right\rangle \right|\\ & \le\norm{x_{n}}{}\norm{f_{n}-f}{}+\left|\left\langle f,x_{n}\right\rangle -\left\langle f,x\right\rangle \right|. \end{align*}

From (ii), \norm{x_{n}}{} is bounded. So letting n\rightarrow\infty, we get the desired result.

Let \left\{ f_{n}\right\} be a sequence in X^{*}. We say \left\{ f_{n}\right\}
converges to f weakly-star (or weakly{*}) if \lim_{n\rightarrow\infty}\left\langle f_{n},x\right\rangle =\left\langle f,x\right\rangle
for any x\in X. 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 X^{*} has a weakly-star convergent sequence.

Proof. For simplicity, we assume X is separable. Then there exists a countable dense subset D=\left\{ x_{1},x_{2},\dots\right\} of X. We may assume that \left\{ f_{n}\right\} \subset X^{*} satisfies \norm{f_{n}}{}\le1 for all n. Now consider

    \[ \begin{array}{cccc} \left\langle f_{1},x_{1}\right\rangle & \left\langle f_{2},x_{1}\right\rangle & \left\langle f_{3},x_{1}\right\rangle & \cdots\\ \left\langle f_{1},x_{2}\right\rangle & \left\langle f_{2},x_{2}\right\rangle & \left\langle f_{3},x_{2}\right\rangle & \cdots\\ \left\langle f_{1},x_{3}\right\rangle & \left\langle f_{2},x_{3}\right\rangle & \left\langle f_{3},x_{3}\right\rangle & \cdots\\ \vdots & \vdots & \vdots \end{array} \]

Then for the first line, by the Bolzano-Weierstrass theorem, there exists a convergent subsequence \left\{ \left\langle f_{1,n},x_{1}\right\rangle \right\} of \left\{ \left\langle f_{n},x_{1}\right\rangle \right\}. For such \left\{ f_{1,n}\right\}, there exists a convergent subsequence of \left\{ \left\langle f_{2,n},x_{2}\right\rangle \right\} of \left\{ \left\langle f_{1,n},x_{2}\right\rangle \right\}. Continuing this process, then we see that

    \[ \lim_{n\rightarrow\infty}f_{j,n}\left(x_{j}\right) \]

exists. Now define

    \[ f_{n_{k}}\left(x\right)=f_{k,k}\left(x\right). \]

Then \lim_{k\rightarrow\infty}f_{n_{k}}\left(x_{j}\right) exists for all j. Since \left\{ f_{n_{k}}\right\} is bounded in X^{*}, \left\{ f_{n_{k}}\left(x\right)\right\} is a Cauchy sequence for each x in D. So from this, \left\{ f_{n_{k}}\left(x\right)\right\} is Cauchy for all x in X. Hence,

    \[ \lim_{k\rightarrow\infty}f_{n_{k}}\left(x\right)=f\left(x\right) \]

exists. Then f is linear and

    \[ \left|f_{n_{k}}\left(x\right)\right|\le\norm x{}\quad\text{for all }k\quad\text{and}\quad\text{all }x\in X \]

and so

    \[ \left|f\left(x\right)\right|=\lim_{k\rightarrow\infty}\left|f_{n_{k}}\left(x\right)\right|\le\norm x{}\quad\text{for all }x\in X. \]

So f is a bounded linear functional and \norm f{}\le1. So f_{n_{k}}\rightharpoonup f weakly star in X^{*}.

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

We say X is \emph{reflexive }if for every F\in X^{**}, there
is u\in X such that \left\langle F,f\right\rangle =\left\langle f,u\right\rangle
for all f\in X^{*}. When X is reflexive, we identify X and
X^{**}.

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

Proof. Let \left\{ x_{n}\right\} be a bounded sequence in X. Since X is reflexive, x_{n} is a bounded linear functional on X^{*} and \norm{x_{n}}{X^{**}}\le C for all n. Hence by Theorem 5, there exists a convergent subsequence \left\{ x_{n_{k}}\right\} of \left\{ x_{n}\right\} in X^{**} which converges to x. So for any f\in X^{*},

    \[ \lim_{k\rightarrow\infty}\left\langle x_{n_{k}},f\right\rangle =\left\langle x,f\right\rangle , \]

i.e.,

    \[ \lim_{k\rightarrow\infty}\left\langle f,x_{n_{k}}\right\rangle =\left\langle f,x\right\rangle \quad\text{for all }f\in X^{*}. \]

This shows x_{n_{k}} converges weakly to x in X.

2. Compact operators and completely continuous

Recall that a mapping T:X\rightarrow Y is said to be continuous if T\left(x_{n}\right)\rightarrow T\left(x\right) in Y for any sequence x_{n}\rightarrow x in X. A mapping T:X\rightarrow Y is weakly continuous if T\left(x_{n}\right)\rightharpoonup T\left(x\right) weakly in Y whenever x_{n}\rightharpoonup x weakly in X. A mapping T:X\rightharpoonup Y is said to be completely continuous if T\left(x_{n}\right)\rightarrow T\left(x\right) strongly in Y whenever x_{n}\rightharpoonup x weakly in X. Finally, a mapping T:X\rightharpoonup Y is compact if T is continuous and for every bounded sequence \left\{ x_{n}\right\} in X, there exists a subsequence \left\{ x_{n_{k}}\right\} such that T\left(x_{n_{k}}\right) converges strongly in Y.

Remark. Let B be a bounded set in X and T:X\rightarrow Y be linear. Suppose \overline{T\left(B\right)} is compact in Y. Then T is bounded. When T is linear, to check the compactness, it suffices to check \overline{T\left(B\right)} is compact in Y whenever B is a bounded set in X.

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

Proposition 7. If X is reflexive and T:X\rightarrow Y is completely continuous, then T is compact.

Proof. Let \left\{ x_{n}\right\} be a bounded sequence in X. Then by Theorem ??, there exists a subsequence \left\{ x_{n_{k}}\right\} for which x_{n_{k}} converges weakly to x in X. Since T is completely continuous, Tx_{n_{k}} converges strongly to Tx. Hence T is compact.

Example 8. Define T:\Leb 2\left(\left[0,2\pi\right]\right)\rightarrow\mathbb{R} by T\left(f\right)=\norm f2. Then T is compact. Let \left\{ x_{n}\right\} be a bounded sequence in \Leb 2\left(\left[0,2\pi\right]\right). Then \left\{ \norm{x_{n}}{}\right\} is bounded. So by the Bolzano-Weierstrass theorem, there exists a subsequence \left\{ x_{n_{k}}\right\}
for which \lim_{k\rightarrow\infty}\norm{x_{n_{k}}}{} converges. So T is compact. Define e_{n}\left(x\right)=\frac{1}{\sqrt{2\pi}}e^{inx}. Then \left\{ e_{n}\right\} forms an orthonormal basis for \Leb 2\left(\left[0,2\pi\right]\right). Then from Parseval’s inequality, we see that \lim_{n\rightarrow\infty}\left(f,e_{n}\right)=0 for any f\in\Leb 2\left(\left[0,2\pi\right]\right). So e_{n} converges weakly to 0. But

    \[ \lim_{n\rightarrow\infty}T\left(e_{n}\right)=1\neq0. \]

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

Example 9. The identity map I:\ell^{1}\left(\mathbb{N}\right)\rightarrow\ell^{1}\left(\mathbb{N}\right)
is completely continuous but not compact. Note that \ell^{1}\left(\mathbb{N}\right)
is non-reflexive.

Theorem 10. Let T:X\rightarrow Y be linear. Then

  1.  if T is continuous, then T is weakly continuous.
  2.  if T is compact, then it is completely continuous.
  3.  if X,Y are Banach spaces, then T is weakly continuous if and only if T is continuous.

Proof. (i) Let x_{n}\rightharpoonup x and let y^{*} be a bounded linear functional on Y. Then y^{*}\circ T:X\rightarrow\mathbb{R} is
linear and

    \begin{align*} \left\langle y^{*}\circ T,x\right\rangle & =\left\langle y^{*},Tx\right\rangle \\ & \le\norm{y^{*}}{}\norm{Tx}{}\\ & \le\norm{y^{*}}{}\norm T{}\norm x{}. \end{align*}

So y^{*}\circ T is a bounded linear functional on X. Hence

    \[ \lim_{n\rightarrow\infty}\left\langle y^{*}\circ T,x_{n}\right\rangle =\left\langle y^{*}\circ T,x\right\rangle , \]

i.e.,

    \[ \lim_{n\rightarrow\infty}\left\langle y^{*},Tx_{n}\right\rangle =\left\langle y,Tx\right\rangle . \]

So Tx_{n}\rightharpoonup Tx weakly in Y.

(ii) Let x_{n}\rightharpoonup x weakly in X. Then by Proposition 4, \left\{ x_{n}\right\} is bounded in X. Since T is compact, there is a subsequence \left\{ x_{n_{k}}\right\}
for which Tx_{n_{k}} converges strongly to y in Y. Then by (i), Tx_{n}\rightharpoonup Tx and so Tx_{n_{k}}\rightharpoonup Tx weakly in Y. It remains to show y=Tx.

Note that Tx_{n_{k}}\rightharpoonup y weakly in Y. So for any bounded linear functional l on Y,

    \[ \left\langle l,y\right\rangle =\lim_{k\rightarrow\infty}\left\langle l,Tx_{n_{k}}\right\rangle =\left\langle l,Tx\right\rangle . \]

So l\left(y-Tx\right)=0 for any l\in Y^{*}. Now by the consequence of Hahn-Banach theorem, there exists a bounded linear functional l such that

    \[ l\left(y-Tx\right)=\norm{y-Tx}{}. \]

So y=Tx in Y. Hence, Tx_{n_{k}}\rightarrow Tx strongly in Y. Now we are left to show that Tx_{n}\rightarrow Tx strongly in Y. Suppose not. Then there exists \varepsilon_{0}>0 and a subsequence \left\{ x_{n_{k}}\right\} such that

(1)   \begin{equation*} \norm{Tx_{n_{k}}-Tx}Y\ge\varepsilon_{0}\quad\text{for all }k. \end{equation*}

Note that \left\{ x_{n_{k}}\right\} is a bounded sequence. So using the compactness of T again, there exists a subsequence \left\{ x_{n_{k_{m}}}\right\} such that Tx_{n_{k_{m}}}\rightarrow Tx in Y. This contradicts (1). Hence, Tx_{n}\rightarrow Tx strongly in Y.

(iii) Suppose T is weakly continuous. Consider

    \[ G\left(T\right)=\left\{ \left(x,Tx\right):x\in X\right\} . \]

Then G\left(T\right) is weakly closed subspace of X\times Y. Hence G\left(T\right) is strongly closed in X\times Y. Thus, T is continuous by the closed graph theorem.

A normed linear space X is continuously embedded in Y if X\subset Y and there exists a constant c>0 such that \norm xY\le c\norm xX for all x\in X. We write X\hookrightarrow Y. We say X is compactly embedded in Y if X is continuously embedded in Y and if \left\{ x_{n}\right\} is bounded in X, then there exists a subsequence \left\{ x_{n_{k}}\right\} which converges strongly in Y.

Example 11. By the Sobolev embedding theorem, the inclusion map I:\Sob 12\left(\mathbb{R}^{3}\right)\rightarrow\Leb 6\left(\mathbb{R}^{3}\right)Ix=x is linear and continuous. But the embedding is not compact. Indeed, take

    \[ f_{k}\left(x\right)=\begin{cases} k^{\frac{1}{2}}\left(1-k\left|x\right|\right) & \text{if }\left|x\right|<\frac{1}{k},\\ 0 & \text{if }\left|x\right|\ge\frac{1}{k}. \end{cases} \]

Then \supp f_{k}\subset\overline{B_{1}} for every k\in\mathbb{N} and \left\{ f_{k}\right\} is bounded in \Sob 12\left(\mathbb{R}^{n}\right).
Note

    \begin{align*} \int_{\left|x\right|<\frac{1}{k}}\left|f_{k}\left(x\right)\right|^{2}dx & =c_{3}\int_{0}^{\frac{1}{k}}k\left(1-k\rho\right)^{2}\rho^{2}d\rho\\ & =c_{3}\int_{0}^{1}\left(1-y\right)^{2}\left(\frac{y}{k}\right)^{2}dy\\ & \le\frac{c}{k^{2}}\le C \end{align*}

and similarly,

    \[ \int_{\mathbb{R}^{3}}\left|\nabla f_{k}\left(x\right)\right|^{2}dx\le C\quad\text{for all }k. \]

But

    \begin{align*} \int_{\left|x\right|<\frac{1}{k}}\left|f_{k}\left(x\right)\right|^{6}dx & =c_{3}\int_{0}^{\frac{1}{k}}k^{3}\left(1-k\rho\right)^{6}\rho^{2}d\rho\\ & =ck^{4}, \end{align*}

which cannot be bounded. So it cannot have a convergent subsequence in \Leb 6\left(\mathbb{R}^{3}\right). Also, I is linear and continuous but not compact. Hence it is not completely continuous.

Corollary 12. Let X\subset Y. If X is compactly embedded in Y and x_{n} converges weakly to x in X, then x_{n} converges strongly to x in Y. In addition, if X is reflexive, then the converse holds.

Proof. Since X is compactly embedded in Y, by (i) I:X\rightarrow Y is completely continuous. Then the corollary is proved by the definition of completely continuous. If X is reflexive, the inclusion map I is compact by Proposition 7.