Continuous, completely continuous and compact mapping

By | January 8, 2018

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.

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