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Let $X$ be a reflexive Banach space. Let $T: X \to Y$ a linear operator. I want to show that:

$$T \in \mathcal{L}(X, Y) \iff ((x_n \stackrel{w}{\rightharpoonup} x) \implies (T(x_n) \stackrel{w}{\rightharpoonup} T(x))) $$

Any help, thoughts, hints, solutions will be greatly appreciated. I am lots of trouble regarding weak convergence.

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Perhaps you should give some of your thoughts on the question. How did you try to start, and what went wrong? – GEdgar Mar 1 '13 at 18:30
Is that $L(X,Y)$ or $L(X,X)$? – Vahid Shirbisheh Mar 1 '13 at 18:46
up vote 7 down vote accepted

Forward direction $ (\Longrightarrow) $

Suppose that $ T \in \mathcal{L}(X,Y) $ and that $ x_{n} \stackrel{\text{wk}}{\longrightarrow} x $. Then for any $ \varphi \in Y^{*} $, we have \begin{align} \lim_{n \to \infty} \varphi(T(x_{n}) - T(x)) &= \lim_{n \to \infty} \varphi(T(x_{n} - x)) \\ &= \lim_{n \to \infty} (\varphi \circ T)(x_{n} - x) \\ &= 0. \quad (\text{As $ \varphi \circ T \in X^{*} $.}) \end{align} Therefore, $ T(x_{n}) \stackrel{\text{wk}}{\longrightarrow} T(x) $.

Backward direction $ (\Longleftarrow) $

Suppose that $ T: X \to Y $ is a linear operator (not assumed to be bounded) and that $ x_{n} \stackrel{\text{wk}}{\longrightarrow} x $ implies $ T(x_{n}) \stackrel{\text{wk}}{\longrightarrow} T(x) $.

For the sake of contradiction, assume that $ T $ is not bounded. Then we can easily construct a sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ X $ that converges in norm to $ 0_{X} $ but for which $ \| T(x_{n}) \|_{Y} \geq n^{2} $ for all $ n \in \mathbb{N} $. As $ \dfrac{1}{n} \cdot x_{n} \stackrel{\| \cdot \|_{X}}{\longrightarrow} 0_{X} $, we obviously have $ \dfrac{1}{n} \cdot x_{n} \stackrel{\text{wk}}{\longrightarrow} 0_{X} $. From our initial hypothesis, it follows that $ T \left( \dfrac{1}{n} \cdot x_{n} \right) \stackrel{\text{wk}}{\longrightarrow} T(0_{X}) = 0_{Y} $.

Now, for each $ n \in \mathbb{N} $, we can view $ T \left( \dfrac{1}{n} \cdot x_{n} \right) $ as an element $ \Psi_{n} $ of $ Y^{**} $, via the canonical embedding $ J: Y \to Y^{**} $. As \begin{align} \forall \varphi \in Y^{*}: \quad \lim_{n \to \infty} {\Psi_{n}}(\varphi) &= \lim_{n \to \infty} \varphi \left( T \left( \frac{1}{n} \cdot x_{n} \right) \right) \\ &= \varphi(0_{Y}) \\ &= 0, \end{align} we see that $ \displaystyle \sup_{n \in \mathbb{N}} |{\Psi_{n}}(\varphi)| < \infty $ for all $ \varphi \in Y^{*} $. Applying the Uniform Boundedness Principle (also known as the Banach-Steinhaus Theorem) to $ Y^{**} $, we obtain $ \displaystyle \sup_{n \in \mathbb{N}} \| \Psi_{n} \|_{Y^{**}} < \infty $. This yields $ \displaystyle \sup_{n \in \mathbb{N}} \left\| T \left( \dfrac{1}{n} \cdot x_{n} \right) \right\|_{Y} < \infty $ as the canonical embedding $ J $ is an isometry. We have thus arrived at a contradiction as our construction of the sequence $ (x_{n})_{n \in \mathbb{N}} $ forces us to have $ \left\| T \left( \dfrac{1}{n} \cdot x_{n} \right) \right\|_{Y} \geq n $ for all $ n \in \mathbb{N} $ instead.

The assumption is therefore false, so we conclude that $ T $ is indeed bounded.

Note: There is no need to assume that $ X $ is reflexive.

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I will only assume that $X$ is a Banach space.

($\Longrightarrow$) Let $T\in\mathcal{B}(X)$. Take arbitrary $\{x_n:n\in\mathbb{N}\}\subset X$ such that $\lim\limits_{n\to\infty}x_n\underset{w}{=}x$. Then $$ \forall f\in X^*\qquad\lim\limits_{n\to\infty}f(x_n)=f(x)\tag{1} $$ Take arbitrary $g\in X^*$, and denote $f=T^*(g)$. Using $(1)$ we get $$ \lim\limits_{n\to\infty} g(T(x_n))=\lim\limits_{n\to\infty} (T^*(g))(x_n)=\lim\limits_{n\to\infty} f(x_n)=f(x)=T^*(g)(x)=g(T(x))\tag{2} $$ Since $g\in X^*$ is arbitrary we conclude $\lim\limits_{n\to\infty}T(x_n)\underset{w}{=}T(x)$. Thus we proved implication $$ (\lim\limits_{n\to\infty}x_n\underset{w}{=}x) \implies (\lim\limits_{n\to\infty}T(x_n)\underset{w}{=}T(x)) $$

($\Longleftarrow$) Take arbitrary $\{x_n:n\in\mathbb{N}\}\subset X$ such that $\lim\limits_{n\to\infty}x_n=x$ and $\lim\limits_{n\to\infty}T(x_n)=y$. Take arbitrary $f\in X^*$. Obviously $\lim\limits_{n\to\infty}f(x_n)=f(x)$. Since $f\in X^*$ is arbitrary, we see that $\lim\limits_{n\to\infty}x_n\underset{w}{=}x$. By assumption this implies that $\lim\limits_{n\to\infty}T(x_n)\underset{w}{=}T(x)$. This means that for all $g\in X^*$ we have $\lim\limits_{n\to\infty}g(T(x_n))=g(T(x))$. In this case $$ g(T(x)-y)=g(T(x)-\lim\limits_{n\to\infty}T(x_n))=g(T(x))-g(\lim\limits_{n\to\infty}T(x_n))= g(T(x))-\lim\limits_{n\to\infty}g(T(x_n))=0 $$ Thus for all $g\in X^*$ we have $g(T(x)-y)=0$. By corollary of Hahn-Banach theorem this means that $T(x)-y=0$, i.e. $T(x)=y$. To summarize we showed that conditions $\lim\limits_{n\to\infty}x_n=x$, $\lim\limits_{n\to\infty}T(x_n)=y$ implies $T(x)=y$. Since $X$ is Banach by closed graph theorem this means that $T\in\mathcal{B}(X)$.

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It's worth noting that your statement "Since $g \in X^*$ is arbitrary we conclude..." is secretly an application of the Hahn-Banach theorem. – Nate Eldredge Mar 1 '13 at 19:34
Strange... in that line I'm talking about weak convergece $T(x_n)$ to $T(x)$ – Norbert Mar 1 '13 at 19:47
I'm sorry, I didn't read carefully enough. – Nate Eldredge Mar 1 '13 at 22:28

For the backward direction (⟸), here's the proof given by Norbert in a simplified form:

Let $x_n \to x$ and $Tx_n \to y$, then we have $x_n \to x$ weakly and $Tx_n \to y$ weakly, now using the assumption we have also $Tx_n \to Tx$ weakly. By the uniqueness of the weak limit we have $y=Tx$. T is then closed and therefore bounded (continuous) by the closed graph theorem.

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