# Weak convergence in reflexive Banach space

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.

• 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 $L(X,X)$? – user56706 Mar 1 '13 at 18:46

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.

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 $T\in\mathcal{B}(X)$.

• 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.