weak convergence of a bounded linear operator

Question

I need help with this problem

Let $X$ be a reflexive Banach space and $T: X \to X$ a linear operator. Show that $T$ belongs to $\mathcal{L}(X,X)$ if and only if whenever $\{x_n \}$ converges weakly to $x$, $\{T(x_n)\}$ converges weakly to $T(x)$.

any hints or suggestions, thank you!

Answer 1

Preliminaries.

1. By $\iota_X$ we denote natural embedding into the second dual.

2. For any $S\in\mathcal{L}(X,Y)$ the operrator $S^*\in\mathcal{L}(Y^*,X^*)$ is weak-${}^*$ continuous.

3. For any $S\in\mathcal{L}(X,Y)$ holds $S^{**}\iota_X=\iota_Y S$

4. For reflexive $X$ operator $\iota_X$ is a isometric weak-to-weak-${}^*$ homeomorphism.

Proof. $(\Rightarrow)$ Assme $x_n\overset{w}{\to}x$, then $\iota_X(x_n)\overset{w^*}{\to}\iota_X(x)$. As any dual operator $T^{**}$ is weak-${}^*$ continuous, so $\iota_X(T(x_n))=T^{**}(\iota_X(x_n))\overset{w^*}{\to} T^{**}(\iota_X(x))=\iota_X(T(x))$. Since $X$ is reflexive, this means that $T(x_n)\overset{w}{\to} T(x)$.

$(\Leftarrow)$ See this post.