Weak convergence in a reflexive, separable and infinite Banach space.

Question

Let $Y$ be a infinite Banach space and $X$ be a reflexive, separable and infinite Banach space. Further let $T ∈\mathcal{L}(X,Y )$ satisies that $$\|Tx_n − Tx\|_{Y} \to 0,~~~~~as~~n\to\infty$$ , as $n → ∞$. whenever $(x_n)_{n≥1}$ is a sequence in $X$ converging weakly to $x ∈ X$ then show that $T ∈\mathcal{K}(X,Y )$.

Answer 1

Since $X$ is reflexive, by Banach-Alaoglu theorem the unit ball of X is weakly compact.

Hence for $(x_n)_n$ a bounded sequence (which is contains in a weak compact ball) there exists a subsequence $(x_{n_j})_j$ that converges weakly in $X$ to some $x$.

By assumption on $T$ we. have $$\|Tx_{n_j} − Tx\|_{Y} \to 0$$ This shows that T is a compact operator that is $T\in\mathcal{K}(X;Y)$

Answer 2

Use the fact that if $X$ is a reflexive normed space, then any bounded sequence in $X$ has a weakly convergent subsequence.

Now let $(\eta_{n})$ be a bounded sequence in $X$, we are to prove that there is a convergent subsequence $(T(\eta_{n_{k}}))$ of $(T(\eta_{n}))$.

Find some weakly convergent subsequence $\eta_{n_{k}}\rightarrow\eta$ as in the first paragraph. So by assumption we have $\|T(\eta_{n_{k}})-T(\eta)\|\rightarrow 0$ as $k\rightarrow\infty$, so $(T(\eta_{n_{k}}))$ is convergent.