I'm trying to solve an exercise in Brezis' Functional Analysis
Let $E$ and $F$ be Banach spaces and $T:E \to F$ a bounded linear operator. Let $B_E$ be the closed unit ball of $E$. If $E$ is reflexive, then $T(B_E)$ is closed in the norm topology of $F$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it? Thank you so much for your help!
Let $(x_n) \subset B_E$ and $y\in F$ such that $Tx_n \to y$. We need to find some $x \in B_E$ such that $y=Tx$. Let $(x_{n_k})_k$ be a subsequence of $(x_n)$. Because $E$ is reflexive, $B_E$ is sequentially compact in the weak topology $\sigma(E, E^*)$. Then there is $x \in B_E$ such that $x_{n_k} \xrightarrow{k \to \infty} x$ in $\sigma(E, E^*)$. By Theorem 3.10 in the same book, $T$ is continuous from $E$ endowed with $\sigma(E, E^*)$ to $F$ endowed with $\sigma(F, F^*)$. Then $Tx_{n_k} \xrightarrow{k \to \infty} Tx$ in $\sigma(F, F^*)$. On the other hand, $Tx_{n_k} \xrightarrow{k \to \infty} y$. So $Tx=y$. This completes the proof.