Let $T:X\longrightarrow Y$ be a weakly compact operator between Banach spaces. Prove range of $T$ is closed iff range of $T$ is reflexive

Question

The proof is from Robert Megginson, Introduction to Banach Spaces (3.5.6)

PROOF. One implication is a trivial consequence of the fact that every reflexive subspace of a Banach space is closed. For the other, suppose that a weakly compact linear operator $T$ from a Banach space $X$ into a Banach space $Y$ has closed range. Then $T$ is an open mapping from $X$ onto $T(X)$, so $T(B_X)$ is a relatively weakly compact subset of $T(X)$ that includes a neighborhood of $0$ in $T(X)$. This implies that $B_{T(X)}$ is weakly compact and therefore that $T(X)$ is reflexive. $\qquad\blacksquare$

However I dont understand why $B_{T(X)}$ is weakly compact, I've been using Eberlein - Smulian Theorem to prove the previos propositions but I still dont get why if $T(B_X)$ is a relatively weakly compact subset of $T(X)$ that includes a neighbornood of 0 in $T(X)$ then $B_{T(X)}$ is weakly compact.

I would appreciate any insight.

Answer 1

So, $T(B_X)$ is weakly relatively compact in $T(X)$ and contains a $T(X)$-open neighborhood $U$ of $0$. Of course, this also holds for $T(cB_X)$ for any $c>0$.

Denote the $T(X)$-closed ball around zero in $T(X)$ with radius $r>0$ by $B_{T(X),r}$. So, one of these guys is contained in $T(B_X)$, i.e., $B_{T(X),r}\subset T(B_X)$. Hence, $B_{T(X)} = r^{-1}B_{T(X),r}\subset r^{-1}T(B_X) = T(r^{-1}B_X)$. Thus, the weakly closed set $B_{T(X)}$ is weakly compact.