# 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

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.

• Is $B_X$ the open or the closed ball in $X$? Commented Oct 7, 2019 at 16:34
• The closed unit ball of $X$. Commented Oct 7, 2019 at 16:37
• Well, by definition $T(B_X)$ is relatively weakly compact in $Y$ and hence also in the closed subspace $T(X)$. Moreover, $\overset{\circ}{B_X}$ is open and so is $T(\overset{\circ}{B_X})$ in $T(X)$. Hence, $T(\overset{\circ}{B_X})$ is an open neighborhood of zero in $T(B_X)$. Commented Oct 7, 2019 at 16:42
• Yeah, I got that, the part that I dont see is the one that implies $B_{T(X)}$ is weakly compact. Commented Oct 7, 2019 at 16:48
• Uhh sorry, I misread your question. But you also misread something. It's not neighbornood, but neighborhood. ;-) Commented Oct 7, 2019 at 16:49

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.