# Totally continuous implies bounded

Consider a separable, reflexive Banach space $V$. We define the mapping $A: V \rightarrow V^{*}$ as totally continuous if it is continuous as a mapping $(V, \text{weak}) \rightarrow (V^{*}, norm)$. I want to show that $A$ is bounded. I am looking for a more direct proof of the boundedness of $A$. I can prove that the mapping $A$ is compact(and therefore bounded) in the following way:

Let $B \subset V$ be a bounded set. Consider a sequence in $A(B)$, say ${A(v_{k})}_{k \in \mathbb{N}} \subset V^{*}$ with $v_{k} \in B$. I will show that it has a convergent subsequence. Since $V$ is separable it follows that the weak topology on the closed unit ball is metrizable. By the Kakutani Theorem we have that the closed unit ball in $V$ is compact in the weak topology, so there exists a weakly convergent subsequence ${v_{k}}_{l} \rightharpoonup v$ in $V$. By total continuity it follows that $A({v_{k}}_{l}) \rightarrow A(v)$ in $V^{*}$. We have shown that $A$ is compact and therefore bounded. $\square$

Is this proof fine and is there a more direct proof of boundedness for a totally continuous mapping $A:V \rightarrow V^{*}$? Maybe missing something trivial.

• That shows that it is norm - norm continuous. Does the equivalence of continuous and bounded operators not only hold for linear operators($A$ is not assumed linear ) on metrizable vector spaces? Is that how you are implying boundedness?
– user100431
Feb 25, 2014 at 13:41
• Sorry. Somehow I read "linear" in there. Feb 25, 2014 at 13:42
• But, couldn't you argue that the image of the weakly compact norm-closed unit ball of $V$ is norm compact, thus bounded, in $V^*$? Feb 25, 2014 at 13:49
• That's what he did @DavidMitra and it seems to me that the proof is good. Feb 25, 2014 at 13:50
• @Tomás Yes, but the only fact you need is that the continuous image of a compact set is compact. Feb 25, 2014 at 13:52

Now you need only use the fact that a continuous image of a compact set is compact. $A$ maps the weakly compact norm-closed unit ball of $V$ onto a norm-compact, thus bounded, subset of $V^*$.
(Note the separability of $V$ is not needed here.)