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.

  • $\begingroup$ 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? $\endgroup$
    – user100431
    Feb 25, 2014 at 13:41
  • $\begingroup$ Sorry. Somehow I read "linear" in there. $\endgroup$ Feb 25, 2014 at 13:42
  • $\begingroup$ But, couldn't you argue that the image of the weakly compact norm-closed unit ball of $V$ is norm compact, thus bounded, in $V^*$? $\endgroup$ Feb 25, 2014 at 13:49
  • $\begingroup$ That's what he did @DavidMitra and it seems to me that the proof is good. $\endgroup$
    – Tomás
    Feb 25, 2014 at 13:50
  • $\begingroup$ @Tomás Yes, but the only fact you need is that the continuous image of a compact set is compact. $\endgroup$ Feb 25, 2014 at 13:52

1 Answer 1


Your proof is fine, though it uses more machinery than necessary (in particular, you need not appeal to sequences):

The norm-closed unit ball of a reflexive normed space is weakly compact. This follows from Alaoglu's theorem and the fact that the natural embedding from such a space onto its second dual is a weak-weak* homeomorphism.

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.)

  • $\begingroup$ What is the difference between Kakutani's Thm and Alaoglu's Thm? From Brezis book Kakutani's states: Let E be a Banach space. Then E is reflexive if and only if the closed unit ball is compact in the weak topology. You refer to this as Alaoglu's theorem? $\endgroup$
    – Alex
    Feb 25, 2014 at 15:50
  • 1
    $\begingroup$ @Alex "Alaoglu's Theorem" states the closed unit ball of a dual space is weak*-compact. of course, what I wrote above is just half of Kakutani's Theorem. $\endgroup$ Feb 25, 2014 at 15:58

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