# completely continuous implies compact

I'm searching for a proof of the fact that if: $T$ is a bounded operator in a reflexive Banach space that maps weakly convergent sequences onto convergent sequences then $T$ is compact.

If we let $\{x_n\}$ be any bounded sequence in $X$, then $|x*(x_n)|\leq ||x*|| ||x_n||\leq K$ for some constant $K$ since $\{x_n\}$ bounded. Then for each $x*$ we have a subsequence $\{x*(x_{nj}\}$ that converges to some real number $y_{x*}$ if $X*$ was countable we could construct a subsequence for which all $x*$ converges. The problem is also to show that there exists some $x_0\in X$ for which $x*(x_0)=y_{x*}$ for all $x*$. I guess there is somewhere in this mess where we use that $X$ is reflexive and Banach?

• I've just learned here that "the term completely-continuous operator originally meant what we now call compact operator, which has sometimes resulted in confusion. – Julien Dec 5 '13 at 23:44

Accepting that as given, from any bounded sequence $(x_n)$ we can extract a weakly convergent subsequence $(x_{n_k})$, and by the complete continuity, $T(x_{n_k})$ is convergent. Thus the image of the closed unit ball is relatively sequentially compact, and since we're dealing with Banach spaces, it is relatively compact.
Now a sketch of a proof of the used half of the Eberlein-Šmulian Theorem. Let $X$ be a reflexive space, and let $(x_n)$ be a sequence in the closed unit ball of $X$. Let $E = \overline{\operatorname{span} \{ x_n : n \in \mathbb{N}\}}$. Then $E$ is a separable reflexive Banach space, hence its dual is also separable, and therefore the closed unit ball of $E$ is metrisable in the weak topology, hence it is weakly sequentially compact. But a subsequence that is weakly convergent in $E$ is also weakly convergent in $X$.