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My professor left for self-convincing the following statement: "If $T: X \to Y$ with $X$ Banach is completely continuous (that is, takes $w$-convergent sequences to norm convergent sequences) and $X$ has a separable dual, then $T$ is compact".

Well, I haven't been able to convince myself so far. Not only that, whenever I think I found a way to prove it, I discover myself in need of reflexivity for $X$. Looking in the literature I found theorems both in Conway and Megginson stating that the condition ensuring this implication is actually reflexivity.

So, should I assume my professor made a mistake, or am I missing some subtle point, perhaps separable $X^{\ast}$ implies reflexive $X$?

ulilaka
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Your professor did not make a mistake.

A diagonalization argument will show that if $X^*$ is separable, then any bounded sequence in $X$ has a weakly Cauchy subsequence (c.f. Joeseph Diestel, Sequences and Series in Banach Spaces, pg. 200) or this answer.

A completely continuous operator maps weakly Cauchy sequences to norm convergent sequences. See this post for a proof of this statement.

(Incidentally, it follows that a completely continuous operator on a Banach space $X$ is compact if $X$ does not contain $\ell_1$. In this case, Rosenthal's $\ell_1$ Theorem insures every bounded sequence has a weakly Cauchy subsequence.)

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David Mitra
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    I presume this "diagonalization argument" is equivalent to invoking the Banach-Alaoglu theorem for the weak-* topology on $X^{}$? If $X^*$ is separable then the closed ball of $X^{}$ is weak-* compact metrizable, so any bounded sequence in $X$ has a subsequence converging weak-* in $X^{**}$; in particular this subsequence must be weakly Cauchy. – Nate Eldredge Nov 27 '13 at 19:13