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Let $ X $ be Banach space, and $X^*$ its dual. A set $ F \subset X ^ * $ is weakly-* compact if and only if $ F $ is closed in the weak* topology and is bounded in norm.

How does one prove this (well-known) fact?

Notes

This characterization of compactness in weak* topology is similar to compactness in finite dimensional spaces, where it is equivalent to being closed and bounded.

The analogous statement for weak topology is false. An example is given in Weakly compact implies bounded in norm: the closed unit ball of $c_0$ is weakly closed and norm bounded, but not weakly compact.

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    $\begingroup$ Hint: Alaoglu-Bourbaki theorem. $\endgroup$ – Daniel Fischer Jul 9 '13 at 19:56
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Daniel Fischer said it all. I would like to make a precision which would have been a comment if I had enough reputation.

The key point is that the closed unit ball of $X^*$ is compact for the weak*-topology. That's called Banach-Alaoglu theorem.

Now to get your equivalence, you also need the following easier facts:

  • the weak*-topology is Hausdorff.

  • a compact in a Hausdorff space is closed (and Hausdorff is needed).

  • a closed subset of a compact space is always compact.

  • a norm bounded set in a normed vector space is contained in a positive-scalar-multiple of the closed unit ball.

  • to show that weak*-compact implies norm bounded, you need the Uniform Boundedness Principle and the fact that the image of a compact space by a continuous complex-valued map (point evaluations in this case, which are continuous by definition of the weak*-topology) is compact in $\mathbb{C}$, whence bounded.

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  • $\begingroup$ Ah, that works too, of course. I'd have said Ascoli-Bourbaki ;) $\endgroup$ – Daniel Fischer Jul 9 '13 at 20:27
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    $\begingroup$ @DanielFischer Or "it follows from Tychonoff..." ;). $\endgroup$ – user85486 Jul 9 '13 at 20:31

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