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Banach–Alaoglu theorem states that:

The closed bounded subsets of $X^*$ is compact with respect to the weak-* topology.

Is closeness and boundedness in weak-* topology on $X^*$? or the norm topology on $X^*$

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    $\begingroup$ Compact isn't equivalent to closed and bounded in general. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Commented Mar 12, 2019 at 20:14

2 Answers 2

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It is norm-closed and norm-bounded. A usual formulation of Banach-Alaoglu states that the closed unit ball of $X^*$ is weak-* compact.

EDIT: this is incorrect. The set must be weak* closed.

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  • $\begingroup$ The usual formulation and the one here are equivalent. Aren't they? $\endgroup$
    – Saj_Eda
    Commented Mar 12, 2019 at 19:39
  • $\begingroup$ Yes. It is a straightforward exercise to prove. $\endgroup$
    – TM Gallagher
    Commented Mar 12, 2019 at 19:42
  • $\begingroup$ Sorry, but this is wrong. $\endgroup$
    – Jochen
    Commented Mar 14, 2019 at 7:56
  • $\begingroup$ Yes, of course you are correct. I had my implications reversed. $\endgroup$
    – TM Gallagher
    Commented Mar 14, 2019 at 21:09
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If $X$ is Banach, then norm-boundedness and weak$^*$-boundedness are equivalent because of the Banach-Steinhaus theorem. However, norm-closed bounded sets needn't be weak$^*$-closed (and thus cannot be weak$^*$-compact). An example is the unit ball of $c_0$ considered as a subset of $\ell_\infty= \ell_1^*$.

EDIT. The usual statement is that the polar $U^\circ=\{\varphi^*\in X^*: |\varphi(x)|\le 1\}$ of any neighbourhood of $0$ of a Banach space (or any topological vector space) is weak$^*$-compact. Hence all weak$^*$-closed and equicontinuous subsetes of $X^*$ are weak$^*$-compact. The example of the unit ball of $c_0$ in $\ell_\infty$ shows that one cannot replace weak$^*$-closedness by norm-closedness.

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  • $\begingroup$ This is different form the above response. Which one is meant in Banach-Alaoglu theorem? $\endgroup$
    – Saj_Eda
    Commented Mar 13, 2019 at 20:20
  • $\begingroup$ The version which is true in the dual of any topological vector space states that equicontinuous weak*-closed sets are weak*-compact. $\endgroup$
    – Jochen
    Commented Mar 13, 2019 at 20:30
  • $\begingroup$ I don't think on Banach spaces "closed-ness" is with respect to weak-* topology. $\endgroup$
    – Saj_Eda
    Commented Mar 13, 2019 at 20:37
  • $\begingroup$ But for norm-closed it is not true. $\endgroup$
    – Jochen
    Commented Mar 13, 2019 at 20:59
  • $\begingroup$ Why the downvote? $\endgroup$
    – Jochen
    Commented Mar 14, 2019 at 8:03

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