# Understanding Banach–Alaoglu theorem

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 $$X^*$$

• Compact isn't equivalent to closed and bounded in general. – Martín-Blas Pérez Pinilla Mar 12 '19 at 20:14

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.

• The usual formulation and the one here are equivalent. Aren't they? – Saj_Eda Mar 12 '19 at 19:39
• Yes. It is a straightforward exercise to prove. – TM Gallagher Mar 12 '19 at 19:42
• Sorry, but this is wrong. – Jochen Mar 14 '19 at 7:56
• Yes, of course you are correct. I had my implications reversed. – TM Gallagher Mar 14 '19 at 21:09

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.

• This is different form the above response. Which one is meant in Banach-Alaoglu theorem? – Saj_Eda Mar 13 '19 at 20:20
• The version which is true in the dual of any topological vector space states that equicontinuous weak*-closed sets are weak*-compact. – Jochen Mar 13 '19 at 20:30
• I don't think on Banach spaces "closed-ness" is with respect to weak-* topology. – Saj_Eda Mar 13 '19 at 20:37
• But for norm-closed it is not true. – Jochen Mar 13 '19 at 20:59
• Why the downvote? – Jochen Mar 14 '19 at 8:03