Let be a Banach space and its continuous dual. By we denote the closed unit ball in centered at zero. We will prove the following metrization theorem and its “dual” form:

Theorem 1. is weak*-metrizable iff is separable.

Theorem 2. is weakly metrizable iff is separable.

We will prove both of them and discuss alternative proofs. We will kill two birds with one stone since proofs are very similar. However, there is one moment when this symmetry breaks, which makes this feat more interesting!

We mostly follow Brezis’ textbook “Functional Analysis, Sobolev Spaces and Partial Differential Equations”. However, proof of the ⇒ part of theorem 2 and alternative, shorter, proofs in the comment section are mine (probably there are similar proofs somewhere, so I guess it belongs to folklore).

Let’s jump straight to the proofs. The reader can find definitions in the last section.

# Proof of ⇐

We first sketch proof of ⇐ part of Theorem 1: Separable implies weak*-metrizable ball. Let be a dense sequence in . Map defines a norm that agrees with the weak* topology on . It’s quite cumbersome to verify this, but it works.

The analogous method (with dense sequence and norm ) works for the “dual” statement.

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The reader might have seen a similar series in when proving that a countable product of metric spaces remains a metric space. There is a much shorter, alternative proof of ⇐ part of theorem 1 described in the “Comments” section, that relies on this exact general fact.

# Proof of ⇒

We first show ⇒ part of Theorem 1. Assume be weak*-metrizable, say with a metric . We will show that is separable.

For each natural let Note that these open balls form a local base of zero. Moreover, for each there is weak* open basic set . That is, there is finite and with Now, let It suffices to show that the linear span of is dense. Indeed, finite linear combinations of elements of with rational coefficients form a countable dense subset of .

To show that subspace is dense in we use the following fact.

*Fact 1.* Linear subspace
of a normed space
is dense iif whenever
disappears on
, then
disappears on the whole space.

Take with . Then for all we have . This means for all . But since , we have And we are done.

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Now we prove the “dual” statement: If is weak-metrizable, then is separable.

Here the symmetry breaks… Nevertheless, we start somehow similarly, modulo a technical caveat: We need to consider as a subset of vie the canonical map where . Why we need to do this? Try to proceed without this step and see where the proof breaks. Anyway, this map is isometry hence Since the canonical injection is a homeomorphism between with weak topology and with weak* topology, we get that is weak*-metrizable.

Now, let ’s form a local base of zero coming from the metric in . Then, we find weak* open basic sets where is finite and . Let Take any with . As before, it suffices to show that . Now, however, we cannot make the analogous conclusion, since might not live in .

WLOG (otherwise scale , do the work, and scale it back). Since is weak*-dense in (Goldstine lemma) we find a net in withThis means that for every we have . In particular, since is finite, for there is such that implies Hence for any there is such that implies . This means in . Recall that by assumption. Since is a net in and weak* topology is Hausdorff (this guarantees unique limits), we conclude that as needed.

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# Comments

- Alternative (shorter but less direct) proof of ⇐ part of theorem 1:
Separable
implies
is weak*-metrizable.
*Proof.*Let be a dense subset in . Consider the following mapIt is injective (functions to a Hausdorff space are uniquely described on a dense subset). It is continuous (in both topologies convergence = pointwise convergence). Since the domain is compact (Banach–Alaoglu theorem) it must be homeomorphism onto its image. - The analogous version of this proof of a “dual” statement has to be modified since does not need to be weakly compact. So extra work, namely continuity of the inverse, has to be done.
- However, we can choose another path using already established theorem 1. Assume is separable. Then is weak*-metrizable by theorem 1. Since with weak topology can be considered as a subset of with weak* topology, we are done.

# Definitions

Let be a Banach space. By we denote a set of continuous linear functionals. By () we denote the closed unit ball (dual) centered at zero.

By weak topology on we mean a weak topology generated by . We denote it by . This is the smallest topology on that makes all functions from continuous. It is generated by a subbasis made of for and . By weak* topology on we mean the weak topology generated by evaluation functionals , that is for each . Since they can be identified with elements of it is sensible to denote weak* topology by . It is formed by subasis made of where and .