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

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

Theorem 2. is weakly metrizable iff is separable.

Why these theorems are interesting? Well, an infinite-dimensional Banach space is never weakly metrizable (same for a dual space with weak* topology). But a natural corollary of the above theorems states that all bounded subsets are weakly metrizable whenever dual is separable (and the analogous version for the weak* topology. By the way, Terence Tao made a mistake about this fact which makes the hypothesis that he is only a semi-god more probable 🙃.

In this post we will prove both theorems and discuss alternative proofs. We will kill two birds with one stone since proofs of Theorem 1 and Theorem 2 are quite similar. However, there is one place where this symmetry breaks, which makes the whole feat more interesting!

We mostly follow Brezis’ textbook “Functional Analysis, Sobolev Spaces and Partial Differential Equations”. However, the proof of the ⇒ part of Theorem 2 and shorter proofs in the Comments section are mine (I guess that they belong to folklore, let me know if you saw them somewhere).

We jump straight to the proofs. If needed, the reader can find necessary definitions in the last section. But a familiarity with basis of functional analysis are assumed.

# Proof of ⇐

Theorem 1: Separable implies a weak*-metrizable dual ball.

*Proof.* Let
be a dense sequence in
. The map defines a norm that agrees with the weak* topology on
. This needs a cumbersome verification which can be found
in the mentioned Brezis’ textbook. □

Theorem 2: Separable implies a weak metrizable ball.

*Proof.* Proceed analogously with a dense sequence
and the norm
.

□

*Remark.* The reader might have seen a similar series while
proving that a countable product of metric spaces remains a metric
space. A shorter proof of the ⇐ part of Theorem 1 that relies on this
general fact is described in Comments.

# Proof of ⇒

Theorem 1. Weak*-metrizable implies separable .

*Proof.* Assume that
is weak*-metrizable with a metric
.

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

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

*Fact 1.* A 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.

□

Theorem 2. If is weak-metrizable then is separable.

The proof is by me but it probably belnogs to folkolore.

*Proof.* Here the symmetry breaks… That is, we cannot simply
rewrite the previous proof exchenging
with
etc. (as it was the case in the ⇐ part of the proof).
Nevertheless, we start somehow similarly, modulo a technical caveat: We
need to consider
as a subset of the bidual
via the canonical map
where
. Why we need to do this? Try to proceed without this step
and see where the proof breaks. Anyway, the canonical map is an isometry
hence The map
is also a homeomorphism between
with the weak topology and
with the weak* topology. Therefore
is metrizable with the subsapce topology inherited from
with some metric
.

Then is a local base of zero. There are weak* open basic sets where is finite and . Let Take any such that the restriction . As before, showing that completes the proof. Now, however, we cannot make proceed analogously because 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 with This 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 the weak* topology is Hausdorff (this guarantees unique limits), we conclude that as needed.

□

# Comments

An alternative (shorter but less direct) proof of the ⇐ part of Theorem 1 by me (but, again, probably belongs to the folkolore): A separable implies that is weak*-metrizable.

*Proof.* Let
be a dense subset of
. Consider the following
map It 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 (by Banach–Alaoglu theorem) it must be homeomorphism onto its
image.

□

The analogous version of this proof for the “dual” statement, namely the ⇐ part of Theorem 2, has to be modified since does not need to be weakly compact (in fact unit ball is weakly compact iff is reflexive). So compared to the ⇐ part of Theorem 1 an extra work has to be done (namely, the continuity of the inverse).

However, we can choose another path. Using the already proven Theorem 1 we show that a separable dual of a Banach space implies that a ball is weakly metrizable.

*Proof.* Assume that
is separable. Then
is weak*-metrizable by Theorem 1. Since
with the weak topology can be considered as a subset of
with the weak* topology on bidual 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 the weak topology on we mean a weak topology generated by the dual . We denote it by . This is the smallest topology on that makes all functions from continuous. A subbasic set is of the form where and . By the weak* topology on we mean the weak topology generated by the set of all evaluation functionals , that is for each . Since they can be identified with elements of it is sensible to denote the weak* topology by . A subbaisc set if of the form where and .