Metrizability of a subset in the weak topology. Let $X$ be a Banach space (not reflexive). It is well-known that $(X,w)$, which is $X$ with its weak topology, is not metrizable if $X$ is infinite dimensional. > I want to know under which condition can a subset $S\subset X$ be given a metric that is compatible with the weak topology $(S,w)$? If $X$ is reflexive then, if I recall correctly, the norm-boundedness of $S$ is enough. However, I am dealing with a non-reflexive space $X=W^{1,1}(\Omega)$ so the previous criterion is not applicable. What if I assume that $S$ is norm-compact? Would that be enough?

Reflexivity isn't the right kind of condition for metrizability of the weak topology. Instead we have the following result.

> **Lemma:** Let $X$ be a normed space. The relative weak topology on the unit ball $B_X$ of $X$ is metrizable if and only if $X^*$ is separable.

In particular, if $X$ is reflexive but has non-separable dual then we can conclude that your first claim about boundedness of $S$ being enough in reflexive spaces is false.

It also follows from the lemma that no set with non-empty interior (in the norm topology) can be weakly metrizable if $X^*$ is not separable. If such a set were metrizable, we could fit a homeomorphic image of $B_X$ into it and then, since a subspace of a metrizable space is metrizable, we would have that $B_X$ is metrizable.

In the particular case $X = W^{1,1}(\Omega)$ we have that $X^* = W^{1,\infty}(\Omega)$ is not separable and so $B_X$ (and hence any set with non-empty interior in the norm topology) is not metrizable for the weak topology.