# Are bounded sets always weakly metrizable in reflexive separable spaces?

It is known that if a Banach space is reflexive and separable, its unit ball is weakly metrizable.

My question is about the generalization of this property :

1) Is it true that for all reflexive separable locally convex space, bounded sets are weakly metrizable?

2) If that's true, is there a way to explicitly construct a distance for the weak-topology on any bounded set?

• You would need that the dual is separable to have the bounded sets weakly metrizable. If your space $X$ is metrizable separability of $(X',\sigma(X',X))$ follows from the separability of $X$, but in general, I doubt. As a reference I recommend the book Barrelled Locally Convex Spaces of Bonet and Perez-Carreras, in particular, section 2.5. – Jochen Wengenroth Oct 11 '16 at 7:56
• Concerning the second question: If $\varphi_n$ is a dense sequence in $X'$ the following defines a metric which gives the weak topology on all bounded sets: $d(x,y)=\sup\lbrace |\varphi_n(x-y)| \wedge 1/n: n\in\mathbb N\rbrace$, where $a\wedge b$ is the minimum of $a,b$. – Jochen Wengenroth Oct 11 '16 at 8:13
• @Jochen : That's very helpful, thank you. – Jon-S Oct 11 '16 at 12:41

No. Let $I$ be an index set with the cardinality of the continuum. Endow $X=\mathbb R^I$ with the product topology. According to (a particular case of) the Hewitt-Marczewski-Pondiczery theorem (which is 2.3.15 in Engelking's General Topology) $X$ is separable. Moreover, it is semi-reflexive (by Tychonov) and barrelled (because barrelledness is stable w.r.t. products, proposition 4.2.5 in Barrelled Locally Convex Spaces of Bonet and Perez Carreras). Therefore, $X$ is reflexive. The set $B=[-1,1]^I$ is bounded. Assume that it is weakly metrizable (by the way, $X$ carries its weak topology). Then it has a countable basis of $0$-neighbourhoods and since every $0$-neighbourhood in $X$ contains one of the form $\lbrace x\in X: |\varphi_i(x)|<\varepsilon, i=1,\ldots,n \rbrace$ for some $n\in \mathbb N$, $\varphi_1,\ldots,\varphi_n \in X'$, and $\varepsilon>0$, we find a sequence of $\phi_n\in X'$ such that $$B\cap \bigcap_{n\in\mathbb N} \text{kern}(\phi_n) = \lbrace 0 \rbrace.$$ But each $\phi_n$ only depends on finitely many coordinates, i.e., it is of the form $\phi_n((x_i)_{i\in I})= \sum_{i\in E_n} a_{n,i} x_i$ for some finite $E_n \subseteq I$, and it is enough to consider $j\in I\setminus \bigcup_{n\in\mathbb N} E_n$ and $b=(\delta_{i,j})_{i\in I} \in B$ to get a contradiction.
No. Let $X$ be any separable Banach space s.t. $X^*$ is non separable and consider the space $X^{**}$ with the weak$^*$ topology.
Edit: As Jochen and vitava point out in the comments below, this only gives a semi-reflexive example even if you use the Mackey topology on $X^{**}$ with respect to the duality $(X^{**}, X^*)$. It is not reflexive under the classical definition of reflexive for LCTVS.
• So that I understand well : how do we know if $X$ is a separable Banach space, that $X^{**}$ is reflexive and separable ? – Jon-S Oct 10 '16 at 16:01
• For separability, the unit ball of $X$ is weak$^*$ dense in the unit ball of $X{^**}$ (Goldstine's theorem). It is obvious that any dual Banach space is reflexive in the weak$^*$ topology. – Bill Johnson Oct 10 '16 at 16:23
• I think that vltava's comment does matter. $X^{**}$ with the Mackey topology from the dual pair $(X^*,X^{**})$ isn't reflexive in the sense that it coincides topologically with its bidual (the latter endowed with the strong topology = uniform convergence on all bounded sets). The strong bidual of $(X^{**}$, Mackey$)$ is the Banach space topology of $X^{**}$ which is strictly finer than the Mackey topology (it has even a larger dual). – Jochen Wengenroth Oct 11 '16 at 7:51