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? 



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


