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?

  • $\begingroup$ 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. $\endgroup$ Oct 11, 2016 at 7:56
  • $\begingroup$ 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 $a,b$. $\endgroup$ Oct 11, 2016 at 8:13
  • $\begingroup$ @Jochen : That's very helpful, thank you. $\endgroup$
    – Jon-S
    Oct 11, 2016 at 12:41

2 Answers 2


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.

  • $\begingroup$ So that I understand well : how do we know if $X$ is a separable Banach space, that $X^{**}$ is reflexive and separable ? $\endgroup$
    – Jon-S
    Oct 10, 2016 at 16:01
  • $\begingroup$ 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. $\endgroup$ Oct 10, 2016 at 16:23
  • 1
    $\begingroup$ It is standard lcs terminology that such a space is reflexive if it is isomorphic as a tvs with its bidual (in the standard way), whereby one uses the strong topology with both dualities. It is semireflexive if it is isomorphic only as a vector space to the bidual. In this sense, the above example is semireflexive, but not reflexive. $\endgroup$
    – vltava
    Oct 10, 2016 at 17:27
  • $\begingroup$ I have looked into this in more detail. There seems to be some discrepancy in how to define biduals of lcs's. According to the definition given by Grothendieck, the above space would be reflexive. I don't have any other sources at hand but wikipedia and my recollection of graduate courses suggest that his is not the standard definition. $\endgroup$
    – vltava
    Oct 10, 2016 at 17:41
  • 2
    $\begingroup$ 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). $\endgroup$ Oct 11, 2016 at 7:51

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