Let $(X,\tau)$ be a topological vector space such that the associated dual space $X^*$ is separable. Can we say that $X$ is separable ?
I know that this property is valid for Banach spaces but for topological vector spaces, I have no idea.
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7$\begingroup$ Just consider a non-separable $X$ with $X^*=\{0\}$. Maybe you want $X$ to be locally convex. $\endgroup$– YCorCommented Mar 31, 2020 at 14:23
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2$\begingroup$ Since we know this is true for Banach spaces and false for a rather exotic lcs example, I am tempted to ask if there is a positive answer for some of the central classes of well-behaved spaces—the obvious candidates are the (lc) Fréchet spaces and $(DF)$-spaces. $\endgroup$– user131781Commented Apr 2, 2020 at 11:21
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2 Answers
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YCor has given a counterexample for topological vector spaces. The statement is still false for locally convex spaces. Consider the space $X$ defined to be a locally convex coproduct of $\newcommand{\R}{\mathbb{R}}\R$, $\R$-many times. So $X$ is the space of functions $\R \rightarrow \R$ that are zero except for on a finite subset of $\R$, equipped with the locally convex coproduct topology.
The dual space $X^* \cong \R^\R$ by the usual pairing of an element of $X$ with an element of $\R^\R$ by summation, and for the topology it doesn't matter whether we take the strong dual topology or the weak-* topology, either way we get the usual product topology on $\R^\R$. This space is separable, because the polynomial functions with rational coefficients $\R \rightarrow \R$ forms a countable dense subset.
However, $X$ itself is not separable. If $X$ were separable, we would have some countable set $D \subseteq X$ such that each $f \in \R^\R$ is determined by its values when paired with each $\phi \in D$. But since each element of $D$ has finite support, $D$ only determines countably many "coordinates" in $\R^\R$, so we could pick two $f_1,f_2 \in \R^\R$ agreeing on those elements of $\R$ but not being equal. This contradicts $D$ being dense.
Instead of an $\R$-fold coproduct we could have taken any $\kappa$-fold coproduct where $\aleph_0 < \kappa \leq 2^{\aleph_0}$, by the Hewitt-Marczewski-Pondiczery theorem.
answered Mar 31, 2020 at 15:35
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$\begingroup$ For $X^*$ to be ${\mathbb R}^{\mathbb R}$, every linear functional on $X$ must be continuous. That is, the initial topology must be locally convex. Is this obvious? $\endgroup$ Commented Mar 31, 2020 at 16:19
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1$\begingroup$ @მამუკაჯიბლაძე That depends on what you mean by "initial topology". It is true that every linear functional is continuous (and it has to be, to be a coproduct). However, the topological vector space initial topology and locally convex initial topology are not identical for uncountable coproducts. This is why I specified the "locally convex coproduct topology" (which should be considered as one word, rather than as saying that the copduct topology is locally convex). See Schaefer's Topological Vector Spaces section II.6. The coproduct is called the "locally convex direct sum" there. $\endgroup$ Commented Mar 31, 2020 at 16:31
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$\begingroup$ Thank you for the explanation. It is subtler than I expected, I thought you meant that the topological vector space coproduct is locally convex in this case, which is presumably not true. $\endgroup$ Commented Mar 31, 2020 at 17:09
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1$\begingroup$ @მამუკაჯიბლაძე In terms of counterexamples, in Schaefer it is just not stated that the two are the same, and the fact that they are the same for a $\aleph_0$-fold coproduct is set as an exercise. But in this answer and this comment by Paul Garrett links are provided to counterexamples for the case of an uncountable coproduct. $\endgroup$ Commented Apr 1, 2020 at 1:34
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Let $\Gamma$ be a set and consider the product measure $\mu^{\otimes \Gamma}$, where $\mu$ is the Lebesgue measure on the unit interval. For any $p\in (0,1)$, the space $L_p(\mu^{\otimes \Gamma})$ has trivial dual because $\mu^{\otimes \Gamma}$ is atomless. However, when $\Gamma$ is uncountable, the space $L_p(\mu^{\otimes \Gamma})$ is non-separable.
answered Apr 2, 2020 at 8:07