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For a Tychonoff space $X$ let $C_k(X)$ denote the space of continuous real-valued functions on $X$, endowed with the compact-open topology.

Problem. Is the space $C_k(X)$ Polish if it is Polishable in the sense that it admits a stronger Polish group topology?

Remark. It is known that $C_k(X)$ is Polish if and only if $X$ is a submetrizable hemicompact $k$-space. On the other hand, a Tychonoff space $X$ is submetrizable and hemicompact if $C_k(X)$ is Polishable.

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  • $\begingroup$ What about $\mathbb{Q}$? At least for the bounded functions: the compact-open topology is not first-countable yet the sup-norm topology is Polish. $\endgroup$ – KP Hart Sep 16 '18 at 11:44
  • $\begingroup$ @KPHart Why the sup-norm on $C_b(\mathbb Q)$ is Polish? That space contains a copy of $\ell_\infty=C_b(\mathbb N)$. Counterexample if exists should be a submetrizable hemicompact space, which is not a $k$-space. But $\mathbb Q$ is a $k$-space and is not hemicompact. $\endgroup$ – Taras Banakh Sep 16 '18 at 12:07
  • $\begingroup$ You're right, I wrote in haste $\endgroup$ – KP Hart Sep 17 '18 at 8:43
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This problem has a partial affirmative answer:

Theorem. For any Tychonoff space $X$ the function space $C_k(X)$ is Polish if and only if it admits a stronger Polish locally convex topology.

Proof: The "only if" part is trivial. To prove the "if'' part, fix a stronger Polish locally convex topology $\tau$ on $C_k(X)$ and denote the Polish locally convex space $(C_k(X),\tau)$ by $C_\tau(X)$.

It follows that the idenity map $C_\tau(X)\to C_k(X)$ is continuous and so is the identity map $C_\tau(X)\to C_p(X)$ where $C_p(X)$ is the space $C(X)$ of all continuous real-valued functions on $X$, endowed with the topology of pointwise convergence. Then the space $C_p(X)$ has countable network of the topology, being a continuous image of the Polish space $C_\tau(X)$. By the Duality Theorem I.1.3 from the book "Topological Function spaces" of Arkhangelskii, the space $X$ has a countable network and hence is Lindelof and a $\mu$-space (= all closed bounded sets in $X$ are compact). Since $X$ is a $\mu$-space, the function space $C_k(X)$ is barrelled, see Theorem 10.1.20 from the book "Barrelled locally convex space" by Carreras and Bonet.

Since the identity map $C_\tau(X)\to C_k(X)$ is continuous, it has closed graph. Then the "inverse" identity map $C_k(X)\to C_\tau(X)$ also has closed graph and hence is continuous by the Closed Graph Theorem 4.1.10 from the book of Carreras and Bonet. Now we see that the identity map $C_\tau(X)\to C_k(X)$ is a topological isomorphism and $C_k(X)$ is Polish.


So it remains to answer the following

Problem. Let $X$ be a Tychonoff space and $C_k(X)$ be the space of continuous real-valued functions endowed with the compact-open topology. Are the following conditions equivalent?

1) $X$ admits a stronger Polish group topology;

2) $X$ admits a stronger Polish locally convex topology.

Remark. The space $\ell_{1/2}:=\{(x_n)\in\ell_1:\sum_{n=1}^\infty|x_n|^{1/2}<\infty\}$ has a Polish group topology but fails to have a Polish locally convex topology.

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