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.