Let $X$ and $Y$ be topological spaces, and consider the compact-open topology on $C(X,Y)$, which is generated by open sets of the form
$$\{\text{continuous }f\colon X\to Y:f(K)\subseteq U\}\text{ for compact }K\subseteq X\text{ and open }U\subseteq Y.$$
Without assuming that $Y$ is a metric space, are there conditions on $X$ and $Y$ that imply that the compact open topology on $C(X,Y)$ is locally compact?
One related result of particular importance is the fact that $\mathrm{Hom}(G,T)$ is locally compact, where $G$ is a locally compact Hausdorff abelian group, and $T$ is the circle group. This is important for Pontryagin duality, because it is one part of knowing that the Pontryagin dual of a locally compact Hausdorff abelian group is still a locally compact Hausdorff abelian group.
For this related result, the standard proof uses equicontinuity and Arzela-Ascoli. But this approach only works if $Y$ is a metric space.
Best Answer
This is not a complete answer, but gives some useful information. Let us assume that $Y$ is Hausdorff.
Necessary condition: If $C(X,Y)$ is locally compact, then $Y$ is locally compact.
It is well-known that $j : Y \to C(Y,X), j(y) =$ constant map with value $y$, is an embedding. If $Y$ is Hausdorff, then $j$ is a closed embedding which proves 1.
Sufficient condition: If $Y$ is locally compact and $X$ is a finite discrete space, then $C(X,Y)$ is locally compact.
This is true because $C(X,Y)$ is homeomorphic to a finite product of copies of $Y$.
If $Y$ is locally compact non-compact and $X$ is an infinite discrete space, then $C(X,Y)$ is not locally compact.
This is true because $C(X,Y)$ is homeomorphic to an infinite product of copies of $Y$.
If $X$ is an infinite compact Hausdorff space, $C(X,\mathbb R)$ is not locally compact.
It is well-known that for compact Hausdorff $X$ the compact open topology on $C(X,\mathbb R)$ agrees with the $\sup$-norm topology. Thus $C(X,\mathbb R)$ is a normed linear space which is locally compact if and only if it is finite-dimensional. But for an infinite $X$ the space $C(X,\mathbb R)$ is not finite-dimensional. To see this, let $i : A \hookrightarrow X$ be the inclusion of a closed subspace $A$. It induces a linear map $i^* : C(X,\mathbb R) \to C(A,\mathbb R), i^*(f) = f \mid_A$. By the Tietze extension theorem this map is onto. But for each $n$ the set $X$ contains finite subsets $A_n$ with $n$ elements, thus we get linear surjections $C(X,\mathbb R) \to C(A_n,\mathbb R) \approx \mathbb R^n$.
This shows that it is quite difficult to get a locally compact $C(X,Y)$.