Sorry for my bad english.
Let $X$ and $Y$ be two topological spaces, and $G$ a topological group, let $\theta : G \times X \to X$ be a continuous action of $G$ on $X$. We defined the action of $G$ on the space of continuous functions from $X$ to $Y$ with compact open topology, $\phi : G \times \mathcal{C} (X,Y) \to \mathcal{C} (X,Y)$, where $[\phi (g,f)](x) = f(g^{-1} x)$ for all $x \in X$.
Now, I have to prove that if $G$ is locally compact then $\phi$ is a continuous action.
I have already seen that $\phi$ is well defined and that $\phi$ is an action, but I can not prove that $\phi$ is continuous. I found in a book the case when $G = X$ and it say it follows from the Theorem that said that $\phi$ is continuous iff the function $\phi *: (G \times \mathcal{C} (X,Y)) \times X \to Y$ is continuous, where $\phi * ((g,f),x) = [\phi (g,f)] (x) = f(g^{-1} x)$, but I can't see how it follows.
What I wanted to do, is use the Theorem that said: The continuity of an action $\phi : G \times \mathcal{C} (X,Y) \to \mathcal{C} (X,Y)$ of a topological group $G$ with identity $e$ on a space $\mathcal{C} (X,Y)$ is equivalent to the continuity of $\phi$ at the points of the set $\lbrace e \rbrace \times \mathcal{C} (X,Y) \subset G \times \mathcal{C} (X,Y)$. So I only need to see that if $f \in \mathcal{C} (X,Y)$ then for every neighborhood $(K,U)$ of $f$, in compact open topology, there exist neighborhoods $O$ of $e \in G$ and $(K',U')$ of $f$ such that $\phi (O \times (K',U')) \subset (K,U)$, but I don't know how to use the hypothesis of G locally compact.