Lemma 3

Let X be a locally compact space, Y and Z topological spaces, $\varphi$ a continuous mapping of $X \times Y$ into Z, and $\varphi_x$ the mapping $y \mapsto \varphi(x,y)$ of Y into Z. The spaces $\mathscr{C}(Y), \mathscr{C}(Z)$ being equipped with the topology of compact convergence, the mapping $(x,f) \mapsto f \circ \varphi_x$ of $X \times \mathscr{C}(Z)$ into $\mathscr{C}(Y)$ is continuous.

It clearly suffices to consider the case that X is compact. Let $(x_0, f_0) \in X \times \mathscr{C}(Z)$, K a compact subset of Y, and $\varepsilon > 0$. Let $K' = \varphi(X \times K)$. Since $f_0 \circ \varphi$ is uniformly continuous in $X \times K$, there exists a neighborhood W of $x_0$ such that $|f_0(\varphi(x,y)) - f_0(\varphi(x_0,y))| \leq \varepsilon$ for $x \in W$ and $y \in K$. On the other hand, if one takes $f \in \mathscr{C}(Z)$ to be such that $|f(z) - f_0(z)| \leq \varepsilon$ for all $z \in K'$, one will have $|f(\varphi(x,y)) - f_0(\varphi(x,y))| \leq \varepsilon$ for $x \in X, y \in K$, and therefore $|f(\varphi(x,y)) - f_0(\varphi(x_0,y))| \leq 2\varepsilon$ for $x \in W, y \in K$. Whence the lemma.

Let us now return to the earlier notations.