Corollary

Let X be a compact topological space, Y a locally compact topological space, $\mu$ a complex measure on X, and $k$ a continuous mapping of $X\times Y$ into $\mathbf{C}$. Then formula (1) defines a compact linear mapping, again denoted by $\widetilde{k}$, of $\mathscr{C}(X)$ endowed with the topology of uniform convergence into $\mathscr{C}(Y)$ endowed with the topology of compact convergence.

Let us verify conditions (i) and (ii) for $k$. For every $y\in Y$, the mapping $k_y$ is continuous on the compact space X, hence $\mu$-integrable. Moreover, the mapping $y\mapsto k_y$ is continuous from Y into $\mathscr{C}(X)$ since X is compact (TG, X, p. 28, th. 3). Since $\|k_y-k_{y'}\|_1\leqslant \|\mu\| \|k_y-k_{y'}\|$ for every $(y, y')\in Y^2$, this mapping of Y into $\mathscr{L}_{\mathbf{C}}^1(X, \mu)$ is continuous. By prop. 2, the mapping $\widetilde{k}$ of $\mathscr{L}_{\mathbf{C}}^{\infty}(X, \mu)$ into $\mathscr{C}(Y)$ is compact. By composition with the canonical mapping $\mathscr{C}(X)\rightarrow \mathscr{L}_{\mathbf{C}}^{\infty}(X, \mu)$, one obtains a compact linear mapping of $\mathscr{C}(X)$ into $\mathscr{C}(Y)$.