proof of composition limit law for uniform convergence

Theorem 1.

Let $X\mathrm{,}Y\mathrm{,}Z$ be metric spaces, with $X$ compact and $Y$ locally compact.If $f_n\mathrm{:}X\mathrm{\to }Y$ is a sequence of functions converging uniformlyto a continuous function $f\mathrm{:}X\mathrm{\to }Y$, and $h\mathrm{:}Y\mathrm{\to }Z$is continuous, then $h\mathrm{\circ }{f}_n$ converge to $h\mathrm{\circ }f$ uniformly.

Proof.

Let $K$ denote the compact set $f(X)\subseteq Y$. By local compactness of $Y$,for each point $y\in K$, there is an open neighbourhood $U_y$ of $y$ such that$\overline{U_y}$ is compact. The neighbourhoods $U_y$ cover $K$, so there is a finitesubcover $U_{y_1},\mathrm{\dots },U_{y_n}$ covering $K$. Let$U={\bigcup }_i{U}_{y_i}\supseteq K$. Evidently$\overline{U}={\bigcup }_i\overline{U_{y_i}}$ is compact.

Next, let $V$ be the ${\delta }_0$-neighbourhood of $K$contained in $U$, for some ${\delta }_0>0$.$\overline{V}$ is compact, since it is contained in $\overline{U}$.

Now let $ϵ>0$ be given.$h$ is uniformly continuous on $\overline{V}$, sothere exists a $\delta >0$ such thatwhen $y,y^{\prime }\in \overline{V}$ and ,we have .

From the uniform convergence of $f_n$, choose $N$ so thatwhen $n\ge N$, for all $x\in X$.Since $f(x)\in K$, it follows that $f_n(x)$ is inside the${\delta }_0$-neighbourhood of $K$, i.e. both $y=f_n(x)$ and $y^{\prime }=f(x)$are both in $V$. Thus when $n\ge N$,uniformly for all $x\in X$.∎