# proof of composition limit law for uniform convergence

###### Theorem 1.

Let $X,Y,Z$ be metric spaces, with $X$ compact and $Y$ locally compact. If $f_{n}\colon X\to Y$ is a sequence of functions converging uniformly to a continuous function $f\colon X\to Y$, and $h\colon Y\to Z$ is continuous, then $h\circ f_{n}$ converge to $h\circ f$ 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 finite subcover $U_{y_{1}},\ldots,U_{y_{n}}$ 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 $\epsilon>0$ be given. $h$ is uniformly continuous on $\overline{V}$, so there exists a $\delta>0$ such that when $y,y^{\prime}\in\overline{V}$ and $d(y,y^{\prime})<\delta$, we have $d(g(y),g(y^{\prime}))<\epsilon$.

From the uniform convergence of $f_{n}$, choose $N$ so that when $n\geq N$, $d(f_{n}(x),f(x))<\min(\delta,\delta_{0})$ 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 $d(g(f_{n}(x)),g(f(x)))<\epsilon$ when $n\geq N$, uniformly for all $x\in X$. ∎

Title proof of composition limit law for uniform convergence ProofOfCompositionLimitLawForUniformConvergence 2013-03-22 15:23:08 2013-03-22 15:23:08 stevecheng (10074) stevecheng (10074) 4 stevecheng (10074) Proof msc 40A30