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 uniformly to 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 finite subcover $U_{y_1},\mathrm{\dots },U_{y_n}$ 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 $\overline{V}$, so there exists a $\delta >0$ such that when $y,y^{\prime }\in \overline{V}$ and , we have .

From the uniform convergence of $f_n$, choose $N$ so that when $n\ge N$, 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 $x\in X$. ∎

Title	proof of composition limit law for uniform convergence
Canonical name	ProofOfCompositionLimitLawForUniformConvergence
Date of creation	2013-03-22 15:23:08
Last modified on	2013-03-22 15:23:08
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	4
Author	stevecheng (10074)
Entry type	Proof
Classification	msc 40A30