# 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 $\u03f5>0$ 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 |