(Theorem) The limit of a uniformly convergent sequence of continuous functions is continuous

Theorem (Munkres 21.6, Rudin 7.12): Uniform limit

Let $f_n:X\to Y$ be a sequence of continuous functions from the topological space $X$ to the metric space $Y$. If $(f_n)$ converges uniformly to $f$, then $f$ is continuous.

wip

Note that the converse is true only under particular conditions.

Theorem (Rudin 7.13): Dini’s convergence theorem

Let $X$ be compact, and suppose $(f_n)$ is a sequence of continuous functions on $X$ such that:

• (i) There exists a continuous function $f$ on $X$ such that $f_n\to f$ pointwise;
• (ii) $f_n(x)\ge f_{n+1}(x)$ for all $x\in X$, $n=1,2,3,\dots$.

Then $f_n\to f$ converges uniformly on $X$.

Proof from Modern Analysis I. Uses:

• (Proof pattern) Defining a new function
• (Proof pattern) Trivial subset

Define a new function $g_n=f_n-f$. We know $g_n$ is continuous for all $n$, and $g_n\to 0$ pointwise. It remains to show that $g_n\to 0$ uniformly on $X$.

Fix $ϵ>0$, and let $K_n$ be the set of all $x\in X$ such that $g_n(x)\ge ϵ$. Since $g_n$ is continuous and the set

C = \{ g_n(x) \ | \ g_n(x) \geq \epsilon \} $$ is finite (❓ is this why?), the pre-image $g^{-1}(C) = K_n$ is closed, hence [[Compactness|compact]] as a subset $K_n \subseteq X$. Given any $x \in X$, the fact that $g_n \to 0$ implies that $x \notin K_n$ for sufficiently large $n$. Thus $x \notin \bigcap K_n$ and $\bigcap K_n = \varnothing$. Further, the fact that $g_n(x) \geq g_{n+1}(x)$ implies $K_{n+1} \subseteq K_n$ is a nested sequence of compact sets. Then we may apply the negation of [[(Theorem) Cantor intersection]] to find some $N$ such that $K_N$ is empty, and it follows that

n \geq N \implies 0 \leq g_n(x) < \epsilon

for all $x \in X$.