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Let $\phi_n\colon [a,b]\to \mathbb R$ be a sequence of continuous functions. Assuming there is an $A\subset [a,b]$ such that $\phi_n|_A$ converges uniformly to $\phi\colon A\to \mathbb R$, I have already proven that for all $x\in \bar A$ the closure of $A$ the sequence $(\phi_n(x))$ converges. Defining $\bar\phi$ on $\bar A$ by the pointwise limit of $\phi_n$ it seems to me, $\bar\phi$ should be continuous on $\bar A$ but I cannot bring a proof together.

My idea: I have already proven, that for any finite set $B\subset \bar A$ the function $\bar\phi$ is continuous on $B\cup A$, and I was now trying to extend this to the entire $\bar A$ with Zorn's Lemma, but showing that, if $A\subset B_1\subset B_2 \subset \dots\subset \bar A$ and $\bar\phi$ is continuous on each $B_k$ then it is also on $\bigcup B_k$, is making some problems. Note that could also assume uniform continuity everywhere.

Note the thoughts of Continuity on a union.

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Show that $\phi_n \to \phi$ uniformly on $\overline{A}$. – Qiaochu Yuan May 25 '11 at 16:38
@Qiaochu: Did you think this to the end or is this only a guess how you would approach it? I will try it myself, but it didn't seem to promising to me on first glance. – Peter Patzt May 25 '11 at 18:33
I don't see why this wouldn't work. $\sup_{x \in A} |\phi_n(x) - \phi_m(x)|$ should be equal to $\sup_{x \in \bar{A}} |\phi_n(x) - \phi_m(x)|$, so one sequence is Cauchy in the sup norm if and only if the other is. – Qiaochu Yuan May 25 '11 at 18:40
@Qiaochu: See and I don't know why I didn't consider that. – Peter Patzt May 25 '11 at 18:44

1 Answer 1

up vote 2 down vote accepted

Since the sequence of continuous functions $(\phi_n)$ converges uniformly in $A$, then it converges uniformly also in $\overline{A}$. This fact can be proved in a moment using the Cauchy criterion for uniform convergence, since by continuity $\sup_{a\in A} |\phi_n(x) - \phi_k(x)| = \sup_{x\in\overline{A}} |\phi_n(x) - \phi_k(x)|$. Now, again by the continuity of the $\phi_n$, we conclude that the uniform limit $\phi$ is continuous.

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