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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
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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

2 Answers 2

up vote 1 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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You want to show that for all $x\in\bar A$, for all $(x_n) \in A^\mathbb{N}$ st. $x_n \rightarrow x$, then $\bar\phi(x_n) \rightarrow \bar\phi(x)$. I suggest you bound $\| \bar\phi(x) - \bar\phi(x_n) \|$ by a sum $\|\bar\phi(x) - \bar\phi_p(x)\| + \|\bar\phi_p(x) - \phi_p(x_n)\| + \|\phi_p(x_n) - \bar\phi(x_n)\|$ for some $p \in \mathbb{N}$, and try to bound them by a certain $\varepsilon$. You've already done most of the work, it should be easy (don't mess up with the order in which you find the bounds!).

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As I stated, I have already proven that $\bar \phi$ is continuous on $\{x\}\cup A$. You didn't prove more. –  Peter Patzt May 25 '11 at 18:31
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I don't believe this is enough: you need this to be true for all sequences which take values in $\bar{A}$. –  Qiaochu Yuan May 25 '11 at 18:37
    
@Qiaochu: Come on! Zulon's bound shows first that $\bar{\phi}(x)$ is well-defined, then using uniform continuity of $\phi_{p}$ and approximating two close enough $x$ and $y$ by sequences $(x_n)$ and $(y_n)$ and applying zulon's suggested bound twice shows immediately that $\bar{\phi}$ is uniformly continuous. –  t.b. May 25 '11 at 19:03
    
@Theo: yes, but the first sentence in this answer is not equivalent to the desired result. –  Qiaochu Yuan May 25 '11 at 19:09
    
Ah, my bad, you're correct. I'm fairly new to math.stackexchange, should I leave my answer as it is or delete it? –  Najib Idrissi May 26 '11 at 10:41

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