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Suppose $\mathcal{F} \subset C(A)$ be a family of continuous functions with domain $A$. If $\mathcal{F}$ is pointwise equicontinuous, is it true that $\cal F$ is uniformly equicontinuous?

I guess the answer is NO. Indeed, if the domain $A$ is compact set, then the $\cal F$ is uniformly equicontinuous. But I'm not sure if the compactness assumption drops.

Here is my thinking about constructing a counterexample: (but not sure...)

Suppose $A:=(0,1]$, I want to say $\cal F$ is NOT uniformly equicontinuous; i.e., there exists $\varepsilon>0$ s.t. $\forall \delta>0$, there exists $x,y \in A$, $f \in \mathcal{F}$ such that $|x-y| < \delta$ but $|f(x) - f(y)| \ge \varepsilon$

Take $\varepsilon=1$, and choose $x=1/n, y=2/n$ (then $x,y \in A$); define a function $f_n(x):= 2n x$ ($f_n \in \mathcal{F} \subset C(A)$ ), then I have $$|x-y| = |1/n - 2/n|<1/n := \delta,$$ but $|f_n(x) - f_n(y)| = |2 -4 |=2 \ge 1$

But I didn't see very clear what the role of compactness played..

Thanks you.

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It may help to consider a special case: $\mathcal F=\{f\}$, the family consists of one function. Then

  • $\mathcal F$ is pointwise equicontinuous $\iff $ $f$ is continuous
  • $\mathcal F$ is uniformly equicontinuous $\iff $ $f$ is uniformly continuous

This both gives you plenty of examples, and indicates the relevance of compactness.

A remark on the latter: pointwise equicontinuity says that for every $\epsilon$, every point has a neighborhood where $\mathcal F$ is under control. In the compact case, Lebesgue's number lemma gives $\delta$ for uniform equicontinuity. In the noncompact case, some of those good neighborhoods may be so small that there is no $\delta$ that fits all of them. Like the union of $(1/(n+2),1/n)$ covering $(0,1)$.

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