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.


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 $(0,1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.