As an example of Arzela-Ascoli theorem in intro. to topology course, the following case was exhibited:
Let $A\subset C[a,b]$ be a set of differentiable functions such that: $\forall f\in A,\ \vert f \vert \leq M_1,\ \vert f' \vert \leq M_2$ thus A is bounded and equicontinuous, thus every sequence $\{f_n\}$ has a converging subsequence.

My question is: The limit function isn't necessarily in $A$, right?
A isn't necessarily a closed set, and the more detaild chain of consequences is: bounded & equicontinuous $\rightarrow$ absolutely bounded $\rightarrow\ \bar A$ is closed and absolutely bounded $\rightarrow\ \bar A$ is compact $\rightarrow$ every sequence $\{f_n\}\subset \bar A$ has a converging sub-sequence.
So a sequence in $A$ might converge to $C[a,b]\supset \bar A \ni f \notin A$, am I right?

  • 3
    $\begingroup$ Right, take $f_n(x) \equiv 1/n$ for example. $\endgroup$ – zhw. Mar 14 '17 at 20:48
  • 2
    $\begingroup$ Another way to think about this is that Arzelà–Ascoli only tells you $A$ is precompact. So if it's not closed (and hence not compact), the sequence need not converge to a limit in $A$. $\endgroup$ – Fimpellizieri Mar 14 '17 at 20:52
  • $\begingroup$ Great, thank you both $\endgroup$ – galra Mar 14 '17 at 21:47

As pointed out in the comments by zhw and Fimpellizieri:

We only know that $A$ is precompact for the metric induced by the uniform norm, but it may not be closed. For example, if $A=\left\{ x\mapsto 1/n,n\in\mathbb N, n\geqslant 1\right\}$, the limit of $\left(1/n\right)$ does not belong to $A$.


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.