0
$\begingroup$

For a closed, bounded interval $[a,b]$, let $\{ f_{n}\}$ be a sequence in $C[a,b]$. If $\{f_{n}\}$ is equicontinuous, does $\{f_{n}\}$ necessarily have a uniformly convergent subsequence?

I would think not, because according to the Arzela-Ascoli Theorem, $\{f_{n} \}$ also needs to be uniformly bounded. Is this all that needs to be violated in order for an equicontinuous sequence of continuous functions on a compact interval to not have a uniformly convergent subsequence?

And if so, what is an example of a sequence that illustrates this, and how to show it does not have a uniformly convergent subsequence? Thank you.

$\endgroup$
1
$\begingroup$

Take $f_n(x) = n$ for $x \in [0,1]$. These functions are all constant, so clearly equicontinuous, but $\| f_n - f_m \|_\infty = \lvert n - m \rvert \ge 1$ for $n \neq m$ so no subsequence can converge since no subsequence is Cauchy.

$\endgroup$
4
  • $\begingroup$ this sequence is also not uniformly bounded, right? $\endgroup$ – user100463 Apr 11 '16 at 23:28
  • 1
    $\begingroup$ Yes of course. If it was uniformly bounded, then we could apply Arzela Ascoli and find a convergent subsequence. A sequence of functions $f_n$ is uniformly bounded if and only if the sequence of real numbers $\| f_n\|_\infty$ is bounded. Here that clearly is not the case because $\| f_n \|_\infty = n$. $\endgroup$ – User8128 Apr 11 '16 at 23:29
  • $\begingroup$ How can we infer that there is no convergent subsequence from proving that no sequence is Cauchy? $\endgroup$ – PLanderos33 Oct 20 '19 at 22:41
  • $\begingroup$ Any convergent sequence in a normed space is also Cauchy. Contrapositively: not Cauchy implies not convergent. Here I’ve shown that each subsequence is not Cauchy. $\endgroup$ – User8128 Oct 21 '19 at 18:01

Your Answer

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