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.


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.

  • $\begingroup$ this sequence is also not uniformly bounded, right? $\endgroup$ – ALannister 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

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