Let $\{f_n\}$ be a sequence of functions which are continuous over $[0, 1]$ and continuously differentiable in $(0, 1)$. Assume that $|f_n(x)| \leq 1$ and that $|f_n’(x)| \leq 1$ for all $x \in(0, 1)$ and for each positive integer $n$. Pick out the true statements.
(a) $f_n$ is uniformly continuous for each $n$.
(b) $\{f_n\}$ is a convergent sequence in $C[0, 1]$.
(c) $\{f_n\}$ contains a subsequence which converges in $C[0, 1]$.

I am totally stuck. How should I solve this? Thanks for your help.

  • $\begingroup$ I think the title of the post is entirely too long for being so nondescript, but I could not come up with a better title. $\endgroup$ – JavaMan Jan 6 '13 at 5:58
  • $\begingroup$ Do you know any theorems about uniform continuity? Or about convergent subsequences? If you know any, you might want to try them and say why they help or don't help. $\endgroup$ – Jesse Madnick Jan 6 '13 at 6:02
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    $\begingroup$ $[0,1]$ is compact. You might know a theorem on continuous functions on a compact space? Also, try a simple alternating sequence of constant functions to kill (b), and look up Arzela-Ascoli to see if it applies to c). $\endgroup$ – Henno Brandsma Jan 6 '13 at 6:18
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Feb 21 '13 at 19:31


(a) $[0,1]$ is compact. Probably you know a theorem like this:

Let $(X,d)$ a compact metric space and $f: X \to \mathbb{R}$ continuous. Then...

(b) What about $f_n(x) := (-1)^n$?

(c) Apply Arzelà-Ascoli. (Use $|f_n(x)| \leq 1$, $|f_n'(x)| \leq 1$. Think about mean-value theorem to prove one of the conditions.)

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  • $\begingroup$ But I can not understand how to prove that ${f_n}$ is closed? $\endgroup$ – sani Jun 16 '17 at 18:45
  • $\begingroup$ @sani What do you mean by closedness of a function? $\endgroup$ – saz Jun 17 '17 at 4:52

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