1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ $[0,1]$ 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
2
$\begingroup$

Hints

(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.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

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.