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Let $f_n:[a,b]\to\mathbb{R}$ be a monotone sequence of continuous functions, pointwise convergent to $f:[a,b]\to\mathbb{R}$. Then $f_n\rightrightarrows f$ on $[a,b]$.

Then I want to use this version of the Arzelà-Ascoli theorem:

Let $\Omega$ be a open set of $R^{m}$, and $\{f_n\}$, $f_n:\Omega \to R^{l}$ a sequence of continous functions in $\Omega$ (not necesarilly bounded) equicontinous and pointwise bounded. Then there exists a subsequence of $\{f_n\}$ that converges uniformly in each compact subset of $\Omega$.

But it is not clear How can I enssure the hypotheses and that the subsequence is in fact the whole $\{f_n\}$. Can someone help me to prove this please?

Thanks a lot in advance.

NOTE:Why Can I apply A-A theorem?

I have the following result:

Let $\mathcal F \subset C(X,Y)$. If the family $\mathcal F$ is totally bounded then it equicontinous in $X$. So I think this result should help but I don't know how to prove totally boundness of the family given in the theorem.

user162343
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    $f_n \rightrightarrows f$ means uniform convergence? The hypotheses don't guarantee uniform convergence. You need to require that $f$ is continuous (or something that after all is equivalent to that). – Daniel Fischer Sep 16 '15 at 17:41
  • Can you clarify please? Thnks :) – user162343 Sep 16 '15 at 17:46
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    You can't use Arzelà-Ascoli with only these hypotheses: first, although it is true that each $f_n$ is bounded, the sequence of functions is not necessarily bounded (i.e. the sequence $\max |f_n|$ may tend to $\infty$); second, the continuity of each $f_n$ does not imply the equicontinuity of the whole family. Finally, and without any connection to the Arzelà-Ascoli theorem, if $f$ is not continuous then your statement is simply not true. – Alex M. Sep 16 '15 at 17:47
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    If the convergence is uniform, the limit function is continuous. There are monotonic sequences of continuous functions that converge pointwise to discontinuous functions. If the limit function is continuous, however, Dini's theorem asserts that the convergence is uniform. – Daniel Fischer Sep 16 '15 at 17:48
  • Ok so let check :), and when I get something, (a little bit later) I edit my post and I comment here for your help please :) right? – user162343 Sep 16 '15 at 17:50
  • @AlexM. The monotonic convergence gives pointwise boundedness. The interesting quirk if we assume that $f$ is continuous is that the easiest method to show equicontinuity is showing the uniform convergence ;) – Daniel Fischer Sep 16 '15 at 17:50
  • And assuming $f$ continous, what can be done? – user162343 Sep 16 '15 at 17:53
  • If $f$ is continuous, just apply Dini's theorem, this has nothing to do with Arzelà-Ascoli. Plus, you get that the whole seuquence converges uniformly, not just a subsequence. – Alex M. Sep 16 '15 at 17:54
  • Invoke or prove Dini's theorem to show that the convergence is uniform. Then, if you insist on using Ascoli, you use that uniformly convergent sequences of continuous functions (on compact domains) are equicontinuous. – Daniel Fischer Sep 16 '15 at 17:56
  • Let me check what does it states properly :) – user162343 Sep 16 '15 at 17:56
  • I am back :), I am sorry for the delay but I had a conference, so the thing is Why do I have a compact space here? – user162343 Sep 16 '15 at 21:08
  • The domain is $[a,b]$. Such intervals are compact. What the best/most appropriate way to prove that is depends on your background. – Daniel Fischer Sep 16 '15 at 21:13
  • Rigth, but what I have seen so far is that then $C([a,b])$ is complete, but then Does the domain $[a,b]$ implies totally boundness? and then conclude that it is fact compact – user162343 Sep 16 '15 at 21:15
  • ajajaja I got it :), the compact space shouldn't be $C$ it should be $X$ jajaja :), let me check the proof again :) – user162343 Sep 16 '15 at 21:31
  • @DanielFischer Have you found something for this question using A-A? – user162343 Sep 17 '15 at 14:07
  • I have edited my post, What do you think ? :) – user162343 Sep 17 '15 at 23:39

1 Answers1

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Notice that the limit $f$ needs to be continuous. You don't need A-A for this.

Some extensive hints:

  1. Without loss of generality let $f_n$ be non-strictly decreasing and $f=0$, in particular $f_n\ge 0$.

  2. $\|f_n\|$ is non-strictly decreasing. Let $c$ be its limit.

  3. Let $x_n$ such that $f_n(x_n) = \|f_n\|$. Without loss of generality assume $x_n\to x$.

  4. Show $f_n(x)\to c$ and thus $c=0$.

user251257
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