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Let $\Omega\subset\mathbb{R^m}$ be open and $(f_n)$ an equicontinuous sequence of functions that converges pointwise in $\Omega$. Then $(f_n)$ converges uniformly on compact subsets of $\Omega$.

So one of the versions of the theorem that I've got is the following

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 the thing is How can I cover all the hypotheses and ensure that the subsequence is the whole sequence $(f_n)$? I think that there is happening the same problem as here How to use Arzelà-Ascoli theorem in this situation? right ?

Thanks a lot in advance I appreciate your help :)

My approach using @Ian's criterion:

We pick a subsequence of $(f_n)$, let say $(f_{n_k})$, then we get that $(f_{n_k})$ is equicontinuous and it converges pointwise to $f$, now since this happens we get:

$$|f_{n_k}-f(x)|<\epsilon \Rightarrow f(x)-\epsilon<f_{n_k}<f(x)+\epsilon$$

Then we only take the ball of radius $f_{n_k}<f(x)+\epsilon$ and with center at $0$, so we get by A-A theorem that $f_{n_k}$ has a subsequence that converges in each compact of $\Omega$, therefore by the criterion given by @Ian we are done.

The only thing I am unsure of is the radius of the ball , Am I right in the above proof?

And How do I prove that the limit should be continuous ?

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user162343
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    A useful fact: if you have a sequence $x_n$ of real numbers such that every subsequence $x_{n_k}$ has a further subsequence $x_{n_{k_\ell}}$ which converges to $x$ (and $x$ doesn't depend on the subsequence) then $x_n$ converges to $x$. You can apply this to $x_n=| f_n - f |_\infty$ (where $f$ is the pointwise limit), $x=0$. – Ian Sep 17 '15 at 00:15
  • Right but the theorem only ensures a sequence not every sequence :) – user162343 Sep 17 '15 at 00:17
  • Strictly speaking, yes. But consider the following: define $x_n$ as I said. Let $x_{n_k}$ be an arbitrary subsequence of $x_n$. Does it satisfy the hypotheses of Arzela-Ascoli? If so, then it has a subsequence $x_{n_{k_\ell}}$ which converges to zero. Now use the result I stated. – Ian Sep 17 '15 at 00:18
  • Well the only thing I need to check then is that the arbitrary sequence is pointwise bounded right? – user162343 Sep 17 '15 at 00:20
  • Also equicontinuous. But you get these basically "for free" from the original sequence – Ian Sep 17 '15 at 00:21
  • Yes jajaj, but then what do you think about the linked question I have posted ? – user162343 Sep 17 '15 at 00:22
  • The subsequence aspect is handled the same way, you just have to check that the monotonicity gets you the boundedness/equicontinuity you need. – Ian Sep 17 '15 at 00:26
  • jajaj right but that is the question :) – user162343 Sep 17 '15 at 00:27
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    The easiest way to get the equicontinuity in the situation of the other question is to show that the sequence converges uniformly (by Dini's theorem). Yes, that kind of defeats the purpose, but I have a hard time thinking of sane ways to show equicontinuity in a different way. – Daniel Fischer Sep 17 '15 at 00:33
  • Ok @DanielFischer :) I appreciate that a lot :), and What do you think about this one?, Can you provide an answer using only Arzelá-Ascoli? – user162343 Sep 17 '15 at 00:40
  • You can't use literally "only Ascoli-Arzelà", you also need some other tools or theorems. This criterion - already mentioned by Ian - is the standard tool to show the convergence of the full sequence in situations like this. – Daniel Fischer Sep 17 '15 at 00:51
  • Ok, but Is that criterion useful in the other question?, Well is because with the things said by @Ian this problem is almost done I only have to verify the pointwise bound but I am don't know it that will be difficult or not jajajaja – user162343 Sep 17 '15 at 00:54
  • I am back, just a queston for @Ian , Why do I have to check equicontinuity of $||f_n-f||_u$? Because this is a sequence of numbers not of functions, I think you meant a subsequence of $(f_n)$ right? – user162343 Sep 17 '15 at 13:50
  • @user162343 My phrasing was imprecise; I meant "if $n_k$ is an arbitrary subsequence of $n$, is $f_{n_k}$ pointwise bounded and equicontinuous? If so, then $x_{n_k}$ has a further subsequence which goes to zero, hence..." – Ian Sep 17 '15 at 13:52
  • Well Is because what I was understanding is that, ok I have the sequence $(f_n)$, so by the criterion you gave I have to check equicontinuity of $f_{n_k}$ and pointwise bounded, so then I get by Arzelà-Ascoli that this subsequence has a subsequence that converges in every compact set, therefore invoking your criterion I get that the whole sequence converges :) right? – user162343 Sep 17 '15 at 13:57
  • @DanielFischer Have you found something for the other question using A-A? – user162343 Sep 17 '15 at 14:06
  • I have edited my post, what do you think about it? – user162343 Sep 17 '15 at 16:18
  • Do I have to ask this question in another post to get help? – user162343 Sep 18 '15 at 01:21
  • @user162343: I don't get your problem. $f_n$ converges to $f$ pointwise. It is trivially pointwise bounded, so does any subsequence of $f_n$. – user251257 Sep 18 '15 at 14:30
  • Right :) but What I want to know is that if the above proof is well writed and performed or there is something wrong or missing in my math :) – user162343 Sep 18 '15 at 14:41
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    @user162343: It is okay. If you want to be pedantic, the bound for $f_{n_k}(x)$ is: $\max{ |f(x)| + \epsilon, |f_{n_1}(x)|, \dotsc, |f_{n_{k_0}}(x)| }$ for some $k_0\in\mathbb N$ depending on $x$ and $\epsilon$. – user251257 Sep 18 '15 at 14:53
  • Ok :) thnks for your verification :), Now what do you think about this one http://math.stackexchange.com/questions/1439619/show-that-the-set-mathcal-l-is-closed, Thnsk in advance :) – user162343 Sep 18 '15 at 14:58

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