0
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ 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), $x=0$. $\endgroup$ – Ian Sep 17 '15 at 0:15
  • $\begingroup$ Right but the theorem only ensures a sequence not every sequence :) $\endgroup$ – user162343 Sep 17 '15 at 0:17
  • $\begingroup$ 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. $\endgroup$ – Ian Sep 17 '15 at 0:18
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Sep 17 '15 at 0:33
  • 1
    $\begingroup$ @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 $\epsilon$. $\endgroup$ – user251257 Sep 18 '15 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.