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Let $\;{f_n}\;$ be a sequence of equicontinuous, real valued, uniformly bounded continuous functions on $\; \mathbb R\;$. Show that $\;{f_n}\;$ has a convergent subsequence which converges uniformly on any bounded subset of $\; \mathbb R\;$.

I read the answer in Applying Arzela-Ascoli to show pointwise convergence on $\mathbb{R}$. and here is my approach:

  • $\;I_1=[-1,1]\;$ There is by Arzela-Ascoli a subsequence $\;\{f_n^1\}\;$ of $\;f_n\;$ which converges uniformly to $\;f^1\;$ on $\;I_1\;$. It's obvious that $\;f^1\;$ is continuous.
  • $\;I_2=[-2,2] \supset [-1,1]=I_1\;$. Since $\;\{f_n^1\}\;$ is convergent there is a subsequence $\;\{f_n^2\}\subset\{f_n^1\}\;$ such that $\;\{f_n^2\}\;$ converges uniformly to $\;f^2\;$ on $\;I_2\;$. $\;f^2\;$ is also continuous.

$\dots \dots \dots \dots \dots$ Continuing this process, one can find $\;\{f_n^m\}\subset\{f_n^{m-1}\}\subset \dots \subset\{f_n^2\}\subset\{f_n^1\}\;$ which converges uniformly to continuous $\;f^m\;$ on $\;I_m=[-m,m]\supset \dots \supset [-1,1]=I_1\;$

Now let $\;F_j:=f_j^j\;$ for $\;1\le j \le m\;$ for some $\;m \in \mathbb N\;$ and define $\;F(x)=\begin{cases} f^1 & x\in I_1 \\ f^2 & x\in I_2\\ . \\ . \\ . \\ f^m & x \in I_m\\ \end{cases}$

From the above it follows $\;F_j\;$ converges uniformly to $\;F\; \;\forall x \in I_j\;$ where $\;1\le j \le m\;$.

Questions:

  1. How do I proceed in order to show $\;F_j \to F\;$ as $\; j \to \infty\;$? Should I show $\;F_j\;$ is Cauchy sequence?
  2. Is the above "structure" of my proof right and formal enough? I haven't used anywhere of the $\;\varepsilon$-definition for convergence and so I believe it's not well written.

It's the first time I use the diagonal argument and I want to be sure I completely understand it. If there are any suggestions on where should I read and learn more about it and how to use it, they would be really welcome.

Any help would be valuable. Thanks in advance!

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The approach is mostly sound, as is your notation. However you need to be a bit more careful with your definition of $F$. As $I_1 \subset I_2 \subset ...$, you got multiple definitions for $F$ at each point. In this case, to have a well defined limit, you will need to show that $f^i(x) = f^j(x)$, where both are defined, that is for $x \in I_i \cap I_j$. This follows from the fact that $(F_n(x))_n$ is a subsequence of both $(f^i_n)_n$ and $(f^j_n)_n$, which both converge, so all three need to have the same limit.

Concerning the convergence, this is nearly trivial: Note that $(g_n)_n$ converges uniformly on $I$, it also converges uniformly on all $J\subset I$ against the same limit, and similarly if $(g_n)_n$ converges to $g$ on $I$, $(g_n(x))_n$ converges to $g(x)$ for all $x\in I$. This should give you enough hints.

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  • $\begingroup$ First of all thanks a lot for your answer! You 're right about the well definition of $\;F\;$, I'll fix it. On convergence, the hint you suggested is that uniform convergence implies pointwise convergence, am I right? But I can't see right now how should I use it. Should I prove $\;F_j\;$ is Cauchy or the above is enough in order to show that $\;F_j \to F\;\;as\;j \to \infty\;$? I feel I'm close but I lose it $\endgroup$ – kaithkolesidou Jul 6 '17 at 9:21
  • $\begingroup$ Yes, uniform convergence implies pointwise convergence. Pointwise convergence on $\mathbb{R}$ just means convergence at every $x \in \mathbb{R}$. However every $x\in\mathbb{R}$ is also in $I_m$ for some $m\in\mathbb{N}$. So uniform convergence on all $I_m$ implies pointwise convergence on all $I_m$ and thus pointwise convergence on $\mathbb{R}$. $\endgroup$ – mlk Jul 6 '17 at 9:24
  • $\begingroup$ Oh I see. If i fix $\;F\;$ is well defined , having shown that $\;F_j\;$ converges uniformly to $\;F\;\;\forall 1\le j \le m\;$, since $\;m\;$ could be arbitary large enough, through pointwise convergence (which is implied by the uniform one), I get convergence on $\;\mathbb R\;$ and the proof is complete. I want to understand it completely so please tell me if I misunderstood something. However you've been very helpful!! $\endgroup$ – kaithkolesidou Jul 6 '17 at 9:34
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    $\begingroup$ Yes you are right. If you want to understand this completely, you may also want to think about why this does not imply uniform convergence on $\mathbb{R}$. (Both where the proof fails and if there is a counterexample.) $\endgroup$ – mlk Jul 6 '17 at 10:39

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