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 $\;1\le j \le m\;$.


  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!


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.

  • $\begingroup$ 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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