One version of Arzela-Ascoli theorem can be stated as follows:

Consider a sequence of real-valued continuous functions $\;\{ f_n \}_{n \in \mathbb N}\;$ defined on a closed and bounded interval $\;[a, b]\;$ of the real line. If this sequence is uniformly bounded and equicontinuous, then there exists a subsequence $\; \{ {f_n}_k \}_{k \in \mathbb N} \;$ that converges uniformly.

I am trying to extend the theorem for sequences of real-valued functions defined on $\; \mathbb R\;$. One hint I've got , is to use a diagonal argument...but I'm not very familiar with this.

EDIT: I need to derive that a sequence of real-valued functions defined on $\; \mathbb R\;$ which satisfy the conditions of boundedness and equicontinuity on a closed interval of $\; \mathbb R\;$, has a uniformly convergent subsequence on compact intervals.

How do I proceed?

I would appreciate if somebody could enlighten me about this. Any help would be valuable!

Thanks in advance!!!

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    $\begingroup$ It can't be true as stated for $\mathbb{R}$. Take $f$ to be, say, continuous and compactly supported (and not the zero function), and set $f_n(x) = f(x-n)$. Then $f_n$ is uniformly bounded and equicontinuous, but no subsequence converges uniformly. $\endgroup$ – Nate Eldredge Apr 19 '17 at 17:18
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    $\begingroup$ You will get locally uniform convergence. $\endgroup$ – Daniel Fischer Apr 19 '17 at 17:21
  • $\begingroup$ @NateEldredge What assumptions should 've been added then? I need to derive a result similar to Arzela-Ascoli for $\;\mathbb R\;$ $\endgroup$ – kaithkolesidou Apr 19 '17 at 17:21
  • $\begingroup$ I think only you can say what "similar" means for you. What are you trying to accomplish? $\endgroup$ – Nate Eldredge Apr 19 '17 at 17:22
  • $\begingroup$ @NateEldredge I just edited my original post. I hope to be more specific this time.. $\endgroup$ – kaithkolesidou May 3 '17 at 15:40