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Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Assume that $f(x) > 0$ for all $x \in \mathbb{R}$, but $f$ decays to $0$ at $\infty, -\infty.$

Let $\{f_n\}$ be a sequence of real-valued equicontinuous functions on $\mathbb{R}.$ Assume that $$|f_n(x)| \leq f(x)$$ for all $n$ and $x.$ Verify that $\{f_n\}$ has a uniformly convergent subsequence.

$\textbf{Proof} \ $ Since $f$ decays at $\infty$ and $-\infty,$ there exists $M > 0$ such that $|f(x)| < 1$ for $|x|>M$. Since $[-M, M]$ is compact and $f$ is continuous, there is $K > 0$such that $|f(x)| \leq K$ with $x \in [-M,M].$

So $|f_n(x)| \leq |f(x)| \leq K+1$ for all $n, x.$ That is, $\{f_n\}$ is bounded under the infinity norm $||\cdot||_\infty.$

So $F = \{f_n : n \in \mathbb{N}\}$ is a bounded and equicontinuous family of function. Clearly, $F \subseteq C(\mathbb{R})$. The original proof of Arzela-Ascoli Theorem will apply only if the space is compact, but $\mathbb{R}$ is not compact. So I try to modify the proof of Arzela-Ascoli Theorem in Carothers book https://archive.org/details/CarothersN.L.RealAnalysisCambridge2000Isbn0521497566416S page 181. However, the proof relies on totally bounded to choose a finite points which their balls will cover the space. This is no where possible for the space that is not compact. I also try to restricted $F$ to $[-M,M]$ first and extract the subsequence. Then I try to extend it on $\mathbb{R}$ with decaying at $\infty, -\infty.$ But still not quite successful. Any help please ?

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For each positive integer $M$ choose a subsequence that converges uniformly on $[-M,M]$. Then take a diagonal subsequence and show that it converges unifomly on the whole line by showing that it is uniformly Cauchy.

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  • $\begingroup$ Chooae it iteratedly or independently ? Like for [-1,1], there is a uniformly Cauchy subsequence $(f_n_{k})_{k=1}^\infty$. Then use this subsequence to obtain $(f_{n_k_l})$ for [-2,2] and $(f_n_k_l_m)$ for [-4,4] ... $\endgroup$ – Both Htob Aug 9 '18 at 19:15
  • $\begingroup$ Or choose independently ? $(f_n_{k1})$ for $[-1,1]$, $(f_n_{k2})$ for $[-2,2]$ , ... then extract the subsequence. But there is no relation between each such subsequences except the equicontinuity. If I construct it iteratedly, each subsequence will relate to each other. But to extract it countably infinite times. Is that possible ? $\endgroup$ – Both Htob Aug 9 '18 at 19:18

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