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 ?
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.