An equicontinuous sequence of functions bounded by a function $\phi$ with $\lim_{x\rightarrow\pm\infty}\phi(x)$ has a uniformly convergent subsequence

Is this proof correct?

Let $$\phi(x) > 0$$ be a continuous function on $$\mathbb{R}$$ with $$\lim_{x\rightarrow\pm\infty} \phi(x) = 0$$ and let $$f_n$$ be an equicontinuous sequence of functions on $$\mathbb{R}$$ satisfying $$\lvert f_n(x)\rvert \leq \phi(x)$$ for all $$x\in \mathbb{R}$$ and $$n\in\mathbb{N}$$. Prove that $$f_n$$ has a uniformly convergent subsequence.

Since $$\lim_{x\rightarrow\pm\infty} \phi(x) = 0$$ there exists an $$R \in \mathbb{R}$$ such that if $$\lvert x \rvert > R$$ then $$\phi(x) < 1$$. Then $$\phi(x)$$ attains a maximum on $$[-R, R]$$, say M, as $$\phi$$ is continuous there. So $$\max_{x\in\mathbb{R}} \phi(x) \leq \max(M, 1) = K$$.

Since $$\lvert f_n(x) \rvert \leq \phi(x) \leq K$$ for all $$x\in \mathbb{R}$$ and $$n\in\mathbb{N}$$, we have that $$f_n$$ is uniformly bounded.

So $$f_n$$ is uniformly bounded and equicontinuous and so has a uniformly convergent subsequence $$g_n$$ on $$[-\alpha, \alpha]$$ for all $$\alpha \in \mathbb{R}$$.

Now, let $$\epsilon > 0$$ and choose $$R \in \mathbb{R}$$ such that $$\phi(x) < \frac{\epsilon}{2}$$ when $$\lvert x\rvert > R$$, and choose an $$N$$ large enough that when $$n,m > N$$ $$\max_{\lvert x\rvert \leq R}\lvert g_n(x) - g_m(x)\rvert < \epsilon$$ as $$g_n$$ is uniformly convergent on $$[-R,R]$$ and hence is uniformly Cauchy there. Then when $$n, m > N$$ we have that $$\max_{x\in\mathbb{R}}\lvert g_n(x) - g_m(x)\rvert = \max(\max_{\lvert x\rvert \leq R}\lvert g_n(x) - g_m(x)\rvert, \max_{\lvert x\rvert>R}\lvert g_n(x) -g_m(x)\rvert) < \max(\epsilon, 2\frac{\epsilon}{2}) = \epsilon$$ and so $$g_m$$ is actually uniformly Cauchy on all of $$\mathbb{R}$$. Hence $$f_n$$ has a uniformly convergent subsequence.

• No. The convergent subsequence $g_n$ depends on $\alpha$, so you have to be more subtle in forming the subsequence for your last paragraph. Probably choose a growing sequence of intervals $[-R_k,R_k]$ so that $\phi<1/k$ outside of them. Aug 23, 2020 at 3:57
• But wouldn't for any given $\alpha$, the uniformly convergent subsequence agree with all $\beta < \alpha$, lest it not be uniformly convergent? (and hence agree for every interval $[-R, R]$) Aug 23, 2020 at 4:03
• Yes, if it converges on a bigger interval it will converge on smaller ones. But what is your choice for the entire subsequence on the whole line? You can not just pick a subsequence that converges on a finite interval, no matter how wide, so you currently do not even have a candidate subsequence $g_{n_k}$ of which to prove that it converges. Aug 23, 2020 at 4:09
• Well then I don't see how your suggestion of using a growing sequence of intervals is any different from what I did. Aug 23, 2020 at 4:19
• You have to select $g_{n_k}$ from the convergent subsequence corresponding to the $k$-th interval and then prove that it converges on the whole line. Aug 23, 2020 at 4:22