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.

  • $\begingroup$ 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. $\endgroup$
    – Conifold
    Aug 23, 2020 at 3:57
  • $\begingroup$ 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 $[-R, R]$) $\endgroup$
    – Aphyd
    Aug 23, 2020 at 4:03
  • $\begingroup$ 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. $\endgroup$
    – Conifold
    Aug 23, 2020 at 4:09
  • $\begingroup$ Well then I don't see how your suggestion of using a growing sequence of intervals is any different from what I did. $\endgroup$
    – Aphyd
    Aug 23, 2020 at 4:19
  • $\begingroup$ 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. $\endgroup$
    – Conifold
    Aug 23, 2020 at 4:22


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