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Let $\langle f_n \rangle$ be sequence of equi-continuous real-valued functions on $\mathbb R$ such that $f_n(0)=0$ for every $n$. Does $\langle f_n\rangle$ have a converging subsequence?

I found that even equi-continuity condition and the value at $x=0$ does not guarantee the uniform convergence, as there is a counter-example of $f_n (x) = x/n$, which does not converge uniformly, but do the conditions of the question guarantee pointwise convergence of some subsequence?

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We can extract a subsequence which convergences uniformly on compact sets.

Indeed, first we notice that for each $N$, the sequence $\left(f_n\right)_n$ is uniformly bounded on $[-N,N]$. Indeed, using the definition of equicontinuity there exists a $\delta\gt 0$ such that if $\left\lvert x-y\right\rvert\lt\delta$ then $\left\lvert f_n(x)-f_n(y)\right\rvert\lt 1$. For a fixed $x\in [-N,N]$, there exists a $k\in\{0,\dots,2N\lfloor \delta^{-1}\rfloor\}$ such that $-N+k\delta\leqslant x\lt -N+(k+1)\delta$. Then $$ \left\lvert f_n(x)\right\rvert\leqslant \left\lvert f_n(x)-f_n\left(-N+k\delta\right)\right\rvert +\left\lvert f_n\left(-N+k\delta\right)\right\rvert\leqslant 1+\sum_{i=1}^k\left\lvert f_n\left(-N+i\delta\right)-f_n\left(-N+(i-1)\delta\right)\right\rvert $$ and all the terms in the sum are smaller than one hence $$ \left\lvert f_n(x)\right\rvert\leqslant 1+2N\lfloor \delta^{-1}\rfloor. $$ By Arzela-Ascoli theorem, there exists a uniformly convergent subsequence.

Now, in order to get a subsequence for which the convergence is uniform on each compact set, we proceed as follows. We construct a non-increasing sequence of subset $\left(I_N\right)_{N\geqslant 1}$ of $\mathbb N$ such that each $I_N$ is infinite and the sequence $\left(f_n\right)_{n\in I_N}$ converges uniformly on $[-N,N]$. Then let $n_k$ be the $k$-element of $I_k$. The subsequence $\left(f_{n_k}\right)_k$ is uniformly convergent on compact sets.

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  • $\begingroup$ How do we know that $(f_n)_n$ is uniformly bounded on $[-N, N]$ for any $N$? $\endgroup$
    – Dasherman
    Mar 29, 2020 at 17:44
  • $\begingroup$ This is was is shown in the first paragraph. $\endgroup$ Mar 29, 2020 at 19:46
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    $\begingroup$ I see now, thank you! $\endgroup$
    – Dasherman
    Mar 30, 2020 at 19:49

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