# Does this function sequence have a convergent subsequence?

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?

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.
• How do we know that $(f_n)_n$ is uniformly bounded on $[-N, N]$ for any $N$? Mar 29, 2020 at 17:44