Suppose that $(f_n)$ is an equicontinuous sequence of functions $f_n : \mathbb{R}\rightarrow \mathbb{R}$, such that $(f_n (0))$ is a bounded sequence in $\mathbb{R}$.
Does there exist a subsequence that converges uniformly? Prove or give a counter-example.
If we limit the functions to a closed interval $[a,b]$, by the following fact and $\mathbb{R}$ is chain-connected, And then by Arzelà–Ascoli theorem, the problem is solved.
$M$ is chain connected if and only if pointwise boundedness of an equicontinuous family at one point of $M$ implies pointwise boundedness at every point of $M$.
But I still don't know the case of $f:\mathbb{R}\rightarrow \mathbb{R}$
What if :
- $\mathbb{R}^m$ replaces $\mathbb{R}$ ?
- $\mathbb{N}$ replaces $\mathbb{R}$ ?
Define $f_n(x):=\min\{|x|,n\}$. We have for all $x,y\in\mathbb N$ and each integer $n$ that $$|f_n(x)-f_n(y)|\leqslant |x-y|,$$ hence $(f_n)_{n\geqslant 1}$ is uniformly equi-continuous. Since $f_n(0)=0$, the all the conditions are fulfilled.
But if $m\neq n$, we have $\sup_{x\in\mathbb R}|f_n(x)-f_m(x)|=|n-m|\geqslant 1$, hence no subsequence can converge uniformly on the real line.