# Almost Arzela Ascoli

Definitions:

We say a sequence of continuous functions $$f_n: [0, 1] \to \mathbb R$$ is equicontinuous on average if for every $$x \in [0, 1]$$ and $$\varepsilon > 0$$ there exists some $$\delta > 0$$ such that $$\limsup_{N \to \infty} \frac{1}{N} \sum_{n = 0} ^{N-1} |f_n (x) - f_n (y)| < \varepsilon$$ whenever $$|x - y| < \delta$$.

Suppose $$f_n$$ are continuous functions that are equicontinuous on average and uniformly bounded - that is, $$\sup_n \sup_x |f_n(x)| < \infty$$.

Question: Does there exist a subsequence $$f_{n_k}$$ that converges in measure to some continuous $$f$$?

• What do you mean by convergence in measure for a sequence of continuous functions? May 25, 2021 at 13:18
• @DieterKadelka The usual notion of convergence in measure of functions with respect to the Lebesgue measure. May 25, 2021 at 13:19
• I guess $f_n(x) = \arctan(n(x-a_n))$ with $a_n$ sampled uniformly from $[\tfrac13, \tfrac23]$ is a counterexample (almost surely), but I did not check the details. May 25, 2021 at 14:24

## 1 Answer


Take now any $$x$$ and $$y$$ such that $$0\le x. Then \begin{align*} \sum_{n=1}^N |f_n(x)-f_n(y)|=\sum_{n=1}^N (f_n(y)-f_n(x))\le S_1+S_2+S_2, \end{align*} where $$$$S_1:=\sum_{n=1}^N 1(a_n\le x\le a_n+1/n)=\sum_{n=1}^N 1(x-1/n\le a_n\le x)=o(N),$$$$ since $$(a_n)$$ is uniformly distributed on $$[1/4,3/4]$$; similarly, $$$$S_2:=\sum_{n=1}^N 1(a_n\le y\le a_n+1/n)=o(N);$$$$ and $$$$S_3:=\sum_{n=1}^N 1(x\le a_n, y\ge a_n+1/n)\le\sum_{n=1}^N 1(x\le a_n\le y)\lesssim 2N(y-x).$$$$ Thus, \begin{align*} \limsup_N\frac1N\sum_{n=1}^N |f_n(x)-f_n(y)|\le2|y-x| \end{align*} for any $$x,y$$ in $$[0,1]$$.

On the other hand, the limit $$f$$ of any subsequence $$f_{n_k}$$ that converges in measure is of the form $$f(x)=1(x>a)$$ for some $$a\in[1/4,3/4]$$ and almost all $$x\in[0,1]$$. So, the limit cannot be continuous.