# Prove that there is a subsequence $f_{nk}$ of $f_n$ converging uniformly on $[0,1]$ to a function $f$

Let $f_n$ be a sequence of functions defined on $[0,1]$. Suppose that there exists an $\alpha\in(0,1)$ and positive number $M$ such that $|f_n(0)|\leq M$ and $|f_n(x)-f_n(y)|\leq|x-y|^\alpha$ for all $n$ and $x,y\in[0,1]$.

Prove that there is a subsequence $f_{nk}$ of $f_n$ converging uniformly on $[0,1]$ to a function $f$ that satisfies $|f(x)-f(y)|\leq|x-y|^\alpha$, $\forall x,y\in[0,1]$.

*I'm trying to apply Arzela Ascoli theorem. Now I have compact domain, so it's totally bounded, I then can have some $\delta$ net covering the domain. I'm given a bounded point at $0$, and I would like to show point wise bounded for some subsequence. Question: Don't I need the sequence of functions be continuous in order to show equicontinuous? Also, does the domain for $\alpha$ matter? Can't just say a fixed number $\alpha$?

Thank you.

the fact that $|f_n(x)-f_n(y)|\leq |x-y|^{\alpha}$ implies that $f_n$ is equicontinuous (thus continuous).
To see this, consider $c>0$, $|x-y|^{\alpha}=exp(\alpha ln(|x-y|)<c$ is equivalent to $ln(|x-y|)<{{ln(c)}\over\alpha}$ and $|x-y|<exp({{ln(c)}\over\alpha})$, so if you take $|x-y|<exp({ln({c)}\over\alpha})$, you obtain that $|f_n(x)-f_n(y)|\leq|x-y|^{\alpha}<c$, this shows that the sequence $f_n$ is equicontinuous and $f_n$ henceforth continuous for every $n$.
• thank you. Does the domain of $\alpha$ matter? – 2ndaccount Aug 2 '17 at 23:59