Is this collection of function uniformly equicontinuous? Hints on the proof.

Question

Let $f_n(x)=\frac{1}{n}\cos(e^{nx})$ for $n\in\mathbb{N}$ be a sequence of functions for $x\in[0,1]$. Is it true that $\{f_n\}$ is a uniformly equicontinuous collection of functions?

My attempt so far before I got stuck:

Let $X=[0,1]$ and $Y=\mathbb{R}$. We need to show that for all $\epsilon>0$, there exists a $\delta>0$ such that for every $x,y\in X$ we have $|x-y|<\delta\implies|f_n(x)-f_n(y)|<\epsilon$ for all $n\in\mathbb{N}$.

Now pick $\epsilon>0$. Then $|f_n(x)-f_n(y)|=|\frac{1}{n}\cos(e^{nx})-\frac{1}{n}\cos(e^{ny})|=|\frac{1}{n}(\cos(e^{nx})-\cos(e^{ny}))|$?

Pick $x_0\in[0,1]$ and let $\delta=|x_0-\epsilon|$??

I could not think of any formula to simplify the expression with cos and I could not find the required $\delta>0$. Could somebody please give a help on how to continue?

Thanks!

Answer 1

Choose $\epsilon>0$. Let $n>2/\epsilon$, then for $m>n$, then $|f_m(z)-f_m(w)|\le 2|f_m|\le 2\frac{1}{n}\le \epsilon$. Thus we only need to show that $f_m$ is uniformly continuous, since then we can take the minimum of the values for $\delta$ corresponding to the given $\epsilon$. But these are continous functions on a compact domain, and thus are indeed uniformly contious. Thus the $f_m$ are uniformly equicontinous as desired.

Answer 2

Alternative twisted proof: $f_n\to 0$ uniformly, so $\{f_n\}_{n\in\Bbb N}\cup\{0\}$ is compact in $C([0,1])$ (why?). By Ascoli-Arzelà is equicontinuous, so $\{f_n\}_{n\in\Bbb N}$ is also equicontinuous.