# The sequence of functions $f_n = \sin{nx}$ is not uniformly equicontinuous on any compact interval.

The sequence of functions $$f_n = \sin{nx}$$ is not uniformly equicontinuous on any compact interval.

I already proved that this sequence is not uniformly equicontinuous on $$[0,1]$$:

Let $$f_n(x) = \sin(nx)$$ be defined on $$[0,1]$$. Put $$\epsilon = \sin(1)$$. Let $$\delta >0$$. Choose $$N \in \mathbb{N}$$ such that $$\frac{1}{N} < \delta$$ and define $$x_N = 1/N$$ and $$y_N = 0$$. Then $$|x_N-y_N| = 1/N < \delta$$ while $$|f_N(x_N) - f_N(y_N)| = |\sin(1) - \sin(0)| = |\sin(1)| \geq \epsilon.$$ This proves that $$\{f_n\}$$ is not uniformly equicontinuous on $$[0,1]$$.

My question is this: is there a way to use the fact that $$[0,1]$$ is homeomorphic to $$[a,b]$$ to prove the claim?

Consider the homeomorphism $$f:[0,1]\to[a,b]$$ defined by $$f(x)=a+x(b-a)$$ and consider the sequence $$\{\sin(n\cdot f)\}$$.
Suppose, $$\{y\mapsto\sin(ny)\}$$ is equicontinuous family on $$[a,b]$$. Then given any $$\epsilon >0$$ we have $$\delta>0$$ such that, $$|\sin(ny_1)-\sin(ny_2)|<\epsilon$$ for all $$y_1,y_2\in [a,b]$$ and for all positive integer $$n$$.
In particular when, $$x_1,x_2\in [0,1]$$ with $$|x_1-x_2|<\frac{\delta}{b-a}$$ we have, $$|f(x_1)-f(x_2)|<\delta$$, so that for any positive integer $$n$$ we have, $$\big|\sin\big(nf(x_1)\big)-\sin\big(nf(x_2)\big)\big|<\epsilon$$. So that, $$\{\sin(n\cdot f)\}$$ is a equicontinuous family on $$[0,1]$$. But you have already proved that it is not possible.