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.

  • $\begingroup$ I am aware of the homeomorphism, but I am more specifically asking how to go about using it. $\endgroup$ – johnny133253 Nov 11 '19 at 18:34
  • $\begingroup$ Consider the compositions. $\endgroup$ – Sumanta Nov 11 '19 at 18:40
  • $\begingroup$ Thanks I will try that. $\endgroup$ – johnny133253 Nov 11 '19 at 18:48
  • $\begingroup$ Check the solution.. $\endgroup$ – Sumanta Nov 11 '19 at 19:38

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