Test whether the sequence of functions $\sin^n(x)$ is equicontinuous or not

Question

Prove that $\{f_n(x)\}=\{\sin^n(x)\}$ in NOT equicontinuous in $[0,\pi]$ (using definition).

I have tried to choose $x,y\in [0,\pi]$ in terms of $n$ so that $|x-y|\to 0$ but $|f_n(x)-f_n(y)|\not \to 0$. But unable to choose such pair $(x,y)$.

Any hint. please?

Answer 1

fix $\epsilon=1/2,\forall \delta>0$, let $\delta_0=\min\lbrace\delta/2,1\rbrace,N=\lfloor \log_{\sin(\pi/2-\delta_0)}\epsilon\rfloor+1,x=\pi/2-\delta_0,y=\pi/2$

then $|f_N(x)-f_N(y)|=1-\sin^N(\pi/2-\delta_0)>1/2=\epsilon$, which means $\lbrace f_n\rbrace$ is not equicontinuous

Answer 2

Maybe this is an overkill but here it goes:

If the sequence of functions $f_n(x)=\sin^nx$ were equicontinuous on $[0,\pi]$, then, as $\{f_n\}$ is uniformly bounded, by Arzella-Ascoli's theorem $\{f_n\}$ admits a uniformly convergent subsequence. Now, $f_n(x)\xrightarrow{n\rightarrow\infty}\mathbb{1}_{\{\pi/2\}}(x)$ which is not a continuous function. Thus no subsequence of $\{f_n\}$ can converge uniformly. Hence, $\{f_n\}$ is not equicontinuous.