Discuss the uniform convergence of $\sin^n(x)$

Question

Discuss the uniform convergence in $\mathbb{R}$ of $f_n(x)=\sin^n(x)$.

I need to find the domain in which this sequence is uniformly convergent. But it seems to me that this limit: $\lim_{n\to \infty} \sin^n(x)$ does not exist. So we should not obtain the point limit for any real value of $x$. So this sequence can not be pointwise convergent. Hence it should not converge uniformly for any real $x$. Is this going right?

Answer 1

Hint. Note that the pointwise limit for $x\in\mathbb{R}$ is $$\lim_{n\to \infty} \sin^n(x)=\begin{cases} 1 \quad\text{if $x=\pi/2+ 2k\pi$ for $k\in\mathbb{Z}$,}\\ \not\exists \quad\text{if $x=3\pi/2+ 2k\pi$ for $k\in\mathbb{Z}$,}\\ 0 \quad\text{otherwise.}\\ \end{cases}$$

Answer 2

No, it is going wrong.

The limit $\lim_{n\to\infty}\sin^n(x)$ is equal to $0$ if $x\notin\frac\pi2+\pi\mathbb Z$. If $x\in\frac\pi2+2\pi\mathbb Z$, then $\lim_{n\to\infty}\sin^n(x)=1$. It is only when $x\in-\frac\pi2+2\pi\mathbb Z$ that the limit does not exist.

The sequence convergs uniformly (to the null function) on $\left[-\frac\pi4,\frac\pi4\right]$, for instance. And it does not convege uniformly on $[0,\pi]$, because if it did, then it would converge to a continuous functions. And the function$$\begin{array}{ccc}[0,\pi]&\longrightarrow&\mathbb{R}\\x&\mapsto&\begin{cases}0&\text{ if }x\neq\frac\pi2\\1&\text{ if }x=\frac\pi2\end{cases}\end{array}$$is discontinuous.