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?

  • $\begingroup$ $x\to\infty$? Should it not be $n\to\infty$? $\endgroup$ – drhab Nov 18 '17 at 9:24
  • $\begingroup$ The sequence is typically defined by $n$, not $x$. $\endgroup$ – John Hughes Nov 18 '17 at 9:24
  • $\begingroup$ Yup. That was a typing error, it is n not x. $\endgroup$ – Shatabdi Sinha Nov 18 '17 at 9:38

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

  • $\begingroup$ So that means the limit wont exist when the value of sin x ranges between [-1,0[. Is it correct to say this sir? $\endgroup$ – Shatabdi Sinha Nov 18 '17 at 11:35
  • $\begingroup$ No. The limit does not exist only when $\sin(x)=-1$. If $\sin(x)\in (-1,1)$ then $\sin^n(x)\to 0$. $\endgroup$ – Robert Z Nov 18 '17 at 11:38
  • $\begingroup$ Yup, my bad, I realized that on my own a little later. Thank you, got it. $\endgroup$ – Shatabdi Sinha Nov 18 '17 at 16:31

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.

  • $\begingroup$ Ok sir got it. It may be silly to ask, still i would; $x \to 0$ when $x$ is not equal to $\pi /2$. Is this because the value of sin x is ranging between [0,1[ all this while and higher powers of such fractions eventually tends to zero? $\endgroup$ – Shatabdi Sinha Nov 18 '17 at 11:33
  • $\begingroup$ @ShatabdiSinha What I wrote was that when $x\notin\frac\pi2+\pi\mathbb Z$, then $\lim_{n\to\infty}\sin^n(x)=0$. That's because $\sin x\in(-1,1)$ and for any number $a\in(-1,1)$, $\lim_{n\to\infty}a^n=0$. $\endgroup$ – José Carlos Santos Nov 18 '17 at 13:19
  • $\begingroup$ Yes got it sir. I understood it. Thank you. $\endgroup$ – Shatabdi Sinha Nov 18 '17 at 16:34
  • $\begingroup$ Please mark one of the answers as your accepted answer. $\endgroup$ – José Carlos Santos Nov 18 '17 at 16:41

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