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Is the sequence of functions $(\sin(x))^n$ uniformly convergent in $[0,\pi]$? Can you give me a hint or solution, I have already already prove that it is UC in $[0,1]$ but I don't know how to proceed in $[1,\pi]$.

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  • $\begingroup$ How did you show it for $[0, 1]$? $\endgroup$
    – Dair
    Aug 16, 2017 at 0:11
  • $\begingroup$ $|(sin(x))^n|\leq |x|^n$ for $x\in [0,1]$ $\endgroup$
    – Ragnar1204
    Aug 16, 2017 at 0:13
  • $\begingroup$ @Ragnar1204 $|x|^n$ is not uniformly convergent on $[0,1]$ or even $[0,1)$. $\endgroup$
    – stewbasic
    Aug 16, 2017 at 3:09

2 Answers 2

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Hint: find the pointwise limit of this sequence of functions on $[0,\pi]$. If a sequence is uniformly convergent, then its limit is a continuous function. Is it continuous in this case?

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It is not uniformly convergent: At $\pi/2$ we have $\sin(\pi/2)=1$ and thus $$ \sin(\pi/2)^n\to 1 $$ What happens everywhere else where $\sin(x)<1$?

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