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As far as I know, $f(x)=\sum\limits_{n=0}^\infty a_n(x-x_0)^n$ is continuous on the whole convergence interval $K:=\{x\in\mathbb R:|x-x_0|<r\}$. Is there anything we could say about uniform continuity?

Added: Would it be correct to claim the following?

Power series is uniformly continuous on $\bar K:=\{x\in\mathbb R:|x-x_0|<\bar r\}$ where $0<\bar r<r$.

I'm asking for the clarification since we haven't defined compactness yet.

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    $\begingroup$ Yes, the series converges uniformly on compact subsets of $K$. $\endgroup$ – copper.hat Apr 4 '13 at 17:25
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    $\begingroup$ And is likewise uniformly continuous on compact subsets of $K$. $\endgroup$ – Cameron Buie Apr 4 '13 at 17:37
  • $\begingroup$ @CameronBuie: Must have been asleep. My comment is irrelevant. $\endgroup$ – copper.hat Apr 4 '13 at 17:59
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    $\begingroup$ It converges normally, hence uniformly, on every compact subset of the interval of convergence. And likewise on compact subsets of the disk of convergence in $\mathbb{C}$. $\endgroup$ – Julien Apr 4 '13 at 18:22
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    $\begingroup$ @mathusiast What you say is correct. It would be slightly more natural to let $\leq \overline{r}$ in the definition of your disk where $f$ is uniformly continuous. Not then that it is nothing but the usual Heine-Cantor: $f$ continuous is uniformly continuous on compact subsets. $\endgroup$ – Julien Apr 8 '13 at 11:02
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Check relevant theorems- Weierstrass M-test can be used to prove the uniform continuity.

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  • $\begingroup$ The set $K$ is not compact. The open cover $\{ B(x_0,r-\frac{1}{n}) \}_n$ has no finite subcover. $\endgroup$ – copper.hat Apr 4 '13 at 17:58

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