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Let $R>0$ be the radius of convergence of a power series $Σa_nx^n$. Is it not uniformly convergent in $(-R,R)$? My book goes out of its way to say that if $[a,b]⊂(-R,R)$, then the power series converges uniformly in $[a,b]$. Can't we just say that it is uniformly convergent in $(-R,R)$?

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  • $\begingroup$ This mode of convergence is sometimes called "uniform convergence on compact subsets", "local uniform convergence", or "normal convergence". As shown below it's not the same as uniform convergence. $\endgroup$ – Nate Eldredge Nov 30 '17 at 22:16
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No. Think about the geometric series, which converges (but not uniformly) to $$ 1 + x + x^2 + \cdots = \frac{1}{1-x} $$ on $(-1,1)$.

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  • $\begingroup$ This seems so obvious now that you've pointed it out. Thank you. $\endgroup$ – Hrit Roy Nov 30 '17 at 13:29
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    $\begingroup$ It's an occupational hazard among mathematicians: things often look obvious after you understand them. That doesn't mean they were obvious beforehand. $\endgroup$ – Ethan Bolker Nov 30 '17 at 13:31
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    $\begingroup$ @HritRoy Yes, pretty much. To be a little more technical, continuity is a property regarding values near a single point, not at a single point. $\endgroup$ – Ethan Bolker Nov 30 '17 at 14:00
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    $\begingroup$ Yes. The theorem you quote in a previous comment implies pointwise convergence on the whole interval (and more, as you note). $\endgroup$ – Ethan Bolker Nov 30 '17 at 14:29
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    $\begingroup$ Also note that if the series converges uniformly in $(-R,R)$, then it in fact converges uniformly in $[-R,R]$, and in particular, it converges to a continuous function on $[-R,R]$. Thus, any series that does not have finite limits at the endpoints of the interval must be a counterexample. $\endgroup$ – Emil Jeřábek Nov 30 '17 at 17:31
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No. Example: $ \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $|x|<1$.

Suppose that $s_n(x):=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}$ converges uniformly on $(-1,1)$ to $\frac{1}{1-x}$.

Then , to $ \epsilon=1$ we get $N \in \mathbb N$ such that

$\frac{|x|^{N+1}}{1-x}=|s_N(x)-\frac{1}{1-x}|<1$ for all $x \in (-1,1)$.

Hence $\lim_{x \to 1-}\frac{|x|^{N+1}}{1-x} \le 1$. But $\lim_{x \to 1-}\frac{|x|^{N+1}}{1-x}= \infty$, a contradiction.

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  • $\begingroup$ Ah, of course. Silly of me. Thanks alot. $\endgroup$ – Hrit Roy Nov 30 '17 at 13:28

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