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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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. – Nate Eldredge Nov 30 '17 at 22:16
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No. Think about the geometric series, which converges (but not uniformly) to $$ 1 + x + x^2 + \cdots = \frac{1}{1x} $$ on $(1,1)$.
Ethan Bolker
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12It's an occupational hazard among mathematicians: things often look obvious after you understand them. That doesn't mean they were obvious beforehand. – Ethan Bolker Nov 30 '17 at 13:31

1+1 for the consolation. The confusion came from the fact that we proved certain properties of a power series in $(R,R)$, from the uniform convergence in a closed subinterval $[R+δ,Rδ]$. For instance, the sum function of the power series is continuous on the subinterval due to uniform convergence in said subinterval. We then proceed to conclude the continuity in $(R,R)$ since $δ$ was arbitrary. I thought if we can do this, why not use the same reasoning for uniform convergence? – Hrit Roy Nov 30 '17 at 13:53

Is this because continuity is a property regarding a single point whereas uniform convergence has to do with the whole interval? – Hrit Roy Nov 30 '17 at 13:56

2@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. – Ethan Bolker Nov 30 '17 at 14:00

True. Thanks for correcting me. But pointwise convergence is also such a property. Then wouldn't the pointwise convergence in the aforementioned subinterval imply the pointwise convergence in $(R,R)$? – Hrit Roy Nov 30 '17 at 14:06

1Yes. The theorem you quote in a previous comment implies pointwise convergence on the whole interval (and more, as you note). – Ethan Bolker Nov 30 '17 at 14:29

3Also 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. – Emil Jeřábek Nov 30 '17 at 17:31
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No. Example: $ \sum_{n=0}^{\infty}x^n=\frac{1}{1x}$ for $x<1$.
Suppose that $s_n(x):=\sum_{k=0}^{n}x^k=\frac{1x^{n+1}}{1x}$ converges uniformly on $(1,1)$ to $\frac{1}{1x}$.
Then , to $ \epsilon=1$ we get $N \in \mathbb N$ such that
$\frac{x^{N+1}}{1x}=s_N(x)\frac{1}{1x}<1$ for all $x \in (1,1)$.
Hence $\lim_{x \to 1}\frac{x^{N+1}}{1x} \le 1$. But $\lim_{x \to 1}\frac{x^{N+1}}{1x}= \infty$, a contradiction.
Fred
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