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I am taking a course in analysis, and I am wondering whether it possible for a power series with radius of convergence $1$ to converge uniformly on $(-1,1)$ but not on $[-1,1]?$

I don't think this is possible, since the power series will define a continuous function over $[-1,1]$ (assuming it is defined at $-1$ and $1$) which drags in $-1,$ and $1$ into the game when considering uniform convergence on $(-1,1)$. I can't decide what happens if the series blows up at $-1$ or $1$. It looks like we cannot have uniform convergence, but I am not sure why.

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There is no such series, and this is not something specific to power series.

Claim

If a sequence of continuous functions on $E$ converges uniformly on a dense subset of set $E$, then it converges uniformly on $E$.

Proof. Uniform convergence is equivalent to being Cauchy in the uniform norm, which means $$ \forall \epsilon\ \exists N \text{ such that }\sup_E |f_n-f_m|<\epsilon\quad \forall m,n\ge N $$ Since $|f_n-f_m|$ is continuous, it has the same supremum over $E$ as over any dense subset of $E$. $\quad\Box$

In your situation, $f_n$ is the $n$th partial sum of the series, $E=[-1,1]$, and the dense subset is $(-1,1)$.

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  • $\begingroup$ I'm not sure, but have you used the assumption here that it converges at ±1? Or does that also follow as a consequence? $\endgroup$ Commented Mar 20, 2023 at 15:05
  • $\begingroup$ Nevermind, I think follows as a consequence since the partial sums are continuous on [-1,1] $\endgroup$ Commented Mar 20, 2023 at 15:45

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