0

A (complex) power series is always (give or take some extra hypotheses) uniformly convergent on the interior of its disc of convergence. How do we prove this? Also, what is the exact statement? Does it converge uniformly on the interior, or just on every closed subset of the interior?

I think I can prove that it's normally convergent on any closed sub-disc of the interior, and since I think I also have that the function space $X\to\mathbb C$ for any set $X$ is complete (under the usual supremum of absolute value metric), I think I can prove that this implies absolute convergence. Is this correct? Can the result be strengthened?

Jack M
  • 27,819
  • 7
  • 63
  • 129

1 Answers1

1

A (complex) power series is always uniformly convergent on the interior of its disc of convergence.

Certainly not. Try to prove the uniform convergence of $\sum\limits_nz^n$ on $\{z\in\mathbb C\mid |z|\lt1\}$ and you should soon run into trouble.

On the other hand, every series $\sum\limits_na_nz^n$ with positive radius of convergence $R$ does converge uniformly on every disk (open or closed) with radius $R'\lt R$.

Did
  • 279,727
  • Oh, I must have been mixing it up with the theorem that it's continuous on the entire open disc of convergence. – Jack M Feb 02 '14 at 11:15
  • Power series are almost uniformly convergent, that is, uniformly convergent on every compact subset of domain. – xyzzyz Feb 02 '14 at 11:31
  • Can this be proven in the way I outlined in my question? – Jack M Feb 02 '14 at 12:20
  • Of course if normal convergence holds everything follows. A direct road to normal convergence is to estimate the $n$th term of the series, using its convergence at some $R''\gt R'$. – Did Feb 02 '14 at 14:13