$\sum_0^\infty a_nz^n$ , $z\in \mathbb{C}$, a power series with $R:=\sup \{ t\ge0 : a_n t^n$ is bounded $\}$ as its convergence radius. I wish to prove that $\sum_0^\infty a_nz^n$ is locally normally convergent over $B(0,R)$.

What I did so far:

  • Let $r_1 <r_2<R$ , then for all $z\in B(0,r_1)$ : $$\sum_0^\infty|a_nz^n| = \sum _0^\infty|a_n|\frac{r_2^n}{r_2^n}|z|^n \le \sum _0^\infty|a_n|r_2^n(\frac{r_1}{r_2})^n .$$
  • Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$\sum _0^\infty|a_n|r_2^n(\frac{r_1}{r_2})^n \le M\sum _0^\infty(\frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$. So any power series for $z\in B(0,r_1)$ is mutually bounded by $M\sum_0^\infty(\frac{r_1}{r_2})^n$.

Yet I don't succeed to make the next step, and show that any $z\in B(0,r_1)$ has a neighborhood $U_z$ such that $\sum_0^\infty \sup_{U_z} |a_n z^n|$ converges.

  • $\begingroup$ @MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it... $\endgroup$ Dec 30, 2018 at 17:36
  • $\begingroup$ Actually what I said is nonsense. You need to show that $\sum \sup_U |a_n z^n| $ converges, not $\sum \sup_U |f|$. $\endgroup$
    – Martin R
    Dec 30, 2018 at 17:41
  • $\begingroup$ @MartinR ,Right! I'm sorry I miss typed, just corrected it. $\endgroup$ Dec 30, 2018 at 17:43
  • $\begingroup$ Or in short, for any $z_0\in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment. $\endgroup$ Dec 31, 2018 at 12:37

1 Answer 1


You have done all the necessary work, you just need to put it together:

For a given $z_0 \in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then $U=B(0, r_1)$ is a neighbourhood of $z_0$, and $ \sum_{n=0}^\infty \sup_U |a_n z^n| $ is convergent because $$ \sup_U |a_n z^n| \le M \left( \frac{r_1}{r_2} \right)^n $$ for some $M > 0$.


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