If $f$ is a holomorphic function on the unit disc $\{z:|z|<1\}$, we can write $f(z)=\sum_{n=0}^{\infty}a_nz^n$. Can we be sure that this series converges uniformly on compact subsets of the disc, or do I need to add assumption that $\sum_{n=0}^{\infty}a_n<\infty$ somewhere?
Uniform convergence of power series expansion of a holomorphic function
complex-analysisconvergence-divergenceuniform-convergence
Best Answer
Yes, we can be sure of that. Note that a compact subset $K$ of the disc is contained in $\{z: |z| \le r\}$ for some $r < 1$ (because the continuous real-valued function $|z|$ attains a maximum on a compact set). Take $r < s < 1$. Now $|a_n| s^n$ must be bounded in order for $\sum_n a_n s^n$ to converge, so there is $B$ such that $|a_n z^n| \le B (|z|/s)^n \le B (r/s)^n$ and the series converges uniformly on $K$.