Topology of compact convergence on space of holomorphic functions

Question

Consider $\mathbb{D}_{R} = \{z \in \mathbb{C}: |z| < R\}$. For given $f \in \mathscr{O}(\mathbb{D}_{R})$ - space of holomorphic functions on the open disk, denote by $c_n$ the $n$-th Taylor's coeffitient of $f$ at $0.$ Hot to show that the topology of compact convergence on $\mathscr{O}(\mathbb{D}_{R})$ is generated by family of seminorms $\{\|\;.\|_{r, \infty}: 0 <r < R\}$ where $\|f\|_{r, \infty} = \sup_{n \geq0}|c_n|r^n$. It's sufficient to show that we can major the norm from one family by finite norms from another family and vice versa, but i failed writing inequalities.

Answer 1

It suffices to prove that, for every $R>r>\delta>0$, if $||f_\lambda-f||_{r, \infty}\to 0$ then $f_\lambda$ converges uniformly to $f$ on $\{z:|z|\le r-\delta\}$ and that if it converges uniformly on $\{z:|z|\le r\}$ then $||f_\lambda-f||_{r,\infty}\to 0$ (where $f_\lambda$ is a net).

Let us first prove that uniform convergence implies $||f_\lambda-f||\to 0$. Recall that, by Cauchy's inequalities we have $$|c_n|r^n\le \max_{|z|=r}|f(z)|$$ So if $f_\lambda\to f$ uniformly on $\{z:|z|\le r\}$, $||f_\lambda-f||_{r,\infty}=\sup_n |c_n|r^n \le \max_{|z|=r}|f(z)-f_n(z)|\to 0$.

On the other hand, suppose that $||f_\lambda-f||_{r,\infty}\to 0$. Then $\sup_k|c_k-c_{k,\lambda}|r^k\to 0$. Take $N$ big enough so that $\forall \lambda\ge N$ we have $|c_{k}-c_{k,\lambda}|r^k<\varepsilon$. Then for every $z:|z|\le r(1-\delta)$ we have $$|f(z)-f_\lambda(z)|\le \sum |c_k-c_{k,\lambda}||z|^k\le \sum \varepsilon(1-\delta)^k= \frac{\varepsilon}{\delta}$$ Thus, fixed $\delta>0$, we have uniform convergence on $\overline{\mathbb{D}_{r-\delta}}$.

This implies the result: its clear that a net converges in the topology iff it converges uniformly on compact subsets of $\mathbb{D}_R$. Actually, the topology is clearly metrizable, as one can take $r\in \mathbb{Q}\cap [0,R]$ to define a countable family of seminorms generating the same topology and so one can dispense from using nets, using sequences instead.

Answer 2

For fixed $R>r+\epsilon>r>\epsilon>0$

If $\sup_{n\ge 0} |c_n| r^n$ is small then $\sup_{|z|\le r-\epsilon}|f(z)|$ is small,

if $\sup_{|z|\le r+\epsilon}|f(z)|$ is small then $r^n c_n = r^n\frac1{2i\pi}\int_{|z|=r+\epsilon} \frac{f(z)}{z^{n+1}}dz$ is small.