I need to show that the operator of taking the derivative is continuous, but at a certain point, I got stuck. This must be a simple problem but I'm a layman in complex analysis.
Denote $\mathbb{D}=\{z\in\mathbb{C}:|z| < 1\}$ and consider vector space $$ H(\mathbb{D})=\{f:\mathbb{D}\to\mathbb{C}: f \mbox{ is analytic}\} $$ We endow this space with a family of seminorms $$ p_q : H(\mathbb{D})\to\mathbb{R}_+:f\mapsto \max\limits_{|z|\leq q}|f(z)|, \quad (0< q < 1) $$ to make it a locally convex space. By the maximum modulus principle $$ p_s(f)=\max\limits_{|z|=s}|f(z)|\tag{1} $$ $$ p_s(f)\leq p_t(f)\tag{2} $$ for any $f\in H(\mathbb{D})$ and $s,t\in(0,1)$ such that $s\leq t$.
I need to show that the linear operator
$$
T:H(\mathbb{D})\to H(\mathbb{D}): f \mapsto f'
$$
is continuous. In other words, I need to show that
$$
\forall q\in(0,1)\quad\exists C>0 \quad \exists r_1,\ldots, r_k\in (0,1)\quad \forall f\in H(\mathbb{D})\quad p_q(f')\leq C \max_{i\in\{1,\ldots,k\}} p_{r_i}(f).
$$
Using $(2)$ the latter definition is equivalent to
$$
\forall q\in(0,1)\quad\exists C>0 \quad \exists r\in (0,1)\quad \forall f\in H(\mathbb{D})\quad p_q(f')\leq C p_r(f).\tag{3}
$$
Consider arbitrary $f\in H(\mathbb{D})$, then for some $(c_n)_{n\in\mathbb{Z}_+}$ and all $z\in \mathbb{D}$ we have
$$
f(z)=\sum_{n=0}^\infty c_n z^n,
\quad\quad
f'(z)=\sum_{n=0}^\infty nc_n z^{n-1}.
$$
Then for any $s\in(0,1)$
$$
p_s(f)=\max\limits_{\phi\in[0,2\pi)}\left|\sum_{n=0}^\infty c_n s^{n} e^{in\phi}\right|
\quad\quad
p_s(f')=\max\limits_{\phi\in[0,2\pi)}\left|\sum_{n=0}^\infty n c_n s^{n-1} e^{in\phi}\right|
$$
At this point I'm stuck.
By Cauchy’s formula, if $|z| \leq s < t<1$, $f(z)=\frac{1}{2i\pi}\int_{C(0,t)}{\frac{f(w)}{w-z}dw}$ so that $f’(z)=\frac{1}{2i\pi}\int_{C(0,t)}{\frac{f(w)}{(w-z)^2}dw}$. It follows that $|f’(z)| \leq p_t(f)\frac{t}{(t-s)^2}$. In other words, $p_s(f’) \leq \frac{t}{(t-s)^2}p_t(f)$ for any $0 < s < t <1$.