I am studying operators on the space of entire functions and I don't know if the following is true.
Let $H$ the space of entire functions $f:\mathbb{C}\to \mathbb{C}$ and $\left\|f\right\|_{R}:=\sup_{|z|\leq R}|f(z)|$ with $R>0$. Then if $f\in H$, $\left\|\partial_{z} f\right\|_{R}\leq C \left\|f\right\|_{S}$, some $S>0$?
I just see that writing $f$ by power series, $f=\sum_{n=0}^{\infty} a_n z^n$ then $\left\|\partial_{z} f\right\|_{R}=\sup_{|z|\leq R} | \sum_{n=1}^{\infty}a_n\cdot n z^{n-1}|$ but I don't know how I could continue to bound
Actualization. I have this: $$\left\|f'\right\|_{R}=\sup_{|z|\leq R} |f'(z)|=|f'(a)|$$ some $|a|\leq R$
Now, by Cauchy's estimate, $$|f'(a)|\leq \frac{M}{S-R}$$ with $|f(z)|\leq M:=\sup_{|z-a|\leq S-R}|f(z)|$ for all $|z-a|\leq S-R,\, S>R$.
Then, $$\left\|f'\right\|_{R}\leq \frac{1}{S-R}sup_{|z-a|\leq S-R}|f(z)|$$
Is it possible to bound $\sup_{|z-a|\leq S-R}|f(z)|\leq \sup_{|z|\leq S} |f(z)|$ noting that $|z|\leq |z-a|+|a|\leq (S-R)+R=S$?