Suppose I have a function $f(z)$ analytic in an annulus $A_r := \{z\in\mathbb{C}\mid 1/r<|z|<r\}$ and I know that $|f(z)|<\epsilon$ on $A_r$. Furthermore, $f(z) = \sum_{k\in\mathbb{Z}}f_k z^k$, and $f(z) = f^+(z)+f^-(z)$, which are defined as $f^+(z) = \frac{1}{2\pi i}\int_{|\zeta| = r} \frac{f(\zeta)}{\zeta-z}\,d\zeta$ and $f^-(z) = -\frac{1}{2\pi i}\int_{|\zeta| = 1/r} \frac{f(\zeta)}{\zeta-z}\,d\zeta$.
Is there anything that can be said about $|f'(z)|$ on a smaller annulus? I know that I can estimate the coefficients as $|f_k|<\epsilon r^{-|k|}$ and I also know that if I could estimate the coefficients of the derivative as $|\tilde{f}_k|<Kr^{-|k|}$, I would have $|f'(z)|<2Kr/\delta$ on the annulus $A_{r-\delta}$. However, since the coefficients of $f'$ are of the form $(k+1)f_k$, I have no idea how to find a bound so that I can use the last estimate. $f_k$ can be written as an integral, but even then, the factor $(k+1)$ would still appear....
Any help would be appreciated!
For $z \in A_r$ is $$ |f'(z)| = \left| \sum_{k=-\infty}^\infty k f_k z^{k-1}\right| \le \sum_{k=-\infty}^\infty |k| \frac{\epsilon}{r^{|k|}} |z|^{k-1} \\ = \epsilon \left( \sum_{k=0}^\infty k \frac{|z|^{k-1}}{r^k} + \sum_{k=1}^\infty k \frac{1}{r^k|z|^{k+1}}\right) \, . $$ Now assume that $z \in A_s$ with $0 <s < r$. Then $|z| < s$ and the first sum can be estimated as $$ \frac 1r\sum_{k=0}^\infty k \left( \frac sr\right)^{k-1} = \frac 1r \frac{1}{(1-s/r)^2} \, . $$ Similarly, the second sum can be estimated using $|z| > 1/s$ as $$ \frac {s^2}r\sum_{k=0}^\infty k \left( \frac sr\right)^{k-1} = \frac {s^2}r \frac{1}{(1-s/r)^2} \, . $$ Adding these estimates one gets $$ |f'(z)| \le \frac{(1+s^2)\epsilon}{r(1-s/r)^2} $$ for $1/r < 1/s < |z| < s < r$.
Another option is to estimate $$ f'(z) = \frac{1}{2\pi i}\int_{|z|=r} \frac{f(\zeta)}{(\zeta-z)^2}\, d\zeta - \frac{1}{2\pi i}\int_{|z|=1/r} \frac{f(\zeta)}{(\zeta-z)^2}\, d\zeta $$ as $$ |f'(z)| \le \frac{r \epsilon}{(r-|z|)^2} + \frac{\epsilon/r}{(|z|-1/r)^2} \, . $$ For $z \in A_s$ with $0 < s < r$ this leads to the same estimate as the above one.