How to bound the $L^\infty$ norm of a holomorphic function by its $L^2$ norm on a larger domain? More speficifically, what I need to prove is that for every positive numbers $r,s$ with $r<s$ there exists a $C_{r,s}$ such that for all holomorphic functions $f(z)$ on a region that contains $\bar B_s$,$\operatorname{Max}_{|z|\leq r} |f(z)|\leq C_{r,s} \left(\int_{\bar B_s}|f(z)|^2\mathrm d x \mathrm d y\right)^{\frac{1}{2}}$. The former is the $||f||_\infty$ on the small disk and the latter is the (multiple of ) $||f||_2$ on the larger disk. (the two disks are concentric)
 A: Two approaches, pick what works better for you. 


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*Since the function $|f|$ is subharmonic, it follows that
$$|f(z)|\le \frac{1}{\pi \rho^2} \iint_{|\zeta-z|< \rho} |f|$$ 
as long as disk of radius $\rho$ centered at $z$ is in the domain of $f$. If $|z|\le r$, you can use $\rho = s-r$ here. Thus, the denominator is bounded from below, while the integral is under control. (If you want to get to the $L^2$ norm quicker, work with $|f|^2$, which is also subharmonic.)

*Express the integral of $|f|^2$ over the disk in terms of the coefficients of $f$, by integrating $|f|^2=f\bar f$ in polar coordinates:  $$\sum_{n=0}^\infty \frac{ s^{2n} |c_n|^2}{n+1} =  \frac{1}{\pi}\iint_{B_s} |f|^2$$ Then use the Cauchy-Schwarz inequality to estimate $|f(z)|$ for $|z|\le r$: $$\sum_{n=0}^\infty r^n |c_n|=\sum_{n=0}^\infty \frac{s^n|c_n|}{\sqrt{n+1}} \frac{r^n\sqrt{n+1}}{s^n}\le \left(\sum_{n=0}^\infty \frac{s^{2n}|c_n|^2}{{n+1}} \right)^{1/2}\left(\sum_{n=0}^\infty  \frac{r^{2n}(n+1)}{s^{2n}}\right)^{1/2} $$
