$L^{\infty}$ norm is bounded by $L^2$ norm for holomorphic functions I want to prove that, given $f$ holomorphic in a disc $D_r(z)$ and $0<s<r$, we have
$$\sup_{D_s(z)}|f| \leq \|f\|_{L^2(D_r(z))}$$
where the $L^2$ norm is defined on an open set $U \subset \mathbb{C}$ as $$\|f\|_{L^2(U)} = \Bigl(\int_U |f(z)|^2\,\mathrm dx\,\mathrm dy\Bigr)^{1/2}$$
I know I should use the mean value principle somehow, but I'm not sure how we can pull out the $L^2$ norm from it.  I can see that by the max modulus principle, we have that 
$$\sup_{D_s(z)} |f| \leq \sup_{C_s(z)}|f| \leq \sup_{C_r(z)}|f|$$
where $C_k(z)$ is the circle of radius $k$ centered on $z$.
After a while of trying to figure out how to apply the mean value principle, I tried working backwards from the $L^2$ norm via polar coordinates to get
$$\frac{r}{\sqrt 2} \Bigl(\int_{0}^{2\pi} |f(z+r\mathrm e^{\mathrm i\theta})|\,\mathrm d\theta\Bigr)^{1/2}.$$
Can we say that the integral is the mean value principle applied to $|f^2|$? I wanted to use max modulus and the fact that $D_s(z_0) \subset D_r(z_0)$ to conclude $\sup_{D_s}|f| \leq |f(z)|_{z \in D_r}$, but this inequality may not be true - it only applies if we consider $z \in D_r \setminus D_s$, correct?
I'd appreciate it if anyone could lead me into the right direction with this. It seems like the related questions on here have answers beyond the scope of this course.
 A: Let $d > 0$. If $g$ is holomorphic on some neighbourhood of $\overline{D_d(z)}$, the mean-value theorem tells us that
$$g(z) = \frac{1}{2\pi} \int_0^{2\pi} g(z + \rho e^{i\varphi})\,d\varphi$$
for $0 \leqslant \rho \leqslant d$, hence
$$| g(z)| \leqslant \frac{1}{2\pi} \int_0^{2\pi} | g(z + \rho e^{i\varphi}|\,d\varphi. \tag{1}$$
Multiplying $(1)$ with $\rho$ and integrating from $0$ to $d$ we obtain
$$|g(z)| \frac{d^2}{2} \le \frac{1}{2\pi} \int_0^d \int_0^{2\pi} | g(z + \rho e^{i\varphi})|\,\rho\,d\varphi\,d\rho = \frac{1}{2\pi} \iint_{\lvert w - z\rvert \leqslant d} | g(w)|\,dx\,dy.\tag{2}$$
Letting $g = f^2$ in $(2)$ and dividing by $d^2/2$, then  taking the square root, we obtain
$$|f(z)|\le \frac{1}{\sqrt{\pi}\,d} \,\lVert f\rVert_{L^2}(D_d(z)). \tag{3}$$
Choose $d = r-s$ to see
$$\lvert f(z)\rvert \le \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_{r-s}(z))} \le \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}\tag{4}$$
for all $z \in D_s(z_0)$, whence
$$\lVert f\rVert_{L^{\infty}(D_s(z_0))} \le \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}.$$
