$L^{\infty}$ norm is bounded by $L^2$ norm for holomorphic functions

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

$$\lVert f\rVert_{L^{\infty}(D_s(z_0))} \le \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}.$$