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