The following is exercise 4.9 in Donald Marshalls Complex analysis.
Show that there is a constant $C < \infty$ so that if $f$ is analytic on $\mathbb{D}$, then
$$|f'(z)| \leq C \int_{\mathbb{D}} |f(x+iy)| dxdy$$
for $|z| \leq 1/2$.
Here is my try: Let $A$ be the closed region bounded between the circles centered at $0$ of radius $3/4$ and $9/10$. It suffices to show that that there is $C < \infty$ such that
$$|f'(z)| \leq C\int_{A} |f(x+iy)| dxdy$$
Indeed, let $|z| \leq 1/2$ and $3/4 \leq r \leq 9/10$. Define $D_r$ to be the disk centered at the origin of radius $r$. Then, by the Cauchy Integral Formula
\begin{align} |f'(z)| & = \left| \frac{1}{2\pi i}\int_{\partial D_r} \frac{f(\zeta)}{(\zeta - z)^2} d\zeta \right| \\ & = \left|\frac{1}{2\pi i} \int_0^{2\pi} \frac{f(re^{i\theta})}{|re^{i\theta} - z|^2} rie^{i\theta} d\theta \right| \\ & \leq \frac{1}{2\pi}\int_0^{2\pi} \frac{|f(re^{i\theta})|}{|re^{i\theta} - z|^2} rd\theta \\ & \leq \frac{1}{2\pi} \int_0^{2\pi} \frac{|f(re^{i\theta})|}{(r-1/2)^2} r d\theta \\ & \leq \frac{1}{2\pi}\int_0^{2\pi} \frac{|f(re^{i\theta})|}{(1/4)^2} r d\theta \\ & \leq \frac{8}{\pi} \int_0^{2\pi} |f(re^{i\theta})| r d\theta \end{align}
Define $C = 8/\pi\cdot\frac{1}{9/10 - 3/4}$. Note $\int_A |f(x+iy)| < \infty$ and so by Fubini's Theorem and a change of variables
\begin{align} 8/\pi\cdot\frac{1}{9/10 - 3/4} \int_A |f(x+iy)| dxdy & = \frac{1}{9/10 - 3/4} \int_{3/4}^{9/10} 8/\pi \int_0^{2\pi} |f(re^{i\theta})|r d\theta dr \\ & \geq \frac{1}{9/10 - 3/4} \int_{3/4}^{9/10} |f'(z)| dr \\ & = |f'(z)| \end{align}
Does this solution seem correct? Any comments or suggestions would be appreciated.