Show that there exists a constant C depending on $U$ and $K$ such that $|f(z)| \leq C(\int_{U} |f|^2)^{1/2}$.

Question

Let $f$ be analytic in an open set $U \subseteq \Bbb C$ and let $K \subseteq U$ be compact. Show that there exists a constant C depending on $U$ and $K$ such that $|f(z)| \leq C(\int_{U} |f|^2)^{1/2}$ for all $z \in K$. I'm not sure what knowledge in complex analysis should be used here. I think $f(z)$ should be equal to an integral then apply Cauchy-Schwarz inequality?

Answer 1

There exists $r = r(U,K)>0$ such that the closed disk $D_r(z)$ of radius $r$ centered at $z$ is contained in $U$ for all $z \in K$. By the mean value property (area version) and triangle inequality for integrals $$ |f(z)| = \frac{1}{\pi r^2}\left|\iint_{|w-z| \le r} f(w) \, du \, dv\right| \le \frac{1}{\pi r^2}\iint_{|w-z| \le r} |f(w)| \, du \, dv $$ Now by Cauchy-Schwarz and monotonicity of the integral this is $$ \le \frac{1}{\pi r^2} \left(\iint_{|w-z| \le r} 1^2 \, du \, dv\right)^{1/2} \left(\iint_{|w-z| \le r} |f(w)|^2 \, du \, dv\right)^{1/2} \le \frac{1}{\sqrt{\pi r^2}} \left(\iint_U |f(w)|^2 \, du \, dv\right)^{1/2} $$