Let $f_n$ be analytic over a region $D$ on the complex plane. If there is an $M>0$ such that $$ \iint_D\left| f_n(z)\right|^2\,dx\,dy\leqslant M $$ for each $n$, then $\{f_n\}$ is a normal family.
It was an exercise in my class of modern complex analysis...I know it could be implied by Montel's theorem by proving that $\{f_n\}$ is uniformly bounded on each compact subset of $D$, but I find it difficult to get the uniform boundedness from the condition...
Let $K$ be a compact subset of $D$. There exists $r>0$ such that a disk $B(a,r)$ centered at $a$ with radius $r$ is contained in $D$ for every $a\in K$. Let $$f_n(z)=c_0+\sum_{n=0}^\infty c_n(z-a)^n\quad (c_0=f_n(a))$$ be its Taylor expansion at $a$. Then \begin{align} M&\ge \iint_D |f_n(z)|^2\,dxdy\ge \iint_{B(a,r)} |f_n(z)|^2\, dxdy\\ &=\int_0^{2\pi}\int_0^r\left(\sum_{n=0}^\infty c_nr^ne^{in\theta }\right)\left(\sum_{m=0}^\infty \overline{c_m}r^me^{-im\theta }\right) rdrd\theta \\ &=\int_0^r\left\{\int_0^{2\pi}\left(\sum_{n,m=0}^\infty c_n\overline{c_m}r^{n+m}e^{i(n-m)\theta }\right)d\theta \right\}rdr\\ &=2\pi\int_0^r\left(\sum_{n=0}^\infty |c_n|^2r^{2n+1}\right)dr\\ &\ge \pi r^2|c_0|^2. \end{align} Thus we have $$ |f_n(a)|\le \frac{\sqrt{M}}{r\sqrt{\pi}}$$ for every $a\in K$.