The Set of Functions satisfying $\int_{D}\vert f(z)\vert (1-\vert z\vert)^2dA(z)\le 1$ is a Normal Family 
Let $\mathcal{F}$ be a family of holomorphic functions on the unit disc so that for any $f\in \mathcal{F}$ one has $$\int_{D}\vert f(z)\vert (1-\vert z\vert)^2dA(z)\le 1$$
  Prove $\mathcal{F}$ is a normal family.

I have also seen a similar problem where the integral was $\int_{D}\vert f \vert^2dA(z)$. I am looking for a solution that can can used in both of these examples. The only technique that comes to mind is the following:
Montel's Theorem: If $U\subset \mathbb{C}$ is an open set and $\mathcal{F}$ is a family of holomorphic functions on $U$ that is bounded on compact sets, then there is a sequence ${f_n}\subset \mathcal{F}$ that converges normally in $U$.
Attempt: For each $n$ let $g_n(z)=\int_{D}\vert f_n(z)\vert (1-\vert z \vert )^2dA(z)$. Then $g_n$ is bounded by $1$ on any compact subset of $D$. So by Montel's Theorem and continuity of the integral, $g_n$ converges normally to some $\int_{D}\vert g(z)\vert (1-\vert z \vert)^2dA(z)$. Hence $f_n$ converges normally to $g$.
Am I on the right track? For the example $\int_{D}\vert f \vert^2 dA(z)$, I am not sure if this method suffices since the hint is to use Cauchy inequalities to show that $\vert f \vert ^2$ does not exceed the mean value of $\vert f\vert ^2$ on a small disc centered at $z$.
 A: Hint:   $ |f|$ does not exceed the mean value of $|f|$ on a small disc centered at $z$, that is, $$
|f(z)|\le \frac{1}{\pi r^2}\int_{D_r(z)} |f(\zeta)| dA(\zeta)\tag{1}$$
where $D_r(z)$ is a disc with radius $r$ centered at $z$.  
Let $\mathcal{F}$ be a family of holomorphic functions on the unit disc so that for any $f\in \mathcal{F}$  one has $$\int_ D |f(z)|(1-|z|)^2 dA(z)\le 1.$$ 
Prove $\mathcal{F} $ is a normal family.
Proof. Let $z_0\in D$ and take a disc $D_r(z_0)$ centered at $z_0$ with radius $r=\frac{1-|z_0|}{3}$.
For every $z\in {D_r(z_0)}$ we have from (1)
$$|f(z)|\le\frac{1}{\pi r^2}\int_{D_r(z)} |f(\zeta)| dA(\zeta)\le \frac{1}{\pi r^2}\int_{D_r(z)}\frac{|f(\zeta)|\cdot 9(1-|\zeta|)^2}{(1-|z_0|)^2} dA(\zeta)$$
since $\frac{3(1-|\zeta|)}{1-|z_0|}> 1$ for all $\zeta \in {D_r(z)}$.
Therefore
$$\begin{align}|f(z)|&\le \frac{9}{(1-|z_0|)^2}\cdot\frac{1}{\pi r^2}\int_{D_r(z)} |f(\zeta)|(1-|\zeta|)^2 dA(\zeta)\\
&\le \left(\frac{3}{1-|z_0|}\right)^4\cdot\frac{1}{\pi}\int_{D} |f(\zeta)|(1-|\zeta|)^2 dA(\zeta)\\
&\le \left(\frac{3}{1-|z_0|}\right)^4\cdot\frac{1}{\pi }\end{align}$$
This means that $\mathcal{F}$ is uniformly bounded in ${D_r(z_0)}$ and hence $\mathcal{F}$ is a normal family by Montel's theorem.
Proof of a similar problem where the integral is $\int_{D}\vert f \vert^2dA(z)$:
It is simpler than arguments above. We use  the same notation. Since 
$$|f(z)|^2\le \frac{1}{\pi r^2}\int_{D_r(z)} |f(\zeta)|^2 dA(\zeta)\le \frac{1}{\pi r^2}\int_{D} |f(\zeta)|^2 dA(\zeta)
\le \frac{1}{\pi r^2},$$
$\mathcal{F}$ is uniformly bounded in ${D_r(z_0)}$.
