TITLE: Proving that $\{ f_a \}_{a \in A}$ satisfying $\int \limits_{0}^{2\pi} |f_a(e^{i \phi})|^{1/2} d\phi \leq 1$ is a normal family in $\mathbb{D}$
QUESTION [7 upvotes]: This is another problem from a complex analysis qualifying exam from last year for the preparation course that I'm teaching right now. The question is the following.


Let $F = \{ f_a \}_{a \in A}$ be a family of holomorphic functions on a neighborhood of the closed unit disk $\overline{\mathbb{D}} = \{ z \in \mathbb{C}  \mid |z| \leq 1 \}$. Suppose also that 
$$
\int \limits_{0}^{2\pi} |f_a(e^{i \phi})|^{1/2} d\phi \leq 1
$$
for every $a \in A$. Prove that $F = \{ f_a \}_{a \in A}$ is a normal family in the unit disk $\mathbb{D} = \{ z \in \mathbb{C}  \mid |z| < 1 \}$.



My Attempt
My idea is to use Montel's theorem, so I want to show that the family $F$ is uniformly bounded on compact subsets of $\mathbb{D}$. Thus letting $K \subset \mathbb{D}$ be compact, then there's a positive constant $0 < M_K < 1$, such that for every $\xi \in K$ we have $|\xi| \leq M_K$. Then using Cauchy's integral formula we obtain, for any $a \in A$ and $\xi \in K$
\begin{align}
f_a(\xi) = \frac{1}{2\pi i} \int \limits_{\partial \mathbb{D}} \frac{f_a(z)}{z - \xi} \, dz = \frac{1}{2\pi i} \int \limits_{0}^{2\pi} \frac{f_a(e^{i \phi})}{e^{i\phi} - \xi} \cdot ie^{i\phi} \, d\phi
\end{align}
Then taking absolute values we get
\begin{align}
|f_a(\xi)| &\leq \frac{1}{2\pi} \int \limits_{0}^{2\pi} \frac{|f_a(e^{i \phi})|}{|e^{i\phi} - \xi|} \cdot |ie^{i\phi}| \, d\phi \\
&\leq  \frac{1}{2\pi} \int \limits_{0}^{2\pi} \frac{|f_a(e^{i \phi})|}{ 1 - |\xi|}  \, d\phi\\
&\leq \frac{1}{2\pi(1 - M_K)}  \int \limits_{0}^{2\pi} |f_a(e^{i \phi})| \, d\phi\\
&= C_K \int \limits_{0}^{2\pi} |f_a(e^{i \phi})| \, d\phi
\end{align}
where we put $C_K := \dfrac{1}{2\pi(1 - M_K)}$, and the reverse triangle inequality was used to bound $\dfrac{1}{|e^{i\phi} - \xi|} \leq \dfrac{1}{1 - |\xi|}$. The constant $C_K$ is independent of $a$ and $\xi$. Hence if I can somehow use the hypothesis that $\int \limits_{0}^{2\pi} |f_a(e^{i \phi})|^{1/2} \, d\phi \leq 1$ to bound the last integral then the problem would be solved because Montel's theorem would apply. But unfortunately I am stuck at this point.

Question
How can I finish the argument? (assuming that what I did is the right way to proceed)
Thanks a lot for any help.

REPLY [5 votes]: I don't see how to conclude with what you have done. However, you can prove the result as follows.
The key fact is that for any holomorphic function $f$, the function $\sqrt{\vert f\vert}$ is $subharmonic$. It follows that for any $a\in A$ and $z\in \mathbb D$, we have 
$$\sqrt{\vert f_a(z)\vert}\leq \int_0^{2\pi} P_z(\theta) \sqrt{\vert f_a(e^{i\theta})\vert}\, \frac{d\theta}{2\pi}\, ,$$
where $P_z$ is the Poisson kernel at $z$, 
$$P_z(\theta)=\frac{1-\vert z\vert^2}{\vert e^{i\theta}-z\vert^2}\cdot $$
Now, if $z$ stays in a compact set $K\subset \mathbb D$, the Poisson kernel $P_z$ remains bounded by a constant $C$ depending only on $K$, so we get that the functions $\sqrt{\vert f_a\vert}$, $a\in A$ are uniformly bounded on $K$. Hence, $(f_a)_{a\in A}$ is a normal family.