I have a question from a past complex analysis qualifying exam that I would like some help on. Let $f(z)$ be an analytic function on the unit disk $\mathbb{D}$ such that $\lvert f(z)\rvert \leq 1$. Suppose that $z_1,z_2,…,z_n\in \mathbb{D}$ are zeroes of $f(z)$.
Show that
\begin{align}\lvert f(0)\rvert \leq \prod_{j=1}^n \lvert z_j\rvert,\end{align}
where each $z_j$ is repeated in the product on the right hand side corresponding to its multiplicity. For instance, if $z_1$ has multiplicity 3, then we would see $\lvert z_1\rvert^3$ on the right hand side.
This is what I have so far. Suppose that $f(z_j)=0$ with an order of $m_j\geq 1$. Letting $g_j:= f\circ \psi_j$, where $\psi_j$ is the FLT $\psi_j(z)=\frac{z+z_j}{1+\overline{z_j}z}$, we have that $g$ is analytic on $\mathbb{D}$ with a zero of order $m_i$ at $z=0$ and satisfies $\lvert g(z)\rvert \leq 1$. Thus, $\lvert g(z)\rvert \leq \lvert z\rvert ^{m_j}$ by the Schwarz Lemma, and in particular $\lvert g(-z_j)\rvert = \lvert f(0)\rvert \leq \lvert z_j\rvert^{m_j}$. However, I don't know how to combine this over all of the $z_j$ to get the desired inequality.
Best Answer
Hint: Use induction on $n$. Let $\phi_j (z)=\frac {z-z_j}{1-\overline {z_j}z} =\psi_j^{-1}(z)$ and consider $\frac {f{(z)}} {\phi_1(z)....\phi_{n-1}(z)}$. Use MMP to show that this is bounded in modulus by $1$.