# For $f$ analytic on $|z|<1$, $|f|\le M$ with $a_1,\ldots,a_n\in \Bbb{D}$ zeros of $f$ show that $|f(0)|\le M \prod |a_j|$

For $f$ analytic in unit disk $\Bbb{D}$ where $|f|\le M$ with $a_1,\ldots,a_n\in \Bbb{D}$ such that $f(a_1)=\cdots=f(a_n)=0$ show that $|f(0)|\le M \prod |a_j|$.

I have tried many approaches including modifying the function using the Cauchy Formula, ML estimate, Maximum Principle and more. nothing seems to get me forward. I am really clueless here and could use a guidance.

• Can you show it when there is exactly one zero? – Jorge Fernández Hidalgo Jul 25 '16 at 1:29

By induction, we only have to show for $n=1$. Let $f(\alpha)=0$ and suppose $|f(z)|\leq 1$, i.e. $M=1$. (Otherwise, replace $f$ to $\frac{1}{M}f$.) Let \begin{align} \varphi_{\alpha}(z)=\frac{\alpha-z}{1-\overline{\alpha}z}, \end{align} which is automorphism on $\mathbb{D}$ with $\varphi_{\alpha}(\alpha)=0, \varphi_{\alpha}(0)=\alpha$. Define \begin{align} g(z)=f(\varphi_{\alpha}(z)). \end{align} where $g:\mathbb{D}\to \mathbb{D}$. Then $g(0)=0$ and $g(\alpha)=f(0)$. By Schwarz lemma, $|f(0)|=|g(\alpha)|\leq |\alpha|$.