An exercise from Conway's Complex Analysis on an application of Schwarz's Lemma

Question

Let $f$ be analytic in $B(0;1)$, and suppose that $|f(z)|\leq M$ for all $z\in B(0,1)$. If $f(z_{k})=0$ for $1\leq k \leq n$. Show that $$|f(z)| \leq M\prod_{k=1}^{n}\frac{|z-z_{k}|}{|1-\bar{z_{k}}z|}$$ for $|z|<1$.

What I have done is proved the result when $k=1$. That is if $f(z_{1})=0$, with $|f(z)|\leq M$ on $B(0;1)$, I have proved that $$|f(z)|\leq M \frac{|z-z_{1}|}{|1-\bar{z_{1}}z|}$$ for $|z|<1$.

Now I tried to consider this function:$g:B(0;1)\to \mathbb{C}$

$$g(z)=\frac{f(z)}{\prod_{k=2}^{n}\frac{|z-z_{k}|}{|1-\bar{z_{k}}z|}}$$

Now $g$ has removable singularity at the points $z_{2}, z_{3},.... z_{n}$. So after removing these singularities we can consider $g$ to be analytic. Now wanted to invoke the result I have proved above since $g(z_{1})=0$. But couldn't do so because $g(z)$ might not be bounded by $M$ on $B(0;1)$. How can I work this problem out.

Thanks in advance!!

Answer 1

As Jack D'Aurizio commented, we must assume $z_j \in B(0,1)$.

It's easier to do if we assume $f$ is continuous on the closed disk $\overline{B(0,1)}$. We can do the question for $f((1-\epsilon) z)$ and then take the limit as $\epsilon \to 0+$.

With this assumption, note that $|g(z)| = |f(z)|$ for $|z|=1$, because $|z - z_k| = |1 - \overline{z_k} z|$ there.

Answer 2

Ok, I think this works:

By the Maximum Modulus Principle, for every $0 < r < 1$ we have that for all $|w|<r$ \begin{equation} \left|\dfrac{f(w)}{M\prod_{k=1}^n\varphi_{a_k}(w)}\right| \leq \sup_{|z|=r} \prod_{k=1}^n |\varphi_{a_k}(z)|^{-1}. \end{equation} The quotient on the LHS is holomorphic on de disc (after removing singularities) and the RHS goes to $1$ when $r \to 1$, which implies the desired inequality.