# Complex Analysis: Exhibiting an upper bound

Let $f:D(0,1) \to \mathbb{C}$ be an analytic function such that $|f(z)| \leq M, ~\forall z \in D(0,1)$ and $f(z_1) = 0.$

Claim: The estimate

\begin{equation*} |f(z)| \leq M \left( \frac{|z-z_1|}{|1-\overline{z_1}z|}\right) \end{equation*}

holds.

Can anyone give suggestions on how to start this?

• Compose $f$ with $(z_1 - z)/(1-\bar {z_1}z).$ – zhw. May 12 '15 at 18:24
• I think want you want is to use a certain Möbius transformation and then apply Schwarz lemma. – Suugaku May 12 '15 at 18:24
• @zhw. Is $f\bigg(\frac{z_1 - z}{1-\overline{z_1}z}\bigg) \leq M$? I don't know where you are going with that. Is the thing I am composing with in D(0,1)? – Mr.Fry May 12 '15 at 18:28

Hint. Set $$\varphi(z)=\frac{z+z_1}{1+\overline{z}_1z}$$ then $$\varphi^{-1}(z)=\frac{z-z_1}{1-\overline{z}_1z}.$$ Both, $\varphi$ and $\varphi^{-1}$ are bijections between $D$ and $D$.
In particular, if $g(z)=f\big(\varphi(z)\big)$, then $g(0)=0$, $\lvert g(z)\rvert\le M\lvert z\rvert$.
Then apply Schwarz lemma on $g$ and obtain that $\lvert g(z)\rvert \le M\lvert z\rvert$, and hence $$\lvert\,f(z) \rvert=\left|g\big(\varphi^{-1}(z)\big)\right| \le M\lvert\varphi^{-1}(z)\rvert =M\left|\frac{z-z_1}{1-\overline{z}_1z}\right|,$$ for all $z\in D$.
• How did $M |\phi^{-1}(z)|$ come about? ( I understand everything else) – Mr.Fry May 12 '15 at 19:02
• Also is $|g| \leq 1$? THis is required for the Lemma. – Mr.Fry May 12 '15 at 19:08
• I review this and schwarz lemma gives $|g| \leq|z|$ how do you get $|g| \leq M|z|$? – Mr.Fry May 13 '15 at 17:40
• Apply Schwarz Lemma to $h(z)=g(z)/M$. – Yiorgos S. Smyrlis May 13 '15 at 17:52