This is taken from an old complex analysis qualifying exam.


Let $\Delta$ denote the unit disc $\{z\in\mathbb{C}:|z|<1\}$.

Suppose $f:\Delta\rightarrow\Delta$ is holomorphic. Show that $$\frac{|f(0)|-|z|}{1-|f(0)||z|}\leq|f(z)|\leq\frac{|f(0)|+|z|}{1+|f(0)||z|}$$ for all $|z|<1$.


Since $f(0)\in\Delta$, define $\phi\in\operatorname{Aut}(\Delta)$ as $$\phi(z)=\frac{f(0)-z}{1-\overline{f(0)}z}.$$ Then $\phi\circ f$ maps the unit disc into the unit disc and fixes zero. Thus, by Schwarz' lemma, we have $$\left|\frac{f(0)-f(z)}{1-\overline{f(0)}f(z)}\right|\leq|z|.$$

But I don't see how this will lead to either of the desired inequalities. Any help would be greatly appreciated.


set $ h = (f(0) - f(z))/ (1- \bar f(0) f(z))$ and solve explicity for f, though I would prefer that the signs in the denominator were reversed, but maybe that's the hard part

  • $\begingroup$ Thanks @mike. Maybe I'm missing something... when I solve for f I get f(z)=(f(0)-h(z))/(1-(f(0)bar)h(z)). How does that help? $\endgroup$ – John Adamski Apr 10 '12 at 19:28
  • $\begingroup$ I think if you let $-h = \phi (f(z))$ you will have slightly nicer signs. The inequality $|h| \leq |z|$ still holds, of course. $\endgroup$ – copper.hat Apr 10 '12 at 19:31
  • $\begingroup$ Sorry @mike for unaccepting your answer. I didn't realize you could only accept one. I've accepted it again. $\endgroup$ – John Adamski Apr 24 '12 at 18:00

Here is a proof that uses only Schwarz' lemma and the triangle inequality.

Set $a=f(0)$ and $$\phi_a(z)=\frac{z-a}{1-\overline{a}z}.$$ Then $\phi_a\circ f$ maps the unit disc to the unit disc and fixes zero. Thus, by Schwarz' lemma, $$|\phi_a(f(z))|=\left|\frac{f(z)-a}{1-\overline{a}f(z)}\right|\leq|z|,$$ and so \begin{equation}|f(z)-a|\leq|z||1-\overline{a}f(z)|\leq|z|+|z||a||f(z)|.\tag{1}\end{equation}

Applying the triangle inequality, we arrive at $$|f(z)|\leq|z|+|z||a||f(z)|+|a|.$$ Then $$|f(z)|-|z||a||f(z)|\leq|z|+|a|$$ $$|f(z)|\leq\frac{|z|+|a|}{1-|a||z|}=\frac{|f(0)|+|z|}{1-|f(a)||z|},$$ Where the final equality uses the fact that $f(0)=a$. This is the second desire inequality.

To obtain the first inequality, we begin with $$|a|=|a-f(z)+f(z)|\leq |a-f(z)|+|f(z)|\leq |z|+|z||a||f(z)|+|f(z)|,$$ where the last inequality follows from (1). Then $$|a|-|z|\leq |z||a||f(z)|+|f(z)|$$ $$\frac{|a|-|z|}{|z||a|+1}=\frac{|f(0)|-|z|}{1+|f(0)||z|}\leq |f(z)|.$$


Expanding on the answer of mike. Set

$$ \frac{f(z) - f(0)}{1-\overline{f(0)}f(z)} = w. $$

Then $|w| \leq |z|$. Solve for $f(z)$:

$$ f(z) = \frac{w + f(0)}{1 + \overline{f(0)}w}. $$

The right hand side is an automorphism of the unit disc as a function in $w$. It maps circles $|w| = r$ to (non concentric) circles around $f(0)$. The following inequalities hold for such a Möbius transformation:

$$ \frac{|f(0)| - |w|}{1 - |f(0)| |w|} \leq \left| \frac{w + f(0)}{1 + \overline{f(0)}w} \right| \leq \frac{|f(0)|+|w|}{1 + |f(0)| |w|}. $$

The left hand side decreases in $|w|$ and the right hand side increases in $|w|$. Since $|w| \leq |z|$, the inequalities for $|f(z)|$ follow.


First we define $$ \rho(z,w)=\left|\dfrac{z-w}{1-\overline w z}\right| $$ Then is easy prove that $$ 1-\rho(z,w)^2=\dfrac{(1-|z|^2)(1-|w|^2)}{|1 - \overline w z|^2} $$ Now you can check that $$ z \overline w+ \overline z w\le 2|z||w| $$ and this inequality help to obtained $$ (1-|w||z|)^2\le |1-\overline w z|^2 $$ So: \begin{align*} \rho(z,w)^2=\left|\dfrac{z-w}{1-\overline w z}\right|^2 =&1- \dfrac{(1-|z|^2)(1-|w|^2)}{|1 - \overline w z|^2}\\ \ge& 1- \dfrac{(1-|z|^2)(1-|w|^2)}{(1 - |\overline w| |z|)^2}\\ =&\dfrac{(1 - |\overline w| |z|)^2- (1-|z|^2)(1-|w|^2)}{(1 - |\overline w||z|)^2}\\ =&\dfrac{ |z|^2-2|z||w|+|w|^2}{(1 - |\overline w||z|)^2}\\ =&\dfrac{ (|z|-|w|)^2}{(1 - |\overline w||z|)^2}\\ \end{align*} Now chaning $z$ by $f(z)$, and $w$ by $f(0)$ we have $$ \left|\dfrac{ (|f(z)|-|f(0)|)}{(1 - |f(0)||f(z)|)}\right|\le\left|\dfrac{f(z)-f(0)}{1-\overline{f(0)}f(z)}\right|\le|z| $$ the last inequality by Schwarz-Pick. Then $$ \left|\dfrac{ (|f(z)|-|f(0)|)}{(1 - |f(0)||f(z)|)}\right|\le|z| $$ we obtained $$ \frac{|f(0)|-|z|}{1-|f(0)||z|}\leq|f(z)|\leq\frac{|f(0)|+|z|}{1+|f(0)||z|} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.