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This is taken from an old complex analysis qualifying exam.

Problem

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 $|z|<1$.

Attempt

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.

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4 Answers 4

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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 $$|a|-|z|\leq |z||a||f(z)|+|f(z)|$$ $$\frac{|a|-|z|}{|z||a|+1}=\frac{|f(0)|-|z|}{1+|f(0)||z|}\leq |f(z)|.$$

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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.

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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 $$ \frac{|f(0)|-|z|}{1-|f(0)||z|}\leq|f(z)|\leq\frac{|f(0)|+|z|}{1+|f(0)||z|} $$

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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

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  • $\begingroup$ 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? $\endgroup$ Commented Apr 10, 2012 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
    Commented Apr 10, 2012 at 19:31
  • $\begingroup$ Sorry @mike for unaccepting your answer. I didn't realize you could only accept one. I've accepted it again. $\endgroup$ Commented Apr 24, 2012 at 18:00

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