# Bound for Analytic Function on Unit Disk

The following is an old qualifying exam problem that I can't seem to piece together:

Suppose we have an analytic function $f$ on the unit disk $\mathbb{D}$ s.t. $|f| \leq 1$. Show $$\frac{|f(0)|-|z|}{1-|f(0)||z|} \leq |f(z)| \leq \frac{|f(0)|+|z|}{1+|f(0)||z|}$$.

I've tried two things. First, decomposing $f$ into its real and imaginary parts and applying the Harnack inequality to each (after adding 1 so that it is nonnegative) and then piecing them together so they say something about $f$. I couldn't get that to look close to the inequality. Second, I define $h(z) = \frac{f(z)-f(0)}{1+|f(0)|}$ and apply the Schwartz lemma. This comes close but I couldn't get it to work.

• You are applying the Schwarz lemma on the wrong function. – Braindead Jan 7 '14 at 9:05

Use a conformal mapping of the unit disk to normalize your function $f$, instead of the linear function. That is, a mapping of the form
$h(z) = \dfrac{z-a}{1-\bar{a}{z}}$, where $|a| < 1$.
Note that for $|z| \le 1$, we have $|h(z)| \ge \left|\dfrac{ |z| - |a| }{1-|a||z|}\right|$.
Let $f(0) = a$ and consider the function $\phi_a(z) = \dfrac{a-z}{1-\bar{a}z}$. Apply Schwarz lemma on $(\phi_a \circ f)(z)$ to use the result that $|(\phi_a \circ f)(z)| \leq |z|$. From this, using triangle and reverse triangle inequality you can arrive at $\dfrac{|f(0)|-|z|}{1-|f(0)||z|} \leq |f(z)|$.
Similarly apply Schwarz lemma on $(\tilde{\phi}_a \circ f)(z)$ where $\tilde{\phi}_a(z) = \dfrac{z-a}{1-\bar{a}z}$.
This will lead you to $|f(z)| \leq \dfrac{|f(0)|+|z|}{1+|f(0)||z|}$