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.
Best Answer
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 $|h(z)| \ge \left|\dfrac{ |z| - |a| }{1-|a||z|}\right|$.