1
$\begingroup$

I encountered the following old qualifier problem:

Suppose $f: \mathbb{D} \to \mathbb{D}$ is holomorphic, $f(0)=0$, and $f'(0)=1$. Show that $$|f(z)-z| \leq 2|z|^2 \, \forall z \in \mathbb{D}. \, \, \, \, \, (1)$$

I believe I can solve this by applying the Schwarz lemma to the function $$g(z)=\frac{f(z)}{2z}-\frac{1}{2}.$$ What seems a little strange is that from the beginning, the Schwarz lemma tells us exactly what $f(z)$ looks like: namely, it is a rotation. Is it possible to prove (1) straight from the fact that $f(z)=az$ for some $|a|=1$?

$\endgroup$
1
$\begingroup$

In fact, it is trivial, since the assumption $f'(0)=1$ implies $a=1$, so $f(z) \equiv z$.

$\endgroup$
  • $\begingroup$ Ha! That makes me feel silly. But from the way the problem is stated, it seems like whoever made it up also overlooked this. Strange... $\endgroup$ – TorsionSquid Jan 7 '15 at 23:38

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.