0
$\begingroup$

I am having difficulties with the following problem:

$\bf Given$: $ f$ is an analytic map from unit disk $D$ to itself and: $f(0)=0$. $\bf To \; prove:$ $|f(z)|\leq |z|$ for $z\in D$; and: $|f'(0)|\leq 1$

What I thought is: $$|f(z)|\leq 1,$$ because $f$ maps to unit disk.

This is apparently wrong. But how then should I approach this problem?

$\endgroup$
  • $\begingroup$ The expression $f(z) \leq |z|$ is not meaningful: $\leq$ is not defined for complex numbers. Do you mean $|f(z)|$? $\endgroup$ – Ulrik Jan 19 '14 at 18:44
  • $\begingroup$ @Svinepels Thank you, yes indeed. I changed it. $\endgroup$ – user104662 Jan 19 '14 at 18:46
  • $\begingroup$ Why do you need to prove $f(0)=0$? You are given that. $\endgroup$ – Robert Israel Jan 19 '14 at 18:51
  • 2
    $\begingroup$ This is en.wikipedia.org/wiki/Schwarz_lemma $\endgroup$ – user110822 Jan 19 '14 at 18:54
  • 1
    $\begingroup$ @user104662 $\lim\limits_{h \to 0} \dfrac{f(h)}{h}=\lim\limits_{h \to 0} \dfrac{f(h)-f(0)}{h-0}=f'(0)$ $\endgroup$ – user110822 Jan 19 '14 at 19:01
3
$\begingroup$

Hint: consider $g(z) = f(z)/z$.

$\endgroup$
  • $\begingroup$ Thanks, that hint is also given in the question :). But I don't know why exactly.. Can you tell me why you propose this? $\endgroup$ – user104662 Jan 19 '14 at 18:51

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.