Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be an analytic function on $|z|\leq 1$ with $f(0)=0$ and let $|f(z)|$ have a maximum for $|z|\leq 1$ at $z_0=1$. Show that $f^{'}(z_0)\neq 0$ unless $f$ is constant.

Here is what I have: If $|f(1)|=0$ then $f$ is constant and we are done. So assume $|f(1)|>0$. Then $f(1)\neq 0$ and the function $g(z)=\frac{f(z)}{f(1)}$ is analytic. Also $g$ satisfies the conditions of Schwarz Lemma.

Here is where I am stuck. I want to get $g(z)=cz$ for some $c$ with $|c|=1$.

share|improve this question
    
Theorem 4.1 in this link may be useful arxiv.org/pdf/1001.1805v1.pdf –  PAD Jul 29 '12 at 21:55

1 Answer 1

By "$f$ is an analytic function on $|z|\le 1$" I assume you mean $f$ is analytic on some neighbourhood of $|z|\le 1$. If $f$ is not constant, there is some smallest positive integer $k$ such that $f^{(k)}(1) \ne 0$. Suppose $f'(1) = 0$, so $k \ne 1$. Thus $f(z) = f(1) + a (z-1)^k + O(|z-1|^{k+1})$ as $z \to 1$, with $a \ne 0$, $f(1) \ne 0$ and $k \ge 2$. So $|f(z)|^2 = |f(1)|^2 + 2 \text{Re}(\overline{f(1)} a (z-1)^k) + O(|z-1|^{k+1})$. As $\omega$ goes around the unit circle, $\overline{f(1)} a \omega^k$ makes $k$ revolutions around the origin. There is some $\omega$ in the open left half plane such that $\text{Re}(\overline{f(1)} a \omega^k) > 0$. Then for $z = 1 + s \omega$, if $s > 0$ is sufficiently small we have $|f(z)| > |f(1)|$ and $|z| < 1$, contradicting the maximality of $f$ at $z=1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.