0
$\begingroup$

Suppose $f$ is continuous and: $$ |f(e^{i \theta})| \leq M $$

Suppose: $$ |\int_{|z|=1}f(z)dz| = 2\pi M $$ I am trying to show that $f(z) = c \bar{z}$ for some constant $c$ with $|c|=M$

enter image description here

$\endgroup$
9
  • $\begingroup$ Liouville's theorem? $\endgroup$ Dec 10 '20 at 21:48
  • 1
    $\begingroup$ Hi Paul, that theorem is valid for entire bounded functions. I think the asker meant that f is bounded on the unit circle. $\endgroup$ Dec 10 '20 at 21:51
  • $\begingroup$ @RedPhoenix That's strange because I have double checked question from the textbook. Which part of the question do you think is wrong? $\endgroup$ Dec 10 '20 at 21:51
  • $\begingroup$ @RedPhoenix: No, the closed integral over a constant function vanishes. $\endgroup$ Dec 10 '20 at 21:53
  • $\begingroup$ I attached the picture of the question in the textbook (Gamelin's) $\endgroup$ Dec 10 '20 at 21:55
0
$\begingroup$

Let $\int_{|z|=1}f(z)dz=\alpha|\int_{|z|=1}f(z)dz|, |\alpha|=1$; also note that $dz=iz|dz|$ on the unit circle

$2\pi M=\int_{|z|=1}\alpha^{-1} f(z)dz= \int_{|z|=1}\alpha^{-1} if(z)z|dz| \le \int_{|z|=1}|\alpha^{-1} if(z)z||dz| = \int_{|z|=1}|f(z)||dz| \le 2\pi M$ so we have equality (ae and then everywhere by continuity).

But $|dz|$ is a positive measure so this first means that $|f(z)|=M$ on the unit circle and then by taking real and imaginary parts that $\Re \alpha^{-1} if(z)z=|f(z)|=M, \Im \alpha^{-1} if(z)z=0$ hence $f(z)=-i\alpha M/z=-i\alpha M \bar z$ so we are done!

$\endgroup$
2
  • $\begingroup$ @Could you explain more why $dz=iz|dz|$ on the unit circle? because I found $dz=izd \theta$ $\endgroup$ Dec 11 '20 at 0:24
  • $\begingroup$ Because on the unit circle $|dz|=d\theta$ the arclength; in general of course $|dz|=Rd\theta$ when the circle has radius $R$ $\endgroup$
    – Conrad
    Dec 11 '20 at 0:30

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