A Bounded Continuous Function on a Circle [duplicate]

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 $$|c|=M$$

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

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