The problem is: Suppose that $f(z)$ is continuous on a domain $D$ that contains the unit circle, and that $f(z)$ satisfies: $$\vert f(e^{i\theta})\vert \leq M \; \forall \theta \, \in [0, 2\pi )$$ and $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M $$ show that there exists a $c\in \mathbb{C}$ such that $f(z)=c \bar{z}$ for all $\vert z\vert=1$.

I've tried lots of different approaches such as defining a function $g(z)=\frac{f(z)}{\bar{z}}$ and trying to show that $g'(z)=0$ so that $g$ is constant on the circle, or trying to use some bounding theorems for analytic functions to show that $g$ is constant but this requires $g$ to be analytic. Some other methods were using the Cauchy-Riemann equations but I couldn't find any reason why $g$ should be analytic. Any help would be greatly appreciated, thanks!


The two conditions imply that $|f(z)|=M$ for all $|z|=1$, because if the inequality is strict on some $\theta$, then it will be so in an interval, and that prevents the equality in the integral (note that $2\pi M$ is an upper bound for the integral).

So $f(e^{i\theta})=Me^{ig(\theta)}$ for some continuous $g$. Then $$ \int_{|z|=1}f(z)dz=M\int_{0}^{2\pi}e^{ig(\theta)}\,ie^{i\theta}\,d\theta=iM\int_0^{2\pi}e^{i(\theta+g(\theta))}\,d\theta. $$ So $$ \left|\int_0^{2\pi}e^{i(\theta+g(\theta))}\,d\theta\right|=2\pi, $$ and then $$ 2\pi=\int_0^{2\pi}e^{i(\theta+g(\theta)+d)}\,d\theta $$ for an appropriate $d$. This last equality shows that the integral of the imaginary part is zero, and the integral of the real part is $2\pi$. If at any point the real part were less than $1$, we would not achieve the $2\pi$: we deduce that the real part of $e^{i(\theta+g(\theta)+d)}$ is $1$, which implies $$ e^{i(\theta+g(\theta)+d)}=1 $$ for all $\theta$. This implies that $g(\theta)=2k(\theta)\pi-\theta-d$, and so $$ f(e^{i\theta})=Me^{-i\theta-id}=Me^{-id}\,e^{-i\theta}, $$ i.e. $$ f(z)=c\overline z $$ where $c=Me^{-id}$.


Unfortunately, my first attempt was incorrect because I messed up the equality case in the triangle inequality. I've corrected my argument, and left my first attempt below.

Current Solution

By the Triangle Inequality, $$2\pi M=\left|\int_{0}^{2\pi}f(e^{i\theta})e^{i\theta}\, d\theta\right|\leq\int_{0}^{2\pi}|f(e^{i\theta})|\, d\theta\leq 2\pi M\qquad (1)$$ so in fact equality holds. Examining a proof of the triangle inequality (for instance Theorem 1.33 in Rudin's Real and Complex Analysis), we see that for equality to hold, there exists some constant $\alpha$ with $|\alpha|=1$ for which $$\arg[\alpha f(e^{i\theta})e^{i\theta}]\equiv 0\qquad(2)$$ [this is a consequence of the first $\leq$ in $(1)$]

As a consequence of the second $\leq$ in $(1)$, we have that $$|f(e^{i\theta})|=M\qquad(3)$$ Combining $(2)$ and $(3)$, we see that $$\alpha f(e^{i\theta})e^{i\theta}=M$$ whereupon $$f(z)=f(e^{i\theta})=\frac{M}{\alpha}e^{-i\theta}=c\bar{z}$$ if we choose $c=\frac{M}{\alpha}$ $\square$

First Attempt

By the Triangle Inequality, $$2\pi M=\left|\int_{|z|=1}f(z)\, dz\right|\leq\int_{|z|=1}|f(z)|\, dz\leq 2\pi M$$ so in fact equality holds in the triangle inequality. Thus, $|f(e^{i\theta})|=M$ and $f(e^{i\theta})$ has constant argument. So $f(z)=Me^{i\theta_0}$ for some constant angle $\theta_0$.

In particular, the statement you are trying to prove appears incorrect.

  • 2
    $\begingroup$ But if $f(z)$ is constant then $\int_{|z|=1} f(z)\,dz = 0$ by Cauchy's theorem, violating the assumption. $\endgroup$ – Antonio Vargas Aug 29 '13 at 2:13
  • 1
    $\begingroup$ I don't see how you deduce that the argument is constant. $\endgroup$ – Martin Argerami Aug 29 '13 at 2:30
  • $\begingroup$ Thanks for your help so far guys! Could this be correct after a few of the steps in pre-kideny's post have been used: $f(e^{i \theta})=Me^{i \phi} \;\: , \phi \in [0,2\pi)$. So $\int_{0}^{2\pi} Me^{i \phi} i e^{i \theta} d\theta = 0 $ if $\theta \neq \phi$ or $2\pi i M$ if $\theta =-\phi$. Since $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M\neq 0 $$, then $\theta =-\phi$ and hence $f(e^{i \theta})=M e^{-i \theta}$. So $f(z)=M \bar z$ because $f$ is continuous on D and hence at $z=0$ so it can't be $f(z)=1/z$. $\endgroup$ – user92110 Aug 29 '13 at 2:47
  • $\begingroup$ @user92110 There's no difference between $f(z)=1/z$ and $f(z) = \bar z$ for $|z|=1$. There are many ways to extend $f$ continuously to the rest of the disk. $\endgroup$ – Erick Wong Aug 29 '13 at 4:43
  • $\begingroup$ @AntonioVargas and MartinArgerami thanks for pointing out the issues with my first attempt. I went back and thought some more about how I was applying the triangle inequality, and have produced a (hopefully) correct proof this time around. $\endgroup$ – pre-kidney Aug 30 '13 at 8:30

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.