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This is a homework problem, so hints or rough outlines are strongly preferred to a full solution.

Problem. Let $C$ be the unit circle. Suppose the continuous function $f : C \rightarrow \mathbb{C}$ on the unit circle satisfies $|f(z)| \leq M$ and $\left|\int_{C} f(z)\;dz\right| = 2\pi M$. Show that $f(z) = c\bar{z}$ for some constant $c$ with modulus $|c| = M$.

I've been able to argue $|f(z)| = M$ for all $z \in C$ by assuming this is not the case and finding a contradiction using continuity, the given identity, and the $ML$-inequality. This is clearly necessary, but may not be useful to solve the problem.

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up vote 3 down vote accepted

Your observation that $|f(z)|=M$ is useful. From the equality we get, if $z=e^{it}$, $$ f(z)=M\,e^{ig(t)},\quad g\in C([0,2\,\pi],\mathbb{R}),\quad g(0)=g(2\,\pi)\pmod {2\,\pi} $$ and $$ \int_0^{2\pi}e^{ig(t)}e^{it}\,dt=2\,\pi\,e^{i\alpha} $$ for some $\alpha\in[0,2\,\pi)$. From this $$ \Re(e^{-i\alpha}\int_0^{2\pi}e^{ig(t)}e^{it}\,dt)=\int_0^{2\pi}\Re(e^{i(g(t)+t-\alpha)})\,dt=2\,\pi. $$ Deduce that $$ \Re(e^{i(g(t)+t-\alpha)})=1\quad \forall t $$ and that $g(t)+t-\alpha=0$ for all $t$. I hope I have not given too much detail.

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This was very helpful and left me with some to verify for myself, but I believe I have a complete solution now. Thank you. – Hans Parshall Jan 31 '12 at 23:47

My idea is to write $$\left| \int_C f(z) dz \right| =\left| \int_0^{2 \pi} f(e^{it}) i e^{it} dt \right|=\left| \int_0^{2 \pi} f(e^{it}) e^{it} dt \right| \le \int_0^{2 \pi} \left| f(e^{it}) \right| dt \le 2 \pi M$$ then reason about when this can be an equality.

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I followed Julián Aguirre's answer as it built on what I had already shown for myself. I do see how the second inequality being an equality forces $|f(e^{it})| = M$ for all $t \in [0,2\pi]$. I do not quite understand what the first inequality being an equality forces. Would you be willing to elaborate on this? – Hans Parshall Jan 31 '12 at 23:48
It forces that the integrand, as a vector over $\mathbb{R}$, must always point to the same direction i.e. have constant argument. – Tib Feb 24 '12 at 18:16

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