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Let $D$ denote the open unit disk around the origin in the complex plane. Let $f:D\rightarrow D$ holomorphic and $f$, $f^\prime$ extend continuously to $\overline{D}$. Let $u$ be the real part of $f$.

If $f$ attains a maximum at $z$ (which must be on the unit circle by the maximum modulus principle), is it true that $u=\mathrm{Re}(f)$ also has a maximum at $z$?

I think yes, but I don't know how to prove it. I started saying that the tangential derivative of $u$ at $z$ must be $0$, but that's not enough..

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up vote 0 down vote accepted

It's false; you should be able to see this by checking some elementary functions.

Even better is to think about why it's false. Write $f = u + i v$, where $u$ and $v$ are real (harmonic) functions, and suppose $f$ is holomorphic in some neighbourhood of $\overline D$. Suppose your proposition is true; then it is also true for $-i f = v - i u$. So both $u$ and $v$ attain their maximum modulus at the same point $z$ on $\overline D$. This means that their directional derivatives along the boundary vanish at this point. But $f$ is holomorphic in some neighbourhood of $\overline D$, so the Cauchy-Riemann equations hold at $z$. From this you can conclude that $f'(z) = 0$, and clearly this is not generally true.

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