Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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