1
$\begingroup$

in a paper I'm reading I stumbled upon the following claim:

The holomorphic functions $v \colon \overline{D(s)} \to C$ which satisfy $v(z) \in z^{1/2}\Bbb R$ for all $z \in \partial \overline{D(s)}$ are $v(z) = s^{−1/2}cz + s^{1/2}\overline{c }$, for $c \in \Bbb C$.

$\overline{D(s)}$ is the closed disk of radius $s>0$.

I thought it was an immediate application of maximum principle and Schwarz Lemma but I'm stuck.

Can someone provide some insight on what I'm missing? I'm especially interested in understanding how to use the condition on the boundary. I'm tempted to "divide" by $z^{1/2}$ and then try to get some mileage thanks to the function being always real but $z^{1/2}$ is not holomorphic everywhere and the function will have a singularity at zero.

$\endgroup$
1
$\begingroup$

To prove this is very easy, to gain insight maybe a little harder.

Assume first $s=1$ and note that $f_c(z)=cz+\bar c$ obviously satisfies the required condition. Now assume $v$ satisfies the condition and $v(0)=0$ and let $h(z)=\frac{v^2(z)}{z}$, $h(0)=0$ and $h$ is holomoprhic in the unit disc with imaginary part zero on the boundary, so it immediately follows (max/min principle for harmonic functions etc) that $h$ is a real constant, hence zero by $h(0)=0$, so $v(z)=0$ too.

But now if $v$ is arbitrary satisfying the condition, and $\bar v(0)=c$, it obviously follows that $v-f_c$ satisfies the condition and $(v-f_c)(0)=0$ hence $v(z)=f_c(z)$

For a general disc of radius $s>0$, by changing variables we get $v(z)=\frac{bz}{s}+\bar b$ and then taking $c=\frac{b}{\sqrt s}$ gives us the symmetric form required.

To gain insight, I would note that the condition immediately implies (by looking at the argument, etc) that (unless $v$ is constant hence zero) $v$ is univalent (injective) and has precisely one zero on the boundary (so it is univalent inside too incidentally as by continuity the image of the boundary is a Jordan curve) and as before it is uniquely determined by $v(0)$ while the condition is real linear (so it scales by real constants), hence it makes sense to look for a family of linear functions that satisfy it and then use the unicity of $v(0)$ to conclude that's all

$\endgroup$
  • $\begingroup$ That's a beautiful answer Conrad, thanks. $\endgroup$ – Riccardo Oct 3 at 15:48
  • $\begingroup$ you are welcome! $\endgroup$ – Conrad Oct 3 at 15:54

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.