0
$\begingroup$

My question was inspired by this question.

Let $\Omega\subset \mathbb C^n$ be an open bounded set, and $f\colon \overline \Omega \to\overline \Omega$ a continous map such that

$$\forall x\in \partial \Omega,\quad f(x)=x.$$

Is $f$ surjective?

We know that if $\Omega\subset \mathbb R^n$, this is the case (you can find a proof here or here).

But I don't think the result holds for $\mathbb C$ instead of $\mathbb R$, though I am not able to find a counterexample.

I also think that if $f$ is differentiable, then $f$ is holomorphic so for all $z\in \Omega$, $f(z)$ is uniquely determined by it's value on $\partial \Omega$ so $f=\mathrm{id}_\Omega$ and the result holds. So a counterexample would require $f$ continuous but not differentiable.

$\endgroup$
4
  • 2
    $\begingroup$ $\mathbb{C}^n =\mathbb{R}^{2n}$ $\endgroup$
    – Tim kinsella
    Jan 9, 2017 at 9:32
  • $\begingroup$ @Timkinsella Yes, but is it sufficient ? I thought the condition $f(x)=x$ for $x\in \partial \Omega$ was weaker in that case, but may be I was wrong. $\endgroup$
    – E. Joseph
    Jan 9, 2017 at 9:38
  • 1
    $\begingroup$ @E.Joseph Why would you think this and what would be the meaning? $\endgroup$
    – Moishe Kohan
    Jan 9, 2017 at 13:09
  • $\begingroup$ @MoisheCohen No, I was wrong, open sets of $\mathbb C$ are exactly the same as open sets of $\mathbb R^2$ so the question is trivial. $\endgroup$
    – E. Joseph
    Jan 9, 2017 at 13:27

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .