Uniqueness of analytic continuation on a domain of C^n.

Question

Hi. I have been struggling with this question for a while now.

Given a domain $\Omega \subset \mathbb{C}^n$, $\Omega^\prime = \Omega \cap \mathbb{R}^n$, and an analytic function $f: \Omega \longrightarrow \Omega$ such that $f_{|\Omega^\prime}$ is a (real) constant, under which assumptions on $\Omega$ and/or $\Omega^\prime$ can one conclude that $f$ is constant on the whole domain $\Omega$?

The case $n = 1$ is quite classical. For $n>1$, assuming $\Omega$ is a pseudoconvex domain does not really help, and I'm not sure of what can be said about real domains of the zeros of an analytical function of several complex variables.

Thanks for any clue about this.

Answer 1

Without loss of generality, assume that your $f_{|\Omega^\prime}\equiv 0$. The knowledge of this restriction is sufficient to conclude that $f$ and all of its derivatives vanish on $\Omega'$. The subset of $\Omega$ on which $f$ and all of its derivatives vanish is both closed and open (clopen). Since $\Omega$ is a domain (it's connected), the only clopen subsets are $\Omega$ and $\emptyset$. If $\Omega'$ is non-empty, it then follows that $f$ must vanish on all of $\Omega$.

In other words, the assumptions you specified are already sufficient.

A couple more details. The clopen property holds because (i) continuity ensures that the set of all points on which some derivative of $f$ vanishes is closed and (ii) the Taylor series with all-zero coefficients has a non-zero radius of convergence.