My question stems from the following result about holomorphic functions on the unit disc:
"A function, continuous on the closed unit disc, holomorphic inside, and vanishing on an open subset of the boundary, vanishes identically. A simple proof of this well-known proposition is obtained by considering its Cauchy integral representation." (Generalized Analytic Functions by Richard Arens and I.M.Singer)
There are other more direct proofs too, including the one by P.R. Chernoff, using a product of finitely many rotations of $f: \Bbb D \to \Bbb C$ and the identity theorem. See details here, for example.
In P.R. Chernoff's article, I found the statement of a stronger result – the set of zeros of $f: \Bbb D \to \Bbb C$ on $∂𝔻$ cannot have positive Lebesgue measure unless $f$ vanishes identically. The proof is more involved and can be found in Chapter 4 of Banach Spaces Of Analytic Functions by Kenneth Hoffman. See the Corollary on Pg. 52 of a Theorem on Pg. 51.
Question: It is natural to ask what happens if $\Bbb D$ is replaced by an arbitrary open (connected?) subset $\Omega \subset \Bbb C$. Does the result continue to hold, i.e., is it true that the set of zeros of a holomorphic function $f:\Omega\to \Bbb C$ on $\partial \Omega$ cannot have positive Lebesgue measure unless $f$ vanishes identically on $\Omega$? Does a weaker result hold? One may require $\Omega$ to be bounded.
Thanks a lot!
Best Answer
It is not clear in your question what "Lebesgue measure on $\partial\Omega$" really means.
Let us begin with the unit disk. In the unit disk, every bounded holomorphic function which is zero on a set $E$ of positive measure, in the sense that $$\limsup_{r\to 1}|f(re^{i\theta})|=0,\quad e^{i\theta}\in E,$$ vanishes. This is the direct consequence of Jensen's inequality $$\log|f(0)|\leq\frac{1}{2\pi}\int_{-\pi}^\pi\log|f(re^{i\theta})|\,d\theta, \quad r\in(0,1).$$ (If $f\neq 0$, you can find an automorphism $\phi$ of the unit disk such that $g=f\circ\phi(0)\neq 0$. Applying Jensen's inequality to this $g$ and letting $r\to 1$ gives a contradiction.)
Now for other regions. If $\Omega$ is simply connected, you can use a conformal map of the unit disk onto $\Omega$. But then the question is what is the correct "measure" on $\partial \Omega$. It is called the harmonic measure. For a simply connected $\Omega$ it can be defined as the image of the Lebesgue measure on the unit circle under the conformal map. But it can be defined for arbitrary region possessing a Green function (in particular for any bounded region, or more generally for any region for which $\partial D$ contains a continuum). Let $G$ be the Green function, with the pole at some point $z_0\in\Omega$. Extend it to the whole plane by setting $G(z)=0,z\not\in\Omega$. The resulting function is subharmonic in ${\mathbf{C}}\backslash\{z_0\}$, and harmonic measure is the Riesz measure, $(2\pi)^{-1}\Delta G$ of this subharmonic function. Here $\Delta$ is in the sense of Schwarz distributions.
Now your question has the following general answer for arbitrary region possessing a Green function: if $f$ is bounded, and tends to zero as $z\to E$, where $E$ is a set of positive harmonic measure then $f=0$.
The question remains how to characterize the sets of positive or zero harmonic measure in terms of some usual metric properties, for example Hausdorff measures, in particular the $1$-dimensional measure when $\partial G$ is rectifiable etc. The answer depends on the class of domains $\Omega$ considered, and this is a subject of a very large body of literature, of which I can recommend the modern book:
MR2150803 Garnett, John B.; Marshall, Donald E. Harmonic measure, Cambridge, 2005.
If $\Omega$ is simply connected, and $\partial\Omega$ is a rectifiable Jordan curve, then, according to a theorem of F. and M. Riesz, the sets of zero harmonic measure are the same as those of zero 1-Hausdorff measure.
F. and M. Riesz (1916), "Über die Randwerte einer analytischen Funktion", Quatrième Congrès des Mathématiciens Scandinaves, Stockholm, pp. 27–44.