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Suppose $u$ is subharmonic, real valued and continuous in the open disk $D\subset\mathbb{C}$ and is non constant on any open set in $D$. By adding a constant to $u$ if necessary, I want to know whether the set $$ \{z\in D: u(z)=0\} $$ has measure zero?

Mathmo
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1 Answers1

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No. Let $K\subset\mathbb{C}$ be a compact set with positive measure, empty interior, and connected complement (e.g., a simple curve of positive area or a "thick" Cantor-type set). Let $\mu$ be the equilibrium measure of $K$, i.e., the measure that maximizes the energy $$ I(\mu) = \iint \log|z-w|\,d\mu(z)\,d\mu(w) $$ among all probability measures supported on $K$. Such a measure exists: see, e.g., section 3.3 of Potential Theory in the Complex Plane by Ransford or these lecture notes by Saff. Moreover (from the same sources), the potential of $\mu$, defined by $$ P_\mu(z) = \int \log|z-w|\,d\mu(w) $$ is

  1. subharmonic on $\mathbb{C}$
  2. harmonic on $\mathbb{C}\setminus K$
  3. equal to $I(\mu)$ a.e. on $K$

Property 2 implies $P_\mu$ is not constant on any open set, since such a set would overlap $\mathbb{C}\setminus K$, and a harmonic function can't be constant on an open subset of a connected open set unless it's identically constant (recall that harmonic functions are real-analytic).

Community
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  • Sorry, was unclear - I actually want $u$ not to be open on any open set. Nice example though. – Mathmo Mar 31 '17 at 14:02
  • Answer is still no, however. –  Mar 31 '17 at 15:01
  • Nice! My original motivation was the study of the resolvent norm of a bounded operator on a separable Hilbert space (for bounded operators the resolvent norm satisfies the above properties). Do you know if this can be converted? I.e. is there an operator such that $P_\mu(z)$ is locally $\left|(z-T)^{-1}\right|$ around K? – Mathmo Apr 01 '17 at 10:29