# Minimum set of subharmonic function in $\mathbb R^n$

Let $f :\mathbb R^n\to \mathbb [0, \infty)$ be a (continuous, $C^2$, or smooth) subharmonic function with minimum value $0$. Then we know the sublevel set $f^{-1}((-\infty, c])$ is mean convex for $c > 0$. The interior minimum set $f^{-1}(0)$ has to be minimum if it's a $C^1$ submanifold. My question is it necessary a manifold?

How about for general open Riemaniann manifold?

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According to the following, no (you can make the function nonnegative by taking maximum of $f$ and the constant 0):
Suppose that $L$ and $M$ are two lines in the plane. There is a nontrivial harmonic function which vanishes on both $L$ and $M$ if and only if the angle between the two lines is a rational multiple of $\pi$. H. S. Shapiro asked which cones in ${\bf R}^3$ have the property that there is a nontrivial harmonic function in ${\bf R}^3$ which vanishes on the cone. The author shows that a cone has this property if and only if the opening of the cone is a zero of a derivative of a Legendre polynomial. The result stated is for cones in ${\bf R}^N$ and then ultraspherical polynomials arise. The proof is elegant and well presented. It uses results of Kuran on homogeneous harmonic polynomials. Reviewed by Tom Carroll
If you are familiar with holomorphic dynamics, the Green function with pole at $\infty$ for the Julia set $J_f$ of a polynomial $f$ is always a nonnegative subharmonic function. Take $f(z)=z^2+3$. Then $J_f$ is a Cantor-like set, which is exactly the zero set of the Green function. So you get another example. – Margaret Friedland Oct 11 '11 at 19:23