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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):

MR1173388 (93h:31003) Armitage, D. H.(4-QUEEN) Cones on which entire harmonic functions can vanish. Proc. Roy. Irish Acad. Sect. A 92 (1992), no. 1, 107–110. 31B05

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

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    $\begingroup$ 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. $\endgroup$
    – Margaret Friedland
    Commented Oct 11, 2011 at 19:23

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