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?



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


