Proving nonexistence of local extrema of harmonic functions using Hessian

Question

Edited I want to prove that solution to Laplace's equation do not support local maximum or minimum using Hessian.

Suppose $f(x,y,z)$ is a real-valued function of three real variables that satisfies the Laplace equation, $\nabla^2f=0$. Consider the Hessian matrix $$ H(x,y,z)= \begin{bmatrix} f_{xx} & f_{xy} & f_{xz}\\ f_{yx} & f_{yy} & f_{yz}\\ f_{zx} & f_{zy} & f_{zz}\\ \end{bmatrix}, $$ which is real symmetric. Therefore, its eigenvalues must all be real. Due to Laplace equation, $${\rm Tr}(H)=\nabla^2f=0.$$ So either (1) all eigenvalues are zero or (2) at least some of them must have mixed signs.

In case (2) i.e., when the eigenvalues have mixed signs, it follows that solutions to Laplace equation do not tolerate local maxima or minima.

In the case (1) i.e., when the eigenvalues are all zero, Hessian cannot tell us anything. Here, the test becomes inconclusive. How can I proceed, in this case, to continue with the proof of the fact that Laplace's equation do not support local maximum or local minimum?

Answer 1

I do not think that your statement about the eigenvalues not being zero holds (consider e.g. constant or linear functions which have vanishing Hessian), but if you only want to exclude local extrema, then you can just use the maximum principle for harmonic functions. One possible formulation is the following:

Let $\Omega\subset\mathbb{R}^n$ be open and connected. Then any harmonic function $u\in C^2(\overline{\Omega},\mathbb{R})$ is either constant or attains neither a local maximum nor a local minimum inside of $\Omega$.