Possible to have only zero eigenvalues of the Hessian of a harmonic function that is neither of the form $ax+by+cz+d$ nor a constant?

Question

(Following an earlier post here) I intuit that if we restrict to functions $f(x,y,z)$ that are harmonic (i.e. satisfying $\nabla^2f=0)$ but neither of the form $ax+by+cz+d$ ($a,b,c,d\in\mathbb{R}$) nor a constant, all the three eigenvalues of the $3\times 3$ Hessian matrix $$ H(x,y,z)= \begin{bmatrix} f_{xx} & f_{xy} & f_{xz}\\ f_{yx} & f_{yy} & f_{yz}\\ f_{zx} & f_{zy} & f_{zz}\\ \end{bmatrix}, $$ cannot all be zero. Of course, I don't have a proof. But does anyone have any counterexample?

Answer 1

The answer depends on the precise formulation of the question.

The Hessian vanishes everywhere: In this case the function must be (locally) affine. Indeed, if the Hessian vanishes, then we have for every $j\in \{1,2,3\}$ that $$\nabla (\partial_j f)(x,y,z)=((\partial_1 \partial_j f)(x,y, z), (\partial_2 \partial_j f)(x,y,z), (\partial_3 \partial_j f)(x,y,z))=(0,0,0).$$ This means that $\partial_j f$ is (locally) constant. This implies that $f$ is (locally) affine.

Hessian is only vanishing at certain points: The function $f(x,y,z) = xyz$ is a counterexample to the claim "if $f$ is harmonic and the Hessian vanishes at least in one point, then $f$ must be an affine function". It is harmonic, we even have $\partial_j^2 f \equiv 0$ for all $j\in \{1,2,3\}$. However, we have $$ Hf(x,y,z) = \begin{pmatrix} 0 & z & y \\ z & 0 & x \\ y &x & 0 \end{pmatrix}. $$

Thus, we get that $Hf(0,0,0)$ is the zero matrix.

Note that we also have $\nabla f(0,0,0)=(0,0,0)$ (it is also a critical point). Thus, if you are trying to show that harmonic function do not have local extrema, then you need to find a way to deal with the second case too.