$\nabla^2u=0$ implies every critical point is a saddle point

Question

Hi everybody I need help with this problem: let $u:R^n \rightarrow R$ be a function so that $\nabla^2u=0$ prove that every critical point of the function is a saddle point.

Answer 1

Hints:

1. The condition $\nabla^2u=0$ is equivalent to the trace of the Hessian being zero
2. The trace of a square matrix is the sum of its eigenvalues
3. A point $x$ is a saddle point if the Hessian matrix of $u$ at $x$ has both positive and negative eigenvalues.

What if all the eigenvalues are zero, you ask? Well, the Hessian is symmetric and hence diagonalizable; if all the eigenvalues are zero, then the Hessian is similar to the zero matrix! (What does this say about the Hessian?)

Edit:

Indeed, as pointed out in a comment below, this argument is not sufficient for the case when the Hessian matrix is zero at a point, for this only tells us that the function $u$ is locally constant, and nothing about the behavior in various directions. The rigorous proof of this saddle-point property is essentially the maximum/minimum principle for harmonic functions. The "strong" version goes like this:

If $u$ is harmonic on an open, bounded, connected set $\Omega\subset\Bbb{R}^n$, and there exists $x_0\in\Omega$ such that $u(x_0)=\sup\{u(x):x\in\bar{\Omega}\},$ then $u(x)$ is constant on $\Omega$. (Similarly if $u(x_0)=\inf\{u(x):x\in\bar{\Omega}\}$, then $u$ is constant on $\Omega$).

So, suppose $\nabla^2u=0$ on $\Bbb{R}^n$ and $x$ is a critical point of $u$. Then, consider $B(x,M)$ for any arbitrary $M>0$. For each $M$ this is a bounded, open, connected set on which $\nabla^2u=0$. Thus if $x$ were a minimum, $u$ would be constant throughout $B(x,M)$. Since $M$ is arbitrary, it must be that $u$ is constant throughout $\Bbb{R}^n$. Similarly if $x$ were a maximum, $u$ must be constant throughout $\Bbb{R}^n$.

Thus, if $u$ is non constant and $x$ is a critical point, it can be neither a maximum nor minimum, and is hence by definition a saddle point.

So, it seems that we need to assume that $u$ is non constant, since constant functions are harmonic and have no saddle points (every point is a local max and local min). Otherwise the proof goes.