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

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.

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but if you use $u(x,y)=x^2+y^2$ then $\nabla^2u=2+2=4$ –  Buddharta Dec 2 '12 at 18:36
I deleted that comment - I read the question wrong (thought it said $\nabla u=0$ instead of $\nabla^2u=0$) –  icurays1 Dec 2 '12 at 18:41
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Julian Kuelshammer Dec 2 '12 at 19:34

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.

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oh thank you! yes I've noticed that the condicion $\nabla^2U=0$ implies $Tr(HessU)=0$ but I wasn't sure if the trace of the matrix was allways the sums of the eigenvalues or if that only aplies when you diagolize. –  Buddharta Dec 2 '12 at 18:57
It always holds. See this –  icurays1 Dec 2 '12 at 19:00
Can you actually conclude that the critical point is a saddle point when the Hessian is $0$? We might need a more general definition than the standard one ... –  Matt Dec 3 '12 at 2:32
Thanks @Matt - completely overlooked that. It seemed too easy... –  icurays1 Dec 3 '12 at 4:07
I hate to do this (mostly because this is really close to being a beautiful way to prove the maximum principle in 2-D which I hadn't thought of before), but can we really conclude that a harmonic function with $0$ Hessian is locally constant? If I'm not mistaken $u(x,y)=xy$ has an isolated critical point at $(0,0)$ with Hessian $0$, but every neighborhood of $(0,0)$ contains both positive and negative (and zero) values. –  Matt Dec 3 '12 at 4:44