# $\nabla^2u=0$ 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 $\nabla^2u=2+2=4$ – user1080987 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

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– Julian Kuelshammer Dec 2 '12 at 19:34

## 1 Answer

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. – user1080987 Dec 2 '12 at 18:57
– icurays1 Dec 2 '12 at 19:00

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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
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