Sign of Laplacian at critical points of $\mathbb R^n$

Suppose we are in $\mathbb R^n$. What can we say about the sign of $\Delta u(\vec x)$ if u($\vec x$) has a local max/min at $\vec x$? I've tried looking at the reverse of the second partial derivative test and it seems to suggest that $\Delta u(\vec x) \leq 0$ if $\vec x$ is a min and $\Delta u(\vec x) \geq 0$ if $\vec x$ is a max. However, I'm not convinced of this result. Can somebody point me towards a statement (theorem, etc.) involving the above. I appreciate the help.

Solution 1:

Your claim is true, I'll give two proofs in case that's helpful.

Proposition: Let $u : U \to \mathbb{R}$ be twice differentiable where $U$ is some open set in $\mathbb{R}^n$. If $u$ has a local max at $a\in U$, then $\Delta u (a) \leqslant 0$.

Proof 1. This is the more direct proof: $\Delta u (a) \leqslant 0$ follows from the fact that all second order derivatives ${\partial^2 u \over \partial x_i^2}$ are $\leqslant 0$ at $a$, let's see why. Let $f(x) = (x, a_2, \dots, a_n)$ for $x$ in some neighborhood of $a_1$ in $\mathbb{R}$ (where $a = (a_1, \dots, a_n)$). Then $f$ is twice differentiable and has a local max at $x = a_1$, so we must have $f''(a) \leqslant 0$ i.e. ${\partial^2 u \over \partial x_1^2}(a) \leqslant 0$. Proceed likewise for the other second derivatives.

Proof 2. Perhaps you are aware that the Hessian of $u$ must be negative semidefinite at a local max $a$. It follows (via the spectral theorem for instance) that the trace of the Hessian must be $\leqslant 0$, but the trace of the Hessian is precisely the Laplacian.

Of course, in case is of a local min you get $\Delta u \geqslant 0$ (either redo the proofs inverting the inequalities, or apply the previous proposition to $-u$).