Inequality between Laplacians at a point of local maximum

Question

I never before worked with maximum principle. Can someone tell me how I proceed with this:

Let $v\in C^{\infty}$ $u$ is continuous and $\Omega\in \mathbf{R}^n$ and suppose that $u − v$ has a strict local maximum at $x = x_0 \in\Omega$.

That is, $u(x_0) − v(x_0)> u(x) − v(x)$ in some neighborhood of $x_0$. For $\xi$ sufficiently small, $u_\xi − v$ has a strict local maximum at $x = x_\xi$ where $x_\xi \to x_0$ as $\xi→ 0$. We can relate the derivatives of $u$ and $v$ through the maximum principle.

$$ Du_\xi(x_\xi) = Dv(x_\xi) \\ -\Delta u_\xi(x_\xi) \geq -\Delta v(x_\xi) $$

How do I get this inequality by the maximum principle?

Answer 1

This is not really about the maximum principle. The general fact is, if $f$ is a $C^2$ function that attains a local maximum at point $p$, then $Df(p)=0$ and $\Delta f(p)\le 0$. The proof is by Taylor expansion.

1. First order expansion: $f(x)=f(p)+\langle Df(p),x-p\rangle +o(|x-p|)$. This implies $Df(p)=0$, for otherwise $f(x)>f(p)$ when $x-p$ is parallel to $Df(p)$ and small enough.

2. Second order expansion: $f(x)=f(p) + (x-p)^TD^2f(p)(x-p)+o(|x-p|^2)$. The maximality of $f(p)$ implies that $D^2f(p)$ is nonnegative semidefinite. Hence, all of its eigenvalues are $\le 0$, and in particular its trace is $\le 0$. But this trace is $\Delta f(p)$.