I am self-studying Evan's PDE, chapter 6.4, which talks about the maximum principle of elliptic equations.
Theorem (Weak maximum principle) Define $L := \sum_{i,j=1}^n a^{ij} u_{x_i x_j} + \sum_{i=1}^n b^i u_{x_i} + cu$ and let $U \subset \mathbb{R}^n$ be a bounded open set. Assume $u \in C^2(U) \cap C(\overline{U})$ and $c \equiv 0$ in $U$. If $Lu \leq 0$ in $U$, then $\max_U u = \max_{\partial U} u$.
Proof Suppose $Lu<0$ in $U$ first and suppose there exists some $x_0 \in U$ such that $u(x_0)$ attains the maximum. Then $Du(x_0) = 0$ and $D^2u(x_0)$ is non-positive definite at $x_0$.
My question: Why $D^2u(x_0) \leq 0$? My attempts are as follows:
The conclusion is true in 1-dimensional case. WLOG let $x_0 = 0$. Fix some $y \in \mathbb{R}^n$, define $f(t) := u(ty)$. Then $f'(t) = Du(ty) \cdot y = \sum_{i=1}^n u_{x_i}(ty)y_i$ and then $f''(t) = \sum_{i=1}^n y_i \nabla (u_{x_i}(ty)) \cdot y = y^T D^2u(ty) y$. Since $u\in C^2(U)$, sending $t \to 0$ shows that $D^2u(0) \leq 0$.
Is my above proof correct? Could anyone give me some hint?
Edit: with the help in comments, the above proof should be correct; and $u \in C^2$ is not necessary, instead, we could directly write $f''(0) = y^T D^2(0) y \leq 0$.
Best Answer
Evans is appealing to a general result coming only from the fact that $u$ is sufficiently smooth (twice differentiable) and has a (local) maximum at the interior point $x_0$.
It is standard that $Du$ vanishes at the maxmimum $x_0$. Suppose that $D^2u(x_0)\not\le 0$. Then there would exist a unit vector $v$ such that $v^T D^2 u(x_0)v=C>0$ (as real numbers). Recall the second order Taylor expansion as $x\to x_0$, (Peano remainder is enough) $$ u(x) = u(x_0) + \underbrace{Du(x_0)\cdot (x-x_0)}_{=0} + (x-x_0)^T Du(x_0)(x-x_0) + o(|x-x_0|^2). $$ Choose $x-x_0 = \lambda v$ where $\lambda\ll1$. Then we see as $\lambda\to 0$, $$ u(x) = u(x_0) + \lambda^2 C + o(\lambda^2).$$ since the $o(\lambda^2)$ term is eventually smaller than $\lambda^2C/2$, we see that $u(x)>u(x_0)$, which contradicts the fact that $u(x_0)$ was a maximum.