Picture below is part of proof of weak maximum principle. On the red line ,I don't know how to use the Taylor expansion to get $-u''(x_0) \le 0$. I think the Taylor expansion of $u(x)$ at $x_0$ is $$ u(x)=u(x_0)+u'(x_0)(x-x_0)+\frac{u''(x_0)}{2}(x-x_0)^2+o((x-x_0)^2) $$ Let $x\in \partial \Omega$ and $u(x)=m$, then $$ m=u(x_0)+\frac{u''(x_0)}{2}(x-x_0)^2+o((x-x_0)^2) $$ So, $$ 0> \frac{u''(x_0)}{2}(x-x_0)^2+o((x-x_0)^2) $$ then $$ u''(x_0) > 2\frac{o((x-x_0)^2)}{(x-x_0)^2} $$ But the $x$ has been choice , how to show $-u''(x_0) \le 0$ ?
Picture below is from the 7th page of this .
