Corollary 2.14: Suppose $\overline{\Omega}$ is compact. If $u$ is harmonic and real-valued on $\Omega$ and continuous on $\overline{\Omega}$, then the maximum value of $u$ on $\overline\Omega$ is achieved on $\partial \Omega$
The image below is an exerise that guides me on how to prove it:
Here is another proof of Corollary (2.14) that works for more general operators. Let $\Omega$ be a bounded domain in $\mathbb{R}^n$, and let $$ L = \sum a_{jk}(x) \partial_j \partial_k + \sum b_j(x) \partial_j $$ where $a_{jk}$ and $b_j$ are continuous functions on $\overline{\Omega}$ and the matrix $(a_{jk})$ is positive definite on $\overline{\Omega}$.
a. Show that if $v \in C^2(\Omega)$ is real-valued and $Lv > 0$ in $\Omega$, then $v$ cannot have a local maximum in $\Omega$. (Hint: Given $x_0 \in \Omega$, by a rotation of coordinates one can assume that the matrix $(a_{jk}(x_0))$ is diagonal [cf. the discussion of coordinate changes in §1A].)
b. Show that if $x_0 \notin C(\overline{\Omega})$ and $M > 0$ is sufficiently large, then $w(x) = \exp[-M|x-x_0|^2]$ satisfies $Lw > 0$ in $\Omega$.
c. Suppose $u \in C^2(\Omega) \cap C(\overline{\Omega})$ is real-valued and $Lu = 0$ in $\Omega$. Show that $\max_{\overline{\Omega}} u = \max_{\partial \Omega} u$. (Hint: Show that this conlusion holds for $v = u + \epsilon w$ where $w$ is as in (b) and $\epsilon > 0$.)
(Original image here.)
a)
$$Lv(x) = \sum a_{jk}(x)\frac{\partial v(x)}{\partial x_j}\frac{\partial v(x)}{\partial x_k} + \sum b_j(x)\frac{\partial v(x)}{\partial x_j}$$
As the exercise suggests, for a point $x_0$, I'll have $Lv(x_0)$ and if I do some coordinate changes, I get a diagonal matrix. So
$$Lv(z) = \sum a_{jj}(z)\frac{\partial^2 v(x)}{\partial x_j^2} + \sum b_j(x)\frac{\partial v(x)}{\partial x_j}$$
That is, the terms $\frac{\partial v(x)}{\partial x_j}\frac{\partial v(x)}{\partial x_k}$ cancel for $k\neq j$, and $z$ is $x_0$ in new coordinates.
From the hypotesis, $a_{jk}$ is positive definite on $\overline\Omega$. Since the matrix depends on $x$, I assume that it means that $xA(z)x>0$ for all $x,z\in\overline\Omega$. I think this is sufficient to say
$$\sum a_{jj}(z)\frac{\partial^2 v(x)}{\partial x_j^2}>0$$
since this is just $\nabla v^T[a_{jk}]\nabla v$. I still need to take care of $\sum b_j(x)\partial_j$. And if I suppose $Lv>0$ in $\Omega$, I should get that there's no local minimum there. I think it has something to do with proving the second derivative is positive or that the first derivative is too, cause they appear in the expression.
b) If $w(x) = e^{-M|x-x_0|^2}$ then:
$$Lw(x) = \sum a_{jk}(x)(-2M)(-2M)x_jx_ke^{-2M|x-x_0|^2} + \sum b_j(x)(-2M)x_je^{-M|x-x_0|^2} = \\ \sum a_{jk}(x)4M^2x_jx_ke^{-2M|x-x_0|^2} + \sum b_j(x)(-2M)e^{-M|x-x_0|^2}$$
I don't know how to pick large $M$ to make this always positive. I think I need to take care of $bj$ but I don't see any relation here.
c) Any ideas here?