Proof of Weak Maximum Principle for Parabolic Equations

Question

I'm trying understand the proof of the Weak Maximum Principle for Parabolic Equations in this presentation (the proof starts on the slide $11$) for the operator

$$Lu \equiv \sum_{i,j = 1}^n a_{ij}(x,t) \frac{\partial^2 u}{\partial x_i \partial x_j} + \sum_{i = 1}^n b_i(x,t) \frac{\partial u}{\partial x_i} + c(x,t)u - \frac{\partial u}{\partial t}$$

when $c \leq 0$.

My doubts are

1) Why $D^2u \leq 0$ on the slide $11$? I know that the operator $Lu$ is evaluated on a maximum point, but how $D^2u$ appears in $Lu$ instead of $\sum_{i,j = 1}^n a_{ij}(x,t) \frac{\partial^2 u}{\partial x_i \partial x_j}$?

2) How exactly appears the inequality on slide $12$ which involves $||b_1||_{\infty}||$ and $||c||_{\infty}$?

3) Why $\sup_{\overline{\Omega_T}} u_{\varepsilon} \leq \max_{\overline{\Omega_T}} u_{\varepsilon}^+$ and not $\sup_{\overline{\Omega_T}} u_{\varepsilon} = \max_{\overline{\Omega_T}} u_{\varepsilon}^+$?

Thanks in advance!

$\textbf{EDIT:}$

2) I understood how appears the inequality. The term $\alpha^2 \lambda$ appears by definition of parabolic operator $L$, the terms $- ||b_1||_{\infty}$ and $- ||c||_{\infty}$ appear because $|b_1(x,t)| \leq ||b_1||_{\infty}$ and $|c(x,t)| \leq ||c||_{\infty}$ for every $(x,t) \in \Omega_T$ by definition of the uniform norm and, by the inequalities involving modulus, we have

$$- ||b_1||_{\infty} \leq b_1(x,t) \leq ||b_1||_{\infty} \hspace{1cm} \text{and} \hspace{1cm} - ||c||_{\infty} \leq c(x,t) \leq ||c||_{\infty}, \forall (x,t) \in \Omega_T.$$

3) I understood why $\sup_{\overline{\Omega_T}} u_{\varepsilon} \leq \max_{\overline{\Omega_T}} u_{\varepsilon}^+$. I saw now that $u = u^+ - u^-$ and $u^- = - \min \{ u, 0 \}$, then the inequality is clear now for me.

Answer 1

The sum $\sum_{i,j=1}^n a_{ij}(x,t)\frac{\partial^2 u}{\partial x_i\partial x_j}$ is exactly the trace of product $AH$ where $H$ is the hessian of $u$ and $A$ is the matrice with coefficients $(a_{ij})_{ij}$. Since $A$ is positive definite and $H$ is negative semidefinite, the trace $AH$ is non positive.