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.