While reading a paper on the topic 'Numerical solutions for generalized Black-Scholes equation', It is given that their numerical scheme can be executed explicitly by solving a linear system $\mathbf A^nU^n=\mathbf b^n$, where $n$ represents the current time $t_n$. ie, $U^n$ is an $M-$ vector whose coordinates are the solutions at each $x_m,~m=1,2,...,M$ along the temporal line $t=t_n$. Furthermore, they showed that the matrix $\mathbf{A}^{n}$ has a positive inverse. Afterwards they given a lemma,
lemma: Assume the conditions in previous lemma for the mesh parameters $h$ and $k$, the operator $L_h^k$ defined by \begin{equation}\label{eq15} L_h^ku_m^n=\mathbf{A}^{n}(m){U}^{n}={\mathbf b}^{n}(m), \end{equation} where $\mathbf{A}^{n}(m)$ is the $m$ th row of $\mathbf{A}^n$ and ${\mathbf b}^{n}(m)$ is the $m$ th coordinate of ${\mathbf b}^{n}$, satisfies the discrete maximum principle. ie., Let $u_m^n,~v_m^n$ be mesh functions satisfying $u_1^n \leq v_1^n,~u_M^n \leq v_M^n,~n=1,2,...,N,~u_m^1 \leq v_m^1,~m=0,1,...,M$ and $L_h^ku_m^n\leq L_h^kv_m^n,~m=0,1,...,M,~n=0,1,...,N$, then $u_m^n \leq v_m^n$ for all $m,n$.
My doubt: Here $L_h^k$ is given by \begin{equation}\label{LBSKD9} \begin{array}{l} L^k_h u_m^n \equiv \left(\alpha_{m,-}^n+\frac{1}{k}\beta_{m,1}^n\right)u_{m-1}^n+\left(\alpha_{m,c}^n+\frac{1}{k}\beta_{m,2}^n\right) u_{m}^n+\left(\alpha_{m,+}^n+\frac{1}{k}\beta_{m,3}^n\right) u_{m+1}^n=\\ \beta_{m,1}^n\left(f_{m-1}^n+\frac{1}{k}u_{m-1}^{n-1}\right)+\beta_{m,2}^n\left(f_{m}^n+\frac{1}{k}u_{m}^{n-1}\right)+\beta_{m,3}^n\left(f_{m+1}^n+\frac{1}{k}u_{m+1}^{n-1}\right)=F^n_m \end{array} \end{equation} How can we say that operator $L_h^k$ satisfies the discrete maximum principle? What is the role of invertibility of $\mathbf{A}^{n}$ in the discrete maximum principle? What happens for the inequality $L_h^ku_m^n\leq L_h^kv_m^n$ to obtain $u_m^n \leq v_m^n$?
