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The displayed estimate requires a nonnegative barrier. As printed, a counterexample is obtained with two interior nodes, each having row $L_h{v}_p=v_a-v_p$ to one boundary node. Set $e_a=0$, $e_p=1$, and ${\varphi }_a=-1$, ${\varphi }_p=-2$. Both interior rows have $|T_p|=1$ and $L_h{\varphi }_p=1$. Put one node in each of $J_1,J_2$ and take $T=S=d_1=d_2=1$. Then the proposed right-hand side is $-1$ while $E=1$.

For the intended use, suppose ${\varphi }_p\ge 0$ and the stencil has the maximum principle established in part (a). Set $C=\max \{T/d_1,S/d_2\}$. On either subregion,At boundary nodes these functions equal $C{\varphi }_a$. Hencewhich proves the intended discrete elliptic barrier estimateFor a normalized stencil with $L_{h}1=0$ and a sign-changing barrier, shifting it by its minimum gives the valid general replacement $E\le C({\max }_a{\varphi }_a-{\min }_p{\varphi }_p)$. The supplied barrier in part (d) is positive, so the printed estimate applies there without a sign problem.

For a normalized two-dimensional discrete Laplacian satisfying the discrete maximum principle, suppose quadratic functions are differentiated exactly and $L_{h}1=0$. In a domain of diameter $D$, choose $x_0$ in the domain and $\varphi (x)=|x-x_0{|}^2$. Then $0\le \varphi \le D^2$ and $L_h\varphi =4$. Applying the discrete elliptic barrier argument to zero-boundary error $L_{h}e=\tau$ gives the displayed bound. A mere sign pattern of the stencil does not determine this constant: rescaling the whole operator changes the error-to-residual ratio without changing the domain diameter.