= Solution
{c}
The displayed estimate requires a nonnegative barrier. As printed, a counterexample is obtained with two interior nodes, each having row $L_hv_p=v_a-v_p$ to one boundary node. Set $e_a=0$, $e_p=1$, and $\phi_a=-1$, $\phi_p=-2$. Both interior rows have $|T_p|=1$ and $L_h\phi_p=1$. 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 $\phi_p\geq0$ and the stencil has the maximum principle established in part (a). Set $C=\max\{T/d_1,S/d_2\}$. On either subregion,
$$
L_h(\pm e+C\phi)\geq-|T_p|+Cd_i\geq0.
$$
At boundary nodes these functions equal $C\phi_a$. Hence
$$
\pm e_p\leq C\left(\max_a\phi_a-\phi_p\right)\leq C\max_a\phi_a,
$$
which proves the intended <discrete elliptic barrier> estimate
$$
\boxed{E\leq\max_{a\in J_{\partial\Omega}}\phi_a\max\{T/d_1,S/d_2\}.}
$$
For a normalized stencil with $L_h1=0$ and a sign-changing barrier, shifting it by its minimum gives the valid general replacement $E\leq C(\max_a\phi_a-\min_p\phi_p)$. The supplied barrier in part (d) is positive, so the printed estimate applies there without a sign problem.
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