 Solution

Assume real bounded coefficients, with a principal part defining a uniformly elliptic operator, say $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ for some $\lambda >0$, and assume $c\le 0$. No smoothness of the boundary is needed for this classical weak maximum principle for elliptic operators. Its conclusion isIf $c=0$, the zero may be omitted. Only the symmetric part of the principal coefficients contributes to the Hessian matrix contraction.

For a strict inequality $Lu>0$, a positive interior maximum is impossible: there $Du=0$, $D^{2}u$ is negative semidefinite, and $cu\le 0$, so $Lu\le 0$. To reduce the non-strict case to this one, let $q(x)=e^{\kappa x_1}$. Boundedness of $b^1,c$ permits choosing $\kappa$ so large thatThus $L(u+\epsilon q)>0$. The preceding maximum argument and continuity on the compact closure giveLet $\epsilon \downarrow 0$; the exponential is bounded on the bounded domain. This is the exponential perturbation proof of the weak maximum principle with drift, with the zeroth-order term included. For $c=0$, any interior maximum, of either sign, is impossible for the perturbed function, giving the sharper boundary bound. The sign restriction on $c$ is essential: a positive first Dirichlet Laplacian eigenfunction violates the conclusion for $L=\Delta +{\lambda }_1$.