 Solution

Assume $c\le 0$, the coefficients are bounded, and $u$ is bounded above. Choose $\kappa$ so large that the bounded positive function $q(y)=e^{\kappa y}$ satisfiesAlso set $\psi (x)=\log (1+|x{|}^2)$. Boundedness of the coefficients, $x\cdot D\psi \le 2$, and $c\le 0$ give a global upper bound $L\psi \le C$.

For $\epsilon >0$ and $0<\delta <\epsilon m/C$, the functiontends to $-\infty$ as $|x|\to \infty$ and satisfies $Lv>0$. If $v$ exceeded both zero and its values on $y=\pm 1$, it would attain a positive interior maximum. At that point $Dv=0$ and $D^{2}v\le 0$, whence $Lv\le cv\le 0$, a contradiction. Letting $\delta \downarrow 0$ and then $\epsilon \downarrow 0$ provesThis is the weak maximum principle for elliptic operators on the slab. The boundedness or a comparable growth condition is necessary because the domain is unbounded.