Maximum principle of p-Laplacian operator

Question

Suppose $\Delta_p u \geq \Delta_p v$ then can it be said that $u \geq v$? The domain considered is a bounded subset of $\mathbb{R}^n$.

Answer 1

No. First, you could of course have $u\equiv1$ and $v\equiv0$. Even if $u$ and $v$ have the same boundary values, this can happen. If, say, the domain is the unit ball and $u\equiv0$ and $v(x)=1-|x|^2$, then $\Delta_pv\leq0$ for any $p$ but still $v\leq u$.

If you reverse one of the inequalities and ask if $\Delta_pu\leq\Delta_pv$ for $u-v\in W^{1,p}_0$ implies $u\geq v$, the question is harder. For $p=2$ this is true and perhaps also in general, but I don't recall seeing such a theorem. It is also true if one of the functions is constant, because there is a maximum principle for $p$-subharmonic functions.