weak version of maximum principle for not-quite-subharmonic functions

Question

For smooth functions $f(x,y)$ in a disk, if $f$ is subharmonic then it satisfies the maximum principle. What happens if we relax the subharmonic condition, by requiring only certain bounds on $\Delta f$? Are there estimates on "how far" $f$ would be from having a maximum on the boundary in such situations?

Answer 1

Let $M\geq 0$ and $u\in C^2(B_1)\cap C^0(\overline{B_1})$ satisfy $-\Delta u \leq M$ in $B_1$.

Then, let $v:\overline {B_1} \to \mathbb R$ be given by $$v(x)= u(x) -\frac{M}{2n} \big (1-\vert x \vert^2 \big ). $$ It follows that $$-\Delta v= -\Delta u -M \leq 0 \text{ in } B_1$$ and $v=u$ on $\partial B_1$. By the weak maximum principle, \begin{align*} \max_{\partial{B_1}}u&= \max_{\partial {B_1}}v \\ &= \max_{\overline {B_1}}v \\ &\geq \max_{\overline {B_1}}u - C M \end{align*} with $C=C(n)>0$. Thus,

$$ \max_{\overline {B_1}}u \leq \max_{\partial{B_1}}u +CM.$$ This is a direct extension of the weak maximum principle. Indeed, when $M=0$ (the case $u$ is subharmonic), we recover the maximum principle since we always have that $ \max_{\partial{B_1}}u \leq \max_{\overline{B_1}}u$.