Growth of nonnegative functions satisfying $\Delta u \geq C>0$

Question

Let $u\in C^2(\Omega)$ with $\Omega = B_1(0)\subseteq \mathbb{R}^d$. Assume that $u$ is subharmonic and satisfies the inequality $$ \Delta u(x) \geq C>0 $$ for all $x\in \Omega$. Furthermore, we know that $u\geq 0$. Is there a reasonable lower bound for $\sup_\Omega u$ in terms of $C$?

By the strong maximum principle for subharmonic functions we get that the supremum of $u$ on $\Omega$ cannot be attained in the interior (the function cannot be constant as this would violate the inequality $\Delta u >0$). In fact, we cannot even have a local maximum by the same argument. So morally I would expect that $u$ has some directions in which is grows quite fast and I was wondering whether there is a quantitative way of describing this (respectively whether it is true to begin with).

Answer 1

Let $a=\sup_\Omega u$ and solve the problem $\Delta v=C$ in $\Omega$ with $v=a$ at the boundary. We get $v(x)= \frac{C}{2d}|x|^2+a-\frac{C}{2d}$. The function $w=u-v$ satisfies $\Delta w \geq 0$ and $w \leq 0$ on $\partial \Omega$ and by the maximum principle $w \leq 0$ in $\Omega$. This gives $0 \leq u(x) \leq \frac{C}{2d}|x|^2+a-\frac{C}{2d}$ and then $a \geq \frac {C}{2d}$.

Answer 2

Assume $C=1$ (by scaling), then $$ \sup_{B_1} u= \sup_{\partial B_1} u \ge \frac{1}{2n}. $$

Assume indeed that $$ \sup_{\partial B_1} u \le \frac{1}{2n}-\varepsilon. $$ The function $$ v(x)=u(x)-\frac{|x|^2}{2n}-\varepsilon $$ satisfies $$ \Delta v \ge 0\quad\text{in $B_1$} \quad \text{and} \quad v\le 0 \quad\text{on $\partial B_1$} $$ Hence, $v\le 0$ in $B_1$ and $u(0)\le -\varepsilon$, a contradiction with $u\ge 0$.