Consider a function $u \in C^\infty(\mathbb{R}^n)$ such that
$$ \begin{cases} u(x) > 0 & \forall x \in \mathbb{R}^n \\ \Delta u(x) \geq u(x) > 0 &\forall x \in \mathbb{R}^n \end{cases} $$
Does there exist a solution to this PDE that is bounded i.e
Does there exist a solution to the above PDE, $ v(x) $, such that
$$ \vert v(x) \vert < M \ \forall x\in \mathbb{R}^n $$
for some $M > 0$?
No, such a function does not exist. If $u-\Delta u \leq 0$ with $u$ bounded, then $u\leq 0$. In fact, assume that $u>0$ somewhere. If $u$ has a maximum in $x_0$, then $u(x_0)>0$ and $\Delta u(x_0) \leq 0$ so that $u(x_0)-\Delta u(x_0) >0$, in contrast with the assumption.
However $u$ is merely bounded and could not have a maximum. Take $v(x)=2n+|x|^2$ which satisfies $v-\Delta v \geq 0$ and $u_\epsilon=u-\epsilon v$. Then $u_\epsilon- \Delta u_\epsilon \leq 0$ and $u_\epsilon$ has a maximum since $u$ is bounded and $v \to \infty$. The argument above now applies and gives $u_\epsilon \leq 0$ and then $u \leq 0$, letting $\epsilon \to 0$.
