Let $(M,g)$ be a complete Riemannian manifold. Suppose that $u$ is a nonnegative solution to $\Delta_gu\ge u^2$. Does it follow that $u$ must be identically 0?
I know that the answer to above question is yes if one assumes that $Ric(g)$ has a lower bound, which allows for a maximum principle argument, using the distance function to cut-off.
I wonder if this is true in general, with no additional assumptions?
Let $(\theta,r)$ be the polar coordinates on the plane. Consider the Riemannian metric of the form $$g=\left(\begin{smallmatrix}1&0\\0&\ell^2(r)\end{smallmatrix}\right).$$
Let $f(\theta,r)=\phi(r)$. Then $\Delta f=\phi''+\tfrac{\ell'}{\ell}\cdot \phi'$, and the inequality becomes $$\phi''+\tfrac{\ell'}{\ell}\cdot \phi'\geqslant \phi^2.$$ So, you may choose $\phi(r)=r^2$ and select a rapidly growing function $\ell$ to satisfy this equation. If $r \approx 0$, you may assume $\ell = r$, making $g$ a Riemannian metric on the plane.
(It is not surprising that the curvature of the resulting metric diverges to $-\infty$ as $r \to \infty$.)
