I am not sure whether I should ask for help here or math stackexchange. I got trouble with an inequality involving the Schrodinger operator on manifolds. Any suggestion is appreciated!


Let $(M,g)$ be a closed Riemannian manifold, and two functions $\phi, \psi\in C^{\infty}(M)$ satisfying 
$$\phi+\psi>0\ \mathrm{on}\ M.$$


Is it possible to find a positive function $u>0$ such that
\begin{equation}
(-\Delta+\phi)u\geq0\ \mathrm{on}\ M,
\end{equation}
or
\begin{equation}
(-\Delta+\psi)u\geq0\ \mathrm{on}\ M?
\end{equation}



A question on Schrodinger operator
Here is an alternative formulation of a question related to yours.
Consider the case when $M=\Omega\subset\mathbb{R}^{n}$, a (smooth)
open bounded domain. Suppose that $u>0$ and 
$$
-\Delta u+\psi u=f \mbox{ in }\Omega.
$$


Integrate against $\frac{w^{2}}{u}$ and rearranging you obtain for
all $w\in H\_{0}^{1}(\Omega),$
$$
\int\nabla w\cdot\nabla w-\int\_{\Omega}\left|\frac{u\left|\nabla w\right|-w\left|\nabla u\right|}{u}\right|^{2}+\int\_{\Omega}\psi w^{2}=\int f\frac{w^{2}}{u}
$$
therefore for all $w\in H\_{0}^{1}(\Omega)$ 
$$
\int\nabla w\cdot\nabla w+\int\_{\Omega}\psi w^{2}\geq\int f\frac{w^{2}}{u}
$$
therefore, if $\lambda\_{\psi}$ is the first dirichlet eigenvalue
of $-\Delta+\psi$, you obtain 
$$
\lambda\_{\psi}\int w^{2}\geq\int f\frac{w^{2}}{u}.
$$
So if there exists a positive solution such that $-\Delta u+\psi u>0$,
then $\lambda\_{\psi}>0$. The converse is obsiously true, as we can
choose the first eigenvector to satisfy $u\_{\psi}>0$ in $\Omega$,
and therefore $-\Delta u+\psi u=\lambda\_{\psi}u>0$ in $\Omega$.
So your question could be written in this case: 



> 
> Is it true that
>  $$
> \mbox{if }\quad \psi+\phi>0,
> \mbox{ then } 
> \lambda\_{\psi}>0 \mbox{ or } \lambda\_{\phi}>0\quad?
> $$
> 
> 
> 


It is not obvious, because $\lambda\_{f}$ is concave in $f$: the fact that $\lambda\_{\frac{1}{2}\psi+\frac{1}{2}\phi}>0$ does not help..


