Let $(M,g)$ be a closed, smooth Riemannian manifold. Let $\Delta = -div\nabla$ be the Laplace-Beltrami operator. Let $h$ be a smooth function on $M$. Is there a condition on $h$ weaker than non-negativity such that $\Delta + h$ is a positive or non-negative operator? 


I'm thinking of something akin to the following: For the conformal Laplacian, non-negativity of the Yamabe constant is sufficient for any $h$ that is a scalar curvature of a metric in the conformal class.



Positivity of Second-Order Elliptic Differential Operators
This can't be right as stated: if $h$ takes a negative value at some point $p$ of $M$ then $\Delta+ch$ has a negative eigenvalue for sufficiently large $c$.


Proof: let $f: M \rightarrow {\bf R}$ be a nonnegative smooth function that's positive at $p$ and supported on a small enough neighborhood of $p$ that $f(q)=0$ whenever $h(q)>0$. Then $\langle f, hf \rangle < 0$. Therefore if $c$ is large enough then $\langle f, (\Delta+ch) f \rangle < 0$, which would be impossible if every eigenvalue of $\Delta + ch$ were nonnegative.


