It is known that the kernel of the (non-negative) Laplacian operator on a closed manifold consists of constant functions. I would like to ask if some similar phenomena happens for the modified operator:
$$ Lu=\Delta u+ fu,$$
where $f$ is a smooth function.


**More specifically:** If $f$ equals to minus an eigenvalue of $\Delta$, then $Lu=0$ has non-trivial solutions. Are these the only $f$ with non-trivial solutions? Can we conclude that $Lu=0$ has only zero (or constant solutions) if we assume $f$ non-constant? Otherwise, can you parametrize its kernel (as you parametrize constant functions by their integrals, or by their value in one point)?


Thank you very much.



Kernel of the Laplacian + a function
**Q:** Can we conclude that $Lu=\Delta u+ fu=0$ has only zero (or constant solutions) if we assume $f$ non-constant?


**A:** No, a counter example in one dimension is the [Mathieu equation](https://en.wikipedia.org/wiki/Mathieu_function), which has non-constant $\pi$-periodic or $2\pi$-periodic solutions $u(x)$ when $f(x)=a-2q\cos 2x$ for any given real $q$ at an infinite sequence of values of $a\_n(q)$, $n=1,2,3,\ldots$.


More generally, $-L$ is the Hamiltonian of a particle in the potential $-f$, and we can readily adjust the potential so that it has a bound state at zero energy – simply by adding a constant to the potential to shift the bound state up or down.


Do note that the answer to the question would be affirmative for any generic $f$. To have a nonzero $u$ with $Lu=0$ requires fine tuning of the function $f$, for a generic $f$ such a solution will not exist.


