Kernel of Laplacian plus a function 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.
 A: As explained at https://mathoverflow.net/a/418867/11260 --- the answer is "no", you cannot conclude from $Lu=0$ that $u=0$. Basically you are asking whether it is possible for a potential $f$ to have a bound state at zero energy (interpreting $-L$ as the Hamiltonian); by adding a constant to $f$ you can always ensure that there is a bound state at zero.
Note that you will need to fine tune parameters to line up a bound state at zero, this will not happen generically. For any random $f$ the kernel of $L$ will be trivial. In physics there is much literature on Hamiltonians where such fine tuning is not needed; one then speaks of a "topologically protected zero-mode"; your $L$ is not of this type.
An example of a physical system where this does happen is graphene in a magnetic field, with Hamiltonian
$$H=\begin{pmatrix}
0&-i\partial/\partial x +\partial/\partial y + cy\\
-i\partial/\partial x-\partial/\partial y+cy&0
\end{pmatrix}.$$
The equation $H{u_1\choose u_2}=0$ has a nonzero solution for any real $c$ (which represents the magnetic field), no fine tuning is needed. Andrei Geim and Kostya Novoselov won the 2010 Nobel prize in physics for the discovery of this "zero-mode" in graphene.
