I am used to deal with second order Sturm-Liouville operator on $L^2(\mathbb{R})$. What is useful for my problems is to identify the kernel, and count the number of zeroes of the eigenfunction. From this, one can count how many negative eigenvalue the operator has. This has repercussion on the stability of solutions to a nonlinear PDE.
I just stumbled of an operator $H$ of the form $$Hv:=Lv+2\phi\star v,$$ where $L$ is a second order Sturm-Liouville operator, $\phi=e^{-|x|}/2$, and $\star$ denotes the convolution operation. Since $\phi$ is the Green's function for $1-\partial_x^2$, we have that $$(1-\partial_x^2)(\phi\star v)=v.(1)$$
I know the eigenfunction of $H$ corresponding to the kernel (eigenvalue zero) has one zero. Is there any way to use that information to infer that there is only one negative eigenvalue for $H$, even in the presence of that extra convolution operator?
The things I pondered with: I tried to take advantage of (1). I also think there must be a way to use the fact that the convolution part of the operator has no point spectrum on $L^2(\mathbb{R})$.