I am stuck with the following problem:
Let $Af = -f'' + uf$ be an unbounded operator, with $u \in \mathcal{S}(\mathbb{R},\mathbb{R})$ in the Schwartz space. The domain of $A$ is the linear subspace $H^2(\mathbb{R})$ of $L^2(\mathbb{R})$.
(i) Show that if $\lambda<0$ is an eigenvalue, then $\lambda > \inf_{x \in \mathbb{R}}\{u(x)\}$.
(ii) Show that if $\lambda_2 < \lambda_1$ are eigenvalues with corresponding eigenfunctions $f_2$ and $f_1$ and if $x_1 > x_2$ are two consecutive zeros of $f_1$ then the function $f_2$ must have at least one zero in the interval $(x_1,x_2)$.
For part (i) I attempted to solve it by proof of contradiction and showing that the corresponding ODE does not have a solution. I get $-f''+(u-\lambda)f =0$ but how do I know that this is not solvable?
I think part (ii) follows from part (i). I would greatly appreciate any input, since I do not know how to approach this. Thank you!
Suppose $-f''(x)+u(x)f(x)=\lambda f(x)$, where $-f''+uf\in L^2(\mathbb{R})$ and $f\in L^2(\mathbb{R})$ is twice absolutely continuous. Let $\mu = \inf u(x)$. Then \begin{align} -\langle f'',f\rangle & = \lambda\|f\|^2-\langle uf,f\rangle \\ \|f'\|^2 & = \lambda \|f\|^2-\langle uf,f) \\ & \le \lambda \|f\|^2-\inf u(x)\|f\|^2 \\ & = (\lambda-\inf u(x))\|f\|^2 \end{align} Therefore $\lambda < \inf u(x)$. (If $\lambda=\inf u(x)$, then $\|f'\|=0$, which forces $f=C$ and $f\in L^2$ forces $C=0$.)