Spectrum below zero for $-\beta(x) \partial^2_x : L^2(\mathbb{R}) \to L^2(\mathbb{R})$

Question

Let $\beta \in L^\infty(\mathbb{R} ; (0, \infty))$ be bounded from above and below by positive constants. Consider the self-adjoint operator $ -\beta^{-1} \partial^2_x : L^2(\mathbb{R}; \beta dx) \to L^2(\mathbb{R}; \beta dx)$ with domain the Sobolev space $H^2(\mathbb{R})$.

If $1 - \beta(x)$ decays sufficiently fast as $|x| \to \infty$, is the spectrum of $-\beta^{-1} \partial^2_x$ discrete in $(-\infty, 0)$? If so, are there only finitely many negative eigenvalues?

This sort of result is well known if we are instead considering the operator $-\partial^2_x + V(x) : L^2(\mathbb{R}) \to L^2(\mathbb{R})$ with potential $V \in L^\infty$ decaying fast enough toward $\pm \infty$. In that case, the standard argument to show discreteness of spectrum uses the resolvent identity:

\begin{equation*} (-\partial^2_x + V - \lambda^2)(\partial^2_x- \lambda^2)^{-1} = ( \text{Id} + V (-\partial^2_x -\lambda^2)^{-1}), \qquad \text{Im}\, \lambda > 0. \end{equation*}

The operator $V (-\partial^2_x -\lambda^2)^{-1}$ is compact, thanks to the sufficient decay of $V$, and has norm less than one for $\text{Im} \lambda \gg 1$. Hence (using a Neumann series) $(-\partial^2_x + V - \lambda^2)^{-1}$ exists for $\text{Im} \, \gg 1$, and (by the Analytic Fredholm Theorem) exists as meromorphic operator in $\text{Im}\, \lambda > 0$.

This argument seems to break down for the higher order perturbation of $-\partial^2_x$. Noting that $(-\beta^{-1} \partial_x^2 - \lambda^2)^{-1}$ exists and is bounded if and only if $(-\partial_x^2 - \beta \lambda^2)^{-1}$ exists and is bounded, the analogous resolvent identity seems to be \begin{equation*} (-\partial^2_x - \beta \lambda^2)(-\partial^2_x - \lambda^2)^{-1} = \text{Id} + (1 - \beta) \lambda^2 (-\partial^2_x - \lambda^2)^{-1}. \end{equation*}

This issue now seems to be that the extra factor of $\lambda^2$ precludes us from concluding that the norm of $(1 - \beta) \lambda^2 (-\partial^2_x - \lambda^2)^{-1}$ is small for $\text{Im} \gg 1$, and hence we cannot construct $(-\beta \partial^2_x - \lambda^2)^{-1}$ by a Neumann series.

Answer 1

Edit (complete rewrite, my first attempt was utter nonsense): In fact, $\sigma(H)=[0,\infty)$, $H=-(1/\beta)d^2/dx^2$.

One quick way to see that there is no negative spectrum is to consider the quadratic form $$ \langle y, Hy \rangle = -\int \overline{y(x)}\frac{1}{\beta(x)}y''(x)\beta(x) \, dx = \int |y'(x)|^2 \, dx \ge 0 . $$

The same conclusion follows from oscillation theory: If $\lambda= 0$, we have the zero free solution $y=1$, so there is no spectrum below zero. If we somewhat more systematically introduce Prufer variables by writing $(y',y)=R(\cos\varphi, \sin\varphi)$, then the Prufer angle $\varphi$ solves $\varphi'=\lambda\beta\sin^2\varphi+\cos^2\varphi$. This can be used to show the stronger statement $\sigma=[0,\infty)$ that I already mentioned above. Here we now assume that $\beta\to 1$.