Spectrum near zero of $-\partial^2_x + V : L^2(\mathbb{R}) \to L^2(\mathbb{R})$, where $V = O(|x|^{-2 - \delta})$

Question

Let $H = -\partial^2_x + V(x) : L^2(\mathbb{R}) \to L^2(\mathbb{R})$ be a one dimensional Schrödinger operator, where the potential $V$ is real-valued, belongs to $L^\infty(\mathbb{R})$, and, as $|x| \to \infty$, decays as $|x|^{-2 - \delta}$ for some $\delta > 0$. Since $V \in L^\infty$, by the Kato-Rellich theorem, the operator $H$ is self-adjoint with respect to the domain the Sobolev space $H^2(\mathbb{R})$.

I would like to show that $0$ cannot be an eigenvalue of $H$, that the spectrum of $H$ on $(-\infty, 0)$ consists only of eigenvalues, and finally that there exists $\mu > 0$ sufficiently small so that $(-\mu, 0) \subseteq \rho(H)$.

These properties of $H$ are asserted but not proved in the article by Jensen and Nenciu, Rev. Math. Phys. 13(6) (2001). It seems that the above properties are quite classical, because I have consulted many of the prominent modern references (the books of Reed--Simon and Yafaev, articles by Jensen, Kato, Gesztesy, etc.), but have not found any concrete discussion about them. Hint, solutions, or pointers to relevant references are greatly appreciated. I would like to establish each of these properties without using too much abstract machinery, if possible.

Letting $H_0 = -\partial^2_x$, I think I have worked out that, due to the decay of $V$, $V(H_0 - \lambda^2)^{-1}: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ is compact in $\text{Im} \, \lambda> 0$. We should also have $(H - \lambda^2)^{-1} = (H_0 - \lambda^2)^{-1}(I + V(H_0 - \lambda^2)^{-1})$ exists for $\text{Im} \, \lambda \gg 1$ and (by the analytic Fredholm theorem) is meromorphic in $\text{Im} \, \lambda> 0$. But this is as far as I have got toward proving any of the above properties.

Answer 1

As a general rule of thumb, it's usually most convenient in one-dimensional problems to work with solutions of the ODE $-y''+Vy=Ey$ rather than operator theoretic methods.

Here, everything follows from the behavior of the $E=0$ solutions. To analyze the spectrum below $E=0$, we can replace the actual potential by a smaller potential (by min-max). The borderline case for finite spectrum below $E=0$ is $V_0=-(1/4)|x|^{-2}$. This we can see by solving $-y''-cx^{-2}y=0$ explicitly (for $x\ge 1$, say), which is possible since this is an Euler equation. For $c<1/4$ the solutions have only finitely many zeros, and then the claim follows from oscillation theory.

Finally, $E=0$ cannot be an eigenvalue because a square integrable solution $y$ would also solve $$ y(x)=1+\int_x^{\infty} (t-x)V(t)y(t)\, dt , $$ but then $y\to 1$ as $x\to\infty$.