# Spectrum of "classical" operators

Lately, I've been reading a couple of papers from different one-dimensional PDE contexts on which operators like $$\mathcal{L}:=-\partial_x^2+c_*+\Phi$$ repeatedly appear. Usually, on these contexts $$\Phi$$ is a smooth exponentially decaying function and $$c_*\in\mathbb{R}$$ is a positive constant.

I am very surprised that in all of these papers the authors, just by knowing these facts, they immediately conclude that the continuous spectrum of $$\mathcal{L}$$ is exactly $$[c_*,+\infty)$$ and the rest of the spectrum consists on a finite number of eigenvalues. My question is, how do they know that the continuous spectrum starts exactly at $$c_*$$? I've seen this at least five times for different values of $$c_*$$ and different functions $$\Phi$$ (all of them smooth with exponential decay). Does anyone have an explanation for this? Or any reference?

A second question (I know that the difficulty of the question can exponentially grow now so I am actually very happy just by understanding the previous part!): What if I consider a non-smooth but still exponentially decaying $$\Phi$$? For instance $$\Phi=e^{-\vert x\vert}$$? The previous "result" still holds?

• The 3D case is treated here around page 88. I can't remember whether they did the 1D case, or how it was proved. Mar 25, 2020 at 3:00

Let $$A$$ be a self-adjoint operator with domain $$D(A)\subset\mathcal H$$ ($$\mathcal H$$ is some Hilbert space). An operator $$C$$ with $$D(A)\subset D(C)$$ is called relatively compact with respect to $$C$$ if $$C(A-zI)^{-1}$$ is compact for some (hence all) $$z\notin\sigma(A)$$. Paraphrasing Corollary 2, page 113 Section XIII.4, in [1], we have

If $$C$$ is relatively compact with respect to $$A$$, then $$B:=A+C$$ is closed with domain $$D(B)=D(A)$$, and $$\sigma_{ess}(B)=\sigma_{ess}(A).$$

In fact, this is elementary once one realises that $$C(A-zI)^{-1}:\mathcal H\to \mathcal H$$ is compact if and only if $$C:D(A)\to\mathcal H$$ is compact.

In your case, setting $$A = -\partial_x^2 + c_*$$ and $$C$$ the multiplication operator by $$\Phi$$, is it not hard to prove that $$C:H^2(\mathbb R)\to L^2(\mathbb R)$$ is compact. We then obtain that $$\sigma_{\mathrm{ess}}(\mathcal{L}) = \sigma_{\mathrm{ess}}(-\partial_x^2 + c_*).$$ Since it is well know that $$\sigma_{\mathrm{ess}}(-\partial_x^2 + c_*) = [c_*,+\infty)$$, the result follows.

For your second question, this answer is yes the result hold for $$\Phi(x) = e^{-|x|}$$. In fact it will hold in dimension $$d\leq 3$$ for any $$\Phi\in L^2(\mathbb R^d)$$. As explained earlier, it is enough to show that multiplication by $$\Phi$$ is (well-defined and) compact from $$H^2(\mathbb R^d)\to L^2(\mathbb R^d)$$; let us prove this result.

Let $$(f_n)_{n\geq0}$$ be a bounded sequence in $$H^2(\mathbb R^d)$$, $$d\leq 3$$. The core of the argument is the following fact.

$$(f_n)_{n\geq0}$$ is bounded in $$L^\infty$$ and, up to extraction, converges $$L^\infty$$-locally to some function $$f$$.

Note that $$f$$ then has to be bounded as well. Once this is established, we see that \begin{align*} \limsup |\Phi f_n-\Phi f|_{L^2}^2 & = \limsup \int|\Phi|^2\cdot|f_n-f|^2 \\\\ & \leq \limsup |f_n-f|_{L^\infty([-k,k]^d)}^2\cdot\int|\Phi|^2{\mathbf 1}_{[-k,k]^d} \\\\ & \quad + \limsup (|f_n|_{L^\infty}+|f|_{L^\infty})^2\cdot\int|\Phi|^2{\mathbf 1}_{([-k,k]^d)^\complement} \\\\ & \leq |\Phi|_{L^2}^2\cdot0 + 4M^2\cdot\limsup \int |\Phi|^2{\mathbf 1}_{([-k,k]^d)^\complement} \end{align*} for $$M$$ a bound on the norms $$|f_n|_{L^\infty}$$, and $$\limsup$$ the limit superior along a convergent subsequence. Because $$\Phi$$ is in $$L^2$$, Lebesgue's dominated convergence theorem guarantees that the limit is zero, and $$\Phi f_n$$ converges to $$\Phi f$$ as expected.

Let us turn to the proof of the fact. The boundedness of $$(f_n)_{n\geq0}$$ on compact sets follows from the continuous embeddings from $$H^2(\ell+[0,1]^d)$$ to $$\mathcal C^0(\ell+[0,1]^d)$$ for all $$\ell\in\mathbb Z^d$$ (because $$d\leq3$$). Since the norm of these injections does not depend on $$\ell$$, $$(f_n)_{n\geq0}$$ is in fact uniformly bounded over $$\mathbb R^d$$

As for the convergence up to extraction, according to the usual Sobolev embeddings/inequalities, the sequence of restrictions $$\big((f_n)_{|[-k,k]^d}\big)_{n\geq0}$$ is relatively compact in $$\mathcal C^0([-k,k]^d)$$ for all $$k\geq1$$ (we use again that $$d\leq3$$). We can then extract diagonally a subsequence $$(f_{\sigma(m)})_{m\geq0}$$ that converges to some continuous function $$f$$ uniformly over the compact sets, concluding the proof of the fact.

[1] Reed, M., Simon, B. (1978). Methods of Modern Mathematical Physics. IV Analysis of Operators. New York: Academic Press.