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?



Spectrum of "classical" operators
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.




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


