For context, this question is related to a previous one of mine.
Consider the operator $H$ acting on $L^2(\mathbb{R})$ and of the form $$Hv:=\phi\ast v,$$ where $\phi=e^{-|x|}/2$ and $\ast$ denotes the convolution operation. Since $\phi$ is the Green's function for $1-\partial_x^2$, we have that $$(1-\partial_x^2)(\phi\star v)=v.\;\;\;\;\;(1)$$
From (1), the eigenvalue problem $Hv=\lambda v$ is equivalent to $$ v''=(1-1/\lambda)\,v.\;\;\;\;(2) $$ Because (2) does not have $L^2$ solutions, $H$ does not have point spectrum. Intuitively/heuristically, I find the essential spectrum by using Fourier analysis on (2), with $v=e^{ikx}$ and find the dispersion relation $$ \lambda=\frac{1}{1+k^2}. $$ Thus I have the following result: The spectrum of $H$ is the interval $[0,1]$ and it entirely consists of essential spectrum.
Are my proof and result valid? I am asking because I find the proof to be heuristic thus have some doubts. Perhaps somebody can help me with the validity (or invalidity) of one of them?