Consider a bounded self-adjoint operator $L$ on $L^2(\mathbb{R})$. I want to study its essential spectrum, thus I need to determine when is $L-\lambda I$ Fredholm. Am I right with the statement $$ L-\lambda I\text{ is Fredholm iff }(1-\partial_x^2)(L-\lambda I)\text{ is.} $$ My proof is based on the fact that $(1-\partial_x^2)$ has a trivial kernel. It thus does not change the property that the kernel must be finite-dimensional. The part where the cokernel is finite-dimensional can be dealt with the fact that $(1-\partial_x^2)$ is surjective and has a trivial kernel.
I am not particularly self-confident in this topic and just want to ask if it makes sense or if there is a result that makes this statement trivial and thus not deserves a proof. There is of course the possibility that I am wrong.
The reason I am interested in this is because I deal with this operator, which looks like an ordinary S-L differential operator after the application of $(1-\partial_x^2)$, thus much more easy to deal with when computing the spectrum.