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In Volume 4 of Reed and Simon on page 83 the authors refer to the space $(L^\infty(\mathbb{R}^3))_\varepsilon$,

and later on page 119 they use $L^\\infty_\varepsilon$. Are these two spaces the same? Is that the set of functions $f$ such that $(1+|x|^2)^{\varepsilon/2} f \in L^\infty$?

Thanks for any help -- I really tried looking this up for a while and turned up nothing, and it didn't seem to be in the index!

PS Just to motivate the question, Theorem XIII.15 of that book says (among other things), that if $V \in L^\infty_\varepsilon$, then the essential spectrum of $-\Delta + V$ is $[0,\infty)$.

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I am pretty sure they are the same. – GH from MO Jul 21 '11 at 16:22

I still haven't found the definition of this notation in the books, but I think $f \in L^\infty_\varepsilon$ means that $f \in L^\infty$ and that for every $\varepsilon > 0$ there exists $R$ such that $|x| > R$ implies $|f(x)| < \varepsilon$ (almost everywhere of course).

Then you can prove the application I mentioned by using the Rellich compactness theorem (for example in the version of -- you need a version for noncompactly supported functions) to show that $V$ is a relatively compact perturbation of $-\Delta$ (in the sense that $V(-\Delta - i)^{-1}$ is compact). Then the statement about essential spectrum can be proved by using the analytic Fredholm theorem to show that $\textrm{Id} + V(-\Delta - z)^{-1}$ is invertible off of a discrete subset of $\mathbb{C} \setminus [0,\infty)$, allowing one to put $(-\Delta + V - z)^{-1} = (-\Delta - z)^{-1}(\textrm{Id} + V(-\Delta - z)^{-1})^{-1}$.

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In the definition of "relatively compact" are you inverting $-\Delta - i$ (with $i=\sqrt{-1}$) or $-\Delta - I$ (with $I$ the identity operator)? – Noam D. Elkies Aug 27 '11 at 19:09
Noam: It should be $\sqrt{-1}$, since that works for any self-adjoint operator. But since the (negative) Laplacian $-\Delta$ is positive, one could also take $-\Delta + Identity$. This choice is sometimes made to preserve reality. – Helge Aug 28 '11 at 7:59
On another note: It seems to me that this question is answered to the authors satisfaction? Is there some way to stop from making it come back to the top? – Helge Aug 28 '11 at 8:00

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