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Let $A$ be an unbounded self-adjoint operator with spectrum $\sigma(A)=\mathbb R$ in a Hilbert space $\mathcal H$. Let $P$ be a bounded operator in $\mathcal H$ satisfying $P\ge1$ and $$ {\rm Domain}(AP) \equiv\big[\varphi\in\mathcal H:P\varphi\in{\rm Domain}(A)\big] ={\rm Domain}(A). $$ Finally, suppose that the operator $$ H:=PAP,\quad{\rm Domain}(H):={\rm Domain}(A), $$ is self-adjoint in $\mathcal H$.

Question: Can we find conditions on $A$ and $P$ guaranteeing that the spectrum of $H$ is the same of that of $A$; that is, guaranteeing that $\sigma(H)=\sigma(A)=\mathbb R$ ?

Even though operators like $H$ appear in Spectral Theory, I haven't been able to find much information on their spectrum in the literature. I am only aware of the paper:

Hladnik, Milan, Omladic, Matjaz, Spectrum of the product of operators. Proc. Amer. Math. Soc. 102 (1988), no. 2, 300–302,

whose results are too general to answer the present question.

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  • $\begingroup$ Shouldn't a polar decomposition help here? $\endgroup$
    – Matthias Ludewig
    Commented Feb 2, 2012 at 15:54
  • $\begingroup$ Simplest condition $P = P^{-1}$. (So $P = g(Q)$ for $g$ taking values in $+1, -1$.) Note, this condition corresponds to preserving eigenvalues & eigenfunctions. Of course one can imagine many other reasons for the claim to hold ... in the form stated it is just TOO general to handle. $\endgroup$
    – Helge
    Commented Feb 2, 2012 at 21:17

2 Answers 2

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The only reasonable condition I can think of is that $P = g(A)$ where $g \ge 1$ is a bounded continuous function on $\mathbb R$.

It is instructive to consider the case where $A$ is the multiplication operator $(A f)(x) = x f(x)$ on $L^2({\mathbb R})$ and $P = g(A)$ for some bounded measurable function $g \ge 1$. Then $\sigma(PAP)$ is the essential range of $\{ x g(x)^2: x \in {\mathbb R}\}$. For suitable discontinuous $g$, this could be any closed subset $S$ of the plane that intersects $(-\infty,N)$, $(-1/N,0)$, $(0,1/N)$ and $(N,\infty)$ for all $N$.

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  • $\begingroup$ Unfortunately, in the example I have in mind the operators $A$ and $P$ do not commute. The typical situation would be the one where $A$ is the operator of differentiation $(Af):=-if'(x)$ in $L^2(\mathbb R)$ and $P=g(X)$ with a bounded $C^1$ function $g\ge1$ and $X$ the multiplication operator $(Xf):=xf(x)$ in $L^2(\mathbb R)$. $\endgroup$
    – Peter F
    Commented Feb 2, 2012 at 19:40
  • $\begingroup$ Have you looked at the results on perturbation of continuous spectrum in Kato, "Perturbation Theory for Linear Operators"? $\endgroup$
    – Robert Israel
    Commented Feb 3, 2012 at 17:47
  • $\begingroup$ I guess that Kato's book is indeed the best we can get. Thanks for the answer. $\endgroup$
    – Peter F
    Commented Feb 20, 2012 at 21:16
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A way to prove that the spectrum are the same, since they are equal to $\mathbb{R}$, is to prove that they have the same essential spectrum. To do so, an easy way is to ensure that $(H+i)^{-1}- (A+i)^{-1}$ is compact.

By using the resolvent formula, one sees that it is enough to suppose that $H - A$ is compact from $D(A)$ to $D(A)^*$ [recall that $H$ and $A$ have the same domain by hypothesis]. Here I used the Riesz isomorphism to identify $\mathcal{H}$ with its antidual.

Finally, since $H-A= (1-P)A + PA(1-P)$ you can suppose for instance $P\in \mathcal{B}(\mathcal{D}(A))$ (which implies $P\in \mathcal{B}(\mathcal{D}(A)^*)$ by duality) and $1-P\in \mathcal{K}(\mathcal{D}(A), \mathcal{H})$ (which implies $1-P\in \mathcal{K}(\mathcal{H}, \mathcal{D}(A)^*)$ by duality).

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