# A.C. spectrum of the non additive perturbation BAB of a self-adjoint operator A where B is strictly positive

If have the following problem:

Let $A : \mathcal{H} \to \mathcal{H}$ be a bounded, self-adjoint operator on some Hilbert space $\mathcal{H}$. Let $B: \mathcal{H} \to \mathcal{H}$ be a bounded, strictly positive operator $B \geq c > 0$. Is the following statement true or false: $$0 \in \sigma_{ac}(A) \Leftrightarrow 0 \in \sigma_{ac}(BAB) \quad ?$$ Or can one say something about the relation of $\lim_{y \searrow 0} \Im \langle f, (A- i y)^{-1} f \rangle$ and $\lim_{y \searrow 0} \Im \langle f, (A- iB y)^{-1} f \rangle$. $A$ and $B$ do not commute.

I'm aware of a similar question: Spectrum of the operator PAP, with A self-adjoint and P strictly positive. I hope that the one I state here is a little more specific as to come up with a counterexample or a proof.

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I do not know if it helps, but at least in this case you can argue that $BAB$ is similar to $B^2A$ or $AB^2$. –  András Bátkai Aug 12 '13 at 10:38

This is false. Let $A=S+S^*$ be the free Jacobi matrix on $\ell^2$; I write $S$ for the shift $(Sy)_n = y_{n+1}$. Then $\sigma_{ac}(A)=[-2,2]$. Let $B$ be multiplication by a bounded sequence $b_n\ge c$. Then $$BAB = b(S+S^*)b = bb_+S + bb_-S^*$$ is another Jacobi matrix, and pretty much any choice of $b$ provides a counterexample as it's so easy to destroy ac spectrum of one-dimensional operators. For example, any $b$ that takes only two values and is not eventually periodic works: $\sigma_{ac}(BAB)=\emptyset$
Not very directly the way I did it, but I could also finish the argument by referring to Anderson localization (pick a random $b$). –  Christian Remling Apr 1 '14 at 19:09