Scattering isometries do not establish asymptotic completeness

By [F3], the bounded linear isometry ${\mathrm{\Omega }}_{+}$ has a bounded Hilbert adjoint. For any $z$ in a Hilbert space, $\mathrm{\parallel }z\mathrm{\parallel }={\sup }_{\mathrm{\parallel }x\mathrm{\parallel }\le 1}\mathrm{\mid }(z,x)\mathrm{\mid }$: Cauchy–Schwarz gives the upper bound, and if $z\mathrm{\ne }0$ the choice $x=z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$ attains it. For $y\in \mathcal{H}$, the adjoint identity and [F2] therefore give $\mathrm{\parallel }{\mathrm{\Omega }}_{+}^{\ast }y\mathrm{\parallel }={\sup }_{\mathrm{\parallel }x\mathrm{\parallel }\le 1}\mathrm{\mid }({\mathrm{\Omega }}_{+}^{\ast }y,x)\mathrm{\mid }={\sup }_{\mathrm{\parallel }x\mathrm{\parallel }\le 1}\mathrm{\mid }(y,{\mathrm{\Omega }}_{+}x)\mathrm{\mid }\le \mathrm{\parallel }y\mathrm{\parallel }$, since $\mathrm{\parallel }{\mathrm{\Omega }}_{+}x\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$. Therefore $\mathrm{\parallel }Sx\mathrm{\parallel }=\mathrm{\parallel }{\mathrm{\Omega }}_{+}^{\ast }{\mathrm{\Omega }}_{-}x\mathrm{\parallel }\le \mathrm{\parallel }{\mathrm{\Omega }}_{-}x\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ for every $x\in {\mathcal{H}}_{\mathrm{a}\mathrm{s}}$, so $S$ is a contraction.

By [F3], the bounded maps ${\mathrm{\Omega }}_{\pm }$ and $S$ have bounded Hilbert adjoints with the stated adjoint identities. For either isometry $\mathrm{\Omega }$, the adjoint identity gives $(x,{\mathrm{\Omega }}^{\ast }\mathrm{\Omega }y)=(\mathrm{\Omega }x,\mathrm{\Omega }y)=(x,y)$ for all $x,y$; nondegeneracy therefore gives ${\mathrm{\Omega }}^{\ast }\mathrm{\Omega }=I$. For $P=\mathrm{\Omega }{\mathrm{\Omega }}^{\ast }$ the same identity gives $(Px,y)=({\mathrm{\Omega }}^{\ast }x,{\mathrm{\Omega }}^{\ast }y)=(x,Py)$, so $P^{\ast }=P$, and ${\mathrm{\Omega }}^{\ast }\mathrm{\Omega }=I$ gives $P^2=P$. Also $P$ fixes each $\mathrm{\Omega }x$ and its image is contained in $\mathrm{Ran}\mathrm{\Omega }$, so $P$ is the orthogonal projection onto the closed range. If $\mathrm{Ran}{\mathrm{\Omega }}_{+}=\mathrm{Ran}{\mathrm{\Omega }}_{-}=:R$, both products are the same projection $P_R$. For all $x,z\in {\mathcal{H}}_{\mathrm{a}\mathrm{s}}$, $(Sx,z{)}_{\mathrm{a}\mathrm{s}}=({\mathrm{\Omega }}_{-}x,{\mathrm{\Omega }}_{+}z{)}_{\mathcal{H}}=(x,{\mathrm{\Omega }}_{-}^{\ast }{\mathrm{\Omega }}_{+}z{)}_{\mathrm{a}\mathrm{s}}$, so uniqueness of the adjoint gives $S^{\ast }={\mathrm{\Omega }}_{-}^{\ast }{\mathrm{\Omega }}_{+}$. It follows that $S^{\ast }S={\mathrm{\Omega }}_{-}^{\ast }{P}_R{\mathrm{\Omega }}_{-}={\mathrm{\Omega }}_{-}^{\ast }{\mathrm{\Omega }}_{-}=I$ and $S{S}^{\ast }={\mathrm{\Omega }}_{+}^{\ast }{P}_R{\mathrm{\Omega }}_{+}={\mathrm{\Omega }}_{+}^{\ast }{\mathrm{\Omega }}_{+}=I$, so $S$ is unitary.

Conversely, suppose $S$ is unitary and put $P_{+}={\mathrm{\Omega }}_{+}{\mathrm{\Omega }}_{+}^{\ast }$. By [F3] and step 1.2 this is a bounded orthogonal projection. For each $x$, $\mathrm{\parallel }{P}_{+}{\mathrm{\Omega }}_{-}x\mathrm{\parallel }=\mathrm{\parallel }{\mathrm{\Omega }}_{+}^{\ast }{\mathrm{\Omega }}_{-}x\mathrm{\parallel }=\mathrm{\parallel }Sx\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }{\mathrm{\Omega }}_{-}x\mathrm{\parallel }$. For an orthogonal projection $P$, the identity $\mathrm{\parallel }Py\mathrm{\parallel }=\mathrm{\parallel }y\mathrm{\parallel }$ implies $\mathrm{\parallel }y{\mathrm{\parallel }}^2=\mathrm{\parallel }Py{\mathrm{\parallel }}^2+\mathrm{\parallel }(I-P)y{\mathrm{\parallel }}^2$ and hence $(I-P)y=0$; therefore ${\mathrm{\Omega }}_{-}x\in \mathrm{Ran}{\mathrm{\Omega }}_{+}$ for every $x$, proving $\mathrm{Ran}{\mathrm{\Omega }}_{-}\subseteq \mathrm{Ran}{\mathrm{\Omega }}_{+}$. Since $S$ is onto, for every $z\in {\mathcal{H}}_{\mathrm{a}\mathrm{s}}$ choose $x$ with $Sx=z$. Then ${\mathrm{\Omega }}_{+}z={\mathrm{\Omega }}_{+}Sx=P_{+}{\mathrm{\Omega }}_{-}x={\mathrm{\Omega }}_{-}x$, which gives the reverse range inclusion. Thus the ranges coincide.