Compact-support asymptotic completeness

Let ${\mathcal{R}}_{\pm }=\mathrm{ran}{\mathrm{\Omega }}_{\pm }$. These ranges are closed because ${\mathrm{\Omega }}_{\pm }$ are isometries. Intertwining with both positive and negative group times makes each range reducing. By [F1], the scalar spectral measure of ${\mathrm{\Omega }}_{\pm }u$ equals the free scalar measure of $u$, which is absolutely continuous. Thus ${\mathcal{R}}_{\pm }\subseteq {\mathcal{H}}_{\mathrm{a}\mathrm{c}}(H)$. Their closed linear span $\mathcal{R}=\overline{{\mathcal{R}}_{+}+{\mathcal{R}}_{-}}$ is also reducing and lies in the closed absolutely continuous subspace. Let $Q$ be the orthogonal projection onto ${\mathcal{R}}^{\perp }$; it commutes with $E_H(B)$ and $f(H)$. Since ${\mathcal{H}}_{\mathrm{p}\mathrm{p}}\perp \mathcal{R}$, $Q$ is the identity on ${\mathcal{H}}_{\mathrm{p}\mathrm{p}}$ and preserves ${\mathcal{H}}_{\mathrm{c}}$.

The negative spectral subspace is pure point, by the following concrete compact-resolvent argument. For a self-adjoint operator with the present separable PVM, a spectral point $\lambda$ has an infinite-rank spectral neighborhood at every radius exactly when there is a weakly null unit sequence $u_n\in D(A)$ with $\mathrm{\parallel }(A-\lambda )u_n\mathrm{\parallel }\to 0$. In the forward direction for each integer $n\ge 1$, choose an orthonormal vector in $E_A((\lambda -1\mathrm{/}n,\lambda +1\mathrm{/}n))\mathcal{H}$ perpendicular to the previous ones: project the fixed countable dense sequence into that closed subspace and select its first nonzero member, then normalize. In the reverse direction, a finite-rank $E_{\epsilon }$ would imply $E_{\epsilon }{u}_n\to 0$ and $\mathrm{\parallel }(I-E_{\epsilon })u_n\mathrm{\parallel }\le \mathrm{\parallel }(A-\lambda )u_n\mathrm{\parallel }\mathrm{/}\epsilon \to 0$, contradicting unit norm. With $K=(H-i{)}^{-1}-(H_0-i{)}^{-1}$ compact, transfer such a sequence for $A=H_0$ to $v_n=(H-i{)}^{-1}(H_0-i)u_n=u_n+K(H_0-i)u_n$: the compact term tends to zero, and $(H-\lambda )v_n=(H_0-\lambda )u_n+(i-\lambda )(v_n-u_n)\to 0$. The same transfer in reverse gives equality of the sets of spectral points having only infinite-rank neighborhoods. Free momentum multiplication has this set $[0,\mathrm{\infty })$, since every positive shell or ball near zero contains infinitely many disjoint positive-measure momentum subsets, and its projections vanish below zero. Hence every negative spectral point of $H$ has a finite-rank neighborhood. The restriction of $H$ to such a bounded neighborhood is a finite-dimensional self-adjoint matrix and its spectral support is a finite set of eigenvalues. Negative spectral points are therefore isolated eigenvalues of finite multiplicity, countable by assigning a distinct least rational isolating interval. Outside their union below zero, every point has a zero-projection neighborhood; rational subintervals provide a countable cover. Countable PVM additivity consequently shows that $E_H((-\mathrm{\infty },0))\mathcal{H}$ is the closed span of these negative eigenspaces.

The singleton spectral projection $E_H(\{0\})$ has range exactly $\ker H$: its vectors belong to $D(H)$ because their spectral measures are supported at zero, and the PVM integral defining $H$ vanishes there; conversely an eigenvector at zero is supported on that singleton. Thus this whole threshold atom, which may be nonzero, is pure point; no absence of a zero eigenvalue or zero resonance is assumed. Together with step 1.2, this gives ${\mathcal{H}}_{\mathrm{c}}(H)\subseteq E_H((0,\mathrm{\infty }))\mathcal{H}$. For integers $n\ge 1$, the positive compact intervals $J_n=[1\mathrm{/}n,n]$ exhaust every vector in that continuous subspace by PVM countable continuity.

Fix a compact interval $J\subset (0,\mathrm{\infty })$ and choose a real $f\in C_c^{\mathrm{\infty }}((0,\mathrm{\infty }))$ equal to one on $J$. Put $f_0=f(H_0)$. By [F2], $C_{+}+C_{-}=({\mathrm{\Omega }}_{+}{f}_0{P}_{+}+{\mathrm{\Omega }}_{-}{f}_0{P}_{-})-f_0$, using $P_{+}+P_{-}=I$ from [F3]; no commutation of $P_{\pm }$ with $f_0$ is needed. Step 1.1 gives $Q{\mathrm{\Omega }}_{\pm }=0$, hence $Q{f}_0=-Q(C_{+}+C_{-})$ is compact. Adding the compact $Q(f(H)-f_0)$ gives compact $Qf(H)$. Consequently $Q{E}_H(J)=Qf(H)E_H(J)$ is compact; since its two projection factors commute, it is an orthogonal projection. Its range is finite-dimensional, reduces $H$, and has spectrum in the bounded interval $J$, so it is spanned by eigenvectors of $H$ and lies in ${\mathcal{H}}_{\mathrm{p}\mathrm{p}}$. For $u\in {\mathcal{H}}_{\mathrm{c}}$, its vector $Q{E}_H(J)u$ also lies in ${\mathcal{H}}_{\mathrm{c}}$ by step 1.1 and [F4], and therefore is zero. It follows that $E_H(J)u\in \ker Q=\mathcal{R}\subseteq {\mathcal{H}}_{\mathrm{a}\mathrm{c}}$. Exhaustion by step 2.1 and closedness of ${\mathcal{H}}_{\mathrm{a}\mathrm{c}}$ give ${\mathcal{H}}_{\mathrm{c}}\subseteq {\mathcal{H}}_{\mathrm{a}\mathrm{c}}$; the reverse inclusion follows from orthogonality to eigenspaces. Thus ${\mathcal{H}}_{\mathrm{c}}={\mathcal{H}}_{\mathrm{a}\mathrm{c}}$. An atomless singular vector is orthogonal to every eigenspace, hence belongs to ${\mathcal{H}}_{\mathrm{c}}$, and its null carrier makes it orthogonal to ${\mathcal{H}}_{\mathrm{a}\mathrm{c}}$; it must therefore be zero. Thus ${\mathcal{H}}_{\mathrm{s}\mathrm{c}}=\{0\}$. This step establishes spectral type and joint coverage only; it does not yet establish either individual range equality.

To prove outgoing completeness individually, let $u\in {\mathcal{H}}_{\mathrm{a}\mathrm{c}}\cap {\mathcal{R}}_{+}^{\perp }$. Both subspaces reduce the PVM, so $u_J=E_H(J)u$ still belongs to this intersection for every compact positive interval $J$. Choose the same real smooth $f=1$ on $J$ and first work with $\psi =u_J$, so $f(H)\psi =\psi$. Let ${\psi }_t=e^{-itH}\psi$; then $f(H){\psi }_t={\psi }_t$, ${\psi }_t\perp {\mathcal{R}}_{+}$, and $\mathrm{\parallel }{\psi }_t\mathrm{\parallel }=\mathrm{\parallel }\psi \mathrm{\parallel }$. By [F5], ${\psi }_t$ is weakly null as $t\to +\mathrm{\infty }$. Compactness [F2] and [F4] imply $\mathrm{\parallel }{\psi }_t-f_0{\psi }_t\mathrm{\parallel }=\mathrm{\parallel }(f(H)-f_0){\psi }_t\mathrm{\parallel }\to 0$ and $\mathrm{\parallel }{C}_{\pm }{\psi }_t\mathrm{\parallel }\to 0$. Thus, using the exact identity of step 3.1, $\mathrm{\parallel }\psi {\mathrm{\parallel }}^2={\lim }_{t\to +\mathrm{\infty }}⟨{\psi }_t,{\mathrm{\Omega }}_{+}{f}_0{P}_{+}{\psi }_t+{\mathrm{\Omega }}_{-}{f}_0{P}_{-}{\psi }_t⟩\mathrm{.}$ The first summand is zero by ${\psi }_t\perp {\mathcal{R}}_{+}$. For the second, adjoints give the bound $\mathrm{\mid }⟨{\psi }_t,{\mathrm{\Omega }}_{-}{f}_0{P}_{-}{\psi }_t⟩\mathrm{\mid }\le \mathrm{\parallel }{P}_{-}{f}_0^{\ast }{\mathrm{\Omega }}_{-}^{\ast }{\psi }_t\mathrm{\parallel }\mathrm{\parallel }\psi \mathrm{\parallel }$. The adjoint intertwining identity is ${\mathrm{\Omega }}_{-}^{\ast }{e}^{-itH}=e^{-it{H}_0}{\mathrm{\Omega }}_{-}^{\ast }$, obtained by taking adjoints of [F1] at time $-t$. Hence the norm in this bound is $\mathrm{\parallel }{P}_{-}\bar{f}(H_0)e^{-it{H}_0}{\mathrm{\Omega }}_{-}^{\ast }\psi \mathrm{\parallel }$, which tends to zero by the incoming opposite-time strong limit in [F3], applied to the fixed vector ${\mathrm{\Omega }}_{-}^{\ast }\psi$. Therefore $\psi =0$. Taking $J=J_n$ and using step 2.1 proves $u=0$. Since the closed range ${\mathcal{R}}_{+}$ lies in ${\mathcal{H}}_{\mathrm{a}\mathrm{c}}$, its orthogonal complement there being zero proves ${\mathcal{R}}_{+}={\mathcal{H}}_{\mathrm{a}\mathrm{c}}$.

For incoming completeness take $u\in {\mathcal{H}}_{\mathrm{a}\mathrm{c}}\cap {\mathcal{R}}_{-}^{\perp }$, localize exactly as in step 4.1, and let $t\to -\mathrm{\infty }$. The ${\mathrm{\Omega }}_{-}$ summand is now zero by orthogonality. Moving the adjoint in the remaining ${\mathrm{\Omega }}_{+}$ term bounds it by $\mathrm{\parallel }{P}_{+}\bar{f}(H_0)e^{-it{H}_0}{\mathrm{\Omega }}_{+}^{\ast }\psi \mathrm{\parallel }\mathrm{\parallel }\psi \mathrm{\parallel }$, which tends to zero by the outgoing opposite-time strong limit in [F3] for the fixed vector ${\mathrm{\Omega }}_{+}^{\ast }\psi$. The weak-null and compact-error arguments hold at negative time by [F5]. Therefore $\psi =0$ on every positive compact spectral interval, exhaustion gives $u=0$, and ${\mathcal{R}}_{-}={\mathcal{H}}_{\mathrm{a}\mathrm{c}}={\mathcal{H}}_{\mathrm{c}}$. This proves (a) and (b) without a stationary-transform argument or an inference of individual completeness from joint coverage.

By steps 4.1–5.1 the isometries ${\mathrm{\Omega }}_{\pm }$ have the same range ${\mathcal{H}}_{\mathrm{a}\mathrm{c}}$. Let $P_{\mathrm{a}\mathrm{c}}$ be its orthogonal projection. Then ${\mathrm{\Omega }}_{\pm }^{\ast }{\mathrm{\Omega }}_{\pm }=I$ and ${\mathrm{\Omega }}_{\pm }{\mathrm{\Omega }}_{\pm }^{\ast }=P_{\mathrm{a}\mathrm{c}}$, so $S^{\ast }S={\mathrm{\Omega }}_{-}^{\ast }{P}_{\mathrm{a}\mathrm{c}}{\mathrm{\Omega }}_{-}=I$ and $S{S}^{\ast }={\mathrm{\Omega }}_{+}^{\ast }{P}_{\mathrm{a}\mathrm{c}}{\mathrm{\Omega }}_{+}=I$. Finally [F1] and its adjoint identity give $S{e}^{-it{H}_0}={\mathrm{\Omega }}_{+}^{\ast }{e}^{-itH}{\mathrm{\Omega }}_{-}=e^{-it{H}_0}S$ for every real $t$. This proves (c).