Concrete three-dimensional Cook existence

[Exact $L^2$ potential bound] For $u\in H^2$ and $R\ge 1$, split the absolutely integrable Fourier inverse into $\mathrm{\mid }\xi \mathrm{\mid }\le R$ and $\mathrm{\mid }\xi \mathrm{\mid }>R$. Cauchy–Schwarz and ${\int }_{\mathrm{\mid }\xi \mathrm{\mid }>R}\mathrm{\mid }\xi {\mathrm{\mid }}^{-4}d\xi =4\pi \mathrm{/}R$ give $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}\le (4\pi \mathrm{/}3{)}^{1\mathrm{/}2}{R}^{3\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_2+(4\pi {)}^{1\mathrm{/}2}{R}^{-1\mathrm{/}2}\mathrm{\parallel }{H}_{0}u{\mathrm{\parallel }}_2\mathrm{/}(4{\pi }^2)$. The same bounds show the inverse is absolutely convergent and continuous. Thus $Vu\in L^2$, and for every $\eta >0$ choosing a sufficiently large explicit $R$ gives $\mathrm{\parallel }Vu{\mathrm{\parallel }}_2\le \eta \mathrm{\parallel }{H}_{0}u{\mathrm{\parallel }}_2+b_{\eta }\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Reality of $V$ gives symmetry of this multiplication pairing on $H^2$; no boundedness or compact support of $V$ is assumed.

[All-vector free density with the exact Jacobian] Fix any $\psi \in L^2$ and a Borel representative $f$ of ${\mathcal{F}}_2\psi$. Such a representative exists under CC: truncate and quantize each real and imaginary component of a Lebesgue measurable representative; each of the countably many measurable level sets differs from a Borel set by a subset of a Borel null set, by the completed Lebesgue measure definition. Choose these countably many replacements using CC. Their Borel simple functions converge to the given representative outside one Borel null set; take their pointwise limit where it exists and zero elsewhere. For every Borel $B$, the explicit free PVM gives ${\mu }_{\psi }(B)={\int }_{{\mathbb{R}}^3}{\mathbf{1}}_B(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2)\mathrm{\mid }f(\xi ){\mathrm{\mid }}^{2}d\xi$. Polar integration and the substitution $E=(2\pi r{)}^2$ give $r^{2}dr=\sqrt{E}dE\mathrm{/}[2(2\pi {)}^3]$, hence ${\mu }_{\psi }(B\cap (0,\mathrm{\infty }))={\int }_{B\cap (0,\mathrm{\infty })}{d}_{\psi }(E)dE$, where $d_{\psi }(E)=\sqrt{E}{\int }_{S^2}\mathrm{\mid }f(\sqrt{E}\mathrm{\Omega }\mathrm{/}(2\pi )){\mathrm{\mid }}^{2}d\sigma (\mathrm{\Omega })\mathrm{/}[2(2\pi {)}^3]$. Tonelli proves measurability and nonnegativity of the angular integral; the total integral is $\int \mathrm{\mid }f{\mathrm{\mid }}^2=\mathrm{\parallel }\psi {\mathrm{\parallel }}_2^2$, so it is finite for almost every $E$ and is an $L^1$ density. For any other Lebesgue measurable representative, the disagreement set lies in a Borel null set $N$. Apply the Borel polar formula and Tonelli to ${\mathbf{1}}_N$: for almost every positive energy, its spherical section has surface measure zero. On those spheres the changed representative equals the Borel one outside a subset of that null section and is measurable for completed surface measure. Thus the density is unchanged for almost every energy; exceptional energies can be assigned density zero. This justifies the formula for any measurable representative without asserting that its restriction to every sphere is measurable. This derives the displayed density for every vector directly, without only proving it on Schwartz functions or making pointwise sphere restrictions of an $L^2$ class.

[Self-adjoint common domain under CC] The free maximal multiplier is self-adjoint by [F1]. Let $R_0(z)={\mathcal{F}}_2^{\ast }{M}_{(q-z{)}^{-1}}{\mathcal{F}}_2$. For $z=\pm i\tau$, $\mathrm{\parallel }{R}_0(z)\mathrm{\parallel }\le 1\mathrm{/}\tau$ and $\mathrm{\parallel }{H}_0{R}_0(z)\mathrm{\parallel }\le 1$. Step 1.1 with $\eta =1\mathrm{/}2$ and $\tau >2b_{1\mathrm{/}2}$ gives $\mathrm{\parallel }{M}_V{R}_0(z)\mathrm{\parallel }<1$. The Neumann series inverts $I+M_V{R}_0(z)$, and the exact factorization $H-z=(I+M_V{R}_0(z))(H_0-z)$ on $H^2$ gives onto inverses $R_0(z)(I+M_V{R}_0(z){)}^{-1}$ at both signs. The densely defined symmetric $H$ is self-adjoint on precisely $H^2$: for $y\in D(H^{\ast })$, solve $(H-i\tau )x=(H^{\ast }-i\tau )y$ with $x\in H^2$; then $y-x\in \ker (H^{\ast }-i\tau )=\mathrm{ran}(H+i\tau {)}^{\perp }=\{0\}$. Hence no adjoint-domain extension remains. The same Neumann argument at $z=-\kappa$, $\kappa >2b_{1\mathrm{/}2}$, uses $\mathrm{\parallel }{H}_0{R}_0(-\kappa )\mathrm{\parallel }\le 1$, $\mathrm{\parallel }{R}_0(-\kappa )\mathrm{\parallel }\le 1\mathrm{/}\kappa$ and shows every sufficiently negative real number is in the resolvent; the self-adjoint calculus [F3] therefore also gives a lower bound. Thus the full generic $L^2$ self-adjoint interface needed here is established under CC independently of Kato–Rellich.

[All-time norm-continuity and integrability] For Schwartz $\psi$, put $u(t)=U_0(t)\psi$. The explicit multiplier gives continuous $u(t)$ in the free graph norm, with $\mathrm{\parallel }u(t){\mathrm{\parallel }}_2=\mathrm{\parallel }\psi {\mathrm{\parallel }}_2$ and $\mathrm{\parallel }{H}_{0}u(t){\mathrm{\parallel }}_2=\mathrm{\parallel }{H}_0\psi {\mathrm{\parallel }}_2$. Step 1.1 at $R=1$ bounds $\mathrm{\parallel }Vu(t){\mathrm{\parallel }}_2$ by $C\mathrm{\parallel }V{\mathrm{\parallel }}_2(\mathrm{\parallel }\psi {\mathrm{\parallel }}_2+\mathrm{\parallel }{H}_0\psi {\mathrm{\parallel }}_2)$ uniformly on $[-1,1]$, and its relative bound applied to differences makes $Vu(t)$ norm-continuous. For $\mathrm{\mid }t\mathrm{\mid }\ge 1$, [F1] gives $\mathrm{\parallel }Vu(t){\mathrm{\parallel }}_2\le \mathrm{\parallel }V{\mathrm{\parallel }}_2(4\pi \mathrm{\mid }t\mathrm{\mid }{)}^{-3\mathrm{/}2}\mathrm{\parallel }\psi {\mathrm{\parallel }}_1$. Both half-line norm integrals are finite, proving (a). This proves continuity directly in the graph norm rather than presupposing a domain product rule.

[Concrete moving-vector differentiation and Cook limits] Let $F(t)=e^{itH}{U}_0(t)$, with the CC group from [F3] and step 2.1. For fixed $t$ and $h\mathrm{\ne }0$, the exact difference quotient equals $e^{itH}\{e^{ihH}[u(t+h)-u(t)]\mathrm{/}h+[(e^{ihH}-I)\mathrm{/}h]u(t)\}$. The first term tends to $-i{H}_{0}u(t)$ by the free graph-domain derivative and strong continuity of $e^{ihH}$; the second tends to $iHu(t)$ because $u(t)\in H^2=D(H)$. Thus $\frac{d}{dt}F(t)\psi =i{e}^{itH}Vu(t)$, continuously by step 3.1. The Banach fundamental theorem and unitarity give $\mathrm{\parallel }F(t)\psi -F(s)\psi \mathrm{\parallel }\le {\int }_s^t\mathrm{\parallel }Vu(r){\mathrm{\parallel }}_{2}dr$ for every $s<t$. Both tails are Cauchy by step 3.1, so ${\mathrm{\Omega }}_{\pm }\psi$ exist. Since every $F(t)$ is unitary, the limits preserve norms and are linear on the dense Schwartz class. For arbitrary $\eta \in L^2$ choose Schwartz ${\psi }_n\to \eta$ using CC; the isometry extends by completeness, and $\mathrm{\parallel }F(t)(\eta -{\psi }_n)\mathrm{\parallel }=\mathrm{\parallel }\eta -{\psi }_n\mathrm{\parallel }$ proves $F(t)\eta \to {\mathrm{\Omega }}_{\pm }\eta$. Hence both limits are bounded linear isometries on the whole space. These are the concrete Cook consequences of the displayed hypotheses.

[Group and generator intertwining] For every real $s$ the operator identity $e^{-isH}F(t)e^{is{H}_0}=F(t-s)$ holds, so the strong limits at either sign give $e^{-isH}{\mathrm{\Omega }}_{\pm }={\mathrm{\Omega }}_{\pm }{e}^{-is{H}_0}$. If $\psi \in D(H_0)$, this identity implies that the strong derivative at $s=0$ of $e^{-isH}{\mathrm{\Omega }}_{\pm }\psi$ is $-i{\mathrm{\Omega }}_{\pm }{H}_0\psi$. For an explicit domain-converse check, write this derivative as $-i\eta$ and test against any $v\in D(H)$: moving the unitary to the second slot gives $⟨\eta ,v⟩=⟨{\mathrm{\Omega }}_{\pm }\psi ,Hv⟩$, or equivalently $⟨Hv,{\mathrm{\Omega }}_{\pm }\psi ⟩=⟨v,\eta ⟩$. Thus ${\mathrm{\Omega }}_{\pm }\psi \in D(H^{\ast })=D(H)$ and $H{\mathrm{\Omega }}_{\pm }\psi =\eta ={\mathrm{\Omega }}_{\pm }{H}_0\psi$. This proves every assertion in (b), including the actual domain inclusion.

[Zero atom and conclusion] The Fourier preimage of $\{0\}$ under $q$ is exactly the Lebesgue-null singleton $\{0\}$, so ${\mu }_{\psi }(\{0\})={\int }_{\{0\}}\mathrm{\mid }f{\mathrm{\mid }}^2=0$ for every $\psi$. Also $q\ge 0$ gives ${\mu }_{\psi }((-\mathrm{\infty },0))=0$. Combined with step 1.2, this proves absolute continuity of each entire scalar measure and $H_{\mathrm{a}\mathrm{c}}(H_0)=L^2$, including all-vector density and direct zero exclusion, rather than inferring an atom from an arbitrary pointwise density value. Mixed scalar measures follow by polarization in the stated first-variable-linear convention if needed. Assertions (a) and (b) follow from the integrability, Cook-limit and intertwining steps above; no range equality or asymptotic completeness is used or asserted.