Cook domain criterion

[Resolvent algebra on the exact domains] For nonreal $z,w$ and $x,y\in \mathcal{H}$, put $u=R_{z}x$, $v=R_{\bar{z}}y\in D(C)$. Symmetry gives $⟨R_{z}x,y⟩=⟨u,(C-\bar{z})v⟩=⟨(C-z)u,v⟩=⟨x,R_{\bar{z}}y⟩$, hence $R_z^{\ast }=R_{\bar{z}}$. Applying the two inverse identities to $R_{w}x\in D(C)$ gives $R_{z}x-R_{w}x=(z-w)R_z{R}_{w}x$; interchanging $z,w$ gives the same left side as $(z-w)R_w{R}_{z}x$, so the resolvents commute (also trivially when $z=w$). On $D(C)$, $C{R}_{z}y=y+z{R}_{z}y=R_{z}Cy$, and on all $\mathcal{H}$, $C{R}_z=I+z{R}_z$ is bounded. Finally closedness follows directly from the adjoint definition: if $u_j\to u$ and $C{u}_j\to v$, then $⟨Cy,u⟩=⟨y,v⟩$ for every $y\in D(C)$, so $u\in D(C^{\ast })=D(C)$ and $Cu=v$. These arguments apply to either $C=A$ or $C=A_0$.

[Actual arbitrary-H autonomous construction] For either $C=A$ or $C=A_0$ and integers $n\ge 1$, define $J_n=n^2{R}_{in}{R}_{-in}$ and the bounded self-adjoint $C_n=C{J}_n=(n^2\mathrm{/}2)(R_{in}+R_{-in})$. Resolvent algebra gives $\mathrm{\parallel }{J}_n\mathrm{\parallel }\le 1$, $\mathrm{\parallel }{C}_n\mathrm{\parallel }\le n$ and, for $y\in D(C)$, $(I-J_n)y=\frac{1}{2}(R_{in}+R_{-in})Cy$, so $\mathrm{\parallel }(I-J_n)y\mathrm{\parallel }\le \mathrm{\parallel }Cy\mathrm{\parallel }\mathrm{/}n$. Density gives $J_n\to I$ strongly on $\mathcal{H}$, and $C_{n}y=J_{n}Cy\to Cy$. The bounded power-series exponentials $E_n(t)=e^{-it{C}_n}$ are unitary, commute with one another and with resolvents, and satisfy $\mathrm{\parallel }(E_n(t)-E_m(t))y\mathrm{\parallel }\le \mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }(C_n-C_m)y\mathrm{\parallel }$ by the bounded Duhamel identity and commutation. Thus they converge strongly, uniformly on compact time intervals, first on $D(C)$ and then by density on all $\mathcal{H}$. The limits $E_C(t)$ are a strongly continuous unitary group: products pass through strong limits of norm-one factors, the opposite-time products are $I$, and each limit preserves norms. Each bounded approximation preserves $D(C)$ and commutes with $C$ there, as follows by graph-norm convergence of its power series and resolvent commutation. Closedness therefore gives $E_C(t)y\in D(C)$ and $C{E}_C(t)y=E_C(t)Cy$. Passing to the bounded integral identities gives $E_C(t)y=y-i{\int }_0^t{E}_C(r)Cydr$, hence $E_C^{\mathrm{\prime }}(t)y=-iC{E}_C(t)y$. In particular, both $E_C(t)y$ and $C{E}_C(t)y$ are norm-continuous: this is actual graph continuity, not domain preservation inferred solely from a strong limit. Write $E_C(t)=e^{-itC}$.

[Exact generator converse by adjoint testing] If ${\lim }_{h\to 0}(e^{ihC}x-x)\mathrm{/}h=g$ exists strongly, then for every $v\in D(C)$, unitarity and step 2.1 give $⟨g,v⟩={\lim }_{h\to 0}⟨x,(e^{-ihC}v-v)\mathrm{/}h⟩=i⟨x,Cv⟩$. Thus $⟨Cv,x⟩=⟨v,-ig⟩$ for every $v\in D(C)$. The exact definition of the unbounded adjoint implies $x\in D(C^{\ast })=D(C)$ and $Cx=-ig$. Conversely step 2.1 gives the derivative $iCx$ for every $x\in D(C)$. This proves the precise generator-domain characterization without spectral Fatou arguments, Stone premises or separability.

[Moving-domain vector differentiation] Fix $\psi \in D$ and put $x(t)=e^{-it{A}_0}\psi$. Step 2.1 gives $x\in C^1(\mathbb{R};\mathcal{H})$, $x^{\mathrm{\prime }}(t)=-i{A}_{0}x(t)$, $x(t)\in D(A_0)\subseteq D(A)$, and norm-continuity of $A_{0}x(t)$. The given difference curve gives $Ax(t)=A_{0}x(t)+b_{\psi }(t)$, hence actual continuity of $x(t)$ in the graph norm of $A$. For the product derivative, use the identity $F(t+h)\psi -F(t)\psi =e^{i(t+h)A}[x(t+h)-x(t)]+e^{itA}[e^{ihA}-I]x(t)$. After division by $h$, the first term tends to $e^{itA}{x}^{\mathrm{\prime }}(t)$ by strong continuity and the uniform norm-one bound; the second tends to $i{e}^{itA}Ax(t)$ by the autonomous derivative on the fixed vector $x(t)\in D(A)$. Therefore $F^{\mathrm{\prime }}(t)\psi =i{e}^{itA}{b}_{\psi }(t)$. This is norm-continuous: subtract its value at a fixed $u$, use the norm-continuity of $b_{\psi }$ for the moving-vector term and strong continuity of the group on the fixed vector $b_{\psi }(u)$ for the remaining term. No general rule about differentiating an unbounded operator on an unspecified moving domain is invoked.

[Both Cook limits and both tail estimates] For every pair of real numbers $r_1<r_2$, step 3.2 and [F2] give $F(r_2)\psi -F(r_1)\psi =i{\int }_{r_1}^{r_2}{e}^{irA}{b}_{\psi }(r)dr$ and hence $\mathrm{\parallel }F(r_2)\psi -F(r_1)\psi \mathrm{\parallel }\le {\int }_{r_1}^{r_2}\mathrm{\parallel }{b}_{\psi }(r)\mathrm{\parallel }dr$. The positive tail is Cauchy by its assumed integrability; completeness gives its limit, which can be constructed from the integer times and then identified with all real times by this estimate. Fixing $r_1=s\ge 0$ and taking $r_2\to +\mathrm{\infty }$ yields the stated positive tail bound. On the negative half-line the same displayed identity holds, without changing the free-group sign. Negative-tail integrability makes it Cauchy as both $r_1,r_2\to -\mathrm{\infty }$; the integer sequence $-n$ and the same estimate give the full negative limit. Fixing $r_2=-s$ and taking $r_1\to -\mathrm{\infty }$ gives exactly the asserted negative tail bound. Linearity of both limits follows from linearity of every $F(t)$, and their norms equal $\mathrm{\parallel }\psi \mathrm{\parallel }$ because $F(t)$ is unitary.

[Unique extension and full strong limits without energy-indexed choice] For each fixed $\eta \in \mathcal{H}$, density of $D$ and unitarity give $\mathrm{\parallel }(F(t)-F(u))\eta \mathrm{\parallel }\le 2\mathrm{\parallel }\eta -\psi \mathrm{\parallel }+\mathrm{\parallel }(F(t)-F(u))\psi \mathrm{\parallel }$ for every $\psi \in D$. Given a tolerance, choose a single sufficiently close $\psi$ and then use step 4.1; this proves Cauchyness on both full tails for $\eta$. Integer-time completeness and the same estimate give unique limits ${\mathrm{\Omega }}_{\pm }\eta$ over all real times. This construction uses uniquely specified limits for each vector, with no simultaneous arbitrary-index choice of approximating sequences. They are linear and norm-preserving by passage from the linear unitary $F(t)$, hence bounded isometries extending their values on $D$. Any bounded extension agreeing on dense $D$ is identical. Polarization gives $⟨{\mathrm{\Omega }}_{\pm }\eta ,{\mathrm{\Omega }}_{\pm }\zeta ⟩=⟨\eta ,\zeta ⟩$, so their actual bounded adjoints satisfy ${\mathrm{\Omega }}_{\pm }^{\ast }{\mathrm{\Omega }}_{\pm }=I$. They are injective because they preserve norms. If ${\mathrm{\Omega }}_{\pm }{\eta }_n$ converges, isometry makes ${\eta }_n$ Cauchy, and completeness followed by continuity gives its limit in the range; the range is therefore closed.

[Intertwining and exact domain on the wave range] For fixed real $\tau$ and either time direction, the identity $F(t)e^{-i\tau A_0}=e^{-i\tau A}F(t+\tau )$ follows from the two group laws. Step 5.1's full strong limits are unchanged by a fixed time shift, so they give ${\mathrm{\Omega }}_{\pm }{e}^{-i\tau A_0}=e^{-i\tau A}{\mathrm{\Omega }}_{\pm }$. If $\eta \in D(A_0)$, differentiate the group identity on its free side: the derivative of $e^{itA}{\mathrm{\Omega }}_{\pm }\eta$ exists and is $i{\mathrm{\Omega }}_{\pm }{A}_0\eta$. Step 3.1 therefore implies ${\mathrm{\Omega }}_{\pm }\eta \in D(A)$ and $A{\mathrm{\Omega }}_{\pm }\eta ={\mathrm{\Omega }}_{\pm }{A}_0\eta$. Conversely, if ${\mathrm{\Omega }}_{\pm }\eta \in D(A)$, multiply the group identity by ${\mathrm{\Omega }}_{\pm }^{\ast }$ to obtain $e^{it{A}_0}\eta ={\mathrm{\Omega }}_{\pm }^{\ast }{e}^{itA}{\mathrm{\Omega }}_{\pm }\eta$. Its derivative exists by boundedness of ${\mathrm{\Omega }}_{\pm }^{\ast }$ and the interacting-domain derivative, so step 3.1 gives $\eta \in D(A_0)$. This proves $D(A)\cap \mathrm{ran}{\mathrm{\Omega }}_{\pm }={\mathrm{\Omega }}_{\pm }D(A_0)$, with no onto or completeness assumption. The original existence, both tails, isometry, extension and closed-range claims and the specified consuming domain argument follow.