Bounded-commutator graph-domain Schrödinger response

Use the graph norm on $D(H)$ fixed in the Statement. If $({\psi }_n)$ is Cauchy in this norm, then ${\psi }_n\to \psi$ and $H{\psi }_n\to \xi$ in $\mathcal{H}$; closedness of $H$ from [F3] gives $\psi \in D(H)$ and $H\psi =\xi$, so the graph norm is complete.

For $B_t=U_0(t{)}^{\ast }B{U}_0(t)$ and $\psi \in D(H)$, invariance and commutation in [F2] give $H{B}_t\psi =B_{t}H\psi +U_0(t{)}^{\ast }C{U}_0(t)\psi$. Hence $B_t$ is uniformly bounded on the graph space by a constant depending only on $\mathrm{\parallel }B\mathrm{\parallel },\mathrm{\parallel }C\mathrm{\parallel },E_0$. To check graph-norm continuity at $t_0$, the Hilbert-norm part follows from strong continuity of $U_0$; for the $H$-part use the displayed identity and continuity of $t\mapsto B_{t}H\psi$ and $t\mapsto U_0(t{)}^{\ast }C{U}_0(t)\psi$. The latter follows from $\mathrm{\parallel }{U}_0(t{)}^{\ast }C{U}_0(t)\psi -U_0(t_0{)}^{\ast }C{U}_0(t_0)\psi \mathrm{\parallel }\le \mathrm{\parallel }C\mathrm{\parallel }\mathrm{\parallel }{U}_0(t)\psi -U_0(t_0)\psi \mathrm{\parallel }+\mathrm{\parallel }(U_0(t{)}^{\ast }-U_0(t_0{)}^{\ast })C{U}_0(t_0)\psi \mathrm{\parallel },$ whose two terms tend to zero by strong continuity of $U_0$ and the group law.

Apply [F4] on the graph Banach space with coefficient $K_t=ih(t)B_t\mathrm{/}\mathrm{\hslash }$; its Volterra series defines $W_{ϵ}(t)$ on $D(H)$. For any $y\in D(H)$, solve the reversed-time graph equation $z^{\mathrm{\prime }}(s)=-ϵK_{t-s}z(s)$ with $z(0)=y$ on $[0,t]$; then $x(s)=z(t-s)$ solves the forward equation and $W_{ϵ}(t)x(0)=y$. Hence $W_{ϵ}(t)D(H)=D(H)$. On the Hilbert space the same coefficient is skew-adjoint, and the Hilbert clause of [F4] gives the unitary propagator used in the Statement; uniqueness identifies both constructions. The domain-preservation conclusion therefore uses the full-Choice spectral group hypothesis [A1] together with [F2], [F4] and the commutator condition.

For $\psi \in D(H)$, the graph-space Volterra solution makes $W_{ϵ}(t)\psi$ continuous and differentiable in graph norm. Also $U_0(t)$ acts continuously on $D(H)$ in graph norm because it is strongly continuous on both $\psi$ and $H\psi$ and commutes with $H$ there. Thus $\mathrm{\Psi }\in C([0,T];D(H))$ in graph norm, and the product rule gives $\mathrm{\Psi }\in C^1([0,T];\mathcal{H})$ with $\mathrm{\Psi }(0)=\psi$. Using $U_0(t)B_t=B{U}_0(t)$ and $i\mathrm{\hslash }{W}_{ϵ}^{\mathrm{\prime }}=-ϵh(t)B_t{W}_{ϵ}$, we obtain $i\mathrm{\hslash }{\mathrm{\Psi }}^{\mathrm{\prime }}(t)=(H-ϵh(t)B)\mathrm{\Psi }(t)$ for every $t\in [0,T]$, with one-sided derivatives at the endpoints. For each fixed $t$, $H-ϵh(t)B$ is self-adjoint on $D(H)$: it is symmetric, and if $y\in D((H-ϵh(t)B{)}^{\ast })$, subtracting the bounded $B$ term in the adjoint functional shows $y\in D(H^{\ast })=D(H)$, proving equality of the adjoint domain.

The trace-class response assertion is exactly [F1] under its stated density, stationarity, bounded-observable and control hypotheses; step 4.1 adds the strong Schrödinger realization only for the graph-domain class in this Statement. The two conclusions therefore retain their separate hypotheses.