Norm Dyson series and uniqueness

[The operator Banach algebra and the bounded coefficient] For an operator-norm Cauchy sequence $(B_j)$, each $(B_j\psi )$ is Cauchy in $H$ and has a unique limit; define $B\psi ={\lim }_j{B}_j\psi$. The limit is linear and $\mathrm{\parallel }B\psi \mathrm{\parallel }\le M\mathrm{\parallel }\psi \mathrm{\parallel }$ for $M={\sup }_j\mathrm{\parallel }{B}_j\mathrm{\parallel }<\mathrm{\infty }$. For sufficiently large $j,k$, $\mathrm{\parallel }(B_j-B_k)\psi \mathrm{\parallel }\le \epsilon \mathrm{\parallel }\psi \mathrm{\parallel }$; letting $k\to \mathrm{\infty }$ gives $\mathrm{\parallel }{B}_j-B\mathrm{\parallel }\le \epsilon$. Thus $\mathcal{B}(H)$ is complete, even for nonseparable or zero-dimensional $H$. This construction uses unique limits for all $\psi$, not a choice of a basis or of unrelated representatives. Norm continuity on compact $I$ makes $K$ finite. The bounded adjoint construction in [F4] applies on arbitrary $H$; $A(t{)}^{\ast }=-A(t)$ follows from self-adjointness of $\mathcal{H}(t)$.

[Oriented nested integrals and their fixed parameter region] For $D=t-s$ and $n\ge 1$, substitution in the defining nested integrals gives $U_n(t,s)=D^n{\int }_0^{1}d{\xi }_1{\int }_0^{{\xi }_1}d{\xi }_2\cdots {\int }_0^{{\xi }_{n-1}}d{\xi }_{n}A(s+D{\xi }_1)\cdots A(s+D{\xi }_n)$. This formula follows by replacing each $r_j$ with $s+D{\xi }_j$ successively; it remains valid for $D<0$ because each oriented differential contributes its signed factor $D$. The integrands are continuous and bounded on the fixed closed region $0\le {\xi }_n\le \cdots \le {\xi }_1\le 1$, and their successive integrals exist by [F2]. Thus every $U_n$ is defined. For $D\ge 0$ relabel $t_j=r_{n+1-j}$ to obtain exactly the latest-time-left simplex in the Statement, without invoking a Banach-valued product-measure Fubini theorem. At $D=0$ all positive-order terms are zero.

[Factorial bound and joint continuity] The repeated integral of the constant $1$ over the fixed region in step 2.1 is $1\mathrm{/}n!$, by successive elementary integration. The norm inequality and submultiplicativity give $\mathrm{\parallel }{U}_n(t,s)\mathrm{\parallel }\le \mathrm{\mid }D{\mathrm{\mid }}^n{K}^n\mathrm{/}n!$; for $n=0$, $\mathrm{\parallel }{I}_H\mathrm{\parallel }\le 1$, including $H=\{0\}$. This proves (a) for both time directions. If $(s^{\mathrm{\prime }},t^{\mathrm{\prime }})\to (s,t)$, the points $s^{\mathrm{\prime }}+(t^{\mathrm{\prime }}-s^{\mathrm{\prime }})\xi$ converge uniformly for $0\le \xi \le 1$ to $s+D\xi$. Uniform continuity of $A$ on $I$ and the telescoping product identity make the products in step 2.1 converge uniformly in all ${\xi }_j$; the scalar factor $(t^{\mathrm{\prime }}-s^{\mathrm{\prime }}{)}^n$ also converges. The norm inequality through the fixed nested integrals proves joint norm continuity of each $U_n$.

[Uniform convergence and the Volterra equation] The uniform bound $(K(b-a){)}^n\mathrm{/}n!$ has summable series, so completeness from step 1.1 gives a uniformly operator-norm convergent sum $U$ on $I^2$. The sum is norm-continuous, and summing the positive-order bounds gives $\mathrm{\parallel }U-I_H\mathrm{\parallel }\le e^{K\mathrm{\mid }t-s\mathrm{\mid }}-1$. The terms at $t=s$ give $U(s,s)=I_H$. Summing the defining recursion and passing the uniform limit through its oriented integral by [F3] gives $U(t,s)=I_H+{\int }_s^{t}A(r)U(r,s)dr$ for all $s,t$. This proves (b) and the identity in (c).

[Derivatives and uniqueness on both sides of the initial time] For fixed $s$, the integrand $A(r)U(r,s)$ is norm-continuous; [F3] applied to step 4.1 gives ${\mathrm{\partial }}_{t}U=A(t)U$ and its endpoint one-sided versions. For uniqueness with arbitrary initial $C$, let $W$ be another continuous solution and $E(t)=W(t)-U(t,s)C$. Then $E(t)={\int }_s^{t}A(r)E(r)dr$ and $M={\sup }_{t\in I}\mathrm{\parallel }E(t)\mathrm{\parallel }<\mathrm{\infty }$. Iterating this identity $N$ times along the oriented segment between $s$ and $t$, and taking norms using its fixed region from step 2.1, yields $\mathrm{\parallel }E(t)\mathrm{\parallel }\le M(K\mathrm{\mid }t-s\mathrm{\mid }{)}^N\mathrm{/}N!$. The bound tends to zero for each $t$; hence $E=0$ on all $I$. This also proves uniqueness for a bounded continuous operator coefficient with any norm bound, and for right multiplication or a sum of left/right multiplication, since these are bounded linear operators on the Banach space $\mathcal{B}(H)$.

[Isometry and surjectivity are separate deductions] Operator-norm differentiability permits the product rule and differentiation of adjoints. For fixed $s$, $G(t)=U(t,s{)}^{\ast }U(t,s)$ has $G^{\mathrm{\prime }}=U^{\ast }(A^{\ast }+A)U=0$, so [F3] gives $G=I_H$. This establishes isometry. To prove surjectivity, put $Q(t)=U(t,s)U(t,s{)}^{\ast }$. Its derivative is $Q^{\mathrm{\prime }}=A(t)Q-QA(t)$ and $Q(s)=I_H$; this derivative is not assumed to vanish by cancellation. The constant function $I_H$ solves the same equation. The linear coefficient $T_t(B)=A(t)B-BA(t)$ has norm at most $2K$ and is norm-continuous. The uniqueness argument in step 5.1, with coefficient $T_t$ and bound $2K$, shows $Q=I_H$. Thus $U$ is unitary for every $s,t$, including $t<s$.

[Cocycle, reverse propagator and lower-start derivative] Fix any $r,s\in I$. Both $t\mapsto U(t,r)$ and $t\mapsto U(t,s)U(s,r)$ satisfy $X^{\mathrm{\prime }}=A(t)X$ throughout $I$ and have value $U(s,r)$ at $t=s$, so step 5.1 gives $U(t,r)=U(t,s)U(s,r)$ for every $t$. Taking $r=t$ gives $U(t,s)U(s,t)=I_H$, and step 6.1 identifies $U(s,t)=U(t,s{)}^{\ast }$. Now fix the reference endpoint $a$. The cocycle gives $U(t,s)=U(t,a)U(s,a{)}^{\ast }$. Differentiating in $s$ using the already proved upper-time derivative of $U(s,a)$ gives ${\mathrm{\partial }}_{s}U(t,s)=U(t,a)(A(s)U(s,a){)}^{\ast }=-U(t,s)A(s)=iU(t,s)\mathcal{H}(s)\mathrm{/}\mathrm{\hslash }$, with the endpoint one-sided interpretations. This proves (d) and completes (c).

[Strong equations, domains and choice ledger] For any $\psi \in H$, applying the operator-norm difference quotients to $\psi$ gives the stated strong derivative equations and their continuity. Every $\mathcal{H}(t)$ is a bounded operator with domain all $H$, so no invariant unbounded domain is hidden in this assertion. All constructions use explicit grid simple approximants, countable series and uniquely defined limits; no basis selection, spectral representation, or Stone theorem is used for arbitrary $H$. The declared CC hypothesis is retained at the Lebesgue/Bochner and FTC interfaces [F2]–[F3], and through Hilbert Riesz for the adjoints in [F4]. If $a=b$ every oriented integral has length zero and the identities reduce to $I_H$ as stated. Steps 3.1, 4.1, 5.1 and 7.1, 6.1 and 7.1 establish all four claims.