Bounded Dyson convergence and temporal factorization

If $a=b$, every integral is zero, $U_0=I$, and $U_n=0$ for $n\ge 1$; all evolution and unitarity claims follow, while no $c<d$ lies in $I$, so the separated-support clause has no instance. Assume henceforth $a<b$. Suppose $U_n(r,s)\psi$ is continuous in $r$. For fixed $r_0$, $\mathrm{\parallel }V(r)U_n(r,s)\psi -V(r_0)U_n(r_0,s)\psi \mathrm{\parallel }\le M\mathrm{\parallel }{U}_n(r,s)\psi -U_n(r_0,s)\psi \mathrm{\parallel }+\mathrm{\parallel }(V(r)-V(r_0))U_n(r_0,s)\psi \mathrm{\parallel },$ which tends to zero by strong continuity of $V$. Thus the recursive integrand is continuous. Apply [F7] to $I=[a,b]$ and the Hilbert space with their norm metrics; the estimate gives uniform continuity of the integrand. On any interval with endpoints $s,t$, the midpoint tagged sums of its equal-mesh partitions are Cauchy by uniform continuity, so completeness of $\mathcal{H}$ gives a norm limit. The same modulus-of-continuity estimate shows that every tagged sum of sufficiently small mesh approaches this limit, proving existence and tag/partition independence of the Riemann integral defined in the Statement; reverse its sign when $t<s$. The tagged-sum triangle inequality and scalar Riemann-sum limit [F5] give $\mathrm{\parallel }\int f\mathrm{\parallel }\le \int \mathrm{\parallel }f\mathrm{\parallel }$. The integral as a function of its upper endpoint is continuous, since its increment has norm at most the interval length times the supremum of the continuous integrand; this supremum is finite by [F3]. This proves strong continuity of $U_{n+1}(\cdot ,s)\psi$ and closes the induction.

The identity operator has norm at most one, including when $\mathcal{H}=\{0\}$. The integral estimate then gives inductively $\mathrm{\parallel }{U}_{n+1}(t,s)\mathrm{\parallel }\le {\int }_{\min (s,t)}^{\max (s,t)}M\frac{M^n\mathrm{\mid }r-s{\mathrm{\mid }}^n}{n!}dr=\frac{M^{n+1}\mathrm{\mid }t-s{\mathrm{\mid }}^{n+1}}{(n+1)!}\mathrm{.}$ This proves the factorial bound for both orientations of the integral.

The bound in step 2.1 makes $\sum z^n{U}_n(t,s)$ absolutely and uniformly convergent in operator norm on $I^2$, with total norm at most ${\sum }_{n\ge 0}(\mathrm{\mid }z\mathrm{\mid }M(b-a){)}^n\mathrm{/}n!=e^{\mathrm{\mid }z\mathrm{\mid }M(b-a)}$ by [F4]. The recursion and this uniform bound permit termwise integration on each vector: the norm of the integral of a uniform tail is at most $(b-a)$ times its supremum norm, which tends to zero. This gives the displayed integral equation.

If $X(t,s)$ and $Y(t,s)$ are two strongly continuous solutions, then for fixed $\psi$, $D(t)=\mathrm{\parallel }(X(t,s)-Y(t,s))\psi \mathrm{\parallel }$ is bounded on $I$ by [F3] and is Riemann integrable by [F5]. It satisfies $D(t)\le \mathrm{\mid }z\mathrm{\mid }M{\int }_{\min (s,t)}^{\max (s,t)}D(r)dr$. Repeated substitution gives an $n$-fold integral over the ordered simplex between $s$ and $t$; induction using [F5] and the polynomial derivative rule [F6] evaluates its volume as $\mathrm{\mid }t-s{\mathrm{\mid }}^n\mathrm{/}n!$, so $D(t)\le C_{\psi }\frac{(\mathrm{\mid }z\mathrm{\mid }M\mathrm{\mid }t-s\mathrm{\mid }{)}^n}{n!}$ with $C_{\psi }={\sup }_{r\in I}D(r)<\mathrm{\infty }$. The right side tends to zero, so the solution is unique. To prove composition, split the equation for $U_z(t,s)$ at $r$: for $t\ge r$, it reads $U_z(t,s)=U_z(r,s)-iz{\int }_r^{t}V(q)U_z(q,s)dq$. Multiplying the equation for $U_z(t,r)$ on the right by $U_z(r,s)$ shows $U_z(t,r)U_z(r,s)$ solves this same initial-value equation and has the same value at $t=r$. The uniqueness estimate just proved, with initial time $r$, gives equality; the reversed orientation proves it for $t<r$.

For a continuous vector-valued integrand, the average over a signed interval of length $h$ converges in norm to its value at the base point: the error is bounded by the supremum of its variation on that interval. Taking difference quotients in the integral equation therefore gives ${\mathrm{\partial }}_t{U}_z(t,s)\psi =-izV(t)U_z(t,s)\psi$, with relative one-sided derivatives at the endpoints. If $u(t)=U_z(t,s)\psi$, then $\frac{\mathrm{\parallel }u(t+h){\mathrm{\parallel }}^2-\mathrm{\parallel }u(t){\mathrm{\parallel }}^2}{h}=2\mathrm{Re}⟨u(t),\frac{u(t+h)-u(t)}{h}⟩+\frac{\mathrm{\parallel }u(t+h)-u(t){\mathrm{\parallel }}^2}{h}\mathrm{.}$ The second term tends to zero and the first tends to $2\mathrm{Re}⟨u(t),-izV(t)u(t)⟩=0$ for real $z$, because self-adjointness makes $⟨u,V(t)u⟩$ real. By the continuity estimate in step 1.1, the scalar function $t\mapsto \mathrm{\parallel }u(t){\mathrm{\parallel }}^2$ has continuous derivative zero. The real fundamental theorem [F5] on each subinterval therefore makes $\mathrm{\parallel }u(t){\mathrm{\parallel }}^2$ constant, so $U_z(t,s)$ is an isometry.

By the composition law in step 4.1, $U_z(t,s)U_z(s,t)=U_z(t,t)=I_{\mathcal{H}}$ and $U_z(s,t)U_z(t,s)=U_z(s,s)=I_{\mathcal{H}}$. Hence $U_z(t,s)$ is surjective. Step 4.2 shows that it is an isometry when $z$ is real. By the unitary convention stated above, $U_z(t,s)$ is unitary.

Let $Y(t)=U_z^{V_1}(t,a)U_z^{V_2}(t,a)$. On $[a,c]$, $U_z^{V_1}(t,a)=I$, so the integral equation for $U_z^{V_2}$ is the equation for $Y$ with generator $V_1+V_2$. On $[c,d]$, both generators vanish on the open gap and, by strong continuity, at its endpoints; thus $Y(t)=U_z^{V_2}(c,a)$ is constant and the same equation continues across the gap. On $[d,b]$, $U_z^{V_2}(t,a)=U_z^{V_2}(c,a)$, while the integral equation for $U_z^{V_1}$ reduces to $U_z^{V_1}(t,a)=I-iz{\int }_d^t{V}_1(q)U_z^{V_1}(q,a)dq\mathrm{.}$ Right multiplication by the fixed bounded operator $U_z^{V_2}(c,a)$ commutes with the vector integral by its tagged-sum definition, so $Y$ satisfies the equation for $V_1+V_2$ on this interval too. The three interval equations agree at their endpoints and combine to the full integral equation on $I$; uniqueness from step 4.1 gives the stated factorization at $b$. Both sides have operator-norm-convergent power series for every $z$. Their equality for real $z$ forces equality of coefficients: a first differing coefficient would remain nonzero after division by its least power of $z$ and passage to $z\to 0$.