Dyson truncation bound

[The first remainder and recursive reduction] The supplier's Volterra identity gives $R_0(t,s)=U(t,s)-I_H={\int }_s^{t}A(r)U(r,s)dr$. For $N\ge 1$, subtract the finite sum of recursions ${\sum }_{n=1}^N{U}_n(t,s)={\int }_s^{t}A(r){\sum }_{n=0}^{N-1}{U}_n(r,s)dr$ from that identity. Integral linearity yields $R_N(t,s)={\int }_s^{t}A(r)R_{N-1}(r,s)dr$, for both $t\ge s$ and $t<s$. This does not interchange an infinite sum with an integral. Each integrand is norm-continuous by [F1] and [F3], so the integrals exist by [F2].

[Exact nesting and the residual endpoint] Apply the recursive identity in step 1.1 successively $N$ times and then insert its formula for $R_0$. Bounded left multiplication passes through each integral by [F2], giving the universal nested expression in (a). At its last variable $r_{N+1}$, the remaining factor is exactly $U(r_{N+1},s)$, because the inserted equation is $R_0(r_N,s)={\int }_s^{r_N}A(r_{N+1})U(r_{N+1},s)d{r}_{N+1}$. For $s\le t$, $s\le r_{N+1}\le \cdots \le r_1\le t$, so renaming $t_j=r_{N+2-j}$ puts the product in latest-time-left order and the residual factor at $t_1$. The asserted simplex means precisely these nested integrals, so no unproved Banach-valued product-measure Fubini theorem is needed. This proves the whole identity (a), including $N=0$.

[Sign and factorial volume for either time direction] Put $m=N+1$ and $D=t-s$. In step 2.1 substitute $r_j=s+D{\xi }_j$ successively. The remainder becomes $D^m{\int }_0^{1}d{\xi }_1{\int }_0^{{\xi }_1}d{\xi }_2\cdots {\int }_0^{{\xi }_{m-1}}d{\xi }_{m}A(s+D{\xi }_1)\cdots A(s+D{\xi }_m)U(s+D{\xi }_m,s)$. This is valid for $D<0$: every integral is oriented and contributes its signed $D$, so the overall sign is $D^m$. For $D>0$ the outermost $r_1$ is the latest physical time and the innermost $r_m$ is earliest; for $D<0$ these physical inequalities reverse, while the recursive operator order stays unchanged. Norm estimates take $\mathrm{\mid }D{\mathrm{\mid }}^m$. The repeated integral of the constant $1$ over $0\le {\xi }_m\le \cdots \le {\xi }_1\le 1$ is $1\mathrm{/}m!$, by integrating one power at each step. At $D=0$ every nested integral and every $R_N$ vanish.

[Both explicit remainder estimates] For every variable in step 3.1, $r_m$ lies between $s$ and $t$, so $\mathrm{\mid }{r}_m-s\mathrm{\mid }\le \tau$. Using [F1], [F2], [F3] and the volume calculation there gives $\mathrm{\parallel }{R}_N(t,s)\mathrm{\parallel }\le {\tau }^m{K}^m{e}^{K\tau }\mathrm{/}m!$. This is (b)'s retained exponential estimate. Using instead the actual supplier's unitarity bound $\mathrm{\parallel }U(r_m,s)\mathrm{\parallel }\le 1$ gives $\mathrm{\parallel }{R}_N(t,s)\mathrm{\parallel }\le (K\tau {)}^m\mathrm{/}m!$. Self-adjointness and unitarity are not new assumptions: they are exactly the family and conclusion in [F1]. The first estimate can also be derived from its factorial series bounds alone, so no sharper norm hypothesis is needed to retain the original bound.

[Scaling and boundary cases] Choose $N=1$ in step 4.1 for the first-order error estimate. A real multiple $\lambda \mathcal{H}$ has the same self-adjoint/norm-continuous class, and its norm bound is $K_{\lambda }=\mathrm{\mid }\lambda \mathrm{\mid }K$. Substituting that bound gives the displayed scaling estimates; when $\mathrm{\mid }\lambda \mathrm{\mid }K\tau \le M$, the original estimate is at most $e^M\mathrm{\mid }\lambda {\mathrm{\mid }}^2{K}^2{\tau }^2\mathrm{/}2$, proving the qualified quadratic assertion with its explicit constant. The $t=s$ case was proved in step 3.1; if $K=0$ the coefficient vanishes identically and all positive-order terms and remainders vanish; if $H=\{0\}$ all operators are zero; and if $a=b$ the sole time pair is diagonal. No construction distinguished positive from negative absolute time. The proof uses only the actual CC bounded Dyson supplier, finite nesting and the Bochner interfaces, with no choice of a Hilbert basis or new AC/spectral/Stone assumption. This completes all the stated claims.