Banach-space Volterra and Dyson series with a norm remainder

If $x:[0,T]\to X$ is continuous, then $s\mapsto K(s)x(s)$ is continuous: for $s\to t$, its difference from $K(t)x(t)$ is bounded by ${\sup }_r\mathrm{\parallel }K(r)\mathrm{\parallel }\mathrm{\parallel }x(s)-x(t)\mathrm{\parallel }+\mathrm{\parallel }K(s)x(t)-K(t)x(t)\mathrm{\parallel }$, and both terms tend to zero. By [F1], the norm-Riemann integral defining ${\mathrm{\Delta }}_{n+1}$ therefore exists whenever ${\mathrm{\Delta }}_n$ is continuous. Moreover, for a continuous $f:[0,t]\to X$, the triangle inequality on each tagged Riemann sum and convergence of the scalar Riemann sums give $\mathrm{\parallel }{\int }_0^{t}f(s)ds\mathrm{\parallel }\le {\int }_0^t\mathrm{\parallel }f(s)\mathrm{\parallel }ds$.

Put $\mathrm{\Theta }(t)={\int }_0^{t}a(s)ds$. By [F1], $\mathrm{\Theta }$ is continuously differentiable and ${\mathrm{\Theta }}^{\mathrm{\prime }}=a$. Inductively the curves ${\mathrm{\Delta }}_n$ are continuous and satisfy $\mathrm{\parallel }{\mathrm{\Delta }}_n(t)\mathrm{\parallel }\le \mathrm{\parallel }{x}_0\mathrm{\parallel }\mathrm{\Theta }(t{)}^n\mathrm{/}n!$: it is equality as a bound for $n=0$, and if it holds at $n$, then $\mathrm{\parallel }{\mathrm{\Delta }}_{n+1}(t)\mathrm{\parallel }\le {\int }_0^{t}a(s)\mathrm{\parallel }{\mathrm{\Delta }}_n(s)\mathrm{\parallel }ds\le \mathrm{\parallel }{x}_0\mathrm{\parallel }{\int }_0^{t}a(s)\mathrm{\Theta }(s{)}^n\mathrm{/}n!ds=\mathrm{\parallel }{x}_0\mathrm{\parallel }\mathrm{\Theta }(t{)}^{n+1}\mathrm{/}(n+1)!$, where the last equality follows by differentiating ${\mathrm{\Theta }}^{n+1}\mathrm{/}(n+1)!$.

Since $\mathrm{\Theta }(t)\le c_T$, the bound in step 2.1 gives uniform absolute convergence of ${\sum }_n{ϵ}^n{\mathrm{\Delta }}_n(t)$. Uniform convergence, the integral bound in step 1.1, and the recursion show $x_{ϵ}(t)=x_0+ϵ{\int }_0^{t}K(s)x_{ϵ}(s)ds$; [F1] then gives the differential equation and initial value. If two solutions differ by $z$, repeated substitution in $z(t)=ϵ{\int }_0^{t}K(s)z(s)ds$ gives $\mathrm{\parallel }z(t)\mathrm{\parallel }\le {\sup }_{[0,T]}\mathrm{\parallel }z\mathrm{\parallel }(\mathrm{\mid }ϵ\mathrm{\mid }{c}_T{)}^n\mathrm{/}n!$ for every $n$, whose right side tends to zero; hence the solution is unique.

With $y=\mathrm{\mid }ϵ\mathrm{\mid }{c}_T$, the tail after its linear term is at most $\mathrm{\parallel }{x}_0\mathrm{\parallel }{\sum }_{n\ge 2}{y}^n\mathrm{/}n!\le \mathrm{\parallel }{x}_0\mathrm{\parallel }{y}^2{e}^y\mathrm{/}2$, because $n!\ge 2(n-2)!$ for every $n\ge 2$. This is the stated remainder.

In the $C^{\ast }$-algebra case, write $W=x_{ϵ}$ and assume $b(t{)}^{\ast }=-b(t)$ and $ϵ\in \mathbb{R}$. The product rule and $W^{\mathrm{\prime }}=ϵbW$ give $(W^{\ast }W{)}^{\mathrm{\prime }}=ϵW^{\ast }(b^{\ast }+b)W=0$, so $W^{\ast }W=1$. Also $Z=W{W}^{\ast }$ satisfies $Z^{\mathrm{\prime }}=ϵ(bZ-Zb)$ and $Z(0)=1$. The uniqueness argument of step 3.1, applied to the bounded operator family $Z\mapsto ϵ(b(t)Z-Zb(t))$, shows that the constant curve $Z=1$ is its only solution. Hence $W{W}^{\ast }=1$ as well, so $W$ is unitary in the unital $C^{\ast }$-algebra.

In the Hilbert-space case, the series is linear in $x_0$ and step 2.1 gives $\mathrm{\parallel }{W}_{ϵ}(t)x_0\mathrm{\parallel }\le e^{\mathrm{\mid }ϵ\mathrm{\mid }{c}_T}\mathrm{\parallel }{x}_0\mathrm{\parallel }$, so it defines a bounded operator. Differentiating inner products of two solutions gives $\frac{d}{dt}⟨W_{ϵ}(t)x,W_{ϵ}(t)y⟩=ϵ⟨K(t)W_{ϵ}(t)x,W_{ϵ}(t)y⟩+ϵ⟨W_{ϵ}(t)x,K(t)W_{ϵ}(t)y⟩=0$, hence $W_{ϵ}(t)$ is an isometry. For a fixed terminal time $t$, solve $z^{\mathrm{\prime }}(s)=-ϵK(t-s)z(s)$ on $[0,t]$ with arbitrary $z(0)=y$ using the first part; then $x(s)=z(t-s)$ solves the forward equation and $W_{ϵ}(t)x(0)=y$. Thus $W_{ϵ}(t)$ is onto and therefore unitary.