A posteriori Schrödinger propagation bound

Fix $s$ in the compact segment joining $t_0$ to $t$, put $W(\tau )=e^{i\tau T}$ and $U(s)=W((s-t_0)\mathrm{/}\mathrm{\hslash })$, and define $w(s)=U(s)v(s)$. For $h$ with $s+h$ in the interval, the group law gives $\frac{w(s+h)-w(s)}{h}=U(s+h)\frac{v(s+h)-v(s)}{h}+U(s)\frac{W(h\mathrm{/}\mathrm{\hslash })-I}{h}v(s)$. By [F1], $U$ is strongly continuous and unitary, the first term tends to $U(s)v^{\mathrm{\prime }}(s)$, and since $v(s)\in D(T)$ the second tends to $\frac{i}{\mathrm{\hslash }}U(s)Tv(s)$ by the derivative of $W$ at zero and the scalar chain rule. Thus $w$ is differentiable and $w^{\mathrm{\prime }}(s)=U(s)v^{\mathrm{\prime }}(s)+\frac{i}{\mathrm{\hslash }}U(s)Tv(s)=-\frac{i}{\mathrm{\hslash }}U(s)r(s)$, after substituting $i\mathrm{\hslash }{v}^{\mathrm{\prime }}=Tv+r$. The derivative is continuous because $v^{\mathrm{\prime }}$, $Tv$, and hence $r$ are continuous and $U$ is strongly continuous.

If $t\ge t_0$, apply [F3] on $[t_0,t]$; if $t<t_0$, apply it on $[t,t_0]$ and negate the resulting identity. In either case $w(t)-w(t_0)=-\frac{i}{\mathrm{\hslash }}{\int }_{t_0}^{t}U(s)r(s)ds$, using step 1.1. Then [F4] gives $\mathrm{\parallel }w(t)-w(t_0)\mathrm{\parallel }\le \frac{1}{\mathrm{\hslash }}{\int }_{\min (t,t_0)}^{\max (t,t_0)}\mathrm{\parallel }U(s)r(s)\mathrm{\parallel }ds=\frac{1}{\mathrm{\hslash }}{\int }_{\min (t,t_0)}^{\max (t,t_0)}\mathrm{\parallel }r(s)\mathrm{\parallel }ds$; the equality uses unitarity of $U(s)$. Since $w(t_0)=v(t_0)$ and $w(t)=U_T(t,t_0{)}^{\ast }v(t)$, unitarity also gives $\mathrm{\parallel }w(t)-w(t_0)\mathrm{\parallel }=\mathrm{\parallel }v(t)-U_T(t,t_0)v(t_0)\mathrm{\parallel }$. This is the required estimate, including $t=t_0$ and either time orientation.