{{\mathcal H}}$ we get $$\begin{aligned}
 \|t\tau H(it\tau)w\|^2
 &\!=\!& \|\,[t\tau B(t\tau) + it\tau A(t\tau)]w\|^2\\
 &\!=\!& \|\,[(I-\cos t\tau H) + i\sin t\tau H]w\|^2\\
 &\!=\!& \langle w, [(I-\cos t\tau H)^2 + \sin^2(t\tau H) ]w \rangle \\
 &\!=\!& 4 \|\sin(t\tau H/2) w\|^2 \to 0\,,
 \end{aligned}$$ uniformly on compact $t$-intervals in ${\mathbb{R}}$. In this way we have proved the claim, noting that $P(\tau)
\stackrel{\scriptsize{s}}{\longrightarrow} P$ holds uniformly on each compact $t$-interval in ${\mathbb{R}}$ as $\tau \to 0$.

Now we employ the following identity, $$\begin{aligned}
&&\bigl(P(1/n)\,e^{-itH/n}\bigr)^n v -
 \bigl(P(1/n)\,e^{-itH/n}P(1/n)\bigr)^n v \label{diff:ns-s}\\
&&= -\bigl(P(1/n)e^{-itH/n}
          P(1/n)\bigr)^{n-1}[\,e^{-itH/n}, P(1/n)\,]
v\,, \nonumber\end{aligned}$$ the right-hand side of which converges by Lemma \[L3\] to zero for all $t \not= 0$ and any $v\in{{\mathcal H}}$, because $\bigl(P(1/n)e^{-itH/n}P(1/n)\bigr)^{n-1}$ is a contraction on ${{\mathcal H}}$, and hence also in $L^2_\mathrm{loc}([0,\infty);\, {\cal
H})$. This yields the formula (\[product-ns\]).

[*Proof of (\[product-ns2\]):*]{} In view of the already proven formula (\[product-s\]) we have for every $f \in{{\mathcal H}}$ and $T>0$ the following chain of relations $$\begin{aligned}
 T\|Pf\|^2
     &\geq& \limsup \int_0^T \|(e^{-itH/n}P(1/n))^n f\|^2\, dt \\
     &=& \limsup
     \Big[\int_0^T \|P(1/n)(e^{-itH/n}P(1/n))^n f\|^2 \, dt\\
     &&\qquad\quad
      + \int_0^T \|Q(1/n)(e^{-itH/n}P(1/n))^n f\|^2 \, dt\Big]\\
     &\geq& \limsup
     \int_0^T \|(P(1/n)e^{-itH/n}P(1/n))^n f\|^2 \, dt \\
     &=& \int_0^T \|e^{-itH_P}P f\|^2 \, dt =T\|Pf\|^2,\end{aligned}$$ with the $\limsup$ taken along $n\rightarrow 