1/n')\Big)^{-1}f \stackrel{\scriptsize{s}}
 {\longrightarrow}Pf \quad  \mathrm{as}\;\; n'\rightarrow \infty,$$ for every $t \in [0,\infty)\setminus M$. Next we use [@Ch1 Lemma 2] which gives for any $g \in {\cal H}$ the inequality $$\left\|F(it, 1/n')^{n'} g - e^{-n'(I-F(it,1/n'))}g\right\|
 \leq \sqrt{n'}\left\|(I-F(it, 1/n'))g\right\|\,.$$ Choosing $g = \left(I+ \frac{1}{\sqrt{n'}} S(it,
1/n')\right)^{-1}\!f\:$ we infer that $$\begin{aligned}
&&\left\|\left[ F(it, 1/n')^{n'} - e^{-S(it,1/n')}\right]
       \left(1+ \frac{1}{\sqrt{n'}} S(it, 1/n')\right)^{-1}f
       \right\| \\
&& \quad \leq
 \left\|\left(I+ \frac{1}{\sqrt{n'}} S(it, 1/n')\right)^{-1}f
 - f\right\|\,,\end{aligned}$$ where the right-hand side tends to zero as $n' \rightarrow \infty$ by (\[L3.3-resolvconv\]). Using (\[L3.3-resolvconv\]) once again we get $$\label{L3.3-semigrconv2}
\left\|F(it, 1/n')^{n'} f - e^{-S(it,1/n')}f \right\|
 \longrightarrow  0\,.$$ The sought relation (\[prodconv\]) immediately follows from (\[L3.3-semigrconv\]) and (\[L3.3-semigrconv2\]), since by (\[red-evol\]) we have $F(it,1/n')^{n'} = [P(1/n')\exp(-itH/n')
P(1/n')]^{n'}$.

The case $f \in Q{{\mathcal H}}$ is easier being independent of the arguments preceding Lemma \[L2\]. We have, along the sequence of [*all*]{} positive integers $n$, $$[P(1/n)\exp(-itH/n) P(1/n)]^n f \rightarrow 0$$ strongly in ${{\mathcal H}}$ and for each $t\in[0, \infty)$, since $P(1/n)f =
P(1/n)Qf$ converges by assumption to $PQf =0$ as $n\to\infty$, while $\exp(-itH_P)Pf = 0$.

This yields the sought result because $\{[P(1/n')\exp(-itH/n')
P(1/n')]^{n'}\}$ is a bounded sequence for any $t\ge 0$ and by Lebesgue dominated-convergence theorem it tends to the expected limit in $L_\mathrm{loc}^2([0,\infty); 