strongly in ${{\mathcal H}}$ provided $\:t\notin M$.

Notice that this claim is in fact the “symmetric” part of Corollary \[cor2\].

[*Proof of Lemma \[L2\]:*]{} We use arguments analogous to those employed in derivation of Chernoff’s theorem – see [@Ch2 Theorem 1.1], [@Ch1] and [@Ka1 Thm IX.3.6]. We divide the proof into two steps referring to $f$ belonging to $P{{\mathcal H}}$ and to its orthogonal complement.

Suppose first that $f \in P{\cal H}$. For $t \notin M$ and $\tau$ fixed, the operator $S(it, \tau)$ generates a strongly continuous semigroup $\{\,e^{-\theta S(it, \tau)}:\: \theta \geq 0\,\} $ on ${{\mathcal H}}$, and the resolvent convergence (\[resolconv\]) implies the convergence of the corresponding semigroups [@Ka1 Thm IX.2.16], so we have $$e^{-\theta S(it, 1/n')}f \stackrel{\scriptsize{s}}{\longrightarrow}
 e^{-i\theta t H_P}f$$ in ${{\mathcal H}}$ as $\:n'\to\infty$ for $\:t \notin M$, uniformly on each compact interval of the variable $\theta \geq 0$. In particular, choosing $\theta = 1$ we get for each $t \in [0,\infty)\setminus
M$ $$\label{L3.3-semigrconv}
 e^{-S(it, 1/n')}f \stackrel{\scriptsize{s}}{\longrightarrow}
 e^{-it H_P}f\,, \quad n'\rightarrow \infty\,.$$ The same equivalence implies for any $\lambda \geq 0$ and $t \in
[0,\infty)\setminus M$ that $$(I+ \lambda S(it, 1/n'))^{-1}f
 \stackrel{\scriptsize{s}}{\longrightarrow}
 (I+i\lambda tH_P)^{-1}Pf\,,$$ in particular, using the diagonal trick we obtain $$\label{L3.3-resolvconv}
 \Big(I+ \frac{1}{\sqrt{n'}} S(it, 