{\cal H})$. Using the standard “subsequence” trick mentioned above we have thus shown that Lemma \[L1\] implies the symmetric product formula (\[product-s\]) of Theorem \[main\].

\(b) Let us turn to the non-symmetric product-formula cases, i.e. to prove that (\[product-s\]) implies (\[product-ns\]) and (\[product-ns2\]).

[*Proof of (\[product-ns\]):*]{} We employ the standard notation, $[U,S] = US - SU$, for the commutator of bounded operators $U$ and $S$. First we observe the following fact.

\[L3\] It holds that $\,[\,e^{-it\tau H}\!, P(\tau)\,]
 \stackrel{\scriptsize{s}}{\longrightarrow} 0$ as $\:\tau\to 0$, uniformly on each compact $t$-interval in ${\mathbb{R}}$.

[*Proof:*]{} By (\[Htau\]) with $\zeta = it\tau$ we have $$[\,e^{-it\tau H}\!, P(\tau )\,]
   = i\Bigl(P(\tau)t\tau H(it\tau)
         - t\tau H(it\tau)P(\tau) \Bigr),$$ and hence for any $v\in {{\mathcal H}}$ we can estimate $$\bigl\|[\,e^{-it\tau H}\!, P(\tau)\,]v \bigr\|
   \leq \|t\tau H(it\tau)v \| + \|t\tau H(it\tau)P(\tau)v \|\,.$$ We rewrite (\[Htau\]) with $\zeta = it\tau$ as $$\label{decomp-H}
 iH(it\tau) = \frac{I- \cos t\tau H}{t\tau} + i\,
 \frac{\sin t\tau H}{t\tau}
=: B(t\tau) + iA(t\tau)\,,$$ where $B(t\tau)$ and $A(t\tau)$ are obviously bounded self-adjoint operators on ${\cal H}$, and $B(t\tau)$ is in addition nonnegative. The definition makes sense if $t\ne0$ but we need not exclude this case because what we really need is the operator $t\tau H(it\tau)$. For any $w\in 