which $H$ is the multiplication operator, $(H\psi)(x)=x\psi(x)$ on $L^2({\mathbb{R}}_+)$, and $P$ is the one-dimensional projection onto the subspace spanned by the vector $\psi_0:\: \psi_0(x)= [(\pi/2)(1\!+\!x^2)]^{-1/2}$. In this case obviously $H_P$ is the zero operator on the domain $D[H_P]=
\{\psi_0\}^\perp$. On the other hand, $P\, e^{-itH}P$ acts on ${\mathrm{Ran}\,}P$ as multiplication by the function $$v(t):= e^{-t} -{i\over\pi} \left\lbrack e^{-t}\,
 \overline\mathrm{Ei}(t) - e^{t}\,\mathrm{Ei}(-t) \right\rbrack =
 1+{2i\over\pi}\, t\ln t + \mathcal{O}(t),$$ where $E_{i}(-t)$ and $\overline{E}_i(t)$ are exponential integrals [@AS]; due to the rapid oscillations of the imaginary part as $t \downarrow 0$ a pointwise limit of $v(t/n)^n$ for $n\to\infty$ does not exist. Notice also that different limits may be obtained in this example along suitably chosen subsequences $\{n'\}.$ ]{}

[In their recent study of Trotter-type formulae involving projections Matolcsi and Shvidkoy [@MaS] presented two examples in which expressions of the type $[\exp(-iH/n) P]^n$ do not converge strongly. This result does not answer the question, however, whether the product expressions considered here converge in the strong topology of $\mathcal{H}$ or not, because our assumptions are not satisfied there. In the first example of [@MaS] the analogue of the operator $H_P$ is not densely defined, in the second one $H$ is not semi-bounded. ]{}

Proof 