To begin with I introduce the basic elements of my question.
Let $\mathcal{K}$ be a real Hilbert space, for each $f\in \mathcal{K}$ there is a densily defined selfadjoint operator $\varphi(f)$ on a Hilbert space $\mathfrak{F}$, also $\varphi(tf)=t\varphi(f)$ for all $t\in \mathbb{R}$. So for $f\in \mathcal{K}$, $e^{i\varphi(f)}$ are unitary operators (bounded).
Then $R(\mathcal{K}):= \{ e^{i\varphi(f)} / f\in \mathcal{K} \}''$ is a Von Neumann algebra.
Lastly, for a fix $f\in \mathcal{K}$ the map $U:\mathbb{R}\to R(\mathcal{K})$ given by $U(t)=e^{i\varphi(tf)}$ is strongly and also weak continuous.
In order to see that $\varphi(f) \eta R(\mathcal{K})$ (is afiliated to) I did the folowing.
Any $A\in R(\mathcal{K})'$ satisfies $AU(t)=U(t)A$, and by deriving this at $t=0$ we have $A\varphi(f)=\varphi(f)A$ in the common domain, i.e. $\varphi(f) \eta R(\mathcal{K})$.
(1) Is the above argument right?
My last question is about the following chain of equalities,
(2) I'm taking limits of elements in $R(\mathcal{K})$, wich is (weak and strongly) closed, so I conclude that $\varphi(f)\in R(\mathcal{K})$, but this is wrong, because $\varphi(f)$ is not bouded. What is the mistake?
$\varphi(f)=w-\lim_{t\to 0}\dfrac{U(t)-U(0)}{it}$