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\begin{document}
\noindent

\vspace{1cm}
\hfill{\it Cracow, le 19 Dec 2009}

\medskip
Dear Guys,

About the equality

\begin{equation}
	\int_{\C} \ket{z}\  W(|z|^2)\ \bra{z}\ dz=I
\end{equation}

At this point of our discussion, one has at hand two definitions. We denote $\HG$ our ``untouchable'' ground Hilbert space with basis of states $(\ket n)_{n\in\N}$.

\bs
{\bf First definition}\pointir\notemg{After our discussion with Kus. In fact the left hand side of \mref{woe} is a weak operator integration.}

\begin{equation}\label{woe}
(\forall n,m\in\N)(\int_{\C} W(|z|^2)\ \scal{n}{z}\scal{z}{m}\ dz)=\scal{n}{m}
\end{equation}
``weak operator equality''.

\bs
{\bf Second definition}\pointir\notemg{From yesterday. In fact the left hand side of \mref{soe} is a strong operator integration.}
\begin{equation}\label{soe}
(\forall \phi,\psi\in\HG)(\int_{\C} W(|z|^2)\ \scal{\phi}{z}\scal{z}{\psi}\ dz)=\scal{\phi}{\psi}
\end{equation}
``strong operator equality''.

Let us prove that, in our case the weak definition implies the strong definition and then what was computed yesterday 

\begin{equation}
\int_{\C} W(|z|^2).Osc(|z|^2)\ \scal{\phi}{z}\scal{z}{\phi}\ dz
\end{equation}

will always amount to zero as $W(|z|^2).Osc(|z|^2)=W_1(|z|^2)-W_2(|z|^2)$.
%\textcolor{inkred}{this is  wise because we have things to check for this new Hopf algebra} 
\end{document} 