I have the following problem;


Fix a Hilbert space $\mathcal{H}$. Let $S \colon \mathrm{Dom}S \subset L\_2(\mathbb{R}\_+, \mathcal{H}) \rightarrow L\_2(\mathbb{R}\_+, \mathcal{H}) $ be a closed densely defined (possibly unbounded) linear operator. 
Do we know when can we find such a family of operators $(T\_t)\_{t \geq 0}$ on $\mathcal{H}$, such that
$$(Sf)(s)= T\_s(f(s)) $$
for all $f \in L\_2(\mathbb{R}\_+, \mathcal{H})$ and $s \geq 0$.
One condition which I observed is to make $S$ of the form $S= I\_{L\_2(\mathbb{R}\_+)} \otimes X$, where X is a closed densely defined linear operator on $\mathcal{H}$. But I was just wondering maybe there are some other special conditions which will guarantee the existence of a 'nice' family $(T\_t)\_{t \geq 0}$ which will satisfy the condition which I mentioned. Thank you for any help. 



Special form of unbounded operators on $L_2(\mathbb{R}_+, \mathcal{H})$
Basically you're asking for the map to take almost every fiber into itself. For bounded operators this is equivalent to commuting with every operator of the form $M\_g$ with $g \in L^\infty({\bf R}\_+)$, defined by $M\_gf(s) = g(s)f(s)$. Or, equivalently, to commuting with all unitaries of this form, and that version of the condition makes sense for unbounded operators too, so I think that's a nice natural characterization of the operators you're looking for.


