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.


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.

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  • $\begingroup$ Do you know any interesting references, which contain the result of the similar nature, that is, a theory of the operators on $L^2(\mathbf{R}_+; \mathcal{K})$ or sth equivalent? Thanks! $\endgroup$ – Chidoru Feb 26 '14 at 17:02

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