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.


1 Answer 1


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.

  • $\begingroup$ 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, 2014 at 17:02

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