# Action of Bochner integral of operator-valued functions on vectors

Consider a separable Hilbert space $$\mathcal H$$ and the bounded linear operators $$B(\mathcal H)$$. Consider a function $$T: [0, \infty) \to B(\mathcal H)$$, under what assumptions on $$T(t)$$ is it true that $$\big(\int_0^c T(t) \, dt \big) (x) = \int_0^c T(t)x \, dt \ , \ \ \ \forall x \in \mathcal H , c \in (0, \infty)$$
for the Bochner integral? I am aware of a similar question for semigroups of BLO on Banach spaces,The Bochner integral about a semigroup of bounded linear operators on a Banach space, but I am interested in general operator-valued functions.

• Either the family of operators should be renamed, or the limit on the integral... :) But/and, for example, continuity $t\to T(t)$ is sufficient, in any of operator norm, strong, or weak topologies on bounded operators, for example. Is this the sort of condition that would be adequate for you? Sep 26, 2021 at 17:37
• Can you provide some references, or is this easy to see? Yes, this is the sort of condition that I'm after! Is it also true if $T(t)$ is piecewise norm-continuous, i.e.\ it has a countable number of discontinuities? Sep 26, 2021 at 17:54
• The result you need is Theorem 6 on p. 47 of the standard "Vector Measures" by Diestel and Uhl (required reading for anybody working with the Bochner integral). This shows, roughly speaking, that a continuous (even closed) linear operator applied to a Bochner integral can be pulled under the integral sign just as one would expect. There is a slight twist in that you must interpret an element $x$ of $E$ as a continuous linear operator from $L(E,F)$ into $F$ (via evaluation). Sep 26, 2021 at 20:48

I understand you assume that $$T:[0,c]\to\mathcal{B(H)}$$ is Bochner integrable in order to write $$\int_0^c T(t)dt$$ as Bochner integral. Then for any $$x\in\mathcal H$$ the map $$[0,c]\ni t\mapsto \mathcal H$$ is also Bochner integrable and the identity you wrote holds. More generally: for a measure situation $$(X,\mathcal S,\mu)$$, a couple of B-spaces $$\mathbb E$$ and $$\mathbb F$$, a bounded linear operator $$L:\mathbb E\to \mathbb F$$, and a Bochner integrable map $$f:X\to \mathbb E$$, the composition $$L\circ f:X\to \mathbb F$$ is Bochner integrable and $$\int_X L f(u) d\mu(u)=L\int_X f(u))d\mu(u)$$ (in your case $$L$$ is the evaluation map $$\mathcal{B(H)}\ni A\mapsto Ax\in\mathcal H$$).
(The proof is immediate if $$f:=v\chi_S$$ with $$v\in\mathbb E$$ and $$S\in\mathcal S$$; by linearity it generalizes immediately to integrable simple functions $$f:X\to \mathbb E$$; it further generalizes immediately to $$f\in \mathcal L^1(\mu,\mathbb E)$$, by the very definition of Bochner integrable function and integral).
In addition to @PietroMajer's good answer, I'd want to make a point about "vector-valued integrals" (and related), that the Bochner integral gives a construction (of something we want, with certain obvious/natural properties), but/and we have to prove that this construction succeeds. Oppositely, we can go the Gelfand-Pettis route, and "define" a "weak" integral on $$V$$-valued functions $$f$$ to be a linear functions $$I$$ such that, for all $$\lambda$$ in the dual space to $$V$$, $$\lambda(I(f))=\int \lambda(f)$$. After all, moving a linear operator inside the integral is the main thing we want to do.
A very convenient sufficient criterion is that the value-space $$V$$ is quasi-complete, locally convex, and that the $$V$$-valued function is continuous and compactly supported. The arguments to prove that various non-Frechet TVS's are quasi-complete (such as spaces of distributions, or continuous operators with strong or weak topologies) are not as well known, but do hold, for robust reasons.