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.

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    $\begingroup$ 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? $\endgroup$ Sep 26, 2021 at 17:37
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    $\begingroup$ 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? $\endgroup$
    – Arbiter
    Sep 26, 2021 at 17:54
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    $\begingroup$ 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). $\endgroup$
    – memorial
    Sep 26, 2021 at 20:48

2 Answers 2


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.

Many sources do treat this issue (Rudin's "Functional Analysis", up to a point, Bourbaki's "Integration theory", and various notes of mine, for example).

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.


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