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.

## 2 Answers

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.

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? $\endgroup$