Proposition 9

Let $U$ be a scalarly essentially $\mu$-integrable mapping of $T$ into $F = \mathscr{L}_s(G; H)$. In order that $\int U d\mu \in F$, it is necessary and sufficient that the following two conditions be satisfied:

a) For every $x \in G$, $\int (U(t) \cdot x) d\mu(t) \in H$.

b) For every equicontinuous subset $B'$ of $H'$, the set of linear forms $u_{y'} : x \mapsto \int \langle U(t) \cdot x, y' \rangle d\mu(t)$, where $y'$ runs over $B'$, is equicontinuous.

The conditions a) and b) are necessary. For, since for every $x \in G$, the mapping $\tilde{x} : V \mapsto V \cdot x$ of $\mathscr{L}_s(G; H)$ into $H$ is linear, one sees (No. 1,

Prop. 1) that $\widetilde{\mathbf{x}} \circ U : t \mapsto U(t) \cdot \mathbf{x}$ is scalarly essentially $\mu$-integrable and that

where $S = \int U\, d\mu \in \mathscr{L}_s(G; H)$. This proves a). Moreover, (1) may also be written

in other words ${}^t S \cdot \mathbf{y}' = u_{\mathbf{y}'}$. Since $S$ is continuous, ${}^t S$ transforms every equicontinuous subset of $H'$ into an equicontinuous subset of $G'$, whence b).

Conversely, suppose a) and b) verified. By virtue of a), the formula (1) defines a linear mapping $S$ of $G$ into $H$, and, for every $\mathbf{y}' \in H'$, this mapping satisfies (2) (No. 1, Prop. 1); but then, condition (b) says that $S$ is continuous (TVS, II, §6, No. 4, Props. 5 and 6, and III, §3, No. 5, Prop. 7), therefore $S \in \mathscr{L}_s(G; H)$. Finally, formula (2) proves that $S = \int U\, d\mu$.