Proposition 10

— Let F be a separable Banach space. For every function $f \in \overline{\mathscr{L}}^1_F$ and every function $g \in \mathscr{L}_{F'_s}^\infty$, the numerical function $\langle f, g \rangle : t \mapsto \langle f(t), g(t) \rangle$ is essentially $\mu$-integrable, and

*For every class $\dot{g} \in L_{F'_s}^\infty$, let $\theta(\dot{g})$ be the continuous linear form on $L^1_F$ deduced from the linear form $f \mapsto \int \langle f, g \rangle \, d\mu$ on $\overline{\mathscr{L}}^1_F$ by passage to the quotient; then $\theta$ is a linear isometry of $L_{F'_s}^\infty$ onto the strong dual $(L^1_F)'$ of the Banach space $L^1_F$.

For every compact subset K of T and every $\varepsilon > 0$, there exists a compact subset $K'$ of K such that $\mu(K - K') \leq \varepsilon$ and the restriction of $f$ (resp. $g$) to $K'$ is a continuous mapping of $K'$ into F (resp. into $F'_s$); consequently $g(K')$ is weakly compact, hence equicontinuous on $F'$ (TVS, III, §4, No. 2, Th. 1 or IV, §1, Exer. 10). Now, the restriction of the canonical bilinear form on $F \times F'$ to the product of F and an equicontinuous subset of $F'$ is continuous for the product topology of the topology of F and $\sigma(F', F)$ (GT, X, §2, No. 1, Cor. 4 of Prop. 1); it follows that the restriction of $\langle f, g \rangle$ to $K'$ is continuous, hence that $\langle f, g \rangle$ is $\mu$-measurable. Moreover,

locally almost everywhere, consequently $\langle f, g \rangle$ is essentially $\mu$-integrable and the inequality (3) holds (Ch. IV, §5, No. 6, Th. 5 and Ch. V, §1, No. 3, Lemma).

It remains to show that $\theta$ is a surjective isometry. Let $u$ be a continuous linear form on $L^1_F$. The mapping $(\tilde{h}, z) \mapsto u(\tilde{h}z)$ is a continuous bilinear form on $L^1 \times F$, because

By Cor. 1 of Th. 1 of No. 5, there exists a mapping $g$ of T into $F'$, belonging to $\mathscr{L}_{F'_s}^\infty$, such that $|g(t)| \leq \|u\|$ for all $t \in T$ and such that $u(\tilde{h}z) = \int \langle hz, g \rangle \, d\mu$ for every function $h \in \mathscr{L}^1$ with class $\tilde{h}$ in $L^1$ and every $z \in F$. In other words, the linear forms $u$ and $\theta(\dot{g})$ coincide on the subspace of $L^1_F$ generated by the elements of the form $\tilde{h}z$ ($\tilde{h} \in L^1, z \in F$). Since this subspace is dense in $L^1_F$ (Ch. IV, §3, No. 5, Prop. 10), it follows that $u = \theta(\dot{g})$, which already proves that $\theta$ is surjective. Moreover, by (3), $\| \theta(\dot{g}) \| \leq N_\infty(g) \leq \| u \| = \| \theta(\dot{g}) \|$, whence $\| \theta(\dot{g}) \| = N_\infty(g)$, and this concludes the proof.