Example

— Let us take for $m$ the identity mapping of $\mathscr{K}(T)$ onto itself. Since the dual of $\mathscr{K}(T)$ is the space $\mathscr{M}(T)$ of scalar measures on $T$, the functions $f \in \mathscr{L}(m)$ are those that are essentially integrable for every scalar measure $\mu$ (cf. Exer. 1), and the integral $\int f dm$ is the linear form $\mu \mapsto \int f d\mu$ on $\mathscr{M}(T)$. One cannot have $\int f d\mu = 0$ for every measure $\mu \in \mathscr{M}(T)$ unless $f = 0$, as one sees on taking $\mu = \varepsilon_t$, where $t$ is arbitrary in $T$; in other words, the mapping $f \mapsto \int f dm$ is an injection of $\mathscr{L}(m)$ into the algebraic dual of $\mathscr{M}(T)$, which extends the identity mapping of $\mathscr{K}(T)$. The relation $\int f dm \in F = \mathscr{K}(T)$ is therefore equivalent to $f \in \mathscr{K}(T)$.

Let $u$ be a continuous linear mapping of $F$ into a Hausdorff locally convex space $G$, and let us denote again by $u$ its extension by bitransposition to a linear mapping of ${F'}^*$ into ${G'}^*$ (§ 1, No. 1). With this convention: