Remark 2

— Let E be a locally convex vector space and let m be a vectorial measure on $X/H$ with values in E. The mapping $f \mapsto m(f^b)$ of $\mathscr{K}(X)$ into E is then a vectorial measure on X, with values in E, which we shall again denote by $m^\#$. The mapping $m \mapsto m^\#$ is again an isomorphism of $\mathscr{L}(\mathscr{K}(X/H); E)$ onto a linear subspace A of $\mathscr{L}(\mathscr{K}(X); E)$ (when these spaces are equipped with the topology of pointwise convergence). Moreover, since the mapping $f \mapsto f^b$ is a surjective strict morphism, the subspace A consists precisely of the vectorial measures n on X that are zero on the kernel N of the mapping $f \mapsto f^b$. In order that $n \in A$, it is therefore necessary and sufficient that the scalar measures $z' \circ n$ be zero on N for every $z' \in E'$. One then deduces from Prop. 3 that $n \in A$ if and only if $\delta(\xi)n = \Delta_H(\xi)n$ for all $\xi \in H$.