Proposition 9

— (i) For a subset A of $(E/M)'$ to be equicontinuous, it is necessary and sufficient that $\pi(A)$ is an equicontinuous subset of $E'$.

(ii) Let $\mathfrak{S}$ be a set of bounded subsets of $E$, and $\mathfrak{S}_1$ the set of the images of subsets $A \in \mathfrak{S}$ in $E/M$. Then $\pi$ is an isomorphism from $(E/M)'_{\mathfrak{S}_1}$ onto $M^\circ$, where $M^\circ$ is assigned the topology induced by that of $E'_{\mathfrak{S}}$.

(iii) Suppose $E$ is a normed space then $\pi$ is an isometry from the normed space $(E/M)'$ onto the normed subspace $M^\circ$ of $E'$.

Let $A$ be a subset of $(E/M)'$ and $B = {}^t p(A) \subset E'$. Put

for all $\xi \in E/M$. In order that $A$ be equicontinuous, it is necessary and sufficient that the mapping $q$ from $E/M$ into $\overline{\mathbf{R}}_+$ is a continuous semi-norm. This implies that $q \circ p$ is a continuous semi-norm on $E$ (II, p. 27, prop. 5, (ii)). Since we have for all $x \in E$, this in turn implies that $B$ is equicontinuous in $E'$, and (i) follows.

Let $A \in \mathfrak{S}$ and let $f$ be a continuous linear form on $E/M$. For every $\lambda \in \mathbf{R}_+$, we have $|f| \leq \lambda$ on $p(A)$ if and only if $|{}^t p(f)| \leq \lambda$ on $A$; hence (ii).

Finally we prove (iii). Let $y'$ be in $(E/M)'$. An element in $E/M$ has norm $< 1$ if and only if it is the image under $p$ of an element of norm $< 1$ in $E$. Hence

and ${}^t p$ induces an isometry from $(E/M)'$ onto $M^\circ$.