Proposition 10

— (i) For a subset $A$ of $M'$ to be equicontinuous, it is necessary and sufficient that it is the image under ${}^t i$ of an equicontinuous subset of $E'$.

(ii) Suppose $M$ is closed in $E$. Let $\mathfrak{S}$ be a covering of $E$ consisting of bounded subsets and let $\mathfrak{S}_1$ be the set of subsets of $M$ of the form $M \cap A$ for $A$ in $\mathfrak{S}$. The bijective linear mapping $i$ from $E'_{\mathfrak{S}}/M^\circ$ onto $M'_{\mathfrak{S}_1}$ is continuous. It is a homeomorphism if $\mathfrak{S}$ is a directed set for the relation $\subset$ and consists of closed convex and compact sets for $\sigma(E, E')$.

(iii) If $E$ is assumed to be normed, then $i$ is an isometry from $E'/M^\circ$ onto $M'$.

The image under ${}^t i$ of an equicontinuous subset of $E'$ is an equicontinuous subset of $M'$ (IV, p. 7, prop. 7). Conversely, let $A$ be an equicontinuous subset of $M'$. The topology of $M$ is defined by the set of restrictions to $M$ of the continuous semi-norms on $E$. Hence there exists a continuous semi-norm $p$ on $E$ such that $|f(x)| \leq p(x)$ for all $f \in A$ and for all $x \in M$. Let $B$ be the set of all linear forms $g$ on $E$ such that $|g| \leq p$ and whose restriction to $M$ belongs to $A$. The set $B$ is equicontinuous in $E'$; by Hahn-Banach theorem (II, p. 23, cor. 1), we have ${}^t i(B) = A$, hence (i) follows.

We now prove (ii). By prop. 6 of IV, p. 6, the linear mapping ${}^t i$ from $E'_{\mathfrak{S}}$ into $M'_{\mathfrak{S}_1}$ is continuous, and defines, by passing to the quotient, a continuous linear mapping $\iota$ from $E'_\mathscr{S}/M^\circ$ onto $M'_{\mathscr{S}_1}$. Let $\mathscr{T}$ be the topology on $M'$ obtained by transferring that of $E'_\mathscr{S}/M^\circ$ by $\iota$; this is finer than the $\mathscr{S}_1$-topology.

Suppose now that $\mathscr{S}$ is a directed set for $\subset$ and consists of closed, convex, balanced and compact sets for $\sigma(E, E')$. To show that $\iota$ is a homeomorphism, i.e. that $\mathscr{T}$ is coarser than the $\mathscr{S}_1$-topology on $M'$, it is enough to prove that $\mathscr{T}$ is compatible with the duality between $M'$ and $M$ and that every equicontinuous set in $M$ (considered as the dual of $M$ with $\mathscr{T}$) is contained in the homothetic of a set belonging to $\mathscr{S}_1$. Since $\mathscr{T}$ is finer than the $\mathscr{S}_1$-topology and $\mathscr{S}_1$ is a covering of $M$, the linear form $y' \mapsto \langle y, y' \rangle$ on $M'$ is continuous for $\mathscr{T}$ for every $y \in M$. Let $f$ be a linear form on $M'$ which is continuous for $\mathscr{T}$; then $f \circ \iota$ is a continuous linear form on $E'_\mathscr{S}$. The $\mathscr{S}$-topology on $E'$ is coarser than the Mackey topology $\tau(E', E)$; for, the mapping $d_B : E \to {E'}^*$ is continuous for the topologies $\sigma(E, E')$ and $\sigma({E'}^*, E')$, and since the latter is Hausdorff, the image under $d_B$ of a set which is compact for $\sigma(E, E')$ is compact for $\sigma({E'}^*, E')$. By lemma 1 of IV, p. 3, there exists $x_0 \in E$ such that $f(\iota(x')) = \langle x_0, x' \rangle$ for all $x' \in E'$. In particular, $\langle x_0, x' \rangle = 0$ for all $x' \in M^\circ$, and since $M$ is closed in $E$, we have $x_0 \in M$ (II, p. 45, cor. 2); and finally, $f(y') = \langle x_0, y' \rangle$ for all $y' \in M'$. This proves that $\mathscr{T}$ is compatible with the duality between $M$ and $M'$.

Now let $A$ be a subset of $M$ equicontinuous for the topology $\mathscr{T}$ on $M'$. By the definition of $\mathscr{T}$, and in view of the hypothesis that $\mathscr{S}$ is directed, this means that there exists a set $B \in \mathscr{S}$ containing 0 and such that the upper bound $\lambda$ of the numbers $|\langle y, x' \rangle|$ for $y \in A$ and $x' \in B^\circ$, is finite (III, p. 19, prop. 7). Since $B$ is closed in $E$, the theorem of bipolars (II, p. 44, th. 1) shows that we have $A \subset \lambda(B \cap M)$; this completes the proof of (ii).

We shall now prove (iii). Let $y' \in M'$. We shall prove the formula

By prop. 8, (ii) of IV, p. 7, we have $\|\iota\| = \|i\|$, and so $\|\iota\| \leq 1$, and

By Hahn-Banach theorem (II, p. 23, cor. 3), there exists a linear form $x'_0$ on $E$ which extends $y'$ and is of the same norm; hence we get the inequality opposite to (7), since $\iota(x'_0) = y'$.