Let $\mathscr{T}$ be a topology on E compatible with the duality between E and F, hence finer than $\sigma(E, F)$. If a convex subset of E is closed for $\mathscr{T}$, it is the intersection of closed, real half-spaces (II, p. 38, cor. 1), hence it is closed for $\sigma(E, F)$. This proves (i). Assertion (ii) was proved in cor. 3 of III, p. 27.

Let $F_\sigma$ denote the vector space F endowed with the weak topology $\sigma(F, E)$. Then the linear mapping $s_B$ maps E onto the dual $(F_\sigma)'$ of $F_\sigma$ (II, p. 43, prop. 3). Let $\mathfrak{S}$ be a family of bounded subsets of $F_\sigma$. By abuse of language, the inverse image under $s_B$ of the $\mathfrak{S}$-topology on $(F_\sigma)'$ is called the $\mathfrak{S}\text{-topology on } E$. It is defined by the family of semi-norms

where A runs through $\mathfrak{S}$. In particular, when $\mathfrak{S}$ is the family of finite subsets of F, the $\mathfrak{S}$-topology is precisely the weak topology $\sigma(E, F)$.

Every element of F is a continuous mapping for $\sigma(E, F)$, hence for $\tau(E, F)$.

Conversely, let $f \in E^*$ be continuous for $\tau(E, F)$. There exists a neighbourhood U of 0 in E (for $\tau(E, F)$), such that $|f| \leq 1$ on U; we can assume that there exists a set $A \in \mathscr{S}_0$ such that $U = A^\circ$. In other words, $f$ belongs to the bipolar $A^{\circ\circ}$ of A for the duality between $E^*$ and E. But the topology $\sigma(F, E)$ on F is induced by $\sigma(E^*, E)$; consequently A is convex, balanced and compact for $\sigma(E^*, E)$, and the theorem of bipolars (II, p. 44, th. 1) implies the equality $A = A^{\circ\circ}$. Therefore we have that $f \in F$, from which the lemma follows.

Let $\mathscr{U}$ be the set of convex, balanced neighbourhoods of 0 for $\mathscr{T}$. Let $\mathscr{S}$ be the set of polars in F of elements of $\mathscr{U}$. By cor. 2 of III, p. 17, we have $\mathscr{S} \subset \mathscr{S}_0$, and by cor. 1 of prop. 7 of III, p. 19, $\mathscr{T}$ is identical with the $\mathscr{S}'$-topology, where $\mathscr{S}'$ is the set of polars of sets of $\mathscr{U}$ in the dual $E'$ of E. But $E' \subset F$, by hypothesis, hence every set of $\mathscr{S}'$ is contained in a set of $\mathscr{S}$; and the lemma follows.

Let $\mathscr{T}$ be a topology on E compatible with the duality between E and F. Then $\mathscr{T}$ is coarser than $\tau(E, F)$ by lemma 2, and evidently $\mathscr{T}$ is finer than $\sigma(E, F)$. Conversely, F is the dual of E for the topology $\tau(E, F)$ (lemma 1) and for the topology $\sigma(E, F)$ (II, p. 43, prop. 3), hence also for every topology intermediate between $\tau(E, F)$ and $\sigma(E, F)$.

(i) p is continuous for the topology $\tau(E, F)$; (ii) every linear form f on E, such that $|f| \leq p$, comes from an element of F.

(i) $\Rightarrow$ (ii) : if p is continuous for $\tau(E, F)$, every linear form f on E such that $|f| \leq p$ is continuous for $\tau(E, F)$, hence comes from an element of F by lemma 1.

(ii) $\Rightarrow$ (i) : let $\mathscr{T}$ be the topology on E defined by the semi-norm p. If condition (ii) is satisfied, the linear forms on E which are continuous for $\mathscr{T}$ belong to F. By lemma 2 $\mathscr{T}$ is coarser than $\tau(E, F)$, hence p is continuous for $\tau(E, F)$.

a) E is barrelled; b) E is bornological; c) E is metrizable.

We note first that the Mackey topology on E is identical with the initial topology if and only if every convex subset of $E'$ which is compact for $\sigma(E', E)$, is equi-continuous. This is certainly the case if E is barrelled (III, p. 24, corollary).

Suppose E is bornological; let V be a convex and balanced neighbourhood of 0 in E for the topology $\tau(E, E')$. Let B be a subset of E, bounded for the initial topology. Since B is bounded for the Mackey topology, V absorbs B, and since E is bornological, V is a neighbourhood of 0 for the initial topology.

In case c), the space E is bornological (III, p. 12, prop. 2).

(i) If $E_1$ and $E_2$ are Hausdorff then $u$ is injective if and only if the image of $^t u$ is dense in $E'_1$ for the weak topology $\sigma(E'_1 : E_1)$.

(ii) For $^t u$ to be injective it is necessary and sufficient that the image of $u$ is dense in $E_2$.

A vector subspace of $E_2$ is dense for the initial topology if and only if it is dense for the weakened topology (IV, p. 4, prop. 2). Prop. 5 then follows from II, p. 47, cor. 2.

This is an immediate consequence of prop. 2 of III, p. 15.

The only point which requires a proof is the case c), when the topologies of $E_1$ and $E_2$ are not necessarily Hausdorff. Then for every linear form $f \in E'_1*$, $f \circ ^t u$ is a linear form on $E'_2$; hence there is a linear mapping $v : E'_1* \to E'_2*$ which is continuous for the topologies $\sigma(E'_1*, E'_1)$ and $\sigma(E'_2*, E'_2)$ and is such that $d_{B_2} \circ u = v \circ d_{B_1}$, where $d_{B_i}$ is the canonical mapping from $E_i$ into $E'_i*$ ($i = 1, 2$). Consequently, if $A$ is a subset of $E_1$ such that $d_{B_1}(A)$ is convex, balanced and compact for $\sigma(E'_1*, E'_1)$ then $d_{B_2}(u(A)) = v(d_{B_1}(A))$ is convex, balanced and compact for $\sigma(E'_2*, E'_2)$ since the topologies $\sigma(E'_1*, E'_1)$ and $\sigma(E'_2*, E'_2)$ are Hausdorff.

(i) The mapping $u$ is continuous if $E_1$ and $E_2$ are assigned their Mackey topologies.

(ii) If $E_1$ is bornological or barrelled, then $u$ is continuous for the initial topologies of $E_1$ and $E_2$.

(iii) In order that $u$ be continuous for the initial topologies of $E_1$ and $E_2$, it is necessary and sufficient that the image under $'u$ of every equicontinuous subset of $E_2'$ be equicontinuous in $E_1'$.

The hypothesis implies that $'u$ is continuous for the weak topologies $\sigma(E_2', E_2)$ and $\sigma(E_1', E_1)$ (II, p. 46, corollary) hence the image under $'u$ of a convex, balanced and compact subset for $\sigma(E_2', E_2)$ is convex, balanced and compact for $\sigma(E_1', E_1)$, the topologies $\sigma(E_2', E_2)$ and $\sigma(E_1', E_1)$ being Hausdorff. Therefore, assertion (i) follows from GT, X, § 1, No. 4, prop. 3, b). Assertion (ii) is a consequence of (i) : for, if $E_1$ is bornological or barrelled, its initial topology is the Mackey topology, and the Mackey topology of $E_2$ is finer than the initial topology of $E_2$. Finally, the initial topology of $E_i$ is that of uniform convergence on equicontinuous subsets of $E_i'$ (III, p. 19, cor. 1 of prop. 7). This proves (iii).

a) $u$ is continuous;

b) $u$ is continuous for the weakened topologies;

c) the image of the unit ball in $E_1$ under $u$ is bounded in $E_2$;

d) for every sequence $(x_n)$ of points of $E_1$ tending to 0 for the initial topology, the sequence $(u(x_n))$ is bounded for the weakened topology of $E_2$.

Since $E_1$ is bornological the equivalence of a) and b) follows from prop. 7; that of a) and c) is immediate. The equivalence of a) and d) follows from prop. 1 of IV, p. 1, and from prop. 1 of III, p. 11.

(ii) Let $E_1$ and $E_2$ be two normed spaces and $u$ a continuous linear mapping from $E_1$ into $E_2$. We have

Let $x \in E$. For every $x' \in E'$ such that $\|x'\| \leq 1$, we have

By Hahn-Banach theorem (II, p. 23, cor. 2), there exists an element $x'$ in $E'$ such that $\|x'\| \leq 1$ and $\langle x, x' \rangle = \|x\|$. This proves (i).

Let us now prove (ii). By formula (3) and the definition of the transpose, we have

for all $x \in E$.

Let B' be the unit ball of the dual $E'$ of E with the weak topology $\sigma(E', E)$ assigned to it. Then B' is a compact metrizable space (III, p. 19, cor. 2); hence there exists a countable dense subset D' in B'. Put $D = D' \cap (E' - \{0\})$. Let $x \in E$; the mapping $x' \mapsto \langle x, x' \rangle$ from B' into K is continuous, therefore

Formula (5) now follows from (3).

(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 $M^\circ$.

(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 $\iota(x'_0) = y'$.

(ii) The Mackey topology on E/M is the quotient of that on E; the Mackey topology on M is finer than the topology induced by τ(E, E').

Assertion (i) follows from prop. 7 of II, p. 48.

The canonical injection $i : M \to E$ is continuous for the weakened topologies, hence for the Mackey topologies $\tau(M, M')$ and $\tau(E, E')$ (IV, p. 7, prop. 7). Similarly, the canonical projection $p : E \to E/M$ is continuous for the Mackey topologies. We see immediately that the quotient topology on $E/M$ obtained from $\tau(E, E')$ is compatible with the duality between $E/M$ and $(E/M)'$, hence is coarser than the Mackey topology on $E/M$, by Mackey's theorem (IV, p. 2, th. 1). This proves (ii).

(ii) For every $i \in I$, let $\mathfrak{J}_i$ be an adapted bornology on the space $E_i$ endowed with the weak topology $\sigma(E_i, F_i)$, none equal to $\{ \varnothing \}$. Let $\mathfrak{J}$ be the family of subsets $A$ of $E = \prod_{i \in I} E_i$ such that $\operatorname{pr}_i(A) \in \mathfrak{J}_i$ for all $i \in I$. Then the $\mathfrak{J}$-topology on $F$ is the direct sum of the $\mathfrak{J}_i$-topologies on the $F_i$.

Let $\mathscr{T}$ be the product of the $\mathfrak{S}_i$-topologies. The sets of the form

where $J \subset I$ is finite and $A_i \in \mathfrak{S}_i$ for all $i \in J$, form a fundamental system of neighbourhoods of 0 for $\mathscr{T}$. We have $A = (\bigcup_{i \in J} A_i)^\circ$, hence $\mathscr{T}$ is identical with the $\mathfrak{S}$-topology. This proves (i).

We assign the $\mathfrak{J}$-topology to $F$ and the $\mathfrak{J}_i$-topology to each $F_i$. For every subset $A$ of $E$, we have $F_i \cap A^\circ = \operatorname{pr}_i(A)^\circ$, hence the injection from $F_i$ into $F$ is continuous. Let $q$ be a semi-norm on $F$; we assume that the restriction $q_i$ of $q$ to $F_i$ is continuous for all $i \in I$. Then we can find non-empty subsets $A_i \in \mathfrak{J}_i$ such that we have

Put $A = \prod_{i \in I} A_i$; then $A \in \mathfrak{J}$. For $y = (y_i)_{i \in I}$ in $F$, we have

where the last equality follows from (8) since the family $(y_i)_{i \in I}$ has finite support and the $A_i$ are non-empty and can be assumed balanced (GT, IV, § 5, No. 7, cor. 2 to prop. 12). This inequality proves that $q$ is continuous on $F$, and hence (ii).