be linear mappings of G in the dual $F^*$ of F and of F in the dual $G^*$ of G, associated respectively to the right and to the left of B (A, IX, § 1.1). Then B puts F and G in a duality separating in G (resp. in F), if and only if $d_B$ (resp. $s_B$) is injective.*

When F and G are put in separating duality by B, we often identify F (resp. G) with a subspace of $G^*$ (resp. $F^*$) by means of $s_B$ (resp. $d_B$). When we consider F (resp. G) as a subspace of $G^*$ (resp. $F^*$) without specifying how this identification is to be made, we are always using the preceding identifications; the bilinear form B is then identified with the restriction to $F \times G$ of the canonical bilinear form :

This is a particular case of II, p. 3, prop. 2.

For, to say that the linear form $f$ on $F$ is continuous for $\sigma(F, G)$ means that there exists a finite set of points $y_i \in G$ ($1 \leq i \leq n$) such that, for all $x$ in $F$, $|f(x)| \leq \sup_{1 \leq i \leq n} |\langle x, y_i \rangle|$ (II, p. 6, prop. 5). The $n$ relations $\langle x, y_i \rangle = 0$ ($1 \leq i \leq n$) imply therefore $f(x) = 0$, and hence (A, II, § 7.5, cor. 1), there exists a linear combination $y = \sum_{i=1}^n \lambda_i y_i$ such that $f(x) = \langle x, y \rangle$ for all $x \in F$. The uniqueness follows from (D$_\text{II}$).

In other words, when the duality is separating in $G$, and $F$ has the topology $\sigma(F, G)$, then we can identify $G$ canonically with the dual of $F$ for this topology (II, p. 42, def. 1).

For using prop. 3 and I, p. 13, th. 1, the property expresses the fact that for $\sigma(F, G)$ no closed hyperplane contains all the $a_i$; the corollary follows therefore from cor. 3 of II, p. 38.

This means, that for all $i$, there exists a closed hyperplane in $\sigma(F, G)$, which contains all the $a_\kappa$ with index $\kappa \neq i$ but does not contain $a_i$.

The condition is obviously sufficient; conversely, if $\sigma(F, G_2)$ is finer than $\sigma(F, G_1)$, then every linear form that is continuous for $\sigma(F, G_1)$ is also continuous for $\sigma(F, G_2)$, hence $G_1 \subset G_2$ by prop. 3.

This follows from cor. 1.

(ii) If M is a cone of vertex 0, then $M^\circ$ is a cone of vertex 0 and it is also the set of $y \in G$ such that $\langle x, y \rangle \geq 0$ for all $x \in M$.

(iii) If M is a vector subspace of F, then $M^\circ$ is a vector subspace of G, and it is also the set of $y \in G$ such that $\langle x, y \rangle = 0$ for all $x \in M$.

(i) Since the linear forms $y \mapsto \langle x, y \rangle$ are continuous for $\sigma(G, F)$ the statement follows immediately from the definitions and the fact that a half-space determined by a hyperplane is convex.

(ii) If M is a cone with vertex 0 and if $x \in M, y \in M^\circ$, then as $\lambda x \in M$, for all $\lambda > 0$, we have $\langle \lambda x, y \rangle \geq -1$, i.e. $\lambda \langle x, y \rangle \geq -1$. Since this holds for all $\lambda > 0$, it follows that $\langle x, y \rangle \geq 0$, and (ii) is proved.

(iii) Similarly, if M is a vector subspace of F, the relations $x \in M, y \in M^\circ$ imply, this time, that $\lambda \langle x, y \rangle \geq -1$ for all real $\lambda$ which is possible only if $\langle x, y \rangle = 0$.

If M is a vector subspace of F we say that $M^\circ$ is the orthogonal of M in G; if $G \subset F^*$, then $M^\circ$ is the intersection of G, and of the subspace orthogonal to M in the algebraic dual $F^*$ of F (A, II, § 2.4, def. 4).

For a vector subspace M of F and a vector subspace N of G we say that M and N are orthogonal if $M \subset N^o$ (or, equivalently, if $N \subset M^\circ$).

We have seen that we need only consider the case where $M$ is convex and $0 \in M$. Denote the closure, in topology $\sigma(F, G)$ of $M$ by $\overline{M}$, then $\overline{M}$ is a convex set in $F$; prop. 4 of II, p. 44 shows that $M^{\circ\circ} \supseteq \overline{M}$. On the other hand if $a \in F$ does not belong to $\overline{M}$ then there exists a closed hyperplane $H$ in $F$ which separates $a$ strictly from $\overline{M}$ (II, p. 38, prop. 4); since $H$ does not contain 0, there exists $y \in G$ such that $H$ has the equation $\langle x, y \rangle = -1$ (II, p. 43, prop. 3); thus $\langle x, y \rangle > -1$ for all $x \in \overline{M}$ and $\langle a, y \rangle < -1$. This implies that $y \in M^\circ$ and $a \notin M^{\circ\circ}$, and the relation $M^{\circ\circ} = \overline{M}$ follows.

For, if $N$ is this convex closed envelope, then

whence $N = N^{\circ\circ} = M^\circ$.

The conclusion of cor. 1 does not necessarily hold if the $M_\alpha$ are not convex.

(i) *The closed convex sets in E are the same for the topology $\mathscr{T}$ and for the weak topology $\sigma(E, E')$.

(ii) *For every subset M of E, the polar set $M^{\circ\circ}$ in E of the polar set $M^\circ$ of M in $E'$, is the convex closed envelope of $M \cup \{0\}$ for the topology $\mathscr{T}$.

Clearly, (ii) follows from (i) and th. 1. From the definition of the dual $E'$, it follows from II, p. 43, prop. 3 that the continuous linear forms on E for the topology $\mathscr{T}$ are the same as the continuous linear forms for $\sigma(E, E')$. The closed half-spaces in E are therefore the same for $\mathscr{T}$ and for $\sigma(E, E')$ (II, p. 15, prop. 17) and the assertion (i) follows therefore from II, p. 38, cor. 1.

a) u is continuous for the weak topologies $\sigma(F, G)$ and $\sigma(F_1, G_1)$;

b) there exists a mapping $v : G_1 \to G$ such that

for all $y \in F$ and $z \in G_1$.

If these properties hold and if the duality between F and G is separating in G, then the mapping v satisfying (1) is unique, and v is linear.

If u is continuous for the weak topologies, then, for all $z_1 \in G_1$, the linear form $y \mapsto \langle u(y), z_1 \rangle$ on F is continuous for $\sigma(F, G)$, thus (II, p. 43, prop. 3) can be written as $y \mapsto \langle y, v(z_1) \rangle$ with $v(z_1) \in G$, which shows that a) implies b). Conversely, if b) is true, for all $z_1 \in G_1$, the linear form

is continuous for $\sigma(F, G)$ : it follows from the definition of weak topologies that u is continuous for $\sigma(F, G)$ and $\sigma(F_1, G_1)$ (I, p. 10, cor. 1). The uniqueness of v follows from (D_{II}) and this uniqueness implies that v is linear.

It is sufficient to exchange $F$ and $F_1$ with $G$ and $G_1$ in prop. 5.

(ii) We have ${}^t u(A_1^{\circ}) \subset (u^{-1}(A_1))^{\circ}$; further, if $A$ is closed, (for $\sigma(F_1, G_1)$) convex, and contains the origin, then we have ${}^t u(A_1^{\circ}) = (u^{-1}(A_1))^{\circ}$.

Let $z_1 \in G_1$, the relation $z_1 \in (u(A))^{\circ}$ is equivalent to $\langle u(y), z_1 \rangle \geq -1$ for all $y \in A$, and the relation ${}^t u(z_1) \in A^{\circ}$ is equivalent to $\langle y, {}^t u(z_1) \rangle \geq -1$ for all $y \in A$ and our assertion (i) follows using (1). Next interchanging $u$ and ${}^t u$ and applying (i) to the set $A_1^{\circ}$ of $G_1$ we get

from which, on taking polars

We have ${}^t u(A_1^{\circ}) \subset ({}^t u(A_1^{\circ}))^{\circ \circ}$ by the bipolar theorem (II, p. 44, th. 1); the final statement follows from (2) and the bipolar theorem since then $A_1^{\circ \circ} = A_1$ and ${}^t u(A_1^{\circ})$ is convex and contains the origin.

In fact, the relation $u(A) \subset A_1$ equivalent to $A \subset u^{-1}(A_1)$, therefore implies

and conversely the relation ${}^t u(A_1^{\circ}) \subset A^{\circ}$ implies

from (2). When $A_1 = A_1^{\circ \circ}$, we deduce that $A \subset u^{-1}(A_1)$.

(3)

(4)

Suppose that the dualities between $F$ and $G$ and between $F_1$ and $G_1$ are separating; then $u(F)$ is dense in $F_1$ (for $\sigma(F_1, G_1)$), if and only if ${}^t u$ is injective.

Apply prop. 6 with $A = F$ and $A_1 = \{0\}$, using the fact that the weak topologies $\sigma(G, F)$ and $\sigma(F_1, G_1)$ are Hausdorff. The last assertion results from (4), interchanging $u$ and $^t u$.

(i) *The topology $\sigma(M, G/N)$ for this duality is induced by $\sigma(F, G)$ (and in particular we have $\sigma(F, G) = \sigma(F, G/F^\circ)$).

(ii) *The topology $\sigma(G/N, M)$ for this duality is coarser than the quotient topology of $\sigma(G, F)$ by $N$; these topologies are identical if and only if $M + G^\circ = N^\circ$.

(i) Every element of $G/N$ is a class mod. $N$ of an element of $G$; if $z_i (1 \leq i \leq n)$ are elements of $G$ and $\dot{z}_i (1 \leq i \leq n)$ is the class of $z_i$ in $G/N$ then the set of $y \in M$ such that $|\langle y, \dot{z}_i \rangle| \leq \alpha$ for $1 \leq i \leq n$ is the trace on $M$ of the set of those $x \in F$ such that $|\langle x, z_i \rangle| \leq \alpha$ for $1 \leq i \leq n$; the conclusion follows from the definition of neighbourhoods of 0 for the weak topology.

(ii) Let $p : G \to G/N$ be the canonical surjection. We show that the quotient topology $\mathscr{T}$ of $\sigma(G, F)$ by $N$ is identical with $\sigma(G/N, N^\circ)$. As, for $z \in G, y \in N^\circ$, we have $\langle y, p(z) \rangle = \langle y, z \rangle$, it follows that every neighbourhood of 0 for $\sigma(G/N, N^\circ)$ is of the form $p(V)$, where $V$ is a neighbourhood of 0 for $\sigma(G, F)$ saturated for the relation $z - z' \in N$, therefore $\mathscr{T}$ is finer than $\sigma(G/N, N^\circ)$. Conversely let $U = W(y_1, ..., y_n; \alpha)$ be a neighbourhood of 0 in $G$ for $\sigma(G, F)$, where $y_i \in F$ for $1 \leq i \leq n$ and $\alpha > 0$; we are going to see that for $1 \leq i \leq n$, there exist elements $t_i \in N^\circ$ such that if one puts $U' = W(t_1, ..., t_n; \alpha)$, then $p(U') \subset p(U)$; this will show that $\sigma(G/N, N^\circ)$ is finer than $\mathscr{T}$ and therefore is actually identical with $\mathscr{T}$. Now, let $L$ be the vector subspace of $F$ generated by $N^\circ$ and the $y_i$, and denote by $P$ the complementary subspace of $N^\circ$ in $L$; it is of finite dimension, say $m$. Let $(x_j)_{1 \leq j \leq m}$ be a basis of $P$; the restrictions to $N$ of the linear forms $x \mapsto \langle x_j, z \rangle$ are linearly independent, since otherwise there exists $x \neq 0$ in $P$ such that $\langle x, z \rangle = 0$ for all $z \in N$, that is to say $x \in N^\circ$, which contradicts the definition of $P$. Thus we conclude that for all $z' \in G$, there exists $s \in N$ such that $\langle x_j, z' \rangle = \langle x_j, s \rangle$ for all $j$; if $z' = z + s$, we have $\langle x, z \rangle = 0$ for all $x \in P$. This being so, put $y_i = t_i + w_i$, where $t_i \in N^\circ$ and $w_i \in P$; we have $\langle y_i, z \rangle = \langle t_i, z \rangle = \langle t_i, z' \rangle$ for $1 \leq i \leq n$; therefore, for all $z' \in U'$, there exists $z \in U$ such that $z' - z \in N$, that is to say we have $p(U') \subset p(U)$.

Returning to the case where M is any subspace of N°, note that evidently $\sigma(G/N, M) = \sigma(G/N, M + G^\circ)$; further, from prop. 3 of II, p. 43, we see that, if $y \in N^\circ$ is such that the linear form $\dot{z} \mapsto \langle y, \dot{z} \rangle$ is continuous for $\sigma(G/N, M)$, then necessarily $y \in M + G^\circ$. We conclude that the condition $M + G^\circ = N^\circ$ is necessary and sufficient for the quotient topology $\mathscr{T}$ to be equal to $\sigma(G/N, M)$.

This follows from prop. 7 putting $N = M^\circ$, and recalling that $M^{\circ \circ}$ is the closure of M for $\sigma(F, G)$ (II, p. 45, cor. 2).

For, $G/M^0$ is in separating duality with M; if M is of dimension n, the same is therefore true of $G/M^\circ$ (II, p. 41, example 1). If M is closed, $F/M = F/M^{\circ \circ}$ is in separating duality with $M^\circ$; if F/M is of dimension n, it is therefore the same for $M^\circ$ (II, p. 41, example 1).

Let $N = \operatorname{Im}(^t u) \subset G$; we know that $N^0 = \operatorname{Ker}(u)$ in F (II, p. 47, formula (3)). Let $p : F \to F/N^\circ$ be the canonical mapping so that u factorises as

where w is injective. The spaces $F/N^\circ$ and N are in separating duality and by formula (1) of II, p. 48, we have $\langle w(\dot{y}), z_1 \rangle = \langle \dot{y}, ^t u(z_1) \rangle$ for all $\dot{y} \in F/N^\circ$ and $z_1 \in G_1$. This relation shows that w is an isomorphism of $F/N^\circ$, carrying the topology $\sigma(F/N^\circ, N)$, on $u(F)$ with the topology induced by $\sigma(F_1, G_1)$. The conclusion results therefore from cor. 1 and the definition of a strict morphism.