Proposition 7

— Let $M$ be a vector subspace of $F$ and $N$ a vector subspace of $G$ where $F$ and $G$ are two vector spaces in duality. Suppose that $M$ and $N$ are orthogonal (which is equivalent to saying that $N \subset M^\circ$, or $M \subset N^\circ$). The vector spaces $M$ and $G/N$ are then in duality by the bilinear form $(x, \dot{y}) \mapsto \langle x, \dot{y} \rangle$.

(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 $\sigma(G/N, M)$.