On account of Remarks 1) and 2) above, we see that a necessary and sufficient condition for the $\mathfrak{S}$*-topology* to be compatible with the vector space structure of H is that the sets $H \cap T(M, V)$ are absorbent in H (I, p. 7, prop. 4); but this implies that for every $u \in H$, every subset $M \in \mathfrak{S}$, and every balanced neighbourhood V of 0 in F, there exists $\lambda \neq 0$ such that $u(M) \subset \lambda V$; that is to say (III, p. 2) that $u(M)$ is bounded in F. Finally, the last assertion of the proposition follows from the fact that if V is convex, so is $T(M, V)$.

It is enough to remark that if $u$ is a continuous linear mapping from E into F and M is a bounded subset of E, then $u(M)$ is bounded in F (III, p. 4, cor. 1).

Given two locally convex vector spaces E and F, and a family $\mathfrak{S}$ of bounded subsets of E, let $\mathscr{L}_\mathfrak{S}(E; F)$ denote the locally convex space obtained by assigning the $\mathfrak{S}$*-topology* to $\mathscr{L}(E; F)$.

1) The $\mathfrak{S}$-topology on $\mathscr{L}(E; F)$ is identical with the $\tilde{\mathfrak{S}}$-topology, where $\tilde{\mathfrak{S}}$ denotes the smallest adapted, bornology (III, p. 3) on E which contains $\mathfrak{S}$.

2) Suppose that $\{0\}$ is not dense in F and let $\mathfrak{S}'$ be another set of bounded subsets of E. Then the $\mathfrak{S}'$-topology is coarser than the $\mathfrak{S}$-topology if and only if $\mathfrak{S}' \subset \tilde{\mathfrak{S}}$.

Let $u \in \mathscr{L}(E; F)$, $M \in \mathfrak{S}$ and let $p$ be a continuous semi-norm on F. Since $p \circ u$ is a continuous semi-norm on E, this is the same as saying that $p \circ u$ is bounded above by 1 on M or on the closed, convex balanced envelope $\tilde{M}$ of M; in other words, we have $p_M = p_{\tilde{M}}$. Moreover, it is clear that we have $p_{\lambda M} = \lambda p_M$ for all $\lambda > 0$ and $p_{M \cup M'} = \sup(p_M, p_{M'})$, from which the first assertion follows, since $\tilde{\mathfrak{S}}$ has the set of homothetics of the closed convex balanced envelopes of finite unions of sets of $\mathfrak{S}$ as a base.

We now prove the second assertion : first, if F is the base field, it follows from the definition that the $\tilde{\mathfrak{S}}$-topology on $E' = \mathscr{L}(E; F)$ has as a fundamental system of neighbourhoods of 0, the set of polars of the sets of $\tilde{\mathfrak{S}}$. Let A be a bounded subset of E, whose polar $A^\circ$ is a neighbourhood of 0 for the $\tilde{\mathfrak{S}}$-topology; then there exists a closed convex balanced set $B \in \tilde{\mathfrak{S}}$ such that $A^\circ \supset B^\circ$, and so $A \subset B^{\circ \circ}$; but by cor. 3 of II, p. 45, we have $B^{\circ \circ} = B$, and hence $A \subset B$ and $A \in \tilde{\mathfrak{S}}$. Therefore if $\mathfrak{S}'$ is a set of bounded subsets of E, the $\mathfrak{S}'$-topology is coarser than the $\mathfrak{S}$-topology on $E'$ if and only if $\mathfrak{S}' \subset \tilde{\mathfrak{S}}$. The general case follows immediately, since if $y \in F$ is not in the closure of 0, we can verify that the mapping which makes $f \in E'$ correspond to the mapping $x \mapsto f(x)\ y$ is an isomorphism of the locally convex spaces $E'_\mathfrak{S}$ onto its image in $\mathscr{L}_{\tilde{\mathfrak{S}}}(E; F)$.

Let $u_0$ be a non-zero element of $\mathscr{L}(E; F)$; since $u_0$ is continuous and A is total in E, there exists an $x_0$ in A such that $u_0(x_0) \neq 0$. Since F is Hausdorff, there exists a neighbourhood V of 0 in F such that $u_0(x_0) \notin V$. Let $M \in \mathfrak{S}$ be such that $x_0 \in M$. Then the set U of all $u \in \mathscr{L}(E; F)$ such that $u(M) \subset V$ is a neighbourhood of 0 in $\mathscr{L}(E; F)$, and we have $u_0 \notin U$, hence $\mathscr{L}(E; F)$ is Hausdorff.

In particular, the following topologies on $\mathscr{L}(E; F)$ are Hausdorff whenever F is Hausdorff : simple convergence, compact convergence, precompact or compact convex, and bounded convergence.

(i) The set of linear mappings from E into F is closed in $F^E$.

(ii) If H is an equicontinuous subset of $\mathscr{L}(E; F)$, the closure $\overline{H}$ of H in $F^E$ is contained in $\mathscr{L}(E; F)$ and is equicontinuous.

We know that $\overline{H}$ is equicontinuous (GT, X, § 2, No. 3, prop. 6). It remains to prove the assertion (i). Let x, y be in E and $\lambda, \mu$ in K, and let $A(x, y, \lambda, \mu)$ be the set of all $u \in F^E$ such that

This set is closed in $F^E$ since the mapping $u \mapsto u(x)$ from $F^E$ into F is continuous for every $x \in E$ and since F is Hausdorff. But the set of linear mappings from E into F is equal to

Thus this set is closed in $F^E$.

In fact, this condition is necessary and sufficient for $\overline{H}$ to be compact in $F^E$ (GT, I, § 9, No. 5, cor.).

For, if H is an equicontinuous subset of $E'$, $\sup_{u \in H} |u|$ is a continuous semi-norm on E; in particular, for every $x \in E$, the set $H(x)$ is bounded, hence relatively compact in the field of scalars.

This ball is also closed for $\sigma(E', E)$.

1) the uniform structure of simple convergence in T; 2) the uniform structure of simple convergence in E; 3) the uniform structure of convergence in the precompact subsets of E.

We recall (III, p. 15, prop. 2) that the $\mathfrak{S}$-topology on $\mathscr{L}(E; F)$ coincides with the $\tilde{\mathfrak{S}}$-topology, where $\tilde{\mathfrak{S}}$ is the smallest bornology adapted to E and containing $\mathfrak{S}$.

In the statement of prop. 5, we can therefore replace the word « total » by « everywhere dense ». The proposition then follows from the general properties of equicontinuous sets (GT, X, § 2, No. 4, th. 1).

The first assertion follows from prop. 4 (III, p. 16) and the second from prop. 5 (III, p. 17).

Let $(a_n)$ be a total sequence in E. Then the mapping $u \mapsto (u(a_n))$ is an isomorphism from $\mathscr{L}(E; F)$, where $\mathscr{L}(E; F)$ has the uniform structure of simple convergence on the set of the $a_n$, onto a uniform subspace of $F^\mathbf{N}$. If F is metrizable (resp. metrizable and satisfies the first axiom of countability) then this is also true for $F^\mathbf{N}$ (GT, IX, § 2, No. 4, cor. 2 and § 2, No. 8, corollary), and the proposition follows from prop. 5 (III, p. 17).

Let B be the unit ball of F and $(V_n)$ a countable fundamental system of neighbourhoods of 0 in E. For every integer n, the set $H_n$ of all $u \in \mathscr{L}(E; F)$ such that $u(V_n) \subset B$ is equicontinuous and $\mathscr{L}(E; F)$ is the union of the $H_n$. The corollary then follows from prop. 6.*

This follows from prop. 6 and from III, p. 17, cor. 3.

(i) $M$ is equicontinuous ; (ii) $M$ is contained in the polar of a neighbourhood of 0 in E ; (iii) the polar of M is a neighbourhood of 0 in E.

If M is equicontinuous, there exists a convex balanced neighbourhood V of 0 such that $|u(x)| \leq 1$ for all $x \in V$ and all $u \in M$; then we have that $M \subset V^\circ$ and (i) implies (ii). With the same notations, if $M \subset V^\circ$ then $V \subset V^{\circ\circ} \subset M^\circ$ and (ii) implies (iii). Finally, if $M^\circ$ contains a convex balanced neighbourhood V of 0, then $M \subset M^{\circ\circ} \subset V^\circ$ and the relations $x \in \varepsilon V,\ u \in M$ imply $|u(x)| \leq \varepsilon$ for all $\varepsilon > 0$, which proves that (iii) implies (i).

We remark that every $x \in E$ defines a mapping $j(x): u \mapsto u(x)$ from $E'$ into K. Hence we can talk of the $\mathscr{S}$-topology on E, where $\mathscr{S}$ is a family of subsets of $E'$: this is the inverse image under $j$ of the $\mathscr{S}$-topology on $K^{E'}$. We verify immediately that if $\mathscr{S}$ is a convex bornology on $E'$, then the polars of sets of $\mathscr{S}$ form a fundamental system of neighbourhoods of 0 for the $\mathscr{S}$-topology on E. This is so, in particular, when $\mathscr{S}$ is the family of equicontinuous subsets of $E'$ and prop. 7 implies :

More generally, let F be a locally convex space; every $u \in \mathscr{L}(E; F)$ defines a map $j(u):(x, f) \mapsto f(u(x))$ from $E \times F'$ into K (i.e. into $\mathbf{R}$ or $\mathbf{C}$). This enables us to define, on the space $\mathscr{L}(E; F)$, the topology of uniform convergence on a set of subsets of $E \times F'$. In particular :

For every $u \in \mathscr{L}(E; F)$, every $A \in \mathscr{S}$ and every closed convex balanced neighbourhood V of 0 in F, the relation $u(A) \subset V$ is equivalent to « $j(u)(A \times V^\circ)$ is contained in the unit ball of K ».

It is obvious that the condition is necessary. Suppose it is verified, and let V be a closed convex balanced neighbourhood of 0 in F. Since V^\circ is equicontinuous, there exists a neighbourhood W of 0 in E such that |f(u(x))| \leq 1 for all x \in W, u \in H and f \in V^\circ; in other words, u(W) \subset V^{\circ\circ} = V for all u \in H, hence H is equicontinuous.

Since every simple limit of linear forms on E is a linear form (III, p. 16, prop. 4) and since the bornology \mathscr{S} on E is covering, it follows from GT, X, § 1, No. 6, cor. 2 that the space F with the \mathscr{S}-topology is Hausdorff and complete. It is clear that E'_\mathscr{S} is a topological vector subspace of F; hence it is enough to prove that E'_\mathscr{S} is everywhere dense in F. This follows from the following lemma:

Let \varepsilon > 0. There exists a closed convex balanced neighbourhood U of 0 in E such that |u(x)| \leq \varepsilon for all x \in U \cap A. We know that the polar U^\circ of U in E* is contained in E' and is compact for the topology \sigma(E^*, E) (III, p. 17, cor. 2). Since the polar A^\circ of A in E* is closed for \sigma(E^*, E), it follows that A^\circ + U^\circ is a closed convex subset of E* (GT, III, § 4, No. 1, cor. 1).

Let C be a closed convex balanced subset of E. Then C is closed for \sigma(E, E') (II, p. 45, cor. 3), hence also for \sigma(E, E^*), and as a consequence, we have C = C^{\circ\circ} (for the duality between E and E*). As a result, we have

from which, we get

Since the linear form \varepsilon^{-1}u belongs to (A \cap U)^\circ, there exist v \in A^\circ and w \in U^\circ such that u = \varepsilon(v + w). Hence x' = \varepsilon w belongs to E' and u - x' = \varepsilon v is bounded above in absolute value by \varepsilon on A; hence the lemma.

Now let E be a locally convex Hausdorff space and \hat{E} its completion. Every continuous linear form f on E extends to \hat{E} by continuity; hence we have (\hat{E})' = E'

(III, p. 16) and every element of $\hat{E}$ defines a linear form on $E'$; that is, an element of the algebraic dual ${E'}^*$ of $E'$. In addition, the duality between $E$ (resp. $\hat{E}$) and $E'$ is separating (II, p. 24, cor. 1). Consequently $E$ and $\hat{E}$ can be identified with vector subspaces of ${E'}^*$.

The space $E$ can be identified with the topological dual of $E'$ when $E'$ is assigned the topology $\sigma(E', E)$ (II, p. 43, prop. 3); on the other hand, if $\mathscr{S}$ is the set of equicontinuous subsets of $E'$, the given topology on $E$ is the $\mathscr{S}$-topology (III, p. 19, cor. 1). Then it follows from III, p. 13, prop. 1, that the sets of $\mathscr{S}$ are bounded for $\sigma(E', E)$ (cf. later on, III, p. 22, prop. 9); in other words, $\mathscr{S}$ is an adapted and covering bornology for the topology $\sigma(E', E)$. Theorem 2 is then a consequence of th. 1 if we replace $E$ by $E'$ and $E'_\mathscr{S}$ by $E$.

We know that the dual of $E$ can be identified with the product of the duals of the $E_i$ (II, p. 30, formula (1)). Let $u \in \hat{E}$, and let $u_i \in {E_i'}^*$ be the restriction of $u$ (considered as an element of ${E'}^*$) to $E_i' \subset E'$. It is immediate that it is enough to prove that $u_i = 0$ except for a finite number of indices $i \in I$. Suppose on the contrary that there exists a sequence $(i_n)_{n \in \mathbf{N}}$ of distinct indices such that $u_{i_n} \neq 0$. Then there exists $x_{i_n} \in E_{i_n}'$ such that $u_{i_n}(x_{i_n}) = n$. The set $H$ of all $x_{i_n}$ is equicontinuous in $E'$ and the restriction of $u$ to $H$ is not bounded, which is impossible.

For, if $H$ is an equicontinuous subset of $\mathscr{L}(E; F)$ and $V$ a neighbourhood of 0 in F, the set $\bigcap_{u \in H} u^{-1}(V)$ is a neighbourhood of 0 in E, hence absorbs every bounded subset of E.

A subset of $\mathscr{L}(E; F)$ which is bounded for a $\mathscr{S}$-topology is not necessarily equicontinuous, even if $\mathscr{S}$ is covering and $\mathscr{S}$ is the canonical bornology on E (IV, p. 50, exerc. 17). In the following paragraph we shall study, under the name barrelled spaces, the spaces E such that every simply bounded subset of $\mathscr{L}(E; F)$ is equicontinuous. For the present note the following result :

For every convex balanced neighbourhood $V$ of 0 in F, the set $\bigcap_{u \in H} u^{-1}(V)$ absorbs every bounded subset of E, hence is a neighbourhood of 0 in E ; this proves that H is equicontinuous.

Since H is bounded in $\mathscr{L}_{\mathscr{S}}(E; F)$ (III, p. 22, prop. 9) and closed in $F^E$ for the $\mathscr{S}$-topology (III, p. 16, prop. 4), this follows from cor. 3 of GT, X, § 1, No. 5.

For, by virtue of prop. 5 (III, p. 17) $\Phi$ is a Cauchy filter for the uniform structure of precompact convergence in E; by prop. 11, the closure $\overline{H}$ of H in $\mathscr{L}_{pc}(E; F)$ is complete and so $\Phi$ converges uniformly on every precompact subset of E to a mapping $u \in \overline{H}$.

Let $\Phi$ be a Cauchy filter in $\mathscr{L}_{\mathfrak{S}}(E; F)$. Then $\Phi$ is a Cauchy filter for the topology of simple convergence, hence converges in $F^E$; moreover, its limit $u$ is a linear mapping from E into F and $\Phi$ converges to $u$ uniformly on every set of $\mathfrak{S}$ (GT, X, § 1, No. 5, prop. 5). It follows that the image under $u$ of a sequence converging to zero is a sequence converging to zero, hence, that $u$ is continuous, since E is bornological (III, p. 11, prop. 1, (iii)).