(i) Let U be a subset of E which absorbs every bounded subset of E, and which is the intersection of a sequence of convex, balanced and closed neighbourhoods of 0 in E. Then U is a neighbourhood of 0 in E.

(ii) For every locally convex space F, every bounded subset of $\mathscr{L}_b(E; F)$ which is the union of a countable family of equicontinuous subsets, is equicontinuous.

(iii) In the strong dual $E'_b$ of E, every bounded subset which is the union of a countable family of equicontinuous subsets, is equicontinuous.

It is clear that (iii) is a particular case of (ii).

(i) $\Rightarrow$ (ii) : let H be a bounded subset of $\mathscr{L}_b(E; F)$, and let $(H_n)$ be a sequence of equicontinuous subsets of $\mathscr{L}_b(E; F)$ such that $H = \bigcup H_n$. Let V be a convex, balanced and closed neighbourhood of 0 in F. For every $n$, the set $W_n = \bigcap_{u \in H_n} u^{-1}(V)$ is a convex, balanced and closed neighbourhood of 0 in E since $H_n$ is equicontinuous. The set $W = \bigcap_{u \in H} u^{-1}(V)$ absorbs every bounded subset of E, since H is bounded in $\mathscr{L}_b(E; F)$ (III, p. 22), and we have $W = \bigcap_n W_n$. If E satisfies (i), then the set W is a neighbourhood of 0 in E, hence H is equicontinuous.

(iii) $\Rightarrow$ (i) : let $(U_n)$ be a sequence of convex, balanced and closed neighbourhoods of 0 in E. We assume that the set $U = \bigcap_n U_n$ absorbs every bounded subset of E, hence that its polar $U^\circ$ is bounded in $E'_b$. Then the set $B = \bigcup_n U_n^\circ$ is contained in $U^\circ$, hence is bounded in $E'_b$. If E satisfies (iii), the set B is equicontinuous in $E'$; consequently, the polar $B^\circ = \bigcap_n (U_n^\circ)^\circ = \bigcap_n U_n = U$ of B in E is a neighbourhood of 0 in E.

The space E is bornological (III, p. 12, prop. 2), hence its strong dual is complete (III, p. 23, cor. 1).

Let $(V_n)_{n \in \mathbf{N}}$ be a decreasing sequence of neighbourhoods of 0 in E, such that every neighbourhood of 0 in E contains one of the $V_n$. Let $A_n$ be the polar of $V_n$ in F. Since E is bornological, every bounded subset of F is equicontinuous (III, p. 22, prop. 10), therefore contained in one of the $A_n$. In other words, the space F satisfies the condition (DB).

We now show that F is semi-barrelled. Let $(U_n)_{n \in \mathbf{N}}$ be a sequence of convex, balanced and closed neighbourhoods of 0 in F. We assume that the set $U = \bigcap_n U_n$ absorbs every bounded subset of F. We shall prove that U is a neighbourhood of 0 in F. For this, we shall construct, by induction on the integer $n \geqslant 0$, real numbers $\lambda_n > 0$ and convex balanced neighbourhoods $W_n$ of 0 in F, whose which are closed for $\sigma(F, E)$, and satisfy the following relations

(1)

(2)

Suppose that the numbers $\lambda_i$ and the sets $W_i$ have been constructed for $0 \leq i < n$. By the hypothesis, the set U absorbs the bounded subsets of F; moreover, for $0 \leq i < n$, $W_i$ is a neighbourhood of 0 in F, hence absorbs the bounded subsets of F. We can therefore find a number $\lambda_n > 0$ satisfying (1). Let C denote the closed convex balanced envelope, for $\sigma(F, E)$, of $\bigcup_{0 \leq i \leq n} \lambda_i A_i$; the set C is equicontinuous, hence compact for $\sigma(F, E)$ (III, p. 17, cor. 2). Since $U_n$ is a neighbourhood of 0 in F, there exists a bounded subset B of E such that $B^\circ \subset \frac{1}{2} U_n$. Put $W_n = C + B^\circ$. Since $B^\circ$ is a neighbourhood of 0 in F, we see that $W_n$ is a convex and balanced neighbourhood of 0 in F. In addition, C is compact and $B^\circ$ closed for $\sigma(F, E)$; by cor. 1 of GT, III, § 4, No. 1, $W_n$ is closed for $\sigma(F, E)$. Finally, we have $C \subset \frac{1}{2} U \subset \frac{1}{2} U_n$ and $B^\circ \subset \frac{1}{2} U_n$, hence $W_n \subset U_n$ since $U_n$ is convex. We have thus established (2).

Put $W = \bigcap_n W_n$, then $W \subset U$. By (1) and (2), we have $\lambda_i A_i \subset W_j$ for all $i$ and $j$ in $\mathbf{N}$, and so $\lambda_i A_i \subset W$ for all $i \in \mathbf{N}$. In particular, W is absorbent, hence is a barrel for $\sigma(F, E)$. By remark 3 of IV, p. 4, W is a neighbourhood of 0 in F. A fortiori, U is a neighbourhood of 0 in F, and F is semi-barrelled.

The following corollary extends the Banach-Steinhaus theorem to the dual of a Fréchet space (cf. III, p. 25, cor. 2).

Since F is complete, the set of all $u_n$, which is bounded for the topology of simple convergence, is bounded in $\mathscr{L}_b(F; G)$ (III, p. 27, cor. 1). Since the space F is semi-barrelled (prop. 2), every countable and bounded subset of $\mathscr{L}_b(F; G)$ is equicontinuous by prop. 1 of IV, p. 21. Therefore the set of the $u_n$ is equicontinuous, and the corollary follows from III, p. 18, corollary.

By prop. 2 (IV, p. 22), there exists a sequence $(A_n)$ of bounded subsets of $E'_b$ such that every bounded subset of $E'_b$ is contained in one of the $A_n$. Let $(V_n)$ be a countable fundamental system of neighbourhoods of 0 in G. Let $H_{mn}$ be the set of linear mappings $u$ from $E'_b$ into G such that $u(A_m) \subset V_n$. Then $(H_{mn})$ is a fundamental system of neighbourhoods of 0 in $\mathscr{L}_b(E'_b ; G)$, and the latter space is then metrizable.

To show that $\mathscr{L}_b(E'_b ; G)$ is complete, it is enough to prove that every Cauchy sequence $(u_n)$ in this space is convergent; since G is complete, there exists a linear mapping $u : E'_b \to G$ such that $(u_n)$ converges simply to u. By IV, p. 23, corollary, we have $u \in \mathscr{L}_b(E'_b ; G)$. It then follows from prop. 5 of GT, X, § 1, No. 5, that $(u_n)$ converges to u in $\mathscr{L}_b(E'_b ; G)$.

Let $(V_n)_{n \in \mathbf{N}}$ be a decreasing sequence of convex, balanced and closed neighbourhoods of 0 in E, such that every neighbourhood of 0 in E contains one of the $V_n$. Let $A_n$ be the polar of $V_n$ in $E'$. Then $A_n$ is convex, balanced and compact for $\sigma(E', E)$; by III, p. 8, corollary the space $E'_{A_n}$ is a Banach space. We shall prove that $E'_b$ is the inductive limit of the spaces $E'_{A_n}$; in other words, that every convex and balanced subset U of $E'$ which absorbs each of the $A_n$ is a neighbourhood of 0 in $E'_b$. For every $n \in \mathbf{N}$, choose a real number $\lambda_n > 0$ such that $\lambda_n A_n \subset U$. Let $B_n$ be the convex envelope of the set $\bigcup_{0 \leq i \leq n} \lambda_i A_i$; put $V = \bigcup_n B_n$, then $V \subset U$. For every $n \in \mathbf{N}$, the set $B_n$ is convex, balanced and compact for $\sigma(E', E)$ (II, p. 14, prop. 15).

Now we shall show that $\frac{1}{2} V^{oo} \subset V$. Let $x \in E'_b - V$; for every $n \in \mathbf{N}$, we have $x \notin B_n$, and since $B_n$ is closed for $\sigma(E', E)$ there exists an element $y_n$ in $B_n^\circ$ such that \langle y_n, x \rangle = 1 (\text{II, p. 38, prop. 4}). Since E is reflexive, every bounded subset of E is relatively compact for $\sigma(E, E')$ (IV, p. 16, th. 2). By the definition of $B_n$, we have

hence the sequence $(y_n)$ is bounded. Let $y$ be a limit point of $(y_n)$ for the topology $\sigma(E, E')$. We have $y \in V^\circ = \bigcap_n B_n^\circ$ and $\langle y, x \rangle = 1$. Hence $x \notin \frac{1}{2}V^{\circ\circ}$, and so we have the inclusion $\frac{1}{2}V^{\circ\circ} \subset V$ and a fortiori, $\frac{1}{2}V^{\circ\circ} \subset U$.

Since every bounded subset of $E'_b$ is contained in one of the sets $A_n$, the set $V = \bigcup_n B_n$ absorbs every bounded subset of $E'_b$. Consequently, $V^\circ$ is bounded in E, hence $\frac{1}{2}V^{\circ\circ}$ is a neighbourhood of 0 in $E'_b$. A fortiori, U is a neighbourhood of in $E'_b$.

An inductive limit of Banach spaces is bornological by definition. Further, a Banach space is barrelled (III, p. 25, corollary) and every inductive limit of barrelled spaces is a barrelled space (III, p. 25, cor. 3).

a) the topology $\mathscr{T}_\mathfrak{N}$ of $\mathfrak{N}$-convergence, where $\mathfrak{N}$ is the family of subsets of E each consisting of points of a sequence converging to 0 ; b) the topology $\mathscr{T}_c$ of uniform convergence on compact subsets of E ; c) the topology $\mathscr{T}_{pc}$ of uniform convergence on precompact subsets of E ; d) the topology $\mathscr{T}_f$ which is the finest topology inducing the same topology as $\sigma(E', E)$ on every equicontinuous subset of $E'$.

First observe that a subset A of $E'$ is closed for $\mathscr{T}_f$ if and only if $A \cap H$ is closed for $\sigma(E', E)$ for every subset H of $E'$ which is equicontinuous and closed for $\sigma(E', E)$. The weak topology $\sigma(E', E)$ and $\mathscr{T}_{pc}$ induce the same topology on every equicontinuous subset of $E'$ (III, p. 17, prop. 5). Consequently each of the topologies $\mathscr{T}_\mathfrak{N}$, $\mathscr{T}_c$, $\mathscr{T}_{pc}$, $\mathscr{T}_f$ is coarser than the one following it. It is therefore enough to prove that $\mathscr{T}_\mathfrak{N}$ is finer than $\mathscr{T}_f$. Moreover, every translation in $E'$ is a homeomorphism for $\mathscr{T}_f$. Hence it is enough to prove that, if F is a subset of $E'$ which is closed for $\mathscr{T}_f$, and does not contain 0, then there exists a set $S \in \mathfrak{N}$ such that $S^\circ \cap F = \varnothing$.

Let $(U_n)_{n \geq 0}$ be a decreasing sequence of neighbourhoods of 0 in E forming a fundamental system of neighbourhoods of 0. We shall construct, by induction on $n \geq 0$, finite sets $X_n$ such that we have

for every integer $n \geqslant 0$. Let $m \geqslant 0$ be an integer such that $X_n$ has been constructed for $0 \leqslant n < m$ and satisfies (4) and (5) for $0 \leqslant n < m$. For every $x \in U_m$, put

Formula (5) with $n = m - 1$ implies that $\bigcap_{x \in U_m} F_x = \varnothing$. Further, the set $U_{m+1}^{\circ}$ is equicontinuous, and compact for $\sigma(E', E)$. In view of the definition of $\mathscr{T}_f$, each of the sets $F_x$ is compact for $\sigma(E', E)$; therefore there exists a finite subset $X_m$ of $U_m$ such that $\bigcap_{x \in X_m} F_x = \varnothing$, i.e. relation (5) is satisfied for $n = m$.

Put $S = \bigcup_{n \geqslant 0} X_n$. We have $X_n \subset U_p$ for $n \geqslant p$, therefore $S$ is the set of points of a sequence which converges to 0 in $E$. From (5) we deduce that $S^{\circ} \cap U_{n+1}^{\circ} \cap F = \varnothing$, and since $E'$ is the union of the sequence of sets $U_{n+1}^{\circ}$, we get $S^{\circ} \cap F = \varnothing$.

This follows from the fact that the topologies $\mathscr{T}_{pc}$ and $\mathscr{T}_{\mathfrak{H}}$ are identical, on account of prop. 2 of III, p. 15.

Since $E$ is complete, the topology $\mathscr{T}_c$ on $E'$ is compatible with the duality between $E'$ and $E$ (IV, p. 3, Example); consequently the closed convex subsets in $E'$ are the same for $\mathscr{T}_c$ and $\sigma(E', E)$ (IV, p. 1, prop. 1). The corollary then follows from the identity of the topologies $\mathscr{T}_c$ and $\mathscr{T}_f$.

Recall (I, p. 13) that the hyperplanes of $E'$ which are closed for $\sigma(E', E)$ are the kernels of linear forms on $E'$ associated with elements of $E$. Cor. 2 therefore gives another proof (for Fréchet spaces) of cor. 1 of III, p. 21.

Let $(U_n)_{n \in \mathbf{N}}$ (resp. $(V_n)_{n \in \mathbf{N}}$) be a fundamental system of neighbourhoods of 0 in E (resp. F). We assume that the sets $U_n$ are balanced and form a decreasing sequence. Since u is continuous, for every $n \in \mathbf{N}$, there exists a bounded set $B_n$ in E such that $u(B_n^\circ) \subset V_n$. Since $B_n$ is bounded, there exists a real number $\lambda_n > 0$ such that $\lambda_n B_n \subset U_n$. Put $B = \bigcup_{n \in \mathbf{N}} \lambda_n B_n$.

We shall prove that the set B is bounded in E, in other words that for every integer $m \geqslant 0$, there exists a real number $\mu > 0$ such that $\mu B \subset U_m$. Since the sets $B_n$ are bounded, there exists a real number $\mu$ such that $0 < \mu \leqslant 1$ and such that $\mu (\lambda_n B_n) \subset U_m$ for $0 \leqslant n \leqslant m$; we have also $\lambda_n B_n \subset U_n \subset U_m$ if $n > m$; hence $\mu B \subset U_m$ since $U_m$ is balanced.

Let U be the polar of B in $E'_b$. This is a neighbourhood of 0 in $E'_b$ and we have $\lambda_n B^\circ \subset B_n^\circ$, hence $\lambda_n u(U) \subset V_n$ for all $n \in \mathbf{N}$. Consequently $u(U)$ is bounded in F.

The space G is isomorphic to a subspace of a product of Banach spaces (II, p. 5, prop. 3). Therefore it is enough to prove the theorem under the additional hypothesis that G is a Banach space. But $F_1$ is barrelled and $F_2$ bornological (IV, p. 24, corollary), and $\mathscr{L}_b(F_2 ; G)$ is a Fréchet space (IV, p. 23, prop. 3). Let v denote the linear mapping from $F_1$ into $\mathscr{L}_b(F_2 , G)$ associated with u by the relation

Since $F_1$ is barrelled and u separately continuous, v is continuous (III, p. 31, prop. 6).

Since v is continuous, lemma 1 implies the existence of a neighbourhood $U_1$ of 0 in $F_1$ whose image under v is bounded in $\mathscr{L}_b(F_2 ; G)$. In other words, for every bounded subset $B_2$ in $F_2$, the set $u(U_1 \times B_2)$ is bounded in the Banach space G. Let $U_2$ be the set of all $x_2 \in F_2$ such that $\|u(x_1, x_2)\| \leqslant 1$ for all $x_1 \in U_1$. The set $U_2$ then absorbs every bounded subset; since $F_2$ is bornological, $U_2$ is a neighbourhood of 0 in $F_2$, and this proves that u is continuous.