Proposition 5

Let E be a vector space, and $(F_\alpha)_{\alpha \in A}$ be a family of topological vector spaces and for each $\alpha \in A$, let $g_\alpha$ be a linear mapping of $F_\alpha$ in E.

(i) Denote by $\mathfrak{B}$ the family of absorbent, symmetric convex subsets V of E such that $g_\alpha^{-1}(V)$ is a neighbourhood of 0 in $F_\alpha$ for every $\alpha$; the family $\mathfrak{B}$ is a fundamental system of neighbourhoods of 0 in E for a topology $\mathscr{T}$ that is compatible with the vector space structure.

(ii) A linear mapping f of E in a locally convex space G (resp. a semi-norm p on E) is continuous for $\mathscr{T}$ if and only if, for every index $\alpha$, $f \circ g_\alpha$ (resp. $p \circ g_\alpha$) is continuous in $F_\alpha$.

(iii) The topology $\mathscr{T}$ is the finest of the locally convex topologies on E for which the $g_\alpha$ are continuous.

Further, the topology $\mathscr{T}$ is the only locally convex topology on E that satisfies condition (ii) for linear mappings (resp. for the semi-norms).

As $\mathfrak{B}$ is a filter base invariant under homotheties of ratio > 0, the assertion (i) follows immediately from II, p. 23, prop. 1. By the definition of $\mathfrak{B}$, the topology $\mathscr{T}$ is the finest of locally convex topologies on E making the $g_\alpha$ continuous; whence (iii). Finally, it is clear that if $f$ is continuous, so is $f \circ g_\alpha$; conversely if the $f \circ g_\alpha$ are continuous for every $\alpha$, then for each symmetric convex neighbourhood W of 0 in G, the set $g_\alpha^{-1}(f^{-1}(W))$ is a neighbourhood of 0 in $F_\alpha$ for each $\alpha$. Now $f^{-1}(W)$ is absorbent, symmetric and convex thus $f^{-1}(W)$ is a neighbourhood of 0 in $\mathscr{T}$, and $f$ is continuous. Similarly if $p$ is a semi-norm on E such that $p \circ g_\alpha$ is continuous for every $\alpha$, and if U is the set of points $x \in E$ such that $p(x) < 1$, then, for every $\alpha$, the set $g_\alpha^{-1}(U)$ is a convex neighbourhood of 0 in $E_\alpha$ that is symmetric and absorbent; thus U is a neighbourhood of 0 in E and $p$ is continuous (II, p. 2, prop. 1).

The last statement follows from S, IV, § 2.5, criterion CST 18.

We say that $\mathscr{T}$ is the locally convex final topology of the family of topologies $\mathscr{T}_\alpha$ of the $F_\alpha$, for the family of linear mappings $g_\alpha$.

It may happen that $\mathscr{T}$ is not the finest of the topologies on E compatible with its vector space structure and making the $f_\alpha$ continuous (II, p. 75, exerc. 15; see also II, p. 75, exerc. 14).

In the most important case $E = \sum_{\alpha \in A} g_\alpha(F_\alpha)$, we get a fundamental system of neighbourhoods of 0 for $\mathscr{T}$ as follows; for each $\alpha \in A$, let $V_\alpha$ be a symmetric neighbourhood of 0 for $\mathscr{T}_\alpha$, form the union of the $g_\alpha(V_\alpha)$ for $\alpha \in A$ and denote the convex envelope in E of this union by $\Gamma((g_\alpha(V_\alpha)))$; since every element of E is of the form $\sum_{\alpha \in J} x_\alpha$, where J is a finite subset of I and $x_\alpha \in g_\alpha(F_\alpha)$, it is immediate that $\Gamma((g_\alpha(V_\alpha)))$ is an absorbent symmetric convex set in E (each of the $V_\alpha$ is absorbent in $F_\alpha$); as $\Gamma((g_\alpha(V_\alpha)))$ contains all the $g_\alpha(V_\alpha)$, it is a neighbourhood of 0 for $\mathscr{T}$. On the other hand, it is clear that for every symmetric convex neighbourhood V of 0 for $\mathscr{T}$, we have $V \supset \Gamma((V \cap g_\alpha(F_\alpha)))$, from which our assertion follows.