Lemma 3

— (i) For every $i \in I$, let $\mathfrak{S}_i$ be a family of subsets of $F_i$, which is bounded for $\sigma(F_i, E_i)$; put $\mathfrak{S} = \bigcup_{i \in I} \mathfrak{S}_i$. Then the $\mathfrak{S}$-topology on $E$ is the product of the $\mathfrak{S}_i$-topologies on the $E_i$.

(ii) For every $i \in I$, let $\mathfrak{J}_i$ be an adapted bornology on the space $E_i$ endowed with the weak topology $\sigma(E_i, F_i)$, none equal to $\{ \varnothing \}$. Let $\mathfrak{J}$ be the family of subsets $A$ of $E = \prod_{i \in I} E_i$ such that $\operatorname{pr}_i(A) \in \mathfrak{J}_i$ for all $i \in I$. Then the $\mathfrak{J}$-topology on $F$ is the direct sum of the $\mathfrak{J}_i$-topologies on the $F_i$.

Let $\mathscr{T}$ be the product of the $\mathfrak{S}_i$-topologies. The sets of the form

where $J \subset I$ is finite and $A_i \in \mathfrak{S}_i$ for all $i \in J$, form a fundamental system of neighbourhoods of 0 for $\mathscr{T}$. We have $A = (\bigcup_{i \in J} A_i)^\circ$, hence $\mathscr{T}$ is identical with the $\mathfrak{S}$-topology. This proves (i).

We assign the $\mathfrak{J}$-topology to $F$ and the $\mathfrak{J}_i$-topology to each $F_i$. For every subset $A$ of $E$, we have $F_i \cap A^\circ = \operatorname{pr}_i(A)^\circ$, hence the injection from $F_i$ into $F$ is continuous. Let $q$ be a semi-norm on $F$; we assume that the restriction $q_i$ of $q$ to $F_i$ is continuous for all $i \in I$. Then we can find non-empty subsets $A_i \in \mathfrak{J}_i$ such that we have

Put $A = \prod_{i \in I} A_i$; then $A \in \mathfrak{J}$. For $y = (y_i)_{i \in I}$ in $F$, we have

where the last equality follows from (8) since the family $(y_i)_{i \in I}$ has finite support and the $A_i$ are non-empty and can be assumed balanced (GT, IV, § 5, No. 7, cor. 2 to prop. 12). This inequality proves that $q$ is continuous on $F$, and hence (ii).