Proposition 12

The topology $\beta(F, E)$ is the direct sum of the topologies $\beta(F_i, E_i)$. The topology $\beta(E, F)$ is the product of the topologies $\beta(E_i, F_i)$.

We shall apply lemma 3 taking for $\mathfrak{S}_i$ the family of all subsets of $F_i$ which are bounded for $\sigma(F_i, E_i)$ and for $\mathfrak{J}_i$ the family of all subsets of $E_i$ which are bounded for $\sigma(E_i, F_i)$.

By cor. 2 of III, p. 4, $\mathfrak{J}$ is the family of all subsets of $E_i$ which are bounded for the product topology of the $\sigma(E_i, F_i)$, which is identical with $\sigma(E, F)$. Hence our assertion on $\beta(F, E)$ follows.

We endow $F = \bigoplus_{i \in I} F_i$ with the topology $\mathscr{T}$ which is the direct sum of the topologies $\sigma(F_i, E_i)$. Then the dual of $F$ consists of the linear forms $y \mapsto B(x, y)$ where $x$ runs through $E$ (II, p. 30, prop. 6). By prop. 1 of IV, p. 1, the topologies $\mathscr{T}$ and $\sigma(F, E)$ have the same bounded sets. Assume first that that the topologies $\sigma(F_i, E_i)$ are Hausdorff. By prop. 5 of III, p. 5, these sets are contained in a subset of the form $\sum_{i \in J} B_i$ with $J \subset I$ finite and $B_i$ bounded in $F_i$ (for $\sigma(F_i, E_i)$) for all $i \in J$. Since $\sum_{i \in J} B_i$ is contained in the convex envelope of $\bigcup_{i \in J} nB_i$, where $n = \mathrm{Card}(J)$, we can apply lemma 3, to prove the assertion on $\beta(E, F)$ in this case.

For the general case, let $N_i$ be the intersection of all neighbourhoods of 0 for $\sigma(F_i, E_i)$, and let $N = \sum_{i \in I} N_i$; then $F/N$ is the topological direct sum of the $F_i/N_i$ (II, p. 31, prop. 8); we deduce from this that every subset of $F$ which is bounded for $\mathscr{T}$ is contained in a set of the form $N + \sum_{i \in J} B_i$ with $J \subset I$ finite and $B_i$ bounded in $F_i$ for all $i \in J$ (III, p. 2, Remark 3); since the polar of this set in $E$ is the same as that of $\sum_{i \in J} B_i$, the result follows as above.