Corollary 2

— Let $(e_i)_{i \in J}$ be a basis of the vector space F with dual $F^*$, and let $u : \mathbf{R}^{(I)} \to F$ be an (algebraic) isomorphism defined by this basis. Then the transposed mapping $^t u : F^* \to \mathbf{R}^I$ is a topological isomorphism when $F^*$ carries the topology $\sigma(F^*, F)$ and $\mathbf{R}^I$ the product topology.

We know (A, II, § 2.6, prop. 10) that $^t u$ is a bijection, and that if for a $x^* \in F^*$, we put $\langle e_i, x^* \rangle = \xi_i^*$ for all $i \in I$, then the image $^t u(x^*)$ is the vector $(\xi_i^*)$ of $\mathbf{R}^I$, so that, for all $x = \sum_i \xi_i e_i$ in F, we have $\langle x, x^* \rangle = \sum_i \xi_i \xi_i^*$. The corollary then follows from this formula and prop. 8.