Example

— 1) Let E be a vector space and let E* be its dual. The canonical bilinear form $(x, x^*) \mapsto \langle x, x^* \rangle$ on $E \times E^*$ (A, II, § 2.3) puts E and E* in separating duality : for (D_{II}) is true because of the definition of the relation $x^* \neq 0$, and we know on the other hand, that for all $x \neq 0$ in E, there exists a linear form $x^* \in E^*$ such that $\langle x, x^* \rangle \neq 0$ (A, II, § 7.5, th. 6), which proves (D_I); the identifications of E with a subspace of $E^{**}$ is made here by the canonical mapping $c_E$ (loc. cit.).

When E is of finite dimension, the only subspace G of E* that is in separating duality by the restriction to $E \times G$ of the canonical bilinear form, is the space E* itself ; for, E being then canonically identified with $E^{**}$ (loc. cit.), if we had $G \neq E^*$, there would exist $a \neq 0$ in E such that $\langle a, x^* \rangle = 0$ for all $x^* \in G$ (A, II, § 7.5, th. 7), which contradicts the hypothesis.

2) When E is an infinite dimensional vector space, and E' is a vector subspace of E*, the duality between E and E' defined by the restriction to $E \times E'$ of the canonical bilinear form is always separating in E' ; it can be separating in E even if $E' \neq E^*$. The most important case occurs where E is a topological vector space.