Corollary 2

Let E and F be two hilbertian spaces. For $u \in \mathscr{L}(E; F)$, $x \in E$ and $y \in F$, put

The mapping $u \mapsto \Phi_u$ is an isomorphism from the Banach space $\mathscr{L}(E; F)$ onto the space of all continuous sesquilinear $^1$ forms on $F \times E$, endowed with the norm

It is clear that $\Phi_u$ is sesquilinear and continuous for all $u \in \mathscr{L}(E; F)$. Conversely, let $f$ be a continuous sesquilinear form on $F \times E$. For every $x \in E$, the mapping $y \mapsto f(y, x)$ is a continuous linear form on the hilbertian space F. By th. 3, for every $x \in E$, there exists a unique element $u(x)$ in F such that $f(y, x) = \langle u(x)|y \rangle$ for all $y \in F$. The mapping $u : x \mapsto u(x)$ from E into F is linear and we have

hence $u$ belongs to $\mathscr{L}(E; F)$, $f = \Phi_u$ and $\|u\| = \|f\|$. This proves cor. 2.

$^1$ Recall (A, IX, § 1, No. 5) that a sesquilinear form (on the left) $f$ on $F \times E$ is a mapping from $F \times E$ into K which satisfies relations (1) and (2) of V, p. 1.

The canonical mapping from E into its bidual E'' (IV, p. 14) maps E onto E'', in other words (IV, p. 16), E is a reflexive Banach space. In fact, if E is a real (resp. complex) hilbertian space, the canonical mapping φ from E' onto E is an isomorphism from the normed space E' onto E (resp. onto the conjugate space $\overline{E}$ of E); applying th. 3 to E (resp. E), we see that every continuous linear form on the normed space E' is of the form $x' \mapsto \langle \phi(x')|x \rangle = \langle x, x' \rangle$ with $x \in E$, hence our assertion follows.

As a consequence (IV, p. 17, prop. 6):