Theorem 1

Let E and F be two hilbertian spaces. (i) The set $\mathscr{L}^2(E, F)$ is a vector subspace of $\mathscr{L}(E; F)$ and $u \mapsto \|u\|_2$ is a hilbertian norm (V, p. 6) on $\mathscr{L}^2(E; F)$. (ii) The isomorphism $\theta$ from $F \otimes E'$ onto $\mathscr{L}_f(E; F)$ characterized by $\theta(y \otimes x^*) = yx^*$ extends to an isomorphism $\hat{\theta}$ from $F \otimes_2 E'$ onto $\mathscr{L}^2(E; F)$. In particular, $\mathscr{L}_f(E; F)$ is dense in $\mathscr{L}^2(E; F)$.

Let $(e_i)_{i \in I}$ (resp. $(f_j)_{j \in J}$) be an orthonormal basis of $E$ (resp. $F$). For every $u \in \mathscr{L}(E; F)$, let $\Lambda(u)$ be the matrix of $u$ with respect to chosen orthonormal bases for $E$ and $F$ (V, p. 22). Let $\|a\|_2$ denote the norm of an element $a$ of the hilbertian space $\ell^2(J \times I)$. By formula (34), $\Lambda$ is a mapping from $\mathscr{L}^2(E; F)$ into $\ell^2(J \times I)$ such that $\|\Lambda(u)\|_2 = \|u\|$; it is clear that $\Lambda$ is injective. To prove (i), it is enough to prove that $\Lambda$ is surjective. Let $a = (a_{ji})$ be an element of $\ell^2(J \times I)$; by Cauchy-Schwarz inequality, we have

for every $\xi = (\xi_i)$ in $\ell^2(I)$ and $\eta = (\eta_j)$ in $\ell^2(J)$. Then there exists a continuous sesquilinear form $\Phi$ on $F \times E$ such that $\Phi(y, x) = \sum_{j,i} \overline{\eta}_j a_{ji} \xi_i$ for $x = \sum_i \xi_i e_i$ in $E$ and $y = \sum_j \eta_j f_j$ in $F$. Let $u \in \mathscr{L}(E; F)$ be such that $\Phi(y, x) = \langle y|u(x)\rangle$ (V, p. 16, cor. 2). We get hence $a = \Lambda(u)$.

Since $\Lambda$ is a hilbertian space isomorphism from $\mathscr{L}^2(E; F)$ onto $\ell^2(J \times I)$ and since $(f_j \otimes e_i^*)$ is an orthonormal basis of $F \otimes_2 E'$, there exists an isomorphism $\hat{\theta}$ from $F \otimes_2 E'$ onto $\mathscr{L}^2(E; F)$ such that

for every $i \in I,\ j \in I$ and $t \in F \otimes_2 E'$. In particular, for $t = y \otimes x^*$, we find

hence $\hat{\theta}(y \otimes x^*) = y x^*$. This proves (ii). Q.E.D.