Remark 1

Suppose $K = \mathbf{C}$. Let $u$ and $v$ be in $\mathscr{L}^2(E; F)$. We have the relation

hence $u^* v$ belongs to $\mathscr{L}^1(E)$. The scalar product in the hilbertian space $\mathscr{L}^2(E; F)$ is given by since this formula defines a hermitian form on $\mathscr{L}^2(E; F)$ and we get $\langle u | u \rangle = \| u \|_2^2$.

If $u \in \mathscr{L}^2(E; F)$ and $v \in \mathscr{L}^2(F; E)$, then $vu$ belongs to $\mathscr{L}^1(E)$ and $uv$ to $\mathscr{L}^1(F)$ by the preceding; moreover, we have

By linearity and continuity, it is enough to verify this formula when $u = y_1 x_1^*$ and $v = x_2 y_2^*$ (with $x_1, x_2$ in E, $y_1, y_2$ in F); but then $uv$ is the mapping $y \mapsto y_1 \langle x_1 | x_2 \rangle \langle y_2 | y \rangle$ and $vu$ the mapping $x \mapsto x_2 \langle y_2 | y_1 \rangle \langle x_1 | x \rangle$, and (41) follows from formula (22) of V, p. 48.

Consequently, if $u_1, u_2$ are two elements of $\mathscr{L}^2(E; F)$, we have, in the hilbertian space $\mathscr{L}^2(F; E)$,

in other words, $u \mapsto u^*$ is an isomorphism from the hilbertian space $\mathscr{L}^2(E; F)$ onto the conjugate (V, p. 6) of the hilbertian space $\mathscr{L}^2(F; E)$. If we identify this conjugate with the dual of $\mathscr{L}^2(F; E)$ (V, p. 15), we see that $\mathscr{L}^2(E; F)$ can be identified with the dual of $\mathscr{L}^2(F; E)$, the canonical bilinear form $(v, u) \mapsto \langle v, u \rangle$ being identified with $(v, u) \mapsto \operatorname{Tr}(vu)$.