Proposition 18

The mapping $\theta$ is an isometric isomorphism.

By lemma 9, the mapping $\theta$ is isometric, and it is enough to prove that $\theta$ is surjective.

Let $\lambda \in \mathscr{L}^c(F; E)'$. Since $\|v\|\leqslant \|v\|_2$ for every $v\in \mathscr{L}_2(F; E)$ (EVT, V, p. 52, formula (33)), the restriction of $\lambda$ to the subspace $\mathscr{L}_2(F; E)$ is a continuous linear form on the Hilbert space $\mathscr{L}_2(F; E)$. There therefore exists an element $u$ of $\mathscr{L}_2(F; E)$ such that $\lambda (v) =\langle u|v\rangle$ for every $v\in \mathscr{L}_2(F; E)$ (EVT, V, p. 15, th. 3).

For every $w\in \mathscr{L}_2(E; F)$ of norm $\leqslant 1$, we have $|\langle w|u\rangle |=|\lambda (w)|\leqslant \|\lambda \|$. Consequently, $u$ is a nuclear mapping from E into F (prop. 14).

The continuous linear forms $\lambda$ and $\theta (u)$ are equal on the dense subspace $\mathscr{L}_2(F,E)$ of $\mathscr{L}^c(F; E)$. It follows that $\lambda =\theta (u)$, which concludes the proof.