Proposition 8

Let E be a Banach space, F a normed space. For every $u\in \mathscr{L}(E; F)$, one has $((u)) = ((^tu))$.

Let $j$ denote the canonical injection of the normed space F into its completion $\widehat{F}$. One has $((u)) = ((j\circ u))$. Since the linear mapping $^tj: (\widehat{F})'\rightarrow F'$ is bijective and isometric, one has likewise $((^tu)) = ((^t(j\circ u)))$. It is therefore enough to prove the proposition when the normed space F is complete, which we shall assume in the rest of the proof.

If $u$ is zero, one has $((u)) = ((^tu)) = +\infty$. If $u$ is not a strict morphism, then $^tu$ is not one either (EVT, IV, p. 29, Cor. 3), and one has $((u)) = ((^tu)) = 0$.

Assume henceforth that $u$ is a strict nonzero morphism. Let N denote the kernel of $u$ and I its image, and consider the canonical decomposition of $u:$

The kernel of $^tu$ is the orthogonal $I^{\circ}$ of I in $F'$ (EVT, IV, p. 27, Prop. 2), and $^ti$ defines by passing to the quotient an isometry $\iota$ of $F'/I^{\circ}$ onto $I'$ (EVT, IV, p. 9, Prop. 10). Moreover, $^tp$ defines an isometry $\pi$ of $(E/N)'$ onto the orthogonal $N^{\circ}$ of N in $E'$ (EVT, IV, p. 8, Prop. 9). The canonical decomposition of $^tu$ is therefore

where $v'=\pi \circ^tv\circ \iota$. One then has

(EVT, IV, p. 7, prop. 8), whence $((u)) = ((^tu))$ after formula (10).