Theorem 3

Let $E$ be a hilbertian space. For every $x \in E$, let $x^*$ be the continuous linear form $y \mapsto \langle x | y \rangle$ on $E$; the mapping $x \mapsto x^*$ is a bijective, semi-linear (for the automorphism $\xi \mapsto \bar{\xi}$) mapping from $E$ onto its dual $E'$, and an isometry from the normed space $E$ onto the normed space $E'$.

The mapping $x \mapsto x^*$ is semi-linear by (2) (V, p. 1) and by virtue of the Cauchy-Schwarz inequality, we have $\|x^*\| = \sup_{\|y\| \leq 1} |\langle x | y \rangle| = \|x\|$, hence $x \mapsto x^*$ is an isometry from $E$ into $E'$, and in particular, is injective. To complete the proof, we need to prove that for all $x' \neq 0$ in $E'$, there exists $x \in E$ such that $x' = x^*$. But the hyperplane $H = \mathrm{Ker}\, x'$ is closed in $E$; its orthogonal is a line $D$. Let $b$ be a non-zero element of $D$; the kernel of the linear form $b^*$ is equal to $H$ and hence there exists a scalar $\lambda \neq 0$ such that $x' = \lambda \cdot b^* = (\overline{\lambda} \cdot b)^*$. Q.E.D.

The mapping $x \mapsto x^*$ from $E$ onto its dual $E'$ is said to be canonical. The inverse mapping from $E'$ onto $E$ is also called canonical and is denoted by $x' \mapsto {x'}^*$. We have

for $x, y$ in $E$ and $x'$ in $E'$. Also $(x^*)^* = x$ for $x \in E$.

When K is $\mathbf{R}$, the mapping $x \mapsto x^*$ is linear. We shall transfer the scalar product of E to E' by this mapping. When $K = \mathbf{C}$, we can consider the mapping $x \mapsto x^*$ as an isomorphism from the vector space $\overline{E}$, the conjugate of E onto E' (V, p. 6). We shall transfer the scalar product of $\overline{E}$ to E' by this mapping.

In the two cases considered, E' is a hilbertian space and we have the formulae

for $x, y$ in E and $x'$ in E'.

To say that the vector $x \in E$ is orthogonal to a vector $y \in E$ is equivalent to saying that the linear form $x^* \in E'$ is orthogonal to $y$ in the sense defined in II, p. 41 (this justifies the use of the word « orthogonal » in the two cases). If M is a closed vector subspace of E, the subspace $M^\circ$ orthogonal to M in E' (II, p. 44) is the image under $x \mapsto x^*$ of the orthogonal of M in E, defined in V, p. 13 (this justifies the use of the notation $M^\circ$ in the two cases).