Example 5

Let $E$ be a real prehilbertian space, $f$ the corresponding symmetric bilinear form on $E$. Let $E_{(c)}$ be the vector space complexification of $E$; we identify $E$ with a subset of $E_{(c)}$ by the map $x \mapsto 1 \otimes x$, in such a way that every element of $E_{(c)}$ can be written uniquely as $x_1 + ix_2$ with $x_1, x_2$ in $E$. The map $f$ extends uniquely to a hermitian form $f_{(c)}$ on $E_{(c)}$; we have,

In particular, we have

hence $f_{(\mathbf{C})}$ is positive. We say that $E_{(\mathbf{C})}$, with $f_{(\mathbf{C})}$ is the prehilbertian space complexification of $E$.

Whenever only one prehilbertian space structure on a vector space $E$ is under consideration, the value, for a pair $(x, y)$ of points of $E$, of the hermitian form which defines the said structure is denoted by $\langle x|y \rangle_E$ or simply $\langle x|y \rangle$, if no confusion is likely to arise. This number is called the scalar product $^1$ of $x$ and $y$ (scalar square of $x$ if $y = x$). Two vectors $x, y$ are said to be orthogonal if $\langle x|y \rangle = 0$. The function $x \mapsto \|x\| = \langle x|x \rangle^{1/2}$ is a semi-norm on the vector space $E$ (V, p. 3); a prehilbertian space is always considered with this semi-norm assigned to it (and consequently also with the corresponding topology and uniform structure).

With these notations, in a prehilbertian space $E$, the Cauchy-Schwarz inequality can be written as

Consequently, the scalar product is a continuous sesquilinear form on $E \times E$ (II, p. 5, prop. 4).

In order that $E$ be Hausdorff, it is necessary and sufficient that $x \mapsto \|x\|$ is a norm on $E$; in other words, that the hermitian form $(x, y) \mapsto \langle x|y \rangle$ is positive and separating; this is equivalent to saying that $0$ is the only vector of $E$, which is orthogonal to itself.

According to general definitions (S, IV, § 1, No. 5), an isomorphism from a prehilbertian space $E$ onto a prehilbertian space $F$ is a bijective linear mapping $u$ from $E$ onto $F$ such that

for every $x$ and $y$ in $E$. We deduce from this that $\|u(x)\| = \|x\|$ for all $x \in E$, and $u$ is evidently an isomorphism for the topological vector space structures of $E$ and of $F$; if $E$ and $F$ are Hausdorff, $u$ is an isometry from $E$ onto $F$. Conversely, if $u$ is a bijective linear mapping from $E$ onto $F$, such that $\|u(x)\| = \|x\|$ for all $x \in E$, the polarization formulas (V, p. 2) show that $u$ is a prehilbertian space isomorphism from $E$ onto $F$.

Let $E$ be a complex prehilbertian space, and $\langle x|y \rangle$ the scalar product in $E$. On the set $E$, we can define a second vector space structure with respect to $\mathbf{C}$, taking the same law of the additive group and for the law of external composition $(\lambda, x) \mapsto \overline{\lambda} x$ (A, II, § 1, No. 13) for this vector space structure, $(x, y) \mapsto \langle y|x \rangle$ is a positive hermitian

1 It may happen sometimes that we write $(x|y)$ for $\langle y|x \rangle$. Observe that the formula (4) of V, p. 2, takes the following equivalent forms:

form. The prehilbertian space $\overline{E}$ obtained by assigning this new vector space structure and the new hermitian form to E, is said to be conjugate to E. An isomorphism $u$ from E onto $\overline{E}$ is a semi-linear mapping from E onto itself (with respect to the automorphism $\xi \mapsto \overline{\xi}$ of $\mathbf{C}$) such that $\langle u(y)|u(x) \rangle = \langle x|y \rangle$ or $\langle u(x)|u(y) \rangle = \overline{\langle x|y \rangle}$ (for $x, y$ in E); such a mapping is said to be a semi-automorphism of the prehilbertian space E.

If E is a prehilbertian space, M a vector subspace of E, the restriction of the scalar product $\langle x|y \rangle$ to $M \times M$ is a positive hermitian form on M, which then defines a prehilbertian space structure on M; we say that this structure is induced by the structure of E, or that M is a prehilbertian subspace of E.