§ 1. Prehilbertian spaces and hilbertian spaces

1. Hermitian forms

We recall the following definition given in Algebra (A, IX, § 3, No. 1):

Definition 1

Let E be a vector space over the field K. A hermitian form (on the left) on E is a map f from E × E into K satisfying the following conditions (for $x_1, x_2, x, y_1, y_2, y$ in E and $\lambda, \mu$ in K):

(1)

\begin{cases} f(x_1 + x_2, y) = f(x_1, y) + f(x_2, y) \\ f(x, y_1 + y_2) = f(x, y_1) + f(x, y_2) \end{cases}

(2)

\begin{cases} f(\lambda x, y) = \overline{\lambda} f(x, y) \\ f(x, \mu y) = \mu f(x, y) \end{cases}

(3)

f(x, y) = \overline{f(y, x)} .

When the field K is R, the notion of hermitian form on E reduces to that of symmetric bilinear form on E × E (A, III, § 6, No. 3).

We note that the second condition (1) and the second condition (2) follow from the other three.

¹ For the reader specially interested in hilbertian spaces, we point out that only No. 7 of § 1 and No. 8 of § 4 depend on results of chapters III and IV. For this the reader can consult « Summary of some important properties of Banach spaces » which appears at the end of this volume. The only references to chapters I and II concern the definition of a convex set and of a semi-norm (II, p. 1 and p. 7), that of a topological direct sum (I, p. 4), of a total family and a topologically independent family (I, p. 12).

From (1) and (2) we deduce immediately that

f(\sum_j \lambda_j x_j, \sum_k \mu_k y_k) = \sum_{j,k} \overline{\lambda}_j \mu_k f(x_j, y_k).

In particular, if E is finite dimensional, and if $(e_j)_{1 \leq j \leq n}$ is a basis of E, then for $x = \sum_{j=1}^n \xi_j e_j$ and $y = \sum_{j=1}^n \eta_j e_j$ , we have,

f(x, y) = \sum_{j,k} \alpha_{jk} \overline{\xi}_j \eta_k

with the notation $\alpha_{jk} = f(e_j, e_k)$ ; moreover, relation (3) amounts to $\alpha_{jk} = \overline{\alpha}_{kj}$ for every pair of indices $j, k$ ; this implies in particular that the numbers $\alpha_{jj}$ are real.

From (3), the number $Q(x) = f(x, x)$ is real for all $x \in E$ . Moreover, we immediately establish the following formulas, known as polarization formulas

(5)

4f(x, y) = \sum_{\varepsilon^2 = 1} \varepsilon Q(x + \varepsilon y) \quad \text{if } K \text{ is } \mathbf{R},

(6)

4f(x, y) = \sum_{\varepsilon^4 = 1, \varepsilon \in \mathbf{C}} \varepsilon Q(x + \overline{\varepsilon} y) \quad \text{if } K \text{ is } \mathbf{C}.

Remark

— We observe that formula (6) is valid for every sesquilinear form on $E \times E$ (that is to say, for every function $f$ satisfying (1) and (2), but not necessarily (3)). This remark shows that, when $K = \mathbf{C}$ , a sesquilinear form $f$ such that $f(x, x)$ is real for all $x \in E$ is necessarily hermitian: relation (6) then gives $\overline{f(y, x)} = f(x, y)$ since we have $y + \varepsilon x = \varepsilon(x + \overline{\varepsilon} y)$ and $Q(\varepsilon z) = Q(z)$ whenever $\varepsilon^4 = 1$ .

From the polarisation formulas, we have in particular,

Proposition 1

If $f$ is a hermitian form on $E$ , and $M$ a vector subspace of $E$ such that $f(x, x) = 0$ for all $x \in M$ , then we also have $f(x, y) = 0$ for every pair of points $x, y$ in $M$ .

Let $f$ be a hermitian form on $E$ ; the set $N$ of all $x \in E$ such that $f(x, y) = 0$ for all $y \in E$ is a vector subspace of $E$ . It follows from (3) that, if $x_1 \equiv x_2$ (mod. $N$ ) and $y_1 \equiv y_2$ (mod. $N$ ), we have $f(x_1, y_1) = f(x_2, y_2)$ ; hence, on the quotient space $E/N$ we define a sesquilinear form $f$ by putting $f(\dot{x}, \dot{y}) = f(x, y)$ for all $x \in \dot{x}$ and all $y \in \dot{y}$ ; it is clear that $f$ is hermitian and that the relation « $f(\dot{x}, \dot{y}) = 0$ for all $\dot{y} \in E/N$ » implies $\dot{x} = 0$ in $E/N$ , in other words (A, IX) $f$ is separating. We say that $f$ is the separating hermitian form associated with $f$ .

2. Positive hermitian forms

Definition 2

Let $E$ be a vector space over the field $K$ . A hermitian form $f$ on $E$ is said to be positive if $f(x, x) \geq 0$ for all $x \in E$ .

It is clear that hermitian forms on a vector space $E$ form a vector space over the field $\mathbf{R}$ (but not over the field $\mathbf{C}$ , when $K$ is $\mathbf{C}$ ) : in this space the positive hermitian forms constitute a pointed convex proper cone (II, p. 10) as a result of def. 2 and prop. 1.

Proposition 2

— *If f is a positive hermitian form, we have

|f(x, y)|^2 \leq f(x, x) \ f(y, y)

for every x and y in E (Cauchy-Schwarz inequality).

First assume that we have $f(y, y) \neq 0$ . For every $\xi \in \mathbf{K}$ , we have

f(y, y) \ f(x + \xi y, x + \xi y) \geq 0

which can be written as

f(x, x) \ f(y, y) - |f(x, y)|^2 + (\xi f(y, y) + \overline{f(x, y)}) (\overline{\xi} f(y, y) + f(x, y)) \geq 0 .

Replacing $\xi$ by $-\overline{f(x, y)}/f(y, y)$ in this inequality, we get (7). If $f(x, x) \neq 0$ , we argue similarly.

Finally, if $f(x, x) = f(y, y) = 0$ , we have $f(x + \xi y, x + \xi y) \geq 0$ for all $\xi \in \mathbf{K}$ , which can be written as

\xi f(x, y) + \overline{\xi} f(x, y) \geq 0 .

Replacing $\xi$ by $-\overline{f(x, y)}$ in this inequality, we get $-2 \ |f(x, y)|^2 \geq 0$ , and therefore $f(x, y) = 0$ ; we again get (7) in this case.

Corollary 1

— *If f is a positive hermitian form, the set N of all x \in E such that f(x, x) = 0 coincides with the vector subspace of all x \in E such that f(x, y) = 0 for all y \in E.

Corollary 2

— *For a positive hermitian form to be separating, it is necessary and sufficient that the relation x \neq 0 implies f(x, x) > 0.

This follows immediately from cor. 1.

For every positive hermitian form f on E, the separating hermitian form associated with f (V, p. 2) is evidently a positive hermitian form on E/N.

Proposition 3

— *Let f be a positive hermitian form on E. Put

p(x) = f(x, x)^{1/2}

for all x \in E. Then p is a semi-norm on E, and is a norm if and only if f is separating.

It is enough to prove the inequality $p(x + y) \leq p(x) + p(y)$ . But we have

f(x + y, x + y) = f(x, x) + f(y, y) + f(x, y) + \overline{f(x, y)}

and, by Cauchy-Schwarz inequality

f(x + y, x + y) \leq f(x, x) + f(y, y) + 2(f(x, x) f(y, y))^{1/2}

= (f(x, x)^{1/2} + f(y, y)^{1/2})^2 .

Remark

— 1) Suppose f is positive and separating, and let x, y be two vectors $\neq 0$ . The proof of Cauchy-Schwarz inequality shows that, if the two members of (7) are equal, then there exists a scalar $\xi$ such that $f(x + \xi y, x + \xi y) = 0$ , hence $x + \xi y = 0$ , in other words, $x$ and $y$ are linearly dependent; the converse is immediate. The proof of inequality (8) shows that the equality $p(x + y) = p(x) + p(y)$ is possible only if $x$ and $y$ are linearly dependent; if $y = \lambda x$ , the preceding equality can be written as $|1 + \lambda| = 1 + |\lambda|$ , and implies that $\lambda$ is real and positive.

2) Let $f$ be a positive hermitian form on $E$ , and let $E$ be assigned the semi-norm $x \mapsto f(x, x)^{1/2}$ ; if $f$ is the positive, separating hermitian form defined on $E/N$ associated with $f$ , then the normed space obtained by assigning the norm $x \mapsto f(\dot{x}, \dot{x})^{1/2}$ to $E/N$ is the normed space associated with $E$ (II, p. 5).

Definition 3

Let $E$ be a vector space over the field $K$ . A semi-norm $p$ on $E$ is said to be prehilbertian if there exists a positive hermitian form $f$ on $E$ such that $p(x) = f(x, x)^{1/2}$ for all $x \in E$ *.

Observe that for a semi-norm $p$ on $E$ , there exists at most one positive hermitian form $f$ such that $p(x) = f(x, x)^{1/2}$ for all $x \in E$ ; this follows from the polarization formulas (V, p. 2).

3. Prehilbertian spaces

Definition 4

A prehilbertian space is a set $E$ with the structure of a vector space over $K$ and with a positive hermitian form. We say that $E$ is a real (resp. complex) prehilbertian space when $K$ is $\mathbf{R}$ (resp. $K$ is $\mathbf{C}$ ).

Example 1

The form $(\lambda, \mu) \mapsto \overline{\lambda} \mu$ defines a prehilbertian structure on $K$ , said to be canonical. When $K$ is considered as a prehilbertian space, we shall always mean, unless otherwise mentioned that it has this structure.

Example 2

Let $I$ be an interval (bounded or not) in $\mathbf{R}$ , and let $E$ be the set of regulated functions (FVR, II, p. 4) defined on $I$ with values in $\mathbf{C}$ , having compact support. It is clear that $E$ is a vector space over $\mathbf{C}$ ; let $f$ be the sesquilinear form $(x, y) \mapsto \int_1^1 \overline{x(t)}\ y(t)\ dt$ ; it is immediate that $f$ is a positive hermitian form on $E$ , and hence defines a prehilbertian structure on this space.

Example 3

Let $n \geqslant 0$ be an integer. We define a prehilbertian space structure on the space $K^n$ , by means of the hermitian form

(x, y) \mapsto \sum_{j=1}^n \overline{x_j} y_j

(for $x = (x_1, ..., x_n)$ and $y = (y_1, ..., y_n)$ ). When $K$ is $\mathbf{R}$ , we see that this is just the scalar product of two vectors of $\mathbf{R}^n$ (GT, VI, § 2, No. 2).

Example 4

Let $\ell^2$ (or $\ell^2(\mathbf{N})$ ) be the set of sequences $x = (x_n)_{n \in \mathbf{N}}$ of elements of $K$ such that $\sum_{n=0}^\infty |x_n|^2$ is finite. One can show that $\ell^2$ is a vector subspace of $K^\mathbf{N}$ and define a prehilbertian space structure on $\ell^2$ by means of the hermitian form $(x, y) \mapsto \sum_{n=0}^\infty \overline{x_n} y_n$ (cf. V, p. 18).

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,

f_{(c)}(x_1 + ix_2, y_1 + iy_2) = f(x_1, y_1) + f(x_2, y_2) + i(f(x_1, y_2) - f(x_2, y_1)) .

In particular, we have

f_{(\mathbf{C})}(x_1 + ix_2, x_1 + ix_2) = f(x_1, x_1) + f(x_2, x_2) \geqslant 0,

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

|\langle x|y \rangle| \leqslant \|x\| \cdot \|y\|.

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

\langle u(x)|u(y) \rangle = \langle x|y \rangle

for every

x

and

y

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:

\begin{align*} \langle \sum_i \lambda_i x_i | \sum_j \mu_j y_j \rangle &= \sum_{i,j} \overline{\lambda}_i \mu_j \langle x_i|y_j \rangle. \\ (\sum_i \lambda_i x_i | \sum_j \mu_j y_j) &= \sum_{i,j} \lambda_i \overline{\mu}_j (x_i|y_j). \end{align*}

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.

4. Hilbertian spaces

Definition 5

A hilbertian space* (or Hilbert space) is a prehilbertian space which is Hausdorff and complete. We say that a norm on a vector space E (over K) is hilbertian if it is prehilbertian, and if the normed space E is complete.

If E is a hilbertian space and M a closed vector subspace of E, the prehilbertian space structure induced on M is in fact a hilbertian space structure. In this case we say that M, with the induced structure is a hilbertian subspace of E.

Example 1

The prehilbertian spaces defined in examples 1, 3, 4 of V, p. 4, are hilbertian spaces. On the other hand, the prehilbertian space defined in example 2 is neither Hausdorff, nor complete. The complexification of a hilbertian space is a hilbertian space.

Example 2

\* Let X be a Hausdorff topological space and let $\mu$ be a positive measure on X. Let $L^2(X, \mu)$ be the space consisting of equivalence classes, for $\mu$ , of all square $\mu$ -integrable functions on X with values in $\mathbf{C}$ . This is a complex hilbertian space, whose scalar product is given by

\langle f|g \rangle = \int_X \overline{f(x)}\ g(x)\ d\mu(x) .

Example 3

\* Let $n \geqslant 1$ be an integer and let U be an open set in $\mathbf{R}^n$ . Let $\mu$ be the measure on U induced by the Lebesgue measure on $\mathbf{R}^n$ , and put $\mathscr{H}^0 = L^2(U, \mu)$ . Let $\mathscr{H}^1$ denote the space of all functions $f \in \mathscr{H}^0$ with the following property; for $1 \leqslant i \leqslant n$ , there exists a function $g_i \in \mathscr{H}^\circ$ such that

\int_U g_i(x)\ h(x)\ d\mu(x) = - \int_U f(x)\ D_i h(x)\ d\mu(x)

for every function $h$ of class $C^1$ with compact support in U. The function $g_i$ is defined uniquely up to equivalence with respect to $\mu$ , and is denoted by $D_i f$ or $\partial f / \partial x_i$ (ith partial derivative). By induction on the integer $s \geqslant 1$ , we define $\mathscr{H}^s$ as the set of all functions $f \in \mathscr{H}^1$ such that $D_i f \in \mathscr{H}^{s-1}$ for $1 \leqslant i \leqslant n$ . We define a scalar product on $\mathscr{H}^s$ by the formula

\langle f|g \rangle = \sum_{k=0}^s \sum_{1 \leqslant i_1 \leqslant \ldots \leqslant i_k \leqslant n} \int \overline{D_{i_1} \ldots D_{i_k} f} \cdot D_{i_1} \ldots D_{i_k} g\ d\mu .

Then $\mathscr{H}^s$ is a complex hilbertian space, called Sobolev space of index s.

Example 4

\* Let X be a differential variety of class $C^r$ (with $r \geqslant 1$ ) pure of finite dimension $n$ .

In the vector fibre space $\Lambda^n T(X)$ , let L be the complement of the zero section. For every real number $\lambda \neq 0$ , the mapping $u \mapsto \lambda u$ from $\Lambda^n T(X)$ into itself leaves L stable.

Let $\alpha$ be a complex number. A complex valued function $\omega$ on L such that $\omega(\lambda u) = |\lambda|^{\alpha} \omega(u)$ for $u \in L$ and any non-zero real number $\lambda$ is called a density of order $\alpha$ on X. We say that a density $\omega$ of order 1 is locally integrable if there exists an open cover $(U_i)_{i \in I}$ of X, and for every $i \in I$ a system of coordinates $\xi_i = (\xi_i^1, ..., \xi_i^n)$ on $U_i$ and a complex valued function $f_i$ on $\xi_i(U_i)$ satisfying the following conditions:

a) The function $f_i$ is locally integrable on the open set $\xi_i(U_i)$ of $\mathbf{R}^n$ with respect to the Lebesgue measure $\mu$ ;

b) Let $x \in U_i$ ; if $(\partial_{1,i,x}, ..., \partial_{n,i,x})$ is the basis of $T_x X$ associated to the system of coordinates $(\xi_i', ..., \xi_i^n)$ in $U_i$ we have

\omega(\partial_{1,i,x} \wedge ... \wedge \partial_{n,i,x}) = f_i(\xi_i^1(x), ..., \xi_i^n(x)) .

Then, there exists one and only one measure $\tilde{\omega}$ on X such that for every $i \in I$ , the image under $\xi_i$ of the restriction of $\tilde{\omega}$ to $U_i$ is equal to the measure $f_i \cdot \mu$ (cf. VAR, R, 10.4.3).

Let $\mathscr{V}$ (resp. $\mathscr{N}$ ) be the vector space of measurable densities $\omega$ of order 1/2 such that the measure associated with the density $|\omega|^2$ of order 1 is bounded (resp. null). Let $\omega_1$ and $\omega_2$ be in $\mathscr{V}$ ; then $\omega = \overline{\omega_1} \omega_2$ is a density of order 1, and the measure $\tilde{\omega}$ associated with $\omega$ is bounded; the number $\int_X \tilde{\omega}$ depends only on the classes $\dot{\omega}_1$ and $\dot{\omega}_2$ of $\omega_1$ and $\omega_2$ modulo $\mathscr{N}$ and is denoted by $\langle \omega_1 | \omega_2 \rangle$ or $\langle \dot{\omega}_1 | \dot{\omega}_2 \rangle$ . Then the mapping $(\dot{\omega}_1, \dot{\omega}_2) \mapsto \langle \dot{\omega}_1 | \dot{\omega}_2 \rangle$ assigns a complex hilbertian space structure to the vector space $\Omega_{1/2}(X) = \mathscr{V}/\mathscr{N}$ .

Example 5

\* Let D be the open disc with centre 0 and radius 1 in $\mathbf{C}$ . The Hardy space $H^2(D)$ consists of all holomorphic functions $f : D \to \mathbf{C}$ for which

\sup_{0 < R < 1} \int_0^1 |f(R \cdot e(\theta))|^2 d\theta < + \infty .

f_1

and

f_2

belong to

H^2(D)

, the limit

\langle f_1 | f_2 \rangle = \lim_{R \to 1} \int_0^1 \overline{f_1(R \cdot e(\theta))} \cdot f_2(R \cdot e(\theta)) \, d\theta

exists; the mapping

(f_1, f_2) \mapsto \langle f_1 | f_2 \rangle

assigns a complex hilbertian space structure to the vector space

H^2(D)

For a function $f : D \to \mathbf{C}$ to belong to $H^2(D)$ , it is necessary and sufficient that there exists a sequence $(a_n)_{n \in \mathbf{N}}$ of complex numbers such that $\sum_{n=0}^\infty |a_n|^2 < + \infty$ and that

f(z) = \sum_{n=0}^\infty a_n z^n

for all

z \in D

. Then we have

\|f\|^2 = \sum_{n=0}^\infty |a_n|^2

which gives an isomorphism from

H^2(D)

onto the hilbertian space

\ell^2

(V, p. 4).

Every Hausdorff prehilbertian space is isomorphic to an everywhere dense subspace of a hilbertian space determined up to an isomorphism. Precisely :

Proposition 4

— Let E be a Hausdorff prehilbertian space, $\hat{E}$ the normed space completion of E(GT, IX, § 3, No. 3). The scalar product* $(x, y) \mapsto \langle x|y \rangle$ extends by continuity to a positive and separating hermitian form on* $\hat{E}$ , and defines a hilbertian space structure on $\hat{E}$ .

The existence of the extension of $(x, y) \mapsto \langle x|y \rangle$ to $\hat{E} \times \hat{E}$ follows from the continuity of this sesquilinear form an $E \times E$ (GT, III, § 6, No. 5, th. 1). Moreover, this extension, which will also be denoted by $(x, y) \mapsto \langle x|y \rangle$ is a hermitian form and satisfies the relation $\langle x|x \rangle = \|x\|^2$ , by virtue of the principle of extension of identities ( $\|x\|$ being the norm on $\hat{E}$ obtained by extending the norm on E by continuity) ; this proves that the relation $\langle x|x \rangle = 0$ implies $x = 0$ in $\hat{E}$ , hence that the form $(x, y) \mapsto \langle x|y \rangle$ is positive and separating, and consequently defines a hilbertian space structure on $\hat{E}$ . Q.E.D.

This hilbertian space is said to be the completion of the Hausdorff prehilbertian space E.

\* Example 6. — Let U be an open subset of $\mathbf{R}^n$ ( $n \geqslant 1$ ). Let $\mathscr{C}_0^1(\mathrm{U})$ be the vector space of all functions of class $\mathrm{C}^1$ with compact support in U. We define a Hausdorff prehilbertian space structure on $\mathscr{C}_0^1(\mathrm{U})$ whose scalar product is given by

\langle f|g \rangle = \sum_{i=1}^{n} \int_{\mathrm{U}} \overline{\mathrm{D}_i f(x)} . \mathrm{D}_i g(x) \, dx .

This prehilbertian space is not complete. Its completion is called the Dirichlet space associated with U. \*

Corollary

— Let V be a vector space over K and f a positive hermitian form on V. a) There exists a Hilbert space E and a linear mapping $u : V \to E$ such that $f(x, y) = \langle u(x)|u(y) \rangle$ for x, y in V, and such that u(V) is dense in E. b) If two pairs $(E_i, u_i)$ satisfy the conditions analogous to a), then there exists a unique isomorphism $\phi$ from the Hilbert space $E_1$ onto the Hilbert space $E_2$ such that $u_2 = \phi \circ u_1$ .

Let N be the set of all $x \in V$ such that $f(x, x) = 0$ . We define a positive and separating hermitian form on the space V/N by $\langle \dot{x}|\dot{y} \rangle = f(x, y)$ for $x \in \dot{x}$ and $y \in \dot{y}$ . Let E be the hilbertian space completion of V/N and $u$ the mapping $x \mapsto x + N$ from V into E. Then the conditions of a) are satisfied.

Under the hypotheses of b), N is equal to the kernel of $u_1$ and to that of $u_2$ . Hence there exists a bijective linear mapping $\phi_0$ from $u_1(V)$ onto $u_2(V)$ such that $u_2(x) = \phi_0(u_1(x))$ for all $x \in V$ . We verify immediately that $\phi_0$ is an isomorphism of prehilbertian spaces, hence an isometry. Since $u_i(V)$ is dense in $E_i$ for $i = 1, 2$ , $\phi_0$ extends uniquely to an isometry $\phi$ from $E_1$ onto $E_2$ , and b) follows.

We say that the hilbertian space E is the separated completion of V (for the form f).

Example 7. — Let G be a group (with unit element 1) and $\pi$ a homomorphism from G into the group of automorphisms of a complex hilbertian space E ; we say that $\pi$ is a unitary representation of G in E. Let $a \in E$ ; we put

\phi(x) = \langle a|\pi(x).a \rangle

for all $x \in G$ . Then $\phi : G \to \mathbf{C}$ is positive definite, in other words satisfies the relation :

(PD) For every $\lambda_1, ..., \lambda_n$ in $\mathbf{C}$ and $x_1, ..., x_n$ in $G$ , we have

\sum_{i,j=1}^n \overline{\lambda_i} \lambda_j \phi(x_i^{-1} x_j) \geqslant 0 .

In fact, the first member of (11) is precisely $\| \sum_{i=1}^n \lambda_i \pi(x_i) . a \| ^2$ .

Conversely, let $\phi$ be a positive definite function on $G$ . Let $C^{(G)}$ be the vector space of all functions with finite support on $G$ . We define a hermitian form $\Phi$ on $G$ by

\Phi(u, v) = \sum_{x,y \in G} \overline{u(x)} \ v(y) \ \phi(x^{-1} y)

and the relation (PD) expresses the fact that $\Phi$ is positive. By the corollary of prop. 4, there exists a hilbertian space $E$ and a linear mapping $\rho : C^{(G)} \to E$ , with a dense image, such that

\Phi(u, v) = \langle \rho(u)|\rho(v) \rangle \quad \text{for} \quad u, v \quad \text{in} \quad C^{(G)} .

For every $x \in G$ , let $\gamma_x$ be the left translation by $x$ in $C^{(G)}$ defined by $\gamma_x u(y) = u(x^{-1} y)$ for $u \in C^{(G)}$ and $y \in G$ . We have $\Phi(\gamma_x u, \gamma_x v) = \Phi(u, v)$ . Now apply assertion b) of the corollary of prop. 4 to $\rho$ and $\rho \circ \gamma_x$ : there exists a unique automorphism $\pi(x)$ of the hilbertian space $E$ such that $\rho \circ \gamma_x = \pi(x) \circ \rho$ . We see immediately that $\pi$ is a homomorphism from $G$ into the group of automorphisms of $E$ .

Let $\delta$ be the element of $C^{(G)}$ defined by $\delta(1) = 1, \delta(x) = 0$ for $x \neq 1$ in $G$ . We have $u = \sum_{x \in G} u(x) . \gamma_x \delta$ for all $u \in C^{(G)}$ , and so $\rho(u) = \sum_{x \in G} u(x) \ \pi(x) . a$ by putting $a = \rho(\delta)$ .

Formulas (12) and (13) imply that $\phi(x) = \langle a|\pi(x).a \rangle$ for all $x \in G$ . We remark that the set of vectors $\pi(x).a$ for all $x \in G$ , is total in $E$ .

5. Convex subsets of a prehilbertian space

\| \frac{1}{2}(x + y) \| ^2 + \| \frac{1}{2}(x - y) \| ^2 = \frac{1}{2} (\| x \| ^2 + \| y \| ^2 ) .

From this identity we deduce the following proposition :

!Diagram showing sets B', B, and A, with x and y marked within A

Fig. 1.

Proposition 5

— Let $E$ be a prehilbertian space. Let $d$ be a real number $> 0$ , $\delta$ a real number such that $0 \leqslant \delta \leqslant d$ . Let $B$ and $B'$ be subsets of $E$ defined by $\| x \| < d$ ,

\|x\| \leq d + \delta \text{ respectively, and let } A \text{ be a convex set contained in } B' - B. \text{ Then for every pair of points } x, y \text{ of } A, \text{ we have } \|x - y\| \leq \sqrt{12d\delta} \text{ (fig. 1).}

In fact, we have

\frac{1}{2}(x + y) \in A

, hence

\left\| \frac{1}{2}(x + y) \right\| \geq d

; hence from (14) we get the inequality

\left\| \frac{1}{2}(x - y) \right\|^2 = \frac{1}{2}(\|x\|^2 + \|y\|^2) - \left\| \frac{1}{2}(x + y) \right\|^2 \leq (d + \delta)^2 - d^2 \leq 3d\delta

from which the proposition follows.

Theorem 1

Let E be a prehilbertian space, and H a non-empty convex subset of E such that H is a Hausdorff and complete uniform subspace of E. For every $x \in E$ , there exists a unique point $p_H(x)$ in H such that $\|x - p_H(x)\| = \inf_{y \in H} \|x - y\|$ . The element $p_H(x)$ of H is also the unique element a of H satisfying the relation*

\Re \langle x - a | y - a \rangle \leq 0

for all

y \in H

!Diagram showing a point x, a point y, and a line segment labeled a + λ(y − a)

Fig. 2.

Put $d = \inf_{y \in H} \|x - y\|$ , and for every integer $n > 0$ , let $H_n$ be the set of points $y$ of H such that $\|x - y\| \leq d + n^{-1}$ . The set $H_n$ is closed in H, is convex and non-empty, and its diameter is bounded by $\sqrt{12 \frac{d}{n}}$ for all large enough $n$ , by prop. 5. The sequence $(H_n)_{n \geq 1}$ being decreasing, and the set H being Hausdorff and complete it follows that the base of the Cauchy filter $(H_n)_{n \geq 1}$ converges to a point $p_H(x)$ of H; we have $\{ p_H(x) \} = \bigcap_{n \geq 1} H_n$ , hence $p_H(x)$ is the unique point $a$ of H such that $\|x - a\| = d$ .

1 We recall (GT, VIII, § 1, No. 1) that $\Re(z)$ denotes the real part of the complex number z; we have $\Re(z) = z$ if z is real.

Let $y \in \mathbf{H}$ ; since $\mathbf{H}$ is convex, the point $z(\lambda) = p_{\mathbf{H}}(x) + \lambda(y - p_{\mathbf{H}}(x))$ of E belongs to $\mathbf{H}$ for every real number $\lambda$ such that $0 < \lambda < 1$ . Hence we have

\| x - z(\lambda) \|^{2} \geq \| x - p_{\mathbf{H}}(x) \|^{2} \quad \text{for} \quad 0 < \lambda < 1 ,

which gives

\mathscr{R} \langle x - p_{\mathbf{H}}(x)|y - p_{\mathbf{H}}(x) \rangle = \lim_{\lambda \to 0} \frac{1}{2\lambda} \left\{ \| x - p_{\mathbf{H}}(x) \|^{2} - \| x - z(\lambda) \|^{2} \right\} \leq 0 .

Conversely, let

a

be a point of

\mathbf{H}

such that

\mathscr{R} \langle x - a|y - a \rangle \leq 0

for all

y \in \mathbf{H}

. For every

y \in \mathbf{H}

, we have

\| x - y \|^{2} = \| x - a \|^{2} + \| y - a \|^{2} - 2 \mathscr{R} \langle x - a|y - a \rangle \geq \| x - a \|^{2} ,

and so

\| x - a \| = d

and finally that

a = p_{\mathbf{H}}(x)

follows from the first part of the proof. Q.E.D.

In what follows the mapping $p_{\mathbf{H}}$ of E in $\mathbf{H}$ will be called the projection from E onto $\mathbf{H}$ . We remark that $p_{\mathbf{H}}(x) = x$ for all $x \in \mathbf{H}$ .

The first part of th. 1 is valid under more general hypotheses on the space E (V, p. 67, exerc. 31).

The proof of th. 1 establishes, among others, the following property :

Corollary 1

Let I be a set directed by a filter $\mathfrak{F}$ and let $(y_{i})_{i \in I}$ be a family of points of $\mathbf{H}$ . Let $x \in \mathbf{E}$ . Suppose that we have*

\lim_{i, \mathfrak{F}} \| x - y_{i} \| = \inf_{z \in \mathbf{H}} \| x - z \| .

*Then

y_{i}

tends to

p_{\mathbf{H}}(x)

with respect to the filter

\mathfrak{F}

Corollary 2

For every $x, y$ in $\mathbf{E}$ , we have*

\| p_{\mathbf{H}}(x) - p_{\mathbf{H}}(y) \| \leq \| x - y \| .

*In particular, the mapping

p_{\mathbf{H}}

from

\mathbf{E}

into

\mathbf{H}

is continuous.

Let $x, y$ be two points of $\mathbf{E}$ . Put $a = p_{\mathbf{H}}(x) - x, b = p_{\mathbf{H}}(y) - p_{\mathbf{H}}(x), c = y - p_{\mathbf{H}}(y)$ . By formula (15) (V, p. 10) we have $\mathscr{R} \langle a|b \rangle \geq 0$ and $\mathscr{R} \langle c|b \rangle \geq 0$ . We also have $a + b + c = y - x$ , which gives,

\| x - y \|^{2} = \| a + b + c \|^{2} = \| b \|^{2} + \| a + c \|^{2} + 2 \mathscr{R} \langle a|b \rangle + 2 \mathscr{R} \langle c|b \rangle

\geq \| b \|^{2} = \| p_{\mathbf{H}}(x) - p_{\mathbf{H}}(y) \|^{2} .

This proves corollary 2.

Proposition 6

Let $\mathbf{E}$ be a prehilbertian space and let $\Phi$ be a non-empty, directed decreasing set of non-empty Hausdorff and complete convex subsets of $\mathbf{E}$ . For every $x \in \mathbf{E}$ and every subset $\mathbf{H}$ of $\mathbf{E}$ , put $d(x, \mathbf{H}) = \inf_{z \in \mathbf{H}} \| x - z \|$ . In order that the intersection M of the sets H belonging to $\Phi$ be non-empty, it is necessary and sufficient that there exists $x_0$ in E such that $\sup_{H \in \Phi} d(x_0, H)$ is finite. For every $x \in E$ we then have

p_M(x) = \lim_{H \in \Phi} p_H(x)

(limit with respect to the directed set

\Phi

If M is non-empty, $d(x, H) \leq d(x, M)$ for all $H \in \Phi$ and all $x \in E$ .

Conversely, suppose that there exists a point $x_0$ in E and a real number $C \geq 0$ such that $d(x_0, H) \leq C$ for all $H \in \Phi$ . Let $x \in E$ ; then

d(x, H) \leq \|x - x_0\| + C \quad \text{for all } H \in \Phi,

hence the number

d = \sup_{H \in \Phi} d(x, H)

is finite. Let B be the set of all

z \in E

such that

\|x - z\| \leq d

. Since B is convex and closed in E, the sets

H \cap B

, for H ranging over

\Phi

, are convex, Hausdorff and complete. Let

\varepsilon > 0

; there exists a set

H \in \Phi

such that

d(x, H) \geq d - \varepsilon

, and if

\varepsilon < d/2

, the diameter of

H \cap B

is bounded by

\sqrt{12 \varepsilon (d - \varepsilon)}

by prop. 5 (V, p. 9). In other words, for all

H_0 \in \Phi

, the closed sets

H \cap B

, for

H \in \Phi

and

H \subset H_0

, form a base of the Cauchy filter on the Hausdorff and complete space

H_0

. Hence the intersection of the sets

H \cap B

(for

H \in \Phi

) reduces to a point y. We get

y \in M

and

\|x - y\| = d = d(x, M)

. Since M is closed in

H_0

, it is a Hausdorff, convex and complete set in E, and so

y = p_M(x)

. For every

H \in \Phi

, we have

p_H(x) \in H \cap B

, from which we get that

p_M(x) = \lim_{H \in \Phi} p_H(x)

Proposition 7

Let E be a Hausdorff prehilbertian space and let $\Psi$ be a non-empty directed increasing set of non-empty, convex, complete subsets of E. Put $A = \bigcup_{H \in \Psi} H$ and suppose that the closure N of A is complete. Then N is convex and we have*

p_N(x) = \lim_{H \in \Psi} p_H(x) \text{ for all } x \in E.

It is clear that A is convex, hence its closure N is convex (II, p. 13). With the notations of prop. 6, $d(x, N) = \inf_{H \in \Psi} d(x, H)$ , and consequently $d(x, N)$ is the limit of $d(x, H)$ with respect to the section filter of $\Psi$ . Since $p_H(x) \in H$ and

\lim_{H \in \Psi} \|x - p_H(x)\| = \lim_{H \in \Psi} d(x, H) = d(x, N),

it follows from cor. 1 of V, p. 11 that

p_H(x)

tends to the projection

p_N(x)

x

onto N with respect to the section filter of

\Psi

6. Vector subspaces and orthoprojectors

Let E be a prehilbertian space. Recall that two vectors $x$ and $y$ of E are said to be orthogonal if $\langle x | y \rangle = 0$ ; then

\|x + y\|^2 = \|x\|^2 + \|y\|^2

(« Pythagoras’ theorem »).

Let A be a subset of E. We say that a vector x in E is orthogonal to A if it is orthogonal to every vector of A. The set of all vectors orthogonal to A is a closed vector subspace of A, denoted by $A^\circ$ and called (by abuse of language) the orthogonal of A.

Let A and B be two subsets of E. We say that A and B are orthogonal if every vector of A is orthogonal to every vector of B. This is equivalent to saying that $A \subset B^\circ$ , or that $B \subset A^\circ$ . If E is Hausdorff and if A and B are orthogonal then $A \cap B$ is empty or reduces to 0 since 0 is the only vector of E orthogonal to itself.

Theorem 2

Let E be a prehilbertian space and M a vector subspace of E, which is Hausdorff and complete. Then E is the topological direct sum of M and of $M^\circ$ the subspace orthogonal to M. The projector from E onto M associated with the decomposition $E = M \oplus M^\circ$ is the projection $p_M$ from E onto M defined in th. 1 (V, p. 10).

We first show that $x - p_M(x)$ belongs to $M^\circ$ for all $x \in E$ . Let $y \in M$ . For every scalar $\lambda \in K$ , the vector $p_M(x) + \lambda y$ belongs to M; hence by formula 15 (V, p. 10) we have,

\mathscr{R}(\lambda \langle x - p_M(x)|y \rangle) \leq 0

for all $\lambda \in K$ . If, in particular we take $\lambda = \overline{\langle x - p_M(x)|y \rangle}$ we conclude that $\langle x - p_M(x)|y \rangle = 0$ , hence our assertion.

Since M is Hausdorff, 0 is the only vector of M, orthogonal to itself, hence $M \cap M^\circ = \{0\}$ . For every $x \in E$ , we have $p_M(x) \in M$ and $x - p_M(x) \in M^\circ$ . Consequently, E is the direct sum of M and $M^\circ$ , and $p_M$ is the projector from E onto M with kernel $M^\circ$ . Since $p_M$ is a continuous mapping from E into M (V, p. 11, cor. 2), if follows from GT, III, § 6, No. 2 that E is the topological direct sum of M and $M^\circ$ .

Corollary

Let E be a Hausdorff prehilbertian space and M a finite dimensional vector subspace of E. Then E is the direct sum of M and $M^\circ$ .

Since E is Hausdorff, so is M; since M is finite dimensional, it is complete (I, p. 13). It is therefore enough to apply th. 2.

With the notations of th. 2, we say that $M^\circ$ is the orthogonal complement of M and that $p_M$ is the orthoprojector (or the orthogonal projector, or by abuse of language, the projector) from E onto M; if x is a vector of E, the vector $p_M(x)$ of M is also called the orthogonal projection of x on M. Note that $p_M$ is a continuous linear mapping from E onto M and that we have $\|p_M\| = 1$ by cor. 2 of V, p. 11, except in the case when $M = \{0\}$ in which case $p_M = 0$ .

It follows immediately from Pythagoras theorem that the canonical mapping $\psi$ from $E/M$ onto $M^\circ$ deduced from the direct sum decomposition $E = M \oplus M^\circ$ is isometric if $E/M$ is assigned the quotient semi-norm from that of E (II, p. 4). We shall always assign that prehilbertian structure to $E/M$ for which $\psi$ is an isomorphism of prehilbertian spaces; the quotient semi-norm on $E/M$ is then deduced from this prehilbertian structure.

We shall often use the preceding results when E is a hilbertian space and M a closed vector subspace of E. In this case, $M^\circ$ is a closed vector subspace of E, and $p_{M^\circ} = 1 - p_M$ , and $(M^\circ)^\circ = M$ .

Proposition 8

— Let E be a hilbertian space, M a closed vector subspace of E, I a non-empty ordered directed set and $(M_i)_{i \in I}$ a family of closed vector subspaces of E. We assume that either the mapping $i \mapsto M_i$ is increasing and that M is the closure of $\bigcup_{i \in I} M_i$ or that the mapping $i \mapsto M_i$ is decreasing and that $M = \bigcap_{i \in I} M_i$ . Then $p_M(x) = \lim_{i \in I} p_{M_i}(x)$ for all $x \in E$ .

Prop. 8 follows immediately from props. 6 (V, p. 11) and 7 (V, p. 12).

Proposition 9

— Let E be a hilbertian space and M, N two closed vector subspaces of E.

a) The following conditions are equivalent : (i) $p_M p_N = p_N p_M$ ; (ii) if $x \in M$ is orthogonal to $M \cap N$ and if $y \in N$ is orthogonal to $M \cap N$ , then x and y are orthogonal; (iii) every vector of M orthogonal to $M \cap N$ is orthogonal to N; (iv) $M = (M \cap N) + (M \cap N^\circ)$ .

b) If the equivalent conditions of a) are satisfied, we have $p_{M \cap N} = p_M p_N$ , the vector subspace $M + N$ of E is closed and we have $p_{M+N} = p_M + p_N - p_M p_N$ .

c) We have $p_M p_N = 0$ if and only if M is orthogonal to N. If this is so, then the vector subspace $M + N$ of E is closed, and $p_{M+N} = p_M + p_N$ .

Put $L = M \cap N$ , $M_1 = M \cap L^\circ$ and $N_1 = N \cap L^\circ$ . Condition (ii) implies that $M_1$ and $N_1$ are orthogonal, and (iii) implies that $M_1$ and N are orthogonal. Since we have $N = N_1 + L$ and $M_1$ is orthogonal to L, we have proved the equivalence of (ii) and (iii). If condition (iii) is satisfied, we have $M_1 = M \cap N^\circ$ and since $M = L + M_1$ , condition (iv) is satisfied. Conversely, from (iv) we conclude that $M_1 = M \cap N^\circ$ since the subspaces $M \cap N$ and $M \cap N^\circ$ of M are orthogonal, and so $M_1 \subset N^\circ$ , that is, the relation (iii).

Assume that condition (iv) is satisfied. It is immediate that $p_N(y) = p_L(y)$ for all $y \in M$ and hence $p_N p_M(x) = p_L p_M(x)$ for all $x \in E$ . But, for every $x \in E$ , the vector $p_L p_M(x)$ belongs to L, and the vector

x - p_L p_M(x) = (x - p_M(x)) + (p_M(x) - p_L(p_M(x)))

belongs to

M^\circ + L^\circ = L^\circ

; hence we have

p_L p_M(x) = p_L(x)

. Finally,

p_N p_M = p_L p_M = p_L

. Since condition (ii) is equivalent to (iv) and is symmetric in M and N, we also have

p_M p_N = p_L

. Finally we get

p_M p_N = p_N p_M = p_{M \cap N}

which gives (i).

Conversely, suppose condition (i) is satisfied. Let $x \in M$ ; we have

p_M(p_N(x)) = p_N(p_M(x)) = p_N(x)

and so

p_N(x) \in M

. We conclude that

x - p_N(x) \in M

, hence

x

is the sum of an element

p_N(x)

M \cap N

and an element

x - p_N(x)

M \cap N^\circ

, which gives (iv).

We have proved a) and the first part of b). Assume now that $p_M$ and $p_N$ commute and put $q = p_M + p_N - p_M p_N$ ; since $p_M$ and $p_N$ are idempotents in the algebra $\mathscr{L}(E)$ , so is q; hence (GT, III, § 6, No. 2) the image of q is a closed vector subspace of E.

It is clear that the image of $q$ is contained in $M + N$ ; however, we have $p_N(x) = x$ , hence $q(x) = x$ for all $x \in N$ ; since we also have $q = p_M + p_N - p_N p_M$ , we get $q(x) = x$ for all $x \in M$ . We conclude that the image of $q$ is equal to $M + N$ . The orthogonal of $M + N$ is equal to $M^\circ \cap N^\circ$ , and the kernel of $q$ obviously contains $M^\circ \cap N^\circ$ , hence $q = p_{M+N}$ . This proves b).

We have $p_M p_N = 0$ if and only if the image $N$ of $p_N$ is contained in the kernel $M^\circ$ of $p_M$ , that is, if and only if $M$ is orthogonal to $N$ . The rest of the assertion c) is then a particular case of b).

Remark

— Let $E$ be a hilbertian space and $M, N$ two closed vector subspaces of $E$ . The relation $M \subset N$ is equivalent to the orthogonality of $M$ and $N^\circ$ , that is to say, to the relation $p_M p_{N^\circ} = 0$ by prop. 9, c). Since we have $p_{N^\circ} = 1 - p_N$ , we conclude that the relations $M \subset N$ and $p_M = p_M p_N$ are equivalent (« the three perpendicular theorem », cf. fig. 3).

!Diagram showing subspaces M, N, and projections p_M(x), p_N(x)

Fig. 3.

7. Dual of a hilbertian space

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

\langle x | y \rangle = \langle y, x^* \rangle, \quad \langle x, x' \rangle = \langle {x'}^* | x \rangle

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

\langle x^*|y^* \rangle = \overline{\langle x|y \rangle}, \quad \langle x'|x' \rangle = \|x'\|^2

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).

Corollary 1

In order that the family $(x_i)_{i \in I}$ of points of a hilbertian space E be total, it is necessary and sufficient that the relations $\langle x_i|y \rangle = 0$ for $y \in E$ and for all indices $i \in I$ imply that $y = 0$ .

In fact, this says that 0 is the only vector of E' which is orthogonal to all the $x_i$ (II, p. 43 and IV, p. 1).

Corollary 2

Let E and F be two hilbertian spaces. For $u \in \mathscr{L}(E; F)$ , $x \in E$ and $y \in F$ , put

\Phi_u(y, x) = \langle y|u(x) \rangle .

The mapping $u \mapsto \Phi_u$ is an isomorphism from the Banach space $\mathscr{L}(E; F)$ onto the space of all continuous sesquilinear $^1$ forms on $F \times E$ , endowed with the norm

\|f\| = \sup_{\substack{x \in E, y \in F \\ \|x\| \leq 1, \|y\| \leq 1}} |f(y, x)| .

It is clear that $\Phi_u$ is sesquilinear and continuous for all $u \in \mathscr{L}(E; F)$ . Conversely, let $f$ be a continuous sesquilinear form on $F \times E$ . For every $x \in E$ , the mapping $y \mapsto f(y, x)$ is a continuous linear form on the hilbertian space F. By th. 3, for every $x \in E$ , there exists a unique element $u(x)$ in F such that $f(y, x) = \langle u(x)|y \rangle$ for all $y \in F$ . The mapping $u : x \mapsto u(x)$ from E into F is linear and we have

\begin{align*} \|f\| &= \sup_{\|x\| \leq 1} \sup_{\|y\| \leq 1} |f(y, x)| = \sup_{\|x\| \leq 1} \sup_{\|y\| \leq 1} |\langle y|u(x) \rangle| \\ &= \sup_{\|x\| \leq 1} \|u(x)\| ; \end{align*}

hence $u$ belongs to $\mathscr{L}(E; F)$ , $f = \Phi_u$ and $\|u\| = \|f\|$ . This proves cor. 2.

$^1$ Recall (A, IX, § 1, No. 5) that a sesquilinear form (on the left) $f$ on $F \times E$ is a mapping from $F \times E$ into K which satisfies relations (1) and (2) of V, p. 1.

The canonical mapping from E into its bidual E'' (IV, p. 14) maps E onto E'', in other words (IV, p. 16), E is a reflexive Banach space. In fact, if E is a real (resp. complex) hilbertian space, the canonical mapping φ from E' onto E is an isomorphism from the normed space E' onto E (resp. onto the conjugate space $\overline{E}$ of E); applying th. 3 to E (resp. E), we see that every continuous linear form on the normed space E' is of the form $x' \mapsto \langle \phi(x')|x \rangle = \langle x, x' \rangle$ with $x \in E$ , hence our assertion follows.

As a consequence (IV, p. 17, prop. 6):

Theorem 4

In a hilbertian space E, the unit ball is weakly compact.

Proposition 10

If, in a hilbertian space E, a filter $\mathfrak{F}$ converges weakly to $x_0$ , and if moreover $\lim_{\mathfrak{F}} \|x\| = \|x_0\|$ , then $\mathfrak{F}$ converges to $x_0$ for the initial topology of E.

In fact, $\|x - x_0\|^2 = \|x\|^2 - 2\Re \langle x|x_0 \rangle + \|x_0\|^2$ . Since $\langle x|x_0 \rangle$ tends to $\|x_0\|^2$ with respect to $\mathfrak{F}$ by hypothesis, and $\|x\|$ tends to $\|x_0\|$ with respect to $\mathfrak{F}$ , $\|x - x_0\|$ tends to 0 with respect to $\mathfrak{F}$ , hence the proposition.

Remark

— If E is a Hausdorff prehilbertian space and $\hat{E}$ the hilbertian space completion of E, we know (III, p. 16) that the dual E' of E can be identified with the dual of $\hat{E}$ ; it then follows from th. 3 (V, p. 15) that every continuous linear form on E can be written in a unique way as $x \mapsto \langle a|x \rangle$ , where $a \in \hat{E}$ .

Exercises

See the exercises for § 1.