a) E is a vector subspace of P.

b) For every $x = (x_i)_{i \in I}$ and $y = (y_i)_{i \in I}$ in E, the family $(\langle x_i|y_i \rangle)_{i \in I}$ is summable. If we put $\langle x|y \rangle = \sum_{i \in I} \langle x_i|y_i \rangle$, we define a positive separating hermitian form on E.

c) For the scalar product so defined, E is a hilbertian space; the direct sum S of the $E_i$ is dense in E.

For $x = (x_i)_{i \in I}$ and $y = (y_i)_{i \in I}$ in E, we have

hence $x + y = (x_i + y_i)_{i \in I}$ belongs to E. This proves a).

By the Cauchy-Schwarz inequality, we have

hence $\sum_{i \in I} |\langle x_i|y_i \rangle| < +\infty$. If $x \neq 0$, we have $\langle x|x \rangle = \sum_{i \in I} \|x_i\|^2 > 0$, hence assertion b) follows.

We recall that S is the subspace of P consisting of all families $x = (x_i)_{i \in I}$ such that the set of all $i \in I$ for which $x_i \neq 0$ is finite. It follows immediately that S is dense in E; hence it remains to prove that E is complete for the topology $\mathscr{T}_1$ obtained by the norm $\|x\| = \langle x|x \rangle^{1/2}$. Let $\mathscr{T}_2$ be the topology induced on E by the product topology on $\prod_{i \in I} E_i$. For every $r > 0$, let $B_r$ be the set of all $x \in E$ such that $\|x\| \leq r$. This relation implies that we have $\sum_{i \in J} \|x_i\|^2 \leq r^2$ for every finite subset J of I, and so $B_r$ is a closed subset of $\prod_{i \in I} E_i$, hence also complete. The fact that E is complete for $\mathscr{T}_1$ now follows from GT, III, § 3, No. 5, cor. 2 to prop. 10.

Let $S \subset F$ be the « algebraic » direct sum of the $E_i$, and let $g$ be the linear mapping $(x_i)_{i \in I} \mapsto \sum_{i \in I} x_i$ from S into E. We shall show that $g$ is an isomorphism from the prehilbertian space S onto the (prehilbertian) subspace $g(S)$ of E, generated by the union of the $E_i$ : for, for two elements $x = (x_i)_{i \in I}$, $y = (y_i)_{i \in I}$, we have

But if $i \neq k$, $\langle x_i|y_k \rangle = 0$ by hypothesis, hence

this proves our assertion. Since S is dense in F and $g(S)$ dense in E, the isomorphism $g$ extends to an isomorphism $\overline{g}$ from F onto E (V, p. 8, cor.). It is clear that the inverse isomorphism $f$ of $\overline{g}$ is the required mapping; its uniqueness follows from the fact that the closed subspace of E generated by the union of the $E_i$ is E itself.

When E is the hilbertian sum of a family $(E_i)_{i \in I}$ of subspaces, we shall often identify E with the external hilbertian sum F of the $E_i$ by means of the isomorphism $f$. If the set I is finite, saying that E is the hilbertian sum of the family $(E_i)_{i \in I}$ means that the $E_i$ are two by two orthogonal and that the vector space E is the direct sum of the family $(E_i)_{i \in I}$ of subspaces.

a) For all $x \in E$, the family $(\|p_{E_i}(x)\|^2)_{i \in I}$ is summable in $\mathbf{R}$, the family $(p_{E_i}(x))_{i \in I}$ is summable in E, and we have

b) Conversely, if $(x_i)_{i \in I}$ is a family of elements of E such that $x_i \in E_i$ for all $i \in I$ and $\sum_{i \in I} \|x_i\|^2 < + \infty$, this family is summable, and the sum $x$ is the only point of E for which $p_{E_i}(x) = x_i$ for all $i \in I$.

c) For every pair of points $x, y$ of E, we have

These properties are in fact obvious for the external hilbertian sum of the $E_i$, and can be transferred to E by isomorphism.

1) We have $\sum_{i \in I} \|p_{E_i}(x)\|^2 \leq \|x\|^2$.

2) The following conditions are equivalent : a) $x \in V$; b) $\sum_{i \in I} \|p_{E_i}(x)\|^2 = \|x\|^2$; c) the family $(p_{E_i}(x))_{i \in I}$ is summable in E, and we have $x = \sum_{i \in I} p_{E_i}(x)$.

3) Suppose V is complete. Then the family $(p_{E_i}(x))_{i \in I}$ is summable in E, and

where $p_V$ denotes the orthoprojector from E onto V.

Let $\hat{E}$ be the hilbertian space completion of E; we identify E with a dense subspace of $\hat{E}$; the $E_i$, being complete, are closed subspaces of $\hat{E}$. The closure $\overline{V}$ of V in $\hat{E}$ is the closed vector subspace of $\hat{E}$ generated by the union of the $E_i$, and $V = \overline{V} \cap E$. The space $\hat{E}$ is the hilbertian sum of the $E_i$ and of the subspace W, the orthogonal complement of $\overline{V}$ in $\hat{E}$; put $x_0 = p_W(x)$ and $x_i = p_{E_i}(x)$ for all $i \in I$. By cor. 1, we have $\|x\|^2 = \|x_0\|^2 + \sum_{i \in I} \|x_i\|^2$, and $x = x_0 + \sum_{i \in I} x_i$ in $\hat{E}$. This implies assertion 1), and the fact that conditions b) and c) of 2) are equivalent to the condition $x_0 = 0$, hence to the condition $x \in V$. Finally, if V is complete, and if we put $x' = p_V(x)$, we have $x' - x_i = (x - x_i) - (x - p_V(x))$, hence $x' - x_i$ is orthogonal to $E_i$, and so $x_i = p_{E_i}(x')$ for all $i \in I$; it is now enough to apply property 2) to the vector $x'$.

To show that the condition is necessary, it is enough to see that $V_\alpha$ and $V_\beta$ are orthogonal if $\alpha \neq \beta$. But, every element of $W_{\alpha \mu}$ ($\mu \in M_\alpha$) is orthogonal to all the $W_{\beta v}$ ($v \in M_\beta$), hence to the closed vector subspace $V_\beta$ which they generate; the same argument then shows that every element of $V_\beta$ is orthogonal to $V_\alpha$, being orthogonal to all the $W_{\alpha \mu}$ ($\mu \in M_\alpha$).

To show that the condition is sufficient, it is enough to verify that, if it is satisfied, E is equal to the closed vector subspace F generated by the union of the $W_{\lambda \mu}$ ($\lambda \in L$, μ ∈ M_λ); but, for each λ ∈ L, F contains the closed vector subspace generated by the union of the W_{λμ} such that μ ∈ M_λ, that is, F contains V_λ; hence F is the closed vector subspace generated by the union of the V_λ, which is E by hypothesis.

We note that this property immediately follows from the characterization of topologically independent families (IV, p. 1 and II, p. 43, cor. 2), on account of the identification of the dual of E with the completion of E or with the space conjugate to E according as K is equal to R or C (V, p. 17, Remark).

1) For every x ∈ E, we have

(Bessel’s inequality); here the set of all $i \in I$ such that $⟨e_i|x⟩ \neq 0$ is countable. Moreover, the following conditions are equivalent : a) $x \in V$; b) $\|x\|^2 = \sum_{i\in I} |⟨e_i|x⟩|^2$; c) the family $⟨e_i|x⟩.e_i$ is summable in E, and $x = \sum_{i\in I} ⟨e_i|x⟩.e_i$.

2) If V is complete, then the family of all $⟨e_i|x⟩.e_i$ is summable in E for all $x \in E$, and $\sum_{i\in I} ⟨e_i|x⟩.e_i = p_V(x), \sum_{i\in I} |⟨e_i|x⟩|^2 = \|p_V(x)\|^2$.

3) Suppose V is complete. For every family $(\lambda_i)_{i\in I}$ of scalars such that $\sum_{i\in I} |\lambda_i|^2 < +\infty$, there exists a unique point $x \in V$ such that $⟨e_i|x⟩ = \lambda_i$ for all $i \in I$. If $(\mu_i)_{i\in I}$ is a second family of scalars such that $\sum_{i\in I} |\mu_i|^2 < +\infty$, and if $y \in V$ is such that $⟨e_i|y⟩ = \mu_i$ for all $i \in I$, then $⟨x|y⟩ = \sum_{i\in I} \overline{\lambda_i} \mu_i$.

(Parseval’s relation).

When E is hilbertian these conditions are also equivalent to : d) the relations $\langle e_i | x \rangle = 0$ for all $i \in I$ imply that $x = 0$.

The equivalence of conditions a), b), c) follows immediately from prop. 4. When E is hilbertian, the equivalence of conditions a) and d) follows from cor. 1 of V, p. 16.

In fact, let $\mathfrak{D}$ be the family of all orthonormal subsets of E, linearly ordered by inclusion; it is immediate that this family has finite character (S, III, § 4, No. 5). Hence there exists a maximal family B in $\mathfrak{D}$ containing L, by th. 1 of S, III, § 4, No. 5. It remains to prove that B is a total set. If not, there will exist a vector $y \neq 0$ which is orthogonal to all the vectors of B (V, p. 22, prop. 5), and multiplying y by a suitable scalar, we can assume that $\|y\| = 1$; then, $B \cup \{y\}$ will be an orthonormal set distinct from B and containing B; this contradicts the definition of B; hence the theorem.

It is enough to apply th. 2 to the case $L = \varnothing$.

More precisely, let $(e_i)_{i \in I}$ be an orthonormal basis of a hilbertian space E. By props. 4 (V, p. 21) and 5 (V, p. 22), the mapping $\phi$ defined by

is a hilbertian space isomorphism from E onto $\ell^2(I)$. The inverse isomorphism $\psi$ is defined by

1) for every integer $p \in I$, the vector subspace of E generated by $e_1, e_2, ..., e_p$ is identical with the vector subspace of E generated by $a_1, a_2, ..., a_p$; 2) for every integer $p \in I$, the number $\langle a_p|e_p \rangle$ is real and $> 0$.

In fact, let $V_n$ be the subspace (of dimension n) generated by $a_1, a_2, ..., a_n$. If $n + 1 \in I$ and $b_{n+1} = a_{n+1} - p_{V_n}(a_{n+1})$ (where $p_{V_n}$ is the orthoprojector onto the complete subspace $V_n$), the line $K b_{n+1}$ is the orthogonal of $V_n$ in $V_{n+1}$. If the $e_n$ satisfy condition 1) of the proposition, we must have $e_{n+1} = \lambda b_{n+1}$; the condition $\|e_{n+1}\| = 1$ then implies $|\lambda|^2 \|b_{n+1}\|^2 = 1$ and the condition $\langle a_{n+1}|e_{n+1} \rangle > 0$ implies $\lambda \langle a_{n+1}|b_{n+1} \rangle > 0$; this completely determines $\lambda$, and we have proved that we can determine, by induction, an orthonormal family $(e_n)_{n \in I}$, and only one, so as to satisfy conditions 1) and 2) of the proposition.

The sequence $(e_n)_{n \in I}$ is said to be obtained by orthonormalisation from the independent family $(a_n)_{n \in I}$. It is clear that the vector subspace generated by the family $(e_n)$ is identical with the vector subspace generated by the family $(a_n)$. In particular, if $(a_n)$ is a total sequence, so is $(e_n)$, which is then an orthonormal basis of E; hence we get:

If E satisfies the first axiom of countability, then there exists a total sequence in E, and we can always extract an independent total family from such a sequence (A, II, § 7, No. 1, th. 2).

We can give examples of Hausdorff prehilbertian spaces not having any orthonormal basis (V, p. 70, exerc. 2). Example. — Let I be the interval $(-1, 1)$ of $\mathbf{R}$ and E the vector space of real valued continuous functions on I. Let $x$ denote the canonical injection from I into $\mathbf{R}$, considered as an element of E. By the Stone-Weierstrass theorem the sequence $(x^n)_{n \in \mathbf{N}}$ is total in E for the topology of uniform convergence GT, X, § 4, No. 2). Consider E as a real Hausdorff prehilbertian space in which the scalar product is given by

The sequence $(x^n)_{n \in \mathbf{N}}$ is then total in the prehilbertian space E. Let $(\Pi_n)_{n \in \mathbf{N}}$ be the sequence obtained by the orthonormalisation of the sequence $(x^n)_{n \in \mathbf{N}}$. We can show that $\Pi_n = (n + \frac{1}{2})^{1/2} P_n$, where the Legendre polynomial $P_n$ is defined by

Let B and C be two orthonormal bases of E. The case when one of the two sets B, C is finite is trivial, since a finite orthonormal basis is an algebraic basis of the space. Suppose therefore that B and C are infinite. For every $x \in B$, let $C_x$ be the subset of C consisting of all $y \in C$ such that $\langle x | y \rangle \neq 0$. The set $C_x$ is countable (V, p. 21, prop. 4). For every $y \in C$, there exists $x \in B$ such that $y \in C_x$, since B is an orthonormal basis and $y \neq 0$; in other words C is the union of the countable sets $C_x$ as $x$ ranges over B. The cardinality of C is hence less than that of $\mathbf{N} \times B$, hence less than that of B (S, III, § 6, No. 3, cor. 4); similarly, the cardinality of B is less than that of C; this completes the proof.

The cardinality of an arbitrary orthonormal basis of a hilbertian space E is called the hilbertian dimension of E.