The formula $\Phi(z, t) = \overline{\Phi(t, z)}$ follows from (3) when $z = x_1 \otimes x_2$ and $t = y_1 \otimes y_2$. The general case is obtained by linearity, hence $\Phi$ is hermitian.

Suppose $E_1$ and $E_2$ are Hausdorff; we shall prove that the hermitian form $\Phi$ is positive and separating. Let $z = \sum_{i=1}^n x_i \otimes y_i$ be a non-zero element of $F = E_1 \otimes E_2$. Let $(e_1, ..., e_n)$ be an orthonormal basis of the subspace of $E_1$ generated by $x_1, ..., x_n$ (V, p. 23, cor. 1). There exist elements $f_1, ..., f_m$ in $E_2$, not all null, such that $z = \sum_{i=1}^n e_i \otimes f_i$, hence

For the general case, we shall now prove that $\Phi$ is positive. Let $\tilde{E}_i$ be the Hausdorff prehilbertian space associated with $E_i$ and let $\pi_i$ be the canonical mapping from $E_i$ onto $\tilde{E}_i$ ($i = 1, 2$). Put $\pi = \pi_1 \otimes \pi_2$. Let $\tilde{\Phi}$ be the hermitian form on $\tilde{E}_1 \otimes \tilde{E}_2$ constructed in the same way as $\Phi$. We have

and since $\tilde{\Phi}$ is positive, so is $\Phi$.

The prehilbertian space defined in prop. 1 is called the tensor product of the prehilbertian spaces $E_1$ and $E_2$ and is written as $E_1 \otimes E_2$. Henceforth we shall write $\langle z|t \rangle$ for $\Phi(z, t)$, and therefore by definition

we shall also write $\|z\|_2$ or $\|z\|$ for $\langle z|z \rangle^{1/2}$. From (4), we get then the bilinear mapping $(x_1, x_2) \mapsto x_1 \otimes x_2$ from $E_1 \times E_2$ into $E_1 \otimes E_2$ is continuous.

For $i = 1, 2$ let $F_i$ be a vector subspace of $E_i$, endowed with the induced prehilbertian structure. Then $F_1 \otimes F_2$ can be identified with a vector subspace of $E_1 \otimes E_2$ (A, II, § 7, No. 7). Formula (4) shows that $F_1 \otimes F_2$ with the prehilbertian space structure induced by that of $E_1 \otimes_2 E_2$, is exactly $F_1 \otimes_2 F_2$. We shall henceforth identify $F_1 \otimes_2 F_2$ with a prehilbertian subspace of $E_1 \otimes_2 E_2$.

Consider the positive hermitian form on $E_1$ given by

By prop. 1 (V, p. 25), there exists a positive hermitian form $\Phi$ on $E_1 \otimes E_2$ such that

for $x_1, y_1$ in $E_1$ and $x_2, y_2$ in $E_2$. By linearity we have for $z, t$ in $E_1 \otimes E_2$. Since $\Phi$ is positive, we get $\Phi(z, z) \geq 0$, that is, $\|(u_1 \otimes 1).z\|_2$

On the other hand, This completes the proof of prop. 2. Q.E.D.

Let $E_1, ..., E_n$ be prehilbertian spaces ($n \geq 2$). We define the tensor product $E_1 \otimes_2 ... \otimes_2 E_n$ (also denoted by $\bigotimes_{i=1}^n E_i$) by induction, by

Hence, by the definition of scalar product, we have and in particular $^1$ for $x_i, y_i$ in $E_i$ ($1 \leq i \leq n$). If the $E_i$ are Hausdorff, then so is $E_1 \otimes_2 ... \otimes_2 E_n$.

Let $F_1, ..., F_n$ be prehilbertian spaces and $u_i \in \mathscr{L}(E_i; F_i)$ for $1 \leq i \leq n$. By induction on $n$, prop. 2 implies that $u_1 \otimes ... \otimes u_n$ is a continuous linear mapping from $E_1 \otimes_2 ... \otimes_2 E_n$ into $F_1 \otimes_2 ... \otimes_2 F_n$ and that

Let $\sigma \in S_n$ be a permutation of the set $\{1, 2, ..., n\}$. Because of (6), the linear mapping $p_\sigma$ from $E_1 \otimes_2 ... \otimes_2 E_n$ onto $E_{\sigma^{-1}(1)} \otimes_2 ... \otimes_2 E_{\sigma^{-1}(n)}$ characterized by is a prehilbertian space isomorphism ("commutativity of tensor product").

Similarly, consider a partition of $\{1, 2, ..., n\}$ into $m$ consecutive intervals $I_1, ..., I_n$ with $I_k = [a_k, a_{k+1} - 1]$ for $1 \leq k \leq m$. Put

$^1$ Here again we put $\|z\|_2 = \langle z|z \rangle^{1/2}$ for $z$ in $E_1 \otimes_2 ... \otimes_2 E_n$.

The canonical isomorphism from $F_1 \otimes ... \otimes F_m$ onto $E_1 \otimes ... \otimes E_n$ which transforms $\bigotimes_{k=1}^{m} \bigotimes_{i=a_k}^{a_{k+1}-1} x_i$ into $x_1 \otimes ... \otimes x_n$ (A, II, § 3, No. 9) is a prehilbertian space isomorphism (« associativity of the tensor product »).

By formula (6) of V, p. 27, the subspaces $E_{1,\alpha_1} \hat{\otimes}_2 ... \hat{\otimes}_2 E_{n,\alpha_n}$ of $E_1 \hat{\otimes}_2 ... \hat{\otimes}_2 E_n$ are mutually orthogonal. For every integer $i$ between 1 and $n$, the set $\bigcup_{\alpha \in A(i)} E_{i,\alpha}$ is total in $E_i$, and the multilinear mapping $(x_1, ..., x_n) \mapsto x_1 \otimes ... \otimes x_n$ is continuous.

It follows that the union of the subspaces $E_{1,\alpha_1} \hat{\otimes}_2 ... \hat{\otimes}_2 E_{n,\alpha_n}$ is total, hence prop. 3.

Let $F_i$ be the line in $E_1$ generated by the $e_i$ ($i \in I$). Then $E_1$ is the hilbertian sum of the family of subspaces $(F_i)_{i \in I}$. By prop. 3, the space $E_1 \hat{\otimes}_2 E_2$ is the hilbertian sum of the family of subspaces $(F_i \hat{\otimes}_2 E_2)_{i \in I}$, hence cor. 2 follows.

Then $(z_{\alpha})_{\alpha \in \mathbf{N}^{(I)}}$ is an orthonormal basis of $\hat{\mathbf{S}}(E)$.

Let $E_0$ be the vector subspace of $E$ generated by the vectors $e_i$ for $i$ ranging over $I$. Then the $z_{\alpha}$ form a basis of the vector space $\mathbf{S}(E_0)$ (A, III, § 6, No. 6). But $E_0$ is dense in $E$, and the multilinear mapping $(x_1, \ldots, x_n) \mapsto x_1 \ldots x_n$ from $E \times \cdots \times E$ into $\mathbf{S}(E)$ is continuous for all $n \geqslant 1$; hence $\mathbf{S}(E_0)$ is dense in $\mathbf{S}(E)$. It is now enough to prove that the family of the $z_\alpha$ is orthonormal. First observe that $\hat{S}^n(E)$ and $\hat{S}^m(E)$ are orthogonal for $n \neq m$. Hence it is enough to prove the formula

when $|\alpha| = \sum_{i \in I} \alpha_i$ and $|\beta| = \sum_{i \in I} \beta_i$ are equal to the same integer $n$.

Consider a partition $(P_i)_{i \in I}$ of the set $\{ 1, 2, ..., n \}$ such that Card $P_i = \alpha_i$ for all $i \in I$. Put $x_k = e_i$ if $k$ belongs to $P_i$, then $x_1 ... x_n = \prod_{i \in I} e_i^{\alpha_i}$. Similarly we define $(Q_i)_{i \in I}$ and $y_k$ in such a way that Card $Q_i = \beta_i$ and $y_1 ... y_n = \prod_{i \in I} e_i^{\beta_i}$. Since the $e_i$ are mutually orthogonal, we have $\langle x_k | y_{\sigma(k)} \rangle = 0$ except if there exists an indice $i \in I$ such that $k \in P_i$ and $\sigma(k) \in Q_i$. By formula (15), we then have $\langle x_1 ... x_n | y_1 ... y_n \rangle = 0$ unless there exists a permutation $\sigma \in \mathfrak{S}_n$ such that $\sigma(P_i) = Q_i$ for all $i \in I$, which implies that $\alpha = \beta$. Then $\langle z_\alpha | z_\beta \rangle = 0$ for $\alpha \neq \beta$. The same argument proves that $\| x_1 ... x_n \| ^2$ is equal to the number of the $\sigma \in \mathfrak{S}_n$ such that $\sigma(P_i) = P_i$ for all $i \in I$, hence equal to $\prod_{i \in I} \alpha_i !$. We get $\| z_\alpha \| = 1$, and the proposition is proved.

Let $(e_i)_{i \in I}$ (resp. $(f_j)_{j \in J}$) be an orthonormal basis of the hilbertian space M (resp. N) and let $M_0$ (resp. $N_0$) be the vector subspace of E generated by the vectors $e_i$ (resp. $f_j$). Put $E_0 = M_0 + N_0$ and let $g_0$ be the canonical isomorphism from $S(M_0) \otimes S(N_0)$ onto $S(E_0)$. Put

for $\alpha \in \mathbf{N}^{(I)}$ and $\beta \in \mathbf{N}^{(J)}$. By prop. 4, we have thus defined the orthonormal bases $(z_\alpha)_{\alpha \in \mathbf{N}^{(I)}}$ for $\hat{S}(M)$, $(t_\beta)_{\beta \in \mathbf{N}^{(J)}}$ for $\hat{S}(N)$ and $(z_\alpha t_\beta)_{\alpha \in \mathbf{N}^{(I)}, \beta \in \mathbf{N}^{(J)}}$ for $\hat{S}(E)$. Since we have $z_\alpha t_\beta = g_0(z_\alpha \otimes t_\beta)$, and since the elements $z_\alpha \otimes t_\beta$ form an orthonormal basis of $\hat{S}(M) \hat{\otimes}_2 \hat{S}(N)$ (V, p. 29, cor. 1), we see that $g_0$ extends to a hilbertian space isomorphism $h : \hat{S}(M) \hat{\otimes}_2 \hat{S}(N) \to \hat{S}(E)$. By the construction, we have

for all vectors $x_1, ..., x_m$ in $M_0$ and $y_1, ..., y_n$ in $N_0$. By continuity, the same relation also holds for the vectors $x_1, ..., x_n$ in M and the vectors $y_1, ..., y_n$ in N; in other words, $h$ extends $g$. The uniqueness of $h$ is clear.

Let E and F be two hilbertian spaces and $u \in \mathscr{L}(E; F)$. The linear mapping $\hat{T}^n(u) = u \hat{\otimes}_2 ... \hat{\otimes}_2 u$ ($n$ factors) from $\hat{T}^n(E)$ into $\hat{T}^n(F)$ is continuous with norm $\|u\|^n$ (V, p. 28, formula (10)). Moreover, formulas (13) and (14) of V, p. 30, show that there exists an isomorphism $\phi_{n,E}$ from $\hat{S}^n(E)$ onto the subspace $\widehat{T S^n}(E)$ of $\hat{T}^n(E)$, and only one, such that

Hence there exists a continuous linear mapping $\hat{S}^n(u)$ from $\hat{S}^n(E)$ into $\hat{S}^n(F)$ and only one which makes the following diagram commutative

We now prove the formula

We clearly have $\| \hat{S}^n(u) \| \leq \| \hat{T}^n(u) \| = \| u \|^{n}$. Further, for all $x \in E$, we have $\hat{S}^n(u)(x^n) = u(x)^n, \| x^n \| = (n!)^{1/2} \| x \|^{n}$ and $\| u(x)^n \| = (n!)^{1/2} \| u(x) \|^{n}$, which gives

it follows immediately that $\| \hat{S}^n(u) \| \geq \| u \|^{n}$, hence formula (19).

It is clear that we have the formulas

Finally, $\hat{S}^n(u)$ coincides on $S^n(E)$ with the linear mapping $S^n(u) : S^n(E) \to S^n(F)$ defined in A, III, § 6, No. 2 since it transforms $x_1 ... x_n$ into $u(x_1) ... u(x_n)$ for every $x_1, ..., x_n$ in $E$.