Proposition 1

— The sesquilinear form $\Phi$ on $E_1 \otimes E_2$ is hermitian and positive, hence assigns the structure of a prehilbertian space to $E_1 \otimes E_2$. This space is Hausdorff if $E_1$ and $E_2$ are Hausdorff.

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 $E_1 \otimes_2 E_2$.