Proposition 14

— Let $E_1, ..., E_n$ be hilbertian spaces, $E = E_1 \hat{\otimes}_2 ... \hat{\otimes}_2 E_n$, and $u_i$ a continuous endomorphism of $E_i$ for $1 \leq i \leq n$. If $u_1, ..., u_n$ are positive, then so is $u = u_1 \hat{\otimes}_2 ... \hat{\otimes}_2 u_n$, and

If $u_i \in \mathscr{L}^1(E_i)$ for all $1 \leq i \leq n$, then $u \in \mathscr{L}^1(E)$ and formula (31) is again valid in this case.

Proceeding by induction on $n$, we immediately reduce to the case $n = 2$.

For $i = 1, 2$, we define a sesquilinear form $\Phi_i$ on $E_i$ by the formula $\Phi_i(x, y) = \langle x | u_i(y) \rangle$ for $x, y$ in $E_i$. If $u_1$ and $u_2$ are positive, the forms $\Phi_1$ and $\Phi_2$ are hermitian and positive. By prop. 1 of V, p. 25 there exists a positive hermitian form $\Phi$ on the vector space $E_1 \otimes E_2$ such that

for $x_1, y_1$ in $E_1$ and $x_2, y_2$ in $E_2$. We verify immediately the relation $\Phi(z, t) = \langle z | u(t) \rangle$ for $z$ and $t$ in $E_1 \otimes E_2$. Since $\Phi$ is positive, we have $\langle z | u(z) \rangle \geq 0$ for all $z$ in $E_1 \otimes E_2$. Since $u$ is continuous and $E_1 \otimes E_2$ is dense in the hilbertian space $E = E_1 \hat{\otimes}_2 E_2$, we conclude that $u$ is a continuous and positive endomorphism of E.

Let $(e_i)_{i \in I}$ be an orthonormal basis of $E_1$ and $(f_j)_{j \in J}$ an orthonormal basis of $E_2$; then the family $(e_i \otimes f_j)_{i \in I, j \in J}$ is an orthonormal basis of $E$ and we have

In particular, if $u_1$ and $u_2$ are positive endomorphisms with finite trace, then so is $u$. By linearity, we deduce that $u$ belongs to $\mathscr{L}^1(E)$ when $K = \mathbf{C}$ and that the $u_i$ belong to $\mathscr{L}^1(E_i)$ for $i = 1, 2$; formula (31) extends to this case by linearity. Finally, the case when $K = \mathbf{R}$ and the $u_i \in \mathscr{L}^1(E_i)$ reduces to the complex case by extension of the scalars.