Proposition 2

For every $u \in \mathscr{L}(E ; F)$, we have

(11)

By prop. 1, $\|u^*\| = \|u\|$, hence $\|u^*u\| \leq \|u^*\|\cdot\|u\| \leq \|u\|^2$. On the other hand,

hence $\|u^*u\| = \|u\|^2$. Replacing $u$ by $u^*$, we get $\|uu^*\| = \|u^*\|^2$, hence (11) follows since $\|u\| = \|u^*\|$.

Let $E_1, ..., E_n$ and $F_1, ..., F_n$ be hilbertian spaces, and for every integer $i$ between 1 and $n$, let $u_i$ be a continuous linear mapping from $E_i$ into $F_i$. Then

(12)

Let $v$ be the continuous linear mapping $u_1 \hat{\otimes}_2 ... \hat{\otimes}_2 u_n$ from

and $w$ the continuous linear mapping $u_1^* \hat{\otimes}_2 ... \hat{\otimes}_2 u_n^*$ from $F$ into $E$. It is enough to prove the equality $\langle y|v(x)\rangle = \langle w(y)|x\rangle$ for $x \in E$ and $y \in F$. By linearity and continuity, we reduce to the case when $x$ and $y$ have the following form

with $x_i \in E_i$ and $y_i \in F_i$ for $1 \leq i \leq n$. From the definition of scalar product in a tensor product (V, p. 27, formula (6)), we then get

This proves our assertion.

Let $E$ and $F$ be two hilbertian spaces, $u \in \mathscr{L}(E ; F)$ and $n$ a positive integer. If we put $u_1 = ... = u_n = u$ in formula (12) we obtain the result that the continuous linear mapping $\hat{T}^n(u^*)$ from $\hat{T}^n(F)$ into $\hat{T}^n(E)$ is the adjoint of the continuous linear mapping $\hat{T}^n(u)$ from $\hat{T}^n(E)$ into $\hat{T}^n(F)$. The formulas

can be established in the same way as formula (12), on account of the definition of the scalar product in $\hat{S}^n(E)$ (V, p. 30, formula (15)) and in $\hat{\Lambda}^n(E)$ (V, p. 33, formula (26)).