Proposition 6

For every $n\in I$, one has

$\lambda_n(u) =$ sup$_{F\in\mathscr{F}_{n+1}}r_F(u) =$ inf$_{F\in\mathscr{F}_n}R_F(u)$.

Let $(e_n)_{n\in I}$ be an orthonormal family having the property of Prop. 3 of IV, p. 152. For every integer such that $1\leqslant n <M + 1$, let $F_n\in \mathscr{F}_n$ be the subspace of dimension $n$ of E generated by $(e_0, . . . , e_{n-1})$; by construction, the space $F_n$ is adapted to $u$. By Prop. 5, one therefore has

Let $n\in I$. Let $F\in \mathscr{F}_{n+1}$. The restriction to F of the orthogonal projector onto $F_n$ is not injective, hence there exists $x\not = 0$ in F orthogonal to $F_n$. Since $x\in F^{\circ}_n$, one then has (Prop. 3 of IV, p. 152)

$m\geqslant n$

This proves that $r_F(u)\leqslant \lambda_n(u)$, whence in particular the inequality

(3) Fsup$_{\in\mathscr{F}_{n+1}}r_F(u)\leqslant \lambda_n(u)$.

Let $F\in \mathscr{F}_n$. The restriction to $F_{n+1}$ of the orthogonal projector onto F is not injective, hence there exists a vector $x\not = 0$ in $F_{n+1}$ orthogonal to F. Since $x\in F_{n+1}$, one has (loc. cit.)

and therefore $R_F(u)\geqslant \lambda_n(u)$. In particular, one obtains

(4) Finf$_{\in\mathscr{F}_n}R_F(u)\geqslant \lambda_n(u)$.

In view of formulas (2), (3) and (4), the proposition is proved.