Proposition 17

a) For every integer $n\in \mathbf{N}$, one has

$\lambda_n(u) =$ sup$_{F\in\mathscr{F}_n}\widetilde{r}_F(u) =$ inf$_{F\in\mathscr{F}_{n+1}^u}\widetilde{R}_F(u)$;

b) For every integer $n <M$ and for every subspace $F\in \mathscr{F}_n^u$ adapted to $u$, one has $\lambda_n(u) =\widetilde{r}_F(u)$;

c) For every integer $n <M$ and for every subspace $F\in \mathscr{F}_{n+1}^u$ adapted to $u$, one has $\lambda_n(u) =\widetilde{R}_F(u)$.

There exists an orthonormal basis $(e_n)_{0\leqslant n<M}$ of the space $E_b$ such that $e_n\in$ dom($u$) and $u(e_n) =\lambda_n(u)e_n$ for every $n$ such that $0\leqslant n <M$. For every $x\in E_b\cap$ dom($u$), one therefore has

For every integer $n$ such that $1\leqslant n <M + 1$, let $F_n$ be the subspace of dimension $n$ of $E_b$ generated by $(e_0, . . . , e_{n-1})$. One has $F_n\subset$ dom($u$) and $F_n$ is adapted to $u$.

Let $n\in \mathbf{N}$ and $F\in \mathscr{F}_n$. Let us prove that $\widetilde{r}_F(u)\leqslant \lambda_n(u)$ and, consequently, that

(19) Fsup$_{\in\mathscr{F}_n}\widetilde{r}_F(u)\leqslant \lambda_n(u)$.

If $0\leqslant n <M$, in particular if M is infinite, then the restriction to $F_{n+1}$ of the orthogonal projection of E onto F is not injective, hence there exists $x\not = 0$ in $F_{n+1}$ orthogonal to F. Since

one has $\widetilde{r}_F(u)\leqslant \lambda_n(u)$.

If $M\in \mathbf{N}$ and $n\geqslant M$, then for every real number $\varepsilon >0$, there exists $x$ of norm 1 in dom($u$) which is orthogonal to F and whose spectral measure $\mu_x$ has support contained in $[\varrho_e, \varrho_e+\varepsilon ]$ (Lemma 11 and Prop. 9, a) of IV, p. 278), whence

by definition. Since $\varepsilon >0$ is arbitrary, one therefore has $\widetilde{r}_F(u)\leqslant \lambda_n(u)$. Inequality (19) is therefore established.

Let $n\in \mathbf{N}$ and $F\in \mathscr{F}_{n+1}^u$. Let us prove that $\widetilde{R}_F(u)\geqslant \lambda_n(u)$ and, consequently, that

(20) Finf$_{\in\mathscr{F}_{n+1}^u}\widetilde{R}_F(u)\geqslant \lambda_n(u)$.

If $0\leqslant n <M$, observe that the restriction to F of the orthogonal projector onto $F_n$ is not injective, hence that there exists a vector $x\not = 0$ in F orthogonal to $F_n$. Put $x_b=p_{Sp_b(u)}(x)$ and $x_h=p_{Sp_h(u)}(x)$. We therefore have $x=x_b+x_h$. The elements $x_b$ and $x_h$ belong to the domain of $u$ (Prop. 9, c) of IV, p. 278) and are orthogonal to $F_n$. We have the lower bound

and $\langle x_h|u(x_h)\rangle \geqslant \varrho_e\|x_h\|^2$ (formula (18)), whence

If M is finite and $n\geqslant M$, there exists $x\not = 0$ in F orthogonal to $E_b$, hence $x\in E_h$, and

by (18). Inequality (20) is therefore proved.

We shall now prove assertions b) and c), which imply that inequalities (19) and (20) are equalities when $n <M$.

Let us prove assertion b). Let $F\in \mathscr{F}_n^u$ be a space adapted to $u$. We have the Hilbert direct sum

$F^{\circ}= E_h\oplus \bigoplus_{\lambda\in Sp_b(u)}$ Ker($u-\lambda \cdot 1_E$)$\oplus (F^{\circ}\cap$ Ker($u-\lambda_{n-1}(u)\cdot 1_E$)).

$\lambda >\lambda_{n-1}(u)$

Let $x\in F^{\circ}-\{0\}$. Write $x=x_h+y+z$, where $x_h\in E_h$ and $y$ (resp. $z$) belongs to the second (resp. third) space of the above decomposition. Using again (18) and the fact that every eigenvalue $\lambda > \lambda_{n-1}(u)$ of $u$ is $\geqslant \lambda_n(u)$, we obtain

If $z\not = 0$, then by Lemma 12, we have $\lambda_n(u) =\lambda_{n-1}(u)$. It follows therefore that $\langle x|u(x)\rangle \geqslant \lambda_n(u)\|x\|^2$, whence $\widetilde{r}_F(u)\geqslant \lambda_n(u)$. Combined with (19), this implies assertion b).

Let us prove assertion c). Let $F\in \mathscr{F}_{n+1}^u$ be a subspace adapted to $u$. Let $(f_i)_{0\leqslant i\leqslant n}$ be an orthonormal basis of F such that $u(f_i) =\lambda_i(u)f_i$ for $0\leqslant i\leqslant n$. For every $x\in F$, we have

with equality if $x=f_n$, hence $\widetilde{R}_F(u) =\lambda_n(u)$.

Let us prove finally that (19) and (20) are equalities when $n\geqslant M$. In this case, M is finite, hence $E_b$ is contained in the domain of $u$; moreover, we have $\lambda_n(u) =\varrho_e$ by definition.

There exists $F\in \mathscr{F}_n$ containing $E_b$. Any element $x\not = 0$ of dom($u$) orthogonal to F is therefore orthogonal to $E_b$, and satisfies

(formula (18)), hence $\widetilde{r}_F(u)\geqslant \varrho_e=\lambda_n(u)$, and consequently

sup$_{F\in\mathscr{F}_n}\widetilde{r}_F(u)\geqslant \lambda_n(u)$.

Let $\varepsilon >0$. Since $E_b$ is finite-dimensional, it follows from lemma 11 that there exists an orthonormal family $(x_i)_{i\in I}$ of elements of E such that the subspace F generated by $E_b$ and $(x_i)_{i\in I}$ is of dimension $n+ 1$ and is contained in dom($u$), and such that $\langle x_i|u(x_i)\rangle \leqslant \varrho_e+\varepsilon$ for all $i\in I$. One then has $\widetilde{R}_F(u)\leqslant \varrho_e+\varepsilon$. Since $\varepsilon >0$ is arbitrary, one concludes that

inf$_{F\in\mathscr{F}_{n+1}^u}\widetilde{R}_F(u)\leqslant \varrho_e=\lambda_n(u)$.

The proposition is proved.