Lemma 12

Let $F\in \mathscr{F}_n^u$ be a subspace adapted to $u$. Every eigenvector of $u$ belonging to the space $F^{\circ}\cap E_b$ has an eigenvalue $\geqslant \lambda_n(u)$.

Let $\ell$ be such that $\lambda_{n-1}(u) =\nu_{\ell}$. Let $x\in F^{\circ}\cap E_b$ be an eigenvector of $u$ for an eigenvalue $\lambda$, and let $k <$ Card(Sp$_b(u)$) be such that $\lambda =\nu_k$.

The condition $k < \ell$ is impossible, for $\nu_k< \nu_{\ell}$, and the space F would then contain the eigensubspace for the eigenvalue $\nu_k$, contradicting the fact that $x$ is orthogonal to F. If $k=\ell$, then $x$ is an eigenvector for the eigenvalue $\lambda_{n-1}(u)$; since $x\in F^{\circ}$, the multiplicity $n_k$ is strictly greater than the number of integers $i < n$ such that $\lambda_i(u) =\nu_k$, which implies that $\lambda_n(u) =\lambda_{n-1}(u)$. Finally, if $k > \ell$, one has $\lambda =\nu_k> \nu_{\ell}=\lambda_{n-1}(u)$, hence $\lambda \geqslant \lambda_n(u)$.

For every subspace F of E, let

$\widetilde{r}_F(u) =$ inf$_{x\in dom(u)\cap F^{\circ}}\frac{\langle x|u(x)\rangle}{\|x\|^2}$,

$x\not =0$

$\widetilde{R}_F(u) =$ sup $\frac{\langle x|u(x)\rangle}{2}$.

$x\not =0$