Exercise 30, § 6

We take up again the hypotheses of the preceding exercise so that $u$ is linear, compact, and that conditions (i) to (v) are assumed to be satisfied. The endomorphism $^tu$ is also compact.

(a) Let $\eta \in C^{\circ}-\{0\}$. For every $\lambda \in \mathbf{R}_+$ such that $\lambda < \lambda_m$, the unique solution $\xi (\lambda )$ of the equation $\xi -\lambda^tu(\xi ) =\eta$ belongs to $C^{\circ}$. (If $x\in C$, let $y\in C$ be such that $y-\lambda u(y) =x$; one has $\langle y, \eta \rangle =\langle x, \xi \rangle .$)

(b) Let $x_0\in \mathring{C}$. The mapping $\lambda \mapsto \langle x_0, \xi (\lambda )\rangle$ tends to infinity when $\lambda$ tends to $\lambda_m$. Now prove that the set of values $\xi (\lambda )/\langle x_0, \xi (\lambda )\rangle$ possesses a value of adherence $\xi_m$ such that :

(i) $\xi_m\in C^{\circ}$;

(ii) $\langle x_0, \xi_m\rangle = 1$;

(iii) $\xi_m=\lambda_m^tu(\xi_m)$.

(c) Let $f\in E$ such that $\langle f, \xi_m\rangle = 0$. For $\lambda \in \mathbf{R}_+$ such that $\lambda < \lambda_m$, denote by $x(\lambda )$ the unique solution $x$ in E of the equation $x-\lambda u(x) =f$. Now prove that the set of the $x(\lambda )$ is bounded in E (let $d$ be a distance defining the topology of E; if there existed a sequence $\lambda_n\rightarrow \lambda_m$ such that $x(\lambda_n)$ tends to infinity, put $\widehat{x}_n=x(\lambda_n)/d(0, x(\lambda_n))$, and show that the sequence $(\widehat{x}_n)$ converges to an element $\widehat{x}$ such that $d(0,\widehat{x}) = 1, \widehat{x}=\lambda u(\widehat{x})$ and $\langle \widehat{x, \xi}_m\rangle = 0$, which contradicts the fact that one must have $\widehat{x}\in \mathring{C}$).

(d) There exists a unique element $x\in E$ such that $f=x-\lambda_mu(x)$ and $\langle x, \xi_m\rangle = 0$. (For the existence, now prove with the aid of the preceding question that there exists a sequence $(\lambda_n)$ converging towards $\lambda_m$ such that $(x(\lambda_n))$ converges towards an element $x$ satisfying the required conditions.)

(e) The endomorphism $1_{E'}-\lambda_m^tu$ of $E'$ induced by passing to the subspaces gives a bijection of the polar $x^{\circ}_m$ of $x_m$.

(f) The eigenspace Ker(1$_E-\lambda_mu$) has dimension 1 and is equal to the polar of the image of $1_E-\lambda_mu$.