Proposition 19

Let $u$ be a self-adjoint partial operator on E. The following conditions are equivalent :

(i) There exists an orthonormal basis $B = (e_j)_{j\in J}$ of E such that $u$ is diagonal in the basis B and the absolute value of the family of eigenvalues of $u$ tends to infinity along the filter of complements of finite subsets of J;

(ii) For every $\lambda$ belonging to the resolvent set of $u$, the resolvent $R(u, \lambda )$ is compact;

(iii) There exists a complex number $\lambda$ belonging to the resolvent set of $u$ such that the resolvent $R(u, \lambda )$ is compact;

(iv) The spectrum of $u$ coincides with the point spectrum of $u$.

Suppose that (i) is satisfied, and let $(\lambda_j)_{j\in J}$ be the family of eigenvalues of $u$. Let $\mu$ be the counting measure on J. The spectrum of $u$ is the $\mu$-essential image of the family $(\lambda_j)$ (Prop. 22 of IV, p. 252); it is the set of values of this family. For every $\lambda \notin$ Sp($u$), the resolvent $R(u, \lambda )$ is diagonal in the basis B and the family of its eigenvalues is $((\lambda -\lambda_j)^{-1})_{j\in J}($loc. cit.). Since this family converges to 0, the endomorphism $R(u, \lambda )$ is compact (Prop. 2, (iii) of IV, p. 148). This proves that (i) implies (ii).

Since the resolvent set of $u$ is nonempty (cf. Prop. 17 of IV, p. 248), properties (ii) and (iii) are equivalent by formula (8) of IV, p. 245 and Proposition 3 of III, p. 5.

Since $R(u, \lambda )$ is normal for $\lambda \notin$ Sp($u$) (Prop. 16 of IV, p. 247), condition (iii) implies (iv) by Prop. 15 of IV, p. 247 and Prop. 5 of III, p. 90.

Let us prove finally that (iv) implies (i). Let $\mathscr{O}$ be the set of orthonormal subsets of E consisting of eigenvectors for $u$. The set $\mathscr{O}$, ordered by inclusion, is of finite character (E, III, p. 34, Def. 2) since O belongs to $\mathscr{O}$ if and only if the sets consisting of at most two elements of O belong to $\mathscr{O}$. By E, III, p. 35, Th. 1, there exists a maximal element O of $\mathscr{O}$. Denote by F the closed subspace of E generated by O. For $e\in O$, there exists a unique $\lambda (e)\in \mathbf{R}$ such that $u(e) =\lambda (e)e$ (Prop. 17 of IV, p. 248).

The set of values of the mapping $\lambda$ of O into $\mathbf{R}$ coincides with the spectrum of $u$. In fact, on the one hand this set is contained in the spectrum of $u$ and on the other hand, if there exists $\lambda_0\in$ Sp($u$) which is not a value of $\lambda$, then $\lambda_0$ is an eigenvalue of $u$ by hypothesis. There exists a vector $e\in$ dom($u$) of norm 1 such that $u(e) =\lambda_0e$ and $e \notin O$; the subset $O\cup \{e\}$ is orthonormal (since eigenspaces of $u$ relative to distinct eigenvalues are orthogonal by assertion b) of the corollary to Prop. 17 of IV, p. 248); it belongs to $\mathscr{O}$, contradicting the maximality of O.

Suppose that $F\not = E$. Then $F^{\circ}$ is nonzero. The endomorphism $R(u, i)$ of E is normal (prop. 16 of IV, p. 247). It leaves stable the subspace $F^{\circ}$ of E (lemma 4 of I, p. 135). Let $v$ be the endomorphism of $F^{\circ}$ deduced from $R(u, i)$ by passing to subspaces. It is a continuous and normal endomorphism of $F^{\circ}($loc. cit.) whose spectrum is contained in that of $R(u, i)$. Since $F^{\circ}$ is nonzero, the spectrum of $v$ is nonempty (cor. 1 of I, p. 26). Moreover, the spectrum of $v$ is not reduced to 0, since the normal endomorphism $v$ is nonzero (example 1 of I, p. 110). Let $s\in$ Sp($v$)$-\{0\}$. Since $s$ belongs to the spectrum of $R(u, i)$, there exists $e\in O$ such that $s= (i-\lambda (e))^{-1}$, and $s$ is an eigenvalue of $R(u, i)$ (prop. 15, a) of IV, p. 247). Since $s$ is nonzero, it is an isolated point of Sp(R($u, i$)) by assumption, hence also of Sp($v$). Thus, $s$ is an eigenvalue of $v$ (prop. 5, c) of I, p. 134). Let $e\in F^{\circ}$ be an eigenvector of norm 1 of $v$; it is also an eigenvector of $u$ (prop. 15, b) of IV, p. 247), and the set $O\cup \{e\}$ contradicts the fact that O is maximal in $\mathscr{O}$. Hence F = E.

The family of elements of O is therefore an orthonormal basis of E formed of eigenvectors of $u$. The spectrum of $u$ is closed in $\mathbf{R}$, and the point spectrum is discrete; since these sets coincide, the set of elements $\lambda \in$ Sp($u$) such that $|\lambda |\leqslant R$ is compact, hence finite, for every $R>0$. Thus the family of absolute values of the eigenvalues of $u$ tends to infinity along the filter of complements of finite subsets of O. Therefore (iv) implies (i).