Proposition 2

Let $u\in \mathscr{D}_B(E)$ and $\lambda = (\lambda_i)_{i\in I}$ be the family of its eigenvalues.

a) The following conditions are equivalent:

(i) One has $\lambda \in \mathscr{C}_0(I; K)$;

(ii) The endomorphism $u$ is compact ;

(iii) The family $(\lambda_ip_i)_{i\in I}$ is summable in the Banach space

$\mathscr{L}(E)$. Its sum is then equal to $u$.

b) Suppose that $u$ is compact and denote by Λ the set of values of $\lambda$. The set of the $i\in I$ such that $\lambda_i\not = 0$ is countable. If E is of infinite dimension, one has Sp$_s(u) = \Lambda -\{0\}$ and Sp($u$) $= \Lambda \cup \{0\}$. If E is of finite dimension, then Sp$_s(u) =$ Sp($u$) $= \Lambda$.

By Lemma 1, one has $\|\sum_{j\in J}\lambda_jp_j\|=$ sup$_{j\in J}|\lambda_j|$ for every finite subset J of I. It follows, by the Cauchy criterion, that the family $(\lambda_ip_i)$ is summable in $\mathscr{L}(E)$ if and only if the family $\lambda$ tends to 0 at infinity, which implies that conditions (i) and (iii) are equivalent.

Condition (iii) implies that $u$ is compact (corollary of Proposition 2 of III, p. 4). Conversely, suppose that the endomorphism $u$ is compact. Since the family B is bounded in E, its image by $u$ in E is relatively compact in E (III, p. 2), and hence precompact. Let $\varepsilon >0$ and J be a finite subset of I such that the image of B by $u$ is contained in the union of the balls of radius $\varepsilon$ and centre $u(e_j)$ for $j\in J$. Let $i\in I$ - J. There exists $j\in J$ such that $\|u(e_i)-u(e_j)\|\leqslant \varepsilon$, hence

Consequently, one has $\lambda \in \mathscr{C}_0(I; K)$, that is, condition (i).

Finally, the spectrum of $u$ and its sensitive spectrum are calculated as functions of $\lambda$ by using prop. 1, c) and Proposition 5 of III, p. 90.