Proposition 14

Let E be a complex Banach space. In order that an element $u$ of $\mathscr{L}(E)$ be a Riesz endomorphism of E, it is necessary and sufficient that one of the following two mutually exclusive conditions be satisfied:

(i) The endomorphism $u$ of E is an automorphism of E;

(ii) The point 0 is an isolated point of Sp($u$) and the projector $e_0(u)$ has finite rank.

When condition (ii) is satisfied, the image of $e_0(u)$ is the nilspace of $u$ and the kernel of $e_0(u)$ is the conilspace of $u$.

Every automorphism of E is a Riesz endomorphism of E (III, p. 47, remark 1). If $u$ satisfies condition (ii), the kernel F of $e_0(u)$ is a closed vector subspace of finite codimension of E, stable under $u$. Let $u_F$ be the endomorphism of F induced by $u$. Its spectrum is contained in $\mathbf{C}-\{0\}$ (No. 3 of I, p. 131), hence $u_F$ is an automorphism of F. It follows that $u$ is a Riesz endomorphism of E (III, p. 48, prop. 8).

Conversely, let $u$ be a Riesz endomorphism of E. Let N denote its nilspace and I its conilspace. By prop. 6 of III, p. 47, the space E is the topological direct sum of N and I, and the mapping $u$ defines, by passing to the subspaces, a nilpotent endomorphism $u_N$ of N and an automorphism $u_I$ of I. In particular, we have Sp($u_N$)$\subset \{0\}$ and $0\not \in$ Sp($u_I$). If N is zero, then I = E and $u$ is an automorphism of E. Otherwise, 0 is an isolated point of Sp($u$) and $e_0(u)$ is the projector with image N and kernel I (prop. 2 of I, p. 129); in particular, it is of finite rank.