Theorem 1

Let E be a complete normable space and $u$ an endomorphism of E.

a) The set Sp$_{\omega}(u)$ is compact. It is non-empty if E is infinite-dimensional;

b) Let $n\in \overline{\mathbf{Z}}$. The set Sp$_n(u)$ is the union of a family of bounded connected components of $\mathbf{C}-$ Sp$_{\omega}(u)$. It is open in $\mathbf{C}$, and its boundary in $\mathbf{C}$ is contained in Sp$_{\omega}(u)$.

The set $\mathbf{C}-$ Sp$_{\omega}(u)$ consists of the complex numbers $\lambda \in \mathbf{C}$ such that $u-\lambda 1_E$ is a strict morphism, whose kernel or cokernel has finite dimension. By props. 11 of III, p. 67 and 13 of III, p. 70, it is open. The set Sp$_{\omega}(u)$ is therefore closed. Since it is bounded, it is compact.

Let us prove b). Let $n\in \overline{\mathbf{Z}}$. The set Sp$_n(u)$ is contained in $\mathbf{C}-$Sp$_{\omega}(u)$. Let U be a connected component of $\mathbf{C}-$ Sp$_{\omega}(u)$ which meets Sp$_n(u)$. The mapping $\lambda \mapsto$ ind($u-\lambda 1_E$) from $\mathbf{C}-$ Sp$_{\omega}(u)$ into $\overline{\mathbf{Z}}$ being locally constant (cor. 1 of prop. 12 of III, p. 68 and cor. 1 of prop. 13 of III, p. 70), the index of $u-\lambda 1_E$ is equal to $n$ for every $\lambda \in U$. If $n\not = 0$, this implies that U is contained in Sp$_n(u)$. If $n= 0$, remark that the set U is a connected component of $\mathbf{C}-$ (Sp$_{\omega}(u)\cup$ Sp$_{-\infty}(u)\cup$ Sp$_{+\infty}(u)$). Since U meets Sp$_0(u)$ and hence Sp$_e(u)$, it follows then from Remark 2 that the set U is contained in Sp$_e(u)$, and consequently in Sp$_0(u)$. In all cases we conclude that Sp$_n(u)$ is the union of the connected components of $\mathbf{C}-$ Sp$_{\omega}(u)$ which meet Sp$_n(u)$. These are necessarily bounded since the set Sp($u$) is bounded. Consequently, Sp$_n(u)$ is open in $\mathbf{C}$ and its boundary is contained in Sp$_{\omega}(u)$. This proves b).

Suppose finally that the set Sp$_{\omega}(u)$ is empty. According to b), each of the sets Sp$_n(u)$, for $n\in \overline{\mathbf{Z}}$, is then empty. We therefore have Sp($u$) $=$ Sp$_s(u)$. The spectrum of $u$ is consequently discrete and compact, hence finite, and since all its points are of finite spectral multiplicity, the vector space E is finite-dimensional (III, p. 83, prop. 2). This completes the proof of a).