Theorem 1

Suppose $K =\mathbf{C}$. Let $u$ be a compact and normal endomorphism of E. There exists an orthonormal basis B of E such that $u$ is diagonal in the basis B.

The set Sp$_s(u)$ is countable, and it does not contain 0 if E is of infinite dimension (III, p. 90, prop. 5, b)). For every element $\lambda \in$ Sp$_s(u)$, denote by $N_{\lambda}$ the nullspace of $u-\lambda 1_E$. It is of finite dimension (loc. cit.) and, since $u$ is normal, it coincides with the eigenspace of $u$ relative to $\lambda$ (EVT, V, p. 43, cor. of prop. 8). The spaces $N_{\lambda}$ are pairwise orthogonal (I, p. 132, n$^o5$).

Let F be the subspace of E Hilbert sum of the spaces $N_{\lambda}$ for $\lambda \in$ Sp$_s(u)-\{0\}$. It is a space of countable type, stable under $u$ since each subspace $N_{\lambda}$ is stable under $u$. Since $N_{\lambda}$ is also the eigenspace of $u^*$ corresponding to $\overline{\lambda}$ (EVT, V, loc. cit.), the endomorphism $u$ induces an endomorphism $\widetilde{u}$ of $F^{\circ}$ by passing to subspaces. The endomorphism $\widetilde{u}$ is compact (Proposition 3 of III, p. 5) and normal (Lemma 4 of I, p. 135). By construction, the point spectrum of $\widetilde{u}$ is contained in $\{0\}$ (for, any eigenvector for $\widetilde{u}$ would be one for $u$, hence would belong to one of the spaces $N_{\lambda}$ if the corresponding eigenvalue were non-zero). Thus the spectral radius of $\widetilde{u}$ is zero, whence $\widetilde{u}= 0$ since $\widetilde{u}$ is normal (Corollary 1 of I, p. 108). Hence $F^{\circ}\subset$ Ker($u$).

For every $\lambda \in$ Sp$_s(u)$, let $B_{\lambda}$ be an orthonormal basis of $N_{\lambda}$ and let $(e_j)_{j\in J}$ be the family union of the $B_{\lambda}$; it is an orthonormal basis of F. Let B be the union of $(e_j)_{j\in J}$ and of an orthonormal basis of $F^{\circ}$. It is an orthonormal basis of E and $u$ is diagonal in the basis B.