Proposition 3

Let $u$ be a compact positive endomorphism of E. There exists an orthonormal family $(e_n)_{n\in I}$ in E such that, for every $x\in E$, one has

Let $B = (f_j)_{j\in J}$ be an orthonormal basis of E in which $u$ is diagonal (cor. 1 of IV, p. 150) and $(\lambda_j)_{j\in J}$ the family of eigenvalues of $u$ in the basis B. Let $J'$ be the set of $j\in J$ such that $\lambda_j$ belongs to the point spectrum of $u$.

For each $\lambda \in$ Sp$_s(u)$, there exists a bijection between the integers $n$ such that $0\leqslant n <M$ and $\lambda_n(u) =\lambda$ and the $j\in J'$ such that $\lambda_j=\lambda$, for these two sets have as cardinal the spectral multiplicity of $\lambda$. A choice of such bijections for each $\lambda$ defines a bijection $\iota$ of the set of integers such that $0\leqslant n <M$ onto the set $J'$. One defines the sequence $(e_n)_{0\leqslant n<M}$ in E by putting $e_n=f_{\iota(n)}$ for $0\leqslant n <M$. It is an orthonormal family in E.

In the case where M $<$ Card(I) $= +\infty$, the space F generated by $\{e_0, . . . , e_{M-1}\}$ is of finite dimension and its orthogonal complement $F^{\circ}$ is of infinite dimension; one chooses for $(e_n)_{n\geqslant M}$ an orthonormal family in $F^{\circ}$.

For every $x\in E$, one has

since $u(f_j) = 0$ when $j\in J$ - $J'$. If $n\in I$ satisfies $n\geqslant M$, one has $\lambda_n(u) = 0$, and one obtains the first formula of the proposition. The second results from it.