Proposition 5

Let E be a complex Hilbert space and $u$ a normal endomorphism of E.

a) For every open and closed subset H of the spectrum of $u$, the spectral projector $e_H(u)$ is an orthogonal projector whose kernel is the image of the spectral projector $e_{Sp(u)-H}(u)$ ;

b) If $H_1$ and $H_2$ are disjoint open and closed subsets of the spectrum of $u$, then the spectral subspaces $E_{H_1}$ and $E_{H_2}$ are orthogonal ;

c) If $\lambda \in \mathbf{C}$ is an isolated point of the spectrum of $u$, then $\lambda$ is an eigenvalue of $u$ and the image of the spectral projector $e_{\lambda}(u)$ is the eigenspace of $u$ relative to $\lambda$.

Let us prove a). Since the holomorphic functional calculus is compatible with the continuous functional calculus (I, p. 111, cor. 1), we have $e_H(u) =\varphi_H(u)$, where $\varphi_H\in \mathscr{C}$ (Sp($u$)) is the characteristic function of H. This implies $e_H(u)^*=\overline{\varphi}_H(u) =\varphi_H(u) =e_H(u)$, hence $e_H(u)$ is an orthogonal projector (lemma 3, (ii)). Its kernel is the image of the projector $1-e_H(u) =e_{Sp(u)-H}(u)$.

Let us prove b). The characteristic functions $\varphi_{H_1}$ and $\varphi_{H_2}$ of $H_1$ and $H_2$ in Sp($u$) are continuous and their product is zero, which implies $e_{H_1}(u)\circ e_{H_2}(u) =e_{H_2}(u)\circ e_{H_1}(u) = 0$. The inclusions $E_{H_2}(u)\subset E_{H_1}(u)^{\circ}$ and $E_{H_1}(u)\subset E_{H_2}(u)^{\circ}$ follow.

Finally, let us prove assertion c). The characteristic function $\varphi_{\lambda}$ of $\{\lambda \}$ is continuous and non-zero on Sp($u$) ; it satisfies $(z-\lambda )\varphi_{\lambda}(z) = 0$ for every $z\in$ Sp($u$). Hence we have $(u-\lambda 1_E)\varphi_{\lambda}(u) = 0$. The image of $\varphi_{\lambda}(u)$, which is non-zero, is therefore contained in the proper subspace of $u$ relative to $\lambda$. Since we have $e_{\lambda}(u) =\varphi_{\lambda}(u)$ and the image of $e_{\lambda}(u)$ contains the proper subspace of $u$ relative to $\lambda$, the assertion follows.