Show that $\lambda$ is an eigenvalue for normal bounded linear operator $N$ on Hilbert space $H$ with spectral measure $E$ iff $E(\{\lambda\})\neq 0$, in which case the range of $E(\{\lambda\})$ is the eigenspace $\ker (N-\lambda)$.
My difficulty for this Exercise is showing that $E(\{\lambda\})H = \ker(N-\lambda)$ when $\lambda$ is an eigenvalue for $N$.I just know that $E(\{\lambda\})H \subset \ker(N-\lambda)$ , but for converse I can not show it. Please give me a hint. Thanks .
For any bounded Borel function $f$, and Borel subset $S\subset\mathbb{C}$, $$ E(S)\int f(s)dE(s)x =\int_{S} f(s)dE(s)x,\;\;\; x \in H. $$ And $\int s dE(s)x = Ax$. Therefore, $\mathcal{R}(E(\{\lambda\}))\subseteq\mathcal{N}(A-\lambda I)$ follows from $$ (A-\lambda I)E(\{\lambda\})x=E(\{\lambda\})(A-\lambda I)x=\int_{\{\lambda\}}(s-\lambda)dE(s)x=0. $$ Conversely, suppose $(A-\lambda I)x=0$. For any $\rho > 0$, let $E_{\rho}=E(\{ \mu : |\mu-\lambda| \ge \rho > 0\})$. Then, using properties of the spectral integral, $$ 0= \|E_{\rho}(A-\lambda I)x\|^{2}=\|(A-\lambda I)E_{\rho}x\|^{2} \ge \rho^{2}\|E_{\rho}x\|^{2}. $$ Hence, $E_{\rho}x=0$ for all $\rho > 0$, which gives $E(\mathbb{C}\setminus\{\lambda\})x=0$. Hence, $E(\{\lambda\})x=x$.