Let $u\in B(H)$ be a normal element with spectral resolution of the identity $E$ and $\lambda$ be an isolated point of spectrum $u$. Show that $E(\lambda)H = \ker(u-\lambda)$ .
I can show that $E(\lambda)H \subset \ker(u-\lambda)$, but I do not have any idea to show $\ker(u-\lambda)\subset E( \lambda)H $. Please give me a hint. Thanks in advance.
Starting with the comment, $$ E(S)\left(\int f(\mu)dE(\mu)\right)=\left(\int f(\mu)dE(\mu)\right) E(S)=\int_{S}f(\mu)dE(\mu). $$ Let $f(\mu)=\mu-\lambda$ and let $S$ be the singleton set $\{\lambda\}$. Then the second equality gives $$ (u-\lambda I)E\{\lambda\} = 0. $$ So, as you deduced, $E\{\lambda\}H \subseteq \mbox{ker}(u-\lambda I)$.
Conversely, suppose that $(u-\lambda I)x=0$. To show that $x \in E\{\lambda\}H$, let $S_{\epsilon} = \{ \mu : |\mu-\lambda| \ge \epsilon \}$. Then $$ \begin{align} \epsilon^{2}\|E(S_{\epsilon})x\|^{2} & \le \int_{S_{\epsilon}}|\mu-\lambda|^{2}d\|E(\mu)x\|^{2} \\ & = \left\|\int_{S_{\epsilon}}(\mu-\lambda)dE(\mu)x\right\|^{2} \\ & = \left\|E(S_{\epsilon})\int(\mu-\lambda)dE(\mu)x\right\|^{2} \\ & = \|E(S_{\epsilon})(u-\lambda I)x\|^{2} = 0. \end{align} $$ Therefore $E(S_{\epsilon})x = 0$ for all $\epsilon > 0$. It follows that $$ E(\mathbb{C}\setminus\{\lambda\})x = 0. $$ Hence, if $x \in \mbox{ker}(u-\lambda I)$, $$ x = E(\mathbb{C}\setminus\{\lambda\})x+E\{\lambda\}x = E\{\lambda\} x \in E\{\lambda\} H \\ \implies \mbox{ker}(u-\lambda I) \subseteq E\{\lambda\}H. $$