Spectral projection and isolated point of spectrum 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.
 A: 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 $x \in \mbox{ker}(u-\lambda I)$,
$$
                 x = E(\mathbb{C}\setminus\{\lambda\})x+E\{\lambda\}x = E\{\lambda\} x \in E\{\lambda\} H \\
                   \implies \mbox{ker}(u-\lambda I) \subseteq E\{\lambda\}H.
$$
