Let $A \in B(H)_{sa}$. For a closed subset $F \subset \sigma (A)$. Let $\lambda \in \sigma (A)$. Show the following statements:
1- $(A- \lambda I) E^A (\{ \lambda \})=0$
2- $E^A (\{ \lambda \})$ is the projection onto $\ker (A- \lambda I)$.
My proof:
1- Define $f_n \in C_{(-\infty,\lambda]}$ (as defined in the lecture notes). Then $E^A_\lambda = \underset{n \to \infty}{\operatorname{s-\lim}} f_n (A)$. Let $f(A) = ( A -\lambda I) $.
$$\| f(A) E^A (\{ \lambda \}) x \| = \underset{n \to \infty}{\lim} \| f(A) f_n (A) x\| \le \underset{n \to \infty}{\lim} \| f(A) f_n (A) \| \|x\|,$$
and
$$ \| f(A) E^A (\{ \lambda \}) x \| \le \underset{n \to \infty}{\lim} \underset{t \in [\lambda, \lambda+ \frac{1}{n})}{\sup} |f(t)| =f(\lambda) = \lambda I - \lambda I =0$$
2- From (1) it is clear that $R(E^A(\{\lambda\})) \subset \ker (A - \lambda I)$. Let $\{x_n\}_{n=1}^\infty$ be an orthonormal basis for $\ker (A- \lambda I)$, then $\| x_n \| =1$, and $(A- \lambda I) x_k =0$ for all $k \in \mathbb{N}$. Then from theorem (3.5-3) we have
\begin{align*} \|(E^A(\{\lambda \}) - \lambda) x_k \| &=\underset{n \to \infty}{s-\lim} \| \underset{g \in C_{[\lambda, \lambda+ \frac{1}{n})}}{\inf}\{g(A)\} (x_k) - \lambda x_k\|=0\\ \end{align*} So, $\underset{n \to \infty}{s-\lim} \underset{g \in C_{[\lambda, \lambda+ \frac{1}{n})}}{\inf}\{g(A)\} (x_k) = \lambda x_k \Rightarrow E^A (\{ \lambda \}) x_k = \lambda x_k \Rightarrow x_k \in R(E^A(\{ \lambda \}))$.