$\lambda\in\sigma(A)\Leftrightarrow P_{(\lambda-\epsilon,\lambda+\epsilon)}\neq 0\;\forall \epsilon>0$ I would be grateful for a proof of a proposition in Reed Simon I, Chapter VII.3. It states that for a bounded self-adjoint operator $A$ a point $\lambda$ is in the spectrum of $A$ if and only if for every $\epsilon>0$ the spectral projection $P_{(\lambda-\epsilon,\lambda+\epsilon)}$ is nonzero.
As a hint they say that the main ingredient in the proof is the equality $\|(A-\lambda)^{-1}\|=\operatorname{dist}(\lambda,\sigma(A))^{-1}$, but I've got no clue how to use this.
 A: For any $x \in H$ and $\epsilon > 0$,
\begin{align}
         \|(A-\lambda I)P(\lambda-\epsilon,\lambda+\epsilon)x\|^2 
    & = \int_{(\lambda-\epsilon,\lambda+\epsilon)}(\mu-\lambda)^2d(P(\mu)x,x) \\
    & \le \epsilon^2\int_{(\lambda-\epsilon,\lambda+\epsilon)}d(P(\mu)x,x) \\
    & = \epsilon^2\|P(\lambda-\epsilon,\lambda+\epsilon)x\|^2.
\end{align}


*

*Suppose $\lambda\notin\sigma(A)$. Then let $x=(A-\lambda I)^{-1}y$, and notice that
\begin{eqnarray*}
  \|(A-\lambda I)P(\lambda-\epsilon,\lambda+\epsilon)(A-\lambda I)^{-1}y\|
    &\le \epsilon\|P(\lambda-\epsilon,\lambda+\epsilon)(A-\lambda I)^{-1}y\| \\
  \|P(\lambda-\epsilon,\lambda+\epsilon)y\|
    &\le \epsilon \|(A-\lambda I)^{-1}P(\lambda-\epsilon,\lambda+\epsilon)y\| \\
  \|P(\lambda-\epsilon,\lambda+\epsilon)y\|
    &\le \epsilon \,\mbox{dist}(\lambda,\sigma(A)^{-1}
          \|P(\lambda-\epsilon,\lambda+\epsilon)y\|.
\end{eqnarray*}
Therefore $P(\lambda-\epsilon,\lambda+\epsilon)y=0$ for $\epsilon\cdot\mbox{dist}(\lambda,\sigma(A))^{-1} < 1$. Because $y$ was arbitrary, then
$$
    P(\lambda-\epsilon,\lambda+\epsilon)=0 \mbox{ for  } \epsilon < \mbox{dist}(\lambda,\sigma(A)).
$$

*Conversely, suppose $P(\lambda-\epsilon,\lambda+\epsilon) = 0$ some $\epsilon > 0$. Then
$$
       \int\frac{1}{\mu-\lambda}dP(\mu)=\int_{|\mu-\lambda| \ge \epsilon}\frac{1}{\mu-\lambda}dP(\mu)
$$
is a bounded operator $R(\lambda)$ such that $(A-\lambda I)R(\lambda)=R(\lambda)(A-\lambda I)=I$, which proves $\lambda\notin\sigma(A)$.
