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Let $A:H\longrightarrow H$ be a self-adjoint operator, where H is an Hilbert space. Let $(E_{\lambda})_{\lambda}$ be the spectral decomposition of $A$ and $\lambda_0$ a regular value of A with finite multiplicity and isolated in the spectrum of $A$. Chosen $\varepsilon$ small enough we have $$A=\int_{\mathbb{R}}\lambda dE_{\lambda}= \lambda_0\Pi_{\lambda_0}+\int_{|\lambda-\lambda_0|>\varepsilon}\lambda dE_{\lambda}$$. Why is it that $$(A-z)^{-1}=\frac{1}{\lambda_0-z}\Pi_{\lambda_0}+\int_{|\lambda-\lambda_0|>\varepsilon}\frac{1}{\lambda-z}dE_{\lambda}$$?. (z is not in the spectrum of A). I know a theorem that states that if $A=\int\lambda dE_{\lambda}$ then $(A-z)^{-1}=\int\frac{1}{\lambda-z}dE_{\lambda}$ but I don't get it in the case above.

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up vote 2 down vote accepted

Since $\lambda_0$ is an eigenvalue of $A$ with finite multiplicity and an isolated point of the spectrum, the projection-valued measure $E_\lambda$ on the interval $$ I := \{\lambda\in \mathbb R : \vert \lambda - \lambda_0 \vert < \epsilon \} $$ is pure point with just one atom, $\lambda_0$. For each Borel set $\Omega\subseteq I$, we have $$ E_\Omega = \begin{cases} E_{\{\lambda_0\}} & \text{if } \lambda_0\in \Omega \\ 0 & \text{if } \lambda_0\notin \Omega \end{cases} $$ and therefore if $f$ is a bounded Borel function $$ \int_I f(\lambda)dE_\lambda = f(\lambda_0)E_{\{\lambda_0\}} $$ Applying the above equality we can write $$ \begin{align} A &= \int_{\mathbb{R}}\lambda dE_{\lambda}\\ &= \int_I\lambda dE_{\lambda} + \int_{\mathbb{R}\setminus I}\lambda dE_{\lambda}\\ &= \lambda_0 E_{\{\lambda_0\}}+\int_{\mathbb R \setminus I}\lambda dE_{\lambda} \end{align} $$ and $$ \begin{align} (A - z)^{-1} &= \int_{\mathbb{R}}\frac{1}{\lambda-z} dE_{\lambda} \\ &= \int_I\frac{1}{\lambda-z} dE_{\lambda} + \int_{\mathbb{R}\setminus I}\frac{1}{\lambda-z} dE_{\lambda} \\ &= \frac{1}{\lambda_0 - z} E_{\{\lambda_0\}}+\int_{\mathbb R \setminus I}\frac{1}{\lambda-z} dE_{\lambda} \\ \end{align} $$

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