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Let $x$ be a self-adjoint operator on $H$. By spectral theorem, there is a spectral measure $\mu$ correspondence to $*-$ homomorphism $\pi:C(\sigma(x)) \to B(H)$ such that $x=\int_{-||x||}^{||x||} \lambda d\mu(\lambda)$. Put $$e_n=\int_{|\lambda|\geq \frac{1}{n}} d\mu(\lambda)$$ Suppose $x_n=e_nx =\int_{|\lambda|\geq \frac{1}{n}}\lambda d\mu(\lambda)$. I want to find $\sigma(x_n) $ and also show that $\mu_{|\sigma-alg(\sigma(x_n))} = \mu_n$ where $\mu_n$ is spectral measure correspondence $*-$homomorphism $\pi':C(\sigma(x_n))\to B(e_nH)$.

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Please help me. Thanks in advance.

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Assume $e_{n} \ne 0$ and $e_{n} \ne 1$; otherwise, the problem is trivial.

The Hilbert space $H$ splits into $e_{n}H\oplus (1-e_{n})H$ and these subspaces are invariant under $x$ and all functions of $x$. Let $x'$ denote the restriction of $x$ to $e_{n}H$ and $x''$ the restriction of $x$ to $(1-e_{n})H$. Then $x$ has the diagonal operator matrix representation $$ x = \left[\begin{array}{cc} x' & 0 \\ 0 & x''\end{array}\right]. $$ It is obvious that $x-\lambda 1$ is invertible iff $x'-\lambda 1'$ and $x''-\lambda 1''$ are invertible on their respective spaces. In other words, $$ \sigma(x) = \sigma(x')\cup\sigma(x''), $$ even though the union may not be disjoint. If $\alpha \notin \sigma(x)\cap\{|\lambda| \ge 1/n\}$, then $x'-\alpha 1'$ is invertible because $\alpha$ must be a positive distance from the compact set $\sigma(x)\cap\{|\lambda|\ge 1/n\}$, which guarantees that the following is in the algebra $$ r_{n}(\alpha) = \int_{|\lambda|\ge 1/n}\frac{1}{\lambda-\alpha}d\mu(\lambda). $$ And $r_{n}(\alpha)(x_{n}-\alpha 1)=(x_{n}-\alpha 1)r_{n}=e_{n}$, which gives $\alpha\in\rho(x')$. Therefore, $$ \sigma(x') \subseteq \sigma(x)\cap\{|\lambda| \ge 1/n\}. $$ Similarly, letting a superscript of 'c' denote topological closure, the same type of argument shows $$ \sigma(x'') \subseteq (\sigma(x)\cap\{|\lambda| < 1/n\})^{c}. $$ So we know that $\sigma(x')\cap\sigma(x'')\subseteq \sigma(x)\cap\{|\lambda|=1/n\}$ because the above closure is contained in (but may not equal) the closed set $\sigma(x)\cap\{|\lambda| \le 1/n\}$. That definitely gives $$ \sigma(x)\cap\{|\lambda| > 1/n\} \subset\sigma(x'),\\ \sigma(x)\cap\{|\lambda| < 1/n\} \subset\sigma(x''). $$ The frontier set $S=\sigma(x)\cap\{|\lambda|=1/n\}$ can be a subset of either spectrum or of both. If $\lambda \in S$ is in the point spectrum, then $\lambda\in\sigma(x')$ definitely holds.

That still doesn't answer your question about $x_{n}$. Note that $x_{n}$ has the matrix representation $$ x_{n}=\left[\begin{array}{cc} x' & 0 \\ 0 & 0\end{array}\right]. $$ Therefore $\sigma(x_{n})=\sigma(x')\cup\{0\}$ because $0$ is definitely in the spectrum and, for $\alpha \ne 0$, the following is invertible iff $x'-\alpha 1'$ is invertible: $$ \left[\begin{array}{cc} x'-\alpha 1' & 0 \\ 0 & -\alpha 1''\end{array}\right]. $$ In fact, if $\alpha \ne 0$ and $x'-\alpha 1'$ is invertible then $$ (x-\alpha 1)^{-1} = \left[\begin{array}{cc} (x'-\alpha 1')^{-1} & 0 \\ 0 & -\frac{1}{\alpha}1'\end{array}\right] $$ So $\sigma(x_{n})=\sigma(x')\cup\{0\}$, even though the question of $\sigma(x')$ cannot be fully answered because of the questionable points $\sigma(x)\cap{|\lambda|=1/n}$.

For the final part, note that spectral representations are unique, and the representation of $x'=\int \lambda d\mu'$, where $\mu'$ is the restriction of $\mu$ to $e_{n}H$, is a spectral representation because $\mu'$ is easily shown to be a spectral measure with values in $\mathcal{L}(H')$.

Added because of your remark: You asked about $x_{n}$ using the functional calculus. The operator matrix representation, being diagonal, is the same as $$ x_{n}=\pi(z\chi_{|\lambda| \ge 1/n}). $$ To look at the resolvent, $$ x_{n}-\lambda 1=\pi(z\chi_{|\lambda| \ge 1/n}-\lambda 1) = \pi((z-\lambda)\chi_{|\lambda| \ge 1/n}-\lambda\chi_{|\lambda| < 1/n}). $$ If $\lambda \ne 0$ and $\lambda\notin\sigma(x)\cap\{ |\lambda| \ge 1/n\}$, then $x_{n}-\lambda 1$ has an inverse $$ (x_{n}-\lambda 1)^{-1}=\pi\left(\frac{1}{z-\lambda}\chi_{|\lambda|\ge 1/n}(z)-\frac{1}{\lambda}\chi_{|\lambda| < 1/n}(z)\right). $$ This is the same as the diagonal operator matrix approach.

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  • $\begingroup$ I agree with you about $\sigma(x_n)$. But my question is a part of discussion about a representation of a compact operator which I added above. In this theorem the arthur claims $x_n$ is invertable while $0\in \sigma(x_n)$!!! $\endgroup$ – niki Nov 2 '14 at 15:08
  • $\begingroup$ @niki : His $x_{n}$ is the restriction to $e_{n}H$, which is my $x'$. If you read his proof carefully, $x_{n}$ is not $xe_{n}$, but the restriction of $xe_{n}$ to $e_{n}H$. $\endgroup$ – DisintegratingByParts Nov 2 '14 at 15:16
  • $\begingroup$ What is the correspondence function of $x_n$ in $B(\sigma(x))$( Bounded Borel-measurable function space)? I think it's $z\chi_{|\lambda|\geq 1/n}$. $\endgroup$ – niki Nov 2 '14 at 15:27
  • $\begingroup$ Yes, the $x_{n}$ you have defined is on $H$ and the correspondence is with $z\chi_{|\lambda|\ge 1/n}(z)$. However, the restriction to $e_{n}H$ corresponds $x_{n}$ with $z$. $\endgroup$ – DisintegratingByParts Nov 2 '14 at 15:28
  • $\begingroup$ Thanks, I think now it's clear for me. $\endgroup$ – niki Nov 2 '14 at 15:48

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