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 $\pi':C(\sigma(x_n))\to B(e_nH)$.

Add: enter image description here

enter image description here

Please help me. Thanks in advance.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.