What is the relationship between spectral resolution and spectral measure?

1k Views Asked by At

In Kadison and Ringrose's book "FUNDAMENTALS OF THE THEORY OF OPERATOR ALGEBRAS", the author gives the following theorem.

Theorem: If $A$ is a self-adjoint operator acting on a Hilbert space $\mathscr{H}$ and $\mathscr{A}$ is an abelian von Neumann algebra containing $A$, there is a family $\{E_\lambda\}$ of projections, indexed by $\mathbb{R}$, in $\scr{A}$ such that

(1) $E_\lambda=0$ if $\lambda<-\|A\|$ and $E_\lambda=I$ if $\lambda\geq\|A\|$;

(2) $E_\lambda\leq E_{\lambda'}$ if $\lambda\leq\lambda'$;

(3) $E_\lambda=\wedge_{\lambda'>\lambda}E_{\lambda'}$;

(4) $AE_\lambda\leq\lambda E_\lambda$ and $A(I-E_\lambda)\leq\lambda(I-E_\lambda)$;

(5) $A=\int^{\|A\|}_{-\|A\|}\lambda dE_\lambda$ in the sense of norm convergence of approximating Riemann sums; and $A$ is the norm limit of finite linear combinations with coefficients in $sp(A)$ of orthogonal projections $E_{\lambda'}-E_\lambda$.

I want to know what (3) means and also what is the relationship between such so called "spectral resolution" $E_\lambda$ and the spectral measure?

1

There are 1 best solutions below

6
On BEST ANSWER

The formula connecting the spectral resolution of $E_{\lambda}$ and the spectral measure $E$ is $E_\lambda=E((-\infty,\lambda]),\ \mbox{for all}\ \lambda\in\mathbb R.$ You can get $E$ from $E_\lambda$ by setting $E((a,b]):=E_b-E_a$ etc.

Condition (3) means $E_\lambda$ is the greatest lower bound of $E_{\lambda'},\ \lambda'>\lambda.$ Equivalently, $E_\lambda$ is the strong limit of $E_{\lambda'},\ \lambda'\to\lambda+0.$