# Problem with spectral theorem and spectral measure.

There is a passage in a book that is not very clear to me:

A is a C*Algebra and $a$ is selfadjoint.

Then

"Indeed identifying A with an algebra of operators on a Hilbert space $\mathcal{H}$, by the spectral theorem there is a projection-valued compactly supported measure $E(\cdot \space; a)$ so that for every continuous function,

$f(a)=\int f(t)dE((-\infty,t],a)$"

I know that the spectral theorem says that if a is selfadjoint then it can be viewed as a multiplication operator . I also know what a spectral measure is.

It's not clear to me why in the dE there are intervals in the form $(-\infty,t]$. I've always seen only dE (without specifications). Maybe is only a matter of notation.

and in addition, $f(a)$ should be itself an operator in A (obtained by the taylorization of f) so why is equal a complex nuber?!

Thank you for your help :-)

## 2 Answers

Spectral theorem for a self-adjoint operator is often formulated as $a=\int\lambda dE_\lambda,$ where $E_\lambda:\mathbb R\to B(H)$ is the spectral resolution of $a=a^*\in B(H).$ In your book $E(\cdot\ ;a)$ is the spectral measure (function of Borel subsets $\mathbb R$) associated with $a.$ The mapping $\lambda\mapsto E_\lambda=E((-\infty,\lambda];a)$ is now the spectral resolution of $a$ and the integral is the projection-valued Riemann-Stieltjes integral.

That has to be some of the most awkward notation for the Spectral Theorem that I've seen. The author could state that $E$ is the spectral measure for $a$, thereby eliminating $a$ from the decorations added to $E$. Who decorates the eigenfunctions or eigenvalues for an operator with the name of that operator, unless there is some special reason to do so? The author appears to mix Riemann-Stieltjes integration with the spectral measure. Otherwise, why would he add $(-\infty,t]$ into the picture. For both Riemann-Stieltjes integrals and for Borel operator measures, it seems to me that one could simply write $$f(a) = \int f(t)dE(t) \mbox{ or } f(a)=\int f dE,$$ with possible upper- and lower-limits or a set of integration. If more than one spectral measure were being used in the discussion, perhaps $$f(a) = \int fdE_{a}.$$ I've never liked "projection-valued measure" either. One rarely writes "complex-valued measure" to mean "complex measure." The adjective describes the values, and "on" refers to the domain. "Spectral measure for a selfadjoint element a" should be clear enough.

• Thanks! And what about $f(a)$? If $f$ is a continuous function on $C(\sigma(a))$, then $f(a)$ shouldn't be an operator? Why is identifyed with a complex number here? – Benzio Dec 5 '13 at 16:30
• Here f(a) is supposed to be an element of the C* algebra. This is a bit confusing because the Spectral theorem for 'a' assumes 'a=a*' is an operator on a Hilbert space. Of course, every C* algebra is such an operator algebra, but I don't know what has been introduced at this point in your reading. – DisintegratingByParts Dec 6 '13 at 1:05