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

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


"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 :-)


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.

See also What is the relationship between spectral resolution and spectral measure?


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.

  • $\begingroup$ 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? $\endgroup$ – Benzio Dec 5 '13 at 16:30
  • $\begingroup$ 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. $\endgroup$ – DisintegratingByParts Dec 6 '13 at 1:05

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