# Showing there is a unique spectral measure

All the books I have seen have proved that, for a normal bounded operator $T$, there is a unique spectral measure $E$ such that $\int_{\sigma(T)}^{}\lambda\,dE=T$ by first proving in it for a general Algebra. I am trying to modify this proof to get straight to $\int_{\sigma(T)}^{}\lambda\,dE=T$ without proving it for a general algebra.

I have almost immediately come up against a problem because it uses the Riesz representation theorem and I can't think how to integrate (pardon the pun) that into my proof. Surely though there is a way to do this since I am trying to prove something easier. Can anyone help?

Note: I also posted this on stack exchange as I am not sure which is more suitable.

Thanks

## 1 Answer

I think most books start with the spectral theorem for a single bounded self-adjoint operator; Reed and Simon vol. 1 is a canonical example. But spectral theory for a single normal operator is essentially the same as spectral theory for a pair of commuting self-adjoint operators (namely, its real and imaginary parts). So at that point it is natural to just give the general construction for algebras.

However, it so happens that in my book Mathematical Quantization I give a proof for a single bounded normal operator. See Theorem 3.5.3.

• thanks for trying but that's not quite approach I want. – user108605 Feb 14 '14 at 8:22
• okay, good luck then – Nik Weaver Feb 14 '14 at 8:39