If you Google something like "Borel functional calculus" or "spectral decomposition" you get plenty of results for operators on Hilbert spaces. What about an (unrepresented) $\mathrm{C}^*$-algebra $A$?
Apologies if this question seems a bit dumb: I am a little unsteady after learning some things about unitary inequivalence of Hilbert space representations of $\mathrm{C}*$-algebras.
We have an isometric *-isomorphism (direct sum of GNS of pure states) of $A$ with $\pi(A)\subset B(\mathsf{H})$. I am happy to use the Gelfand Naimark *-isomorphism to say that when a finite-spectrum self-adjoint element of $A$, say $f$, we can look at it in $\pi(A)$, and then give it a spectral decomposition back in $A$. Briefly: $\pi(a)$ is a self-adjoint operator with finite spectrum (the spectrum of an element is a unitary invariant), and so it has a spectral decomposition:
$$\pi(f)=\sum_{f_i\in\sigma(f)}f_i\,p_i,$$
and the $(p_i)_{i}$ form a partition of unity. I can take the inverse isomorphism then and so $f$ has a spectral decomposition back in $A$:
$$f=\sum_{f_i\in\sigma(f)}f_i\,\pi^{-1}(p_i).$$
From Murphy:
Let $u$ be a normal operator on a Hilbert space $\mathsf{H}$. Then there is a unique spectral measure relative to $(\sigma(u),\mathsf{H})$ such that $u=\int z\,dE$ where $z$ is the inclusion of $\sigma(u)$ in $\mathbb{C}$
Question 1: So the question is, if I have a self-adjoint bounded operator $f\in A$, no longer with a finite spectrum, do I still have a spectral decomposition back in $A$?
$$f=\pi^{-1}(\pi(f))=\pi^{-1}\left(\int_{\sigma(\pi(f))}z\,dE\right)\overset{?}{=}\int_{\sigma(f)}z\,\pi^{-1}(dE).$$
If Question 2 is false, what I really want is the following. The spectrum of $f$ (not e.g. the point spectrum but the full spectrum) is the same in every faithful representation ($\sigma(f)=\sigma(\pi(f))$). In $\pi(f)$ I can certainly use Borel functional calculus to talk about for a Borel set $E\subset \sigma(f)$, the projection $\mathbf{1}_{E}(\pi(f))\in B(\mathsf{H})$, can I safely define:
$$\mathbf{1}_E(f):=\pi^{-1}(\mathbf{1}_E(\pi(f))).$$
In other words:
Question 2: Do I have Borel functional calculus on 'unrepresented' $A$?
Best Answer
In general, you don't have a spectral decomposition in $A$ because the projection $1_E(\pi(x))$ does not necessarily belong to $\pi(A)$. For example, if $X$ is a connected compact space, the only projections in $C(X)$ are $0$ and $1$, and the only elements you can write as limits of linear combinations of $0$ and $1$ are multiples of the identity.