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For any normal element $a \in A$ for a $C^*$-algebra $A$, we can easily define a continuous functional calculus using $\varphi:C(\sigma(a))\to C^*[1,a]$. We can also define a spectral measure using $a = \int_{\sigma(a)} \lambda \ dE(\lambda)$ by considering $C^*[1,a] \subset B(H)$ where $H$ is the universal representation of $A$. Most of the books I've sorted through only really care about the the spectral measure for normal elements in $B(H)$, but I'm wondering how this matches back up with $A$?

In particular, let's call $\overline A$ the weak closure of $A \subset B(H)$. Let's also explicitly call $E_a$ the spectral measure associated to $a \in A$. It's clear that for $f \in C_0(\mathbb C)$ we have that $\int_{\sigma(a)} f(\lambda) \ dE_a(\lambda) \in A$ for all normal $a \in A$.

Is it the case that for borel measurable $f$ that $\int_{\sigma(a)} f(\lambda) \ dE_a(\lambda) \in \overline{C^*[1,a]}$ for all normal $a \in A$? Moreover when is it the case that for all borel measurable $f$ and normal $a \in A$ that $\int_{\sigma(a)} f(\lambda) \ dE_a(\lambda) \in A$? From this answer, Is a von Neumann algebra just a C*-algebra which is generated by its projections? I'm guessing that this is not strong enough to characterize von neumann algebras, but what sorts of spaces is this true for? I think this last condition should be equivalent to $E_a(S) = \chi_S (a) \in A$ for all normal $a$, but I'm not sure. If the first question is true, then in this case we would have $\overline{C^*[1,a]} \subset A$ for all normal $a \in A$, but I don't know what sort of algebra that is!

Addendum: For the first question it seems like one could use the fact that borel sets appear at stage $\omega_1$ in the hierarchy? And so taking all $\omega_1$ length nets of functions it should be enough that $\langle \varphi(f_n)x,y \rangle \to \langle \varphi(f)x,y \rangle$. I'm not so sure about this $\omega_1$ net argument though.

Ice Tea
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Opisthokont
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  • The compact operators are not a von Neumann algebra, but they are generated by projections. – J. De Ro Oct 10 '21 at 07:07
  • Isn't the borel functional calculus a stronger condition than being generated by projections? Because being generated by projections is a norm-closure condition and the functional calculus is a weak-closure condition? – Opisthokont Oct 10 '21 at 07:24

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Regarding your second paragraph, you can only have a meaningful spectral measure for normal elements.

The integral over a spectral measure is a norm limit, so it will stay in any C$^*$-algebra that contains the values of $E_a$. One can also show, via the Double Commutant Theorem, that any von Neumann algebra that contains the normal element $a$, also contains the range of the spectral measure $E_a$. So if $a\in A$, with $a$ normal and $A$ a von Neumann algebra, then $\int_{\sigma(a)}f(\lambda)\,dE_a(\lambda)\in A$ for all bounded Borel functions $f$.

The algebra you denote by $\overline{C^*(1,a)}$ is the von Neumann subalgebra generated by $a$, and the above applies. Said algebra contains, sometimes strictly the algebra $$ \Big\{\int_{\sigma(a)}f(\lambda)\,dE_a(\lambda):\ f\ \text{bounded Borel}\Big\}. $$

Martin Argerami
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  • Forgot to include the word "normal" there, typed it so much I forgot (edited it in now)! And that's a very easy proof, it should have occurred to me right away. Is it also true that all normal elements in the vNa generated by $a$ are achieved as the spectral integral of a borel function? Since the other way around is now fairly clear. And what precisely are the $C^$ algebras where $\overline{C^(1,a)} \subset A$ for all normal $a$, the algebras that have an internal spectral calculus? – Opisthokont Oct 10 '21 at 22:37
  • To clarify I mean the algebra where $\overline{C^(1,a)} \subset A$ for all normal $a$. Which is really asking what the enveloping $C^$ algebra of $\cup_i \overline{C^*(1,a_i)}$ looks like for a set of normal elements $a_i \in B(H)$. – Opisthokont Oct 10 '21 at 22:47
  • Ah sorry for the repeated comments, just realized that the first part is also obvious, since for a net $\langle a_n x,y \rangle \to \langle a x,y \rangle$, we can apply the functional calculus for each $a_n$ and that's clear. It's interesting to note, baire-1 consists of all the functions that are pointwise limits of continuous functions, but it seems like this sort of weak limit of continuous functions in fact results in all borel functions. Which is something to keep in mind! Thank you for all the help. I need to think more about the question of that algebra, maybe a separate question later? – Opisthokont Oct 10 '21 at 23:11
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    I included mention of the characterization of the algebra. As soon as you put more than one normal element in play, there is absolutely nothing you can say; it is still an open problem whether every separable von Neumann algebra is generated by two normals, which means that it is perfectly possible (and likely) that any separable von Neumann algebra is $\overline{C^*(a_1,a_2)}$ for two normals. – Martin Argerami Oct 11 '21 at 01:38