How to think of functional calculus for C*-algebras I was wondering if it were possible to think about functional calculus for C*-algebras in terms of actual "extension of functions".
More specifically, let $A$ be a C*-algebra, $x\in A$ a normal element.
Then Gelfand duality gives a *-morphism $\Phi_x: C(\mathrm{Sp}_A(x)) \to A$, such that $\Phi_x(\mathrm{Sp}_A(x) \hookrightarrow \mathbb{C}) = x$.
One then usually write $f(x) := \Phi_x(f)$ for a continous map $f: \mathrm{Sp}_A(x) \to \mathbb{C}$.
I have no intuition how to think of this element $f(x)$, both abstractly and geometrically. In particular, I have two questions:

*

*For such a given $f$, what is the relation between the actual continous map $f$ and the element $f(x) \in A$. I have no clue how one would think of it as "extending" $f$. Would it maybe possible to promote this $f$ to a *-morphism $\tilde{f}: A[x] \to A$ such that $\left. \tilde{f}\right|_{\mathrm{Sp}_A(x)} = f$ or something like this holds, where $A[x]$ is the subalgebra generated by $x$ (This would maybe be the algebraic way of thinking of this process as extending.)

*How does all of this work in open neighborhoods of $x$? $A$ is a Banach algebra after all, so I would hope to get some reasonable behavior of these constructions with respect to the topology. For example, how do the maps $\Phi_x$ and $\Phi_y$ relate for $y$ in a sufficiently small neighborhood of $x$? I'm not even sure how one would compare these two maps, given that they have different codomains so there isn't any good shared topological space of maps they would live in. (This would maybe be a prerequisite to think about the functional calculus in a topological way as a form of extending.)

As you might see, I don't have a particular good intuition when it comes to these things, which is probably due to the fact that I don't really know a lot of good examples. So I would very much appreciate these as answers as well.
 A: Let a unital $C^*$-algebra $A$ be given and $x$ a normal element in it, i.e. $x$ commutes with its adjoint. The functional calculus is a $*$-isomorphism
$$\Phi_x: C(\sigma_A(x)) \to C^*(1,x): f \mapsto f(x)$$
such that $\Phi(z) = x$ and $\Phi(1)=1$, where $z: \sigma_A(x)\hookrightarrow \mathbb{C}$ is the inclusion mapping. Note that the notation $f \mapsto f(x)$ is purely notational, but it conveys the right intuition.

The idea of functional calculus is simple: given a normal element $x$, we want to be able to apply a continuous function on it because this allows us to construct particular nice elements with imposed properties. In this way, many properties that hold for functions in $C(X)$ for a compact Hausdorff space $X$ also hold for normal elements in an abstract $C^*$-algebra.
For example, given a positive element $x\in A$, we want to be able to talk about its square root. So simply define $\sqrt{x}:= \Phi_x(\sqrt{\cdot})$! Since $\Phi_x$ preserves the $*$-operations, the element $\sqrt{x}$ satisies $$\sqrt{x}^2 = \Phi_x(\sqrt{\cdot})^2 = \Phi_x((\sqrt{\cdot})^2) = \Phi_x(z) = x$$
which is the relation we hoped to be true. Note also that $\sqrt{x}$ is positive since $x$ is positive.

But how would you come up with the functional calculus yourself? Given a normal element $x \in A$, everything sensible thing you can do with this element without involving other elements is encoded in the $C^*$-subalgebra $C^*(1,x)$. Since $x$ commutes with $x^*$, this is a commutative $C^*$-algebra, so we know that it is of the form $C(X)$ where $X$ is the space of characters of $C^*(1,x)$. It is not so hard to see that the space of characters can be identified with $\sigma_A(x)$, so we are naturally led to investigate the isomorphism $C(\sigma_A(x)) \cong C^*(1,x)$. In some sense, this already tells us that many properties of a normal element are encoded by its  spectral data. This is for example illustrated by the spectral theorem for normal compact operators.
Investigating the isomorphism $C(\sigma_A(x))\to C^*(1,x)$, we see that $z \mapsto x, 1 \mapsto 1$. And this is really all we need to know! Indeed, consider a polynomial $p \in C(\sigma_A(x))$. Then the image of this polynomial under our map is simply the element $p(a)$ as you would naively define. But every element in $C(\sigma_A(x))$ is a uniform limit of such polynomials, so our isomorphism transfers the non-polynomials nicely as well so that they behave like we would expect.
I hope this answers helped a bit.
