# What Mathematical Object Does a “Functional Calculus” Correspond to Precisely?

I'm somewhat confused on what specific mathematical object a/the Borel/Continuous/Holomorphic functional calculus corresponds to. Based on my understanding so far, I would say that it corresponds to some sort of mapping between a Banach algebra to the space of eigenfunctions of the operator in question. However, I haven't seen that definition in a text and everything I've read makes it seem more like a concept than an actual mathematical object. So what exactly is it? Also is there only one "Borel functional calculus" for a given class of operators or does each operator define it's own?

In the three cases you ask about, a functional calculus of an element $x$ in some algebra $B$ can be seen as a algebra homomorphism $\Phi_x \colon A \rightarrow B$ between some algebra of functions $A$ (that contains complex polynomials) and $B$. When you impose more restrictions on $B$ and $x$, you get a larger functional calculus. More explicitly,
1. If $B$ is a Banach algebra and $x \in B$, the holomorphic functional calculus is an algebra homomorphism $\Phi_x \colon \mathcal{A}(\sigma(x)) \rightarrow B$ between the algebra $\mathcal{A}(\sigma(x))$ of holomorphic functions defines on a neighborhood of $\sigma(x)$ and $B$. This map extends the polynomial functional calculus in the sense that if $p \in \mathbb{C}[z]$ is a polynomial, then $\Phi_x(p) = p(x)$ where the right hand side is defined by plugging in $x$ into the polynomial $p$. In addition, if $f_n,f \in \mathcal{A}(\sigma(x))$ and $f_n \rightarrow f$ uniformly on compact subsets then $\Phi_x(f_n) \rightarrow \Phi_x(f)$ with respect to the norm topology on $B$. These properties determine $\Phi_x$ uniquely and we have a spectral mapping theorem that states that $$\sigma(\Phi_x(f)) = \sigma(f(x)) = f(\sigma(x)).$$
2. If $B$ is a $C^{*}$-algebra and $x \in B$ is a normal element, the continuous functional calculus is a $C^{*}$-algebra homomorphism $\Phi_x \colon C(\sigma(x)) \rightarrow B$ between the $C^{*}$-algebra $C(\sigma(x))$ of continuous complex-valued functions on $\sigma(x)$ (endowed with the supremum norm and the obvious conjugation) and $B$. Again, this map extends the polynomial functional calculus and also the holomorphic functional calculus and the spectral mapping theorem continues to hold.
3. If $B = \mathcal{B}(H)$ is the $C^{*}$-algebra of bounded operators on a Hilbert space $H$ (or, more generally, a von Neumann algebra) and $x \in B$ is a normal element, the Borel functional calculus is an algebra homomorphism $\Phi_x \colon B(\sigma(x)) \rightarrow B$ between the $C^{*}$-algebra $B(\sigma(x))$ of bounded complex-valued measurable Borel functions on $\sigma(x)$ (endowed with the supremum norm and the obvious conjugation) and $B$. Again, this map extends the polynomial functional calculus and also the continuous functional calculus. It preserves the involution $((f^{*})(x) = \overline{f}(x) = (f(x))^{*})$ and is continuous but not an isometry and the spectral mapping theorem fails to hold. It also satisfies a stronger continuity property analogous to the dominated convergence theorem
From the point of view I described above, each element $x \in B$ (for example, a normal operator in $\mathcal{B}(H)$) defines a functional calculus of its own. However, there are properties that connect the functional calculi of different elements. For example, if $x,y$ are normal commuting elements in a $C^{*}$-algebra, you can define a joint continuous and Borel calculus on the space of continuous/Borel measurable and bounded functions on the joint spectrum of $x$ and $y$ which is a subspace of $\sigma(x) \times \sigma(y)$. In this way you can show for example that if $x,y$ commute then so does $f(x),g(y)$ and so on.
• @Pseudo-IntellectualBS: You're welcome! Note that I've edited slightly slightly my answer for part $(3)$ as you need some more conditions on the algebra $B$ for the Borel functional calculus to be well-defined. It works if $B = \mathcal{B}(H)$ or a von Neumann algebra but not in general. – levap Oct 22 '16 at 15:54