Defining holomorphic functions in terms of Banach algebras, and similarly for C*-algebras Let $C$ be the category of commutative Banach algebras and let $U : C \to \text{Set}$ be the usual forgetful functor. The holomorphic functional calculus guarantees that every holomorphic function $f : \mathbb{C}^n \to \mathbb{C}$ defines a natural transformation $U^n \to U$ (in a way which is compatible with composition, etc.). Does every natural transformation $U^n \to U$ have this form?
Similarly, let $C$ be the category of commutative C*-algebras and let $U : C \to \text{Set}$ be the usual forgetful functor. The continuous functional calculus guarantees that every continuous function $f : \mathbb{C}^n \to \mathbb{C}$ defines a natural transformation $U^n \to U$. Does every natural transformation $U^n \to U$ have this form? 
In both cases the first thing to try would be to check if $U$ is representable, and in both cases I think it can't be. 
 A: I'll just consider a natural map $\phi:U\to U$; the multivariate case should be similar.
Let $D_n$ be the closed disk of radius $n$ centred at the origin, and let $A_n$ be the Banach algebra of functions that are continuous on $D_n$ and holomorphic on the interior.  Let $\iota_n\in A_n$ be the identity map.  By naturality for the restriction maps $A_{n+1}\to A_n$, the functions $\phi(\iota_n)\in A_n$ will fit together to give an entire function $f$.  Now consider an arbitrary Banach algebra $B$ and an element $b\in B$.  Choose $n>\|b\|$, so functional calculus gives a homomorphism $\beta:A_n\to B$ with $\beta(\iota_n)=b$.  (Here we can use the simplest version of functional calculus using convergent power series; we do not need need anything about the spectrum of $b$, or $B$-valued contour integrals.)  Naturality of $\phi$ with respect to $\beta$ tells us that $\phi(b)=f(b)$, as required.
I think that $A_n$ represents the functor $U_n(B)=\{b\in B:\|b\|\leq n\}$ (or should that be $\rho(b)\leq n$?), and $U$ is the colimit of these. 
UPDATE:
I didn't read Simon Henry's comment properly before; now I see that the argument above is more or less what he proposed.
