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. 



Defining holomorphic functions in terms of Banach algebras, and similarly for C*-algebras
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.


