For (suitable) real- or complex-valued functions $f$ and $g$ on a (suitable) abelian group $G$, we have two bilinear operations: multiplication -
$$(f\cdot g)(x) = f(x)g(x),$$


and convolution -
$$(f\*g)(x) = \int\_{y+z=x}f(y)g(z)$$


Both operations define commutative ring structures (possibly without identity) with the usual addition. (For that to make sense, we have to find a subset of functions that is closed under addition, multiplication, and convolution. If $G$ is finite, this is not an issue, and if G is compact, we can consider infinitely differentiable functions, and if $G$ is $\mathbb R^d$, we can consider the Schwarz class of infinitely differentiable functions that decay at infinity faster than all polynomials, etc. As long as our class of functions doesn't satisfy any additional nontrivial algebraic identities, it doesn't matter what it is precisely.) 


My question is simply: **do these two commutative ring structures satisfy any additional nontrivial identities?** 


A "trivial" identity is just one that's a consequence of properties mentioned above: e. g., we have the identity
$$f\*(g\cdot h) = (h\cdot g)\*f,$$


but that follows from the fact that multiplication and convolution are separately commutative semigroup operations. 


**Edit:** to clarify, an "algebraic identity" here must be of the form $A(f\_1, ..., f\_n) = B(f\_1, ..., f\_n)$," where $A$ and $B$ are composed of the following operations: 


* addition
* negation
* additive identity (0)
* multiplication
* convolution


(Technically, a more correct phrasing would be "for all $f\_1, ..., f\_n$: $A(f\_1, ..., f\_n) = B(f\_1, ..., f\_n)$," but the universal quantifier is always implied.) While it's true that the Fourier transform exchanges convolution and multiplication, that doesn't give valid identities unless you could somehow write the Fourier transform as a composition of the above operations, since I'm not giving you the Fourier transform as a primitive operation. 


**Edit 2**: Apparently the above is still pretty confusing. This question is about identities in the sense of universal algebra. I *think* what I'm really asking for is the variety generated by the set of abelian groups endowed with the above five operations. Is it different from the variety of algebras with 5 operations (binary operations +, \*, .; unary operation -; nullary operation 0) determined by identities saying that $(+, -, 0, \*)$ and $(+, -, 0, \cdot)$ are commutative ring structures? 



Do convolution and multiplication satisfy any nontrivial algebraic identities?
I think the answer to the original question (i.e. are there any universal algebraic identities relating convolution and multiplication over arbitrary groups, beyond the "obvious" ones?) is negative, though establishing it rigorously is going to be tremendously tedious.


There are a couple steps involved. To avoid technicalities let's restrict attention to discrete finite fields G (so we can use linear algebra), and assume the characteristic of G is very large.


Firstly, given any purported convolution/multiplication identity relating a bunch of functions, one can use homogeneity and decompose that identity into homogeneous identities, in which each function appears the same number of times in each term. (For instance, if one has an identity involving both cubic expressions of a function f and quadratic expressions of f, one can separate into a cubic identity and a quadratic identity.) So without loss of generality one can restrict attention to homogeneous identities.


Next, by [depolarisation](http://en.wikipedia.org/wiki/Polarization_identity), one should be able to reduce further to the case of multilinear identities: identities involving a bunch of functions $f\_1, f\_2, ..., f\_n$, with each term being linear in each of the functions. (I haven't checked this carefully but it should be true, especially since we can permit the functions to be complex valued.)


It is convenient just to consider evaluation of these identities at the single point 0 (i.e. scalar identities rather than functional identities). One can actually view functional identities as scalar identities after convolving (or taking inner products of) the functional identity with one additional test function.


Now (after using the distributive law as many times as necessary), each term in the multilinear identity consists of some sequence of applications of the pointwise product and convolution operations (no addition or subtraction), evaluated at zero, and then multiplied by a scalar constant. When one expands all of that, what one gets is a sum (in the discrete case) of the tensor product $f\_1 \otimes ... \otimes f\_n$ of all the functions over some subspace of $G^n$. The exact subspace is given by the precise manner in which the pointwise product and convolution operators are applied.


The only way a universal identity can hold, then, is if the weighted sum of the indicator functions of these subspaces (counting multiplicity) vanishes. (Note that finite linear combinations of tensor products span the space of all functions on $G^n$ when $G$ is finite.)
But when the characteristic of $G$ is large enough, the only way that can happen is if each subspace appears in the identity with a net weight of zero. (Indeed, look at a subspace of maximal dimension in the identity; for $G$ large enough characteristic, it contains points that will not be covered by any other subspace in the identity, and so the only way the identity can hold is if the net weight of that subspace is zero. Now remove all terms involving this subspace and iterate.)


So the final thing to do is to show that a given subspace can arise in two different ways from multiplication and convolution only in the "obvious" manner, i.e. by exploiting associativity of pointwise multiplication and of convolution. This looks doable by an induction argument but I haven't tried to push it through.


