I'm trying to make the following theorem as blindingly obvious to a layperson as possible. We have a periodic function f(x)=f(x+P), e.g., P=2pi. We'd like to cyclicly convolve this periodic function f(x) over the period P with a constant function c0(x). It is obvious to most mathematicians/engineers that the result is another constant function c1(x). One way to see this is that a constant function factors out of the convolution integral, leaving the integral to simply compute the mean of f(x) over the period. Another way to see this uses Fourier transforms: Let F(z) be the Fourier transform of f(x), and let C0(Z) be the Fourier transform of c0(x). But the Fourier transform C0(z) of a constant function c0(x) is an impulse at the origin, whose coefficient value is the mean (or "DC") value of c0(x) over one period. When we point-wise multiply F(z)*C0(z), every coefficient but the DC coefficient of F(z) is multiplied by zero, so we end up with F(0)*C0(0) as the only coefficient in the result, so once we inverse transform this result, we once again get a constant function c1(x) -- whose value is now mean(f(x))*mean(c0(x)), all over one period. QED Since the audience isn't composed of mathematicians, is there any way to make this obvious point "trivially obvious" ? The best intuition I can come up with is that if f(x) averages to 1 -- i.e., F(0)=1 -- and if c0(x)=1 -- i.e., C0(0)=1 -- then c0(x) acts like a kind of "zero" for the convolution operation, in that c0(x) (*) f(x) = f(x) (*) c0(x) = c0(x). (Here, we denote convolution by "(*)".) ------- If we consider a discrete domain, then we can represent a function by a Z-transform: i.e., f(i) by f(i)*z^i, and c0(i)=c*z^i. Let F(z)=sum(f(i)*z^i,i,0,N) and C0(z)=sum(c*z^i,i,0,N), so to perform a convolution, we compute F(z)*C0(z). Finally, to get the cyclic convolution we do (mod (Z^N)-1). Once again, I'd love to make this idea blindingly obvious to a lay audience.