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Knowing that the convolution operation (*), when defined, is commutative, associative and distributive with respect to addition, what happens with multiplication (.) in both time and frequency domains? Example:

y(t) = [a(t).b(t)]*h(t)


Y(jw) = [A(jw)*B(jw)].H(jw)

Is it possible to perform any operation to a(t), b(t) and/or h(t), so that:

y(t) = [a'(t)*h'(t)].[b'(t)*h'(t)]


Y(jw) = [A'(jw).H'(jw)]*[B'(jw).H'(jw)]

Thanks, Pires

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Could you please try and formulate your question more precisely? In particular, your third equation always defines a function $y(t)$, so presumably you are asking if that function is related to some other one in some way? – Yemon Choi Mar 9 '10 at 22:00
I think he wants the y(t) in equation 3 to be the same as the one in equation 1 (and the same for eqs 2 and 4) – Suresh Venkat Mar 10 '10 at 6:14
Hi, sorry if I did not explain properly. Suresh is right, I want that: [a(t).b(t)]*h(t) = [a'(t)*h'(t)].[b'(t)*h'(t)] and/or [A(jw)*B(jw)].H(jw) = [A'(jw).H'(jw)]*[B'(jw).H'(jw)] if possible, of course. The x' should represent some sort of modification of the original function and not a differentiation. I hope this clarifies, but if not, please let me know. – Pires Mar 10 '10 at 9:06

I don't have a complete answer yet, but it seems highly implausible such a thing exists. Assume there exists some map $T: \mathcal{S}\to\mathcal{S}$ on Schwarz space such that $ (f\cdot g) * h = (Tf*Th) \cdot(Tg*Th) $, it is clear that $T$ cannot be a linear map (else the left and right sides scale differently when $h$ is replaced by $\lambda h$).

Now say we want $T$ to map real-valued functions to real-valued ones, then we run into a bit of a problem: let $h$ be negative of the normal Gaussian. Then for any real-valued $f$, $f^2 \geq 0$, so the LHS $(f\cdot f) * h$ is non-positive. On the other hand, the RHS becomes $(Tf * Th)^2$ is non-negative. And we get a contradiction.

How about translation invariance? Suppose $(Tf)(x-s) = T(f(x-s))$. Now take $f$ to be some bump function with support in the unit interval. And take $g$ to be an arbitrary translate of $f$. In the case where $g = f$, we have that the left hand side does not always vanish, which implies that $(Tf*Th)$ cannot vanish identically. But when $g$ is a translate by some large (compared to 1) value, the LHS becomes 0. However, writing $f_t(x) = f(x-t)$, we have that $Tf_t*Th = Tf * Th_t$, so this implies that $(f\cdot f_t)*h = (h\cdot h_t)*f = 0$ for any $h$, and this is obviously nonsense. So $T$ cannot be translation invariant.

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Thank you for the response. – Pires Mar 15 '10 at 14:22

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