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I know that according to the Convolution theorem the Fourier transform of the convolution of two functions $f$ and $g$ is equal to the product of their Fourier spectra:

$\mathcal{F}\{f*g\} = \mathcal{F}\{f\} \times \mathcal{F}\{g\} $

and the other way round:

$\mathcal{F}\{f\} * \mathcal{F}\{g\} = \mathcal{F}\{f\times g\}$.

Is there any relation between the cross product of two functions and the cross product of their Fourier transforms?

$f*g \sim \mathcal{F}\{f\} * \mathcal{F}\{g\} $

My own intuition is that there are none, as when a function is defined in a limited domain and is 0 elsewhere (which guarantees that the convolution exists), it has an infinite domain spectrum and there is no guarantee that the convolution integral for the spectra converges and vice versa, a function with a limited domain spectrum will have an infinite domain and again it should satisfy special conditions for the convolution integral to exist.

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  • $\begingroup$ In general, why a convolution of two infinite-supported functions should not converge? As an counterexample, take two gaussians y=e⁽⁻ˣ⁾ˣ, which have an integral when convolved or multiplied, and their FT is again another gaussian... $\endgroup$ – dominecf Feb 8 '16 at 17:17
  • $\begingroup$ Yes, in this particular case the convolution converges. But for a general function with an infinite domain there is no guarantee that it does. It should satisfy special conditions. Probably it would be better to say that in general there is no guarantee that both convolution of the functions and the convolution of the spectra converge. So probably for a limited set of functions both of these conditions are satisfied, but I am interested in a general relation $\endgroup$ – Ilya Feb 9 '16 at 9:46
  • $\begingroup$ What do you mean by "cross product" ? on my browser, at least, the math is not rendering correcctly, I cannot see the formulas. Do you simply mean the pointwise product of their values? $\endgroup$ – joseph f. johnson Feb 9 '16 at 10:22
  • $\begingroup$ No i mean the Convolution taken from -infinity to +infinity. $\endgroup$ – Ilya Feb 9 '16 at 13:53
  • $\begingroup$ Okay, right, I see that now. I mean, I see it in my mind's eye---for some ungodly reason, the math formulas in your post have completely disappeared from my browser. $\endgroup$ – joseph f. johnson Feb 10 '16 at 18:07
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You are right, there is no relation. And your intuition gives one way to see why. If you take an average function f and convolve it with a nice, smooth, function g, this smooths f more but also spreads it out more: the smoothness becomes better, but the support of f, the domain where it is non-zero, becomes "worse", so to speak. So the convolution is smoother but broader than f. But that means that its F.T. has better decay properties at infinity than did the FT of f, because the F.T. transforms decay properties into smoothness properties and vice versa. Now on the other hand, you are convolving the F.T. of f with a reasonably nice function (the F.T. of g). So the same line of reasoning says that the convolution of the F.T. of f with the F.T. of g is more spread out, i.e., worse decay properties, than the F.T. of f (and is smoother). This is a contradiction. Another way to phrase this is to see that the F.T. itself is a kind of convolution: f is being convolvoed with the exponential function. If your hypothetical formula were true, you would be saying that convolution was "distributive", which is obviously false: it is associative, not distributive.

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