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I have two functions $f(x)$ and $g(x)$ where $f(x)=g(x)=0$ for $x<0$. I want to use the convolution theorem to get $\tilde{f}(\omega)\tilde{g}(\omega)$, and for some reason I only want the imaginary part of $\tilde{g}(\omega)$, i.e. $\tilde{f}(\omega)iIm\big(\tilde{g}(\omega)\big)$. To do that I use the fact that $FT(g(x)-g(-x))=2iIm\big(\tilde{g}(\omega)\big)$, which means that the convolution is now between $f(x)$ and $\frac{1}{2}(g(x) - g(-x))$. But since $g(-x)=0$ for $x>0$ now and $f(x)=0$ for $x>0$, this means that we can simply discard $g(-x)$ in the convolution. So we are back to square one.

What is the reason behind the inconsistency?

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  • $\begingroup$ You can't discard $g(-x)$ in the convolution. With $h(x) = g(-x)$, we have $$(f\ast h)(x) = \int_{\mathbb{R}} f(t)h(x-t)\,dt = \int_{\mathbb{R}} f(t)g(t-x)\,dt = \int_{\max \{0,x\}}^{+\infty} f(t)g(t-x)\,dt.$$ $\endgroup$ – Daniel Fischer Mar 1 '17 at 14:05
  • $\begingroup$ I see, then since I'm always evaluating the convolution at $x=0$, then in fact $g(x)$ can be discarded, instead of $g(-x)$ right? $\endgroup$ – David Young Mar 1 '17 at 14:34
  • $\begingroup$ Well, $(f\ast g)(0) = 0$, but why are you "always evaluating the convolution at $x = 0$"? $\endgroup$ – Daniel Fischer Mar 1 '17 at 14:41
  • $\begingroup$ $f(t)$ can be considered to be a "force" as a function of time $t$, and it is only defined for $t<t_0$ where $t_0$ is the present time which I can always set to be zero (so $f(t)$ is like a "record" of the force in the past). Also I am not sure if my problem is resolved? $\endgroup$ – David Young Mar 1 '17 at 15:00

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