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Consider two functions $f(x),g(x) = 0 \forall x<0$, I'd like to know if we can always find an $h(x)$ which satisfies the integral equation $$f(x)g(x) = h(x)*f(x)$$ where '$*$' is the convolution operator.

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  • $\begingroup$ are you looking for the Fourier transform? $\endgroup$ – John Dvorak Dec 31 '13 at 5:33
  • $\begingroup$ @JanDvorak : No, I am interested in a function, not transform. I know Fourier transform converts product into convolution, but thats not what I want. $\endgroup$ – Rajesh Dachiraju Dec 31 '13 at 5:34
  • $\begingroup$ Well, then there's $h(x) = 0$, but I guess you're not looking for that either. I'm afraid there's no function that satisfies your needs. $\endgroup$ – John Dvorak Dec 31 '13 at 5:39
  • $\begingroup$ @JanDvorak : How does $h(x) = 0$ satisfy? looks like there is some misterpretation. $\endgroup$ – Rajesh Dachiraju Dec 31 '13 at 5:42
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    $\begingroup$ For one, $g(x)$ appears nowhere on the RHS of that formula, meaning that $h$ should be dependent on $g$. $\endgroup$ – John Dvorak Dec 31 '13 at 5:49
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Simple answer is no except for some trivial cases such as $g(x) \equiv 0$. The equation on the left depends only on instantaneous value of $f$ while the one on the right depends on all the values of $f$. There is no way the two can be equal as an identity.

More to the point you want

$$f(x) g(x) = \int_0^{\infty} f(\tau) h(x-\tau) d \tau$$

Even if you make $h$ causal, the right hand side will still depend on $f(\tau), 0\le \tau \le x$ but the left hand side depends only on $f(x)$

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