Is there a function that replaces a product by convolution?

Question

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.

Answer 1

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 $f(x)$