Convolution division exists or not?

Question

Can we solve a convolution equation $a(t)*x(t)=b(t)$ for $x(t)$, where a(t),b(t) are "nice" distribution functions (whatever nice means)? So a(t),b(t) are non-negative, continuous, and having integral 1.

Answer 1

The Fourier transform can be used for functions that have a Fourier transform.

$$F\{a(t)*x(t)\} = F\{b(t)\}$$ $$F\{a(t)\}F\{x(t)\} = F\{b(t)\}$$ $$A(s)X(s) = B(s)$$ $$X(s) = B(s)/A(s)$$ $$x(t) = F^{-1}\{B(s)/A(s)\}$$

if that inverse transform exists.