Let $X$ be a random variable that has full support and is continuously distributed on $\mathbb{R}$ according to the density $f$. I want to "deconvolve" $f$ with a kernel that has also full support and is continuously distributed on $\mathbb{R}$ according to density $g$.
I am looking for sufficient (necessary would also help) conditions such that $f$ can be deconvolved with $g$ as a kernel.
Example
I know that if both $f$ and $g$ are Gaussian with variances $\sigma_f^2$ and $\sigma_g^2$, respectively, the only requirement is that $\sigma_f^2>\sigma_g^2$ and the result of the deconvolution is a Gaussian with a variance of $\sigma^2=\sigma^2_f-\sigma^2_g$ . But if we get rid of the particular functional form, I am pretty sure that $\mathop{\rm Var}\nolimits [f]>\mathop{\rm Var}\nolimits [g]$ is not enough.
Attempts
I have tried going through the Fourier transform path and find that this should be possible iff $$\frac{\hat f}{\hat g}$$ is the Fourier transform of a distribution (density) function. But this is not easy at all to check. What would be "enough" to impose on $f$ and $g$ to ensure that this is true?
In particular, does $f$ being a mean-preserving spread of $g$ ensure that the deconvolution is possible? That is, if there exist random variables $Y$ and $Z$ with $Y$ being distributed according to $g$ and $\mathbb{E}[Z|Y=y]=0$ for all $y\in\mathbb{R}$ such that $$Z+Y\overset{d}{=}X$$ (where $\overset{d}{=}$ means equal in distribution), is it true that there exists some other random variable $W$ such that $$W+Y\overset{d}{=}X$$ with $W$ being independent of $Y$?