Suppose that $f$ is a positive-definite Schwartz distribution, that is, $$\langle\phi,f*\phi\rangle\geq0\qquad\text{for every }\phi\in C_0^\infty(\mathbb R^n).$$ By the Bochner-Schwartz theorem, there exists a unique tempered measure $\mu$ such that $\hat f=\mu$, where $\hat\cdot$ denotes the Fourier transform. I'm interested in the following:
Question. Are there simple conditions that guarantee that $\mu$ is absolutely continuous, i.e., $d\mu(x)=g(x)dx$ for a measurable $g$? What can we say about $g$'s regularity?
If $f$ were continuous, then we could use for example the fact that decay of $f$ translates into regularity for $\hat f$, but since $f$ is only a Schwartz distribution, this seems difficult.
For example, using the formal correspondence $$\sup_{x\in K}f(x)"="\sup_{\|\phi\|_1=1,~\mathrm{supp}(\phi)\subset K}\langle f,\phi\rangle$$ for any $K\subset\mathbb R^n$, I suppose one could somehow access the "decay" of a distribution by looking at quantities like $$\sup_{\|\phi\|_1=1,~\mathrm{supp}(\phi)\subset B(0,t)^c}\langle f,\phi\rangle$$ for large $t>0$, where $B(0,t)^c=\{x\in\mathbb R^d:|x|>t\}$ is the complement of the unit ball in $t>0$. It seems to me like results regarding the absolute continuity of $f$'s Fourier transform should be classical, but I've not been able to find any.
