Given are $f\in L^1(\mathbb R^n)$, $f>0$, such that $\log f\in L^1_{\mathrm{loc}}(\mathbb R^n)$ and $\nabla \log f = g$ in the sense of distributions, with $g\in L^1_{\mathrm{loc}}(\mathbb R^n)\cap L^1(\mathbb R^n,fdx)$. Is it true that $$ f\nabla \log f = \nabla f, $$ again in the sense of distributions?

Obviously the result is true if $f\in C^1(\mathbb R^n)$; it would also be true if $\log f $ were a Sobolev function with bounded range, since then the mapping $\log f \mapsto f$ (i.e. the exponential function $s\mapsto e^s$) can be considered $C^1$ and Lipschitz, and usual theorems on composition apply (e.g. Brezis, *Functional Analysis, Sobolev Spaces, and PDEs*, Corollary 8.11). In homogeneous space, i.e. without the weight in the $L^1$-space, the result would follow from regularization by convolution.

Without these helping properties, does anyone know how to prove this?