# Chain rule for weakly differentiable functions

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?

• What do you mean by $\nabla f$ if $f$ is merely in $L^1$? Or is it that you want to prove that under your assumptions $f$, in fact, has a Sobolev gradient and that it satisfies this formula? If $f \nabla \log f$ is well-defined under your assumptions, you could define that $\nabla f := f \nabla \log f$, if the Sobolev gradient does not exist otherwise. Such procedures can be found in the literature; of course, whether this makes any sense, depends on what you want to do,. Jun 17, 2015 at 15:06
• $\nabla f = h$ is intended in the sense of distributions, i.e. $\int f \mathrm{div}\, \phi = -\int h \phi$ for all smooth compactly supported $\phi$. Jun 17, 2015 at 15:31

I think it works by cutting-off as follows: Fix $\epsilon>0$, let $f_\epsilon=\min\{1/\epsilon,\max\{\epsilon,f\}\}$, and observe that $f_\epsilon\to f$ in $L^1_{loc}$.
Then obviously $\log f_\epsilon=\min\{\log(1/\epsilon),\max\{\log(\epsilon),\log f\}\}$ and $\nabla \log f_\epsilon =(\nabla\log f)\chi_{[\epsilon<f<1/\epsilon]}$. Since after truncation we made sure that $\log f_\epsilon$ has bounded range we can use the composition with the exponential to conclude that $$\nabla f_\epsilon=\nabla(\exp(\log f_\epsilon))=f_\epsilon\nabla \log f_\epsilon=(f\nabla\log f)\chi_{[\epsilon<f<1/\epsilon]}.$$ Because $f_\epsilon\to f$ in $L^1_{loc}$ we have $$\int f_\epsilon\, div(\phi)\to \int f \,div(\phi)$$ for all test functions when $\epsilon\to 0$. On the other hand since $\nabla f_\epsilon=f_\epsilon\nabla\log f_\epsilon$ we can write $$-\int f_\epsilon\, div(\phi)=\int f_\epsilon\nabla\log f_\epsilon\cdot \phi=\int (f\nabla\log f)\cdot(\phi\chi_{[\epsilon<f<1/\epsilon]}).$$ With your assumption that $\nabla\log f\in L^1(f\, dx)$ we can apply Lebesgue's dominated convergence to the last term (with the uniform bound almost everywhere $|(f\nabla\log f)\cdot(\phi\chi_{[\epsilon<f<1/\epsilon]})|\leq |f\nabla\log f|.\|\phi\|_{\infty}\in L^1$) and conclude that $$\int (f\nabla\log f)\cdot(\phi\chi_{[\epsilon<f<1/\epsilon]})\to \int f\nabla\log f \cdot\phi.$$ Thus $$\int f\,div(\phi)=-\int f\nabla\log f \cdot\phi$$ for all test-functions $\phi\in C^\infty_c$.