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?

  • $\begingroup$ 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,. $\endgroup$ Jun 17, 2015 at 15:06
  • $\begingroup$ $\nabla f = h$ is intended in the sense of distributions, i.e. $\int f \mathrm{div}\, \phi = -\int h \phi$ for all smooth compactly supported $\phi$. $\endgroup$ Jun 17, 2015 at 15:31

1 Answer 1


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 $\phi\in C^\infty_c$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.