Let $\varphi:\mathbb R^d\to\mathbb R_+$ be given as $$ \varphi(x) := \begin{cases} c\exp\big(1/(|x|^2-1)\big) & \mbox{if } |x|\le 1 \\ 0 & \mbox{otherwise}, \end{cases} $$ where $c>0$ is chosen such that $\int_{\mathbb R^d}\varphi(x)dx=1$. Could we prove the following convergence for $\varphi_t(x):=\varphi(x/t)/t^d$ $$\lim_{t\to 0+} \int_{\mathbb R^d} \big|{\nabla (\varphi_t\ast f)(x)-\nabla f(x)\big|}dx ~=~0?$$ Here $f:\mathbb R^d\to\mathbb R$ is a fixed Lipschitz function and $\varphi_t\ast f$ denotes the convolution of $\varphi_t$ and $f$, i.e. $$(\varphi_t\ast f)(x):=\int_{\mathbb R^d}\varphi_t(y)f(x-y)dy.$$ Indeed, by means of a change of variable, it follows that $$ \int_{\mathbb R^d} \big|{\nabla (\varphi_t\ast f)(x)-\nabla f(x)\big|}dx ~=~\int_{\mathbb R^d} \left(\int_{\mathbb R^d}\varphi(y)\big |\nabla f(x-ty)-\nabla f(x)\big |dy\right)dx.$$ If we know $\nabla f$ is a.e. continuous, then we may conclude using the dominated convergence theorem. But I can not find any reference on the continuity of $\nabla f$. Any proof, comments or references are highly appreciated!


1 Answer 1


Yes and no.

No, because this would imply that $\nabla f$ is in $L^1(\mathbb R^d)$, but you didn't assume that ("only" $\nabla f\in L^\infty$, which seems of course much better but does not control decay at infinity in the whole space).

Yes, because if you assume indeed $\nabla f\in L^1$ then this is a classical exercise: It is well-known that $\nabla (\phi_t * f)=\phi_t*(\nabla f)$ (and actually that's the whole point of mollifying sequences). So forget that you're dealing with gradients, the question becomes: if $g$ (here $g=\nabla f$) is an $L^1$ function, is it true that $\phi_t* g\to g$ in $L^1$? This is of course true, you can find the proof in any basic textbook.


  1. Even without the assumption that $\nabla f\in L^1$ you can conclude that $\nabla(\phi_t*f)\to \nabla f$ a.e. (and therefore in $L^1_{loc}$). See theorem 8.15 in Folland's book "Real Analysis, modern Techniques and their applications". Actually the pointwise convergence holds at any Lebesgue point of $\nabla f$, which are of full measure by Lebesgue's differentiation theorem.
  2. It is not true that, if all you know is that $\nabla f\in L^\infty$, then $\nabla f$ is continuous a.e. This is why you didn't find any reference to help you conclude the proof!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy