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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!

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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.


Observations:

  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!
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