Convergence of the regularized gradient of a Lipschitz function 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!
 A: 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:


*

*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.

*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!

