# 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!

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!