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This is in response to a claim made in the second line of the question here, namely:

Given the standard mollifier $\eta$ and a locally integrable function $f:U \rightarrow \mathbb{R}^n$, by defining the mollification of $f^{\epsilon}$ in the usual way ($f^{\epsilon}:=f*\eta_{\epsilon}$), (where $\eta_{\epsilon}:=\epsilon^{-n}\eta(x/\epsilon))$, if $f\in W^{1,p}(U)$ then $D(f^{\epsilon}) = Df * \eta_{\epsilon}$.

(Where $Df$ is, of course, the weak derivative of $f$.)

I can show this works for the other side (i.e. that $D(f^{\epsilon}) = f * D(\eta_{\epsilon})$), but don't know how to incorporate the definition of the weak derivative to show the claim. Does anyone know a nice proof?

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  • $\partial^\alpha(uv)=(\partial^\alpha u)v$ is true for all Schwartz distributions such that one of them has compact support. – Jochen Aug 25 '15 at 10:26
  • Is there a way to do it without referring to distributions? – Cookie Monster Aug 25 '15 at 13:52
  • If you shrink the set $U$ suitably, this is true, I.e. your equality holds on $U_\epsilon ={x \mid dist(x,U^c)>\epsilon}$. I don't think (at least I never saw the claim) that this also holds on all of $U$. – PhoemueX Aug 25 '15 at 16:08
  • You have correctly deduced that $D(f^{\epsilon})(x)=\int f(y)D(\eta_{\epsilon})(x-y)dy$. How are you having trouble applying the definition of the weak derivative to this integral expression? – Disintegrating By Parts Aug 26 '15 at 15:51
  • Never mind, I see the answer to my question is in Evans. – Cookie Monster Sep 13 '15 at 14:22
  • This is why I hate this website. – Martin Erhardt Oct 29 '20 at 15:46
  • It's on page 250, for future readers of this thread. Basically the idea is that $$\int f(y) D_x \eta(x-y) dy = - \int f(y) D_y \eta(x-y) dy $$ and then applying the definition. – Niklas Nov 15 '23 at 18:16

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