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Let $f \in C^1_c(\Omega)$ where $\Omega \subset \mathbb{R}^d$ is a bounded domain. Let $\phi \in C^1_c(\mathbb{R}^d)$ be an approximation of the identity (i.e. $\int_{\mathbb{R}^d} \phi=1$, $\phi \geq 0$, $\phi_\epsilon := \frac{1}{\epsilon^d} \phi(\frac{x}{\epsilon})$.

How would you prove that

$\int_\Omega |f(x) - f \ast \phi_\epsilon(x)| dx \leq \epsilon \int_\Omega |\nabla f| dx?$

I'm trying to show that the family of $C^1_c$ functions convolved with a mollifier is uniformly close to the function in $L^1$ (which would be true after having this result if we assume something like the family of functions being bounded in $W^{1,1}(\Omega)$).

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    Well, you clearly have to be able to bound $|f-f*\phi_\epsilon|$ in terms of the derivatives of $f$. When $\epsilon$ is small, $|f-f*\phi_\epsilon|$ is a difference between the value of $f$ and the values nearby it averaged according to $\phi$...2011-03-09

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WLOG assume $\phi(x)$ supported in the unit ball (it has compact support, so it is supported in some ball). First look at

$ |f(x) - f*\phi_\epsilon(x)| \leq \int_{|z|\leq 1} \phi(z) |f(x) - f(x-\epsilon z)| dz $

replace the integrand

$ f(x) - f(x-\epsilon z) = - \int_0^{\epsilon|z|} D_rf(x - r\omega) dr $

with $\omega = z / |z|$ and using the fundamental theorem of calculus. So

Integrating the whole thing over $x$, and changing the order of integration, you have

$ \int_{\Omega}|f(x) - f*\phi_\epsilon(x)|dx \leq \int_{|z|\leq 1} \phi(z) \int_0^{\epsilon|z|} \int_{\Omega} |D_rf(x - r\omega)| dx~ dr~ dz $

The inside most integral for fixed $r\omega$ gives you $\int_\Omega |\nabla f| dx$. The integral over $r$ gives you the factor of $\epsilon$. And integrating $\phi(z)$ over the ball of radius 1 gives you 1.

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    @Willie, yup that's what $I$ expected regarding the scaling. Thanks again for all your help! $Y$our notes also look like a fun read.2011-03-10