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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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Does Taylor's theorem ( for the function $f$ around $x$ in the convolution help? – Fabian Mar 9 '11 at 18:11
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 $\phi$... – Mariano Suárez-Alvarez Mar 9 '11 at 18:32
up vote 7 down vote accepted

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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An alternative approach, if you know about difference quotients, is to note $\int_\Omega|f(x) - f*\phi_\epsilon(x)|dx \leq \sup_{|z| < \epsilon} \int_\Omega |f(x - z) - f(x) | dx$ The integral on the right is bounded by $|z|$ times the $L^1$ norm of the $|z|$-difference quotient, which the standard theory of difference quotients tells you is bounded by the homogeneous ${W}^{1,1}$ norm. – Willie Wong Mar 9 '11 at 18:45
@Willie: Don't you mean $f$ instead of $u$ in the first formula? Cool stuff nevertheless. – Jonas Teuwen Mar 9 '11 at 19:40
@Jonas: thanks, corrected. – Willie Wong Mar 9 '11 at 22:43
@Willie: Thanks a lot! Do you have any suggestions on where to read up on the "standard theory of difference quotients" (including the result about the $W^{1,1}$ norm that you mentioned)? Also, regarding the initial assumptions, we can always choose $\epsilon$ small enough to be supported in the unit ball eventually anyway right? If support is outside of the unit ball, is the bound of $\epsilon$ still guaranteed? – user1736 Mar 10 '11 at 0:57
@user1736: If the support of $\phi$ is outside of the unit ball, then directly following the above argument you will lose a constant; that is, the RHS instead of being $\epsilon \|\nabla f\|_{1}$ will be $R\epsilon \|\nabla f\|_{1}$ where $R$ is the radius of the support set of $\phi$. Which as long as you fix your mollifier function is insignificant in the larger setting. – Willie Wong Mar 10 '11 at 12:14

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