TITLE: Bounding the integral of a $C^1$ function using its gradient
QUESTION [5 upvotes]: 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 $W^{1,1}(\Omega)$).

REPLY [7 votes]: 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.