Total variation and Lipschitz continuity Let $f:B_R(0) \subset \mathbb{R}^N \to \mathbb{R}$ be a $L$-Lipschitz continuous function.
Is it true that the total variation $|Df|(B_R(0))$ is controlled by the Lipschitz constant $L$? How?
 A: The total variation of $f$ on $B_R(0)$ is given by
$$\|Df\| = \sup \left\{ \int_{B_R(0)} u\, \mathrm{div\,} \varphi \, dx : \varphi \in C_0^\infty (B_R(0);\mathbb R^n),\, \|\phi\|_\infty \le 1. \right\}.$$
You can define the sequence $\{f_\epsilon\}$ of regularizations of $f$ in the usual manner. Each belongs to $C^\infty(B_{R-\epsilon}(0))$.
Select $\varphi \in C_0^\infty (B_R(0);\mathbb R^n)$. For all sufficiently small $\epsilon > 0$ you have that $\mathrm{supp\, } \varphi \subset B_{R-\epsilon}(0)$ so that
$$\int f\, \mathrm{div\,} \varphi \, dx = \int_{\mathrm{supp\, \varphi }} f\, \mathrm{div\,} \varphi \, dx = \lim_{\epsilon \to 0^+} \int_{B_{R - \epsilon}(0)} f_\epsilon\, \mathrm{div\,} \varphi \, dx$$
Since $\varphi$ vanishes on $\partial B_{R - \epsilon}(0)$ you can integrate by parts to get
$$ \int_{B_{R - \epsilon}(0)} f_\epsilon\, \mathrm{div\,} \varphi \, dx = - \int_{B_{R - \epsilon}(0)} Df_\epsilon \cdot \phi \, dx \le \|Df_\epsilon\|_{L^1(B_{R-\epsilon}(0))} \|\varphi\|_\infty.$$
To complete the argument you need to bound the $L^1$ norm of $\|Df_\epsilon\|$ in a manner independent of $\epsilon$ in terms of the Lipschitz constant $L$.
