## BV function and its property involves translation

Theorem 1 Suppose ${u\in L^1(\mathbb{R})}$, then ${u\in \text{BV}}$ if and only if ${\exists\, C}$ such that

$\displaystyle ||\tau_hu-u||_{L^1(\mathbb{R})}\leq C|h|,\quad \forall\, h$

Moreover, one can take ${C=|u|_{BV}}$. Here ${\tau_hu(\cdot)=u(\cdot+h)}$ is the tanslation operator.

Proof: Firstly suppose ${u\in \text{BV}}$. Let us prove

$\displaystyle \left|\int_{\mathbb{R}}(\tau_hu(x)-u(x))\phi(x)dx\right|\leq |u|_{BV}|\phi|_{L^\infty}|h|,\quad \forall \phi\in C^\infty_c(\mathbb{R}) \ \ \ \ \ (1)$

To show that

$\displaystyle LHS=\left|\int_{\mathbb{R}}u(x)(\phi(x-h)-\phi(x))dx\right|$

$\displaystyle =\left|\int_{\mathbb{R}}u(x)\psi(x)hdx\right|$

$\displaystyle \leq |u|_{BV}|\psi|_{L^\infty}|h|$

where

$\displaystyle \psi(x)=\int^x_{-\infty}\frac{\phi(s-h)-\phi(s)}{h}ds\in C_c^\infty(\mathbb{R})$

it is easy to verify ${|\psi|_\infty=|\phi|_\infty}$ therefore (1) is proved. Next one can choose such ${\phi_n\rightarrow sign(\tau_hu-u)\in L^1}$ with ${|\phi_n|\leq 1}$(it is easy to show by mollification). By dominating theorem, one get

$\displaystyle \int_{\mathbb{R}}|\tau_hu(x)-u(x)|dx\leq |u|_{BV}|h|.$

The other direction need more analysis. $\Box$