Let $u(x) \in L^1( \mathbb{R}^n) \cap BV(\mathbb{R}^n)$ and let $\rho\ge 0$ be the standard mollifier on $\mathbb{R}^d$, supported in unit ball with $\int_{\mathbb{R}^d}\rho \,dx=1$ and define $\rho_\delta= \frac{1}{\delta^n}\rho(\frac{x}{\delta})$, then $\int_{\mathbb{R}^d}\rho_\delta \,dx=1$ . How shall I prove that $$\int_{\mathbb{R}^n} \int_{\mathbb{R}^n}| u(y)-u(x)| \rho_\delta(y-x) \,dx\,dy \le \delta |u|_{BV(\mathbb{R}^d)}$$
I wrote this as after putting $z=\frac{y-x}{\delta}$, $$\int_{|z|\le 1} \int_{\mathbb{R}^n}| u(x+\delta z)-u(x)| \rho(z) \,dx\,dz$$ $$= \int_{|z|\le 1}\int_{\mathbb{R}^n}\Big|\int^1_0 \nabla u( x+\lambda \delta z) \cdot \delta z \,d\lambda\Big| \rho(z)\,dx\,dz$$ $$\le \delta \int_{|z|\le 1} \int^1_0 \int_{\mathbb{R}^n}| \nabla u(x+\lambda \delta z)|\,dx\rho(z)\,d\lambda\,dz$$ Now if I can prove that $\int_{\mathbb{R}^n}| \nabla u(x+\lambda \delta z)|\,dx\ \le |u|_{BV(\mathbb{R}^n)}$ , then I will be done. But I'm not sure how to do this.
