Is the translation of a BV(R^n) function, regarded as a function from R^n into L1, Lipschitz-continuous?

Question

Suppose $f:\mathbb R^n\to\mathbb R$ is a $BV$ function, in the sense that $$\sup\Big\{\int_U f(\nabla\cdot\phi)\,dx\,|\,\phi\in C_c^1(\mathbb R^n;\mathbb R^n), |\phi|\leq 1\Big\}<\infty$$ Now, consider the translation $\tau:\mathbb R^n\to L^1(\mathbb R^n)$, given by $$\tau(t)= f(x- t)$$ Does it hold that $\tau$ is Lipschitz continuous?

Answer 1

Take $\phi\in C_c^1(\mathbb R^n,\mathbb R)$ with $|\phi|\le 1$. Let $\|e\|=1$, $t>0$. Define $\tilde\phi(x) = e\cdot \phi(x)$. Then $\nabla\cdot \tilde \phi = e\cdot \nabla \phi$. Now consider $$ \int_{\mathbb R^n} (f(x+te)-f(x))\phi(x) dx= \int_{\mathbb R^n} f(x)(\phi(x+te)-\phi(x)) dx = \int_{\mathbb R^n} f(x) \int_0^t \nabla\phi(x+se)e\ ds\ dx\\ =\int_0^t\int_{\mathbb R^n} f(x) \nabla\phi(x+se)e\ dx\ ds =\int_0^t\int_{\mathbb R^n} f(x) \nabla\cdot \tilde\phi(x+se)\ dx\ ds \le \|f\|_{BV} \cdot t. $$ By continuity, this inequality holds for all $\phi\in C_c(\mathbb R^n,\mathbb R)$, and for all $\phi\in C_b(\mathbb R^n,\mathbb R)$. By the uniform boundedness principle, the functional $$ \phi\mapsto \int_{\mathbb R^n} (f(x+te)-f(x))\phi(x) dx $$ is bounded in $C_b(\mathbb R^n,\mathbb R)^*$ with norm less than $t\|f\|_{BV}$. Since this functional is realized by integration against a $L^1$-function, this implies $$ \|\tau(te)-\tau(0)\|_{L^1} \le t\|f\|_{BV}. $$ Since this estimate is independent of the unit vector $e$ and the BV-norm is translation invariant, the claim follows.