Is the convolution by BV function a BV function? Let $BV(\mathbb R^n)$ be the set of functions $f\in L^1(\mathbb R^n)$ such that
$$\sup\Bigl\{\int_U f(\nabla\cdot\phi)\,dx\Bigm|\phi\in C_c^1(\mathbb R^n;\mathbb R^n), |\phi|\leq 1\Bigr\}<\infty$$
Suppose $f\in BV(\mathbb R^n)$ and $g\in L^1(\mathbb R^n)$. Is it true that $f\ast g\in BV(\mathbb R^n)$?
 A: Denote the supremum in the question as $V_f$. 
You should be able to show that if $f \in BV$ then the translate $f_y(x) = f(x-y)$ is also $BV$ and $V_{f_y} = V_f$.
Let $f \in BV(\mathbb R^n)$ and $g \in L^1(\mathbb R^n)$. It is a standard result that $f \ast g \in L^1(\mathbb R^n)$. Assume that $g \ge 0$.
Now let $\phi \in C_c^\infty(\mathbb R^n;\mathbb R^n)$, $|\phi| \le 1$. Then \begin{align*}\int_{\mathbb R^n} f \ast g (\nabla \cdot \phi) \,dx &= \int_{\mathbb R^n} \int_{\mathbb R^n} f(x-y) g(y) \, dy (\nabla \cdot \phi)(x) \, dy dx \\
&= \int_{\mathbb R^n} \left( \int_{\mathbb R^n} f(x-y) (\nabla \cdot \phi)(x) \, dx \right) g(y) \, dy \end{align*}
Now,
$$\int_{\mathbb R^n} f(x-y) (\nabla \cdot \phi)(x) \, dx = \int_{\mathbb R^n} f_y(x) (\nabla \cdot \phi)(x) \, dx \le V_{f_y} = V_f$$
and since $g \ge 0$
$$ \int_{\mathbb R^n}  \left( \int_{\mathbb R^n} f(x-y) (\nabla \cdot \phi)(x) \, dx \right) g(y)  \, dy \le V_f \int_{\mathbb R^n} g(y) \, dy = V_f \|g\|_{L^1}.$$
Now take the sup over all admissible $\phi$ to obtain $$V_{f \ast  g} \le V_f \|g\|_{L^1}.$$
You can remove the requirement that $g \ge 0$ by considering $g^+$ and  $g^-$ separately.
