Is the convolution by BV function a BV function?

Question

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 $f\ast g\in BV(\mathbb R^n)$?

Answer 1

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.