Let $\mathbf{f} \in W^{1,q}_{loc}(\mathbb{R}^n)$, $J \in L^p (\mathbb{R}^n)$, with $1 \leq p,q \leq \infty$, and $\frac{1}{p}+\frac{1}{q} = 1$, $\rho_{\epsilon}$ be a regularizing kernel for $\epsilon > 0$, $ \rho_{\epsilon} = \frac{1}{\epsilon ^n} \rho(\frac{\cdot}{\epsilon}), $ with $\rho \in C^{\infty}_0 (\mathbb{R}^n)$, $\rho (\cdot) \geq 0$, $\int_{\mathbb{R}^n} \rho dx = 1$ and the compact support $supp~ \rho = B_1(0)$, where $B_1 (0) \subset \mathbb{R}^n$ is the ball with the center $0$ and the radius $1$.
Prove the inequality \begin{eqnarray*} && \int_{B_R} | \int_{\mathbb{R}^n} J(y) (\mathbf{f} (y) - \mathbf{f} (x)) \cdot \nabla \rho_{\epsilon}(x-y) dy | dx \\ & \leq & C \| J \|_{L^p (B_{R+1})} \Big \{ \int_{B_{R+1}} dx \int_{|x-y| \leq \epsilon } \{ \frac{|\mathbf{f} (y) - \mathbf{f} (x)|}{\epsilon} \}^q dy \Big \}^{\frac{1}{q}} \end{eqnarray*}
I know the Holder's inequality should be used. But I don't know how to eliminate the item $\nabla \rho_{\epsilon}(x-y) $.
Could someone give help? Any suggestion will be appreciated.
The inequality should be \begin{eqnarray*} && \int_{B_R} | \int_{\mathbb{R}^n} J(y) (\mathbf{f} (y) - \mathbf{f} (x)) \cdot \nabla \rho_{\epsilon}(x-y) dy | dx \\ & \leq & C \| J \|_{L^p (B_{R+1})} \Big \{ \int_{B_{R+1}} dx \int_{|x-y| \leq \epsilon } \{ \frac{|\mathbf{f} (y) - \mathbf{f} (x)|}{\epsilon} \}^q \frac{1}{\epsilon^n} dy \Big \}^{\frac{1}{q}} \end{eqnarray*}