Consider the following integral expression: $$\mathcal I :=\iint_{\epsilon \leq|x-y| \leq 1/2} f(x) f(y) \frac{\langle g(x)-g(y), x-y\rangle}{|x-y|^{n+2}} d x d y $$ for $\epsilon>0$, $f \in L^\infty(\mathbb R^n;\mathbb R)$ and $f$ compactly supported.
If $g \in BV(\mathbb R^n;\mathbb R^n)$, is it true that $$\mathcal I \lesssim TV(g)$$ or something of this nature (possibly adding the $\epsilon$ somewhere)?
Related questions are Bounding integral expression with Sobolev norm of integrand and Bounding integral expression with total variation of integrand
Info on BV functions: Let $\Omega \subset \mathbb R^n$. We say that $u \in L^{1}(\Omega; \mathbb R)$ is a function of bounded variation in $\Omega$ if the distributional derivative of $u$ is representable by a finite Radon measure in $\Omega$: \begin{align*} \int_{\Omega} u \frac{\partial \phi}{\partial x_{i}} d x=-\int_{\Omega} \phi d D_{i} u \quad \forall \phi \in C_{c}^{\infty}(\Omega), \quad i=1, \ldots, N \end{align*} for some $\mathbb{R}^{n}$-valued measure $D u=\left(D_{1} u \ldots . D_{n} u\right)$ in $\Omega$.
For functions $u \in B V(\Omega; \mathbb R^m)$, $D u$ is an $m \times n$ matrix of measures $D_{i} u^{\alpha}$ in $\Omega$ satisfying \begin{align*} \int_{\Omega} u^{\alpha} \frac{\partial \phi}{\partial x_{i}} d x=-\int_{\Omega} \phi d D_{i} u^{\alpha} \quad \forall \phi \in C_{c}^{\infty}(\Omega), i=1 \ldots, n, \alpha=1 \ldots m \end{align*}
Note that Sobolev space $W^{1,1}(\Omega)$ is contained in $B V(\Omega)$: indeed, for any $u \in W^{1,1}(\Omega)$ the distributional derivative is given by $\nabla u \mathcal{L}^{n}$.
For $u \in L_{\text {loc }}^{1}(\Omega; \mathbb R^m)$. The total variation $TV(u, \Omega)$ is defined by \begin{align*} TV(u, \Omega):=\sup \left\{\sum_{\alpha=1}^{m} \int_{\Omega} u^{\alpha} \operatorname{div} \varphi^{\alpha} d x: \varphi \in C_{c}^{\infty}(\Omega; \mathbb R^{m n}), \|\varphi\|_{\infty} \leq 1\right\} \end{align*} Note that $TV(u, \Omega)=\int_{\Omega}|\nabla u| d x$ if $u$ is continuously differentiable in $\Omega$.
We remark that $u \in L^{1}((\Omega); \mathbb R^{m})$ belongs to $BV(\Omega; \mathbb R^m)$ if and only if $TV(u, \Omega)<\infty$.