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* (NB compact support assumption added later).


If $g \in W^{1,p}(\mathbb R^n;\mathbb R^n)$ (for any $p \in (1,\infty]$), is it true that $$\mathcal I \lesssim \|\nabla g\|\_{L^p(\mathbb R^n)}$$
or something similar (possibly adding the $\epsilon$ somewhere)?


Related questions are [Bounding integral expression with total variation of integrand](https://mathoverflow.net/questions/425400/bounding-integral-expression-with-total-variation-of-integrand?rq=1) and [Bounding integral expression with BV norm of integrand](https://mathoverflow.net/questions/425441/bounding-integral-expression-with-bv-norm-of-integrand)



Bounding integral expression with Sobolev norm of integrand
Such a bound is impossible already for $n=1$. Indeed, suppose that $f=1$ and
$$g'(x)=\frac1{1+|x|}$$
for all real $x$. Then $TV(g)=\infty$ and hence, according to the [previous answer](https://mathoverflow.net/a/425408/36721), $\mathcal I=\infty$, whereas $\|\nabla g\|\_{L^p(\mathbb R^n)}<\infty$ for any $p\in(1,\infty]$.


