Consider the following integral expression:
$$\mathcal I :=\iint\_{\epsilon \leq|x-y| \leq 1/2} f(x) f(y) \frac{(g(x)-g(y))(x-y)}{|x-y|^{3}} d x d y $$
for $\epsilon>0$, $f \in L^\infty(\mathbb R)$, and $g \in BV(\mathbb R)$.
Is it true that
$$\mathcal I \lesssim TV(g)$$
or something of this nature (possibly adding the $\epsilon$ somewhere)?




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Added later: does the dependence on $\epsilon$ in the answer below improve if we further assume $f$ to be compactly supported?




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This is motivated by a question related to approximate differentiability.



Bounding integral expression with total variation of integrand
$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$Yes, this is true. Indeed, for $\ep\in(0,1/2]$ we have
\begin{equation}
 \mathcal I\le2\|f\|\_\infty^2\, J, \tag{1}\label{1}
\end{equation}
where
\begin{equation}
\begin{aligned}
 J&:=\iint\_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,|g(x)-g(y)|\,1(\ep\le y-x\le1/2) \\ 
 &\le\iint\_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,\int\_x^y|dg(z)|\,1(\ep\le y-x\le1/2) \\ 
 &=\int\_\R |dg(z)|\iint\_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,1(\ep\le y-x\le1/2,x\le z\le y) \\ 
 &=\int\_\R |dg(z)|\int\_{z-1/2}^z dx\,\int\_{\max(z,x+\ep)}^{x+1/2}\frac{dy}{(x-y)^2} \\ 
 &=\Big(\ln\frac1{2\ep}\Big)\,\int\_\R |dg(z)|=\Big(\ln\frac1{2\ep}\Big)\,TV(g). 
\end{aligned}
\tag{2}\label{2}
\end{equation}
Thus,
\begin{equation}
 \mathcal I\le2\Big(\ln\frac1{2\ep}\Big)\|f\|\_\infty^2\, TV(g). \tag{3}\label{3}
\end{equation}


The bound in \eqref{3} is exact. Indeed, the ineqialities in \eqref{1} and \eqref{2}, and hence
in \eqref{3}, turn into the equalities if $f$ is a constant and $g$ is nondecreasing.


