Bounding integral expression with total variation of integrand 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)?

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

This is motivated by a question related to approximate differentiability.
 A: $\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$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.
A: $\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$Added later by the OP:

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

The answer to this additional question is: of course, not.
Indeed, suppose that $f=1_{[a,b]}$ for some real $a$ and $b$ such that $b-a>1$. Let $g$ be any nondecreasing function such that $g$ is constant on $(-\infty,a+1/2)$ and on $(b-1/2,\infty)$, so that
$$TV(g)=\int_{[a+1/2,\,b-1/2]}dg(z).$$
Then for $\ep\in(0,1/2]$ we have
\begin{equation}
    \mathcal I=2\|f\|_\infty^2\, J, 
\end{equation}
where
\begin{equation}
\begin{aligned}
    J  
&:=\iint_{[a,b]^2}\,\frac{dx\, dy}{(x-y)^2}\,|g(x)-g(y)|\,1(\ep\le y-x\le1/2) \\ 
    &=\iint_{[a,b]^2}\,\frac{dx\, dy}{(x-y)^2}\,\int_x^y dg(z)\,1(\ep\le y-x\le1/2) \\ 
    &=\int_{[a,b]} dg(z) \\ 
&\quad\times\iint_{[a,b]^2}\,\frac{dx\, dy}{(x-y)^2} \,1(x\le z\le y,\,\ep\le y-x\le1/2)\\ 
&\ge\int_{[a+1/2,\,b-1/2]} 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_{[a+1/2,\,b-1/2]} dg(z)=\Big(\ln\frac1{2\ep}\Big)\,TV(g).  
\end{aligned}
\end{equation}
Thus,
\begin{equation}
    \mathcal I\ge2\Big(\ln\frac1{2\ep}\Big)\|f\|_\infty^2\, TV(g). \tag{4}\label{4}
\end{equation}
(In view of inequality (3) in the previous answer, inequality \eqref{4} is actually the equality.)
Thus, the form of the exact bound on $\mathcal I$ with the additional restriction that $f$ be compactly supported is exactly the same as the exact bound on $\mathcal I$ without this restriction.
