# 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.

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

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.

• Thank you very much. Do you see any way to refine the estimate so that it doesn't blow up as $\epsilon \to 0$?
– Dal
Jun 24 at 18:31
• @Dal : No, you cannot avoid a blow-up. Now the bound is exact, as it is attained. Jun 24 at 18:51
• Thank you. Where does the factor 2 in formula (1) come from?
– Dal
Jun 24 at 19:03
• @Dal : The factor $2$ arises because (i) the integrand and the region over which the integration is done are invariant with respect to the interchange of $x$ and $y$ and (ii) the restriction $\epsilon\le|x-y|\le1/2$ was replaced by $\epsilon\le y-x\le1/2$. Jun 24 at 19:25
• Thank you. Does the same hold in higher dimension: namely for $$\mathcal I :=\iint_{\epsilon \leq|x-y| \leq 1/2} f(x) f(y) \frac{\langle g(x)-g(y), x-y\rangle}{|x-y|^{n+2}} d x d y$$ for $\epsilon>0$, $f \in L^\infty(\mathbb R^n;\mathbb R)$, and $g \in BV(\mathbb R^n;\mathbb R^n)$?
– Dal
Jun 24 at 22:33

$$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$$Added later by the OP:

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

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 $$$$\mathcal I=2\|f\|_\infty^2\, J,$$$$ where \begin{aligned} J &:=\iint_{[a,b]^2}\,\frac{dx\, dy}{(x-y)^2}\,|g(x)-g(y)|\,1(\ep\le y-x\le1/2) \\ &=\iint_{[a,b]^2}\,\frac{dx\, dy}{(x-y)^2}\,\int_x^y dg(z)\,1(\ep\le y-x\le1/2) \\ &=\int_{[a,b]} dg(z) \\ &\quad\times\iint_{[a,b]^2}\,\frac{dx\, dy}{(x-y)^2} \,1(x\le z\le y,\,\ep\le y-x\le1/2)\\ &\ge\int_{[a+1/2,\,b-1/2]} dg(z)\int_{z-1/2}^z dx\,\int_{\max(z,x+\ep)}^{x+1/2}\frac{dy}{(x-y)^2} \\ &=\Big(\ln\frac1{2\ep}\Big)\,\int_{[a+1/2,\,b-1/2]} dg(z)=\Big(\ln\frac1{2\ep}\Big)\,TV(g). \end{aligned} Thus, $$$$\mathcal I\ge2\Big(\ln\frac1{2\ep}\Big)\|f\|_\infty^2\, TV(g). \tag{4}\label{4}$$$$ (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.

• Thank you so much. That's enlightening.
– Dal
Jul 3 at 21:05
• The answer though begs an additional question: it seems that in your construction is important that $g$ is not compactly supported. What if we assume it is? Would there be any hope?
– Dal
Jul 3 at 21:08
• @Dal : My guess is that it would not significantly help if $g$ is assumed to be compactly supported. You can try to do a more or less explicit calculation/bounding of $\mathcal I$ with $f$ as in this answer and a compactly supported $g$ with just two nonzero values of the slope, one of them much greater in absolute value than the other. I guess that would be an example showing that you cannot get rid of $\epsilon$. Or maybe try $g(x)=e^{-a x}\,1(x>0)$ with $a>0$. Jul 3 at 21:24