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.


2 Answers 2


$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$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.

  • $\begingroup$ Thank you very much. Do you see any way to refine the estimate so that it doesn't blow up as $\epsilon \to 0$? $\endgroup$
    – Dal
    Jun 24 at 18:31
  • $\begingroup$ @Dal : No, you cannot avoid a blow-up. Now the bound is exact, as it is attained. $\endgroup$ Jun 24 at 18:51
  • $\begingroup$ Thank you. Where does the factor 2 in formula (1) come from? $\endgroup$
    – Dal
    Jun 24 at 19:03
  • $\begingroup$ @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 $\epsilon\le y-x\le1/2$. $\endgroup$ Jun 24 at 19:25
  • $\begingroup$ 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 $g \in BV(\mathbb R^n;\mathbb R^n)$? $\endgroup$
    – 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?

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.

  • $\begingroup$ Thank you so much. That's enlightening. $\endgroup$
    – Dal
    Jul 3 at 21:05
  • $\begingroup$ 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? $\endgroup$
    – Dal
    Jul 3 at 21:08
  • $\begingroup$ @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 $a>0$. $\endgroup$ Jul 3 at 21:24

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