Prove or disprove: if $\mu(E) > 0$, then $\int_E \int_E \frac{dx\,dy}{|x-y|} = \infty.$

Question

Prove or disprove: If $E \subseteq \mathbb{R}$ has positive Lebesgue measure, then $\displaystyle\int_E \int_E \frac{dx\,dy}{|x-y|} = \infty$.

I believe this is true. Since $\mu(E) > 0$, $E-E$ contains a neighborhood of 0, so $\int_{E-E} \frac{dx}{|x|} = \infty$, but I am having difficulty formally justifying the inequality $$\int_E \int_E \frac{dx\,dy}{|x-y|} \geq \int_{E-E} \frac{dx}{|x|}$$

Intuitively, it seems obvious since the values $|x-y|$ range over $E-E$, but I would like to see the fully justified intermediate steps of reducing this double integral to a single integral.

Answer 1

By choosing an appropriate subset, we may assume that $\mu(E) \in (0, \infty)$. Then

\begin{align*} \int_{E}\int_{E} \frac{dxdy}{\lvert x-y \rvert} &= \int_{\mathbb{R}}\int_{\mathbb{R}} \frac{\mathbf{1}_E(x)\mathbf{1}_E(y)}{\lvert x-y \rvert} \, dxdy \\ &= \int_{\mathbb{R}}\int_{\mathbb{R}} \frac{\mathbf{1}_E(x+y)\mathbf{1}_E(y)}{\lvert x \rvert} \, dxdy \\ &= \int_{\mathbb{R}} \frac{\phi(x)}{\lvert x \rvert}\,dx \end{align*}

where $\phi$ is defined by

$$\phi(x) := \int_{\mathbb{R}} \mathbf{1}_E(x+y)\mathbf{1}_E(y) \, dy.$$

Notice that $\phi(0) = \mu(E) > 0$ and $\phi$ is continuous by $L^p$-continuity of translation. So there exist $c > 0$ and $\delta > 0$ such that $\phi(x) \geq c$ on $[-\delta, \delta]$, and therefore

$$ \int_{E}\int_{E} \frac{dxdy}{\lvert x-y \rvert} \geq \int_{-\delta}^{\delta} \frac{c}{\lvert x \rvert}\,dx = \infty. $$

Answer 2

Let $\Omega$ be an open set in $E$ $$\int_E \int_E \frac{\mathrm d x\mathrm dy}{|x-y|}\ge \int_{\Omega} \int_{\Omega} \frac{\mathrm d x\mathrm dy}{|x-y|} =_{u= x-y, v = x} \int_\Omega \int_{\Omega-\Omega} \frac{\mathrm d u\mathrm d v}{|u|} = \mu(\Omega) \int_{\Omega-\Omega} \frac{\mathrm d u}{|u|}$$

Answer 3

Stronger result: For a.e $y\in E,$

$$\int_E \frac{1}{|x-y|}\, dx = \infty.$$

Proof: A.e. $y\in E$ is a point of density of $E$ in the Lebesgue sense. Fix such a $y.$ Note that

$$\frac{1}{|x-y| }= \int_{|x-y|}^\infty\frac{1}{t^2}\, dt,$$

which is a little devious. So the integral in question equals

$$\int_E \int_{|x-y|}^\infty\frac{1}{t^2}\, dt.$$

Switching the order of integration gives

$$\int_0^\infty \frac{1}{t^2}\int_{E\cap(y-t,y+t)} dx \, dt=\int_0^\infty m({E\cap(y-t,y+t)})\, \frac{1}{t^2}\, dt.$$

Since $y$ is a point of density of $E,$ there exists $r >0$ such that $m({E\cap(y-t,y+t)})>t$ for $0<t<r.$ Thus the last integral is at least

$$\int_0^r \frac{t}{t^2}\, dt = \infty.$$