Vtyjkyuk

Here's the problem:

Let $fin L^1$ satisfy
$$limsup_epsilon to 0 int_mathbbRint_mathbbR fracx-y dxdy < infty$$
Show that $f=0$ almost everywhere.

I've got what I think to be a solution to this problem, but I don't find it to be very elegant. Denote the integral by $I(epsilon)$. It involves restricting to a strip of width $h$ along the line $y=x$ and using Lebesgue Differentiation. I have to do a bunch of basic-type estimates on the integral, and what I finally get is that $I(epsilon) geq C frac1h int_K |f|^2$, where $K_n$ is a sequence of compact set which increase to $mathbbR$, minus perhaps a null set. Letting $h to 0$ shows that $int_K |f|^2 = 0$, so that $int_K |f| = 0$.

So I effectively hit the integral with a bunch of rather crude estimates. But this is a qual problem, whose solutions I'm used to being a little more clean. Is there any easier way? I can provide details to my solution, but if you have anything that looks better than what I've outlined above, I'd like to see it.

Thank you!

Some ideas: 1. By the monotone convergence theorem, the $limsup$ equals

$$int_mathbbRint_mathbbR frac^2 dxdy.$$

2. Suppose $f= 0$ a.e. fails. Then there is a set $E$ of positive measure and a constant $c>0$ such that $|f|>c$ on $E.$




3. A.e. point of $E$ is a point of density of $E.$ If $x$ is a point of density of $E$, then it seems probable to me that

$$int_E frac1^2 dy =infty.$$