A problem concerning a divergent function on $[0, 1]$ This problem was posted on another forum and was given at the 1992 Miklós Schweitzer Competition. This competition is known for its very difficult problems and this one seems no exception. I also can't find a solution anywhere online. Here is the problem:
Let $E \subset [0,1]$ be Lebesgue measurable with measure $\lvert E\rvert < 1/2$. Define
$$h (s) = \int _ {E^c} \frac{dt}{{(s-t)}^2}$$
where $E^c = [0, 1]\setminus E$. Prove that there exists $t \in E^c$ such that
$$\int_E \frac {ds} {h (s) {(s-t)} ^ 2} \leq c {\lvert E \rvert} ^ 2$$
with some constant $c$ that is independent of $E$.
The tricky part of the problem is the quadratic convergence to zero with the measure of $E$. It is easy to see that the left hand side of the inequality is linearly bound with $\lvert E\rvert$.  Any ideas (or links to sites with solutions to Miklós Schweitzer problems?)
 A: I asked a colleague in Hungary, and he found the solution here (on page 170):
http://real-j.mtak.hu/9393/1/MTA_MatematikaiLapok_1992.pdf
It is in Hungarian, but with some effort and Google translator, finally I understood.
EDIT This is the translation of the argument above. I keep the same notation but write $E^c$ for $[0,1] \setminus E$ insetad of $\overline E$.
a) Let $$F=\{t  \in E^c: |I \cap E| \leq K|E||I|\}$$ for every interval $t \in I$. In other words, $F$ consists of all points for which the maximal function of $\chi_E$ is less than $K|E|$. Using the maximal inequality, one selects $K$ (independent of $E$) such that $|F| \approx 1$, hence  $|E_1|:=|E^c \cap F| \geq 1/2$.
b) Given $s \in E$ let $I_s$ be an interval containing $s$ such that $|I_s \cap E|=(1/2)|I_s|$. Such a $I_s$ exists in all Lebesgue points of $E$, again using that $|E| <1/2$ and a continuity argument.
c) If $s \in E$ and $t \in E_1$, then applying a) to $J=I_s \cup [s,t]$ (or the other way around) we get $(1/2)|I_s|=|I_s \cap E| \leq |J\cap E| \leq K|E|(|t-s|+|I_s|)$. Then $|t-s| \geq \frac{|I_s|}{4K|E|}$ if $K|E| \leq 1/4$.
d) Suppose that $K|E| \leq 1/4$. Then
$$
\int_{E_1} \frac{dt}{(s-t)^2} \leq 2 \int_{\frac{|I_s|}{4K|E|}}^\infty \frac{du}{u^2}=\frac{8K|E|}{|I_s|}$$ but also, using b),
$$h(s)=\int_{E^c} \frac{dt}{(s-t)^2}\geq \int_{I_s \cap  \geq E^c} \frac{dt}  {(s-t)^2} \geq \frac{1}{|I_s|^2} \frac 12 |I_s|=\frac{1}{2|I_s|}.$$ Summing up,
$$
\int_{E_1} \frac{dt}{(s-t)^2} \leq 16 h(s) K|E|.$$
e) If $K|E| \geq 1/4$ then
$$
\int_{E_1} \frac{dt}{(s-t)^2} \leq 
\int_{E^c} \frac{dt}{(s-t)^2} =h(s) \leq 4h(s) K|E| \leq 16 h(s) K|E|.$$
f) Finally,
$$
\int_{E_1}dt\int_{E}\frac{ds }{h(s)(s-t)^2}=\int_{E}\frac{ds}{h(s)}\int_{E_1}\frac{ dt}{(s-t)^2} \leq 16K|E|^2 $$ and there is a point $t \in E_1$ for which the statement holds with $c=32 K$ since $|E_1| \geq 1/2$.
PS This does not deserve badges for maths...maybe for translating from hungarian!
