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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?)

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  • $\begingroup$ TeX note: for set difference use $[0, 1] \setminus E$ [0, 1] \setminus E, not $[0, 1]\backslash E$ [0, 1]\backslash E. I have edited accordingly. $\endgroup$
    – LSpice
    Dec 2, 2021 at 20:30
  • $\begingroup$ Have you tried artofproblemsolving.com ? $\endgroup$ Dec 2, 2021 at 22:24
  • $\begingroup$ That's where the problem came from (still no solution). I also suspect that this problem is beyond what is typically discussed over there. $\endgroup$
    – Ivan
    Dec 3, 2021 at 0:06
  • $\begingroup$ I think it is true, but not obvious, that $h <\infty$ almost everywhere in $E$. Am I right? $\endgroup$ Dec 4, 2021 at 12:12
  • $\begingroup$ You could have put that information in your original post, Ivan. $\endgroup$ Dec 8, 2021 at 0:35

1 Answer 1

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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!

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    $\begingroup$ Could you describe the solution, now that you have understood it? $\endgroup$
    – LSpice
    Dec 8, 2021 at 2:15
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    $\begingroup$ @L.Spice Done, see the Edit to the answer. $\endgroup$ Dec 8, 2021 at 14:55

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