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Let $A$ be a nonempty measurable subset of $\mathbb R$, with Lebesgue measure $|A|=1$, and let $c>0$. Define the scalar $I(A)$ by

$$ I(A) := \int_{-c}^c |A \cap (x + A)|\, dx, $$ where $x+A := \{x + a \mid b \in A\}$.

Question. Under the constraint $|A| = 1$, is it true that $I(A)$ is maximized when $A$ is an interval centered at the origin, i.e $[-1/2,1/2]$ ?

In the case where $A$ is restricted to an interval, then the answer to the question is affirmative, and has been provided here https://mathoverflow.net/a/282941/78539 (under the name "Fedja's lemma"). Unfortunately, the solution is somewhat complicate and I don't see how to modify it so that it applies without convexity (i.e non-interval sets).

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  • $\begingroup$ (i) "a nonempty measurable subsets"? (ii) "respectively"? (iii) "centered at the origin"? $I(A)$ is shift-invariant. (iv) The answer at mathoverflow.net/a/282941/78539 is to a substantially different question. (v) Is your $c$ fixed? (vi) Overall, are you sure this is the question you really wanted to ask? $\endgroup$ Commented Feb 9, 2022 at 1:06
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    $\begingroup$ This is immediate from the Riesz rearrangement inequality, since $I(A) = \int_{\bf R} \int_{\bf R} 1_A(x) 1_A(x-y) 1_{[-c,c]}(y)\ dx dy$. en.wikipedia.org/wiki/Riesz_rearrangement_inequality $\endgroup$
    – Terry Tao
    Commented Feb 9, 2022 at 2:05
  • $\begingroup$ @TerryTao Indeed, this solves the problem. Thanks. $\endgroup$
    – dohmatob
    Commented Feb 9, 2022 at 9:03
  • $\begingroup$ @IosifPinelis Sorry for the typos. Fixed. $\endgroup$
    – dohmatob
    Commented Feb 9, 2022 at 9:04
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    $\begingroup$ @dohmatob Feel free to give an answer to this question to close it off. Incidentally you may be interested in some recent work on questions of this type in arxiv.org/abs/2106.13873 and arxiv.org/abs/1903.08731 . $\endgroup$
    – Terry Tao
    Commented Feb 9, 2022 at 19:32

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Disclaimer: Upon recommendation (in the comments), this post is just to close off the question, since it has been completely answered in the comments.


  • Indeed, as observed by user @Terry Tao in the comments, the question has a simple affirmative answer via the Riez rearrangment inequality.

  • Using this machinery, a more general question in $n$ dimensions is answered here https://mathoverflow.net/a/415754/78539, and the answer to my original question (on the real line) follows as an immediate corollary.

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