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Problem. Is there a partition $\mathbb R^2=A\sqcup B$ of the Euclidean plane into two Lebesgue measurable sets such that for any disk $D$ of the unit radius we get $\lambda(A\cap D)=\lambda(B\cap D)=\frac12\lambda(D)$?

(I.V.Protasov called such partitions kaleidoscopic).

Observe that for the $\ell_1$- or $\ell_\infty$-norms on the plane such partitions exist: just take a suitable chessboard coloring.

The problem can be reformulated in terms of convolutions: Is there a measurable function $f:\mathbb R^2\to\{1,-1\}$ such that its convolution with the characteristic function $\chi_D$ of the unit disk $D$ is identically zero?

(The problem was posed 08.11.2015 by T.Banakh and I.Protasov on page 19 of Volume 0 of the Lviv Scottish Book).

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    $\begingroup$ About the tag scottish-book, see the meta discussion: meta.mathoverflow.net/questions/3948/tag-referring-to-a-book/… $\endgroup$
    – YCor
    Commented Oct 28, 2018 at 19:37
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    $\begingroup$ I'll record the basic observations in the Fourier domain: Let $f$ be the stated function, let $\hat{f}$ be its Fourier transform and let $\hat{\chi_D}$ be the Fourier transform of $\chi_D$. Then $\hat{f} \hat{\chi_D}$ is orthogonal to every character of $\mathbb{R}^2$. In my naive nonanalyst way, I think this means $\hat{f}$ is supported on the zeroes of $\hat{\chi_D}$. According to Mathematica, $\hat{\chi_D}(k_1, k_2) = 2 \pi \tfrac{J_1(k)}{k}$ where $J_1$ is the first Bessel function and $k = \sqrt{k_1^2+k_2^2}$. Unfortunately, $J_1$ has infinitely many zeroes. $\endgroup$
    – David E Speyer
    Commented Oct 29, 2018 at 14:40
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    $\begingroup$ I seem to see a 3-color kaleidoscope for the perfect hexagonal ball. $\endgroup$
    – Wlod AA
    Commented Oct 30, 2018 at 9:21
  • $\begingroup$ Do you know any plane figure (maybe nonconvex) without kaleidoscopic colorings? (wrt parallel translations) $\endgroup$
    – Anton Petrunin
    Commented Jan 5, 2021 at 3:39
  • $\begingroup$ @AntonPetrunin No, I do not know such a figure. It seems that for any triangle a kaleidoscopic coloring does exist. $\endgroup$
    – Taras Banakh
    Commented Jan 5, 2021 at 12:41

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