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Problem: Let $E \subset \mathbb{R}^N$ be a connected, bounded, open and smooth (or just $N$-measurable) set and denote with $\mathcal{L}^N$ the Lebesgue measure. Define $\Omega_i = \{ x \in \mathbb{R}^N: x_i > 0 \}$ and for any $J \subset \{1,\dots,N\}$ define: $$\Omega_J= \bigcap\limits_{i \in J} \Omega_i \cap \bigcap\limits_{i \in \{1,\dots,N\} \smallsetminus J} \Omega_i ^c$$

I am asking if there is $z \in \mathbb{R}^N$ and a rotation $R \in \mathcal{SO}(n)$ such that, defining $\tau_z(x)=x+z$ for all $x \in \mathbb{R}^N$, the following property holds:

$$\forall I,J \subset \{1,\dots,N\} : \mathcal{L}^N ((\tau_z \circ R) (E) \cap \Omega_J)=\mathcal{L}^N ((\tau_z \circ R) (E) \cap \Omega_I)$$

I do not even know which is the answer.

My attempt:

I proved the following:

Lemma: In the assumption as before we can prove that: $$\mathcal{L}^N ((\tau_z \circ R) (E) \cap \Omega_i)=\mathcal{L}^N ((\tau_z \circ R) (E) \cap \Omega_i^c)$$

Proof: We do not need rotations. Just consider that $z \to \mathcal{L}^N (\tau_z (E) \cap \Omega_i)$ is continuous and assume the values $0$ and $\mathcal{L}^N(E)>0$, thus there is a $z$ such that $\mathcal{L}^N (\tau_z (E) \cap \Omega_i)=\frac{\mathcal{L}^N(E)}{2}$. Then proceeding this way for all the components we obtain what we want. $\Box$

Background: I am studying BV functions and in particular the proof of isoperimetric inequality for finite perimeter sets of $\mathbb{R}^N$.

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    $\begingroup$ Off hand the answer should be "no" if $n$ is not too small because you have roughly speaking $n^2$ free parameters and $2^n$ independent equations to satisfy. Things normally just do not work this way. $\endgroup$
    – fedja
    Nov 29, 2021 at 23:41

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