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What probability measure(s) maximize the quantity $\iiint_{\mathbb{S}^1}|(x-z)\times(y-z)|d\mu(x)d\mu(y)d\mu(z)$?

The answer appears to be uniform measure, since informally it appears better to have more triangles in the support of $\mu$ which the function $|(x-z)\times(y-z)|$ computes the area of.

Is there an argument that shows this is true generally for $\mathbb{S}^{d-1}$?

Edit: Thanks to @fedja for the answer to the main question and the request for clarification. While it could be possible to replace the area of triangles with volumes of simplices (and so on), what was meant by generalization was most immediately the question of whether uniform measure maximizes $$\iiint_{\mathbb{S}^2}|(x-z)\times(y-z)|d\mu(x)d\mu(y)d\mu(z).$$

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    $\begingroup$ Probably not. Let's take a small disk of radius $r$ on the sphere and replace the area measure in it by the delta-measure at the center. Then it looks like you'll gain $cr^2$ for a typical height of the triangle with one vertex in the disk, so the total gain will be $cr^4$ while the loss will come from the triangles with 2 vertices in the disk (order $r^5$) and three vertices in the disk (order $r^8$). It is not a 2-dimensional equator either because in high dimensions for the uniform measure most triangles are about equilateral with side $\sqrt 2$ and $\frac{\sqrt 3}2>\frac 3{2\pi}$. $\endgroup$
    – fedja
    Commented Feb 3, 2020 at 3:56
  • $\begingroup$ Ok. There is a semidefinite programming argument which shows that surface measure is optimal (among other optimizers) for $|(x-z)\times(y-z)|^2$ and $\mathbb{S}^2$. It is curious for $|(x-z)\times(y-z)|$ what should be the optimizer in the case uniform measure is not optimal. $\endgroup$
    – Josiah Park
    Commented Feb 3, 2020 at 4:18
  • $\begingroup$ Interesting. To be honest, I have no idea what to suggest for the optimizer since it looks like we have a break of symmetry here (an asymmetric solution of a symmetric problem) and such things are pretty hard to guess in general... $\endgroup$
    – fedja
    Commented Feb 3, 2020 at 4:27
  • $\begingroup$ Arghhh... I made a (very) stupid mistake in my computations. Just forget everything I said about the sphere for now. $\endgroup$
    – fedja
    Commented Feb 3, 2020 at 12:45
  • $\begingroup$ Follow-up question, mathoverflow.net/questions/352236/… $\endgroup$
    – Gerry Myerson
    Commented Feb 8, 2020 at 21:49

1 Answer 1

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Yes, it is true for the circle (though the reason is not quite the one you suggested). We shall consider the discrete version of the problem, which is to put some odd number $n\ge 3$ of points (think of them as of being assigned the probability of $1/n$ each) on the circle so that the sum of triangle areas is maximized. Then an optimal configuration exists by compactness. We just want to show that it should be an equispaced distribution. Since you can obtain any measure on the circle as a weak limit of such discrete measures (and the integrand is continuous, so a small displacement of the measure does not change the integral much), you'll get the required optimality of the uniform measure as a limiting statement.

Take any of the points $e$ and enumerate all other points by the numbers $1,\dots,n-1$ clockwise starting from $e$ (so the whole system is $e,e_1,\dots,e_{n-1}$. Fix all points except $e$ and think a bit of how the total double area depends on the position of $e$. You'll realize quite soon that it is just some constant plus $v_e\times e$ where $$v_e=\sum_{1\le k< m \le n-1}(e_m-e_k).$$

If you now consider for fixed $\ell<\frac n2$ all vectors $e_m-e_k$ with [$(\ell\le k< m\le n-\ell$) and ($k=\ell$ or $m=n-\ell$)], you'll see that $v_e=\sum_{\ell<\frac n2}A_\ell(e_{n-\ell}-e_\ell)$ with some $A_\ell>0$. You will be able to increase the sum of (even oriented) areas if $e$ is not perpendicular to $v_e$, so in the optimal configuration we must always have $\langle e,v_e\rangle=0$ for every choice of $e$.

Now recall the famous "drive around the circle without running out of gas" problem (more precisely, its particular case when all gas stations have the same amount of fuel sufficient for driving $\frac 1n$ of the entire circle). The conclusion of it is that you can choose $e$ so that the arc distance to $e_k$ is at most $\frac k n$ of the whole circle for every $k=1,\dots,n-1$, which makes all scalar products $\langle e,e_{n-\ell}-e_\ell\rangle$ with $\ell<\frac n2$ non-positive with the only chance of the equality $\langle e,v_e\rangle=0$ when the arc distance from $e$ to $e_k$ is exactly $\frac kn$ for all $k=1,\dots,n-1$.

As to higher dimensions, can you, please, specify first what you mean by "this"? (there are several ways to try to generalize the setup, so the word seems rather ambiguous to me :-) )

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