# The product of distance between $n$ points in $[-1,1]$

calculusgeometry

$$a_1,a_2,\cdots,a_n$$ are real numbers between $$-1$$ and $$1$$, i.e., for each $$1\leq i \leq n$$, $$-1 \leq a_i \leq 1$$.

Define
$$f(n):= \max_{-1\leq a_1,\cdots,a_n \leq 1} \min_{1\leq p \leq n} \prod_{q \neq p} (a_q – a_p)^2.$$
My question is that if $$f(n)$$ decreases exponentially as $$n$$ increases.

I guess that $$f(n) \leq 16\left(\frac{3}{4}\right)^n$$.

Yes, $$f(n)$$ decreases exponentially and decreases faster than any $$(\frac14 + \varepsilon)^n$$.

For compact $$E \subset \mathbb{C}$$, let $$\mathcal{E}_n = \{ A \subset E : |A| = n \}$$ be those subsets of $$E$$ with $$n$$ elements.

For $$A = \{ z_1, \ldots, z_n \} \in \mathcal{E}_n$$, let $$\Delta_A$$ be product of distances among its elements:

$$\Delta_A = \prod_{1\le i < j \le n}|z_i - z_j|$$

Among all $$A \in \mathcal{E}_n$$, those maximizes $$\Delta_A$$ is called an $$n$$-point Fekete set of $$E$$.

For $$E = [-1,1]$$, its $$n$$-point Fekete set $$\mathcal{F}_n$$ is known to be unique and

$$z \in \mathcal{F}_n \iff (1-z^2)P'_\ell(z) = 0$$

where $$\ell = n-1$$ and $$P_\ell(z)$$ is the Legendre polynomial of order $$\ell$$.

From now on, we will assume $$E = [-1,1]$$.

For $$k = 1,\ldots, n$$, let $$\displaystyle\;\varphi(A,z_k) = \prod\limits_{i=1,\ne k}^n |z_i - z_k|$$. It is easy to see

$$\prod_{k=1}^n \varphi(A,z_k) = \Delta_A^2 \implies \min_{1\le k \le n}\varphi(A,z_k) \le \Delta_A^{2/n}$$

Compare this with definition of $$f(n)$$, we find

$$f(n) \le \sup \{ \Delta_A^{4/n} : A \in \mathcal{E}_n \} = \Delta_{\mathcal{F}_n}^{4/n}$$

Let $$\alpha_1,\ldots,\alpha_n$$ be elements of $$\mathcal{F}_n$$ and $$\psi(z) = \prod\limits_{i=1}^n(z - \alpha_k)$$. Since $$\alpha_k$$ are roots of $$(1-z^2)P'_\ell(z)$$, by comparing leading coefficients, we find

$$\psi(z) = \frac{2^\ell}{\ell}\binom{2\ell}{\ell}^{-1}(z^2-1) P'_\ell(z)$$

In term of $$\psi(z)$$, we have

$$\varphi(\mathcal{F}_n,\alpha_k) = \prod_{i=1,\ne k}^n |\alpha_k - \alpha_i| = |\psi'(\alpha_k)| = (\ell+1)2^\ell\binom{2\ell}{\ell}^{-1}|P_{\ell}(\alpha_k)|$$

$$\Delta_{\mathcal{F}_n}^{2/n} = \prod_{k=1}^n \varphi(\mathcal{F}_n,\alpha_k)^{1/n} = (\ell+1)2^\ell\binom{2\ell}{\ell}^{-1}\left|\prod_{k=1}^n P_\ell(\alpha_k)\right|^{1/n}$$
Since $$|P(z)| \le 1$$ for $$z \in [-1,1]$$, we can bound $$f(n)$$ from above as $$f(n) \le \Delta_{\mathcal{F}_n}^{4/n} \le \left((\ell+1) 2^\ell\binom{2\ell}{\ell}^{-1}\right)^2 = (2n-1)^24^n \binom{2n}{n}^{-2}$$ Notice $$\displaystyle\;\frac{4^n}{\sqrt{\pi(n + \frac12)}} \le \binom{2n}{n}$$ for all $$n \ge 1$$, we have
$$f(n) \le \frac{(2n-1)^2\pi(n + \frac12)}{4^n} \le \frac{4\pi n^3}{4^n}$$
This means $$f(n)$$ decreases faster than $$(\frac14+\varepsilon)^n$$ for any $$\varepsilon > 0$$.