The product of distance between $n$ points in $[-1,1]$ $a_1,a_2,\cdots,a_n$ are real numbers between $-1$ and $1$, i.e., for each $1\leq i \leq n$, $-1 \leq a_i \leq 1$.
Define
\begin{equation}
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.
\end{equation}
My question is that if $f(n)$ decreases exponentially as $n$ increases.
I guess that $f(n) \leq 16\left(\frac{3}{4}\right)^n$.
 A: 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
$$\varphi(\mathcal{F}_n,\alpha_k) =
\prod_{i=1,\ne k}^n |\alpha_k - \alpha_i|
= |\psi'(\alpha_k)| = (\ell+1)2^\ell\binom{2\ell}{\ell}^{-1}|P_{\ell}(\alpha_k)|$$
This leads to
$$\Delta_{\mathcal{F}_n}^{2/n} = \prod_{k=1}^n \varphi(\mathcal{F}_n,\alpha_k)^{1/n}  = (\ell+1)2^\ell\binom{2\ell}{\ell}^{-1}\left|\prod_{k=1}^n P_\ell(\alpha_k)\right|^{1/n}$$
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 $\varepsilon > 0$.
