Consider the Hilbert space $H=L^2_w(I)$ as the weighted $L^2$ space, where $I\subseteq\mathbb{R}$: $$ L_w^2(I)=\{\phi:I\rightarrow\mathbb{R}:\,\|\phi\|^2=\int_I \phi(x)^2w(x)\,dx<\infty\}. $$ In particular, let $w$ be a probability density function with support in $I$. For example, $I=\mathbb{R}$ and $w(x)=(1/\sqrt{2\pi})e^{-x^2/2}$, $I=(-1,1)$ and $w(x)=\frac12 1_{(-1,1)}(x)$, etc.
Assume that $x^i\in H$ for all $i\geq0$ (that is, the probability distribution has moments of any order). Consider the linearly independent set $\mathcal{B}=\{\phi_i\}_{i=1}^\infty\subseteq H$ defined as $\phi_i(x)=x^{i-1}/\|x^{i-1}\|$, $i\geq1$. Define $A^p=(\langle \phi_i,\phi_j\rangle )_{1\leq i,j\leq p}$ as the Gram matrix of $\phi_1,\ldots,\phi_p$, which is symmetric and positive definite. Let $\lambda_p>0$ be the minimum eigenvalue of $A^p$, for $p\geq1$. I would like to ask two questions:
The first question is whether $\lim_{p\rightarrow\infty} \lambda_p=0$ at polynomial or exponential rate. I did some numerical experiments with $w(x)=\frac12 1_{(-1,1)}(x)$ and obtained: $\lambda_6=0.000627468$, $\lambda_7=0.000117839$, $\lambda_8=0.0000218272$, $\lambda_9=4.00222\cdot 10^{-6}$, $\lambda_{10}=7.28163\cdot 10^{-7}$, etc. For $w(x)=(1/\sqrt{2\pi})e^{-x^2/2}$, I got: $\lambda_6=0.195204$, $\lambda_7=0.158382$, $\lambda_8=0.10844$, $\lambda_9=0.0852584$, $\lambda_{10}=0.0628251$, etc.
Another question is whether there is a simple lower bound for each $\lambda_p$ in this case, in terms of the entries of $A^p$, but which does not involve $\det(A^p)$. Usually, lower bounds depend on $\det(A^p)$, see for instance here.
