Let $H$ be a Hilbert space with inner product $\langle,\rangle$. Let $\mathcal{B}=\{\phi_i\}_{i=1}^\infty\subseteq H$ be a linearly independent set with $\|\phi_i\|=1$. Consider the matrices $A^p=(\langle \phi_i,\phi_j\rangle )_{1\leq i,j\leq p}$ of size $p\times p$, for $p\geq1$. Each matrix $A^p$ is symmetric and positive definite, and it corresponds to the Gram matrix of the set $\{\phi_i\}_{i=1}^p$. Let $\lambda_p>0$ be the minimum eigenvalue of $A^p$, for $p\geq1$. My question is whether $\lim_{p\rightarrow\infty} \lambda_p=0$ at polynomial or exponential rate.
In particular, take $H=L^2_w(I)$ as the weighted $L^2$ space, where $I\subseteq\mathbb{R}$: $$ L_w^2(I)=\{\phi:I\rightarrow\mathbb{R}:\,\|\phi\|^2=\int_I \phi(x)^2w(x)\,dx<\infty\}. $$ It is assumed that $x^i\in H$ for all $i\geq0$. 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 $A^p=(\langle \phi_i,\phi_j\rangle )_{1\leq i,j\leq p}$. My question is whether $\lim_{p\rightarrow\infty} \lambda_p=0$ at polynomial or exponential rate.
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 $\det(A^p)$.
This depends on your choice of $(\phi_i)$. If $(\phi_i)$ are orthonormal, then $\lambda_p=1$ for all $p$.
Take $H=l^2$. Define $\phi_1:=e_1$, $\phi_k:=\frac1{\sqrt{1+a_k^2}}e_1 + a_ke_k$. Then the eigenvalues of $$ \pmatrix{ 1 & \frac1{\sqrt{1+a_k^2}} \\ \frac1{\sqrt{1+a_k^2}} & \frac{a_k^2}{\sqrt{1+a_k^2}}} $$ are upper bounds of $\lambda_p$ for $k\le p$. The smallest eigenvalue of this matrix is bounded above by $\frac{a_k^2}{\sqrt{1+a_k^2}}$. This follows from Cauchy eigenvalue interlacing theorem. Hence the decay rate of $\lambda_p$ can be arbitrarily fast.
In fact, constructing basis like $$ \phi_{2k}=e_{2k}, \quad \phi_{2k+1} = \frac1{\sqrt{1+a_k^2}} (e_{2k} + a_k e_{2k+1}), $$ you can prescribe any decay rate of $\lambda_p$.