Let $C$ be a PSD $n\times n$ matrix with unit diagonals $C_{ii}=1,\forall i$. What is a tight lower bound for $\det(C)^{\odot k}$ based on the value of $\det(C)$?
Below, I have some proof that shows the sequence is non-decreasing, bounded, and has a simple limiting case, which suggests a convergence. But it doesn't lead to a convergence rate.
Limit case If $\det(C)=0$ it's easy to show that sequence will simply remain zero. Otherwise, $\det(C)>0$ implies that all off-diagonals of $C$ are strictly smaller than $1.$ Therefore, $C^{\odot k}\to I_n,$ and consequently $\det(C^{\odot k})\to 1$ when $k$ goes to infinity.
Bounded range By GM vs. AM inequality and the fact that $C$ has unit diagonals, we can conclude that $\det(C^{\odot k})\in[0,1]$ $$\det(C^{\odot k})^{1/n}=\left(\prod_i\lambda_i(C^{\odot k})\right)^{1/n}\le\frac1n\sum_i\lambda_i(C^{\odot k}) =\frac1n tr(C^{\odot k}) = 1$$
Monotonicity
Theorem 3.4 here attributed to Schur states that when $A$ and $B$ are positive semidefinite, $$ch_p(A) . b_{min} \le ch_i(A \odot B) \le ch_1(A) b_{max}, \qquad \text{for } i = 1,\dots, p $$ where $ch_j$ denotes the $j$th largest characteristic root, and $b_{min}$ and $b_{max}$ are the smallest and largest diagonal elements of $B$.
By setting $A = C^{\odot k}, B=C$ we have $b_{min}=b_{max}=1$, and characteristic roots are equal to the eigenvalues of $C$ and $C^\odot k$, leading to the conclusion that $$ \lambda_{p}(C^{\odot}) \le \lambda_i(C^{\odot (k+1)}) \le \lambda_1(C^{\odot k}), \qquad i=1,\dots,p $$
In particular, we can conclude that the eigenspectrum of $C^{\odot k}$ is non-decreasing in $k$, therefore, we can conclude that sequence $\det(C^{\odot k})$ is non-decreasing as well.
Question The conclusions above suggest that if $\det(C)>0$, there might be a strong convergence of the determinant sequence to $1$ as $k$ tends to infinity. Is there a way to analyse the convergence rate?