Uniform convergence under matrix inverse

55 Views Asked by At

Let $A_n\in\mathbb{R}^{d\times d}$ be a sequence of positive definite matrices such that $\sigma_{\text{min}}(A_n) = 1/n$ and $\sigma_{\text{max}}(A_n) = 1$ for all $n\in\mathbb{N}$. Let $X\in\mathbb{R}^{N\times d}$, where $N<d$ be such that $\lambda_{\text{min}}(XX^T)>0$ and $X$ does not depend on $n\in\mathbb{N}$. Consider a sequence of functions $f_{n}: [0,1]\to \mathbb{R}$ defined by $$ f_n (\lambda) = \text{Tr}\big[\big(X(A_n + \lambda I_d)^{-1}X^T\big)^{-1}\big] - \text{Tr}\big[\big(XA_n^{-1}X^T\big)^{-1}\big].$$ It holds that $f_n(\lambda) \to 0$ as $\lambda \to 0$, for all $n\in\mathbb{N}$. Is this convergence uniform in $n\in\mathbb{N}$?

For example, when $N=d$, $X(A_n + \lambda I_d)^{-1} = (A_n + \lambda I_d)^{-1}X$ and $XA_n^{-1} = A_n^{-1}X$, then the convergence is uniform.