$\newcommand{\A}{\mathcal A}\newcommand{\Tr}{\operatorname{tr}}$For $c$ and $C$ such that $0<c<C<\infty$, let $\A_{d;c,C}$ denote the set of all symmetric positive-definite real $d\times d$ matrices $A$ such that $$cI\le A\le CI,$$ where $I$ is the $d\times d$ identity matrix and $A\le B$ for $d\times d$ matrices $A$ and $B$ means that $B-A$ is positive semidefinite. Let $|A|$ denote the determinant of a square matrix $A$.

Proposition 1:For any $A_0$ and $A_1$ in $\A_{d;c,C}$ \begin{equation} |A_1|-|A_0|\le L\|A_1-A_0\|_F\le L\sqrt d\,\|A_1-A_0\|, \end{equation} where $L:=C^d\sqrt d/c$, $\|\cdot\|_F$ is the Frobenius norm, and $\|\cdot\|$ is the spectral norm.

A proof of Proposition 1 will be given at the end of this post.

Question 1:Is Proposition 1 known?

Question 2:Can Proposition 1 be improved?

A correct and complete answer to either one of these questions will be considered a correct and complete answer to this entire post.

*Proof of Proposition 1:* In view of the inequality $\|B\|_F\le\sqrt d\,\|B\|$ for any matrix $B$, it is enough to prove the first inequality in Proposition 1.

Note that the set $\A_{d;c,C}$ is convex. For $A\in\A_{d;c,C}$, \begin{equation} f(A):=\sqrt{|A|}=(2\pi)^{-d/2}\int_{\mathbb R^d}dx\,e^{-x^\top A^{-1}x/2}. \end{equation} Let $X_A$ stand for any zero-mean Gaussian random vector in $\mathbb R^d$ with covariance matrix $A\in\A_{d;c,C}$. Let $\Tr A$ denote the trace of a square matrix $A$. Then the derivative of $f$ at $A$ applied to any $d\times d$ real matrix $D$ is \begin{equation} \begin{aligned} f'(A)(D)&=(2\pi)^{-d/2}\int_{\mathbb R^d}dx\,e^{-x^\top A^{-1}x/2}x^\top A^{-1}DA^{-1}x/2 \\ &=\frac{f(A)}2\,E(X_A^\top A^{-1}DA^{-1}X_A) \\ &=\frac{f(A)}2\,E\Tr(X_AX_A^\top A^{-1}DA^{-1}) \\ &=\frac{f(A)}2\,\Tr(EX_AX_A^\top A^{-1}DA^{-1}) \\ &=\frac{f(A)}2\,\Tr(DA^{-1}) \\ &\le\frac{f(A)}2\,\|D\|_F \|A^{-1}\|_F \\ &\le\frac{f(A)}2\,\|D\|_F \sqrt d\,\|A^{-1}\| \\ &\le\frac{C^{d/2}}2\,\|D\|_F \sqrt d\,/c=K\|D\|_F, \end{aligned} \end{equation} with $K:=\frac{C^{d/2}}2\,\sqrt d\,/c$. So, for any $A_0$ and $A_1$ in $\A_{d;c,C}$ \begin{equation} \sqrt{|A_1|}-\sqrt{|A_0|}=f(A_1)-f(A_0) \le K\|A_1-A_0\|_F, \end{equation} whence \begin{equation} |A_1|-|A_0|=(\sqrt{|A_1|}+\sqrt{|A_0|})(\sqrt{|A_1|}-\sqrt{|A_0|}) \\ \le 2C^{d/2}K\|A_1-A_0\|_F=L\|A_1-A_0\|_F, \end{equation} which proves the first inequality in Proposition 1. $\quad\Box$