$\newcommand{\A}{\mathcal A}\newcommand{\Tr}{\operatorname{tr}}$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. $\quad\Box$



Lipschitz property of the determinant
The best constant is $C^{d-1}\sqrt{d}$.


Write $D(A)=\det A$. We can rephrase the inequality as the claim that $\|D'\|\_F\le L$ for $c\le A\le C$. (It's perhaps best to think of the matrices as long column vectors and the Frobenius norm as the Euclidean norm, and then $D'$ can be viewed as the gradient of $D$.)


By [Jacobi's formula](https://en.wikipedia.org/wiki/Jacobi%27s_formula) for the derivative of a determinant, we have
$$
\frac{\partial D}{\partial a\_{jk}}= D(A)\;\textrm{tr}\: (A^{-1}E\_{jk}) = D(A) (A^{-1})\_{kj} ;
$$
here $E\_{jk}$ is the matrix with a $1$ in the $jk$ slot and zero entries otherwise.


So $\|D'(A)\|\_F= D(A) \|A^{-1}\|\_F$. (This immediately recovers the original inequality since $D\le C^d$, $\|A^{-1}\|^2\_F\le d/c^2$.) To find the optimal constant, we write
$$
D^2\|A^{-1}\|^2\_F= \prod \lambda\_j^2\cdot \sum \lambda\_j^{-2} = \lambda\_2^2\cdots\lambda\_d^2 +\lambda\_1^2\lambda\_3^2\cdots \lambda^2\_d +\ldots +\lambda\_1^2\cdots \lambda^2\_{d-1}
$$
in terms of the eigenvalues of $A$. Clearly this is maximized at $A=C$, with value $C^{d-1}\sqrt{d}$, so this constant works.


It is also optimal as we can confirm by simply taking $A=C$, $B=C-\epsilon$: then $\|A-B\|^2\_F=d\epsilon^2$, $D(A)-D(B)=C^d-(C-\epsilon)^d=dC^{d-1}\epsilon + O(\epsilon^2)$.


(The fact that the derivative does have this value $C^{d-1}\sqrt{d}$ does not immediately rule out still smaller constants since the derivative controls movements in arbitrary directions while we are dealing with positive definite matrices only. However, we can also observe, in more abstract style, that $D'(A)$ is symmetric, so we do stay inside positive definite matrices when following the direction of the largest change of $D(A)$.)


