Is the determinant of a matrix Lipschitz continuous? I want to know if the determinant of a matrix is Lipschitz continuous or not.
To be precise, does there exist a constant $K$ such that 
$|\det(A)-\det(B)|\leq K||A-B||_F$,
for all matrices $A,B\in \mathcal{C}^{n\times n}$? 
If the answer is no, then what about being Hölder continuous?
Does $|\det(A)-\det(B)|\leq K||A-B||_F^\alpha$ hold for some constant $K$ and $\alpha$?
Can anyone help me on this? Thank you in advance!
 A: In the case $n = 1$, the determinant is the identity, and hence globally Lipschitz continuous.
For $n > 1$, the determinant is not globally $\alpha$-Hölder continuous for any $\alpha \in (0,1]$, since
$$\lvert \det (r\cdot I) - \det (0\cdot I)\rvert = \lvert r^n\rvert = \lVert I\rVert_F^{-1}\cdot\lvert r\rvert^{n-\alpha}\cdot \lVert r\cdot I - 0\rVert_F^\alpha,$$
and $\lvert r\rvert^{n-\alpha}$ is unbounded.
The determinant is however a polynomial in the entries of the matrix, and hence continuously differentiable everywhere, and that implies that it is locally $\alpha$-Hölder continuous for all $\alpha\in (0,1]$, in particular locally Lipschitz continuous.
A: A connection between determinant and Frobenius norm results from the fact that $|\det A|$ is the volume of the spat $A[0,1]^n$ and that this is smaller than the volume of a rectangular box with the same side lengths.
If $A$ has columns $A=(a_1,a_2,..., a_n)$ then 
$$
|\det(A)|
\le 
\|a_1\|\,\|a_2\|\,....\,\|a_n\|
\le 
\left(\frac{\|a_1\|^2+\|a_2\|^2+....+\|a_n\|^2}{n}\right)^{\frac{n}{2}}
=
\left(\frac{\|A\|_F^2}{n}\right)^{\frac{n}{2}}
=\frac{\|A\|_F^n}{n^{n/2}}
$$
