Is it true that \begin{equation} \exists L>0, \forall m,n\in\mathbb{N}, \forall A,B \in \mathbb{R}^{n\times m}, \|\sqrt{AA^T}-\sqrt{BB^T}\| \le L \|A-B\|, \end{equation} where $\|\cdot\|$ is the operator norm with respect to Euclidean norms in the domain and in the codomain?
If not, is it true that \begin{equation} \forall m,n\in\mathbb{N}, \exists L>0, \forall A,B \in \mathbb{R}^{n\times m}, \|\sqrt{AA^T}-\sqrt{BB^T}\| \le L \|A-B\|? \end{equation}
If $m = 1 = n$, the previous inequality holds trivially with $L =1$, since for each $A,B \in \mathbb{R}$ we have \begin{equation*} \|\sqrt{AA^T}-\sqrt{BB^T}\|= |\sqrt{A^2}-\sqrt{B^2}| = \big||A|-|B|\big| \le |A-B| = \|A-B\|. \end{equation*} If $m,n \in \mathbb{N}$, we can use:
- $\forall A,B \in \mathbb{R}^{n\times m}, \|AA^T-B B^T\| \le \big(\|A\|+\|B\|\big) \|A-B\|$
- $\forall A,B \in \mathbb{R}^{n\times n}, \big((A^T = A) \land (B^T = B) \land (A,B\succeq0)\big) \implies\big(\lambda_{\operatorname{min}}(\sqrt{A}+\sqrt{B})\|\sqrt A- \sqrt B\| \le \|A-B\|\big)$
to get for example that, for $A,B \in \mathbb{R}^{n\times m}$ such that $\operatorname{rank}(A)=n=\operatorname{rank}(B)$, it holds \begin{equation*} \|\sqrt{AA^T}-\sqrt{BB^T}\| \le \frac{\sigma_{\operatorname{max}}(A)+\sigma_{\operatorname{max}}(B)}{\sigma_{\operatorname{min}}(A)+\sigma_{\operatorname{min}}(B)} \|A-B\|, \end{equation*} where $\lambda_{\operatorname{min}}(\cdot)$ is the minimum eigenvalue, $\sigma_{\operatorname{max}}(\cdot)$ is the maximum singular value and $\sigma_{\operatorname{min}}(\cdot)$ is the minimum singular value.
However, this inequality seems somewhat underwhelming, at least compared to the one we have in the $1$-dimensional case (i.e., the triangle inequality), where $A \mapsto \sqrt{AA^T}$ is globally $1$-Lipschitz. Can we do better in the general case?