Let $U \subset\mathbb{R}^n$ be an open set, $\mathbb{S}^n$ be the set of all $n\times n$ symmetric real matrices, $A:U\to \mathbb{S}^n$ be a uniformly Lipschitz continuous function.
Suppose $\exists \lambda>0$ s.t. $\forall x\in U, \xi\in\mathbb{R}^n$ $\ ^t\xi A(x)\xi\geq\lambda|\xi|^2$. Then we can define $A^{1/2}:U\to \mathbb{S}^n$, which is the nonnegative sqrt of A at each point.
Then is $A^{1/2}$ also uniformly Lipschitz continuous? If it holds, how to prove it?
If $a,b$ were scalars, $\sqrt a - \sqrt b = (a-b)/(\sqrt a + \sqrt b)$ is the common strategy to prove Lipschitzness of the square root.
Here similarly, if vector $x$ with $\|x\|=1$ is an eigenvector of $\sqrt A - \sqrt B$ with eigenvalue $\mu$ then \begin{align*} x^T(A-B)x &= x^T(\sqrt A - \sqrt B)\sqrt A x + x^T\sqrt B (\sqrt A - \sqrt B) x \\&= \mu x^T(\sqrt A + \sqrt B)x. \end{align*} As $\mu$ can be chosen as $\pm\|\sqrt A - \sqrt B\|_{op}$, this implies $$\|\sqrt A - \sqrt B\|_{op} \le \|A-B\|_{op} \lambda_{\min}(\sqrt A + \sqrt B)^{-1} \le \|A - B\|_{op} /(2\sqrt \lambda) $$ and Lipschitzness with respect to the operator norm if the eigenvalues of $A$ and $B$ belong to $[\lambda,+\infty)$.
On the set of positive definite matrices, the matrix square root function $f:P\mapsto\sqrt{P}$ is differentiable. If, in a suitable orthonormal basis, the eigenvalues of $P$ are $\lambda_1,\lambda_2,\ldots,\lambda_n$, then the derivative of $f$ evaluated at $P$ is given by $Df(P):H\mapsto K$, where $k_{ij}=\frac{h_{ij}}{\sqrt{\lambda_i}+\sqrt{\lambda_j}}$ for every $H\in\mathbb S$. Therefore, when the eigenvalues of $P$ are bounded away from zero, $f$ has a bounded first derivative and hence it is Lipschitz. As compositions of Lipschitz functions are Lipschitz, $f(A(x))=\sqrt{A(x)}$ is also Lipschitz on $U$.
