This question is in part related to a [question that I have already posed][1].

Say I have two symmetric positive definite matrices and their respective Cholesky decompositions $\mathbf{A} = \mathbf{L}_A \mathbf{L}_A^{\top}$ and $\mathbf{B} = \mathbf{L}_B \mathbf{L}_B^{\top}$, where $\mathbf{L}_A$ and $\mathbf{L}_B$ have positive diagonal entries. Furthermore, let $\sqrt{\mathbf{A}}$ and $\sqrt{\mathbf{B}}$ denote the unique symmetric positive definite square roots of $\mathbf{A}$ and $\mathbf{B}$, respectively, i.e., $\sqrt{\mathbf{A}}^{\top} = \sqrt{\mathbf{A}}$ and $\sqrt{\mathbf{A}}\sqrt{\mathbf{A}} = \mathbf{A}$.

I would like to know whether there is an inequality relating the operator norms of $\mathbf{L}_A - \mathbf{L}_B$ and $\sqrt{\mathbf{A}}- \sqrt{\mathbf{B}}$ up to a positive constant $C>0$, i.e.,
\begin{equation}
\max_{\mathbf{x}} \frac{\Vert (\mathbf{L}_A - \mathbf{L}_B)\mathbf{x}\Vert}{\Vert\mathbf{x}\Vert} \leq C \max_{\mathbf{x}} \frac{\Vert(\sqrt{\mathbf{A}} - \sqrt{\mathbf{B}})\mathbf{x}\Vert}{\Vert\mathbf{x}\Vert}.
\end{equation}

Here $\Vert \cdot \Vert$ denotes the Euclidean norm.

**Edit**: As pointed out by Denis Serre, the inequality does not hold in general. However, I can also include the assumption that the entries of $\mathbf{A}$ and $\mathbf{B}$ are bounded (and therefore also their square-roots and Cholesky decompositions). From Denis' answer, it seems to me that Lipschitz continuity from $\sqrt{\mathbf{A}}$ to $\mathbf{L}_A$ would hold, since the derivative of $\mathbf{L}_A$ with respect to the entries of $\sqrt{\mathbf{A}}$ cannot become unbounded.

  [1]: https://mathoverflow.net/questions/390293/uniform-continuity-of-cholesky-decomposition