This question is in part related to a [question that I have already posed](https://mathoverflow.net/questions/390293/uniform-continuity-of-cholesky-decomposition).


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.



Operator norm of difference of matrix decompositions
The answer is **negative**, and this happens as soon as $n=2$. The question is whether the composition $X\mapsto L:=L\_{X^2}$ is globally Lipschitz over ${\bf SPD}\_n$. Let $x\_j\in{\mathbb R}^n$ denote the $j$th column of $X$. Then we have the following formulae
$$\ell\_{11}=\|x\_1\|,\quad \ell\_{j1}=\frac{\langle x\_1,x\_j\rangle}{\|x\_1\|}\,,$$
and so on, the expression getting more and more complicated as the column index increases.


The first entry $l\_{11}$ is obviously globally Lipschitz. But this is not the case for the next ones. We have
$$d\ell\_{j1}=\frac{x\_1}{\|x\_1\|}\cdot dx\_j+x\_j\cdot d\frac{x\_1}{\|x\_1\|}.$$
Because the entries $x\_{kj}$ are independent from $x\_1$ if $k\ge2$, the global Lipschitz property can hold only if
$$d\frac{x\_1}{\|x\_1\|}\parallel \vec e\_1,$$
which is not true in general.


**Edit**. The obstacle discussed above occurs at infinity, because there is no *a priori* bound of $x\_{22}$, hence $d\ell\_{21}$ is not uniformly bounded as $x\_{22}\rightarrow+\infty$. But because $X\mapsto L$ is homogeneous of degree $1$, this has the counterpart that the Lipschitz property also fails near the origin. Actually the norm of
$$d\frac{x\_1}{\|x\_1\|}=\frac1{\|x\_1\|}\left(I\_n-\frac{x\_1\otimes x\_1}{\|x\_1\|^2}\right)dx\_1$$
is $1/\|x\_1\|$, which can be arbitrarily large. Thus even if we impose an *a priori* bound on $X$, the map $X\mapsto L$ is not Lipschitzian.


