Hölder continuity for operators Let $x,y$ be positive real numbers then 
$$|\sqrt{x}-\sqrt{y}|=\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}=\sqrt{|x-y|}\cdot \dfrac{\sqrt{|x-y|}}{\sqrt{x}+\sqrt{y}}\leq 1\cdot |x-y|^{\frac{1}{2}}$$
we obtain $1/2$-Hölder continuity for the square-root.
I would like to know if $x,y$ are positive Hilbert-Schmidt operators. Does it follow then that for some $C>0$
$$\left\lVert \sqrt{x}-\sqrt{y} \right\rVert_{HS} \le C \left\lVert x-y\right\rVert_{HS}^{\frac{1}{2}}.$$
Sounds natural, but on the other hand, it is less obvious to me how this should follow.
One remark however is that if it would hold for finite-rank operators, then a density argument yields the claim.
 A: Your proposed inequality doesn't even work on diagonal matrices.  Let $x\in\mathbb M_n$ be diagonal with entries $x_i\geq 0$.  Then $x^{1/2}$ is diagonal with entries $\sqrt{x_i}$.  Thus
$$ \|x^{1/2}\|_{HS}^2 = \sum_i x_i, \qquad
\|x\|_{HS} = \Big( \sum_i x_i^2 \Big)^{1/2}. $$
Taking e.g. $x_i=1/\sqrt n$ we obtain $n/\sqrt n \leq C^2$ so $C\geq n^{1/4}$.
The Powers-Stormer inequality (see section 4 of https://projecteuclid.org/euclid.cmp/1103842028) says that
$$ \|x^{1/2} - y^{1/2}\|_{HS}^2 \leq \|x - y\|_1 $$
for all $x,y$ positive and Hilbert Schmidt.  Here $\|\cdot\|_1$ is the trace-class norm (which takes the value $\infty$ if $x-y$ is not trace-class).
In finite-dimensions, we have that $\|x\|_1 \leq n^{1/2} \|x\|_{HS}$ (reduce to the diagonal case and apply Cauchy-Schwarz) and so we see that $C=n^{1/4}$ works in general.
A: For the Hilbert-Schmidt norm, the inequality $\|X^{1/2}-Y^{1/2}\|_2 \le C\|X-Y\|_2^{1/2}$ is false in general. Consider for that the case of $n\times n$ positive definite matrices. Put e.g., $Y=0$, then asking for a $C>0$ amounts to requiring
\begin{equation*}
\text{tr}\,X \le C'\sqrt{\text{tr}\,X^2}
\end{equation*}
which in general cannot hold for a constant independent of the dimension of $X$.
However, for all positive operators $X$ and $Y$, and for any unitarily invariant norm $|\!|\!| \cdot |\!|\!|$
\begin{equation*}
|\!|\!|X^p - Y^p|\!|\!| \le |\!|\!| |X-Y|^p |\!|\!|,\quad 0 \le p \le 1.
\end{equation*}
Thus, in particular for the operator norm, this inequality implies $p$-Hölder continuity (with $C=1$). 
