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.



Hölder continuity for operators
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.


