mjhjmtu

Let $x,y$ be positive real numbers then
$$|sqrtx-sqrty|=dfracsqrtx+sqrty=sqrtcdot dfracsqrtsqrtx+sqrtyleq 1cdot |x-y|^frac12$$

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 $C>0$

$$leftlVert sqrtx-sqrty rightrVert_HS le C leftlVert x-yrightrVert_HS^frac12.$$

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.

Your proposed inequality doesn't even work on diagonal matrices. Let $xinmathbb M_n$ be diagonal with entries $x_igeq 0$. Then $x^1/2$ is diagonal with entries $sqrtx_i$. 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 $Cgeq n^1/4$.

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.

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 $ntimes n$ positive definite matrices. Put e.g., $Y=0$, then asking for a $C>0$ amounts to requiring
beginequation*
texttr,X le C'sqrttexttr,X^2
endequation*
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 $|!|!| cdot |!|!|$
beginequation*
|!|!|X^p - Y^p|!|!| le |!|!| |X-Y|^p |!|!|,quad 0 le p le 1.
endequation*
Thus, in particular for the operator norm, this inequality implies $p$-Hölder continuity (with $C=1$).