A square root inequality for symmetric matrices?

Question

In this post all my matrices will be $\mathbb R^{N\times N}$ symmetric positive semi-definite (psd), but I am also interested in the Hermitian case. In particular the square root $A^{\frac 12}$ of a psd matrix $A$ is defined unambiguously via the spectral theorem. Also, I use the conventional Frobenius scalar product and norm $$ \newcommand\abs[1]{\lvert#1\rvert}\newcommand\pair[2]{\langle#1, #2\rangle}\pair A B:=\operatorname{Tr}(A^tB), \qquad \abs A^2:=\pair A A. $$

Question: is the following inequality true $$ \abs{A^{\frac 12}-B^{\frac 12}}^2\leq C_N \abs{A-B}\quad ??? $$ for all psd matrices $A,B$ and a positive constant $C_N$ depending on the dimension only.

For non-negative scalar number (i-e $N=1$) this amounts to asking whether $\abs{\sqrt a-\sqrt b}^2\leq C\abs{a-b}$, which of course is true due to $\abs{\sqrt a-\sqrt b}^2=\abs{\sqrt a-\sqrt b}\times \abs{\sqrt a-\sqrt b}\leq \abs{\sqrt a-\sqrt b} \times \abs{\sqrt a+\sqrt b}=\abs{a-b}$.

$\DeclareMathOperator\diag{diag}$If $A$ and $B$ commute then by simultaneous diagonalisation we can assume that $A=\diag(a_i)$ and $B=\diag(b_i)$, hence from the scalar case $$ \abs{A^\frac 12-B^\frac 12}^2 =\sum\limits_{i=1}^N \abs{\sqrt a_i-\sqrt b_i}^2 \leq \sum\limits_{i=1}^N \abs{a_i-b_i} \leq \sqrt N \left(\sum\limits_{i=1}^N \abs{a_i-b_i}^2\right)^\frac 12=\sqrt N \abs{A-B}. $$

Some hidden convexity seems to be involved, but in the general (non diagonal) case I am embarrassingly not even sure that the statement holds true and I cannot even get started. Since I am pretty sure that this is either blatantly false, or otherwise well-known and referenced, I would like to avoid wasting more time reinventing the wheel than I already have.

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Suvrit's answer to Subadditivity of the square root for matrices and answer to Ratio sum comparison on operators seem to be related but do not quite get me where I want (unless I missed something?)

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Context: this question arises for technical purposes in a problem I'm currently working on, related to the Bures distance between psd matrices, defined as $$ d(A,B)=\min\limits_U \abs{A^\frac 12-B^\frac 12U} $$ (the infimum runs over unitary matrices $UU^t=\operatorname{Id}$).

Answer 1

The classical operator generalization of the scalar inequality $|\sqrt{a}-\sqrt{b}|^2 \leq |a-b|$ is the Powers-Størmer inequality, which involves two different norms : the trace norm $\|X\|_1 = \operatorname{Tr}|X|$ and the Froebenius norm $\|X\|_2 = (\operatorname{Tr}(X^* X))^{\frac 1 2}$, where $|X| = (X^* X)^{\frac 1 2}$ is the usual absolute value of matrices. It says that for all positive matrices $A,B$ (or operators on a Hilbert space), $$ \|\sqrt{A} - \sqrt{B}\|_2^2 \leq \|A-B\|_1.$$

It implies a positive answer to your question with $C_N = \sqrt{N}$, because by Hoelder's inequality, $\|A-B\|_1 \leq \sqrt{N} \|A-B\|_2$. The constant is optimal (take $A=\operatorname{Id},B=0$).

The Powers-Størmer surprisingly does not have a wikipedia page yet, but it probably appears in most textbooks on operator algebras or matrix analysis. The original reference is R. T. Powers, E. Størmer, Free states of canonical anticommutation relations, Commun. Math. Phys. 16, 1-33 (1970).