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Given two symmetric, positive definite matrices $A$ and $B$, let $$ d(A, B) = \textrm{tr}(A) + \textrm{tr}(B) - 2 \, \textrm{tr} \, \left((A^{1/2} B A^{1/2})^{1/2}\right). $$ This function coincides with the square of the 2-Wasserstein distance between Gaussians with equal means and with covariance matrices given by $A$ and $B$, respectively. Is $d(\cdot, \cdot)$ bounded from above by a matrix norm? That is to say, is there a constant $C$ such that $$ d(A, B) \leq C \|A - B\| \qquad \forall A, B \in \mathbb R^{n\times n} \, s.p.d., $$ where $\|\cdot\|$ is any matrix norm?

  • In dimension one, this is true: $$ d(A, B) = (\sqrt A - \sqrt B)^2 \leq |A-B|, $$ because $|\sqrt{A} - \sqrt{B}| \leq \sqrt{|A - B|}$ by concavity.

  • More generally, if $A$ and $B$ commute, i.e. if there exists $P$ such that $A = P D_A P^T$ and $B = P D_B P^T$, $$ d(A, B) = \mathrm{tr}(D_A) + \mathrm{tr}(D_B) - 2 \, \mathrm{tr}((D_A D_B)^{1/2}) = \mathrm{tr}(|D_A^{1/2} - D_B^{1/2}|^2) \leq \mathrm{tr} (|D_A - D_B|) \leq n\|A - B\|_2. $$

  • What about the general case? A friend pointed out to me that the Araki–Lieb–Thirring inequality, with $r=1/2$ and $q=1$, could be employed to obtain $$ \mathrm{tr}((A^{1/2}BA^{1/2})^{1/2}) \geq \mathrm{tr}(A^{1/4}B^{1/2}A^{1/4}) = \mathrm{tr}(A^{1/2}B^{1/2}), $$ which implies that $$ d(A, B) \leq \mathrm{tr}((A^{1/2} - B^{1/2})^2) = \|A^{1/2} - B^{1/2}\|_F^2. $$
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    $\begingroup$ Just a comment: the distance can be written $tr(A)+tr(B)-2\|B^\frac12 A^\frac12\|_*$ where $\|\cdot\|_*$ is the nuclear norm. Don't know if that helps. $\endgroup$ – Keaton Jul 24 at 15:21
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    $\begingroup$ Isn't $d$ just given by $d(A,B) = \|A^{\frac{1}{2}}-B^{\frac{1}{2}}\|_F^2$ ? $\endgroup$ – Hyperplane Jul 26 at 23:32
  • $\begingroup$ @Hyperplane Could you explain why you think this is the case? $\endgroup$ – Roberto Rastapopoulos Jul 29 at 16:02
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    $\begingroup$ Apparently it is only the case when $A$ and $B$ commute. Now I think you could build on that by deploying an upper bound on the commutator, cf. the links given in this answer mathoverflow.net/a/50870/107094 $\endgroup$ – Hyperplane Jul 29 at 18:10
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If $A,B$ are positive semidefinite Hermitian matrices with trace 1 then $$ d(A, B) = 2 - 2 \, \textrm{tr}\left((A^{1/2} B A^{1/2})^{1/2}\right)\leq \textrm{tr}\left(((B-A)^2)^{1/2}\right)=\|B-A\|_1.$$

This inequality follows from equation 24 (page 13) and inequality 46 (page 17) of this paper.

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From the answer of this question, and in view of the fact that $$d(A, B) \leq \|A^{1/2} - B^{1/2}\|_F^2$$ by the Araki-Lieb-Thirring inequality, the answer to my question is yes. Quoting from the linked page:

In fact, we can say much more: every $α$-Holder continuous function $F$ is operator Holder continuous ($0<α<1$) on the space of self-adjoint matrices.

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