# Is this matrix function bounded from above by a norm

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 $$d(A, B) \leq \mathrm{tr}((A^{1/2} - B^{1/2})^2) = \|A^{1/2} - B^{1/2}\|_F^2.$$
• 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. – Keaton Jul 24 at 15:21
• Isn't $d$ just given by $d(A,B) = \|A^{\frac{1}{2}}-B^{\frac{1}{2}}\|_F^2$ ? – Hyperplane Jul 26 at 23:32
• @Hyperplane Could you explain why you think this is the case? – Roberto Rastapopoulos Jul 29 at 16:02
• 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 – Hyperplane Jul 29 at 18:10

If $$A,B$$ are positive semidefinite Hermitian matrices with trace 1 then $$d(A, B) = 2 - 2 \, \textrm{tr}\left((A^{1/2} B A^{1/2})^{1/2}\right)\leq \textrm{tr}\left(((B-A)^2)^{1/2}\right)=\|B-A\|_1.$$
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.