TITLE: Prove that the Log-Euclidean distance is negative-definite
QUESTION [5 upvotes]: Let $\Bbb{S}_{++}^n$ be the $\frac{n(n+1)}{2}$-dimensional Riemannian manifold of the symmetric positive definite (SPD) $n\times n$ real matrices.
The Log-Euclidean distance between two points of $\Bbb{S}_{++}^n$, i.e. between two SPD matrices $A,B\in\Bbb{S}_{++}^n$, is given by
$$
d(A,B)=\lVert\log(A)-\log(B)\rVert_{F},
$$
where $\|\cdot\|_{F}$ denotes the regular Frobenius norm.
We want to prove that $d\colon\Bbb{S}_{++}^n\times\Bbb{S}_{++}^n\to\Bbb{R}$ is a conditionally negative definite function, as defined below. 
For a topological space $\mathcal{X}$, the function $f\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$ is called conditionally negative definite if for any $m\in\Bbb{N}$, $x_1,\ldots,x_m\in\mathcal{X}$, and any real numbers $c_1,\ldots,c_m$ for which $\sum_{i=1}^{m}c_i=0$, the following holds true
$$
\sum_{i,j=1}^{m}c_ic_jf(x_i,x_j)\leq0.
$$
In our case, we want to prove that for any $m\in\Bbb{N}$, $X_1,\ldots,X_m\in\Bbb{S}_{++}^n$, and any real numbers $c_1,\ldots,c_m$ for which $\sum_{i=1}^{m}c_i=0$, the following holds true
$$
\sum_{i,j=1}^{m}c_ic_j\lVert\log(X_i)-\log(X_j)\rVert_{F}\leq0.
$$
Thank you very much in advance!
Edit: In case of the squared Frobenius norm, I think it would be proven more easily, but I still need some help...

REPLY [3 votes]: First, recall that if $\psi$ is negative definite, then $\exp(-\gamma \psi)$ is positive definite for all $\gamma >0$. Now, from Corollary 2.10 of Harmonic analysis on Semigroups (Berg, Christensen, Ressel) we also know that if $\psi$ is negative definite, then $\psi^\alpha$ is also negative definite for $0 < \alpha < 1$ as long as $\psi(x,x) \ge 0$. 
Consider now some map $f: \mathcal{X} \to \mathbb{R}^n$, and set $\psi(x,y) = \|f(x)-f(y)\|^2$. Clearly, $\psi$ is negative definite. 
In particular, choosing $\mathcal{X} = \mathbb{S}^n_{++}$,  $f \equiv \log$, and $\alpha=1/2$, from the above corollary it follows that $\|\log X - \log Y\|_\text{F}$ is negative definite.