Let $(X, d)$ be an ultrametric space. In particular, X satisfies the strong triangle inequality: for any $x, y, z \in X$, we have $$d(x,y) \leq \max\{d(x,z), d(y,z)\}.$$
I want to show that $X$ satisfies the Gromov four-point condition: $$ (x \cdot y)_w \geq \min\{(x\cdot z)_w, (z\cdot y)_w\} $$
where $x,y,z,w \in X$ and $(x \cdot y)_w = \frac{1}{2} (d(x,w) + d(y,w) – d(x,y)) $.
I feel this should be relatively straightforward, but I have not been successful in using the strong triangle inequality to somehow deduce the four-point condition. I would appreciate any hints.
Best Answer
This is true, but the proof that I know is tedious.
First, observe that if $x, y, z$ are points in an ultrametric space $(X,d)$ then the largest two of the distances $d(x,y), d(y,z), d(z,x)$ are equal.
Now, consider a quadruple $x, y, z, w$ in an ultrametric space $(X,d)$. WLOG,
$$ d(x,y)=a\le d(x,z)=b \le d(x,w)=c. $$ Our first observation then implies that $$ d(y,z)=b, d(z,w)=c, d(w,y)=c. $$ The rest is a straightforward computation of 12 Gromov-products between points in $\{x, y, z, w\}$ and checking the 4-point inequalities. (With 4 exceptions, these will be equalities.) I leave these computations to you.