In Géométrie et théorie des groupes : les groupes hyperboliques de Gromov, ch. 8, the authors describe how one can approximate a hyperbolic space that satisfy some finiteness condition with a metric tree. Then they note how this fact can be used to translate finite properties of trees (i.e. metric properties that involve only a finite amount of points, eg. convexity of the distance function, which only involves three points at a time) into a properties of $\delta$-hyperbolic spaces.
I would like to use this lemma to obtain a hyperbolicity criterion for an isometry $g$ of a $\delta$-hyperbolic space $X$, namely '' If there exist a point $x\in X$ s.t. $d(g^2(x),x)>d(g(x),x)+2\delta$ then $g$ is hyperbolic. '' from the analogue criterion for trees '' If there exist a point $x\in X$ s.t. $d(g^2(x),x)>d(g(x),x)$ then $g$ is hyperbolic. ''
I have seen that same lemma applyied to prove a different result, but I dont see how to do it in this case. I would like to have some direction on how to do it, and to understand if this is some kind of procedure that can be done in many cases in more or less the same fashion, or if one has to invent something different every time.
Suppose $g$ has a fixed point $F$. By $\delta$-hyperbolicity, there exist points $Y$ on $Xg(X)$, $M$ on $XF$ and $N$ on $g(X)F$ such that $d(Y,M), d(Y,N) \leq \delta$. By symmetry via $g$, $d(N,g(M))$ is less than $2\delta$.
Now $$d(X,g^2(X)) \leq d(X,Y) + d(Y,N) + d(N,g(M)) + d(g(M),g(Y)) + d(g(Y),g^2(X))$$ $$\leq d(X,Y) + \delta + 2\delta + \delta + d(Y,g(X)) = 4\delta + d(X,g(X)).$$
If your space is sufficiently symmetrical [e.g. by having a symmetry $h$ fixing $F$ and exchanging $X$ and $g(X)$], you can ensure $N = g(M)$ and improve the estimate to $2\delta$. I'm almost certain there's a trick that lets you choose $N = g(M)$ in the less-symmetrical case too.