# Gromov hyperbolic metric spaces are quasi-convex

I'm aware about the fact stated above, but I'm not able to find some references or proofs besides Gromov's Hyperbolic Groups - Essays in Group Theory. I'll state things precisely. I will consider a complete geodesic metric space $(X,d)$. Given $x,y\in X$, we denote any short geodesic joinning $x$ and $y$ by $[x,y]$.

Definition: a complete geodesic metric space $(X,d)$ is said to be Gromov $\delta$-hyperbolic if every geodesic triangle whose vertices are $x_1,x_2,x_3$ satisfy the following condition: the distance between any $x\in [x_i,x_j]$ and $[x_i,x_j]\cup[x_j,x_k]$ is less than $\delta$, where we take index $i,j,k$ mod.3

Definition: A complete geodesic metric space $(X,d)$ is (A,B)-quasi-convex if any two geodesic segments $\alpha,\beta:[0,1]\rightarrow X$ satisfy the following condition: $$d^H(\alpha([0,1]),\beta([0,1]))\leqslant A\max\{d(\alpha(0),\beta(0)),d(\alpha(1),\beta(1))\}+B$$ where $d^H$ is the Hausdorff distance.

Theorem: If $(X,d)$ is a complete geodesic metric space and is $\delta$-hyperbolic, then $(X,d)$ is $(A,B)$-quasi convex for some $A,B$.

Does anyone know some reference for the proof? It seems to be a well-known fact on the context. Any help will be really appreciated.

• Brigson and Haefliger discuss quasiconvexity in their book Metric spaces of nonpositive curvature (pp 460-464), but only in the context of hyperbolic groups. Their definition of quasiconvexity is similar to what you have, but it only has the upper bound. – user147263 Aug 7 '14 at 0:23
• Thanks again for helping. That is what was missing for me to change the statement of the question. – matgaio Aug 11 '14 at 1:56

## 1 Answer

The upper bound, though I have not seen it written in precisely this form, is a quick consequence of the thinness of triangles. Here is something from A Course in Metric Geometry by Burago, Burago, Ivanov:

Lemma 8.4.2 A shortest path $[ab]$ belongs to the $(d(b,c)+\delta)$-neighborhood of a shortest path $[ac]$.

The authors omit the proof, which is easy: $[ab]$ is in the $\delta$-neighborhood of $[ac]\cup [bc]$, and the latter set is in the $d(b,c)$-neighborhood of $[ac]$.

Applying the above lemma twice (moving endpoints of geodesic one at a time), you'll get $$d^H(\alpha([0,1]),\beta([0,1])) \leqslant d(\alpha(0),\beta(0)) + d(\alpha(1),\beta(1)) +2\delta$$ which gives the right hand side with $A=2$, $B=2\delta$.

The left hand side cannot hold as stated: what if $\alpha,\beta$ are the same long geodesic path traveled in different directions?

• Perhaps I'm not figuring out what you mean by "moving endpoints of geodesics one at a time". Could you, if it doesn't bother you, perhaps fill out a little bit more details? Thank you very much again. – matgaio Aug 5 '14 at 20:38
• @matgaio Say, your geodesics are $[ab]$ and $[cd]$. Then you first estimate the Hausdorff distance from $[ab]$ to $[ad]$, and then the distance from $[ad]$ to $[cd]$. Then use the triangle inequality. – user147263 Aug 5 '14 at 20:44
• Ok! Now I've figured out! Thank you again! – matgaio Aug 5 '14 at 20:48
• @matgaio But what about the lower bound? Do you have a form of it that works, perhaps as stated in Gromov's paper? – user147263 Aug 5 '14 at 20:56
• @matgaio Sure, if you bring the question to the form that better matches the answer, the whole thing will make more sense to those reading it in the future. – user147263 Aug 5 '14 at 21:38