I am reading the book by Clara Löh on Geometric Group theory and stumbled on the following exercise:
Let $\delta,D\in\mathbb{R}_{\geq 0}$ and let $(X,d)$ be a $\delta$-hyperbolic space. Let $\gamma:[0,L]\rightarrow X$ and $\gamma':[0,L']\rightarrow X$ be geodesics in $X$ with $$\gamma(0)=\gamma'(0)\quad\text{and}\quad d(\gamma(L),\gamma'(L'))\leq D$$
Then $\gamma,\gamma'$ are uniformly ($2\delta + D$)-close, i.e.:
$$\forall_{t\in [0,\min(L,L')]} \,d(\gamma(t),\gamma'(t))\leq 2\delta +D\quad \text{and}\quad \lvert L-L'\rvert\leq D$$
A geodesic is an isometric embedding $\gamma$ from an interval $[0,L]$ into the space $X$, i.e. such that $d(\gamma(t),\gamma(t'))=\lvert t-t'\rvert$.
A $\delta$-hyperbolic space $(X,d)$ is a space which is geodesic, i.e. between any two points there exists a geodesic and which is $\delta$-slim, i.e. any geodesic triangle (three geodesics each one starting at the end point of the previous one and the final one ending at the start point of the first one) has the following property: for a point that lies on a geodesic of the triangle, there exists another point on another geodesic of the triangle, which is at most a distance $\delta$ away or equivalently, any geodesic of the triangle is contained in a $\delta$-ball around the other two geodesics.
My attempt so far:
Since $X$ is geodesic, we can easily create a geodesic triangle by taking $\gamma$, a geodesic of length at most $D$ starting at $\gamma(L)$ and ending at $\gamma'(L')$ and the inverted geodesic $\tilde{\gamma}'$ of $\gamma'$ for which $\tilde{\gamma}'(t)=\gamma'(L'-t)$. A visualisation is give by the following image:
Now suppose that $\gamma(t)$ lies a distance less than $\delta$ from some $\gamma'(t')$. I then need to show that $\lvert t-t'\rvert\leq \delta+D$. I am also unsure why $\lvert L-L'\rvert\leq D$. But under this assumption of we moreover assume that $\gamma(t),\gamma'(t)$ lie a distance less than $\delta$ from points on this right geodesic, then their distance is easily seen to be less than $2\delta+D$. I am unsure on how to show all the possible cases.
Best Answer
Consider the triangle with the vertices $\gamma(0), \gamma(L), \gamma'(L')$. Since $d(\gamma(L), \gamma'(L'))\le D$, the triangle inequality implies that $$ |L-L'|=|d(\gamma(0), \gamma(L)) - d(\gamma(0), \gamma'(L'))|\le D $$
Other proofs are similar, just triangle inequalities. Suppose that $\gamma(t)$ is within distance $\delta$ from $\gamma'(t')$. Then you consider the triangle with the vertices $\gamma(0), \gamma(t), \gamma'(t')$ and observe that, by the triangle inequality, $$ d(\gamma'(t), \gamma'(t')) = |t'-t|= |d(\gamma(0), \gamma'(t'))- d(\gamma(0), \gamma(t))|\le \delta $$ Hence, applying the triangle inequality one more time: $$ d(\gamma(t), \gamma'(t))\le d(\gamma'(t), \gamma'(t'))+ d(\gamma(t), \gamma'(t'))\le 2\delta. $$
The case when $\gamma(t)$ is within distance $\delta$ from a point in the segment between $\gamma(L), \gamma'(L')$ you already know how to handle.