Are local quasi-geodesics already quasi-geodesics in hyperbolic spaces? Recall the following definitions
1) A $(\lambda, \varepsilon)$-quasi-isometric embedding $f$ between metric spaces $X$ and $Y$ is a map $X \to Y$ such that
$\frac{1}{\lambda} d_X(x,y) - \varepsilon \leq d_Y(f(x), f(y)) \leq \lambda d_X(x,y) + \varepsilon$
holds for all $x,y \in X$. (of course $\lambda \geq 1$ and $\varepsilon \geq 0$)
2) A $(\lambda, \varepsilon)$-quasi-geodesic in a metric space $X$ is a $(\lambda, \varepsilon)$-quasi-isometric embedding $c: I \to X$. 
3) Let $c: [a,b] \to X$ be a path and $k > 0$ be some constant. Then $c$ is said to be a $k$-local geodesic if $d_X(c(s), c(t)) = |t - s|$ for all $s,t \in [a,b]$ with $|s - t| \leq k$.
4) Define a $k$-local-$(\lambda, \varepsilon)$-quasi-geodesic in the obvious way.
We have the following well known Theorems
T1) For all $\delta > 0, \lambda \geq 1, \varepsilon \geq 0$ there exists a constant $R = R(\delta, \lambda, \varepsilon)$ with the following property:
If $X$ is a $\delta$-hyperbolic geodesic space, $c$ is a $(\lambda, \varepsilon)$-quasi-geodesic in $X$ and $[p,q]$ is some geodesic segment joining the endpoints of $c$, then the Hausdorff distance between $[p,q]$ and the image of $c$ is less than $R$. (Hence there is some constant such that $[p,q]$ is contained in the neighbourhood of $c$ and vice versa)
T2) Let $X$ be a $\delta$-hyperbolic geodesic space and let $c: [a,b] \to X$ be a $k$-local geodesic, where $k > 8\delta$. Then:
(i) im(c) is contained in the $2 \delta$-neighbourhood of any geodesic segment connecting the endpoints of $c$.
(ii) $[c(a),c(b)]$ is contained the $3 \delta$-neighbourhood of im(c)
(iii) $c$ is a $(\lambda, \varepsilon)$-quasi-geodesic, where $\varepsilon = 2 \delta$ and $\lambda = (k + 4 \delta)/(k - 4 \delta)$.
My question is if Theorem 2 (T2) is also true (in an apropriate way) for $k$-local-$(\lambda, \varepsilon)$-quasi-geodesic, i.e. that such local quasi geodesics are actually quasi-geodesics.
In the book of Bridson and Haefliger (Metric spaces of non-positive curvature) this should follow in the 'obvious' way of Theorem 1 (T1) and Theorem 2 (T2) above. However I have sme troubles writing this down explicitly.
EDIT: In the book of Bridson and Haefliger (Metric spaces of non-positive curvature) on p. 407 the following is written
'By combining (1.7) [our T1] and (1.13) [our T2] in the obvious way one gets a criterion for seeing that paths which are locally quasi-geodesic (with suitable parameters) are actually quasi-geodesics.'
This quote is exactly the content of my question.
 A: Let $c:[a,b]\to X$ be a $k$-local $(L,A)$ quasi-geodesic in a uniquely geodesic $\delta$-hyperbolic space $X$. I will prove a local-to-global principle based on the proofs in Bridson-Haefliger, and want to thank Jeff Danciger who assisted me greatly in the proof.
I will need the following lemma, which may be proved by techniques similar to the proof of the theorem.
Lemma. If $k>2L(2D+4\delta+A)$ then $c$ lies within $D+2\delta$ of the geodesic $[c(a),c(b)]$ joining its endpoints.
Theorem. If $k>2L(3D+4\delta+A)$ then $c$ is an $(L',A')$-quasigeodesic.
Proof: Let $x,y,z$ be three points on $c$ such that $x=c(t-k/2)$, $y=c(t)$, and $z=c(t+k/2)$. Let $x',y',z'$ be points on $[c(a),c(b)]$ within $D+2\delta$ of $x,y,z$ respectively. Join $x$ and $z$ by a geodesic, and let $y_0$ be a point on $[x,z]$ within $D$ of $y$. I claim that $y_0$ is within $2\delta$ of a point $y''$ between $x'$ and $z'$.
To see the claim, draw the quadrilateral $(x',x,z,z')$ and cut it into two triangles. By $\delta$-hyperbolicity, $y_0$ lies within $2\delta$ of a point $w$ on an edge of the quadrilateral (beside the one it started on). Assume for the sake of contradiction that $w$ is on $[x,x']$. Then for large $k$ we have that $y_0$ is closer to $x'$ than $x$ because
$$ k/2L-A-D \le d(x,y_0) \le \delta + d(x,w) $$
so
$$ d(y_0,x') - d(x,x') \le \delta+d(w,x')-d(w,x')-d(x,w) \le -k/2L +A+D+2\delta $$
while we also have that $y_0$ is farther from $x'$ than $x$ because
$$ d(y_0,x')-d(x,x') \ge d(y,x) - D -(D+2 \delta)-(D+2 \delta) \ge k/2L-A-3D-4\delta $$
So we get a contradiction. 
Now assume that $y_0$ is within $2\delta$ of a point $w$ on $[z,z']$. Then we have
\begin{align*}
d(y_0,z')-d(z,z') & \le 2 \delta + d(w,z') - d(z,w) - d(w,z') \\
& \le -\frac{k}{2L} + D+3 \delta +A
\end{align*}
is negative for large $k$, while 
$$ d(y_0,z') -d(z,z') \ge d(y,z) - D-(D+2\delta) - (D+2 \delta) \ge \frac{k}{2L}-3D-4\delta-A $$
so its also positive for large $k$, again yielding a contradiction.
We conclude that $y_0$ is within $2\delta$ of a point $y''$ on $[x',z']$. 
Thus $y$ is within $D+2\delta$ of a point $y''$ on the geodesic $[x',z']$ and by $\delta$-hyperbolicity and because $k$ is large, $y'$ is on $[x',z']$ as well. Hence suffiently spaced point projections are monotonic.
Let $t_n=a+nk/2$ and set $x_n=c(t_n)$ and let $x_n'$ be the nearest point projections to $[c(a),c(b)]$. We have 
$$ d(x_n',x_{n+1}') \ge \frac{k}{2L} -A-2D-4\delta $$
and
$$ d(x_0,x_N') \ge N \left(\frac{k}{2L} -A-2D-4\delta \right) .$$
Now let $t,t'\in [a,b]$ and let $t_n,t_n'$ be their nearest points among $\{t_n\}$. For now let's put aside the case when $t'$ is close to $b$ and $b$ is nearly but not quite $a+Nk/2$ for some $N$. Then 
$$ |t-t'|<  2 k/4 + |n-n'|(k/2) $$
$$ d(c(t),c(t_n)) \le kL/4+A $$
and likewise for $t',t_n'$. So
\begin{align*}
d(c(t),c(t'))  & \ge d(c(t_n),c(t_n')) - d(c(t),c(t_n))-d(c(t'),c(t_n')) \\
& \ge |n-n'|(k/2L -A-2D-4\delta) - kL/2 -2A \\
& \ge (|t-t'|-1)(2/k)(k/2L -A-2D-4\delta) - kL/2 -2A \\ 
& \ge |t-t'|(2/k)(k/2L-A-2D-4\delta)- (2/k)(k/2L -A-2D-4\delta)-kL/2-2A \\
& =|t-t'|/L' -A'
\end{align*}
where $L'=(1/L -(2/k)(A+2D+4\delta))^{-1}$ and $A'=(2/k)(k/2L -A-2D-4\delta)+kL/2+2A$. To cover the case we excluded above, we should use something like $A'=3/2L+3kL/4+2A$, or for simplicity you can just use $(3/2)$ times the old bound.  
We conclude that $c$ is a global $(L',A')$-quasigeodesic. Notice that we could have picked any $k_0$ bigger than $2L(3D+4\delta+A)$ and less than $k$, so you have some flexibility in obtaining $L'$ and $A'$. This may let you trade some badness of $L'$ for some goodness of $A'$, if you are interested.
Remark: For my personal recordkeeping, the best $A'$ which follows from my work is $A'=\frac{3}{2L}+\frac{3kL}{4}+2A-\frac{3}{k}(A+2D+4\delta)$
