Hg toolkit polygonal interpolation of quasi geodesics

If $\epsilon =0$, set $p=q$: the upper inequality makes it $\lambda$-Lipschitz and continuous, and all approximation errors vanish. If $\epsilon >0$ and the endpoint distance is less than $2\epsilon$, the lower inequality gives $b-a<3\lambda \epsilon$, and the upper inequality gives $d(q(t),q(a))\le \lambda (b-a)+\epsilon <3{\lambda }^2\epsilon +\epsilon$. This includes a one-point domain. Henceforth assume positive $\epsilon$ and separated endpoints. Write $h=\epsilon \mathrm{/}\lambda$. The upper endpoint inequality implies $b-a\ge h>0$.

If $h\le b-a\le 2h$, use a single marked interval $[a,b]$. If $b-a>2h$, put $N=\lfloor (b-a)\mathrm{/}h\rfloor -2\ge 0$, mark $a,a+h,\dots ,a+Nh$, and then $(a+Nh+b)\mathrm{/}2,b$. The last two gaps lie in $[h,3h\mathrm{/}2)$; earlier gaps equal $h$. In particular all gaps are between $h$ and $2h$. Choose one geodesic for each consecutive pair of marked images, and interpolate it at constant speed to define $p$. At a zero image distance use the constant map. Endpoints of adjacent pieces agree. On a marked interval of length $H$, its speed is at most $(\lambda H+\epsilon )\mathrm{/}H\le 2\lambda$. Splitting a parameter interval at its finitely many marks proves the global $2\lambda$-Lipschitz bound and hence continuity.

For any parameter $t$, the closer endpoint $v$ of its marked interval satisfies $\mathrm{\mid }t-v\mathrm{\mid }\le h$ (including the single-interval case). Since $p(v)=q(v)$, the quasi-geodesic upper bound gives $d(q(t),p(v))\le \lambda h+\epsilon =2\epsilon$, while Lipschitz control gives $d(p(t),q(v))\le 2\lambda h=2\epsilon$. These prove the two Hausdorff inclusions individually. Adding the same two bounds gives $d(q(t),p(t))\le 4\epsilon$.

Within one marked interval of length $H$, constant speed gives $d(p(s),p(t))\le (\lambda +\epsilon \mathrm{/}H)\mathrm{\mid }s-t\mathrm{\mid }\le \lambda \mathrm{\mid }s-t\mathrm{\mid }+\epsilon$. In distinct intervals, join $p(s)$ to its right marked endpoint, then to the left marked endpoint for $t$, then to $p(t)$. The middle pair are original $q$ values. Adding their three upper estimates gives $d(p(s),p(t))\le \lambda (t-s)+3\epsilon \le \lambda (t-s)+4\epsilon$ for $s<t$.

For the lower bound, the single-interval case satisfies $t-s\le 2h$, so $(t-s)\mathrm{/}\lambda -4\epsilon \le 0$. In the longer construction, parameters in the same or adjacent marked intervals satisfy $t-s\le 3h$, again making that lower bound nonpositive. For separated intervals write $s\in [u,u+h]$ and $t\in [v,w]$, $u+h<v$, $w-v\le 3h\mathrm{/}2$. The first interval has length exactly $h$: the two exceptional intervals are the final adjacent ones, so neither can be the first of a separated pair. Put $A=(s-u)\mathrm{/}h\in [0,1]$ and $B=(t-v)\mathrm{/}h\in [0,3\mathrm{/}2]$.

In the separated case select marked endpoints $U,V$ as follows. If $A\le 3\mathrm{/}5$ take $U=u$, otherwise take $U=u+h$; if $B\le 3\mathrm{/}5$ take $V=v$, otherwise take $V=w$. The four cases give respectively the following upper bounds for $E=(\mathrm{\mid }s-U\mathrm{\mid }+\mathrm{\mid }t-V\mathrm{\mid })\mathrm{/}h$ and for $J=\max \{0,(t-s)-(V-U)\}\mathrm{/}h$: $(E,J)\le (6\mathrm{/}5,3\mathrm{/}5),(1,1),(3\mathrm{/}2,0),(13\mathrm{/}10,2\mathrm{/}5),$ in the order $(A\le 3\mathrm{/}5,B\le 3\mathrm{/}5)$, $(A>3\mathrm{/}5,B\le 3\mathrm{/}5)$, $(A\le 3\mathrm{/}5,B>3\mathrm{/}5)$, $(A>3\mathrm{/}5,B>3\mathrm{/}5)$. For example the last case has $u+h-s\le 2h\mathrm{/}5$ and $w-t\le 9h\mathrm{/}10$, while $(t-s)-(w-u-h)\le u+h-s\le 2h\mathrm{/}5$. In every case $1+2E+J\le 4$. Since $U<V$, the original lower inequality and the $2\lambda$ Lipschitz bound give $d(p(s),p(t))\ge (V-U)\mathrm{/}\lambda -\epsilon -2\lambda hE\ge (t-s)\mathrm{/}\lambda -\epsilon (1+2E+J\mathrm{/}{\lambda }^2)\ge (t-s)\mathrm{/}\lambda -4\epsilon$.

Combining the lower cases, upper estimate and approximation estimates proves all assertions. Equality $s=t$ is immediate and reversed parameter order follows by symmetry. The construction uses finitely many chosen geodesics, and requires no continuity of the original map.