 Solution

SetThe triangle inequality makes $f$ Lipschitz continuous and hence a continuous function. Since $\gamma$ is an isometric embedding,so $f(t)\to \infty$ as $|t|\to \infty$. The function is therefore coercive, and the extreme value theorem on a sufficiently large compact interval gives a minimizing parameter.

Suppose $t_1,t_2$ both minimize $f$, put $p=\gamma (t_1)$ and $q=\gamma (t_2)$, and letThe restriction of $\gamma$ between the two parameters is a geodesic from $p$ to $q$. Let $m$ be its midpoint. In the geodesic triangle with vertices $x,p,q$, the thin geodesic triangle condition gives a point $y$ on one of the other two sides with $d(m,y)\le \delta$. By symmetry suppose $y\in [x,p]$. ThensoBut $m$ lies on $\gamma (\mathbb{R})$ and $p$ is a closest point, so $R\le d(x,m)$. Therefore $L\le 4\delta$, which is stronger than the requiredThis is the closest point on a geodesic line in a hyperbolic metric space estimate.