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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.

The defining formulas imply the matching-distance identitiesFor example,and the other two follow cyclically. Thus $p,q,r$ are the tripod points of a geodesic triangle.

By the thin geodesic triangle condition, $p$ lies within $\delta$ of some point $p^{\prime }$ on $[x,y]$ or $[y,z]$. If $p^{\prime }\in [x,y]$, then the triangle inequality givesSince $q$ is the point of $[x,y]$ at distance $d(x,p)$ from $x$, the geodesic parametrization gives $d(p^{\prime },q)\le \delta$, and hence $d(p,q)\le 2\delta$. If instead $p^{\prime }\in [y,z]$, comparison of distances from $z$ gives $d(p,r)\le 2\delta$.

Applying the same argument cyclically, each of $p,q,r$ lies within $2\delta$ of at least one of the other two. The graph on these three points whose edges join pairs at distance at most $2\delta$ therefore has no isolated vertex, so it is connected. Any two vertices are joined by at most two edges, and the triangle inequality yields