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We will always assume that $X$ is a Hadamard manifold, that is, a simply connected complete Riemannian manifold of nonpositive curvature. We consider the topology of uniform convergence on bounded subsets.

Recall again that a function $f\colon X\to\mathbb R$ is convex if for any geodesic $c\colon X\to\mathbb R$ the composition $f\circ c\colon \mathbb R\to\mathbb R$ is convex. A set $B\subset X$ is convex if for any pair $p,q\in B$ the connecting geodesic is in $B$.

Note that the pointwise limit of convex functions is convex.

Example 1 The distance function $d\colon X\times X\to\mathbb R$ is convex.

Example 2 Given $y\in X$, $x\mapsto d_y(x)\eqdef d(y,x)$ is convex and Lipschitz with Lipschitz constant 1. Indeed, $$ d(y,x)-d(y,z)\le d(x,z), \quad d(y,z)-d(y,x)\le d(x,z). $$

Example 3 If $\sigma_1,\sigma_2\colon I\to X$ are geodesics, then $t\in I\mapsto d(\sigma_1(t),\sigma_2(t))$ is convex.

Example 4 If $\sigma_{x,x'},\sigma_{y,y'}\subset X$ are geodesics rays connecting $x$ and $x'$ and $y$ and $y'$, respectively, then $d_H(\sigma_{x,x'},\sigma_{y,y'})\le \max\{d(x,y),d(x',y')\}$.

Example 5 Given $y\in X$, the function $x\mapsto d(x,y)$ is convex and 1-Lipschitz. Hence, given some geodesic $\sigma\colon\bR\to X$, for every $t\ge0$, $$ x\mapsto d(\sigma(t),x)-t $$ is convex. It follows that (using the existence of the limit, shown below) the function $$ x\mapsto h(x)\eqdef\lim_{t\to\infty}d(x,\sigma(t))-t $$ is convex and 1-Lipschitz. For every $x\in X$, consider the geodesic connecting $x$ with $\sigma(t)$, its intersection points with $\partial B_r(x)$ contains exactly two points $x_1^t,x_2^t$ satisfying $d(x,x_1^t)=r=d(x,x_2^t)$ and hence $$ \Big(d(x_1^t,\sigma(t))-t\Big)-\Big(d(x_2^t,\sigma(t))-t\Big) = \Big(d(x,\sigma(t))+r-t\Big)-\Big(d(x,\sigma(t))-r-t\Big) = 2r $$ Taking accumulation points $x_i$ of $x_i^t$ as $t\to\infty$, $i=1,2$, for every $x\in X$ there are points $x_1,x_2\in\partial B_r(x)$ with $$ \lvert h(x_1)-h(x_2)\rvert =2r. $$

Recall that two geodesics $\sigma_1,\sigma_2\colon\bR\to X$ are (forward) asymptotic if $$ d_H(\sigma_1(\bR^+),\sigma_2(\bR^+))<\infty, $$ where $d_H$ denotes the Hausdorff distance. In other words, this defines a relation between geodesic rays. Note that being forward asymptotic is an equivalence relation. The set of equivalence classes of asymptotic geodesics is called sphere at infinity or also boundary at infinity and denoted by $X(\infty)$. Given a geodesic ray $\sigma\colon\bR^+\to X$, denote by $\sigma(\infty)$ the equivalence class it is contained in.

Lemma 1

For every $\xi\in X(\infty)$ and $x\in X$ there is at most one geodesic ray such that $\sigma(0)=x$ and $\sigma(\infty)=\xi$.

Proof.

Recall that $t\mapsto d(\sigma_1(t),\sigma_2(t))$ is convex. Hence, $\sigma_1(0)=\sigma_2(0)$ and $\sigma_1(\infty)=\sigma_2(\infty)$ implies $\sigma_1=\sigma_2$. $\qed$

We will use the following lemma.

Lemma 2 (see [Lemma 2.1, Ballmann])

Given a geodesic ray $\sigma\colon\mathbb R^+\to X$ and $x\in X$, for $n\in\mathbb N$ denote by $\sigma_n\colon[0,d(x,\sigma(n)]\to X$  the geodesic segment from $x$ to $\sigma(n)$. Then for every $R>0$ and $\varepsilon>0$, for $n,m\in\mathbb N$ sufficiently large, it holds

$$ d(\sigma_n(t),\sigma_m(t))<\varepsilon \quad\text{ for all }t\in[0,R]. $$ Moreover, $\sigma_n$ converges to a geodesic ray $\sigma_{x,\xi}\colon\mathbb R^+\to X$ which is asymptotic to $\sigma$: for every $R>0$ and $\varepsilon>0$, for $n$ sufficiently large $$ d(\sigma_{x,\xi}(t),\sigma_n(t))<\varepsilon \quad\text{ for all }t\in[0,R]. $$

The following is now an immediate consequence.

Lemma 3.

For every $\xi\in X(\infty)$ and $x\in X$ there is a unique geodesic ray $\sigma_{x,\xi}\colon\mathbb R^+\to X$ such that $\sigma_{x,\xi}(0)=x$ and $\sigma_{x,\xi}(\infty)=\xi$.

Lemma 4 Let $\sigma\colon\mathbb R\to\infty$ be a geodesic, $y:=\sigma(0)$, and $\xi=\sigma(\infty)$. Then for every $x\in X$ the limit $$ b_y(x,\xi) \eqdef \lim_{t\to\infty}d(x,\sigma(t))-t $$ exists.

Proof. Check that the triangle inequality and $d(y,\sigma(t))=t$ together imply $$ 0 = d(y,\sigma(t))-t \le d(y,x)+d(x,\sigma(t))-t, $$ which implies $$ -d(y,x) \le d(x,\sigma(t))-t. $$ On the other hand, $$ d(x,\sigma(t)) \le d(x,y)+d(y,\sigma(t)) = d(x,y)+t $$ and for $t\le s$ $$\begin{split} d(x,\sigma(s))-s &\le d(x,\sigma(t))+d(\sigma(t),\sigma(s))-s = d(x,\sigma(t))+(s-t)-s \\ &= d(x,\sigma(t)) -t \le d(x,y). \end{split}$$ Hence, the sequence is nonincreasing and bounded from below, hence converging.