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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 $f\colon X\to\bR$ is \emph{convex} if for any geodesic $c\colon X\to\bR$ the composition $f\circ c\colon \bR\to\bR$ is convex. A set $B\subset X$ is \emph{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 $d\colon X\times X\to\bR$ is convex.
	
	Given $y\in X$, $x\mapsto d_y(x)\eqdef d(y,x)$ is convex and Lipschitz with Lipschitz constant 1. Indeed,
$$
	d(y,x)-d(y,z)\le d(x,z), \quad
	d(y,z)-d(y,x)\le d(x,z).
$$	



	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.	

	If $\sigma_{x,x'},\sigma_{y,y'}\subset X$ are geodesics rays connecting $x$ and $x'$ and $y$ and $y'$, respectively, then $d_H(\sigma_{x,x'},\sigma_{y,y'})\le \max\{d(x,y),d(x',y')\}$.
	%BBallmann 5.4

	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 \emph{(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 \emph{sphere at infinity} or also \emph{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.

\begin{lemma}[{\cite[Lemma 2.1]{Bal:95}}]\label{lem:rays}
	Given a geodesic ray $\sigma\colon\bR^+\to X$ and $x\in X$, for $n\in\bN$ 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 $n,m\in\bN$ 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 $\sigma_{x,\xi}\colon\bR^+\to X$ which is asymptotic to $\sigma$: for every $R>0$ and $\varepsilon>0$, for $n$ sufficiently large
$$
	d(\sigma_{x,\xi}(t),\sigma_n(t))<\varepsilon
	\quad\text{ for all }t\in[0,R].
$$	
\end{lemma}

The following is now an immediate consequence.

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


%Ballmann Lemma 2.2, topologia em X(infty)

\begin{lemma}
	Let $\sigma\colon\bR\to\infty$ be a geodesic, $y\eqdef\sigma(0)$, 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.
\end{lemma}

\begin{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.
\end{proof}