[proofplan] For each fixed point $x \in M$, we study the real-valued function $t \mapsto t - d_g(x,\rho(t))$. The triangle inequality and the ray property show that this function is monotone increasing and bounded above, so the Busemann limit exists. Each approximating function $x \mapsto t - d_g(x,\rho(t))$ is $1$-Lipschitz by the [reverse triangle inequality](/theorems/2300), and passing to the limit preserves the Lipschitz bound. Finally, evaluating on the ray uses exactly the minimizing identity $d_g(\rho(s),\rho(t)) = t-s$ for $t \ge s$. [/proofplan]

[step:Prove that the defining limit exists at every point]Fix $x \in M$. Define \begin{align*} F_x: [0,\infty) &\to \mathbb{R} \\ t &\mapsto t - d_g(x,\rho(t)). \end{align*} Let $0 \le t_1 \le t_2$. Since $\rho$ is a ray, \begin{align*} d_g(\rho(t_1),\rho(t_2)) = t_2 - t_1. \end{align*} By the triangle inequality for $d_g$, \begin{align*} d_g(x,\rho(t_2)) &\le d_g(x,\rho(t_1)) + d_g(\rho(t_1),\rho(t_2)) \\ &= d_g(x,\rho(t_1)) + t_2 - t_1. \end{align*} Rearranging gives \begin{align*} t_1 - d_g(x,\rho(t_1)) \le t_2 - d_g(x,\rho(t_2)), \end{align*} so $F_x$ is monotone increasing. The reverse triangle inequality applied to the points $x$, $\rho(0)$, and $\rho(t)$ gives \begin{align*} d_g(x,\rho(t)) \ge d_g(\rho(0),\rho(t)) - d_g(x,\rho(0)). \end{align*} Since $\rho$ is a ray, $d_g(\rho(0),\rho(t)) = t$, hence \begin{align*} F_x(t) = t - d_g(x,\rho(t)) \le d_g(x,\rho(0)). \end{align*} Thus $F_x$ is monotone increasing and bounded above in $\mathbb{R}$, so the limit \begin{align*} \lim_{t \to \infty} F_x(t) \end{align*} exists and is finite. Therefore $b_\rho(x)$ is well-defined for every $x \in M$.[/step]

[guided]Fix a point $x \in M$. The definition of $b_\rho(x)$ requires a limit as the parameter point $\rho(t)$ moves out along the ray, so we first prove that the real-valued function being limited has a finite limit. Define \begin{align*} F_x: [0,\infty) &\to \mathbb{R} \\ t &\mapsto t - d_g(x,\rho(t)). \end{align*} The key point is that increasing $t$ moves farther along a minimizing ray. Let $0 \le t_1 \le t_2$. Since $\rho$ is a ray, the segment from $\rho(t_1)$ to $\rho(t_2)$ has distance exactly equal to its parameter length: \begin{align*} d_g(\rho(t_1),\rho(t_2)) = t_2 - t_1. \end{align*} The triangle inequality gives \begin{align*} d_g(x,\rho(t_2)) &\le d_g(x,\rho(t_1)) + d_g(\rho(t_1),\rho(t_2)) \\ &= d_g(x,\rho(t_1)) + t_2 - t_1. \end{align*} After moving terms to the other side, this becomes \begin{align*} t_1 - d_g(x,\rho(t_1)) \le t_2 - d_g(x,\rho(t_2)). \end{align*} Thus $F_x(t_1) \le F_x(t_2)$ whenever $t_1 \le t_2$, so $F_x$ is monotone increasing. A monotone increasing function has a finite limit if it is bounded above. The upper bound comes from the reverse triangle inequality. Applied to $x$, $\rho(0)$, and $\rho(t)$, it gives \begin{align*} d_g(x,\rho(t)) \ge d_g(\rho(0),\rho(t)) - d_g(x,\rho(0)). \end{align*} Because $\rho$ is a ray starting at $\rho(0)$, \begin{align*} d_g(\rho(0),\rho(t)) = t. \end{align*} Therefore \begin{align*} F_x(t) = t - d_g(x,\rho(t)) \le t - \bigl(t - d_g(x,\rho(0))\bigr) = d_g(x,\rho(0)). \end{align*} So $F_x$ is monotone increasing and bounded above by the finite number $d_g(x,\rho(0))$. Hence \begin{align*} \lim_{t \to \infty} F_x(t) \end{align*} exists in $\mathbb{R}$. This proves that $b_\rho(x)$ is well-defined for the chosen point $x$, and since $x \in M$ was arbitrary, $b_\rho$ is defined on all of $M$.[/guided]

[step:Pass the Lipschitz bound from the distance approximants to the limit] For each $t \ge 0$, define \begin{align*} B_t: M &\to \mathbb{R} \\ x &\mapsto t - d_g(x,\rho(t)). \end{align*} Let $x,y \in M$. The reverse triangle inequality for $d_g$ gives \begin{align*} |d_g(x,\rho(t)) - d_g(y,\rho(t))| \le d_g(x,y). \end{align*} Therefore \begin{align*} |B_t(x) - B_t(y)| &= |d_g(y,\rho(t)) - d_g(x,\rho(t))| \\ &\le d_g(x,y). \end{align*} Taking the limit as $t \to \infty$ and using the already established pointwise convergence $B_t(z) \to b_\rho(z)$ for each $z \in M$, we obtain \begin{align*} |b_\rho(x) - b_\rho(y)| \le d_g(x,y). \end{align*} Since $x,y \in M$ were arbitrary, $b_\rho$ is $1$-Lipschitz. [/step]

[step:Evaluate the Busemann function along the ray] Fix $s \ge 0$. For every $t \ge s$, the ray property gives \begin{align*} d_g(\rho(s),\rho(t)) = t-s. \end{align*} Hence, for every $t \ge s$, \begin{align*} t - d_g(\rho(s),\rho(t)) = t - (t-s) = s. \end{align*} Taking the limit as $t \to \infty$ yields \begin{align*} b_\rho(\rho(s)) = s. \end{align*} Because $s \ge 0$ was arbitrary, the identity holds for all $s \ge 0$. This completes the proof. [/step]