How does $\operatorname{Ric} \ge 0$ guarentee the Busemann function is regular in the splitting theorem? Cheeger-Gromoll's famous splitting theorem says

If $(M,g)$ contains a line and $\operatorname{Ric} \ge 0$. Then $(M,g)$ is isometric to a product.

I want to know how does $\operatorname{Ric} \ge 0$ guarentee that the Busemann function is regular, since it's regular then by Morse theory $M$ is homeomorphic to a product.
 A: Let $\gamma: \mathbb{R}\to M$ be a line and let $\beta_{+}$ and $\beta_{-}$ be the Busemann functions associated with the rays $\gamma_{+}(t)=\gamma(t)$ for $t>0$ and $\gamma_{-}(t)=\gamma(-t)$ for $t>0$. In general,  $\beta_{+}$ and $\beta_{-}$ are Lipschitz functions so they are differentiable almost everywhere. But with $\mathrm{Ric}\geq 0$ you can do better, they are actually smooth. By the Laplacian comparison theorem we see that $\Delta\beta_{+}\geq 0$ and similarly $\Delta\beta_{-}\geq 0$. Therefore, the function $\beta_{+}+\beta_{-}$ is subharmonic on $M$. Now observe that by the triangle inequality we get $\beta_{+}+\beta_{-}\leq 0$ with equality achieved on $\gamma$ so the strong maximum principle tells us that $\beta_{+}+\beta_{-}\equiv 0$. In particular, $\Delta (\beta_{+}+\beta_{-})\equiv 0$ and it follows that $\Delta \beta_{+}=\Delta\beta_{-}=0$. Hence, by regularity theory, $\beta_{+}$ and $\beta_{-}$ are smooth. Finally, if you look into the definition of Busemman function and you use that $|\nabla d_p|=1$, you get $|\nabla\beta_{+}|=|\nabla\beta_{-}|=1$. 
Edit: This is basically from the proof found on Chapter 4 of Peter Li's lecture notes.
