Buseman function hyperbolic plane (Y)

Question

Fix $z_0 \in \mathbb{H}$ and $\alpha \in \mathbb{R} \cup \infty$. Show for $q \in H$ $\text{lim}_{q \rightarrow \xi} (d_\mathbb{H}(q,z) -d_\mathbb{H}(q,z_0))$ exists $\forall z \in \mathbb{H}$.

Case: $\xi \neq \infty$

Since we are in the hyperbolic plane, we can assume that $z_0 = i$ and $z = r \cdot i$ for some $r \in \mathbb{R}_{>0}$ by using Möbius maps and the fact that Möbius maps send $\mathbb{R}$ to $\mathbb{R}$.

So far I could show that $\text{lim}_{q \rightarrow \xi'} (d_\mathbb{H}(q,ri) -d_\mathbb{H}(q,i)) = \text{lim}_{q \rightarrow \xi'}(log(\frac{|q+ri|+|q-ri|}{|q+i|+|q-i|})+ log(\frac{|q+ri-|q-ri|}{|q+i|-|q-i|}))$. Everything is clearly okay with $log(\frac{|q+ri|+|q-ri|}{|q+i|+|q-i|})$. But how to proceed with $log(\frac{|q+ri|-|q-ri|}{|q+i|-|q-i|})$?

Can anyone give me a hint? Or is there a better way to solve this problem?

Answer 1

I think what you are asking is the following:

Fix $z_0, z \in \mathbb{H}$ and $\alpha \in \partial\mathbb{H} = \mathbb{R} \cup \{\infty \}$. Show that for $q \in \mathbb{H}$, $\lim_{q \rightarrow \alpha} (d_\mathbb{H}(q,z) > -d_\mathbb{H}(q,z_0))$ exists.

One may take $$ d_{\mathbb{H}}(q,z) = \log \left(\frac{|q-\bar{z}|+|q-z|}{|q-\bar{z}|-|q-z|}\right) $$ for the distance.

Your choice of $z_0=i$ and $z=r \cdot i$ is a great start. The point $q$ tends to $\alpha \in \partial \mathbb{H}$, which can be described as a sequence $q_n=u_n + i v_n$ for $n\in \mathbb{N}$, such that $u_n \to \alpha$ and $v_n \to 0$ as $n\to \infty$. Where you got stuck, you can evaluate the expression with $q_n = u_n + i v_n$.

We calculate \begin{align} \lim_{q \to \alpha} (d_{\mathbb{H}}(q,ri)-d_{\mathbb{H}}(q,i)) &= \lim_{n\to \infty} \left( \log \left(\frac{|q_n+ri|+|q_n-ri|}{|q_n+ri|-|q_n-ri|}\right) - \log \left(\frac{|q_n+i|+|q_n-i|}{|q_n+i|-|q_n-i|}\right) \right) \\ &= \lim_{n\to \infty} \left( \log \left(\frac{|q_n+ri|+|q_n-ri|}{|q_n+i|+|q_n-i|}\right) + \log \left(\frac{|q_n+i|-|q_n-i|}{|q_n+ri|-|q_n-ri|}\right) \right) \\ &= \lim_{n \to \infty} \log \frac{\sqrt{u_n^2+(v_n+r)^2}+\sqrt{u_n^2+(v_n-r)^2}}{\sqrt{u_n^2+(v_n+1)^2}+\sqrt{u_n^2+(v_n-1)^2}} \\&+ \lim_{n \to \infty} \log \frac{\sqrt{u_n^2+(v_n+1)^2}-\sqrt{u_n^2+(v_n-1)^2}}{\sqrt{u_n^2+(v_n+r)^2}-\sqrt{u_n^2+(v_n-r)^2}} \\ &= \log \frac{\sqrt{\alpha^2+r^2}}{\sqrt{\alpha^2+1}} + \lim_{v \to 0} \log \frac{\sqrt{\alpha^2+(v+1)^2}-\sqrt{\alpha^2+(v-1)^2}}{\sqrt{\alpha^2+(v+r)^2}-\sqrt{\alpha^2+(v-r)^2}} . \end{align} and then use l'Hospital's rule as follows in the spoiler:

\begin{align} f(v) &= \sqrt{\alpha^2+(v+r)^2}-\sqrt{\alpha^2+(v-r)^2} ,\\ f'(v) &= \frac{2r}{\sqrt{\alpha^2+(v+r)^2}} , \\ g(v) &= \sqrt{\alpha^2+(v+1)^2}-\sqrt{\alpha^2+(v-1)^2} ,\\ g'(v) &= \frac{2}{\sqrt{\alpha^2+(v+1)^2}} . \end{align} Thus \begin{align} \lim_{v \to 0} &\frac{\sqrt{\alpha^2+(v+1)^2}-\sqrt{\alpha^2+(v-1)^2}}{\sqrt{\alpha^2+(v+r)^2}-\sqrt{\alpha^2+(v-r)^2}} = \lim_{v \to 0}\frac{f(v)}{g(v)} = \lim_{v \to 0}\frac{f'(v)}{g'(v)} \\ &= \lim_{v \to 0} r \frac{\sqrt{\alpha^2+(v+1)^2}}{\sqrt{\alpha^2+(v+r)^2}} = r \frac{\sqrt{\alpha^2+1}}{\sqrt{\alpha^2+r^2}} . \end{align}

We get the result

\begin{align} \lim_{q \to \alpha} (d_{\mathbb{H}}(q,ri)-d_{\mathbb{H}}(q,i)) = \log \frac{\sqrt{\alpha^2+r^2}}{\sqrt{\alpha^2+1}} + \log r \frac{\sqrt{\alpha^2+1}}{\sqrt{\alpha^2+r^2}} =\log (r) . \end{align}