Update
I'm trying to show the Corollary, but I have stuck...That is:
For any complex space $X$, we have: $$\begin{align} (1).\ c_X(p,q) &\le d_X(p,q),\ \text{for}\ p, q \in X \\ (2).\ c_X(p,q)&\le \delta_X(p,q) \le d_X(p,q),\ \text{for}\ p, q \in X\end{align}$$
with
- $c_X$: the Carathéodory pseudo - distance: $$c_X=\sup_{f\in \text{Hol}(X,\Delta)} \{\rho \left(f(p),f(q)\right) \}$$
- $d_X$: the Kobayashi pseudo - distance: $$d_X=\inf_{\alpha\in \Omega_{x,y}}\{ \sum_{i=1}^{k}\rho(0,a_i) \},\ \forall a_i \in \Delta=\{z \in \Bbb C:|z|<1\}$$
- $\rho$: Bergman - Poincaré distance: $$\rho_\Delta=\log \left(\dfrac{1+|z|}{1-|z|}\right),\ \forall z \in \Delta$$
- $\delta_X$: is a pseudo - distance on $X$.
I think we can apply Proposition:
I can show that $c_\Delta \le d_\Delta $ But How to prove that $c_X \le d_X $?
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Any help will be appreciated. Thanks!
Let $x, y\in X$. Then, by the definition of $d_C$, for every $\epsilon>0$ we can find $g\in Hol(X, \Delta)$ such that $$ d_C(x,y)-\epsilon< \rho(g(x), g(y))\le d_C(x,y). $$ By definition of $d_X$, for every $\epsilon>0$ we can also find a "holomorphic chain" $$ f_i: \Delta\to X, f_i(p_i)=x_i, i=1,...,n, f_1(p_0)=x, f_n(p_n)=x_n=y, $$ such that $$ d_X(x,y)\le \sum_{i=1}^n \rho(p_{i-1}, p_i) < d_X(x,y)+\epsilon. $$ Set $h_i=g\circ f_i$. Then, by Schwartz lemma, for every $i$, $$ \rho(g(x_{i-1}), g(x_i))= \rho(h_i(p_{i-1}), h(p_i))\le \rho(p_{i-1}, p_i) $$ and in view of the triangle inequalities for the metric $\rho$, $$ \rho(g(x), g(y))\le \sum_{i=1}^n \rho(g(x_{i-1}), g(x_i)) \le \sum_{i=1}^n \rho(p_{i-1}, p_i). $$ Thus, $$ d_C(x,y)-\epsilon< \rho(g(x), g(y))< d_X(x,y)+\epsilon. $$ Since $\epsilon>0$ was arbitrary, $$ d_C(x,y)\le d_X(x,y) $$ as required. qed