Kobayashi Pseudo-distance I was reading Kobayashi's book: hyperbolic complex spaces and when he talks about the prototype of his Kobayashi metric, he gave the following definition which he claims that this 'metric' fail to satisfy the triangle's inequality:
Let $X$ be a complex manifold, $p,q\in X$. Let $\text{Hol($D,X$)}$ be the set of holomorphic mappings from the unit disc equipped with Poincare metric $\rho$ to $X$. Now we define 
$$
d_X(p,q)=\inf_{f\in \text{Hol}(D,X)} \rho(a,b)\,\,\,\,\,\,\text{where  $f(a)=p,f(b)=q$}
$$
Does anyone have any counterexample to show this?
 A: I think the standard counterexample is this: $\newcommand{\eps}{\varepsilon}$
Let $\Omega_\eps = \{ (z,w) : |z| < 1, |w| < 1, |zw| < \eps \}$ and let $P = (0,1/2)$, $Q = (1/2, 0)$.
Then (using the mapping $f(\zeta) = (0,\zeta)$), it follows that $d_{\Omega_\eps}(P,0) \le \rho(1/2,0)$. Similarly $d_{\Omega_\eps}(0,Q) \le \rho(1/2,0)$. 
On the other hand, I claim that $d_{\Omega_\eps}(P,Q) \to \infty$ as $\eps \to 0$. Assume that it does not. Then there is a constant $M$ and a sequence $\eps_j \to 0$ and mappings $f_j = (g_j,h_j) : D \to \Omega_j$ with $f_j(0) = (1/2, 0)$ and $f_j(\sigma_j) = (0,1/2)$ such that $\rho(0, \sigma_j) = \tanh^{-1}|\sigma_j| \le M$.
By Montel's theorem, the components $g_j$ and $h_j$ converge locally uniformly (after passing to a subsequence if necessary) to $g$ and $h$ respectively. By passing to a further subseqence, we may also assume that $\sigma_j \to \sigma$ for some $\sigma$ with $|\sigma| < 1$.
Now, $\| g_j h_j \| < \eps_j$, so in fact $g_j h_j$ converges to $0$ locally uniformly. Hence by the identity principle, either $g$ or $h$ is identically $0$. However, $g_j(0) = 1/2$ for all $j$, so $g$ cannot be identically zero. Also $h_j(\sigma_j) = 1/2$, so $h(\sigma) = 1/2$ and $h$ also cannot be identically $0$. 
This contradiction shows that for $\eps$ small enough, $d_{\Omega_\eps}$ does not satisfy the triangle inequality.
