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I'm working through the book Pick Interpolation and Hilbert Function Spaces by Jim Agler and John E. McCarthy.

They define the Schwarz-Pick Lemma as follows: Suppose $ \lambda_1, \lambda_2, w_1, w_2 $ are points in $ \mathbb{D} $ [the unit disk of the complex plane]. Then there exists a holomorphic function $ \phi : \mathbb{D} \to \mathbb{D}$ that maps $\lambda_1$ to $w_1$ and $\lambda_2$ to $w_2$ if and only if $$ \rho(w_1, w_2) \leq \rho(\lambda_1, \lambda_2). $$ Moreover, if equality holds in the above equation, then the function $ \phi $ is unique and is a Mobius transformation.

[$\rho (w_1, w_2)$ is the pseudo-hyperbolic distance $\rho (w_1, w_2) := |\frac{w_1-w_2}{1-\bar{w_1}w_2}|$ ]

Most other sources only state the lemma as being in one direction and so I've found plenty of ways to show that $\phi$ holomorphic implies the inequality holds, but I've not been able to find any proofs of the converse. The closest I've found was a question asked here titled "The converse to Schwarz Pick lemma?", however I'm not familiar with a lot of the mathematics used and can't grasp the concept of the answer given.

Would anyone be able to give me a push in the right direction to prove this direction of the lemma? i.e. that $ \rho(w_1, w_2) \leq \rho(\lambda_1, \lambda_2) \implies$ there exists some holomorphic $\phi$.

Community
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Harry
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  • You cannot hope to prove that any map $\phi:\mathbb D\to\mathbb D$ sending $\lambda_i$ to $w_i$ is holomorphic, since this is obvioulsly not true. The converse part of the lemma says that if $\rho(w_1,w_2)\leq \rho(\lambda_1,\lambda_2)$, then there exists some holomorphic $\phi:\mathbb D\to\mathbb D$ sending $\lambda_i$ to $w_i$. – Etienne Apr 04 '14 at 18:11
  • Ahh yes, thanks for clearing that up, but I'm still having issues trying to find some function that works for phi. – Harry Apr 04 '14 at 18:33

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Hint: You first prove the equality case of your question: if $\rho(z_1, z_2)=\rho(w_1,w_2)$ there there exists a linear-fractional transformation of the unit disk sending $z_i$ to $w_i$, $i=1,2$. If you can prove this, you are half-way towards the general case. For the general case, you again use linear-fractional transformations of the complex plane sending $D$ to itself but not necessarily surjectively. For this, it suffices to assume that $z_1=w_1=0$.

Moishe Kohan
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  • Thanks for the reply, and sorry to be a bother, but I'm still a bit stuck. I've tried substituting $w_i=f(\lambda_i)=\frac{a\lambda_i+b}{c\lambda_i+d}$ into the equation, but I seem to get a horrible expression in the denominator which doesn't help much. Is this the right way of doing it, and I'm making a mistake or missing something? \ The other way I've tried to do it is by using a process called the Schur algorithm, which seems to give a fairly messy expression as well. I'd go into more detail, but if you don't know what the Schur algorithm is, it's probably not the right way to go anyway. – Harry Apr 04 '14 at 23:44
  • No, you do not need any algorithms for this. First, you use the fact that the linear-fractional group acts transitively on $D$. Then you use the fact that the stabilizer of the origin is the group of rotations and that the set of points within $\rho$-distance $r$ from $0$ is just a round circle. – Moishe Kohan Apr 05 '14 at 00:03