TITLE: pick interpolation -- why is it symmetric? $\left[\frac{1 - w_i \overline{w_j}}{1 - z_i \overline{z_j}} \right]_{i,j=1}^n \geq 0$
QUESTION [2 upvotes]: I am reading notes on a complex interpolation problem:

Let $z_1, \dots, z_n \in \mathbb{D}$ and $w_1, \dots, w_n \in \mathbb{C}$. There exists (bounded holomorphic?) $f \in H^\infty(\mathbb{D})$ with $||f||_\infty \leq 1$ taking $$ f: z_1 \mapsto w_1 \; \dots \; z_n \mapsto w_n$$
  If and only if the following matrix inequality holds: $$ \left[\frac{1 - w_i \overline{w_j}}{1 - z_i \overline{z_j}} \right]_{i,j=1}^n \geq 0 $$

I am not a complex analyst, but I have been examining interpolation problems of this kind.  Here is the caricature I currently have for this result:

Let $z_1, \dots, z_n \in \mathbb{D}$ and $w_1, \dots, w_n \in \mathbb{C}$. There exists  holomorphic $f \in H^2(\mathbb{D})$ taking $$ f: z_1 \mapsto w_1 \; \dots \; z_n \mapsto w_n$$
  If and only if the the matrix is positive definite: $$ \left[\frac{1 - w_i \overline{w_j}}{1 - z_i \overline{z_j}} \right]_{i,j=1}^n \geq 0 $$

And it may in fact be false when stated this way.  That's issue #1.

I know that the function $ \frac{1}{1 - z \overline{z}}$ is the reproducing kernel for the Hardy space $H^2(\mathbb{D})$ -- the Szego kernel.  I could imagine trying to solve the system of equations:
$$ \begin{array}[ccccc]
 ff(z) &=& a_0 + a_1 z + \dots + a_n z^n + \dots &=& w \\
&\dots & \\
f(z) &=& a_0 + a_1 z + \dots + a_n z^n + \dots &=& w 
 \end{array}$$
and hopefully this matrix occurs somehow.  Instead of generalizing to other kernels, I would like to know why this condition is symmetric in $z$ and $w$, especially since $f(z)$ is not generally invertible but should have many branch points.  
And I don't really understand why positive-definiteness and boundedness are equivalent in the first place.

REPLY [1 votes]: Your "caricature" is trivially true without any condition of positivity.
(That is when you remove the second sentence it becomes true). Just take
the interolation polynomial. Polynomials are evidently in $H^2$. And also in $H^\infty$.
The main condition in the Pick theorem for which positivity is needed is the
condition that $\| f\|_\infty\leq 1$.
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