2
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ Could you explain why the Pick condition "is symmetric in z and w"? $\endgroup$
    – Yemon Choi
    Jun 28, 2014 at 14:20
  • $\begingroup$ the numerator and denominator have the same form. $\endgroup$ Jun 28, 2014 at 17:43
  • 1
    $\begingroup$ Also, as already suggested by the formulae, Blaschke products play a prominent role here, not polynomials. $\endgroup$ Jun 28, 2014 at 19:03
  • 1
    $\begingroup$ @johnmangual Sure, but I would usually interpret "Condition X is symmetric in z and w" to mean "(z,w) satisfies Condition X if and only if (w,z) does". The operation "entrywise reciprocal" doesn't preserve the property of being positive (semi-)definite. $\endgroup$
    – Yemon Choi
    Jun 28, 2014 at 20:55
  • 1
    $\begingroup$ Re: your final question/comment: if you can get access to a copy, Agler and McCarthy's book books.google.co.uk/books/about/… has a good explanation/proof $\endgroup$
    – Yemon Choi
    Jun 28, 2014 at 21:01

1 Answer 1

1
$\begingroup$

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 $\| f\|_\infty\leq 1$.

$\endgroup$
2
  • $\begingroup$ right. Lagrange interpolation gives us at least one polynomial for any collections of $z_1, \dots, z_n$ and $w_1, \dots, w_n$. $\endgroup$ Jun 29, 2014 at 11:39
  • 2
    $\begingroup$ So what is your question, then? $\endgroup$ Jun 29, 2014 at 15:18

Not the answer you're looking for? Browse other questions tagged or ask your own question.