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On the wikipedia page of the Nevanlinna-Pick theorem the following claim appears:

Let $\lambda_1,\lambda_2,f(\lambda_1),f(\lambda_2)\in\mathbb{D}$. The matrix $P_{ij}:=\frac{1-f(\lambda_i)\overline{f(\lambda_j)}}{1-\lambda_i\overline{\lambda_j}}$ is positive semidefinite if and only if the following inequality holds: $\frac{|f(\lambda_1)-f(\lambda_2)|}{|\lambda_1-\lambda_2|}\leq \frac{|1-\overline{f(\lambda_1)}f(\lambda_2)|}{|1-\overline{\lambda_1}\lambda_2|}$.

My first question is how to see this equivalence? In particular the Sylvester criterion implies that $P$ is positive semi definite if and only if $\mathrm{det}(P)\geq 0$, which I was hoping would clarify this, but it didn’t.

My second question is- Does a similar equivalence holds for a larger set of points? That is, let $\{\lambda_i\}_{i=1}^n, \{f(\lambda_i)\}_{i=1}^n\subseteq \mathbb{D}$, then the matrix $P_{ij}:=\frac{1-f(\lambda_i)\overline{f(\lambda_j)}}{1-\lambda_i\overline{\lambda_j}}$ is positive semidefinite if and only if for every $i\neq j\in\{1,\ldots, n\}$,$\frac{|f(\lambda_i)-f(\lambda_j)|}{|\lambda_i-\lambda_j|}\leq \frac{|1-\overline{f(\lambda_i)}f(\lambda_j)|}{|1-\overline{\lambda_i}\lambda_j|}$ ?

Many thanks to all the helpers!

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About your first question this is exactly the positivity of the determinant of the Pick matrix. That is because if $\lambda_1,\lambda_2 \in \mathbb{D}$ then \begin{align*} \det(P) & = \begin{bmatrix} \frac{1-|f(\lambda_1)|^2}{1-|\lambda_1|^2} & \frac{1-\overline{f(\lambda_1)}f(\lambda_2)}{1-\overline{\lambda_1}\lambda_2} \\\frac{1-\overline{f(\lambda_2)}f(\lambda_1)}{1-\overline{\lambda_2}\lambda_1} & \frac{1-|f(\lambda_2)|^2}{1-|\lambda_2|^2} \end{bmatrix} \\ & = \frac{1-|f(\lambda_1)|^2}{1-|\lambda_1|^2}\frac{1-|f(\lambda_2)|^2}{1-|\lambda_2|^2} - \Big| \frac{1-\overline{f(\lambda_2)}f(\lambda_1)}{1-\overline{\lambda_2}\lambda_1} \Big|^2 \\ & = \frac{|1-\overline{f(\lambda_1)}f(\lambda_2)|^2-|f(\lambda_1)-f(\lambda_2)|^2}{|1-\overline{\lambda_1}\lambda_2|^2-|\lambda_1-\lambda_2|^2} -\Big| \frac{1-\overline{f(\lambda_2)}f(\lambda_1)}{1-\overline{\lambda_2}\lambda_1} \Big|^2. \end{align*} The key is to use the identity $$ 1-\Big| \frac{a-b}{1-\overline{a}b} \Big|^2 = \frac{(1-|a|^2)(1-|b|^2)}{|1-\overline{a}b|^2}. $$

About your second question, it is not true that if $f$ is a contraction on the finite set then the Pick matrix is positive. (This is asking that all $2\times2$ principle submatrices are positive semidefinite, implying that the matrix is positive semidefinite). For a concrete counterexample take $\lambda_1=0,\lambda_2=-r,\lambda_3=r, f(0)=0, f(-r)=-r, f(r)=r/2$. Then the pick matrix is $$ \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & \frac{1+r^2/2}{1+r^2} \\ 1 & \frac{1+r^2/2}{1+r^2} & \frac{1-r^2/4}{1-r^2} \end{bmatrix} $$ which has determinant $-\frac{r^4}{4(1+r^2)^2} <0 $ but all $2\times2$ principle submatrices have non negative determinant.

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  • $\begingroup$ Thank you! This is a very illuminating example. $\endgroup$ Commented Mar 27, 2024 at 21:06

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