You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'We use reproducing kernel methods to study various rigidity problems. The methods and setting allow us to also consider the non-positive case.'
address:
- |
    (DA) Department of Mathematics\
    Ben-Gurion University of the Negev\
    POB 653\
    84105 Beer-Sheva, Israel
- |
    (SR) Department of Mathematics\
    The Technion - Israel Institute of Technology\
    32000 Haifa, Israel
- |
    (DS) Department of Mathematics\
    Ort Braude College\
    POB 78, 21982 Karmiel, Israel
author:
- Daniel Alpay
- Simeon Reich
- David Shoikhet
title: 'Rigidity, boundary interpolation and reproducing kernels'
---

Introduction
============

We use reproducing kernel methods to study various rigidity problems. First, recall that a function $s$ is analytic and contractive in the open unit disk if and only if the kernel $$k_s(z,w)=\dfrac{1-s(z)\overline{s(w)}}{1-z\overline{w}}$$ is positive in the open unit disk ${\mathbb D}$. Such functions have various names in the literature. We will use the term [*Schur functions*]{} in the present paper, and will denote by ${\mathcal S}_0$ the family of Schur functions. We note that constants in the closed unit disk, and, in particular, unimodular constants are Schur functions. See [@Alpay01] for a review of Schur functions. Krein and Langer introduced in [@kl1] the class of [*generalized Schur functions*]{} as the set of functions $s$ meromorphic in the open unit disk and such that the kernel $$k_s(z,w)=\dfrac{1-s(z)\overline{s(w)}}{1-z\overline{w}}$$ has a finite number (say, $\kappa$) of negative squares for $z$ and $w$ in the domain of analyticity $\Omega(s)$ of $s$. This means that for every choice of an integer $n$ and of points $w_1,\ldots, w_n$ in $\Omega(s)$, the $n\times n$ hermitian matrix the $(\ell,j)$-th entry of which is equal to $k_s(w_\ell, w_j)$, has at most $\kappa$ strictly negative eigenvalues, and exactly $\kappa$ strictly negative eigenvalues for some choice of $n,w_1,\ldots ,
w_n$.\
We denote by ${\mathcal S}_{\kappa}$ the family of [*generalized Schur functions*]{} with $\kappa$ negative squares, and set $${\rm sq}_-(s)=\kappa.$$ By a result of Krein and Langer [@kl1], a function $s\in{\mathcal S}_\kappa$ if and only if it can be written as $$s(z)=\dfrac{s_0(z)}{b(z)},$$ where $s_0\in{\mathcal S}_0$ and $b$ is a Blaschke product of order $\kappa$, and where, moreover, $s_0$ and $b$ have no common zeros. For example, the function $s(z)=1/z$ belongs to ${\mathcal
S}_1$. See, for instance, [@adrs] for more information on generalized Schur functions. It is well to recall at this point that the positivity of the kernel $k_s(z,w)$ implies the analyticity of $s$. On the other hand, a function $s$ can be such that the kernel $k_s(z,w)$ has a finite number of negative squares in the open unit disk without being meromorphic there, as is illustrated by the example $$s(z)=\begin{cases}0,\quad z\not= 0,\\
                    1,\quad z=0.

                    \end{cases}$$ The corresponding kernel $k_s(z,w)$ has one negative square, but $s$ is not meromorphic in ${\mathbb D}$. See [@adrs p. 82]. Thus, meromorphicity is an essential part of the definition of a generalized Schur function.\
The methods and setting of the present paper allow us to also study rigidity problems in the non-positive case, where Schur functions are replaced by generalized Schur functions.\
We now turn to the outline of the paper. The paper consists of six sections, including this introduction. In Section 2, we review the main results on boundary interpolation which we will need. Section 3 contains Theorem \[Malakoff\], which is the main result of this paper. It is, to the best of our knowledge, the first rigidity theorem where meromorphic functions are considered. We illustrate Theorem \[Malakoff\] in two cases, in Sections 4 and 5: in Section 4 we recover, in particular, a well-known theorem of Burns and Krantz, while in Section 5 we consider an example in the non-positive case. Finally, in Section 6, we consider a case where, in general, there is no rigidity result.

Boundary interpolation problem: a quick review
==============================================

We review in this section some of the results proved in [@ADLW] and which we will use in the sequel. We denote by ${\mathbb T}$ the unit circle and by $z \hat\rightarrow z_1$ the nontangential (or angular) convergence of $z$ to $z_1\in{\mathbb T}$. This means (see, for instance, [@sarason94 p. 47]) that $z$ stays in a region of the form $$\label{julia}
\{ z \in \mathbb{D}: |z-z_1| <  K(|1-|z|)\}$$ for some $K>1$.

\[pb1\] Let $z_1\in{\mathbb T}$, an integer $k\geq 1$, and complex numbers $\tau_0$, $\tau_k$, $\tau_{k+1}, \dots ,
\tau_{2k-1}$ with $|\tau_0|=1$, $\tau_k\ne 0$, be given. Find all generalized Schur functions $s$ such that $$\label{exp}
s(z)=\tau_0+\sum_{i=k}^{2k-1}\tau_i(z-z_1)^i + {\rm
O}((z-z_1)^{2k}), \quad z \hat\rightarrow z_1.$$

Before presenting the solution to this problem we introduce some notations. First, we define the matrices $$\label{Bk}
 T=\begin{pmatrix}
  \tau_k& 0 & \cdots&0 &0\\
\tau_{k+1} & \tau_k & \dots  &0&0  \\
\vdots  &    \vdots  & \ddots&\vdots & \vdots \\
\tau_{2k-2}&\tau_{2k-3}&\cdots&\tau_k&0\\
\tau_{2k-1} &\tau_{2k-2}& \cdots &\tau_{k+1}& \tau_k
\end{pmatrix}$$ and $$\label{bbb}
B=\begin{pmatrix}
0&0&\cdots&              0  &  (-1)^{k-1}\binom{k-1}{0} z_1^{2k-1} \\[2mm]
0     & 0  &\cdots        &(-1)^{k-2}\binom{k-2}{0} z_1^{2k-3}
&  (-1)^{k-1}\binom{k-1}{1} z_1^{2k-2}     \\[2mm]
\vdots  & \vdots          & \vdots        &    \vdots        &   \vdots      \\[2mm]
0      &-\binom10 z_1^3 &\cdots &            (-1)^{k-2}
\binom{k-2}{k-3} z_1^{k} &(-1)^{k-1}\binom{k-1}{k-2} z_1^{k+1}\\[2mm]
 z_1 &-\binom11 z_1^2& \cdots&(-1)^{k-2}\binom{k-2}{k-2} z_1^{k-1}
 &(-1)^{k-1}\binom{k-1}{k-1}
 z_1^k
\end{pmatrix}.$$ We note that $B$ is a right lower triangular matrix, which is invertible because $z_1\not=0$, and since $\tau_k,\,\tau_0\not=0$, the matrix $$\label{P} \mathbb P:=\overline{\tau_0}TB$$ is also invertible. We also define the vector function $$R(z)=\begin{pmatrix} \dfrac{1}{1-z\overline{z_1}} & \dfrac{z}{(1-z\overline{z_1})^2}
 & \ldots
& \dfrac{z^{k-1}}{(1-z\overline{z_1})^k}\end{pmatrix},$$ fix some $z_0\in \mathbb T$, $z_0\not =z_1$, and define the polynomial $$\label{polynomial} p(z)=(1-z\overline{z_1})^k \,R(z)\mathbb P^{-1}
R(z_0)^*.$$ Its degree is at most $k-1$ and $p(z_1)\not =0$. See [@ADLW p. 13] for this last fact.

\[bbip\] Let $z_1 \in \mathbb T$ and $\tau_0, \tau_k, \ldots , \tau_{2k-1}$ be as in Problem $\ref{pb1}$, assume that the matrix $\mathbb P$ in is hermitian, and let $\Theta$ be the $J$–unitary rational matrix function $$\label{theta} \Theta(z)= \begin{pmatrix}
a(z)&b(z)\\c(z)&d(z)\end{pmatrix}= I_2-
\dfrac{(1-z\overline{z_0})p(z)}{(1-z\overline{z_1})^k}
{\mathbf u}
{\mathbf u}^*J
,\quad J=\begin{pmatrix}1&0\\0&-1\end{pmatrix},
\quad \mathbf u=
\begin{pmatrix} 1 \\ \overline{\tau_0} \end{pmatrix},$$ with $p$ defined by and a fixed $z_0 \in
\mathbb T$, $z_0\neq z_1$. Then the fractional-linear transformation $$\label{lft} s(z)=T_{\Theta(z)}(s_1(z))
=\dfrac{a(z)s_1(z)+b(z)}{c(z)s_1(z)+d(z)}$$ establishes a bijective correspondence between all solutions $s$ of Problem $\ref{pb1}$ and all $s_1\in\cup_{\kappa=0}^\infty{\mathcal S}_\kappa$ with the property $$\label{liminf}
\liminf _{z \hat{\rightarrow}
z_1}|s_1(z)-\tau_0|>0.$$ Moreover, if $s$ and $s_1$ are related by , then $$\label{eqq}
{\rm sq}_-(s)={\rm sq}_-(s_1)+{\rm ev}_-({\mathbb P}),$$ where ${\rm ev}_-({\mathbb P})$ denotes the number of strictly negative eigenvalues of ${\mathbb P}$.

This is [@ADLW Theorem 3.2, p. 43]. The proof there is based on the theory of reproducing kernel Pontryagin spaces of the kind introduced by L. de Branges and J. Rovnyak. Having in mind the proofs of the forthcoming theorems, it is well to recall that one can write $$\Theta(z)=\begin{pmatrix}1-\theta(z)&\tau_0\theta(z)\\
-\overline{\tau_0}\theta(z)&1+\theta(z)\end{pmatrix},$$ where $$\theta(z)=\frac{(1-z\overline{z_0})p(z)}{(1-z\overline{z_1})^k}.$$ See [@ADLW (3.18), p. 16].\
We have, in particular (see the formula on the sixth line from the top of page 16 of [@ADLW]), $$c(z)s_1(z)+d(z)=-\overline{\tau_0}(s_1(z)-\tau_0)\theta(z)+1,$$ and hence $$\label{opera_bastille}
c(z)s_1(z)+d(z)
=\frac{(1-z\overline{z_1})^k-\overline{\tau_0}(1-z\overline{z_0})p(z)(s_1(z)-\tau_0)}
{(1-z\overline{z_1})^k}.
$$ This expression also makes clear why property is important. When combined with the fact that $p(z_1)\not=0$, it yields the conclusion that the numerator in stays bounded away from $0$ as $z$ tends nontangentially to $z_1$.\
Finally, we recall that the description of all the solutions of the interpolation problem is independent of the choice of $z_0\in{\mathbb T}\setminus\{z_1\}$. This follows from Theorem \[bbip\] since the problem itself does not depend on $z_0$. For the convenience of the reader we present a direct argument. Let $\Theta$ be the matrix function corresponding to $z_0$ and $\widehat{\Theta}$ the matrix function corresponding to a different point $\widehat{z_0}$ on the unit circle. Then, see [@ADLW (3.22), p. 17], the corresponding functions $\theta$ and $\widehat{\theta}$ are related by $$\theta(z)-\widehat{\theta}(z)=-
\widehat{\theta}(z_0).$$ It follows that $$I_2-\theta(z){\mathbf u} {\mathbf u}^*J=\left(
I_2-\widehat{\theta}(z){\mathbf u} {\mathbf u}^*J\right) \left(
I_2+\widehat{\theta}(z_0){\mathbf u} {\mathbf u}^*J\right),$$ that is, $$\label{la_defense}
\Theta(z)=\widehat{\Theta}(z)U,$$ where $U$ is the $J$–unitary matrix $$U=I_2+\widehat{\theta}(z_0){\mathbf u} {\mathbf u}^*J=
\begin{pmatrix}1+\widehat{\theta}(z_0)&-\tau_0\widehat{\theta}(z_0)\\
\overline{\tau_0}\widehat{\theta}(z_0)&1-\widehat{\theta}(z_0)\end{pmatrix}.$$ Let us use the notation $$T_{\Theta(z)}(s_1(z))=\frac{a(z)s_1(z)+b(z)}{c(z)s_1(z)+d(z)}.$$ From it follows that $$T_{\Theta(z)}(s_1(z))=T_{\widehat{\Theta}(z)}\left(T_U(s_1(z))\right).$$ Hence, to show that the description of the set of solutions does not depend on the normalization point, we have to show that condition is invariant under the linear-fractional transformation defined by $U$, that is, that the function $$\widehat{s_1}(z)=
\frac{(1+\widehat{\theta}(z_0))s_1(z)-\tau_0\widehat{\theta}(z_0)}
{\overline{\tau_0}\widehat{\theta}(z_0)s_1(z)+1-\widehat{\theta}(z_0)}$$ satisfies if and only if $s_1$ does. But this is clear from the equality $$\widehat{s_1}(z)-\tau_0=\frac{s_1(z)-\tau_0}{1+\widehat{\theta}(z_0)\overline{\tau_0}
(s_1(z)-\tau_0)}.$$

A general rigidity theorem
==========================

As we have already mentioned, the theorem to be proved in this section is, to the best of our knowledge, the first rigidity theorem where meromorphic functions are considered. Let us first explain in words this theorem. Consider the boundary interpolation problem \[pb1\], and fix $z_1=1$ to lighten the notation. One shows that any two solutions of this problem, $s$ and $\sigma$, satisfy $$s(z)-\sigma(z)=((z-1)^{2k})g(z),$$ where the function $g$ stays bounded as $z$ tends nontangentially to $1$. Using Julia’s lemma, we then show that for a special choice of $\sigma$, and if $s$ and $\sigma$ satisfy the stronger requirement $$s(z)-\sigma(z)=O((z-1)^{2k+2}),$$ then they must coincide: $s=\sigma$.

\[Malakoff\] Consider the boundary interpolation problem \[pb1\] with $z_1=1$, and assume that the associated matrix ${\mathbb P}$ is hermitian. Let $x\in{\mathbb
T}\setminus\left\{\tau_0\right\}$ and let $b(z)=T_{\Theta(z)}(x)$. Let $s\in{\mathcal S}_{{\rm
ev}_-({\mathbb P})}$ be a solution of Problem \[pb1\] such that $$\label{Luxembourg} s(z)-b(z)=O((z-1)^{2k+2}).$$ Then $$s(z)\equiv b(z).$$

Note that $x\in{\mathbb T}\setminus\left\{\tau_0\right\}$ obviously has property .\
Theorem \[Malakoff\] includes a number of results, and, in particular, the well-known Burns-Krantz rigidity theorem. We note that $b$ is a rational function which takes unitary values on the unit circle, that is, $b$ is a quotient of two finite Blaschke products.\
Before proving Theorem \[Malakoff\], we present a preliminary lemma.

\[Richard-Lenoir\] Let $x\in{\mathbb T}$ and let $\sigma$ be a Schur function such that $$\label{220707}
\sigma(z)-x=O((z-1)^2).$$ Then $\sigma\equiv x$.

[**Proof:**]{} Indeed, by , we have that $$\liminf_{z \hat\rightarrow 1}
\Big|\dfrac{\sigma(z)-x}{1-z}\Big|=0,$$ and hence $$c\,:\,\,=\liminf_{z \hat\rightarrow 1} \dfrac{1-|\sigma(z)|}{1-|z|}=0,$$ when $z$ is restricted to a region of the form . Assume now by contradiction that $\sigma(z)\not \equiv x$. It then follows from Julia’s lemma (see, for instance, [@sarason94 p. 51]) that $$\dfrac{|\sigma(z)-x|^2}{1-|\sigma(z)|^2}\le
c\dfrac{|z-1|^2}{1-|z|^2}=0$$ for any $z\in{\mathbb D}$. This contradicts the hypothesis $\sigma(z)\not \equiv x$.\
[**Proof of Theorem \[Malakoff\]:**]{} Using the formula $$\det (I_p-AB)=\det(I_q-BA),$$ where $A\in{\mathbb C}^{p\times q}$ and $B\in{\mathbb C}^{q\times
p}$, and since $${\mathbf u}^*J{\mathbf u}=0,$$ we obtain from that $$\det
\Theta(z)=1-\dfrac{(1-z\overline{z_0})p(z)}{(1-z\overline{z_1})^k}{\mathbf
u}^*J{\mathbf u}=1.$$ Let $s$ be a solution to Problem \[pb1\], which belongs to ${\mathcal S}_{\nu_-({\mathbb P})}$. There is a Schur function $s_1$ subject to such that $s=T_\Theta(s_1)$. Thus, with $b(z)=T_{\Theta(z)}(x)$, we have $$\begin{split}
s(z)-b(z)&=\dfrac{(\det\Theta(z))(s_1(z)-x)}{(c(z)s_1(z)+d(z))(c(z)x+d(z))}\\
&=\dfrac{s_1(z)-x}{(c(z)s_1(z)+d(z))(c(z)x+d(z))}.
\end{split}$$ Using for $s_1$ and for $x$, we have $$s(z)-b(z)=\dfrac{(1-z)^{2k}(s_1(z)-x)}{
((1-z)^k-\overline{\tau_0}(1-z\overline{z_0})p(z)(s_1(z)-\tau_0))
((1-z)^k-\overline{\tau_0}(1-z\overline{z_0})p(z)(x-\tau_0))}.
$$ Now condition and the fact that $p(z_1)\not =0$ come into play. If holds, then it follows that $$s_1(z)-x=O((z-1)^2),$$ and hence $s_1(z)\equiv x$ by Lemma \[Richard-Lenoir\].\

Recovering a theorem of Burns and Krantz
========================================

\[lala\] Theorem \[Malakoff\] specialized to $k=1, \;
\tau_0=\tau_1=1$, and $x=-1$ reduces to Theorem \[Chatelet\] presented in this section. When the angular convergence is replaced by the unrestricted one, Theorem \[Chatelet\] is due to D. M. Burns and S. G. Krantz [@MR1242454]. For a survey of related results, see [@ELRS].

Assume that a Schur function $s$ satisfies $$s(z)=z+O((1-z)^4),\quad z\hat{\rightarrow} 1,
\label{Crimee, ligne 7}$$ where $\hat{\rightarrow}$ denotes nontangential convergence. Then $s(z)\equiv
z$. \[Chatelet\]

We can recover this result as a special case of Theorem \[Malakoff\]. We present the argument for completeness. To begin with, $s$ is, in particular, a solution of the boundary interpolation problem $$\label{voltaire} s(z)=z+O((1-z)^2),\quad z\hat{\rightarrow} 1.$$ To solve this last problem we use Theorem \[bbip\]. In the notation of the theorem, we have $$z_1=1,\quad k=1\quad{\rm and}\quad\tau_0=\tau_1=1.$$ The matrix ${\mathbb P}$ given by reduces to a number: ${\mathbb P}=\overline{\tau_0}\tau_1z_1=1$.\
To compute the coefficient matrix–function (which will allow us to describe the set of all solutions to ), we need to fix a point $z_0\in{\mathbb T}\setminus\{1\}$. We will choose $z_0=-1$. With this choice of $z_0$, the polynomial $p$ given in [@ADLW (3.2), p. 11], the definition of which is recalled in , is then: $$p(z)=(1-z)\frac{1}{1-z}1^{-1}\frac{1}{2}=\frac{1}{2}.$$ The matrix function $\Theta$ in Theorem \[bbip\] is then equal to $$\begin{split}
\Theta(z)&=\begin{pmatrix}a(z)&b(z)\\
c(z)&d(z)\end{pmatrix}
=\begin{pmatrix}1&0\\0&1\end{pmatrix}-\frac{1+z}{2(1-z)}\begin{pmatrix}1\\1\end{pmatrix}
\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\\
&=\frac{1}{2(1-z)}\begin{pmatrix}1-3z&1+z \\-1-z&3-z
\end{pmatrix}.
\end{split}$$ Thus, Theorem \[bbip\] becomes here:

The fractional-linear transformation $$\label{lumbago1}
s(z)=\dfrac{(1-3z)s_1(z)+1+z}{3-z-(1+z)s_1(z)}$$ gives a one-to-one correspondence between Schur functions satisfying , $$s(z)=z+O((1-z)^2),\quad z\hat{\rightarrow} 1,$$ and Schur functions $s_1$ which satisfy condition .

The choice $x=-1$ corresponds to $s(z)=z$.\
We now compute $s(z)-z$: $$\begin{split}
s(z)-z&=\dfrac{(1-3z)s_1(z)+1+z}{3-z-(1+z)s_1(z)}-z\\
&=\dfrac{(1-3z)s_1(z)+1+z-3z+z^2+(z^2+z)s_1(z)}{3-z-(1+z)s_1(z)}\\
&=(1-z)^2\dfrac{s_1(z)+1}{3-z-(1+z)s_1(z)}.
\end{split}
\label{Corentin Cariou, ligne 7}$$ We intend to show that condition forces $s_1(z)\equiv -1$. It will then follow from that $s(z)\equiv z$.\
Combining and , we obtain $$\dfrac{s_1(z)+1}{3-z-(1+z)s_1(z)}=O((1-z)^2),\quad
z\hat{\rightarrow} 1.$$ But $|3-z-(1+z)s_1(z)|\le 6$ in the open unit disk, and therefore $$|s_1(z)+1|=|3-z-(1+z)s_1(z)|\cdot\left|\dfrac{s_1(z)+1}{3-z-(1+z)s_1(z)}\right|\le
K|1-z|^2,\quad z\hat{\rightarrow} 1,$$ for some positive constant $K$. Thus $$s_1(z)+1=O((1-z)^2),\quad z\hat{\rightarrow} 1.$$ By Lemma \[Richard-Lenoir\], $s_1(z)+1\equiv 0$, and this ends the proof.

An example in the indefinite case
=================================

Let $s$ be a generalized Schur function with one negative square and assume that $$s(z)-\dfrac{1}{z}=O((1-z)^4).$$ Then $$s(z)\equiv\dfrac{1}{z}.$$

[**Proof:**]{} The function $s$ is, in particular, a solution to Problem \[pb1\] with $$\label{lumbago11}
 z_1=1,\quad k=1\quad{\rm
and}\quad\tau_0=1,\quad \tau_1=-1.$$ The matrix ${\mathbb P}$ reduces to a strictly negative number: ${\mathbb P}=\overline{\tau_0}\tau_1z_1=-1$. Thus there are solutions to the interpolation problem in the class of generalized Schur functions ${\mathcal S}_1$. The polynomial $p$ is now equal to $$p(z)=(1-z)\frac{1}{1-z}\frac{1}{(-1)}\frac{1}{2}=-\frac{1}{2},$$ and we have, with $z_0=-1$, $$\Theta(z)=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\frac{1}{2}
\frac{1+z}{1-z}\begin{pmatrix}1\\
1\end{pmatrix}\begin{pmatrix}1&-1\end{pmatrix}.$$ Thus $$2(1-z)\Theta(z)=\begin{pmatrix}2(1-z)+(1+z)&-1-z\\
1+z&2(1-z)-1-z\end{pmatrix} =
\begin{pmatrix}
3-z&-1-z\\1+z&1-3z\end{pmatrix}.$$ By Theorem \[bbip\], a generalized Schur function $s$ with one negative square satisfies the interpolation problem with data if and only if it is of the form $$\label{lumbago33} s(z)=\dfrac{(3-z)s_1(z)-1-z}{1-3z+(1+z)s_1(z)},$$ where the parameter $s_1$ is any Schur function satisfying .\
We now compute the difference $s(z)-\dfrac{1}{z}$:

$$\begin{split}
s(z)-\dfrac{1}{z}&=\dfrac{(3-z)s_1(z)-1-z}{1-3z+(1+z)s_1(z)}-\dfrac{1}{z}\\
&=\dfrac{((3-z)z-1-z)s_1(z)-z^2-z-1+3z}{z(1-3z+(1+z)s_1(z))}\\
&=-(1-z)^2\dfrac{s_1(z)+1}{z(1-3z+(1+z)s_1(z))}.
\end{split}$$ Assume now that $$s(z)-\dfrac{1}{z} =O((1-z)^4).$$ Then $$\dfrac{s_1(z)+1}{z(1-3z+(1+z)s_1(z))}=O((1-z)^2).$$ In particular, $s_1$ satisfies the interpolation condition $$\label{lumbago44}
s_1(1)=-1,\quad s_1^{\prime}(1)=0,$$ and hence, by Lemma \[Richard-Lenoir\], $s_1(z)\equiv -1$. The corresponding $s(z)=\frac{1}{z}$.

The case where $s^\prime (1)=\alpha$
====================================

In this section we replace the condition $s^\prime(1)=1$ (see ) by $s^\prime(1)=\alpha$, where $\alpha\in[0,1)$. Note that if $\alpha=0$, then it follows immediately from Lemma \[Richard-Lenoir\] that $s(z)\equiv 1$. However, the fact that a Schur function $s$ satisfies the interpolation conditions $$s(1)=1,\quad s^\prime(1)=\alpha,\quad s^{(2)}(1)=s^{(3)}(1)=0
\label{lumbago}$$ does not imply that $s(z)=\alpha z+1-\alpha$. Actually, for each $\alpha\in(0,1)$, there exists a sufficiently small $\beta>0$ such that the function $$s(z)=\alpha z+ 1-\alpha+\beta(1-z)^4$$ is still a Schur function. For instance, for $\alpha=1/2$, the following example is given in [@ELRS]: $$s(z)=\frac{1+z}{2}+\frac{(z-1)^4}{20}.
\label{Daumesnil}$$

To study the case $\alpha\in(0,1)$, we first give, as in the previous section, a description of all Schur functions $s$ such that $$\label{parmentier}
s(z)=1+\alpha(z-1)+O((z-1)^4).$$ We now have $$z_1=1,\quad k=1\quad{\rm and}\quad\tau_0=1,\quad \tau_1=\alpha.$$ The matrix ${\mathbb P}$ given in [@ADLW (1.6), p. 3] reduces to a number: ${\mathbb P}=\overline{\tau_0}\tau_1z_1=\alpha>0$. The polynomial $p$ is now equal to $$p(z)=(1-z)\frac{1}{1-z}\frac{1}{\alpha}\frac{1}{2}=\frac{1}{2\alpha},$$ and we have $$\Theta_\alpha(z)=\begin{pmatrix}1&0\\0&1\end{pmatrix}-\frac{1}{2\alpha}
\frac{1+z}{1-z}\begin{pmatrix}1\\
1\end{pmatrix}\begin{pmatrix}1&-1\end{pmatrix}.$$ Thus $$\begin{split}
2\alpha(1-z)\Theta_\alpha(z)&=\begin{pmatrix}2\alpha(1-z)-(1+z)&1+z\\
-(1+z)&2\alpha(1-z)+1+z\end{pmatrix}\\
&=
\begin{pmatrix}
2\alpha-1-z(2\alpha+1)&1+z\\-1-z&2\alpha+1-z(1+2\alpha)\end{pmatrix}.
\end{split}$$ A Schur function $s$ satisfies the interpolation conditions if and only if it is of the form $$\label{lumbago3} s(z)= \dfrac{(2\alpha-1-z(2\alpha+1))s_1(z)+1+z}
{2\alpha+1-z(2\alpha-1)-(1+z)s_1(z)},$$ where, as in , the parameter $s_1$ is any Schur function satisfying .\
The parameter $s_1(z)=1-2\alpha$ satisfies , and corresponds to the solution $$s(z)=\dfrac{(2\alpha-1-z(2\alpha+1))(1-2\alpha)+1+z}
{2\alpha+1-z(2\alpha-1)-(1+z)(1-2\alpha)}=\alpha z+1-\alpha.$$ Furthermore, with $s$ of the form , $$s(z)-(\alpha
z+1-\alpha)=\alpha(z-1)^2\dfrac{s_1(z)+2\alpha-1}{2\alpha+1-z(2\alpha-1)-(1+z)s_1(z)}.$$ Assume that holds. Then $$\dfrac{s_1(z)+2\alpha-1}{2\alpha+1-z(2\alpha-1)-(1+z)s_1(z)}=O((1-z)^2).$$ In particular, $s_1$ satisfies the interpolation conditions $$\label{lumbago4} s_1(1)=1-2\alpha,\quad s_1^{\prime}(1)=0.$$ However, as we have already mentioned above, these conditions do not imply that $s_1(z)\equiv 1-2\alpha$. For instance, for the function defined by , we have $$s_1(z)=\frac{\dfrac{(z-1)^4}{10}}{2z+(1+z)\left(\dfrac{1+z}{2}
+\dfrac{(z-1)^4}{20}\right)},$$ which indeed satisfies with $\alpha = 1/2$, but is not identically equal to $0$.\
At this stage we see the difference between the cases $\alpha=1$ and $\alpha\in(0,1)$. When $\alpha=1$, conditions force $s_1(z)\equiv
-1$. When $\alpha\in(0,1)$, we need to impose more conditions on $s$ in order to force $s_1(z)\equiv 1-2\alpha$. These conditions are spelled out in the following theorem.

Let $\alpha \in (0,1)$, assume that $s$ is a Schur function which admits the representation $$s(z)=\alpha z+1-\alpha+O((1-z)^4)$$ as $z$ tends nontangentially to $1$, and let $s_1$ be defined by . The following assertions are equivalent:\
$(i)$ $s(z)=\alpha z +1-\alpha$.\
$(ii)$ $s_1(z)\equiv 1-2\alpha$.\
$(iii)$ For every $z\in{\mathbb D}$, it holds that $|s_1(z)|\le
|1-2\alpha|$.\
$(iv)$ For every $z\in{\mathbb D}$, it holds that $$|(2\alpha+1-z(2\alpha-1))s(z)-1-z|\le|1-2\alpha|\cdot|
(2\alpha-1-z(2\alpha+1)-((2\alpha+1)s(z)-z(2\alpha-1))|.$$ $(v)$ For every $z\in{\mathbb D}$, it holds that $$\frac{|1-s(z)|^2}{1-|s(z)|^2} <  \frac{\alpha}{1-\alpha}.$$

[**Proof:**]{} It is clear that $$(i)\iff(ii)\Rightarrow (iii)\iff (iv).$$ Assume now that $(iv)$ is in force. If $\alpha =1/2$, then $(ii)$ and $(i)$ are clear. If $\alpha \not=1/2$, consider the function $\sigma
=s_{1}/(1-2\alpha )$. By hypothesis, $\sigma $ is a Schur function. If its value at $z=1$ is $1$ and its angular derivative at $z=1$ is $0$, then $\sigma (z)\equiv 1$ by Lemma \[Richard-Lenoir\], that is, $(ii)$ holds.

It remains to be shown that $(i)\iff (v)$. One direction is clear by a direct computation. To prove the reverse assertion, we consider the holomorphic function $f:{\mathbb{D}}\rightarrow \Pi
_{+}=\{w\in
\mathbb{C}
:\mathrm{Re}~w>0\}$ defined by $$f(z)=C(s(z)),  \label{N5}$$where $C(z)=\frac{1+z}{1-z}$ is the Cayley transform. Calculations show that $f$ admits the representation $$f(z)=\frac{1}{\alpha }\cdot \frac{1+z}{1-z}+\frac{1-\alpha }{\alpha }+r_{f}(z),  \label{N5'}$$where $$r_{f}(z)=O((1-z)^{2}).$$In addition, condition $(v)$ implies that $\mathrm{Re}~f(z)>\frac{1-\alpha }{\alpha }$. So the function $f_{1}$, defined by $$f_{1}(z)=f(z)-\frac{1-\alpha }{\alpha },$$is of positive real part and $$(1-z)f_{1}\left( z\right) \rightarrow \frac{2}{\alpha },$$when $z\hat{\rightarrow}1$. Therefore, if we define a self-mapping $h$ of $\mathbb D $ by $h=C^{-1}\left( f_{1}\right),
$ then we have $$(1-z)\frac{1+h(z)}{1-h(z)}\rightarrow \frac{2}{\alpha },$$when $z\hat{\rightarrow}1,$ which means that $h\left( 1\right) =1$ and $h^{\prime }\left( 1\right) =\alpha .$ Applying now the Julia-Wolff-Carathéodory theorem [@sarason94; @RS] to the function $h$, we get that$$\frac{|1-h(z)|^{2}}{1-|h(z)|^{2}}\leq \alpha
\frac{|1-z|^{2}}{1-|z|^{2}}$$or, equivalently, $$\mathrm{Re}~f_{1}(z)\geq \frac{1}{\alpha
}\mathrm{Re}~\frac{1+z}{1-z},$$ because $$\mathrm{Re}~f_{1}(z) =\frac{1-|h(z)|^{2}}{|1-h(z)|^{2}}\quad{\rm
and}\quad \mathrm{Re}~\frac{1+z}{1-z}=\frac{1-|z|^{2}}{
|1-z|^{2}}.$$ Thus we obtain that $$\mathrm{Re}~r_{f}(z)=\mathrm{Re}~\left( f_{1}(z)-\frac{1}{\alpha }\frac{1+z}{1-z}\right) \geq 0.$$Therefore the function $g$ defined by $$g\left( z\right) =\frac{1-r_{f}(z)}{1+r_{f}(z)}$$is a Schur function. In addition, $g\left(
1\right) =1$ and the angular derivative $g^{\prime }\left( 1\right) =0.$ Therefore Lemma [Richard-Lenoir]{} implies that $$g\left( z\right)  \equiv 1$$or $$r_{f}(z)\equiv 0.$$

Consequently, $f(z)=\frac{1}{\alpha }\cdot \frac{1+z}{1-z}+\frac{1-\alpha }{\alpha },$ or $s\left( z\right) =C^{-1}\left( f\left( z\right)
\right) =\alpha z+1-\alpha $, as asserted.\
**Remark:** Condition $(v)$ has a nice geometric interpretation: the image $s({\mathbb{D}})$ of the open unit disk $\mathbb{D}$ lies inside the horocycle $${D(1, K)=\left\{z\in{\mathbb{D}} \, : \, \frac{|1-z|^2}{1-|z|^2} < K\right\}},$$ of “size" $K=\alpha/(1-\alpha)$, which is internally tangent to the unit circle ${\mathbb{T}}$ at the point $z=1$. See, for example, [@RS p. 121]\
[**Acknowledgments:**]{} the first author thanks the Earl Katz family for endowing the chair which supports his research. The second author was partially supported by the Fund for the Promotion of Research at the Technion and by the Technion President’s Research Fund.

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