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: 'It is well known that the solutions of a (relaxed) commutant lifting problem can be described via a linear fractional representation of the Redheffer type. The coefficients of such Redheffer representations are analytic operator-valued functions defined on the unit disc ${{\mathbb D}}$ of the complex plane. In this paper we consider the converse question. Given a Redheffer representation, necessary and sufficient conditions on the coefficients are obtained guaranteeing the representation to appear in the description of the solutions to some relaxed commutant lifting problem. In addition, a result concerning a form of non-uniqueness appearing in the Redheffer representations under consideration and an harmonic maximal principle, generalizing a result of A. Biswas, are proved. The latter two results can be stated both on the relaxed commutant lifting as well as on the Redheffer representation level.'
address: |
    Department of Mathematics\
    Utrecht University\
    Budapestlaan 6\
    3584 CD Utrecht\
    The Netherlands
author:
- 'S. ter Horst'
title: Redheffer representations and relaxed commutant lifting
---

Introduction {#S:intro}
============

Linear fractional representations have been used for describing solutions to metric constrained interpolation problems since the work of Adamyan-Arov-Kreĭn [@AAK68; @AAK71], and later appeared in the encompassing commutant lifting theory of Sz.-Nagy-Foiaş [@NF68] and D. Sarason [@S67]; cf., [@FF90; @FFGK98]. They also play an important role in linear system theory [@ZDG96]. The topic of the present paper is the class of linear fractional representation that appears in the context of the description of the solutions to the relaxed commutant lifting problem obtained in [@tH07]. Such representations are called Redheffer representations, after the work of R. Redheffer [@R60; @R62].

Before we can introduce this class of representations, we require some definitions and notations. Let ${{\mathcal U}}$ and ${{\mathcal Y}}$ be Hilbert spaces. With the symbol ${{\mathbf H}}^\infty({{\mathcal U}},{{\mathcal Y}})$ we denote the set of uniformly bounded analytic functions on the open unit disc ${{\mathbb D}}$ with values in ${{\mathscr L}}({{\mathcal U}},{{\mathcal Y}})$. Here, as usual, we write ${{\mathscr L}}({{\mathcal U}},{{\mathcal Y}})$ for the space of operators from ${{\mathcal U}}$ into ${{\mathcal Y}}$. By definition, an operator is assumed to be linear and bounded. The set ${{\mathbf H}}^\infty({{\mathcal U}},{{\mathcal Y}})$ is a Banach space with respect to the supremum norm $\|\ \|_\infty$; its closed unit ball is denoted by ${{\mathbf S}}({{\mathcal U}},{{\mathcal Y}})$ and called the [*Schur class*]{} associated with ${{\mathcal U}}$ and ${{\mathcal Y}}$. Functions in ${{\mathbf S}}({{\mathcal U}},{{\mathcal Y}})$ are called [*Schur class functions*]{}. It is well known that a function $F$ in ${{\mathbf H}}^\infty({{\mathcal U}},{{\mathcal Y}})$ defines a multiplication operator $M_F$ between the Hardy spaces $H^2({{\mathcal U}})$ and $H^2({{\mathcal Y}})$, and that $\|F\|_\infty$ is equal to the operator norm of $M_F$ in ${{\mathscr L}}(H^2({{\mathcal U}}),H^2({{\mathcal Y}}))$. We further define ${{\mathbf H}}^2({{\mathcal U}},{{\mathcal Y}})$ to be the set of analytic functions $H$ on ${{\mathbb D}}$ whose values are in ${{\mathscr L}}({{\mathcal U}},{{\mathcal Y}})$ with the property that the formula $$\label{GaH}
({\Gamma}_H u)({\lambda})=H({\lambda})u\quad(u\in{{\mathcal U}},{\lambda}\in{{\mathbb D}})$$ defines an operator from ${{\mathcal U}}$ into the Hardy space $H^2({{\mathcal Y}})$. The set ${{\mathbf H}}^2({{\mathcal U}},{{\mathcal Y}})$ is a Banach space with norm $\|H\|=\|{\Gamma}_{H}\|$, i.e., the norm of $H\in{{\mathbf H}}^2({{\mathcal U}},{{\mathcal Y}})$ is equal to the operator norm of the associated operator ${\Gamma}_H$. We write ${{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal Y}})$ for the closed unit ball of ${{\mathbf H}}^2({{\mathcal U}},{{\mathcal Y}})$. Note that, conversely, any operator ${\Gamma}\in{{\mathscr L}}({{\mathcal U}},H^2({{\mathcal Y}}))$ defines a function $H\in{{\mathbf H}}^2({{\mathcal U}},{{\mathcal Y}})$ via the same identity $H({\lambda})u=({\Gamma}u)({\lambda})$.

Now let ${{\mathcal U}}$, ${{\mathcal Y}}$, ${{\mathcal E}}$ and ${{\mathcal E}}'$ be Hilbert spaces and $\Psi$ an operator-valued function on ${{\mathbb D}}$ that decomposes as $\Psi={\left[\begin{smallmatrix} \Psi_{1,1}&\Psi_{1,2}\\ \Psi_{2,1}&\Psi_{2,2}\end{smallmatrix}\right]}$ so that $\Psi_{1,1}(0)=0$ and $$\label{RedCoefs}
\Psi_{1,1}\in{{\mathbf S}}({{\mathcal E}}',{{\mathcal E}}),\ \
\Psi_{1,2}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal E}}),\ \
\Psi_{2,1}\in{{\mathbf S}}({{\mathcal E}}',{{\mathcal Y}}),\ \
\Psi_{2,2}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal Y}}),$$ The [*Redheffer representation*]{} associated with $\Psi$ is the map $V\mapsto{{\mathscr R}}_\Psi[V]$ given by $$\label{RedTrans}
{{\mathscr R}}_\Psi[V]({\lambda})=
\Psi_{2,2}({\lambda})+\Psi_{2,1}({\lambda})V({\lambda})
(I-\Psi_{1,1}({\lambda})V({\lambda}))^{-1}\Psi_{1,2}({\lambda})$$ and defined for $V\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$. The functions $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ are referred to as the associated [*Redheffer coefficients*]{}. In addition, we shall assume the [*coefficient matrix*]{} $$\label{RedMat}
{\ensuremath{\left[\begin{array}{cc}M_{\Psi_{1,1}}&{\Gamma}_{\Psi_{1,2}}\\M_{\Psi_{2,1}}&{\Gamma}_{\Psi_{2,2}}\end{array} \right]}}
{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal E}}')\\{{\mathcal U}}\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal E}})\\H^2({{\mathcal Y}})\end{array} \right]}}$$ to be a contraction, or, more often a co-isometry. It is then the case that the Redheffer representation ${{\mathscr R}}_\Psi$ maps ${{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ into ${{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal Y}})$ (see Proposition \[P:Redheffer\]). Note that this includes the more conventional class of Redheffer representations where the function $\Psi$ is a Schur class function itself, in which case ${{\mathcal R}}_\Psi$ maps ${{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ into ${{\mathbf S}}({{\mathcal U}},{{\mathcal Y}})$.

The relaxed commutant lifting problem was introduced in [@FFK02]; the developed theory extends the classical commutant lifting theory [@NF68], as well as the Treil-Volberg commutant lifting setting [@TV94] and the weighted version of [@BFF99]. See [@DBF] for an application in filterbank design. In [@FFK02] a particular (central) solution is given; while descriptions of all solutions are obtained in [@LT06; @FtHK06a; @FtHK06b; @tH07; @tH1]. The starting point for the relaxed commutant lifting problem is a [*lifting data set*]{} ${\{A,T',U',R,Q\}}$ consisting of five Hilbert space operators: the operator $A$ is a contraction mapping ${{\mathcal H}}$ into ${{{\mathcal H}}}'$, the operator $U'$ on ${{\mathcal K}}'$ is a minimal isometric lifting of the contraction $T'$ on ${{\mathcal H}}'$, i.e., $U'$ is an isometry on ${{\mathcal K}}'$ with ${{\mathcal H}}'\subset{{\mathcal K}}'$ being cyclic for $U'$ and $\Pi_{{{\mathcal H}}'}U'=T'\Pi_{{{\mathcal H}}'}$, and $R$ and $Q$ are operators from ${{\mathcal H}}_0$ to ${{\mathcal H}}$, satisfying $$\label{intertw}
T'AR=AQ{\quad\mbox{and}\quad}R^*R\leq Q^*Q.$$ Given this data set the [*relaxed commutant lifting problem*]{} is to describe a (all) contraction(s) $B$ from ${{\mathcal H}}$ to ${{\mathcal K}}'$ such that $$\label{rclt}
\Pi_{{{\mathcal H}}'}B=A{\quad\mbox{and}\quad}U'BR=BQ.$$ Here we follow the convention that for a subspace ${{\mathcal V}}$ of a Hilbert space ${{\mathcal W}}$ the symbol $\Pi_{{\mathcal V}}$ stands for the orthogonal projection from ${{\mathcal W}}$ onto ${{\mathcal V}}$ viewed as an operator from ${{\mathcal W}}$ onto ${{\mathcal V}}$, whereas $P_{{\mathcal V}}$ is used for the orthogonal projection from ${{\mathcal W}}$ onto ${{\mathcal V}}$ as an operator from ${{\mathcal W}}$ into ${{\mathcal W}}$. A contraction $B$ from ${{\mathcal H}}$ into ${{\mathcal K}}'$ that satisfies is called a [*contractive interpolant for ${\{A,T',U',R,Q\}}$*]{}. The Treil-Volberg version appears when $R$ is the identity operator on ${{\mathcal H}}$, and thus ${{\mathcal H}}_0={{\mathcal H}}$; for classical commutant lifting it is assumed in addition that $Q$ is an isometry.

Without loss of generality we may, and will, assume that the isometric lifting $U'$ in ${\{A,T',U',R,Q\}}$ is the Sz.-Nagy Schäffer isometric lifting of $T'$, i.e., $$U'={\ensuremath{\left[\begin{array}{cc}T'&0\\E_{{{\mathcal D}}_{T'}}D_{T'}&S_{{{\mathcal D}}_{T'}}\end{array} \right]}}{\mbox{ on }}{\ensuremath{\left[\begin{array}{c}{{\mathcal H}}'\\H^2({{\mathcal D}}_{T'})\end{array} \right]}},$$ where we use the general notation $S_{{\mathcal U}}$ for the forward shift on the Hardy space $H^2({{\mathcal U}})$, and $E_{{{\mathcal U}}}$ for the canonical embedding of ${{\mathcal U}}$ into the subspace of constant functions in $H^2({{\mathcal U}})$, that is, $(E_{{\mathcal U}}u)({\lambda})=u$ for all ${\lambda}\in{{\mathbb D}}=\{{\lambda}\in{{\mathbb C}}\colon |{\lambda}|<1\}$ and $u\in{{\mathcal U}}$. Furthermore, as usual, given a contraction $N$, we write $D_N$ for the defect operator and ${{\mathcal D}}_N$ for the defect space of $N$, that is, $D_N$ is the positive square root of $I-N^*N$ and ${{\mathcal D}}_N$ is the closure of the range of $D_N$.

An initial step in the process of describing the solutions to a commutant lifting problem is the extraction of an operator from the data. In the case of the relaxed commutant lifting problem we obtain a contraction ${\omega}$ defined by $$\label{om}
{\omega}:{{\mathcal F}}={{\overline{D_AQ{{\mathcal H}}_0}}}\to{\ensuremath{\left[\begin{array}{c}{{\mathcal D}}_{T'}\\{{\mathcal D}}_A\end{array} \right]}},\quad {\omega}D_AQ={\ensuremath{\left[\begin{array}{c}D_{T'}AR\\D_AR\\\end{array} \right]}},$$ which we refer to as the [*underlying contraction*]{} of ${\{A,T',U',R,Q\}}$ (see [@FFK02]). We write ${\omega}_1$ for the component of ${\omega}$ that maps ${{\mathcal F}}$ into ${{\mathcal D}}_{T'}$ and ${\omega}_2$ for the component of ${\omega}$ that maps ${{\mathcal F}}$ into ${{\mathcal D}}_{A}$. In the classical commutant lifting setting, as well as in many of the examples considered in [@FFK02], we have $R^*R=Q^*Q$, which is equivalent to ${\omega}$ being an isometry.

Once the underlying contraction ${\omega}$ is obtained from the data, we set ${{\mathcal G}}={{\mathcal D}}_A\ominus{{\mathcal F}}$ and write $\Pi_{{\mathcal F}}$ and $\Pi_{{\mathcal G}}$ for the orthogonal projections in ${{\mathscr L}}({{\mathcal D}}_A,{{\mathcal F}})$ and ${{\mathscr L}}({{\mathcal D}}_A,{{\mathcal G}})$, respectively. The symbols $\Pi_{{{\mathcal D}}_A}$ and $\Pi_{{{\mathcal D}}_{T'}}$ will be used for the orthogonal projections from ${{\mathcal D}}_A\oplus{{\mathcal D}}_{T'}$ onto ${{\mathcal D}}_A$, respectively ${{\mathcal D}}_{T'}$. Next we define operator-valued functions $\Phi_{1,1}$, $\Phi_{1,2}$, $\Phi_{2,1}$ and $\Phi_{2,2}$ on ${{\mathbb D}}$ by $$\label{RCLcoefs}
\begin{array}{rl}
&\Phi_{1,1}({\lambda})={\lambda}\Pi_{{\mathcal G}}(I_{{{\mathcal D}}_A}-{\lambda}{\omega}_2\Pi_{{\mathcal F}})^{-1}
\Pi_{{{\mathcal D}}_A} D_{{\omega}^*},\\[.1cm]
&\Phi_{1,2}({\lambda})=\Pi_{{\mathcal G}}(I_{{{\mathcal D}}_A}-{\lambda}{\omega}_2\Pi_{{\mathcal F}})^{-1},\\[.1cm]
&\Phi_{2,1}({\lambda})=\Pi_{{{\mathcal D}}_{T'}} D_{{\omega}^*}+{\lambda}{\omega}_1\Pi_{{\mathcal F}}(I_{{{\mathcal D}}_A}-{\lambda}{\omega}_2\Pi_{{\mathcal F}})^{-1}\Pi_{{{\mathcal D}}_A} D_{{\omega}^*},\\[.1cm]
&\Phi_{2,2}({\lambda})={\omega}_1\Pi_{{\mathcal F}}(I_{{{\mathcal D}}_A}-{\lambda}{\omega}_2\Pi_{{\mathcal F}})^{-1}.
\end{array}\quad({\lambda}\in{{\mathbb D}})$$ and put $$\label{PhiRCL}
\Phi={\ensuremath{\left[\begin{array}{cc}\Phi_{1,1},&\Phi_{1,2}\\\Phi_{2,1}&\Phi_{2,2}\end{array} \right]}}.$$ The set of all contractive interpolants is then described by the following theorem; see Theorem 5.1.1 in [@tH07].

\[T:RCLRedheffer\] Let ${\{A,T',U',R,Q\}}$ be a lifting data set with underlying contraction ${\omega}$. Set ${{\mathcal G}}={{\mathcal D}}_A\ominus D_AQ{{\mathcal H}}_0$ and define $\Phi_{1,1}$, $\Phi_{1,2}$, $\Phi_{2,1}$ and $\Phi_{2,2}$ by . Then $$\Phi_{1,1}\!\in{{\mathbf S}}({{\mathcal D}}_{{\omega}^*},{{\mathcal G}}),\
\Phi_{1,2}\!\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal D}}_A,{{\mathcal G}}),\
\Phi_{2,1}\!\in{{\mathbf S}}({{\mathcal D}}_{{\omega}^*},{{\mathcal D}}_{T'}),\
\Phi_{2,2}\!\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal D}}_A,{{\mathcal D}}_{T'}),$$ $\Phi_{1,1}(0)=0$ and the coefficient matrix $$\label{CoefMat}
K_0={\ensuremath{\left[\begin{array}{cc}M_{\Phi_{1,1}}&{\Gamma}_{\Phi_{1,2}}\\M_{\Phi_{2,1}}&{\Gamma}_{\Phi_{2,2}}\end{array} \right]}}
:{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal D}}_{{\omega}^*})\\{{\mathcal D}}_A\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{cc}H^2({{\mathcal G}})\\ H^2({{\mathcal D}}_{{\omega}^*})\end{array} \right]}}$$ is a co-isometry. Moreover, set $H_V={{\mathscr R}}_\Phi[V]\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal D}}_A,{{\mathcal D}}_{T'})$ for any $V$ in ${{\mathbf S}}({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ and define $$\label{RCLsols2}
B_V={\ensuremath{\left[\begin{array}{c}A\\{\Gamma}_{H_V} D_A\end{array} \right]}}:{{\mathcal H}}\to{\ensuremath{\left[\begin{array}{c}{{\mathcal H}}'\\ H^2({{\mathcal D}}_{T'})\end{array} \right]}}.$$ Then $B_V$ is a contractive interpolant for ${\{A,T',U',R,Q\}}$, and all contractive interpolants are obtained in this way. Finally, $K_0$ is unitary if and only if ${\omega}$ is an isometry and ${\omega}_2\Pi_{{\mathcal F}}$ on ${{\mathcal D}}_A$ is strongly stable (i.e., $\lim_{n\to\infty}({\omega}_2\Pi_{{\mathcal F}})^n u=0$ for each $u\in{{\mathcal D}}_A$).

When we take for $V$ the constant function whose value is the zero operator, we see that a particular contractive interpolant is obtained by taking $H_V=\Phi_{2,2}$ in ; this contractive interpolant is called the [*central contractive interpolant*]{} and was already obtained in [@FFK02].

One point where the Redheffer representation of Theorem \[T:RCLRedheffer\] is distinctively different from the one in the classical commutant lifting setting is that the map ${{\mathscr R}}_\Phi$ is, in general, not one-to-one, and thus the same holds true for the map $V\mapsto B_V$. This property is inherited from the Schur representation of [@FtHK06b] (from which the Redheffer representation of Theorem \[T:RCLRedheffer\] is deduced in [@tH07]) where the same phenomenon occurs. We return to this issue later in the introduction.

Theorem \[T:RCLRedheffer\] raises the question how the set of Redheffer representations ${{\mathscr R}}_\Phi$ with $\Phi$ obtained from a relaxed commutant lifting problem is situated in the set of all Redheffer representations ${{\mathscr R}}_\Psi$ with coefficients as in . We further restrict the set of Redheffer representations ${{\mathscr R}}_\Psi$ by demanding that $\Psi_{1,1}(0)=0$ and that the coefficient matrix is a co-isometry, as this is a natural restriction based on the result of Theorem \[T:RCLRedheffer\].

To be precise, we will consider the following question: Given a Redheffer representation ${{\mathcal R}}_\Psi$ with coefficients as a in , with $\Psi_{1,1}(0)=0$ and a co-isometry, when does there exist a lifting data set ${\{A,T',U',R,Q\}}$ so that the set of all contractive interpolants for the associated relaxed commutant lifting problem is given by $$\label{VtoB}
\left\{ B_V={\ensuremath{\left[\begin{array}{c}A\\{\Gamma}_{H_V} D_A\end{array} \right]}}\mid H_V={{\mathscr R}}_\Psi[V]\text{ for some }V\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}') \right\}.$$ If ${\{A,T',U',R,Q\}}$ has this property, we say it is a lifting data set [*associated with the Redheffer representation ${{\mathscr R}}_\Psi$*]{}. Note that the lifting data set ${\{A,T',U',R,Q\}}$ should satisfy ${{\mathcal D}}_A={{\mathcal U}}$ and ${{\mathcal D}}_{T'}={{\mathcal Y}}$.

There is a different kind of inverse problem for interpolation and commutant lifting problems that goes back to the work of Adamjan-Arov-Kreĭn [@AAK68; @AAK71] on the Nehari problem, and has since been considered in various settings [@A89; @Arov90; @K95a; @K95b; @VY; @BK]. In its original form, for the Nehari problem, this inverse problem takes the form of the question which $\gamma$-generating pairs are also Nehari pairs. An extension to the classical commutant lifting setting is posed and solved by J.A. Ball and A. Kheifets in [@BK]. For such problems the data of the considered problem is known, and a Redheffer representation is given for which it is also known that its range forms a subset of the set of all solutions for the problem defined by the given data. The inverse problem is then to determine if the solutions generated by the Redheffer transformation are in fact all solutions. In the present paper, the inverse problem is to construct a data set so that the given Redheffer representation provides all solutions. As a consequence, the solution criteria we find here looks quite different from the one obtained in [@BK].

Somewhat surprisingly, it turns out that for the problem considered here, it is always possible to find an associated lifting data set.

\[T:inverse\] Let ${{\mathscr R}}_\Psi$ be a Redheffer representation with coefficients as a in so that $\Psi_{1,1}(0)=0$ and the coefficient matrix is a co-isometry. Then there exists a lifting data set ${\{A,T',U',R,Q\}}$ associated with ${{\mathscr R}}_\Psi$. More precisely, if $\Phi_{1,1}$, $\Phi_{1,2}$, $\Phi_{2,1}$ and $\Phi_{2,2}$ are the Redheffer coefficients associated with ${\{A,T',U',R,Q\}}$ via , then there exist a unitary operator $\psi$ mapping ${{\mathcal E}}$ onto ${{\mathcal G}}$ and a co-isometry ${\varphi}$ from ${{\mathcal E}}'$ onto ${{\mathcal D}}_{{\omega}^*}$ such that for each ${\lambda}\in{{\mathbb D}}$ $$\label{CoefsRelation}
{\ensuremath{\left[\begin{array}{cc}\psi&0\\0&I_{{{\mathcal Y}}}\end{array} \right]}}
{\ensuremath{\left[\begin{array}{cc}\Psi_{1,1}({\lambda})&\Psi_{1,2}({\lambda})\\\Psi_{2,1}({\lambda})&\Psi_{2,2}({\lambda})\end{array} \right]}}=
{\ensuremath{\left[\begin{array}{cc}\Phi_{1,1}({\lambda})&\Phi_{1,2}({\lambda})\\\Phi_{2,1}({\lambda})&\Phi_{2,2}({\lambda})\end{array} \right]}}{\ensuremath{\left[\begin{array}{cc}{\varphi}&0\\0&I_{{{\mathcal U}}}\end{array} \right]}}.$$ Moreover, if in addition the coefficient matrix is unitary, then ${\varphi}$ is unitary.

Since the Redheffer coefficients are completely determined by the underlying contraction ${\omega}$, we see that the construction goes along the following path.

\[path\]

(1,1) (-26,0)[$\begin{array}{c}\text{Lifting data set}\\{\{A,T',U',R,Q\}}\end{array}$]{} (-13,0)[$\boldsymbol\Longrightarrow$]{} (-10,0)[$\begin{array}{c}\text{Underlying contraction}\\{\omega}\end{array}$]{} (7,0)[$\boldsymbol\Longrightarrow$]{} (10,0)[$\begin{array}{c}\text{Redheffer coefficients}\\
\Phi_{1,1},\ \Phi_{1,2},\ \Phi_{2,1},\ \Phi_{2,2}\end{array}$]{}

It was already observed in [@FtHK08] that any contraction ${\omega}$ of the form $$\label{genom}
{\omega}={\ensuremath{\left[\begin{array}{c}{\omega}_1\\{\omega}_2\end{array} \right]}}:{{\mathcal F}}\to{\ensuremath{\left[\begin{array}{c}{{\mathcal Y}}\\{{\mathcal U}}\end{array} \right]}},\quad{{\mathcal F}}\subset{{\mathcal U}}$$ appears as the underlying contraction of some lifting data set. We prove Theorem \[T:inverse\] in Section \[S:inverseT\] below by extracting a contraction ${\omega}$ of the form from the Redheffer coefficients.

Theorem \[T:inverse\] guarantees the existence of a lifting data set with the required properties. Moreover, as we will see in the course of the proof in Section \[S:inverseT\] below, the lifting data set can in fact be constructed explicitly from the given coefficients. We already observed that for an associated lifting data set ${\{A,T',U',R,Q\}}$ we have ${{\mathcal D}}_A={{\mathcal U}}$ and ${{\mathcal D}}_{T'}={{\mathcal Y}}$. In addition, the space ${{\mathcal F}}$ on which the underlying contraction ${\omega}$ associated with the lifting data set is defined and the unitary operator $\psi$ can also be extracted from the given coefficients. Indeed, from the definition of $\Phi_{1,2}$ in and the relation it follows that ${{\mathcal F}}={\textup{Ker\,}}\Psi_{1,2}(0)$, while $\psi^*=\Psi_{1,2}(0)|_{{{\mathcal U}}\ominus{{\mathcal F}}}$. Hence the Hilbert spaces between which the contraction ${\omega}$ underlying ${\{A,T',U',R,Q\}}$ acts are determined by the Redheffer coefficients.

Note that, as one can see from , $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ are, in general, not the actual coefficients associated with the constructed data set, not even up to unitary transformations on the spaces ${{\mathcal E}}$ and ${{\mathcal E}}'$, unless the operator ${\varphi}$ is unitary. If ${\varphi}$ is not unitary and ${{\mathcal E}}_0'={\textup{Ker\,}}{\varphi}$, then $\Psi_{1,1}({\lambda})|_{{{\mathcal E}}_0'}=0$ and $\Psi_{2,1}({\lambda})|_{{{\mathcal E}}_0'}=0$ for each ${\lambda}\in{{\mathbb D}}$, which only adds to the non-uniqueness in the Redheffer representation: For $V,V'\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ with $P_{{{\mathcal E}}'\ominus{{\mathcal E}}_0'}V({\lambda})=P_{{{\mathcal E}}'\ominus{{\mathcal E}}_0'}V'({\lambda})$ for all ${\lambda}\in{{\mathbb D}}$ we have $H_V=H_{V'}$.

The two steps in the construction, lifting data set to underlying contraction and underlying contraction to Redheffer coefficients, are in general not one-to-one. Many lifting data sets can have the same underlying contraction and for each admissible set of Redheffer coefficients there can be more than one underlying contraction. However, it turns out that under some additional assumptions all contractions ${\omega}$ that define a fixed set of Redheffer coefficients via are unique up to a unitary transformation; see Proposition \[P:unitRedmat\] below.

Let ${\omega}'$ be the underlying contraction of another lifting data set associated with the Redheffer coefficients , and assume that the conditions of Theorem \[T:inverse\] are met. Then ${\omega}'$ also maps ${{\mathcal F}}={\textup{Ker\,}}\Psi_{1,2}(0)$ into ${{\mathcal Y}}\oplus{{\mathcal U}}$. We say that ${\omega}$ and ${\omega}'$ are [*unitarily equivalent*]{} if there exists a unitary operator ${\Theta}$ on ${{\mathcal U}}$ for which ${{\mathcal F}}$ is a reducing subspace (${\Theta}{{\mathcal F}}\subset{{\mathcal F}}$ and ${\Theta}^*{{\mathcal F}}\subset{{\mathcal F}}$) such that $${\omega}_1\Pi_{{\mathcal F}}{\Theta}={\omega}_1'\Pi_{{\mathcal F}}{\quad\mbox{and}\quad}{\omega}_2\Pi_{{\mathcal F}}{\Lambda}={\Lambda}{\omega}_2'\Pi_{{\mathcal F}}.$$ It is not difficult to see that unitarily equivalent underlying contractions define the same Redheffer coefficients via . If, in addition, the coefficient matrix is unitary, then the converse is also true.

\[P:unitRedmat\] Let ${\{A,T',U',R,Q\}}$ and $\{{\widetilde{A}},{\widetilde{T}}',{\widetilde{U}}',{\widetilde{R}},{\widetilde{Q}}\}$ be lifting data sets associated with Redheffer coefficients that satisfy the conditions of Theorem \[T:inverse\]. Assume that the coefficient matrix is unitary. Then the underlying contractions of ${\{A,T',U',R,Q\}}$ and $\{{\widetilde{A}},{\widetilde{T}}',{\widetilde{U}}',{\widetilde{R}},{\widetilde{Q}}\}$ are unitarily equivalent.

Since the steps in the construction are in general not one-to-one, it is of interest which properties are invariant under these steps. Two invariants are known:

-   $R^*R=Q^*Q$ $\quad{\Longleftrightarrow}\quad$ ${\omega}$ is an isometry;\

-   $
    \begin{array}{c}
    {\omega}\text{ is an isometry and}\\ {\omega}_2\Pi_{{\mathcal F}}\text{ is strongly stable}\end{array}$ $\ {\Longleftrightarrow}\ $ $
    \begin{array}{c}
    \text{the coefficient matrix}\\ \text{is unitary.}\end{array}$

If the operators in the lifting data set satisfy: $R^*R=Q^*Q$, $\|A\|<1$ and $R$ is left invertible (or equivalently $R^*R$ is invertible), then we are in case (ii), but these conditions on the lifting data set are only sufficient; in fact, any ${\omega}$ of the form is the underlying contraction of a lifting data set with $A$ a co-isometry (see the proof of Theorem \[T:inverse\] below).

We now return to the topic of the non-uniqueness in the Redheffer representation of Theorem \[T:RCLRedheffer\]. Let $B$ be a contractive interpolant for the lifting data set ${\{A,T',U',R,Q\}}$, and define $$\label{nonuniqueSet}
{{\mathscr V}}_{B}=\{V\in{{\mathbf S}}({{\mathcal G}},{{\mathcal D}}_{{\omega}^*}) \mid B=B_V\}.$$ In [@FtHK06b] necessary and sufficient conditions were obtained that guarantee that the set ${{\mathscr V}}_B$ consists of one element only. To state the criteria some additional notation is required. Assume that $B$ is given in the form , i.e., we have a contraction ${\Gamma}\in{{\mathscr L}}({{\mathcal D}}_A,H^2({{\mathcal D}}_{T'}))$ such that $$B={\ensuremath{\left[\begin{array}{c}A\\{\Gamma}D_A\end{array} \right]}}.$$ As in [@FtHK06b] we define a contraction ${\omega}_B$ by $$\label{omB}
{\omega}_B:{{\mathcal F}}_B={{\overline{D_{{\Gamma}}{{\mathcal F}}}}}\to{{\mathcal D}}_{{\Gamma}},\quad {\omega}_B D_{{\Gamma}}|_{{\mathcal F}}=D_{{\Gamma}}{\omega}_2.$$ It turns out that ${\omega}_B$ is an isometry if and only if ${\omega}$ is an isometry, i.e., if and only if $R^*R=Q^*Q$. Hence the statement “${\omega}_B$ is an isometry” holds true, or not, independent of the contractive interpolant $B$ in question. Set ${{\mathcal G}}_B={{\mathcal D}}_{\Gamma}\ominus{{\mathcal F}}_B$. Using this contraction ${\omega}_B$ a one-to-one map from the Schur class ${{\mathbf S}}({{\mathcal G}}_B,{{\mathcal D}}_{{\omega}_B^*})$ onto the set ${{\mathscr V}}_B$ can be constructed; see Theorem 1.2 and the subsequent paragraph in [@FtHK06b]. Since ${{\mathcal G}}_B=\{0\}$ if and only if ${{\mathcal F}}_B={{\mathcal D}}_{{\Gamma}_H}$ and ${{\mathcal D}}_{{\omega}_B^*}=\{0\}$ if and only if ${\omega}_B$ is a co-isometry, we have the following criterion for the set ${{\mathscr V}}_B$ to be a singleton.

\[P:NonProper1\] Let $B$ be a contractive interpolant for ${\{A,T',U',R,Q\}}$. Let ${\omega}_B$ be the contraction from ${{\mathcal F}}_{H}$ into ${{\mathcal D}}_{{\Gamma}_H}$ defined by . Then the set ${{\mathscr V}}_B$ is a singleton if and only if ${{\mathcal F}}_B={{\mathcal D}}_{{\Gamma}_H}$ or ${\omega}_B$ is a co-isometry.

Proposition \[P:NonProper1\] gives necessary and sufficient conditions under which the set ${{\mathscr V}}_B$ consists of just one element for each contractive interpolant $B$ individually. It is not clear how to derive necessary and sufficient conditions for the map $V\mapsto B_V$ to be one-to-one from this result; sufficient conditions obtained in the literature are: ${{\mathcal F}}={{\mathcal D}}_A$, ${\omega}$ is a co-isometry or ${\omega}$ is an isometry and ${{\overline{{\omega}_2{{\mathcal F}}}}}={{\mathcal D}}_A$; see Proposition 4.2.8 in [@tH07] and the proof of Theorem 1.3 in [@FtHK06b]. However, the fact that the contraction ${\omega}_B$ is an isometry either for all contractive interpolants $B$ or for none gives hope that a global result may be obtainable.

The next theorem presents a global result when we restrict to the set of contractive interpolants $B_V$ where $V$ is from the open ball ${{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ of ${{\mathbf H}}^\infty({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$, that is, from the set $${{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*}):=\{V\in{{\mathbf S}}({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})\colon \|V\|_\infty<1\},$$ and under the additional assumptions the underlying contraction ${\omega}$ is an isometry and ${\omega}_2\Pi_{{\mathcal F}}$ is strongly stable. The latter happens in many of the examples considered in [@FFGK98; @FFK02] if the operator $A$ is a strict contraction.

\[T:NonProper2\] Let ${\{A,T',U',R,Q\}}$ be a lifting data set such that the underlying contraction ${\omega}$ is an isometry and ${\omega}_2\Pi_{{\mathcal F}}$ on ${{\mathcal D}}_A$ is strongly stable. Then for each $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ we have ${{\mathscr V}}_{B_V}=\{V\}$ if and only if $${\Gamma}_{\Phi_{1,2}}{{\mathcal D}}_A=\{\Pi_{{{\mathcal G}}}(I-{\lambda}{\omega}_2\Pi_{{\mathcal F}})u\mid u\in{{\mathcal D}}_A\}$$ is dense in $H^2({{\mathcal G}})$. In particular, if ${{\mathscr V}}_{B_V}=\{V\}$ for some $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$, then ${{\mathscr V}}_{B_V}=\{V\}$ for all $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$.

Given a lifting data set ${\{A,T',U',R,Q\}}$ that satisfies the requirement of Theorem \[T:NonProper2\], it follows that the Redheffer representation of Theorem \[T:RCLRedheffer\] cannot be one-to-one in case ${{\mathcal D}}_{A}$ is finite dimensional, unless ${{\mathcal G}}=\{0\}$. This occurs, for instance, in the examples of the relaxations of classical interpolation problems introduced in [@FFK02], where the common ingredient is that $R$ and $Q$ are operators from ${{\mathcal V}}^{n-1}$ into ${{\mathcal V}}^n$ of the form $$R={\ensuremath{\left[\begin{array}{c}I_{{{\mathcal V}}^{n-1}}\\0\end{array} \right]}}{\quad\mbox{and}\quad}Q={\ensuremath{\left[\begin{array}{c}0\\I_{{{\mathcal V}}^{n-1}}\end{array} \right]}},$$ assuming that ${\textup{dim}}\, {{\mathcal V}}<\infty$ and $\|A\|<1$.

Another examples of a global result for the set of contractive interpolants $B_V$ associated with ${{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ is the harmonic maximum principle for the classical commutant lifting problem due to A. Biswas [@B97]. As a result of our investigations of Redheffer representations in Section \[S:Redheffer\] below we obtain the following extension of Theorem 3.1 from [@B97] to the relaxed commutant lifting setting.

\[T:MaxPrinciple\] Given a lifting data set ${\{A,T',U',R,Q\}}$ we have:

-   If there exists a ${\widetilde{V}}\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ so that $\|B_{{\widetilde{V}}}\|<1$, then $\|B_V\|<1$ for all $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$.

-   If there exists a ${\widetilde{V}}\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ so that $\|B_{{\widetilde{V}}}\|=1$, then $\|B_V\|=1$ for all $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$.

-   Moreover, for $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ and ${\widetilde{V}}\in{{\mathbf S}}({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ we have ${\textup{Ker\,}}D_{B_V}\subset{\textup{Ker\,}}D_{B_{{\widetilde{V}}}}$. In particular, ${\textup{Ker\,}}D_{B_V}={\textup{Ker\,}}D_{B_{{\widetilde{V}}}}$ in case ${\widetilde{V}},V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$.

When restricting to the classical commutant lifting setting, Part 2 coincides with Theorem 3.1 from [@B97]. Part 1 is just the contrapositive of Part 2. The techniques from [@B97] can be used to prove Part 1 as well, but in the present paper we give a shorter, more intuitive proof. The last part of Theorem \[T:MaxPrinciple\] seems to be a new result, even for classical commutant lifting. Part 3 and the fact that a contractive interpolant $B$ with $\|B\|=\|A\|$ always exists imply the following corollary.

\[C:nonnormattaining\] Let ${\{A,T',U',R,Q\}}$ be a lifting data set with $\|A\|<1$. Then ${\textup{Ker\,}}D_{B_V}=\{0\}$ for each $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$, i.e., for any $h\in{{\mathcal H}}$ and $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ we have $\|B_Vh\|<\|h\|$.

The paper consists of 6 sections, not counting the present introduction. We start with a section where some preliminary system theory results that will be used throughout the paper is presented. In Section \[S:inverseT\] we prove Proposition \[P:unitRedmat\] and Theorem \[T:inverse\]. The following section contains a proof of Theorem \[T:NonProper2\] restricted to the case that the contractive interpolant in question is the central contractive interpolant. We then proceed with an intermezzo about general Redheffer representations in Section \[S:Redheffer\]. Finally, Theorems \[T:NonProper2\] and \[T:MaxPrinciple\] are proved in Sections \[S:nonproper\] and \[S:maxprinciple\], respectively. The order in the proofs goes in the reversed direction: we first prove analogous results on the level of Redheffer representations before proving Theorems \[T:NonProper2\] and \[T:MaxPrinciple\].

System theory preliminaries {#S:Systems}
===========================

In this section we review some results from linear system theory that will be useful in the sequel. See [@F87; @F81] for a general overview.

#### **contractive systems.**

A [*contractive system*]{} is a quadruple $\{{X},B,C,D\}$ consisting of Hilbert space operators: ${X}$ on ${{\mathcal X}}$, $B$ from ${{\mathcal U}}$ to ${{\mathcal X}}$, $C$ from ${{\mathcal X}}$ to ${{\mathcal Y}}$ and $D$ mapping ${{\mathcal U}}$ into ${{\mathcal Y}}$ such that the [*system matrix*]{} $$\label{SysMat}
{\ensuremath{\left[\begin{array}{cc}{X}&B\\C&D\end{array} \right]}}:{\ensuremath{\left[\begin{array}{c}{{\mathcal X}}\\{{\mathcal U}}\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}{{\mathcal X}}\\{{\mathcal Y}}\end{array} \right]}}$$ is a contraction. Of particular interest will be the case when the system matrix is a co-isometry; we then say that $\{{X},B,C,D\}$ is a [*co-isometric system*]{}. The system $\{{X},B,C,D\}$ is said to be [*strongly stable*]{} in case the state operator ${X}$ is strongly stable, that is, if for each $x\in{{\mathcal X}}$ the sequence ${X}^nx$ converges to zero as $n\to\infty$. Since ${X}$ is contractive, we can define analytic functions $F$ and $W$ on ${{\mathbb D}}$ by $$\label{MultObs}
F({\lambda})=D+{\lambda}C(I-{\lambda}{X})^{-1}B{\quad\mbox{and}\quad}W({\lambda})=C(I-{\lambda}{X})^{-1}\quad\quad({\lambda}\in{{\mathbb D}}).$$ We refer to $F$ and $W$ as the [*transfer function*]{} and [*observability function*]{} associated with $\{{X},B,C,D\}$, respectively. From the fact that the system matrix is a co-isometry it follows that $F\in{{\mathbf S}}({{\mathcal U}},{{\mathcal Y}})$ and $W\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal X}},{{\mathcal Y}})$. Hence $F$ defines a contractive multiplication operator $M_F$ from $H^2({{\mathcal U}})$ to $H^2({{\mathcal Y}})$, and $W$ defines a contraction ${\Gamma}_W$ from ${{\mathcal X}}$ into $H^2({{\mathcal Y}})$ via . The system $\{{X},B,C,D\}$ is said to be [*observable*]{} in case ${\textup{Ker\,}}{\Gamma}_W=\{0\}$.

Two co-isometric systems $\{{X}_1,B_1,C_1,D_1\}$ and $\{{X}_2,B_2,C_2,D_2\}$ with the same input spaces ${{\mathcal U}}$ and output spaces ${{\mathcal Y}}$ but possibly different state spaces ${{\mathcal X}}_1$ respectively ${{\mathcal X}}_2$ are said to be [*unitarily equivalent*]{} if $D_1=D_2$ and there exists a unitary operator ${\Theta}$ mapping ${{\mathcal X}}_1$ onto ${{\mathcal X}}_2$ such that $${\Theta}{X}_1={X}_2{\Theta},\quad C_1=C_2{\Theta},\quad {\Theta}B_1=B_2.$$ One easily verifies that unitarily equivalent systems have the same transfer function. The converse statement is true if the co-isometric systems are also observable.

\[T:CoOpMat\] Let $\{{X},B,C,D\}$ be a co-isometric system with transfer function $F\in{{\mathbf S}}({{\mathcal U}},{{\mathcal Y}})$ and observability function $W\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal X}},{{\mathcal Y}})$. Then the operator $$\label{opmat}
{\ensuremath{\left[\begin{array}{cc}\!\!M_F&{\Gamma}_W\!\!\end{array} \right]}}:{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal U}})\\{{\mathcal X}}\end{array} \right]}}\to H^2({{\mathcal Y}})$$ is a co-isometry. Moreover, the operator is unitary if and only if the system $\{{X},B,C,D\}$ is strongly stable and the system matrix is unitary.

The statement for the case that is unitary follows from Theorem III.10.4 in [@FFGK98] and the fact that the positive operator ${\Delta}$ on ${{\mathcal X}}$ defined by $${\Delta}^2={\textup{strong--}}\!\!\lim_{n\to\infty} {X}^{*n}{X}^n$$ is zero if and only if ${X}$ is strongly stable. For the statement on co-isometric systems see Theorem 1.3 in [@tH2].

The first part of Theorem \[T:CoOpMat\] has the following converse result.

\[T:CoOpMatInv\] Let $F\in{{\mathbf S}}({{\mathcal U}},{{\mathcal Y}})$ and $W\in{{\mathbf H}}^2({{\mathcal X}},{{\mathcal Y}})$ be functions such that the operator matrix is a co-isometry. Then there is a co-isometric system $\{{X},B,C,D\}$ so that $F$ and $W$ are the associated transfer and observability functions.

It follows from [@Ando90] that the function $F$ is the transfer function of an observable co-isometric system $\{{\widetilde{X}},{\widetilde{B}},{\widetilde{C}},{\widetilde{D}}\}$, say with state space $\widetilde{{\mathcal X}}$. From Theorem \[T:CoOpMat\] we know that the operator matrix ${\ensuremath{\left[\begin{array}{cc}M_{F}&{\Gamma}_{{\widetilde{W}}}\end{array} \right]}}$ is a co-isometry. Hence $$M_FM_{F^*}+{\Gamma}_{{\widetilde{W}}}{\Gamma}_{{\widetilde{W}}}^*=I=M_FM_{F^*}+{\Gamma}_{W}{\Gamma}_{W}^*,\quad\text{ and thus }\quad
{\Gamma}_{{\widetilde{W}}}{\Gamma}_{{\widetilde{W}}}^*={\Gamma}_{W}{\Gamma}_{W}^*.$$ By Douglas factorization lemma [@D66] there exists a unitary operator ${\Lambda}$ mapping ${{\overline{{\textup{Im\,}}{\Gamma}_{{\widetilde{W}}}^*}}}=\widetilde{{\mathcal X}}$ onto ${{\mathcal N}}={{\overline{{\textup{Im\,}}{\Gamma}_{W}^*}}}$ defined by the identity ${\Gamma}_{{\widetilde{W}}}{\Lambda}^*={\Gamma}_{W}$. Now set ${{\mathcal M}}={{\mathcal X}}\ominus{{\mathcal N}}$ and define $\{{X},B,C,D\}$ to be the co-isometric system given by the system matrix $${\ensuremath{\left[\begin{array}{cc}{X}&B\\C&D\end{array} \right]}}={\ensuremath{\left[\begin{array}{cc|c}{\Lambda}&0&0\\0&I_{{{\mathcal M}}}&0\\\hline0&0&I_{{\mathcal Y}}\end{array} \right]}}
{\ensuremath{\left[\begin{array}{cc|c}{\widetilde{X}}&0&{\widetilde{B}}\\0&I_{{\mathcal M}}&0\\\hline{\widetilde{C}}&0&{\widetilde{D}}\end{array} \right]}}
{\ensuremath{\left[\begin{array}{cc|c}{\Lambda}^*&0&0\\0&I_{{{\mathcal M}}}&0\\\hline0&0&I_{{\mathcal Y}}\end{array} \right]}}.$$ Then $\{{X},B,C,D\}$ and $\{{\widetilde{X}},{\widetilde{B}},{\widetilde{C}},{\widetilde{D}}\}$ are unitarily equivalent, and thus $F$ is also the transfer function of $\{{X},B,C,D\}$. Moreover, the observability function $\widehat W$ of $\{{X},B,C,D\}$ satisfies $\widehat
W({\lambda})={\ensuremath{\left[\begin{array}{cc}{\widetilde{W}}({\lambda}){\Lambda}^*&0\end{array} \right]}}$ for each ${\lambda}\in{{\mathbb D}}$. The latter implies that $${\Gamma}_{\widehat W}={\ensuremath{\left[\begin{array}{cc}{\Gamma}_{{\widetilde{W}}}{\Lambda}^*&0\end{array} \right]}}={\Gamma}_W.$$ But then $\widehat W=W$, which proves our claim.

#### **Redheffer cascading systems.**

Suppose we are given Hilbert spaces ${{\mathcal U}}_1$, ${{\mathcal U}}_2$, ${{\mathcal Y}}_1$, ${{\mathcal Y}}_2$, ${{\mathcal X}}$ and ${{\mathcal X}}'$ and two operators $M_1$ mapping ${{\mathcal X}}\oplus{{\mathcal U}}_1$ into ${{\mathcal X}}'\oplus{{\mathcal Y}}_1$ and $M_2$ mapping ${{\mathcal X}}'\oplus{{\mathcal U}}_2$ into ${{\mathcal X}}\oplus{{\mathcal Y}}_2$ with operator matrix decompositions $$\label{M1M2}
M_1={\ensuremath{\left[\begin{array}{cc}{X}_1&B_1\\C_1&D_1\end{array} \right]}}:{\ensuremath{\left[\begin{array}{c}{{\mathcal X}}\\{{\mathcal U}}_1\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}{{\mathcal X}}'\\{{\mathcal Y}}_1\end{array} \right]}},\ \
M_2={\ensuremath{\left[\begin{array}{cc}{X}_2&B_2\\C_2&D_2\end{array} \right]}}:{\ensuremath{\left[\begin{array}{c}{{\mathcal X}}'\\{{\mathcal U}}_2\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}{{\mathcal X}}\\{{\mathcal Y}}_2\end{array} \right]}}.$$ Furthermore, assume that $I_{{\mathcal X}}-{X}_2{X}_1$ is invertible. Then $I_{{{\mathcal X}}'}-{X}_1{X}_2$ is invertible, and we can form the [*Redheffer product*]{} $M_1\circ M_2$ of $M_1$ and $M_2$, which is the operator from ${{\mathcal U}}_1\oplus{{\mathcal U}}_2$ into ${{\mathcal Y}}_1\oplus{{\mathcal Y}}_2$ given by the operator matrix decomposition $$\label{FeedbackMat}
{\ensuremath{\left[\begin{array}{cc}D_1+C_1{X}_2(I_{{{\mathcal X}}'}-{X}_1{X}_2)^{-1}B_1&C_1(I_{{{\mathcal X}}}-{X}_2{X}_1)^{-1}B_2\\
C_1(I_{{{\mathcal X}}'}-{X}_1{X}_2)^{-1}B_1&D_2+C_2{X}_1(I_{{{\mathcal X}}}-{X}_2{X}_1)^{-1}B_2\end{array} \right]}}.$$ For an elaborate discussion on Redheffer products we refer to [@FF90 Chapter XIV]. The next proposition shows how norm properties of $M_1$ and $M_2$ carry over to $M_1\circ M_2$.

\[P:CoOpMatInv\] In case the block operator matrices $M_1$ and $M_2$ in are both contractive, isometric, co-isometric or unitary, and $I-X_2X_1$ is invertible, the Redheffer product $M_1\circ M_2$ is also contractive, isometric, co-isometric or unitary, respectively.

Since we assume $I_{{\mathcal X}}-{X}_2{X}_1$ to be invertible, the case where $M_1$ and $M_2$ are both contractive follows from Lemma XIV.1.2 in [@FF90]. The case where $M_1$ and $M_2$ are isometries follows from Lemma XIV.1.2 in [@FF90] as well, using identity (1.15) on Page 433 of [@FF90]. The co-isometric case is obtained by applying the statement for the isometric case to $M_1^*$ and $M_2^*$, and finally the unitary case follows from the result when $M_1$ and $M_2$ are both isometric and co-isometric.

Proofs of Proposition \[P:unitRedmat\] and Theorem \[T:inverse\] {#S:inverseT}
================================================================

We start this section with a proof of Theorem \[T:inverse\], followed by a proof of Proposition \[P:unitRedmat\].

Assume that ${{\mathcal U}}$, ${{\mathcal Y}}$, ${{\mathcal E}}$ and ${{\mathcal E}}'$ are Hilbert spaces, and $$\Psi_{1,1}\in{{\mathbf S}}({{\mathcal E}}',{{\mathcal E}}),\ \
\Psi_{1,2}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal E}}),\ \
\Psi_{2,1}\in{{\mathbf S}}({{\mathcal E}}',{{\mathcal Y}}),\ \
\Psi_{2,2}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal Y}}),$$ with $\Psi_{1,1}(0)=0$ such that the coefficient matrix is a co-isometry. Set $$F({\lambda})={\ensuremath{\left[\begin{array}{c}\Psi_{1,1}({\lambda})\\\Psi_{2,1}({\lambda})\end{array} \right]}}\in{{\mathbf S}}({{\mathcal E}}',{{\mathcal E}}\oplus{{\mathcal Y}}){\quad\mbox{and}\quad}W({\lambda})={\ensuremath{\left[\begin{array}{c}\Psi_{1,2}({\lambda})\\\Psi_{2,2}({\lambda})\end{array} \right]}}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal E}}\oplus{{\mathcal Y}}).$$ Then ${\ensuremath{\left[\begin{array}{cc}M_F&{\Gamma}_W\end{array} \right]}}$ is a co-isometry from $H^2({{\mathcal E}}')\oplus{{\mathcal U}}$ into $H^2({{\mathcal E}}\oplus{{\mathcal Y}})$. According to Theorem \[T:CoOpMatInv\] there exists a co-isometric system $\{{X},B,{\left[\begin{smallmatrix} C_1\\C_2\end{smallmatrix}\right]},{\left[\begin{smallmatrix} D_1\\D_2\end{smallmatrix}\right]}\}$ with system matrix $$N={\ensuremath{\left[\begin{array}{cc}{X}&B\\C_1&D_1\\C_2&D_2\end{array} \right]}}:{\ensuremath{\left[\begin{array}{c}{{\mathcal U}}\\{{\mathcal E}}'\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}{{\mathcal U}}\\{{\mathcal E}}\\{{\mathcal Y}}\end{array} \right]}}$$ such that $F$ and $W$ are the corresponding transfer function and observability function, respectively. The fact that $\Psi_{1,1}(0)=0$ implies that $D_1=0$, and thus, since the system matrix $N$ is a co-isometry, that $C_1=\Psi_{1,2}(0)$ is a co-isometry. Set ${{\mathcal F}}={\textup{Ker\,}}C_1$ and ${{\mathcal G}}={{\mathcal U}}\ominus{{\mathcal F}}$. Then $C_1$ maps ${{\mathcal G}}$ isometrically onto ${{\mathcal E}}'$. Define ${\omega}_1=C_2|_{{\mathcal F}}$ and ${\omega}_2={X}|_{{\mathcal F}}$, i.e., $$N={\ensuremath{\left[\begin{array}{cc}{\omega}_2\Pi_{{\mathcal F}}&B\\C_1P_{{\mathcal G}}&0\\{\omega}_1\Pi_{{\mathcal F}}&D_1\end{array} \right]}}.$$ By a special case of Parrot’s lemma [@P78], again using that $N$ is a co-isometry, there exists a co-isometry ${\varphi}$ mapping ${{\mathcal E}}'$ onto ${{\mathcal D}}_{{\omega}^*}$ such that $$\label{Defectom*}
{\ensuremath{\left[\begin{array}{cc}B\\D_1\end{array} \right]}}=D_{{\omega}^*}{\varphi},\quad\text{where}\quad
{\omega}={\ensuremath{\left[\begin{array}{c}{\omega}_1\\{\omega}_2\end{array} \right]}}:{{\mathcal F}}\to{\ensuremath{\left[\begin{array}{cc}{{\mathcal Y}}\\{{\mathcal U}}\end{array} \right]}}.$$ Let $\psi$ be the unitary map from ${{\mathcal G}}$ onto ${{\mathcal E}}$ given by $\psi=C_1|_{{\mathcal G}}$. Then $${\ensuremath{\left[\begin{array}{ccc}I_{{\mathcal U}}&0&0\\0&\psi&0\\0&0&I_{{\mathcal Y}}\end{array} \right]}}N{\ensuremath{\left[\begin{array}{cc}I_{{\mathcal U}}&0\\0&{\varphi}^*\end{array} \right]}}
={\ensuremath{\left[\begin{array}{cc}{\omega}_2\Pi_{{\mathcal F}}&\Pi_{{\mathcal U}}D_{{\omega}^*}\\\Pi_{{\mathcal G}}&0\\{\omega}_1\Pi_{{\mathcal F}}&\Pi_{{\mathcal Y}}D_{{\omega}^*}\end{array} \right]}}.$$ Thus $\{{X},B,{\left[\begin{smallmatrix} C_1\\C_2\end{smallmatrix}\right]},{\left[\begin{smallmatrix} D_1\\D_2\end{smallmatrix}\right]}\}$ is unitarily equivalent to the system $\{\widehat{{X}},\widehat{B},\widehat{C},\widehat{D}\}$ given by $$\label{REDsysmat}
{\ensuremath{\left[\begin{array}{cc}\widehat{{X}}&\widehat{B}\\\widehat{C}&\widehat{D}\end{array} \right]}}
={\ensuremath{\left[\begin{array}{c|c}{\omega}_2\Pi_{{\mathcal F}}&\Pi_{{\mathcal U}}D_{{\omega}^*}\\\hline\Pi_{{\mathcal G}}&0\\{\omega}_1\Pi_{{\mathcal F}}&\Pi_{{\mathcal Y}}D_{{\omega}^*}\end{array} \right]}}
:{\ensuremath{\left[\begin{array}{c}{{\mathcal U}}\\{{\mathcal D}}_{{\omega}^*}\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}{{\mathcal U}}\\{{\mathcal G}}\\{{\mathcal Y}}\end{array} \right]}}.$$ One easily verifies that $${\ensuremath{\left[\begin{array}{c}\Phi_{1,1}({\lambda})\\\Phi_{2,1}({\lambda})\end{array} \right]}}
=\widehat{D}+{\lambda}\widehat{C}(I-{\lambda}\widehat{{X}})^{-1}\widehat{B}{\quad\mbox{and}\quad}{\ensuremath{\left[\begin{array}{c}\Phi_{1,2}({\lambda})\\\Phi_{2,2}({\lambda})\end{array} \right]}}=\widehat{C}(I-{\lambda}\widehat{{X}})^{-1},$$ where $\Phi_{1,1}$ and $\Phi_{2,1}$ are defined by with ${\omega}$ as in . A straightforward computation shows that the transfer function $\widehat{F}$ and observability function $\widehat{W}$ of the co-isometric system $\{\widehat{X},\widehat{B},\widehat{C},\widehat{D}\}$ given in relate to $F$ and $W$ via $${\ensuremath{\left[\begin{array}{cc}\psi&0\\0&I_{{\mathcal Y}}\end{array} \right]}}{\ensuremath{\left[\begin{array}{cc}F({\lambda})&W({\lambda})\end{array} \right]}}
={\ensuremath{\left[\begin{array}{cc}\widehat{F}({\lambda})&\widehat{W}({\lambda})\end{array} \right]}}{\ensuremath{\left[\begin{array}{cc}{\varphi}&0\\0&I_{{\mathcal U}}\end{array} \right]}}\quad({\lambda}\in{{\mathbb D}}).$$ Thus, to prove Theorem \[T:inverse\] it remain to show that the contraction ${\omega}$ appear as the underlying contraction of a lifting data set. However, it is known from [@FtHK08] that any contraction ${\omega}$ of the form in with ${{\mathcal F}}\subset{{\mathcal U}}$, is the underlying contraction of some lifting data set. In fact, one easily verifies that the operators $$A={\ensuremath{\left[\begin{array}{cc}I_{{\mathcal Y}}&0\end{array} \right]}}:{\ensuremath{\left[\begin{array}{c}{{\mathcal Y}}\\{{\mathcal U}}\end{array} \right]}}\to{{\mathcal Y}},\ T'=0{\mbox{ on }}{{\mathcal Y}},\ R={\omega},Q=\Pi_{{\mathcal F}}^*:{{\mathcal F}}\to{{\mathcal Y}}\oplus{{\mathcal U}}$$ define a lifting data set ${\{A,T',U',R,Q\}}$, with $U'$ the Sz.-Nagy-Schäffer isometric lifting of $T'$, which has ${\omega}$ as its underlying contraction.

To complete the proof, assume that the coefficient matrix $K_0$ in is unitary. Then $K_0|_{H^2({{\mathcal E}}')}$ is an isometry, and thus, in particular, for each $e\in{{\mathcal E}}'$ there exists a ${\lambda}\in{{\mathbb D}}$ such that $\Psi_{1,1}({\lambda})e\not=0$ or $\Psi_{2,1}({\lambda})e\not=0$. So, by and the fact that $\psi$ is unitary, ${\varphi}e\not=0$ for all $e\in{{\mathcal E}}'$, and thus ${\varphi}$ is, in fact, unitary.

Let ${\omega}$ and ${\omega}'$ denote the underlying contractions of the data sets ${\{A,T',U',R,Q\}}$ and $\{{\widetilde{A}},{\widetilde{T}}',{\widetilde{U}}',{\widetilde{R}},{\widetilde{Q}}\}$, respectively.

Since the coefficient matrix is assumed to be unitary, we may without loss of generality assume that the Redheffer coefficients defined by ${\omega}$ and ${\omega}'$ via are equal to the given Redheffer coefficients . Indeed, this is the case because, according to the last part of Theorem \[T:inverse\], the three sets of Redheffer coefficients are the same up to unitary transformations between ${{\mathcal E}}$ and ${{\mathcal G}}={{\mathcal U}}\ominus{{\mathcal F}}$ and between ${{\mathcal E}}'$ and ${{\mathcal D}}_{{\omega}^*}$ and ${{\mathcal D}}_{{\omega}^{'*}}$, and these unitary transformations do not affect whether ${\omega}$ and ${\omega}'$ are unitarily equivalent, or not.

Now let $\{\widehat{{X}},\widehat{B},\widehat{C},\widehat{D}\}$ be the co-isometric system given by and define $\{\widehat{{X}}',\widehat{B}',\widehat{C}',\widehat{D}'\}$ accordingly with ${\omega}$ replaced by ${\omega}'$. Then both systems have the same transfer function $F$ and observability function $W$ that are given by $$F({\lambda})={\ensuremath{\left[\begin{array}{c}\Psi_{1,1}({\lambda})\\\Psi_{2,1}({\lambda})\end{array} \right]}}
{\quad\mbox{and}\quad}W({\lambda})={\ensuremath{\left[\begin{array}{c}\Psi_{1,2}({\lambda})\\\Psi_{2,2}({\lambda})\end{array} \right]}}.$$ By assumption, the coefficient matrix is unitary. This implies that the operator ${\Gamma}_W$, defined via , is an isometry. In particular, ${\textup{Ker\,}}{\Gamma}_W=\{0\}$. Hence the two co-isometric systems $\{\widehat{{X}},\widehat{B},\widehat{C},\widehat{D}\}$ and $\{\widehat{{X}}',\widehat{B}',\widehat{C}',\widehat{D}'\}$ are observable and have the same transfer function. This implies that they are unitarily equivalent. It then follows from the definition of the systems $\{\widehat{{X}},\widehat{B},\widehat{C},\widehat{D}\}$ and $\{\widehat{{X}}',\widehat{B}',\widehat{C}',\widehat{D}'\}$ that the operator ${\Theta}$ on ${{\mathcal U}}$ that establishes this unitary equivalence is also the operator that establishes the unitary equivalence of ${\omega}$ and ${\omega}'$.

Proof of Theorem \[T:NonProper2\] for the central solution
==========================================================

Let $B_c$ denote the central contractive interpolant, that is, $B_c=B_{{\underline{0}}}$, where ${\underline{0}}$ denotes the constant Schur class function whose value is the zero operator in ${{\mathscr L}}({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$. In this section we prove Theorem \[T:NonProper2\] restricted to the case that the contractive interpolant $B$ is equal to $B_c$. To be more precise, we prove the following proposition.

\[P:CentralNonProper\] Let ${\{A,T',U',R,Q\}}$ be a lifting data set such that the underlying contraction ${\omega}$ is an isometry and ${\omega}_2\Pi_{{\mathcal F}}$ on ${{\mathcal D}}_A$ is strongly stable. Then the set ${{\mathscr V}}_{B_c}$ in consists of just one element if and only if ${\Gamma}_{\Phi_{1,2}}{{\mathcal D}}_A$ is dense in $H^2({{\mathcal G}})$.

In order to prove Proposition \[P:CentralNonProper\] it is convenient to first prove the next lemma. Recall that $E_{{\mathcal G}}\in{{\mathscr L}}({{\mathcal G}}, H^2({{\mathcal G}}))$ stands for the canonical embedding of ${{\mathcal G}}$ into the set of constant functions in $H^2({{\mathcal G}})$.

\[L:DenseRange\] Let ${{\mathcal K}}\subset H^2({{\mathcal G}})$ be an invariant subspace of the backward shift $S_{{\mathcal G}}^*$. Then ${{\mathcal G}}_0:=E_{{\mathcal G}}{{\mathcal G}}\subset{{\mathcal K}}$ and $S_{{{\mathcal G}}}^*{{\mathcal K}}={{\mathcal K}}$ if and only if ${{\mathcal K}}=H^2({{\mathcal G}})$.

If ${{\mathcal K}}=H^2({{\mathcal G}})$, then obviously ${{\mathcal G}}_0\subset{{\mathcal K}}$ and $S_{{{\mathcal G}}}^*{{\mathcal K}}={{\mathcal K}}$. Conversely, assume that ${{\mathcal G}}_0\subset{{\mathcal K}}$ and $S_{{{\mathcal G}}}^*{{\mathcal K}}={{\mathcal K}}$. Then $S_{{{\mathcal G}}}^n{{\mathcal G}}_0\subset
S_{{{\mathcal G}}}^n{{\mathcal K}}=S_{{{\mathcal G}}}^nS_{{{\mathcal G}}}^{*n}{{\mathcal K}}$ for $n=0,1,2,\ldots$. Since ${{\mathcal G}}_0=S_{{{\mathcal G}}}^0{{\mathcal G}}_0\subset{{\mathcal K}}$, we obtain that $S_{{\mathcal G}}{{\mathcal G}}_0\subset S_{{\mathcal G}}S_{{\mathcal G}}^*{{\mathcal K}}={{\mathcal K}}\ominus{{\mathcal G}}_0\subset{{\mathcal K}}$, and with a similar argument it follows, recursively, that $$S_{{{\mathcal G}}}^n{{\mathcal G}}_0\subset S_{{\mathcal G}}^nS_{{\mathcal G}}^{*n}{{\mathcal K}}={{\mathcal K}}\ominus(\oplus_{k=0}^{n-1}S_{{{\mathcal G}}}^k{{\mathcal G}}_0)\subset{{\mathcal K}}$$ for each nonnegative integer $n$. The latter inclusions imply that ${{\mathcal K}}=H^2({{\mathcal G}})$.

Note that $B_c$ is of the form with $H_V$ equal to $H_c:=\Phi_{2,2}$. Let ${\omega}_{B_c}$ be the contraction defined according to . It follows from the last statement of Theorem \[T:RCLRedheffer\] and the remark in the sentence after that ${\omega}_{B_c}$ is an isometry. So, by Proposition \[P:NonProper1\], it suffices to prove that ${\Gamma}_{\Phi_{1,2}}{{\mathcal D}}_A$ is dense in $H^2({{\mathcal G}})$ if and only if ${{\mathcal F}}_{B_c}={{\mathcal D}}_{{\Gamma}_{\Phi_{2,2}}}$ or ${\omega}_{B_c}{{\mathcal F}}_{B_c}={{\mathcal D}}_{{\Gamma}_{\Phi_{2,2}}}$.

Since, by Theorem \[T:RCLRedheffer\], the coefficient matrix $K_0$ is unitary we have $$D_{{\Gamma}_{H_c}}^2=D_{{\Gamma}_{\Phi_{2,2}}}^2=I-{\Gamma}_{\Phi_{2,2}}^*{\Gamma}_{\Phi_{2,2}}
={\Gamma}_{\Phi_{1,2}}^*{\Gamma}_{\Phi_{1,2}}.$$ Hence there exists a unique unitary operator $\rho$ mapping ${{\mathcal D}}_{H_c}$ onto ${{\mathcal K}}:={{\overline{{\Gamma}_{\Phi_{1,2}}{{\mathcal D}}_A}}}$ defined by $\rho D_{{\Gamma}_{\Phi_{2,2}}}={\Gamma}_{\Phi_{1,2}}$. The fact that ${\Gamma}_{\Phi_{1,2}}$ is contractive and $\Phi_{1,2}(0)=\Pi_{{\mathcal G}}$ implies that ${{\mathcal G}}_0:=E_{{\mathcal G}}{{\mathcal G}}\subset{{\mathcal K}}$ and that $\rho$ maps ${{\mathcal F}}_{B_c}=D_{{\Gamma}_{\Phi_{2,2}}}{{\mathcal F}}$ onto ${\widetilde{{\mathcal F}}}_{B_c}:={{\mathcal K}}\ominus{{\mathcal G}}_0$. In particular, ${{\mathcal F}}_{B_c}={{\mathcal D}}_{{\Gamma}_{\Psi_{2,2}}}$ holds if and only if ${{\mathcal G}}=\{0\}$, from which the denseness of ${\Gamma}_{\Phi_{1,2}}{{\mathcal D}}_A$ in $H^2({{\mathcal G}})$ obviously follows.

Let ${\widetilde\omega}_{B_c}$ be the isometry from ${\widetilde{{\mathcal F}}}_{B_c}$ to ${{\mathcal K}}$ defined by ${\widetilde\omega}_{B_c}\!=\!\rho{\omega}_{B_c}\rho^*|_{{\widetilde{{\mathcal F}}}_{B_c}}$. Note that ${\widetilde\omega}_{B_c}\Pi_{{\widetilde{{\mathcal F}}}_{B_c}}=\rho{\omega}_{B_c}\Pi_{{{\mathcal F}}_{B_c}}\rho^*$. We claim that ${{\mathcal K}}$ is invariant under $S_{{{\mathcal G}}}^*$ and ${\widetilde\omega}_{B_c}\Pi_{{\widetilde{{\mathcal F}}}_{B_c}}=S_{{\mathcal G}}^*|_{{\mathcal K}}$. The latter, which implies the former, follows since $$\begin{aligned}
{\widetilde\omega}_{B_c}\Pi_{{\widetilde{{\mathcal F}}}_{B_c}}{\Gamma}_{\Phi_{1,2}}
&=&\rho{\omega}_{B_c}\Pi_{{{\mathcal F}}_{B_c}}\rho^*{\Gamma}_{\Phi_{1,2}}
=\rho{\omega}_{B_c}\Pi_{{{\mathcal F}}_{B_c}}D_{{\Gamma}_{\Phi_{2,2}}}
=\rho{\omega}_{B_c}D_{{\Gamma}_{\Phi_{2,2}}}P_{{{\mathcal F}}}\\
&=&\rho D_{{\Gamma}_{\Phi_{2,2}}}{\omega}_2\Pi_{{{\mathcal F}}}
={\Gamma}_{\Phi_{1,2}}{\omega}_2\Pi_{{{\mathcal F}}}
=S^*_{{\mathcal G}}{\Gamma}_{\Phi_{1,2}}P_{{{\mathcal F}}}
=S^*_{{\mathcal G}}{\Gamma}_{\Phi_{1,2}}.\end{aligned}$$ The third identity follows from the fact that ${\Gamma}_{\Phi_{2,2}}|_{{\mathcal G}}=0$, and thus $D_{{\Gamma}_{\Phi_{2,2}}}{{\mathcal G}}={{\mathcal G}}$. For the last identity it is again used that ${\Gamma}_{\Phi_{1,2}}$ is a contraction and $\Phi_{1,1}(0)=\Pi_{{\mathcal G}}$, which implies that ${\Gamma}_{\Phi_{1,2}}{{\mathcal G}}={{\mathcal G}}_0={\textup{Ker\,}}S^*_{{{\mathcal G}}}$.

We have thus shown that ${{\mathcal K}}$ is a backward shift invariant subspace of $H^2({{\mathcal G}})$ with ${{\mathcal G}}_0\subset{{\mathcal K}}$. It then follows from Lemma \[L:DenseRange\] that ${{\mathcal K}}=H^2({{\mathcal G}})$ if and only if ${\widetilde\omega}_{B_c}{\widetilde{{\mathcal F}}}_{B_c}=S^*_{{{\mathcal G}}}{{\mathcal K}}={{\mathcal K}}$. This is the same as saying that ${\Gamma}_{\Phi_{1,2}}{{\mathcal D}}_A$ is dense in $H^2({{\mathcal G}})$ if and only if ${\omega}_{B_c}{{\mathcal F}}_{B_c}={{\mathcal D}}_{{\Gamma}_{\Phi_{2,2}}}$.

As a result of Theorem \[T:inverse\] we obtain the following analogue of Proposition \[P:CentralNonProper\] in the setting of general Redheffer representations.

\[P:CentralNonProper2\] Let ${{\mathscr R}}_\Psi$ be a Redheffer representation with Redheffer coefficients $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ as in such that the coefficient matrix is unitary. Then the set $${{\mathscr W}}_{H_{{\underline{0}}}}:=\{V\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')\mid H_{{\underline{0}}}=H_V\}$$ is equal to $\{{\underline{0}}\}$ if and only if ${\Gamma}_{\Psi_{1,2}}{{\mathcal U}}$ is dense in $H^2({{\mathcal E}})$.

The result follows immediately from Theorem \[T:inverse\] and Proposition \[P:CentralNonProper\].

Intermezzo: Redheffer representations {#S:Redheffer}
=====================================

Assume we are given Hilbert spaces ${{\mathcal U}}$, ${{\mathcal Y}}$, ${{\mathcal E}}$ and ${{\mathcal E}}'$ and operator-valued functions $$\label{RedCoefs2}
\Psi_{1,1}\in{{\mathbf S}}({{\mathcal E}}',{{\mathcal E}}),\ \
\Psi_{1,2}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal E}}),\ \
\Psi_{2,1}\in{{\mathbf S}}({{\mathcal E}}',{{\mathcal Y}}),\ \
\Psi_{2,2}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal Y}})$$ with $\Psi_{1,1}(0)=0$. In the present section we derive some results for the Redheffer representation ${{\mathscr R}}_\Psi$ with coefficients $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$.

In case the coefficient matrix $$\label{RedMat2}
K_0={\ensuremath{\left[\begin{array}{cc}M_{\Psi_{1,1}}&{\Gamma}_{\Psi_{1,2}}\\M_{\Psi_{2,1}}&{\Gamma}_{\Psi_{2,2}}\end{array} \right]}}
{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal E}}')\\{{\mathcal U}}\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal E}})\\H^2({{\mathcal Y}})\end{array} \right]}}$$ is a co-isometry, it follows from Theorems \[T:inverse\] and \[T:RCLRedheffer\] that ${{\mathscr R}}_\Psi$ maps ${{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ into ${{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal Y}})$. The next proposition shows that this is still true if the coefficient matrix is a contraction.

\[P:Redheffer\] Let $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ be as in with $\Psi_{1,1}(0)=0$ and assume that is contractive. Then ${{\mathscr R}}_\Psi$ maps ${{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ into ${{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal Y}})$.

For the case that the coefficient matrix $K_0$ is a co-isometry, the statement follows right away from Theorems \[T:RCLRedheffer\] and \[T:inverse\]. Now assume that $K_0$ is contractive. By Parrot’s lemma [@P78] there exist a Hilbert space ${{\mathcal V}}$ and functions $\Psi_{1,3}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal V}},{{\mathcal E}})$ and $\Psi_{2,3}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal V}},{{\mathcal Y}})$ such that $${\widetilde{K}}_0={\ensuremath{\left[\begin{array}{ccc}M_{\Psi_{1,1}}&{\Gamma}_{\Psi_{1,2}}&{\Gamma}_{\Psi_{1,3}}\\
M_{\Psi_{2,1}}&{\Gamma}_{\Psi_{2,2}}&{\Gamma}_{\Psi_{2,3}}\end{array} \right]}}:
{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal E}}')\\{{\mathcal U}}\\{{\mathcal V}}\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal E}})\\H^2({{\mathcal Y}})\end{array} \right]}}$$ is a co-isometry. Set $\widetilde\Psi_{1,2}({\lambda})={\ensuremath{\left[\begin{array}{cc}\Psi_{1,2}({\lambda})&\Psi_{1,3}({\lambda})\end{array} \right]}}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}}\oplus{{\mathcal V}},{{\mathcal E}})$ and $\widetilde\Psi_{2,2}({\lambda})={\ensuremath{\left[\begin{array}{cc}\Psi_{2,2}({\lambda})&\Psi_{2,3}({\lambda})\end{array} \right]}}\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}}\oplus{{\mathcal V}},{{\mathcal Y}})$. It follows that the Redheffer representation ${{\mathscr R}}_{\widetilde\Psi}$ with Redheffer coefficients $\Psi_{1,1}$, $\widetilde\Psi_{1,2}$, $\Psi_{2,1}$ and $\widetilde\Psi_{2,2}$ has a co-isometric coefficient matrix, and thus maps ${{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ into ${{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}}\oplus{{\mathcal V}},{{\mathcal Y}})$. The statement now follows because ${{\mathscr R}}_\Psi[V]({\lambda})={{\mathscr R}}_{\widetilde\Psi}[V]({\lambda})|_{{{\mathcal U}}}$ for each ${\lambda}\in{{\mathbb D}}$ and $V\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$. Hence $H_V\in{{\mathbf H}}^2_{\textup{ball}}({{\mathcal U}},{{\mathcal Y}})$.

As in the relaxed commutant lifting setting, the map ${{\mathscr R}}_\Psi$ is in general not one-to-one. This is the underlying obstruction preventing the obviously sufficient condition “${{\mathcal E}}=\{0\}$ or ${{\mathcal E}}'=\{0\}$” for ${{\mathscr R}}_\Psi[{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')]$ to be a singleton from also being necessary. Necessary and sufficient conditions under which the range of ${{\mathscr R}}_\Psi$ is a singleton, in the relaxed commutant lifting setting, were obtained in [@LT06] for the case that $R^*R=Q^*Q$ and in [@tH2] for the general case. An adaptation of the proof in [@tH2] gives the following result.

\[P:UniqueSolution\] Let $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ be as in with $\Psi_{1,1}(0)=0$ and assume that the coefficient matrix is contractive. Then the range of ${{\mathscr R}}_\Psi$ is a singleton if and only if $\Psi_{1,2}({\lambda})=0$ for each ${\lambda}\in{{\mathbb D}}$ or $\Psi_{2,1}({\lambda})=0$ for each ${\lambda}\in{{\mathbb D}}$. In case the coefficient matrix is a co-isometry, the range of ${{\mathscr R}}_\Psi$ is a singleton if and only if ${{\mathcal E}}=\{0\}$ or ${\Gamma}_{\Psi_{2,2}}$ is a co-isometry.

The statement for the case that the coefficient matrix is a co-isometry follows from Theorem \[T:inverse\] above and Theorem 0.2 in [@tH2]. The result for the case that the coefficient matrix is contractive can easily be deduced from the proof of Theorem 0.2 in [@tH2].

We now restrict our attention to the Schur class functions in the open unit ball ${{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ of ${{\mathbf H}}^\infty({{\mathcal E}},{{\mathcal E}}')$, i.e., $V\in{{\mathbf H}}^\infty({{\mathcal E}},{{\mathcal E}}')$ with $\|V\|_\infty<1$. For $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ the multiplication operator $M_V$ is a strict contraction, which enables us to express the operator ${\Gamma}_{H_V}$ associated with $H_V={{\mathscr R}}_\Psi[V]$ via explicitly as $$\label{GaHVexplicit}
{\Gamma}_{H_V}={\Gamma}_{\Psi_{2,2}}+M_{\Psi_{2,1}}M_V(I-M_{\Psi_{1,1}}M_V)^{-1}{\Gamma}_{\Psi_{1,2}}.$$ We write $R_V$ for the rotation operator associated with the strict contraction $M_V$: $$\label{RotOperV2}
R_V:={\ensuremath{\left[\begin{array}{cc}M_V&D_{M_V^*}\\-D_{M_V}&M_V^*\end{array} \right]}}
:{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal E}})\\H^2({{\mathcal E}}')\end{array} \right]}}\to{\ensuremath{\left[\begin{array}{c}H^2({{\mathcal E}}')\\H^2({{\mathcal E}})\end{array} \right]}}.$$ Then $R_V$ is unitary, and, since $I-M_{\Psi_{1,1}}M_V$ is invertible, we can form the Redheffer product $K_V=R_V\circ K_0$ of $R_V$ with the coefficient matrix $K_0$ in , which gives us the following block operator matrix $$\label{KV2}
K_V={\ensuremath{\left[\begin{array}{cc}M_V^*+D_{M_V}(I-M_{\Psi_{1,1}}M_V)^{-1}M_{\Psi_{1,1}}&
-D_{M_V}(I-M_{\Psi_{1,1}}M_V)^{-1}{\Gamma}_{\Psi_{1,2}}\\
M_{\Psi_{2,1}}(I-M_VM_{\Psi_{1,1}})^{-1}D_{M_V^*}&
{\Gamma}_{H_V}\end{array} \right]}}$$ that maps $H^2({{\mathcal E}}')\oplus{{\mathcal U}}$ into $H^2({{\mathcal E}})\oplus H^2({{\mathcal Y}})$. Note that in case $V=\underline{0}$, the constant function with value $0\in{{\mathscr L}}({{\mathcal E}},{{\mathcal E}}')$, $K_V$ is equal to the coefficient matrix $K_0$.

\[P:KVMat2\] For each $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ the operator matrix $K_V$ is contractive, co-isometric, isometric or unitary whenever the coefficient matrix $K_0$ in is contractive, co-isometric, isometric or unitary, respectively.

The result follows right away from Proposition \[P:CoOpMatInv\] and the fact that the rotation operator $R_V$ is unitary.

Proof of Theorem \[T:NonProper2\] {#S:nonproper}
=================================

Before giving the proof of Theorem \[T:NonProper2\], we first prove the following Redheffer representation analogue.

\[T:NonProper2RED\] Let $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ be as in with $\Psi_{1,1}(0)=0$ and assume that the coefficient matrix $K_0$ in is unitary. For each $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ the set $$\label{nonuniqueSet2}
{{\mathscr W}}_{H_V}=\{{\tilde{V}}\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}') \mid H_V=H_{{\tilde{V}}}\}$$ is equal to $\{V\}$ if and only if ${\Gamma}_{\Psi_{1,2}}{{\mathcal U}}$ is dense in $H^2({{\mathcal E}})$. In particular, if ${{\mathscr W}}_{H_V}=\{V\}$ for some $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$, then ${{\mathscr W}}_{H_V}=\{V\}$ for all $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$.

By Theorem \[T:inverse\] there exist a lifting data set ${\{A,T',U',R,Q\}}$ with underlying contraction ${\omega}$ as in and associated Redheffer coefficients $\Phi_{1,1}$, $\Phi_{1,2}$, $\Phi_{2,1}$ and $\Phi_{2,2}$ given by , and unitary operators $\psi$ mapping ${{\mathcal E}}$ onto ${{\mathcal G}}={{\mathcal D}}_A\ominus{{\mathcal F}}$ and ${\varphi}$ from ${{\mathcal E}}'$ onto ${{\mathcal D}}_{{\omega}^*}$ such that holds. Thus $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ can be identified with coefficients $\Phi_{1,1}$, $\Phi_{1,2}$, $\Phi_{2,1}$ and $\Phi_{2,2}$, up to the two unitary maps $\psi$ and ${\varphi}$. In particular, the coefficient matrix is unitary as well, which, according to Theorem \[T:RCLRedheffer\], implies that ${\omega}$ is an isometry.

Now fix a $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$. The relation between the two sets of Redheffer coefficients and the lifting data set ${\{A,T',U',R,Q\}}$ implies that we can define a contraction ${\omega}_{H_V}$ by $${\omega}_{H_V}:{{\mathcal F}}_{H_V}={{\overline{D_{{\Gamma}_{H_V}}{{\mathcal F}}}}}\to{{\mathcal D}}_{{\Gamma}_{H_V}},\quad
{\omega}_{H_V}D_{{\Gamma}_{H_V}}|_{{\mathcal F}}=D_{{\Gamma}_{H_V}}{\omega}_2.$$ In fact, ${\omega}_{H_V}$ is an isometry, because ${\omega}$ is an isometry; see the paragraph preceding Proposition \[P:NonProper1\]. Moreover, by Proposition \[P:NonProper1\] and from the fact that ${\omega}_{H_V}$ is an isometry, it follows that ${{\mathscr W}}_{H_V}=\{V\}$ if and only if ${{\mathcal F}}_{H_V}={{\mathcal D}}_{{\Gamma}_{H_V}}$ or ${\omega}_{H_V}{{\mathcal F}}_{H_V}={{\mathcal D}}_{{\Gamma}_{H_V}}$.

Since the coefficient matrix is unitary, we obtain from Proposition \[P:KVMat2\] that the operator $K_V$ in is unitary. In particular, $$D_{{\Gamma}_{H_V}}^2=I-{\Gamma}_{H_V}^*{\Gamma}_{H_V}=C_V^*C_V,$$ where we define $C_V$ by $$C_V=Y_V{\Gamma}_{\Psi_{1,2}}\quad\text{with}\quad Y_V=D_{M_V}(I-M_{\Psi_{1,1}}M_V)^{-1}.$$ Set ${{\mathcal K}}_V={{\overline{C_V{{\mathcal U}}}}}$. According to Douglas’ factorization lemma [@D66], the identity $\rho D_{{\Gamma}_{H_V}}=C_V$ defines a unitary map $\rho$ from ${{\mathcal D}}_{{\Gamma}_{H_V}}$ onto ${{\mathcal K}}_V$. Set $${\widetilde\omega}_{H_V}=\rho{\omega}_{H_V}\rho^*|_{{\widetilde{{\mathcal F}}}_{H_V}},\quad\text{where}\quad{\widetilde{{\mathcal F}}}_{H_V}=\rho{{\mathcal F}}_{H_V}.$$ Then ${\widetilde\omega}_{H_V}$ is an isometry from ${\widetilde{{\mathcal F}}}_{H_V}$ into ${{\mathcal K}}_V$, and ${{\mathscr W}}_{H_V}=\{V\}$ if and only if $$\label{aaa}
{\widetilde{{\mathcal F}}}_{H_V}={{\overline{Y_V{\Gamma}_{\Psi_{1,2}}{{\mathcal F}}}}}={{\mathcal K}}_V\quad\text{or}\quad
{\widetilde\omega}_{H_V}{\widetilde{{\mathcal F}}}_{H_V}={{\overline{Y_V{\Gamma}_{\Psi_{1,2}}{\omega}_2{{\mathcal F}}}}}={{\mathcal K}}_V.$$ Since $\|M_V\|=\|V\|_\infty<1$, $D_{M_V}$ is invertible, and hence $Y_V$ is invertible. But then holds if and only if $$\label{bbb}
{{\overline{{\Gamma}_{\Psi_{1,2}}{{\mathcal F}}}}}={{\overline{{\Gamma}_{\Psi_{1,2}}{{\mathcal U}}}}}\quad\text{or}\quad
{{\overline{{\Gamma}_{\Psi_{1,2}}{\omega}_2{{\mathcal F}}}}}={{\overline{{\Gamma}_{\Psi_{1,2}}{{\mathcal U}}}}}.$$ The conditions in are equal to those in in case $V={\underline{0}}$, the constant function in ${{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ whose value is the zero operator. This case was resolved in Proposition \[P:CentralNonProper2\], from which we infer that ${{\mathscr W}}_{H_V}=\{V\}$ if and only if ${\Gamma}_{\Psi_{1,2}}{{\mathcal U}}$ is dense in $H^2({{\mathcal E}})$.

Let ${\{A,T',U',R,Q\}}$ be a lifting data set such that the underlying contraction ${\omega}$ is an isometry and ${\omega}_2\Pi_{{\mathcal F}}$ on ${{\mathcal D}}_A$ is strongly stable. Then the coefficient matrix for the Redheffer coefficients associated with the lifting data set ${\{A,T',U',R,Q\}}$ is unitary according to Theorem \[T:RCLRedheffer\].

Moreover, for any $V\in{{\mathbf S}}({{\mathcal D}}_A,{{\mathcal D}}_{T'})$, since, by Parrott’s lemma [@P78], the operators $B_V$ and ${\Gamma}_V$ in determine each other uniquely, it follows that ${{\mathscr V}}_{B_V}={{\mathscr W}}_{H_V}$, where ${{\mathscr V}}_{B_V}$ and ${{\mathscr W}}_{H_V}$ are given by and , respectively. Therefore, we obtain Theorem \[T:NonProper2\] by applying Theorem \[T:NonProper2RED\] to the Redheffer representation ${{\mathscr R}}_\Phi$.

Proof of Theorem \[T:MaxPrinciple\] {#S:maxprinciple}
===================================

In this section we derive Theorem \[T:MaxPrinciple\] from the following harmonic maximal principle for Redheffer representations.

\[T:MaximumPrinciple\] Let $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ be as in with $\Psi_{1,1}(0)=0$ and assume that the coefficient matrix is contractive. Then, for the Redheffer representation ${{\mathscr R}}_\Psi$ we have:

-   If there exists a ${\widetilde{V}}\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ with $\|{\Gamma}_{H_{{\widetilde{V}}}}\|<1$, then $\|{\Gamma}_{H_{V}}\|<1$ for any $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$.

-   If there exists a ${\widetilde{V}}\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ with $\|{\Gamma}_{H_{{\widetilde{V}}}}\|=1$, then $\|{\Gamma}_{H_{V}}\|=1$ for any $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$.

-   Moreover, for $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ and ${\widetilde{V}}\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ we have ${\textup{Ker\,}}D_{{\Gamma}_{H_V}}\subset{\textup{Ker\,}}D_{{\Gamma}_{H_{{\widetilde{V}}}}}$. In particular, ${\textup{Ker\,}}D_{{\Gamma}_{H_V}}={\textup{Ker\,}}D_{{\Gamma}_{H_{{\widetilde{V}}}}}$ in case ${\widetilde{V}},V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$.

We first prove (3). Let $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ and $u\in{\textup{Ker\,}}D_{{\Gamma}_{H_V}}$, i.e., $\|{\Gamma}_{H_V}u\|=\|u\|$. Since the operator matrix $K_V$ of Proposition \[P:KVMat2\] is contractive, it follows that $-D_{M_V}(I-M_{\Psi_{1,1}}M_V)^{-1}{\Gamma}_{\Psi_{1,2}}u=0$. The operator $D_{M_V}(I-M_{\Psi_{1,1}}M_V)^{-1}$ is invertible, because $\|M_V\|<1$. Thus ${\Gamma}_{\Psi_{1,2}}u=0$. In other words, $\Psi_{1,2}({\lambda})u=0$ for each ${\lambda}\in{{\mathbb D}}$. This implies that $H_{{\widetilde{V}}}({\lambda})u=\Psi_{2,2}({\lambda})u$ for any ${\widetilde{V}}\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$ and ${\lambda}\in{{\mathbb D}}$. Then ${\Gamma}_{H_{{\widetilde{V}}}}u={\Gamma}_{\Psi_{2,2}}u={\Gamma}_{H_{V}}u$, and thus $\|{\Gamma}_{H_{{\widetilde{V}}}}u\|=\|{\Gamma}_{H_{V}}u\|=\|u\|$. Hence $u\in{\textup{Ker\,}}D_{{\Gamma}_{H_{{\widetilde{V}}}}}$ for any ${\widetilde{V}}\in{{\mathbf S}}({{\mathcal E}},{{\mathcal E}}')$.

Note that (1) is just the contrapositive of (2). To see that (2) holds, we use a similar argument as in the proof of (3). Assume we have a $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ with $\|{\Gamma}_{H_{V}}\|=1$. Let $u_n$, $n=0,1,\ldots$, be a sequence with $\|u_n\|=1$ and $\|{\Gamma}_{H_{V}}u_n\|\to1$. Since $K_V$ in Proposition \[P:KVMat2\] is contractive, $-D_{M_V}(I-M_{\Psi_{1,1}}M_V)^{-1}{\Gamma}_{\Psi_{1,2}}u_n\to0$, and thus ${\Gamma}_{\Psi_{1,2}}u_n\to0$; again because $D_{M_V}(I-M_{\Psi_{1,1}}M_V)^{-1}$ is invertible. Now let ${\widetilde{V}}$ be an arbitrary function in ${{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$. Using that ${\Gamma}_{H_{{\widetilde{V}}}}$ and ${\Gamma}_{H_V}$ can be written explicitly as in , it follows that $\lim_{n\to\infty} {\Gamma}_{H_{{\widetilde{V}}}}u_n
=\lim_{n\to\infty} {\Gamma}_{\Psi_{2,2}}u_n=\lim_{n\to\infty} {\Gamma}_{H_{V}}u_n$. In particular, $\|{\Gamma}_{H_{{\widetilde{V}}}}u_n\|\to1$. Thus $\|{\Gamma}_{H_{{\widetilde{V}}}}\|=1$.

Assume there exists a ${\widetilde{V}}\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ with $\|B_{{\widetilde{V}}}\|<1$. Let $H_{{\widetilde{V}}}={{\mathscr R}}_\Phi[{\widetilde{V}}]$, with $\Phi$ defined by . Then both $A$ and ${\Gamma}_{H_{{\widetilde{V}}}}$ are strict contractions. By Theorem \[T:MaximumPrinciple\] it then follows for any $V\in{{\mathbf S}}_0({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ that the operator ${\Gamma}_{H_{V}}$ is a strict contraction, and thus that $B_V$ is a strict contraction. This proves Part 1. Part 2 follows because it is the contrapositive of Part 1.

To see that the third statement holds, note that for any $V\in{{\mathbf S}}({{\mathcal G}},{{\mathcal D}}_{{\omega}^*})$ the contractive interpolant $B_V$ in satisfies $$\|D_{V}h\|=\|D_{{\Gamma}_{H_V}}D_A h\|\quad(h\in{{\mathcal H}}).$$ Thus $h\in{{\mathcal H}}$ is in ${\textup{Ker\,}}D_{V}$ if and only if $D_A h$ is in ${\textup{Ker\,}}D_{{\Gamma}_V}$. The statement then follows by applying Part 3 of Theorem \[T:MaximumPrinciple\] to the Redheffer representation ${{\mathscr R}}_\Phi$.

We conclude with the some bounds on $\|{\Gamma}_{H_V}\|$ for $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$.

\[P:bounds\] Let $\Psi_{1,1}$, $\Psi_{1,2}$, $\Psi_{2,1}$ and $\Psi_{2,2}$ be as in with $\Psi_{1,1}(0)=0$ and assume that the coefficient matrix is contractive. For any $V\in{{\mathbf S}}_0({{\mathcal E}},{{\mathcal E}}')$ and $u\in{{\mathcal U}}$ we have $$\label{bound1}
\|{\Gamma}_{H_V}u\|^2\leq\|u\|^2-\frac{1-\|M_V\|^2}{\|I-M_{\Psi_{1,1}}M_V\|^2}\|{\Gamma}_{\Psi_{1,2}}u\|^2.$$ In particular, $$\label{bound2}
\|{\Gamma}_{H_V}u\|^2\leq\|u\|^2-\frac{1-\|M_V\|}{1+\|M_V\|}\|{\Gamma}_{\Psi_{1,2}}u\|^2.$$ If, in addition, the coefficient matrix is an isometry, then $$\label{bound3}
\|{\Gamma}_{H_V}u\|^2\leq\frac{2\|M_V\|}{1+\|M_V\|}\|u\|^2
+\frac{1-\|M_V\|}{1+\|M_V\|}\|{\Gamma}_{\Psi_{2,2}}u\|^2.$$

From the fact that the operator matrix $K_V$ in is contractive we obtain $$\|{\Gamma}_{H_V}u\|^2\leq \|u\|^2-\|D_V(I-M_{\Psi_{11}}M_V)^{-1}{\Gamma}_{\Psi_{1,2}}u\|^2.$$ Using that for an invertible operator $T\in{{\mathcal L}}({{\mathcal H}})$ and $h\in{{\mathcal H}}$ we have the inequality $\|Th\|\geq\|T^{-1}\|^{-1}\|h\|$, it follows that $$\|{\Gamma}_{H_V}u\|^2\leq
\|u\|^2-\|D_V^{-1}\|^{-2}\|(I-M_{\Psi_{11}}M_V)\|^{-2}\|{\Gamma}_{\Psi_{1,2}}u\|^2.$$ The first inequality for $\|{\Gamma}_{H_V}u\|$ holds because $\|D_{M_V}^{-1}\|\leq(1-\|M_V\|^2)^{-{\frac{1}{2}}}$. Indeed, for any $h\in H^2({{\mathcal G}})$ we have $\|D_{M_V}h\|^2=\|h\|^2-\|M_Vh\|^2\geq(1-\|M_V\|^2)\|h\|^2$. So for $h=D_{M_{V}}^{-1}k$ we obtain $\|k\|^2\geq (1-\|M_V\|^2)\|D_{M_{V}}^{-1}k\|^2$, which proves our claim.

To see that the second inequality holds, we use $$\|I-M_{\Psi_{1,1}}M_V\|\leq 1+\|M_{\Psi_{1,1}}M_V\|\leq1+\|M_V\|,$$ in order to obtain $$\frac{1-\|M_V\|^2}{\|I-M_{\Psi_{1,1}}M_V\|^2}\geq
\frac{1-\|M_V\|^2}{(1+\|M_V\|)^2}=\frac{1-\|M_V\|}{1+\|M_V\|}.$$

In case the coefficient matrix $K_0$ is an isometry, $\|{\Gamma}_{\Psi_{1,2}}u\|^2=\|u\|^2-\|{\Gamma}_{\Psi_{2,2}}u\|^2$. Hence, starting from the second inequality, $$\begin{aligned}
\|{\Gamma}_{H_V}u\|^2&\leq&
\|u\|^2-\frac{1-\|M_V\|}{1+\|M_V\|}(\|u\|^2-\|{\Gamma}_{\Psi_{2,2}}u\|^2)\\
&=&\frac{2\|M_V\|}{1+\|M_V\|}\|u\|^2+\frac{1-\|M_V\|}{1+\|M_V\|}\|{\Gamma}_{\Psi_{2,2}}u\|^2.\end{aligned}$$

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