A Schwarz Lemma for the Pentablock

Abstract

In this paper, we prove a Schwarz lemma for the pentablock. The pentablock $\mathcal {P}$ is defined by

$$\begin{aligned} \mathcal {P}=\{(a_{21}, {\text {tr}}A, \det A) : A=[a_{ij}]_{i,j=1}^2 \in \mathbb {B}^{2\times 2}\} \end{aligned}$$

where $\mathbb {B}^{2\times 2}$ denotes the open unit ball in the space of $2\times 2$ complex matrices. The pentablock is a bounded non-convex domain in $\mathbb {C}^3$ which arises naturally in connection with a certain problem of $\mu$-synthesis. We develop a concrete structure theory for the rational maps from the unit disc $\mathbb {D}$ to the closed pentablock $\overline{\mathcal {P}}$ that map the unit circle $\mathbb {T}$ to the distinguished boundary $b\overline{\mathcal {P}}$ of $\overline{\mathcal {P}}$. Such maps are called rational ${\overline{\mathcal {P}}}$-inner functions. We give relations between ${\overline{\mathcal {P}}}$-inner functions and inner functions from $\mathbb {D}$ to the symmetrized bidisc. We describe the construction of rational ${\overline{\mathcal {P}}}$-inner functions $x = (a, s, p) : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ of prescribed degree from the zeroes of a, s and $s^2-4p$. The proof of this theorem is constructive: it gives an algorithm for the construction of a family of such functions x subject to the computation of Fejér–Riesz factorizations of certain non-negative trigonometric functions on the circle. We use properties and the construction of rational ${\overline{\mathcal {P}}}$-inner functions to prove a Schwarz lemma for the pentablock.

1 Introduction

An unsolved problem in $H^\infty$ control theory led us to consider inner rational mappings from $\mathbb {D}$ to certain domains in $\mathbb {C}^d$ which arise in connection with the $\mu$-synthesis problem. One such domain is the pentablock. Other well-known examples of such domains are the symmetrized bidisc $\Gamma$ and the tetrablock. We should mention papers on the construction of rational $\Gamma$-inner functions [5, 6] and rational tetra-inner functions [15, 16] for the symmetrized bidisc $\Gamma$ and the tetrablock, respectively. The pentablock $\mathcal {P}$ was introduced by Agler, Lykova and Young in [4] in 2015. It was shown there that $\mathcal {P}$ arises naturally in the context of $\mu$-synthesis.

Definition 1.1

[4] The open pentablock is the domain defined by

$$\begin{aligned} \mathcal {P}=\{(a_{21}, {\text {tr}}A, \det A) : A=[a_{ij}]_{i,j=1}^2 \in \mathbb {B}^{2\times 2}\}, \end{aligned}$$

(1.1)

where $\mathbb {B}^{2\times 2}$ denotes the open unit ball in the space of $\;\;2\times 2$ complex matrices with respect to the operator norm arising from the standard inner product on $\mathbb {C}^2$.

Recall [20] that the structured singular value $\mu _E$ of $A\in \mathbb {C}^{m\times n}$ corresponding to subspace E of $\mathbb {C}^{n\times m}$ is defined by

$$\begin{aligned} \frac{1}{\mu _E(A)} = \inf \{\Vert X\Vert : X\in E \text{ and } \det (1-AX)=0\}. \end{aligned}$$

(1.2)

The cost function $\mu _E$ plays a central role in the “$H^\infty$ approach” to the problem of stabilizing a linear system in a way that is maximally robust with respect to structured uncertainty. This approach, developed and promoted by Doyle and Stein [21], reduces the “robust stabilization problem” to the solution of a variant of the classical Nevanlinna–Pick problem for matrix-valued functions, in which the cost function to be minimized is given by $\mu _E$ for some uncertainty space E, in place of the usual operator norm.

To date there is not a satisfactory mathematical treatment of this “$\mu$-synthesis problem” in general, and so mathematicians have studied some special cases, such as for $2\times 2$-matrix-valued functions and for some natural choices of the space E. In particular the authors of [4] investigated the following special case of $\mu _E$.

Definition 1.2

Let

$$\begin{aligned} E=\mathrm {span~}\left\{ 1,\begin{bmatrix} 0&{}1\\ 0&{}0 \end{bmatrix}\right\} \subset \mathbb {C}^{2\times 2}, \end{aligned}$$

$\mathcal {P}_\mu$ is the domain in $\mathbb {C}^3$ given by

$$\begin{aligned} \mathcal {P}_\mu = \{(a_{21},{\text {tr}}A,\det A): A\in \mathbb {C}^{2\times 2}, \, \mu _E(A)<1\} \subset \mathbb {C}^3. \end{aligned}$$

(1.3)

It was proved in [4] that $\mathcal {P}=\mathcal {P}_\mu$.

The pentablock $\mathcal {P}$ is a region in 3-dimensional complex space which intersects $\mathbb {R}^3$ in a convex body bounded by five faces, comprising two triangles, an ellipse and two curved surfaces [4]. The closure of $\mathcal {P}$ is denoted by $\overline{\mathcal {P}}$.

In this paper, we study rational $\overline{\mathcal {P}}$-inner functions. We define a rational $\overline{\mathcal {P}}$-inner function to be a rational analytic function from $\mathbb {D}$ into $\overline{\mathcal {P}}$ which maps $\mathbb {T}$ into $b\overline{\mathcal {P}}$, where $b\overline{\mathcal {P}}$ is the distinguished boundary of $\mathcal {P}$. The distinguished boundary $b\overline{\mathcal {P}}$ of $\mathcal {P}$ is

$$\begin{aligned} b\overline{\mathcal {P}} = \bigg \{(a, s, p) \in \mathbb {C}^3 : |s| \le 2, \ |p|=1, \ s = \overline{s}p \; \text {and} \; |a| = \sqrt{1-\frac{1}{4}|s|^2}\bigg \}, \end{aligned}$$

see [4]. The degree of a rational $\overline{\mathcal {P}}$-inner function $x = (a, s, p)$ is defined to be the pair of numbers $(deg \; a, deg \;p)$. We say that deg $x \le (m, n)$ if deg $a \le m$ and deg $p \le n$. The group of automorphisms of the pentablock was studied in [4] and [26].

Recall that a classical rational inner function is a rational map f from the unit disc $\mathbb {D}$ to its closure ${\overline{\mathbb {D}}}$ with the property that f maps the unit circle $\mathbb {T}$ into itself. A survey of results connecting inner functions and operator theory is given in [17]. All rational inner functions from the unit disc $\mathbb {D}$ to its closure ${\overline{\mathbb {D}}}$ are finite Blaschke products.

Definition 1.3

[9, p. 2] A finite Blaschke product is a function of the form

$$\begin{aligned} B(z) = c\prod ^n_{i=1}B_{\alpha _i}(z) \ \ \ \ \text {for} \ z\in \mathbb {C}\setminus \{1/\overline{\alpha _1},\dots ,1/\overline{\alpha _n}\}, \end{aligned}$$

(1.4)

where $B_{\alpha _i}(z) = \frac{z-\alpha _i}{1-\overline{\alpha _i}z}$, $|c|=1$ and $\alpha _1, \dots , \alpha _n \in \mathbb {D}$.

We have proved several results on the description and the construction of rational $\overline{\mathcal {P}}$-inner functions and on the connections between rational $\Gamma$-inner functions and rational $\overline{\mathcal {P}}$-inner functions.

One of our main results is the construction of a rational $\overline{\mathcal {P}}$-inner function $x = (a, s, p)$ of prescribed degree from the zeros of a and s and $s^2 -4p$. The zeros of $s^2 -4p$ in $\overline{\mathbb {D}}$ are called the royal nodes of (s, p). One can consider this result as an analogue of the expression (1.4) for a finite Blaschke product in terms of its zeros. Concretely, the following result is a corollary of Theorem 8.6.

Theorem 1.4

Let n, m be positive integers and suppose the following points are given

• (1) $\alpha _1, \alpha _2, \dots , \alpha _{k_0} \in \mathbb {D}$ and $\eta _1, \eta _2, \dots , \eta _{k_1} \in \mathbb {T}$, where $2k_0+k_1=n$;

• (2) $\beta _1, \beta _2, \dots , \beta _m \in \mathbb {D}$;

• (3) $\sigma _1, \dots , \sigma _n$ in $\overline{\mathbb {D}}$ which are distinct from $\eta _1, \dots , \eta _{k_1}$.

Then there exists a rational $\overline{\mathcal {P}}$-inner function $x = (a, s, p)$ of degree $\le (m+n, n)$ such that the zeros of a in $\mathbb {D}$ are $\beta _1, \beta _2, \dots , \beta _m$, the zeros of s in $\overline{\mathbb {D}}$ are $\alpha _1, \alpha _2, \dots , \alpha _{k_0}$, $\eta _1, \eta _2, \dots , \eta _{k_1}$, and the royal nodes of (s, p) are $\sigma _1, \dots , \sigma _n$.

There is a well-developed theory of Schwarz lemmas for various domains by many authors, including Dineen and Harris [19, 23]. In particular, for the symmetrized bidisc and the tetrablock, see [1, 11, 22]. Connections established in this paper between ${\overline{\mathcal {P}}}$-inner functions and $\Gamma$-inner functions (especially Theorem 7.8) and a Schwarz lemma for the symmetrized bidisc due to Agler and Young [11] allow us to prove a Schwarz lemma for the pentablock (Theorem 11.3).

Theorem 1.5

Let $\lambda _0 \in \mathbb {D}\setminus \{0\}$ and $(a_0, s_0, p_0) \in \overline{\mathcal {P}}$. Then the following conditions are equivalent:

• (i) there exists a rational $\overline{\mathcal {P}}$-inner function $x=(a, s, p)$, $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$;

• (ii) there exists an analytic function $x=(a, s, p)$, $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$, and $|a_0| \; \le \; |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$;

• (iii)
  $$\begin{aligned} \dfrac{2|s_0 - p_0\overline{s}_0| + |s_{0}^{2} - 4p_0|}{4 - |s_0|^2} \le |\lambda _0|\;\; \text {and} \;\;|s_0| <2, \end{aligned}$$

  (1.5)

  and

  $$\begin{aligned} |a_0| \; \le \; |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}. \end{aligned}$$

  (1.6)

The construction of an interpolating function $x=(a, s, p)$, $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$ is given in Theorems 11.2 and 11.3.

The authors are grateful to Nicholas Young for some helpful suggestions.

2 The Pentablock $\mathcal {P}$ and the Symmetrized Bidisc $\Gamma$

In 1999, Agler and Young introduced the symmetrized bidisc in [10]. Following [10], we shall often use the co-ordinates (s, p) for points in the symmetrized bidisc $\mathbb {G}$, chosen to suggest ‘sum’ and ‘product’.

Definition 2.1

The symmetrized bidisc is the set

$$\begin{aligned} \mathbb {G} {\mathop {=}\limits ^{\textrm{def}}} \{(z+w,zw):|z| \;<\; 1, |w| \; <\; 1\}, \end{aligned}$$

(2.1)

and its closure is

$$\begin{aligned} \Gamma {\mathop {=}\limits ^{\textrm{def}}} \{(z+w,zw):|z| \le 1, |w| \le 1\}. \end{aligned}$$

The following results from [2] give useful criteria for membership of $\mathbb {G}$, of the distinguished boundary $b\Gamma$ of $\Gamma$ and of the topological boundary $\partial \Gamma$ of $\Gamma$.

Proposition 2.2

[2, Proposition 3.2] Let (s, p) belong to $\mathbb {C}^2$. Then

• (1) (s, p) belongs to $\mathbb {G}$ if and only if

  $$\begin{aligned} |s- \overline{s}p| \; <\; 1-|p|^2; \end{aligned}$$

• (2) (s, p) belongs to $\Gamma$ if and only if

  $$\begin{aligned} |s| \le 2 \textit{ and } |s- \overline{s}p| \le 1-|p|^2; \end{aligned}$$

• (3) (s, p) lies in $b\Gamma$ if and only if

  $$\begin{aligned} |p|=1, |s| \le 2 \textit{ and } s- \overline{s}p = 0. \end{aligned}$$

The following terminology was introduced in [12].

Definition 2.3

The royal variety $\mathcal {R}_{\Gamma }$ of the symmetrized bidisc is

$$\begin{aligned} \mathcal {R}_{\Gamma } = \{(s, p) \in \mathbb {C}^2 : s^2 = 4p\}. \end{aligned}$$

Lemma 2.4

[14, Lemma 4.3] Every automorphism of $\mathbb {G}$ maps the royal variety $\mathcal {R}_{\Gamma } \cap \mathbb {G}$ onto itself.

The royal variety is the only complex geodesic in the symmetrized bidisc that is invariant under all automorphisms of $\mathbb {G}$ [8].

Remark 2.5

The pentablock is closely related to the symmetrized bidisc. Indeed, Definition 1.1 shows that $\mathcal {P}$ is fibred over $\mathbb {G}$ by the map $(a, s, p) \mapsto (s, p)$, since if $A \in \mathbb {B}^{2\times 2}$ then the eigenvalues of A lie in $\mathbb {D}$ and so $({\text {tr}}A, \det A) \in \mathbb {G}$. Thus, for every point $(a, s, p) \in \mathcal {P}$, the point $(s,p) \in \mathbb {G}$.

Remark 2.6

In [26] Kosinski commented that the pentablock is a Hartogs Domain. It follows from the descriptions of the pentablock $\mathcal {P}$ in [4] that $\mathcal {P}$ can be seen as a Hartogs domain in $\mathbb {C}^3$ over the symmetrized bidisc $\mathbb {G}$, that is,

$$\begin{aligned} \mathcal {P} = \Big \{(a, s, p) \in \mathbb {D} \times \mathbb {G} : |a|^2 \; <\; e^{-\varphi (s, p)}\Big \}, \end{aligned}$$

where

$$\begin{aligned} \varphi (s, p) = -2\log \left| 1-\frac{\frac{1}{2}s{\overline{\beta }}}{1+\sqrt{1-|\beta |^2}}\right| , \end{aligned}$$

$(s, p) \in \mathbb {G}$ and $\beta = \frac{s-\overline{s}p}{1-|p|^2}$.

Hartogs domains are important objects in several complex variables.

Definition 2.7

[24, p. 259] A domain $D \subset \mathbb {C}^n$ is called $\mathbb {C}$-convex if for any complex line $\ell = a+b\mathbb {C}, \; 0 \ne a, b \in \mathbb {C}^n$ such that $\ell \cap D \ne \emptyset$, this intersection $\ell \cap D$ is connected and simply connected.

It is known that the pentablock is polynomially convex and starlike, see [4]. It was shown in [31] that the pentablock $\mathcal {P}$ is hyperconvex and that $\mathcal {P}$ cannot be exhausted by domains biholomorphic to convex ones. Later in [30, Theorem 1.1] it was proved that $\mathcal {P}$ is a $\mathbb {C}$-convex domain.

The following results from [4] give useful criteria for membership of $\mathcal {P}$.

Definition 2.8

[4, Definition 4.1] For $z \in \mathbb {D}$ and $(a, s, p) \in \mathbb {C}^3$ define $\Psi _z(a, s, p)$ by

$$\begin{aligned} \Psi _z(a, s, p) = \frac{a(1-|z|^2)}{1-sz+pz^2} \ \ \text { whenever } \ 1-sz+pz^2 \ne 0. \end{aligned}$$

(2.2)

The polynomial map implicit in the definition (1.1) can be written as

$$\begin{aligned} \pi (A)=(a_{21}, {\text {tr}}A, \det A) \; \text {for} \; A=[a_{ij}]_{i,j=1}^2 \in \mathbb {C}^{2\times 2}. \end{aligned}$$

(2.3)

Thus $\mathcal {P} = \pi (\mathbb {B}^{2\times 2})$.

Theorem 2.9

[4, Theorem 1.1] Let

$$\begin{aligned} (s,p)=(\lambda _1+\lambda _2,\lambda _1\lambda _2) \end{aligned}$$

here $\lambda _1$, $\lambda _2$ $\in$ $\mathbb {D}$. Let a $\in \mathbb {C}$ and let

$$\begin{aligned} \beta =\frac{s-\overline{s}p}{1-|p|^2}. \end{aligned}$$

Then $|\beta | \; <\; 1$ and the following statements are equivalent:

• (1) (a,s,p) $\in \mathcal {P}$, that is, there exists $A \in \mathbb {C}^{2\times 2}$ such that $\Vert A\Vert < 1$ and $\pi (A) = (a, s, p)$;

• (2) $|a| \; <\; |1-\frac{\frac{1}{2}s {\overline{\beta }}}{1+\sqrt{1-|\beta |^2}}|$;

• (3) $|a| \; <\; \frac{1}{2}|1-\overline{\lambda _2}\lambda _1|+\frac{1}{2}(1-|\lambda _1|^2)^\frac{1}{2}(1-|\lambda _2|^2)^\frac{1}{2}$;

• (4) $\sup _{z\in \mathbb {D}}|\Psi _{z} (a,s,p)| \; <\; 1$.

Theorem 2.10

[4, Theorem 5.3] Let

$$\begin{aligned} (s, p) = (\beta +{\overline{\beta }}p, p) = (\lambda _1+\lambda _2, \lambda _1\lambda _2)\in \Gamma , \end{aligned}$$

where $|\beta | \le 1$ and if $|p| = 1$ then $\beta = \frac{1}{2}s$. Let $a \in \mathbb {C}$. The following statements are equivalent:

• (1) $(a, s, p) \in \overline{\mathcal {P}}$;

• (2) $|a| \le |1-\frac{\frac{1}{2}s{\overline{\beta }}}{1+\sqrt{1-|\beta |^2}}|$;

• (3) $|a| \le \frac{1}{2}|1-{\overline{\lambda }}_2\lambda _1| + \frac{1}{2}(1-|\lambda _1|^2)^{\frac{1}{2}}(1-|\lambda _2|^2)^{\frac{1}{2}}$;

• (4) $|\Psi _z(a, s, p)| \le 1$ for all $z \in \mathbb {D}$, where $\Psi _z$ is defined by Eq. (2.2);

• (5) there exists $A \in \mathbb {C}^{2\times 2}$ such that $\Vert A\Vert \le 1$ and $\pi (A) = (a, s, p).$

3 The Distinguished Boundary of $\mathcal {P}$

Let $\Omega$ be a domain in $\mathbb {C}^n$ with closure ${\overline{\Omega }}$ and let $A(\Omega )$ be the algebra of continuous scalar functions on ${\overline{\Omega }}$ that are holomorphic on $\Omega$. A boundary for $\Omega$ is a subset C of ${\overline{\Omega }}$ such that every function in $A(\Omega )$ attains its maximum modulus on C. Since $\overline{\mathcal {P}}$ is polynomially convex, there is a smallest closed boundary of $\mathcal {P}$, contained in all the closed boundaries of $\mathcal {P}$, called the distinguished boundary of $\mathcal {P}$ and denoted by $b\overline{\mathcal {P}}$. If there is a function $g \in A(\mathcal {P})$ and a point $u \in \overline{\mathcal {P}}$ such that $g(u)=1$ and |g(x)| 1 for all $x \in \overline{\mathcal {P}} \backslash \{u\}$, then u must belong to $b\overline{\mathcal {P}}$. Such a point u is called a peak point of $\overline{\mathcal {P}}$ and the function g a peaking function for u. Define

$$\begin{aligned} K_0 {\mathop {=}\limits ^{def }}\bigg \{(a, s, p) \in \mathbb {C}^3 : (s, p) \in b\Gamma , \ |a| = \sqrt{1-\frac{1}{4}|s|^2}\bigg \}. \end{aligned}$$

and

$$\begin{aligned} K_1 {\mathop {=}\limits ^{def }}\bigg \{(a, s, p) \in \mathbb {C}^3 : (s, p) \in b\Gamma , \ |a| \le \sqrt{1-\frac{1}{4}|s|^2}\bigg \}. \end{aligned}$$

(3.1)

Proposition 3.1

[4, Proposition 8.3] The subsets $K_0$ and $K_1$ of $\overline{\mathcal {P}}$ are closed boundaries for $A(\mathcal {P})$.

Theorem 3.2

[4, Theorem 8.4] For $x \in \mathbb {C}^3$, the following are equivalent:

• (1) $x \in K_0$;

• (2) x is a peak point of $\overline{\mathcal {P}}$;

• (3) $x \in b\overline{\mathcal {P}}$, the distinguished boundary of $\mathcal {P}$. Therefore,

  $$\begin{aligned} b\overline{\mathcal {P}} = \bigg \{(a, s, p) \in \mathbb {C}^3 : (s, p) \in b\Gamma , \ |a| = \sqrt{1-\frac{1}{4}|s|^2}\bigg \} \end{aligned}$$

  and so

  $$\begin{aligned} b\overline{\mathcal {P}} = \bigg \{(a, s, p) \in \mathbb {C}^3 : |s| \le 2, \ |p|=1, \ s = \overline{s}p \ and \ |a| = \sqrt{1-\frac{1}{4}|s|^2}\bigg \}.\nonumber \\ \end{aligned}$$

  (3.2)

Theorem 3.3

[4, Theorem 8.5] The distinguished boundary $b\overline{\mathcal {P}}$ is homeomorphic to

$$\begin{aligned} \{(\sqrt{1-x^2}w, x, \theta ) : -1 \le x \le 1, \ 0 \le \theta \le 2\pi , \ w \in \mathbb {T}\} \end{aligned}$$

with the two points $(\sqrt{1-x^2}w, x, 0)$ and $(\sqrt{1-x^2}w, -x, 2\pi )$ identified for every $w \in \mathbb {T}$ and $x \in [-1, 1]$.

4 The Royal Variety of $\mathcal {P}$ and Aut $\mathcal {P}$

Recall that $\mathcal {P}=\pi (\mathbb {B}^{2\times 2})$ where $\pi :\mathbb {C}^{2\times 2}\rightarrow \mathbb {C}^3$ is defined as

$$\begin{aligned} \pi :A \mapsto (a_{21}, tr A, det A). \end{aligned}$$

(4.1)

We define the singular set of $\mathcal {P}=\pi (\mathbb {B}^{2\times 2})$ to be the image under $\pi$ of the set of critical points of $\pi$.

Proposition 4.1

The singular set of the pentablock is $\mathcal {R}_{\mathcal {P}}=\{(0, s, p )\in \mathcal {P}: s^2=4p\}$.

Proof

The set of critical points of $\pi$ is the set $\pi (\{A \in \mathbb {B}^{2\times 2}: \textbf{J}_{\pi }(A)$ is not of full rank}), where $\textbf{J}_{\pi }(A)$ is the Jacobian matrix of $\pi$.

The Jacobian matrix of $\pi$, $\textbf{J}_{\pi }(A)$, is defined by

$$\begin{aligned} \textbf{J}_{\pi }(A)= \begin{bmatrix} \dfrac{\partial \pi _1}{\partial a_{11}} &{} \dfrac{\partial \pi _1}{\partial a_{12}} &{} \dfrac{\partial \pi _1}{\partial a_{21}} &{} \dfrac{\partial \pi _1}{\partial a_{22}} \\ \\ \dfrac{\partial \pi _2}{\partial a_{11}} &{} \dfrac{\partial \pi _2}{\partial a_{12}} &{} \dfrac{\partial \pi _2}{\partial a_{21}} &{} \dfrac{\partial \pi _2}{\partial a_{22}} \\ \\ \dfrac{\partial \pi _3}{\partial a_{11}} &{} \dfrac{\partial \pi _3}{\partial a_{12}} &{} \dfrac{\partial \pi _3}{\partial a_{21}} &{} \dfrac{\partial \pi _3}{\partial a_{22}} &{} \end{bmatrix}\;\; \text {for} \;\; A= \begin{bmatrix} a_{11} &{} a_{12} \\ a_{21} &{} a_{22} \end{bmatrix} \in \mathbb {C}^{2 \times 2}. \end{aligned}$$

Thus,

$$\begin{aligned} \textbf{J}_\pi (A)= \begin{bmatrix} 0 &{} 0 &{} 1 &{} 0 \\ 1 &{} 0 &{} 0 &{} 1 \\ a_{22} &{} -a_{21} &{} -a_{12} &{} a_{11} \end{bmatrix}. \end{aligned}$$

Note that $\textbf{J}_{\pi }(A)$ is not of full rank if and only if rank $\textbf{J}_{\pi }(A) \le 2$. That means all $3\times 3$ minors of $\textbf{J}_{\pi }(A)$ are zero. Let us find all $3\times 3$ minors of $\textbf{J}_\pi (A)$.

$$\begin{aligned} \begin{vmatrix} 0&0&1\\ 1&0&0 \\ a_{22}&-a_{21}&-a_{12} \end{vmatrix}=-a_{21},&~~&\begin{vmatrix} 0&1&0\\0&0&1 \\ -a_{21}&-a_{12}&a_{11} \end{vmatrix}=-a_{21}, \\ \begin{vmatrix} 0&0&0\\1&0&1 \\ a_{22}&-a_{21}&a_{11} \end{vmatrix}=0,&~~&\begin{vmatrix} 0&1&0\\1&0&1 \\ a_{22}&-a_{12}&a_{11} \end{vmatrix}=-a_{11}+a_{22}. \end{aligned}$$

Thus $\textbf{J}_\pi (A)$ is not of full rank if and only if $a_{21}=0$ and $a_{11}=a_{22}$. Therefore,

$$\begin{aligned} \mathcal {R}_{\mathcal {P}} = \pi (\{A \in \mathbb {B}^{2\times 2}:\textbf{J}_\pi (A) \text { is not of full rank} \})= & {} \pi \Bigg ( A = \begin{bmatrix} a &{} * \\ 0 &{} a \end{bmatrix} \in \mathbb {B}^{2\times 2} \Bigg )\\= & {} \bigg \{ \Big (0, 2a, a^2 \Big ): a \in \mathbb {D}\bigg \}\\= & {} \bigg \{ \Big (0, s, p \Big ): (s, p)\in \mathbb {G}, s^2=4p\bigg \}. \end{aligned}$$

here $s=$ tr A and $p=$ det A. $\square$

Remark 4.2

The singular set of the pentablock can be presented as

$$\begin{aligned} \mathcal {R}_{\mathcal {P}} = \{(0, s, p) \in \mathcal {P} : (s, p) \in \mathcal {R}_\Gamma \cap \mathcal {G}\}. \end{aligned}$$

By analogy with the established terminology for the symmetrized bidisc, we shall call the set

$$\begin{aligned} \mathcal {R}_{\overline{\mathcal {P}}}= \{(0, s, p) \in \mathbb {C}^3 : s^2= 4p\}. \end{aligned}$$

the royal variety of the pentablock.

Lemma 4.3

Let $(a, s, p) \in b\overline{\mathcal {P}}$. Then the following conditions are equivalent:

• (i) $a = 0$;

• (ii) $(a, s, p) \in b\overline{\mathcal {P}} \cap \mathcal {R}_{\overline{\mathcal {P}}}$;

• (iii) $|s|=2$.

Proof

It easily follows from the definition of $\mathcal {R}_{\overline{\mathcal {P}}}$ and the formula (3.2) for $b\overline{\mathcal {P}}$. $\square$

The automorphism group of $\mathcal {P}$. Recall the known information on the automorphism group Aut $\mathcal {P}$ of $\mathcal {P}$ from [4]. For $w \in \mathbb {T}$ and $v \in Aut \ \mathbb {D}$, let

$$\begin{aligned} f_{wv}(a, s, p) = \Big (\frac{w\eta (1-|\alpha |^2)a}{1-{\overline{\alpha }}s+{\overline{\alpha }}^2p}, \tau _v(s, p) \Big ), \end{aligned}$$

(4.2)

where $v=\eta B_\alpha$ for $\alpha \in \mathbb {D}$, $\eta \in \mathbb {T}$, $B_\alpha (z) = \dfrac{z-\alpha }{1-{\overline{\alpha }}z}$ is a Blaschke factor and $\tau _v \in Aut \ \mathbb {G}$ is defined by

$$\begin{aligned} \tau _v(z+w, zw) = \big (v(z)+v(w), v(z)v(w)\big ). \end{aligned}$$

Theorem 4.4

[4, Theorem 7.1] The maps $f_{wv}$, for $w \in \mathbb {T}$ and $v \in Aut \ \mathbb {D}$, constitute a group of automorphisms of $\mathcal {P}$ under composition. Each automorphism $f_{wv}$ extends analytically to a neighbourhood of $\overline{\mathcal {P}}$.

Moreover, for all $w_1, w_2 \in \mathbb {T}$, $v_1, v_2 \in Aut \ \mathbb {D}$,

$$\begin{aligned} f_{w_1v_1} \circ f_{w_2v_2} = f_{(w_1w_2)(v_1\circ v_2)}, \end{aligned}$$

and, for all $w \in \mathbb {T}$, $v \in Aut \ \mathbb {D}$,

$$\begin{aligned} (f_{wv})^{-1} = f_{\overline{w}v^{-1}}. \end{aligned}$$

Kosiński proved in [26] that the set $\{f_{wv}:w \in \mathbb {T}, v \in Aut \ \mathbb {D}\}$ is the full group of automorphisms of $\mathcal {P}$.

Lemma 4.5

$\mathcal {R}_{\overline{\mathcal {P}}} \cap \mathcal {P}$ is invariant under Aut $\mathcal {P}$.

Proof

Every element of $\mathcal {R}_{\overline{\mathcal {P}}} \cap \mathcal {P}$ is of the form $(0, s, p) \in \mathcal {P}$, where $s^2 = 4p$. It is easy to see that, for every element $f_{wv}$ of Aut $\mathcal {P}$ given by Eq. (4.2),

$$\begin{aligned} f_{wv}(0, s, p) = \Big (0, \tau _v(s, p) \Big ). \end{aligned}$$

Since $\tau _v \in Aut \ \mathbb {G}$ and $(s, p)\in \mathcal {R}_\Gamma \cap \mathbb {G}$, by Lemma 2.4, $\tau _v(s, p) \in \mathcal {R}_\Gamma \cap \mathbb {G}$. Therefore, $f_{wv}(0, s, p) \in \mathcal {R}_{\overline{\mathcal {P}}}.$ $\square$

For any domain U in $\mathbb {C}^{n}$, $\text {Hol}(\mathbb {D}, U)$ denotes the space of analytic functions from $\mathbb {D}$ to U.

Definition 4.6

Let U be a domain in $\mathbb {C}^n$ and let $\mathcal {D} \subset U$. We say $\mathcal {D}$ is a complex geodesic in U if there exists a function $k \in Hol (\mathbb {D}, U)$ and a function $C \in Hol (U, \mathbb {D})$ such that $C \circ k = \mathrm id_{\mathbb {D}}$ and $\mathcal {D} = k(\mathbb {D})$.

For a geometric classification of complex geodesics in the symmetrized bidisc $\mathbb {G}$, see [7]. We define, for $\omega \in \mathbb {T}$, the rational function $\Phi _\omega$ of two variables by

$$\begin{aligned} \Phi _\omega (s, p) = \frac{2\omega p-s}{2-\omega s}, \; \; \text {where} \;\; \omega s \ne 2 . \end{aligned}$$

In the function theory and geometry of $\mathbb {G}$ much depends on the properties of these functions $\Phi _\omega ,\; \omega \in \mathbb {T}$, see [14].

Lemma 4.7

$\mathcal {R}_{\overline{\mathcal {P}}} \cap \mathcal {P}$ is a complex geodesic in $\mathcal {P}$.

Proof

Define the analytic functions k and c by

$$\begin{aligned} k:\mathbb {D}\rightarrow \mathcal {P}, \; k(\lambda ) = (0, -2\lambda , \lambda ^2) \end{aligned}$$

and

$$\begin{aligned} c:\mathcal {P} \rightarrow \mathbb {D}, \; c(a, s, p) = \Phi _\omega (s, p) = \frac{2\omega p-s}{2-\omega s}, \; \; \text {where} \;\; \omega \in \mathbb {T}. \end{aligned}$$

For $\lambda \in \mathbb {D}$,

$$\begin{aligned} (c \; \circ \; k)(\lambda ) = c(k(\lambda )) = c(0, -2\lambda , \lambda ^2) = \frac{2\omega \lambda ^2+2\lambda }{2+2\omega \lambda } = \lambda , \end{aligned}$$

which means $c \; \circ \; k = \mathrm id_{\mathbb {D}}.$ By definition of $\mathcal {R}_{\overline{\mathcal {P}}}$, it is easy to see that $\mathcal {R}_{\overline{\mathcal {P}}} \cap \mathcal {P} = k(\mathbb {D})$. Therefore, $\mathcal {R}_{\overline{\mathcal {P}}} \cap \mathcal {P}$ is a complex geodesic in $\mathcal {P}$. $\square$

5 Examples of $\overline{\mathcal {P}}$-Inner Functions

Descriptions of inner and outer functions in $H^\infty (\mathbb {D})$ and properties of inner and outer functions can be found [29, Chapter III]. Here $H^\infty (\mathbb {D})$ is the space of holomorphic functions u on $\mathbb {D}$ such that the corresponding norm

$$\begin{aligned} \Vert u\Vert _\infty = \sup \limits _{\lambda \in \mathbb {D}} |u(\lambda )| \end{aligned}$$

is finite.

Definition 5.1

An inner function is an analytic map $f:\mathbb {D} \rightarrow \overline{\mathbb {D}}$ such that the radial limit

$$\begin{aligned} \lim _{r \rightarrow 1^{-}}f(r\lambda ) \end{aligned}$$

exists and belongs to $\mathbb {T}$ for almost all $\lambda \in \mathbb {T}$ with respect to Lebesgue measure.

It is well known that the rational inner functions on $\mathbb {D}$ are precisely the finite Blaschke products. One can see that the only functions which are at the same time inner and outer are the constant functions of modulus 1.

Definition 5.2

A $\overline{\mathcal {P}}$-inner or penta-inner function is an analytic map $f:\mathbb {D}\rightarrow \overline{\mathcal {P}}$ such that the radial limit

$$\begin{aligned} \lim _{r \rightarrow 1^{-}}f(r\lambda ) \end{aligned}$$

exists and belongs to $b\overline{\mathcal {P}}$ for almost all $\lambda \in \mathbb {T}$ with respect to Lebesgue measure.

By Fatou’s Theorem, the $\lim _{r \rightarrow 1^{-}}f(r\lambda )$ exists for almost all $\lambda \in \mathbb {T}$.

Remark 5.3

Let $f:\mathbb {D}\rightarrow \overline{\mathcal {P}}$ be a rational $\overline{\mathcal {P}}$-inner function. Since f is rational and bounded on $\mathbb {D}$ it has no poles in $\overline{\mathbb {D}}$ and hence f is continuous on $\overline{\mathbb {D}}$. Thus one can consider the continuous function

$$\begin{aligned} {\tilde{f}}:\mathbb {T}\rightarrow b\overline{\mathcal {P}}, \text {where} {\tilde{f}}(\lambda )=\lim _{r \rightarrow 1^-}f(r\lambda ) \text { for all } \lambda \in \mathbb {T}. \end{aligned}$$

Example 5.4

Let us consider an example of an analytic function $f:\mathbb {D}\rightarrow \overline{\mathcal {P}}$. Consider the analytic map $h:\mathbb {D}\rightarrow \mathbb {B}^{2\times 2}$ defined by

$$\begin{aligned} h(\lambda )= \begin{bmatrix} \varphi (\lambda ) &{} 0 \\ 0 &{} \psi (\lambda ) \end{bmatrix} \ \ \ \text {for} \ \lambda \in \mathbb {D}, \end{aligned}$$

(5.1)

where $\varphi , \psi \in H^\infty (\mathbb {D})$ are non-constant inner functions. Note that

$$\begin{aligned} \Vert h(\lambda )\Vert = \max \{|\varphi (\lambda )|,|\psi (\lambda )|\} \; <\; 1\; \text { for} \; \lambda \in \mathbb {D}. \end{aligned}$$

For all $\lambda \in \mathbb {D}$, let

$$\begin{aligned} f(\lambda ) = \pi (h(\lambda ))=(0, {\text {tr}}h(\lambda ), \det h(\lambda )). \end{aligned}$$

Let us show that the function f is a $\overline{\mathcal {P}}$ -inner function only when $\varphi = \psi$.

Let, for $\lambda \in \mathbb {D}, \; a(\lambda ) = 0, \; s(\lambda ) = {\text {tr}}h(\lambda ) = \varphi (\lambda )+\psi (\lambda )$ and $p(\lambda ) = \det h(\lambda ) = \varphi (\lambda )\psi (\lambda )$. Clearly $f=(a, s, p):\mathbb {D}\rightarrow \overline{\mathcal {P}}$ is an analytic function.

To prove that f is $\overline{\mathcal {P}}$-inner, we need to check that $f(\lambda ) \in b\overline{\mathcal {P}}$ for almost every $\lambda \in \mathbb {T}$, that is, $\big (s(\lambda ), p(\lambda )\big ) \in b\Gamma$ and $\sqrt{1-\frac{1}{4}|s(\lambda )|^2} = 0$ for almost every $\lambda \in \mathbb {T}$. Note that, for almost every $\lambda \in \mathbb {T}$,

$$\begin{aligned} |p(\lambda )| = |\varphi (\lambda )\psi (\lambda )| = |\varphi (\lambda )||\psi (\lambda )| = 1, \ \text { since } \ |\varphi (\lambda )| = 1 \text { and } \ |\psi (\lambda )| = 1, \\ |s(\lambda )| = |\varphi (\lambda )+\psi (\lambda )| \le |\varphi (\lambda )|+|\psi (\lambda )| = 2, \end{aligned}$$

and

$$\begin{aligned} (\overline{s}p)(\lambda )= & {} \big (\overline{\varphi (\lambda )+\psi (\lambda )}\big )\big (\varphi (\lambda ) \psi (\lambda )\big ) = \varphi (\lambda )\overline{\varphi (\lambda )}\psi (\lambda ) + \varphi (\lambda )\psi (\lambda )\overline{\psi (\lambda )} \\= & {} |\varphi (\lambda )|^2 \psi (\lambda ) + \varphi (\lambda )|\psi (\lambda )|^2 = \varphi (\lambda )+\psi (\lambda ) = s(\lambda ). \end{aligned}$$

Hence for almost every $\lambda \in \mathbb {T}, \ |p(\lambda )| = 1, \ |s(\lambda )|\le 2$ and $(\overline{s}p)(\lambda ) = s(\lambda )$, and so $\big ({\text {tr}}h(\lambda ), \det h(\lambda )\big ) \in b\Gamma$. Now, for almost every $\lambda \in \mathbb {T}$,

$$\begin{aligned} 1-\frac{1}{4}|s(\lambda )|^2= & {} 1-\frac{1}{4}|\varphi (\lambda )+\psi (\lambda )|^2 = 1-\frac{1}{4}\Big (\big (\varphi (\lambda )+\psi (\lambda )\big )\big (\overline{\varphi (\lambda )+\psi (\lambda )}\big )\Big ) \\= & {} 1-\frac{1}{4}\Big (1+1+2 \textrm{Re}\big (\varphi (\lambda )\overline{\psi (\lambda )}\big )\Big ) = \frac{1}{2}-\frac{1}{2}\textrm{Im}\big (i\varphi (\lambda )\overline{\psi (\lambda )}\big ). \end{aligned}$$

Hence $|a|=\sqrt{1-\frac{1}{4}|s|^2}$ almost everywhere on $\mathbb {T}$ if and only if

$$\begin{aligned} \frac{1}{2}-\frac{1}{2} \textrm{Im} \big (i\varphi (\lambda )\overline{\psi (\lambda )}\big ) = 0, \text { for almost every }\lambda \in \mathbb {T}, \end{aligned}$$

if and only if $\textrm{Im}(i\varphi (\lambda )\overline{\psi (\lambda )}) = 1 \text { for almost every }\lambda \in \mathbb {T}.$ Therefore, $|a|=\sqrt{1-\frac{1}{4}|s|^2}$ almost everywhere on $\mathbb {T}$ if and only if $\varphi (\lambda )\overline{\psi (\lambda )} = 1$ almost everywhere on $\mathbb {T}$, and so, $\varphi (\lambda ) = \psi (\lambda )$ almost everywhere on $\mathbb {T}$. Thus the function f is a $\overline{\mathcal {P}}$-inner function only when $\varphi = \psi$, and so $f = (0, 2 \varphi , \varphi ^2)$.

Example 5.5

Let $h_1 : \mathbb {D} \rightarrow \mathbb {C}^{2 \times 2}$ be defined by $h_1=Uh$, where h is defined by Eq. (5.1) and

$$\begin{aligned} U= \begin{bmatrix} \dfrac{1}{\sqrt{2}} &{} \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}}i &{} -\dfrac{1}{\sqrt{2}}i \end{bmatrix}. \end{aligned}$$

Note that U is a unitary matrix. Then, for $\lambda \in \mathbb {D}$,

$$\begin{aligned} h_1(\lambda )= & {} Uh(\lambda ) \\= & {} \frac{1}{\sqrt{2}} \begin{bmatrix} \varphi (\lambda ) &{} \psi (\lambda ) \\ i\varphi (\lambda ) &{} -i\psi (\lambda ) \end{bmatrix}, \end{aligned}$$

and, for all $\lambda \in \mathbb {D}$,

$$\begin{aligned} \Vert h_1(\lambda )\Vert \le \Vert U\Vert \Vert h(\lambda )\Vert =\Vert h(\lambda )\Vert =\max \{|\varphi (\lambda )|, |\psi (\lambda )|\} < 1. \end{aligned}$$

Hence $h_1(\lambda ) \in \mathbb {B}^{2\times 2}$ for all $\lambda \in \mathbb {D}$. Define $f_1=\pi \circ h_1$ on $\mathbb {D}$. Then, for $\lambda \in \mathbb {D}$,

$$\begin{aligned} f_1(\lambda )&= \pi (h_1(\lambda )) \nonumber \\&= \pi \bigg (\frac{1}{\sqrt{2}} \begin{bmatrix} \varphi (\lambda ) &{} \psi (\lambda ) \\ i\varphi (\lambda ) &{} -i\psi (\lambda ) \end{bmatrix}\bigg ) \nonumber \\&= \bigg (\frac{i\varphi (\lambda )}{\sqrt{2}}, \frac{\varphi (\lambda )-i\psi (\lambda )}{\sqrt{2}}, -i\varphi (\lambda )\psi (\lambda )\bigg ). \end{aligned}$$

(5.2)

Clearly, $f_1:\mathbb {D}\rightarrow \overline{\mathcal {P}}$ is an analytic function since $\varphi , \psi$ are analytic on $\mathbb {D}$.

Let us show that $f_1$ is a $\overline{\mathcal {P}}$ -inner function if and only if $\varphi = i\psi$.

Let us check when the function $f_1$ is $\overline{\mathcal {P}}$-inner. We need to find conditions when $f_1$ maps $\mathbb {T}$ into the distinguished boundary $b\overline{\mathcal {P}}$ of $\mathcal {P}$. Since $\varphi$, $\psi$ are inner functions, they have unit modulus almost everywhere on $\mathbb {T}$. Thus, for $s = \dfrac{\varphi -i\psi }{\sqrt{2}}$, $p = -i\varphi \psi$ and for almost every $\lambda \in \mathbb {T}$,

$$\begin{aligned} |p(\lambda )|= & {} |-i\varphi (\lambda )\psi (\lambda )|=|\varphi (\lambda )||\psi (\lambda )|=1.\\ |s(\lambda )|= & {} \bigg |\frac{\varphi (\lambda )-i\psi (\lambda )}{\sqrt{2}}\bigg | \le \frac{|\varphi (\lambda )|+|-i\psi (\lambda )|}{\sqrt{2}}=\frac{2}{\sqrt{2}}.\\ (\overline{s}p)(\lambda )= & {} \frac{\overline{\varphi (\lambda )-i\psi (\lambda )}}{\sqrt{2}}(-i\varphi (\lambda )\psi (\lambda )) \\= & {} \frac{-i|\varphi (\lambda )|^2 \psi (\lambda )+\varphi (\lambda )|\psi (\lambda )|^2}{\sqrt{2}} = \frac{\varphi (\lambda )-i\psi (\lambda )}{\sqrt{2}}=s(\lambda ). \end{aligned}$$

Therefore, for almost every $\lambda \in \mathbb {T}, |p(\lambda )| = 1$, $|s(\lambda )|\le 2$ and $(\overline{s}p)(\lambda ) = s(\lambda )$ and so $(s(\lambda ), p(\lambda )) \in b\Gamma$. Finally,

$$\begin{aligned} \sqrt{1-\frac{1}{4}|s(\lambda )|^2}= & {} \sqrt{1-\frac{1}{4} \bigg (\frac{1}{2}|\varphi (\lambda )-i \psi (\lambda )|^2\bigg )} \\= & {} \sqrt{1-\frac{1}{8}\bigg (1+1+2 \textrm{Re} (\varphi (\lambda ) i \overline{\psi (\lambda )})\bigg )} = \frac{1}{2} \sqrt{3+\textrm{Im} (\varphi (\lambda )\overline{\psi (\lambda )})}. \end{aligned}$$

We want $|a|=\sqrt{1-\frac{1}{4}|s|^2}$ almost everywhere on $\mathbb {T}$, that is, for almost every $\lambda \in \mathbb {T}$,

$$\begin{aligned} \dfrac{1}{\sqrt{2}} = \dfrac{|\varphi (\lambda )|}{\sqrt{2}} = \dfrac{1}{2} \sqrt{3+\textrm{Im} (\varphi (\lambda )\overline{\psi (\lambda )})}. \end{aligned}$$

Hence $|a|=\sqrt{1-\frac{1}{4}|s|^2}$ almost everywhere on $\mathbb {T}$ if and only if $\sqrt{3+\textrm{Im} (\varphi (\lambda )\overline{\psi (\lambda )})} = \sqrt{2}$ for almost every $\lambda \in \mathbb {T}$, or equivalently, $\varphi (\lambda ) = -i\psi (\lambda )$ for almost every $\lambda \in \mathbb {T}$. Thus $f_1$ given by Eq. (5.2) is a $\overline{\mathcal {P}}$-inner function if and only if $\varphi = -i\psi$. In this case $f_1= \bigg (\frac{i\varphi }{\sqrt{2}}, \sqrt{2} \varphi , \varphi ^2\bigg )$.

Example 5.6

Let v, $\varphi$ and $\psi$ be inner functions on $\mathbb {D}$. Consider the functions

$$\begin{aligned} V(\lambda ) = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 &{} v \\ -1 &{} v \end{bmatrix}(\lambda ) \ \text { and } \ h(\lambda ) = \begin{bmatrix} \varphi &{} 0 \\ 0 &{} \psi \end{bmatrix}(\lambda ),\; \text { for }\; \lambda \in \mathbb {D}. \end{aligned}$$

Define

$$\begin{aligned} U(\lambda )= & {} (V^* h V)(\lambda )= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 &{} -1 \\ \overline{v} &{} \overline{v} \end{bmatrix} \begin{bmatrix} \varphi &{} 0 \\ 0 &{} \psi \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 &{} v \\ -1 &{} v \end{bmatrix}(\lambda ) \\= & {} \frac{1}{2} \begin{bmatrix} \varphi + \psi &{} (\varphi - \psi )v \\ (\varphi - \psi )\overline{v} &{} \varphi + \psi \end{bmatrix}(\lambda ), \ \text { for } \lambda \in \mathbb {D}. \end{aligned}$$

Note that, for all $\lambda \in \mathbb {D}$,

$$\begin{aligned} \Vert V^*(\lambda )\Vert = \left\| \begin{bmatrix} 1 &{} 0 \\ 0 &{} \overline{v}(\lambda ) \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 &{} -1 \\ 1 &{} 1 \end{bmatrix} \right\| \le 1, \end{aligned}$$

since $\frac{1}{\sqrt{2}}\begin{bmatrix} 1 &{} -1 \\ 1 &{} 1 \end{bmatrix}$ is unitary and $| \overline{v}(\lambda ) | \le 1$ on $\mathbb {D}$. Hence

$$\begin{aligned} \Vert U(\lambda )\Vert \le \Vert V(\lambda )\Vert ^2 \Vert h(\lambda )\Vert < 1 \ \text {for } \lambda \in \mathbb {D}. \end{aligned}$$

Define $f:\mathbb {D} \rightarrow \overline{\mathcal {P}}$ by $f=\pi \circ U$. Then, for $\lambda \in \mathbb {D}$,

$$\begin{aligned} f(\lambda ) = \pi \circ U(\lambda ) = \Big (\frac{1}{2}(\varphi - \psi )\overline{v}, \varphi + \psi , \frac{1}{4}\big ((\varphi + \psi )^2-(\varphi - \psi )^2|v|^2\big ) \Big )(\lambda ).\nonumber \\ \end{aligned}$$

(5.3)

Note that f is analytic on $\mathbb {D}$ if and only if v is constant or $\varphi = \psi$.

Let us show that f is a $\overline{\mathcal {P}}$-inner function if and only if v is constant or $\varphi = \psi$.

Case 1 Suppose v is constant. As v is inner, $|v|=1$. Let us check that $f:\mathbb {D} \rightarrow \overline{\mathcal {P}}$ is $\overline{\mathcal {P}}$-inner, that is, $f(\mathbb {T}) \subset b\overline{\mathcal {P}}$. Note, for almost all $\lambda \in \mathbb {T}$, $\det U(\lambda ) = (\varphi \psi )(\lambda )$ and $(\varphi +\psi , \varphi \psi )(\lambda ) \in b\Gamma$ as in Example 5.4.

$$\begin{aligned} |a|^2 = \dfrac{1}{4}|\varphi - \psi |^2 = \dfrac{1}{4} \Big (1+1-2 \textrm{Re} ({\overline{\varphi }}\psi ) \Big ) = \dfrac{1}{2}-\dfrac{1}{2} \textrm{Re} ({\overline{\varphi }}\psi ) \text { almost everywhere on } \mathbb {T}. \\ 1 - \frac{1}{4} |s|^2 = 1 - \frac{1}{4}|\varphi + \psi |^2 = 1 - \frac{1}{4} \Big (1 + 1 + 2 \textrm{Re}({\overline{\varphi }}\psi ) \Big ) = 1-\frac{1}{2} \textrm{Re} ({\overline{\varphi }}\psi ) = |a|^2. \end{aligned}$$

Thus $|a|^2 = 1-\frac{1}{4}|s|^2$ almost everywhere on $\mathbb {T}$, and so, f given by Eq. (5.3) is a $\overline{\mathcal {P}}$-inner function if v is constant. Since $|v|=1$, $f = \Big (\frac{1}{2}(\varphi - \psi )\overline{v}, \varphi + \psi , \varphi \psi \Big )$.

Case 2 Suppose $\varphi = \psi$. Then

$$\begin{aligned} f = \Big (0, 2\varphi , \frac{1}{4}(2\varphi )^2\Big ) = (0, 2\varphi , \varphi ^2). \end{aligned}$$

We have shown in Example 5.4 that $f= (0, 2\varphi , \varphi ^2)$ is a $\overline{\mathcal {P}}$-inner function.

Example 5.7

Define the function $x(\lambda ) = (\lambda ^m, 0, \lambda ): \mathbb {D}\rightarrow \overline{\mathcal {P}}$. First we need to show that for all $\lambda \in \mathbb {D}, \; x(\lambda ) \in \overline{\mathcal {P}}$. Let us show that x is a rational $\overline{\mathcal {P}}$-inner function.

By Proposition 2.2, $(s, p) \in \Gamma$ if and only if

$$\begin{aligned} |s|\le 2 \ \text { and } |s-\overline{s}p|\le 1-|p|^2. \end{aligned}$$

It is easy to see that $(0, \lambda ) \in \Gamma$. By Theorem 2.10, if $(s, p) \in \Gamma$, then for $a \in \mathbb {C}, (a, s, p) \in \overline{\mathcal {P}}$ if and only if

$$\begin{aligned} |a| \le \left| 1-\frac{\frac{1}{2}s{\overline{\beta }}}{1+\sqrt{1-|\beta |^2}}\right| , \ \text { where } \ \beta =\frac{s-\overline{s}p}{1-|p|^2}. \end{aligned}$$

(5.4)

In the case $s = 0$, Eq. (5.4) is equivalent to $|a| \le 1$. Note that $x(\lambda ) = \big (a(\lambda ), s(\lambda ), p(\lambda )\big ) = (\lambda ^m, 0, \lambda ), \; \lambda \in \mathbb {D}, \text { is analytic in } \mathbb {D}$, and

$$\begin{aligned} |a(\lambda )| = |\lambda ^m| \le 1, \ \text { for all } \ \lambda \in \mathbb {D}. \end{aligned}$$

Thus for $\lambda \in \mathbb {D}, \ x(\lambda ) \in \overline{\mathcal {P}}$.

Now, let us check that x maps $\mathbb {T}$ into the distinguished boundary $b\overline{\mathcal {P}}$ of $\mathcal {P}$. For all $\lambda \in \mathbb {T}$,

$$\begin{aligned} |p(\lambda )| = |\lambda | = 1, \ |s(\lambda )| = |0| \le 2, \\ (\overline{s}p)(\lambda ) = 0 = s(\lambda ) \ \text { and } \\ |a(\lambda )| = |\lambda ^m| = \sqrt{1-\frac{1}{4}|s(\lambda )|^2} = 1. \end{aligned}$$

Therefore, for every $\lambda \in \mathbb {T}$, $x(\lambda ) \in b\overline{\mathcal {P}}$ and hence x is a rational $\overline{\mathcal {P}}$-inner function.

Example 5.8

For $\lambda \in \mathbb {D}$, define the function $x(\lambda ) = \big (a(\lambda ), s(\lambda ), p(\lambda )\big )= (\lambda , 0, \lambda ^n)$. As in the previous example, for all $\lambda \in \mathbb {D}$, by Proposition 2.2, $(0, \lambda ^n) \in \Gamma$. For $a \in \mathbb {C}$, we want

$$\begin{aligned} |a| \le |1-\frac{\frac{1}{2}s{\overline{\beta }}}{1+\sqrt{1-|\beta |^2}}|. \end{aligned}$$

(5.5)

Since $s = 0$, the condition (5.5) is equivalent to $|a| \le 1$. Note that

$$\begin{aligned} |a(\lambda )| = |\lambda | \le 1, \text { for all } \ \lambda \in \mathbb {D}. \end{aligned}$$

Thus, by Theorem 2.10, for $\lambda \in \mathbb {D}, \ x(\lambda ) \in \overline{\mathcal {P}}$. Let us show that x is a rational $\overline{\mathcal {P}}$- inner function.

Let us check that x maps $\mathbb {T}$ into the distinguished boundary $b\overline{\mathcal {P}}$ of $\mathcal {P}$. For all $\lambda \in \mathbb {T}$,

$$\begin{aligned} |p(\lambda )| = |\lambda ^n| = 1, \ |s(\lambda )| = |0| \le 2, \\ (\overline{s}p)(\lambda ) = 0 = s(\lambda ) \ \ \text {and } \\ |a(\lambda )| = |\lambda | = \sqrt{1-\frac{1}{4}|s(\lambda )|^2} = 1. \end{aligned}$$

Therefore, for every $\lambda \in \mathbb {T}$, $x(\lambda ) \in b\overline{\mathcal {P}}$ and hence x is a rational $\overline{\mathcal {P}}$-inner function.

6 Some Properties of Analytic Functions $x:\mathbb {D}\rightarrow \overline{\mathcal {P}}$

In this section and in Sect. 7, we will show that there are close relations between $\overline{\mathcal {P}}$-inner functions and $\Gamma$-inner functions. Recall that the $\Gamma$-inner functions were first mentioned in [13]. A good understanding of rational $\Gamma$-inner functions will play an important part in any future solution of the finite interpolation problem for $\text {Hol}(\mathbb {D}, \Gamma )$, since such a problem has a solution if and only if it has a rational $\Gamma$-inner solution (see, for example, [18, Theorem 4] and [3, Theorem 8.1]). The rational $\Gamma$-inner functions were classified in [2]. Algebraic and geometric aspects of rational $\Gamma$-inner functions were presented in [6].

Definition 6.1

A $\Gamma$-inner function is an analytic function $h:\mathbb {D} \rightarrow \Gamma$ such that the radial limit

$$\begin{aligned} \lim _{r \rightarrow 1^-} h(r \lambda ) \end{aligned}$$

(6.1)

exists and belongs to $b\Gamma$ for almost all $\lambda \in \mathbb {T}$ with respect to Lebesgue measure.

Lemma 6.2

• (i) Let $x = (a, s, p) : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ be an analytic function. Then $h = (s, p) : \mathbb {D} \rightarrow \Gamma$ is an analytic function.

• (ii) Let $x = (a, s, p) : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ be a $\overline{\mathcal {P}}$-inner function. Then $h = (s, p) : \mathbb {D} \rightarrow \Gamma$ is a $\Gamma$-inner function.

Proof

• (i) By assumption, $x = (a, s, p)$ is analytic on $\mathbb {D}$ and for all $\lambda \in \mathbb {D}$, $x(\lambda ) = (a(\lambda ), s(\lambda ), p(\lambda )) \in \overline{\mathcal {P}}$. By Remark 2.5, for all $\lambda \in \mathbb {D}$, $(s(\lambda ), p(\lambda )) \in \Gamma$. Thus $h = (s, p):\mathbb {D} \rightarrow \Gamma$, where $h(\lambda ) = (s(\lambda ), p(\lambda ))$, for $\lambda \in \mathbb {D}$, is well defined and analytic from $\mathbb {D}$ to $\Gamma$.

• (ii) By assumption $x = (a, s, p):\mathbb {D} \rightarrow \overline{\mathcal {P}}$ is a penta-inner function, and so, for almost all $\lambda \in \mathbb {T}$, $x(\lambda ) \in b\overline{\mathcal {P}}$. Recall $b\overline{\mathcal {P}} = \Big \{(a, s, p) \in \mathbb {C}^3 : (s, p) \in b\Gamma , \; |a| = \sqrt{1-\frac{1}{4}|s|^2}\Big \}$. By Theorem 3.2, for almost all $\lambda \in \mathbb {T}$, $h(\lambda ) = \big (s(\lambda ), p(\lambda )\big ) \in b\Gamma$. Hence h is a $\Gamma$-inner function.

$\square$

Recall that, by Proposition 3.1, $K_1 = \bigg \{(a, s, p) \in \overline{\mathcal {P}} : (s, p) \in b\Gamma , \ |a| \le \sqrt{1-\frac{1}{4}|s|^2}\bigg \}$ is a closed boundary of $A(\mathcal {P})$.

Proposition 6.3

• (i) Let $h = (s, p) : \mathbb {D} \rightarrow \Gamma$ be an analytic function. Then $x = (0, s, p)$ is an analytic function from $\mathbb {D}$ to $\overline{\mathcal {P}}$. (ii) Let $h = (s, p) : \mathbb {D} \rightarrow \Gamma$ be a $\Gamma$-inner function. Then $x = (0, s, p):\mathbb {D} \rightarrow \overline{\mathcal {P}}$ is an analytic function such that, for almost all $\lambda \in \mathbb {T}, \; x(\lambda ) \in K_1$.

Proof

• (i) It follows from Theorem 2.10 that, for all $\lambda \in \mathbb {D}, \big (0, s(\lambda ), p(\lambda )\big ) \in \overline{\mathcal {P}}$.

• (ii) Suppose that h is a $\Gamma$-inner function. By Proposition 2.2, $|p(\lambda )| = 1, \ |s(\lambda )| \le 2 \ \text { and } \ (\overline{s}p)(\lambda ) = s(\lambda )$, for almost all $\lambda \in \mathbb {T}$. Since $a = 0 \text { and } \sqrt{1-\frac{1}{4}|s(\lambda )|^2} \ge 0$ for almost all $\lambda \in \mathbb {T}$, $x(\mathbb {T}) \subset K_1$.

$\square$

Proposition 6.4

Let $x = (a, s, p)$ be a $\overline{\mathcal {P}}$-inner function. Let $a_{\textrm{in}} \; a_\textrm{out}$ be the inner-outer factorization of a. Then ${\widetilde{x}} = (a_\textrm{out}, s, p)$ is a $\overline{\mathcal {P}}$-inner function.

Proof

By assumption $x = (a, s, p)$ is a $\overline{\mathcal {P}}$-inner function. Hence, for each $\lambda \in \mathbb {D}$, $\;(a(\lambda ), s(\lambda ), p(\lambda )) \in \overline{\mathcal {P}}$. By Theorem 2.10, for all $\lambda \in \mathbb {D}$, $|\Psi _z(a(\lambda ), s(\lambda ), p(\lambda ))| \le 1$ for all $z \in \mathbb {D}.$ Thus

$$\begin{aligned} \left| \frac{a(\lambda )(1-|z|^2)}{1-s(\lambda )z+p(\lambda )z^2}\right| \le 1, \; \text { for all } \lambda , z \in \mathbb {D}. \end{aligned}$$

Recall that $a=a_{\textrm{in}}a_{\textrm{out}}$, where $a_{\textrm{in}}$ is inner, and so $|a_{\textrm{in}}(\lambda )| = 1$ for almost all $\lambda \in \mathbb {T}$. Therefore, for every $z \in \mathbb {D}$, and, for almost all $\lambda \in \mathbb {T}$,

$$\begin{aligned} \left| a_{\textrm{in}}(\lambda )\frac{a_{\textrm{out}}(\lambda )(1-|z|^2)}{1-s(\lambda )z+p(\lambda )z^2}\right| = \left| \frac{a_{\textrm{out}}(\lambda )(1-|z|^2)}{1-s(\lambda )z+p(\lambda )z^2}\right| \le 1. \end{aligned}$$

Note that, for every $z \in \mathbb {D}$, the function

$$\begin{aligned} \lambda \mapsto \frac{a_{\textrm{out}}(\lambda )(1-|z|^2)}{1-s(\lambda )z+p(\lambda )z^2} \end{aligned}$$

is analytic on $\mathbb {D}.$ By the maximum principle for analytic functions, for every $z \in \mathbb {D}$,

$$\begin{aligned} \left| \frac{a_{\textrm{out}}(\lambda )(1-|z|^2)}{1-s(\lambda )z+p(\lambda )z^2}\right| \le 1, \; \text { for all } \lambda \in \mathbb {D}. \end{aligned}$$

Hence, by Theorem 2.10, for each $\lambda \in \mathbb {D}, \; (a_{\textrm{out}}(\lambda ), s(\lambda ), p(\lambda )) \in \overline{\mathcal {P}}$. Therefore, ${\widetilde{x}} = (a_{\textrm{out}}, s, p) \in$ Hol$(\mathbb {D}, \overline{\mathcal {P}})$.

To prove the statement of the proposition, we must show that, for almost all $\lambda \in \mathbb {T}$, ${\widetilde{x}}(\lambda ) = (a_{\textrm{out}}(\lambda ), s(\lambda ), p(\lambda )) \in b\overline{\mathcal {P}}$. Recall that

$$\begin{aligned} b\overline{\mathcal {P}} = \left\{ (a, s, p) \in \mathbb {C}^3 : (s, p) \in b\Gamma , \; |a|=\sqrt{1-\frac{1}{4}|s|^2}\right\} . \end{aligned}$$

By Lemma 6.2, for almost all $\lambda \in \mathbb {T}$, $(s(\lambda ), p(\lambda )) \in b\Gamma$. Since $x = (a, s, p)$ is a $\overline{\mathcal {P}}$-inner function, we have, for almost all $\lambda \in \mathbb {T}$,

$$\begin{aligned} |a(\lambda )| = \sqrt{1-\frac{1}{4}|s(\lambda )|^2}. \end{aligned}$$

Since $a_{\textrm{in}}a_{\textrm{out}} = a$ is the inner-outer factorization of a and $|a_{\textrm{in}}(\lambda )| = 1$ for almost all $\lambda \in \mathbb {T}$,

$$\begin{aligned} |a_{\textrm{out}}(\lambda )| = \sqrt{1-\frac{1}{4}|s(\lambda )|^2} \; \text { for almost all } \lambda \in \mathbb {T}. \end{aligned}$$

Therefore, ${\widetilde{x}} = (a_{\textrm{out}}, s, p)$ is a $\overline{\mathcal {P}}$-inner function. $\square$

7 Connections Between Rational $\Gamma$-Inner and Rational $\overline{\mathcal {P}}$-Inner Functions

Theorem 7.1

(Fejér–Riesz theorem) [27, Sect. 53] If $f(\lambda ) = \sum ^{n}_{i=-n}a_i\lambda ^i$ is a trigonometric polynomial of degree n such that $f(\lambda ) \ge 0$ for all $\lambda \in \mathbb {T}$, then there exists an analytic polynomial $D(\lambda ) = \sum ^n_{i=0}b_i\lambda ^i$ of degree n such that D is outer (that is, $D(\lambda )\ne 0$ for all $\lambda \in \mathbb {D}$) and

$$\begin{aligned} f(\lambda )=|D(\lambda )|^2 \end{aligned}$$

for all $\lambda \in \mathbb {T}$.

Recall that for every $a \ne 0$ in $H^\infty (\mathbb {D})$ there is an outer-inner factorization. Rational inner functions can be written in the form $c\prod \limits _{i=1}^{n}B_{\alpha _{i}}$ for some $n \ge 1$ and $\alpha _1, \dots , \alpha _n \in \mathbb {D}$ and $c \in \mathbb {C}$.

Definition 7.2

[6, Definition 3.1] The degree deg(h) of a rational $\Gamma$-inner function h is defined to be $h_*(1)$, where $h_* : \mathbb {Z}=\pi _1(\mathbb {T}) \rightarrow \pi _1(b\Gamma )$ is the homomorphism of fundamental groups induced by h when it is regarded as a continuous map from $\mathbb {T}$ to $b\Gamma$.

Recall that, by [6, Proposition 3.3], for any rational $\Gamma$-inner function $h=(s, p)$, deg(h) is the degree deg(p) (in the usual sense) of the finite Blaschke product p.

Definition 7.3

Let g be a polynomial of degree less than or equal to n, where $n \ge 0$. Then we define the polynomial $g^{\sim n}$ by

$$\begin{aligned} g^{\sim n}(\lambda )= \lambda ^n \overline{g(1/{\overline{\lambda }})}. \end{aligned}$$

Definition 7.4

The degree of a rational $\overline{\mathcal {P}}$-inner function $x = (a, s, p)$ is defined to be the pair of numbers $(deg \;a, deg \;p)$. We say that deg $x \le (m, n)$ if deg $a \le m$ and deg $p \le n$.

The next theorem provides a description of the structure of rational penta-inner functions of prescribed degree.

Theorem 7.5

Let $x=(a, s, p) : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ be a rational penta-inner function of degree (m, n). Let $a \ne 0$ and let an inner-outer factorization of a be given by $a = a_{\textrm{in}} a_{\textrm{out}}$, where $a_{\textrm{in}}$ is an inner function and $a_{\textrm{out}}$ is an outer function. Then there exist polynomials A, E, D such that

• (1) $deg(A), deg(E), deg(D) \le n$,

• (2) $E^{\sim n} = E$,

• (3) $D(\lambda ) \ne 0$ on $\overline{\mathbb {D}}$,

• (4) $|E(\lambda )| \le 2|D(\lambda )|$ on $\overline{\mathbb {D}}$,

• (5) A is an outer polynomial such that $|A(\lambda )|^2 = |D(\lambda )|^2 - \frac{1}{4}|E(\lambda )|^2$ on $\mathbb {T}$,

• (6) $a = a_{\textrm{in}} \dfrac{A}{D}$ on $\overline{\mathbb {D}}$,

• (7) $s = \dfrac{E}{D}$ on $\overline{\mathbb {D}}$,

• (8) $p = \dfrac{D^{\sim n}}{D}$ on $\overline{\mathbb {D}}$.

Proof

Suppose that $x=(a, s, p)$ is a rational penta-inner function. By Lemma 6.2, $h=(s, p)$ is a rational $\Gamma$-inner function. By [2, Corollary 6.10], p can be written in the form

$$\begin{aligned} p(\lambda ) = c\frac{\lambda ^kD^{\sim (n-k)}(\lambda )}{D(\lambda )}, \end{aligned}$$

where $|c|=1$, $0 \le k \le n$ and D is a polynomial of degree $n-k$ such that $D(0)=1$. Therefore, by [6, Proposition 2.2], there exist polynomials E and D such that

$$\begin{aligned} \text {(i)}&\text {deg}(E), \text {deg}(D) \le n, \nonumber \\ \text {(ii)}&E^{\sim n}=E,\nonumber \\ \text {(iii)}&D(\lambda ) \ne 0 \textit{ on } \overline{\mathbb {D}},\nonumber \\ \text {(iv)}&|E(\lambda )| \le 2 |D(\lambda )| \textit{ on } \overline{\mathbb {D}},\nonumber \\ \text {(v)}&s=\dfrac{E}{D} \textit{ on } \overline{\mathbb {D}},\nonumber \\ \text {(vi)}&p = \dfrac{D^{\sim n}}{D} \text { on } \overline{\mathbb {D}}. \end{aligned}$$

(7.1)

By assumption $x = (a, s, p)$ is a $\overline{\mathcal {P}}$-inner function, and so, for almost all $\lambda \in \mathbb {T}, \; (a(\lambda ), s(\lambda ), p(\lambda )) \in b\overline{\mathcal {P}}$, which implies

$$\begin{aligned} |a_{\textrm{out}}(\lambda )|^2 = 1-\frac{1}{4}|s(\lambda )|^2, \; \text { since } |a_{\textrm{in}}(\lambda )| = 1 \text { almost everywhere on } \mathbb {T}. \end{aligned}$$

Thus

$$\begin{aligned} |a_{\textrm{out}}(\lambda )|^2 = 1-\frac{1}{4}\frac{|E(\lambda )|^2}{|D(\lambda )|^2} \ \ \text { since } s(\lambda ) = \frac{E(\lambda )}{D(\lambda )}, \end{aligned}$$

and so,

$$\begin{aligned} |a_{\textrm{out}}(\lambda )|^2|D(\lambda )|^2 = |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2. \end{aligned}$$

(7.2)

By [6, Proposition 2.2], $|E(\lambda )| \le 2|D(\lambda )|$. By the Fejér–Riesz Theorem, since $|D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2 \ge 0$, there exists an analytic polynomial A of degree $\le n$ such that A is outer and

$$\begin{aligned} |A(\lambda )|^2=|D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2 \end{aligned}$$

(7.3)

for all $\lambda \in \mathbb {T}$.

From Eqs. (7.2) and (7.3) we have, $|A(\lambda )|^2 = |a_{\textrm{out}}(\lambda )|^2 |D(\lambda )|^2$. Note that $D(\lambda )\ne 0$ on $\overline{\mathbb {D}}$. Thus $|a_{\textrm{out}}(\lambda )| = \left| \dfrac{A}{D}(\lambda )\right|$ for $\lambda \in \mathbb {T}$, and so $\dfrac{A}{D}$ is an outer function such that $|a(\lambda )| = \left| \dfrac{A}{D}(\lambda )\right|$ for almost all $\lambda \in \mathbb {T}$. Since outer factors are unique up to unimodular constant multiples, there exists $\omega \in \mathbb {T}$ such that

$$\begin{aligned} a_{\textrm{out}}(\lambda ) = \omega \dfrac{A(\lambda )}{D(\lambda )}. \end{aligned}$$

Therefore, $a = a_{\textrm{in}} \dfrac{A}{D}$ on $\overline{\mathbb {D}}$, after replacement of A by $\omega A$. $\square$

Remark 7.6

Results similar to our Theorem 7.5 were announced on ArXiv in [25].

Example 7.7

Consider a rational $\overline{\mathcal {P}}$-inner function $x(\lambda ) = (\lambda ^m, 0, \lambda )$ for $\lambda \in \mathbb {D}$. It is easy to see that polynomials described in Theorem 7.5 for this function are the following: $E(\lambda ) = 0$, $D(\lambda ) = 1$, $D^{\sim 1}(\lambda ) = \lambda$, $A(\lambda ) = 1$. Since $a(\lambda ) = \lambda ^m$ is inner, $a_{\textrm{in}}= a$ and so $a_{\textrm{in}}(\lambda ) = \lambda ^m$.

Theorem 7.8

Let $h = (s, p) : \mathbb {D} \rightarrow \Gamma$ be a rational $\Gamma$-inner function of degree n. Let E, D be defined by equations (7.1) ( [6, Proposition 2.2]). Let A be an outer polynomial such that

$$\begin{aligned} |A(\lambda )|^2 = |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2. \end{aligned}$$

(7.4)

Then, for every finite Blaschke product B and $|c|=1$, $x=\left( c B \dfrac{A}{D}, \dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right)$ is a rational $\overline{\mathcal {P}}$-inner function.

Proof

Let a, s, p be defined by

$$\begin{aligned} a = c B \dfrac{A}{D}, \; \; \; s = \dfrac{E}{D} \; \; \; \text { and } p = \dfrac{D^{\sim n}}{D}. \end{aligned}$$

Let us show that $x = (a, s, p)$ is a rational $\overline{\mathcal {P}}$-inner function. We have to prove that $x : \mathbb {D}\rightarrow \overline{\mathcal {P}}$ and, for almost all $\lambda \in \mathbb {T}, \; x(\lambda ) \in b\overline{\mathcal {P}}$.

By assumption $h = (s, p) : \mathbb {D} \rightarrow \Gamma$ is a rational $\Gamma$-inner function, which means $|p(\lambda )| = 1, \ |s(\lambda )| \le 2 \ \text { and } \ (\overline{s}p)(\lambda ) = s(\lambda )$, for almost all $\lambda \in \mathbb {T}$. Now we need to show that for almost all $\lambda \in \mathbb {T}, \; |a(\lambda )| = \sqrt{1-\frac{1}{4}|s(\lambda )|^2}$. For almost all $\lambda \in \mathbb {T}$,

$$\begin{aligned} |a(\lambda )|^2&= \left| c B(\lambda ) \dfrac{A(\lambda )}{D(\lambda )}\right| ^2 = \dfrac{|A(\lambda )|^2}{|D(\lambda )|^2} \; \; \; \text { (since } |c| = 1 \text { and } |B(\lambda )| = 1 \text { on } \mathbb {T}) \\&= \dfrac{|D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2}{|D(\lambda )|^2} = 1-\frac{1}{4}\left| \frac{E(\lambda )}{D(\lambda )}\right| ^2 \\&= 1-\frac{1}{4}|s(\lambda )|^2. \end{aligned}$$

Let us show that $x = (a, s, p) = \left( c B \dfrac{A}{D}, \dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right)$ maps $\mathbb {D}$ to $\overline{\mathcal {P}}$, that is, $x(\lambda ) = (a(\lambda ), s(\lambda ), p(\lambda )) \in \overline{\mathcal {P}}$ for all $\lambda \in \mathbb {D}$. By the construction, $D(\lambda ) \ne 0$ on $\overline{\mathbb {D}}$, and so $(a(\lambda ), s(\lambda ), p(\lambda ))$ is analytic on $\mathbb {D}$. By Theorem 2.10, for each $\lambda \in \mathbb {D}, \; x(\lambda ) \in \overline{\mathcal {P}}$ if and only if $|\Psi _z(x(\lambda ))| \le 1 \text { for all } z \in \mathbb {D},$ where

$$\begin{aligned} \Psi _z(x(.)) :&\; \mathbb {D} \rightarrow \mathbb {C} \\&\; \lambda \mapsto (1-|z|^2) \; \frac{a(\lambda )}{1-s(\lambda )z+p(\lambda )z^2}. \end{aligned}$$

For all $\lambda \in \mathbb {D}$, $(s(\lambda ), p(\lambda )) \in \Gamma$, and so $1-s(\lambda )z+p(\lambda )z^2\ne 0$ for all $z \in \mathbb {D}$. Hence, for every $z \in \mathbb {D},\; \Psi _z(x(.))$ is analytic on $\mathbb {D}$. For fixed $z \in \mathbb {D}$, by the maximum principle, to prove that $|\Psi _z(x(\lambda ))| \le 1$ for all $\lambda \in \mathbb {D}$, it suffices to show that $|\Psi _z(x(\lambda ))| \le 1$ for all $\lambda \in \mathbb {T}$. We have shown above that, for almost all $\lambda \in \mathbb {T}, \; (a(\lambda ), s(\lambda ), p(\lambda )) \in b\overline{\mathcal {P}}$. Thus, for all $\lambda \in \mathbb {T}$, $|a(\lambda )| = \sqrt{1-\frac{1}{4}|s(\lambda )|^2}, \; |p(\lambda )| = 1, \; |s(\lambda )| \le 2$ and $s(\lambda ) = \overline{s(\lambda )}p(\lambda )$, and so $(s(\lambda ), p(\lambda )) = (\beta +{\overline{\beta }}p, p)(\lambda ) \in b\Gamma$, where $\beta (\lambda )=\frac{1}{2}s(\lambda )$. One can see that, for all $\lambda \in \mathbb {T}$,

$$\begin{aligned} \left| 1-\dfrac{\frac{1}{2}s(\lambda ){\overline{\beta }}(\lambda )}{1+\sqrt{1-|\beta (\lambda )|^2}}\right|&= \left| 1-\frac{\frac{1}{4}|s(\lambda )|^2}{1+\sqrt{1-\frac{1}{4}|s(\lambda )|^2}}\right| \\&=\left| \frac{1+\sqrt{1-\frac{1}{4}|s(\lambda )|^2}-\frac{1}{4}|s(\lambda )|^2}{1+\sqrt{1-\frac{1}{4}|s(\lambda )|^2}}\right| \\&= \left| \frac{\sqrt{1-\frac{1}{4}|s(\lambda )|^2} \left( 1 +\sqrt{1-\frac{1}{4}|s(\lambda )|^2} \right) }{1+\sqrt{1-\frac{1}{4}|s(\lambda )|^2}}\right| \\&= \sqrt{1-\frac{1}{4}|s(\lambda )|^2} = |a(\lambda )|. \end{aligned}$$

By Theorem 2.10 (3) $\Leftrightarrow$ (5), for each $\lambda \in \mathbb {T}$,

$$\begin{aligned} |a(\lambda )| \le \left| 1-\dfrac{\frac{1}{2}s(\lambda ){\overline{\beta }}(\lambda )}{1+\sqrt{1-|\beta (\lambda )|^2}}\right| \text { if and only if } |\Psi _z(a(\lambda ), s(\lambda ), p(\lambda ))| \le 1 \; \text {for all} \; z \in \mathbb {D}. \end{aligned}$$

Hence, by the maximum principle, for all $z, \lambda \in \mathbb {D}$, $|\Psi _z(a(\lambda ), s(\lambda ), p(\lambda ))| \le 1$. Thus, by Theorem 2.10, $x(\lambda ) = (a(\lambda ), s(\lambda ), p(\lambda )) \in \overline{\mathcal {P}}$ for all $\lambda \in \mathbb {D}$. $\square$

Theorem 7.9

(Converse to Theorem 7.5) Suppose polynomials A, E, D satisfy

• (1) $deg(A), deg(E), deg(D) \le n$,

• (2) $E^{\sim n} = E$,

• (3) $D(\lambda ) \ne 0$ on $\overline{\mathbb {D}}$,

• (4) $|E(\lambda )| \le 2|D(\lambda )|$ on $\overline{\mathbb {D}}$,

• (5) A is an outer polynomial such that $|A(\lambda )|^2 = |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2$ for $\lambda \in \mathbb {T}$,

• (6) $a_{\textrm{in}}$ is a rational inner function on $\mathbb {D}$ of degree $\le m$.

Let a, s, p be defined by

$$\begin{aligned} a = a_{\textrm{in}}\dfrac{A}{D}, \; \; \; s = \dfrac{E}{D} \; \; \; \text { and } \; \; \; p = \dfrac{D^{\sim n}}{D} \; \; \; \text { on } \overline{\mathbb {D}}. \end{aligned}$$

Then

$$\begin{aligned} x = \left( a_{\textrm{in}} \dfrac{A}{D}, \dfrac{E}{D},\dfrac{D^{\sim n}}{D}\right) \end{aligned}$$

is a rational $\overline{\mathcal {P}}$-inner function of degree less than or equal $(m+n, n)$.

Proof

By the converse of [6, Proposition 2.2], $h = (s, p)$, where

$$\begin{aligned} s = \dfrac{E}{D} \; \; \; \text { and } \; \; \; p = \dfrac{D^{\sim n}}{D}, \end{aligned}$$

is a rational $\Gamma$-inner function of degree at most n. Since the rational inner functions on $\mathbb {D}$ are precisely the finite Blaschke products, the statement of the theorem follows from Theorem 7.8. $\square$

8 Construction of Rational $\overline{\mathcal {P}}$-Inner Functions

In this section, we describe an algorithm for the construction of rational $\overline{\mathcal {P}}$-inner function from certain interpolation data. Firstly we recall some notions and statements from [6] which were useful for the construction of rational $\Gamma$-inner functions.

Definition 8.1

[6, p. 140] Let $h =(s, p)$ be a rational $\Gamma$-inner function of degree n. Let E and D be as in Eq. (7.1) ([6, Proposition 2.2]). The royal polynomial $R_h$ of h is defined by

$$\begin{aligned} R_h(\lambda ) = 4D(\lambda )D^{\sim n}(\lambda )-E(\lambda )^2. \end{aligned}$$

(8.1)

We call the points $\lambda \in \overline{\mathbb {D}}$ such that $h(\lambda ) \in \mathcal {R}_{\Gamma }$ the royal nodes of h and, for such $\lambda$, we call $h(\lambda )$ a royal point of h, that is, $4p(\lambda )-s(\lambda )^2=0$. Since $D(\lambda )\ne 0$ on $\overline{\mathbb {D}}$, the royal nodes of h exactly correspond to the zeros of the royal polynomial $R_h$. Hence, $\lambda \in \overline{\mathbb {D}}$ is a royal node of h if and only if $R_h(\lambda ) = 0$.

Definition 8.2

[6, Definition 3.4] We say that a polynomial f is n-symmetric if deg$(f)\le n$ and $f^{\sim n} = f$. For any set $E \subset \mathbb {C}$, ord$_E(f)$ will denote the number of zeros of f in E, counted with multiplicity, and ord$_0(f)$ will mean the same as ord$_{\{0\}}(f)$.

Definition 8.3

[6, Definition 4.1] A non-zero polynomial R is n-balanced if deg$(R) \le 2n, \; R$ is 2n-symmetric and $\lambda ^{-n}R(\lambda ) \ge 0$ for all $\lambda \in \mathbb {T}$.

Proposition 8.4

[6, Proposition 3.5] Let h be a rational $\Gamma$-inner function of degree n and let $R_h$ be the royal polynomial of h as defined by Eq. (8.1). Then $R_h$ is 2n-symmetric and the zeros of $R_h$ that lie on $\mathbb {T}$ have either even or infinite order.

Definition 8.5

[6, Definition 3.6] Let h be a rational $\Gamma$-inner function such that $h(\overline{\mathbb {D}})\nsubseteq \mathcal {R}_\Gamma \cap \Gamma$ and let $R_h$ be the royal polynomial of h. If $\sigma$ is a zero of $R_h$ of order $\ell$, we define the multiplicity $\#\sigma$ of $\sigma$ (as a royal node of h) by

$$\begin{aligned} \#\sigma = {\left\{ \begin{array}{ll} \ell &{} \text { if } \sigma \in \mathbb {D} \\ \frac{1}{2}\ell &{} \text { if } \sigma \in \mathbb {T}. \end{array}\right. } \end{aligned}$$

We next present a description of rational penta-inner functions (a, s, p) in terms of the zeros of a, s and $s^2 -4p$.

Theorem 8.6

Suppose that $\alpha _1, \alpha _2, \dots , \alpha _{k_0} \in \mathbb {D}$ and $\eta _1, \eta _2, \dots , \eta _{k_1} \in \mathbb {T}$, where $2k_0+k_1=n$ and suppose that $\beta _1, \beta _2, \dots , \beta _m \in \mathbb {D}$. Suppose that $\sigma _1, \dots , \sigma _n$ in $\overline{\mathbb {D}}$ are distinct from $\eta _1, \dots , \eta _{k_1}$. Then there exists a rational $\overline{\mathcal {P}}$-inner function $x = (a, s, p)$ of degree less than or equal $(m+n, n)$ such that

• (1) the zeros of a in $\mathbb {D}$, repeated according to multiplicity, are $\beta _1, \beta _2, \dots , \beta _m$,

• (2) the zeros of s in $\overline{\mathbb {D}}$, repeated according to multiplicity, are $\alpha _1, \alpha _2, \dots , \alpha _{k_0}$ and $\eta _1, \eta _2, \dots , \eta _{k_1}$,

• (3) the royal nodes of (s, p) are $\sigma _1, \dots , \sigma _n$.

Such a function x can be constructed as follows. Let $t_+ \; \>\; 0$ and let $t \in \mathbb {R} \setminus \{0\}$. Let R and E be defined by

$$\begin{aligned} R(\lambda ) = t_{+} \prod _{j=1}^{n} (\lambda -\sigma _j)(1-\overline{\sigma _j}\lambda ), \\ E(\lambda ) = t\prod _{j=1}^{k_0} (\lambda -\alpha _j)(1-\overline{\alpha _j}\lambda ) \prod _{j=1}^{k_1} ie ^{-i \theta _j/2}(\lambda -\eta _j), \end{aligned}$$

where $\eta _j = e ^{i\theta _j}, \; 0 \le \theta _j \; <\; 2\pi$. Let $a_{\textrm{in}} : \mathbb {D} \rightarrow \overline{\mathbb {D}}$ be defined by

$$\begin{aligned} a_{\textrm{in}}(\lambda ) = c\prod _{i=1}^{m} B_{\beta _i}(\lambda ), \end{aligned}$$

(8.2)

where $|c|=1$ and $\beta _i \in \mathbb {D}, \; i=1, \dots , m$.

• (i) There exist outer polynomials D and A of degree at most n such that

  $$\begin{aligned} \lambda ^{-n}R(\lambda )+|E(\lambda )|^2 = 4|D(\lambda )|^2 \end{aligned}$$

  (8.3)

  and

  $$\begin{aligned} \lambda ^{-n}R(\lambda ) = 4|A(\lambda )|^2 \end{aligned}$$

  (8.4)

  for all $\lambda \in \mathbb {T}$.

• (ii) The function x defined by

  $$\begin{aligned} x=(a, s, p)=\Big (a_{\textrm{in}}\frac{A}{D}, \frac{E}{D}, \frac{D^{\sim n}}{D}\Big ) \end{aligned}$$

  (8.5)

  is a rational $\overline{\mathcal {P}}$-inner function such that deg$(x) \le (m+n, n)$ and conditions (1), (2) and (3) hold. The royal polynomial of (s, p) is R.

Proof

• (i) By [6, Lemma 4.4], R is n-balanced, and so $\lambda ^{-n}R(\lambda ) \ge 0$ for all $\lambda \in \mathbb {T}$. Therefore,

  $$\begin{aligned} \lambda ^{-n}R(\lambda )+|E(\lambda )|^2 \ge 0 \text { for all } \lambda \in \mathbb {T}. \end{aligned}$$

  By the Fejér–Riesz theorem, there exist outer polynomials A and D of degree at most n such that

  $$\begin{aligned} \lambda ^{-n}R(\lambda ) = 4|A(\lambda )|^2 \; \text { for all } \lambda \in \mathbb {T} \end{aligned}$$

  and

  $$\begin{aligned} \lambda ^{-n}R(\lambda )+|E(\lambda )|^2 = 4|D(\lambda )|^2 \; \text { for all } \lambda \in \mathbb {T}. \end{aligned}$$

• (ii) By [6, Theorem 4.8], the function h defined by

  $$\begin{aligned} h=(s, p)=\Big (\frac{E}{D}, \frac{D^{\sim n}}{D}\Big ) \end{aligned}$$

  is a rational $\Gamma$-inner function such that deg$(h)=n$ and conditions (2) and (3) hold. The royal polynomial of h is R. By Eqs. (8.3) and (8.4),

  $$\begin{aligned} |A(\lambda )|^2= & {} |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2. \end{aligned}$$

  Therefore, by Proposition 7.8,

  $$\begin{aligned} x = \Big (a_{\textrm{in}}\frac{A}{D}, \frac{E}{D}, \frac{D^{\sim n}}{D}\Big ) \end{aligned}$$

  is a rational $\overline{\mathcal {P}}$- inner function. By the definition (8.2) of $a_{\textrm{in}}$, the zeros of $a_{\textrm{in}}$ are $\beta _1, \dots , \beta _m$, while, since A is an outer polynomial, A has no zeros in $\mathbb {D}$. Hence the zeros of $a = a_{\textrm{in}}\dfrac{A}{D}$ in $\mathbb {D}$ are $\beta _1, \dots , \beta _m$, as required for (1).

$\square$

Theorem 8.7

Let $x=(a, s, p)$ be a rational $\overline{\mathcal {P}}$-inner function of degree $(m+n, n)$ such that

• (1) the zeros of a, repeated according to multiplicity, are $\beta _1, \beta _2, \dots , \beta _m \in \mathbb {D}$,

• (2) the zeros of s, repeated according to multiplicity, are $\alpha _1, \alpha _2, \dots , \alpha _{k_0} \in \mathbb {D}$ and $\eta _1, \eta _2, \dots , \eta _{k_1} \in \mathbb {T}$, where $2k_0+k_1=n$,

• (3) the royal nodes of (s, p) are $\sigma _1, \dots , \sigma _n \in \overline{\mathbb {D}}$.

There exists some choice of $c \in \mathbb {T}, \; t_{+} \; \>\; 0, \; t \in \mathbb {R} \setminus \{0\}$ and $\omega \in \mathbb {T}$ such that the recipe in Theorem 8.6 with these choices produces the function x.

Proof

By Lemma 6.2, $h=(s, p)$ is a rational $\Gamma$-inner function of degree n. As in [6, Proposition 4.9], there exists some choice of $t_{+} \; \>\; 0, \; t \in \mathbb {R} \setminus \{0\}$ and $\omega \in \mathbb {T}$ such that the recipe of [6, Theorem 4.8] produces the function h. Let us give those steps.

By [6, Proposition 2.2], there exist polynomials $E_1$ and $D_1$ such that deg$(E_1)$, deg$(D_1) \le n$, $E_1$ is n-symmetric, $D_1(\lambda ) \ne 0$ on $\overline{\mathbb {D}}$, and

$$\begin{aligned} s = \dfrac{E_1}{D_1} \; \text { and } \; \; p = \dfrac{D_{1}^{\sim n}}{D_1} \text { on } \overline{\mathbb {D}}. \end{aligned}$$

By hypothesis, the zeros of s, repeated according to multiplicity, are $\alpha _1, \alpha _2, \dots , \alpha _{k_0}$ and $\eta _1, \eta _2, \dots , \eta _{k_1}$, where $2k_0+k_1=n$. Since $E_1$ is n-symmetric, by [6, Lemma 4.6], there exists $t \in \mathbb {R} \setminus \{0\}$ such that

$$\begin{aligned} E_1(\lambda ) = t\prod _{j=1}^{k_0} (\lambda -\alpha _j)(1-\overline{\alpha _j}\lambda ) \prod _{j=1}^{k_1} ie ^{-i \theta _j/2}(\lambda -\eta _j), \end{aligned}$$

where $\eta _j = e^{i \theta _j}$ for $j =1, \dots , k_1.$ The royal nodes of h are assumed to be $\sigma _1, \dots , \sigma _n$. By [6, Proposition 4.5], for the royal polynomial $R_1$ of h, there exists $t_{+} \; \>\; 0$ such that

$$\begin{aligned} R_1(\lambda ) = t_{+} \prod _{j=1}^{n} Q_{\sigma _j}(\lambda ), \end{aligned}$$

where $Q_{\sigma _j}(\lambda )= (\lambda -\sigma _j)(1-\overline{\sigma _j}\lambda )$, $j= 1, \dots ,n$.

Since $E_1$ and $R_1$ coincide with E and R in the construction of Theorem 8.6, for a suitable choice of $t_{+} \; \>\; 0$ and $t \in \mathbb {R} \setminus \{0\}$, $D_1$ is a permissible choice for $\omega D$ for some $\omega \in \mathbb {T}$, as a solution of Eq. (8.3). By assumption the zeros of a, repeated according to multiplicity, are $\beta _1, \beta _2, \dots , \beta _m \in \mathbb {D}$. Then the inner part of a will be equal to $a_{\textrm{in}}^{1} = c_1 \prod \limits _{i=1}^{m}B_{\beta _i}$, where $|c_1|=1$. For the outer part of a there is an outer polynomial $A_1$ such that

$$\begin{aligned} |A_1(\lambda )|^2= & {} |D_1(\lambda )|^2-\frac{1}{4}|E_1(\lambda )|^2 \\= & {} |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2 \\= & {} \lambda ^{-n}R(\lambda ), \end{aligned}$$

for $\lambda \in \mathbb {T}$. By Eq. (8.4), $A_1 = c_2 A$ up to a constant $c_2$ such that $|c_2|=1$. Also, $a_{\textrm{in}}^{1}$ coincides with $a_{\textrm{in}}$ for a suitable choice of $c \in \mathbb {T}$. Hence the construction of Theorem 8.6 yields $x=(a, s, p)$ for the appropriate choices of $t_{+} \; \>\; 0$, $t \in \mathbb {R} \setminus \{0\}$, $\omega$ and $c \in \mathbb {T}$. $\square$

9 A Special Case of Schwarz Lemma for $\mathcal {P}$

The classical Schwarz lemma gives a solvability criterion for a two-point interpolation problem in $\mathbb {D}$. In [4] a simple analogue of Schwarz lemma the for two-point $\mu$-synthesis was given. We consider a general linear subspace E of $\mathbb {C}^{n\times m}$ and the corresponding $\mu _E$ on $\mathbb {C}^{m\times n}$, as in Eq. (1.2). We shall denote by N the Nevanlinna class of functions on the disc [28] and if F is a matrix function on $\mathbb {D}$ then we write $F\in N$ to mean that each entry of F belongs to N. It then follows from Fatou’s Theorem that if $F\in N$ is an $m\times n$-matrix-valued function then

$$\begin{aligned} \lim _{r\rightarrow 1-} F(r\lambda ) \text{ exists } \text{ for } \text{ almost } \text{ all } \lambda \in \mathbb {T}. \end{aligned}$$

The following Schwarz lemma was proved in [4, Proposition 10.3].

Proposition 9.1

Let $\lambda _0 \in \mathbb {D}\setminus \{0 \}$, let $W \in \mathbb {C}^{m\times n}$ and let E be a subset of $\mathbb {C}^{n\times m}$. There exists $F \in N\cap {{\,\textrm{Hol}\,}}(\mathbb {D},\mathbb {C}^{m\times n})$ such that

• 1. $F(0) =0$ and $F(\lambda _0)= W$,

• 2. $\mu _E(F(\lambda )) < 1$ for all $\lambda \in \mathbb {D}$

if and only if $\mu _E(W) \le |\lambda _0|$.

In this section, we consider a simple case of a Schwarz lemma for the pentablock. We will need the following elementary technical lemma.

Lemma 9.2

Let $A =$ $\begin{bmatrix} \lambda _1 &{} 0 \\ a &{} \lambda _2 \end{bmatrix}$, where $\lambda _1, \; \lambda _2, \; a \in \mathbb {C}$. Then the following conditions are equivalent:

• (i) $\lambda _1, \lambda _2 \in \overline{\mathbb {D}}, \; |a| \le (1 - |\lambda _1|^2)^{\frac{1}{2}}(1 - |\lambda _2|^2)^{\frac{1}{2}}$,

• (ii) $\Vert A\Vert \le 1$,

• (iii) $1-A^*A \ge 0$.

Definition 9.3

$H^{\infty }(\mathbb {D}, \mathbb {C}^{2\times 2})$ denotes the space of bounded analytic $2\times 2$ matrix-valued functions on $\mathbb {D}$ with the supremum norm:

$$\begin{aligned} \Vert f\Vert _{H^\infty } = \sup \limits _{z \in \mathbb {D}}\Vert f(z)\Vert _{\mathbb {C}^{2\times 2}}. \end{aligned}$$

Definition 9.4

$L^{\infty }(\mathbb {T}, \mathbb {C}^{2\times 2})$ denotes the space of essentially bounded Lebesgue-measurable $2\times 2$ matrix-valued functions on $\mathbb {T}$ with the essential supremum norm:

$$\begin{aligned} \Vert f\Vert _{L^\infty } = ess \sup \limits _{|z|=1}\Vert f(z)\Vert _{\mathbb {C}^{2\times 2}}. \end{aligned}$$

Lemma 9.5

If $g \in H^{\infty }(\mathbb {D}, \mathbb {C}^{2\times 2})$ and $\lambda _0 \in \mathbb {D}$ then $\Vert g(\lambda _0)\Vert _{\mathbb {C}^{2\times 2}} \le \Vert g\Vert _{L^\infty }$.

Proof

Consider any unit vectors $x, y \in \mathbb {C}^{2}$ and the scalar function

$$\begin{aligned} f&:&\mathbb {D} \rightarrow \mathbb {C} \\&:&\lambda \longmapsto \langle g(\lambda )x, y \rangle _{\mathbb {C}^2}. \end{aligned}$$

Note that, for every $\lambda \in \mathbb {D}$, since $\Vert x\Vert _{\mathbb {C}^2} = \Vert y\Vert _{\mathbb {C}^2} = 1$

$$\begin{aligned} |f(\lambda )| = |\left\langle g(\lambda )x, y \right\rangle _{\mathbb {C}^2}|\le & {} \Vert g(\lambda )x\Vert _{\mathbb {C}^2} \; \Vert y\Vert _{\mathbb {C}^2} \; \; \; \; \; \text {(Cauchy--Schwarz inequality)}\\\le & {} \Vert g(\lambda )\Vert _{\mathbb {C}^{2\times 2}} \; \Vert x\Vert _{\mathbb {C}^2} \; \Vert y\Vert _{\mathbb {C}^2} \\\le & {} \Vert g\Vert _{H^\infty }. \end{aligned}$$

Thus f is bounded on $\mathbb {D}$. Since g is analytic on $\mathbb {D}$, it is easy to show that f is analytic on $\mathbb {D}$ and, for every $z_0 \in \mathbb {D}, \; f'(z_0) = \langle g'(z_0)x, y \rangle _{\mathbb {C}^2}$. By the maximum principle for scalar analytic functions, for every $\lambda _0 \in \mathbb {D}, \; |f(\lambda _0)| \le ess \sup \limits _{z \in \mathbb {T}} |f(z)|$, and so

$$\begin{aligned} \left| \left\langle g(\lambda _0)x, y \right\rangle _{\mathbb {C}^2}\right|\le & {} ess \sup \limits _{z \in \mathbb {T}} \left| \left\langle g(z)x, y \right\rangle \right| \\\le & {} ess \sup \limits _{z \in \mathbb {T}} \Vert g(z)\Vert _{\mathbb {C}^{2\times 2}} = \Vert g\Vert _{L^{\infty }}. \end{aligned}$$

Take the supremum of both sides in this inequality over unit vectors x, y to get

$$\begin{aligned} \Vert g(\lambda _0)\Vert _{\mathbb {C}^{2\times 2}} \le \Vert g\Vert _{L^{\infty }}. \end{aligned}$$

$\square$

The following statement is known and follows easily from Lemma 9.5.

Corollary 9.6

If $F \in H^{\infty }(\mathbb {D}, \mathbb {C}^{2\times 2})$ and $F(0) = 0$ then, for any $\lambda _0 \in \mathbb {D}$,

$$\begin{aligned} \Vert F(\lambda _0)\Vert _{\mathbb {C}^{2\times 2}} \le |\lambda _0|\;\Vert F\Vert _{H^{\infty }}. \end{aligned}$$

We next describe a special case of a Schwarz lemma for the pentablock.

Theorem 9.7

Let $\lambda _0 \in \mathbb {D} \setminus \{0\}$, and $(a_0, s_0, p_0) \in \overline{\mathcal {P}}$, where $s_0 = \lambda _1+\lambda _2, \; p_0 = \lambda _1\lambda _2$, for some $\lambda _1, \lambda _2 \in \mathbb {D}$. Then the following conditions are equivalent:

• (i) $|\lambda _1| \le |\lambda _0|, \; |\lambda _2| \le |\lambda _0|$, and

  $$\begin{aligned} |a_0| \le |\lambda _0|\left( 1-\left| \dfrac{\lambda _1}{\lambda _0}\right| ^2\right) ^{\frac{1}{2}} \left( 1-\left| \dfrac{\lambda _2}{\lambda _0}\right| ^2\right) ^{\frac{1}{2}}. \end{aligned}$$

  (9.1)

• (ii) There exists an analytic map $F : \mathbb {D} \rightarrow \overline{\mathbb {B}^{2\times 2}}$ such that

  $$\begin{aligned} F(0) = 0 \text { and } F(\lambda _0) = \begin{bmatrix} \lambda _1 &{} 0 \\ a_0 &{} \lambda _2 \end{bmatrix}. \end{aligned}$$

  Furthermore, if (i) holds and $x = \pi \circ F$, then x is an analytic map from $\mathbb {D}$ to $\overline{\mathcal {P}}$ such that $x(0) = (0,0,0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$.

Proof

(i) $\Rightarrow$ (ii) By assumption, $|\lambda _1| \le |\lambda _0|, \; |\lambda _2| \le |\lambda _0|$ and

$$\begin{aligned} |a_0| \le |\lambda _0|\left( 1-\left| \dfrac{\lambda _1}{\lambda _0}\right| ^2\right) ^{\frac{1}{2}} \left( 1-\left| \dfrac{\lambda _2}{\lambda _0}\right| ^2\right) ^{\frac{1}{2}}. \end{aligned}$$

Define

$$\begin{aligned} F(\lambda ) = \frac{\lambda }{\lambda _0}\begin{bmatrix} \lambda _1 &{} 0 \\ a_0 &{} \lambda _2 \end{bmatrix} = \lambda \begin{bmatrix} \lambda _1 / \lambda _0 &{} 0 \\ a_0 / \lambda _0 &{} \lambda _2 / \lambda _0 \end{bmatrix}. \end{aligned}$$

(9.2)

By Lemma 9.2,

$$\begin{aligned} \left\| \begin{bmatrix} \dfrac{\lambda _1}{\lambda _0} &{} 0 \\ \dfrac{a_0}{\lambda _0} &{} \dfrac{\lambda _2}{\lambda _0} \end{bmatrix}\right\| _{\mathbb {C}^{2\times 2}} \le 1. \end{aligned}$$

Hence $\Vert F(\lambda )\Vert \le |\lambda |$ for all $\lambda \in \mathbb {D}$, and so $\Vert F\Vert _{\infty } \le 1$. From the definition (9.2) of F, we have

$$\begin{aligned} F(0) = \begin{bmatrix} 0 &{} 0 \\ 0 &{} 0 \end{bmatrix} \text { and } F(\lambda _0) = \begin{bmatrix} \lambda _1 &{} 0 \\ a_0 &{} \lambda _2 \end{bmatrix}. \end{aligned}$$

(ii) $\Rightarrow$ (i) Suppose (ii) is satisfied. By Corollary 9.6,

$$\begin{aligned} \left\| \begin{bmatrix} \lambda _1 &{} 0 \\ a_0 &{} \lambda _2 \end{bmatrix}\right\| _{\mathbb {C}^{2\times 2}} = \Vert F(\lambda _0)\Vert _{\mathbb {C}^{2\times 2}} \le |\lambda _0| \; \Vert F\Vert _{H^\infty }. \end{aligned}$$

By assumption $\Vert F\Vert _{H^\infty } \le 1$, and so $\left\| \begin{bmatrix} \lambda _1 &{} 0 \\ a_0 &{} \lambda _2 \end{bmatrix}\right\| _{\mathbb {C}^{2\times 2}} \le |\lambda _0|$, hence, $\left\| \begin{bmatrix} \dfrac{\lambda _1}{\lambda _0} &{} 0 \\ \dfrac{a_0}{\lambda _0} &{} \dfrac{\lambda _2}{\lambda _0} \end{bmatrix}\right\| _{\mathbb {C}^{2\times 2}} \le 1. \;$ By Lemma 9.2,

$$\begin{aligned}&\left\| \begin{bmatrix} \dfrac{\lambda _1}{\lambda _0} &{} 0 \\ \dfrac{a_0}{\lambda _0} &{} \dfrac{\lambda _2}{\lambda _0} \end{bmatrix}\right\| _{\mathbb {C}^{2\times 2}} \le 1 \text { if and only if } \\&\quad |a_0| \le |\lambda _0|\left( 1-\left| \dfrac{\lambda _1}{\lambda _0}\right| ^2\right) ^{\frac{1}{2}} \left( 1-\left| \dfrac{\lambda _2}{\lambda _0}\right| ^2\right) ^{\frac{1}{2}}, \; \; |\lambda _1| \le |\lambda _0| \text { and } |\lambda _2| \le |\lambda _0|. \end{aligned}$$

Let us consider $x = \pi \circ F$ on $\mathbb {D}$. By assumption $(a_0, s_0, p_0) \in \overline{\mathcal {P}}$. By Theorem 2.10 (6), since $\Vert F(\lambda )\Vert \le 1$ for each $\lambda \in \mathbb {D}$, $x(\lambda ) = \pi (F(\lambda )) = \dfrac{\lambda }{\lambda _0}(a_0, s_0, p_0)$ maps $\mathbb {D}$ to $\overline{\mathcal {P}}$. Therefore, $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ is analytic on $\mathbb {D}$ and maps 0 to (0, 0, 0) and $\lambda _0$ to $(a_0, s_0, p_0)$. $\square$

10 A Schwarz Lemma for the Symmetrized Bidisc $\Gamma$

In [11], Agler and Young proved the following theorems.

Theorem 10.1

[11, Theorem 1.1] Let $\lambda _0 \in \mathbb {D}$ and $(s_0, p_0) \in \Gamma$. The following conditions are equivalent:

• (1) There exists an analytic function $\varphi : \mathbb {D} \rightarrow \Gamma$ such that $\varphi (0) = (0, 0)$ and $\varphi (\lambda _0) = (s_0, p_0)$;

• (2) $|s_0| \; <\; 2$ and

  $$\begin{aligned} \dfrac{2|s_0 - p_0\overline{s_0}| + |s_{0}^{2} - 4p_0|}{4 - |s_0|^2} \le |\lambda _0|; \end{aligned}$$
• (3)
  $$\begin{aligned} \big | |\lambda _0|^2s_0 - p_0\overline{s_0}\big | + |p_0|^2 + (1 - |\lambda _0|^2) \dfrac{|s_0|^2}{4} - |\lambda _0|^2 \le 0; \end{aligned}$$

  (10.1)
• (4)
  $$\begin{aligned} |s_0| \le \frac{2}{1 - |\lambda _0|^2}(|\lambda _0| |1 - p_0{\overline{\omega }}^2| - \big | |\lambda _0|^2 - p_0 {\overline{\omega }}^2\big |), \end{aligned}$$

  where $\omega$ is a complex number of unit modulus such that $s_0 = |s_0|\omega$.

Moreover, for any analytic function $\varphi = (\varphi _1, \varphi _2) : \mathbb {D} \rightarrow \Gamma$ such that $\varphi (0) = (0, 0)$,

$$\begin{aligned} \frac{1}{2}|\varphi _{1}^{'}(0)| + |\varphi _{2}^{'}(0)| \le 1. \end{aligned}$$

The following theorem shows the construction of an interpolating function $\varphi$ satisfying the inequalities of Theorem 10.1 with equality.

Theorem 10.2

[11, Theorem 1.4] Let $\lambda _0 \in \mathbb {D}$, and $(s_0, p_0) \in \Gamma$ be such that $\lambda _0 \ne 0$, $|s_0| <2$ and

$$\begin{aligned} \dfrac{2|s_0 - p_0\overline{s}_0| + |s_{0}^{2} - 4p_0|}{4 - |s_0|^2} = |\lambda _0|. \end{aligned}$$

Then there exists an analytic function $\varphi : \mathbb {D} \rightarrow \Gamma$ such that $\varphi (0) = (0, 0)$ and $\varphi (\lambda _0) = (s_0, p_0)$, given explicitly as follows:

If $|p_0| = |\lambda _0|$, then

$$\begin{aligned} \varphi (\lambda ) = (0, \omega \lambda ), \end{aligned}$$

(10.2)

where $\omega$ is a complex number of unit modulus such that $\omega \lambda _0 = p_0$.

If $|p_0| \; <\; |\lambda _0|$, then $\varphi = (\varphi _1, \varphi _2)$ where

$$\begin{aligned} \varphi _1(\lambda )= & {} \frac{c \zeta \lambda }{(1 - {\overline{\lambda }}_{0}\lambda ) (1 + \overline{p}_1 \zeta ^2 v(\lambda ))}, \end{aligned}$$

(10.3)

$$\begin{aligned} v(\lambda )= & {} \frac{\lambda - \lambda _0}{1 - {\overline{\lambda }}_0 \lambda }, \; \; \; \; \; \zeta \lambda _0|s_0| = |\lambda _0| s_0, \; \; \; \; \; |\zeta | = 1,\nonumber \\ p_1= & {} \frac{p_0}{\lambda _0}, \; \; \; \; \; c = \frac{2}{|\lambda _0|}\{|{\overline{\lambda }}_0 - \overline{p}_0 \lambda _0 \zeta ^2| - |\lambda _0^2 \zeta ^2 - p_0|\},\nonumber \\ \varphi _2(\lambda )= & {} \frac{\lambda (\zeta ^2 v(\lambda ) + p_1)}{1 + \overline{p}_1 \zeta ^2 v(\lambda )}. \end{aligned}$$

(10.4)

Lemma 10.3

Consider the rational function $\varphi = (\varphi _1, \varphi _2): \mathbb {D} \rightarrow \Gamma$, where $\varphi _1, \varphi _2$ are defined as in Eqs. (10.3) and (10.4) above. Define the polynomials E and D by the equations:

$$\begin{aligned} E(\lambda )= & {} c \lambda ,\\ D(\lambda )= & {} {\overline{\zeta }}\{(1-\overline{\lambda _0}\lambda ) + \overline{p_1}\zeta ^2 (\lambda - \lambda _0)\}, \end{aligned}$$

where

$$\begin{aligned} |\zeta | = 1, \; \; \; p_1 = \dfrac{p_0}{\lambda _0}, \; \; \; \; \; c = \dfrac{2}{|\lambda _0|}\{|{\overline{\lambda }}_0 - \overline{p}_0 \lambda _0 \zeta ^2| - |\lambda _0^2 \zeta ^2 - p_0|\}. \end{aligned}$$

Then $\varphi _1 = \dfrac{E}{D}$ and $\varphi _2 = \dfrac{D^{\sim 2}}{D}$. Moreover, $E^{\sim 2} = E$ and $|E(\lambda )| \le 2 |D(\lambda )|$ on $\overline{\mathbb {D}}$.

Proof

Let us check that $\varphi _1(\lambda ) = \dfrac{E(\lambda )}{D(\lambda )}$.

$$\begin{aligned} \frac{E(\lambda )}{D(\lambda )}= & {} \frac{c\lambda }{{\overline{\zeta }}\{(1-\overline{\lambda _0}\lambda ) + \overline{p_1}\zeta ^2 (\lambda - \lambda _0)\}} \times \frac{\zeta }{\zeta } \\= & {} \frac{c \zeta \lambda }{(1-\overline{\lambda _0}\lambda ) + \overline{p_1}\zeta ^2 (\lambda - \lambda _0)} \\= & {} \frac{c \zeta \lambda }{(1-\overline{\lambda _0}\lambda ) (1 + \overline{p_1}\zeta ^2 v(\lambda ))} = \varphi _1(\lambda ). \end{aligned}$$

To check that $\varphi _2(\lambda ) = \dfrac{D^{\sim 2}(\lambda )}{D(\lambda )}$, we need to find $D^{\sim 2}(\lambda )$.

$$\begin{aligned} D^{\sim 2}(\lambda ) = \lambda ^2 \overline{D(1 / {\overline{\lambda }})}= & {} \lambda ^2 \overline{{\overline{\zeta }}\left\{ \left( 1- \frac{\overline{\lambda _0}}{{\overline{\lambda }}}\right) + \overline{p_1}\zeta ^2\left( \frac{\ 1 \ }{\ {\overline{\lambda }} \ }-\lambda _0\right) \right\} } \\= & {} \lambda ^2 \zeta \left\{ \left( 1- \frac{\lambda _0}{\lambda }\right) + p_1 {\overline{\zeta }}^2 \left( \frac{1}{\lambda } - \overline{\lambda _0}\right) \right\} \\= & {} \lambda \zeta (\lambda - \lambda _0) + \lambda p_1 {\overline{\zeta }}(1-\overline{\lambda _0}\lambda ). \end{aligned}$$

Now,

$$\begin{aligned} \frac{D^{\sim 2}(\lambda )}{D(\lambda )}= & {} \frac{\lambda \zeta (\lambda - \lambda _0) + \lambda p_1 {\overline{\zeta }}(1-\overline{\lambda _0}\lambda )}{{\overline{\zeta }}\{(1-\overline{\lambda _0}\lambda ) + \overline{p_1}\zeta ^2 (\lambda - \lambda _0)\}} \times \frac{\left( \dfrac{\zeta }{1-\overline{\lambda _0}\lambda }\right) }{\left( \dfrac{\zeta }{1-\overline{\lambda _0}\lambda }\right) } \\= & {} \frac{\lambda \zeta ^2 \left( \dfrac{\lambda - \lambda _0}{1-\overline{\lambda _0}\lambda }\right) + \lambda p_1}{1+\overline{p_1}\zeta ^2 \left( \dfrac{\lambda - \lambda _0}{1-\overline{\lambda _0}\lambda }\right) } \\= & {} \frac{\lambda (\zeta ^2 v(\lambda )+ p_1)}{1+\overline{p_1} \zeta ^2 v(\lambda )} = \varphi _2(\lambda ), \end{aligned}$$

where $v(\lambda ) = \dfrac{\lambda - \lambda _0}{1-\overline{\lambda _0}\lambda }$.

We would like to show that $E^{\sim 2} = E$. For $\lambda \in \mathbb {D}$,

$$\begin{aligned} E^{\sim 2}(\lambda ) = \lambda ^2 \overline{E(1/{\overline{\lambda }})}= & {} \lambda ^2 \overline{\left( c \; \frac{\ 1 \ }{\ {\overline{\lambda }} \ }\right) } \\= & {} c \; \lambda = E(\lambda ), \;\; \text {since} \; c \in \mathbb {R}. \end{aligned}$$

By assumption, $\varphi = (\varphi _1, \varphi _2): \mathbb {D} \rightarrow \Gamma$, and we have proved that $\varphi _1 = \dfrac{E}{D}$, thus $|E(\lambda )| \le 2 |D(\lambda )|$ on $\overline{\mathbb {D}}$. $\square$

Proposition 10.4

Let $h = (s, p)$ be the function from $\mathbb {D}$ to $\Gamma$ defined by

$$\begin{aligned} s(\lambda ) = \varphi _1(\lambda ), \; p(\lambda ) = \varphi _2(\lambda ), \; \lambda \in \mathbb {D}, \end{aligned}$$

as in Eqs. (10.3) and (10.4). Then h is a rational $\Gamma$-inner function of degree 2.

Proof

By Lemma 10.3 and by the converse part of [6, Proposition 2.2], h is a rational $\Gamma$-inner function.

Another way to prove that h is a rational $\Gamma$-inner function is as follows. One can easily see that $h = (s, p)$ is a rational function, and so there are only finitely many singularities of this function. Hence we can extend h continuously to almost all points in $\mathbb {T}$.

Let us show that for almost all $\lambda \in \mathbb {T}, \; h(\lambda ) \in b\Gamma$. We need to show that, for almost all $\lambda \in \mathbb {T}, \; |p(\lambda )| = 1, \; |s(\lambda )| \le 2$ and $s(\lambda ) = \overline{s(\lambda )} p(\lambda )$. Since $v(\lambda ) = \dfrac{\lambda -\lambda _0}{1-\overline{\lambda _0}\lambda }$ is an inner function from $\mathbb {D}$ to $\overline{\mathbb {D}}$, for almost all $\lambda \in \mathbb {T}, \; |v(\lambda )| = 1$.

$$\begin{aligned} |p(\lambda )|= & {} \frac{|\lambda | |\zeta ^2 v(\lambda ) + p_1|}{|1 + \overline{p_1} \zeta ^2 v(\lambda )|} = \frac{|\zeta ^2 v(\lambda ) + p_1|}{|\zeta ^2 \overline{\zeta ^2} (v \overline{v})(\lambda ) + \overline{p_1} \zeta ^2 v(\lambda )|} \\= & {} \frac{|\zeta ^2 v(\lambda ) + p_1|}{|\zeta ^2 v(\lambda ) \overline{(\zeta ^2 v(\lambda ) + p_1)}|} = \frac{|\zeta ^2 v(\lambda ) + p_1|}{|\zeta ^2 v(\lambda ) + p_1|} = 1, \end{aligned}$$

for almost all $\lambda \in \mathbb {T}$.

In [11, Theorem 1.5] it was proved that, for all $\lambda \in \mathbb {D}$,

$$\begin{aligned} \big | |\lambda |^2 s(\lambda ) - \overline{s(\lambda )}p(\lambda )\big | + |p(\lambda )|^2 + (1-|\lambda |^2)\dfrac{|s(\lambda )|^2}{4} - |\lambda |^2 = 0. \end{aligned}$$

(10.5)

Choose a sequence $(r_n)_{n \ge 1}$ such that $0< r_n < 1$ for each n and $\lim \limits _{n \rightarrow \infty } r_n = 1$. Consider $\mu \in \mathbb {T}$ and let $\lambda = r_n \mu$ in Eq. (10.5). On letting $n \rightarrow \infty$ we find that, for almost all $\mu \in \mathbb {T}$,

$$\begin{aligned} \big | |\mu |^2 s(\mu ) - \overline{s(\mu )}p(\mu )\big | + |p(\mu )|^2 + (1-|\mu |^2)\dfrac{|s(\mu )|^2}{4} - |\mu |^2 = 0. \end{aligned}$$

(10.6)

Note that $|\mu | = 1$ and $|p(\mu )|^2=1$, and so Eq. (10.6) is equivalent to

$$\begin{aligned} |s(\mu ) - \overline{s(\mu )}p(\mu )| = 0. \end{aligned}$$

Hence $s(\mu ) = \overline{s(\mu )}p(\mu )$ for almost all $\mu \in \mathbb {T}$. Note that for all $\lambda \in \mathbb {D}, \; h(\lambda ) = (s(\lambda ), p(\lambda )) \in \Gamma$, which means $|s(\lambda )| \le 2$ for all $\lambda \in \mathbb {D}$. Thus, for almost all $\mu \in \mathbb {T}, \; |s(\mu )| \le 2$. By [6, Proposition 3.3], $\text {deg}(h) = \text {deg}(p)$. By Lemma 10.3,

$$\begin{aligned} p(\lambda ) = \varphi _2(\lambda ) = \frac{D^{\sim 2}(\lambda )}{D(\lambda )} = \dfrac{\lambda \zeta (\lambda -\lambda _0) + \lambda p_1 {\overline{\zeta }} (1-\overline{\lambda _0}\lambda )}{{\overline{\zeta }}\{(1-\overline{\lambda _0}\lambda ) + \overline{p_1}\zeta ^2 (\lambda - \lambda _0)\}}, \end{aligned}$$

where $D(\lambda ) = {\overline{\zeta }}\{(1-\overline{\lambda _0}\lambda ) + \overline{p_1}\zeta ^2 (\lambda - \lambda _0)\}$. Since $\text {deg}(D^{\sim 2}) = 2$ and $\text {deg}(D) = 1$, then $\text {deg}(p) = 2$. Therefore, $\text {deg}(h) = 2$. $\square$

11 Sharpness of the Schwarz Lemma for $\mathcal {P}$

Recall necessary conditions for a Schwarz lemma for $\mathcal {P}$ which were given in [4, Proposition 11.1].

Proposition 11.1

[4, Proposition 11.1] Let $\lambda _0 \in \mathbb {D}\setminus \{0\}$ and $(a_0, s_0, p_0) \in \overline{\mathcal {P}}$. If $x \in Hol (\mathbb {D}, \mathcal {P})$ satisfies $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$ then $|s_0| <2$,

$$\begin{aligned} \frac{2|s_0-\overline{s}_0 p_0|+|s_0^2-4p_0|}{4-|s_0|^2} \le |\lambda _0| \end{aligned}$$

(11.1)

and

$$\begin{aligned} |a_0|\Big /\left| 1-\frac{\frac{1}{2}s_0{\overline{\beta }}}{1+\sqrt{1-|\beta |^2}}\right| \le |\lambda _0|, \end{aligned}$$

(11.2)

where $\beta = (s_0-\overline{s}_0 p_0)/(1-|p_0|^2)$ when $|p_0| < 1$ and $\beta = \frac{1}{2}s_0$ when $|p_0| = 1$.

We prove the following result.

Theorem 11.2

Let $\lambda _0 \in \mathbb {D}\setminus \{0\}$, and $(a_0, s_0, p_0) \in \overline{\mathcal {P}}$ be such that $|s_0| <2$,

$$\begin{aligned} \dfrac{2|s_0 - p_0\overline{s}_0| + |s_{0}^{2} - 4p_0|}{4 - |s_0|^2} = |\lambda _0| \end{aligned}$$

(11.3)

and

$$\begin{aligned} |a_0| \; \le \; |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}. \end{aligned}$$

(11.4)

Then there exists a rational $\overline{\mathcal {P}}$-inner function $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$ given explicitly as follows:

• (i) If $|p_0| = |\lambda _0|$, then $s_0= 0$ and $x(\lambda ) = (a(\lambda ), 0, \omega \lambda ),$ where $\omega \lambda _0 = p_0, \; \omega \in \mathbb {T}$ and

  • (a) if $|a_0| = |\lambda _0|$, then, for $\lambda \in \mathbb {D}$, $a (\lambda ) = \kappa \lambda , \;$ where $|\kappa |=1$ and $\kappa \lambda _0 = a_0$;

  • (b) if $|a_0| < |\lambda _0|$, then

    $$\begin{aligned} a(\lambda ) = \lambda \dfrac{\lambda - \lambda _0+\eta _0(1-\overline{\lambda _0}\lambda )}{1-\overline{\lambda _0}\lambda +\overline{\eta _0}(\lambda - \lambda _0)}, \; \lambda \in \mathbb {D}, \end{aligned}$$

  and $\eta _0 = \dfrac{a_0}{\lambda _0}$.

• (ii) If $|p_0| \; <\; |\lambda _0|$, then $x(\lambda ) = (a(\lambda ), \varphi _1(\lambda ), \varphi _2(\lambda )),$ where $\varphi _1$ is defined by Eq. (10.3), $\varphi _2$ is defined by Eq. (10.4) and

  • (a) if $|a_0| = |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$, then, for $\lambda \in \mathbb {D}$,

    $$\begin{aligned} a(\lambda ) =\gamma \lambda \frac{A(\lambda )}{D(\lambda )}, \end{aligned}$$

    where $|\gamma |=1$ such that $\;\gamma \lambda _0 \sqrt{1-\frac{1}{4}|s_0|^2} = a_0$, $A(\lambda ) = b_0(1+b_1\lambda )$ is an outer polynomial of degree 1 such that

    $$\begin{aligned} |A(\lambda )|^2 = |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2, \end{aligned}$$

    (11.5)

    and

    $$\begin{aligned} E(\lambda ) = c \lambda ,\; \; D(\lambda ) = {\overline{\zeta }}\{(1-\overline{\lambda _0}\lambda ) + \overline{p_1}\zeta ^2 (\lambda - \lambda _0)\}, \end{aligned}$$

    (11.6)

    with

    $$\begin{aligned} |\zeta | = 1, \; \; \; p_1 = \dfrac{p_0}{\lambda _0}, \; \; \; \; \; c = \dfrac{2}{|\lambda _0|}\{|{\overline{\lambda }}_0 - \overline{p}_0 \lambda _0 \zeta ^2| - |\lambda _0^2 \zeta ^2 - p_0|\}. \end{aligned}$$

  • (b) if $|a_0| < |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$, then, for $\lambda \in \mathbb {D}$,

    $$\begin{aligned} a(\lambda ) = \lambda \left( \frac{\lambda - \lambda _0+\mu _0(1-\overline{\lambda _0}\lambda )}{1-\overline{\lambda _0}\lambda +\overline{\mu _0}(\lambda -\lambda _0)} \right) \frac{A(\lambda )}{D(\lambda )}, \end{aligned}$$

    where $\mu _0 = \dfrac{a_0}{\lambda _0 \sqrt{1-\frac{1}{4}|s_0|^2}}$ and polynomials A, E, D are defined by Eqs. (11.6) and (11.5).

Proof

Since $(a_0, s_0, p_0) \in \overline{\mathcal {P}}$, we have $(s_0, p_0) \in \Gamma$. By assumption the equality (11.3) holds. Hence, by Theorem 10.2, there exists a rational analytic function

$$\begin{aligned} \varphi : \mathbb {D} \rightarrow \Gamma : \lambda \longmapsto (s(\lambda ), p(\lambda )) \end{aligned}$$

such that $\varphi (0) = (0, 0)$ and $\varphi (\lambda _0) = (s_0, p_0)$. It is easy to see that the function $\varphi = (s, p)$ from $\mathbb {D}$ to $\Gamma$ defined as in Eq. (10.2) is a rational $\Gamma$-inner function of degree 1. By Proposition 10.4, the function $\varphi = (s, p)$ from $\mathbb {D}$ to $\Gamma$ defined by

$$\begin{aligned} s(\lambda ) = \varphi _1(\lambda ), \; p(\lambda ) = \varphi _2(\lambda ), \; \lambda \in \mathbb {D}, \end{aligned}$$

as in Eqs. (10.3) and (10.4) is a rational $\Gamma$-inner function of degree 2. By Theorem 10.1, it follows from Eq. (10.1) that $|p_0| \le |\lambda _0| \; <\; 1$. Let us consider two cases.

Case (i) Let $|p_0| = |\lambda _0|$. By Theorem 10.2, if $|p_0|=|\lambda _0|$, then $s_0= 0$ and the function $\varphi = (0, \omega \lambda )$ from $\mathbb {D}$ to $\Gamma$, where $\omega$ is a complex number of unit modulus such that $\omega \lambda _0 = p_0$, is a rational $\Gamma$-inner function of degree 1. Since $s_0=0$, the assumption (11.4) $|a_0| \le |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$ is equivalent to $|a_0| \le |\lambda _0|$. It is easy to see that, for $\lambda \in \mathbb {D}$, $s(\lambda ) = \dfrac{E(\lambda )}{D(\lambda )}$ and $p(\lambda ) = \dfrac{D^{\sim n}}{D}(\lambda )$, where polynomials $E(\lambda )=0$ and $D(\lambda ) = \overline{\omega _1}$ and $\omega _1^2=\omega$. By Theorem 7.8, we can construct a rational $\overline{\mathcal {P}}$-inner function

$$\begin{aligned} x = \left( cB \dfrac{A}{D}, 0, \omega \lambda \right) , \text { for an arbitrary finite Blaschke product } B \text { and } |c| = 1, \end{aligned}$$

where A is a non-zero constant such that

$$\begin{aligned} |A(\lambda )|^2 = |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2= |D(\lambda )|^2= | \overline{\omega _1}|^2 =1. \end{aligned}$$

It is sufficient to construct $a : \mathbb {D} \rightarrow \mathbb {C}$ of the form $a = c B$ with $|c|=1$ such that $a(0) = 0$ and $a(\lambda _0) = a_0$.

If $|a_0| = |\lambda _0|$, then, for $\lambda \in \mathbb {D}$, we define $a(\lambda ) = \kappa \lambda \; \; \text {for} \;\;\lambda \in \mathbb {D},$ where $|\kappa |=1$ and $\kappa \lambda _0 = a_0$. It is easy to see that $a(0) = 0$ and $a(\lambda _0) = a_0$.

Let $|a_0| \; <\; |\lambda _0|$, and let $\eta _0 {\mathop {=}\limits ^{def }} \frac{a_0}{\lambda _0},$ it is clear that $\eta _0 \in \mathbb {D}$. Let us define a by the formula

$$\begin{aligned} a(\lambda ) = \lambda B_{\eta _0}^{-1} \circ B_{\lambda _0}(\lambda ) \; \; \text {for} \;\;\lambda \in \mathbb {D}. \end{aligned}$$

Here the Blaschke factors are defined by

$$\begin{aligned} B_{\eta _0}^{-1}(z) = \frac{z+\eta _0}{1+\overline{\eta _0}z} \text { and } B_{\lambda _0}(z) = \frac{z-\lambda _0}{1-\overline{\lambda _0}z} \; \text {for} \; z \in \mathbb {D}. \end{aligned}$$

It is easy to see that $a(0) = 0$ and $a(\lambda _0) = a_0$.

Define a rational $\overline{\mathcal {P}}$-inner function $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ by $x(\lambda ) = (a(\lambda ), 0, \omega \lambda ),$ where $\omega \lambda _0 = p_0, \; \omega \in \mathbb {T}$ and

• (a) if $|a_0| = |\lambda _0|$, then, for $\lambda \in \mathbb {D}$, $a (\lambda ) = \kappa \lambda , \;$ where $|\kappa |=1$ and $\kappa \lambda _0 = a_0$;

• (b) if $|a_0| < |\lambda _0|$, then

  $$\begin{aligned} a(\lambda ) = \lambda \dfrac{\lambda - \lambda _0+\eta _0(1-\overline{\lambda _0}\lambda )}{1-\overline{\lambda _0}\lambda +\overline{\eta _0}(\lambda - \lambda _0)}, \; \lambda \in \mathbb {D}, \end{aligned}$$

  and $\eta _0 = \dfrac{a_0}{\lambda _0}$. This function x satisfies the conditions $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$.

Case (ii) Let $|p_0| \; <\; |\lambda _0|$. By Lemma 10.3, for the rational $\Gamma$-inner function $\varphi = (s, p)$ from $\mathbb {D}$ to $\Gamma$ defined by

$$\begin{aligned} s(\lambda ) = \varphi _1(\lambda ), \; p(\lambda ) = \varphi _2(\lambda ), \; \lambda \in \mathbb {D}, \end{aligned}$$

as in Eqs. (10.3) and (10.4), there exist polynomials E and D

$$\begin{aligned} E(\lambda ) = c \lambda ,\; \; D(\lambda ) = {\overline{\zeta }}\{(1-\overline{\lambda _0}\lambda ) + \overline{p_1}\zeta ^2 (\lambda - \lambda _0)\}, \end{aligned}$$

(11.7)

where

$$\begin{aligned} |\zeta | = 1, \; \; \; p_1 = \dfrac{p_0}{\lambda _0}, \; \; \; \; \; c = \dfrac{2}{|\lambda _0|}\{|{\overline{\lambda }}_0 - \overline{p}_0 \lambda _0 \zeta ^2| - |\lambda _0^2 \zeta ^2 - p_0|\}, \end{aligned}$$

such that $s = \dfrac{E}{D}$ and $p= \dfrac{D^{\sim 2}}{D}$. Moreover, $E^{\sim 2} = E$ and $|E(\lambda )| \le 2 |D(\lambda )|$ on $\overline{\mathbb {D}}$.

Then, by Theorem 7.8, we can construct a rational $\overline{\mathcal {P}}$-inner function

$$\begin{aligned} x = \left( cB \dfrac{A}{D}, \dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right) , \text { for an arbitrary finite Blaschke product } B \text { and } |c| = 1, \end{aligned}$$

where $A= b_0(1+b_1 \lambda )$ is an outer polynomial of degree 1 such that

$$\begin{aligned} |A(\lambda )|^2 = |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2. \end{aligned}$$

(11.8)

We would like to construct $a : \mathbb {D} \rightarrow \mathbb {C}$ of the form $a = c B \dfrac{A}{D}$ such that $a(0) = 0$ and $a(\lambda _0) = a_0$.

Since, for $\lambda \in \mathbb {D}$,

$$\begin{aligned} \frac{|A(\lambda )|^2}{|D(\lambda )|^2} = 1-\frac{1}{4}|s(\lambda )|^2, \end{aligned}$$

we get $\big |\dfrac{A}{D}(\lambda _0) \big | = \sqrt{1-\frac{1}{4}|s_0|^2}$. Let $B(\lambda ) = \lambda \tilde{B}(\lambda ),$ with some finite Blaschke product $\tilde{B}$. Then $B(0) = 0$, and so $a(0) = 0$. Recall we require $a(\lambda _0) = a_0$,

$$\begin{aligned} a_0= & {} a(\lambda _0) = c B(\lambda _0) \dfrac{A}{D}(\lambda _0)\\= & {} c \lambda _0 \tilde{B}(\lambda _0) \sqrt{1-\frac{1}{4}|s_0|^2}, \end{aligned}$$

for some $|c| = 1$. By assumption (11.4),

$$\begin{aligned} |a_0| \; \le \; |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2} \;\text {and} \; \; |s_0| <2, \,\; \text { thus } \frac{|a_0|}{|\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}} \; \le \; 1. \end{aligned}$$

Suppose $|a_0| < |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$, and let

$$\begin{aligned} \mu _0 {\mathop {=}\limits ^{def }} \frac{a_0}{\lambda _0 \sqrt{1-\frac{1}{4}|s_0|^2}}. \end{aligned}$$

It is clear that $\mu _0 \in \mathbb {D}$. We need to find $\tilde{B} : \mathbb {D} \rightarrow \mathbb {D}$ such that $\tilde{B}(\lambda _0) = \mu _0$. Let $\tilde{B}= B_{\mu _0}^{-1} \circ B_{\lambda _0}$, where

$$\begin{aligned} B_{\mu _0}^{-1}(z) = \frac{z+\mu _0}{1+\overline{\mu _0}z} \text { and } B_{\lambda _0}(z) = \frac{z-\lambda _0}{1-\overline{\lambda _0}z}. \end{aligned}$$

Then, for all $z \in \mathbb {D}$,

$$\begin{aligned} \tilde{B}(z)= & {} B_{\mu _0}^{-1} \circ B_{\lambda _0}(z) = \frac{\dfrac{z-\lambda _0}{1-\overline{\lambda _0}z}+\mu _0}{1+\overline{\mu _0}\left( \dfrac{z-\lambda _0}{1-\overline{\lambda _0}z}\right) }\\= & {} \frac{z-\lambda _0+\mu _0(1-\overline{\lambda _0}z)}{1-\overline{\lambda _0}z+\overline{\mu _0}(z-\lambda _0)}. \end{aligned}$$

Let us define $a : \mathbb {D} \rightarrow \mathbb {C}$, for $\lambda \in \mathbb {D}$, by

$$\begin{aligned} a(\lambda ) = \lambda \tilde{B}(\lambda ) \dfrac{A(\lambda )}{D(\lambda )}= \lambda \left( \frac{\lambda - \lambda _0+\mu _0(1-\overline{\lambda _0}\lambda )}{1-\overline{\lambda _0}\lambda +\overline{\mu _0}(\lambda -\lambda _0)} \right) \frac{A(\lambda )}{D(\lambda )}. \end{aligned}$$

Note that $a(0) = 0$ and $\tilde{B}(\lambda _0) = \dfrac{\mu _0(1-\overline{\lambda _0}\lambda _0)}{1-\overline{\lambda _0}\lambda _0} = \mu _0$. Therefore,

$$\begin{aligned} a(\lambda _0)= & {} \lambda _0 \mu _0 \frac{A(\lambda _0)}{D(\lambda _0)} \\= & {} \lambda _0 \frac{a_0}{\lambda _0 \sqrt{1-\frac{1}{4}|s_0|^2}} \sqrt{1-\frac{1}{4}|s_0|^2} \\= & {} a_0. \end{aligned}$$

Hence, in the case when $|p_0| \; <\; |\lambda _0|$, we define a rational $\overline{\mathcal {P}}$-inner function $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ by $x(\lambda ) = (a(\lambda ), \varphi _1(\lambda ), \varphi _2(\lambda )), \;\ \text {for all } \;\; \lambda \in \mathbb {D},$ where $\varphi _1$ is defined by Eq. (10.3), $\varphi _2$ is defined by Eq. (10.4) and

• (a) if $|a_0| = |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$, then, for $\lambda \in \mathbb {D}$,

  $$\begin{aligned} a(\lambda ) = \gamma \lambda \frac{A(\lambda )}{D(\lambda )}, \end{aligned}$$

  where $|\gamma |=1$ such that $\;\gamma \lambda _0 \sqrt{1-\frac{1}{4}|s_0|^2} = a_0$.

• (b) if $|a_0| < |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$, then

  $$\begin{aligned} a(\lambda ) = \lambda \left( \frac{\lambda - \lambda _0+\mu _0(1-\overline{\lambda _0}\lambda )}{1-\overline{\lambda _0}\lambda +\overline{\mu _0}(\lambda -\lambda _0)} \right) \frac{A(\lambda )}{D(\lambda )}, \end{aligned}$$

  where $\mu _0 = \dfrac{a_0}{\lambda _0 \sqrt{1-\frac{1}{4}|s_0|^2}}$, and polynomials A, E and D are defined by Eqs. (11.7) and (11.8).

One can verify that a suitable choice of A is $A(\lambda ) = b_0(1+b_1\lambda )$, where

$$\begin{aligned} |b_0|^2 = |1-\overline{p_1}\zeta ^2 \lambda _0|^2 \;\; \text {and} \;\; |b_1|^2 = 2 \; \frac{|\lambda _0 \zeta ^2 - p_1|}{|1-\overline{p_1}\zeta ^2 \lambda _0|} - 1. \end{aligned}$$

$\square$

Theorem 11.3

Let $\lambda _0 \in \mathbb {D}\setminus \{0\}$ and $(a_0, s_0, p_0) \in \overline{\mathcal {P}}$. Then the following conditions are equivalent:

• (i) there exists a rational $\overline{\mathcal {P}}$-inner function $x=(a, s, p)$, $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$;

• (ii) there exists an analytic function $x=(a, s, p)$, $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x(0) = (0, 0, 0)$ and $x(\lambda _0) =(a_0, s_0, p_0)$, and $|a_0| \; \le \; |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$;

• (iii)
  $$\begin{aligned} \dfrac{2|s_0 - p_0\overline{s}_0| + |s_{0}^{2} - 4p_0|}{4 - |s_0|^2} \le |\lambda _0|, \;\;|s_0| <2, \end{aligned}$$

  (11.9)

  and

  $$\begin{aligned} |a_0| \; \le \; |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}. \end{aligned}$$

  (11.10)

Proof

We shall prove below that (i) $\Longleftrightarrow$ (iii), from which it will follow trivially that (i) $\Rightarrow$ (ii).

(ii) $\Rightarrow$ (i) Suppose (ii) holds, that is, there exists an analytic function $x_1=(a', s', p')$, $x_1 : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x_1(0) = (0, 0, 0)$ and $x_1(\lambda _0) = (a_0, s_0, p_0)$. By Lemma 6.2, $h_1=(s',p'): \mathbb {D} \rightarrow \Gamma$ is an analytic function such that $h_1(0) = (0, 0)$ and $h_1(\lambda _0) = (s_0, p_0)$.

By [18, Theorem 4] (see also [3, Theorem 8.1]), there exists a rational $\Gamma$-inner function $h: \mathbb {D} \rightarrow \Gamma$ satisfying $h(0) = (0, 0)$ and $h(\lambda _0) = (s_0, p_0)$. Let E and D be polynomials as in Eq. (7.1) (see [6, Proposition 2.2]) with $h= \left( \dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right)$ on $\mathbb {D}$, where $n = \deg h$. By Theorem 7.8, we can construct a rational $\overline{\mathcal {P}}$-inner function

$$\begin{aligned} x = \left( a,\dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right) =\left( cB \dfrac{A}{D},\dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right) , \end{aligned}$$

for any finite Blaschke product B and $c \in \mathbb {C}$ with $|c| = 1,$ where A is an outer polynomial such that

$$\begin{aligned} |A(\lambda )|^2 = |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2, \;\; \text {for} \;\;\lambda \in \mathbb {T}. \end{aligned}$$

Hence, for $\lambda \in \mathbb {D}$,

$$\begin{aligned} \frac{|A(\lambda )|^2}{|D(\lambda )|^2} = 1-\frac{1}{4}|s(\lambda )|^2. \end{aligned}$$

Thus we get $\big |\dfrac{A}{D}(\lambda _0) \big | = \sqrt{1-\frac{1}{4}|s_0|^2}$. By assumption, $|a_0| \le |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$. As in Theorem 11.2, to satisfy conditions $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$, we define a function a the following way:

• (a) if $|a_0| = |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$, then, for $\lambda \in \mathbb {D}$,

  $$\begin{aligned} a(\lambda ) = \gamma \lambda \frac{A(\lambda )}{D(\lambda )}, \end{aligned}$$

  where $|\gamma |=1$ is such that $\;\gamma \lambda _0 \sqrt{1-\frac{1}{4}|s_0|^2} = a_0$.

• (b) if $|a_0| < |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}$, then

  $$\begin{aligned} a(\lambda ) = \lambda \left( \frac{\lambda - \lambda _0+\mu _0(1-\overline{\lambda _0}\lambda )}{1-\overline{\lambda _0}\lambda +\overline{\mu _0}(\lambda -\lambda _0)} \right) \frac{A(\lambda )}{D(\lambda )}, \end{aligned}$$

  where $\mu _0 = \dfrac{a_0}{\lambda _0 \sqrt{1-\frac{1}{4}|s_0|^2}}$. Therefore, condition (i) holds.

(iii) $\Rightarrow$ (i) By Theorem 10.1, since condition (iii) holds, there exists an analytic function $h_1: \mathbb {D} \rightarrow \Gamma$ such that $h_1(0) = (0, 0)$ and $h_1(\lambda _0) = ( s_0, p_0)$. By [18, Theorem 4] (see also [3, Theorem 8.1]), there exists a rational $\Gamma$-inner function $h=(s, p): \mathbb {D} \rightarrow \Gamma$ satisfying $h(0) = (0, 0)$ and $h(\lambda _0) = (s_0, p_0)$. Let E and D be polynomials as in Eq. (7.1) (see [6, Proposition 2.2]) with $h= \left( \dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right)$ on $\mathbb {D}$, where $n = \deg h$. By Theorem 7.8, we can construct a rational $\overline{\mathcal {P}}$-inner function

$$\begin{aligned} x = \left( a,\dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right) =\left( cB \dfrac{A}{D},\dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right) , \end{aligned}$$

for any finite Blaschke product B and $c \in \mathbb {C}$ with $|c| = 1,$ where A is an outer polynomial such that

$$\begin{aligned} |A(\lambda )|^2 = |D(\lambda )|^2-\frac{1}{4}|E(\lambda )|^2. \end{aligned}$$

To satisfy the conditions $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$, we define a function a as in Part (ii) $\Rightarrow$ (i).

Note that in the case when

$$\begin{aligned} \dfrac{2|s_0 - p_0\overline{s}_0| + |s_{0}^{2} - 4p_0|}{4 - |s_0|^2} = |\lambda _0|,\;\;|s_0| <2, \end{aligned}$$

and

$$\begin{aligned} |a_0| \; \le \; |\lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2} \end{aligned}$$

Theorem 11.2 gives the construction of an interpolating rational $\overline{\mathcal {P}}$-inner function $x=(a, s, p)$, $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$.

(i) $\Rightarrow$ (iii) Suppose (i) holds, that is, there exists a rational $\overline{\mathcal {P}}$-inner function $x=(a, s, p)$, $x : \mathbb {D} \rightarrow \overline{\mathcal {P}}$ such that $x(0) = (0, 0, 0)$ and $x(\lambda _0) = (a_0, s_0, p_0)$. Let $\deg x = (m, n)$.

By Lemma 6.2, $h=(s,p)$ is a rational $\Gamma$-inner function of degree n. Note that $h(0) = (0,0)$ and $h(\lambda _0) = (s_0, p_0)$. By Theorem 10.1, $|s_0| \; <\; 2$ and

$$\begin{aligned} \dfrac{2|s_0 - p_0\overline{s_0}| + |s_{0}^{2} - 4p_0|}{4 - |s_0|^2} \le |\lambda _0|. \end{aligned}$$

By Theorem 7.5, there exist polynomials A, E, D such that $E^{\sim n} = E$, $D(\lambda ) \ne 0$ on $\overline{\mathbb {D}}$, A is an outer polynomial such that $|A(\lambda )|^2 = |D(\lambda )|^2 - \frac{1}{4}|E(\lambda )|^2$ on $\mathbb {T}$, $|E(\lambda )| \le 2|D(\lambda )|$ on $\overline{\mathbb {D}}$ and

$$\begin{aligned} x=\left( c B \dfrac{A}{D}, \dfrac{E}{D}, \dfrac{D^{\sim n}}{D}\right) \;\; \text {on} \;\;\overline{\mathbb {D}} \end{aligned}$$

for some finite Blaschke product B and $|c|=1$. The function

$$\begin{aligned} \lambda \mapsto a(\lambda ) = c B (\lambda )\dfrac{A}{D}(\lambda ) \end{aligned}$$

is an analytic map from $\mathbb {D}$ to $\overline{\mathbb {D}}$ such that $a(0) = 0$ and $a(\lambda _0) = a_0$. Note that A and D are outer polynomials on $\overline{\mathbb {D}}$, and so

$$\begin{aligned} f(\lambda ) =\frac{a(\lambda )}{\left( \dfrac{A}{D}(\lambda ) \right) }= c B (\lambda ) \end{aligned}$$

is an analytic map from $\mathbb {D}$ to $\overline{\mathbb {D}}$ such that $f(0) = 0$. By the classical Schwarz lemma we have

$$\begin{aligned} |f(\lambda )| = \left| \frac{a(\lambda )}{\dfrac{A}{D}(\lambda )} \right| \le | \lambda | \;\; \text {for}\;\;\lambda \in \mathbb {D}. \end{aligned}$$

Since

$$\begin{aligned} \big |\dfrac{A}{D}(\lambda ) \big |^2{} & {} = 1- |s (\lambda )|^2\;\; \text {for} \;\; \lambda \in \mathbb {D},\\ |f(\lambda )|{} & {} = \left| \frac{a(\lambda )}{\sqrt{1-\frac{1}{4}|s (\lambda )|^2}} \right| \;\; \text {for} \;\; \lambda \in \mathbb {D}. \end{aligned}$$

Thus

$$\begin{aligned} |a_0| = |a(\lambda _0)| =|f(\lambda _0)| \sqrt{1-\frac{1}{4}|s (\lambda _0)|^2} \le | \lambda _0| \sqrt{1-\frac{1}{4}|s_0|^2}. \end{aligned}$$

Therefore, condition (iii) holds. $\square$