1 Introduction

The quantum walks are certain unitary operators, and they are sometimes regarded as quantum counterparts of the classical random walks. The homogeneous two-state quantum walks (in one dimension with a constant coin matrix) are well understood fundamental models (see, for example, [6, 9, 25]), and recently the scattering-theoretical aspects, as a perturbation of homogeneous walks, are intensively investigated (see [15,16,17, 20, 21]). The Schrödinger operators in one dimension are often called the Sturm–Liouville operators and they are well studied. Thus, it would be rather natural to understand resemblance between one-dimensional quantum walks and Sturm–Liouville operators. The purpose of the present paper is to give a proof of an eigenfunction expansion formula for one-dimensional two-state quantum walks which is analogous to classical formulas of Weyl [27, 28], Stone [24], Titchmarsh [26], and Kodaira [10] for Sturm–Liouville operators. The theory of eigenfunction expansions for Sturm–Liouville operators are discussed, for example, in [12, 14, 18, 19] and a short review can be found in [23]. Probabilistic aspects of one-dimensional quantum walks are also intensively investigated. The notion of transfer matrix is introduced in [11] to construct stationary measures from eigenfunctions for quantum walks, and it is suitable for our analysis. Then, our basic idea in this paper is to use the transfer matrix to develop a theory analogous to that for Sturm–Liouville operators.

Before going to explain our setting-up, we should mention about the work by Gesztesy–Zinchenko [8] on Weyl–Titchmarsh theory for CMV matrices with Verblunsky coefficients in the unit disk. The notion of CMV matrices has been introduced by Cantero–Moral–Velázquez [2] and has further developed and deepened by Simon [22, 23]. The one-dimensional two-state quantum walks are special CMV matrices and the theory of CMV matrices applied to this class of quantum walks in [1, 5] and other works. Therefore, many of our results in this paper are essentially contained in [8]. However, since our presentations and proofs are direct without using the theory of CMV matrices, and formulas are given in usual representations of unitary evolutions for quantum walks. Although our approach only works for the class of quantum walks mentioned above, the setting of our presentation could have advantageous aspect when the quantum walks are applied and used in areas different from pure mathematics such as information science or quantum physics. Furthermore, it seems that a property of the Green function, Theorem 1.4 below, is new, and it can be used to give a concrete formula, Theorem 5.7, to compute the positive-matrix-valued measure, which gives directly the spectral resolution, for a certain special simplest case of the so-called two-phase model [3].

Now, let us prepare notation to mention some of results in the paper. All the inner products in the paper are complex linear in the first variable and anti-complex linear in the second. We denote the standard Hermitian inner product of the two-dimensional complex vector space \({\mathbb{C}}^{2}\) by \(\langle \,\cdot ,\cdot \,\rangle _{{\mathbb{C}}^{2}}\) and the standard orthonormal basis of \({\mathbb{C}}^{2}\) by \(\{e_{L}, e_{R}\}\)

$$\begin{aligned} e_{L}= \left[ \begin{array}{cc} 1 \\ 0 \end{array}\right] ,\quad e_{R}= \left[ \begin{array}{c} 0 \\ 1 \end{array}\right] . \end{aligned}$$

The orthogonal projection onto the one-dimensional subspaces, \({\mathbb{C}}e_{L},\) \({\mathbb{C}}e_{R},\) are denoted by \(\pi _{L},\) \(\pi _{R}.\) In general, the set of maps from a set X to another set Y is denoted by \({\mathrm{Map}}\,(X,Y).\) We fix \({\mathcal{C}} \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathrm{U}}(2)),\) where \({\mathrm{U}}(2)\) is the group of unitary \(2 \times 2\) matrices, and define a linear map

$$\begin{aligned} U({\mathcal{C}}) :{\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}) \end{aligned}$$
(1)

by the following formula:

$$\begin{aligned} U({\mathcal{C}})\Psi (n)=\pi _{L} {\mathcal{C}}(n+1) \Psi (n+1) +\pi _{R} {\mathcal{C}}(n-1) \Psi (n-1), \end{aligned}$$
(2)

where \(\Psi \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}),\) \(n \in {\mathbb{Z}}.\) Let \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\) be the Hilbert space of \(\ell ^{2}\)-functions whose inner product is given by

$$\begin{aligned} \langle \,f,g \,\rangle =\sum _{n \in {\mathbb{Z}}} \langle \,f(n),g(n) \,\rangle _{{\mathbb{C}}^{2}} \quad (f,g \in \ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})). \end{aligned}$$

The linear map \(U({\mathcal{C}})\) defined in (2) becomes a unitary operator on \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\) when it is restricted to \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}),\) and it preserves the space \(C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) of finitely supported \({\mathbb{C}}^{2}\)-valued functions. In this paper, we call the linear map defined in (1), (2) the quantum walk with the coin matrix \({\mathcal{C}}:{\mathbb{Z}} \rightarrow {\mathrm{U}}(2).\) We write

$$\begin{aligned} {\mathcal{C}}(n)= \left[ \begin{array}{cc} a_{n} &{} b_{n} \\ c_{n} &{} d_{n} \end{array}\right] , \quad \triangle _{n}=\det {\mathcal{C}}(n)=a_{n}d_{n}-b_{n}c_{n} \quad (n \in {\mathbb{Z}}). \end{aligned}$$
(3)

Throughout this paper, we assume the following:

$$\begin{aligned} a_{n} \ne 0 \quad \text{for all} \ n \in {\mathbb{Z}}. \end{aligned}$$
(4)

Under the assumption (4), the unitarity of the matrix \({\mathcal{C}}(n)\) causes \(d_{n} \ne 0\) for any \(n \in {\mathbb{Z}}.\)

Theorem 1.1

[11, 13, 17] Suppose that the coin matrix \({\mathcal{C}}\) satisfies the assumption (4). For any \(n \in {\mathbb{Z}}\) and any \(\lambda \in {\mathbb{C}} {\setminus } \{0\},\) we define a \(2 \times 2\) matrix \(T_{\lambda }(n)\) by

$$\begin{aligned} T_{\lambda }(n)= \left[ \begin{array}{cc} { \frac{1}{a_{n+1}} (\lambda -\lambda ^{-1}c_{n} b_{n+1}) }&{} { -\lambda ^{-1} \frac{b_{n+1}d_{n}}{a_{n+1}} }\\ { \lambda ^{-1} c_{n} }&{}{ \lambda ^{-1} d_{n}} \end{array}\right] , \end{aligned}$$
(5)

and a \(2 \times 2\) matrix \(F_{\lambda }(n)\) by

$$\begin{aligned} F_{\lambda }(n) = \left\{ \begin{array}{ll} I &{}\quad (n=0), \\ { T_{\lambda }(n-1) T_{\lambda }(n-2) \cdots T_{\lambda }(1) T_{\lambda }(0) }&{}\quad (n \ge 1), \\ { T_{\lambda }(n)^{-1} T_{\lambda }(n+1)^{-1} \cdots T_{\lambda }(-2)^{-1} T_{\lambda }(-1)^{-1} }&{}\quad (n \le -1). \end{array}\right. \end{aligned}$$
(6)

For any \(u \in {\mathbb{C}}^{2},\) we define \(\Phi _{\lambda }(u) \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) by

$$\begin{aligned} \Phi _{\lambda }(u)(n) = F_{\lambda }(n)u \quad (n \in {\mathbb{Z}}). \end{aligned}$$
(7)

Then, the map \(\Phi _{\lambda } :{\mathbb{C}}^{2} \rightarrow {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) is injective and the eigenspace \({\mathcal{M}}^{\lambda }\) of \(U({\mathcal{C}})\) in \({\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) with an eigenvalue \(\lambda \in {\mathbb{C}} {\setminus } \{0\}\) coincides with the image of \(\Phi _{\lambda }.\) Hence, \(\dim {\mathcal{M}}^{\lambda }=2\) for each such \(\lambda .\) Furthermore, we have \(\dim {\mathcal{M}}^{\lambda } \cap \ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}) \le 1\) for any \(\lambda \in S^{1}.\)

The matrix \(T_{\lambda }(n)\) is called the transfer matrix,Footnote 1 and it is useful to describe various functions and quantities related to the quantum walk \(U({\mathcal{C}}).\) For example, the Green function can be expressed in terms of the transfer matrix. To be more precise, let

$$\begin{aligned} R(\lambda )=(U-\lambda )^{-1}\quad (\lambda \in {\mathbb{C}} {\setminus } \sigma (U)) \end{aligned}$$
(8)

be the resolvent of the restriction U of \(U({\mathcal{C}})\) to \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}),\) where \(\sigma (U)\) denotes the spectrum of the operator U. For any \(u \in {\mathbb{C}}^{2}\) and \(k \in {\mathbb{Z}},\) we define a function \(\delta _{k} \otimes u \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) by

$$\begin{aligned} (\delta _{k} \otimes u)(n)= \left\{ \begin{array}{ll} u &{}\quad (n=k), \\ 0 &{}\quad (n \ne k). \end{array}\right. \end{aligned}$$
(9)

The Green function \(R_{\lambda } \in {\mathrm{Map}}\,({\mathbb{Z}}^{2}, {\mathrm{M}}_{2}({\mathbb{C}})),\) where \({\mathrm{M}}_{2}({\mathbb{C}})\) denotes the space of complex \(2 \times 2\) matrices, is defined as

$$\begin{aligned} R_{\lambda }(n,m)u=[R(\lambda ) (\delta _{m} \otimes u)](n) \quad ((n,m) \in {\mathbb{Z}}^{2},\, u \in {\mathbb{C}}^{2},\, \lambda \in {\mathbb{C}} {\setminus } \sigma (U)). \end{aligned}$$
(10)

Then, the Green function \(R_{\lambda }(n,m)\) is expressed in terms of the matrix \(F_{\lambda }(n)\) as in the following.

Theorem 1.2

We define \({\mathrm{z}}_{L},{\mathrm{z}}_{R} \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathrm{M}}_{2}({\mathbb{C}}))\) by

$$\begin{aligned} {\mathrm{z}}_{L}(n)= \left[ \begin{array}{cc} 1 &{} 0 \\ -c_{n}/d_{n} &{} 0 \end{array}\right] ,\quad {\mathrm{z}}_{R}(n) = \left[ \begin{array}{cc} 0 &{} -b_{n} /a_{n} \\ 0 &{} 1 \end{array}\right] . \end{aligned}$$
(11)

We set \({\mathrm{x}}_{0}(\lambda )=R_{\lambda }(0,0).\) Then, we have

$$\begin{aligned} R_{\lambda }(n,m)= \left\{ \begin{array}{ll} F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda )+\lambda ^{-1} {\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }} (m)^{*} &{}\quad (m>n), \\ F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda )+\lambda ^{-1} {\mathrm{z}}_{L}(0)] F_{1/\overline{\lambda }} (m)^{*} &{}\quad (m<n), \end{array}\right. \end{aligned}$$
(12)

and for each \(n \in {\mathbb{Z}},\) we have

$$\begin{aligned} R_{\lambda }(n,n)&= F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{L}(0)]F_{1/\overline{\lambda }} (n)^{*} -\lambda ^{-1} {\mathrm{z}}_{L}(n) \\ &= F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{R}(0)]F_{1/\overline{\lambda }} (n)^{*} -\lambda ^{-1} {\mathrm{z}}_{R}(n). \end{aligned}$$
(13)

As in Theorem 1.2, the matrix-valued holomorphic function \({\mathrm{x}}_{0}(\lambda )=R_{\lambda }(0,0)\) plays one of the central roles in the present paper. Therefore, it is important to develop methods to compute \({\mathrm{x}}_{0},\) one of which is given by the following two theorems.

Theorem 1.3

Let \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}).\) Let \(A_{L}(\lambda )\) (resp. \(A_{R}(\lambda ))\) be the vector subspace in \({\mathbb{C}}^{2}\) consisting of all vectors \(w \in {\mathbb{C}}^{2}\) satisfying

$$\begin{aligned} \sum _{n \ge 1} \Vert F_{\lambda }(n)w\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty \quad \left( {\text{resp. }} \sum _{n \le -1} \Vert F_{\lambda }(n)w\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty \right) . \end{aligned}$$

Then, we have \(\dim A_{L}(\lambda )=\dim A_{R}(\lambda )=1.\) In particular, we have

$$\begin{aligned} {\mathrm{rank}}\, [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{L}(0)]= {\mathrm{rank}}\, [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{R}(0)]={\mathrm{rank}}\,R_{\lambda }(n,m)=1 \end{aligned}$$

for any \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1})\) and \(n,m \in {\mathbb{Z}}\) with \(n \ne m.\)

For \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}),\) let \({\mathrm{v}}_{\pm }(\lambda )\) be unit vectors satisfying

$$\begin{aligned} \sum _{n \ge 1} \Vert F_{\lambda }(n) {\mathrm{v}}_{+}(\lambda ) \Vert _{{\mathbb{C}}^{2}}^{2}<+\infty ,\quad \sum _{n \le -1} \Vert F_{\lambda }(n) {\mathrm{v}}_{-}(\lambda ) \Vert _{{\mathbb{C}}^{2}}^{2}<+\infty . \end{aligned}$$
(14)

The existence of these unit vectors is assured by Theorem 1.3.

Theorem 1.4

For \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1})\) the unit vectors \({\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda )\) are linearly independent. The matrix-valued holomorphic function \({\mathrm{x}}_{0}(\lambda )=R_{\lambda }(0,0)\) is given by

$$\begin{aligned} {\mathrm{x}}_{0}(\lambda )e_{L}&=-\lambda ^{-1} \frac{\langle \,{\mathrm{z}}_{L}(0)e_{L}, {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}}{\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}} {\mathrm{v}}_{-}(\lambda ),\\ {\mathrm{x}}_{0}(\lambda )e_{R}&= -\lambda ^{-1} \frac{\langle \,{\mathrm{z}}_{R}(0)e_{R}, {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}}{\langle \,{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}}{\mathrm{v}}_{+}(\lambda ), \end{aligned}$$
(15)

where \({\mathrm{v}}_{\pm }(\lambda )^{\perp }\) denotes any unit vector in \({\mathbb{C}}^{2}\) perpendicular to \({\mathrm{v}}_{\pm }(\lambda ),\) respectively.

The eigenfunction expansion theorem due to Weyl [27, 28], Stone [24], Titchmarsh [26], and Kodaira [10]Footnote 2 is regarded as an inversion formula for a generalized Fourier transform defined by eigenfunctions for Sturm–Liouville operators. Let us state an eigenfunction expansion formula for the quantum walk \(U({\mathcal{C}})\) defined by the coin matrix \({\mathcal{C}}\) satisfying (4). For any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}),\) we define a function \({\mathcal{F}}_{{\mathcal{C}}}[f]\) on \({\mathbb{C}} {\setminus } \{0\}\) by

$$\begin{aligned} {\mathcal{F}}_{{\mathcal{C}}}[f](\lambda )=\widehat{f}^{{\mathcal{C}}}(\lambda ):=\sum _{n \in {\mathbb{Z}}} F_{1/\overline{\lambda }}(n)^{*}f(n),\qquad \lambda \in {\mathbb{C}} {\setminus } \{0\}. \end{aligned}$$
(16)

The sum in (16) is finite, because the support of f is finite. Therefore, the function \({\mathcal{F}}_{{\mathcal{C}}}[f](\lambda )\) is a Laurent polynomial in \(\lambda \in {\mathbb{C}} {\setminus } \{0\}.\) We call \({\mathcal{F}}_{{\mathcal{C}}}[f]=\widehat{f}^{{\mathcal{C}}}\) the QW-Fourier transform of f.

Theorem 1.5

There exists a positive-matrix-valued measure \(\Sigma\) on \(S^{1}\), such that we have the following.

  1. (1)

    The resolvent \(R(\lambda )\) is written as

    $$\begin{aligned} \langle \,R(\lambda )f,g \,\rangle =\int _{S^{1}} \frac{1}{\zeta -\lambda } \langle \,{\mathrm{d}}\Sigma (\zeta ){\mathcal{F}}_{{\mathcal{C}}}[f](\zeta ),{\mathcal{F}}_{{\mathcal{C}}}[g](\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
    (17)

    The positive-matrix-valued measure \(\Sigma\) satisfying (17) is unique.

  2. (2)

    For any \(f,g \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}),\) we have

    $$\begin{aligned} \langle \,f,g \,\rangle =\int _{S^{1}} \langle \,{\mathrm{d}}\Sigma (\zeta ){\mathcal{F}}_{{\mathcal{C}}}[f](\zeta ),{\mathcal{F}}_{{\mathcal{C}}}[g](\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
    (18)
  3. (3)

    Let

    $$\begin{aligned} U=\int _{S^{1}} \lambda \,{\mathrm{d}}E(\lambda ) \end{aligned}$$

    be the spectral resolution of the unitary operator Uwhere E is a projection-valued measure on \(S^{1}.\) Then, for each Borel set A in \(S^{1},\) the projection E(A) on \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\) is written as

    $$\begin{aligned}{}[E(A)f](n)=\int _{A} F_{\zeta }(n) {\mathrm{d}}\Sigma (\zeta ) {\mathcal{F}}_{{\mathcal{C}}}[f](\zeta ),\quad f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}). \end{aligned}$$
    (19)

    In particular, the following inversion formula holds for \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}):\)

    $$\begin{aligned} f(n)=\int _{S^{1}} F_{\zeta }(n) {\mathrm{d}}\Sigma (\zeta ) {\mathcal{F}}_{{\mathcal{C}}}[f](\zeta ). \end{aligned}$$
    (20)

Corollary 1.6

The following holds.

  1. (1)

    The spectrum \(\sigma (U)\) coincides with the support of \(\Sigma .\)

  2. (2)

    \(\lambda \in S^{1}\) is an eigenvalue of U if and only if \(\Sigma (\{\lambda \}) \ne 0.\) When \(\lambda\) is an eigenvalue of Uthe projection \(E(\{\lambda \})\) onto the eigenspace of \(\lambda\) is given by

    $$\begin{aligned}{}[E(\{\lambda \})f](n)=F_{\lambda }(n)\Sigma (\{\lambda \}) {\mathcal{F}}_{{\mathcal{C}}}[f](\lambda ). \end{aligned}$$

We refer the readers to [7, 8] for properties of positive-matrix-valued measures. We note that the matrix-valued function \({\mathrm{x}}_{0}(\lambda )\) is not an m-Carathéodory function in the sense of [22], because our operator is unitary. Instead, we use the matrix

$$\begin{aligned} {\mathrm{x}}(\lambda )=I+2\lambda {\mathrm{x}}_{0}(\lambda ), \end{aligned}$$
(21)

which is indeed an m-Carathéodory function. The positive-matrix-valued measure \(\Sigma\) is then a boundary value of the function \({\mathrm{x}}(\lambda )\) in the sense that \(\Sigma\) satisfies

$$\begin{aligned} {\mathrm{x}}(\lambda )=\int _{S^{1}} \frac{\zeta +\lambda }{\zeta -\lambda }\,{\mathrm{d}}\Sigma (\zeta )\quad (|\lambda | <1), \end{aligned}$$
(22)

and \(\Sigma\) is characterized as

$$\begin{aligned} {\mathrm{d}}\Sigma (\zeta )=\mathop {\mathrm{w*-lim}}\limits _{r \uparrow 1} {\mathrm{Re}}\,{\mathrm{x}}(r\zeta )= \mathop {\mathrm{w*-lim}}\limits _{r \uparrow 1}[r \zeta {\mathrm{x}}_{0}(r\zeta ) -r^{-1}\zeta {\mathrm{x}}_{0}(r^{-1}\zeta )]. \end{aligned}$$
(23)

Let \(C(S^{1}\!,{\mathbb{C}}^{2})\) be the space of continuous \({\mathbb{C}}^{2}\)-valued functions on \(S^{1}.\) For any \(k, l \in C(S^{1},{\mathbb{C}}^{2}),\) we define

$$\begin{aligned} \langle \,k,l \,\rangle _{\Sigma }=\int _{S^{1}} \langle \,{\mathrm{d}}\Sigma (\zeta )k(\zeta ),l(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(24)

This is a positive semi-definite Hermitian sesquilinear form on \(C(S^{1}\!,{\mathbb{C}}^{2})\), and hence, it defines an inner product on the quotient space \(C(S^{1},{\mathbb{C}}^{2})/N\) of \(C(S^{1},{\mathbb{C}}^{2})\) by the subspace \(N=\{k \in C(S^{1}\!,{\mathbb{C}}^{2}) \mid \Vert k\Vert _{\Sigma }=0\},\) where we set

$$\begin{aligned} \Vert k\Vert _{\Sigma }=\sqrt{\langle \,k,k \,\rangle _{\Sigma }} \quad (k \in C(S^{1},{\mathbb{C}}^{2})). \end{aligned}$$
(25)

We denote \(L^{2}(S^{1}\!,{\mathbb{C}}^{2})_{\Sigma }\) the completion of \(C(S^{1}\!,{\mathbb{C}}^{2})/N\) by the norm on \(C(S^{1},{\mathbb{C}}^{2})/N\) naturally induced by (25). The QW-Fourier transform (16) induces a map from \(C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) to \(L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }\) which we denote \(\mathfrak {F}_{{\mathcal{C}}}.\)

Theorem 1.7

The map \(\mathfrak {F}_{{\mathcal{C}}}\) extends to a unitary operator from \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\) to \(L^{2}(S^{1}\!, {\mathbb{C}}^{2})_{\Sigma }.\) The quantum walk U on \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\) is unitarily equivalent to the operator defined by the multiplication by \(\lambda \in S^{1}\) on \(L^{2}(S^{1}\!, {\mathbb{C}}^{2})_{\Sigma },\) namely, we have

$$\begin{aligned} \mathfrak {F}_{{\mathcal{C}}}[U({\mathcal{C}})f](\lambda )=\lambda \mathfrak {F}_{{\mathcal{C}}}[f](\lambda ) \end{aligned}$$
(26)

for any \(f \in \ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}).\)

The organization of the paper is as follows. In Sect. 2, we solve two equations, an inhomogeneous eigenvalue equation and its conjugate. The definition of the QW-Fourier transform (16) comes from the fact that it gives a defect of the left-inverse of \(U({\mathcal{C}})-\lambda\) obtained by solving a conjugate equation to an inhomogeneous eigenvalue equation to be the right-inverse. See Theorem 2.6 for a precise statement. The solutions to these equations are used to prove Theorem 1.2 in Sect. 3. Some of the properties of the Green functions, such as Theorem 1.3, are proved also in Sect. 3. In Sect. 4, we give proofs of Theorems 1.5 and 1.7. We calculate the positive-matrix-valued measure \(\Sigma\) in two examples, homogeneous quantum walks, and a simplest case of the two-phase models in Sect. 5.

2 Inhomogeneous eigenvalue equations and its conjugate

Let \(f \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}),\) \(\lambda \in {\mathbb{C}} {\setminus } \{0\}\) and \(w \in {\mathbb{C}}^{2}.\) We consider the following initial value problem:

$$\begin{aligned} (U({\mathcal{C}})-\lambda )\Psi =f,\quad \Psi (0)=w. \end{aligned}$$
(27)

Any \(\Psi (a)\) with a fixed integer \(a \in {\mathbb{Z}}\) can be chosen for an initial value, but we have chosen \(a=0\) for simplicity of notation. To prove Theorem 1.2, it is important to construct solutions to the problem (27) and its conjugate problem

$$\begin{aligned} (U({\mathcal{C}})^{*} -\overline{\lambda })\Psi =f,\quad \Psi (0)=w, \end{aligned}$$
(28)

where the map

$$\begin{aligned} U({\mathcal{C}})^{*} :{\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}) \end{aligned}$$
(29)

is the extension of the formal adjoint operator (on \(C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\)) of \(U({\mathcal{C}})\) given by

$$\begin{aligned} (U({\mathcal{C}})^{*}\Psi )(n)={\mathcal{C}}(n)^{*}\pi _{L} \Psi (n-1)+{\mathcal{C}}(n)^{*}\pi _{R} \Psi (n+1) \quad (n \in {\mathbb{Z}}). \end{aligned}$$
(30)

A brief account on the formal adjoint operator for a linear map \(A :C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) is given in Appendix B. It is well known that \(U({\mathcal{C}})\) defines a unitary operator on \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}).\) This property comes from the following lemma.

Lemma 2.1

As linear maps on \({\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}),\) we have \(U({\mathcal{C}})^{*}=U({\mathcal{C}})^{-1}.\)

Proof

Let \(f \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}).\) Then, we have

$$\begin{aligned}&[U({\mathcal{C}})U({\mathcal{C}})^{*}f](n) \\ &\quad = \pi _{L} {\mathcal{C}}(n+1) [U({\mathcal{C}})^{*}f](n+1) +\pi _{R} {\mathcal{C}}(n-1) [U({\mathcal{C}})^{*} f] (n-1) \\ &\quad = \pi _{L} {\mathcal{C}}(n+1) \left[ {\mathcal{C}}(n+1)^{*} \pi _{L} f(n) +{\mathcal{C}}(n+1)^{*} \pi _{R} f(n+2) \right] \\ &\qquad + \pi _{R} {\mathcal{C}}(n-1) \left[ {\mathcal{C}}(n-1)^{*} \pi _{L} f(n-2) +{\mathcal{C}}(n-1)^{*} \pi _{R} f(n) \right] \\ &\quad = \pi _{L} \pi _{L} f(n) + \pi _{L} \pi _{R} f(n+2) + \pi _{R} \pi _{L} f(n-2) +\pi _{R} \pi _{R} f(n) =f(n), \\ &\qquad [U({\mathcal{C}})^{*}U({\mathcal{C}})f](n) \\ &\quad = {\mathcal{C}}(n)^{*} \pi _{L} [U({\mathcal{C}})f] (n-1) +{\mathcal{C}}(n)^{*} \pi _{R} [U({\mathcal{C}})f](n+1) \\ &\quad = {\mathcal{C}}(n)^{*} \pi _{L} \left[ \pi _{L} {\mathcal{C}}(n) f(n) +\pi _{R} {\mathcal{C}}(n-2) f(n-2) \right] \\ &\qquad + {\mathcal{C}}(n)^{*} \pi _{R} \left[ \pi _{L} {\mathcal{C}}(n+2) f(n+2) +\pi _{R} {\mathcal{C}}(n) f(n) \right] \\ &\quad = {\mathcal{C}}(n)^{*}\pi _{L} {\mathcal{C}}(n)f(n)+{\mathcal{C}}(n)^{*}\pi _{R} {\mathcal{C}}(n) f(n) \\ &\quad = {\mathcal{C}}(n)^{*}[\pi _{L} +\pi _{R}] {\mathcal{C}}(n)f(n)={\mathcal{C}}(n)^{*}{\mathcal{C}}(n) f(n)=f(n), \end{aligned}$$

which shows the assertion. \(\square\)

In Proposition 2.3 below, we give formulas for the solutions to the initial value problems (27) and (28). Before proceed to Proposition 2.3 and its proof, we give recurrence equations equivalent to the equations in (27) and (28).

Lemma 2.2

The initial value problem (27) is equivalent to the equation

$$\begin{aligned} \Psi (n+1)=T_{\lambda }(n) \Psi (n) +\frac{1}{a_{n+1}} \pi _{L}f(n) -\frac{1}{d_{n}} T_{\lambda }(n) \pi _{R} f(n+1) \end{aligned}$$
(31)

with \(\Psi (0)=w.\) The initial value problem (28) is equivalent to the equation

$$\begin{aligned} \Psi (n+1)&= T_{1/\overline{\lambda }}(n) \Psi (n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L} {\mathcal{C}}(n+1) f(n+1) \\ &\quad +\overline{\lambda }^{-1}\left( T_{1/\overline{\lambda }}(n)-\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L} \right) f(n) \end{aligned}$$
(32)

with \(\Psi (0)=w.\)

Proof

We suppose that \(\Psi \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) is a solution to Eq. (27) with \(\Psi (0)=w \in {\mathbb{C}}^{2}.\) We set

$$\begin{aligned} \Psi (n)= \left[ \begin{array}{c} \psi _{L}(n) \\ \psi _{R}(n) \end{array}\right] , \quad f(n)= \left[ \begin{array}{c} f_{L}(n) \\ f_{R}(n) \end{array}\right] , \end{aligned}$$
(33)

so that Eq. (27) is written as

$$\begin{aligned} \left\{ \begin{aligned} a_{n+1} \psi _{L}(n+1) +b_{n+1} \psi _{R}(n+1)&= \lambda \psi _{L}(n) +f_{L}(n), \\ c_{n-1} \psi _{L}(n-1) +d_{n-1} \psi _{R}(n-1)&= \lambda \psi _{R}(n) +f_{R}(n). \end{aligned} \right. \end{aligned}$$
(34)

By the assumption (4), we can rewrite (34) as

$$\begin{aligned} \left\{ \begin{aligned} \psi _{L}(n+1)&= -\frac{b_{n+1}}{a_{n+1}} \psi _{R}(n+1) +\frac{\lambda }{a_{n+1}} \psi _{L}(n) +\frac{1}{a_{n+1}} f_{L}(n), \\ \psi _{R}(n-1)&= -\frac{c_{n-1}}{d_{n-1}} \psi _{L}(n-1) +\frac{\lambda }{d_{n-1}} \psi _{R}(n) +\frac{1}{d_{n-1}} f_{R}(n). \end{aligned} \right. \end{aligned}$$
(35)

Shifting the variable n in (34), we have

$$\begin{aligned} \left\{ \begin{aligned} a_{n} \psi _{L}(n) +b_{n} \psi _{R}(n)&= \lambda \psi _{L}(n-1) +f_{L}(n-1), \\ c_{n} \psi _{L}(n) +d_{n} \psi _{R}(n)&= \lambda \psi _{R}(n+1) +f_{R}(n+1). \end{aligned} \right. \end{aligned}$$
(36)

Solving the second equation of (36) in \(\psi _{R}(n+1)\), we have

$$\begin{aligned} \psi _{R}(n+1) = \lambda ^{-1} c_{n} \psi _{L}(n) +\lambda ^{-1} d_{n} \psi _{R}(n) - \lambda ^{-1}f_{R} (n+1). \end{aligned}$$
(37)

Substituting (37) into the first equation of (35), we see

$$\begin{aligned} \psi _{L}(n+1)&=\frac{1}{a_{n+1}}(\lambda -\lambda ^{-1}b_{n+1}c_{n}) \psi _{L}(n) -\lambda ^{-1} \frac{b_{n+1}d_{n}}{a_{n+1}} \psi _{R}(n) \\ &\quad +\lambda ^{-1}\frac{b_{n+1}}{a_{n+1}} f_{R}(n+1)+\frac{1}{a_{n+1}}f_{L}(n). \end{aligned}$$
(38)

Equations (37) and (38) give Eq. (31). Next, we suppose that \(\Psi \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) satisfies (31) and \(\Psi (0)=w.\) We write \(\Psi\) as in (33). Then, \(\psi _{L},\) \(\psi _{R}\) satisfy (37) and (38). Solving (37) in \(f_{R}(n+1)\) and substituting the result into (38), we obtain the first equation in (35). Shifting the variable n to \(n-1\) in (37), we have the second equation in (35). Therefore, \(\Psi\) satisfies (35). Since (35) is equivalent to the equation in (27), \(\Psi\) solves the initial value problem (27).

Next, we assume that \(\Psi \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) solves the initial value problem (28), and we write \(\Psi\) as in (33). Equation (28) is equivalent to the equation

$$\begin{aligned} \pi _{L} \Psi (n-1) +\pi _{R} \Psi (n+1) ={\mathcal{C}}(n) \left[ \overline{\lambda } \Psi (n) +f(n) \right] \end{aligned}$$

with \(\Psi (0)=w,\) which is written as

$$\begin{aligned} \left\{ \begin{aligned} \psi _{L}(n-1)&= \overline{\lambda } a_{n} \psi _{L}(n) +\overline{\lambda } b_{n} \psi _{R}(n) +a_{n} f_{L}(n) +b_{n} f_{R}(n), \\ \psi _{R}(n+1)&= \overline{\lambda } c_{n} \psi _{L}(n) +\overline{\lambda } d_{n} \psi _{R}(n) +c_{n} f_{L}(n) +d_{n}f_{R}(n). \end{aligned} \right. \end{aligned}$$
(39)

Shifting the variable n in (39), we see

$$\begin{aligned} \left\{ \begin{aligned} \psi _{L}(n)&= \overline{\lambda } a_{n+1} \psi _{L}(n+1) +\overline{\lambda } b_{n+1} \psi _{R}(n+1) \\ &\quad +a_{n+1} f_{L}(n+1) +b_{n+1} f_{R}(n+1), \\ \psi _{R}(n)&= \overline{\lambda } c_{n-1} \psi _{L}(n-1) +\overline{\lambda } d_{n-1} \psi _{R}(n-1) \\ &\quad +c_{n-1} f_{L}(n-1) +d_{n-1}f_{R}(n-1). \end{aligned} \right. \end{aligned}$$
(40)

Solving the first equation of (40) in \(\psi _{L}(n+1),\) we see

$$\begin{aligned} \psi _{L}(n+1)&=\frac{\overline{\lambda }^{-1}}{a_{n+1}} \psi _{L}(n) -\frac{b_{n+1}}{a_{n+1}} \psi _{R}(n+1) \\ &\quad -\overline{\lambda }^{-1} f_{L}(n+1) -\overline{\lambda }^{-1} \frac{b_{n+1}}{a_{n+1}} f_{R}(n+1). \end{aligned}$$
(41)

Substituting the second equation of (39) into (41) shows

$$\begin{aligned}&\psi _{L}(n+1) \\ &\quad = \frac{1}{a_{n+1}} (\overline{\lambda }^{-1} -\overline{\lambda } b_{n+1}c_{n}) \psi _{L}(n) -\overline{\lambda } \frac{b_{n+1}d_{n}}{a_{n+1}} \psi _{R}(n) -\frac{b_{n+1}c_{n}}{a_{n+1}} f_{L}(n) \\ &\qquad -\frac{b_{n+1}d_{n}}{a_{n+1}} f_{R}(n) -\overline{\lambda }^{-1} f_{L}(n+1) -\overline{\lambda }^{-1} \frac{b_{n+1}}{a_{n+1}}f_{R}(n+1). \end{aligned}$$
(42)

Since the combination of the two equations, (42) and the second line of (39), is equivalent to Eq. (32), \(\Psi\) solves (32). Conversely, we suppose that \(\Psi \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) satisfies (32) with \(\Psi (0)=w.\) Thus, \(\Psi\) satisfies (42) and the second line of (39). From these two equations, we have (41). Shifting the variable n in (41) to \(n-1\) and solving it in \(\psi _{L}(n-1)\), we have the first line of (39). Therefore, \(\Psi\) solves Eq. (28). \(\square\)

Lemma 2.2 is used to deduce the concrete formulas of the functions \(v_{\lambda }(n,m),\) \(w_{\lambda }^{o}(n,m)\) defined in (44), (47) below which give solutions to the problems (27), (28). Indeed, if we give an initial value \(\Psi (0)=w,\) then the function \(\Psi\) on the set of non-negative integers satisfying Eq. (32) is automatically determined. Shifting the variable n to \(n-1\) in (32) and solving it in \(\Psi (n-1),\) we see

$$\begin{aligned} \Psi (n-1)&= T_{1/\overline{\lambda }}(n-1)^{-1} \Psi (n) + \frac{\overline{\lambda }^{-1}}{a_{n}} T_{1/\overline{\lambda }}(n-1)^{-1} \pi _{L} {\mathcal{C}}(n)f(n) \\ &\quad +\overline{\lambda }^{-1} \left( \frac{\overline{\lambda }^{-1}}{a_{n}} T_{1/\overline{\lambda }} (n-1)^{-1} \pi _{L} -I \right) f(n-1). \end{aligned}$$
(43)

Thus, once we give \(\Psi (0)=w,\) the function \(\Psi\) on the set of non-positive integers satisfying Eq. (43) is also automatically determined. Therefore, in principle, we can solve the Eq. (28). Similar discussion is also applicable for (27) using (31). The formulas of the solutions given in Proposition 2.3 are deduced from the concrete form of \(\Psi (n)\) obtained using Eqs. (31) and (32) for several small \(n \in {\mathbb{Z}}\) in the absolute value.

Proposition 2.3

  1. (1)

    We define a function \(v_{\lambda } \in {\mathrm{Map}}\,({\mathbb{Z}}^{2},{\mathrm{M}}_{2}({\mathbb{C}}))\) by the following formula:

    $$\begin{aligned}&v_{\lambda }(n,0) = \left\{ \begin{array}{ll} \lambda ^{-1} F_{\lambda }(n) {\mathrm{z}}_{L}(0) &{}\quad (n \ge 1), \\ \lambda ^{-1}F_{\lambda }(n) {\mathrm{z}}_{R}(0) &{}\quad (n \le -1), \end{array}\right. \\ &v_{\lambda }(n,m) \\ &\quad = \left\{ \begin{array}{ll} \lambda ^{-1}F_{\lambda }(n)F_{\lambda }(m)^{-1}({\mathrm{z}}_{L}(m)-{\mathrm{z}}_{R}(m)) &{}\quad (1 \le m \le n-1), \\ -\lambda ^{-1} {\mathrm{z}}_{R}(n) &{}\quad (1 \le m =n), \\ \lambda ^{-1} F_{\lambda }(n) F_{\lambda }(m)^{-1} ({\mathrm{z}}_{R}(m) -{\mathrm{z}}_{L}(m)) &{}\quad (n+1 \le m \le -1), \\ -\lambda ^{-1} {\mathrm{z}}_{L}(n) &{}\quad (n=m \le -1), \\ 0 &{}\quad ({\mathrm{otherwise}}). \end{array}\right. \end{aligned}$$
    (44)

    Then, for each \(f \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}),\) \(\lambda \in {\mathbb{C}} {\setminus } \{0\}\) and \(w \in {\mathbb{C}}^{2},\) Eq. (27) has a unique solution given by

    $$\begin{aligned} \Psi =\Phi _{\lambda }(w)+V_{\lambda }f, \end{aligned}$$
    (45)

    where \(\Phi _{\lambda }(w)\) is defined in (7) and \(V_{\lambda }f\) is defined by

    $$\begin{aligned} V_{\lambda }f(n)= \sum _{m \in {\mathbb{Z}}} v_{\lambda }(n,m)f(m). \end{aligned}$$
    (46)
  2. (2)

    We define a function \(w_{\lambda }^{o} \in {\mathrm{Map}}\,({\mathbb{Z}}^{2},{\mathrm{M}}_{2}({\mathbb{C}}^{2}))\) by the following formula:

    $$\begin{aligned}&w_{\lambda }^{o}(n,0) = \left\{ \begin{array}{ll} \overline{\lambda }^{-1}F_{1/\overline{\lambda }}(n){\mathrm{z}}_{R}(0)^{*} &{}\quad (n \ge 1), \\ \overline{\lambda }^{-1}F_{1/\overline{\lambda }}(n){\mathrm{z}}_{L}(0)^{*} &{}\quad (n \le -1), \end{array}\right. \\ &w_{\lambda }^{o}(n,m) \\ &\quad = \left\{ \begin{array}{ll} \overline{\lambda }^{-1}F_{1/\overline{\lambda }}(n)F_{1/\overline{\lambda }}(m)^{-1}({\mathrm{z}}_{R}(m)^{*} -{\mathrm{z}}_{L}(m)^{*}) &{}\quad (1 \le m \le n-1), \\ -\overline{\lambda }^{-1}{\mathrm{z}}_{L}(n)^{*} &{}\quad (1 \le m =n), \\ \overline{\lambda }^{-1} F_{1/\overline{\lambda }}(n)F_{1/\overline{\lambda }}(m)^{-1} ({\mathrm{z}}_{L}(m)^{*}-{\mathrm{z}}_{R}(m)^{*}) &{}\quad (n+1 \le m \le -1), \\ -\overline{\lambda }^{-1}{\mathrm{z}}_{R}(n)^{*} &{}\quad (n=m \le -1), \\ 0 &{} ({\mathrm{otherwise}}). \end{array}\right. \end{aligned}$$
    (47)

    Then, for each \(f \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}),\) \(\lambda \in {\mathbb{C}} {\setminus } \{0\}\) and \(w \in {\mathbb{C}}^{2},\) Eq. (28) has a unique solution given by

    $$\begin{aligned} \Psi =\Phi _{1/\overline{\lambda }}(w)+W_{\lambda }^{o}f, \end{aligned}$$
    (48)

    where \(W_{\lambda }^{o}f\) is defined by

    $$\begin{aligned} W_{\lambda }^{o}f(n)= \sum _{m \in {\mathbb{Z}}} w_{\lambda }^{o}(n,m)f(m). \end{aligned}$$
    (49)

We need some of the following formulas to prove Proposition 2.3.

Lemma 2.4

For any \(\lambda \in {\mathbb{C}} {\setminus } \{0\},\) \(n \in {\mathbb{Z}},\) we have the following:

  1. (1)

    \(F_{\lambda }(n+1)=T_{\lambda }(n)F_{\lambda }(n),\)

  2. (2)

    \(\pi _{L}{\mathcal{C}}(n) T_{\lambda }(n-1)=\lambda \pi _{L},\)

  3. (3)

    \(\pi _{R} {\mathcal{C}}(n) T_{\lambda }(n)^{-1} =\lambda \pi _{R},\)

  4. (4)

    \({\mathrm{z}}_{L}(n)=\frac{\lambda }{a_{n+1}} T_{\lambda }(n)^{-1}\pi _{L} =\frac{\triangle _{n}}{d_{n}}{\mathcal{C}}(n)^{*}\pi _{L}=\frac{1}{\overline{a_{n}}}{\mathcal{C}}(n)^{*}\pi _{L},\)

  5. (5)

    \({\mathrm{z}}_{R}(n)=\frac{\lambda }{d_{n-1}} T_{\lambda }(n-1) \pi _{R} =\frac{\triangle _{n}}{a_{n}} {\mathcal{C}}(n)^{*} \pi _{R}=\frac{1}{\overline{d_{n}}} {\mathcal{C}}(n)^{*}\pi _{R},\)

  6. (6)

    \({\mathrm{z}}_{L}(n)^{*}+{\mathrm{z}}_{R}(n)={\mathrm{z}}_{L}(n)+{\mathrm{z}}_{R}(n)^{*}=I,\)

  7. (7)

    \(a_{n}{\mathrm{z}}_{L}(n)^{*}+d_{n}{\mathrm{z}}_{R}(n)^{*}=\pi _{L} {\mathcal{C}}(n) {\mathrm{z}}_{L}(n)^{*}+\pi _{R}{\mathcal{C}}(n){\mathrm{z}}_{R}(n)^{*}={\mathcal{C}}(n),\)

  8. (8)

    \(T_{\lambda }(n)[{\mathrm{z}}_{L}(n)-{\mathrm{z}}_{R}(n)]T_{1/\overline{\lambda }}(n)^{*}=[{\mathrm{z}}_{L}(n+1)-{\mathrm{z}}_{R}(n+1)].\)

Proof

(1) follows from (6), (2) and the first two equalities in (5) follow from (5), (11) and the unitarity of \({\mathcal{C}}(n).\) The inverse \(T_{\lambda }(n)^{-1}\) of the matrix \(T_{\lambda }(n)\) is given by

$$\begin{aligned} T_{\lambda }(n)^{-1}= \left[ \begin{array}{cc} { \lambda ^{-1}a_{n+1} }&{}{ \lambda ^{-1}b_{n+1}}\\ { -\lambda ^{-1} \frac{c_{n}a_{n+1}}{d_{n}} }&{}{ \frac{1}{d_{n}}(\lambda -\lambda ^{-1}b_{n+1}c_{n})} \end{array}\right] . \end{aligned}$$
(50)

From this and the unitarity of \({\mathcal{C}}(n),\) (3) and the first two equalities in (4) follow. Since \({\mathcal{C}}(n)\) is unitary, we have

$$\begin{aligned} \triangle _{n} \overline{a_{n}} =d_{n},\quad \triangle _{n} \overline{b_{n}}=-c_{n},\quad \triangle _{n} \overline{c_{n}}=-b_{n},\quad \triangle _{n} \overline{d_{n}}=a_{n}. \end{aligned}$$
(51)

This and (11) show (6). By a direct computation using (11) and (51), we see

$$\begin{aligned} a_{n}{\mathrm{z}}_{L}(n)^{*}&=\pi _{L}{\mathcal{C}}(n) {\mathrm{z}}_{L}(n)^{*}= \left[ \begin{array}{cc} a_{n} &{} b_{n} \\ 0 &{} 0 \end{array}\right] ,\\ d_{n}{\mathrm{z}}_{R}(n)^{*}&=\pi _{R}{\mathcal{C}}(n) {\mathrm{z}}_{R}(n)^{*}= \left[ \begin{array}{cc} 0 &{} 0 \\ c_{n} &{} d_{n} \end{array}\right] , \end{aligned}$$

from which the item (7) follows. To prove (8), we first note that, by (51), the matrix \(T_{1/\overline{\lambda }}(n)^{*}\) is written as

$$\begin{aligned} T_{1/\overline{\lambda }}(n)^{*}= \left[ \begin{array}{cc} { \frac{\triangle _{n+1}}{d_{n+1}} \left( \lambda ^{-1} -\lambda \frac{b_{n} c_{n+1}}{\triangle _{n}\triangle _{n+1}} \right) }&{}{ -\lambda \frac{b_{n}}{\triangle _{n}} }\\ { \lambda \frac{a_{n} c_{n+1}}{\triangle _{n} d_{n+1}} }&{}{ \lambda \frac{a_{n}}{\triangle _{n}} } \end{array}\right] . \end{aligned}$$
(52)

From this and the item (4), we have

$$\begin{aligned}&T_{\lambda }(n) {\mathrm{z}}_{L}(n) T_{1/\overline{\lambda }}(n)^{*} = \frac{\lambda }{a_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*} \\ &\quad = \left[ \begin{array}{cc} { \frac{\triangle _{n+1}}{a_{n+1}d_{n+1}} -\lambda ^{2} \frac{b_{n} c_{n+1}}{a_{n+1}d_{n+1} \triangle _{n}} }&{} { - \lambda ^{2} \frac{b_{n}}{a_{n+1}\triangle _{n}} }\\ { 0 }&{}{ 0 } \end{array}\right] . \end{aligned}$$
(53)

Using the concrete form for \(T_{1/\overline{\lambda }}(n)^{*}\) mentioned above, we also have

$$\begin{aligned} {\mathrm{z}}_{R}(n)T_{1/\overline{\lambda }}(n)^{*} = \frac{\lambda }{\triangle _{n}d_{n+1}} \left[ \begin{array}{cc} -b_{n} c_{n+1} &{} -b_{n} d_{n+1} \\ a_{n} c_{n+1} &{} a_{n} d_{n+1} \end{array}\right] . \end{aligned}$$

By a direct computation using the definition of \(T_{\lambda }(n)\) and the above formula for \({\mathrm{z}}_{R}(n)T_{1/\overline{\lambda }}(n)^{*},\) we have

$$\begin{aligned}&T_{\lambda }(n) {\mathrm{z}}_{R}(n) T_{1/\overline{\lambda }}(n)^{*} \\ &\quad = \left[ \begin{array}{cc} { -\lambda ^{2} \frac{b_{n}c_{n+1}}{a_{n+1} d_{n+1} \triangle _{n}} - \frac{b_{n+1} c_{n+1}}{a_{n+1}d_{n+1}} }&{} { -\lambda ^{2} \frac{b_{n}}{a_{n+1}\triangle _{n}} -\frac{b_{n+1}}{a_{n+1}} }\\ { \frac{c_{n+1}}{d_{n+1}} }&{}{ 1} \end{array}\right] . \end{aligned}$$
(54)

Subtracting (54) from (53), we conclude (8). \(\square\)

Proof of Proposition 2.3

According to (11), the matrices \({\mathrm{z}}_{L}(n),\) \({\mathrm{z}}_{R}(n),\) \({\mathrm{z}}_{L}(n)-{\mathrm{z}}_{R}(n),\) and their adjoints are all nonzero for any \(n \in {\mathbb{Z}}.\) Since the matrix \(F_{\lambda }(n)\) is non-singular for any \(\lambda \in {\mathbb{C}} {\setminus } \{0\}\) and \(n \in {\mathbb{Z}},\) \(v_{\lambda }(n,m)\) and \(w_{\lambda }^{o}(n,m)\) are nonzero if and only if \(n \ne 0\) and m lies between 0 and n. In particular, the sums in (46) and (49) are finite. Thus, \(V_{\lambda }f\) and \(W_{\lambda }^{o}f\) are well defined as elements in \({\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) for any \(f \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}).\)

First, we show the uniqueness of the solution to each of the initial value problems (27) and (28). Suppose that \(\Psi _{1}, \Psi _{2} \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) solve the initial value problem (27). Then, we have \(\Psi _{1}(0)=\Psi _{2}(0)=w \in {\mathbb{C}}^{2}.\) We set \(\Psi =\Psi _{1}-\Psi _{2}.\) Then, \(\Psi\) satisfies \((U({\mathcal{C}})-\lambda )\Psi =0\) with \(\Psi (0)=0.\) According to Theorem 1.1, \(\Psi\) is in the image of the map \(\Phi _{\lambda } :{\mathbb{C}}^{2} \rightarrow {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) defined in (7). Thus, we can write \(\Psi =\Phi _{\lambda }(u)\) with some \(u \in {\mathbb{C}}^{2}.\) Since \(F_{\lambda }(0)=I,\) we see \(0=\Psi (0)=\Phi _{\lambda }(u)(0)=F_{\lambda }(0)u=u,\) and hence, \(\Psi =\Phi _{\lambda }(0)=0.\) This shows that the solution to the initial value problem (27) is unique. Next, suppose that \(\Psi _{1},\Psi _{2}\) are solutions to the initial value problem (28). Then, we have \(\Psi _{1}(0)=\Psi _{2}(0)=w.\) We set \(\Psi =\Psi _{1}-\Psi _{2}.\) Then, \(\Psi\) satisfies \((U({\mathcal{C}})^{*}-\overline{\lambda }) \Psi =0\) with \(\Psi (0)=0.\) Applying \(U({\mathcal{C}})\) to the equation \(U({\mathcal{C}})^{*}\Psi =\overline{\lambda } \Psi\) and using Lemma 2.1, we have \(\Psi =\overline{\lambda } U({\mathcal{C}}) \Psi\) or, what is the same to say, \(U({\mathcal{C}})\Psi =\overline{\lambda }^{-1} \Psi\), because \(\lambda \ne 0\) as assumed. Therefore, by Theorem 1.1, \(\Psi\) is in the image of the map \(\Phi _{1/\overline{\lambda }} :{\mathbb{C}}^{2} \rightarrow {\mathrm{Map}}\,({\mathbb{Z}}, {\mathbb{C}}^{2}),\) and hence, we can write \(\Psi =\Phi _{1/\overline{\lambda }}(u)\) with some \(u \in {\mathbb{C}}^{2}.\) Then, we have \(0=\Psi (0)=\Phi _{1/\overline{\lambda }}(u)(0)=u\) by exactly the same discussion as before. Therefore, we see \(\Psi =\Phi _{1/\overline{\lambda }}(0)=0,\) which shows that the solution to (28) is unique.

Next, we check that the function defined by (45) (resp. (48)) solves the initial value problem (27) (resp. (28)). Since, by Theorem 1.1, we have \((U({\mathcal{C}})-\lambda ) \Phi _{\lambda }(w)=(U({\mathcal{C}})^{*}-\overline{\lambda })\Phi _{1/\overline{\lambda }}(w)=0\) and \(\Phi _{\lambda }(w)(0)=\Phi _{1/\overline{\lambda }}(w)(0)=w,\) it is enough to check that the function \(\Psi =V_{\lambda }f\) (resp. \(\Psi =W_{\lambda }^{o}f\)) solves the initial value problem (27) (resp. (28)) with the initial value \(w=0.\) The definitions (44) and (47) show that \(v_{\lambda }(0,m)=w_{\lambda }^{o}(0,m)=0\) for any \(m \in {\mathbb{Z}},\) and hence we see \(V_{\lambda }f(0)=W^{o}_{\lambda }f(0)=0.\) Thus, by Lemma 2.2, we need only to show that \(\Psi =V_{\lambda }f\) (resp. \(\Psi =W_{\lambda }^{o}f\)) satisfies the recurrence equation (31) (resp. (32)). Before proceeding to the proof that \(V_{\lambda }f\) satisfies (31), it is useful to give concrete formulas for \(T_{\lambda }v_{\lambda }(n,m).\) Namely, we have

$$\begin{aligned}&T_{\lambda }(n) v_{\lambda }(n,0) = \left\{ \begin{array}{ll} v_{\lambda }(n+1,0) &{}\quad (n \ge 1), \\ 0 &{}\quad (n=0), \\ \lambda ^{-1}{\mathrm{z}}_{R}(0) &{} \quad (n=-1), \\ v_{\lambda }(n+1,0) &{}\quad (n \le -2), \\ \end{array}\right. \\ &T_{\lambda }(n) v_{\lambda }(n,m) \\ &\quad = \left\{ \begin{array}{ll} v_{\lambda }(n+1,m) &{}\quad (1 \le m \le n-1), \\ -\lambda ^{-1} T_{\lambda }(n) {\mathrm{z}}_{R}(n) &{}\quad (1 \le m =n), \\ \lambda ^{-1} ({\mathrm{z}}_{R}(n+1) -{\mathrm{z}}_{L}(n+1)) &{}\quad (n+1 = m \le -1), \\ v_{\lambda }(n+1,m) &{}\quad (n+2 \le m \le -1), \\ { -\frac{1}{a_{n+1}} \pi _{L} }&{}\quad (n=m \le -1), \\ 0 &{}\quad (\text{otherwise}). \end{array}\right. \end{aligned}$$
(55)

These can be obtained directly from the definition (44) and the items (1), (4) in Lemma 2.4. (The item (4) in Lemma 2.4 is used only to obtain the formula when \(n=m \le -1.\)) In what follows, we use the notation, for example, ‘\(\overset{(1)}{=}\)’ to indicate that the item (1) in Lemma 2.4 is used to show the equality. We now show that \(\Psi =V_{\lambda }f\) satisfies (31). We have

$$\begin{aligned} (V_{\lambda }f) (1)&= v_{\lambda }(1,0)f(0) +v_{\lambda }(1,1)f(1) \\ &\overset{(1)}{=} \lambda ^{-1} T_{\lambda }(0) {\mathrm{z}}_{L}(0) f(0) -\lambda ^{-1} {\mathrm{z}}_{R}(1)f(1) \\ &\overset{(4)}{=} \frac{1}{a_{1}} \pi _{L} f(0) - \lambda ^{-1} {\mathrm{z}}_{R}(1) f(1) \overset{(5)}{=} \frac{1}{a_{1}} \pi _{L} f(0) - \frac{1}{d_{0}} T_{\lambda }(0) \pi _{R}f(1), \end{aligned}$$

which shows that \(V_{\lambda }f\) satisfies (31) when \(n=0\), because \(V_{\lambda }f(0)=0.\) To prove (31) with \(n=-1\) for \(\Psi =V_{\lambda }f,\) using (55), we see

$$\begin{aligned}&T_{\lambda } (-1) (V_{\lambda }f)(-1) +\frac{1}{a_{0}} \pi _{L}f(-1) -\frac{1}{d_{-1}} T_{\lambda }(-1) \pi _{R}f (0) \\ &\quad = T_{\lambda }(-1) v_{\lambda }(-1,-1)f(-1) +T_{\lambda }(-1) v_{\lambda }(-1,0)f(0) \\ &\qquad +\frac{1}{a_{0}} \pi _{L}f(-1) -\frac{1}{d_{-1}} T_{\lambda }(-1) \pi _{R}f (0) \\ &\quad \overset{(2.29)}{=} -\frac{1}{a_{0}} \pi _{L} f(-1) +\lambda ^{-1} {\mathrm{z}}_{R}(0) f(0) +\frac{1}{a_{0}} \pi _{L} f(-1) -\frac{1}{d_{-1}} T_{\lambda }(-1) \pi _{R} f(0) \\ &\quad \overset{(5)}{=} \frac{1}{d_{-1}} T_{\lambda }(-1) \pi _{R}f(0) -\frac{1}{d_{-1}} T_{\lambda }(-1)\pi _{R} f(0) = 0, \end{aligned}$$

which coincides with \(V_{\lambda }f(0)=0,\) and hence, \(V_{\lambda }f\) satisfies (31) when \(n=-1.\) Next, we consider the case \(n \ge 1.\) We have

$$\begin{aligned}&(V_{\lambda }f)(n+1) \\ &\quad = \sum _{m=0}^{n-1} v_{\lambda }(n+1,m) f(m)+v_{\lambda }(n+1,n)f(n) +v_{\lambda }(n+1,n+1)f(n+1) \\ &\quad \overset{(2.29), (1)}{=} \sum _{m=0}^{n-1} T_{\lambda }(n) v_{\lambda }(n,m)f(m) +\lambda ^{-1}T_{\lambda }(n) ({\mathrm{z}}_{L}(n) -{\mathrm{z}}_{R}(n)) f(n) \\ &\qquad - \lambda ^{-1} {\mathrm{z}}_{R}(n+1) f(n+1) \\ &\quad \overset{(2.29)}{=} \sum _{m=0}^{n-1} T_{\lambda }(n) v_{\lambda }(n,m)f(m) +T_{\lambda }(n)v_{\lambda }(n,n)f(n) +\lambda ^{-1}T_{\lambda }(n) {\mathrm{z}}_{L}(n) f(n) \\ &\qquad - \lambda ^{-1} {\mathrm{z}}_{R}(n+1) f(n+1) \\ &\quad \overset{(4)}{=} T_{\lambda }(n)V_{\lambda }f (n) + \frac{1}{a_{n+1}}\pi _{L}f(n) - \lambda ^{-1} {\mathrm{z}}_{R}(n+1) f(n+1) \\ &\quad \overset{(5)}{=} T_{\lambda }(n)V_{\lambda }f (n) + \frac{1}{a_{n+1}}\pi _{L}f(n) -\frac{1}{d_{n}} T_{\lambda }(n) \pi _{R} f(n+1), \end{aligned}$$

which shows that the function \(V_{\lambda }f\) satisfies (31) for \(n \ge 1.\) For \(n \le -3,\) we calculate the right-hand side of (31) using (55) as follows:

$$\begin{aligned}&T_{\lambda } (n) V_{\lambda }f(n) +\frac{1}{a_{n+1}}\pi _{L}f(n) -\frac{1}{d_{n}}T_{\lambda }(n)\pi _{R} f(n+1) \\ &\quad = \left[ T_{\lambda }(n) v_{\lambda }(n,n) +\frac{1}{a_{n+1}} \pi _{L}\right] f(n) -\frac{1}{d_{n}} T_{\lambda }(n) \pi _{R}f(n+1) \\ &\qquad + T_{\lambda }(n)v_{\lambda }(n,n+1)f(n+1) +\sum _{m=n+2}^{0} T_{\lambda }(n)v_{\lambda }(n,m)f(m) \\ &\quad \overset{(2.29)}{=} -\frac{1}{d_{n}} T_{\lambda }(n) \pi _{R}f(n+1) + \lambda ^{-1} ({\mathrm{z}}_{R}(n+1) -{\mathrm{z}}_{L}(n+1)) f(n+1) \\ &\qquad +\sum _{m=n+2}^{0} v_{\lambda }(n+1,m)f(m) \\ &\quad \overset{(2.18)}{=} -\frac{1}{d_{n}} T_{\lambda }(n) \pi _{R}f(n+1) + \lambda ^{-1} {\mathrm{z}}_{R}(n+1) f(n+1) \\ &\qquad +\sum _{m=n+1}^{0} v_{\lambda }(n+1,m)f(m) \overset{(5)}{=} V_{\lambda }f(n+1). \end{aligned}$$

For \(n=-2,\) also by (55), we see

$$\begin{aligned}&T_{\lambda } (-2) V_{\lambda }f(-2) +\frac{1}{a_{-1}}\pi _{L}f(-2) -\frac{1}{d_{-2}}T_{\lambda }(-2)\pi _{R} f(-1) \\ &\quad = \sum _{m=-2}^{0}T_{\lambda }(-2)v_{\lambda }(-2,m)f(m) +\frac{1}{a_{-1}}\pi _{L}f(-2)-\frac{1}{d_{-2}}T_{\lambda }(-2)\pi _{R}f(-1) \\ &\quad \overset{(2.29)}{=} v_{\lambda }(-1,0)f(0)+\lambda ^{-1}({\mathrm{z}}_{R}(-1)-{\mathrm{z}}_{L}(-1))f(-1) -\frac{1}{d_{-2}}T_{\lambda }(-2) \pi _{R} f(-1) \\ &\quad \overset{(5)}{=} v_{\lambda }(-1,0)f(0) - \lambda ^{-1}{\mathrm{z}}_{L}(-1)f(-1) \\ &\quad \overset{(2.18)}{=} v_{\lambda }(-1,0)f(0)+v_{\lambda }(-1,-1)f(-1)=V_{\lambda }f(-1). \end{aligned}$$

This shows that \(V_{\lambda }f\) satisfies (31) for all \(n \in {\mathbb{Z}}.\) Next, to show that the function \(\Psi =W_{\lambda }^{o}f\) satisfies (32), we prepare the formulas for \(T_{1/\overline{\lambda }} w_{\lambda }^{o}(n,m)\) as follows:

$$\begin{aligned}&T_{1/\overline{\lambda }}(n)w_{\lambda }^{o}(n,0) = \left\{ \begin{array}{ll} w_{\lambda }^{o}(n+1,0) &{}\quad (n \ge 1), \\ 0 &{}\quad (n=0), \\ \overline{\lambda }^{-1} {\mathrm{z}}_{L}(0)^{*} &{}\quad (n=-1), \\ w_{\lambda }^{o}(n+1,0) &{}\quad (n \le -2). \end{array}\right. \\ &T_{1/\overline{\lambda }}(n) w_{\lambda }^{o}(n,m) \\ &\quad = \left\{ \begin{array}{ll} w_{\lambda }^{o}(n+1,m) &{}\quad (1 \le m \le n-1), \\ -\overline{\lambda }^{-1} T_{1/\overline{\lambda }} (n) {\mathrm{z}}_{L}(n)^{*} &{}\quad (1 \le m=n), \\ w_{\lambda }^{o}(n+1,m) &{}\quad (n+2 \le m \le -1), \\ \overline{\lambda }^{-1} ({\mathrm{z}}_{L}(n+1)^{*} -{\mathrm{z}}_{R}(n+1)^{*}) &{}\quad (n+1=m \le -1), \\ -\overline{\lambda }^{-1}T_{1/\overline{\lambda }}(n) {\mathrm{z}}_{R}(n)^{*} &{}\quad (n=m \le -1), \\ 0 &{}\quad (\text{otherwise}). \end{array}\right. \end{aligned}$$
(56)

For \(n=0,\) we have

$$\begin{aligned} W_{\lambda }^{o}f(1)&=w_{\lambda }^{o}(1,1)f(1) +w_{\lambda }^{o}(1,0)f(0) \\ &= -\overline{\lambda }^{-1} {\mathrm{z}}_{L}(1)^{*}f(1) +\overline{\lambda }^{-1} F_{1/\overline{\lambda }}(1) {\mathrm{z}}_{R}(0)^{*} f(0) \\ &\overset{(4),(1)}{=} -\frac{\overline{\lambda }^{-1}}{a_{1}} \pi _{L} {\mathcal{C}}(1) f(1) + \overline{\lambda }^{-1} T_{1/\overline{\lambda }} (0) {\mathrm{z}}_{R}(0)^{*} f(0) \\ &\overset{(6)}{=} -\frac{\overline{\lambda }^{-1}}{a_{1}} \pi _{L} {\mathcal{C}}(1) f(1) + \overline{\lambda }^{-1} T_{1/\overline{\lambda }} (0) (I-{\mathrm{z}}_{L}(0)) f(0) \\ &\overset{(4)}{=} -\frac{\overline{\lambda }^{-1}}{a_{1}} \pi _{L} {\mathcal{C}}(1) f(1) + \overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(0) -\frac{\overline{\lambda }^{-1}}{a_{1}} \pi _{L} \right) f(0), \end{aligned}$$

which shows that \(W_{\lambda }^{o}f\) satisfies (32) for \(n=0.\) For \(n=-1,\) we calculate the right-hand side of (32) with \(n=-1,\) by (56), as

$$\begin{aligned}&T_{1/\overline{\lambda }} (-1) W_{\lambda }^{o}f(-1) -\frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} {\mathcal{C}}(0) f(0) +\overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(-1) - \frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} \right) f(-1) \\ &\quad = T_{1/\overline{\lambda }} (-1) \left( w_{\lambda }^{o}(-1,-1)f(-1) +w_{\lambda }^{o}(-1,0)f(0) \right) \\ &\qquad - \frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} {\mathcal{C}}(0) f(0) +\overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(-1) - \frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} \right) f(-1) \\ &\quad \overset{(2.30)}{=} -\overline{\lambda }^{-1} T_{1/\overline{\lambda }}(-1) {\mathrm{z}}_{R}(-1)^{*} f(-1) +\overline{\lambda }^{-1} {\mathrm{z}}_{L}(0)^{*} f(0) \\ &\qquad -\frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} {\mathcal{C}}(0) f(0) +\overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }} (-1) -\frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} \right) f(-1) \\ &\quad = \overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(-1) (I-{\mathrm{z}}_{R}(-1)^{*}) -\frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} \right) f(-1) \\ &\qquad +\overline{\lambda }^{-1}\left( {\mathrm{z}}_{L}(0)^{*} -\frac{1}{a_{0}} \pi _{L} {\mathcal{C}}(0) \right) f(0). \end{aligned}$$
(57)

By the item (4) in Lemma 2.4, we have \({\mathrm{z}}_{L}(0)^{*}=\frac{1}{a_{0}}\pi _{L} {\mathcal{C}}(0).\) By the item (6) in Lemma 2.4, we have \(I-{\mathrm{z}}_{R}(-1)^{*}={\mathrm{z}}_{L}(-1).\) Substituting these formulas into (57), we obtain

$$\begin{aligned}&T_{1/\overline{\lambda }} (-1) W_{\lambda }^{o}f(-1) -\frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} {\mathcal{C}}(0) f(0) + \overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(-1) - \frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} \right) f(-1) \\ &\quad = \overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(-1) {\mathrm{z}}_{L}(-1) -\frac{\overline{\lambda }^{-1}}{a_{0}} \pi _{L} \right) f(-1) \overset{(4)}{=} 0. \end{aligned}$$

This shows that \(W_{\lambda }^{o}f\) satisfies (32) for \(n=-1\), because \(W_{\lambda }^{o}f(0)=0.\) Let \(n \ge 1.\) We have

$$\begin{aligned}&W_{\lambda }^{o} f(n+1) \\ &\quad = \sum _{m=0}^{n-1} w_{\lambda }^{o}(n+1,m)f(m) +w_{\lambda }^{o}(n+1,n)f(n) + w_{\lambda }^{o}(n+1,n+1)f(n+1) \\ &\quad \overset{(2.30), (1)}{=} \sum _{m=0}^{n-1} T_{1/\overline{\lambda }}(n) w_{\lambda }^{o}(n,m)f(m) \\ &\qquad + \overline{\lambda }^{-1} T_{1/\overline{\lambda }}(n) ({\mathrm{z}}_{R}(n)^{*} -{\mathrm{z}}_{L}(n)^{*}) f(n) -\overline{\lambda }^{-1} {\mathrm{z}}_{L}(n+1)^{*}f(n+1) \\ &\quad \overset{(2.21)}{=}T_{1/\overline{\lambda }} (n) W_{\lambda }^{o} f(n) +\overline{\lambda }^{-1} T_{1/\overline{\lambda }}(n) {\mathrm{z}}_{R}(n)^{*}f(n) -\overline{\lambda }^{-1} {\mathrm{z}}_{L}(n+1)^{*}f(n+1) \\ &\quad \overset{(6), (4)}{=} T_{1/\overline{\lambda }} (n) W_{\lambda }^{o} f(n) +\overline{\lambda }^{-1} T_{1/\overline{\lambda }}(n) (I-{\mathrm{z}}_{L}(n)) f(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L}{\mathcal{C}}(n+1)f(n+1) \\ &\quad \overset{(4)}{=} T_{1/\overline{\lambda }} (n) W_{\lambda }^{o} f(n) +\overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(n)-\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L} \right) f(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L}{\mathcal{C}}(n+1)f(n+1), \end{aligned}$$

which shows that \(W_{\lambda }^{o}f\) satisfies (32) for \(n \ge 1.\) When \(n \le -2,\) we calculate the right-hand side of (32) for \(\Psi =W_{\lambda }^{o}f\) as

$$\begin{aligned}&T_{1/\overline{\lambda }} (n) W_{\lambda }^{o}f(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L}{\mathcal{C}}(n+1) f(n+1) +\overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}}\pi _{L} \right) f(n) \\ &\quad = \overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}}\pi _{L} \right) f(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L}{\mathcal{C}}(n+1) f(n+1) \\ &\qquad + \sum _{m=n}^{0} T_{1/\overline{\lambda }}(n) w_{\lambda }^{o}(n,m)f(m) \\ &\quad \overset{(2.30)}{=} \overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}}\pi _{L} \right) f(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L}{\mathcal{C}}(n+1) f(n+1) \\ &\qquad + \sum _{m=n+2}^{0} w_{\lambda }^{o}(n+1,m)f(m) -\overline{\lambda }^{-1} T_{1/\overline{\lambda }}(n) {\mathrm{z}}_{R}(n)^{*} f(n) \\ &\qquad +\overline{\lambda }^{-1} ({\mathrm{z}}_{L}(n+1)^{*} -{\mathrm{z}}_{R}(n+1)^{*}) f(n+1) \\ &\overset{(2.21), (6)}{=} \overline{\lambda }^{-1} \left( T_{1/\overline{\lambda }}(n) -\frac{\overline{\lambda }^{-1}}{a_{n+1}}\pi _{L} \right) f(n) -\overline{\lambda }^{-1} T_{1/\overline{\lambda }}(n) (I-{\mathrm{z}}_{L}(n)) f(n) \\ &\qquad + \sum _{m=n+1}^{0} w_{\lambda }^{o}(n+1,m)f(m) \\ &\qquad -\frac{\overline{\lambda }^{-1}}{a_{n+1}} \pi _{L}{\mathcal{C}}(n+1) f(n+1) +\overline{\lambda }^{-1} {\mathrm{z}}_{L}(n+1)^{*} f(n+1) \\ &\quad \overset{(4)}{=} W_{\lambda }^{o}f(n+1). \end{aligned}$$

This shows that \(W_{\lambda }^{o}f\) satisfies (32) for any \(n \in {\mathbb{Z}}.\) \(\square\)

Corollary 2.5

For \(\lambda \in {\mathbb{C}}{\setminus } \{0\}\), we define \(w_{\lambda } \in {\mathrm{Map}}\,({\mathbb{Z}}^{2},{\mathrm{M}}_{2}({\mathbb{C}}))\) by

$$\begin{aligned} w_{\lambda }(n,m)=w_{\lambda }^{o}(m,n)^{*} \quad (n,m \in {\mathbb{Z}}). \end{aligned}$$
(58)

Let \(W_{\lambda } :C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) be a map defined by

$$\begin{aligned} (W_{\lambda }f)(n)=\sum _{m \in {\mathbb{Z}}} w_{\lambda }(n,m)f(m)\quad (f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})). \end{aligned}$$
(59)

Then, \(W_{\lambda }f\) satisfies

$$\begin{aligned} W_{\lambda }(U({\mathcal{C}})-\lambda )f=f \end{aligned}$$
(60)

for any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\)

Proof

We first note that \(W_{\lambda }\) defined in (59) is the formal adjoint operator of the linear map \(W_{\lambda }^{o} :C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\).Footnote 3 We take \(f,g \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Then, \((U({\mathcal{C}})-\lambda )f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) By Proposition 2.3, (2) and taking \(w=0\) in (48), we have \((U({\mathcal{C}})^{*} -\overline{\lambda })W_{\lambda }^{o}g=g.\) Since \(W_{\lambda }\) is the formal adjoint operator of \(W_{\lambda }^{o} :C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}),\) we have

$$\begin{aligned} \langle \,W_{\lambda }(U({\mathcal{C}})-\lambda )f,g \,\rangle =\langle \,(U({\mathcal{C}})-\lambda )f,W_{\lambda }^{o}g \,\rangle = \langle \,f,(U({\mathcal{C}})^{*}-\overline{\lambda })W_{\lambda }^{o}g \,\rangle =\langle \,f,g \,\rangle . \end{aligned}$$

Since \(f,g \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) is arbitrary, we have (60). \(\square\)

The function \(w_{\lambda }\) is given explicitly by the following:

$$\begin{aligned}&w_{\lambda }(0,m) = \left\{ \begin{array}{ll} \lambda ^{-1} {\mathrm{z}}_{R}(0) F_{1/\overline{\lambda }}(m)^{*} &{}\quad (m \ge 1), \\ \lambda ^{-1} {\mathrm{z}}_{L}(0) F_{1/\overline{\lambda }}(m)^{*} &{}\quad (m \le -1), \end{array}\right. \\ &w_{\lambda }(n,m) \\ &\quad = \left\{ \begin{array}{ll} \lambda ^{-1} ({\mathrm{z}}_{R}(n)-{\mathrm{z}}_{L}(n)) F_{1/\overline{\lambda }}(n)^{*-1} F_{1/\overline{\lambda }}(m)^{*} &{}\quad (1 \le n \le m-1), \\ -\lambda ^{-1} {\mathrm{z}}_{L}(n) &{}\quad (1 \le n=m), \\ \lambda ^{-1} ({\mathrm{z}}_{L}(n)-{\mathrm{z}}_{R}(n)) F_{1/\overline{\lambda }}(n)^{*-1} F_{1/\overline{\lambda }}(m)^{*} &{}\quad (m+1 \le n \le -1), \\ -\lambda ^{-1} {\mathrm{z}}_{R} (n) &{}\quad (m =n \le -1), \\ 0 &{}\quad ({\mathrm{otherwise}}). \end{array}\right. \end{aligned}$$
(61)

One of the most important properties of the operator \(W_{\lambda }\) is the following.

Theorem 2.6

For any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}),\) we have \(W_{\lambda }f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) and

$$\begin{aligned} (U -\lambda ) W_{\lambda }f (n) =f(n) -(\delta _{0} \otimes \widehat{f}^{{\mathcal{C}}}(\lambda ))(n) \quad (n \in {\mathbb{Z}}), \end{aligned}$$
(62)

where the QW-Fourier transform \(\widehat{f}^{{\mathcal{C}}}\) of f is defined in (16), and the function \(\delta _{m} \otimes u \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) with \(m \in {\mathbb{Z}},\) \(u \in {\mathbb{C}}^{2}\) is defined in (9).

Proof

For fixed \(m \in {\mathbb{Z}},\) the function \(w_{\lambda }(n,m)\) in n can be nonzero only when n lies between 0 and m. Therefore, we have \(W_{\lambda }f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) for any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Let \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) and \(n \ge 2.\) By (4) in Lemma 2.4 and (61), we have

$$\begin{aligned} \pi _{R}{\mathcal{C}}(n-1)w_{\lambda }(n-1,n-1)&\overset{(2.35)}{=} -\lambda ^{-1} \pi _{R} {\mathcal{C}}(n-1) {\mathrm{z}}_{L}(n-1) \\ &\overset{(4)}{=} -\lambda ^{-1} \frac{\triangle _{n-1}}{d_{n-1}} \pi _{R}\pi _{L}=0, \\ \pi _{L}{\mathcal{C}}(n+1) w_{\lambda }(n+1,n+1)&\overset{(2.35)}{=} -\lambda ^{-1} \pi _{L}{\mathcal{C}}(n+1) {\mathrm{z}}_{L}(n+1) \\ &\overset{(4)}{=} -\lambda ^{-1} \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L}. \end{aligned}$$
(63)

By (1), (4), (5) in Lemma 2.4, we also have for \(m \ge n\)

$$\begin{aligned}&\pi _{R} {\mathcal{C}}(n-1) w_{\lambda }(n-1,m) \\ &\quad \overset{(2.35)}{=} \lambda ^{-1} \pi _{R} {\mathcal{C}}(n-1) ({\mathrm{z}}_{R}(n-1)-{\mathrm{z}}_{L}(n-1)) F_{1/\overline{\lambda }}(n-1)^{*-1} F_{1/\overline{\lambda }}(m)^{*} \\ &\quad \overset{(4), (5)}{=} \lambda ^{-1} \pi _{R} \left( \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} -\frac{\triangle _{n-1}}{d_{n-1}} \pi _{L} \right) F_{1/\overline{\lambda }}(n-1)^{*-1} F_{1/\overline{\lambda }}(m)^{*} \\ &\quad \overset{(1)}{=} \lambda ^{-1} \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*} F_{1/\overline{\lambda }}(n)^{*-1} F_{1/\overline{\lambda }}(m)^{*}, \end{aligned}$$
(64)

and for \(m \ge n+2\)

$$\begin{aligned}&\pi _{L} {\mathcal{C}}(n+1) w_{\lambda }(n+1,m) \\ &\quad \overset{(2.35)}{=} -\lambda ^{-1} \pi _{L} {\mathcal{C}}(n+1) \left( {\mathrm{z}}_{R}(n+1) -{\mathrm{z}}_{L}(n+1) \right) \\ &\qquad \times F_{1/\overline{\lambda }}(n+1)^{*-1} F_{1/\overline{\lambda }}(m)^{*} \\ &\quad \overset{(4), (5)}{=} -\lambda ^{-1} \pi _{L} \left( \frac{\triangle _{n+1}}{a_{n+1}} \pi _{R} -\frac{\triangle _{n+1}}{d_{n+1}} \pi _{R} \right) F_{1/\overline{\lambda }} (n+1)^{*-1}F_{1/\overline{\lambda }}(m)^{*} \\ &\quad \overset{(1)}{=} -\lambda ^{-1} \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*-1}F_{1/\overline{\lambda }}(n)^{*-1} F_{1/\overline{\lambda }}(m)^{*}. \end{aligned}$$
(65)

Thus, the formula for \(m=n+1\) in (63) can be regarded as a special case of (64) and (65). We have

$$\begin{aligned}&\pi _{L} {\mathcal{C}}(n+1) W_{\lambda } f(n+1) \\ &\quad = -\lambda ^{-1} \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*-1} F_{1/\overline{\lambda }}(n)^{*-1} \sum _{m \ge n+1} F_{1/\overline{\lambda }}(m)^{*}f(m), \\ &\pi _{R} {\mathcal{C}}(n-1) W_{\lambda } f(n-1) \\ &\quad = \lambda ^{-1} \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*}F_{1/\overline{\lambda }}(n)^{*-1} \sum _{m \ge n} F_{1/\overline{\lambda }}(m)^{*}f(m), \end{aligned}$$

where we have used (1) in Lemma 2.4. Thus, \(U({\mathcal{C}}) W_{\lambda }f(n)\) can be calculated as follows:

$$\begin{aligned}&U({\mathcal{C}}) W_{\lambda } f (n) \\ &\quad = \pi _{L} {\mathcal{C}}(n+1) W_{\lambda } f(n+1) + \pi _{R} {\mathcal{C}}(n-1) W_{\lambda } f(n-1) \\ &\quad = -\lambda ^{-1} \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*-1} F_{1/\overline{\lambda }}(n)^{*-1} \sum _{m \ge n+1} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\qquad + \lambda ^{-1} \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*}F_{1/\overline{\lambda }}(n)^{*-1} \sum _{m \ge n} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad = \lambda ^{-1} \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*} f(n) \\ &\qquad +\lambda ^{-1} \left( \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }} (n-1)^{*} -\frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }} (n)^{*-1} \right) \\ &\qquad \times F_{1/\overline{\lambda }} (n)^{*-1} \sum _{m \ge n+1} F_{1/\overline{\lambda }} (m)^{*}f(m). \end{aligned}$$
(66)

We calculate this expression further. By Lemma 2.4, we see

$$\begin{aligned} \begin{gathered} \lambda ^{-1}\frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*-1} \overset{(4)}{=} \frac{1}{a_{n}} \pi _{L} {\mathcal{C}}(n) \overset{(4)}{=} {\mathrm{z}}_{L}(n)^{*},\\ \lambda ^{-1}\frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*} \overset{(5)}{=} \frac{1}{d_{n}} \pi _{R} {\mathcal{C}}(n) \overset{(5)}{=} {\mathrm{z}}_{R}(n)^{*} \end{gathered} \end{aligned}$$
(67)

for all \(n \in {\mathbb{Z}}.\) Using these formulas for \(n \ge 2,\) (66) can be further calculated as

$$\begin{aligned}&U({\mathcal{C}}) W_{\lambda } f (n) \\ &\quad = \lambda ^{-1} \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*} f(n) \\ &\qquad +\lambda ^{-1} \left( \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }} (n-1)^{*} -\frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }} (n)^{*-1} \right) \\ &\qquad \times F_{1/\overline{\lambda }} (n)^{*-1} \sum _{m \ge n+1} F_{1/\overline{\lambda }} (m)^{*}f(m) \\ &\quad \overset{(2.41)}{=} {\mathrm{z}}_{R}(n)^{*}f(n) + ({\mathrm{z}}_{R}(n)^{*} -{\mathrm{z}}_{L}(n)^{*}) F_{1/\overline{\lambda }} (n)^{*-1} \sum _{m \ge n+1} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad \overset{(6)}{=} f(n)-{\mathrm{z}}_{L}(n)f(n) +({\mathrm{z}}_{R}(n) -{\mathrm{z}}_{L}(n)) F_{1/\overline{\lambda }} (n)^{*-1} \sum _{m \ge n+1} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad \overset{(2.35)}{=} f(n) +\lambda w_{\lambda }(n,n)f(n) + \lambda \sum _{m \ge n+1}w_{\lambda }(n,m)f(m) \\ &\quad = f(n) +\lambda W_{\lambda }f(n), \end{aligned}$$

which shows (62) for \(n \ge 2.\) Next, we consider the case \(n \le -2.\) By (5) in Lemma 2.4, we have

$$\begin{aligned}&\pi _{L} {\mathcal{C}}(n+1) w_{\lambda }(n+1,n+1) \\ &\quad \overset{(2.35)}{=} -\lambda ^{-1} \pi _{L} {\mathcal{C}}(n+1) {\mathrm{z}}_{R}(n+1) \overset{(5)}{=} -\lambda ^{-1} \frac{\triangle _{n+1}}{a_{n+1}} \pi _{L}\pi _{R}=0, \\ &\pi _{R} {\mathcal{C}}(n-1) w_{\lambda }(n-1,n-1) \\ &\quad \overset{(2.35)}{=} -\lambda ^{-1} \pi _{R} {\mathcal{C}}(n-1) {\mathrm{z}}_{R}(n-1) \overset{(5)}{=} -\lambda ^{-1} \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R}. \end{aligned}$$
(68)

By (1), (4), (5) in Lemma 2.4, we also have for \(m \le n\)

$$\begin{aligned}&\pi _{L} {\mathcal{C}}(n+1) w_{\lambda }(n+1,m) \\ &\quad \overset{(2.35)}{=} \lambda ^{-1} \pi _{L} {\mathcal{C}}(n+1) \left( {\mathrm{z}}_{L}(n+1) - {\mathrm{z}}_{R}(n+1) \right) F_{1/\overline{\lambda }}(n+1)^{*-1}F_{1/\overline{\lambda }}(m)^{*} \\ &\quad \overset{(4),(5)}{=} \lambda ^{-1} \pi _{L} \left( \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} - \frac{\triangle _{n+1}}{a_{n+1}} \pi _{R} \right) F_{1/\overline{\lambda }}(n+1)^{*-1}F_{1/\overline{\lambda }}(m)^{*} \\ &\quad \overset{(1)}{=} \lambda ^{-1} \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*-1} F_{1/\overline{\lambda }}(n)^{*-1}F_{1/\overline{\lambda }}(m)^{*}, \end{aligned}$$

and for \(m \le n-2\)

$$\begin{aligned}&\pi _{R} {\mathcal{C}}(n-1) w_{\lambda } (n-1,m) \\ &\quad \overset{(2.35)}{=} \lambda ^{-1}\pi _{R} {\mathcal{C}}(n-1) ({\mathrm{z}}_{L}(n-1) -{\mathrm{z}}_{R}(n-1)) F_{1/\overline{\lambda }} (n-1)^{*-1} F_{1/\overline{\lambda }} (m)^{*} \\ &\quad \overset{(4), (5)}{=} \lambda ^{-1} \pi _{R} \left( \frac{\triangle _{n-1}}{d_{n-1}} \pi _{L} -\frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} \right) F_{1/\overline{\lambda }} (n-1)^{*-1} F_{1/\overline{\lambda }} (m)^{*} \\ &\quad \overset{(1)}{=} - \lambda ^{-1}\frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*} F_{1/\overline{\lambda }} (n)^{*-1} F_{1/\overline{\lambda }} (m)^{*}. \end{aligned}$$

Thus, the second line in (68) is a special case of the formula for \(\pi _{R} {\mathcal{C}}(n-1) w_{\lambda } (n-1,m)\) in the case \(m \le n-2\) given above, and we have

$$\begin{aligned}&\pi _{L} {\mathcal{C}}(n+1) W_{\lambda }f (n+1) \\ &\quad = \lambda ^{-1} \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*-1} F_{1/\overline{\lambda }}(n)^{*-1} \sum _{m \le n} F_{1/\overline{\lambda }} (m)^{*} f(m), \\ &\pi _{R} {\mathcal{C}}(n-1) W_{\lambda }f(n-1) \\ &\quad = -\lambda ^{-1} \frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*} F_{1/\overline{\lambda }} (n)^{*-1} \sum _{m \le n-1} F_{1/\overline{\lambda }}^{*}f(m). \end{aligned}$$

Therefore, \(U({\mathcal{C}})W_{\lambda }f\) can be calculated as

$$\begin{aligned}&U({\mathcal{C}})W_{\lambda }f(n) = \pi _{L} {\mathcal{C}}(n+1) W_{\lambda }f (n+1) + \pi _{R} {\mathcal{C}}(n-1) W_{\lambda }f(n-1) \\ &\quad = \lambda ^{-1} \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*-1} f(n) \\ &\qquad +\lambda ^{-1} \left( \frac{\triangle _{n+1}}{d_{n+1}} \pi _{L} T_{1/\overline{\lambda }}(n)^{*-1} -\frac{\triangle _{n-1}}{a_{n-1}} \pi _{R} T_{1/\overline{\lambda }}(n-1)^{*} \right) \\ &\qquad \times F_{1/\overline{\lambda }}(n)^{*-1} \sum _{m \le n-1} F_{1/\overline{\lambda }} (m)^{*} f(m) \\ &\quad \overset{(2.41)}{=} {\mathrm{z}}_{L}(n)^{*}f(n) \\ &\qquad + \left( {\mathrm{z}}_{L}(n)^{*} -{\mathrm{z}}_{R}(n)^{*} \right) F_{1/\overline{\lambda }}(n)^{*-1} \sum _{m \le n-1} F_{1/\overline{\lambda }} (m)^{*} f(m) \\ &\quad \overset{(6)}{=} f(n) -{\mathrm{z}}_{R}(n) f(n) \\ &\qquad +({\mathrm{z}}_{L}(n) -{\mathrm{z}}_{R}(n)) F_{1/\overline{\lambda }}(n)^{*-1} \sum _{m \le n-1} F_{1/\overline{\lambda }} (m)^{*} f(m) \\ &\quad \overset{(2.35)}{=} f(n) +\lambda W_{\lambda }f (n), \end{aligned}$$
(69)

which shows (62) for \(n \le -2.\) Since \(w_{\lambda }(0,0)=0,\) we have

$$\begin{aligned}&U({\mathcal{C}}) W_{\lambda }f (1) = \sum _{m \ge 2} \pi _{L} {\mathcal{C}}(2) w_{\lambda } (2,m)f(m) +\sum _{m \ge 1} \pi _{R} {\mathcal{C}}(0) w_{\lambda }(0,m)f(m) \\ &\quad \overset{(2.35)}{=} -\lambda ^{-1} \pi _{L} {\mathcal{C}}(2) {\mathrm{z}}_{L}(2) f(2) \\ &\quad +\lambda ^{-1} \pi _{L} {\mathcal{C}}(2) ({\mathrm{z}}_{R}(2) -{\mathrm{z}}_{L}(2)) F_{1/\overline{\lambda }}(2)^{*-1} \sum _{m \ge 3} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\qquad +\lambda ^{-1} \pi _{R} {\mathcal{C}}(0) {\mathrm{z}}_{R}(0) \sum _{m \ge 1} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad \overset{(1), (4), (5)}{=} -\lambda ^{-1} \frac{\triangle _{2}}{d_{2}} \pi _{L} F_{1/\overline{\lambda }} (2)^{*-1} \sum _{m \ge 2} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\qquad +\lambda ^{-1} \frac{\triangle _{0}}{a_{0}} \pi _{R} T_{1/\overline{\lambda }} (0)^{*} F_{1/\overline{\lambda }}(1)^{*-1} \sum _{m \ge 1} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad \overset{(2.41)}{=} -{\mathrm{z}}_{L}(1)^{*}F_{1/\overline{\lambda }}(1)^{*-1} \sum _{m \ge 2} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\qquad +{\mathrm{z}}_{R}(1)^{*} F_{1/\overline{\lambda }}(1)^{*-1} \sum _{m \ge 1} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad = {\mathrm{z}}_{R}(1)^{*}f(1) +({\mathrm{z}}_{R}(1)^{*} -{\mathrm{z}}_{L}(1)^{*}) F_{1/\overline{\lambda }}(1)^{*-1} \sum _{m \ge 2} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad \overset{(6), (2.35)}{=} f(1) -{\mathrm{z}}_{L}(1)f(1) + \lambda \sum _{m \ge 2} w_{\lambda }(1,m)f(m) \\ &\quad \overset{(2.35)}{=} f(1) + \lambda W_{\lambda }f(1). \end{aligned}$$

For \(n=-1,\) we have

$$\begin{aligned}&U({\mathcal{C}}) W_{\lambda }f (-1) \\ &\quad = \sum _{m \le -1} \pi _{L}{\mathcal{C}}(0) w_{\lambda }(0,m) +\sum _{m \le -2} \pi _{R} {\mathcal{C}}(-2) w_{\lambda }(-2,m)f(m) \\ &\quad \overset{(2.35)}{=} \lambda ^{-1}\sum _{m \le -1} \pi _{L} {\mathcal{C}}(0) {\mathrm{z}}_{L}(0) F_{1/\overline{\lambda }}(m)^{*} f(m) - \lambda ^{-1} \pi _{R} {\mathcal{C}}(-2) {\mathrm{z}}_{R}(-2)f(-2) \\ &\qquad +\sum _{m \le -3} \pi _{R} {\mathcal{C}}(-2) ({\mathrm{z}}_{L}(-2) -{\mathrm{z}}_{R}(-2)) F_{1/\overline{\lambda }}^{*-1} F_{1/\overline{\lambda }}(m)^{*} f(m) \\ &\quad \overset{(4), (5)}{=} \lambda ^{-1} \frac{\triangle _{0}}{d_{0}} \pi _{L} F_{1/\overline{\lambda }}(-1)^{*}f(-1) +\lambda ^{-1} \frac{\triangle _{0}}{d_{0}} \pi _{L} \sum _{m \le -2} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\qquad -\lambda ^{-1} \frac{\triangle _{-2}}{a_{-2}} \pi _{R} F_{1/\overline{\lambda }}(-2)^{*-1} \sum _{m \le -2} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad \overset{(1)}{=} \lambda ^{-1} \frac{\triangle _{0}}{d_{0}} \pi _{L} T_{1/\overline{\lambda }} (-1)^{*-1} f(-1) \\ &\qquad +\lambda ^{-1} \left( \frac{\triangle _{0}}{d_{0}} \pi _{L} T_{1/\overline{\lambda }} (-1)^{*-1} -\frac{\triangle _{-2}}{a_{-2}} \pi _{R} T_{1/\overline{\lambda }} (-2)^{*} \right) \\ &\qquad \times F_{1/\overline{\lambda }}(-1)^{*-1} \sum _{m \le -2} F_{1/\overline{\lambda }} (m)^{*}f(m) \\ &\quad \overset{(2.41)}{=} {\mathrm{z}}_{L}(-1)^{*}f(-1) +\left( {\mathrm{z}}_{L}(-1)^{*} -{\mathrm{z}}_{R}(-1)^{*} \right) \\ &\qquad \times F_{1/\overline{\lambda }}(-1)^{*-1} \sum _{m \le -2} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad \overset{(6)}{=} f(-1) -{\mathrm{z}}_{R}(-1)f(-1) +({\mathrm{z}}_{L}(-1) -{\mathrm{z}}_{R}(-1)) \\ &\qquad \times F_{1/\overline{\lambda }}(-1)^{*-1} \sum _{m \le -2} F_{1/\overline{\lambda }}(m)^{*}f(m) \\ &\quad \overset{(2.35)}{=} f(-1) + \lambda W_{\lambda }f(-1). \end{aligned}$$

These calculations show that (62) holds also for \(n=\pm 1.\) Finally, we calculate \(U({\mathcal{C}})W_{\lambda }f (0).\) By Lemma 2.4 again, we see for \(m \ge 1\)

$$\begin{aligned}&\pi _{L}{\mathcal{C}}(1) w_{\lambda }(1,m) = -z_{L}(0)^{*} F_{1/\overline{\lambda }}(m)^{*} \\ &\quad =(-I+{\mathrm{z}}_{R}(0))F_{1/\overline{\lambda }}(m)^{*} =-F_{1/\overline{\lambda }}(m)^{*}+\lambda w_{\lambda }(0,m), \end{aligned}$$

and for \(m \le -1\)

$$\begin{aligned}&\pi _{R}{\mathcal{C}}(-1) w_{\lambda }(-1,m) = -{\mathrm{z}}_{R}(0)^{*} F_{1/\overline{\lambda }}(m)^{*} \\ &\quad =(-I+{\mathrm{z}}_{L}(0)) F_{1/\overline{\lambda }}(m)^{*} =-F_{1/\overline{\lambda }}(m)^{*}+\lambda w_{\lambda }(0,m). \end{aligned}$$

From this, we conclude

$$\begin{aligned} U({\mathcal{C}})W_{\lambda }f(0)&= \sum _{m \ge 1} \pi _{L} {\mathcal{C}}(1) w_{\lambda }(1,m) f(m) + \sum _{m \le -1} \pi _{R} {\mathcal{C}}(-1) w_{\lambda }(-1,m) f(m) \\ &= -\sum _{m \ne 0} F_{1/\overline{\lambda }}(m)^{*}f(m) +\sum _{m \ne 0} \lambda w_{\lambda }(0,m)f(m) \\ &= \lambda W_{\lambda }f(0) -\widehat{f}^{{\mathcal{C}}}(\lambda ) +f(0), \end{aligned}$$

which shows that (62) holds for all \(n \in {\mathbb{Z}}.\) \(\square\)

Before proceeding to the proof of Theorem 1.2, we prove the Eq. (26) in Theorem 1.7 for \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\)

Proof of (26)

for \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) The concrete form of the matrix \(T_{1/\overline{\lambda }}(n)^{*}\) is given in (52) and that of the matrix \(T_{1/\overline{\lambda }}(n)^{*-1}\) is given as

$$\begin{aligned} T_{1/\overline{\lambda }}(n)^{*-1}= \left[ \begin{array}{cc} { \lambda \frac{d_{n+1}}{\triangle _{n+1}} }&{}{ \lambda \frac{d_{n+1}b_{n}}{a_{n} \triangle _{n+1}} }\\ { -\lambda \frac{c_{n+1}}{\triangle _{n+1}} }&{}{ \frac{\triangle _{n}}{a_{n}} \left( \lambda ^{-1} -\lambda \frac{b_{n}c_{n+1}}{\triangle _{n}\triangle _{n+1}} \right) } \end{array}\right] . \end{aligned}$$
(70)

Using (52) and (70), it can be shown directly that

$$\begin{aligned} T_{1/\overline{\lambda }}(m-1)^{*-1} \pi _{L} {\mathcal{C}}(m)&=\lambda {\mathcal{C}}(m)^{*}\pi _{L} {\mathcal{C}}(m) \\ &= \frac{\lambda }{\triangle _{m}} \left[ \begin{array}{cc} a_{m}d_{m} &{} b_{m}d_{m} \\ -a_{m} c_{m} &{} -b_{m}c_{m} \end{array}\right] , \\ T_{1/\overline{\lambda }}(m)^{*} \pi _{R} {\mathcal{C}}(m)&=\lambda {\mathcal{C}}(m)^{*}\pi _{R} {\mathcal{C}}(m) \\ &= \frac{\lambda }{\triangle _{m}} \left[ \begin{array}{cc} -b_{m}c_{m} &{} -b_{m}d_{m} \\ a_{m} c_{m} &{} a_{m} d_{m} \end{array}\right] \end{aligned}$$
(71)

for all \(m \in {\mathbb{Z}}.\) From these formulas and (1) in Lemma 2.4, we have for \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\)

$$\begin{aligned}&{\mathcal{F}}_{{\mathcal{C}}}[U({\mathcal{C}})f](\lambda ) \\ &\quad = \sum _{m \in {\mathbb{Z}}} F_{1/\overline{\lambda }}(m)(U({\mathcal{C}})f)(m) \\ &\quad = \sum _{m \in {\mathbb{Z}}} F_{1/\overline{\lambda }}(m)^{*} \left[ \pi _{L}{\mathcal{C}}(m+1)f(m+1) +\pi _{R}{\mathcal{C}}(m-1) f(m-1) \right] \\ &\quad =\sum _{m \in {\mathbb{Z}}} F_{1/\overline{\lambda }}(m-1)^{*} \pi _{L} {\mathcal{C}}(m)f(m) +\sum _{m \in {\mathbb{Z}}} F_{1/\overline{\lambda }} (m+1)^{*} \pi _{R} {\mathcal{C}}(m) f(m)\\ &\quad \overset{(1)}{=} \sum _{m \in {\mathbb{Z}}} F_{1/\overline{\lambda }}(m)^{*} T_{1/\overline{\lambda }}(m-1)^{*-1} \pi _{L}{\mathcal{C}}(m)f(m) \\ &\qquad +\sum _{m \in {\mathbb{Z}}} F_{1/\overline{\lambda }}(m)^{*}T_{1/\overline{\lambda }}(m)^{*} \pi _{R} {\mathcal{C}}(m)f(m) \\ &\quad \overset{(2.45)}{=} \lambda \sum _{m \in {\mathbb{Z}}} F_{1/\overline{\lambda }}(m)^{*} \left( {\mathcal{C}}(m)^{*} \pi _{L} {\mathcal{C}}(m)+ {\mathcal{C}}(m)^{*} \pi _{R} {\mathcal{C}}(m) \right) f(m) \\ &\quad = \lambda \sum _{m \in {\mathbb{Z}}} F_{1/\overline{\lambda }}(m)^{*}f(m)=\lambda {\mathcal{F}}_{{\mathcal{C}}}[f], \end{aligned}$$

which shows (26). \(\square\)

3 Green function and its properties

In this section, we investigate properties of the Green function (10) and give proofs of Theorems 1.21.3 and 1.4. By the definition (10) of the Green function and (188), (189) in Appendix B, we see

$$\begin{aligned} R(\lambda )f (n)=\sum _{m \in {\mathbb{Z}}} R_{\lambda }(n,m) f(m) \quad (f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})). \end{aligned}$$

It is well known that the resolvent \(R(\lambda )\) is holomorphic as a bounded operator-valued function in \(\lambda \in {\mathbb{C}} {\setminus } \sigma (U),\) and thus, \(R_{\lambda }(n,m)\) is holomorphic as an \({\mathrm{M}}_{2}({\mathbb{C}})\)-valued function in \(\lambda \in {\mathbb{C}} {\setminus } \sigma (U)\) for each fixed \(m,n \in {\mathbb{Z}}.\) For \(\lambda \in {\mathbb{C}} {\setminus } \sigma (U),\) we set

$$\begin{aligned} {\mathrm{x}}_{0}(\lambda )=R_{\lambda }(0,0). \end{aligned}$$
(72)

Using the concrete formula (30) of \(R(0)=U^{*},\) we have

$$\begin{aligned} {\mathrm{x}}_{0}(0)u=U^{*}(\delta _{0} \otimes u)(0) ={\mathcal{C}}(0)^{*}\pi _{L} (\delta _{0} \otimes u)(-1) +{\mathcal{C}}(0)^{*} \pi _{R} (\delta _{0} \otimes u)(1)=0 \end{aligned}$$

for \(u \in {\mathbb{C}}^{2},\) and hence, \({\mathrm{x}}_{0}(0)=0.\) For any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) and \(\lambda \in {\mathbb{C}} {\setminus } \sigma (U),\) the difference \(R(\lambda )f-V_{\lambda }f \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2}),\) where \(V_{\lambda }f\) is defined in (46), satisfies

$$\begin{aligned} (U({\mathcal{C}})-\lambda ) (R(\lambda )f-V_{\lambda }f) = (U-\lambda )R(\lambda )f -(U({\mathcal{C}}) -\lambda ) V_{\lambda }f=0 \end{aligned}$$

by Proposition 2.3. Therefore, by Theorem 1.1, there exists a unique vector \({\mathrm{r}}_{\lambda }(f) \in {\mathbb{C}}^{2}\), such that

$$\begin{aligned} R(\lambda )f(n) =V_{\lambda }f(n) +\Phi _{\lambda }({\mathrm{r}}_{\lambda }(f))(n) \quad (n \in {\mathbb{Z}}). \end{aligned}$$
(73)

Since \(V_{\lambda }f(0)=0\) and \(\Phi _{\lambda }({\mathrm{r}}_{\lambda }(f))(0)={\mathrm{r}}_{\lambda }(f),\) we have

$$\begin{aligned} R(\lambda )f(0)=V_{\lambda }f(0) +\Phi _{\lambda }({\mathrm{r}}_{\lambda }(f))(0)={\mathrm{r}}_{\lambda }(f), \end{aligned}$$

and for any \(u \in {\mathbb{C}}^{2}\)

$$\begin{aligned} {\mathrm{x}}_{0}(\lambda )u=R_{\lambda }(0,0)u=R(\lambda )(\delta _{0} \otimes u)(0) ={\mathrm{r}}_{\lambda }(\delta _{0} \otimes u). \end{aligned}$$
(74)

More generally, we have

$$\begin{aligned} {\mathrm{r}}_{\lambda } (\delta _{m} \otimes u) = R(\lambda )(\delta _{m} \otimes u)(0)=R_{\lambda }(0,m)u \quad (u \in {\mathbb{C}}^{2}). \end{aligned}$$
(75)

To prove Theorem 1.2, we need the following.

Lemma 3.1

For any \(n \in {\mathbb{Z}}\) and \(\lambda \in {\mathbb{C}} {\setminus } \{0\}\), we have

$$\begin{aligned} F_{\lambda }(n)[{\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0)]F_{1/\overline{\lambda }}(n)^{*}=[{\mathrm{z}}_{L}(n)-{\mathrm{z}}_{R}(n)]. \end{aligned}$$
(76)

Proof

Since \(F_{\lambda }(0)=I\) for any \(\lambda \in {\mathbb{C}} {\setminus } \{0\},\) (76) for \(n=0\) is obvious. By the item (8) in Lemma 2.4, and the definition (6) of the matrix \(F_{\lambda }(n),\) we see

$$\begin{aligned}&F_{\lambda } (1) [{\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(1)^{*} \\ &\quad = T_{\lambda }(0) [{\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0)] T_{1/\overline{\lambda }}(0)^{*} = [{\mathrm{z}}_{L}(1)-{\mathrm{z}}_{R}(1)]. \end{aligned}$$

Let \(k \ge 1\) and suppose that (76) holds for \(n=k.\) By the items (1), (8) in Lemma 2.4, we see

$$\begin{aligned}&F_{\lambda } (k+1) [{\mathrm{z}}_{L}(0) -{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(k+1)^{*} \\ &\quad \overset{(1)}{=} T_{\lambda }(k) F_{\lambda }(k) [{\mathrm{z}}_{L}(0) -{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(k)^{*} T_{1/\overline{\lambda }}(k)^{*} \\ &\quad \overset{(3.5)}{=} T_{\lambda }(k) [{\mathrm{z}}_{L}(k) -{\mathrm{z}}_{R}(k)] T_{1/\overline{\lambda }}(k)^{*} \\ &\quad \overset{(8)}{=} [{\mathrm{z}}_{L}(k+1) -{\mathrm{z}}_{R}(k+1)]; \end{aligned}$$

hence, (76) holds for \(k+1.\) By induction, (76) holds for all \(n \ge 1.\) By the item (8) in Lemma 2.4, we see

$$\begin{aligned}{}[{\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0)]=T_{\lambda }(-1) [{\mathrm{z}}_{L}(-1) -{\mathrm{z}}_{R}(-1)] T_{1/\overline{\lambda }}(-1)^{*}, \end{aligned}$$

and hence, we have

$$\begin{aligned}&F_{\lambda }(-1) [{\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(-1)^{*} \\ &\quad =T_{\lambda }(-1)^{-1} [{\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0)] T_{1/\overline{\lambda }}(-1)^{*-1} \\ &\quad =[{\mathrm{z}}_{L}(-1)-{\mathrm{z}}_{R}(-1)]. \end{aligned}$$

This shows that (76) holds for \(n=-1.\) Let \(k \le -1\) and suppose that (76) holds for \(n=k.\) Then, we have

$$\begin{aligned}&F_{\lambda } (k-1) [{\mathrm{z}}_{L}(0) -{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(k-1)^{*} \\ &\quad \overset{(1)}{=} T_{\lambda }(k-1)^{-1} F_{\lambda }(k) [{\mathrm{z}}_{L}(0) -{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(k)^{*} T_{1/\overline{\lambda }}(k-1)^{*-1} \\ &\quad \overset{(3.5)}{=} T_{\lambda }(k-1)^{-1}[{\mathrm{z}}_{L}(k) -{\mathrm{z}}_{R}(k)] T_{1/\overline{\lambda }}(k-1)^{*-1} \\ &\quad \overset{(8)}{=} [{\mathrm{z}}_{L}(k-1) -{\mathrm{z}}_{R}(k-1)], \end{aligned}$$

which shows that (76) holds also for \(n=k-1.\) By induction, (76) holds for all \(n \le -1\) and, hence, for all \(n \in {\mathbb{Z}}.\) \(\square\)

Proof of Theorem 1.2

By the definition (46) of the operator \(V_{\lambda },\) we have

$$\begin{aligned} v_{\lambda }(n,m)u=V_{\lambda }(\delta _{m} \otimes u)(n) \end{aligned}$$

for any \(n,m \in {\mathbb{Z}}\) and \(u \in {\mathbb{C}}^{2}.\) Thus, by setting \(f=\delta _{m} \otimes u\) with \(m \in {\mathbb{Z}},\) \(u \in {\mathbb{C}}^{2}\) in (73), we have

$$\begin{aligned}&R_{\lambda }(n,m)u = R(\lambda ) (\delta _{m} \otimes u)(n) \\ &\quad = V_{\lambda }(\delta _{m} \otimes u)(n) +\Phi _{\lambda } ({\mathrm{r}}_{\lambda }(\delta _{m} \otimes u)) (n) \\ &\quad =v_{\lambda }(n,m)u+F_{\lambda }(n){\mathrm{r}}_{\lambda }(\delta _{m} \otimes u) \\ &\quad \overset{(3.4)}{=} v_{\lambda }(n,m)u+F_{\lambda }(n)R_{\lambda }(0,m)u. \end{aligned}$$
(77)

Applying \(R(\lambda )\) to Eq. (62) in Theorem 2.6, we have

$$\begin{aligned} R(\lambda )f(n)=W_{\lambda }f(n) +R(\lambda ) (\delta _{0} \otimes \widehat{f}^{{\mathcal{C}}}(\lambda ))(n) =W_{\lambda }f(n) +R_{\lambda }(n,0)\widehat{f}^{{\mathcal{C}}}(\lambda ) \end{aligned}$$
(78)

for \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) and \(n \in {\mathbb{Z}}.\) We set \(f=\delta _{m} \otimes u\) in the above. By the definition (16) of \({\mathcal{F}}_{{\mathcal{C}}},\) we have \({\mathcal{F}}_{{\mathcal{C}}}[\delta _{m} \otimes u](\lambda )=F_{1/\overline{\lambda }}(m)^{*}u.\) By the definition (59) of \(W_{\lambda },\) we see

$$\begin{aligned} W_{\lambda }(\delta _{m} \otimes u)(n)=w_{\lambda }(n,m)u \quad (u \in {\mathbb{C}}^{2}). \end{aligned}$$

Hence, substituting \(f=\delta _{m} \otimes u\) into (78), we obtain

$$\begin{aligned} R_{\lambda }(n,m)u=w_{\lambda }(n,m)u +R_{\lambda }(n,0)F_{1/\overline{\lambda }}(m)^{*}u. \end{aligned}$$
(79)

Setting \(m=0\) in (77) gives

$$\begin{aligned} R_{\lambda }(n,0)=v_{\lambda }(n,0) +F_{\lambda }(n) {\mathrm{x}}_{0}(\lambda ), \end{aligned}$$

and substituting this into (79) yields

$$\begin{aligned}&R_{\lambda }(n,m) \\ &\quad = w_{\lambda }(n,m) +v_{\lambda }(n,0)F_{1/\overline{\lambda }}(m)^{*} +F_{\lambda }(n) {\mathrm{x}}_{0}(\lambda ) F_{1/\overline{\lambda }}(m)^{*} \\ &\quad = \left\{ \begin{array}{ll} { w_{\lambda }(n,m)+F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{L}(0)] F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (n \ge 1), \\ { w_{\lambda }(n,m)+F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (n \le -1), \\ { w_{\lambda }(0,m)+{\mathrm{x}}_{0}(\lambda )F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (n=0), \end{array}\right. \end{aligned}$$
(80)

where we used the definition (44) of \(v_{\lambda }.\) For \(n=0,\) it follows from (80) and (61) that:

$$\begin{aligned} R_{\lambda }(0,m) = \left\{ \begin{array}{ll} { [{\mathrm{x}}_{0}(\lambda )+\lambda ^{-1}{\mathrm{z}}_{R}(0)]F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (m \ge 1), \\ { [{\mathrm{x}}_{0}(\lambda )+\lambda ^{-1}{\mathrm{z}}_{L}(0)]F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (m \le -1). \end{array}\right. \end{aligned}$$
(81)

From (61), we have \(w_{\lambda }(n,m)=0\) for the two cases \((n \le -1,\, m \ge n+1)\) and \((n \ge 1,\, m \le n-1).\) Hence, we have

$$\begin{aligned}&R_{\lambda }(n,m) \\ &\quad = \left\{ \begin{array}{ll} { F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (n \le -1,\, m \ge n+1), \\ { F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{L}(0)] F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (n \ge 1,\, m \le n-1). \end{array}\right. \end{aligned}$$
(82)

To consider the other cases, we note that by Lemma 3.1

$$\begin{aligned}&w_{\lambda }(n,m) \\ &\quad = \left\{ \begin{array}{ll} { \lambda ^{-1} F_{\lambda }(n)[{\mathrm{z}}_{R}(0)-{\mathrm{z}}_{L}(0)]F_{1/\overline{\lambda }}(m)^{*} }&{}\quad (1 \le n \le m-1), \\ { \lambda ^{-1} F_{\lambda }(n)[{\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0)]F_{1/\overline{\lambda }}(m)^{*} }&{}\quad (m+1 \le n \le -1). \end{array}\right. \end{aligned}$$
(83)

Therefore, (80) gives

$$\begin{aligned}&R_{\lambda }(n,m) \\ &\quad = \left\{ \begin{array}{ll} { F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (1 \le n \le m-1), \\ { F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{L}(0)] F_{1/\overline{\lambda }}(m)^{*}}&{}\quad (m+1 \le n \le -1). \end{array}\right. \end{aligned}$$
(84)

The formula (12) in Theorem 1.2 is a rewritten form of (81), (82), and (84). For \(n=m \ge 1,\) the formulas (80), (61) and Lemma 3.1 show

$$\begin{aligned}&R_{\lambda }(n,n) \\ &\quad = w_{\lambda }(n,n) + F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{L}(0)] F_{1/\overline{\lambda }}(n)^{*} \\ &\quad =-\lambda ^{-1}{\mathrm{z}}_{L}(n)+F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda )+\lambda ^{-1}{\mathrm{z}}_{L}(0)]F_{1/\overline{\lambda }}(n)^{*} \\ &\quad = -\lambda ^{-1} {\mathrm{z}}_{L}(n)+F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda )+ \lambda ^{-1}{\mathrm{z}}_{R}(0)+\lambda ^{-1}({\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0))]F_{1/\overline{\lambda }}(n)^{*} \\ &\quad = -\lambda ^{-1} {\mathrm{z}}_{L}(n)+\lambda ^{-1}({\mathrm{z}}_{L}(n)-{\mathrm{z}}_{R}(n)) + F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda )+\lambda ^{-1}{\mathrm{z}}_{R}(0)]F_{1/\overline{\lambda }}(n)^{*} \\ &\quad = -\lambda ^{-1} {\mathrm{z}}_{R}(n)+ F_{\lambda }(n)[{\mathrm{x}}_{0}(\lambda )+\lambda ^{-1}{\mathrm{z}}_{R}(0)]F_{1/\overline{\lambda }}(n)^{*}, \end{aligned}$$

which implies (13) for \(n =m\ge 1.\) Similarly, for \(n \le -1,\) we have

$$\begin{aligned}&R_{\lambda }(n,n) \\ &\quad = w_{\lambda }(n,n) + F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(n)^{*} \\ &\quad = -\lambda ^{-1}{\mathrm{z}}_{R}(n) +F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{R}(0)] F_{1/\overline{\lambda }}(n)^{*} \\ &\quad = -\lambda ^{-1}{\mathrm{z}}_{R}(n) +F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{L}(0)+ \lambda ^{-1} ({\mathrm{z}}_{R}(0)-{\mathrm{z}}_{L}(0))] F_{1/\overline{\lambda }}(n)^{*} \\ &\quad = -\lambda ^{-1}{\mathrm{z}}_{R}(n) + \lambda ^{-1}({\mathrm{z}}_{R}(n) -{\mathrm{z}}_{L}(n)) + F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{L}(0) ] F_{1/\overline{\lambda }}(n)^{*} \\ &\quad = -\lambda ^{-1}{\mathrm{z}}_{L}(n) + F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1}{\mathrm{z}}_{L}(0) ] F_{1/\overline{\lambda }}(n)^{*}, \end{aligned}$$

which implies (13) for \(n=m \le -1\) and, hence, for all \(n \in {\mathbb{Z}}.\) \(\square\)

The Green function \(R_{\lambda }(n,m)\) is, as above, expressed in terms of the products \(F_{\lambda }(n)\) of the transfer matrices \(T_{\lambda }(n)\) and the special value \({\mathrm{x}}_{0}(\lambda )=R_{\lambda }(0,0)\) of the Green function. Therefore, we will face a computation of the matrix-valued function \({\mathrm{x}}_{0}(\lambda )\) when we apply results in this paper. Theorems 1.3 and 1.4 give us one of the methods to calculate \({\mathrm{x}}_{0}(\lambda )\) whose proof is given in the rest of this section. The following is one of the most important facts in the proof of Theorem 1.3.

Lemma 3.2

Let \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}).\) Then, we have

$$\begin{aligned} \sum _{n \ge 1} \Vert F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2}=+\infty , \quad \sum _{n \le -1} \Vert F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2}=+\infty . \end{aligned}$$
(85)

In (85), the norm \(\Vert A\Vert _{{\mathrm{HS}}}\) of a \(2 \times 2\) matrix

$$\begin{aligned} A=\left[ \begin{array}{cc} a &{} b \\ c &{} d \end{array}\right] \end{aligned}$$

is the Hilbert–Schmidt norm \(\Vert A\Vert _{{\mathrm{HS}}}\) defined by

$$\begin{aligned} \Vert A\Vert _{{\mathrm{HS}}}^{2}=|a|^{2}+|b|^{2}+|c|^{2}+|d|^{2}=\Vert Ae_{L}\Vert _{{\mathbb{C}}^{2}}^{2}+\Vert Ae_{R}\Vert _{{\mathbb{C}}^{2}}^{2}. \end{aligned}$$

Lemma 3.2 could be proved by a method in [17]. However, we give a proof different from it.

Proof

Let \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}).\) First, we show the following formula:

$$\begin{aligned} {\mathcal{C}}(n) F_{\lambda }(n)=\lambda \pi _{L}F_{\lambda }(n-1) + \lambda \pi _{R}F_{\lambda }(n+1) \quad (n \in {\mathbb{Z}}). \end{aligned}$$
(86)

Indeed, by (1), (2), (3) in Lemma 2.4, we have

$$\begin{aligned} {\mathcal{C}}(n) F_{\lambda }(n)&= \pi _{L} {\mathcal{C}}(n) F_{\lambda }(n) +\pi _{R} {\mathcal{C}}(n) F_{\lambda }(n) \\ &\overset{(1)}{=} \pi _{L} {\mathcal{C}}(n) T_{\lambda }(n-1) F_{\lambda } (n-1) + \pi _{R} {\mathcal{C}}(n) T_{\lambda }(n)^{-1} F_{\lambda }(n+1) \\ &\overset{(2), (3)}{=} \lambda \pi _{L} F_{\lambda }(n-1) +\lambda \pi _{R} F_{\lambda }(n+1), \end{aligned}$$

which gives (86). To prove that

$$\begin{aligned} D:=\sum _{n \ge 1} \Vert F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} \end{aligned}$$

is infinity, we assume \(D<+\infty\) and deduce a contradiction. We note that \(D>0.\) In general, for \(C \in {\mathrm{U}}(2)\) and \(A, B \in {\mathrm{M}}_{2}({\mathbb{C}}),\) we have

$$\begin{aligned} \Vert CA\Vert _{{\mathrm{HS}}}=\Vert A\Vert _{{\mathrm{HS}}},\quad \Vert \pi _{L}A+\pi _{R}B\Vert _{{\mathrm{HS}}}^{2}=\Vert \pi _{L}A\Vert _{{\mathrm{HS}}}^{2}+\Vert \pi _{R}B\Vert _{{\mathrm{HS}}}^{2}. \end{aligned}$$
(87)

Equation (86) and the identities in (87) imply

$$\begin{aligned} \Vert F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2}&=\Vert {\mathcal{C}}(n) F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} \\ &=|\lambda |^{2} \Vert \pi _{L} F_{\lambda }(n-1)\Vert _{{\mathrm{HS}}}^{2} +|\lambda |^{2} \Vert \pi _{R} F_{\lambda }(n+1)\Vert _{{\mathrm{HS}}}^{2}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} D&= \sum _{n \ge 1} \Vert {\mathcal{C}}(x) F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} = \sum _{n \ge 1} \Vert \lambda \pi _{L}F_{\lambda }(n-1) +\lambda \pi _{R} F_{\lambda }(n+1)\Vert _{{\mathrm{HS}}}^{2} \\ &= |\lambda |^{2} \sum _{n \ge 1} \Vert \pi _{L} F_{\lambda }(n-1)\Vert _{{\mathrm{HS}}}^{2} +|\lambda |^{2} \sum _{n \ge 1}\Vert \pi _{R} F_{\lambda }(n+1)\Vert _{{\mathrm{HS}}}^{2} \\ &= |\lambda |^{2} \sum _{n \ge 0} \Vert \pi _{L} F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} +|\lambda |^{2} \sum _{n \ge 2} \Vert \pi _{R} F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} \\ &= |\lambda |^{2} \Vert \pi _{L} F_{\lambda }(0)\Vert _{{\mathrm{HS}}}^{2} -|\lambda |^{2} \Vert \pi _{R} F_{\lambda }(1)\Vert _{{\mathrm{HS}}}^{2} \\ &\quad +|\lambda |^{2} \sum _{n \ge 1} \left( \Vert \pi _{L} F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} +\Vert \pi _{R} F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} \right) \\ &= |\lambda |^{2} \Vert \pi _{L} F_{\lambda }(0)\Vert _{{\mathrm{HS}}}^{2} -|\lambda |^{2} \Vert \pi _{R} F_{\lambda }(1)\Vert _{{\mathrm{HS}}}^{2} +|\lambda |^{2}D. \end{aligned}$$

By definition, we have \(F_{\lambda }(0)=I\) and \(F_{\lambda }(1)=T_{\lambda }(0).\) These and the definition of \(T_{\lambda }(n)\) show

$$\begin{aligned} \Vert \pi _{L} F_{\lambda }(0) \Vert _{{\mathrm{HS}}}^{2} =1,\quad \Vert \pi _{R} F_{\lambda }(1) \Vert _{{\mathrm{HS}}}^{2} =|\lambda |^{-2} (|c_{0}|^{2} +|d_{0}|^{2})=|\lambda |^{-2}. \end{aligned}$$

By substituting these into the above equation and by the assumption \(|\lambda | \ne 1,\) we conclude

$$\begin{aligned} D=|\lambda |^{2} -1 +|\lambda |^{2}D, \quad \text{hence} \quad D=-1, \end{aligned}$$

which is a contradiction. Therefore, \(D=+\infty .\) Next, suppose that

$$\begin{aligned} E:=\sum _{n \le -1} \Vert F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} <+\infty . \end{aligned}$$

We note that \(E>0.\) We have

$$\begin{aligned} E&=\sum _{n \le -1} \Vert {\mathcal{C}}(n)F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} = \sum _{n \le -1} \Vert \lambda \pi _{L} F_{\lambda }(n-1) + \lambda \pi _{R} F_{\lambda }(n+1)\Vert _{{\mathrm{HS}}}^{2} \\ &= \sum _{n \le -1} |\lambda |^{2} \Vert \pi _{L}F_{\lambda }(n-1)\Vert _{{\mathrm{HS}}}^{2} + \sum _{n \le -1} |\lambda |^{2} \Vert \pi _{R} F_{\lambda }(n+1) \Vert _{{\mathrm{HS}}}^{2} \\ &= \sum _{n \le -2} |\lambda |^{2} \Vert \pi _{L}F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} + \sum _{n \le 0} |\lambda |^{2} \Vert \pi _{R} F_{\lambda }(n) \Vert _{{\mathrm{HS}}}^{2} \\ &= -|\lambda |^{2} \Vert \pi _{L}F_{\lambda }(-1)\Vert _{{\mathrm{HS}}}^{2}+|\lambda |^{2} \Vert \pi _{R}F_{\lambda }(0)\Vert _{{\mathrm{HS}}}^{2} \\ &\quad +|\lambda |^{2} \sum _{n \le -1} \left( \Vert \pi _{L}F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2}+\Vert \pi _{R} F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2} \right) \\ &=-|\lambda |^{2} \Vert \pi _{L}F_{\lambda }(-1)\Vert _{{\mathrm{HS}}}^{2}+|\lambda |^{2} \Vert \pi _{R}F_{\lambda }(0)\Vert _{{\mathrm{HS}}}^{2} +|\lambda |^{2}E. \end{aligned}$$

Since \(F_{\lambda }(0)=I,\) we have \(\Vert \pi _{R}F_{\lambda }(0)\Vert _{{\mathrm{HS}}}^{2}=1.\) By the definition of \(F_{\lambda }(-1)=T_{\lambda }(-1)^{-1}\) and its concrete form (50), we see

$$\begin{aligned} \Vert \pi _{L}F_{\lambda }(-1)\Vert _{{\mathrm{HS}}}^{2}=\Vert \pi _{L} T_{\lambda }(-1)^{-1}\Vert _{{\mathrm{HS}}}^{2} =|\lambda |^{-2} (|a_{0}|^{2}+|b_{0}|^{2}) =|\lambda |^{-2}. \end{aligned}$$

Hence, we have

$$\begin{aligned} E=-1+|\lambda |^{2}+|\lambda |^{2}E, \quad \text{hence} \quad E=-1, \end{aligned}$$

which is a contradiction. Therefore, we conclude \(E=+\infty .\) \(\square\)

Lemma 3.3

Let \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}).\) Then, the dimensions of the subspaces \(A_{L}(\lambda )\) and \(A_{R}(\lambda )\) in Theorem 1.3are less than or equal to 1.

Proof

Suppose that \(\dim A_{L}(\lambda )=2.\) Then, we have

$$\begin{aligned} \sum _{n \ge 1} \Vert F_{\lambda }(n)e_{L}\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty , \quad \sum _{n \ge 1} \Vert F_{\lambda }(n)e_{R}\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty . \end{aligned}$$

Summing these two sums gives \(\sum _{n \ge 1} \Vert F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2}<+\infty ,\) which contradicts Lemma 3.2. Thus, we conclude \(\dim A_{L}(\lambda ) \le 1.\) Next, suppose that \(\dim A_{R}(\lambda )=2.\) Then, we see

$$\begin{aligned} \sum _{n \le -1} \Vert F_{\lambda }(n)e_{L}\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty , \quad \sum _{n \le -1} \Vert F_{\lambda }(n)e_{R}\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty . \end{aligned}$$

Summing these two sums gives \(\sum _{n \le -1} \Vert F_{\lambda }(n)\Vert _{{\mathrm{HS}}}^{2}<+\infty ,\) which contradicts Lemma 3.2. Thus, we conclude \(\dim A_{R}(\lambda ) \le 1.\) \(\square\)

The function \(R(\lambda )(\delta _{0} \otimes u)\) is in \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\) for any \(u \in {\mathbb{C}}^{2}\) and we have by Theorem 1.2

$$\begin{aligned}&\Vert R(\lambda )(\delta _{0} \otimes u)\Vert ^{2} \\ &\quad =\sum _{n \in {\mathbb{Z}}} \Vert R(\lambda )(\delta _{0} \otimes u)(n)\Vert _{{\mathbb{C}}^{2}}^{2} = \sum _{n \in {\mathbb{Z}}} \Vert R_{\lambda }(n,0)u\Vert _{{\mathbb{C}}^{2}}^{2} \\ &\quad = \Vert {\mathrm{x}}_{0}(\lambda )u\Vert _{{\mathbb{C}}^{2}}^{2} + \sum _{n \ge 1} \Vert F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{L}(0) ] u\Vert _{{\mathbb{C}}^{2}}^{2} \\ &\qquad +\sum _{n \le -1} \Vert F_{\lambda }(n) [{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{R}(0) ] u\Vert _{{\mathbb{C}}^{2}}^{2} <+\infty . \end{aligned}$$
(88)

Since \(\dim A_{L} \le 1\) and \(\dim A_{R} \le 1\) by Lemma 3.3, we see

$$\begin{aligned} {\mathrm{rank}}[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{L}(0) ] \le 1,\quad {\mathrm{rank}}[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{R}(0) ] \le 1. \end{aligned}$$

Proof of Theorem 1.3

It is left to prove that, for \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}),\) the following holds:

$$\begin{aligned} {\mathrm{rank}}[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{L}(0) ] ={\mathrm{rank}}[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{R}(0) ]=1. \end{aligned}$$
(89)

To prove (89), suppose that \({\mathrm{rank}}[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{L}(0) ]=0.\) Then, \({\mathrm{x}}_{0}(\lambda )=-\lambda ^{-1} {\mathrm{z}}_{L}(0).\) By (88), we see

$$\begin{aligned} \sum _{n \le -1} \Vert F_{\lambda }(n) [{\mathrm{z}}_{L}(0) -{\mathrm{z}}_{R}(0)]u\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty \end{aligned}$$

for any \(u \in {\mathbb{C}}^{2}.\) However, since \(\det ({\mathrm{z}}_{L}(0) -{\mathrm{z}}_{R}(0))= -\frac{\triangle _{0}}{a_{0}d_{0}},\) the matrix \({\mathrm{z}}_{L}(0) -{\mathrm{z}}_{R}(0)\) is non-singular. Thus, for any \(w \in {\mathbb{C}}^{2},\) there exists u, such that \(({\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0))u=w.\) Therefore, we have \(\sum _{n \le -1} \Vert F_{\lambda }(n)w\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty\) for any \(w \in {\mathbb{C}}^{2}.\) This contradicts the fact that \(\dim A_{R}(\lambda )=1\) in Lemma 3.3. Hence, \({\mathrm{rank}}[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{L}(0) ]=1.\) Next, suppose that \({\mathrm{rank}}[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{R}(0) ]=0.\) Then, \({\mathrm{x}}_{0}(\lambda )=-\lambda ^{-1} {\mathrm{z}}_{R}(0).\) By (88), we see

$$\begin{aligned} \sum _{n \ge 1} \Vert F_{\lambda }(n) [{\mathrm{z}}_{L}(0) -{\mathrm{z}}_{R}(0)]u\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty \end{aligned}$$

for any \(u \in {\mathbb{C}}^{2}.\) As above, for any \(w \in {\mathbb{C}}^{2},\) there exists u, such that \(({\mathrm{z}}_{L}(0)-{\mathrm{z}}_{R}(0))u=w.\) Thus, we have \(\sum _{n \ge 1} \Vert F_{\lambda }(n)w\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty\) for any \(w \in {\mathbb{C}}^{2}.\) This contradicts the fact that \(\dim A_{L}(\lambda )=1\) in Lemma 3.3. Hence, \({\mathrm{rank}}[{\mathrm{x}}_{0}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{R}(0) ]=1.\) \(\square\)

To prove Theorem 1.4, we use the following lemma.

Lemma 3.4

Let \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1})\) and \(u \in {\mathbb{C}}^{2}\) be arbitrary. Then, a vector \(w \in {\mathbb{C}}^{2}\) satisfies

$$\begin{aligned}&\sum _{n \ge 1} \Vert F_{\lambda }(n) [w +\lambda ^{-1} {\mathrm{z}}_{L}(0)u] \Vert _{{\mathbb{C}}^{2}}^{2}<+\infty , \\ &\quad {\mathrm{and}} \quad \sum _{n \le -1} \Vert F_{\lambda }(n) [w +\lambda ^{-1} {\mathrm{z}}_{R}(0)u] \Vert _{{\mathbb{C}}^{2}}^{2}<+\infty \end{aligned}$$
(90)

if and only if \(w={\mathrm{x}}_{0}(\lambda )u.\)

Proof

Let \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1})\) and \(u,w \in {\mathbb{C}}^{2}.\) We define \(\Psi \in {\mathrm{Map}}\,({\mathbb{Z}},{\mathbb{C}}^{2})\) by

$$\begin{aligned} \Psi (n)= \left\{ \begin{array}{ll} w &{} (n=0), \\ F_{\lambda }(n)[w +\lambda ^{-1}{\mathrm{z}}_{L}(0) u] &{}\quad (n \ge 1), \\ F_{\lambda }(n)[w+\lambda ^{-1}{\mathrm{z}}_{R}(0) u] &{}\quad (n \le -1). \end{array}\right. \end{aligned}$$
(91)

By Proposition 2.3, we have \(\Psi =V_{\lambda }(\delta _{0} \otimes u)+\Phi _{\lambda } (w).\) Therefore, again by Proposition 2.3, we have

$$\begin{aligned} (U({\mathcal{C}})-\lambda )\Psi =\delta _{0} \otimes u. \end{aligned}$$
(92)

Now, suppose that w satisfies (90). Then, \(\Psi\) defined above is in \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Therefore, by applying \(R(\lambda )\) to (92), we have

$$\begin{aligned} \Psi (n)=R(\lambda )(\delta _{0} \otimes u)(n)=R_{\lambda }(n,0)u. \end{aligned}$$

In particular, we have \(w=\Psi (0)=R_{\lambda }(0,0)u={\mathrm{x}}_{0}(\lambda )u\) by the definition (72) of \({\mathrm{x}}_{0}(\lambda ).\) Conversely, if \(w={\mathrm{x}}_{0}(\lambda )u,\) we have \(\Psi (n)=R_{\lambda }(n,0)u=R(\lambda )(\delta _{0} \otimes u)(n)\) by Theorem 1.2 which is in \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Hence, \(w={\mathrm{x}}_{0}(\lambda )u\) satisfies (90). \(\square\)

Proof of Theorem 1.4

Let \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}).\) We first show that the unit vectors \({\mathrm{v}}_{+}(\lambda ),\) \({\mathrm{v}}_{-}(\lambda )\) form a basis in \({\mathbb{C}}^{2}.\) Indeed, suppose that \(a {\mathrm{v}}_{+}(\lambda )+b {\mathrm{v}}_{-}(\lambda )=0\) with \(a,b \in {\mathbb{C}}\) not simultaneously zero. If \(a =0\), then \(b{\mathrm{v}}_{-}(\lambda )=0\), and hence \(b=0\), because \({\mathrm{v}}_{-}(\lambda )\) is a nonzero vector in \({\mathbb{C}}^{2}.\) Thus, \(a \ne 0\), and thus, we can write \({\mathrm{v}}_{+}(\lambda )=c {\mathrm{v}}_{-}(\lambda )\) with \(c=-b/a.\) Then, by the property (14), the function \(\Phi _{\lambda }({\mathrm{v}}_{+}(\lambda ))\) is in \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\), and hence, \(\lambda\) is in the spectrum \(\sigma (U),\) an eigenvalue, of the unitary operator \(U=U({\mathcal{C}})|_{\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})}.\) This is a contradiction, since we have assumed \(\lambda \not \in S^{1}.\) Thus, \({\mathrm{v}}_{\pm }(\lambda )\) forms a basis of \({\mathbb{C}}^{2}\), and hence, the matrix

$$\begin{aligned} \left[ \begin{array}{cc} 1 &{} - \overline{m(\lambda )} \\ m(\lambda ) &{} -1 \end{array}\right] , \quad m(\lambda )=\langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{aligned}$$

is non-singular. We define \(a_{L}(\lambda ), b_{L}(\lambda ), a_{R}(\lambda ),b_{R}(\lambda ) \in {\mathbb{C}}\) by

$$\begin{aligned} \left[ \begin{array}{c} a_{L}(\lambda ) \\ b_{L}(\lambda ) \end{array}\right]&=\lambda ^{-1} \left[ \begin{array}{cc} 1 &{} - \overline{m(\lambda )} \\ m(\lambda ) &{} -1 \end{array}\right] ^{-1} \left[ \begin{array}{c} \langle \,{\mathrm{z}}_{L}(0)e_{L}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ \langle \,{\mathrm{z}}_{L}(0)e_{L}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{array}\right] , \\ &= -\frac{1}{\lambda (1-|m(\lambda )|^{2})} \\ &\quad \times \left[ \begin{array}{c} \overline{m(\lambda )} \langle \,{\mathrm{z}}_{L}(0)e_{L},{\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} -\langle \,{\mathrm{z}}_{L}(0)e_{L},{\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ \langle \,{\mathrm{z}}_{L}(0)e_{L},{\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} -m(\lambda )\langle \,{\mathrm{z}}_{L}(0)e_{L},{\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{array}\right] , \\ \left[ \begin{array}{c} a_{R}(\lambda ) \\ b_{R}(\lambda ) \end{array}\right]&=-\lambda ^{-1} \left[ \begin{array}{cc} 1 &{} - \overline{m(\lambda )} \\ m(\lambda ) &{} -1 \end{array}\right] ^{-1} \left[ \begin{array}{c} \langle \,{\mathrm{z}}_{R}(0)e_{R}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ \langle \,{\mathrm{z}}_{R}(0)e_{R}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{array}\right] \\ &= \frac{1}{\lambda (1-|m(\lambda )|^{2})} \\ &\quad \times \left[ \begin{array}{c} \overline{m(\lambda )} \langle \,{\mathrm{z}}_{R}(0)e_{R},{\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} -\langle \,{\mathrm{z}}_{R}(0)e_{R},{\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ \langle \,{\mathrm{z}}_{R}(0)e_{R},{\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} - m(\lambda )\langle \,{\mathrm{z}}_{R}(0)e_{R},{\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{array}\right] . \end{aligned}$$
(93)

Then, we claim that

$$\begin{aligned} {\mathrm{x}}_{0}(\lambda )e_{L}&=b_{L}(\lambda ){\mathrm{v}}_{-}(\lambda )= -\lambda ^{-1} {\mathrm{z}}_{L}(0)e_{L} +a_{L}(\lambda ){\mathrm{v}}_{+}(\lambda ),\\ {\mathrm{x}}_{0}(\lambda )e_{R}&= a_{R}(\lambda ) {\mathrm{v}}_{+}(\lambda ) = -\lambda ^{-1} {\mathrm{z}}_{R}(0)e_{R}+b_{R}(\lambda ) {\mathrm{v}}_{-}(\lambda ). \end{aligned}$$
(94)

From the definition (93) and that \(\Vert {\mathrm{v}}_{\pm }(\lambda )\Vert _{{\mathbb{C}}^{2}}=1,\) we have

$$\begin{aligned}&\left[ \begin{array}{c} \langle \,{\mathrm{z}}_{L}(0)e_{L}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ \langle \,{\mathrm{z}}_{L}(0)e_{L}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{array}\right] \\ &\quad = \lambda \left[ \begin{array}{cc} 1 &{} - \overline{m(\lambda )} \\ m(\lambda ) &{} -1 \end{array}\right] \left[ \begin{array}{c} a_{L}(\lambda ) \\ b_{L}(\lambda ) \end{array}\right] \\ &\quad = \left[ \begin{array}{c} \lambda a_{L}(\lambda ) -\lambda \langle \,{\mathrm{v}}_{-}(\lambda ),{\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} b_{L}(\lambda ) \\ \lambda \langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} a_{L}(\lambda ) -\lambda b_{L}(\lambda ) \end{array}\right] \\ &\quad = \left[ \begin{array}{c} \langle \,\lambda a_{L}(\lambda ) {\mathrm{v}}_{+}(\lambda ) -\lambda b_{L}(\lambda ) {\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ \langle \,\lambda a_{L}(\lambda ) {\mathrm{v}}_{+}(\lambda ) -\lambda b_{L}(\lambda ) {\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{array}\right] , \\ &\left[ \begin{array}{c} \langle \,{\mathrm{z}}_{R}(0)e_{R}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ \langle \,{\mathrm{z}}_{R}(0)e_{R}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{array}\right] \\ &\quad = -\lambda \left[ \begin{array}{cc} 1 &{} - \overline{m(\lambda )} \\ m(\lambda ) &{} -1 \end{array}\right] \left[ \begin{array}{c} a_{R}(\lambda ) \\ b_{R}(\lambda ) \end{array}\right] \\ &\quad = \left[ \begin{array}{c} -\lambda a_{R}(\lambda ) +\lambda \langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} b_{R}(\lambda ) \\ -\lambda \langle \,{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}}a_{R}(\lambda ) +\lambda b_{R}(\lambda ) \end{array}\right] \\ &\quad = \left[ \begin{array}{c} \langle \,-\lambda a_{R}(\lambda ) {\mathrm{v}}_{+}(\lambda ) +\lambda b_{R}(\lambda ) {\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ \langle \,-\lambda a_{R}(\lambda ) {\mathrm{v}}_{+}(\lambda ) +\lambda b_{R}(\lambda ) {\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \end{array}\right] . \end{aligned}$$
(95)

Therefore, we have

$$\begin{aligned}&\langle \,\lambda a_{L}(\lambda ) {\mathrm{v}}_{+}(\lambda ) -\lambda b_{L}(\lambda ) {\mathrm{v}}_{-}(\lambda ) -{\mathrm{z}}_{L}(0) e_{L}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = \langle \,\lambda a_{L}(\lambda ) {\mathrm{v}}_{+}(\lambda ) -\lambda b_{L}(\lambda ) {\mathrm{v}}_{-}(\lambda ) -{\mathrm{z}}_{L}(0) e_{L}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} =0, \\ &\langle \,-\lambda a_{R}(\lambda ) {\mathrm{v}}_{+}(\lambda ) +\lambda b_{R}(\lambda ) {\mathrm{v}}_{-}(\lambda ) -{\mathrm{z}}_{R}(0) e_{R}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = \langle \,-\lambda a_{R}(\lambda ) {\mathrm{v}}_{+}(\lambda ) +\lambda b_{R}(\lambda ) {\mathrm{v}}_{-}(\lambda ) -{\mathrm{z}}_{R}(0) e_{R}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} =0. \\ \end{aligned}$$

Since \(\{{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda )\}\) is a basis of \({\mathbb{C}}^{2},\) we obtain

$$\begin{aligned} b_{L}(\lambda ) {\mathrm{v}}_{-}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{L}(0)e_{L}&=a_{L}(\lambda ) {\mathrm{v}}_{+}(\lambda ), \\ a_{R}(\lambda ) {\mathrm{v}}_{+}(\lambda ) +\lambda ^{-1} {\mathrm{z}}_{R}(0)e_{R}&=b_{R}(\lambda ) {\mathrm{v}}_{-}(\lambda ). \end{aligned}$$
(96)

Therefore, by (96), the vector \(w =b_{L}(\lambda ){\mathrm{v}}_{-}(\lambda )\) satisfies

$$\begin{aligned}&\sum _{n \ge 1}\Vert F_{\lambda }(n)[w+\lambda ^{-1}{\mathrm{z}}_{L}(0)e_{L}]\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty , \\ &\sum _{n \le -1}\Vert F_{\lambda }(n)[w+\lambda ^{-1}{\mathrm{z}}_{R}(0)e_{L}\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty , \end{aligned}$$

since we have \({\mathrm{z}}_{R}(0)e_{L}=0\) and the vector \({\mathrm{v}}_{-}(\lambda )\) satisfies (14). By Lemma 3.4, we have \(b_{L}(\lambda ){\mathrm{v}}_{-}(\lambda )={\mathrm{x}}_{0}(\lambda )e_{L}.\) This and (96) show the first line of (94). Similarly, by (96), the vector \(w=a_{R}(\lambda ) {\mathrm{v}}_{+}(\lambda )\) satisfies

$$\begin{aligned}&\sum _{n \ge 1} \Vert F_{\lambda }(n) [w +\lambda ^{-1} {\mathrm{z}}_{L}(0)e_{R}]\Vert _{{\mathbb{C}}^{2}}^{2}<+\infty ,\\ &\sum _{n \le -1} \Vert F_{\lambda }(n) [w+\lambda ^{-1} {\mathrm{z}}_{R}(0)e_{R}] \Vert _{{\mathbb{C}}^{2}}^{2}<+\infty , \end{aligned}$$

since we have \({\mathrm{z}}_{L}(0)e_{R}=0\) and the vector \({\mathrm{v}}_{+}(\lambda )\) satisfies (14). This and (96) show the second line of (94). For a unit vector \(u \in {\mathbb{C}}^{2},\) we take any unit vector \(u^{\perp }\) perpendicular to u. Since \(\{{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp }\},\) \(\{{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{-}(\lambda )^{\perp }\}\) are orthonormal bases of \({\mathbb{C}}^{2},\) we see

$$\begin{aligned} 1&=\Vert {\mathrm{v}}_{-}(\lambda )\Vert _{{\mathbb{C}}^{2}}^{2}=|\langle \,{\mathrm{v}}_{-}(\lambda ),{\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}}|^{2} +|\langle \,{\mathrm{v}}_{-}(\lambda ),{\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2} \\ &= \Vert {\mathrm{v}}_{+}(\lambda )\Vert _{{\mathbb{C}}^{2}}^{2}= |\langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}}|^{2} +|\langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2}, \end{aligned}$$

and hence

$$\begin{aligned} 1- |m(\lambda )|^{2}=|\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2} =|\langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2}. \end{aligned}$$

Using this and (93), we have

$$\begin{aligned}&\lambda b_{L}(\lambda ) \\ &\quad = - \frac{1}{|\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2}} \\ &\qquad \times \left[ \langle \,{\mathrm{z}}_{L}(0) e_{L}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} -\langle \,{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \langle \,{\mathrm{z}}_{L}(0) e_{L}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathrm{C}}^{2}} \right] \\ &\quad = -\frac{1}{|\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2}} \\ &\qquad \times \langle \,{\mathrm{z}}_{L}(0)e_{L} -\langle \,{\mathrm{z}}_{L}(0) e_{L}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} {\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = -\frac{1}{|\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2}} \langle \,{\mathrm{z}}_{L}(0)e_{L}, {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}} \langle \,{\mathrm{v}}_{+}(\lambda )^{\perp }, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = -\frac{\langle \,{\mathrm{z}}_{L}(0)e_{L}, {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}}{\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}}. \end{aligned}$$
(97)

Similarly, we have

$$\begin{aligned}&\lambda a_{R}(\lambda ) \\ &\quad = \frac{1}{|\langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2}} \\ &\qquad \times \left[ \langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \langle \,{\mathrm{z}}_{R}(0) e_{R}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} -\langle \,{\mathrm{z}}_{R}(0)e_{R}, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \right] \\ &\quad = -\frac{1}{|\langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2}} \\ &\qquad \langle \,{\mathrm{z}}_{R}(0) e_{R} -\langle \,{\mathrm{z}}_{R}(0) e_{R}, {\mathrm{v}}_{-}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}}{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = -\frac{1}{|\langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}|^{2}} \langle \,{\mathrm{z}}_{R}(0) e_{R}, {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}} \langle \,{\mathrm{v}}_{-}(\lambda )^{\perp }, {\mathrm{v}}_{+}(\lambda ) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = -\frac{\langle \,{\mathrm{z}}_{R}(0) e_{R}, {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}} }{\langle \,{\mathrm{v}}_{+}(\lambda ),{\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}}. \end{aligned}$$
(98)

Equations (94), (97) and (98) complete the proof of (15) in Theorem 1.4. \(\square\)

4 Integral representation for the Green function

As is seen in the previous sections, the function \({\mathrm{x}}_{0} :{\mathbb{C}} {\setminus } \sigma (U) \rightarrow {\mathrm{M}}_{2}({\mathbb{C}})\) introduced in (72) plays one of the central roles in the series of our results. However, since

$$\begin{aligned} {\mathrm{x}}_{0}(\lambda )^{*}=-\overline{\lambda }^{-1} (I+\overline{\lambda }^{-1} {\mathrm{x}}_{0}(1/\overline{\lambda })) \quad (\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup \sigma (U)), \end{aligned}$$
(99)

it does not seem that the real part \([{\mathrm{x}}_{0}(\lambda ) +{\mathrm{x}}_{0}(\lambda )^{*}]/2\) has nice properties as the m-Carathéodory functions have. The definition of m-Carathéodory functions is given in the statement of Lemma 4.1 below. Instead of using \({\mathrm{x}}_{0},\) we use the following function:

$$\begin{aligned} {\mathrm{x}}(\lambda )=I+2\lambda {\mathrm{x}}_{0}(\lambda ). \end{aligned}$$
(100)

By (99), we have

$$\begin{aligned} {\mathrm{Re}}\,{\mathrm{x}}(\lambda ) :=\frac{1}{2} [{\mathrm{x}}(\lambda )+{\mathrm{x}}(\lambda )^{*}]= \lambda {\mathrm{x}}_{0}(\lambda ) - \overline{\lambda }^{-1} {\mathrm{x}}_{0}(1/\overline{\lambda }), \end{aligned}$$
(101)

which shows the second equality in (23).

Lemma 4.1

The function \({\mathrm{x}} :{\mathbb{C}} {\setminus } \sigma (U) \rightarrow {\mathrm{M}}_{2}({\mathbb{C}})\) is an m-Carathéodory function in the sense that it is a holomorphic function on the unit disc with a positive real part and \({\mathrm{x}}(0)=I.\)

Proof

Since \({\mathrm{x}}_{0}\) is holomorphic on \({\mathbb{C}} {\setminus } \sigma (U),\) \({\mathrm{x}}\) is also holomorphic on \({\mathbb{C}} {\setminus } \sigma (U),\) and in particular, on the unit disc. By definition, we see \({\mathrm{x}}(0)=I.\) We set

$$\begin{aligned} K(\lambda )=(U-\lambda )^{-1}(U+\lambda )=R(\lambda )(U-\lambda +2\lambda )=I+2\lambda R(\lambda ) \quad (\lambda \in {\mathbb{C}} {\setminus } \sigma (U)). \end{aligned}$$
(102)

For \(\lambda \in {\mathbb{C}} {\setminus } \sigma (U)\) and \(u ,v \in {\mathbb{C}}^{2},\) we have

$$\begin{aligned} \langle \,{\mathrm{x}}_{0}(0)u,v \,\rangle _{{\mathbb{C}}^{2}} =\langle \,R_{\lambda }(0,0)u,v \,\rangle _{{\mathbb{C}}^{2}} =\langle \,R(\lambda )(\delta _{0} \otimes u),\delta _{0} \otimes v \,\rangle , \end{aligned}$$

and hence

$$\begin{aligned} \langle \,{\mathrm{x}}(\lambda )u,v \,\rangle _{{\mathbb{C}}^{2}}&=\langle \,[I+2\lambda R(\lambda )] \delta _{0} \otimes u, \delta _{0} \otimes v \,\rangle \\ &=\langle \,K(\lambda ) \delta _{0} \otimes u,\delta _{0} \otimes v \,\rangle . \end{aligned}$$
(103)

This shows that

$$\begin{aligned} \langle \,{\mathrm{x}}(\lambda )^{*} u,v \,\rangle _{{\mathbb{C}}^{2}}&= \langle \,u,{\mathrm{x}}(\lambda )v \,\rangle _{{\mathbb{C}}^{2}} =\overline{\langle \,{\mathrm{x}}(\lambda )v,u \,\rangle _{{\mathbb{C}}^{2}}} \\ &= \overline{\langle \,K(\lambda ) \delta _{0} \otimes v, \delta _{0} \otimes u \,\rangle } = \langle \,\delta _{0} \otimes u, K(\lambda ) \delta _{0} \otimes v \,\rangle \\ &= \langle \,K(\lambda )^{*}(\delta _{0} \otimes u),\delta _{0} \otimes v \,\rangle \\ &= \langle \,[(U^{*}+\overline{\lambda })(U^{*}-\overline{\lambda })^{-1}] \delta _{0} \otimes u,\delta _{0} \otimes v \,\rangle . \end{aligned}$$

We note that

$$\begin{aligned}&\frac{1}{2}(K(\lambda ) +K(\lambda )^{*}) \\ &\quad = \frac{1}{2} [(U-\lambda )^{-1}(U+\lambda ) +(U^{*} +\overline{\lambda }) (U^{*} -\overline{\lambda })^{-1}] \\ &\quad = \frac{1}{2} (U-\lambda )^{-1} [(U+\lambda )(U^{*}-\overline{\lambda }) +(U-\lambda ) (U^{*} +\overline{\lambda })](U^{*}-\overline{\lambda })^{-1} \\ &\quad = (1-|\lambda |^{2}) (U-\lambda )^{-1}(U^{*}-\overline{\lambda })^{-1} = (1-|\lambda |^{2}) R(\lambda )^{*}R(\lambda ). \end{aligned}$$
(104)

Therefore, the real part \({\mathrm{Re}}\,{\mathrm{x}}(\lambda )\) of \({\mathrm{x}}(\lambda )\) satisfies

$$\begin{aligned}&{\mathrm{Re}}\,{\langle \,{\mathrm{x}}(\lambda )u,u \,\rangle _{{\mathbb{C}}^{2}}} \\ &\quad = \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(\lambda )]u,u \,\rangle _{{\mathbb{C}}^{2}} = \langle \,[(K(\lambda )+K(\lambda )^{*})/2](\delta _{0} \otimes u), \delta _{0} \otimes u \,\rangle \\ &\quad = (1-|\lambda |^{2}) \langle \,R(\lambda )^{*}R(\lambda ) \delta _{0} \otimes u,\delta _{0} \otimes u \,\rangle \\ &\quad = (1-|\lambda |^{2}) \Vert R(\lambda )(\delta _{0} \otimes u)\Vert ^{2} >0 \end{aligned}$$

if \(|\lambda |<1\) and \(u \ne 0.\) This completes the proof. \(\square\)

Therefore, there exists a unique positive \(2\times 2\)-matrix-valued measure

$$\begin{aligned} \Sigma = \left[ \begin{array}{cc} \mu _{L} &{} \overline{\alpha } \\ \alpha &{} \mu _{R} \end{array}\right] , \end{aligned}$$
(105)

on \(S^{1},\) where \(\mu _{L}\) and \(\mu _{R}\) are probability measures and \(\alpha\) is a complex measure on \(S^{1},\) such that we have the following Herglotz representation:

$$\begin{aligned} {\mathrm{x}}(\lambda )=\int _{S^{1}} \frac{\zeta +\lambda }{\zeta -\lambda }\,{\mathrm{d}}\Sigma (\zeta ),\quad \Sigma (S^{1}) =I. \end{aligned}$$
(106)

We refer the readers to [8, 12] for the proof of the above fact. We denote C(X) the space of complex-valued continuous functions on a compact topological space X with the norm defined by \(\Vert f\Vert _{C(X)}=\sup _{x \in X}|f(x)|.\) The positive-matrix-valued measure \(\Sigma\) is characterized by

$$\begin{aligned} \int _{S^{1}} h(\zeta ) \langle \,{\mathrm{d}}\Sigma (\zeta )u,u \,\rangle _{{\mathbb{C}}^{2}}&=\lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] u,u \,\rangle _{{\mathbb{C}}^{2}} \,{\mathrm{d}}\ell (\zeta ) \end{aligned}$$
(107)

for any \(h \in C(S^{1})\) and \(u \in {\mathbb{C}}^{2},\) where \({\mathrm{d}}\ell (\zeta )\) denotes the Lebesgue measure on \(S^{1}\) with the unit total mass. In what follows, we choose the counterclockwise orientation on the unit circle \(S^{1}.\) Then, we can write

$$\begin{aligned} \int _{S^{1}} h(\zeta )\, {\mathrm{d}}\ell (\zeta )=\frac{1}{2\pi i}\int _{S^{1}} h(\zeta )\,\frac{{\mathrm{d}}\zeta }{\zeta }=\frac{1}{2\pi } \int _{-\pi }^{\pi } h({\mathrm{e}}^{i\theta })\,{\mathrm{d}}\theta . \end{aligned}$$

To prove Theorem 1.5, (1), we consider, for a fixed \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}),\) the following function:

$$\begin{aligned} H_{f}(\lambda ) =\langle \,K(\lambda )f,f \,\rangle \quad (\lambda \in {\mathbb{C}}^{2}, \, |\lambda |<1). \end{aligned}$$
(108)

The function \(H_{f}\) is holomorphic on the unit disc and, by (104), its real part is

$$\begin{aligned} {\mathrm{Re}}\,H_{f}(\lambda )=(1-|\lambda |^{2}) \Vert R(\lambda )f\Vert ^{2}. \end{aligned}$$
(109)

Hence, \(H_{f}\) is a scalar Carathéodory function with \(H_{f}(0)=\Vert f\Vert ^{2}.\) Therefore, there exists a unique positive measure \(\nu _{f}\) on \(S^{1}\), such that

$$\begin{aligned} H_{f}(\lambda )=\int _{S^{1}} \frac{\zeta +\lambda }{\zeta -\lambda }\,{\mathrm{d}}\nu _{f}(\zeta ), \quad \nu _{f}(S^{1}) =\Vert f\Vert ^{2}, \end{aligned}$$
(110)

and the measure \(\nu _{f}\) is the unique measure satisfying

$$\begin{aligned} \int _{S^{1}} h(\zeta )\,{\mathrm{d}}\nu _{f}(\zeta ) =\lim _{r \uparrow 1}\int _{S^{1}} h(\zeta ) {\mathrm{Re}}\,H_{f}(r\zeta )\, {\mathrm{d}}\ell (\zeta ) \end{aligned}$$
(111)

for any \(h \in C(S^{1}).\) In what follows, we prove the formula (17) with the matrix-valued measure appeared in (105) and (106). We need the following.

Lemma 4.2

The matrix-valued functions \(F_{\lambda }(n),\) \(F_{1/\overline{\lambda }}(n)^{*}\) are holomorphic in \(\lambda \in {\mathbb{C}} {\setminus } \{0\}\) for any fixed \(n \in {\mathbb{Z}}.\)

Proof

By (5) and (52), the matrix-valued functions \(T_{\lambda }(n),\) \(T_{1/\overline{\lambda }}(n)^{*}\) are non-singular and holomorphic in \(\lambda \in {\mathbb{C}} {\setminus } \{0\}\) for any \(n \in {\mathbb{Z}}.\) Therefore, their inverses are also holomorphic. By the definition (6), \(F_{\lambda }(n)\) is defined as a product of some \(T_{\lambda }(m)\)’s or their inverses. Thus, \(F_{\lambda }(n)\) is holomorphic in \(\lambda \in {\mathbb{C}} {\setminus } \{0\}.\) Similarly, \(F_{1/\overline{\lambda }}(n)^{*}\) is defined as a product of some \(T_{1/\overline{\lambda }}(m)^{*}\)’s or their inverses. Therefore, \(F_{1/\overline{\lambda }}(n)^{*}\) is also holomorphic. \(\square\)

Lemma 4.3

Let \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) We set

$$\begin{aligned} h_{f}(\lambda ):=H_{f}(\lambda ) -2\lambda \langle \,{\mathrm{x}}_{0}(\lambda )\widehat{f}^{{\mathcal{C}}}(\lambda ), \widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

Then, \(h_{f}(\lambda )\) is holomorphic on \({\mathbb{C}} {\setminus } \{0\}\), and we have \({\mathrm{Re}}\,h_{f}(\zeta )=\Vert \widehat{f}^{{\mathcal{C}}}(\zeta )\Vert _{{\mathbb{C}}^{2}}^{2}\) for \(\zeta \in S^{1}.\)

Proof

We first note that the statement on the analyticity of the function \(h_{f}(\lambda )\) is non-trivial, because it involves, at first glance, the matrix-valued function \({\mathrm{x}}_{0}(\lambda )\) which is holomorphic only on \({\mathbb{C}} {\setminus } \sigma (U).\) By the identity \(K(\lambda )=I+2\lambda R(\lambda )\) and the definition (16) of the QW-Fourier transform \(\widehat{f}^{{\mathcal{C}}},\) we see

$$\begin{aligned}&\frac{1}{2}h_{f}(\lambda ) \\ &\quad = \frac{1}{2}\Vert f\Vert ^{2} +\lambda \sum _{m,n \in {\mathbb{Z}}} \langle \,R_{\lambda }(n,m)f(m),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\qquad -\lambda \sum _{m,n \in {\mathbb{Z}}} \langle \,{\mathrm{x}}_{0}(\lambda ) F_{1/\overline{\lambda }}(m)^{*}f(m),F_{\lambda }(n)^{*}f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = \frac{1}{2}\Vert f\Vert ^{2} \\ &\qquad +\sum _{m,n \in {\mathbb{Z}}} \langle \,\lambda [R_{\lambda }(n,m)-F_{\lambda }(n) {\mathrm{x}}_{0}(\lambda ) F_{1/\overline{\lambda }}(m)^{*}] f(m),f(n) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(112)

The sums appeared in (112) are all finite sums, because f has a finite support. By Theorem 1.2, we have

$$\begin{aligned}&\lambda [R_{\lambda }(n,m) -F_{\lambda }(n) {\mathrm{x}}_{0}(\lambda ) F_{1/\overline{\lambda }}(m)^{*}] \\ &\quad = \left\{ \begin{array}{ll} F_{\lambda }(n) {\mathrm{z}}_{L}(0) F_{1/\overline{\lambda }} (m)^{*} &{}\quad (m \le n-1), \\ F_{\lambda }(n) {\mathrm{z}}_{R}(0) F_{1/\overline{\lambda }} (m)^{*} &{}\quad (m-1 \ge n), \\ F_{\lambda }(n) {\mathrm{z}}_{L}(0) F_{1/\overline{\lambda }} (n)^{*} - {\mathrm{z}}_{L}(n) &{} \\ \quad =F_{\lambda }(n) {\mathrm{z}}_{R}(0) F_{1/\overline{\lambda }} (n)^{*} - {\mathrm{z}}_{R}(n) &{}\quad (n=m). \end{array}\right. \end{aligned}$$

Substituting these formulas into (112), we have

$$\begin{aligned} \frac{1}{2} h_{f}(\lambda )&= \sum _{n \in {\mathbb{Z}}} \sum _{m \le n-1} \langle \,F_{\lambda }(n) {\mathrm{z}}_{L}(0) F_{1/\overline{\lambda }}(m)^{*}f(m),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad + \sum _{n \in {\mathbb{Z}}} \sum _{m \ge n+1} \langle \,F_{\lambda }(n) {\mathrm{z}}_{R}(0) F_{1/\overline{\lambda }}(m)^{*} f(m), f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad + \sum _{n \in {\mathbb{Z}}} \langle \,F_{\lambda }(n) {\mathrm{z}}_{L}(0) F_{1/\overline{\lambda }}(n)^{*}f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad -\sum _{n \in {\mathbb{Z}}} \langle \,{\mathrm{z}}_{L}(n) f(n), f(n) \,\rangle _{{\mathbb{C}}^{2}} +\frac{1}{2} \Vert f\Vert ^{2}. \end{aligned}$$
(113)

From (113) and Lemma 4.2, the function \(h_{f}(\lambda )\) is holomorphic in \(\lambda \in {\mathbb{C}} {\setminus }\{0\}.\) By taking the complex conjugate of (113), we see

$$\begin{aligned} \frac{1}{2} \overline{h_{f}(\lambda )}&= \sum _{n \in {\mathbb{Z}}} \sum _{m \le n-1} \langle \,F_{1/\overline{\lambda }}(m) {\mathrm{z}}_{L}(0)^{*} F_{\lambda }(n)^{*}f(n),f(m) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +\sum _{n \in {\mathbb{Z}}} \sum _{m \ge n+1} \langle \,F_{1/\overline{\lambda }}(m) {\mathrm{z}}_{R}(0)^{*}F_{\lambda }(n)^{*} f(n), f(m) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +\sum _{n \in {\mathbb{Z}}} \langle \,F_{1/\overline{\lambda }}(n) {\mathrm{z}}_{L}(0)^{*} F_{\lambda }(n)^{*} f(n), f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad -\sum _{n \in {\mathbb{Z}}} \langle \,{\mathrm{z}}_{L}(n)^{*}f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}} +\frac{1}{2}\Vert f\Vert ^{2}. \end{aligned}$$
(114)

By the item (6) in Lemma 2.4, we have \({\mathrm{z}}_{L}(0)^{*}=I-{\mathrm{z}}_{R}(0),\) \({\mathrm{z}}_{R}(0)^{*}=I-{\mathrm{z}}_{L}(0),\) and \({\mathrm{z}}_{L}(n)^{*}=I-{\mathrm{z}}_{R}(n).\) Substituting these identities into (114), we obtain

$$\begin{aligned} \frac{1}{2} \overline{h_{f}(\lambda )}&= \sum _{n \in {\mathbb{Z}}} \sum _{m \le n-1} \langle \,F_{\lambda }(n)^{*}f(n),F_{1/\overline{\lambda }}(m)^{*}f(m) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \sum _{m \le n-1} \langle \,F_{1/\overline{\lambda }}(m){\mathrm{z}}_{R}(0) F_{\lambda }(n)^{*}f(n),f(m) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +\sum _{n \in {\mathbb{Z}}} \sum _{m \ge n+1} \langle \,F_{\lambda }(n)^{*}f(n),F_{1/\overline{\lambda }}(m)^{*}f(m) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \sum _{m \ge n+1} \langle \,F_{1/\overline{\lambda }}(m){\mathrm{z}}_{L}(0) F_{\lambda }(n)^{*}f(n),f(m) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +\sum _{n \in {\mathbb{Z}}} \langle \,F_{\lambda }(n)^{*}f(n),F_{1/\overline{\lambda }}(n)^{*}f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \langle \,F_{1/\overline{\lambda }}(n){\mathrm{z}}_{R}(0) F_{\lambda }(n)^{*}f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad + \sum _{n \in {\mathbb{Z}}} \langle \,{\mathrm{z}}_{R}(n) f(n), f(n) \,\rangle _{{\mathbb{C}}^{2}} -\frac{1}{2}\Vert f\Vert ^{2}. \end{aligned}$$
(115)

Gathering the first, third, and fifth lines in (115) and using (16), we obtain

$$\begin{aligned} \frac{1}{2} \overline{h_{f}(\lambda )}&= \langle \,\widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }), \widehat{f}^{{\mathcal{C}}}(\lambda ) \,\rangle _{{\mathbb{C}}} -\frac{1}{2}\Vert f\Vert ^{2} +\sum _{n \in {\mathbb{Z}}} \langle \,{\mathrm{z}}_{R}(n) f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \sum _{m \le n-1} \langle \,F_{1/\overline{\lambda }}(m){\mathrm{z}}_{R}(0) F_{\lambda }(n)^{*}f(n),f(m) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \sum _{m \ge n+1} \langle \,F_{1/\overline{\lambda }}(m){\mathrm{z}}_{L}(0) F_{\lambda }(n)^{*}f(n),f(m) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \langle \,F_{1/\overline{\lambda }}(n){\mathrm{z}}_{R}(0) F_{\lambda }(n)^{*}f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(116)

In the second and the third lines in (116), we change the order of the summations in n and m,  and exchange the role of the letters mn;  we see

$$\begin{aligned} \frac{1}{2} \overline{h_{f}(\lambda )}&= \langle \,\widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }), \widehat{f}^{{\mathcal{C}}}(\lambda ) \,\rangle _{{\mathbb{C}}} -\frac{1}{2}\Vert f\Vert ^{2} +\sum _{n \in {\mathbb{Z}}} \langle \,{\mathrm{z}}_{R}(n) f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \sum _{m \ge n+1} \langle \,F_{1/\overline{\lambda }}(n){\mathrm{z}}_{R}(0) F_{\lambda }(m)^{*}f(m),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \sum _{m \le n-1} \langle \,F_{1/\overline{\lambda }}(n){\mathrm{z}}_{L}(0) F_{\lambda }(m)^{*}f(m),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \langle \,F_{1/\overline{\lambda }}(n){\mathrm{z}}_{R}(0) F_{\lambda }(n)^{*}f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(117)

In the last line of (117), we use Lemma 3.1, namely the formula

$$\begin{aligned} F_{1/\overline{\lambda }}(n){\mathrm{z}}_{R}(0) F_{\lambda }(n)^{*}= F_{1/\overline{\lambda }}(n){\mathrm{z}}_{L}(0) F_{\lambda }(n)^{*} +{\mathrm{z}}_{R}(n) -{\mathrm{z}}_{L}(n), \end{aligned}$$

and we finally obtain

$$\begin{aligned} \frac{1}{2} \overline{h_{f}(\lambda )}&= - \sum _{n \in {\mathbb{Z}}} \sum _{m \le n-1} \langle \,F_{1/\overline{\lambda }}(n){\mathrm{z}}_{L}(0) F_{\lambda }(m)^{*}f(m),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \sum _{m \ge n+1} \langle \,F_{1/\overline{\lambda }}(n){\mathrm{z}}_{R}(0) F_{\lambda }(m)^{*}f(m),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad - \sum _{n \in {\mathbb{Z}}} \langle \,F_{1/\overline{\lambda }}(n){\mathrm{z}}_{L}(0) F_{\lambda }(n)^{*}f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad + \langle \,\widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }), \widehat{f}^{{\mathcal{C}}}(\lambda ) \,\rangle _{{\mathbb{C}}} -\frac{1}{2}\Vert f\Vert ^{2} \\ &\quad +\sum _{n \in {\mathbb{Z}}} \langle \,{\mathrm{z}}_{L}(n) f(n),f(n) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(118)

Now, we take \(\lambda =\zeta \in S^{1}.\) Since \(1/\overline{\zeta }=\zeta ,\) all of the terms in (113) and (118) are canceled each other after taking the sum of them except the term

$$\begin{aligned} \langle \,\widehat{f}^{{\mathcal{C}}}(1/\overline{\zeta }), \widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}= \Vert \widehat{f}^{{\mathcal{C}}}(\zeta )\Vert ^{2}_{{\mathbb{C}}^{2}} \end{aligned}$$

in (118), and this shows the assertion. \(\square\)

We prepare an estimate of an integral which will be used to prove Proposition 4.5 below.

Lemma 4.4

Let \(u \in {\mathbb{C}}^{2}.\) Then, there exists a positive constant \(C_{u}\), such that, for any real number r satisfying \(2/3 \le r <1,\) we have

$$\begin{aligned} \int _{S^{1}} \Vert {\mathrm{x}}(r\zeta ) u\Vert _{{\mathbb{C}}^{2}} \,{\mathrm{d}}\ell (\zeta ) \le C_{u} (1-r)^{-1/2}. \end{aligned}$$

Proof

We take \(\zeta \in S^{1}\) and \(r \in [2/3,1).\) By definition, we have \({\mathrm{x}}(r\zeta )=I+2r\zeta {\mathrm{x}}_{0}(r\zeta ).\) This shows that

$$\begin{aligned} \Vert {\mathrm{x}}(r\zeta ) u\Vert _{{\mathbb{C}}^{2}} \le \Vert u\Vert _{{\mathbb{C}}^{2}} +2r \Vert {\mathrm{x}}_{0}(r\zeta ) u\Vert _{{\mathbb{C}}^{2}} \le \Vert u\Vert _{{\mathbb{C}}^{2}}+2 \Vert {\mathrm{x}}_{0}(r\zeta ) u\Vert _{{\mathbb{C}}^{2}} \end{aligned}$$
(119)

for any \(u \in {\mathbb{C}}^{2}.\) Taking \(f=\delta _{0} \otimes u\) in (109), we see

$$\begin{aligned} \Vert R(r\zeta ) (\delta _{0} \otimes u)\Vert&= (1-r^{2})^{-1/2} \left[ {\mathrm{Re}}\,H_{\delta _{0} \otimes u} (r\zeta ) \right] ^{1/2} \\ &= (1+r)^{-1/2}(1-r)^{-1/2} \left[ {\mathrm{Re}}\,H_{\delta _{0} \otimes u} (r\zeta )\right] ^{1/2} \\ &\le (1-r)^{-1/2} \left[ {\mathrm{Re}}\,H_{\delta _{0} \otimes u} (r\zeta )\right] ^{1/2}. \end{aligned}$$
(120)

It follows from (119), (120) and the definition of \({\mathrm{x}}_{0}(\lambda )\) in (72) that:

$$\begin{aligned} \Vert {\mathrm{x}}(r\zeta ) u\Vert _{{\mathbb{C}}^{2}}&\le \Vert u\Vert _{{\mathbb{C}}^{2}}+2 \Vert {\mathrm{x}}_{0}(r\zeta ) u\Vert _{{\mathbb{C}}^{2}} \\ &= \Vert u\Vert _{{\mathbb{C}}^{2}}+2\Vert R(r\zeta )(\delta _{0} \otimes u)(0)\Vert _{{\mathbb{C}}^{2}} \\ &\le \Vert u\Vert _{{\mathbb{C}}^{2}}+2\Vert R(r\zeta )(\delta _{0} \otimes u) \Vert \\ &\le \Vert u\Vert _{{\mathbb{C}}^{2}}+2(1-r)^{-1/2} \left[ {\mathrm{Re}}\,H_{\delta _{0} \otimes u} (r\zeta ) \right] ^{1/2}. \end{aligned}$$

Integrating this inequality over \(S^{1}\) with respect to the normalized Lebesgue measure \({\mathrm{d}}\ell ,\) and using the Cauchy–Schwarz inequality, we see

$$\begin{aligned} \int _{S^{1}} \Vert {\mathrm{x}}(r\zeta ) u\Vert _{{\mathbb{C}}^{2}} \,{\mathrm{d}}\ell (\zeta )&\le \Vert u\Vert _{{\mathbb{C}}^{2}}+2(1-r)^{-1/2} \int _{S^{1}} {\mathrm{Re}}\,H_{\delta _{0} \otimes u}(r\zeta )^{1/2}\,{\mathrm{d}}\ell (\zeta ) \\ &\le \Vert u\Vert _{{\mathbb{C}}^{2}}+2(1-r)^{-1/2} \left( \int _{S^{1}} {\mathrm{Re}}\,H_{\delta _{0} \otimes u}(r\zeta ) \,{\mathrm{d}}\ell (\zeta ) \right) ^{1/2}. \end{aligned}$$

Now, from (111), we have

$$\begin{aligned} \Vert u\Vert ^{2}_{{\mathbb{C}}^{2}}=\nu _{\delta _{0} \otimes u}(S^{1}) =\lim _{r \uparrow 1} \int _{S^{1}} {\mathrm{Re}}\,H_{\delta _{0} \otimes u} (r\zeta )\,{\mathrm{d}}\ell (\zeta ), \end{aligned}$$

which shows that the integral \(\int _{S^{1}} {\mathrm{Re}}\,H_{\delta _{0} \otimes u}(r\zeta ) \,{\mathrm{d}}\ell (\zeta )\) is continuous in \(r \in [3/2,1].\) Hence, we can take a positive constant \(A_{u}\), such that

$$\begin{aligned} \int _{S^{1}} {\mathrm{Re}}\,H_{\delta _{0} \otimes u}(r\zeta ) \,{\mathrm{d}}\ell (\zeta ) \le A_{u}^{2} \end{aligned}$$

for any \(r \in [2/3,1].\) Therefore, we have

$$\begin{aligned} \int _{S^{1}} \Vert {\mathrm{x}}(r\zeta ) u\Vert _{{\mathbb{C}}^{2}}\,{\mathrm{d}}\ell (\zeta )&\le [(1-r)^{1/2}\Vert u\Vert _{{\mathbb{C}}^{2}}+2A_{u}] (1-r)^{-1/2} \\ &\le [\Vert u\Vert _{{\mathbb{C}}^{2}}/\sqrt{3}+3A_{u}] (1-r)^{-1/2}. \end{aligned}$$

The assertion follows by setting \(C_{u}=\Vert u\Vert _{{\mathbb{C}}^{2}}/\sqrt{3}+3A_{u}.\) \(\square\)

The following proposition plays a central role in the proof of Theorem 1.5.

Proposition 4.5

For any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}),\) the measure \(\nu _{f}\) on \(S^{1}\) given in (110), (111) is written as

$$\begin{aligned} {\mathrm{d}}\nu _{f}(\zeta )=\langle \,{\mathrm{d}}\Sigma (\zeta )\widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}, \end{aligned}$$
(121)

where \({\mathrm{d}}\Sigma\) is the positive-matrix-valued measure given in (106), (107).

Proof

We use the characterization (111) of the measure \({\mathrm{d}}\nu _{f}.\) By Lemma 4.3, the function \(H_{f}(\lambda )\) can be written as

$$\begin{aligned} H_{f}(\lambda )&= 2\lambda \langle \,{\mathrm{x}}_{0}(\lambda ) \widehat{f}^{{\mathcal{C}}}(\lambda ), \widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }) \,\rangle _{{\mathbb{C}}^{2}} +h_{f}(\lambda ) \\ &=\langle \,{\mathrm{x}}(\lambda )\widehat{f}^{{\mathcal{C}}}(\lambda ), \widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }) \,\rangle _{{\mathbb{C}}^{2}} -\langle \,\widehat{f}^{{\mathcal{C}}}(\lambda ), \widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }) \,\rangle _{{\mathbb{C}}^{2}} +h_{f}(\lambda ), \end{aligned}$$

where \(h_{f}\) is defined in Lemma 4.3. Taking the real part, we see

$$\begin{aligned} {\mathrm{Re}}\,H_{f}(\lambda )&= {\mathrm{Re}}\,\langle \,{\mathrm{x}}(\lambda ) \widehat{f}^{{\mathcal{C}}}(\lambda ), \widehat{f}^{{\mathcal{C}}} (1/\overline{\lambda }) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad -{\mathrm{Re}}\,\langle \,\widehat{f}^{{\mathcal{C}}}(\lambda ), \widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }) \,\rangle _{{\mathbb{C}}^{2}} +{\mathrm{Re}}\,h_{f}(\lambda ). \end{aligned}$$

By Lemmas 4.2 and 4.3, the functions \(\widehat{f}^{{\mathcal{C}}}(\lambda ),\) \(h_{f}(\lambda )\) are holomorphic on \({\mathbb{C}} {\setminus } \{0\}.\) Thus, by Lemma 4.3, we see

$$\begin{aligned} \lim _{r \uparrow 1} \left[ {\mathrm{Re}}\,\langle \,\widehat{f}^{{\mathcal{C}}}(r \zeta ), \widehat{f}^{{\mathcal{C}}}(r^{-1}\zeta ) \,\rangle _{{\mathbb{C}}^{2}} -{\mathrm{Re}}\,h_{f}(r \zeta ) \right] =\Vert \widehat{f}^{{\mathcal{C}}}(\zeta )\Vert _{{\mathbb{C}}^{2}}^{2} -{\mathrm{Re}}\,h_{f}(\zeta )=0. \end{aligned}$$

Therefore, (111) shows that

$$\begin{aligned} {\mathrm{d}}\nu _{f}(\zeta ) = \mathop {\mathrm{w*-lim}}\limits _{r \uparrow 1} {\mathrm{Re}}\,\langle \,{\mathrm{x}}(r\zeta ) \widehat{f}^{{\mathcal{C}}} (r\zeta ), \widehat{f}^{{\mathcal{C}}}(r^{-1}\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \, {\mathrm{d}}\ell (\zeta ). \end{aligned}$$
(122)

We calculate the right-hand side of (122) using (107) as follows. For simplicity of notation, we set

$$\begin{aligned} F_{L}(\lambda )=\langle \,\widehat{f}^{{\mathcal{C}}}(\lambda ),e_{L} \,\rangle _{{\mathbb{C}}^{2}},\quad F_{R}(\lambda )=\langle \,\widehat{f}^{{\mathcal{C}}}(\lambda ),e_{R} \,\rangle _{{\mathbb{C}}^{2}}, \end{aligned}$$

which are holomorphic on \({\mathbb{C}} {\setminus } \{0\}.\) We write

$$\begin{aligned}&\langle \,{\mathrm{x}}(\lambda ) \widehat{f}^{{\mathcal{C}}}(\lambda ), \widehat{f}^{{\mathcal{C}}}(1/\overline{\lambda }) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = F_{L}(\lambda ) \overline{F_{L}(1/\overline{\lambda })} \langle \,{\mathrm{x}}(\lambda ) e_{L},e_{L} \,\rangle _{{\mathbb{C}}^{2}} + F_{L}(\lambda ) \overline{F_{R}(1/\overline{\lambda })} \langle \,{\mathrm{x}}(\lambda ) e_{L},e_{R} \,\rangle _{{\mathbb{C}}^{2}} \\ &\qquad +F_{R}(\lambda ) \overline{F_{L}(1/\overline{\lambda })} \langle \,{\mathrm{x}}(\lambda ) e_{R},e_{L} \,\rangle _{{\mathbb{C}}^{2}} + F_{R}(\lambda ) \overline{F_{R}(1/\overline{\lambda })} \langle \,{\mathrm{x}}(\lambda ) e_{R},e_{R} \,\rangle _{{\mathbb{C}}^{2}}, \end{aligned}$$

and hence

$$\begin{aligned} 2 {\mathrm{Re}}\,\langle \,{\mathrm{x}}(r\zeta )\widehat{f}^{{\mathcal{C}}}(r\zeta ), \widehat{f}^{{\mathcal{C}}}(r^{-1}\zeta ) \,\rangle =g_{1}(r\zeta ) +g_{2}(r\zeta )+g_{3}(r\zeta )+g_{4}(r\zeta ), \end{aligned}$$
(123)

where we set

$$\begin{aligned} g_{1}(\lambda )&=F_{L}(\lambda ) \overline{F_{L}(1/\overline{\lambda })} \langle \,{\mathrm{x}}(\lambda ) e_{L},e_{L} \,\rangle _{{\mathbb{C}}^{2}} +F_{L}(1/\overline{\lambda }) \overline{F_{L} (\lambda )} \langle \,{\mathrm{x}}(\lambda )^{*}e_{L},e_{L} \,\rangle _{{\mathbb{C}}^{2}}, \\ g_{2}(\lambda )&= F_{R}(\lambda ) \overline{F_{L}(1/\overline{\lambda })} \langle \,{\mathrm{x}}(\lambda )e_{R},e_{L} \,\rangle _{{\mathbb{C}}^{2}} +F_{R}(1/\overline{\lambda }) \overline{F_{L}(\lambda )} \langle \,{\mathrm{x}}(\lambda )^{*}e_{R},e_{L} \,\rangle _{{\mathbb{C}}^{2}}, \\ g_{3}(\lambda )&= F_{L}(\lambda ) \overline{F_{R} (1/\overline{\lambda })} \langle \,{\mathrm{x}}(\lambda )e_{L},e_{R} \,\rangle _{{\mathbb{C}}^{2}} + F_{L}(1/\overline{\lambda })\overline{F_{R}(\lambda )} \langle \,{\mathrm{x}}(\lambda )^{*}e_{L},e_{R} \,\rangle _{{\mathbb{C}}^{2}}, \\ g_{4}(\lambda )&= F_{R}(\lambda ) \overline{F_{R}(1/\overline{\lambda })} \langle \,{\mathrm{x}}(\lambda ) e_{R}, e_{R} \,\rangle _{{\mathbb{C}}^{2}} + F_{R}(1/\overline{\lambda }) \overline{F_{R}(\lambda )} \langle \,{\mathrm{x}}(\lambda )^{*}e_{R},e_{R} \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

For \(0<r<1\) and \(\zeta \in S^{1},\) we set

$$\begin{aligned} k_{L}(r,\zeta )=F_{L}(r^{-1}\zeta )-F_{L}(r\zeta ),\quad k_{R}(r,\zeta )=F_{R}(r^{-1}\zeta ) -F_{R}(r\zeta ), \end{aligned}$$

so that

$$\begin{aligned} g_{1}(r\zeta )&= F_{L}(r\zeta ) \overline{k_{L}(r,\zeta )} \langle \,{\mathrm{x}}(r\zeta ) e_{L},e_{L} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +\overline{F_{L}(r\zeta )} k_{L}(r,\zeta ) \langle \,{\mathrm{x}}(r\zeta )^{*}e_{L},e_{L} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad + 2|F_{L}(r\zeta )|^{2} \langle \, [{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] e_{L},e_{L} \,\rangle _{{\mathbb{C}}^{2}}, \\ g_{2}(r\zeta )&= F_{R}(r\zeta ) \overline{k_{L}(r,\zeta )} \langle \,{\mathrm{x}}(r\zeta ) e_{R},e_{L} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +\overline{F_{L}(r\zeta )} k_{R}(r,\zeta ) \langle \,{\mathrm{x}}(r\zeta )^{*}e_{R},e_{L} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad + 2F_{R}(r\zeta ) \overline{F_{L}(r\zeta )} \langle \, [{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] e_{R},e_{L} \,\rangle _{{\mathbb{C}}^{2}}, \\ g_{3}(r\zeta )&= F_{L}(r\zeta ) \overline{k_{R}(r,\zeta )} \langle \,{\mathrm{x}}(r\zeta ) e_{L},e_{R} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +\overline{F_{R}(r\zeta )} k_{L}(r,\zeta ) \langle \,{\mathrm{x}}(r\zeta )^{*}e_{L},e_{R} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad + 2F_{L}(r\zeta ) \overline{F_{R}(r\zeta )} \langle \, [{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] e_{L},e_{R} \,\rangle _{{\mathbb{C}}^{2}}, \\ g_{4}(r\zeta )&= F_{R}(r\zeta ) \overline{k_{R}(r,\zeta )} \langle \,{\mathrm{x}}(r\zeta ) e_{R},e_{R} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +\overline{F_{R}(r\zeta )} k_{R}(r,\zeta ) \langle \,{\mathrm{x}}(r\zeta )^{*}e_{R},e_{R} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad + 2|F_{R}(r\zeta )|^{2} \langle \, [{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] e_{R},e_{R} \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(124)

Since the functions \(F_{L},\) \(F_{R}\) are holomorphic on \({\mathbb{C}} {\setminus } \{0\},\) we have, for \(2/3 \le r <1\)

$$\begin{aligned} |k_{L}(r,\zeta )|&= |F_{L}(r^{-1}\zeta ) -F_{L}(r\zeta )| \\ &\le \Vert F_{L}'\Vert _{C(A)}|r^{-1}-r| \le 3C(1-r), \\ |k_{R}(r,\zeta )|&=|F_{R}(r^{-1}\zeta ) -F_{R}(r\zeta )| \\ &\le \Vert F_{R}'\Vert _{C(A)}|r^{-1}-r| \le 3C(1-r), \end{aligned}$$
(125)

where we set

$$\begin{aligned} A=\{\lambda \in {\mathbb{C}} \mid 2/3 \le |\lambda | \le 3/2\},\quad C=\max \{\Vert F_{L}'\Vert _{C(A)}, \Vert F_{R}'\Vert _{C(A)}\}. \end{aligned}$$

For simplicity, we denote \(k(r,\zeta )\) one of \(k_{L}(r,\zeta ),\) \(\overline{k_{L}(r,\zeta )},\) \(k_{R}(r,\zeta )\) and \(\overline{k_{R}(r,\zeta )},\) and we denote \(F(\lambda )\) one of \(F_{L}(\lambda ),\) \(\overline{F_{L}(\lambda )},\) \(F_{R}(\lambda )\) and \(\overline{F_{R}(\lambda )}.\) Then, for any \(h \in C(S^{1}),\) any \(u,v \in {\mathbb{C}}^{2},\) and any \(r \in [2/3, 1),\) Lemma 4.4 and the estimate (125) show

$$\begin{aligned}&\left| \int _{S^{1}} h(\zeta ) F(r\zeta ) k(r,\zeta ) \langle \,{\mathrm{x}}(r\zeta )u,v \,\rangle _{{\mathbb{C}}^{2}}\, {\mathrm{d}}\ell (\zeta ) \right| \\ &\quad \le \int _{S^{1}} |h(\zeta )| |F(r\zeta )| |k(r,\zeta )| \Vert {\mathrm{x}}(r\zeta )u\Vert _{{\mathbb{C}}^{2}} \Vert v\Vert _{{\mathbb{C}}^{2}} \,{\mathrm{d}}\ell (\zeta ) \\ &\quad \le 3 C \Vert v\Vert _{{\mathbb{C}}^{2}} \Vert h\Vert _{C(S^{1})} \Vert F\Vert _{C(A)} (1-r) \int _{S^{1}} \Vert {\mathrm{x}}(r\zeta )u\Vert _{{\mathbb{C}}^{2}} \,{\mathrm{d}}\ell (\zeta ) \\ &\quad \le 3CC_{u} \Vert v\Vert _{{\mathbb{C}}^{2}} \Vert h\Vert _{C(S^{1})} \Vert F\Vert _{C(A)} (1-r)^{1/2}, \end{aligned}$$

which tends to zero as \(r \uparrow 1.\) This shows that, for any \(h \in C(S^{1}),\) in the integral

$$\begin{aligned} \int _{S^{1}} h(\zeta ) g_{i}(r\zeta )\,{\mathrm{d}}\ell (\zeta ) \quad (i=1,2,3,4), \end{aligned}$$

the contributions from the first and the second term of \(g_{i}(r\zeta )\) in (124) tend to zero as \(r \uparrow 1.\) To handle the contributions from the last term involving \({\mathrm{Re}}\,{{\mathrm{x}}(r\zeta )}\) of \(g_{i}(r\zeta )\) in (124), let \(G(\lambda )\) be a smooth function on A and we take \(u,v \in {\mathbb{C}}^{2}.\) We write \(G(\lambda )=G(x,y)\) for \(\lambda =x+iy.\) Then, for \(\zeta \in S^{1},\) we have

$$\begin{aligned} |G(\zeta ) -G(r\zeta )| \le B (1-r),\quad B=\Vert G_{x}\Vert _{C(A)}+\Vert G_{y}\Vert _{C(A)}, \end{aligned}$$

where \(G_{x}\) (resp. \(G_{y}\)) is the partial derivative of G with respect to x (resp. y). Hence, for any \(h \in C(S^{1}),\) we see by Lemma 4.4

$$\begin{aligned}&\left| \int _{S^{1}} h(\zeta ) (G(r\zeta )-G(\zeta )) \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] u,v \,\rangle _{{\mathbb{C}}^{2}}\,{\mathrm{d}}\ell (\zeta ) \right| \\ &\quad \le \Vert h\Vert _{C(S^{1})} B(1-r) \int _{S^{1}} |\langle \,[{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] u,v \,\rangle _{{\mathbb{C}}^{2}}| \,{\mathrm{d}}\ell (\zeta ) \\ &\quad \le \Vert h\Vert _{C(S^{1})} B(1-r) \int _{S^{1}} \left( \Vert {\mathrm{x}}(r\zeta )u\Vert _{{\mathbb{C}}^{2}} \Vert v\Vert _{{\mathbb{C}}^{2}} +\Vert {\mathrm{x}}(r\zeta ) v\Vert _{{\mathbb{C}}^{2}} \Vert u\Vert _{{\mathbb{C}}^{2}} \right) \,{\mathrm{d}}\ell (\zeta ) \\ &\quad \le \Vert h\Vert _{C(S^{1})} B(1-r) \left( C_{u}\Vert v\Vert _{{\mathbb{C}}^{2}}+C_{v}\Vert u\Vert _{{\mathbb{C}}^{2}} \right) (1-r)^{-1/2} \\ &\quad = \Vert h\Vert _{C(S^{1})} B \left( C_{u}\Vert v\Vert _{{\mathbb{C}}^{2}}+C_{v}\Vert u\Vert _{{\mathbb{C}}^{2}} \right) (1-r)^{1/2}, \end{aligned}$$

which tends to zero as \(r \uparrow 1.\) This and (107) show that

$$\begin{aligned}&\lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) G(r\zeta ) \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] u,v \,\rangle _{{\mathbb{C}}^{2}}\,{\mathrm{d}}\ell (\zeta ) \\ &\quad = \lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) G(\zeta ) \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] u,v \,\rangle _{{\mathbb{C}}^{2}}\,{\mathrm{d}}\ell (\zeta ) \\ &\qquad +\lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) (G(r\zeta )-G(\zeta )) \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] u,v \,\rangle _{{\mathbb{C}}^{2}}\,{\mathrm{d}}\ell (\zeta ) \\ &\quad = \lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) G(\zeta ) \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] u,v \,\rangle _{{\mathbb{C}}^{2}}\,{\mathrm{d}}\ell (\zeta ) \\ &\quad = \int _{S^{1}} h(\zeta ) G(\zeta ) \langle \,{\mathrm{d}}\Sigma (\zeta )u,v \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(126)

Therefore, by taking \(G(\lambda )\) as one of functions \(|F_{L}(\lambda )|^{2},\) \(F_{R}(\lambda )\overline{F_{L}(\lambda )},\) \(F_{L}(\lambda )\overline{F_{R}(\lambda )},\) \(|F_{R}(\lambda )|^{2},\) we obtain

$$\begin{aligned} \lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) g_{1}(r\zeta )\,{\mathrm{d}}\ell (\zeta )&= 2\int _{S^{1}} h(\zeta ) |F_{L}(\zeta )|^{2} \langle \,{\mathrm{d}}\Sigma (\zeta )e_{L},e_{L} \,\rangle _{{\mathbb{C}}^{2}}, \\ \lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) g_{2}(r\zeta )\,{\mathrm{d}}\ell (\zeta )&= 2\int _{S^{1}} h(\zeta ) F_{R}(\zeta ) \overline{F_{L}(\zeta )} \langle \,{\mathrm{d}}\Sigma (\zeta )e_{R},e_{L} \,\rangle _{{\mathbb{C}}^{2}}, \\ \lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) g_{3}(r\zeta )\,{\mathrm{d}}\ell (\zeta )&= 2\int _{S^{1}} h(\zeta ) F_{L}(\zeta ) \overline{F_{R}(\zeta )} \langle \,{\mathrm{d}}\Sigma (\zeta )e_{L},e_{R} \,\rangle _{{\mathbb{C}}^{2}}, \\ \lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) g_{4}(r\zeta )\,{\mathrm{d}}\ell (\zeta )&= 2\int _{S^{1}} h(\zeta ) |F_{R}(r\zeta )|^{2} \langle \,{\mathrm{d}}\Sigma (\zeta )e_{R},e_{R} \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

Therefore, by (122), (123), we conclude

$$\begin{aligned}&\int _{S^{1}} h(\zeta ) \,{\mathrm{d}}\nu _{f}(\zeta ) \\ &\quad = \lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) {\mathrm{Re}}\,\langle \,{\mathrm{x}}(r\zeta ) \widehat{f}^{{\mathcal{C}}} (r\zeta ), \widehat{f}^{{\mathcal{C}}}(r^{-1}\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \, {\mathrm{d}}\ell (\zeta ) \\ &\quad = \lim _{r \uparrow 1} \frac{1}{2}\int _{S^{1}} h(\zeta ) [g_{1}(r\zeta ) +g_{4}(r\zeta )+g_{3}(r\zeta )+g_{4}(r\zeta )] \,{\mathrm{d}}\ell (\zeta ) \\ &\quad = \int _{S^{1}} h(\zeta ) |F_{L}(\zeta )|^{2} \langle \,{\mathrm{d}}\Sigma (\zeta )e_{L},e_{L} \,\rangle _{{\mathbb{C}}^{2}} \\ &\qquad +\int _{S^{1}} h(\zeta ) F_{R}(\zeta ) \overline{F_{L}(\zeta )} \langle \,{\mathrm{d}}\Sigma (\zeta )e_{R},e_{L} \,\rangle _{{\mathbb{C}}^{2}} \\ &\qquad +\int _{S^{1}} h(\zeta ) F_{L}(\zeta ) \overline{F_{R}(\zeta )} \langle \,{\mathrm{d}}\Sigma (\zeta )e_{L},e_{R} \,\rangle _{{\mathbb{C}}^{2}} \\ &\qquad + \int _{S^{1}} h(\zeta ) |F_{R}(r\zeta )|^{2} \langle \,{\mathrm{d}}\Sigma (\zeta )e_{R},e_{R} \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = \int _{S^{1}} h(\zeta ) \langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}, \end{aligned}$$

which completes the proof. \(\square\)

Proof of Theorem 1.5

The formula (121) in Proposition 4.5 is equivalent to the formula

$$\begin{aligned} \langle \,K(\lambda )f,f \,\rangle =\int _{S^{1}} \frac{\zeta +\lambda }{\zeta -\lambda } \langle \,{\mathrm{d}}\Sigma (\zeta )\widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(127)

Since \(H_{f}(\lambda )=\langle \,K(\lambda )f,f \,\rangle ,\) by setting \(\lambda =0\) in (127), we obtain

$$\begin{aligned} \langle \,f,f \,\rangle =\int _{S^{1}}\langle \,{\mathrm{d}}\Sigma (\zeta )\widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}, \end{aligned}$$
(128)

which shows (18) for \(f=g \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Since \(K(\lambda )=I+2\lambda R(\lambda ),\) we see

$$\begin{aligned} \langle \,R(\lambda )f,f \,\rangle&= \frac{1}{2\lambda } \langle \,K(\lambda )f,f \,\rangle -\frac{1}{2\lambda } \langle \,f,f \,\rangle \\ &= \int _{S^{1}} \left( \frac{1}{2\lambda } \frac{\zeta +\lambda }{\zeta -\lambda } -\frac{1}{2\lambda } \right) \,\langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \\ &= \int _{S^{1}} \frac{1}{\zeta -\lambda } \,\langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}, \end{aligned}$$

which shows (17) for \(f=g \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) The polarization identity

$$\begin{aligned} \langle \,f,g \,\rangle =\frac{1}{4} \sum _{k=0}^{3} i^{k} \langle \,f+i^{k}g,f+i^{k}g \,\rangle \end{aligned}$$
(129)

shows (17) and (18) for any \(f,g \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Hence, the assertion (2) in Theorem 1.5 has been proved. To prove the assertion (1) in Theorem 1.5, suppose that the positive-matrix-valued measure \(\Omega\) also satisfies the identity

$$\begin{aligned} \langle \,R(\lambda )f,g \,\rangle =\int _{S^{1}} \frac{1}{\zeta -\lambda } \,\langle \,{\mathrm{d}}\Omega (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \end{aligned}$$
(130)

for any \(f,g \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) We note that \(Uf \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) for \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) The formula (26) shows that for \(\lambda ,\mu \in {\mathbb{C}} {\setminus } \{0\}\)

$$\begin{aligned} {\mathcal{F}}_{{\mathcal{C}}}[(U-\lambda )f](\mu )=(\mu -\lambda ) \widehat{f}^{{\mathcal{C}}}(\mu ). \end{aligned}$$

Thus, replacing f by \((U-\lambda )f\) in (130) shows

$$\begin{aligned} \langle \,f,g \,\rangle&=\langle \,R(\lambda )(U-\lambda )f,g \,\rangle \\ &= \int _{S^{1}} \frac{1}{\zeta -\lambda } \,\langle \,{\mathrm{d}}\Omega (\zeta ) {\mathcal{F}}_{{\mathcal{C}}}[(U-\lambda ) f](\zeta ),\widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \\ &= \int _{S^{1}} \frac{1}{\zeta -\lambda } \,\langle \,{\mathrm{d}}\Omega (\zeta ) (\zeta -\lambda ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \\ &= \int _{S^{1}} \,\langle \,{\mathrm{d}}\Omega (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

Then, using the identity \(K(\lambda )=I+2\lambda R(\lambda )\) again, we have

$$\begin{aligned} \langle \,K(\lambda )f,g \,\rangle&= \int _{S^{1}} \,\langle \,{\mathrm{d}}\Omega (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad +2 \lambda \int _{S^{1}} \frac{1}{\zeta -\lambda } \, \langle \,{\mathrm{d}}\Omega (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \\ &= \int _{S^{1}} \frac{\zeta -\lambda +2\lambda }{\zeta -\lambda } \,\langle \,{\mathrm{d}}\Omega (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \\ &= \int _{S^{1}} \frac{\zeta +\lambda }{\zeta -\lambda } \,\langle \,{\mathrm{d}}\Omega (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ),\widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

From this, (103), and the identity \({\mathcal{F}}_{{\mathcal{C}}}[\delta _{0} \otimes u](\lambda )=u\) for any \(u \in {\mathbb{C}}^{2},\) we obtain the following Herglotz representation of the matrix \({\mathrm{x}}(\lambda ):\)

$$\begin{aligned} \langle \,{\mathrm{x}}(\lambda )u,v \,\rangle _{{\mathbb{C}}^{2}} =\int _{S^{1}} \frac{\zeta +\lambda }{\zeta -\lambda } \,\langle \,{\mathrm{d}}\Omega (\zeta )u,v \,\rangle _{{\mathbb{C}}^{2}} \quad (u,v \in {\mathbb{C}}^{2}), \end{aligned}$$

which shows \(\Sigma =\Omega .\) Thus, the uniqueness of the positive-matrix-valued measure in (17) and the assertion (1) in Theorem 1.5 have been proved. We remark that

$$\begin{aligned} \langle \,h(U)f,g \,\rangle =\int _{S^{1}} h(\zeta ) \langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \end{aligned}$$
(131)

for any \(h \in C(S^{1})\) and any \(f,g \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Let us prove (131) before proceeding to the proof of the assertion (3) in Theorem 1.5. By the formula (26), we have

$$\begin{aligned} \widehat{f}^{{\mathcal{C}}}(\lambda )={\mathcal{F}}_{{\mathcal{C}}}[UU^{-1}f](\lambda ) =\lambda {\mathcal{F}}_{{\mathcal{C}}}[U^{-1}f](\lambda ),\quad {\mathcal{F}}_{{\mathcal{C}}}[U^{-1}f](\lambda )=\lambda ^{-1}\widehat{f}^{{\mathcal{C}}}(\lambda ) \end{aligned}$$

for any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Iterated use of this and the formula (26) show that, for any integer n,  positive or negative

$$\begin{aligned} {\mathcal{F}}_{{\mathcal{C}}}[U^{n}f](\lambda )=\lambda ^{n} \widehat{f}^{{\mathcal{C}}}(\lambda ). \end{aligned}$$

By this formula and the item (2) in Theorem 1.5, we have

$$\begin{aligned} \langle \,U^{n}f,g \,\rangle&=\int _{S^{1}} \langle \,{\mathrm{d}}\Sigma (\zeta ) {\mathcal{F}}_{{\mathcal{C}}}[U^{n}f](\zeta ), \widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \\ &=\int _{S^{1}} \zeta ^{n} \langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

Therefore, for any Laurent polynomial p on \({\mathbb{C}} {\setminus } \{0\},\) we see

$$\begin{aligned} \langle \,p(U)f,g \,\rangle = \int _{S^{1}} p(\zeta ) \langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

Since the set of Laurent polynomials in \(\zeta \in S^{1}\) is dense in \(C(S^{1})\) with respect to the norm \(\Vert \cdot \Vert _{C(S^{1})},\) and since \(\Vert h(U)\Vert _{{\mathrm{op}}} \le \Vert h\Vert _{C(S^{1})},\) where \(\Vert h(U)\Vert _{{\mathrm{op}}}\) denotes the operator norm of h(U) on \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}),\) we have (131). We now proceed to the proof of the assertion (3) in Theorem 1.5. For any \(h \in C(S^{1})\) and \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}),\) we have

$$\begin{aligned} \langle \,h(U)f,f \,\rangle =\int _{S^{1}} h(\zeta )\,{\mathrm{d}}\Vert E(\zeta )f\Vert ^{2}= \int _{S^{1}} h(\zeta ) \langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

Since the spectral measure \(\Vert E(\cdot )f\Vert ^{2}\) is the unique Borel measure on \(S^{1}\) satisfying the above with the total mass \(\Vert f\Vert ^{2},\) we see

$$\begin{aligned} \Vert E(A)f\Vert ^{2}=\langle \,E(A)f,f \,\rangle =\int _{A} \langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{f}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \end{aligned}$$

for any Borel set A in \(S^{1},\) and hence, by the polarization identity (129), we see

$$\begin{aligned} \langle \,E(A)f,g \,\rangle =\int _{A} \langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), \widehat{g}^{{\mathcal{C}}}(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$

Taking \(g=\delta _{n} \otimes v\) with \(n \in {\mathbb{Z}}\) and \(v \in {\mathbb{C}}^{2},\) and noting \({\mathcal{F}}_{{\mathcal{C}}}[\delta _{n} \otimes v](\lambda )=F_{1/\overline{\lambda }}(n)^{*}v,\) we have

$$\begin{aligned} \langle \,[E(A)f](n),v \,\rangle _{{\mathbb{C}}^{2}}&= \langle \,E(A)f,\delta _{n} \otimes v \,\rangle = \int _{A} \langle \,{\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), F_{\zeta }(n)^{*}v \,\rangle _{{\mathbb{C}}^{2}} \\ &= \int _{A} \langle \,F_{\zeta }(n){\mathrm{d}}\Sigma (\zeta ) \widehat{f}^{{\mathcal{C}}}(\zeta ), v \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(132)

Since (132) holds for any \(v \in {\mathbb{C}}^{2},\) we have (19). By setting \(A=S^{1}\) in (19), we conclude (20). \(\square\)

Proof of Corollary 1.6

By taking \(f=\delta _{0} \otimes u\) and \(n=0\) in (132), we have

$$\begin{aligned} \langle \,[E(A)\delta _{0} \otimes u](0), v \,\rangle _{{\mathbb{C}}^{2}} = \int _{A} \langle \,{\mathrm{d}}\Sigma (\zeta ) u, v \,\rangle _{{\mathbb{C}}^{2}}=\langle \,\Sigma (A)u,v \,\rangle _{{\mathbb{C}}^{2}}. \end{aligned}$$
(133)

We remark that the support, \({{\text{supp}}\,}(\Sigma ),\) of the positive-matrix-valued measure \(\Sigma\) on \(S^{1}\) is defined as the complement of the open set

$$\begin{aligned} \bigcup _{U\text{: open in }S^{1}, \, \Sigma (U)=0} U. \end{aligned}$$

Now, suppose that \(\zeta _{o} \in S^{1}\) is not contained in \(\sigma (U).\) Then, since \({{\text{supp}}\,}(E)=\sigma (U),\) there exists an open neighborhood U of \(\zeta _{o}\) in \(S^{1}\), such that \(E(U)=0.\) The identity (133) shows \(\Sigma (U)=0.\) Therefore, \(\zeta _{o} \not \in {{\text{supp}}\,}(\Sigma ).\) Conversely, suppose that \(\zeta _{o} \in S^{1}\) is not contained in \({{\text{supp}}\,}(\Sigma ),\) and let U be an open neighborhood of \(\zeta _{o}\) satisfying \(\Sigma (U)=0.\) This means that each of the entry of \(\Sigma\) is zero on U. Thus, the right-hand side of (132) with \(A=U\) is zero for any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}),\) \(n \in {\mathbb{Z}}\) and \(v \in {\mathbb{C}}^{2}.\) This means that \(E(U)=0.\) Thus, we conclude that \(\zeta _{o} \not \in {{\text{supp}}\,}(E),\) which proves the assertion (1) in Corollary 1.6. Next, to prove the assertion (2) in Corollary 1.6, suppose that \(\lambda \in S^{1}\) satisfies \(\Sigma (\{\lambda \}) \ne 0.\) Then, by (133), we see \(\langle \,[E(\{\lambda \})(\delta _{0} \otimes u)](0),v \,\rangle _{{\mathbb{C}}^{2}} \ne 0\) for some \(u,u \in {\mathbb{C}}^{2}.\) Hence. \(E(\{\lambda \}) \ne 0\) and \(\lambda\) is an eigenvalue of U. Conversely, suppose that \(\lambda \in S^{1}\) is an eigenvalue of U. Let \(f \in \ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\) be an eigenfunction of U with the eigenvalue \(\lambda .\) Since \(E(\{\lambda \})\) is the projection onto the eigenspace of U with the eigenvalue \(\lambda ,\) we see \(E(\{\lambda \})f=f.\) By (132), we have

$$\begin{aligned} \Vert f(n)\Vert _{{\mathbb{C}}^{2}}^{2} =\langle \,[E(\{\lambda \})f](n),f(n) \,\rangle _{{\mathbb{C}}^{2}} =\langle \,F_{\lambda }(n) \Sigma (\{\lambda \}) \widehat{f}^{{\mathcal{C}}}(\lambda ), f(n) \,\rangle _{{\mathbb{C}}^{2}} \end{aligned}$$

for any \(n \in {\mathbb{Z}}.\) The left-hand side of the above expression is nonzero for some \(n \in {\mathbb{Z}}\), because f is not identically zero. Therefore, \(\Sigma (\{\lambda \}) \ne 0,\) which proves the assertion (2) in Corollary 1.6. \(\square\)

Before proceeding to the proof of Theorem 1.7, we give, based on [7], some accounts on the sesquilinear form \(\langle \,\cdot ,\cdot \,\rangle _{\Sigma }\) defined in (24). In particular, we show that the positivity of the measure \(\langle \,{\mathrm{d}}\Sigma (\zeta )k(\zeta ),k(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}\) for any \(k \in C(S^{1},{\mathbb{C}}^{2}).\) This fact can be proved using the characterization (107) of \(\Sigma\) and the non-negativity of \({\mathrm{Re}}\,{\mathrm{x}}(\lambda ).\) However, we give here an alternate proof.

We recall that, about the entries of positive-matrix-valued measure \(\Sigma ,\) the diagonals \(\mu _{L}\) and \(\mu _{R}\) are Borel probability measures, while \(\alpha =\langle \,\Sigma e_{L},e_{R} \,\rangle\) is a complex Borel measure satisfying \(|\alpha (A)|^{2} \le \mu _{L}(A) \mu _{R}(A)\) for any Borel set A, because \(\det \Sigma (A) \ge 0.\) From this, it follows that the total variation measure \(|\alpha |\) of \(\alpha\) also satisfies the inequality \(|\alpha |(A)^{2} \le \mu _{L}(A) \mu _{R}(A)\) for any Borel set A in \(S^{1}.\) This shows that \(\alpha ,\) \(|\alpha |\) is absolutely continuous with respect to both of \(\mu _{L}\) and \(\mu _{R}.\) We define a probability measure \(\mu\) by \(\mu (A)=(\mu _{L}(A)+\mu _{R}(A))/2,\) that is \(\mu ={\text{Tr}}(\Sigma )/2.\) Then, we have \(|\alpha |(A) \le \mu (A)\), and thus, \(\alpha ,\) \(\mu _{L}\) and \(\mu _{R}\) are all absolutely continuous with respect to the measure \(\mu ={\text{Tr}}(\Sigma )/2.\) We can write \(\alpha =\rho \mu ,\) \(\mu _{L}=\rho _{L} \mu ,\) \(\mu _{R}=\rho _{R} \mu\) with \(\rho _{L},\rho _{R},\rho \in L^{1}(S^{1},\mu ),\) where \(\rho _{L}\) and \(\rho _{R}\) are non-negative \(\mu\)-a.e. Then, we have \(|\alpha |=|\rho |\mu .\) For any \(\xi \in S^{1}\) and \(\varepsilon >0,\) we denote \(A(\xi ,\varepsilon )\) the arc in \(S^{1}\) centered at \(\xi\) and length \(\varepsilon .\) Then, by the Lebesgue differentiation theorem (see, for example, [4], 2.9.8 Theorem), we have

$$\begin{aligned} 0&\le \frac{1}{\mu (A(\xi ,\varepsilon ))^{2}} \left( \mu _{L}(A(\xi ,\varepsilon ))\mu _{R}(A(\xi ,\varepsilon )) -|\alpha |(A(\xi ,\varepsilon ))^{2} \right) \\ &=\frac{1}{A(\xi ,\varepsilon )} \int _{A(\xi ,\varepsilon )} \rho _{L} \,{\mathrm{d}}\mu \times \frac{1}{A(\xi ,\varepsilon )} \int _{A(\xi ,\varepsilon )} \rho _{R} \,{\mathrm{d}}\mu -\left( \int _{A(\xi ,\varepsilon )} |\rho |\,{\mathrm{d}}\mu \right) ^{2} \\ &\rightarrow \rho _{L}(\xi ) \rho _{R}(\xi ) -|\rho (\xi )|^{2} \end{aligned}$$

as \(\varepsilon \downarrow 0\) for \(\mu\)-a.e \(\xi \in S^{1}.\) Let X be a Borel set, such that \(\mu (X)=1\) and the above holds for any \(\xi \in X.\) Then, \(|\rho |^{2} \le \rho _{L} \rho _{R}\) on X. Let \(k \in C(S^{1},{\mathbb{C}}^{2})\) and we write \(k=\,\!^{t}(k_{L},k_{R})\) with \(k_{L},k_{R} \in C(S^{1}).\) Then, for any Borel set A on \(S^{1}\), we have

$$\begin{aligned}&\int _{A} \langle \,{\mathrm{d}}\Sigma (\zeta ) k(\zeta ),k(\zeta ) \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = \int _{A} \left( |k_{L}|^{2}\,{\mathrm{d}}\mu _{L}+ k_{L} \overline{k_{R}}\, {\mathrm{d}}\alpha +\overline{k_{L}} k_{R}\, {\mathrm{d}}\overline{\alpha } +|k_{R}|^{2}\,{\mathrm{d}}\mu _{R} \right) \\ &\quad = \int _{A} \left( |k_{L}|^{2}\rho _{L} +k_{L} \overline{k_{R}} \rho +\overline{k_{L}} k_{R} \overline{\rho } +|k_{R}|^{2} \rho _{R} \right) \,{\mathrm{d}}\mu . \end{aligned}$$
(134)

Since the integrand in (134) can be written as

$$\begin{aligned} \rho _{L}\left| k_{L}+\frac{k_{R} \overline{\rho }}{\rho _{L}} \right| ^{2} +|k_{R}|^{2} \left( \rho _{R} -\frac{|\rho |^{2}}{\rho _{L}} \right) , \end{aligned}$$

it is non-negative at points where \(\rho _{L}>0\) in X. At a point \(\xi \in X\) satisfying \(\rho _{L}(\xi )=0,\) the integrand in (134) becomes \(|k_{R}(\xi )|^{2} \rho _{R}(\xi ) \ge 0.\) Thus, the integral in (134) is non-negative. Hence, the measure \(\langle \,{\mathrm{d}}\Sigma (\zeta )k(\zeta ), k(\zeta ) \,\rangle _{{\mathbb{C}}^{2}}\) is indeed a positive measure.

Proof of Theorem 1.7

For \(k \in C(S^{1},{\mathbb{C}}^{2}),\) we define \(\Vert k\Vert _{\infty }\) by the formula \(\Vert k\Vert _{\infty }=\sup _{\zeta \in S^{1}} \Vert k(\zeta )\Vert _{{\mathbb{C}}^{2}}.\) We first show the inequality

$$\begin{aligned} \Vert p\Vert _{\Sigma }^{2} \le 4\Vert p\Vert _{\infty }^{2}. \end{aligned}$$
(135)

Indeed, using (134) for \(A=S^{1},\) we see

$$\begin{aligned} \Vert k\Vert _{\Sigma }^{2}&=\int _{S^{1}} \left( |k_{L}|^{2}\rho _{L} +k_{L} \overline{k_{R}} \rho +\overline{k_{L}} k_{R} \overline{\rho } +|k_{R}|^{2} \rho _{R} \right) \,{\mathrm{d}}\mu \\ &= \int _{S^{1}} \Vert k\Vert _{\infty }^{2} \mu _{L}(S^{1}) +2\int _{S^{1}} |k_{L}||k_{R}||\rho | \,{\mathrm{d}}\mu +\Vert k\Vert _{\infty }^{2}\mu _{R}(S^{1}) \\ &\le 2\Vert k\Vert _{\infty }^{2} +2\Vert k\Vert _{\infty }^{2} |\alpha |(S^{1}) \le 4\Vert k\Vert _{\infty }^{2}, \end{aligned}$$

which shows (135). Let \(\pi\) be the natural projection from \(C(S^{1},{\mathbb{C}}^{2})\) to \(C(S^{1},{\mathbb{C}}^{2})/N.\) Then, the inner product \(\langle \,\cdot ,\cdot \,\rangle _{\Sigma }\) on \(C(S^{1},{\mathbb{C}}^{2})/N\) is defined as \(\langle \,\pi (k),\pi (l) \,\rangle _{\Sigma }=\langle \,k,l \,\rangle _{\Sigma }\) for \(k,l \in C(S^{1},{\mathbb{C}}^{2}),\) and \(L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }\) is the completion of \(C(S^{1},{\mathbb{C}}^{2})/N\) by the norm \(\Vert \pi (k)\Vert _{\Sigma }=\sqrt{\langle \,\pi (k),\pi (k) \,\rangle _{\Sigma }}.\) The inner product \(\langle \,\cdot , \cdot \,\rangle _{\Sigma }\) and its norm \(\Vert \cdot \Vert _{\Sigma }\) is defined on the whole of \(L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }\) in a standard manner. The map

$$\begin{aligned} \mathfrak {F}_{{\mathcal{C}}} :C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma } \end{aligned}$$
(136)

is defined by the composition of \({\mathcal{F}}_{{\mathcal{C}}},\) \(\pi\) and the inclusion \(C(S^{1},{\mathbb{C}}^{2})/N \hookrightarrow L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }.\) For any \(u \in {\mathbb{C}}^{2}\) and \(n \in {\mathbb{Z}},\) we have, by (26), \({\mathcal{F}}_{{\mathcal{C}}}[U^{n} (\delta _{0} \otimes u)](\lambda )=\lambda ^{n}u.\) Therefore, the subspace \({\mathcal{F}}_{{\mathcal{C}}}[C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})]\) of \(C(S^{1},{\mathbb{C}}^{2})\) contains the space of \({\mathbb{C}}^{2}\)-valued Laurent polynomials, and hence, it is dense in \(C(S^{1},{\mathbb{C}}^{2})\) with respect to the supremum norm \(\Vert \cdot \Vert _{\infty }.\) We take \(s \in L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }\) and \(\varepsilon >0.\) Then, we can take a function \(k \in C(S^{1},{\mathbb{C}}^{2})\), such that \(\Vert \pi (k) -s\Vert _{\Sigma }<\varepsilon .\) Since \({\mathcal{F}}_{{\mathcal{C}}}[C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})]\) is dense in \(C(S^{1},{\mathbb{C}}^{2}),\) we can take a function \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\), such that \(\Vert {\mathcal{F}}_{{\mathcal{C}}}(f) -k\Vert _{\infty }<\varepsilon .\) Therefore

$$\begin{aligned} \Vert \mathfrak {F}_{{\mathcal{C}}}(f)-s\Vert _{\Sigma }&\le \Vert \pi {\scriptstyle \circ }{\mathcal{F}}_{{\mathcal{C}}}(f) -\pi (k)\Vert _{\Sigma }+\Vert \pi (k) -s\Vert _{\Sigma } \\ &= \Vert {\mathcal{F}}_{{\mathcal{C}}}(f) -k\Vert _{\Sigma } + \Vert \pi (k) -s\Vert _{\Sigma } \\ &\le 4 \Vert {\mathcal{F}}_{{\mathcal{C}}}(f) -k\Vert _{\infty } +\varepsilon \le 5\varepsilon . \end{aligned}$$

This shows that \(\mathfrak {F}_{{\mathcal{C}}}[C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})]\) is dense in \(L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }.\) For any \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}),\) (18) shows that \(\Vert \mathfrak {F}_{{\mathcal{C}}}(f)\Vert _{\Sigma }=\Vert f\Vert\) for \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}).\) To extend the map \(\mathfrak {F}_{{\mathcal{C}}}\) to the whole space \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}),\) we take \(f \in \ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Noting that \(C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) is dense in \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}),\) we take a sequence \(\{f_{n}\}_{n=1}^{\infty }\) of functions \(f_{n} \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\), such that \(\Vert f-f_{n}\Vert \rightarrow 0\) as \(n \rightarrow \infty .\) Then, we have \(\Vert \mathfrak {F}_{{\mathcal{C}}}(f_{n}) -\mathfrak {F}_{{\mathcal{C}}}(f_{m})\Vert _{\Sigma } =\Vert f_{n}-f_{m}\Vert \rightarrow 0\) as \(n,m \rightarrow 0,\) and thus, the limit \(s=\lim _{n \rightarrow \infty }\mathfrak {F}_{{\mathcal{C}}}(f_{n})\) exists in \(L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }.\) We have

$$\begin{aligned} \Vert s\Vert _{\Sigma }=\lim _{n \rightarrow \infty } \Vert \mathfrak {F}_{{\mathcal{C}}}(f_{n})\Vert _{\Sigma }=\lim _{n \rightarrow \infty } \Vert f_{n}\Vert =\Vert f\Vert . \end{aligned}$$
(137)

We take another sequence \(\{g_{n}\}_{n=1}^{\infty }\) of functions \(g_{n} \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) converging f. Then \(\Vert \mathfrak {F}_{{\mathcal{C}}}(g_{n}) -\mathfrak {F}_{{\mathcal{C}}}(f_{n})\Vert _{\Sigma }=\Vert g_{n}-f_{n}\Vert \rightarrow 0\) as \(n \rightarrow \infty .\) Hence

$$\begin{aligned} \Vert \mathfrak {F}_{{\mathcal{C}}}(g_{n})-s\Vert _{\Sigma } \le \Vert \mathfrak {F}_{{\mathcal{C}}}(g_{n}) -\mathfrak {F}_{{\mathcal{C}}}(f_{n})\Vert _{\Sigma } +\Vert \mathfrak {F}_{{\mathcal{C}}}(f_{n}) -s\Vert _{\Sigma } \rightarrow 0 \end{aligned}$$

as \(n \rightarrow \infty .\) This shows that \(s=\lim _{n \rightarrow \infty }\mathfrak {F}_{{\mathcal{C}}}(f_{n})\) does not depend on the choice of sequences of functions in \(C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) converging f. Thus, we can define \(\mathfrak {F}_{{\mathcal{C}}}(f)=\lim _{n \rightarrow \infty }\mathfrak {F}_{{\mathcal{C}}}(f_{n}).\) We prove that the map \(\mathfrak {F}_{{\mathcal{C}}} :\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }\) is unitary. That the map \(\mathfrak {F}_{{\mathcal{C}}}\) preserves the norm has proved in (137). To prove the surjectivity, we take \(s \in L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }.\) For each positive integer n,  we take \(k_{n} \in C(S^{1},{\mathbb{C}}^{2})\) such that \(\Vert \pi (k_{n}) -s\Vert _{\Sigma }<1/n.\) Since \({\mathcal{F}}_{{\mathcal{C}}}[C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})]\) is dense in \(C(S^{1},{\mathbb{C}}^{2}),\) we can take \(f_{n} \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\), such that \(\Vert {\mathcal{F}}_{{\mathcal{C}}}(f_{n}) -k_{n}\Vert _{\infty }<1/n.\) Then, we have

$$\begin{aligned} \Vert \mathfrak {F}_{{\mathcal{C}}}(f_{n})-s\Vert _{\Sigma }&\le \Vert \mathfrak {F}_{{\mathcal{C}}}(f_{n}) - \pi (k_{n})\Vert _{\Sigma } +\Vert \pi (k_{n}) -s\Vert _{\Sigma } \\ &\le 4 \Vert {\mathcal{F}}_{{\mathcal{C}}}(f_{n}) -k_{n}\Vert _{\infty } +\Vert \pi (k_{n}) -s\Vert _{\Sigma }<5/n, \\ \Vert f_{n} -f_{m}\Vert&=\Vert \mathfrak {F}_{{\mathcal{C}}}(f_{n}) -\mathfrak {F}_{{\mathcal{C}}}(f_{m})\Vert _{\Sigma } \\ &\le \Vert \mathfrak {F}_{{\mathcal{C}}}(f_{n}) -s\Vert _{\Sigma } +\Vert s-\mathfrak {F}_{{\mathcal{C}}}(f_{m})\Vert _{\Sigma } <5/n+5/m. \end{aligned}$$

This shows that the sequence \(\{f_{n}\}\) converges to a function \(f \in \ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2})\) and \(\mathfrak {F}_{{\mathcal{C}}}(f)=\lim _{n \rightarrow \infty } \mathfrak {F}_{{\mathcal{C}}}(f_{n})=s.\) This shows that \(\mathfrak {F}_{{\mathcal{C}}}\) is surjective. Therefore, \(\mathfrak {F}_{{\mathcal{C}}} :\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}) \rightarrow L^{2}(S^{1},{\mathbb{C}}^{2})_{\Sigma }\) is a unitary operator. This completes the proof of Theorem 1.7. \(\square\)

5 Examples

In this section, we consider two examples, homogeneous quantum walks and certain quantum walks with non-constant coin matrix. The quantum walk with a non-constant coin matrix considered here is a special case of the so-called two phase models discussed originally in [3]. In these examples, we use Theorem 1.4 to compute concretely the matrix-valued function \({\mathrm{x}}_{0}(\lambda )\) and the positive-matrix-valued measure \(\Sigma .\)

5.1 Homogeneous quantum walks

First of all, let us consider the fundamental example, namely the case where the coin matrix \({\mathcal{C}}\) is constant, say \({\mathcal{C}}(n)=C \in {\mathrm{U}}(2)\) for any \(n \in {\mathbb{Z}}.\) For simplicity of notation, we assume

$$\begin{aligned} C \in {\mathrm{SU}}(2), \end{aligned}$$

so that C can be written as

$$\begin{aligned} C= \left[ \begin{array}{cc} \alpha &{} \beta \\ -\overline{\beta } &{} \overline{\alpha } \end{array}\right] , \quad |\alpha |^{2}+|\beta |^{2}=1,\quad \alpha \ne 0. \end{aligned}$$

The transfer matrix \(T_{\lambda }(n)\) does not depend on \(n \in {\mathbb{Z}}\) and we denote it by \(T_{C}(\lambda ),\) which and whose inverse are given by

$$\begin{aligned} T_{C}(\lambda )&= \left[ \begin{array}{cc} { \frac{1}{\alpha }(\lambda +\lambda ^{-1}|\beta |^{2}) }&{}{ -\lambda ^{-1}\frac{\beta \overline{\alpha }}{\alpha } }\\ { -\lambda ^{-1}\overline{\beta } }&{}{ \lambda ^{-1}\overline{\alpha } } \end{array}\right] , \\ T_{C}(\lambda )^{-1}&= \left[ \begin{array}{cc} { \lambda ^{-1} \alpha }&{}{ \lambda ^{-1}\beta }\\ { \lambda ^{-1} \frac{\alpha \overline{\beta }}{\overline{\alpha }} } &{}{ \frac{1}{\overline{\alpha }} (\lambda +\lambda ^{-1}|\beta |^{2})} \end{array}\right] . \end{aligned}$$

The matrix \(F_{\lambda }(n)\) is then given by

$$\begin{aligned} F_{\lambda }(n)=T_{C}(\lambda )^{n} \quad (n \in {\mathbb{Z}}). \end{aligned}$$

When \(\beta =0,\) we have \(|\alpha |=1\) and the transfer matrix \(T_{C}(\lambda )\) is a diagonal matrix, and the matrices \(F_{\lambda }(n),\) \(F_{1/\overline{\lambda }}(n)^{*}\) are given by

$$\begin{aligned} F_{\lambda }(n)&=T_{C}(\lambda )^{n}=\alpha ^{-n} \left[ \begin{array}{cc} \lambda ^{n} &{} 0 \\ 0 &{} \lambda ^{-n} \end{array}\right] , \\ F_{1/\overline{\lambda }}(n)^{*}&= \alpha ^{n} \left[ \begin{array}{cc} \lambda ^{-n} &{} 0 \\ 0 &{} \lambda ^{n} \end{array}\right] . \end{aligned}$$
(138)

In this case, we can take

$$\begin{aligned} \left\{ \begin{array}{ll} {\mathrm{v}}_{+}(\lambda ) = e_{L},\quad {\mathrm{v}}_{-}(\lambda ) = e_{R} &{}\quad (|\lambda | <1), \\ {\mathrm{v}}_{+}(\lambda ) = e_{R},\quad {\mathrm{v}}_{-}(\lambda ) = e_{L}&{}\quad (|\lambda |>1). \end{array}\right. \end{aligned}$$

for the unit vectors \({\mathrm{v}}_{+}(\lambda ),\) \({\mathrm{v}}_{-}(\lambda )\) in Theorem 1.4. By the definition (11) of \({\mathrm{z}}_{L},\) \({\mathrm{z}}_{R},\) we have \({\mathrm{z}}_{L}(n)=\pi _{L},\) \({\mathrm{z}}_{R}(n)=\pi _{R}.\) By Theorem 1.4 and (21), we have

$$\begin{aligned} {\mathrm{x}}_{0}(\lambda )= \left\{ \begin{array}{ll} 0 &{}\quad (|\lambda |<1), \\ -\lambda ^{-1} I &{}\quad (|\lambda |>1), \end{array}\right. \quad {\mathrm{x}}(\lambda )= \left\{ \begin{array}{ll} I &{}\quad (|\lambda |<1), \\ -I &{}\quad (|\lambda |>1). \end{array}\right. \end{aligned}$$

Therefore, using the characterization (107) of \(\Sigma ,\) it is concluded that the positive-matrix-valued measure \(\Sigma\) is given by \({\mathrm{d}}\Sigma (\zeta )=I {\mathrm{d}}\ell (\zeta ),\) the identity matrix times the normalized Lebesgue measure \({\mathrm{d}}\ell (\zeta ).\) The corresponding QW-Fourier transform \({\mathcal{F}}_{C}\) is given by

$$\begin{aligned} {\mathcal{F}}_{C}[f](\lambda )=\sum _{n \in {\mathbb{Z}}} \alpha ^{n} \left[ \begin{array}{c} \lambda ^{-n}f_{L}(n) \\ \lambda ^{n} f_{R}(n) \end{array}\right] ,\quad f(n)= \left[ \begin{array}{c} f_{L}(n) \\ f_{R}(n) \end{array}\right] \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2}). \end{aligned}$$

This is basically a usual Fourier series expansion. To be more precise, we define the Fourier series \({\mathcal{F}}[f]\) with \(f \in C_{0}({\mathbb{Z}},{\mathbb{C}}^{2})\) by

$$\begin{aligned} {\mathcal{F}}[f](\lambda )=\sum _{n \in {\mathbb{Z}}} \lambda ^{n} f(n) \quad (\lambda \in {\mathbb{C}} {\setminus } \{0\}). \end{aligned}$$
(139)

The function \({\mathcal{F}}[f]\) is a \({\mathbb{C}}^{2}\)-valued Laurent polynomial. We introduce a map \({\mathcal{J}}:C(S^{1},{\mathbb{C}}^{2}) \rightarrow C(S^{1},{\mathbb{C}}^{2})\) by

$$\begin{aligned} ({\mathcal{J}}k)(\zeta ) = \left[ \begin{array}{c} k_{L}(\zeta ^{-1}) \\ k_{R}(\zeta ) \end{array}\right] ,\quad k(\zeta )= \left[ \begin{array}{c} k_{L}(\zeta ) \\ k_{R}(\zeta ) \end{array}\right] \in C(S^{1},{\mathbb{C}}^{2}). \end{aligned}$$

Then, we have

$$\begin{aligned} {\mathcal{J}}{\mathcal{F}}_{{\mathcal{C}}} ={\mathcal{F}}. \end{aligned}$$

Next, we consider the case \(\beta \ne 0.\) In this case, we have \(0<|\alpha |<1.\) The characteristic equation for \(T_{C}(\lambda )\) is

$$\begin{aligned} z^{2}-\frac{2}{\alpha } J(\lambda ) z +\frac{\overline{\alpha }}{\alpha } =0,\quad J(\lambda )=\frac{\lambda +\lambda ^{-1}}{2}. \end{aligned}$$
(140)

Let ST be subsets of \(S^{1}\) defined by

$$\begin{aligned} S=\{\zeta \in S^{1} \mid |{\mathrm{Re}}\,(\zeta )| \le |\alpha |\},\quad T=\{\zeta \in S^{1} \mid |{\mathrm{Re}}\,(\zeta )| \ge |\alpha |\}. \end{aligned}$$
(141)

To calculate \(\Sigma ,\) we prepare the following lemma.

Lemma 5.1

The following holds.

  1. (1)

    The matrix \(T_{C}(\lambda )\) has an eigenvalue in \(S^{1}\) if and only if \(\lambda \in S.\)

  2. (2)

    \(T_{C}(\lambda )\) has an eigenvalue with multiplicity two if and only if \(\lambda\) is one of the following four points:

$$\begin{aligned} |\alpha | +i |\beta |,\quad -|\alpha |+i|\beta |,\quad -|\alpha |-i|\beta |,\quad |\alpha |-i|\beta |. \end{aligned}$$

Proof

Suppose that \(T_{C}(\lambda )\) has an eigenvalue \(z_{o}\) satisfying \(|z_{o}|=1.\) Then, by (140), we have

$$\begin{aligned} J(\lambda ) =\frac{\alpha z_{o} +\overline{\alpha }\overline{z_{o}}}{2}. \end{aligned}$$

This shows that \(J(\lambda )\) is real and \(|J(\lambda )| \le |\alpha |.\) The parameter \(\lambda\) satisfies the equation

$$\begin{aligned} \lambda ^{2}-2r \lambda +1=0,\quad r=J(\lambda ). \end{aligned}$$

Since r is real and \(|r| \le |\alpha |<1,\) we have \(\lambda \in S^{1}\) and \({\mathrm{Re}}\,(\lambda )=J(\lambda ).\) Therefore, \(|{\mathrm{Re}}\,(\lambda )| \le |\alpha |.\) Conversely, suppose that \(\lambda \in S^{1}\) and \(|{\mathrm{Re}}\,(\lambda )| \le |\alpha |.\) Since \(|\lambda |=1,\) we see \(J(\lambda )={\mathrm{Re}}\,(\lambda )\) and \(|J(\lambda )| \le |\alpha | <1.\) We write \(\lambda ={\mathrm{e}}^{i\theta }\) with \(\theta \in (0,\pi ).\) Then, \(\cos \theta =J(\lambda )\) and the solution to Eq. (140) is

$$\begin{aligned} z=\frac{\cos \theta \pm i \sqrt{|\alpha |^{2} -\cos ^{2}\theta }}{\alpha }. \end{aligned}$$
(142)

Then, a direct calculation shows \(|z|=1,\) and which proves (1). Next, suppose that \(\lambda\) is one of the four points in (2). Then, \(J(\lambda )=\pm |\alpha |,\) and the discriminant of (140) is zero. Hence, \(T_{C}(\lambda )\) has an eigenvalue with multiplicity two. Conversely, suppose that \(T_{C}(\lambda )\) has an eigenvalue \(z_{o}\) with multiplicity two. Then, by (140), we have \(J(\lambda )^{2}=|\alpha |^{2}.\) Therefore, we see \(\lambda =J(\lambda ) \pm i \sqrt{1-|\alpha |^{2}}\) which coincides with one of the four points in the statement. \(\square\)

Thus, for \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}),\) the matrix \(T_{C}(\lambda )\) has two mutually different eigenvalues, and, by (140), the absolute value of one of them is less than one and that of another is greater than one. We remark that, since \(T_{C}(\lambda )\) is holomorphic and its eigenvalues are simple for \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}),\) the eigenvalues can be labelled, so that they are holomorphic there. Let \(z_{\pm }(\lambda )\) be the eigenvalues of \(T_{C}(\lambda )\) satisfying \(|z_{+}(\lambda )|<1<|z_{-}(\lambda )|\) for \(\lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1})\) and holomorphic there.

Lemma 5.2

The function \(z_{+}(\lambda )\) is given by

$$\begin{aligned}&\alpha z_{+}(\lambda ) \\ &\quad = \left\{ \begin{array}{ll} { J(\lambda )-\sqrt{J(\lambda )^{2}-|\alpha |^{2}} }&{}\quad (|\lambda |<1,\, {\mathrm{Re}}\,\lambda>0) \ {\mathrm{or}} \ (|\lambda |>1,\, {\mathrm{Re}}\,\lambda>0), \\ { J(\lambda )+\sqrt{J(\lambda )^{2}-|\alpha |^{2}} }&{}\quad (|\lambda |<1,\, {\mathrm{Re}}\,\lambda<0) \ {\mathrm{or}} \ (|\lambda |>1,\, {\mathrm{Re}}\,\lambda<0), \\ { J(\lambda )+i \sqrt{|\alpha |^{2}-J(\lambda )^{2}} }&{}\quad (|\lambda |<1,\, \mathrm{Im}\,\lambda>0) \ {\mathrm{or}} \ (|\lambda |>1,\, \mathrm{Im}\,\lambda<0), \\ { J(\lambda )-i \sqrt{|\alpha |^{2}-J(\lambda )^{2}} }&{}\quad (|\lambda |<1,\, \mathrm{Im}\,\lambda <0) \ {\mathrm{or}} \ (|\lambda |>1,\, \mathrm{Im}\,\lambda >0), \end{array}\right. \end{aligned}$$
(143)

where we denote \(\sqrt{z}\) for \(z \in {\mathbb{C}} {\setminus } (-\infty ,0]\) the branch of the square root whose values are positive for positive real numbers.

Proof

These expressions are solutions to Eq. (140). We also remark that the function \(\sqrt{J(\lambda )^{2}-|\alpha |^{2}}\) is holomorphic on \({\mathbb{C}} {\setminus } (S \cup i{\mathbb{R}})\) and the function \(\sqrt{|\alpha |^{2}-J(\lambda )^{2}}\) is holomorphic on \({\mathbb{C}} {\setminus } (T \cup {\mathbb{R}}).\) For a nonzero real number \(\lambda ,\) the function \(z_{+}(\lambda )\) is given by

$$\begin{aligned} \alpha z_{+}(\lambda ) = \left\{ \begin{array}{ll} J(\lambda ) -\sqrt{J(\lambda )^{2}-|\alpha |^{2}} &{}\quad (0<\lambda ), \\ J(\lambda ) +\sqrt{J(\lambda )^{2}-|\alpha |^{2}} &{}\quad (0>\lambda ), \end{array}\right. \end{aligned}$$
(144)

which can be checked by the requirement \(|z_{+}(\lambda )|<1.\) The function in the first line of (144) is holomorphic on \(\{ \lambda \in {\mathbb{C}} {\setminus } \{0\} \mid {\mathrm{Re}}\,(\lambda )>0\} {\setminus } S,\) and that in the second line is holomorphic on \(\{ \lambda \in {\mathbb{C}} {\setminus } \{0\} \mid {\mathrm{Re}}\,(\lambda )<0\} {\setminus } S.\) Thus, these expressions of \(z_{+}(\lambda )\) still hold on the respective regions. We note that

$$\begin{aligned} \sqrt{z}= \left\{ \begin{array}{ll} i\sqrt{-z} &{}\quad (\mathrm{Im}\,z>0), \\ -i \sqrt{-z} &{}\quad (\mathrm{Im}\,z <0). \end{array}\right. \end{aligned}$$

A direct calculation shows

$$\begin{aligned} \mathrm{Im}\,(J(\lambda )^{2} -|\alpha |^{2}) =-\frac{1}{2|\lambda |^{4}}(1-|\lambda |^{4}) {\mathrm{Re}}\,(\lambda ) \mathrm{Im}\,(\lambda ). \end{aligned}$$

Thus, we see

$$\begin{aligned}&\sqrt{J(\lambda )^{2}-|\alpha |^{2}} \\ &\quad = \left\{ \begin{array}{ll} -i \sqrt{|\alpha |^{2}-J(\lambda )^{2}} &{}\quad (0<|\lambda |<1,\, {\mathrm{Re}}\,(\lambda )>0,\,\mathrm{Im}\,(\lambda )>0), \\ i \sqrt{|\alpha |^{2}-J(\lambda )^{2}} &{}\quad (0<|\lambda |<1,\, {\mathrm{Re}}\,(\lambda )<0,\, \mathrm{Im}\,(\lambda )>0), \end{array}\right. \end{aligned}$$

and hence, we have

$$\begin{aligned} \alpha z_{+}(\lambda )= J(\lambda ) +i \sqrt{|\alpha |^{2}-J(\lambda )^{2}} \quad (0<|\lambda |<1,\, \mathrm{Im}\,(\lambda )>0). \end{aligned}$$

Other expressions of \(z_{+}(\lambda )\) are obtained in the same way. \(\square\)

Remark 5.3

As in the proof of Lemma 5.2, the function \(z_{+}(\lambda )\) is analytically continued through the relative interior of the subset T in \(S^{1}.\)

For the case of the constant coin, it is rather easy to use the usual Fourier series (139) to calculate the matrix \({\mathrm{x}}_{0}(\lambda ).\) Indeed, we have

$$\begin{aligned} R_{\lambda }(n,m)=\int _{S^{1}} z^{m-n} \left( \widehat{U}(z) -\lambda \right) ^{-1}\,{\mathrm{d}}\ell (z), \end{aligned}$$
(145)

where \(\widehat{U}(z)\) is the matrix-valued function given by

$$\begin{aligned} \widehat{U}(z)= \left[ \begin{array}{cc} \alpha z^{-1} &{} \beta z^{-1} \\ -\overline{\beta } z &{} \overline{\alpha } z \end{array}\right] . \end{aligned}$$

Calculating the integral (145) for \(n=m=0\) using the residue formula and the fact that \(z_{\pm }(\lambda )\) are the solutions to Eq. (140), we see

$$\begin{aligned} {\mathrm{x}}_{0}(\lambda ) = \frac{1}{\lambda \alpha (z_{+}(\lambda ) -z_{-}(\lambda ))} \left[ \begin{array}{cc} { \lambda -\alpha z_{+}(\lambda ) } &{}{ \beta z_{+}(\lambda ) }\\ -{ \frac{\overline{\beta }}{\overline{\alpha }} \alpha z_{+}(\lambda ) } &{} { \lambda -\alpha z_{+}(\lambda ) } \end{array}\right] . \end{aligned}$$
(146)

The formula (146) can also be obtained using Corollary 1.4, although it needs somehow complicated calculation. We give an outline of a calculation. We take the vectors \({\mathrm{v}}_{+}(\lambda ),\) \({\mathrm{v}}_{-}(\lambda )\) as eigenvectors corresponding to the eigenvalues \(z_{\pm }(\lambda ).\) Explicitly, we set

$$\begin{aligned} {\mathrm{v}}_{+}(\lambda )&=\frac{1}{\sqrt{|\lambda z_{+}(\lambda ) -\overline{\alpha }|^{2}+|\beta |^{2}}} \left[ \begin{array}{c} \lambda z_{+}(\lambda ) -\overline{\alpha } \\ -\overline{\beta } \end{array}\right] , \\ {\mathrm{v}}_{-}(\lambda )&=\frac{1}{\sqrt{|\lambda z_{-}(\lambda ) -\overline{\alpha }|^{2}+|\beta |^{2}}} \left[ \begin{array}{c} \lambda z_{-}(\lambda ) -\overline{\alpha } \\ -\overline{\beta } \end{array}\right] . \end{aligned}$$
(147)

Suppose that \(0<\lambda <1.\) Then, \(\alpha z_{\pm }(\lambda )\) is real. Using the fact that \(z_{\pm }(\lambda )\) are the roots of (140), we find that the vectors \({\mathrm{v}}_{+}(\lambda )\) and \({\mathrm{v}}_{-}(\lambda )\) form an orthonormal basis in \({\mathbb{C}}^{2}\) for \(\lambda >0.\) By Theorem 1.4, we see

$$\begin{aligned} {\mathrm{x}}_{0}(\lambda )e_{L}&= -\frac{\lambda \alpha z_{-}(\lambda ) -1}{\lambda \overline{\alpha } (|\lambda z_{-}(\lambda ) -\overline{\alpha }|^{2}+|\beta |^{2})} \left[ \begin{array}{c} \lambda z_{-}(\lambda ) -\overline{\alpha } \\ -\overline{\beta } \end{array}\right] , \\ {\mathrm{x}}_{0}(\lambda ) e_{R}&= \frac{\beta z_{+}(\lambda )}{\overline{\alpha } (|\lambda z_{+}(\lambda ) -\overline{\alpha }|^{2}+|\beta |^{2})} \left[ \begin{array}{c} \lambda z_{+}(\lambda ) -\overline{\alpha } \\ -\overline{\beta } \end{array}\right] . \end{aligned}$$
(148)

That the two formulas (148) and (146) are identical for \(0<\lambda <1\) can be verified by the formula

$$\begin{aligned} |\lambda z_{+}(\lambda ) -\overline{\alpha }|^{2}+|\beta |^{2}&= \frac{\lambda \alpha }{\overline{\alpha }} (z_{+}-z_{-}) (\lambda z_{+} -\overline{\alpha } ), \\ |\lambda z_{-}(\lambda ) -\overline{\alpha }|^{2}+|\beta |^{2}&= \frac{\lambda \alpha }{\overline{\alpha }} (z_{+}-z_{-}) (\overline{\alpha } - \lambda z_{-} ), \end{aligned}$$

which hold for \(0<\lambda <1.\) Then, one can use the coincidence theorem for holomorphic function \({\mathrm{x}}_{0}(\lambda )\) to show that (146) holds also on the region \({\mathrm{Re}}\,(\lambda )>0,\) \(0<|\lambda |<1.\) The same discussion works well for other region in \({\mathbb{C}} {\setminus } (\{0\} \cup S^{1}).\) We note that from the formula (145), the Green function must satisfy

$$\begin{aligned} R_{\lambda }(n,n)=R_{\lambda }(0,0)={\mathrm{x}}_{0}(\lambda ), \end{aligned}$$

which is, according to (13) in Theorem 1.2, equivalent to

$$\begin{aligned} T_{C}(\lambda ) [{\mathrm{x}}_{0}(\lambda )+{\mathrm{z}}_{L}(0)] T_{C}(1/\overline{\lambda })^{*}=[{\mathrm{x}}_{0}(\lambda )+{\mathrm{z}}_{L}(0)], \end{aligned}$$
(149)

because \({\mathrm{z}}_{L}(n)={\mathrm{z}}_{L}(0).\) Equation (149) can be verified directly by (146) and the definition of \(T_{C}(\lambda ).\) A straightforward calculation shows

$$\begin{aligned} {\mathrm{x}}(\lambda )&=I +2 \lambda {\mathrm{x}}_{0}(\lambda ) \\ &=\frac{2}{\alpha (z_{+}(\mu ) -z_{-}(\lambda ))} \left[ \begin{array}{cc} { K(\lambda ) }&{}{ \frac{\beta }{\alpha } \alpha z_{+}(\lambda ) } \\ { -\frac{ \overline{\beta }}{\overline{\alpha }} \alpha z_{+}(\lambda ) }&{}{ K(\lambda ) } \end{array}\right] , \\ K(\lambda )&=\frac{\lambda -\lambda ^{-1}}{2}. \end{aligned}$$
(150)

The positive-matrix-valued measure \(\Sigma\) is described as follows.

Theorem 5.4

The positive-matrix-valued measure \({\mathrm{d}}\Sigma (\zeta )\) is given by

$$\begin{aligned} {\mathrm{d}}\Sigma (\zeta )=\frac{1}{\sqrt{|\alpha |^{2}-{\mathrm{Re}}\,\zeta ^{2}}} \left[ \begin{array}{cc} { |\mathrm{Im}\,(\zeta )| }&{}{ - i \frac{\beta }{\alpha } {\mathrm{Re}}\,\zeta }\\ { i \frac{\overline{\beta }}{\overline{\alpha }} {\mathrm{Re}}\,\zeta }&{}{ |\mathrm{Im}\,(\zeta )| } \end{array}\right] \chi _{\sigma (U)} \,{\mathrm{d}}\ell (\zeta ), \end{aligned}$$

where \(\chi _{\sigma (U)}\) is the characteristic function of the spectrum \(\sigma (U)=\{\zeta \in S^{1} \mid {\mathrm{Re}}\,\zeta \le |\alpha |\}\) of U.

Proof of Theorem 5.4

Let \(u \in {\mathbb{C}}^{2}\) and write it as

$$\begin{aligned} u= \left[ \begin{array}{c} u_{L} \\ u_{R} \end{array}\right] ,\quad u_{L},\, u_{R} \in {\mathbb{C}}. \end{aligned}$$

We need to calculate the limit

$$\begin{aligned} \lim _{r \uparrow 1} \int _{S^{1}} h(\zeta ) \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(r\zeta )] u,u \,\rangle _{{\mathbb{C}}^{2}}\,{\mathrm{d}}\ell (\zeta ) \end{aligned}$$
(151)

for \(h \in C(S^{1}).\) We have the expression

$$\begin{aligned} {\mathrm{Re}}\,{\mathrm{x}}(\lambda ) = \left[ \begin{array}{cc} { 2 {\mathrm{Re}}\,\left( \frac{K(\lambda )}{\alpha (z_{+}(\lambda ) - z_{-}(\lambda ))} \right) } &{} { 2i\frac{\beta }{\alpha } \mathrm{Im}\,\left( \frac{z_{+}(\lambda )}{z_{+}(\lambda ) - z_{-}(\lambda )} \right) } \\ { - 2i\frac{\overline{\beta }}{\overline{\alpha }} \mathrm{Im}\,\left( \frac{z_{+}(\lambda )}{z_{+}(\lambda ) - z_{-}(\lambda )} \right) } &{} { 2 {\mathrm{Re}}\,\left( \frac{K(\lambda )}{\alpha (z_{+}(\lambda ) - z_{-}(\lambda ))} \right) } \end{array}\right] , \end{aligned}$$

and hence

$$\begin{aligned} \langle \,[{\mathrm{Re}}\,{\mathrm{x}}(\lambda )] u, u \,\rangle _{{\mathbb{C}}^{2}}&= 2\Vert u\Vert _{{\mathbb{C}}^{2}}^{2} {\mathrm{Re}}\,\left[ \frac{K(\lambda )}{\alpha (z_{+}(\lambda ) -z_{-}(\lambda ))} \right] \\ &\quad -4\mathrm{Im}\,\left( \frac{\beta }{\alpha } \overline{u_{L}} u_{R}\right) \mathrm{Im}\,\left[ \frac{z_{+}(\lambda )}{z_{+}(\lambda ) -z_{-}(\lambda )} \right] . \end{aligned}$$
(152)

Thus, we need to calculate the limits of the integrals

$$\begin{aligned}&\int _{S^{1}} h(\zeta ) \, {\mathrm{Re}}\,\left[ \frac{K(r\zeta )}{\alpha (z_{+}(r\zeta ) -z_{-}(r\zeta ))} \right] \,{\mathrm{d}}\ell (\zeta ), \\ &\int _{S^{1}} h(\zeta ) \, \mathrm{Im}\,\left[ \frac{z_{+}(r\zeta )}{z_{+}(r\zeta ) -z_{-}(r\zeta )} \right] \,{\mathrm{d}}\ell (\zeta ) \end{aligned}$$

as \(r \uparrow 1.\) We only consider the integral on the positive orthant \(\Delta =\{\zeta \in S^{1} \mid {\mathrm{Re}}\,(\zeta ) \ge 0, \mathrm{Im}\,(\zeta ) \ge 0\}\), because the other parts of the integral can be handled similarly. Thus, we set

$$\begin{aligned} I_{r}&=\int _{\Delta } h(\zeta ) \, {\mathrm{Re}}\,\left[ \frac{K(r\zeta )}{\alpha (z_{+}(r\zeta ) -z_{-}(r\zeta ))} \right] \,{\mathrm{d}}\ell (\zeta ) \\ &= \frac{1}{2\pi }\int _{0}^{\pi /2} h({\mathrm{e}}^{it}) \, {\mathrm{Re}}\,\left[ \frac{K(r{\mathrm{e}}^{it})}{\alpha (z_{+}(r{\mathrm{e}}^{it}) -z_{-}(r{\mathrm{e}}^{it}))} \right] \,{\mathrm{d}}t, \\ J_{r}&= \int _{\Delta } h(\zeta ) \, \mathrm{Im}\,\left[ \frac{z_{+}(r\zeta )}{z_{+}(r\zeta ) -z_{-}(r\zeta )} \right] \,{\mathrm{d}}\ell (\zeta ) \\ &=\frac{1}{2\pi } \int _{0}^{\pi /2} h({\mathrm{e}}^{it}) \, \mathrm{Im}\,\left[ \frac{z_{+}(r{\mathrm{e}}^{it})}{z_{+}(r{\mathrm{e}}^{it}) -z_{-}(r{\mathrm{e}}^{it})} \right] \,{\mathrm{d}}t. \end{aligned}$$

Let \(\psi \in (0,\pi /2)\) be the real number satisfying \(\cos \psi =|\alpha |,\) and we divide the arc \(\Delta\) into two parts, \(D_{1}=\{0 \le t \le \psi \}\) and \(D_{2}=\{\psi \le t \le \pi /2\}.\) From Lemma 5.2, we have

$$\begin{aligned} \frac{z_{+}(\lambda )}{z_{+}(\lambda ) -z_{-}(\lambda )} =\frac{J(\lambda )}{2i \sqrt{|\alpha |^{2}-J(\lambda )^{2}}} +\frac{1}{2} \quad (0<|\lambda |<1,\ \mathrm{Im}\,(\lambda )>0). \end{aligned}$$

By a direct calculation, we see

$$\begin{aligned} \left| |\alpha |^{2}-J(r{\mathrm{e}}^{it})^{2} \right| ^{2}&= D(t)^{2} +K(r)^{2}[J(r)^{2}\sin ^{2}(2t)-2D(t) \cos (2t)] \\ &\quad +K(r)^{4} \cos ^{2}(2t), \end{aligned}$$

where we set

$$\begin{aligned} D(t)=|\alpha |^{2} -\cos ^{2}t. \end{aligned}$$

Since \(J(r) \ge 1,\) \(\sin (2t) \ne 0\) and \(D(\psi )=0,\) we can take a sufficiently small \(\varepsilon >0\), such that

$$\begin{aligned} J(r)^{2} \sin ^{2}(2t) -2 D(t) \cos 2t >0 \quad (t \in [\psi ,\psi +\varepsilon ],\ r \in [2/3,1]). \end{aligned}$$

Then, we see

$$\begin{aligned}&\left| |\alpha |^{2}-J(r{\mathrm{e}}^{it})^{2} \right| ^{2}>D(t)^{2},\quad \text{hence} \\ &\quad \frac{1}{\left| \sqrt{|\alpha |^{2} -J(r{\mathrm{e}}^{it})^{2}}\right| } \le \frac{1}{\sqrt{D(t)}} =\frac{1}{\sqrt{|\alpha |^{2}-\cos ^{2}t }} \end{aligned}$$
(153)

for \(t \in [\psi ,\psi +\varepsilon ]\) and \(r \in [2/3,1].\) For \(t \in (\psi ,\pi /2],\) we have

$$\begin{aligned} \lim _{r \uparrow 1} \mathrm{Im}\,\left[ \frac{z_{+}(r{\mathrm{e}}^{it})}{z_{+}(r{\mathrm{e}}^{it}) -z_{-}(r{\mathrm{e}}^{it})} \right] = -\frac{\cos t}{2D(t)^{1/2}}=-\frac{{\mathrm{Re}}\,(\zeta )}{2 \sqrt{|\alpha |^{2}-{\mathrm{Re}}\,(\zeta )^{2}}}. \end{aligned}$$

Since \(1/D(t)^{1/2}\) is integrable on \([\psi ,\pi /2],\) the Lebesgue convergence theorem shows

$$\begin{aligned}&\lim _{r \uparrow 1} \int _{\psi }^{\psi +\varepsilon } h({\mathrm{e}}^{it}) \mathrm{Im}\,\left[ \frac{z_{+}(r{\mathrm{e}}^{it})}{z_{+}(r{\mathrm{e}}^{it}) -z_{-}(r{\mathrm{e}}^{it})} \right] \,{\mathrm{d}}t \\ &\quad =- \int _{\psi }^{\psi +\varepsilon } h({\mathrm{e}}^{it}) \frac{\cos t}{2\sqrt{|\alpha |^{2} -\cos ^{2}t}}\,{\mathrm{d}}t. \end{aligned}$$

On the interval \([\psi +\varepsilon ,\pi /2],\) D(t) is bounded, and thus, the Lebesgue convergence theorem is also applicable. Therefore, we see

$$\begin{aligned}&\lim _{r \uparrow 1} \int _{D_{2}} h(\zeta ) \mathrm{Im}\,\left[ \frac{z_{+}(r\zeta )}{z_{+}(r\zeta ) -z_{-}(r\zeta )} \right] \,{\mathrm{d}}\ell (\zeta ) \\ &\quad =- \int _{D_{2}} h(\zeta ) \frac{{\mathrm{Re}}\,(\zeta )}{2\sqrt{|\alpha |^{2} -{\mathrm{Re}}\,(\zeta )^{2}}}\,{\mathrm{d}}\ell (\zeta ). \end{aligned}$$

On the arc \(D_{1},\) we can use the first expression of \(z_{+}(\lambda )\) in (143). We can still use the estimate (153) around \(t=\psi .\) However, this time, the imaginary part of \(z_{+}(r\zeta )/(z_{+}(r\zeta ) -z_{-}(r\zeta ))\) tends to zero as \(r \uparrow 1.\) Therefore, again, the Lebesgue convergence theorem shows

$$\begin{aligned} \lim _{r \uparrow 1} \int _{D_{1}} h(\zeta ) \mathrm{Im}\,\left[ \frac{z_{+}(r\zeta )}{z_{+}(r\zeta ) -z_{-}(r\zeta )} \right] \,{\mathrm{d}}\ell (\zeta ) =0. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \lim _{r \uparrow 1} J_{r}=- \int _{D_{2}} h(\zeta ) \frac{{\mathrm{Re}}\,(\zeta )}{2\sqrt{|\alpha |^{2} -{\mathrm{Re}}\,(\zeta )^{2}}}\,{\mathrm{d}}\ell (\zeta ). \end{aligned}$$
(154)

The limit of the integral \(I_{r}\) also handled in the same way. This time, on the arc \(D_{1},\) the real part of \(K(\zeta )\) equals zero. On the arc \(D_{2},\) the real part of \(K(r\zeta )/\alpha (z_{+}(r\zeta ) -z_{-}(r\zeta ))\) converges, as \(r \uparrow 1,\) to \(\mathrm{Im}\,(\zeta ) /2\sqrt{|\alpha |^{2}-{\mathrm{Re}}\,(\zeta )^{2}}.\) Then, we obtain

$$\begin{aligned} \lim _{r \uparrow 1} I_{r}= \int _{D_{2}} h(\zeta ) \frac{\mathrm{Im}\,(\zeta )}{2\sqrt{|\alpha |^{2}-{\mathrm{Re}}\,(\zeta )^{2}}}\,{\mathrm{d}}\ell (\zeta ). \end{aligned}$$
(155)

The same calculations on other orthants combined with (152) show the assertion. \(\square\)

5.2 Simplest two-phase model

Let \(C_{0},\) \(C_{\pm }\) be three \(2 \times 2\) special unitary matrices, and we write them as

$$\begin{aligned}&C_{0} = \left[ \begin{array}{cc} \alpha _{0} &{} \beta _{0} \\ -\overline{\beta }_{0} &{} \overline{\alpha _{0}} \end{array}\right] , \quad C_{\pm } = \left[ \begin{array}{cc} \alpha _{\pm } &{} \beta _{\pm } \\ -\overline{\beta _{\pm }} &{} \overline{\alpha _{\pm }} \end{array}\right] , \\ &|\alpha _{0}|^{2}+|\beta _{0}|^{2}=|\alpha _{\pm }|^{2}+|\beta _{\pm }|^{2}=1. \end{aligned}$$

Let \({\mathcal{C}}:{\mathbb{Z}} \rightarrow {\mathrm{U}}(2)\) be a coin matrix defined as

$$\begin{aligned} {\mathcal{C}}(n)= \left\{ \begin{array}{ll} C_{+} &{}\quad (n \ge 1), \\ C_{0} &{}\quad (n=0), \\ C_{-} &{}\quad (n \le -1). \end{array}\right. \end{aligned}$$

The quantum walk \(U({\mathcal{C}})\) defined by the coin matrix \({\mathcal{C}}\) is called a two-phase model with one defect. Since the calculation is a bit complicated, we impose the following strong assumptions:

$$\begin{aligned} \begin{gathered} 0<\rho :=|\alpha _{+}|=|\alpha _{-}|<1,\quad \beta :=\beta _{+}=\beta _{-}, \quad |{\mathrm{Re}}\,(\beta _{0}\beta )| < |\beta |^{2}. \end{gathered} \end{aligned}$$
(156)

Therefore, \(C_{+}\) and \(C_{-}\) differs only by the phase of the diagonals. We note that the first two assumptions are for simplifying the calculation, but the last assumption is imposed for U to have eigenvalues. Eigenvalues of much general two-phase models are studied in [13]. In this case, the transfer matrix \(T_{\lambda }(n)\) satisfies

$$\begin{aligned} T_{\lambda }(n)= \left\{ \begin{array}{ll} T_{C_{+}}(\lambda ) &{}\quad (n \ge 1), \\ T_{C_{-}}(\lambda ) &{}\quad (n \le -2), \end{array}\right. \end{aligned}$$

and the matrix-valued function \(F_{\lambda }(n)\) satisfies

$$\begin{aligned} F_{\lambda }(n)= \left\{ \begin{array}{ll} T_{C_{+}}(\lambda )^{n-1} T_{\lambda }(0) &{}\quad (n \ge 1), \\ T_{C_{-}}(\lambda )^{n+1} T_{\lambda }(-1)^{-1} &{}\quad (n \le -1). \end{array}\right. \end{aligned}$$
(157)

As in the case of constant coins, we denote \(z_{\pm }(C_{\pm }, \lambda )\) the eigenvalues of \(T_{C_{\pm }}(\lambda )\) satisfying

$$\begin{aligned} |z_{+}(C_{\pm }, \lambda )|<1<|z_{-}(C_{\pm }, \lambda )|,\quad \lambda \in {\mathbb{C}} {\setminus } (\{0\} \cup S^{1}), \end{aligned}$$
(158)

and we define vectors \(w_{\pm }(C_{\pm }, \lambda )\) as

$$\begin{aligned}&w_{\pm }(C_{+}, \lambda ) = \left[ \begin{array}{c} { \overline{\alpha _{0}} \left( \lambda z_{\pm }(C_{+}, \lambda ) -\overline{\alpha _{+}} \right) }\\ - \overline{\alpha _{0}} \overline{\beta } \end{array}\right] , \\ &w_{\pm }(C_{-}, \lambda ) = \left[ \begin{array}{c} { \alpha _{0} \alpha _{-} \left( z_{\pm }(C_{-}, \lambda ) -\overline{\alpha _{-}} \lambda ^{-1} \right) }\\ - \alpha _{0} \alpha _{-} \overline{\beta } \lambda ^{-1} \end{array}\right] . \end{aligned}$$
(159)

The vectors \(w_{\pm }(C_{\pm }, \lambda )\) are eigenvectors of \(T_{C_{\pm }}(\lambda )\) with the eigenvalues \(z_{\pm }(C_{\pm }, \lambda ),\) respectively. We define unit vectors \({\mathrm{v}}_{\pm }(\lambda )\) by

$$\begin{aligned} {\mathrm{v}}_{+}(\lambda )&=\frac{1}{\Vert T_{\lambda }(0)^{-1}w_{+}(C_{+}, \lambda )\Vert _{{\mathbb{C}}^{2}}} T_{\lambda }(0)^{-1}w_{+}(C_{+}, \lambda ), \\ {\mathrm{v}}_{-}(\lambda )&=\frac{1}{\Vert T_{\lambda }(-1) w_{-}(C_{-}, \lambda )\Vert _{{\mathbb{C}}^{2}} } T_{\lambda }(-1) w_{-}(C_{-}, \lambda ). \end{aligned}$$
(160)

These vectors satisfy

$$\begin{aligned} F_{\lambda }(n)&{\mathrm{v}}_{+}(\lambda ) =z_{+}(C_{+}, \lambda )^{n-1} T_{\lambda }(0) {\mathrm{v}}_{+}(\lambda ) \quad (n \ge 1), \\ F_{\lambda }(n)&{\mathrm{v}}_{-}(\lambda ) =z_{-}(C_{-}, \lambda )^{n+1} T_{\lambda }(-1)^{-1} {\mathrm{v}}_{-}(\lambda ) \quad (n \le -1). \end{aligned}$$
(161)

From (161), it follows that we can use the unit vectors \({\mathrm{v}}_{+}(\lambda ),\) \({\mathrm{v}}_{-}(\lambda )\) to apply Theorem 1.4. We set

$$\begin{aligned} S=\{\zeta \in S^{1} \mid |{\mathrm{Re}}\,(\zeta )| \le \rho \}. \end{aligned}$$
(162)

Then, by Lemma 5.1, for \(\lambda \in {\mathbb{C}} {\setminus } \{0\},\) we have \(|z_{\pm }(C_{\pm },\lambda )|=1\) if and only if \(\lambda \in S.\)

Lemma 5.5

There are no eigenvalues of U in S.

Proof

This is proved in [17], but we give a proof for completeness. Suppose contrary that U has an eigenvalue \(\lambda\) in S. Then, there exists a nonzero vector \(u \in {\mathbb{C}}^{2}\), such that the function \(\Phi _{\lambda }(u)\) defined in (7) is in \(\ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}).\) Since \(\lambda \in S,\) all the eigenvalues of \(T_{C_{\pm }}(\lambda )\) are in \(S^{1}\) by Lemma 5.1. By an argument using the Jordan normal form of \(T_{C_{\pm }}(\lambda ),\) we can find a positive constant c, such that

$$\begin{aligned} \Vert T_{C_{\pm }}(\lambda )^{n} w\Vert _{{\mathbb{C}}^{2}} \ge c\Vert w\Vert _{{\mathbb{C}}^{2}} \end{aligned}$$

for any \(w \in {\mathbb{C}}^{2}\) and any positive integer n. By (157), we have

$$\begin{aligned} \Vert \Phi _{\lambda }(u)(n)\Vert _{{\mathbb{C}}^{2}}^{2} \ge c^{2}\Vert T_{\lambda }(0)u\Vert _{{\mathbb{C}}^{2}}^{2}, \quad \Vert \Phi _{\lambda }(u)(-n)\Vert _{{\mathbb{C}}^{2}}^{2} \ge c^{2}\Vert T_{\lambda }(-1)^{-1}u\Vert _{{\mathbb{C}}^{2}}^{2}. \end{aligned}$$

Since \(\Phi _{\lambda }(u) \in \ell ^{2}({\mathbb{Z}},{\mathbb{C}}^{2}),\) we must have \(T_{\lambda }(0)u=T_{\lambda }(-1)^{-1}u=0.\) This implies that \(u=0\) which is a contradiction. \(\square\)

The matrices \(T_{\lambda }(0),\) \(T_{\lambda }(-1)\) and their inverses are given by

$$\begin{aligned}&T_{\lambda }(0)= \left[ \begin{array}{cc} { \frac{1}{\alpha _{+}} (\lambda +\lambda ^{-1}\overline{\beta _{0}}\beta ) }&{}{ -\lambda ^{-1} \frac{\overline{\alpha _{0}}\beta }{\alpha _{+}} }\\ { -\lambda ^{-1} \overline{\beta _{0}} }&{}{ \lambda ^{-1} \overline{\alpha _{0}} } \end{array}\right] , \\ &T_{\lambda }(0)^{-1} = \left[ \begin{array}{cc} { \lambda ^{-1} \alpha _{+} }&{}{ \lambda ^{-1} \beta }\\ { \lambda ^{-1} \frac{\overline{\beta _{0}} \alpha _{+}}{ \overline{\alpha _{0}}} }&{} { \frac{1}{\overline{\alpha _{0}}} (\lambda +\lambda ^{-1} \overline{\beta _{0}} \beta ) } \end{array}\right] , \\ &T_{\lambda }(-1) = \left[ \begin{array}{cc} { \frac{1}{\alpha _{0}} (\lambda + \lambda ^{-1} \overline{\beta } \beta _{0}) }&{}{ -\lambda ^{-1} \frac{\beta _{0} \overline{\alpha _{-}}}{\alpha _{0}} }\\ { -\lambda ^{-1} \overline{\beta } }&{}{ \lambda ^{-1} \overline{\alpha _{-}} } \end{array}\right] , \\ &T_{\lambda }(-1)^{-1} = \left[ \begin{array}{cc} { \lambda ^{-1} \alpha _{0} }&{}{ \lambda ^{-1} \beta _{0} }\\ { \lambda ^{-1} \frac{\overline{\beta }\alpha _{0}}{\overline{\alpha _{-}}} }&{} { \frac{1}{\overline{\alpha _{-}}} (\lambda +\lambda ^{-1} \overline{\beta } \beta _{0}) } \end{array}\right] . \end{aligned}$$

Then, the concrete form of the vectors \({\mathrm{v}}_{\pm }(\lambda )\) is given by

$$\begin{aligned} {\mathrm{v}}_{+}(\lambda )&=\frac{1}{D_{+}(\lambda )} \left[ \begin{array}{c} { \overline{\alpha _{0}} \left( Z_{+}(\lambda ) -\lambda ^{-1}\right) }\\ { \overline{\beta _{0}} \left( Z_{+}(\lambda ) -\lambda ^{-1} \right) -\overline{\beta } \lambda } \end{array}\right] , \\ {\mathrm{v}}_{-}(\lambda )&=\frac{1}{D_{-}(\lambda )} \left[ \begin{array}{c} { \left( \lambda +\beta _{0} \overline{\beta } \lambda ^{-1} \right) Z_{-}(\lambda ) -\rho ^{2} }\\ { - \alpha _{0} \overline{\beta } \lambda ^{-1} Z_{-}(\lambda ) } \end{array}\right] , \end{aligned}$$

where \(D_{\pm }(\lambda )\) are normalization terms and \(Z_{\pm }(\lambda )\) are given by

$$\begin{aligned} Z_{+}(\lambda )=\alpha _{+} z_{+}(C_{+},\lambda ),\quad Z_{-}(\lambda ) =\alpha _{-} z_{-}(C_{-},\lambda ). \end{aligned}$$
(163)

By the assumption (156), \(z=Z_{\pm }(\lambda )\) are the solutions to the equation

$$\begin{aligned} z^{2}-2J(\lambda ) z +\rho ^{2}=0. \end{aligned}$$
(164)

In the case of constant coins, the denominators \(\langle \,{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}\) and \(\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}\) in (15) are constant functions in \(\lambda\) and they do not contribute the asymptotic behavior of \({\mathrm{x}}_{0}(\lambda )\) as \(|\lambda | \rightarrow 1.\) However, in the two-phase model, there are points \(\zeta \in S^{1}\) where these denominators are asymptotically zero as \(\lambda \rightarrow \zeta \in S^{1}.\) We need to describe these points and analyze the behavior of \({\mathrm{x}}_{0}(\lambda )\) near these points. We set

$$\begin{aligned} s={\mathrm{Re}}\,(\beta _{0}\overline{\beta }),\quad t=\mathrm{Im}\,(\beta _{0} \overline{\beta }). \end{aligned}$$
(165)

In what follows, we use the unit vector \(u^{\perp }\) perpendicular to a unit vector u defined by

$$\begin{aligned} u^{\perp }= \left[ \begin{array}{c} -\overline{b} \\ \overline{a} \end{array}\right] \quad \text{when} \quad u= \left[ \begin{array}{c} a \\ b \end{array}\right] . \end{aligned}$$
(166)

Lemma 5.6

The denominators are given by

$$\begin{aligned} \langle \,{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}&=-\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}} \\ &=\frac{2\overline{\beta } }{D_{+}(\lambda )D_{-}(\lambda )} Z_{-}(\lambda )\left( 1-s -J(\lambda ) Z_{-}(\lambda ) \right) . \end{aligned}$$

Proof

Since \(Z_{\pm }(\lambda )\) are the solution to Eq. (164), they satisfy \(Z_{+}(\lambda )-\lambda ^{-1}=Z_{-}(\lambda )-\lambda\) and \(\rho ^{2}=2J(\lambda )Z_{-}(\lambda )-Z_{-}(\lambda )^{2}.\) Thus, we see

$$\begin{aligned} {\mathrm{v}}_{+}(\lambda )&=\frac{1}{D_{+}(\lambda )} \left[ \begin{array}{c} \overline{\alpha _{0}} (\lambda -Z_{-}(\lambda )) \\ \overline{\beta _{0}} (\lambda -Z_{-}(\lambda )) -\overline{\beta } \lambda \end{array}\right] , \\ {\mathrm{v}}_{-}(\lambda )&= \frac{Z_{-}(\lambda )}{D_{-}(\lambda )} \left[ \begin{array}{c} (\beta _{0} \overline{\beta }-1) \lambda ^{-1} +Z_{-}(\lambda ) \\ -\alpha _{0} \overline{\beta } \lambda ^{-1} \end{array}\right] . \end{aligned}$$
(167)

Therefore, a direct calculation using the property \(|\alpha _{0}|^{2}+|\beta _{0}|^{2}=1\) shows

$$\begin{aligned}&\langle \,{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}} \\ &\quad = \frac{Z_{-}(\lambda )}{D_{+}(\lambda )D_{-}(\lambda )} \\ &\qquad \times \left( 2\overline{\beta } -2J(\lambda ) \overline{\beta } Z_{-}(\lambda ) +2J(\lambda ) \overline{\beta _{0}}Z_{-}(\lambda ) -\overline{\beta _{0}}Z_{-}(\lambda )^{2} -\overline{\beta _{0}} -\beta _{0} \overline{\beta }^{2} \right) . \end{aligned}$$

Substituting \(2J(\lambda )Z_{-}(\lambda )-Z_{-}(\lambda )^{2}=\rho ^{2}=1-\beta \overline{\beta }\) into the above, we see

$$\begin{aligned} \langle \,{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}= \frac{2\overline{\beta }Z_{-}(\lambda )}{D_{+}(\lambda )D_{-}(\lambda )} \left( 1-s-J(\lambda ) Z_{-}(\lambda ) \right) . \end{aligned}$$

By the choice of the vector \({\mathrm{v}}_{\pm }(\lambda )^{\perp }\) perpendicular to \({\mathrm{v}}_{\pm }(\lambda )\) described in (166), we have

$$\begin{aligned} \langle \,{\mathrm{v}}_{+}(\lambda ), {\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}= -\langle \,{\mathrm{v}}_{-}(\lambda ), {\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}}, \end{aligned}$$

which completes the proof. \(\square\)

In what follows, we sometimes write

$$\begin{aligned} G(\zeta _{(-)})=\lim _{r \uparrow 1}G(r\zeta ),\quad G(\zeta _{(+)})=\lim _{r \uparrow 1}G(r^{-1}\zeta ) \quad (\zeta \in S^{1}) \end{aligned}$$

for a function G on \({\mathbb{C}} {\setminus } S^{1},\) if the limit exists. We write \(\zeta =z+iy \in S^{1}.\) We define a function \(f(\lambda )\) by

$$\begin{aligned} f(\lambda )=1-s-J(\lambda ) Z_{-}(\lambda ). \end{aligned}$$
(168)

Then, by Lemma 5.2, for \(\zeta \not \in S,\) we have

$$\begin{aligned} f(\zeta ):=f(\zeta _{(+)})=f(\zeta _{(-)})=1-s-x(x +{\text{sgn}}(x)\sqrt{x^{2}-\rho ^{2}}), \end{aligned}$$

where

$$\begin{aligned} {\text{sgn}}(x)= \left\{ \begin{array}{ll} +1 &{}\quad (x>0), \\ -1 &{}\quad (x<0). \end{array}\right. \end{aligned}$$

Thus, \(f(\zeta )=0\) if and only if \(\zeta\) is one of the following four points:

$$\begin{aligned} \zeta _{*}={\mathrm{e}}^{i\theta _{*}}:=x_{*}+iy_{*},\quad -\overline{\zeta _{*}},\quad -\zeta _{*},\quad \overline{\zeta _{*}}, \end{aligned}$$
(169)

where positive real numbers \(x_{*},\) \(y_{*}\) are given by

$$\begin{aligned} x_{*}=\frac{1-s}{\sqrt{2-2s -\rho ^{2}}},\quad y_{*}=\sqrt{1-x_{*}^{2}} =\sqrt{\frac{1-\rho ^{2}-s^{2}}{2-2s-\rho ^{2}}}. \end{aligned}$$
(170)

It is well known (see [13]) that these points are actually the eigenvalues of U,  which will also be shown as point masses of \(\Sigma\) in the following theorem.

Theorem 5.7

The positive-matrix-valued measure \(\Sigma\) for the two-phase model \(U({\mathcal{C}})\) with the assumption (156) is given by the following:

$$\begin{aligned} {\mathrm{d}}\Sigma (\zeta )&= \frac{\sqrt{\rho ^{2}-x^{2}}}{(1-s)^{2}-(2-2s -\rho ^{2}) x^{2}} \\ &\quad \times \left[ \begin{array}{cc} (1-s)|y| + {\text{sgn}}(y)tx &{} -{\text{sgn}}(y) i \overline{\alpha _{0}} \beta x \\ i {\text{sgn}}(y) \alpha _{0} \overline{\beta } x &{} (1-s) |y| - {\text{sgn}}(y) tx \end{array}\right] \chi _{S}(\zeta )\, {\mathrm{d}}\ell (\zeta ) \\ &\quad + \frac{x_{*}\sqrt{x_{*}^{2}-\rho ^{2}}}{2y_{*}(1-s)^{2}} \left[ \begin{array}{cc} [(1-s) y_{*}+tx_{*}] &{} -i\overline{\alpha _{0}} \beta x_{*} \\ i \alpha _{0} \overline{\beta } x_{*} &{} [(1-s) y_{*}-tx_{*}] \end{array}\right] \left( \delta _{\zeta _{*}} +\delta _{-\zeta _{*}} \right) \\ &\quad + \frac{x_{*}\sqrt{x_{*}^{2}-\rho ^{2}}}{2y_{*}(1-s)^{2}} \left[ \begin{array}{cc} [(1-s) y_{*}-tx_{*}] &{} i\overline{\alpha _{0}} \beta x_{*} \\ -i \alpha _{0} \overline{\beta } x_{*} &{} [(1-s) y_{*}+tx_{*}] \end{array}\right] \left( \delta _{\overline{\zeta _{*}}} + \delta _{-\overline{\zeta _{*}}} \right) , \end{aligned}$$

where we write \(\zeta =x+iy \in S^{1}\) with \(x,y \in {\mathbb{R}},\) and \(\zeta _{*}=x_{*}+iy_{*}\) is defined in (170).

Proof of Theorem 5.7

By (167), we have

$$\begin{aligned} \langle \,{\mathrm{z}}_{L}(0)e_{L},{\mathrm{v}}_{+}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}} =\frac{\overline{\beta } \lambda }{D_{+}},\quad \langle \,{\mathrm{z}}_{R}(0)e_{R},{\mathrm{v}}_{-}(\lambda )^{\perp } \,\rangle _{{\mathbb{C}}^{2}} =\frac{Z_{-}}{D_{-}} (Z_{-}-\lambda ^{-1}). \end{aligned}$$

Therefore, by Theorem 1.4, we have

$$\begin{aligned} \lambda {\mathrm{x}}_{0}(\lambda ) = \frac{1}{2 f(\lambda )} \left[ \begin{array}{cc} \beta _{0} \overline{\beta } -1 + \lambda Z_{-}(\lambda ) &{} \overline{\alpha _{0}} \beta \\ -\alpha _{0} \overline{\beta } &{} \overline{\beta _{0}} \beta -1 + \lambda Z_{-}(\lambda ) \end{array}\right] , \end{aligned}$$
(171)

where the holomorphic function \(f(\lambda )\) is given in (168). We first calculate the matrix-valued measure \(\Sigma\) on S. We only consider the upper part \(S_{+}=\{\zeta \in S \mid \mathrm{Im}\,(\zeta )>0\}\) of S. The measure \(\Sigma\) on the lower part of S can be calculated in the same way. For \(\zeta \in S_{+},\) by Lemma 5.2, we have

$$\begin{aligned} Z_{-}(r\zeta )&=J(r\zeta ) -i\sqrt{\rho ^{2}-J(r\zeta )},\\ Z_{-}(r^{-1}\zeta )&=J(r^{-1}\zeta ) +i\sqrt{\rho ^{2}-J(r^{-1}\zeta )}. \end{aligned}$$

Therefore, writing \(\zeta =x+iy\) as before

$$\begin{aligned} f(\zeta _{(-)})=1-s-x^{2}+ixw,\quad f(\zeta _{(+)})=1-s-x^{2}-ixw,\quad w=\sqrt{\rho ^{2}-x^{2}}. \end{aligned}$$

By the assumption (156), we have \(|s|<1-\rho ^{2}\), and hence, \(f(\zeta _{(\pm )})\) are nonzero. Since both of \(\pm \sqrt{\rho ^{2}-J(\lambda )^{2}}\) are continuous on S,  the Lebesgue convergence theorem shows

$$\begin{aligned}&\mathop {\mathrm{w*-lim}}\limits _{r \uparrow 1} [r\zeta {\mathrm{x}}_{0}(r\zeta ) -r^{-1}\zeta {\mathrm{x}}_{0}(r^{-1}\zeta )]\chi _{S}(\zeta ) \\ &= \frac{1}{2f(\zeta _{(-)})} \\ &\qquad \times \left[ \begin{array}{cc} \beta _{0} \overline{\beta } -1 +\zeta (x-i\sqrt{\rho ^{2}-x^{2}}) &{} \overline{\alpha _{0}} \beta \\ -\alpha _{0} \overline{\beta } &{} \overline{\beta _{0}} \beta -1+\zeta (x-i\sqrt{\rho ^{2}-x^{2}}) \end{array}\right] \chi _{S}(\zeta )\,{\mathrm{d}}\ell (\zeta ) \\ &\qquad - \frac{1}{2f(\zeta _{(+)})} \\ &\qquad \times \left[ \begin{array}{cc} \beta _{0} \overline{\beta } -1 +\zeta (x+i\sqrt{\rho ^{2}-x^{2}}) &{} \overline{\alpha _{0}} \beta \\ -\alpha _{0} \overline{\beta } &{} \overline{\beta _{0}} \beta -1+\zeta (x+i\sqrt{\rho ^{2}-x^{2}}) \end{array}\right] \chi _{S}(\zeta )\,{\mathrm{d}}\ell (\zeta ) \\ &\quad = \frac{w}{(1-s)^{2}-[2(1-s) -\rho ^{2}]x^{2}} \left[ \begin{array}{cc} (1-s)y+tx &{} -i\overline{\alpha _{0}} \beta x \\ i \alpha _{0} \overline{\beta } x &{} (1-s)y -tx \end{array}\right] \chi _{S}(\zeta )\,{\mathrm{d}}\ell (\zeta ). \end{aligned}$$
(172)

Next, we calculate the matrix-valued measure \(\Sigma\) on \(S^{1} {\setminus } S.\) We have

$$\begin{aligned} \lim _{r \uparrow 1} (Z_{-}(r\zeta ) -Z_{-}(r^{-1} \zeta )) =0 \quad (\zeta \in S^{1} {\setminus } S \cup \{\pm \zeta _{*}, \pm \overline{\zeta _{*}} \}). \end{aligned}$$
(173)

Therefore, we only need to calculate the behavior of \({\mathrm{x}}_{0}(\lambda )\) near \(\{\pm \zeta _{*}, \pm \overline{\zeta _{*}} \}.\) On the domains

$$\begin{aligned} \Omega _{+}:=\{\lambda \in {\mathbb{C}} \mid {\mathrm{Re}}\,(\lambda ) >0\} {\setminus } (S \cup i{\mathbb{R}}), \quad \Omega _{-}:=\{\lambda \in {\mathbb{C}} \mid {\mathrm{Re}}\,(\lambda ) <0\} {\setminus } (S \cup i{\mathbb{R}}), \end{aligned}$$

the function \(Z_{-}(\lambda )\) is given as

$$\begin{aligned} Z_{-}(\lambda )= \left\{ \begin{array}{ll} J(\lambda ) +\sqrt{J(\lambda )^{2}-\rho ^{2}} &{}\quad (\lambda \in \Omega _{+}), \\ J(\lambda ) -\sqrt{J(\lambda )^{2} -\rho ^{2}} &{}\quad (\lambda \in \Omega _{-}). \end{array}\right. \end{aligned}$$

We still use the notation \(f(\lambda )=1-s-J(\lambda )Z_{-}(\lambda )\) on the domains \(\Omega _{\pm }.\) We denote \(\zeta _{o}=x_{o}+iy_{o}\) one of the points \(\{\pm \zeta _{*},\pm \overline{\zeta _{*}}\}\) and write \({\mathrm{e}}^{i\theta _{o}}=\zeta _{o}.\) For any small \(\varepsilon >0\) and \(r \in (2/3,1)\) close to 1,  we define

$$\begin{aligned} J_{r,\varepsilon }&=\{t {\mathrm{e}}^{i\theta } \mid r \le t \le r^{-1},\,|\theta -\theta _{o}| \le \varepsilon \}, \\ A_{\varepsilon }&=J_{r,\varepsilon } \cap S^{1}=\{{\mathrm{e}}^{i\theta } \mid |\theta -\theta _{o}| \le \varepsilon \}, \end{aligned}$$

where \(\varepsilon >0\) is chosen, so that \(J_{r,\varepsilon }\) is contained in one of \(\Omega _{\pm }\) containing \(\zeta _{o}\) and \(J_{r,\varepsilon }\) contains only \(\zeta _{o}\) among \(\{\pm \zeta _{*},\pm \overline{\zeta _{*}}\}.\) We note that, by Lemma 5.6, \(f(\lambda )\) is nonzero for \(|\lambda | \ne 1\) since, for \(|\lambda | \ne 1,\) \(\{{\mathrm{v}}_{+}(\lambda ),\, {\mathrm{v}}_{-}(\lambda )\}\) is a basis of \({\mathbb{C}}^{2}.\) Then, \(\zeta _{o}\) is a zero of \(f(\lambda )\) of order 1 and there are no zeros of \(f(\lambda )\) on a neighborhood of \(J_{r,\varepsilon },\) where the function \(f(\lambda )\) is defined in (168) which is holomorphic on \(\Omega _{\pm }.\) Thus, for any holomorphic function \(h(\lambda )\) near \(J_{r,\varepsilon },\) Cauchy’s integral formula gives

$$\begin{aligned} \frac{1}{2\pi i} \int _{\partial J_{r,\varepsilon }} \frac{h(\lambda )}{f(\lambda )}\,\frac{{\mathrm{d}}\lambda }{\lambda } = \frac{h(\zeta _{o})}{\zeta _{o} f'(\zeta _{o})}, \end{aligned}$$
(174)

where \(\partial J_{r,\varepsilon }\) is the boundary of \(J_{r,\varepsilon }\) with the counterclockwise direction. Since \(Z_{-}(\lambda )\) satisfies the Eq. (164), we have

$$\begin{aligned} f(\lambda )=1-s-J(\lambda )Z_{-}(\lambda )=1-s-\frac{1}{2}(Z_{-}(\lambda )^{2}+\rho ^{2}). \end{aligned}$$

Thus, we have \(f'(\lambda )=-Z_{-}(\lambda )Z_{-}'(\lambda ).\) Substituting this into (174) yields

$$\begin{aligned} \frac{1}{2\pi i} \int _{\partial J_{r,\varepsilon }} \frac{h(\lambda )}{f(\lambda )}\,\frac{{\mathrm{d}}\lambda }{\lambda } = \frac{h(\zeta _{o})}{\zeta _{o} f'(\zeta _{o})}=-\frac{h(\zeta _{o})}{\zeta _{o} Z_{-}(\zeta _{o}) Z_{-}'(\zeta _{o})}. \end{aligned}$$
(175)

By calculating the contour integral in the left-hand side of (175), we see

$$\begin{aligned} \lim _{r \uparrow 1} \left[ \int _{A_{\varepsilon }} \frac{h(r\zeta )}{f(r\zeta )} \,{\mathrm{d}}\ell (\zeta ) - \int _{A_{\varepsilon }} \frac{h(r^{-1}\zeta )}{f(r^{-1}\zeta )} \,{\mathrm{d}}\ell (\zeta ) \right] = \frac{h(\zeta _{o})}{\zeta _{o} Z_{-}(\zeta _{o}) Z_{-}'(\zeta _{o})}. \end{aligned}$$
(176)

The set of all Laurent polynomials is dense in the space of continuous functions on \(A_{\varepsilon }.\) Thus, the formula (176) still holds for any continuous function h on \(A_{\varepsilon }.\) Let \(u=\,\!^{t}[u_{L}, u_{R}] \in {\mathbb{C}}^{2}\) and \(h \in C(A_{\varepsilon }).\) According to the formula (171), we set

$$\begin{aligned} G(\lambda )=(\beta _{0}\overline{\beta } -1+\lambda Z_{-}(\lambda ))|u_{L}|^{2} + \overline{\alpha _{0}} \beta \overline{u_{L}} u_{R} -\alpha _{0} \overline{\beta } u_{L} \overline{u_{R}} + (\overline{\beta _{0}} \beta -1 +\lambda Z_{-}(\lambda )) |u_{R}|^{2}, \end{aligned}$$

so that

$$\begin{aligned} \langle \,[r\zeta {\mathrm{x}}_{0}(r\zeta ) -r^{-1}\zeta {\mathrm{x}}_{0}(r^{-1}\zeta )]u,u \,\rangle _{{\mathbb{C}}^{2}} =\frac{G(r\zeta )}{2f(r\zeta )} -\frac{G(r^{-1}\zeta )}{2f(r^{-1}\zeta )}. \end{aligned}$$

From (171) and (176), we have

$$\begin{aligned}&\lim _{r \uparrow 1} \int _{A_{\varepsilon }} h(\zeta ) \langle \,[r\zeta {\mathrm{x}}_{0}(r\zeta ) -r^{-1}\zeta {\mathrm{x}}_{0}(r^{-1}\zeta )]u,u \,\rangle _{{\mathbb{C}}^{2}}\,{\mathrm{d}}\ell (\zeta ) \\ &\quad = \lim _{r \uparrow 1} \left( \int _{A_{\varepsilon }} \frac{h(\zeta ) G(r\zeta )}{2f(r\zeta )}\,{\mathrm{d}}\ell (\zeta ) - \int _{A_{\varepsilon }} \frac{h(\zeta ) G(r^{-1}\zeta )}{2f(r^{-1}\zeta )}\,{\mathrm{d}}\ell (\zeta ) \right) \\ &\quad =\frac{h(\zeta _{o})G(\zeta _{o})}{2\zeta _{o} Z_{-}(\zeta _{o})Z_{-}'(\zeta _{o})}. \end{aligned}$$
(177)

A direct calculation shows

$$\begin{aligned} \lambda Z_{-}'(\lambda )=\lambda \frac{J'(\lambda )}{\sqrt{J(\lambda )^{2}-\rho ^{2}}} \left( J(\lambda ) +\sqrt{J(\lambda )^{2}-\rho ^{2}} \right) =\frac{K(\lambda )}{\sqrt{J(\lambda )^{2}-\rho ^{2}}} Z_{-}(\lambda ), \end{aligned}$$

where, as before, \(K(\lambda )=(\lambda -\lambda ^{-1})/2.\) Since \(0=f(\zeta _{o})=1-s-x_{o}Z_{-}(\zeta _{o}),\) when \(x_{o}={\mathrm{Re}}\,(\zeta _{o})>0\) (that is, \(x_{o}=x_{*}\)), we have

$$\begin{aligned} \zeta _{o}Z_{-}(\zeta _{o})Z_{-}'(\zeta _{o}) = \frac{K(\zeta _{o})}{\sqrt{x_{o}^{2}-\rho ^{2}}} Z_{-}(\zeta _{o})^{2}= \frac{iy_{o}(1-s)^{2} }{x_{o}^{2} \sqrt{x_{o}^{2} -\rho ^{2}}}. \end{aligned}$$

Then, we also have

$$\begin{aligned} \beta _{0} \overline{\beta } -1+\zeta _{o} Z_{-}(\zeta _{o})=s+it -1 + \frac{1-s}{x_{o}}(x_{o}+iy_{o}) =\frac{i}{x_{o}} ((1-s)y_{o}+t x_{o}). \end{aligned}$$

Therefore, from (177), we obtain, when \(x_{o}={\mathrm{Re}}\,(\zeta _{o})>0\)

$$\begin{aligned} \Sigma (\{\zeta _{o}\})&= \frac{1}{2\zeta _{o} Z_{-}(\zeta _{o}) Z_{-}'(\zeta _{o})} \\ &\quad \times \left[ \begin{array}{cc} [\beta _{o} \overline{\beta } -1 +\zeta _{o} Z_{-}(\zeta _{o})] \delta _{\zeta _{o}} &{} \overline{\alpha _{0}} \beta \delta _{\zeta _{o}} \\ -\alpha _{0} \overline{\beta } \delta _{\zeta _{o}} &{} [\overline{\beta _{o}} \beta -1 +\zeta _{o} Z_{-}(\zeta _{o})] \delta _{\zeta _{o}} \end{array}\right] \\ &= \frac{x_{o}\sqrt{x_{o}^{2}-\rho ^{2}}}{2y_{o}(1-s)^{2}} \left[ \begin{array}{cc} [(1-s) y_{o}+tx_{o}] \delta _{\zeta _{o}} &{} -i\overline{\alpha _{0}} \beta x_{o} \delta _{\zeta _{o}} \\ i \alpha _{0} \overline{\beta } x_{o} &{} [(1-s) y_{o}-tx_{o}] \delta _{\zeta _{o}} \end{array}\right] , \end{aligned}$$
(178)

where \(\delta _{\zeta _{o}}\) is the delta measure at the point \(\zeta _{o},\) and for \({\mathrm{Re}}\,(\zeta _{o})<0\)

$$\begin{aligned} \Sigma (\{\zeta _{o}\})&= -\frac{x_{o}\sqrt{x_{o}^{2}-\rho ^{2}}}{2y_{o}(1-s)^{2}} \\ &\quad \times \left[ \begin{array}{cc} [(1-s) y_{o}+tx_{o}] \delta _{\zeta _{o}} &{} -i\overline{\alpha _{0}} \beta x_{o} \delta _{\zeta _{o}} \\ i \alpha _{0} \overline{\beta } x_{o} &{} [(1-s) y_{o}-tx_{o}] \delta _{\zeta _{o}} \end{array}\right] . \end{aligned}$$
(179)

From this, we conclude the assertion in Theorem 5.7. \(\square\)