# On expansive operators that are quasisimilar to the unilateral shift of
finite multiplicity

Maria F. Gamal  St. Petersburg Branch
V. A. Steklov Institute of Mathematics
Russian Academy of Sciences
Fontanka 27, St. Petersburg
191023, Russia<EMAIL_ADDRESS>
###### Abstract.

An operator $T$ on a Hilbert space $\mathcal{H}$ is called expansive, if
$\|Tx\|\geq\|x\|$ ($x\in\mathcal{H}$). Expansive operators $T$ quasisimilar to
the unilateral shift $S_{N}$ of finite multiplicity $N$ are studied. It is
proved that $I-T^{*}T$ is of trace class for such $T$. Also the lattice
$\mathrm{Lat}T$ of invariant subspaces of an expansive operator $T$
quasisimilar to $S_{N}$ is studied. It is proved that $\dim\mathcal{M}\ominus
T\mathcal{M}\leq N$ for every $\mathcal{M}\in\mathrm{Lat}T$. It is shown that
if $N\geq 2$, then there exist $\mathcal{M}_{j}\in\mathrm{Lat}T$
($j=1,\ldots,N$) such that the restriction $T|_{\mathcal{M}_{j}}$ of $T$ on
$\mathcal{M}_{j}$ is similar to the unilateral shift $S$ of multiplicity $1$
for every $j=1,\ldots,N$, and $\mathcal{H}=\vee_{j=1}^{N}\mathcal{M}_{j}$. For
$N=1$, that is, for $T$ quasisimilar to $S$, there exist two spaces
$\mathcal{M}_{1}$, $\mathcal{M}_{2}\in\mathrm{Lat}T$ such that
$T|_{\mathcal{M}_{j}}$ is similar to $S$ for $j=1,2$, and
$\mathcal{H}=\mathcal{M}_{1}\vee\mathcal{M}_{2}$. Example of an expansive
operator $T$ quasisimilar to $S$ is given such that intertwining
transformations do not give an isomorphism of $\mathrm{Lat}T$ and
$\mathrm{Lat}S$.

2020 _Mathematics Subject Classification_. 47A45, 47A15, 47A55.

###### Key words and phrases:

Expansive operator, contraction, quasisimilarity, similarity, unilateral
shift, invariant subspaces, unitary asymptote, intertwining relation.

##  1\. Introduction

Let $\mathcal{H}$ be a (complex, separable) Hilbert space, and let
$\mathcal{L}(\mathcal{H})$ be the algebra of all (bounded, linear) operators
acting on $\mathcal{H}$. A (closed) subspace $\mathcal{M}$ of $\mathcal{H}$ is
called _invariant_ for an operator $T\in\mathcal{L}(\mathcal{H})$, if
$T\mathcal{M}\subset\mathcal{M}$. The complete lattice of all invariant
subspaces of $T$ is denoted by $\operatorname{Lat}T$.

The _multiplicity_ $\mu_{T}$ of an operator $T\in\mathcal{L}(\mathcal{H})$ is
the minimum dimension of its reproducing subspaces:

(1.1) $\mu_{T}=\min\\{\dim E:E\subset\mathcal{H},\ \
\vee_{n=0}^{\infty}T^{n}E=\mathcal{H}\\}.$

For Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$, let
$\mathcal{L}(\mathcal{H},\mathcal{K})$ denote the space of (bounded, linear)
transformations acting from $\mathcal{H}$ to $\mathcal{K}$. For
$A\in\mathcal{L}(\mathcal{H})$ and $B\in\mathcal{L}(\mathcal{K})$ set

$\mathcal{I}(A,B)=\\{X\in\mathcal{L}(\mathcal{H},\mathcal{K})\ :\ XA=BX\\}.$

Then $\mathcal{I}(A,B)$ is the set of all transformations $X$ which
_intertwine_ $A$ and $B$. Let $X\in\mathcal{I}(A,B)$. If $X$ is unitary, then
$A$ and $B$ are called _unitarily equivalent_ , in notation: $A\cong B$. If
$X$ is invertible, that is, $X^{-1}\in\mathcal{L}(\mathcal{K},\mathcal{H})$,
then $A$ and $B$ are called _similar_ , in notation: $A\approx B$. If $X$ is a
_quasiaffinity_ , that is, $\ker X=\\{0\\}$ and
$\operatorname{clos}X\mathcal{H}=\mathcal{K}$, then $A$ is called a
_quasiaffine transform_ of $B$, in notation: $A\prec B$. If $\ker X=\\{0\\}$,
we write $A\buildrel i\over{\prec}B$. If
$\operatorname{clos}X\mathcal{H}=\mathcal{K}$, we write $A\buildrel
d\over{\prec}B$. If $A\prec B$ and $B\prec A$, then $A$ and $B$ are called
_quasisimilar_ , in notation: $A\sim B$.

An operator $T\in\mathcal{L}(\mathcal{H})$ is called _expansive_ if
$\|Tx\|\geq\|x\|$ for every $x\in\mathcal{H}$.

An operator $T$ is called _power bounded_ , if $\sup_{n\geq
0}\|T^{n}\|<\infty$. An operator $T$ is called a _contraction_ , if $\|T\|\leq
1$. Clearly, a contraction is power bounded.

Let $T\in\mathcal{L}(\mathcal{H})$ be a power bounded operator. It is easy to
see that the space

(1.2) $\mathcal{H}_{T,0}=\\{x\in\mathcal{H}\ :\ \|T^{n}x\|\to 0\\}$

is invariant for $T$ (cf. [NFBK, Theorem II.5.4]). Classes $C_{ab}$, $a$,
$b=0,1,\cdot$, of power bounded operators are defined as follows (see [NFBK,
Sec. II.4] and [K89]). If $\mathcal{H}_{T,0}=\mathcal{H}$, then $T$ is _of
class_ $C_{0\cdot}$, while if $\mathcal{H}_{T,0}=\\{0\\}$, then $T$ is _of
class_ $C_{1\cdot}$. Furthermore, $T$ is _of class_ $C_{\cdot a}$, if
$T^{\ast}$ is of class $C_{a\cdot}$, and $T$ is _of class_ $C_{ab}$, if $T$ is
of classes $C_{a\cdot}$ and $C_{\cdot b}$, $a$, $b=0,1$.

For a power bounded operator $T\in\mathcal{L}(\mathcal{H})$

$\text{ the \emph{isometric asymptote} }(X_{+,T},T_{+}^{(a)})$

can be defined using a Banach limit $\mathop{\text{\rm Lim}}$, see [K89]. (For
the isometric and unitary asymptotes of a contraction $T$ see also [NFBK, Sec.
IX.1]). Here $T_{+}^{(a)}$ is an isometry on a Hilbert space
$\mathcal{H}_{+}^{(a)}$, and $X_{+,T}$ is the _canonical intertwining mapping_
: $X_{+,T}T=T_{+}^{(a)}X_{+,T}$. Recall that the range of $X_{+,T}$ is dense.
Thus, $X_{+,T}$ realizes the relation $T\buildrel d\over{\prec}T_{+}^{(a)}$.
We do not recall the construction of the canonical intertwining mapping from
[K89] here. We recall only that $\|X_{+,T}x\|^{2}=\mathop{\text{\rm
Lim}}_{n}\|T^{n}x\|^{2}$ for every $x\in\mathcal{H}$. It easy follows from
this relation that an operator $T\in\mathcal{L}(\mathcal{H})$ is similar to an
isometry if and only if $T$ is power bounded and there exists $c>0$ such that
$\|T^{n}x\|\geq c\|x\|$ for every $x\in\mathcal{H}$ and $n\in\mathbb{N}$. In
this case, $X_{+,T}$ is invertible and realizes the relation $T\approx
T_{+}^{(a)}$.

The _unitary asymptote_ $(X_{T},T^{(a)})$ of a power bounded operator
$T\in\mathcal{L}(\mathcal{H})$ is a pair where
$T^{(a)}\in\mathcal{L}(\mathcal{H}^{(a)})$ (here $\mathcal{H}^{(a)}$ is a some
Hilbert space) is the minimal unitary extension of $T_{+}^{(a)}$, and $X_{T}$
is a natural extension of $X_{+,T}$. _The isometry $T_{+}^{(a)}$ and the
unitary operator $T^{(a)}$ will also be called the isometric and unitary
asymptotes of $T$, respectively._

Let $S$ be the simple unilateral shift, that is, the multiplication by the
independent variable on the Hardy space $H^{2}$ on the unit circle
$\mathbb{T}$. A particular case of [K07] is the following (see also [NFBK,
Sec. IX.3]). Let $T\in\mathcal{L}(\mathcal{H})$ be an absolutely continuous
(a.c.) contraction (the definition is recalled in Sec. 2 of the present
paper), and let $T^{(a)}$ contain the bilateral shift as an orthogonal
summand. Then

(1.3) $\mathcal{H}=\vee\\{\mathcal{M}\ :\ \mathcal{M}\in\operatorname{Lat}T,\
T|_{\mathcal{M}}\approx S\\}.$

In [G22] this result is generalized to a.c. polynomially bounded operators
(the definition can be found, for example, in [Pe, Ch. 15], see also [NFBK,
Ch. I.13] where other terminology is used; see references therein). Also it is
shown in [G22] that the quantity of subspaces $\mathcal{M}$ in (1.3) can be
equal to $\mu_{T}$, if $\mu_{T}\geq 2$, and to $2$, if $\mu_{T}=1$. On the
other hand, there exists power bounded operator $T$ such that $T^{(a)}_{+}=S$
and there is no $\mathcal{M}\in\operatorname{Lat}T$ such that
$T|_{\mathcal{M}}\approx S$ [G16, Sec. 5]. The purpose of this paper is to
show that (1.3) is fulfilled for expansive operators $T$ which are
quasisimilar to the unilateral shift $S_{N}=\oplus_{j=1}^{N}S$ of finite
multiplicity $N\in\mathbb{N}$, and the quantity of subspaces $\mathcal{M}$ in
(1.3) is as decribed above (Theorems 4.12 and 4.13). Expansive operators are
right inverses for contractions. The proof is based on the result for
contractions from [K07] (see also [NFBK, Sec. IX.3]) and on representations of
unimodular functions on $\mathbb{T}$ given in [B] and developed in [H]. Some
other properties of an expansive operator $T$ such that $T\sim S_{N}$, where
$N\in\mathbb{N}$, are studied. In particular, it is proved that
$\dim(\mathcal{M}\ominus T\mathcal{M})\leq N$ for every
$\mathcal{M}\in\operatorname{Lat}T$, and $I-T^{*}T\in\mathfrak{S}_{1}$, where
$\mathfrak{S}_{1}$ is the trace class operators (Theorems 4.10 and 4.14).

The paper is organized as follows. In Sec. 2 some simple observations are
collected, some of them are of own interest, and some of them will be used in
the sequel. In Sec. 3 a special kind of finite perturbations of $S_{N}$
($N\in\mathbb{N}$) that are expansive operators is considered. Sec. 4 is the
main part of the paper. In Sec. 5 the relationship between similarity to an
isometry of expansive operator and its Cauchy dual (adjoint of the standard
left inverse) is studied. In Sec. 6 it is shown that there exist expansive
operators $T$ such that $T\sim S$, but the intertwining quasiffinities do not
give the isomorphism of $\operatorname{Lat}T$ and $\operatorname{Lat}S$ (in
contrast with the case when $T$ is a contraction).

The following notation will be used. For a (closed) subspace $\mathcal{M}$ of
a Hilbert space $\mathcal{H}$, by $P_{\mathcal{M}}$ and $I_{\mathcal{M}}$ the
orthogonal projection from $\mathcal{H}$ onto $\mathcal{M}$ and the identity
operator on $\mathcal{M}$ are denoted, respectively. By $\mathbb{O}$ the zero
transformation acting between (maybe nonzero) spaces is denoted.

Symbols $\mathbb{D}$ and $\mathbb{T}$ denote the open unit disc and the unit
circle, respectively. The normalized Lebesgue measure on $\mathbb{T}$ is
denoted by $m$. Set $L^{p}=L^{p}(\mathbb{T},m)$. For $0<p\leq\infty$ by
$H^{p}$ the Hardy space on $\mathbb{T}$ is denoted. Set $\chi(\zeta)=\zeta$
and $\mathbf{1}(\zeta)=1$ for $\zeta\in\mathbb{T}$. The simple unilateral $S$
is the operator of multiplication by $\chi$ on $H^{2}$. Set
$H^{2}_{-}=L^{2}\ominus H^{2}$. For a measurable set $\sigma\subset\mathbb{T}$
denote by $U_{\sigma}$ the operator of multiplication by $\chi$ on
$L^{2}(\sigma,m)$. Then $U_{\mathbb{T}}$ is the simple bilateral shift.

For $N\in\mathbb{N}\cup\\{\infty\\}$ denote by $H^{2}_{N}$, $L^{2}_{N}$,
$(H^{2}_{-})_{N}$ the orthogonal sum of $N$ copies of $H^{2}$, $L^{2}$,
$H^{2}_{-}$, respectively. For $N\in\mathbb{N}$, vectors from $H^{2}_{N}$,
$L^{2}_{N}$, $(H^{2}_{-})_{N}$ are columns of functions from $H^{2}$, $L^{2}$,
$H^{2}_{-}$, respectively. For $1\leq k\leq N$ denote by $e_{k}$ the vector
from $H^{2}_{N}$ with $\mathbf{1}$ on $k$-th place and zeros on all other
places. Then $\\{e_{k}\\}_{k=1}^{N}$ is an orthonormal basis of $\ker
S_{N}^{*}$. By $P_{+}$ and $P_{-}$ the orthogonal projections from $L^{2}_{N}$
onto $H^{2}_{N}$ and $(H^{2}_{-})_{N}$ are denoted, respectively (they depend
on $N$, but it will not be mentioned in notation). Set
$S_{*}=P_{-}U_{\mathbb{T}}|_{H^{2}_{-}}$. By $S_{N}$, $S_{*,N}$, and
$U_{\mathbb{T},N}$ the orthogonal sum of $N$ copies of $S$, $S_{*}$, and
$U_{\mathbb{T}}$ are denoted, respectively. Recall that
$\mu_{S_{N}}=\mu_{U_{\mathbb{T},N}}=N$, and
$\mu_{U_{\mathbb{T},N}|_{\mathcal{M}}}\leq N$ for every
$\mathcal{M}\in\operatorname{Lat}U_{\mathbb{T},N}$.

For a matrix $F=[f_{jk}]_{j,k}$ whose elemets are functions $f_{jk}$ set
$\overline{F}=[\overline{f}_{jk}]_{j,k}$.

##  2\. General observations

The following lemma is well known and can be proved easily, so its proof is
omitted.

######  Lemma 2.1.

Let $A$, $B\in\mathcal{L}(\mathcal{H})$ be such that $BA=I_{\mathcal{H}}$ and
$\dim\ker A^{*}<\infty$. Then the following are equivalent: $\mathrm{(i)}$
$I_{\mathcal{H}}-A^{*}A\in\mathfrak{S}_{1}$; $\mathrm{(ii)}$
$I_{\mathcal{H}}-AA^{*}\in\mathfrak{S}_{1}$; $\mathrm{(iii)}$
$I_{\mathcal{H}}-B^{*}B\in\mathfrak{S}_{1}$; $\mathrm{(iv)}$
$I_{\mathcal{H}}-BB^{*}\in\mathfrak{S}_{1}$.

Recall that $A\in L(\mathcal{H})$ is called a Fredholm operator, if
$A\mathcal{H}$ is closed, $\dim\ker A<\infty$, and $\dim\ker A^{*}<\infty$.
Denote by $\operatorname{ind}$ the Fredholm index of a Fredholm operator $A$,
that is, $\operatorname{ind}A=\dim\ker A-\dim\ker A^{*}$. See, for example,
[Co, Ch. XI].

######  Lemma 2.2.

Suppose that $N\in\mathbb{N}$, $A\in L(\mathcal{H})$, $\ker A=\\{0\\}$,
$\dim\ker A^{*}=N$, and $Y\in\mathcal{I}(S_{N},A)$ is such that
$\operatorname{clos}YH^{2}_{N}=\mathcal{H}$. Then $\ker Y=\\{0\\}$.

###### Proof.

We have

$S_{N}=\left[\begin{matrix}S_{N}|_{\ker Y}&*\\\
\mathbb{O}&R\end{matrix}\right],$

and $Y|_{H^{2}_{N}\ominus\ker Y}$ realizes the relation $R\prec A$. This
relation implies that $\ker R=\\{0\\}$ and $\dim\ker R^{*}\geq N$. By [Co,
Theorem XI.3.7],

$-N=\operatorname{ind}S_{N}=\operatorname{ind}S_{N}|_{\ker
Y}+\operatorname{ind}R\leq\operatorname{ind}S_{N}|_{\ker Y}-N.$

This means that $\operatorname{ind}S_{N}|_{\ker Y}=0$. Therefore, $\ker
Y=\\{0\\}$. ∎

For $A\in\mathcal{L}(\mathcal{H})$ set
$\mathcal{R}^{\infty}(A)=\cap_{n\in\mathbb{N}}A^{n}\mathcal{H}$. If
$\mathcal{R}^{\infty}(A)=\\{0\\}$, then $A$ is called _analytic_ [Sh] or
_pure_ [O]. The following simple lemma is given for convenience of references;
its proof is evident and omitted.

######  Lemma 2.3.

Let $A$ and $B$ be operators, and let $X\in\mathcal{I}(A,B)$. Then
$X\mathcal{R}^{\infty}(A)\subset\mathcal{R}^{\infty}(B)$.

Let $A$ be left-invertible, equivalently, let $A$ be bounded below: there
exists $c>0$ such that $\|Ax\|\geq c\|x\|$ for every $x\in\mathcal{H}$. Then
$\mathcal{R}^{\infty}(A)\in\operatorname{Lat}A$,
$A|_{\mathcal{R}^{\infty}(A)}$ is invertible, and if
$\mathcal{M}\in\operatorname{Lat}A$ is such that $A\mathcal{M}=\mathcal{M}$,
then $\mathcal{M}\subset\mathcal{R}^{\infty}(A)$. Consequently,
$P_{\mathcal{H}\ominus\mathcal{R}^{\infty}(A)}A|_{\mathcal{H}\ominus\mathcal{R}^{\infty}(A)}$
is left-invertible, and

(2.1)
$\mathcal{R}^{\infty}(P_{\mathcal{H}\ominus\mathcal{R}^{\infty}(A)}A|_{\mathcal{H}\ominus\mathcal{R}^{\infty}(A)})=\\{0\\}.$

For a left-invertible $A\in\mathcal{L}(\mathcal{H})$ the operator
$L_{A}=(A^{*}A)^{-1}A^{*}$ is the standard left inverse for $A$:
$L_{A}A=I_{\mathcal{H}}$, and $\ker L_{A}=\ker A^{*}$. Set
$A^{\prime}=L_{A}^{*}=A(A^{*}A)^{-1}$. The operator $A^{\prime}$ is called the
_Cauchy dual_ to $A$ ([Sh], [O]). Note that $A^{\prime}$ is left-invertible
and $A^{\prime\prime}=A$.

######  Lemma 2.4.

  1. (i)

[Sh, Prop. 2.7] Let $A$, $B\in\mathcal{L}(\mathcal{H})$ be such that
$BA=I_{\mathcal{H}}$. Then
$\mathcal{H}=\mathcal{R}^{\infty}(A)\oplus\vee_{n=0}^{\infty}B^{*n}\ker
A^{*}$.

  2. (ii)

[Sh, Lemma 2.1] Let $A\in\mathcal{L}(\mathcal{H})$ be left-invertible.

$\text{ Let }\mathcal{H}=\vee_{n=0}^{\infty}A^{n}\ker A^{*}.\text{ Then
}\mathcal{H}=\vee_{n=0}^{\infty}\ker L_{A}^{n}.$

######  Lemma 2.5.

Let $T\in\mathcal{L}(\mathcal{H})$ be expansive. Then
$P_{\mathcal{H}\ominus\mathcal{R}^{\infty}(T)}T|_{\mathcal{H}\ominus\mathcal{R}^{\infty}(T)}$
is expansive.

###### Proof.

Let $x\in\mathcal{H}\ominus\mathcal{R}^{\infty}(T)$. Since
$T\mathcal{R}^{\infty}(T)=\mathcal{R}^{\infty}(T)$, there exists
$y\in\mathcal{R}^{\infty}(T)$ such that $Ty=P_{\mathcal{R}^{\infty}(T)}Tx$. We
have

$\|P_{\mathcal{H}\ominus\mathcal{R}^{\infty}(T)}Tx\|^{2}=\|T(x-y)\|^{2}\geq\|x-y\|^{2}=\|x\|^{2}+\|y\|^{2}\geq\|x\|^{2}.\qed$

Let $R$ be a contraction. Then $R=U_{s}\oplus U_{a}\oplus R_{1}$, where
$U_{s}$ and $U_{a}$ are singular and absolutely continuous unitary operators
(that is, their spectral measures are singular and absolutely continuous with
respect to $m$), respectively, and $R_{1}$ is a completely nonunitary
contraction (that is, there is no
$\\{0\\}\neq\mathcal{M}\in\operatorname{Lat}R_{1}$ such that
$T|_{\mathcal{M}}$ is unitary). If $U_{s}$ acts on the zero space $\\{0\\}$,
then $R$ is called an _absolutely continuous (a.c.)_ contraction. If $U$ is a
singular unitary operator and $R$ is an a.c. contraction, then
$\mathcal{I}(R,U)=\mathbb{O}$. For an a.c. contraction $R$ the
$H^{\infty}$-functional calculus is defined. If there exists
$0\not\equiv\varphi\in H^{\infty}$ such that $\varphi(R)=\mathbb{O}$, then $R$
is called a _$C_{0}$ -contraction_. $C_{0}$-contractions are of class
$C_{00}$. For references, see [NFBK, Theorems I.3.2, II.2.3, II.6.4, and Secs.
III.2, III.4].

######  Lemma 2.6.

Let $T\in\mathcal{L}(\mathcal{H})$ be expansive. Then $T^{\prime}$ is a
contraction. Furthermore, the following statements hold true.

  1. (i)

Suppose that $U$ is a singular unitary operator,
$\mathcal{M}\in\operatorname{Lat}T$, and $T|_{\mathcal{M}}\approx U$. Then
$T|_{\mathcal{M}}\cong U$ and
$\mathcal{H}\ominus\mathcal{M}\in\operatorname{Lat}T$. Also $\mathcal{M}$,
$\mathcal{H}\ominus\mathcal{M}\in\operatorname{Lat}T^{\prime}$ and
$T^{\prime}|_{\mathcal{M}}\cong U$.

  2. (ii)

If $R$ is an a.c. contraction such that $R\prec T$, then $T^{\prime}$ is an
a.c. contraction.

  3. (iii)

If $\mathcal{R}^{\infty}(T)=\\{0\\}$, then $T^{\prime}$ is a completely non-
unitary contraction;

  4. (iv)

If $\mathcal{H}=\vee_{n=0}^{\infty}T^{n}\ker T^{*}$, then $T^{\prime}$ is a
contraction of class $C_{\cdot 0}$.

###### Proof.

The estimate $\|T^{\prime}\|\leq 1$ easy follows from the relations
$T^{\prime*}T=I$ and $\ker T^{\prime*}=\ker T^{*}$.

(i) Since $T^{\prime*}T=I$ and $T|_{\mathcal{M}}$ is invertible, we have
$\mathcal{M}\in\operatorname{Lat}T^{\prime*}$ and
$T^{\prime*}|_{\mathcal{M}}=(T|_{\mathcal{M}})^{-1}\approx U^{-1}$. Since
$T^{\prime*}$ is a contraction and $U$ is a singular unitary operator, we have
$T^{\prime*}|_{\mathcal{M}}\cong U^{-1}$ and
$\mathcal{H}\ominus\mathcal{M}\in\operatorname{Lat}T^{\prime*}$. The
conclusion of part (i) of the lemma follows from these relations.

(ii) Assume that $T^{\prime}$ is not an a.c. contraction. Therefore, there
exist a singular unitary operator $U$ and
$\\{0\\}\neq\mathcal{M}\in\operatorname{Lat}T^{\prime}$ such that
$\mathcal{H}\ominus\mathcal{M}\in\operatorname{Lat}T^{\prime}$ and
$T^{\prime}|_{\mathcal{M}}\cong U$. Consequently, $\mathcal{M}$,
$\mathcal{H}\ominus\mathcal{M}\in\operatorname{Lat}T$ and
$T|_{\mathcal{M}}\cong U$. Let $Y$ be a quasiaffinity such that $YR=TY$. The
transformation $P_{\mathcal{M}}Y$ realizes the relation $R\buildrel
d\over{\prec}T|_{\mathcal{M}}$. Thus, $R\buildrel d\over{\prec}U$, a
contradiction.

(iii) Assume that there exists $\mathcal{K}\in\operatorname{Lat}T^{\prime}$
such that $U:=T^{\prime}|_{\mathcal{K}}$ is unitary. Then $T^{\prime}=U\oplus
R$ for some $R\in\mathcal{L}(\mathcal{H}\ominus\mathcal{K})$. We have

$\displaystyle T^{\prime*}T^{\prime}=I_{\mathcal{K}}\oplus R^{*}R,\ \
(T^{\prime*}T^{\prime})^{-1}=I_{\mathcal{K}}\oplus(R^{*}R)^{-1},$
$\displaystyle\text{ and }T=T^{\prime\prime}=U\oplus R(R^{*}R)^{-1}.$

Consequently, $\mathcal{K}\subset\mathcal{R}^{\infty}(T)$. Thus,
$\mathcal{K}=\\{0\\}$.

(iv) This is a straightforward corollary of Lemma 2.4(ii), because
$T^{\prime}=L_{T}^{*}$. ∎

######  Lemma 2.7.

Let $R\in\mathcal{L}(\mathcal{H})$ be a contraction, and let $1\leq N=\dim\ker
R^{*}\leq\infty$. Then there exists $Y\in\mathcal{I}(S_{N},R)$ such that
$Y\ker S_{N}^{*}=\ker R^{*}$ and
$\operatorname{clos}YH^{2}_{N}=\vee_{n=0}^{\infty}R^{n}\ker R^{*}$.
Furthermore, if $R$ is left-invertible, then there exists
$X\in\mathcal{I}(R^{\prime},S_{N})$ such that
$\operatorname{clos}X\mathcal{H}=H^{2}_{N}$, $X\ker R^{\prime*}=\ker
S_{N}^{*}$ and $\ker X=\mathcal{R}^{\infty}(R^{\prime})$.

###### Proof.

By [NFBK, Theorem I.4.1], there exists a Hilbert space $\mathcal{K}$ and an
isometry $V\in\mathcal{L}(\mathcal{K})$ such that
$\mathcal{H}\subset\mathcal{K}$,
$\mathcal{K}\ominus\mathcal{H}\in\operatorname{Lat}V$, and
$R=P_{\mathcal{H}}V|_{\mathcal{H}}$. Set $E=\ker R^{*}$. Then $E=\ker
V^{*}\cap\mathcal{H}$. Set $\mathcal{M}=\oplus_{n=0}^{\infty}V^{n}E$. Then
$\ker(V|_{\mathcal{M}})^{*}=E$ and $V|_{\mathcal{M}}\cong S_{N}$. Set
$Y=P_{\mathcal{H}}|_{\mathcal{M}}$. Then $Y$ satisfies the conclusion of the
lemma.

Set $X=P_{\mathcal{M}}|_{\mathcal{H}}$. If $R$ is left-invertible, then
$R^{\prime}=P_{\mathcal{H}}V|_{\mathcal{H}}(V^{*}|_{\mathcal{H}}P_{\mathcal{H}}V|_{\mathcal{H}})^{-1}$
and $\ker R^{\prime*}=\ker R^{*}=E$. Let $x\in\mathcal{H}$, $v\in\mathcal{M}$
and $u\in E$. Then $(R^{\prime}x,u)=0$ and $V^{*}(Vv+u)=v$. Therefore,

$\displaystyle(XR^{\prime}x,Vv+u)$
$\displaystyle=(P_{\mathcal{M}}R^{\prime}x,Vv+u)=(R^{\prime}x,Vv+u)=(V^{*}R^{\prime}x,v)$
$\displaystyle=(V^{*}P_{\mathcal{H}}V|_{\mathcal{H}}(V^{*}|_{\mathcal{H}}P_{\mathcal{H}}V|_{\mathcal{H}})^{-1}x,v)=(x,v)=(P_{\mathcal{M}}x,v)$
$\displaystyle=(Xx,v)=(Xx,V^{*}(Vv+u))=(VXx,Vv+u).$

Since $\mathcal{M}=V\mathcal{M}\oplus E$, we conclude that $XR^{\prime}=VX$.
Clearly, $XE=E$. Since
$E\subset\operatorname{clos}X\mathcal{H}\in\operatorname{Lat}V$, we have
$\operatorname{clos}X\mathcal{H}=\mathcal{M}$.

Set $\mathcal{F}=\ker X$. By Lemma 2.3,
$\mathcal{R}^{\infty}(R^{\prime})\subset\mathcal{F}$. Also,
$\mathcal{F}\in\operatorname{Lat}R^{\prime}$. Since $\mathcal{F}=\ker
P_{\mathcal{M}}|_{\mathcal{H}}=\mathcal{M}^{\perp}\cap\mathcal{H}$, we have
$\mathcal{F}\in\operatorname{Lat}V^{*}$. Consequently,
$\mathcal{F}\in\operatorname{Lat}R^{*}$. The equality
$R^{*}R^{\prime}=I_{\mathcal{H}}$ implies that
$R^{*}|_{\mathcal{F}}R^{\prime}|_{\mathcal{F}}=I_{\mathcal{F}}$. Therefore,
$R^{*}\mathcal{F}=\mathcal{F}$. Furthermore,

$\ker R^{*}|_{\mathcal{F}}=E\cap\mathcal{F}\subset
E\cap\mathcal{M}^{\perp}=\\{0\\}.$

Thus, $R^{*}|_{\mathcal{F}}$ is invertible, and
$(R^{*}|_{\mathcal{F}})^{-1}=R^{\prime}|_{\mathcal{F}}$. Thus,
$\mathcal{F}\subset\mathcal{R}^{\infty}(R^{\prime})$. ∎

######  Corollary 2.8.

Suppose that $T$ is expansive, $1\leq N=\dim\ker T^{*}\leq\infty$, and
$\mathcal{R}^{\infty}(T)=\\{0\\}$. Then there exists a quasiaffinity
$X\in\mathcal{I}(T,S_{N})$ such that $X\ker T^{*}=\ker S_{N}^{*}$.

###### Proof.

Set $R=T^{\prime}$ and apply Lemma 2.7 to $R$.∎

######  Lemma 2.9.

Let $A\in\mathcal{L}(\mathcal{H})$ and $B\in\mathcal{L}(\mathcal{K})$ be power
bounded operators, and let $Y\in\mathcal{L}(\mathcal{H},\mathcal{K})$ be such
that $BYA=Y$. Then

(2.2) $\mathcal{H}_{A,0}\subset\ker Y,$

where $\mathcal{H}_{A,0}$ is defined by (1.2). Consequently,

  1. (i)

if $A$ is of class $C_{0\cdot}$, then $Y=\mathbb{O}$;

  2. (ii)

if $\ker Y=\\{0\\}$, then $A$ is of class $C_{1\cdot}$.

Moreover, there exists
$X_{+}\in\mathcal{L}(\mathcal{H}_{+}^{(a)},\mathcal{K})$ such that
$\|X_{+}\|\leq\sup_{n\in\mathbb{N}}\|B^{n}\|\|Y\|$, $X_{+}X_{+,A}=Y$ and
$X_{+}=BX_{+}A_{+}^{(a)}$.

###### Proof.

For every $n\in\mathbb{N}$ we have $B^{n}YA^{n}=Y$. Set
$C=\sup_{n\in\mathbb{N}}\|B^{n}\|$. Then $\|Yx\|\leq C\|Y\|\|A^{n}x\|$ for
every $x\in\mathcal{H}$ and every $n\in\mathbb{N}$. Consequently, (2.2) is
fulfilled.

Set $X_{+}(X_{+,A}x)=Yx$ for $x\in\mathcal{H}$. Inclusion (2.2) implies that
the definition is correct. We have

$\|X_{+}X_{+,A}x\|^{2}=\|Yx\|^{2}\leq
C^{2}\|Y\|^{2}\mathop{\mathrm{Lim}}_{n}\|A^{n}x\|^{2}=C^{2}\|Y\|^{2}\|X_{+,A}x\|^{2},$

where $\mathop{\mathrm{Lim}}$ is a Banach limit which is used in the
construction of $(X_{+,A},A_{+}^{(a)})$, and

$X_{+}X_{+,A}=Y=BYA=BX_{+}X_{+,A}A=BX_{+}A_{+}^{(a)}X_{+,A}.$

Since the range of $X_{+,A}$ is dense, we conclude that $X_{+}$ can be
extended as (linear, bounded) transformation onto $\mathcal{H}_{+}^{(a)}$, and
$X_{+}=BX_{+}A_{+}^{(a)}$. ∎

######  Corollary 2.10.

Suppose that $T\in\mathcal{L}(\mathcal{H})$ is an expansive operator,
$R\mathcal{\in}\mathcal{L}(\mathcal{K})$ is a power bounded operator, and
$Z\in\mathcal{I}(R,T)$. Then $\mathcal{K}_{R,0}\subset\ker Z$, where
$\mathcal{K}_{R,0}$ is defined by (1.2). Moreover, there exists
$Y\in\mathcal{I}(R_{+}^{(a)},T)$ such that $\|Y\|\leq\|Z\|$ and $Z=YX_{+,R}$.

###### Proof.

Since $ZR=TZ$ and $T^{\prime*}T=I$, we have $T^{\prime*}ZR=Z$. By Lemma 2.9,
there exists $Y\in\mathcal{L}(\mathcal{K}_{+}^{(a)},\mathcal{H})$ such that
$Z=YX_{+,R}$, and $\|Y\|\leq\|Z\|$, because $\|T^{\prime}\|\leq 1$.
Furthermore,

$TYX_{+,R}=TZ=ZR=YX_{+,R}R=YR_{+}^{(a)}X_{+,R}.$

Since the range of $X_{+,R}$ is dense, we conclude that $TY=YR_{+}^{(a)}$. ∎

######  Corollary 2.11.

Suppose that $N\in\mathbb{N}$, $T$ is expansive, $R$ is a contraction, and
$R\prec T\prec S_{N}$. Then $T\sim S_{N}$.

###### Proof.

Since $R\prec S_{N}$, we have $R_{+}^{(a)}\cong S_{N}$ by [G12, Lemma 2.1].
Denote by $Z$ and $X$ the quasiaffinities such that $ZR=TZ$ and $XT=S_{N}X$.
By Corollary 2.10, there exists $Y\in\mathcal{I}(S_{N},T)$ such that
$Z=YX_{+,R}$. The latest equality implies that $Y$ has dense range. Applying
Lemma 2.2 to $XY$ with $A=S_{N}$, we obtain that $\ker XY=\\{0\\}$.
Consequently, $\ker Y=\\{0\\}$. ∎

######  Corollary 2.12.

Suppose that $N\in\mathbb{N}$, $T$ is expansive, and $S_{N}\buildrel
d\over{\prec}T$. Then there exists $M\in\mathbb{N}$ such that $M\leq N$ and
$S_{M}\prec T$.

###### Proof.

Let $Y_{0}$ realize the relation $S_{N}\buildrel d\over{\prec}T$. Set

$\mathcal{N}=\ker Y_{0},\ \ \
R=P_{H^{2}_{N}\ominus\mathcal{N}}S_{N}|_{H^{2}_{N}\ominus\mathcal{N}}\ \
\text{and }\ \ Z=Y_{0}|_{H^{2}_{N}\ominus\mathcal{N}}.$

Then $Z$ realizes the relation $R\prec T$. By Corollary 2.10, $R\in
C_{1\cdot}$. By [U83] or [T84], and [K89] or [NFBK, Sec. IX.1], and [G12,
Lemma 2.1], there exists $M\leq N$ such that $R_{+}^{(a)}\cong S_{M}$. By
Corollary 2.10, there exists $Y\in\mathcal{I}(S_{M},T)$ such that
$Z=YX_{+,R}$. The range of $Y$ is dense, because the range of $Z$ is dense.
Set $\mathcal{E}=\ker Y$. Then $\mathcal{E}\in\operatorname{Lat}S_{M}$. By
[G02], there exists $\mathcal{M}\in\operatorname{Lat}R$ such that
$\mathcal{E}=\operatorname{clos}X_{+,R}\mathcal{M}$. Consequently,
$\mathcal{M}\subset\ker Z=\\{0\\}$. Thus, $\mathcal{E}=\\{0\\}$. ∎

##  3\. Intertwining by Toeplitz operators

Let $\theta\in H^{\infty}$ be an inner function. Set
$\mathcal{K}_{\theta}=H^{2}\ominus\theta H^{2}$. Then

(3.1)
$\mathcal{K}_{\theta}=\theta\overline{\chi}\overline{\mathcal{K}_{\theta}}=P_{+}\theta
H^{2}_{-},$

and $\mathcal{K}_{\theta}\in\operatorname{Lat}S^{*}$. Set
$S(\theta)=P_{\mathcal{K}_{\theta}}S|_{\mathcal{K}_{\theta}}$. Then

(3.2) $S(\theta)f=\chi f-(f,P_{+}\overline{\chi}\theta)\theta\ \
(f\in\mathcal{K}_{\theta}).$

For an inner function $\theta$ such that $\theta(0)=0$ there exists a singular
(with respect to $m$) positive Borel measure $\nu$ on $\mathbb{T}$ such that
$\nu(\mathbb{T})=1$ and

(3.3)
$\frac{1}{1-\theta(z)}=\int_{\mathbb{T}}\frac{1}{1-z\overline{\zeta}}\mathrm{d}\nu(\zeta)\
\ (z\in\mathbb{D}),$

which is called the _Clark measure_ of $\theta$. Every function
$f\in\mathcal{K}_{\theta}$ has nontangential boundary values $f(\zeta)$ for
a.e. $\zeta\in\mathbb{T}$ with respect to $\nu$, and

(3.4)
$f(z)=(1-\theta(z))\int_{\mathbb{T}}\frac{f(\zeta)}{1-z\overline{\zeta}}\mathrm{d}\nu(\zeta)\
\ (z\in\mathbb{D}).$

The relation (3.4) and F. and M. Riesz theorem (see, for example, [GMR,
Theorem 1.21] or [Ga, Theorem II.3.8]) imply that

(3.5) $(1-\theta)H^{2}\cap\mathcal{K}_{\theta}=\\{0\\}.$

For an inner function $\theta$ such that $\theta(0)=0$ set

(3.6) $U(\theta)=S(\theta)+\mathbf{1}\otimes\overline{\chi}\theta.$

For a singular positive Borel measure $\nu$ on $\mathbb{T}$ such that
$\nu(\mathbb{T})=1$ denote by $U_{\nu}$ the operator of multiplication by the
independent variable on $L^{2}(\mathbb{T},\nu)$. If $\nu$ is the Clark measure
for $\theta$, then

(3.7) $U(\theta)\cong U_{\nu}.$

Conversely, if $\nu$ is a singular positive Borel measure on $\mathbb{T}$ such
that $\nu(\mathbb{T})=1$ and $\theta$ is defined by (3.3), then $\theta$ is an
inner function and $\theta(0)=0$.

For references, see [Cl] and [Po93] or [GR], [GMR].

If $\theta$ is an inner function such that $\theta(0)\neq 0$, then $S(\theta)$
is invertible and it is follows from (3.2) that

(3.8)
$(S(\theta)^{*})^{-1}=S(\theta)+(\theta-\frac{1}{\overline{\theta(0)}})\otimes
P_{+}\overline{\chi}\theta.$

The _Toeplitz operator_ $T_{\psi}$ with the symbol $\psi\in L^{2}$ acts by the
formula $T_{\psi}h=P_{+}\psi h$ for $h\in H^{\infty}$. It can be extended as a
_bounded_ operator on $H^{2}$ if and only if $\psi\in L^{\infty}$, and then it
acts by the formula $T_{\psi}h=P_{+}\psi h$ ($h\in H^{2}$). The following
lemma can be found, for example, in [Pe, Theorem 3.1.2].

######  Lemma 3.1.

Let $T\in\mathcal{L}(H^{2})$. Then $T=T_{\psi}$ for some $\psi\in L^{\infty}$
if and only if $S^{*}TS=T$.

It can be checked by the straightforward calculation that

(3.9) $T_{\psi}S-ST_{\psi}=\mathbf{1}\otimes P_{+}\overline{\chi\psi}.$

Let $0\not\equiv g\in H^{2}$, and let $f\in H^{2}$ be such that $|f|\leq|g|$
$m$-a.e. on $\mathbb{T}$. Using the equality
$H^{1}\cap\overline{\chi}\overline{H}^{1}=\\{0\\}$ it is easy to see that
$\ker T_{\frac{f}{\overline{g}}}=\\{0\\}$. If $g$, $1/g\in H^{2}$, then
$T_{\frac{g}{\overline{g}}}$ is a quasiaffinity. A description of functions
$g$ such that $\ker T_{\frac{\overline{g}}{g}}=\\{0\\}$ is given in [Sa89]. A
necessary condition for $\ker T_{\frac{\overline{g}}{g}}=\\{0\\}$ is that $g$
is outer, and $g\neq(1-\theta)f$ for any inner function $\theta\in H^{\infty}$
and any $f\in H^{2}$. But this condition is not sufficient [I]. With respect
to this subject see also [Po01].

For $g\in H^{2}$ such that $\|g\|=1$ define the function $\omega\in
H^{\infty}$ as follows:

(3.10)
$\frac{1}{1-\omega(z)}=\int_{\mathbb{T}}\frac{|g(\zeta)|^{2}\mathrm{d}m(\zeta)}{1-z\overline{\zeta}}\
\ \ (z\in\mathbb{D}).$

Then $1-\omega$ is an outer function. The following theorem is proved in
[Sa88].

######  Theorem A.

[Sa88, Lemma 2] Let $g\in H^{2}$, and let $\|g\|=1$. Define $\omega$ by
(3.10). Then $T_{1-\omega}T_{\overline{g}}\in\mathcal{L}(H^{2})$.

A simplest example of expansive operator is one-dimensional perturbation of
$S$ from Lemma 3.2 below. Other one-dimensional perturbations of $S$ was
considered in [N] and [CT].

######  Lemma 3.2.

Let $g\in H^{2}$, and let $g(0)=1$. Set $T=S-\mathbf{1}\otimes S^{*}g$. Set
$g=\theta f$, where $\theta$ is inner and $f$ is outer. Then the following are
fulfilled.

  1. (i)

$\mathcal{K}_{\theta}=\mathcal{R}^{\infty}(T)$, and
$T|_{\mathcal{K}_{\theta}}=(S(\theta)^{*})^{-1}$.

  2. (ii)

$T_{\frac{f}{\overline{g}}}S=TT_{\frac{f}{\overline{g}}}$, and there exists an
outer function $\varphi\in H^{\infty}$ such that
$T_{\varphi}T_{\overline{g}}\in\mathcal{L}(H^{2})$, and
$T_{\varphi}T_{\overline{g}}T=ST_{\varphi}T_{\overline{g}}$.

  3. (iii)

The following are equivalent:

$\mathrm{(a)}$ $T$ is similar to an isometry;

$\mathrm{(b)}$ $T\approx S$;

$\mathrm{(c)}$ $T_{\frac{g}{\overline{g}}}$ is invertible.

###### Proof.

Straightforward computation shows that
$\mathcal{K}_{\theta}\in\operatorname{Lat}T$ and
$T|_{\mathcal{K}_{\theta}}=(S(\theta)^{*})^{-1}$. Thus,
$\mathcal{K}_{\theta}\subset\mathcal{R}^{\infty}(T)$. Existence of $\varphi$
from (ii) follows from Theorem A. Intertwining relation from (ii) can be
checked by straightforward computation. Set $X=T_{\varphi}T_{\overline{g}}$.
By Lemma 2.3, $\ker X\supset\mathcal{R}^{\infty}(T)$, because $XT=SX$. Since
$\ker X=\mathcal{K}_{\theta}$, we have
$\mathcal{K}_{\theta}\supset\mathcal{R}^{\infty}(T)$. Thus, (i) and (ii) are
proved.

It follows from (i) that if $T$ is similar to isometry, then $g$ is outer.
Indeed, if $T$ is similar to an isometry, then $(S(\theta)^{*})^{-1}\approx U$
for some unitary $U$. Consequently, $S(\theta)^{*}\approx U^{-1}$. But
$S(\theta)$ is a $C_{0}$-contraction. Therefore,
$\mathcal{K}_{\theta}=\\{0\\}$. Also, it is easy to see that (c) implies that
$g$ is outer.

Suppose that $g$ is outer. Then $X$ is a quasiaffinity. Therefore,
(a)$\Rightarrow$(b). The relation (c)$\Rightarrow$(b) follows from (ii), and
the relation (b)$\Rightarrow$(a) is evident.

Let $Y\in\mathcal{L}(H^{2})$ be such that $YS=TY$. Then $S^{*}YS=Y$. By Lemma
3.1, there exists $\psi\in L^{\infty}$ such that $Y=T_{\psi}$. The equality
$T_{\psi}S-ST_{\psi}=-\mathbf{1}\otimes T_{\psi}^{*}S^{*}g$ and (3.9) imply
that $\psi=\frac{h}{\overline{g}}$ for some $h\in H^{2}$.

If $T_{\psi}$ is invertible, then by [Pe, Lemma 3.1.10] $1/\psi\in
L^{\infty}$. Therefore, $h=\vartheta\eta g$, where $\vartheta$ is inner and
$\eta$, $1/\eta\in H^{\infty}$. Since $T_{\psi}=T_{\frac{\vartheta
g}{\overline{g}}}T_{\eta}$ and $T_{\eta}$ is ivertible, we conclude that
$T_{\frac{\vartheta g}{\overline{g}}}$ is invertible. If $\vartheta$ is not a
constant, then $0\not\equiv gS^{*}\vartheta\in\ker T_{\frac{\vartheta
g}{\overline{g}}}^{*}$. Thus, $\vartheta$ is a constant, and
(b)$\Rightarrow$(c) is proved. ∎

######  Example 3.3.

Let $\theta\in H^{\infty}$ be an inner function, and let $\theta(0)=0$. Let
$\nu$ be the Clark measure for $\theta$. Set

$T=S+\mathbf{1}\otimes\overline{\chi}\theta.$

Then $\theta H^{2}\in\operatorname{Lat}T$, $T|_{\theta H^{2}}\cong S$ and
$P_{\mathcal{K}_{\theta}}T|_{\mathcal{K}_{\theta}}\cong U_{\nu}$ (see (3.6)
and (3.7)). Set $Y=T_{\theta}$ and $X=T_{1-\overline{\theta}}$. It is easy to
see that $YS=TY$ and $XT=SX$. Since $X$ is a quasiaffinity, we have $T\prec
S$. Since $\mathcal{I}(U_{\nu}^{*},S^{*})=\\{0\\}$, we conclude that
$S\not\prec T$.

######  Example 3.4.

Let $\theta\in H^{\infty}$ be an inner function, and let $\theta(0)\neq 0$.
Set

$T=S+\mathbf{1}\otimes\overline{\chi}(1-\theta/\theta(0)).$

By Lemma 3.2, $\mathcal{R}^{\infty}(T)=\mathcal{K}_{\theta}$ and
$T|_{\mathcal{K}_{\theta}}=(S(\theta)^{*})^{-1}$. Furthermore, $\theta
H^{2}\in\operatorname{Lat}T$ and $T|_{\theta H^{2}}\cong S$. Let
$X\in\mathcal{I}(T,S)$. By Lemma 2.3, $X|_{\mathcal{K}_{\theta}}=\mathbb{O}$.
Let $Y\in\mathcal{I}(S,T)$. Set $Y_{0}=P_{\mathcal{K}_{\theta}}Y$. Then
$Y_{0}S=(S(\theta)^{*})^{-1}Y_{0}$. Consequently,
$Y_{0}^{*}=S^{*}Y_{0}^{*}S(\theta)$. By Lemma 2.9 (i), $Y_{0}=\mathbb{O}$.
Thus, $S\not\prec T\not\prec S$.

In the proof of the next lemma, Toeplitz operators with matrix-valued symbols
are used. Namely, let $N\in\mathbb{N}$, and let $\Psi$ be an $N\times N$
matrix whose elements are functions from $L^{2}$. The _Toeplitz operator_
$T_{\Psi}$ with the symbol $\Psi$ acts by the formula $T_{\Psi}h=P_{+}\Psi h$
for $h\in H^{2}_{N}$ such that their elements are functions from $H^{\infty}$.
It can be extended as a _bounded_ operator on $H^{2}_{N}$ if and only if all
elements of the matrix $\Psi$ are functions from $L^{\infty}$, and then it
acts by the formula $T_{\Psi}h=P_{+}\Psi h$ ($h\in H^{2}_{N}$). See, for
example, [Pe, Sec. 3.4].

######  Lemma 3.5.

Let $N\in\mathbb{N}$, and let $\\{f_{k}\\}_{k=1}^{N}\subset H^{2}_{N}$. Set
$T=S_{N}+\sum_{k=1}^{N}e_{k}\otimes f_{k}$. Then $S_{N}\buildrel
i\over{\prec}T\buildrel d\over{\prec}S_{N}$.

###### Proof.

The case $N=1$ is considered in Lemma 3.2. Consider the case $N\geq 2$. Recall
that $f_{k}$ are columns of $N$ functions from $H^{2}$. Denote by $f_{kj}$ the
element of $f_{k}$ on $j$-th place. Set

$F=[f_{kj}]_{k,j=1}^{N}=[f_{1},\ldots,f_{N}]\ \ \text{ and }\ \
\psi=\det(I_{N\otimes N}-\chi F).$

Denote by $f_{\mathrm{Ad}kj}$ ($k,j=1,\ldots,N$) the elements of (algebraic)
adjoint matrix of $I_{N\otimes N}-\chi F$. Since $f_{kj}\in H^{2}$, we have
$\psi\in H^{\frac{2}{N}}$ and $f_{\mathrm{Ad}kj}\in H^{\frac{2}{N-1}}$
($k,j=1,\ldots,N$). Since $\psi(0)=1$, we have $\psi\not\equiv 0$. Therefore,
$\log|\psi|$, $\log|f_{\mathrm{Ad}kj}|\in L^{1}$ ($k,j=1,\ldots,N$), and the
elements of the matrix $(I_{N\otimes N}-\chi F)^{-1}$ are functions defined
$m$-a.e. on $\mathbb{T}$. Furthermore, there exists an outer function $\eta\in
H^{\infty}$ such that

$|\eta|=\begin{cases}1,&\text{ if }|f_{\mathrm{Ad}kj}|\leq|\psi|\text{ for all
}k,j=1,\ldots,N,\\\
\displaystyle{\frac{|\psi|}{\max_{k,j=1,\ldots,N}|f_{\mathrm{Ad}kj}|}},&\text{
if }|f_{\mathrm{Ad}kj}|\geq|\psi|\text{ for some }k,j=1,\ldots,N\end{cases}$

$m$-a.e. on $\mathbb{T}$. Set $\Psi=\eta\bigl{(}(I_{N\otimes
N}-\overline{\chi}\overline{F})^{-1}\bigr{)}^{\mathrm{T}}$. Then the elements
$\psi_{kj}$ (where $k$ is the number of row and $j$ is the number of column
($k,j=1,\ldots,N$)) of $\Psi$ are functions from $L^{\infty}$. Set
$Y=T_{\Psi}$ and
$\psi_{k}=[P_{+}\overline{\chi}\overline{\psi}_{kj}]_{j=1}^{N}$
($k=1,\ldots,N$). Then $YS_{N}-S_{N}Y=\sum_{k=1}^{N}e_{k}\otimes\psi_{k}$.
Since $\psi_{k}=Y^{*}f_{k}$ ($k=1,\ldots,N$), we have $YS_{N}=TY$.

Denote by $\varphi_{kj}$ ($k,j=1,\ldots,N$) the outer functions from Theorem A
applied to the elements of $(I_{N\otimes
N}-\overline{\chi}\overline{F})^{\mathrm{T}}$ (which multiplied by appropriate
constants). Set $\varphi=\prod_{1\leq k,j\leq N}\varphi_{kj}$. Set
$X=T_{\varphi I_{N\otimes N}}T_{(I_{N\otimes
N}-\overline{\chi}\overline{F})^{\mathrm{T}}}$. Then
$X\in\mathcal{L}(H^{2}_{N})$. Straightforward calculation shows that
$XT=S_{N}X$.

If $g\in H^{2}$ and $\gamma\in L^{\infty}$, then
$P_{+}\overline{g}P_{+}\gamma=P_{+}\overline{g}\gamma$. Therefore, if $h\in
H^{2}_{N}$ is such that its elements are functions from $H^{\infty}$, then

$XYh=\varphi P_{+}(I_{N\otimes
N}-\overline{\chi}\overline{F})^{\mathrm{T}}P_{+}\Psi h=\varphi
P_{+}(I_{N\otimes N}-\overline{\chi}\overline{F})^{\mathrm{T}}\Psi
h=\varphi\eta h.$

Since $X$ and $Y$ are bounded, we conclude that $XYh=\varphi\eta h$ ($h\in
H^{2}_{N}$). Consequently, $\ker XY=\\{0\\}$ and
$\operatorname{clos}XYH^{2}_{N}=H^{2}_{N}$ (since $\varphi$ and $\eta$ are
outer). Therefore, $\ker Y=\\{0\\}$ and
$\operatorname{clos}XH^{2}_{N}=H^{2}_{N}$. ∎

##  4\. Expansive operators for which the unilateral shift of finite
multiplicity is their quasiaffine transform

###  4.1. Preliminaries

In this subsection, some relationships between isometries are studies, which
will be used in the sequel. Also, Theorem B from [H] is formulated in the end
of this subsection.

######  Lemma 4.1.

Let an isometry $V$ have the representation

$V=\left[\begin{matrix}V_{1}&*\\\ \mathbb{O}&V_{0}\end{matrix}\right],$

where $V_{0}$ is of class $C_{00}$. Then $V\cong V_{1}$.

###### Proof.

Let $V_{1}=U\oplus S_{N}$ be the Wold decomposition of the isometry $V_{1}$,
where $U$ is unitary and $0\leq N\leq\infty$. Then

$V=U\oplus V_{10},\ \ \ \text{ where }\ V_{10}=\left[\begin{matrix}S_{N}&*\\\
\mathbb{O}&V_{0}\end{matrix}\right].$

Since $S_{N}$ and $V_{0}$ are of class $C_{\cdot 0}$, then $V_{10}$ is of
class $C_{\cdot 0}$, too, by [K89, Theorem 3] or [NFBK, Theorem IX.1.6]
(applied to adjoint). Since $V_{0}$ is of class $C_{0\cdot}$, by [K89, Theorem
3] or [NFBK, Theorem IX.1.6], $V_{10}^{(a)}\cong
S_{N}^{(a)}=U_{\mathbb{T},N}$. Since $V_{10}$ is an isometry, we conclude that
$V_{10}\cong S_{N}$. ∎

######  Lemma 4.2.

Suppose that a power bounded operator $R$ has the form

$R=\left[\begin{matrix}R_{1}&*\\\ \mathbb{O}&R_{0}\end{matrix}\right],$

and there exists a $C_{0}$-contraction $A$ such that $A\buildrel
d\over{\prec}R_{0}$. Then $R_{+}^{(a)}\cong(R_{1})_{+}^{(a)}$.

###### Proof.

Denote by $\mathcal{K}$ the space on which $R$ acts. Let
$\mathcal{K}=\mathcal{K}_{1}\oplus\mathcal{K}_{0}$ be the decomposition of
$\mathcal{K}$ such that $R_{1}=R|_{\mathcal{K}_{1}}$ and
$R_{0}=P_{\mathcal{K}_{0}}R|_{\mathcal{K}_{0}}$. Set
$\mathcal{G}_{1}=\operatorname{clos}X_{+,R}\mathcal{K}_{1}$,
$\mathcal{G}_{0}=\mathcal{K}_{+}^{(a)}\ominus\mathcal{G}_{1}$, and
$V=R_{+}^{(a)}$. Then

$V=\left[\begin{matrix}V_{1}&*\\\ \mathbb{O}&V_{0}\end{matrix}\right]$

with respect to the decomposition
$\mathcal{K}_{+}^{(a)}=\mathcal{G}_{1}\oplus\mathcal{G}_{0}$. By [K89],
$X_{+,R_{1}}=X_{+,R}|_{\mathcal{K}_{1}}$ and
$(R_{1})_{+}^{(a)}=V|_{\mathcal{G}_{1}}=V_{1}$.

We have $A\buildrel d\over{\prec}R_{0}\buildrel d\over{\prec}V_{0}$. Since $A$
is a $C_{0}$-contraction, $V_{0}$ is a $C_{0}$-contraction, too. In
particular, $V_{0}$ is of class $C_{00}$ [NFBK, Prop. III.4.2]. By Lemma 4.1,
$V\cong V_{1}$. ∎

######  Lemma 4.3.

Suppose that $\sigma\subset\mathbb{T}$,
$X\in\mathcal{I}(U_{\sigma}^{-1},S^{*})$, and there exists $f_{1}\in
L^{2}(\sigma,m)$ such that $Xf_{1}=\mathbf{1}$. Then $\sigma=\mathbb{T}$ and

$U_{\mathbb{T}}|_{\vee_{n=0}^{\infty}U_{\mathbb{T}}^{n}f_{1}}\cong S.$

###### Proof.

We have $X^{*}S=U_{\sigma}X^{*}$. Set $X^{*}\mathbf{1}=\psi$, then $\psi\in
L^{\infty}(\sigma,m)$ and $X^{*}h=\psi h$ for every $h\in H^{2}$. Therefore,
$Xf=P_{+}\overline{\psi}f$ for every $f\in L^{2}(\sigma,m)$. Since
$\mathbf{1}=P_{+}\overline{\psi}f_{1}$, there exists $h\in H^{2}$ such that
$1+\overline{\chi}\overline{h}=\overline{\psi}f_{1}$ $m$-a.e. on $\mathbb{T}$.
Since $\psi=0$ $m$-a.e. on $\mathbb{T}\setminus\sigma$ and $1+\chi h\in
H^{2}$, we conclude that $m(\mathbb{T}\setminus\sigma)=0$. Furthermore,

$\int_{\mathbb{T}}\log(|\psi||f_{1}|)\mathrm{d}m=\int_{\mathbb{T}}\log|1+\chi
h|\mathrm{d}m>-\infty.$

Since $\psi\in L^{\infty}$, we conclude that
$\int_{\mathbb{T}}\log|f_{1}|\mathrm{d}m>-\infty$. The conclusion of the lemma
follows from this relation and well-known description of
$\operatorname{Lat}U_{\mathbb{T}}$. ∎

######  Lemma 4.4.

Suppose that $N\in\mathbb{N}$, $V_{+}\in\mathcal{L}(\mathcal{K}_{+})$ is an
isometry, $\dim\ker V_{+}^{*}<\infty$,
$X_{+}\in\mathcal{L}(\mathcal{K}_{+},H_{N}^{2})$, and
$S_{N}^{*}X_{+}V_{+}=X_{+}$. Let $V\in\mathcal{L}(\mathcal{K})$ be the minimal
unitary extension of $V_{+}$. Then there exists
$X\in\mathcal{L}(\mathcal{K},H_{N}^{2})$ such $S_{N}^{*}XV=X$ and
$X|_{\mathcal{K}_{+}}=X_{+}$.

###### Proof.

Using the Wold decomposition and appropriate unitary equivalence, we may
assume that $V_{+}=S_{M}\oplus U$, where $U\in\mathcal{L}(\mathcal{G})$ is
unitary and $1\leq M\leq\dim\ker V_{+}^{*}$. Then $V=U_{\mathbb{T},M}\oplus
U$. Set $X_{1}=X_{+}|_{H_{M}^{2}\oplus\\{0\\}}$ and
$X_{0}=X_{+}|_{\\{0\\}\oplus\mathcal{G}}$. Then $S_{N}^{*}X_{1}S_{M}=X_{1}$
and $S_{N}^{*}X_{0}U=X_{0}$. Writing $S_{N}^{*}$ and $S_{M}$ as $N\times N$
and $M\times M$ diagonal matrices, whose elements on the main diagonal are
$S^{*}$ and $S$, respectively, and $X_{1}$ as a $N\times M$ matrix:
$X_{1}=[X_{+jk}]_{j=1,\ldots,N\atop k=1,\ldots,M}$, we have
$S^{*}X_{+jk}S=X_{+jk}$ for all $j=1,\ldots,N$, $k=1,\ldots,M$. By Lemma 3.1,
there exist $\psi_{jk}\in L^{\infty}$ such that $X_{+jk}=T_{\psi_{jk}}$.
Define $X_{jk}\in\mathcal{L}(L^{2},H^{2})$ by the formula $X_{jk}f=P_{+}\psi
f$ ($f\in L^{2}$). Set

$X=\left[[X_{jk}]_{j=1,\ldots,N\atop k=1,\ldots,M},\ X_{0}\right].$

It is easy to see that $X$ satisfies the conclusion of the lemma. ∎

######  Lemma 4.5.

Let $N\in\mathbb{N}$. Write $L^{2}_{N+1}=H^{2}_{N}\oplus(H^{2}_{-})_{N}\oplus
L^{2}$. Let $h_{0}\in H^{2}_{N}$, and let $f\in L^{2}$ be such that
$\int_{\mathbb{T}}\log|f|\mathrm{d}m>-\infty$. Set

$\mathcal{M}=H^{2}_{N}\vee\vee_{n=0}^{\infty}U_{\mathbb{T},N+1}^{n}(\overline{\chi}\overline{h}_{0}\oplus
f).$

Then $U_{\mathbb{T},N+1}|_{\mathcal{M}}\cong S_{N+1}$.

###### Proof.

Set
$\mathcal{N}=\vee_{n=0}^{\infty}(S_{*,N}^{n}\overline{\chi}\overline{h}_{0}\oplus
U_{\mathbb{T}}^{n}f)$. Then $\mathcal{M}=H^{2}_{N}\oplus\mathcal{N}$. We show
that

(4.1) $\mathcal{N}\cap((H^{2}_{-})_{N}\oplus\\{0\\})=\\{0\\}.$

Indeed, assume that $\\{p_{n}\\}_{n}$ is a sequence of analytic polynomials,
$h\in H^{2}_{N}$,

$p_{n}(S_{*,N})\overline{\chi}\overline{h}_{0}\to\overline{\chi}\overline{h}\
\ \text{ and }\ \ \ p_{n}(U_{\mathbb{T}})f\to 0.$

Let $h_{0}=[h_{j}]_{j=1}^{N}$, where $h_{j}\in H^{2}$ ($j=1,\ldots,N$). Set
$s(\zeta)=\max_{j=1,\ldots,N}|h_{j}(\zeta)|$ for $m$-a.e.
$\zeta\in\mathbb{T}$. Since $\int_{\mathbb{T}}\log|f|\mathrm{d}m>-\infty$,
there exists an outer function $\varphi\in H^{\infty}$ such that

$|\varphi|=\begin{cases}\frac{|f|}{s},&\ \ \text{ if }|f|\leq s,\\\ 1,&\ \
\text{ if }|f|\geq s.\end{cases}$

We have

$\varphi(S_{*,N})p_{n}(S_{*,N})\overline{\chi}\overline{h}_{0}=[P_{-}\varphi
P_{-}p_{n}\overline{\chi}\overline{h}_{j}]_{j=1}^{N}=[P_{-}\varphi
p_{n}\overline{\chi}\overline{h}_{j}]_{j=1}^{N}\to\varphi(S_{*,N})\overline{\chi}\overline{h}.$

But

$\displaystyle\|\varphi(S_{*,N})p_{n}(S_{*,N})\overline{\chi}\overline{h}_{0}\|^{2}$
$\displaystyle\leq\sum_{j=1}^{N}\|\varphi
p_{n}\overline{\chi}\overline{h}_{j}\|^{2}\leq\sum_{j=1}^{N}\int_{\mathbb{T}}|\varphi|^{2}s^{2}|p_{n}|^{2}\mathrm{d}m$
$\displaystyle\leq N\int_{\mathbb{T}}|f|^{2}|p_{n}|^{2}\mathrm{d}m\to 0.$

We obtain that $\varphi(S_{*,N})\overline{\chi}\overline{h}=0$. Since
$\varphi$ is outer, [NFBK, Prop. III.3.1] implies that
$\overline{\chi}\overline{h}=0$. Thus, (4.1) is proved.

Set $R=(S_{*,N}\oplus U_{\mathbb{T}})|_{\mathcal{N}}$. There exist $u\in
L^{\infty}$ and $g\in H^{2}$ such that $|u|=1$ $m$-a.e. on $\mathbb{T}$, $g$
is outer, and $f=ug$. We have $\mathcal{N}\subset(H^{2}_{-})_{N}\oplus
uH^{2}$. By (4.1), $P_{\\{0\\}\oplus uH^{2}}|_{\mathcal{N}}$ realizes the
relation $R\prec U_{\mathbb{T}}|_{uH^{2}}$. Since
$U_{\mathbb{T}}|_{uH^{2}}\cong S$ and $R$ is a contraction, we have
$\operatorname{ind}R=-1$ [T87]. Since

$U_{\mathbb{T},N+1}|_{\mathcal{M}}=\left[\begin{matrix}S_{N}&*\\\
\mathbb{O}&R\end{matrix}\right],$

[Co, Theorem XI.3.7] implies that
$\operatorname{ind}U_{\mathbb{T},N+1}|_{\mathcal{M}}=\operatorname{ind}S_{N}+\operatorname{ind}R=-N-1$.
Since $\mu_{U_{\mathbb{T},N+1}|_{\mathcal{M}}}\leq N+1$ (where $\mu_{T}$ for
an operator $T$ is defined in (1.1)), we conclude that
$U_{\mathbb{T},N+1}|_{\mathcal{M}}\cong S_{N+1}$. ∎

Recall that the multiplicity $\mu_{T}$ for an operator $T$ is defined in
(1.1).

######  Theorem 4.6.

Suppose than $N\in\mathbb{N}$, $V_{+}\in\mathcal{L}(\mathcal{K}_{+})$ is an
a.c. isometry, $\mu_{V_{+}}\leq N$,
$X_{+}\in\mathcal{L}(\mathcal{K}_{+},H^{2}_{N})$, and
$S_{N}^{*}X_{+}V_{+}=X_{+}$. Suppose that there exist
$\\{f_{j}\\}_{j=1}^{N}\subset\mathcal{K}_{+}$ such that $X_{+}f_{j}=e_{j}$
($j=1,\ldots,N$). Then

$V_{+}|_{\vee_{j=1}^{N}\vee_{n=0}^{\infty}V_{+}^{n}f_{j}}\cong S_{N}.$

###### Proof.

The theorem will be proved using induction. Let $N=1$. Since there exists
$f_{1}\in\mathcal{K}_{+}$ such that $Xf_{1}=e_{1}=\mathbf{1}$, we have
$\mathcal{K}_{+}\neq\\{0\\}$, and $V_{+}\cong S$ or $V_{+}\cong U_{\sigma}$
for some $\sigma\subset\mathbb{T}$. If $V_{+}\cong S$, the conclusion of the
theorem is fulfilled for every $0\not\equiv f_{1}\in\mathcal{K}_{+}$. If
$V_{+}\cong U_{\sigma}$, Lemma 4.3 is applied. Thus, if $N=1$, then the
theorem is proved.

If $N\geq 1$, assume that the theorem is proved for all $1\leq k\leq N$. We
will to prove the theorem for $N+1$. Let $X$ and $V$ be from Lemma 4.4 applied
to $V_{+}$ and $X_{+}$. Then $V$ is unitary, and $S_{N+1}^{*}XV=X$. Set

(4.2) $\mathcal{M}_{k}=\vee_{j=1}^{k}\vee_{n\in\mathbb{Z}}V^{n}f_{j}\ \ \
\text{ and }X_{k}=P_{H_{k}^{2}\oplus\\{0\\}}X|_{\mathcal{M}_{k}}\ \ \
(k=1,\ldots,N+1).$

Then $S_{k}^{*}X_{k}V|_{\mathcal{M}_{k}}=X_{k}$ and $X_{k}f_{j}=e_{j}$ for
$j=1,\ldots,k$ ($k=1,\ldots,N+1$). Thus, $X_{k}$ and $V|_{\mathcal{M}_{k}}$
satisfy the assumption of the theorem. By the inductive hypothesis,
$V|_{\vee_{j=1}^{k}\vee_{n=0}^{\infty}V^{n}f_{j}}\cong S_{k}$ for
$k=1,\ldots,N$. Consequently,

(4.3) $V|_{\mathcal{M}_{k}}\cong U_{\mathbb{T},k}\ \ \ (k=1,\ldots,N).$

Taking into account relations (4.3) and the estimate $\mu_{V}\leq N+1$, and
using appropriate unitary equivalence, we may assume that
$V=U_{\mathbb{T},N}\oplus U_{\sigma}$ for some $\sigma\subset\mathbb{T}$, and
$\mathcal{M}_{k}=L^{2}_{k}\oplus\\{0\\}\subset L^{2}_{N}$ ($k=1,\ldots,N$).

Write $S_{N+1}^{*}$ as $(N+1)\times(N+1)$ diagonal matrix, whose elements on
the main diagonal are $S^{*}$. Write $V$ as $(N+1)\times(N+1)$ diagonal
matrix, whose $N$ elements on the main diagonal are $U_{\mathbb{T}}$ and the
ending element is $U_{\sigma}$. Write $X$ as a $(N+1)\times(N+1)$ matrix:
$X=[X_{jk}]_{j,k=1,\ldots,N+1}$. Then $S^{*}X_{jk}U_{\mathbb{T}}=X_{jk}$ and
$S^{*}X_{j,N+1}U_{\sigma}=X_{j,N+1}$ for all $j=1,\ldots,N+1$, $k=1,\ldots,N$.
Therefore, there exist $\psi_{jk}\in L^{\infty}$ such that
$X_{jk}f=P_{+}\psi_{jk}f$ for every $f\in L^{2}$, and
$X_{j,N+1}f=P_{+}\psi_{j,N+1}f$ for every $f\in L^{2}(\sigma,m)$ and for all
$j=1,\ldots,N+1$, $k=1,\ldots,N$.

Set $\Psi=[\psi_{jk}]_{j,k=1,\ldots,N+1}$. For $k=1,\ldots,N$ write $f_{k}\in
L^{2}_{k}\oplus\\{0\\}$ as a column whose first $k$ elements are functions
from $L^{2}$ and other are zeros functions. Write $f_{N+1}$ as a column whose
first $N$ elements are functions from $L^{2}$ and $(N+1)$th element is a
function from $L^{2}(\sigma,m)$. Set $F=[f_{1},\ldots,f_{N+1}]$. Then $F$ is a
upper-triangular $(N+1)\times(N+1)$ matrix, whose elements are functions from
$L^{2}$ and $L^{2}(\sigma,m)$. Denote the elements from the main diagonal of
$F$ by $f_{0k}$ ($k=1,\ldots,N+1$). Then $f_{0k}\in L^{2}$ for $k=1,\ldots,N$,
$f_{0,N+1}\in L^{2}(\sigma,m)$, and $\det F=\prod_{k=1}^{N+1}f_{0k}$.

Since $Xf_{j}=e_{j}$ ($j=1,\ldots,N+1$), we have $P_{+}\Psi
F=I_{(N+1)\times(N+1)}$. Therefore, there exists $(N+1)\times(N+1)$ matrix
$G$, whose elements are functions from $H^{2}$, such that $\Psi
F=I_{(N+1)\times(N+1)}+\overline{\chi}\overline{G}$. Set
$h=\det(I_{(N+1)\times(N+1)}+\chi G)$. Then $h\in H^{\frac{2}{N+1}}$, and
$h(0)=1$. Therefore, $\int_{\mathbb{T}}\log|h|\mathrm{d}m>-\infty$. Set
$\psi=\det\Psi$. Then $\psi\in L^{\infty}$. We have

$\overline{h}=\det(\Psi F)=\det\Psi\det F=\psi\prod_{k=1}^{N+1}f_{0k}.$

Therefore,

$\int_{\mathbb{T}}\log|\psi|\mathrm{d}m+\sum_{k=1}^{N+1}\int_{\mathbb{T}}\log|f_{0k}|\mathrm{d}m=\int_{\mathbb{T}}\log|h|\mathrm{d}m>-\infty.$

We obtain that $\int_{\mathbb{T}}\log|f_{0k}|\mathrm{d}m>-\infty$ for all
$k=1,\ldots,N+1$. In particular, $\sigma=\mathbb{T}$ and
$V=U_{\mathbb{T},N+1}$.

By the inductive hypothesis,
$V|_{\vee_{j=1}^{N}\vee_{n=0}^{\infty}V^{n}f_{j}}\cong S_{N}$. We may assume
that

$\vee_{j=1}^{N}\vee_{n=0}^{\infty}V^{n}f_{j}=H^{2}_{N}\oplus\\{0\\}\oplus\\{0\\}\subset
H^{2}_{N}\oplus(H^{2}_{-})_{N}\oplus L^{2}=L^{2}_{N+1}.$

Note that $f_{0,N+1}=P_{\\{0\\}\oplus\\{0\\}\oplus L^{2}}f_{N+1}$. Set
$\overline{\chi}\overline{h}_{0}=P_{\\{0\\}\oplus(H^{2}_{-})_{N}\oplus\\{0\\}}f_{N+1}$,
where $h_{0}\in H^{2}_{N}$. Then

$\vee_{j=1}^{N+1}\vee_{n=0}^{\infty}V^{n}f_{j}=H^{2}_{N}\vee\vee_{n=0}^{\infty}V^{n}(\overline{\chi}\overline{h}_{0}\oplus
f_{0,N+1}).$

By Lemma 4.5,

$V|_{\vee_{j=1}^{N+1}\vee_{n=0}^{\infty}V^{n}f_{j}}\cong S_{N+1}.$

Since $V$ is a unitary extension of $V_{+}$, the theorem is proved. ∎

Let $\psi\in L^{\infty}$. The _Hankel operator_
$H_{\psi}\in\mathcal{L}(H^{2},H^{2}_{-})$ with the symbol $\psi$ acts by the
formula $H_{\psi}h=P_{-}\psi h$ ($h\in H^{2}$). By [Pe, formula (1.1.9)],
$\|H_{\psi}\|=\operatorname{dist}(\psi,H^{\infty})$. If $\theta_{k}$ ($k=1,2)$
are inner functions, then

(4.4)
$\|P_{\mathcal{K}_{\theta_{1}}}|_{\theta_{2}H^{2}}\|=\|H_{\overline{\theta_{1}}\theta_{2}}\|=\operatorname{dist}(\theta_{1},\theta_{2}H^{\infty})\leq\|\theta_{1}-\theta_{2}\|_{\infty}.$

For an inner function $\theta\in H^{\infty}$ and $0\neq a\in\mathbb{D}$ set

(4.5) $\theta_{a}=\frac{\theta-a}{1-\overline{a}\theta}.$

Then $\theta_{a}$ is an inner function, $\theta$ and $\theta_{a}$ are
relatively prime, and

(4.6) $\|\theta-\theta_{a}\|_{\infty}\leq\frac{2|a|}{1-|a|}.$

######  Lemma 4.7.

Suppose that $N\in\mathbb{N}$, $N\geq 2$, $\delta_{0}>0$, $\theta\in
H^{\infty}$ is an inner function, $\mathcal{H}$ is a Hilbert space, and
$Z\in\mathcal{L}(H^{2}_{N},\mathcal{H})$ is such that

$\|Z(\theta h\oplus\\{0\\})\|\geq\delta_{0}\|h\|\ \ \text{ for every }\ h\in
H^{2}.$

Then for every $0<\delta<\delta_{0}$ there exist
$\\{\mathcal{N}_{j}\\}_{j=1}^{N}\subset\operatorname{Lat}S_{N}$ such that
$S_{N}|_{\mathcal{N}_{j}}\cong S$, $\|Zh\|\geq\delta\|h\|$ for every
$h\in\mathcal{N}_{j}$ and $j=1,\ldots,N$, and
$\vee_{j=1}^{N}\mathcal{N}_{j}=H^{2}_{N}$.

###### Proof.

Let $0\neq a\in\mathbb{D}$, and let $0<\varepsilon<1$. Define $N\times N$
matrix $\Theta$ as follows:

$\Theta=\left[\begin{matrix}(1-\varepsilon^{2})^{\frac{1}{2}}\theta_{a}&(1-\varepsilon^{2})^{\frac{1}{2}}\theta&(1-\varepsilon^{2})^{\frac{1}{2}}\theta&\ldots&(1-\varepsilon^{2})^{\frac{1}{2}}\theta\\\
\varepsilon&\varepsilon&0&\ldots&0\\\ 0&0&\varepsilon&\ldots&0\\\
\ldots&\ldots&\ldots&\ldots&\ldots\\\
0&0&0&\ldots&\varepsilon\end{matrix}\right].$

Then

$\displaystyle\det\Theta$
$\displaystyle=\det\left[\begin{matrix}(1-\varepsilon^{2})^{\frac{1}{2}}\theta_{a}&(1-\varepsilon^{2})^{\frac{1}{2}}\theta\\\
\varepsilon&\varepsilon\end{matrix}\right]\det\left[\begin{matrix}\varepsilon&\ldots&0\\\
\ldots&\ldots&\ldots\\\ 0&\ldots&\varepsilon\end{matrix}\right]$
$\displaystyle=(1-\varepsilon^{2})^{\frac{1}{2}}\varepsilon^{N-1}(\theta_{a}-\theta)=-(1-\varepsilon^{2})^{\frac{1}{2}}\varepsilon^{N-1}\frac{a(1-\frac{\overline{a}}{a}\theta^{2})}{1-\overline{a}\theta}.$

Therefore, $\det\Theta$ is an outer function. By [NFBK, Prop. V.6.1 and
Theorem V.6.2], $\Theta$ is an outer function. The columns $\Theta_{j}$
($j=1,\ldots,N$) of the matrix $\Theta$ are inner functions from
$H^{\infty}(\mathbb{C},\mathbb{C}^{N})$. Set $\mathcal{N}_{j}=\Theta_{j}H^{2}$
($j=1,\ldots,N$). Then $\mathcal{N}_{j}\in\operatorname{Lat}S_{N}$,
$S_{N}|_{\mathcal{N}_{j}}\cong S$, ($j=1,\ldots,N$), and
$\vee_{j=1}^{N}\mathcal{N}_{j}=H^{2}_{N}$.

Let $h\in H^{2}$. For $2\leq j\leq N$ we have

$\displaystyle\|Z\Theta_{j}h\|$
$\displaystyle\geq\|Z(1-\varepsilon^{2})^{\frac{1}{2}}(\theta
h\oplus\\{0\\})\|-\|Z(0\oplus\ldots\oplus\varepsilon h\oplus\ldots\oplus 0)\|$
$\displaystyle\geq(1-\varepsilon^{2})^{\frac{1}{2}}\delta_{0}\|h\|-\|Z\|\varepsilon\|h\|$
$\displaystyle=\bigl{(}(1-\varepsilon^{2})^{\frac{1}{2}}\delta_{0}-\|Z\|\varepsilon\bigr{)}\|h\|=\bigl{(}(1-\varepsilon^{2})^{\frac{1}{2}}\delta_{0}-\|Z\|\varepsilon\bigr{)}\|\Theta_{j}h\|.$

By (4.4) and (4.6),
$\|P_{\mathcal{K}_{\theta}}\theta_{a}h\|\leq\frac{2|a|}{1-|a|}\|h\|$.
Therefore,

$\|P_{\theta
H^{2}}\theta_{a}h\|^{2}=\|h\|^{2}-\|P_{\mathcal{K}_{\theta}}\theta_{a}h\|^{2}\geq\frac{1-2|a|-3|a|^{2}}{(1-|a|)^{2}}\|h\|^{2}.$

For $j=1$ we have

$\displaystyle\|Z\Theta_{1}h\|$
$\displaystyle\geq\|Z(1-\varepsilon^{2})^{\frac{1}{2}}(P_{\theta
H^{2}}\theta_{a}h\oplus\\{0\\})\|$
$\displaystyle\quad-\|Z(1-\varepsilon^{2})^{\frac{1}{2}}(P_{\mathcal{K}_{\theta}}\theta_{a}h\oplus\\{0\\})\|-\|Z(0\oplus\varepsilon
h\oplus\ldots\oplus 0)\|$
$\displaystyle\geq(1-\varepsilon^{2})^{\frac{1}{2}}\delta_{0}\frac{(1-2|a|-3|a|^{2})^{\frac{1}{2}}}{1-|a|}\|h\|$
$\displaystyle\quad-\|Z\|(1-\varepsilon^{2})^{\frac{1}{2}}\frac{2|a|}{1-|a|}\|h\|-\|Z\|\varepsilon\|h\|$
$\displaystyle=\Bigl{(}\frac{(1-\varepsilon^{2})^{\frac{1}{2}}}{1-|a|}(\delta_{0}(1-2|a|-3|a|^{2})^{\frac{1}{2}}-2|a|\|Z\|)-\|Z\|\varepsilon\Bigr{)}\|h\|$
$\displaystyle=\Bigl{(}\frac{(1-\varepsilon^{2})^{\frac{1}{2}}}{1-|a|}(\delta_{0}(1-2|a|-3|a|^{2})^{\frac{1}{2}}-2|a|\|Z\|)-\|Z\|\varepsilon\Bigr{)}\|\Theta_{1}h\|.$

When $0<\delta<\delta_{0}$ is given, the conclusion of the lemma is fulfilled
for sufficiently small $|a|$ and $\varepsilon$. ∎

######  Theorem B.

[H] Let $u\in L^{\infty}$, and let $|u|=1$ $m$-a.e. on $\mathbb{T}$. Then for
every $\varepsilon>0$ there exist $\alpha$, $\beta$, $\varphi\in H^{\infty}$
such that $\alpha$ and $\beta$ are inner, $\frac{1}{\varphi}\in H^{\infty}$,
$\|\varphi\|_{\infty}\leq 1+\varepsilon$, $\|\frac{1}{\varphi}\|_{\infty}\leq
1+\varepsilon$, and

$u=\frac{\overline{\varphi}}{\varphi}\alpha\overline{\beta}.$

###  4.2. Resuts

In this subsection main results of the paper are proved.

######  Lemma 4.8.

Suppose that $T$ is an expansive operator, $N=\dim\ker T^{*}<\infty$, and
$S_{N}\prec T$. Set $\mathcal{H}_{1}=\vee_{n=0}^{\infty}T^{\prime n}\ker
T^{*}$. Then $T^{\prime}$ is an a.c. contraction of class $C_{1\cdot}$, and
$(T^{\prime}|_{\mathcal{H}_{1}})_{+}^{(a)}\cong S_{N}$.

###### Proof.

By Lemma 2.6(ii), $T^{\prime}$ is an a.c. contraction. Denote by $Y$ a
quasiaffinity such that $YT^{*}=S_{N}^{*}Y$. Then $Y\ker T^{*}=\ker S_{N}^{*}$
and $Y=S_{N}^{*}YT^{\prime}$. By Lemma 2.9(ii), $T^{\prime}$ is a contraction
of class $C_{1\cdot}$.

Set $V_{+}=(T^{\prime})^{(a)}_{+}$. By [NFBK, Sec. IX.1], $V_{+}$ is an a.c.
isometry. Let $X_{+}$ be from Lemma 2.9. Then $Y=X_{+}X_{+,T^{\prime}}$ and
$X_{+}=S_{N}^{*}X_{+}V_{+}$. Set $\mathcal{F}=X_{+,T^{\prime}}\ker T^{*}$.
Then $\ker S_{N}^{*}=Y\ker T^{*}=X_{+}\mathcal{F}$. By Theorem 4.6,

$V_{+}|_{\vee_{n=0}^{\infty}V_{+}^{n}\mathcal{F}}\cong S_{N}.$

By [K89],
$(T^{\prime}|_{\mathcal{H}_{1}})_{+}^{(a)}=V_{+}|_{\operatorname{clos}X_{+,T^{\prime}}\mathcal{H}_{1}}$.
Furthermore,

$\operatorname{clos}X_{+,T^{\prime}}\mathcal{H}_{1}=\operatorname{clos}X_{+,T^{\prime}}(\vee_{n=0}^{\infty}T^{\prime
n}\ker T^{*})=\vee_{n=0}^{\infty}V_{+}^{n}\mathcal{F}.$

Thus, $(T^{\prime}|_{\mathcal{H}_{1}})_{+}^{(a)}\cong S_{N}$. ∎

######  Lemma 4.9.

Suppose that $T$ is an expansive operator, $N=\dim\ker T^{*}<\infty$, and
$S_{N}\prec T$. Then $T^{\prime}$ is a contraction of class $C_{10}$, and
$(T^{\prime})_{+}^{(a)}\cong S_{N}$.

###### Proof.

Denote by $\mathcal{H}$ the space on which $T$ acts. Set
$\mathcal{H}_{1}=\vee_{n=0}^{\infty}T^{\prime n}\ker T^{*}$,
$T_{1}=T^{\prime}|_{\mathcal{H}_{1}}$,
$\mathcal{H}_{0}=\mathcal{H}\ominus\mathcal{H}_{1}$. Then

$T^{\prime}=\left[\begin{matrix}T_{1}&T_{2}\\\
\mathbb{O}&T_{0}\end{matrix}\right]$

with respect to the decomposition
$\mathcal{H}=\mathcal{H}_{1}\oplus\mathcal{H}_{0}$. Note that $\ker
T^{\prime*}=\ker T_{1}^{*}$. We will to prove that $T_{0}$ is a
$C_{0}$-contraction.

Let $Z_{0}\in\mathcal{I}(S,T_{0})$. By [T93, Lemma 1], there exists
$Z_{2}\in\mathcal{L}(H^{2},\mathcal{H}_{1})$ such that

$\left[\begin{matrix}I_{\mathcal{H}_{1}}&Z_{2}\\\
\mathbb{O}&Z_{0}\end{matrix}\right]\left[\begin{matrix}T_{1}&\mathbb{O}\\\
\mathbb{O}&S\end{matrix}\right]=\left[\begin{matrix}T_{1}&T_{2}\\\
\mathbb{O}&T_{0}\end{matrix}\right]\left[\begin{matrix}I_{\mathcal{H}_{1}}&Z_{2}\\\
\mathbb{O}&Z_{0}\end{matrix}\right].$

Let $Z_{1}\in\mathcal{I}(S_{N},T_{1})$ be from Lemma 2.7 applied to $T_{1}$.
Then $Z_{1}\ker S_{N}^{*}=\ker T_{1}^{*}=\ker T^{*}$. Since
$\operatorname{clos}Z_{1}H^{2}_{N}=\mathcal{H}_{1}$, Lemma 2.2 implies that

(4.7) $\ker Z_{1}=\\{0\\}.$

It is easy to see that

$\left[\begin{matrix}Z_{1}&Z_{2}\\\
\mathbb{O}&Z_{0}\end{matrix}\right]\left[\begin{matrix}S_{N}&\mathbb{O}\\\
\mathbb{O}&S\end{matrix}\right]=\left[\begin{matrix}T_{1}&T_{2}\\\
\mathbb{O}&T_{0}\end{matrix}\right]\left[\begin{matrix}Z_{1}&Z_{2}\\\
\mathbb{O}&Z_{0}\end{matrix}\right].$

Let $Y$ be a quasiaffinity such that $Y^{*}S_{N}=TY^{*}$. Since $\dim\ker
T^{*}=N$, we have $Y\ker T^{*}=\ker S_{N}^{*}$.

Set

$Z=\left[\begin{matrix}Z_{1}&Z_{2}\\\ \mathbb{O}&Z_{0}\end{matrix}\right]$

and $Z_{+}=YZ$. We have $S_{N}^{*}Z_{+}S_{N+1}=Z_{+}$. Since $Z_{1}\ker
S_{N}^{*}=\ker T^{*}$, we have $Z_{+}(\ker S_{N}^{*}\oplus\\{0\\})=\ker
S_{N}^{*}$. Therefore,

$\displaystyle S_{N}Z_{+}=S_{N}S^{*}_{N}Z_{+}S_{N+1}=(I_{H^{2}_{N}}-P_{\ker
S_{N}^{*}})Z_{+}S_{N+1}$ $\displaystyle=(I_{H^{2}_{N}}-P_{Z_{+}(\ker
S_{N}^{*}\oplus\\{0\\})})Z_{+}S_{N+1}=Z_{+}\Bigl{(}S_{N+1}+\sum_{k=1}^{N}e_{k}\otimes
f_{k}\Bigr{)}$

for some $\\{f_{k}\\}_{k=1}^{N}\subset H^{2}_{N}$. Set
$A=S_{N+1}+\sum_{k=1}^{N}e_{k}\otimes f_{k}$. By Lemma 3.5, $S_{N+1}\buildrel
i\over{\prec}A$. If $\ker Z_{+}=\\{0\\}$, then $S_{N+1}\buildrel
i\over{\prec}A\buildrel i\over{\prec}S_{N}$, a contradiction. Thus, $\ker
Z_{+}\neq\\{0\\}$. Consequently, $\ker Z\neq\\{0\\}$. From this relation,
(4.7) and the definition of $Z$ we conclude that $\ker Z_{0}\neq\\{0\\}$. By
[T93, Introduction], $T_{0}$ is a $C_{0}$-contraction.

By Lemmas 4.2 and 4.8, $(T^{\prime})_{+}^{(a)}\cong S_{N}$. Since $T^{\prime}$
is of class $C_{1\cdot}$, we have $T^{\prime}\prec S_{N}$. Therefore,
$T^{\prime}$ is of class $C_{10}$. ∎

######  Theorem 4.10.

Suppose that $T$ is an expansive operator, $N=\dim\ker T^{*}<\infty$, and
$S_{N}\buildrel d\over{\prec}T$. Then $I-T^{*}T\in\mathfrak{S}_{1}$, $T\sim
S_{N}$, and $T^{\prime}\sim S_{N}$.

###### Proof.

By Lemma 2.2, there exists a quasiaffinity $Y$ such that $YT^{*}=S_{N}^{*}Y$.
Consequently, $Y\ker T^{*}=\ker S_{N}^{*}$. Furthermore,
$Y=S_{N}^{*}YT^{\prime}$. By Lemma 4.9, $(T^{\prime})_{+}^{(a)}\cong S_{N}$.
By Lemma 2.9, there exists $X_{+}\in\mathcal{L}(H^{2}_{N})$ such that
$Y=X_{+}X_{+,T^{\prime}}$ and $S_{N}^{*}X_{+}S_{N}=X_{+}$. By Lemma 3.1, there
exists an $N\times N$ matrix $\Psi$ whose elements are functions from
$L^{\infty}$ such that $X_{+}=T_{\Psi}$.

Set $\mathcal{H}_{1}=\vee_{n=0}^{\infty}T^{\prime n}\ker T^{*}$ and
$T_{1}=T^{\prime}|_{\mathcal{H}_{1}}$. Note that $\ker T^{*}=\ker T_{1}^{*}$.
Let $Z\in\mathcal{I}(S_{N},T_{1})$ be from Lemma 2.7 applied to $T_{1}$. Then
$Z\ker S_{N}^{*}=\ker T^{*}$. By Lemma 2.2, $Z$ is a quasiaffinity.

Denote by $\mathcal{H}$ the space on which $T$ acts, and by $J$ the natural
imbedding of $\mathcal{H}_{1}$ into $\mathcal{H}$. Since
$X_{+,T^{\prime}}JZS_{N}=S_{N}X_{+,T^{\prime}}JZ$, there exists an $N\times N$
matrix $\Phi$ whose elements are functions from $H^{\infty}$ such that
$X_{+,T^{\prime}}JZ=T_{\Phi}$. Let
$\Phi=\Theta_{0}\Phi_{0}=\Phi_{1}\Theta_{1}$ be the canonical and
$*$-canonical factorizations of operator-valued function $\Phi$ [NFBK, Sec.
V.4.3]. Namely, $\Theta_{0}$ is inner, $\Phi_{0}$ is outer, $\Phi_{1}$ is
$*$-outer, and $\Theta_{1}$ is $*$-inner. We have

(4.8)
$\operatorname{clos}X_{+,T^{\prime}}\mathcal{H}_{1}=\operatorname{clos}X_{+,T^{\prime}}JZH^{2}_{N}=\operatorname{clos}\Phi
H^{2}_{N}=\Theta_{0}H^{2}_{M}$

for some $M\leq N$. By Lemma 4.8,
$(T^{\prime}|_{\mathcal{H}_{1}})_{+}^{(a)}\cong S_{N}$. Consequently, $M=N$.
By [NFBK, Secs. V.6.1, V.6.2], $\Theta_{0}$ and $\Theta_{1}$ are inner from
both sides, $\Phi_{0}$ and $\Phi_{1}$ are outer from both sides, and
$\varphi:=\det\Phi_{0}=\det\Phi_{1}$ is outer. Clearly, $\varphi\in
H^{\infty}$. Furthermore, $\Phi$ is outer if and only if both $\Theta_{0}$ and
$\Theta_{1}$ are unitary constant functions. Assume that $\Theta_{1}$ is a
non-constant inner function. Set

$\mathcal{K}_{\Theta_{1}}=H^{2}_{N}\ominus\Theta_{1}H^{2}_{N}=\Theta_{1}(H^{2}_{-})_{N}\cap
H^{2}_{N}.$

The equalities $YJZ\ker S_{N}^{*}=\ker S_{N}^{*}$ and $YJZ=T_{\Psi\Phi}$ imply
that $\Psi\Phi=\overline{G}$, where $G$ is an $N\times N$ matrix whose
elements are functions from $H^{\infty}$. Consequently,

(4.9) $\Phi_{1}\mathcal{K}_{\Theta_{1}}\subset\Phi(H^{2}_{-})_{N}\cap
H^{2}_{N}\subset\ker X_{+}.$

Furthermore,

(4.10) $X_{+,T^{\prime}}\mathcal{H}\cap\Phi_{1}H^{2}_{N}\subset\Phi
H^{2}_{N}.$

Indeed, let $x\in\mathcal{H}$ be such that
$X_{+,T^{\prime}}x\in\Phi_{1}H^{2}_{N}$. Then there exist $h\in H^{2}_{N}$ and
$f\in\mathcal{K}_{\Theta_{1}}$ such that
$X_{+,T^{\prime}}x=\Phi_{1}\Theta_{1}h+\Phi_{1}f=\Phi h+\Phi_{1}f$. Since
$\Phi h=X_{+,T^{\prime}}JZh$, we have $\Phi_{1}f=X_{+,T^{\prime}}(x-JZh)$. By
(4.9), $Y(x-JZh)=0$. Since $\ker Y=\\{0\\}$, we have $\Phi_{1}f\equiv 0$.
Since $\Phi_{1}$ is an outer $N\times N$ matrix-valued function, we conclude
that $f\equiv 0$. The inclusion (4.10) is proved.

Let $\Phi_{1}^{\mathrm{Ad}}$ be the (algebraic) adjoint of $\Phi_{1}$. Then
$\Phi_{1}\Phi_{1}^{\mathrm{Ad}}=\varphi I_{N\times N}$. Consequently,

$\Phi_{1}\Phi_{1}^{\mathrm{Ad}}X_{+,T^{\prime}}\mathcal{H}=\varphi(S_{N})X_{+,T^{\prime}}\mathcal{H}=X_{+,T^{\prime}}\varphi(T^{\prime})\mathcal{H}\subset
X_{+,T^{\prime}}\mathcal{H}\cap\Phi_{1}H^{2}_{N}\subset\Phi H^{2}_{N}$

by (4.10). Since $\varphi$ is outer, we have

$H^{2}_{N}=\operatorname{clos}\varphi(S_{N})H^{2}_{N}=\operatorname{clos}\varphi(S_{N})X_{+,T^{\prime}}\mathcal{H}\subset\operatorname{clos}\Phi
H^{2}_{N}.$

The latest inclusion and (4.8) imply that
$\operatorname{clos}X_{+,T^{\prime}}\mathcal{H}_{1}=H^{2}_{N}$. Since the
mapping $\mathcal{M}\mapsto\operatorname{clos}X_{+,T^{\prime}}\mathcal{M}$
($\mathcal{M}\in\operatorname{Lat}T^{\prime}$) is a lattice-isomorphism
between $\operatorname{Lat}T^{\prime}$ and $\operatorname{Lat}S_{N}$ [G02], we
conclude that $\mathcal{H}_{1}=\mathcal{H}$. Thus, $T^{\prime}=T_{1}$ and the
relation $T^{\prime}\sim S_{N}$ is proved. Furthermore, by Lemma 2.4(i),
$\mathcal{R}^{\infty}(T)=\\{0\\}$. By Corollary 2.8, $T\sim S_{N}$.

By [G08], $I-T^{\prime*}T^{\prime}\in\mathfrak{S}_{1}$. By Lemma 2.1,
$I-T^{*}T\in\mathfrak{S}_{1}$. ∎

######  Lemma 4.11.

Suppose that $T$ is an expansive operator, $\dim\ker T^{*}=1$, and $S\buildrel
d\over{\prec}T$. Then there exists a quasiaffinity $Z_{1}\in\mathcal{I}(S,T)$
such that $\|Z_{1}\|=1$ and for every $0<\delta<1$ there exists an inner
function $\vartheta$ such that $\|Z_{1}\vartheta h\|\geq\delta\|h\|$ for every
$h\in H^{2}$.

###### Proof.

We repeat the part of the proof of Theorem 4.10.

By Lemma 2.2, there exists a quasiaffinity $Y$ such that $YT^{*}=S^{*}Y$.
Consequently, there exists $x_{0}\in\ker T^{*}$ such that $Yx_{0}=\mathbf{1}$.
Furthermore, $Y=S^{*}YT^{\prime}$. By Lemma 4.9, $(T^{\prime})_{+}^{(a)}\cong
S$. By Lemma 2.9, there exists $X_{+}\in\mathcal{L}(H^{2})$ such that
$Y=X_{+}X_{+,T^{\prime}}$ and $S^{*}X_{+}S=X_{+}$. By Lemma 3.1, there exists
$\psi\in L^{\infty}$ such that $X_{+}=T_{\psi}$.

Denote by $\mathcal{H}$ the space on which $T$ acts. By Lemma 2.4(i) and
Theorem 4.10,

(4.11) $\mathcal{H}=\vee_{n=0}^{\infty}T^{\prime n}x_{0}.$

Let $Z\in\mathcal{I}(S,T^{\prime})$ be from Lemma 2.7 applied to $T^{\prime}$.
Multiplying $Z$ by an appropriate constant, we may assume that
$Z\mathbf{1}=x_{0}$. By Lemma 2.2, $Z$ is a quasiaffinity.

Set $\varphi_{0}=X_{+,T^{\prime}}x_{0}$. By (4.11), $\varphi_{0}\in H^{2}$ is
an outer function. Since $X_{+,T^{\prime}}x_{0}=X_{+,T^{\prime}}Z\mathbf{1}$,
we conclude that $X_{+,T^{\prime}}Z=\varphi_{0}(S)$. Therefore,
$\varphi_{0}\in H^{\infty}$. Furthermore,

$0=S^{*}\mathbf{1}=S^{*}Yx_{0}=S^{*}X_{+}X_{+,T^{\prime}}x_{0}=S^{*}T_{\psi}\varphi_{0}.$

Consequently, $\overline{\psi\varphi_{0}}\in H^{\infty}$. Therefore, there
exists $\eta\in H^{\infty}$ such that
$\psi=\overline{\eta}\frac{\overline{\varphi}_{0}}{\varphi_{0}}$.

We prove that $\eta$ is outer. Indeed, assume that $\eta=\theta g$, where
$1\not\equiv\theta$ is inner. Let $0\not\equiv f\in\mathcal{K}_{\theta}$. Set
$y=Zf$. Then

$Yy=X_{+}X_{+,T^{\prime}}y=X_{+}X_{+,T^{\prime}}Zf=T_{\psi}\varphi_{0}f=P_{+}\overline{\theta
g}\frac{\overline{\varphi}_{0}}{\varphi_{0}}\varphi_{0}f=P_{+}\overline{\theta
g\varphi_{0}}f=0,$

because $f\in\mathcal{K}_{\theta}$. This contradicts with the equality $\ker
Y=\\{0\\}$. Thus, $\eta$ is outer.

Set
$Z_{1}=X_{+,T^{\prime}}^{*}T_{\frac{\varphi_{0}}{\overline{\varphi}_{0}}}$.
Then $\|Z_{1}\|\leq 1$ and $Z_{1}\eta(S)=Y^{*}$. Therefore, $Z_{1}S=TZ_{1}$
and $\operatorname{clos}Z_{1}H^{2}=\mathcal{H}$. By Lemma 2.2, $\ker
Z_{1}=\\{0\\}$.

Let $0<\delta<1$. Take $0<\delta_{1}<1$ and $\varepsilon_{1}>0$ such that
$\frac{\delta_{1}}{(1+\varepsilon_{1})^{2}}\geq\delta$. By [NFBK, Ch. IX.3] or
[K07], there exists $\mathcal{M}\in\operatorname{Lat}T^{\prime}$ such that
$\|X_{+,T^{\prime}}x\|\geq\delta_{1}\|x\|$ for every $x\in\mathcal{M}$.
Therefore, there exists an inner function $\vartheta_{0}$ such that
$X_{+,T^{\prime}}{\mathcal{M}}=\vartheta_{0}H^{2}$. Consider
$X_{+,T^{\prime}}|_{\mathcal{M}}$ as a transformation from
$\mathcal{L}(\mathcal{M},\vartheta_{0}H^{2})$. Then
$X_{+,T^{\prime}}|_{\mathcal{M}}$ is invertible, and

$\|((X_{+,T^{\prime}}|_{\mathcal{M}})^{*})^{-1}\|=\|(X_{+,T^{\prime}}|_{\mathcal{M}})^{-1}\|\leq
1/\delta_{1}.$

Consequently,

$\|X_{+,T^{\prime}}^{*}h\|\geq\|P_{\mathcal{M}}X_{+,T^{\prime}}^{*}h\|=\|P_{\mathcal{M}}X_{+,T^{\prime}}^{*}P_{\vartheta_{0}H^{2}}h\|\geq\delta_{1}\|P_{\vartheta_{0}H^{2}}h\|$

for every $h\in H^{2}$.

Let $\alpha$, $\beta$, $\varphi$ be from Theorem B applied to
$\frac{\varphi_{0}}{\overline{\varphi}_{0}}$ and $\varepsilon_{1}$. Let $h\in
H^{2}$. Then

$\displaystyle\|Z_{1}\vartheta_{0}\beta
h\|=\|X_{+,T^{\prime}}^{*}T_{\frac{\varphi_{0}}{\overline{\varphi}_{0}}}\vartheta_{0}\beta
h\|\geq\delta_{1}\|P_{\vartheta_{0}H^{2}}T_{\frac{\overline{\varphi}}{\varphi}\alpha\overline{\beta}}\vartheta_{0}\beta
h\|=\delta_{1}\|T_{\overline{\vartheta}_{0}}T_{\frac{\overline{\varphi}}{\varphi}\alpha\overline{\beta}}\vartheta_{0}\beta
h\|$ $\displaystyle=\delta_{1}\|P_{+}\frac{\overline{\varphi}}{\varphi}\alpha
h\|\geq\delta_{1}\frac{1}{\|\varphi\|_{\infty}\|\frac{1}{\varphi}\|_{\infty}}\|\alpha
h\|\geq\frac{\delta_{1}}{(1+\varepsilon_{1})^{2}}\|\vartheta_{0}\beta
h\|\geq\delta\|\vartheta_{0}\beta h\|.$

Setting $\vartheta=\vartheta_{0}\beta$, we conclude that $Z_{1}$ satisfies the
conclusion of the lemma. ∎

######  Theorem 4.12.

Suppose that $T\in\mathcal{L}(\mathcal{H})$ is expansive, $\dim\ker T^{*}=1$,
and $S\buildrel d\over{\prec}T$. Then for every $\varepsilon>0$ there exist
$\mathcal{M}_{1}$, $\mathcal{M}_{2}\in\operatorname{Lat}T$ and invertible
transformations $Y_{1}$, $Y_{2}$ such that

$Y_{k}S=T|_{\mathcal{M}_{k}}Y_{k},\ \ \|Y_{k}\|\|Y_{k}^{-1}\|\leq
1+\varepsilon\ (k=1,2),\ \text{ and }\
\mathcal{M}_{1}\vee\mathcal{M}_{2}=\mathcal{H}.$

###### Proof.

Let $Z_{1}$ be a quasiaffinity from Lemma 4.11. Let $\vartheta_{1}$ be an
inner function from Lemma 4.11 applied with $\delta_{1}>1/(1+\varepsilon)$.
Let $\varepsilon_{1}>0$ be such that
$\delta_{1}(1-\varepsilon_{1}^{2})^{\frac{1}{2}}-\varepsilon_{1}\geq
1/(1+\varepsilon)$. Take an inner function $\vartheta_{2}$ such that
$\vartheta_{1}$ and $\vartheta_{2}$ are relatively prime and
$\|\vartheta_{1}-\vartheta_{2}\|_{\infty}\leq\varepsilon_{1}$ (for example,
use (4.5)). By (4.4),
$\|P_{\mathcal{K}_{\vartheta_{1}}}|_{\vartheta_{2}H^{2}}\|\leq\varepsilon_{1}$.

Let $h\in H^{2}$. Then
$\|P_{\vartheta_{1}H^{2}}\vartheta_{2}h\|^{2}=\|h\|^{2}-\|P_{\mathcal{K}_{\vartheta_{1}}}\vartheta_{2}h\|^{2}\geq(1-\varepsilon_{1}^{2})\|h\|^{2}$.
Therefore,

$\displaystyle\|Z_{1}\vartheta_{2}h\|$
$\displaystyle\geq\|Z_{1}P_{\vartheta_{1}H^{2}}\vartheta_{2}h\|-\|Z_{1}P_{\mathcal{K}_{\vartheta_{1}}}\vartheta_{2}h\|\geq\delta_{1}\|P_{\vartheta_{1}H^{2}}\vartheta_{2}h\|-\varepsilon_{1}\|h\|$
$\displaystyle\geq\delta_{1}(1-\varepsilon_{1}^{2})^{\frac{1}{2}}\|h\|-\varepsilon_{1}\|h\|=\bigl{(}\delta_{1}(1-\varepsilon_{1}^{2})^{\frac{1}{2}}-\varepsilon_{1})\|h\|\geq\|h\|/(1+\varepsilon).$

Since $\|Z_{1}\vartheta_{1}h\|\geq\delta_{1}\|h\|$, we obtain that

$\|Z_{1}\vartheta_{k}h\|\geq\|h\|/(1+\varepsilon)\ \ \text{ for every }h\in
H^{2}\ \text{ and }k=1,2.$

Set $\mathcal{M}_{k}=Z_{1}\vartheta_{k}H^{2}$ ($k=1,2$). Consider
$Y_{k}=Z_{1}|_{\vartheta_{k}H^{2}}$ as the transformations from
$\mathcal{L}(\vartheta_{k}H^{2},\mathcal{M}_{k})$. Then
$T|_{\mathcal{M}_{k}}=Y_{k}S|_{\vartheta_{k}H^{2}}Y_{k}^{-1}$ and
$\|Y_{k}\|\|Y_{k}^{-1}\|\leq 1+\varepsilon$. Clearly,
$S|_{\vartheta_{k}H^{2}}\cong S$. Thus, $\mathcal{M}_{k}$ and $Y_{k}$ (up to
appropriate unitary equivalence) ($k=1,2$) satisfy the conclusion of the
theorem. ∎

######  Theorem 4.13.

Suppose that $N\in\mathbb{N}$, $N\geq 2$, $T\in\mathcal{L}(\mathcal{H})$ is
expansive, $\ker T^{*}\neq\\{0\\}$, and $S_{N}\buildrel d\over{\prec}T$. Then
for every $\varepsilon>0$ there exist
$\\{\mathcal{M}_{j}\\}_{j=1}^{N}\subset\operatorname{Lat}T$ and invertible
transformations $Y_{j}$ such that

$Y_{j}S=T|_{\mathcal{M}_{j}}Y_{j},\ \ \|Y_{j}\|\|Y_{j}^{-1}\|\leq
1+\varepsilon\ (j=1,\ldots,N),\ \text{ and }\
\vee_{j=1}^{N}\mathcal{M}_{j}=\mathcal{H}.$

###### Proof.

Let $Y$ realize the relation $S_{N}\buildrel d\over{\prec}T$. For $1\leq k\leq
N$ set $\mathcal{H}_{k}=\operatorname{clos}Y(\\{0\\}\oplus\ldots\oplus
H^{2}\oplus\ldots\oplus\\{0\\})$ (where a unique nonzero summand $H^{2}$ is on
$k$th place), and $T_{k}=T|_{\mathcal{H}_{k}}$. If $\ker T_{k}^{*}=\\{0\\}$,
then $\mathcal{H}_{k}\subset\mathcal{R}^{\infty}(T)$. If $\ker
T_{k}^{*}=\\{0\\}$ for all $k=1,\ldots,N$, then
$\mathcal{H}=\vee_{k=1}^{N}\mathcal{H}_{k}\subset\mathcal{R}^{\infty}(T)$, a
contradiction with the assumption $\ker T^{*}\neq\\{0\\}$. Consequently, there
exists $1\leq k\leq N$ such that $\ker T_{k}^{*}\neq\\{0\\}$. Without loss of
generality we may assume that $k=1$. Then $T_{1}$ satisfies the assumptions of
Lemma 4.11, because the relation $S\buildrel d\over{\prec}T_{1}$ implies
$\dim\ker T_{1}^{*}\leq 1$.

Let $Z_{1}$ be a quasiaffinity from Lemma 4.11 applied to $T_{1}$. Take
$\varepsilon_{1}>0$ such that $(1+\varepsilon_{1})^{2}\leq 1+\varepsilon$.
Define $Z\in\mathcal{L}(H^{2}_{N},\mathcal{H})$ as follows:

$Z|_{H^{2}\oplus\\{0\\}}=Z_{1},\ \ \ Z|_{\\{0\\}\oplus
H^{2}_{N-1}}=\frac{\varepsilon_{1}}{\|Y\|}Y|_{\\{0\\}\oplus H^{2}_{N-1}}.$

Then $ZS_{N}=TZ$, $\|Z\|\leq 1+\varepsilon_{1}$,
$\operatorname{clos}ZH^{2}_{N}=\mathcal{H}$, and $Z$ satisfies the assumption
of Lemma 4.7 for every $0<\delta_{0}<1$ with some inner function $\theta$
(which depends on $\delta_{0}$). Let
$\\{\mathcal{N}_{j}\\}_{j=1}^{N}\subset\operatorname{Lat}S_{N}$ be from Lemma
4.7 applied with $\delta\geq 1/(1+\varepsilon_{1})$. Set
$\mathcal{M}_{j}=Z\mathcal{N}_{j}$ ($j=1,\ldots,N$). Consider
$Y_{j}=Z|_{\mathcal{N}_{j}}$ as the transformations from
$\mathcal{L}(\mathcal{N}_{j},\mathcal{M}_{j})$. Then
$T|_{\mathcal{M}_{j}}=Y_{j}S_{N}|_{\mathcal{N}_{j}}Y_{j}^{-1}$ and
$\|Y_{j}\|\|Y_{j}^{-1}\|\leq(1+\varepsilon_{1})^{2}\leq 1+\varepsilon$. By
Lemma 4.7, $S_{N}|_{\mathcal{N}_{j}}\cong S$. Thus, $\mathcal{M}_{j}$ and
$Y_{j}$ (up to appropriate unitary equvalence) ($j=1,\ldots,N$) satisfy the
conclusion of the theorem. ∎

######  Theorem 4.14.

Suppose that $T$ is an expansive operator, $N=\dim\ker T^{*}<\infty$,
$I-T^{*}T\in\mathfrak{S}_{1}$, and $\mathcal{R}^{\infty}(T)=\\{0\\}$. Then
$T^{\prime}\sim S_{N}$, $S_{N}\buildrel i\over{\prec}T\prec S_{N}$ and for
every $\mathcal{M}\in\operatorname{Lat}T$ $\dim(\mathcal{M}\ominus
T\mathcal{M})\leq N$.

###### Proof.

By Corollary 2.8, $T\prec S_{N}$. Denote by $\mathcal{H}$ the space in which
$T$ acts. By Lemma 2.4(i),

$\mathcal{H}=\vee_{n=0}^{\infty}T^{\prime n}\ker T^{\prime*}.$

Therefore, $\mu_{T^{\prime}}=N$ (where $\mu_{T}$ for an operator $T$ is
defined in (1.1)), and $S_{N}\prec T^{\prime}$ by Lemmas 2.7 and 2.2. In
particular, $T^{\prime}$ is an a.c. contraction.

Furthermore, $T^{\prime}$ is left-invertible, $\dim\ker T^{\prime*}=N<\infty$,
and $I-T^{\prime*}T^{\prime}\in\mathfrak{S}_{1}$ by Lemma 2.1. By [U84] and
[U83] or [T84], $T^{\prime}$ has the form

$T^{\prime}=\left[\begin{matrix}T_{0}&*\\\
\mathbb{O}&T_{1}\end{matrix}\right],$

where $T_{0}$ is a weak contraction (see [NFBK, Ch. VIII] for definition) and
$T_{1}\prec S_{N}$. By [G12, Lemma 2.1], $(T_{1})^{(a)}_{+}\cong S_{N}$. By
[NFBK, Ch. IX.1], $T^{\prime(a)}\cong T_{0}^{(a)}\oplus
T_{1}^{(a)}=T_{0}^{(a)}\oplus U_{\mathbb{T},N}$, and $T_{0}^{(a)}$ is a.c.
unitary. Therefore,

$\mu_{T_{0}^{(a)}}+N=\mu_{T^{\prime(a)}}\leq\mu_{T^{\prime}}=N.$

Consequently, $\mu_{T_{0}^{(a)}}=0$. This means that $T_{0}$ is a
$C_{0}$-contraction. By [G03, Theorem 0.1], $T^{\prime}\prec T_{0}\oplus
S_{N}$. By [VK],

$\mu_{T_{0}}+N=\mu_{T_{0}\oplus S_{N}}\leq\mu_{T^{\prime}}=N.$

This means that $T_{0}$ acts on the zero space, that is,
$T^{\prime}=T_{1}\prec S_{N}$.

Let $Y$ be a quasiaffinity such that $YS_{N}^{*}=T^{\prime*}Y$. Then $Y\ker
S_{N}^{*}=\ker T^{*}$. Furthermore,

$\displaystyle TY$ $\displaystyle=TT^{\prime*}YS_{N}=(I-P_{\ker
T^{*}})YS_{N}=YS_{N}-P_{Y\ker S_{N}^{*}}YS_{N}$
$\displaystyle=Y\Bigl{(}S_{N}+\sum_{k=1}^{N}e_{k}\otimes f_{k}\Bigr{)}$

for some $\\{f_{k}\\}_{k=1}^{N}\subset H^{2}_{N}$. By Lemma 3.5,
$S_{N}\buildrel i\over{\prec}(S_{N}+\sum_{k=1}^{N}e_{k}\otimes f_{k})$. Thus,
$S_{N}\buildrel i\over{\prec}T$.

Let $\mathcal{M}\in\operatorname{Lat}T$. If $\dim(\mathcal{M}\ominus
T\mathcal{M})>N$, take a subspace $E\subset(\mathcal{M}\ominus T\mathcal{M})$
such that $\dim E=N+1$ and set $\mathcal{N}=\vee_{n=0}^{\infty}T^{n}E$. Then
$\dim\ker(T|_{\mathcal{N}})^{*}=N+1$. Applying to $T|_{\mathcal{N}}$ already
proved part of the theorem, we obtain that $S_{N+1}\buildrel
i\over{\prec}T|_{\mathcal{N}}$. Thus, $S_{N+1}\buildrel
i\over{\prec}T|_{\mathcal{N}}\buildrel i\over{\prec}T\prec S_{N}$, a
contradiction. ∎

##  5\. Similarity to isometry

In this section, the relationship between similarity to isometry of an
operator $T$ and its Cauchy dual $T^{\prime}$ is studies.

######  Proposition 5.1.

Suppose that $V$ and $V_{1}$ are isometries, $T$ is a left-invertible
operator, $T\approx V$ and $T^{\prime}\approx V_{1}$. Then $V\cong V_{1}$.

###### Proof.

Since $\dim\ker V^{*}=\dim\ker T^{*}=\dim\ker T^{\prime*}=\dim\ker V_{1}^{*}$,
we conclude that there exist $0\leq N\leq\infty$ and unitaries
$U\in\mathcal{L}(\mathcal{K})$ and $U_{1}\in\mathcal{L}(\mathcal{K}_{1})$ such
that $V\cong U\oplus S_{N}$ and $V_{1}\cong U_{1}\oplus S_{N}$. Since
$T\approx U\oplus S_{N}$, there exists $\mathcal{M}\in\operatorname{Lat}T$
such that $T|_{\mathcal{M}}\approx U$. Since $T^{\prime*}T=I$, we have
$\mathcal{M}\in\operatorname{Lat}T^{\prime*}$ and
$T^{\prime*}|_{\mathcal{M}}\approx U^{-1}$. Therefore, there exists
$\mathcal{N}\in\operatorname{Lat}(U_{1}\oplus S_{N})^{*}$ such that
$(U_{1}\oplus S_{N})^{*}|_{\mathcal{N}}\approx U^{-1}$. Since

$\ker P_{\mathcal{K}_{1}}|_{\mathcal{N}}=\mathcal{N}\cap H^{2}_{N}=\\{0\\}\ \
\text{ and }\ \ P_{\mathcal{K}_{1}}|_{\mathcal{N}}\in\mathcal{I}((U_{1}\oplus
S_{N})^{*}|_{\mathcal{N}},U_{1}^{-1}),$

we obtain that $U^{-1}\buildrel i\over{\prec}U_{1}^{-1}$.

Since $T^{\prime\prime}=T$, we can apply already proved result and obtain that
$U_{1}^{-1}\buildrel i\over{\prec}U^{-1}$. Consequently, $U\cong U_{1}$. ∎

######  Proposition 5.2.

Suppose that $V$ is an isometry, $T$ is expansive, and $T\approx V$. Then
$T^{\prime}\approx V$.

###### Proof.

Since $T\approx V$, we have $C=\sup_{n\in\mathbb{N}}\|T^{n}\|<\infty$. Denote
by $\mathcal{H}$ the space on which $T$ acts. For $x\in\mathcal{H}$ and
$n\in\mathbb{N}$ we have

$\|x\|=\|T^{*n}T^{\prime n}x\|\leq\|T^{*n}\|\|T^{\prime n}x\|\leq C\|T^{\prime
n}x\|.$

Since $T^{\prime}$ is a contraction, the estimate
$\inf_{n\in\mathbb{N}}\|T^{\prime n}x\|\geq\frac{1}{C}\|x\|$
($x\in\mathcal{H}$) implies that $T^{\prime}$ is similar to an isometry. By
Proposition 5.1, $T^{\prime}\approx V$. ∎

The following two examples show that an expansive operator $T$ in Proposition
5.2 cannot be replaced by contraction. In Example 5.3 $T$ is an expansive
operator such that $T^{\prime}\approx S$, $T\sim S$ and $T\not\approx S$. In
Example 5.4 $T$ is an expansive operator such that $T^{\prime}\approx S$ and
$\mathcal{R}^{\infty}(T)\neq\\{0\\}$. The following result from [N] will be
used.

Let $g\in H^{2}$ be such that $\|g\|=1$ and $0<|g(0)|<1$. Set

(5.1) $T=S-\mathbf{1}\otimes S^{*}\frac{g}{g(0)}.$

Then $T^{\prime}=S-g\otimes S^{*}g$. Let $\omega$ be defined by (3.10) applied
to $g$. By [N, Theorem 5],

$\left[\begin{matrix}\omega\\\ (1-\omega)g\end{matrix}\right]$

is the characteristic function of $T^{\prime}$ (see [NFBK, Ch. VI] for the
characteristic function of a contraction).

######  Example 5.3.

Let $g\in H^{2}$ be such that $\|g\|=1$, $|g(0)|<1$, and $1/g\in H^{\infty}$.
Define $\omega$ by (3.10) and $T$ by (5.1). Then

$\omega+\frac{1}{g}(1-\omega)g=1.$

By [NF73] or [NF76], $T^{\prime}\approx S$. Furthermore, by Lemma 3.2(ii),
$T\sim S$. Indeed, $T_{\frac{\overline{g(0)}g}{g(0)\overline{g}}}$ is a
quasiaffinity, because $g$, $1/g\in H^{2}$, and
$T_{\varphi}T_{\frac{\overline{g}}{\overline{g(0)}}}$ is a quasiaffinity for
some approriate $\varphi$, because $g$ is outer.

By Lemma 3.2(iii), $T$ is similar to an isometry if and only if
$T_{\frac{\overline{g(0)}g}{g(0)\overline{g}}}$ is invertible, what is of
course equivalent that $T_{\frac{g}{\overline{g}}}$ is invertible. If
$T_{\frac{g}{\overline{g}}}$ is invertible, then by [Pe, Corollary 3.2.2]
there exists $p>2$ and $f\in H^{p}$ such that $1/f\in H^{p}$ and
$\frac{g}{\overline{g}}=\frac{\overline{f}}{f}$. Since $gf\in H^{1}$ and
$gf=\overline{gf}$, we conclude that $gf\equiv c$ for some $c\in\mathbb{C}$.
Thus, if $T_{\frac{g}{\overline{g}}}$ is invertible, then there exists $p>2$
such that $g$, $1/g\in H^{p}$. Consequently, if $g\not\in H^{p}$ for any
$p>2$, then $T$ is not similar to an isometry. The function $g$ such that
$1/g\in H^{\infty}$ and $g\not\in H^{p}$ for any $p>2$ is given in [Z, Sec. 7,
Example]. Namely, let $g_{0}$ be the outer function such that

$|g_{0}(\mathrm{e}^{\mathrm{i}\pi
t})|=\frac{1}{|t|^{\frac{1}{2}}\log\frac{2}{|t|}},\ \ \ t\in(-1,0)\cup(0,1),$

and $g=g_{0}/\|g_{0}\|$.

######  Example 5.4.

Let $f\in H^{2}$ be a nonconstant function such that $\|f\|=1$, $|f|^{2}\in
L^{2}$, and $1/f$, $P_{+}|f|^{2}\in H^{\infty}$. For example, it is sufficient
to take $f$ which is analytic on $\mathbb{D}$, continuous on
$\overline{\mathbb{D}}$, and such that $f(z)\neq 0$ and $|f(z)-f(w)|\leq
C|z-w|$ for every $z$, $w\in\overline{\mathbb{D}}$ and some constant $C$. Let
$\omega$ be defined by (3.10) applied to $f$. Then
$\frac{1}{1-\omega}=P_{+}|f|^{2}$. Since $\omega\not\equiv 0$, there exist
$\varphi_{1}$, $\varphi_{2}$, $\theta\in H^{\infty}$ such that
$1\not\equiv\theta$ is inner and $\varphi_{1}\theta+\varphi_{2}\omega=1$ (see,
for example, the proof of [G22, Prop. 5.3]). Since $\omega(0)=0$, we have
$\theta(0)\neq 0$. Note that $\theta$ can be chosen such that
$\dim\mathcal{K}_{\theta}=\infty$. Set $g=\theta f$ and define $T$ by (5.1).
Then

$\varphi_{2}\omega+\frac{1}{1-\omega}\frac{1}{f}\varphi_{1}(1-\omega)g=1.$

By [NF73] or [NF76], $T^{\prime}\approx S$. By Lemma 3.2(i),
$\mathcal{R}^{\infty}(T)=\mathcal{K}_{\theta}\neq\\{0\\}$.

The following example shows that an expansive operator $T$ in Proposition 5.2
cannot be replaced by an operator similar to expansive one.

######  Example 5.5.

Suppose that $g\in H^{2}$, $g(0)=1$, and $S^{*}g\not\equiv 0$. Set
$E=\mathbf{1}\vee g$, $d_{1}=(g-1)/\|S^{*}g\|$ and $d_{2}=\mathbf{1}$. Then
$\\{d_{1},d_{2}\\}$ is an orthonormal basis of $E$. Take $a>\|S^{*}g\|^{2}$.
Let

$Y_{0}=\left[\begin{matrix}a&\|S^{*}g\|\\\ \|S^{*}g\|&1\end{matrix}\right]$

be the matrix of the positive invertible operator $Y_{0}\in\mathcal{L}(E)$ in
the basis $\\{d_{1},d_{2}\\}$. Set $Y=I_{H^{2}\ominus E}\oplus Y_{0}$. Then
$Y\in\mathcal{L}(H^{2})$ is a positive invertible operator, and
$Y\mathbf{1}=g$.

Set $X=Y^{-\frac{1}{2}}$ and $T=XSX^{-1}$. Then
$T^{\prime}=X^{-1}(S-\mathbf{1}\otimes S^{*}g)X$. Indeed,

$T^{\prime*}T=X(S^{*}-S^{*}g\otimes\mathbf{1})X^{-1}XSX^{-1}=X(S^{*}-S^{*}g\otimes\mathbf{1})SX^{-1}=XX^{-1}=I,$
$\ker T^{*}=\\{h\in H^{2}:Xh=c\text{ for some }c\in\mathbb{C}\\},$

and

$\displaystyle\ker T^{\prime*}$ $\displaystyle=\\{h\in H^{2}:X^{-1}h=cg\text{
for some }c\in\mathbb{C}\\}$ $\displaystyle=\\{h\in
H^{2}:Y^{\frac{1}{2}}h=cY\mathbf{1}\text{ for some }c\in\mathbb{C}\\}$
$\displaystyle=\\{h\in H^{2}:h=cY^{\frac{1}{2}}\mathbf{1}\text{ for some
}c\in\mathbb{C}\\}=\ker T^{*}.$

If $T_{\frac{g}{\overline{g}}}$ is not invertible, then, by Lemma 3.2,
$T^{\prime}$ is not similar to an isometry.

##  6\. Quasisimilarity of expansive operator to the unilateral shift does not
preserve the lattice of invariant subspaces

If $T$ is a contraction and $T\sim S$, then the intertwining quasiaffinities
give a bijection between $\operatorname{Lat}T$ and $\operatorname{Lat}S$ (see
[G02] for more general result, see also references therein). In particular,
for every quasiaffinity $Y\in\mathcal{I}(S,T)$ and every
$\\{0\\}\neq\mathcal{M}\in\operatorname{Lat}T$ there exists an inner function
$\vartheta$ such that $\mathcal{M}=\operatorname{clos}Y\vartheta H^{2}$. In
this section, it is shown that a contraction $T$ cannot be replaced by an
expansive operator (Corollary 6.10).

Recall that for an inner function $\vartheta$ the space
$\mathcal{K}_{\vartheta}$ is defined in (3.1).

######  Lemma 6.1.

Suppose that $\theta$ and $\beta$ are inner functions. Then

(6.1)
$\mathcal{K}_{\theta\beta}=\theta\mathcal{K}_{\beta}\oplus\mathcal{K}_{\theta}$

and

(6.2) $(\theta-1)\mathcal{K}_{\beta}\subset\mathcal{K}_{\theta\beta}.$

Moreover,

(6.3)
$\operatorname{clos}(\theta-1)\mathcal{K}_{\beta}=\mathcal{K}_{\theta\beta}$

and if and only if

(6.4) $(\mathcal{K}_{\theta}+(\theta-1)\mathcal{K}_{\beta})\cap\beta
H^{2}=\\{0\\}.$

###### Proof.

The equality (6.1) follows from the definition of the space
$\mathcal{K}_{\vartheta}$ for an inner function $\vartheta$ (see (3.1)), and
the inclusion (6.2) easy follows from (6.1). Let $f\in\mathcal{K}_{\theta}$
and $h\in\mathcal{K}_{\beta}$ be such that $0\not\equiv\theta
h+f\perp(\theta-1)\mathcal{K}_{\beta}$. Then there exist $h_{1}\in H^{2}$ and
$h_{2}\in H^{2}_{-}$ such that

$(1-\theta)h+(\overline{\theta}-1)f=\beta h_{1}+h_{2}.$

By (3.1), $\overline{\theta}f\in H^{2}_{-}$. Consequently,
$(1-\theta)h-f=\beta h_{1}$. By (3.5), the relation $(1-\theta)h-f\equiv 0$
implies $h\equiv 0$ and $f\equiv 0$, a contradiction with the assumption on
$h$ and $f$. Thus, if (6.3) is not fulfilled, then (6.4) is not fulfilled.

Conversely, let $f\in\mathcal{K}_{\theta}$, $h\in\mathcal{K}_{\beta}$, and
$h_{1}\in H^{2}$ be such that $0\not\equiv(1-\theta)h-f=\beta h_{1}$. Then

$(\overline{\theta}-1)\theta h+(\overline{\theta}-1)f=\beta
h_{1}+\overline{\theta}f.$

By (3.1), $(\overline{\theta}-1)\theta
h+(\overline{\theta}-1)f\perp\mathcal{K}_{\beta}$. Consequently, $\theta
h+f\perp(\theta-1)\mathcal{K}_{\beta}$. If $\theta h+f\equiv 0$, then $h\equiv
0$ and $f\equiv 0$, a contradiction with the assumption on $h$ and $f$. Thus,
if (6.4) is not fulfilled, then (6.3) is not fulfilled. ∎

For $\zeta\in\mathbb{T}$ and $t_{0}>0$ set

(6.5) $\Delta(\zeta,t_{0})=\\{\zeta\mathrm{e}^{\mathrm{i}t}\ :\ |t|\leq
t_{0}\\}.$

For $\zeta\in\mathbb{T}$ and $0<s_{0}<\pi/2$ denote by
$\mathcal{S}(\zeta,s_{0})$ the Stolz angle, that is, the closed sector with
vertex $\zeta$ of angle $2s_{0}$ symmetric with respect to the radius
$\\{r\zeta\ :r\in[0,1]\\}$. (Usually, the Stolz angle assumed to be an open
set, but it is convenient to consider closed set here.) For $0<r_{0}<1$
sufficiently close to $1$ both rays which form the boundary of
$\mathcal{S}(\zeta,s_{0})$ intersect the circle $\\{|z|=r_{0}\\}$ in two
points. Denote by $z_{\pm}$ two points from this intersection closest to
$\zeta$. Define $t(s_{0},r_{0})$ as follows:
$z_{\pm}=r_{0}\zeta\mathrm{e}^{\pm\mathrm{i}t(s_{0},r_{0})}$. Then

(6.6) $\tan s_{0}=\frac{r_{0}\sin t(s_{0},r_{0})}{1-r_{0}\cos
t(s_{0},r_{0})}.$

Set

(6.7)
$\mathcal{S}(\zeta,s_{0},r_{0})=\mathcal{S}(\zeta,s_{0})\cap\\{r\zeta\mathrm{e}^{\mathrm{i}t}\
:\ |t|\leq t(s_{0},r_{0}),\ r_{0}\leq r\leq 1\\}.$

Then $\mathcal{S}(\zeta,s_{0},r_{0})$ is a “triangle” with vertices $\zeta$
and $r_{0}\zeta\mathrm{e}^{\pm\mathrm{i}t(s_{0},r_{0})}$; its two edges are
segments and one edge is a subarc of the circle $\\{|z|=r_{0}\\}$.

The following simple lemma is given by convenience of references; its proofs
is omitted.

######  Lemma 6.2.

Let $0<\varepsilon<1$, and let $\zeta_{0}\in\mathbb{T}$. Set
$\lambda_{0}=(1-\varepsilon)\zeta_{0}$ and define $0<s(\varepsilon)<\pi/2$ by
the formula

$\tan
s(\varepsilon)=\frac{(1-\varepsilon)\sin\varepsilon}{1-(1-\varepsilon)\cos\varepsilon}.$

Then $s(\varepsilon)\to\pi/4$ as $\varepsilon\to 0$. Furthermore, if
$\zeta\in\Delta(\zeta_{0},\varepsilon)$, then
$|\zeta-\lambda_{0}|^{2}\leq\varepsilon^{2}+2(1-\varepsilon)(1-\cos\varepsilon)$
and $\lambda_{0}\in\mathcal{S}(\zeta,s(\varepsilon))$.

For $\lambda\in\mathbb{D}$, $\lambda\neq 0$, a Blaschke factor is
$b_{\lambda}(z)=\frac{|\lambda|}{\lambda}\frac{\lambda-z}{1-\overline{\lambda}z}$
($z\in\mathbb{D}$). If $\Lambda\subset\mathbb{D}$ satisfies the Blaschke
condition

(6.8) $\sum_{\lambda\in\Lambda}(1-|\lambda|)<\infty,$

then the Blaschke product $\beta=\prod_{\lambda\in\Lambda}b_{\lambda}$
converges and $\beta$ is an inner function.

For $\zeta\in\mathbb{T}$ and $0<s<1$ set

$Q(\zeta,s)=\\{z\in\mathbb{D}\ :\ 1-s\leq|z|<1,\
\frac{z}{|z|}\in\Delta(\zeta,s)\\}.$

The set $Q(\zeta,s)$ is called the Carleson box or the Carleson window. Let
$\Lambda\subset\mathbb{D}$ satisfy (6.8). A particular case of the Carleson
embedding theorem (see, for example, [GMR, Theorem 11.22]) is the following:
the relations

(6.9) $\sum_{\lambda\in\Lambda}|h(\lambda)|^{2}(1-|\lambda|)<\infty\ \text{
for every }h\in H^{2}$

and

(6.10) $\sup_{\zeta\in\mathbb{T}\atop
0<s<1}\frac{1}{s}\sum_{\lambda\in\Lambda\cap Q(\zeta,s)}(1-|\lambda|)<\infty$

are equivalent.

######  Lemma 6.3.

Suppose that $0<s_{0}<\pi/2$, $\Lambda\subset\mathbb{D}$, $\Lambda$ satisfies
(6.9), $\nu$ is a singular positive Borel measure on $\mathbb{T}$, and

$\nu(\\{\zeta\in\mathbb{T}\ :\
\zeta\in\operatorname{clos}(\Lambda\cap\mathcal{S}(\zeta,s_{0}))\\}=\nu(\mathbb{T})=1.$

Set $\beta=\prod_{\lambda\in\Lambda}b_{\lambda}$, and define $\theta$ by
(3.3). Then $\theta$ and $\beta$ satisfy (6.4).

###### Proof.

Let $C>0$, and let $\\{t_{j}\\}_{j=1}^{\infty}$ be such that $t_{j}>0$ and
$t_{j}\to 0$. Set

$\sigma_{j}=\\{\zeta\in\mathbb{T}\ :\ \nu(\Delta(\zeta,t))\geq Ct\ \text{ for
all }0<t\leq t_{j}\\}\ \ (j\geq 1).$

Since $\nu$ is a singular measure, we have
$\nu(\mathbb{T})=\nu(\cup_{j=1}^{\infty}\sigma_{j})$ (see, for example, [GMR,
Theorem 1.2] or [Ga, formula (II.6.3)]).

Let $0\not\equiv f\in\mathcal{K}_{\theta}$. By (3.4), $f$ has nontangential
boundary values $f(\zeta)$ for $\nu$-a.e. $\zeta\in\mathbb{T}$, and
$\nu(\\{\zeta\in\mathbb{T}\ :\ f(\zeta)\neq 0\\})>0$. Therefore, there exist
$\delta>0$ and $0<r_{0}<1$ such that $\nu(\tau_{0})>0$, where

$\tau_{0}=\\{\zeta\in\mathbb{T}\ :\ |f(z)|\geq\delta\ \text{ for every
}z\in\mathcal{S}(\zeta,s_{0},r_{0})\\}$

and $\mathcal{S}(\zeta,s_{0},r_{0})$ is defined in (6.7). Indeed, let
$\\{\delta_{n}\\}_{n=1}^{\infty}$ and $\\{r_{n}\\}_{n=1}^{\infty}$ be such
that $\delta_{n}>0$, $0<r_{n}<1$, $\delta_{n}\to 0$, and $r_{n}\to 1$. Set

$\tau_{nk}=\\{\zeta\in\mathbb{T}\ :\ |f(z)|\geq\delta_{n}\ \text{ for every
}z\in\mathcal{S}(\zeta,s_{0},r_{k})\\}.$

Then $\\{\zeta\in\mathbb{T}\ :\ f(\zeta)\neq
0\\}=\cup_{n,k=1}^{\infty}\tau_{nk}$. Consequently, there exist $n$ and $k$
such that $\nu(\tau_{nk})>0$. Set $\delta=\delta_{n}$, $r_{0}=r_{k}$, and
$\tau_{0}=\tau_{nk}$. Furthermore, there exists $j$ such that
$\nu(\tau_{0}\cap\sigma_{j})>0$. Set $t_{0}=t_{j}$ and

$\tau=\tau_{0}\cap\sigma_{j}\cap\\{\zeta\in\mathbb{T}\ :\
\zeta\in\operatorname{clos}(\Lambda\cap\mathcal{S}(\zeta,s_{0}))\\}.$

Then $\nu(\tau)>0$.

Let $h\in H^{2}$ be such that $f+(\theta-1)h\in\beta H^{2}$. Then
$h(\lambda)=f(\lambda)/(1-\theta(\lambda))$ for every $\lambda\in\Lambda$. The
equality (3.3) implies that

$\frac{1}{|\theta(z)-1|}\geq\int_{\mathbb{T}}\mathrm{Re}\frac{1}{1-z\overline{\zeta}}\mathrm{d}\nu(\zeta)\geq\frac{1-|z|\cos
t}{1-2|z|\cos
t+|z|^{2}}\nu\Bigl{(}\Delta\Bigl{(}\frac{z}{|z|},t\Bigr{)}\Bigr{)}$

for every $z\in\mathbb{D}$ and $0<t<\pi/2$.

Let $\lambda\in\Lambda\cap\mathcal{S}(\zeta,s_{0},r_{0})$ for some
$\zeta\in\tau$. Then

$|h(\lambda)|=\frac{|f(\lambda)|}{|\theta(\lambda)-1|}\geq\delta\frac{1-|\lambda|\cos
t}{1-2|\lambda|\cos
t+|\lambda|^{2}}\nu\Bigl{(}\Delta\Bigl{(}\frac{\lambda}{|\lambda|},t\Bigr{)}\Bigr{)}$

for every $0<t<\pi/2$. Set $t(\lambda)=2t(s_{0},|\lambda|)$, where
$t(s_{0},r)$ is defined before (6.6) for $0<s_{0}<\pi/2$ and $0<r<1$
sufficiently close to $1$. Then

(6.11) $\displaystyle t(\lambda)$ $\displaystyle\sim 2(1-|\lambda|)\tan s_{0}\
\text{ and }$ $\displaystyle\frac{1-|\lambda|\cos t(\lambda)}{1-2|\lambda|\cos
t(\lambda)+|\lambda|^{2}}$
$\displaystyle\sim\frac{1}{1+4\tan^{2}s_{0}}\frac{1}{1-|\lambda|}\ \text{ as
}|\lambda|\to 1$

(where $a(t)\sim b(t)$ as $t\to c$ means that $\lim_{t\to c}a(t)/b(t)=1$).
Furthermore,

$\Delta(\zeta,t(s_{0},|\lambda|))\subset\Delta\Bigl{(}\frac{\lambda}{|\lambda|},t(\lambda)\Bigr{)}.$

It follows from this inclusion and the construction of $\tau$ that
$\nu(\Delta(\frac{\lambda}{|\lambda|},t(\lambda)))\geq
Ct(s_{0},|\lambda|)=Ct(\lambda)/2$, if $t(s_{0},|\lambda|)\leq t_{0}$. By
(6.11), there exists $0<c<1$ (which does not depend on $\lambda$) such that

$\displaystyle|h(\lambda)|^{2}(1-|\lambda|)$ $\displaystyle\geq
c\delta^{2}C(1-|\lambda|)\tan
s_{0}\frac{1}{(1+4\tan^{2}s_{0})^{2}}\frac{1}{(1-|\lambda|)^{2}}(1-|\lambda|)\nu\Bigl{(}\Delta\Bigl{(}\frac{\lambda}{|\lambda|},t(\lambda)\Bigr{)}\Bigr{)}$
$\displaystyle=c\delta^{2}C\tan
s_{0}\frac{1}{(1+4\tan^{2}s_{0})^{2}}\nu\Bigl{(}\Delta\Bigl{(}\frac{\lambda}{|\lambda|},t(\lambda)\Bigr{)}\Bigr{)}.$

Set $C_{1}=c\delta^{2}C\tan s_{0}\frac{1}{(1+4\tan^{2}s_{0})^{2}}$. Let
$0<\varepsilon<1-r_{0}$. Then

$\displaystyle\sum_{\lambda\in\Lambda:|\lambda|\geq 1-\varepsilon}$
$\displaystyle|h(\lambda)|^{2}(1-|\lambda|)\geq\sum_{\lambda\in\Lambda\cap(\cup_{\zeta\in\tau}\mathcal{S}(\zeta,s_{0},r_{0})):\atop|\lambda|\geq
1-\varepsilon}|h(\lambda)|^{2}(1-|\lambda|)$
$\displaystyle\geq\sum_{\lambda\in\Lambda\cap(\cup_{\zeta\in\tau}\mathcal{S}(\zeta,s_{0},r_{0})):\atop|\lambda|\geq
1-\varepsilon}C_{1}\nu\Bigl{(}\Delta\Bigl{(}\frac{\lambda}{|\lambda|},t(\lambda)\Bigr{)}\Bigr{)}$
$\displaystyle\geq
C_{1}\nu\Bigl{(}\bigcup_{\lambda\in\Lambda\cap(\cup_{\zeta\in\tau}\mathcal{S}(\zeta,s_{0},r_{0})):\atop|\lambda|\geq
1-\varepsilon}\Delta\Bigl{(}\frac{\lambda}{|\lambda|},t(\lambda)\Bigr{)}\Bigr{)}\geq
C_{1}\nu(\tau),$

because for every $\zeta\in\tau$ there exists
$\lambda\in\Lambda\cap\mathcal{S}(\zeta,s_{0},r_{0})$ with $|\lambda|\geq
1-\varepsilon$. Consequently,

$\lim_{\varepsilon\to 0}\sum_{\lambda\in\Lambda:|\lambda|\geq
1-\varepsilon}|h(\lambda)|^{2}(1-|\lambda|)\geq C_{1}\nu(\tau)>0,$

a contradiction with (6.9). Thus, if $f\in\mathcal{K}_{\theta}$ and $h\in
H^{2}$ are such that $f+(\theta-1)h\in\beta H^{2}$, then $f\equiv 0$ and
$(\theta-1)h\in\beta H^{2}$. Since $\theta-1$ is outer, we have $h\in\beta
H^{2}$. If $h\in\mathcal{K}_{\beta}$, then $h\equiv 0$. ∎

The following simple lemma is given by convenience of references; its proofs
is omitted.

######  Lemma 6.4.

Let $K\subset\mathbb{T}$ be a compact, and let $\delta>0$. If $m(K)=0$, then
there exist $N\in\mathbb{N}$ and nonempty closed subarcs
$\\{\Delta_{k}\\}_{k=1}^{N}$ such that $\Delta_{k}\subset\mathbb{T}$,
$\Delta_{k}\cap\Delta_{j}=\emptyset$, if $k\neq j$ $(1\leq k,j\leq N)$,
$K\subset\cup_{k=1}^{N}\Delta_{k}$ and $\sum_{k=1}^{N}\pi
m(\Delta_{k})<\delta$.

######  Lemma 6.5.

Suppose that $\\{K_{n}\\}_{n=1}^{\infty}$ is a sequence of compact subsets of
$\mathbb{T}$ such that $K_{n}\subset K_{n+1}$ and $m(K_{n})=0$ for all
$n=1,2,\ldots$. Then there exists $\Lambda\subset\mathbb{D}$ which satisfies
(6.8) and (6.10) and such that

(6.12) $\zeta\in\operatorname{clos}(\Lambda\cap\mathcal{S}(\zeta,s))$

for every $\zeta\in\cup_{n=1}^{\infty}K_{n}$ and $\pi/4<s<\pi/2$.

###### Proof.

Take $\\{\delta_{k}\\}_{k=1}^{\infty}$ such that $0<\delta_{k}<1$ for all
$k=1,2,\ldots$, and $\sum_{k=1}^{\infty}\delta_{k}<\infty$. Let
$\\{\Delta_{1k}\\}_{k=1}^{N_{1}}$ are subarcs from Lemma 6.4 applied to
$K_{1}$ and $\delta_{1}$. Set $\varepsilon_{1k}=\pi m(\Delta_{1k})$
($k=1,\ldots,N_{1}$). We may assume that

$\varepsilon_{1N_{1}}=\min_{k=1,\ldots,N_{1}}\varepsilon_{1k}.$

There exists $M_{1}>1$ such that
$\sum_{k=M_{1}}^{\infty}\delta_{k}<\varepsilon_{1N_{1}}$. Let
$\\{\Delta_{2k}\\}_{k=1}^{N_{2}}$ are subarcs from Lemma 6.4 applied to
$K_{2}$ and $\delta_{M_{1}}$. Set $\varepsilon_{2k}=\pi m(\Delta_{2k})$
($k=1,\ldots,N_{2}$). We may assume that

$\varepsilon_{2N_{2}}=\min_{k=1,\ldots,N_{2}}\varepsilon_{2k}.$

There exists $M_{2}>M_{1}$ such that
$\sum_{k=M_{2}}^{\infty}\delta_{k}<\varepsilon_{2N_{2}}$. Let
$\\{\Delta_{3k}\\}_{k=1}^{N_{3}}$ are subarcs from Lemma 6.4 applied to
$K_{3}$ and $\delta_{M_{2}}$, and so on. Set $M_{0}=1$. We obtain the sequence
$\\{M_{n}\\}_{n=0}^{\infty}\subset\mathbb{N}$, the closed subarcs
$\Delta_{nk}$ and the quantities $\varepsilon_{nk}>0$ ($n=1,2,\ldots$,
$k=1,\ldots,N_{n}$). Note that

$\displaystyle\varepsilon_{nN_{n}}\leq\varepsilon_{nk}<\varepsilon_{n-1,N_{n-1}}\
\ (k=1,\ldots,N_{n})\ \ \text{and }$
$\displaystyle\sum_{k=1}^{N_{n}}\varepsilon_{nk}<\delta_{M_{n-1}}\ \
(n=1,2,\ldots).$

Define $\lambda_{nk}\in\mathbb{D}$ such that

$\Delta_{nk}=\Delta\Bigl{(}\frac{\lambda_{nk}}{|\lambda_{nk}|},1-|\lambda_{nk}|\Bigr{)}\
\ (n=1,2,\ldots,\ k=1,\ldots,N_{n}),$

where $\Delta(\zeta,s)$ is defined by (6.5). Then
$1-|\lambda_{nk}|=\varepsilon_{nk}$ ($n=1,2,\ldots$, $k=1,\ldots,N_{n}$). Set
$\Lambda=\\{\lambda_{nk},\ n=1,2,\ldots,\ k=1,\ldots,N_{n}\\}$. We will to
prove that $\Lambda$ satisfies the conclusion of the lemma.

The relation (6.8) easy follows from the construction.

Let $\zeta\in\cup_{n=1}^{\infty}K_{n}$. Then there exists $q\in\mathbb{N}$
such that $\zeta\in K_{n}$ for all $n\geq q$. By construction, for every
$n\geq q$ there exists $1\leq k\leq N_{n}$ such that $\zeta\in\Delta_{nk}$. By
Lemma 6.2, $|\zeta-\lambda_{nk}|\to 0$ when $n\to\infty$. Define
$s(\varepsilon_{nk})$ as in Lemma 6.2. Then
$\lambda_{nk}\in\mathcal{S}(\zeta,s(\varepsilon_{nk}))$ and
$s(\varepsilon_{nk})\to\pi/4$ when $n\to\infty$. Consequently, the relation
(6.12) is fulfilled.

Let $\zeta\in\mathbb{T}$, and let $0<s<1$ be sufficiently close to $0$. Then
there exists $q\in\mathbb{N}$ such that $\varepsilon_{qN_{q}}\leq
s<\varepsilon_{q-1,N_{q-1}}$. Let $\lambda_{nk}\in Q(\zeta,s)$. Then
$|\lambda_{nk}|\geq 1-s$. Consequently, $n\geq q$. We have

(6.13)
$\displaystyle\frac{1}{s}\sum_{n=q+1}^{\infty}\sum_{k=1}^{N_{n}}(1-|\lambda_{nk}|)$
$\displaystyle=\frac{1}{s}\sum_{n=q+1}^{\infty}\sum_{k=1}^{N_{n}}\varepsilon_{nk}$
$\displaystyle\leq\frac{1}{s}\sum_{n=q+1}^{\infty}\delta_{M_{n-1}}\leq\frac{1}{s}\sum_{k=M_{q}}^{\infty}\delta_{k}\leq\frac{1}{s}\varepsilon_{qN_{q}}\leq
1.$

Clearly, if $\lambda_{qk}\in Q(\zeta,s)$, then
$\frac{\lambda_{qk}}{|\lambda_{qk}|}\in\Delta(\zeta,s)$. Since
$\Delta_{qk}\cap\Delta_{qj}=\emptyset$, if $k\neq j$ ($1\leq k,j\leq N_{q}$),
we have

$\operatorname{card}\\{k\ :\ 1\leq k\leq N_{q},\lambda_{qk}\in Q(\zeta,s),\
\Delta_{qk}\not\subset\Delta(\zeta,s)\\}\leq 2.$

Therefore,

(6.14) $\displaystyle\frac{1}{s}$ $\displaystyle\sum_{1\leq k\leq
N_{q}:\atop\lambda_{qk}\in
Q(\zeta,s)}(1-|\lambda_{qk}|)\leq\frac{1}{s}\Bigl{(}2s+\sum_{1\leq k\leq
N_{q}:\atop\Delta_{qk}\subset\Delta(\zeta,s)}(1-|\lambda_{qk}|)\Bigr{)}$
$\displaystyle=\frac{1}{s}\Bigl{(}2s+\sum_{1\leq k\leq
N_{q}:\atop\Delta_{qk}\subset\Delta(\zeta,s)}\pi
m(\Delta_{qk})\Bigr{)}\leq\frac{1}{s}\Bigl{(}2s+\pi
m(\Delta(\zeta,s))\Bigr{)}=3.$

The relation (6.10) follows from (6.13) and (6.14). ∎

######  Remark 6.6.

In Lemma 6.5 it is possible that $K_{n}=K_{n+1}$ for all $n=1,2,\ldots$.

######  Theorem 6.7.

Let $\theta$ be an inner function such that $\theta(0)=0$. Then there exists a
Blaschke product $\beta$ with simple zeroes such that $\theta$ and $\beta$
satisfy (6.3), and the set $\Lambda$ of zeroes of $\beta$ satisfies (6.9).

###### Proof.

Let $\nu$ be defined by (3.3). Since $\nu$ is singular with respect to $m$,
there exists a sequence $\\{K_{n}\\}_{n=1}^{\infty}$ of compact subsets of
$\mathbb{T}$ such that $K_{n}\subset K_{n+1}$, $m(K_{n})=0$ for all
$n=1,2,\ldots$, and $\nu(\cup_{n=1}^{\infty}K_{n})=\nu(\mathbb{T})$. Let
$\Lambda$ be the set from Lemma 6.5 applied to $\\{K_{n}\\}_{n=1}^{\infty}$.
Then $\Lambda$ satisfies (6.9), because (6.9) and (6.10) are equivalent when
(6.8) is fulfilled (see the reference before (6.9)). Set
$\beta=\prod_{\lambda\in\Lambda}b_{\lambda}$. By Lemma 6.3, $\theta$ and
$\beta$ satisfy (6.4). By Lemma 6.1, $\theta$ and $\beta$ satisfy (6.3). ∎

The following simple lemma is given for convenience of references; its proof
is omitted.

######  Lemma 6.8.

Let $X\in\mathcal{L}(\mathcal{H},\mathcal{K})$ have the form

$X=\left[\begin{matrix}X_{1}&*\\\ \mathbb{O}&X_{0}\end{matrix}\right]$

with respect to some decompositions
$\mathcal{H}=\mathcal{H}_{1}\oplus\mathcal{H}_{0}$ and
$\mathcal{K}=\mathcal{K}_{1}\oplus\mathcal{K}_{0}$. Then

  1. (i)

if $\ker X_{0}=\\{0\\}$ and $\ker X_{1}=\\{0\\}$, then $\ker X=\\{0\\}$;

  2. (ii)

if $\operatorname{clos}X_{0}\mathcal{H}_{0}=\mathcal{K}_{0}$ and
$\operatorname{clos}X_{1}\mathcal{H}_{1}=\mathcal{K}_{1}$, then
$\operatorname{clos}X\mathcal{H}=\mathcal{K}$.

The following theorem is the main result of this section.

######  Theorem 6.9.

Suppose that $\theta$, $\beta\in H^{\infty}$ are inner functions,
$\theta(0)=0$, and $a$, $b\in\mathbb{C}$. Set $\varphi=a+b(\theta-1)$. Suppose
that $\varphi\not\equiv 0$. Define $T$, $X$, $Y\in\mathcal{L}(H^{2})$ as
follows:

$T=S+(1+(a-1)\theta)\beta\otimes\beta\overline{\chi}\theta+b\theta\beta\otimes
P_{+}\overline{\chi}\beta,$ $X(\theta\beta h+\beta f+g)=(\theta-1)\beta
h+a\beta f+\varphi g\ \ \ (h\in H^{2},\ f\in\mathcal{K}_{\theta},\
g\in\mathcal{K}_{\beta}),$ $Y(\beta h+g)=\theta\beta\varphi h+\theta P_{\beta
H^{2}}\varphi g+(\theta-1)g\ \ \ (h\in H^{2},\ g\in\mathcal{K}_{\beta}).$

Then $\theta\beta H^{2}$, $\beta H^{2}\in\operatorname{Lat}T$,

(6.15) $P_{\beta\mathcal{K}_{\theta}}T|_{\beta\mathcal{K}_{\theta}}\cong
U(\theta),$

where $U(\theta)$ is defined in (3.6), $YS=TY$, $XT=SX$, and $\ker Y=\\{0\\}$.
Furthermore,

  1. (i)

if $\varphi$ is outer and $a\neq 0$, then $X$ is a quasiaffinity;

  2. (ii)

if $\varphi$ is outer and (6.3) is fulfilled for $\theta$ and $\beta$, then
$Y$ is a quasiaffinity;

  3. (iii)

$T$ is expansive if and only if $2\operatorname{Re}\overline{a}b\leq-1$, and
then $1/\varphi\in H^{\infty}$.

###### Proof.

Many statements of the theorem can be checked directly. Unitary equivalence in
(6.15) is given by the operator of multiplication by $\beta$.

The definition of $Y$ implies that $Y$ has the form

$Y=\left[\begin{matrix}Y_{1}&*\\\ \mathbb{O}&Y_{0}\end{matrix}\right]$

with respect to the decompositions $H^{2}=\beta
H^{2}\oplus\mathcal{K}_{\beta}$ and $H^{2}=\theta\beta
H^{2}\oplus\mathcal{K}_{\theta\beta}$, where $Y_{1}$ and $Y_{0}$ are the
operators of multiplication by $\theta\varphi$ and $\theta-1$, respectively.
The equality $\ker Y=\\{0\\}$ and the statement (ii) follow from Lemma 6.8.

The definition of $X$ implies that $X$ has the form

$X=\left[\begin{matrix}X_{1}&*\\\ \mathbb{O}&X_{0}\end{matrix}\right]$

with respect to the decomposition $H^{2}=\beta
H^{2}\oplus\mathcal{K}_{\beta}$. The operator
$X_{0}\in\mathcal{L}(\mathcal{K}_{\beta})$ acts by the formula
$X_{0}g=P_{\mathcal{K}_{\beta}}\varphi g$ ($g\in\mathcal{K}_{\beta}$), that
is, $X_{0}=\varphi(S(\beta))$ (see [NFBK, Theorem III.2.1]). If $\varphi$ is
outer, then $\varphi(S(\beta))$ is a quasiaffinity by [NFBK, Prop. III.3.1].
Furthermore, $X_{1}(\theta\beta h+\beta f)=(\theta-1)\beta h+a\beta f$ ($h\in
H^{2}$, $f\in\mathcal{K}_{\theta}$). Consequently,
$\operatorname{clos}X_{1}\theta\beta H^{2}=\beta H^{2}$, because $\theta-1$ is
outer. If $a\neq 0$, then $\ker X_{1}=\\{0\\}$ by (3.5). The statement (i)
follows from Lemma 6.8.

A computation shows that $\\{\beta\overline{\chi}\theta,\
P_{+}\overline{\chi}\beta/\|P_{+}\overline{\chi}\beta\|\\}$ is an orthonormal
basis of the range of $T^{*}T-I$, and

$A:=\left[\begin{matrix}|a|^{2}&(\overline{a}b+1)\|P_{+}\overline{\chi}\beta\|\\\
(a\overline{b}+1)\|P_{+}\overline{\chi}\beta\|&|b|^{2}\|P_{+}\overline{\chi}\beta\|^{2}\end{matrix}\right]$

is the matrix of the restriction of $T^{*}T-I$ on its range in this basis.
Thus, $T$ is expansive if and only if $A\geq 0$. Furthermore, $A\geq 0$ if and
only if $|a|^{2}|b|^{2}\geq|\overline{a}b+1|^{2}$, which is equivalent to the
inequality $2\operatorname{Re}\overline{a}b\leq-1$. On the other hand, if
$|a-b|>|b|$, then $1/\varphi\in H^{\infty}$. Clearly, $|a-b|>|b|$ if and only
if $|a-b|^{2}>|b|^{2}$, which is equivalent to the inequality
$2\operatorname{Re}\overline{a}b<|a|^{2}$, which follows from the inequality
$2\operatorname{Re}\overline{a}b\leq-1$. The statement (iii) is proved. ∎

######  Corollary 6.10.

There exists an expansive operator $T$ with the following properties: $T\sim
S$, and there exists $\mathcal{M}\in\operatorname{Lat}T$ such that
$\mathcal{M}\neq\operatorname{clos}Y\vartheta H^{2}$ for every
$Y\in\mathcal{I}(S,T)$ and inner function $\vartheta$.

###### Proof.

Take $a$, $b\in\mathbb{C}$ such that $2\operatorname{Re}\overline{a}b\leq-1$,
and inner functions $\theta$, $\beta\in H^{\infty}$ such that $\theta(0)=0$
and (6.3) is fulfilled for $\theta$ and $\beta$. (Such inner functions exist
by Theorem 6.7.) Define $T$ as in Theorem 6.9 and set $\mathcal{M}=\beta
H^{2}$. By Theorem 6.9, $T\sim S$. If
$\mathcal{M}=\operatorname{clos}Y\vartheta H^{2}$ for some
$Y\in\mathcal{I}(S,T)$, then $S\buildrel d\over{\prec}T|_{\mathcal{M}}$.
Consequently, $(T|_{\mathcal{M}})^{*}\buildrel i\over{\prec}S^{*}$. By (6.15),
$U(\theta)^{*}\buildrel i\over{\prec}S^{*}$, a contradiction (see (3.7)). ∎

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