# Factoring Non-negative Operator Valued Trigonometric Polynomials in Two
Variables

Michael A. Dritschel  School of Mathematics, Statistics and Physics, Herschel
Building, University of Newcastle, Newcastle upon Tyne NE1 7RU, UK
<EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

Using Schur complement techniques, it is shown that a non-negative operator
valued trigonometric polynomial in two variables with degree $(d_{1},d_{2})$
can be written as a finite sum of hermitian squares of at most $2d_{2}$
analytic polynomials with degrees at most $(d_{1},2d_{2}-1)$. In analogy with
the Tarski transfer principle in real algebra, when the coefficient space is
finite dimensional, the proof lifts the problem to an ultraproduct, solves it
there, and then shows that this implies the existence of a solution in the
original context. The general result is obtained through a compactness
argument. While the proof is non-constructive, it nevertheless leads to a
concrete algorithm for the factorization.

###### Key words and phrases:

Trigonometric polynomial, Fejér-Riesz theorem, Schur complement, Toeplitz
operator, shift operator, two variables, ultraproducts, transfer principle

###### 1991 Mathematics Subject Classification:

Primary 47A68; Secondary 60G25, 47A56, 47B35, 42A05, 32A70, 30E99

##  1\. Introduction

The Fejér-Riesz theorem on the factorization of a non-negative trigonometric
polynomial in one variable as the hermitian square of an analytic polynomial
is now over $100$ years old. It has since become an essential tool in both
pure mathematics and engineering, especially in signal processing. There have
been numerous generalizations, and there has been an especially keen interest
in finding an analogue in two variables. This is provided here, where it is
proved that operator valued non-negative trigonometric polynomials in two
variables can be factored as a finite sum of hermitian squares of analytic
polynomials, with tight control over the number and degrees of the polynomials
in the factorization.

_Trigonometric polynomials_ are Laurent polynomials in commuting variables
$z=(z_{1},\dots,z_{r})$ on the $r$-torus $\mathbb{T}^{r}$ ($\mathbb{T}$ the
unit circle in the complex plane). This paper is concerned with polynomials
that take their values in the bounded operators
${\mathfrak{L}}({\mathfrak{H}})$ on a Hilbert space. If
$n=(n_{1},\dots,n_{r})$ is a $d$-tuple of integers, the shorthand $z^{n}$ is
used for $z_{1}^{n_{1}}\cdots z_{r}^{n_{r}}$, and $-n$ for
$(-n_{1},\dots,-n_{r})$. Write $z_{j}^{*}$ for $z_{j}^{-1}$. Then any
trigonometric polynomial has the form $p(z)=\sum_{n}a_{n}z^{n}$, where the sum
has only finitely many non-zero integers, and
$p^{*}(z):=\sum_{n}a_{n}^{*}z^{*\,n}=\sum_{n}a_{n}^{*}z^{-n}$. Denote by
$\mathrm{deg}\,p=d=(d_{1},\dots,d_{r})$ the _degree_ of a trigonometric
polynomial $p$; that is, $d_{j}$ is the largest value of $|n_{j}|$ for which
$a_{n}\neq 0$. A trigonometric polynomial where all terms have powers $n$
lying in the positive orthant (that is, each $n_{j}\geq 0$) is called an
_analytic polynomial_.

The ${\mathfrak{L}}({\mathfrak{H}})$ valued trigonometric polynomials have a
natural operator system structure. Those polynomials which satisfy $p^{*}=p$
are termed _hermitian_. The positive cone $\mathcal{C}$ consists of those
hermitian polynomials $p$ for which $p(z)\geq 0$ for all $z\in\mathbb{T}^{d}$;
that is, such that $p(z)$ is a positive operator for all $z$. Thus, for all
$z$ and for all $f\in{\mathfrak{H}}$, $\left<p(z)f,f\right>\geq 0$. If instead
$\left<p(z)f,f\right>>0$ for all $z$ and all $f\neq 0$, $p$ is said to be
_strictly positive_. If $p=q^{*}q$, $q$ an analytic polynomial, then $p$ is
called an _hermitian square_. Finite sums of such squares are positive, and it
is natural to wonder if this describes all elements of $\mathcal{C}$. For
polynomials in one variable, the Fejér-Riesz theorem gives a positive answer.

In this paper, we look at operator valued polynomials in two variables, and
show that once again there is a description of each element in the positive
cone as a finite sum of squares of analytic polynomials. This extends known
results for strictly positive polynomials, where the polynomials evaluate to
operators with spectrum bounded away from $0$ (or equivalently, are non-
singular in the matrix context). It is worth noting that requiring $q_{j}$ to
be analytic when $p=\sum q_{j}^{*}q_{j}$ is not particularly restrictive,
since $z^{N_{j}}q_{j}$ will be an analytic polynomial for sufficiently large
$N_{j}$ and $p=\sum q_{j}^{*}z^{N_{j}\,*}z^{N_{j}}q_{j}$. This sort of
cancellation is the source of many of the challenges in solving this problem.

For ${\mathfrak{L}}({\mathfrak{H}})$ valued trigonometric polynomials in $d$
variables, the cone $\mathcal{C}$ is archimedean with the polynomial $1$,
which has the value of the identity $1\in{\mathfrak{L}}({\mathfrak{H}})$ for
all $z$, as the order unit — that is, for any hermitian $p$, there is a
positive constant $\alpha$ such that $\alpha 1\pm p\in\mathcal{C}$. There is
the obvious generalization to
$M_{n}(\mathbb{C})\otimes{\mathfrak{L}}({\mathfrak{H}})$ valued polynomials.
The cone is in general not closed, and so attention is usually restricted to
the set $\mathcal{P}_{N}$ of hermitian polynomials of degree less than or
equal to some fixed $N=(n_{1},\dots,n_{d})$ with norm closed positive cone
$\mathcal{C}_{N}$. However, since hermitian squares can have reduced degree,
it might not be possible to factor an element of $\mathcal{C}_{N}$ using
analytic polynomials from $\mathcal{P}_{N}$.

For polynomials taking values in $\mathbb{R}$, such factorization problems are
central to real algebraic geometry. It is always possible to express a scalar
valued hermitian trigonometric polynomial in terms of real polynomials,

$p=\sum_{n}a_{n}z^{n}=\sum_{n}(\mathrm{Re}\,a_{n})(\mathrm{Re}\,(x+iy)^{n}),\qquad
x_{j}=\mathrm{Re}\,z_{j},\ y_{j}=\mathrm{Im}\,z_{j}.$

Since $p$ is hermitian, $\mathrm{Re}\,a_{n}\in\mathbb{R}$ and
$\mathrm{Re}\,(x+iy)^{n}$ is a real polynomial in $x$ and $y$, this is a real
polynomial in $2r$ variables. The $r$-torus $\mathbb{T}^{r}$ is a set
consisting of those points in $\mathbb{R}^{2r}$ satisfying the $2r$
constraints, ${\\{\pm(x_{j}^{2}+y_{j}^{2})\geq 0\\}}_{j=1}^{r}$. These
describe a compact _semialgebraic set_ ; that is, a compact set given in terms
of a finite collection of polynomial inequalities. A fundamental problem then
is; given a semialgebraic set such as $\mathbb{T}^{r}$, succinctly
characterize the elements of the positive cone over this set (generally in
terms of “sums of squares”). Such a description is termed a
_Positivstellensatz_. See, for example, [15] and [13].

Recall that in one variable, the Hardy space of functions over the open unit
disk $\mathbb{D}$ with values in a Hilbert space ${\mathfrak{H}}$, is
$H^{2}_{{\mathfrak{H}}}(\mathbb{D})=\\{\sum_{n=0}^{\infty}c_{n}z^{n}:\sum_{n}\|c_{n}\|^{2}<\infty\\}$.
An ${\mathfrak{L}}({\mathfrak{H}})$ valued function $f$ over $\mathbb{D}$ is
called a _multiplier_ if for all $g\in H^{2}_{{\mathfrak{H}}}(\mathbb{D})$,
$f\cdot g\in H^{2}_{{\mathfrak{H}}}(\mathbb{D})$, where the product is taken
pointwise. A standard result is that the space of multipliers of
$H^{2}_{{\mathfrak{H}}}(\mathbb{D})$ equals
$H^{\infty}_{{\mathfrak{L}}({\mathfrak{H}})}(\mathbb{D})$, the bounded
${\mathfrak{L}}({\mathfrak{H}})$ valued analytic functions on the disk.
Obviously, this space contains the analytic polynomials. A multiplier $f$ is
said to be _outer_ if the closure of $f\cdot
H^{2}_{{\mathfrak{H}}}(\mathbb{D})$ equals
$H^{2}_{{\mathfrak{L}}}(\mathbb{D})$, where ${\mathfrak{L}}$ is a closed
subspace of ${\mathfrak{H}}$. In the scalar setting, this is equivalent to
none of the zeros of $f$ lying in $\mathbb{D}$.

As originally formulated, the Fejér-Riesz theorem concerns the factorization
of positive scalar valued trigonometric polynomials in one complex variable
[6]. It was later proved by Rosenblum [18] (see also [20, 19]) that the
theorem remains true for ${\mathfrak{L}}({\mathfrak{H}})$ valued trigonometric
polynomials, again in one variable.

######  Theorem 1.1 (Fejér-Riesz Theorem).

A positive ${\mathfrak{L}}({\mathfrak{H}})$ valued trigonometric polynomial
$p$ in a single variable $z$ of degree $d$ can be factored as $p=q^{*}q$,
where $q$ is an outer analytic ${\mathfrak{L}}({\mathfrak{H}})$ valued
polynomial of degree $d$.

That is, in one variable, the cone $\mathcal{C}_{d}$ in $\mathcal{P}_{d}$
equals the set of hermitian squares of (outer) analytic polynomials in
$\mathcal{P}_{d}$.

There is also a weaker multivariable version of the Fejér-Riesz theorem for
strictly positive trigonometric polynomials [4] (see also [21]).

######  Theorem 1.2 (Multivariable Fejér-Riesz Theorem).

A strictly positive ${\mathfrak{L}}({\mathfrak{H}})$ valued trigonometric
polynomial $p$ of degree $d$ in $r$ variables $z=(z_{1},\dots,z_{r})$, can be
factored as a finite sum $p=\sum q_{i}^{*}q_{i}$, where each $q_{i}$ is an
analytic ${\mathfrak{L}}({\mathfrak{H}})$ valued polynomial. If $p\geq\epsilon
1$, $\epsilon>0$, then the number of polynomials in the sum and their degrees
can be bounded in terms of $\epsilon$.

While there is some control over the degrees of the analytic polynomials in
Theorem 1.2, the bound on the degrees goes to infinity as $\epsilon$ goes to
$0$. This is not surprising because of the cancellations mentioned above.
Examination of the proof yields no obvious choice of closed cone containing
all strictly positive polynomials of a fixed degree, even in two variables.

Suppose that $\mathcal{S}={\\{x\in\mathbb{R}^{d}:r_{j}(x)\geq 0,\
j=1,\dots,m\\}}$, is a semialgebraic set. Let $\Delta$ be the set of functions
from $\\{1,\dots,m\\}$ to $\\{0,1\\}$. On $\mathcal{S}$, there are various
collections of polynomials which are non-negative. Beside the (finite) sums of
squares of polynomials $\sum_{i}q_{i}^{2}$, which are non-negative everywhere,
one also has

  * $\bullet$

the _quadratic module_ : $\sum_{j}r_{j}\sum_{i}q_{ji}^{2}$; and

  * $\bullet$

the _preordering_ :
$\sum_{\delta\in\Delta}\prod_{j}r_{j}^{\delta(j)}\sum_{i}q_{\delta i}^{2}$.

In both cases the sums are assumed to be finite.

In the scalar setting, the multivariable Fejér-Riesz theorem is a special case
of Schmüdgen’s theorem [24] (see also [1]).

######  Theorem 1.3 (Schmüdgen’s theorem).

Let $\mathcal{S}$ be a compact semialgebraic set over $\mathbb{R}^{d}$
described by a finite set of polynomials in $\mathcal{R}$. Any polynomial
which is strictly positive over $\mathcal{S}$ is in the preordering over
$\mathcal{R}$.

There are variations and refinements of this result. For example, if the set
of polynomials includes for each coordinate a constant multiple of $1$ minus
the coordinate squared (that is, disks in each coordinate), then a theorem due
to Putinar [16] says that strictly positive polynomials are in the quadratic
module over $\mathcal{R}$. Also, if the set $\mathcal{R}$ contains at most two
polynomials, the preordering can be replaced by the quadratic module in
Schmüdgen’s theorem [15]. Cimprič [2] and Hol and Scherer [23] have extended
some of these results to matrix valued polynomials.

Matters become more complicated when positivity replaces strict positivity.
Both Schmüdgen’s and Putinar’s theorems are known to be generally false then.
Indeed, Scheiderer has shown that if the dimension of a compact semialgebraic
set is $3$ or more, there will always be positive polynomials which are not in
the preordering [22]. On the other hand, he also proved that under mild
restrictions, for a compact two dimensional semialgebraic set, all positive
scalar valued polynomials are in the preordering [22]. This last result
implies in particular that in two variables, any real valued positive
trigonometric polynomial in two variables is a sum of squares of analytic
polynomials.

The results of Schmüdgen, Putinar and Scheiderer (as well as the various
generalizations mentioned) are intimately tied to the study of real fields,
which has no known analogue when the polynomial coefficients are allowed to
come from a ring of operators. Even the matrix valued results in [2] and [23]
are likewise connected, since they are proved by reducing to the scalar valued
case and applying the known theorems in this setting. For this reason, it is
particularly noteworthy that most proofs of the operator Fejér-Riesz theorems
take a purely analytic approach to the problem. On the other hand, these
theorems are in a sense incomplete, in that they say nothing about polynomials
vanishing on $\mathbb{T}^{r}$ (that is, there is no _Nullstellensatz_).

Motivated by these considerations, we address a generalization of part of
Scheiderer’s work by proving an hermitian factorization theorem for operator
valued positive trigonometric polynomials in two variables.

######  Theorem.

Let ${\mathfrak{K}}$ be a Hilbert space. Given a positive
${\mathfrak{L}}({\mathfrak{K}})$ valued trigonometric polynomial $Q$ in a two
variables $z=(z_{1},z_{2})$ of degree $(d_{1},d_{2})$, $Q$ can be factored as
a sum of hermitian squares of at most $2d_{2}$,
${\mathfrak{L}}({\mathfrak{K}})$ valued, analytic polynomials of degree at
most $(d_{1},2d_{2}-1)$.

The principle techniques are an elaboration of the Schur complement approach
to the operator Fejér-Riesz theorems found in [3] and [4]. Though these
methods are now well known, there are several new ideas appearing here when
tackling a two variable factorization theorem. As usual, a trigonometric
polynomial is associated with a Toeplitz operator. In the two variable
setting, this becomes a Toeplitz operator of Toeplitz operators. Positive
trigonometric polynomials then correspond to finite degree positive Toeplitz
operators (that is, having only finitely many nonzero diagonals). These can be
viewed as either being doubly infinite or singly infinite since the
polynomials are over the $2$-torus, or even some mixture of these. In the
singly infinite case, the multiplication operators corresponding to the
variables are commuting isometries, while in the doubly infinite case they are
commuting unitaries. The interplay between these turns out to be important
here.

By collecting entries of a selfadjoint finite degree Toeplitz operator into
large enough blocks, one gets a tridiagonal Toeplitz operator. Calling the
main diagonal entry $A$ and the off-diagonal entries $B$ and $B^{*}$, it was
observed in [3] that this Toeplitz operator is positive if and only if there
is a positive operator $M$ such that $\begin{pmatrix}A-M&B^{*}\\\
B&M\end{pmatrix}\geq 0$, and in this case the set of all such $M$ forms a norm
closed compact, convex set $\mathcal{M}$. These play a similar role to the
Schur complement supported on the $(1,1)$ entry of the Toeplitz operator,
which happens to be the largest element of $\mathcal{M}$. The Schur complement
gives rise to the outer factorization (unique up to multiplication by a
unitary) of the positive Toeplitz matrix. The smallest element of
$\mathcal{M}$ gives the “co-outer” factorization. Of particular interest will
be those extremal polynomials for which the set is a singleton.

In two variables, a complication arises in that the elements of the set
$\mathcal{M}$ may not consist entirely of Toeplitz operators, and in
particular, the largest element will in general not be Toeplitz. Despite this,
there will always be a closed convex subset $\mathcal{M}_{T}$ of $\mathcal{M}$
consisting of Toeplitz operators. The extremal case when $\mathcal{M}_{T}$ is
a singleton plays a central role. It is proved that $A$ can always be replaced
by a Toeplitz $\hat{A}\leq A$ so that $\mathcal{M}_{T}=\\{\hat{M}\\}$ is a
singleton. However, while $\hat{A}$ and $\hat{M}$ will correspond to bounded
trigonometric functions, it is not evident that they need necessarily have
finite degree, and so correspond to polynomials.

Hence another difficulty is encountered. While the Fejér-Riesz theorem
guarantees the existence of a factorization for positive single variable
trigonometric _polynomials_ , is well known that not all positive
trigonometric _functions_ can be factored. Some restriction, as in for example
Szegő’s theorem, is needed. This problem is finessed by using ultraproduct
methods to dilate to a function which can be factored as an hermitian square.
While it is not clear that the dilation of an extremal is extremal, it
nevertheless retains certain important characteristics, and this is enough to
enable a proof that this function is a polynomial, which in turn implies that
$\hat{A}$ and $\hat{M}$ correspond to polynomials. The procedure is in essence
a sort of _transfer principle_. The single variable Fejér-Riesz theorem is
then used to finish off the factorization.

One further twist is that the dilation argument requires that the coefficient
space of the original trigonometric polynomial be finite dimensional (that is,
the coefficients are matrices). However, since for all finite dimensions and
for all positive polynomials of degree at most $(d_{1},d_{2})$ in
$(z_{1},z_{2})$, there is a uniform bound for the degrees and numbers of
polynomials needed for a factorization, a compactness argument allows the
result to be extended to polynomials with operator coefficients.

Why is it not possible to use the same ideas to factor positive trigonometric
polynomials in three or more variables? While the theorem presented here is
about trigonometric polynomials in two variables, frequent use is made of
results about polynomials in one variable, and in particular, the Fejér-Riesz
factorization in terms of outer polynomials, which plays a fundamental role in
showing that the degrees of $\hat{A}$ and $\hat{M}$ mentioned above are of
finite degree. Outerness (interpreted in the appropriate manner) does not
necessarily apply to factorizations of polynomials in two variables. Indeed,
since there are examples of positive polynomials in three or more variables
which cannot be factored as a sum of square of polynomials, this indirectly
indicates that there will exist positive polynomials in two variables without
an outer factorization. Outer factorizations of multivariable trigonometric
polynomials are explored further in [4].

##  2\. Toeplitz and analytic operators, and their relation to trigonometric
polynomials

A _shift operator_ is an isometry $S$ on a Hilbert space ${\mathfrak{H}}$ with
trivial unitary component in its Wold decomposition. It is then natural to
write for some Hilbert space ${\mathfrak{G}}$,
${\mathfrak{H}}={\mathfrak{G}}\oplus{\mathfrak{G}}\oplus\cdots$ (identified
with $H^{2}({\mathfrak{G}})$), and

$S(h_{0},h_{1},\dots)^{t}=(0,h_{0},h_{1},\dots)^{t}$

when the elements of ${\mathfrak{H}}$ are written as column vectors (here
“$t$” indicates transpose). Fix a shift $S$. If
$T,A\in{\mathfrak{L}}({\mathfrak{H}})$, say that $T$ is _Toeplitz_ if
$S^{*}TS=T$, and that $A$ is _analytic_ if $AS=SA$. To distinguish this case
from Toeplitz operators on $L^{2}$ spaces, which is introduced below, such
operators will be referred to as being _singly infinite_ Toeplitz operators.
Viewed as an operator on $H^{2}({\mathfrak{G}})$, pre-multiplication of $T$ by
$S^{*}$ has the effect of deleting the first row of $T$ and shifting $T$
upwards by one row. Likewise, post-multiplication by $S$ deletes the first
column and shifts left by one column. Hence as matrices with entries in
${\mathfrak{L}}({\mathfrak{G}})$, such Toeplitz and analytic operators have
the forms

$T=\begin{pmatrix}T_{0}&T_{-1}&T_{-2}&\cdots\\\ T_{1}&T_{0}&T_{-1}&\ddots\\\
T_{2}&T_{1}&T_{0}&\ddots\\\ \vdots&\ddots&\ddots&\ddots\end{pmatrix},\qquad
A=\begin{pmatrix}A_{0}&0&0&\cdots\\\ A_{1}&A_{0}&0&\ddots\\\
A_{2}&A_{1}&A_{0}&\ddots\\\ \vdots&\ddots&\ddots&\ddots\end{pmatrix}.$ (2.1)

An analytic operator $A$ is termed _outer_ if $\overline{\mathrm{ran}}\,A$ is
a subspace of ${\mathfrak{H}}$ of the form $H^{2}({\mathfrak{F}})$ for some
closed subspace ${\mathfrak{F}}$ of ${\mathfrak{G}}$.

Now consider Laurent and analytic polynomials $Q(z)=\sum_{k=-d}^{d}Q_{k}z^{k}$
and $P(z)=\sum_{k=0}^{d}P_{k}z^{k}$ on $\mathbb{T}$ with coefficients in
${\mathfrak{L}}({\mathfrak{G}})$. Refer to $d$ as the _degree_ of these
polynomials, assuming that either $Q_{d}$ or $Q_{-d}$ and $P_{d}$ are nonzero.
Set $Q_{j}=0$ for $|j|>d$ and $P_{j}=0$ for $j>d$ or $j<0$. Then the formulas

$T_{Q}=\begin{pmatrix}Q_{0}&Q_{-1}&Q_{-2}&\cdots\\\
Q_{1}&Q_{0}&Q_{-1}&\ddots\\\ Q_{2}&Q_{1}&Q_{0}&\ddots\\\
\vdots&\ddots&\ddots&\ddots\end{pmatrix},\qquad
T_{P}=\begin{pmatrix}P_{0}&0&0&\cdots\\\ P_{1}&P_{0}&0&\ddots\\\
P_{2}&P_{1}&P_{0}&\ddots\\\ \vdots&\ddots&\ddots&\ddots\end{pmatrix}$ (2.2)

define bounded linear operators on ${\mathfrak{H}}$. The operator $T_{Q}$ is
Toeplitz, while $T_{P}$ is analytic, and $d$ is likewise called the _degree_
of $T_{Q}$ and $T_{P}$. Even if $Q$ and $P$ are not polynomials, but are
nevertheless bounded functions, the operators $T_{Q}$ and $T_{P}$ will be
bounded. Moreover,

  1. $\bullet$

$Q(z)\geq 0$ for all $z\in\mathbb{T}$ if and only if $T_{Q}\geq 0$;

  2. $\bullet$

$Q(z)=P(z)^{*}P(z)$ for all $z\in\mathbb{T}$ if and only if
$T_{Q}=T_{P}^{*}T_{P}$.

Recall that $Q(z)\geq 0$ means that for all $g\in{\mathfrak{G}}$,
$\left<Q(z)g,g\right>\geq 0$ and $Q(z)>0$ if the inequality is strict for
$g\neq 0$. Write $Q\geq 0$ and $Q>0$ for $Q(z)\geq 0$, respectively $Q(z)>0$,
for all $z\in\mathbb{T}$.

An analytic function $P(z)$ is _outer_ if the analytic Toeplitz operator
$T_{P}$ is outer. The Fejér-Riesz theorem (Theorem 1.1) can be restated in
terms of Toeplitz operators: A positive Toeplitz operator
$T\in{\mathfrak{L}}(H^{2}({\mathfrak{G}}))$ of finite degree $d$ has the form
$F^{*}F$, where $F\in{\mathfrak{L}}(H^{2}({\mathfrak{G}}))$ is an outer
analytic operator of the same degree as $T$.

So far it has been assumed that the entries $T_{j}$ of the Toeplitz operators
$T$ are in ${\mathfrak{L}}({\mathfrak{G}})$, and while this is necessary if
$T\geq 0$, it is also natural to more generally consider Toeplitz and analytic
operators where
$T_{j}\in{\mathfrak{L}}({\mathfrak{G}}_{1},{\mathfrak{G}}_{2})$,
${\mathfrak{G}}_{1}$, ${\mathfrak{G}}_{2}$, Hilbert spaces.

One can extend the above Toeplitz operators to
${\tilde{\mathfrak{H}}}=\dots\oplus{\mathfrak{G}}\oplus{\mathfrak{G}}\oplus{\mathfrak{G}}\oplus\dots=L^{2}({\mathfrak{G}})$
simply by continuing each of the diagonals. The shift operator is now the
bilateral shift, and is a unitary operator. The resulting _doubly infinite_
Toeplitz operator is positive if and only if the same is true for the
corresponding singly infinite Toeplitz operator.

Since every doubly infinite Toeplitz operator commutes with the bilateral
shift, this is not a useful way of distinguishing those which are analytic.
Instead, the doubly infinite Toeplitz operators with no nonzero entries above
the main diagonal are called _analytic_. It is a simple yet useful observation
that if $A\in{\mathfrak{L}}({\mathfrak{H}})$ is a singly infinite Toeplitz
operator and $\tilde{A}\in{\mathfrak{L}}({\tilde{\mathfrak{H}}})$ is the
corresponding doubly infinite Toeplitz operator, $A$ is analytic if and only
if $\tilde{A}$ is analytic, if and only if ${\mathfrak{H}}$ is an invariant
subspace for $\tilde{A}$, if and only if
$P_{\mathfrak{H}}\tilde{A}|{\mathfrak{H}}$ is an analytic, singly infinite
Toeplitz operator. Likewise, a doubly infinite Toeplitz operator $A$ is
_outer_ if and only if $P_{\mathfrak{H}}\tilde{A}|{\mathfrak{H}}$ is outer.

All other notions considered so far carry over naturally to the multi-index /
multivariable setting. Only the two index / variable case is examined, the
version for three or more then being evident. Suppose that $S_{2}$ is a shift
operator on ${\mathfrak{H}}=\bigoplus_{j_{2}=0}^{\infty}{\mathfrak{G}}$, and
that $S_{1}$ is a shift operator on
${\mathfrak{G}}=\bigoplus_{j_{1}=0}^{\infty}{\mathfrak{K}}$. If $T$ is a
Toeplitz operator on ${\mathfrak{H}}$ with the property that each $T_{j_{2}}$
is a Toeplitz operator on ${\mathfrak{G}}$, say that $T$ is a _bi-Toeplitz_
operator (or _multi-Toeplitz_ more generally). Call $T$ _bi-analytic_
(respectively, _multi-analytic_) if $T$ is analytic and each $T_{j}$ is
analytic. It is sometimes convenient to shift back and forth to the doubly
infinite Toeplitz setting in one of the variables. If $T$ is a bi-Toeplitz
operator on
$\bigoplus_{j_{2}=0}^{\infty}\bigoplus_{j_{1}=0}^{\infty}{\mathfrak{K}}$, the
entries are naturally labeled by two indices, $(j_{2},j_{1})$. Let
${\tilde{\mathfrak{G}}}=\bigoplus_{j_{2}=0}^{\infty}{\mathfrak{K}}$ and
${\tilde{\mathfrak{H}}}=\bigoplus_{j_{1}=0}^{\infty}\bigoplus_{j_{2}=0}^{\infty}{\mathfrak{K}}=\bigoplus_{j_{1}=0}^{\infty}{\tilde{\mathfrak{G}}}$.
The indices of $T$ can be interchanged to get another operator $\tilde{T}$ on
${\tilde{\mathfrak{H}}}$. The exchange is implemented via a permutation of
rows and columns corresponding to conjugation with the unitary operator
$W:{\tilde{\mathfrak{H}}}\to{\mathfrak{H}}$ having the identity $1$ in the
entries labeled with $((j_{1},j_{2}),(j_{2},j_{1}))$ and $0$ elsewhere.

As in the single variable setting, there are Laurent and analytic polynomials
with coefficients in ${\mathfrak{L}}({\mathfrak{K}})$, but now in
$z=(z_{1},z_{2})$. These look like

$Q(z)=\sum_{k_{2}=-m_{2}}^{m_{2}}\left(\sum_{k_{1}=-m_{1}}^{m_{1}}Q_{k_{2},k_{1}}z_{1}^{k_{1}}\right)z_{2}^{k_{2}}\quad\text{and}\quad
P(z)=\sum_{k_{2}=0}^{m_{2}}\left(\sum_{k_{1}=0}^{m_{1}}P_{k_{2},k_{1}}z_{1}^{k_{1}}\right)z_{2}^{k_{2}}.$

Set $Q_{j_{1},j_{2}}=0$ whenever $j_{1}\notin[-m_{1},m_{1}]$ or
$m_{2}\notin[-m_{2},m_{2}]$, and set $P_{j_{1},j_{2}}=0$ whenever
$j_{1}\notin[0,m_{1}]$ or $j_{2}\notin[0,m_{2}]$. This results in
trigonometric polynomials in the variable $z_{2}$ with coefficients which are
trigonometric polynomials in the variable $z_{1}$. Much as before, the
formulas

$T_{Q}=(Q_{j_{2}-k_{2},j_{1}-k_{1}})_{(j_{2},j_{1}),(k_{2},k_{1})\in\mathbb{N}\times\mathbb{N}}\qquad
T_{P}=(Q_{j_{2}-k_{2},j_{1}-k_{1}})_{(j_{2},j_{1}),(k_{2},k_{1})\in\mathbb{N}\times\mathbb{N}}$
(2.3)

define bounded operators on ${\mathfrak{H}}$, the first being bi-Toeplitz and
the second bi-analytic. If indices are interchanged and ${\tilde{T}}_{Q}$ and
${\tilde{T}}_{P}$ are viewed as operators on ${\tilde{\mathfrak{H}}}$, this
amounts to taking $Q$ and $P$ as polynomials in $z_{1}$ with coefficients
which are polynomials in $z_{2}$. The pair $(m_{1},m_{2})$ is referred to as
the _degree_ of the $Q$ (equivalently, degree of $T_{Q}$), if one of the
coefficients of the form $Q_{\pm m_{1},\pm m_{2}}$ is nonzero, while
$Q_{j_{1},j_{2}}=0$ if $j_{1}\notin[-m_{1},m_{1}]$ or
$j_{2}\notin[-m_{2},m_{2}]$,

There are further Toeplitz operators which can be associated with $Q$ and $P$.
Write this as $Q=\sum_{j_{2}}{\hat{Q}}_{j_{2}}(z_{1})z_{2}^{j_{2}}$, where
each ${\hat{Q}}_{j}$ is a trigonometric polynomial in $z_{1}$. Then

$T_{Q_{1}}(z_{1}):=({\hat{Q}}_{j_{2}-k_{2}})_{j_{2},k_{2}\in\mathbb{N}}$

is Toeplitz, with trigonometric polynomial entries. Likewise define
$T_{P_{1}}$, and if the roles of $z_{1}$ and $z_{2}$ are exchanged, denote the
resulting Toeplitz operators as $T_{Q_{2}}$ and $T_{P_{2}}$, respectively. The
functions $T_{Q_{1}}$ and $T_{P_{1}}$ are ${\tilde{\mathfrak{G}}}$ valued
trigonometric polynomials in the variable $z_{1}$, while $T_{Q_{2}}$ and
$T_{P_{2}}$ are ${\mathfrak{G}}$ valued trigonometric polynomials in $z_{2}$.

In analogy with the one variable case,

  1. $\bullet$

$Q(z)\geq 0$ for all $z\in\mathbb{T}^{2}$ if and only if $T_{Q}\geq 0$;

  2. $\bullet$

$Q(z)=\sum_{j}P_{j}(z)^{*}P_{j}(z)$ for all $z\in\mathbb{T}^{2}$ if and only
if $T_{Q}=\sum_{j}T_{P_{j}}^{*}T_{P_{j}}$;

as well as

  1. $\bullet$

$T_{Q_{1}}(z_{1})\geq 0$ for all $z_{1}\in\mathbb{T}$ if and only if
$T_{Q_{2}}(z_{2})\geq 0$ for all $z_{2}\in\mathbb{T}$ if and only if
$T_{Q}\geq 0$;

  2. $\bullet$

$Q(z)=\sum_{j}P_{j}(z)^{*}P_{j}(z)$ for all $z\in\mathbb{T}^{2}$ if and only
if $T_{Q_{1}}(z_{1})=\sum_{j}T_{P_{j,1}}(z_{1})^{*}T_{P_{j,1}}(z_{1})$ if and
only if $T_{Q_{2}}(z_{2})=\sum_{j}T_{P_{j,2}}(z_{2})^{*}T_{P_{j,2}}(z_{2})$.

##  3\. Hyper-Toeplitz operators

Not all of the single variable Toeplitz operators dealt with here are _a
priori_ known to have finite degree, and as a consequence it may be difficult
to determine whether a given non-negative Toeplitz operator is the hermitian
square of an analytic operator. While the Fejér-Riesz theorem implies that
every non-negative Toeplitz operator of finite degree can be factored as an
hermitian square of an outer operator of the same degree, some additional
constraints, such as those found in Szegő’s theorem [19, Section 6.1], are
needed to ensure the existence of a factorization of a non-negative Toeplitz
operator of infinite degree (see [19, Section 6.5] for the operator valued
case). It is shown here that if the coefficient space is finite dimensional,
it is possible to dilate the Toeplitz operator to a factorable Toeplitz
operator on a Hilbert space ultraproduct, referred to as a hyper-Toeplitz
operator, in analogy with terminology used in non-standard analysis.

Let $\mathcal{U}$ be a free ultrafilter over $\mathbb{N}$, and suppose that
${\mathfrak{H}}_{n}$, $n\in\mathbb{N}$, are separable infinite dimensional
Hilbert spaces. Let $a=(a_{n}),b=(b_{n})\in\prod{\mathfrak{H}}_{n}$ (so
$\sup_{n}\|a_{n}\|,\sup_{n}\|b_{n}\|<\infty)$). There is an equivalence
relation $\sim$ whereby $a\sim b$ if $\lim_{\mathcal{U}}\|a_{n}-b_{n}\|=0$.
Denote the completion of $\prod{\mathfrak{H}}_{n}/\sim\,$ by
${\mathfrak{H}}_{\mathcal{U}}$, the _ultraproduct_ of
$\\{{\mathfrak{H}}_{n}\\}$ (with respect to the ultrafilter $\mathcal{U}$).
When all the spaces are the same, the ultraproduct is referred to as an
_ultrapower_. The ultraproduct is a Hilbert space with scalar multiplication
and vector addition defined entry-wise, and with inner product given by
$\left<a,b\right>_{\mathcal{U}}=\lim_{\mathcal{U}}\left<a_{n},b_{n}\right>$
for $a,b\in\prod{\mathfrak{H}}_{n}/\sim$. In case of an ultrapower with
${\mathfrak{H}}_{n}={\mathfrak{H}}_{0}$ for all $n$, the space
${\mathfrak{H}}_{0}$ is isometrically isomorphic to the subspace of
${\mathfrak{H}}_{\mathcal{U}}$ of equivalence classes of constant sequences
(that is, having the same vector in each entry), and the same notation is used
for both ${\mathfrak{H}}_{0}$ and its image in the ultrapower.

A uniformly bounded sequence
$A=(A_{n})\in\prod{\mathfrak{L}}({\mathfrak{H}}_{n},{\mathfrak{K}}_{n})$
defines a bounded operator from ${\mathfrak{H}}_{\mathcal{U}}$ to
${\mathfrak{K}}_{\mathcal{U}}$. Again, addition and multiplication are defined
entry-wise (where they make sense), as are adjoints. When
${\mathfrak{H}}_{n}={\mathfrak{H}}_{0}$ for all $n$,
${\mathfrak{L}}({\mathfrak{H}}_{0})$ is isometrically isomorphic to the
subspace of sequences with the same operator in every entry (here again, the
same notation is used for both). Two operators $A$ and $B$ in
${\mathfrak{L}}({\mathfrak{H}}_{\mathcal{U}},{\mathfrak{K}}_{\mathcal{U}})$
are equivalent if for all $a\in{\mathfrak{H}}_{\mathcal{U}}$,
$\lim_{\mathcal{U}}(A-B)a=0$, the equivalence class of the vector with all
entries equal to $0$. The identity operator
$1\in{\mathfrak{L}}({\mathfrak{H}}_{\mathcal{U}})$ is the equivalence class of
$(1_{n})$, where $1_{n}$ is the identity on ${\mathfrak{H}}_{n}$.

From here on in the context of ultraproducts, when writing vectors and
operators, they are understood as such equivalence classes, or representatives
of these equivalence classes. All of the usual notions for operators continue
hold. So for example,
$A\in{\mathfrak{L}}({\mathfrak{H}}_{\mathcal{U}},{\mathfrak{K}}_{\mathcal{U}})$
is isometric means that $\|Aa\|=\|a\|$ for all
$a\in{\mathfrak{H}}_{\mathcal{U}}$. It is worth emphasizing that this does not
mean that every representative of $A$ acts isometrically entry-wise. However,
by cofinality, if $(A_{n})$ is in the equivalence class of an isometric $A$,
given $a\in{\mathfrak{H}}_{\mathcal{U}}$ and $\epsilon>0$, there is an
$N\in\mathbb{N}$ such that for all $n\geq N$,
$\left|\,\|A_{n}a_{n}\|-\|a_{n}\|\,\right|<\epsilon\|a\|$.

######  Lemma 3.1.

Let ${\mathfrak{H}}_{\mathcal{U}}$, ${\mathfrak{K}}_{\mathcal{U}}$ be
ultrapowers, $T=(T_{0},T_{0},\dots)$, $\hat{T}=(T_{1},T_{2},\dots)$ bounded
operators from ${\mathfrak{H}}_{\mathcal{U}}$ to
${\mathfrak{K}}_{\mathcal{U}}$, and suppose that $\lim_{j}\|T_{0}-T_{j}\|=0$.
Then $T=\hat{T}$.

###### Proof.

This is a consequence of the fact that the elements of an ultrapower are
uniformly bounded sequences. ∎

For $T_{0}\in{\mathfrak{L}}({\mathfrak{H}},{\mathfrak{K}})$, the operator
$T=(T_{0})$ on
${\mathfrak{L}}({\mathfrak{H}}_{\mathcal{U}},{\mathfrak{K}}_{\mathcal{U}})$ is
termed a _dilation_ of $T_{0}$.

Let $S:=(S_{1},S_{2},\dots)$, where $S_{j}$ is the shift operator on
${\mathfrak{H}}_{n}$. This operator is isometric and has no unitary component
as an operator on ${\mathfrak{H}}_{\mathcal{U}}$. So by the Wold
decomposition, $S$ is a shift operator. Of course, not all representatives of
a shift operator $S$ need have this form. A subspace ${\mathfrak{K}}$ with the
property that ${\mathfrak{K}}$ reduces $S$ has the form
$H_{\mathcal{U}}^{2}({\mathfrak{F}})$, where
${\mathfrak{F}}\subseteq\overline{\mathrm{ran}}\,(1-SS^{*})$. The dimension of
${\mathfrak{F}}$ is in general much larger than that of any of the spaces
${\mathfrak{H}}_{n}$. For example, if ${\mathfrak{H}}_{n}={\mathfrak{H}}_{0}$
for all $n$, then as noted, $\overline{\mathrm{ran}}\,(1-SS^{*})$ is the space
of all bounded sequences from ${\mathfrak{H}}_{0}$ (modulo the equivalence
relation).

Letting $S$ refer generically to a shift operator on a $H_{\mathcal{U}}^{2}$
space, an operator
$T\in{\mathfrak{L}}(H_{\mathcal{U}}^{2}({\mathfrak{D}}),H_{\mathcal{U}}^{2}({\mathfrak{E}}))$
is _hyper-Toeplitz_ if $S^{*}TS=T$ and _hyper-analytic_ if $TS=ST$ (which
spaces the shifts act on should be clear from the context). A hyper-analytic
operator $T$ is _hyper-outer_ if
$\overline{\mathrm{ran}}\,T=H_{\mathcal{U}}^{2}({\mathfrak{G}})$. Of course,
since $S$ is a shift operator, these operators are Toeplitz, analytic, or
outer in the usual sense, though entry-wise, equivalence class representatives
may not be. The emphasis here though is that these operators are viewed as
acting on an ultraproduct of $H^{2}$ spaces.

The proof of the following lemma is an easy exercise.

######  Lemma 3.2.

Let $T_{0}\in{\mathfrak{L}}(H^{2}({\mathfrak{H}}),H^{2}({\mathfrak{K}}))$, and
let ${\mathfrak{H}}_{\mathcal{U}}$, ${\mathfrak{K}}_{\mathcal{U}}$ be the
ultrapowers of ${\mathfrak{H}}$ and ${\mathfrak{K}}$, respectively. Define
$T\in{\mathfrak{L}}(H_{\mathcal{U}}^{2}({\mathfrak{H}}),H_{\mathcal{U}}^{2}({\mathfrak{K}}))$
to be the dilation of $T_{0}$ to the ultrapower. The operator $T_{0}$ is
Toeplitz (respectively, analytic / outer) operator if and only if $T$ is a
hyper-Toeplitz (respectively, hyper-analytic / hyper-outer) operator.
Furthermore, $\deg T=\deg T_{0}$.

Additionally, the algebraic structures carry over, in that for example, sums
of operators go to sums of their dilations / restrictions.

While not all positive Toeplitz functions can be factored as hermitian
squares, the next result shows that if $\dim{\mathfrak{H}}<\infty$ and
$T_{0}\in{\mathfrak{L}}(H^{2}({\mathfrak{H}}))$ is a positive Toeplitz
operator, the hyper-Toeplitz dilation of $T_{0}$ can be factored.

######  Theorem 3.3.

Let $T_{0}\in{\mathfrak{L}}(H^{2}({\mathfrak{H}}))$ be a positive Toeplitz
operator, $\dim{\mathfrak{H}}<\infty$, and let
$T\in{\mathfrak{L}}(H_{\mathcal{U}}^{2}({\mathfrak{H}}))$ be a hyper-Toeplitz
operator dilating $T_{0}$. Then there exists a hyper-outer operator $F$ such
that $T=F^{*}F$.

Because of the definition of _outer_ (and hence, hyper-outer) in the doubly
infinite setting, the theorem remains valid if $H^{2}({\mathfrak{H}})$ is
replaced by $L^{2}({\mathfrak{H}})$ and $H_{\mathcal{U}}^{2}({\mathfrak{H}})$
by $L_{\mathcal{U}}^{2}({\mathfrak{H}})$.

###### Proof of Theorem 3.3.

Let $T_{0}\in{\mathfrak{L}}(H^{2}({\mathfrak{H}}))$ be a non-negative Toeplitz
operator, $T\in{\mathfrak{L}}(H_{\mathcal{U}}^{2}({\mathfrak{H}}))$ be a
hyper-Toeplitz operator extending $T_{0}$. Define a positive Toeplitz operator
operator with scalar valued entries by

$J_{n}=\begin{pmatrix}1&\tfrac{n}{n+1}&\tfrac{n-1}{n+1}&\cdots&\tfrac{1}{n+1}&0&\cdots\\\
\tfrac{n}{n+1}&\ddots&\ddots&\ddots&&\ddots&\ddots\\\
\tfrac{n-1}{n+1}&\ddots&&&&&\\\ \vdots&\ddots&&&&&\\\ \tfrac{1}{n+1}&&&&&&\\\
0&\ddots&&&&&\\\ \vdots&\ddots&&&&&\\\ \end{pmatrix}.$ (3.1)

The Schur (that is, entry-wise) product with $T_{0}$ gives $T_{n}$ which is
non-negative and finitely supported. As such, by the operator Fejér-Riesz
theorem, $T_{n}$ factors as $F_{n}^{*}F_{n}$, where $F_{n}$ is outer. Set
$\hat{T}=(T_{n}),F=(F_{n})\in H_{\mathcal{U}}^{2}({\mathfrak{H}})$. Then
$\hat{T}=F^{*}F$, and since each $F_{n}$ is outer, $S_{n}$, the shift operator
on the $n$th copy of $H^{2}({\mathfrak{H}})$, reduces
$\overline{\mathrm{ran}}\,F_{n}$. Hence the shift $S=(S_{n})$ on
$H_{\mathcal{U}}^{2}({\mathfrak{H}})$ reduces $\overline{\mathrm{ran}}\,F$;
that is, $F$ is hyper-outer.

It is left to prove that $T=\hat{T}$. Let $\varphi_{n}$ be the
${\mathfrak{L}}({\mathfrak{H}})$, valued trigonometric function associated to
$T_{n}$, $j\geq 0$, and write
$\varphi_{n}=\varphi_{n}^{+}+\varphi_{n}^{-\,*}$, where $\varphi_{n}^{\pm}$
are analytic. By assumption, $\dim{\mathfrak{H}}=d<\infty$. Let $\epsilon>0$
and fix an orthonormal basis $\\{e_{j}\\}$ for ${\mathfrak{H}}$. By
analyticity, for every $z\in\mathbb{T}$, there exists an $n_{z}$ such that
$\left|\left<(\varphi_{0}^{\pm}(z)-\varphi_{n_{z}}^{\pm}(z))e_{j},e_{k}\right>\right|<\epsilon/(2d^{2})$,
and consequently, $\|\varphi_{0}(z)-\varphi_{n_{z}}(z)\|<\epsilon$. Since
$\mathbb{T}$ is compact, by continuity there is an integer $n$ such that
$\|\varphi_{0}-\varphi_{n}\|<\epsilon$, and hence $\|T_{0}-T_{n}\|<\epsilon$,
thus implying that $(T_{n})$ converges in norm to $T_{0}$. It then follows
from Lemma 3.1 that $T=\hat{T}$. ∎

##  4\. Schur complements

Schur complements play an important role in several proofs of the operator
Fejér-Riesz theorem [3, 4]. A survey of their use in this way can be found in
[5]. Here is the definition.

######  Definition 4.1.

Let ${\mathfrak{H}}$ be a Hilbert space and $0\leq
T\in{\mathfrak{L}}({\mathfrak{H}})$. Let ${\mathfrak{K}}$ be a closed subspace
of ${\mathfrak{H}}$, and
$P_{\mathfrak{K}}\in{\mathfrak{L}}({\mathfrak{H}},{\mathfrak{K}})$ the
orthogonal projection of ${\mathfrak{H}}$ onto ${\mathfrak{K}}$. Then there is
a unique operator $0\leq
M=M(T,{\mathfrak{K}})\in{\mathfrak{L}}({\mathfrak{K}})$ called the _Schur
complement of $T$ supported on ${\mathfrak{K}}$_, such that

  1. (i)

$T-P_{\mathfrak{K}}^{*}MP_{\mathfrak{K}}\geq 0$;

  2. (ii)

if $\widetilde{M}\in{\mathfrak{L}}({\mathfrak{K}})$, $\widetilde{M}\geq 0$,
and $T-P_{\mathfrak{K}}^{*}\widetilde{M}P_{\mathfrak{K}}\geq 0$, then
$\widetilde{M}\leq M$.

There are several equivalent ways of obtaining the Schur complement. For
example, if $T=\begin{pmatrix}A&B^{*}\\\ B&C\end{pmatrix}$ on
${\mathfrak{K}}\oplus{\mathfrak{K}}^{\bot}$, $M$ is defined by

$\left<Mf,f\right>=\inf_{g\in{\mathfrak{K}}^{\bot}}\left<\begin{pmatrix}A&B^{*}\\\
B&C\end{pmatrix}\begin{pmatrix}f\\\ g\end{pmatrix},\begin{pmatrix}f\\\
g\end{pmatrix}\right>,\qquad f\in{\mathfrak{K}}.$ (4.1)

Suppose that $T$ is a Toeplitz operator of finite degree $d$ on
${\mathfrak{H}}=\bigoplus_{0}^{\infty}{\mathfrak{K}}$. By grouping the entries
into $d\times d$ sub-matrices, $T$ can be taken to be tridiagonal. Write
${\tilde{\mathfrak{K}}}$ for $\bigoplus_{0}^{d-1}{\mathfrak{K}}$ and view
${\mathfrak{H}}=\bigoplus_{0}^{\infty}{\tilde{\mathfrak{K}}}$. Then

$T=\begin{pmatrix}A&B^{*}&0&\cdots\\\ B&A&B^{*}&\ddots\\\ 0&B&A&\ddots\\\
\vdots&\ddots&\ddots&\ddots\end{pmatrix}.$ (4.2)

Let $M_{+}$ denote the Schur complement supported on the first copy of
${\tilde{\mathfrak{K}}}$. Then by (4.1) and the Toeplitz structure of $T$ (see
[3] or [5]),

$\hat{A}:=\begin{pmatrix}A-M_{+}&B^{*}\\\ B&M_{+}\end{pmatrix}\geq 0.$ (4.3)

In fact, $T\geq 0$ if and only if there is some $M\geq 0$ such that the
inequality (4.3) holds with $M$ in place of $M_{+}$. In this case, write
$\mathcal{M}$ for the set of positive operators $M$ satisfying (4.3). This set
is norm closed and convex with maximal element equal to the Schur complement
$M_{+}$. There is also a minimal element $M_{-}$ which is constructed by
finding the maximal element $N_{+}$ such that

$\begin{pmatrix}N_{+}&B^{*}\\\ B&A-N_{+}\end{pmatrix}\geq 0$

and setting $M_{-}=A-N_{+}$. Evidently, $N_{+}$ is the Schur complement
supported on the first copy of ${\tilde{\mathfrak{K}}}$ in

$\begin{pmatrix}A&B&0&\cdots\\\ B^{*}&A&B&\ddots\\\ 0&B^{*}&A&\ddots\\\
\vdots&\ddots&\ddots&\ddots\end{pmatrix}.$

In the context of positive Toeplitz operators, Schur complements have a
certain inheritance property, in that if $T_{Q}$ is Toeplitz on
$H^{2}(\mathcal{H})$ and $M_{n}(T_{Q})$ is the Schur complement supported on
the upper left $n\times n$ corner of $T_{Q}$, then
$M_{n}(M_{n+1}(T_{Q}))=M_{n}(T_{Q})$; that is, the Schur complement on the
upper left $n\times n$ corner of the Schur complement on the upper left
$(n+1)\times(n+1)$ corner of $T_{Q}$ is the same as the Schur complement on
the upper left $n\times n$ corner of $T_{Q}$. In addition, if $\deg T_{Q}\leq
n$, then

$M_{n+1}(T_{Q})=\begin{pmatrix}Q_{0}&Q_{1}^{*}&\cdots&Q_{n}^{*}\\\ Q_{1}&&&\\\
\vdots&&M_{n}(T_{Q})&\\\ Q_{n}&&&\end{pmatrix},$ (4.4)

where some $Q_{j}$ may be $0$ if $\deg T_{Q}<n$. This consequently enables the
construction of the Fejér-Riesz factorization in the one variable case, via
the factorization

$\begin{split}&R_{n}(T)=M_{n+1}(T_{Q})-\begin{pmatrix}&&&0\\\
&M_{n}(T_{Q})&&\vdots\\\ &&&0\\\ 0&\cdots&0&0\end{pmatrix}\\\\[3.0pt]
=&\begin{pmatrix}P_{0}^{*}&\cdots&P_{d}^{*}&0&\cdots&\cdots&0\\\
0&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&0\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&P_{d}^{*}\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
0&\cdots&\cdots&\cdots&\cdots&0&P_{0}^{*}\end{pmatrix}\begin{pmatrix}P_{0}&0&\cdots&\cdots&\cdots&\cdots&0\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
P_{d}&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
0&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\ddots&\ddots&\ddots&0\\\
0&\cdots&\cdots&0&P_{d}&\cdots&P_{0}\end{pmatrix}.\end{split}$

See [4] or [5] for more details.

A particularly interesting situation occurs when $\mathcal{M}=\\{M\\}$, a
singleton, which is termed _extremal_. In this case, if there are
factorizations $A-M=E^{*}E$ and $M=F^{*}F$, then $B=F^{*}UE$, where $U$ is
unitary from $\overline{\mathrm{ran}}\,E$ to $\overline{\mathrm{ran}}\,F$.
However such a factorization of $B$ with a unitary for a single element
$M\in\mathcal{M}$ does not necessarily guarantee extremality. In order to
examine this more carefully, the following test is introduced.

######  Lemma 4.2.

Suppose that

$\mathcal{M}=\left\\{M:\begin{pmatrix}A-M&B^{*}\\\ B&M\end{pmatrix}\geq
0\right\\}$

is non-empty with maximal element $M_{+}$, minimal element $M_{-}$, and
$M_{*}=\tfrac{1}{2}(M_{+}+M_{-})$. Let $A-M_{\pm}=E_{\pm}^{*}E_{\pm}$ and
$A-M_{*}=E_{*}^{*}E_{*}$, $M_{\pm}=F_{\pm}^{*}F_{\pm}$ and
$M_{*}=F_{*}^{*}F_{*}$, and $B=F_{\pm}^{*}G_{\pm}E_{\pm}=F_{*}^{*}G_{*}E_{*}$,
where
$G_{\pm}:\overline{\mathrm{ran}}\,E_{\pm}\to\overline{\mathrm{ran}}\,F_{\pm}$
and $G_{*}:\overline{\mathrm{ran}}\,E_{*}\to\overline{\mathrm{ran}}\,F_{*}$
are contractions. The set $\mathcal{M}$ is a singleton if and only if the
operator $G_{+}$, $G_{-}$ and $G_{*}$ are unitary.

###### Proof.

Let $M_{+}$, $M_{-}$ be the maximal and minimal elements of $\mathcal{M}$.
Suppose that $G_{+}$, $G_{-}$ and $G_{*}$ are the corresponding contractions
as in the statement of the lemma. It is straightforward to verify that $G_{+}$
is an isometry and $G_{-}$ is a co-isometry. Hence if $\mathcal{M}$ is a
singleton, $G_{+}=G_{-}=G_{*}$ is unitary.

Conversely, assume that for every $M\in\\{M_{+},M_{-},M_{*}\\}$, there are
factorizations as in the statement of the lemma, where the operators $G_{+}$,
$G_{-}$ and $G_{*}$ are unitary. Without loss of generality, by absorbing
$G_{\pm}$ into $E_{\pm}$ or $F_{\pm}$, it is possible to take $G_{\pm}=1$ on
$\overline{\mathrm{ran}}\,E_{\pm}=\overline{\mathrm{ran}}\,F_{\pm}$.

Since $M_{-}\leq M_{+}$, there exist contractions
$H_{+}:\overline{\mathrm{ran}}\,E_{-}\to\overline{\mathrm{ran}}\,E_{+}$,
$H_{-}:\overline{\mathrm{ran}}\,F_{-}\to\overline{\mathrm{ran}}\,F_{+}$ with
dense ranges such that $E_{+}=H_{-}E_{-}$ and $F_{-}=H_{+}F_{+}$. Let
$H_{-}=V_{-}|H_{-}|$ be the polar decomposition. Then
$E_{+}^{*}E_{+}=E_{+}^{*}V_{-}V_{-}^{*}E_{+}$, so replacing $E_{+}$ by
$V_{-}^{*}E_{+}$ if necessary, take $H_{-}$ to be a positive contraction.
Similarly, assume that $H_{+}$ is also a positive contraction. Since
$H_{\pm}:\overline{\mathrm{ran}}\,E_{-}\to\overline{\mathrm{ran}}\,F_{+}$ and

$\begin{split}B&=F_{-}^{*}E_{-}=F_{+}^{*}H_{+}E_{-}\\\
&=F_{+}^{*}E_{+}=F_{+}^{*}H_{-}E_{-},\end{split}$

it must be that $H_{+}=H_{-}$. Denote this operator by $H$.

The set $\mathcal{M}$ is convex, so

$\begin{split}0&\leq\begin{pmatrix}A-\tfrac{1}{2}(M_{+}+M_{-})&B^{*}\\\
B&\tfrac{1}{2}(M_{+}+M_{-})\end{pmatrix}=\frac{1}{2}\left[\begin{pmatrix}E_{+}^{*}E_{+}&B^{*}\\\
B&F_{+}^{*}F_{+}\end{pmatrix}+\begin{pmatrix}E_{-}^{*}E_{-}&B^{*}\\\
B&F_{-}^{*}F_{-}\end{pmatrix}\right]\\\
&=\frac{1}{2}\left[\begin{pmatrix}E_{-}^{*}H^{2}E_{-}&E_{-}^{*}HF_{+}\\\
F_{+}^{*}HE_{-}&F_{+}^{*}F_{+}\end{pmatrix}+\begin{pmatrix}E_{-}^{*}E_{-}&E_{-}^{*}HF_{+}\\\
F_{+}^{*}HE_{-}&F_{+}^{*}H^{2}F_{+}\end{pmatrix}\right]\\\
&=\frac{1}{2}\begin{pmatrix}E_{-}^{*}&0\\\
0&F_{+}^{*}\end{pmatrix}\begin{pmatrix}1+H^{2}&2H\\\
2H&1+H^{2}\end{pmatrix}\begin{pmatrix}E_{-}&0\\\
0&F_{+}\end{pmatrix}.\end{split}$

Recall that $M_{*}=\tfrac{1}{2}(M_{+}+M_{-})$, $A-M_{*}=E_{*}^{*}E_{*}$,
$M_{*}=F_{*}^{*}F_{*}$, and $B=F_{*}^{*}G_{*}E_{*}$, where by assumption,
$G_{*}:\overline{\mathrm{ran}}\,E_{*}\to\overline{\mathrm{ran}}\,F_{*}$ is
unitary. Then the Schur complement supported on the top left corner of
$\begin{pmatrix}A-M_{*}&B^{*}\\\ B&M_{*}\end{pmatrix}$ is zero. Thus the Schur
complement of the top left corner of $\begin{pmatrix}1+H^{2}&2H\\\
2H&1+H^{2}\end{pmatrix}$ must also be zero. Since $1+H^{2}$ is invertible, it
is a standard fact that this Schur complement equals
$(1+H^{2})-4H(1+H^{2})^{-1}H$, and so $1-2H^{2}+H^{4}=0$. Since $H\geq 0$,
this implies that the only point in the spectrum of $H$ is $\\{1\\}$, and so
$H=1$. From this it follows that $M_{+}=M_{-}$; that is, $\mathcal{M}$ is a
singleton. ∎

Even outside of the extremal situation, it is possible to find $\hat{A}\leq A$
so that by replacing $A$ by $\hat{A}$, the result is extremal.

######  Theorem 4.3.

Suppose that the set

$\mathcal{M}=\left\\{M:\begin{pmatrix}A-M&B^{*}\\\ B&M\end{pmatrix}\geq
0\right\\}$

is not empty. Then there exists $\hat{A}\leq A$ such that the set

$\hat{\mathcal{M}}=\left\\{M:\begin{pmatrix}\hat{A}-M&B^{*}\\\
B&M\end{pmatrix}\geq 0\right\\}$

is a singleton.

###### Proof.

If the set $\mathcal{M}$ is a singleton there is nothing to prove, so assume
this is not the case. Then $\mathcal{M}$ has maximal and minimal elements
$M_{+}$ and $M_{-}$. Set $M_{*}=\tfrac{1}{2}(M_{+}+M_{-})$. By Lemma 4.2, for
$M_{0}$ one of these three, and with $A-M_{0}=E^{*}E$, $M_{0}=F^{*}F$ and
$B=F^{*}GE$, $G:\overline{\mathrm{ran}}\,E\to\overline{\mathrm{ran}}\,F$ a
contraction, the operator $G$ is not unitary. In particular, $G$ is either not
isometric or not co-isometric.

If $G$ is not isometric, set $M_{1}=M_{0}$, and $A_{1}=E^{*}G^{*}GE+M_{0}\geq
0$. Then $A-A_{1}=(A-M_{0})-(A_{1}-M_{1})=E^{*}(1-G^{*}G)E\geq 0$, and is not
$0$, and

$\begin{pmatrix}A_{1}-M_{1}&B^{*}\\\
B&M_{1}\end{pmatrix}=\begin{pmatrix}E^{*}G^{*}GE&E^{*}G^{*}F\\\
F^{*}GE&F^{*}F\end{pmatrix}=\begin{pmatrix}E^{*}G^{*}\\\
F^{*}\end{pmatrix}\begin{pmatrix}GE&F\end{pmatrix}\geq 0.$

If $G$ is isometric but not co-isometric, then choose $M_{1}=F^{*}GG^{*}F$,
and

$A_{1}=A-(M_{0}-M_{1})=A-F^{*}(1-GG^{*})F\geq A-F^{*}F=A-M\geq 0.$

Hence $A\geq A_{1}$ and they are not equal. Also,

$\begin{pmatrix}A_{1}-M_{1}&B^{*}\\\
B&M_{1}\end{pmatrix}=\begin{pmatrix}E^{*}E&E^{*}G^{*}F\\\
F^{*}GE&F^{*}GG^{*}F\end{pmatrix}=\begin{pmatrix}E^{*}\\\
F^{*}G\end{pmatrix}\begin{pmatrix}E&G^{*}F\end{pmatrix}\geq 0.$

Set

$\mathcal{M}_{1}=\left\\{M:\begin{pmatrix}A_{1}-M&B^{*}\\\
B&M\end{pmatrix}\geq 0\right\\},$

and construct $A_{2}$ in an identical manner. Continue in this fashion to
obtain a decreasing net of operators, $A_{0}=A\geq A_{1}\geq
A_{2}\geq\cdots\geq 0$ which then converges strongly to an operator $\hat{A}$.
The set $\\{A_{\alpha}\\}$ may not be countable, but since
$\mathcal{M}\supseteq{\mathcal{M}}_{1}\supseteq{\mathcal{M}}_{2}\supseteq\cdots$,
its cardinality is limited by the cardinality of the set of positive operators
on the space on which $A$ acts. A Zorn’s lemma argument shows that if
$A^{\prime}\leq\hat{A}$ and there is a positive $M$ such that
$\begin{pmatrix}A^{\prime}-M&B^{*}\\\ B&M\end{pmatrix}\geq 0$, then
$A^{\prime}=\hat{A}$. Consequently by the construction given above, for every
$M\in\hat{\mathcal{M}}$ the operator $G$ as in the statement of Lemma 4.2 is
unitary, and so by that lemma, $\hat{\mathcal{M}}$ is a singleton. ∎

Now consider the two index / variable case. Let
$Q(z)=\sum_{j_{2}=-d_{2}}^{d_{2}}{\hat{Q}}_{j_{2}}(z_{1})z_{2}^{j_{2}}$ be an
${\mathfrak{L}}({\mathfrak{K}})$ valued trigonometric polynomial. As in the
last section, there is an associated Toeplitz operator
$T_{Q_{1}}(z_{1}):=({\hat{Q}}_{j_{2}-k_{2}})_{j_{2},k_{2}=0}^{\infty}$, the
entries of which are ${\mathfrak{L}}({\mathfrak{K}})$ valued trigonometric
polynomials in $z_{1}$ of degree at most $d_{1}$. Following the example of the
single variable case, group these into $d_{2}\times d_{2}$ submatrices. Then
with ${\tilde{\mathfrak{K}}}=M_{d_{2}}(\mathbb{C})\otimes{\mathfrak{K}}$, the
entries are ${\mathfrak{L}}({\tilde{\mathfrak{K}}})$ valued trigonometric
polynomials in $z_{1}$ of degree at most $d_{1}$. This in turn is equivalent
to a tridiagonal Toeplitz operator $T$ as in (4.2), with $A$ and $B$ Toeplitz
operators with entries in
${\mathfrak{L}}({\mathfrak{G}})={\mathfrak{L}}(\bigoplus_{0}^{\infty}{\tilde{\mathfrak{K}}})$
of degree at most $d_{1}$. Just to emphasize, the bi-Toeplitz operator
obtained in this way has outer level corresponding to the variable $z_{2}$ and
inner level corresponding to $z_{1}$. In other words, the Toeplitz operators
which are the entries of the tridiagonal Toeplitz operator correspond to
functions in the variable $z_{1}$.

It has been assumed that the operators $A$ and $B$ are singly infinite
Toeplitz operators. If they are instead replaced with the corresponding doubly
infinite Toeplitz operators (so acting on
${\mathfrak{L}}({\mathfrak{G}})=L^{2}({\tilde{\mathfrak{K}}})$) — call them
$\tilde{A}$ and $\tilde{B}$ and the resulting tridiagonal operator $\tilde{T}$
— then $T$ is positive if and only if $\tilde{T}$ is positive. This is
therefore a singly infinite Toeplitz operator with coefficients which are
doubly infinite Toeplitz operators. Write $\tilde{S}$ for the bilateral shift
on ${\tilde{\mathfrak{K}}}$ and suppose that ${\tilde{M}}_{+}$ is the Schur
complement appearing in the resulting version of (4.3). Since $\tilde{A}$ and
$\tilde{B}$ are doubly infinite Toeplitz operators, they are invariant under
conjugation with either $\tilde{S}$ or ${\tilde{S}}^{*}$. Consequently,

$\begin{pmatrix}\tilde{A}-\tilde{S}{\tilde{M}}_{+}{\tilde{S}}^{*}&{\tilde{B}}^{*}\\\
\tilde{B}&\tilde{S}{\tilde{M}}_{+}{\tilde{S}}^{*}\end{pmatrix}\geq 0.$ (4.5)

Hence $\tilde{S}{\tilde{M}}_{+}{\tilde{S}}^{*}\leq{\tilde{M}}_{+}$. On the
other hand, ${\tilde{S}}^{*}\tilde{S}=1$, so conjugating both sides of this
inequality by $\tilde{S}$ gives
${\tilde{M}}_{+}\leq{\tilde{S}}^{*}{\tilde{M}}_{+}\tilde{S}$, and so equality
holds. In other words, the Schur complement in this case is Toeplitz. The same
argument works with the minimal element ${\tilde{M}}_{-}$. In neither case is
it necessary to assume that $\tilde{A}$ and $\tilde{B}$ have finite degree.
There is no immediate guarantee that the degree of $\tilde{M}$ is finite, even
if the degrees of $\tilde{A}$ and $\tilde{B}$ are.

The discussion is summarized in the next lemma.

######  Lemma 4.4.

Suppose that $A$ and $B$ are bounded doubly infinite Toeplitz operators with
entries in ${\mathfrak{L}}({\tilde{\mathfrak{K}}})$ (they need not be of
finite degree). Let
$\tilde{\mathcal{M}}=\\{M:\begin{pmatrix}\tilde{A}-\tilde{M}&{\tilde{B}}^{*}\\\
\tilde{B}&\tilde{M}\end{pmatrix}\geq 0\\}$, and assume that
$\tilde{\mathcal{M}}$ is non-empty. Let $\tilde{\mathcal{M}}_{T}$ be the
subset of elements of $\tilde{\mathcal{M}}$ which are Toeplitz. Then the
maximal and minimal elements of $\tilde{\mathcal{M}}$, ${\tilde{M}}_{+}$ and
${\tilde{M}}_{-}$ are in $\tilde{\mathcal{M}}_{T}$ as maximal and minimal
elements. This set is a closed and convex.

There is a refined version of Theorem 4.3 for doubly infinite Toeplitz
operators.

######  Theorem 4.5.

Let $\tilde{A}$ and $\tilde{B}$ be doubly infinite Toeplitz operators (they
need not be of finite degree), and suppose that the set

$\tilde{\mathcal{M}}=\left\\{M:\begin{pmatrix}\tilde{A}-M&{\tilde{B}}^{*}\\\
\tilde{B}&M\end{pmatrix}\geq 0\right\\}$

is not empty. Then there exists a doubly infinite Toeplitz operator
$\hat{A}\leq\tilde{A}$ such that the set

$\hat{\mathcal{M}}=\left\\{M:\begin{pmatrix}\hat{A}-M&B^{*}\\\
B&M\end{pmatrix}\geq 0\right\\}=\\{\hat{M}\\},$

a singleton. In this case, $\hat{M}$ is a bounded doubly infinite Toeplitz
operator.

###### Proof.

This is essentially a repeat of the proof of Theorem 4.3, taking into account
Lemma 4.4. Let ${\tilde{M}}_{+}$ and ${\tilde{M}}_{-}$ be the maximal and
minimal elements of $\tilde{\mathcal{M}}$, and set
$M_{*}=\tfrac{1}{2}({\tilde{M}}_{+}+{\tilde{M}}_{-})$. All three of these are
doubly infinite Toeplitz operators. If $\tilde{\mathcal{M}}$ is not a
singleton, then for $\tilde{M}_{0}$ equal to one of these,
$\tilde{A}-\tilde{M}_{0}={\tilde{E}}^{*}\tilde{E}$,
$\tilde{M}_{0}={\tilde{F}}^{*}\tilde{F}$. The operators $\tilde{E}$ and
$\tilde{F}$ could for example be square roots. Then
$\tilde{B}={\tilde{F}}^{*}\tilde{G}\tilde{E}$,
$\tilde{G}:\overline{\mathrm{ran}}\,\tilde{E}\to\overline{\mathrm{ran}}\,\tilde{F}$
a contraction, the operator $\tilde{G}$ is not unitary.

If $\tilde{G}$ is not isometric, choose $\tilde{A}_{1}$ so that
$\tilde{A}-\tilde{A}_{1}$ is the Schur complement of the upper left corner of
$\begin{pmatrix}\tilde{A}-\tilde{M}_{0}&{\tilde{B}}^{*}\\\
\tilde{B}&\tilde{M}_{0}\end{pmatrix}$. As in the proof of Lemma 4.4, this
Schur complement is a doubly infinite Toeplitz operator, and so $A_{1}$ has
this form. Here, set $\tilde{M}_{1}=\tilde{M}_{0}$. On the other hand, if
$\tilde{G}$ is isometric but not co-isometric, set $A-A_{1}$ to the Schur
complement of the lower right corner. Again, $A_{1}$ is a doubly infinite
Toeplitz operator. In this case, $\tilde{M}_{1}=M_{0}-(A-A_{1})$.

Consequently, the proof of Theorem 4.3 carries over identically, and the
operators $A\geq A_{1}\geq A_{2}\geq\cdots$ are all doubly infinite Toeplitz
operators, as is the strong limit, $\hat{A}$, and $\hat{\mathcal{M}}$ is a
singleton consisting of a doubly infinite Toeplitz operator. ∎

Of course at this point it is not clear that the degree of $\hat{A}$ in the
theorem is finite, even when the degrees of $\tilde{A}$ and $\tilde{B}$ are.

The arguments just used with doubly infinite Toeplitz operators do not work in
the singly infinite setting with the unilateral shift. Indeed, the Schur
complement in this case will generally not be Toeplitz. If it is, it can be
shown by arguments to follow that it is possible to factor the bivariate
trigonometric polynomial with analytic polynomials of the same degree, and
there are well known examples for which this is not possible [3].
Nevertheless, restricting back to singly infinite operators, the following is
obtained.

######  Lemma 4.6.

Suppose that $A$ and $B$ are bounded (singly infinite) Toeplitz operators with
entries in ${\mathfrak{L}}({\tilde{\mathfrak{K}}})$ (they need not be of
finite degree). Let $\mathcal{M}=\\{M:\begin{pmatrix}A-M&B^{*}\\\
B&M\end{pmatrix}\geq 0\\}$, and assume that $\mathcal{M}$ is non-empty. Then
there exists a closed, convex subset $\mathcal{M}_{T}$ of $\mathcal{M}$, all
the elements of which are Toeplitz. Furthermore, there exists a Toeplitz
operator $\hat{A}\leq A$ such that the set

$\hat{\mathcal{M}}_{T}=\left\\{\hat{M}:\begin{pmatrix}\hat{A}-\hat{M}&B^{*}\\\
B&\hat{M}\end{pmatrix}\geq 0\right\\}$

is a singleton.

The elements of $\mathcal{M}_{T}$ come from restricting the set
$\tilde{\mathcal{M}}_{T}$ of operators $\tilde{M}$ in the doubly infinite
setting satisfying

$\begin{pmatrix}\tilde{A}-\tilde{M}&{\tilde{B}}^{*}\\\
\tilde{B}&\tilde{M}\end{pmatrix}\geq 0.$

The set $\mathcal{M}_{T}$ will therefore have maximal and minimal elements
since $\tilde{\mathcal{M}}$ does, and will be a singleton when
$\tilde{\mathcal{M}}$ is.

As already noted in Lemma 3.2, various properties of Toeplitz operators carry
over to hyper-Toeplitz operators. While it is not clear that extremality
should be preserved upon dilation, one crucial property is retained.

######  Lemma 4.7.

Suppose that $\tilde{A}_{0}$, $\tilde{B}_{0}$, $\tilde{M}_{0}$ are doubly
infinite Toeplitz operators on $L^{2}({\mathfrak{H}})$, the space
${\mathfrak{H}}$ finite dimensional, and that the set

$\left\\{{\tilde{M}_{0}}^{\prime}:\begin{pmatrix}{\tilde{A}}_{0}-{\tilde{M}_{0}}^{\prime}&{\tilde{B}}_{0}^{*}\\\
{\tilde{B}}_{0}&{\tilde{M}_{0}}^{\prime}\end{pmatrix}\geq
0\right\\}=\\{\tilde{M}_{0}\\}.$

Let $\tilde{A}$, $\tilde{B}$, $\tilde{M}$ be the hyper-Toeplitz dilations of
$\tilde{A}_{0}$, $\tilde{B}_{0}$, $\tilde{M}_{0}$, respectively. Let
$\tilde{A}-\tilde{M}={\tilde{E}}^{*}\tilde{E}$,
$\tilde{M}={\tilde{F}}^{*}\tilde{F}$ be hyper-outer factorizations,
$\tilde{B}={\tilde{F}}^{*}\tilde{W}\tilde{E}$, where
$\tilde{W}:\overline{\mathrm{ran}}\,\tilde{E}\to\overline{\mathrm{ran}}\,\tilde{F}$
is hyper-Toeplitz. The operator $\tilde{W}$ is unitary.

###### Proof.

Let $\tilde{A}_{0}-\tilde{M}_{0}={\tilde{E}}_{0}^{*}\tilde{E}_{0}$,
$\tilde{M}_{0}={\tilde{F}}_{0}^{*}\tilde{F}_{0}$ be any factorizations (for
example, they might be in terms of square roots). By Lemma 4.2,
$\tilde{B}_{0}={\tilde{F}}_{0}^{*}\tilde{W}_{0}\tilde{E}_{0}$, where
$\tilde{W}_{0}:\overline{\mathrm{ran}}\,\tilde{E}_{0}\to\overline{\mathrm{ran}}\,\tilde{F}_{0}$
is unitary. The operators $\tilde{E}_{0}$, $\tilde{F}_{0}$ and $\tilde{W}_{0}$
dilate to operators ${\tilde{E}}^{\prime}$, ${\tilde{F}}^{\prime}$ and
${\tilde{W}}^{\prime}$ on $L_{\mathcal{U}}^{2}({\mathfrak{H}})$ given by
constant sequences $(\tilde{E}_{0})$, $(\tilde{F}_{0})$ and $(\tilde{W}_{0}$).
The operator
${\tilde{W}}^{\prime}:\overline{\mathrm{ran}}\,{\tilde{E}}^{\prime}\to\overline{\mathrm{ran}}\,{\tilde{F}}^{\prime}$
is obviously unitary and
$\tilde{B}={\tilde{F}}^{\prime}\,{}^{*}{\tilde{W}}^{\prime}{\tilde{E}}^{\prime}$.

By Theorem 3.3, there are hyper-outer factorizations
$\tilde{A}-\tilde{M}={\tilde{E}}^{*}\tilde{E}$,
$\tilde{M}={\tilde{F}}^{*}\tilde{F}$. By Douglas’ lemma, there are isometries
$U^{\prime}:\overline{\mathrm{ran}}\,\tilde{E}\to\overline{\mathrm{ran}}\,{\tilde{E}}^{\prime}$,
$V^{\prime}:\overline{\mathrm{ran}}\,\tilde{F}\to\overline{\mathrm{ran}}\,{\tilde{F}}^{\prime}$
which are surjective and satisfying
${\tilde{E}}^{\prime}=\tilde{E}U^{\prime}$,
${\tilde{F}}^{\prime}=\tilde{F}V^{\prime}$. Thus
$\tilde{W}:=V^{\prime*}{\tilde{W}}^{\prime}U^{\prime}:\overline{\mathrm{ran}}\,\tilde{E}\to\overline{\mathrm{ran}}\,\tilde{F}$
is unitary. Since $\tilde{B}={\tilde{F}}^{*}\tilde{W}\tilde{E}$, and
$\tilde{B}$, $\tilde{E}$ and $\tilde{F}$ are hyper-Toeplitz, where the
closures of the ranges of $\tilde{E}$ and $\tilde{F}$ are $L^{2}$ spaces,
${\tilde{S}}^{*}\tilde{W}\tilde{S}=\tilde{W}$, as claimed. ∎

##  5\. A two variable factorization theorem

This section contains the proof of the main result, which proceeds by first
dealing with the case where the coefficient space is finite dimensional, and
then extending to positive polynomials with operator coefficients.

######  Theorem 5.1.

Let ${\mathfrak{K}}$ be a Hilbert space. Given a positive
${\mathfrak{L}}({\mathfrak{K}})$ valued trigonometric polynomial $Q$ in a two
variables $z=(z_{1},z_{2})$ of degree $(d_{1},d_{2})$, $Q$ can be factored as
a sum of hermitian squares of at most $2d_{2}$,
${\mathfrak{L}}({\mathfrak{K}})$ valued, analytic polynomials of degree at
most $(d_{1},2d_{2}-1)$.

Of course the roles of $z_{1}$ and $z_{2}$ can be reversed, so as to factor
such a polynomial as a sum of hermitian squares of at most $2d_{1}$,
${\mathfrak{L}}({\mathfrak{K}})$ valued, analytic polynomials of degree at
most $(2d_{1}-1,d_{2})$.

###### Proof of Theorem 5.1.

To begin with, assume that $\dim{\mathfrak{K}}<\infty$. The general case is
dealt with at the end.

As usual, the trigonometric polynomial $Q$ is associated to a bi-Toeplitz
operator $T_{Q}$ in such a way that $T_{Q}$ is Toeplitz of degree $d_{2}$ with
coefficients $R_{0},\dots,R_{d_{2}}$ which are Toeplitz of degree at most
$d_{1}$ acting on ${\mathfrak{G}}=\bigoplus_{0}^{\infty}{\mathfrak{K}}$. These
are collected into $d_{2}\times d_{2}$ blocks, resulting in a tridiagonal
Toeplitz operator on ${\tilde{\mathfrak{G}}}$, which is isomorphic to the
direct sum of $d_{2}$ copies of ${\mathfrak{G}}$. There is then a unitary
operator, call it $V$, which conjugates the entries of this block Toeplitz
operator into Toeplitz operators of degree at most $d_{1}$ with entries acting
on ${\tilde{\mathfrak{K}}}=\bigoplus_{1}^{d_{2}}{\mathfrak{K}}$. The result is
then a bi-Toeplitz operator on $\bigoplus_{0}^{\infty}{\tilde{\mathfrak{K}}}$
with degree $(d_{1},1)$; that is, a Toeplitz operator of degree $1$ with
entries which are Toeplitz operators of degree at most $d_{1}$. Write this
positive operator as

$\begin{pmatrix}A&B^{*}&0&\cdots\\\ B&A&B^{*}&\ddots\\\ 0&B&\ddots&\ddots\\\
\vdots&\ddots&\ddots&\ddots\end{pmatrix}.$

Extend $A$ and $B$ to doubly infinite Toeplitz operators $\tilde{A}$ and
$\tilde{B}$. By Theorem 4.5, there is a doubly infinite Toeplitz operator
$\hat{A}\leq A$ such that the set

$\left\\{M:\begin{pmatrix}\hat{A}-M&{\tilde{B}}^{*}\\\
\tilde{B}&M\end{pmatrix}\geq 0\right\\}=\\{\hat{M}\\},$

a singleton. Dilate $\hat{A}$, $\hat{M}$ and $\tilde{B}$ to hyper-Toeplitz
operators ${\hat{A}}^{\prime}$, ${\hat{M}}^{\prime}$ and
${\tilde{B}}^{\prime}$. The coefficient space for these is
${\tilde{\mathfrak{K}}}_{\mathcal{U}}$. By Theorem 3.3, there are hyper-outer
factorizations ${\hat{A}}^{\prime}-{\hat{M}}^{\prime}=E^{*}E$ and
${\hat{M}}^{\prime}=F^{*}F$. Also, by Lemma 4.7,
${\tilde{B}}^{\prime}=F^{*}\tilde{W}E$, where
$\tilde{W}:\overline{\mathrm{ran}}\,E\to\overline{\mathrm{ran}}\,F$ is hyper-
Toeplitz and unitary.

Let

$C=\begin{pmatrix}{\hat{A}}^{\prime}-{\hat{M}}^{\prime}&0\\\
0&{\hat{M}}^{\prime}\end{pmatrix},\qquad
D=\begin{pmatrix}0&{{\tilde{B}}^{\prime}}{}^{*}\\\
{\tilde{B}}^{\prime}&0\end{pmatrix}.$

Then $D$ is selfadjoint and $\begin{pmatrix}C&D\\\ D&C\end{pmatrix}\geq 0$.
Factor

$D=\begin{pmatrix}E^{*}&0\\\ 0&F^{*}\end{pmatrix}U\begin{pmatrix}E&0\\\
0&F\end{pmatrix},\quad\text{where }U=\begin{pmatrix}0&{\tilde{W}}^{*}\\\
\tilde{W}&0\end{pmatrix}$

is selfadjoint and unitary. Conjugating with unitaries so as to group terms so
that the $(i,j)$ entries of Toeplitz operators in the $2\times 2$ blocks are
placed together in $2\times 2$ matrices, $D$ can be viewed as a hyper-Toeplitz
operator $\tilde{D}$ on
$\bigoplus_{-\infty}^{\infty}({\tilde{\mathfrak{K}}}_{\mathcal{U}}\oplus{\tilde{\mathfrak{K}}}_{\mathcal{U}})$,
$\begin{pmatrix}E&0\\\ 0&F\end{pmatrix}$ can be seen as a hyper-outer operator
$G$ from
$\bigoplus_{-\infty}^{\infty}(\overline{\mathrm{ran}}\,E\oplus\overline{\mathrm{ran}}\,F)$
to
$\bigoplus_{-\infty}^{\infty}({\tilde{\mathfrak{K}}}_{\mathcal{U}}\oplus{\tilde{\mathfrak{K}}}_{\mathcal{U}})$,
and $U$ becomes a hyper-Toeplitz, selfadjoint and unitary operator $\tilde{U}$
on
$\bigoplus_{-\infty}^{\infty}(\overline{\mathrm{ran}}\,E\oplus\overline{\mathrm{ran}}\,F)$.
Furthermore, since the degree of $B$ is at most $d_{1}$, by Lemma 3.2, the
same is true for ${\tilde{B}}^{\prime}$, and hence for $D$ and $\tilde{D}$.

Being selfadjoint and unitary, $\tilde{U}$ has subspaces $\mathcal{P}_{\pm}$
which reduce it and such that $\tilde{U}|\mathcal{P}_{\pm}=\pm
1_{\mathcal{P}_{\pm}}$. Since $U^{2}$ is the identity operator on
$\bigoplus_{-\infty}^{\infty}(\overline{\mathrm{ran}}\,E\oplus\overline{\mathrm{ran}}\,F)$,
and since $\tilde{U}$ is hyper-Toeplitz, $S^{\prime}$, the shift on
$\bigoplus_{-\infty}^{\infty}(\overline{\mathrm{ran}}\,E\oplus\overline{\mathrm{ran}}\,F)$
commutes with the orthogonal projections onto $\mathcal{P}_{\pm}$. In other
words, these spaces reduce $S^{\prime}$. Thus there are orthogonal subspaces
${\mathfrak{L}}_{\pm}\subseteq\overline{\mathrm{ran}}\,E\oplus\overline{\mathrm{ran}}\,F$
such that $\mathcal{P}_{\pm}=\bigoplus_{-\infty}^{\infty}{\mathfrak{L}}_{\pm}$
and
${\mathfrak{L}}_{+}\oplus{\mathfrak{L}}_{-}=\overline{\mathrm{ran}}\,E\oplus\overline{\mathrm{ran}}\,F$.
As a result, $\tilde{U}$ is diagonal; that is, all entries outside of those on
the main diagonal are zero. Hence both $\tilde{U}$ and ${\tilde{U}}^{*}$ are
analytic.

Compress $\tilde{D}$, $G$ and $\tilde{U}$ to their singly infinite variants,
and refer to these with the same symbols. Write $S$ for the unilateral shift
on either
$\bigoplus_{0}^{\infty}(\overline{\mathrm{ran}}\,E\oplus\overline{\mathrm{ran}}\,F)$
or
$\bigoplus_{0}^{\infty}({\tilde{\mathfrak{K}}}_{\mathcal{U}}\oplus{\tilde{\mathfrak{K}}}_{\mathcal{U}})$,
where the space will be clear from the context. Since the degree of
$\tilde{D}$ is at most $d_{1}$, $\tilde{D}S^{d_{1}}$ is analytic. By the
analyticity of $G$ and $\tilde{U}$,

$SG^{*}S^{d_{1}}\tilde{U}G=SG^{*}\tilde{U}GS^{d_{1}}=S\tilde{D}S^{d_{1}}=\tilde{D}S^{d_{1}}S=G^{*}S^{d_{1}}S\tilde{U}G.$

Since $\tilde{U}$ is unitary and $G$ is outer, it follows that
$G^{*}S^{d_{1}}$ is analytic. This implies that the degree of $G$ is at most
$d_{1}$.

Conjugating with the adjoints of the unitaries with which the terms were
grouped above shows that $E$ and $F$ are of degree at most $d_{1}$. Hence
${\hat{A}}^{\prime}$, ${\hat{M}}^{\prime}$ are both of degree at most $d_{1}$.
By Lemma 3.2, the degree of $\hat{A}$ and $\hat{M}$ are at most $d_{1}$. Thus

$H:=\begin{pmatrix}A-\hat{M}&B^{*}\\\
B&\hat{M}\end{pmatrix}=\begin{pmatrix}\hat{A}-\hat{M}&B^{*}\\\
B&\hat{M}\end{pmatrix}+\begin{pmatrix}A-\hat{A}&0\\\ 0&0\end{pmatrix}\geq 0$

has entries which are Toeplitz operators of degree at most $d_{1}$.

Conjugate with an appropriate unitary to make $H$ a Toeplitz operator of
degree at most $d_{1}$ on
$\bigoplus_{0}^{\infty}({\tilde{\mathfrak{K}}}\oplus{\tilde{\mathfrak{K}}})$.
The one variable Fejér-Riesz theorem (Theorem 1.1) implies that this can be
factored as the hermitian square of a single analytic outer operator of the
same degree. Conjugate this by the adjoint of the unitary just used to get

$H=\begin{pmatrix}A-\hat{M}&B^{*}\\\
B&\hat{M}\end{pmatrix}=(F_{ij})^{*}_{i,j=1,2}(F_{ij})_{i,j=1,2},$

where each $F_{i,j}$ is analytic and of degree at most $d_{1}$.

Finally, conjugate the terms of $H$ with the adjoint of the operator $V$ from
the first paragraph of the proof. The operator $F_{ij}$ becomes a $d_{2}\times
d_{2}$ operator $N_{ij}$ with entries that are analytic and of degree at most
$d_{1}$ on ${\mathfrak{G}}$. Write $M=V\hat{M}V^{*}$. Then

$\frac{1}{\sqrt{d_{2}}}\begin{pmatrix}\begin{pmatrix}R_{0}&R_{1}^{*}&\cdots&R_{d_{2}-1}^{*}\\\
R_{1}&\ddots&\ddots&\vdots\\\ \vdots&\ddots&\ddots&R_{1}^{*}\\\
R_{d_{2}-1}&\cdots&R_{1}&R_{0}\end{pmatrix}-M&\begin{pmatrix}R_{d_{2}}^{*}&0&\cdots&0\\\
R_{d_{2}-1}^{*}&\ddots&\ddots&\vdots\\\ \vdots&\ddots&\ddots&0\\\
R_{1}^{*}&\ddots&R_{d_{2}-1}^{*}&R_{d_{2}}^{*}\end{pmatrix}\\\
\begin{pmatrix}R_{d_{2}}&R_{d_{2}-1}&\cdots&R_{1}\\\ 0&\ddots&\ddots&\vdots\\\
\vdots&\ddots&\ddots&R_{d_{2}-1}\\\
0&\ddots&0&R_{d_{2}}^{*}\end{pmatrix}&M\end{pmatrix}=(N_{ij})^{*}_{i,j=1,2}(N_{ij})_{i,j=1,2}.$

For $m=0,\dots,2d_{2}-1$, define polynomials in $z_{2}$ of degree at most
$2d_{2}-1$, $\tilde{F}_{m}$, for which the coefficient of $z_{2}^{k}$ is the
$k$th entry of the $m$th column of $(N_{ij})$. Then

$\tilde{Q}:=\sum_{-d_{2}}^{d_{2}}R_{k}z^{k}=\sum_{0}^{2d_{2}-1}\tilde{F}_{k}^{*}\tilde{F}_{k}.$

Each $R_{j}$ corresponds to an analytic polynomial in $z_{1}$, as do all of
the entries $n_{ijk\ell}$ of each $N_{ij}$ (here $k,\ell$ run from $1$ to
$d_{2}$, and the polynomials have degree at most $d_{1}$ in $z_{1}$). Replace
the entries of $N_{ij}$ by the appropriate polynomials in $z_{1}$, and write
$F_{m}$ for the resulting $2d_{2}$ polynomials of degrees at most
$(d_{1},2d_{2}-1)$ in $(z_{1},z_{2})$. It follows from the discussion towards
the end of Section 2 that $Q=\sum_{k}F_{k}^{*}F_{k}$.

Now suppose that ${\mathfrak{K}}$ is infinite dimensional. Define
$\mathcal{F}$ to be the set of Toeplitz operators $T_{\varphi}$, where
$\varphi$ is a two variable trigonometric polynomial of degree at most
$(d_{1},d_{2})$ and coefficients in ${\mathfrak{L}}({\mathfrak{K}})$ such that
$\|T_{\varphi}\|\leq c_{Q}:=\|T_{Q}\|$. The set $\mathcal{F}$ is norm closed
and convex, and hence weakly closed. Define a subset
$\mathcal{S}\subseteq\mathcal{F}$ of those elements for which
$\varphi=\sum_{m=0}^{2d_{2}}F_{\varphi m}^{*}F_{\varphi m}$, where each
$F_{\varphi m}$ is an analytic ${\mathfrak{L}}({\mathfrak{K}})$ valued
polynomial of degree at most $(d_{1},2d_{2}-1)$.

A weak neighborhood of $T_{Q}$ consists of those operators $R$ on
$H^{2}({\mathfrak{K}})$ such that for a finite set
$\\{(\epsilon_{i},f_{i},g_{i})\in\mathbb{R}^{+}\times
H^{2}({\mathfrak{K}})\times H^{2}({\mathfrak{K}})\\}$,
$\left|\left<(T_{Q}-R)f_{i},g_{i}\right>\right|<\epsilon_{i}$. So to show that
any weak neighborhood of $T_{Q}$ contains an element of $\mathcal{S}$, it
suffices to show that given $\epsilon>0$ and $f,g\in H^{2}({\mathfrak{K}})$,
there is a $T_{\varphi}\in\mathcal{S}$ such that
$\left|\left<(T_{Q}-T_{\varphi})f,g\right>\right|<\epsilon$. Since the
operators $T_{Q}$ and $T_{\varphi}$ are selfadjoint, by polarization it
suffices to show that given $\epsilon>0$ and $f\in H^{2}({\mathfrak{K}})$,
$\left|\left<(T_{Q}-T_{\varphi})f,f\right>\right|<\epsilon$.

Let ${\mathfrak{H}}$ be any finite dimensional subspace of ${\mathfrak{K}}$,
and define $P\in{\mathfrak{L}}(H^{2}({\mathfrak{K}}))$ to be the orthogonal
projection of $H^{2}({\mathfrak{K}})$ onto $H^{2}({\mathfrak{H}})$. By what
was proved above, the operator $PT_{Q}P\in\mathcal{S}$. Conclude that given
any weak neighborhood $\mathcal{V}$ of $T_{Q}$, there is some
$T_{\varphi}\in\mathcal{S}\cap\mathcal{V}$.

Suppose that $(T_{\varphi})$ is a bounded net of elements from $\mathcal{S}$,
and that this net converges weakly to $T_{Q}$. For any $\varphi$, the
coefficients of $\varphi$ have norm bounded by the norm of $T_{\varphi}$, and
so by $c_{Q}$. In particular, the norm of the constant coefficient in any
$\varphi$ is bounded by $c_{Q}$. Since this coefficient equals the sum of the
squares of the coefficients the $F_{\varphi m}$s, where
$\varphi=\sum_{m=0}^{2d_{2}}F_{\varphi m}^{*}F_{\varphi m}$, it follows that
the coefficients of all the $F_{\varphi m}$s lie in the norm closed ball of
radius $c_{Q}$ in $L({\mathfrak{K}})$, which is weakly compact.

Starting with the first coefficient of the first polynomial $F_{\varphi 0}$,
there is a subnet $(T_{\varphi_{1}})$ such that this coefficient converges
weakly to an operator in $L({\mathfrak{K}})$. Then there is a further subnet
of this subnet on which the second coefficient of $F_{\varphi 0}$ converges.
Working through the finite set of polynomials and coefficients, the result is
that there is a subnet $(T_{\varphi_{\beta}})\subset(T_{\varphi})$ in which
for all $m$, $(T_{F_{\beta m}})$ converges weakly to $T_{F_{m}}$. Hence
$Q=\sum_{m=0}^{2d_{2}}F_{m}^{*}F_{m}$. ∎

##  6\. Conclusion

Because of its reliance on ultrapower dilations, the argument used in the
proof of Theorem 5.1 to show that certain Toeplitz operators have finite
degree is far from explicit. However, once it is known that this is the case,
the construction of these operators can be carried out concretely.
Undoubtedly, the coding would involve programming challenges. Roughly speaking
the algorithm begins by following the proof of Theorem 4.5 to determine the
Toeplitz operators $\hat{A}$ and $\hat{M}$. Once this is accomplished, the
proof of Theorem 5.1 gives the steps needed to construct the factorization.

There are numerous applications of Theorem 5.1. For example, following [3],
via a Cayley transform it is possible to construct a rational factorization of
positive operator valued polynomials on $\mathbb{R}^{2}$. For strictly
positive polynomials over $\mathbb{R}^{n}$, such a Cayley transform gives
factorizations involving a restricted class of denominators, although it is
known that for positive semidefinite polynomials, this class may fail to be a
finite [17]. The arguments from [3], imply that not only strictly positive,
but also non-negative operator valued polynomials over $\mathbb{R}^{2}$ can be
factored using a restricted class of denominators, and that a finite set of
denominators works for all polynomials of degree bounded by $(d_{1},d_{2})$.

A number of papers have looked at the problem of factorization for non-
negative trigonometric polynomials in two variables, chiefly in the context of
engineering problems such as filter design. These have tended to restrict to
polynomials having factorizations from the class of stable polynomials; that
is, polynomials with no zeros in the closed bidisk [7, 8, 12], or no zeros in
the closed disk crossed with the open disk [9], or no zeros on a face of the
bidisk [10]. Some deal with the scalar case, and others the operator case.

As noted in the comment following the statement of Theorem 5.1, there at least
two factorizations possible in the two variable setting. Presumably this is
part of some larger family of factorizations. Perhaps there is some special
“central” factorization, though it is unclear by how much, if any, the bounds
on the number of polynomials and their degrees can be improved. The technique
used in [4] to construct outer factorizations, and hence factorizations of
minimal degree, does not generally work in the context of the proof of Theorem
5.1, since in most cases the operator $M$ constructed there is not a Schur
complement.

Finally, there are other related _Positivstellensätze_ in the non-commutative
setting which have not been touched upon (see, for example, [14, 11, 25]). The
lack of a Tarski transfer principle in the non-commutative context has often
been seen as a major stumbling block. The ultrapower construction used here
may offer a way around this.

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