[1]Yash Rastogi

[1]Department of Mathematics, University of Chicago, 5734 S. University
Avenue, Chicago, 60637, IL, USA

# Extremal Growth of Multiple Toeplitz Operators and Implications for
Numerical Stability of Approximation Schemes

<EMAIL_ADDRESS>*

###### Abstract

This paper pioneers the application of reproducing kernel spaces in proving
the stability of numerical methods for approximating linear differential
equations. We analyze families of products of Hardy space Toeplitz operators
that do not fall into a well-understood class of operators through explicit
techniques, many of which are based upon exploiting the reproducing kernel
Hilbert space structure of the Hardy space on the unit disk and properties of
Toeplitz operators. This results in bounds that would not be feasible through
more general functional analytic techniques. In particular, for these
operators, we relate the power bound and a resolvent condition of Kreiss-Ritt
type since it is well-known that such a relationship implies the stability of
certain methods for the numerical solution of linear differential equations.
Our methods apply mutatis mutandis to operators of the form
$T_{g(z)}^{-1}T_{f(z)}T_{g(z)}$ where $f(z)$ is a polynomial in $z$ and
$\overline{z}$ and $g(z)$ is a polynomial in $z$, which arise frequently in
the numerical solution of the Cauchy problem for linear ordinary, partial, and
delay differential equations used as models for processes in science and
engineering.

###### keywords:

Toeplitz operator, Hardy space, Kreiss Matrix Theorem, Numerical stability

###### pacs:

[

MSC Classification]46E22, 47B35, 65M12, 65L20

## Statements and Declarations

  * •

Funding: This work was supported by NSF grant DMS-1851843.

  * •

Competing interests: The author has no financial or proprietary interests in
any material discussed in this article.

##  1 Introduction

A key problem in the stability theory of numerical methods for the solution of
hyperbolic partial differential equations is deriving semigroup estimates from
resolvent-type estimates [13]. The stability theory, due to Gustafsson,
Kreiss, and Sundstrom (GKS), relies on the Laplace transform with respect to
the time variable and consequently the associated stability estimates are
limited to no initial data [11]. Therefore, a major thrust of research is to
incorporate nonzero initial data into these GKS estimates, which are
resolvent-type estimates, to produce semigroup estimates [21]. A fundamental
tool in this analysis, and therefore well-posedness for Cauchy problems in the
theory of partial differential equations, is the Kreiss Matrix Theorem, which
provides the equivalence of uniform boundedness of semigroups and a particular
resolvent estimate for a family of matrices. This issue is subtle as the GKS
stability estimate is equivalent to a slightly stronger version of the Kreiss
resolvent condition, which is the following:

$\sup_{z\in\mathbf{C},|z|>1}(|z|-1)\|(zI-A)^{-1}\|_{\mathcal{L}(\ell^{2}(\mathbf{N}))}<+\infty$
(1)

where T is some bounded operator on $\ell^{2}(\mathbf{N})$ that incorporates
both the discretization of the hyperbolic equation and the numerical boundary
conditions [5]. In finite dimensions, the Kreiss Matrix Theorem asserts the
equivalence of equation 1 and power-boundedness of $A$, which is equivalent to
deriving an optimal semigroup estimate. However, in infinite dimensions, this
equivalence does not generally hold. In fact, the Hille-Yosida resolvent
condition, stated below, is clearly stronger than that in 1, but is still
insufficient to guarantee power-boundedness [5]:

$\sup_{n\geq
1}\sup_{z\in\mathbf{C},|z|>1}(|z|-1)^{n}\|(zI-A)^{-n}\|_{\mathcal{L}(\ell^{2}(\mathbf{N}))}<+\infty$
(2)

An infinite-dimensional analogue of the Kreiss Matrix Theorem, proven by El-
Fallah and Ransford as Corollary 1.3 of [10] is the following:

######  Theorem 1.1.

Let $X$ be a complex Banach space and $A$ be a bounded linear operator on $X$.
Suppose that $A$ satisfies a resolvent condition of the form

$\|(\lambda I-A)^{-1}\|\leq\frac{C}{\operatorname{dist}(\lambda,E)}$

for some compact subset $E$ of the unit circle and some positive scalar $C$.
If $E$ is finite, then

$\sup_{n\geq 0}\|A^{n}\|\leq\frac{e}{2}C^{2}\\#E$

Thus, in infinite dimensions, an operator for which a resolvent condition is
finite is not necessarily power-bounded. To our knowledge, a resolvent
condition that implies power-boundedness in infinite dimensions has not been
found. All that has been established are relationships between the power bound
and a resolvent condition such as that in equation 2. Given the lack of a
general criterion for power-boundedness in the infinite dimensional case, and
motivated by the many relationships between power-bounedness and a particular
resolvent condition being satisfied, we establish such a relationship for
particular families of operators.

In this paper, we prove estimates for two families of multiple Toeplitz
operators on the Hardy space, specifically, composites of the forward and
backward shift operators and the projection onto the Hardy space,
substantially improving upon El-Fallah and Ransford’s general relationship for
infinite-dimensional Banach spaces for these specific cases. Our methods apply
mutatis mutandis to operators of the form $T_{g(z)}^{-1}T_{f(z)}T_{g(z)}$
where $f(z)$ is a polynomial in $z$ and $\overline{z}$ and $g(z)$ is a
polynomial in $z$. This work is of interest since such operators arise in the
numerical solution of the Cauchy problem for linear ordinary, partial, and
delay differential equations that are frequently used as models for processes
in the sciences and engineering. Furthermore, proving the stability of
numerical methods for approximating linear ordinary, partial, and delay
differential equations requires constructing families of operators that
satisfy a particular resolvent condition and have uniformly bounded powers [1,
3, 20].

We investigate families of operators that do not fall into a well-understood
class of operators. In particular, they are not normal, not Toeplitz, and not
contractions. Consequently, standard results on special types of operators are
inapplicable. We utilize the reproducing kernel Hilbert space structure of the
Hardy space on the unit disk as well as properties of Toeplitz operators and
operator norms to compute explicit estimates. This allows us to obtain bounds
that would not be possible through more general functional analytic
techniques. Therefore, this paper pioneers a framework for using reproducing
kernel spaces as a source for new power-bounded operators, which in turn can
be used for approximating broader classes of linear differential equations and
developing novel numerical approximation methods.

The structure of this paper is as follows. In Section 2, we provide background
and a precise formulation of the research problem. In Section 3, we state the
main results of the paper. Sections 4, 5, and 6 are devoted to the proofs of
Theorems 3.1, 3.2, and 3.3, respectively. In Section 7, we detail the
application of this work to the stability analysis of methods for the
numerical solution of the Cauchy problem for linear ordinary, partial, and
delay differential equations.

##  2 Background

###  2.1 Power-Boundedness and the Resolvent Condition

We define the power bound $M(A)$ and a resolvent condition $P(A)$ below. In
this paper, we investigate the relation between $M(A)$ and $P(A)$ for two
classes of operators.

######  Definition 2.1.

Let $A$ be a bounded operator with spectrum $\sigma(A)$ contained within the
closed unit disk $\overline{\mathbf{D}}$ of $\mathbf{C}$.

  1. [a)]

  2. 1.

Set

$M(A):=\sup_{n\geq 0}\|A^{n}\|$

If $M(A)<+\infty$, then $A$ is power-bounded.

  3. 2.

We set

$P(A):=\sup_{|\lambda|>1}\operatorname{dist}(\lambda,\sigma(A))\|(\lambda
I-A)^{-1}\|$

All operators $A$ considered in the remainder of this paper will satisfy the
condition $\sigma(A)\subseteq\overline{\mathbf{D}}$.

######  Remark 2.2.

Diagonal operators illustrate the motivation for the expression $P(A)$. Let
$D=\text{diag}(d_{1},...,d_{n})$ with
$\sigma(D)\subseteq\overline{\mathbf{D}}$. Then,

$\|(zI-D)^{-1}\|=\max_{j}|(z-d_{j})^{-1}|=\frac{1}{\min_{j}|z-d_{j}|}=\frac{1}{\operatorname{dist}(z,\sigma(D))}$

This shows that $\|(zI-A)^{-1}\|$ and $\operatorname{dist}(z,\sigma(A))$
counteract each other and that they do so equally for diagonal operators, as
$P(D)=1$. The association between these quantities can be seen in several
papers such as [8], [17], [19], and [22].

######  Remark 2.3.

For all operators $A$, note that $A^{0}$ is the identity operator, so
$M(A)\geq\|A^{0}\|=1$. Furthermore, all operators $A$ also satisfy $P(A)\geq
1$. We prove this below in the manner of Corollary VII.3.3 of Dunford and
Schwartz [9].

######  Proposition 2.1.

$1\leq P(A)$

###### Proof.

For $|\lambda|>1$, we have that

$\displaystyle P(A)$
$\displaystyle\geq\operatorname{dist}(\lambda,\sigma(A))\|(\lambda
I-A)^{-1}\|$
$\displaystyle\geq\operatorname{dist}(\lambda,\sigma(A))\rho((\lambda
I-A)^{-1})$
$\displaystyle=\operatorname{dist}(\lambda,\sigma(A))\cdot\sup_{\varphi\in\mathfrak{M}}\left|\frac{1}{z-\varphi(A)}\right|$
$\displaystyle=\operatorname{dist}(\lambda,\sigma(A))\cdot\frac{1}{\operatorname{dist}(\lambda,\sigma(A))}$
$\displaystyle=1$

∎

###  2.2 Hardy Space Toeplitz Operators

######  Definition 2.4.

Let $\mathbf{D}\subseteq\mathbf{C}$ denote the open unit disk and let
$\,d\theta$ denote the normalized arc length measure on the unit circle
$\partial\mathbf{D}$. The Hardy space $H^{2}$ is the closed linear span in
$L^{2}(\partial\mathbf{D},\theta)$ of $\\{z^{n}:n\in\mathbf{N}\\}$ where $z$
is the coordinate function on $\partial\mathbf{D}$, meaning
$z(e^{it})=e^{it}$, and $\mathbf{N}$ is the set of nonnegative integers. For
$f\in L^{\infty}(\partial\mathbf{D},\theta)$, the Toeplitz operator with
symbol $f$, denoted $T_{f}$, is the operator on $H^{2}$ defined by

$T_{f}h=P(fh)$

where $P$ denotes the orthogonal projection of
$L^{2}(\partial\mathbf{D},\theta)$ onto $H^{2}$.

The set $\\{z^{n}:n\in\mathbf{N}\\}$ forms an orthonormal basis for $H^{2}$
and the matrix of a Toeplitz operator with respect to it. If $f$ is a function
in $L^{\infty}(\partial\mathbf{D})$ with Fourier coefficients
$\hat{f}(n)=\int_{\partial\mathbf{D}}f\chi_{-n}\,d\theta$, then the matrix
$\\{a_{m,n}\\}_{m,n\in\mathbf{N}}$ for $T_{f}$ with respect to the basis
$\\{\chi_{n}:n\in\mathbf{N}\\}$ is

$a_{m,n}=(T_{f}\chi_{n},\chi_{m})=\int_{\partial\mathbf{D}}f\chi_{n-m}\,d\theta=\hat{f}(m-n)$

The matrix coefficients satisfy $a_{m,n}=a_{m-n,0}$. Such a matrix is called a
Toeplitz matrix.

###  2.3 Reproducing Kernel Hilbert Space

The exposition in this section is based on [15].

######  Definition 2.5.

Given a set $X$, we call $\mathcal{H}$ a reproducing kernel Hilbert space
(RKHS) over $\mathbf{C}$ provided the following hold.

  1. [a)]

  2. 1.

$\mathcal{H}$ is a vector subspace of $\mathcal{F}(X,\mathbf{C})$.

  3. 2.

$\mathcal{H}$ is endowed with an inner product, making it into a Hilbert
space.

  4. 3.

For every $y\in X$, the linear evaluation functional
$E_{y}:\mathcal{H}\to\mathbf{C}$ defined by $E_{y}(f)=f(y)$ is bounded. The
vector that corresponds to this evaluation map is called the kernel vector
corresponding to $g$. An inner product with a kernel vector provides point
evaluation in a RKHS.

The Hardy space $H^{2}$ is unitarily equivalent to the RKHS
$H^{2}(\mathbf{D})$ over the open disk $\mathbf{D}$ in such a way that
coordinate multiplication is preserved. Because of this equivalence, we
identify the spaces $H^{2}(\mathbf{D})$ and $H^{2}$ and carry over the kernel
vectors of $H^{2}(\mathbf{D})$ to $H^{2}$. The kernel vector for
$H^{2}(\mathbf{D})$ for a point $\omega\in\mathbf{D}$ is

$k_{\omega}(z)=\sum_{n=0}^{\infty}\overline{\omega^{n}}z^{n}=\frac{1}{1-\overline{\omega}z}$

because for any $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H^{2}(\mathbf{D})$, the
statement $\langle f,k_{\omega}\rangle=f(\omega)$ holds. For Toeplitz
operators, we have the following property.

######  Proposition 2.2.

If $q$ is analytic and bounded on $\mathbf{D}$, then
$T_{q}^{\ast}k_{\omega}=\overline{q(\omega)}k_{\omega}$ where $k_{\omega}$ is
the kernel vector corresponding to $\omega\in\mathbf{D}$.

###  2.4 Research Problem

Although it is known that $M(A)$ is not controlled by $P(A)$, a relationship
of the form $M(A)\leq CP(A)^{k}$ where $C$ and $k$ are numerical constants
holds for some classes of operators $A$. In this paper, we provide explicit
bounds that relate $M(A)$ and $P(A)$ for the following families of operators
$A$:

  * •

$T_{1+\beta z}^{-1}T_{\overline{z}}T_{1+\beta z}$

  * •

$T_{1+\beta z}^{-1}T_{\frac{z+\overline{z}}{2}}T_{1+\beta z}$

The choice of $1+\beta z$ is due to the following.

######  Remark 2.6.

By linearity of Toeplitz operators,

$T_{a+bz}^{-1}T_{f}T_{a+bz}=T_{1+\frac{b}{a}z}^{-1}T_{f}T_{1+\frac{b}{a}z}$

The two parameters $a$ and $b$ can be encapsulated into a single parameter,
call it $\beta:=\frac{b}{a}$. Thus, we use $1+\beta z$ in the remainder of
this paper for simplicity of notation. The operators we consider have spectrum
in the closed unit disk, so $|\beta|<1$. Furthermore, the case $\beta=0$ is
uninteresting since its undesirable consequence is that $A$ is a Toeplitz
operator. Hence, we study our two families of operators when the parameter
$\beta$ is such that $0<|\beta|<1$.

These two families of operators were selected since they contain infinitely
many points in their spectra and do not fall into a well-understood class of
operators. More precisely, our operators take the form
$A=T_{g}^{-1}T_{f}T_{g}$, where $f$ and $g$ are selected such that $A$ is not
a contraction, not normal, and not Toeplitz. In particular, if $f$ extends to
be analytic on $\mathbf{D}$, then $A$ equals $T_{f}$. Thus, we are interested
in the case where $f$ does not extend analytically to $\mathbf{D}$. For
certain operators of the form of $A$, we have the following result.

######  Proposition 2.3.

Let $A=T_{g}^{-1}T_{f}T_{g}$, $\|f\|_{\infty}\leq 1$, and $g$ analytic in
$\overline{\mathbf{D}}$ such that $\sigma(A)=\overline{\mathbf{D}}$. Then,
$P(A)\leq M(A)$.

###### Proof.

Observe that

$\|(\lambda
I-A)^{-1}\|=\left\|\sum_{j=0}^{\infty}\frac{A^{j}}{\lambda^{j+1}}\right\|\leq\sum_{j=0}^{\infty}\frac{\|A^{j}\|}{|\lambda|^{j+1}}\leq\sum_{j=0}^{\infty}\frac{M(A)}{|\lambda|^{j+1}}=\frac{M(A)}{|\lambda|-1}$

Note that $\sigma(A)=\overline{\mathbf{D}}$, so
$\operatorname{dist}(\lambda,\sigma(A))=|\lambda|-1$. Hence,

$\operatorname{dist}(\lambda,\sigma(A))\|(\lambda I-A)^{-1}\|\leq M(A)$

for all $|\lambda|>1$. In particular,

$P(A)=\sup_{|\lambda|>1}\operatorname{dist}(\lambda,\sigma(A))\|(\lambda
I-A)^{-1}\|\leq M(A)$

∎

###  2.5 Notational Conventions

In this section, we define for the reader some of the conventions employed in
the remainder of the paper.

######  Definition 2.7.

The commutator of two operators $X$ and $Y$ is defined by $[X,Y]=XY-YX$.

The commutator is an indicator of how significantly the operators $X$ and $Y$
fail to commute. It equals zero if and only if $X$ and $Y$ do commute.

######  Definition 2.8.

Given $u$ and $v$ in a Hilbert space $\mathcal{H}$, define the operator
$uv^{\ast}$ on $\mathcal{H}$ by $uv^{\ast}h=\langle h,v\rangle u$.

##  3 Main Results

The main results of this paper are the following three theorems.

######  Theorem 3.1.

Let $A=T_{g}^{-1}T_{\overline{z}}T_{g}$. Then,

$\max\left\\{1,\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}\right\\}\leq M(A)\leq
1+\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$

and

$\max\left\\{1,\frac{c_{0}|\beta|}{\sqrt{1-|\beta|^{2}}}\right\\}\leq
P(A)\leq\min\left\\{M(A),1+\frac{c_{0}|\beta|}{\sqrt{1-|\beta|^{2}}}\right\\}$

where $c_{0}=\frac{\sqrt{2}(\sqrt{5}-1)}{(1+\sqrt{5})^{\frac{3}{2}}}$.

######  Theorem 3.2.

Set

$P_{S}(A):=\sup_{|\lambda|>1}\operatorname{dist}(\lambda,S)\|(\lambda
I-A)^{-1}\|$

where $S\subseteq\sigma(A)=[-1,1]$. Then,

  * •

$P_{\sigma(A)}(A)\leq P_{\\{-1,1\\}}(A)$

  * •

$M(A)\leq eP_{\\{-1,1\\}}(A)^{2}\leq 2eP_{\sigma(A)}(A)^{2}$

######  Theorem 3.3.

Let $A=T_{g}^{-1}T_{\frac{z+\overline{z}}{2}}T_{g}$.

$\max\left\\{\frac{\left|\beta-\overline{\beta}(1-|\beta|^{2})\right|}{2\sqrt{1-|\beta|^{2}}},\frac{\sqrt{1-|\beta|^{2}}}{8}\left|\frac{2+\overline{\beta}^{2}}{1-|\beta|^{2}}-2\overline{\beta}-\overline{\beta}^{3}\right|\right\\}\leq
M(A)\\\ \leq\frac{1+|\beta|}{1-|\beta|}$
$\max\left\\{1,\sqrt{\frac{M(A)}{2e}}\right\\}\leq
P(A)\leq\frac{1+|\beta|}{1-|\beta|}$

The following are corollaries of Theorems 3.1 and 3.3, respectively. They show
that the two families of operators considering in this paper satisfy the
desired form of relationship between the power bound and resolvent condition,
namely, a relationship of the form $M(A)\leq C\cdot P(A)^{k}$, for numerical
constants $C$ and $k$.

######  Corollary 3.1.

Set $A_{\beta}=T_{1+\beta z}^{-1}T_{\overline{z}}T_{1+\beta z}$. As
$|\beta|\to 1^{-}$, $M(A_{\beta})$ and $P(A_{\beta})$ both grow in proportion
to $\frac{1}{\sqrt{1-|\beta|}}$. Hence, for a subsequence of operators in this
family for which $|\beta|\to 1^{-}$, we have
$M(A_{\beta})\to\text{constant}\cdot P(A_{\beta})$.

######  Corollary 3.2.

Set $A_{\beta}=T_{1+\beta z}^{-1}T_{\frac{z+\overline{z}}{2}}T_{1+\beta z}$.
As $|\beta|\to 1^{-}$, we see that $M(A)$ grows in proportion of
$\frac{1}{(1-|\beta|)^{j}}$ for $j\in[\frac{1}{2},1]$, so $P(A)$ grows in
proportion to $\frac{1}{(1-|\beta|)^{\ell}}$ for $\ell\in[\frac{1}{4},1]$.
Therefore, for a subsequence of operators in this family for which $|\beta|\to
1^{-}$, we have that $M(A_{\beta})\to\text{constant}\cdot P(A_{\beta})^{k}$
for some constant $k\in[\frac{1}{2},4]$.

##  4 Proof of Theorem 3.1

###  4.1 Upper Bounds for the Power Bound

######  Proposition 4.1.

If $A=S^{-1}BS$ where $B$ is a contraction, then
$M(A)\leq\|S^{-1}\|\cdot\|S\|$.

###### Proof.

By submultiplicativity of the operator norm,

$M(A)\leq\sup_{n\geq 0}\|S^{-1}\|\cdot\|B^{n}\|\cdot\|S\|\leq\sup_{n\geq
0}\|S^{-1}\|\cdot\|B\|^{n}\cdot\|S\|\leq\|S^{-1}\|\cdot\|S\|$

∎

######  Corollary 4.1.

For $A=T_{g}^{-1}T_{\overline{z}}T_{g}$,

$M(A)\leq\frac{1+|\beta|}{1-|\beta|}$

###### Proof.

Because $\|T_{g}^{-1}\|=\frac{1}{1-|\beta|}$ and $\|T_{g}\|=1+|\beta|$, we
have that $M(A)\leq\frac{1+|\beta|}{1-|\beta|}$. ∎

######  Proposition 4.2.

$M(A)\leq 1+\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$

###### Proof.

Observe that

$\displaystyle[(T_{z}^{\ast})^{n},T_{z}]$
$\displaystyle=T_{\overline{z}}^{n}T_{z}-T_{z}T_{\overline{z}}^{n}$
$\displaystyle=T_{\overline{z}}^{n-1}-(I-e_{0}e_{0}^{\ast})T_{\overline{z}}^{n-1}$
$\displaystyle=e_{0}e_{0}^{\ast}(T_{z}^{n-1})^{\ast}$
$\displaystyle=e_{0}(T_{z}^{n-1}e_{0})^{\ast}$
$\displaystyle=e_{0}e_{n-1}^{\ast}$

Therefore,

$\displaystyle A^{n}$ $\displaystyle=T_{g}^{-1}T_{f}^{n}T_{g}$
$\displaystyle=T_{f}^{n}+T_{g}^{-1}(T_{f}^{n}T_{g}-T_{g}T_{f}^{n})$
$\displaystyle=T_{f}^{n}+T_{g}^{-1}[T_{\overline{z}},T_{g}]$
$\displaystyle=T_{f}^{n}+T_{g}^{-1}([T_{\overline{z}},T_{1}]+\beta[T_{\overline{z}},T_{z}])$
$\displaystyle=(T_{z}^{\ast})^{n}+\beta T_{g}^{-1}[T_{\overline{z}},T_{z}]$
$\displaystyle=(T_{z}^{\ast})^{n}+\beta T_{g}^{-1}e_{0}(e_{n-1}^{\ast})$
$\displaystyle=(T_{z}^{\ast})^{n}+\frac{\beta}{1+\beta z}(e_{n-1}^{\ast})$
$\displaystyle=(T_{z}^{\ast})^{n}+\beta k_{-\overline{\beta}}(e_{n-1}^{\ast})$

By the Triangle Inequality and submultiplicativity,

$\displaystyle\|A^{n}\|$
$\displaystyle\leq\|(T_{z}^{\ast})^{n}\|+|\beta|\cdot\|k_{-\overline{\beta}}(e_{n-1}^{\ast})\|$
$\displaystyle\leq
1+|\beta|\cdot\|k_{-\overline{\beta}}\|\cdot\|(e_{n-1}^{\ast})\|$
$\displaystyle=1+|\beta|\cdot\|k_{-\overline{\beta}}\|$
$\displaystyle=1+\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$

Substituting for $\beta$ gives the result. ∎

###  4.2 Lower Bounds for the Power Bound

######  Proposition 4.3.

$\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}\leq M(A)$

###### Proof.

We consider $\langle T_{\frac{1}{1+\beta z}}T_{\overline{z}}^{n}T_{1+\beta
z}z^{n-1},\widehat{k_{\omega}}\rangle$ where $\omega\in\mathbf{D}$ and
$\widehat{k_{\omega}}=\frac{k_{\omega}}{\|k_{\omega}\|}$ because

$\left|\langle T_{\frac{1}{1+\beta z}}T_{\overline{z}}^{n}T_{1+\beta
z}z^{n-1},\widehat{k_{\omega}}\rangle\right|=\left|\frac{\langle
T_{\frac{1}{1+\beta z}}T_{\overline{z}}^{n}T_{1+\beta
z}z^{n-1},\widehat{k_{\omega}}\rangle}{\|z^{n-1}\|\cdot\|\widehat{k_{\omega}}\|}\right|\leq\|A^{n}\|$

Observe that

$\displaystyle\langle T_{\frac{1}{1+\beta z}}T_{\overline{z}}^{n}T_{1+\beta
z}z^{n-1},\widehat{k_{\omega}}\rangle$ $\displaystyle=\langle
T_{\overline{z}}^{n}T_{1+\beta z}z^{n-1},T_{\frac{1}{1+\beta
z}}^{\ast}\widehat{k_{\omega}}\rangle$ $\displaystyle=\langle
T_{\overline{z}}^{n}T_{1+\beta
z}z^{n-1},\frac{1}{\overline{1+\beta\omega}}\widehat{k_{\omega}}\rangle$
$\displaystyle=\frac{1}{1+\beta\omega}\langle T_{\overline{z}}^{n}T_{1+\beta
z}z^{n-1},\widehat{k_{\omega}}\rangle$
$\displaystyle=\frac{1}{1+\beta\omega}\langle\beta,\widehat{k_{\omega}}\rangle$

Therefore, for any $\omega\in\mathbf{D}$,

$|\langle T_{\frac{1}{1+\beta z}}T_{\overline{z}}^{n}T_{1+\beta
z}z^{n-1},\widehat{k_{\omega}}\rangle|=\frac{|\beta|\sqrt{1-|\omega|^{2}}}{|1+\beta\omega|}$

Now, for $\omega=re^{i\theta}$ we define $\text{sgn}(\omega)=e^{i\theta}$.
Pick $\text{sgn}(\omega)=-\text{sgn}\left(\frac{1}{\beta}\right)$ to obtain

$|\langle T_{\frac{1}{1+\beta z}}T_{\overline{z}}^{n}T_{1+\beta
z}z^{n-1},\widehat{k_{\omega}}\rangle|=\frac{\sqrt{1-|\omega|^{2}}}{\left|\frac{1}{\beta}\right|-|\omega|}$

Using calculus of a real variable, the above expression is maximized at
$\omega=|\beta|$. Substituting this into the above gives
$\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$. Therefore,

$M(A)\geq\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$

∎

######  Proposition 4.4.

$1\leq M(A)$

###### Proof.

This follows from $A$ being a contraction. ∎

###  4.3 Upper Bounds for the Resolvent Condition

######  Proposition 4.5.

$P(A)\leq M(A)$

###### Proof.

See proposition 2.2. ∎

######  Lemma 4.1.

For bounded operators $B$ and $S$,

$(\lambda I-S^{-1}BS)^{-1}=(\lambda I-B)^{-1}+S^{-1}[(\lambda I-S)^{-1},S]$

###### Proof.

Observe that

$\displaystyle(\lambda I-S^{-1}BS)^{-1}$ $\displaystyle=(\lambda
S^{-1}S-S^{-1}BS)^{-1}$ $\displaystyle=[S^{-1}(\lambda I-B)S]^{-1}$
$\displaystyle=S^{-1}(\lambda I-B)^{-1}S$ $\displaystyle=(\lambda
I-B)^{-1}+S^{-1}[(\lambda I-B)^{-1},S]$

∎

######  Lemma 4.2.

$(\lambda
I-A)^{-1}=T_{\frac{1}{\lambda-\overline{z}}}+\frac{\beta}{\lambda^{2}}k_{-\overline{\beta}}k_{\frac{1}{\lambda}}^{\ast}$

###### Proof.

By lemma 4.1,

$(\lambda
I-A)^{-1}=T_{\frac{1}{\lambda-\overline{z}}}+T_{g}^{-1}[T_{\frac{1}{\lambda-\overline{z}}},T_{g}]$

Observe that

$\displaystyle[T_{\frac{1}{\lambda-\overline{z}}},T_{g}]$
$\displaystyle=\beta[T_{\frac{1}{\lambda-\overline{z}}},T_{z}]$
$\displaystyle=\beta\sum_{k=0}^{\infty}\frac{1}{\lambda^{k+1}}[(T_{z}^{\ast})^{k},T_{z}]$
$\displaystyle=\beta\sum_{k=1}^{\infty}\frac{1}{\lambda^{k+1}}e_{0}(e_{k-1}^{\ast})$
$\displaystyle=\beta
e_{0}\left(\sum_{k=1}^{\infty}\frac{1}{\overline{\lambda}^{k+1}}e_{k-1}\right)^{\ast}$
$\displaystyle=\beta
e_{0}\left(\frac{1}{\overline{\lambda}^{2}}\sum_{k=1}^{\infty}\frac{1}{\overline{\lambda}^{k-1}}e_{k-1}\right)^{\ast}$
$\displaystyle=\beta
e_{0}\left(\frac{1}{\overline{\lambda}^{2}}\cdot\frac{1}{1-\frac{z}{\overline{\lambda}}}\right)^{\ast}$

Substituting, we get that

$\displaystyle(\lambda I-A)^{-1}$
$\displaystyle=T_{\frac{1}{\lambda-\overline{z}}}+T_{g}^{-1}\beta
e_{0}\left(\frac{1}{\overline{\lambda}^{2}}\cdot\frac{1}{1-\frac{z}{\overline{\lambda}}}\right)^{\ast}$
$\displaystyle=T_{\frac{1}{\lambda-\overline{z}}}+\frac{\beta}{\lambda^{2}}\left(\frac{1}{1+\beta
z}\right)k_{\frac{1}{\lambda}}^{\ast}$
$\displaystyle=T_{\frac{1}{\lambda-\overline{z}}}+\frac{\beta}{\lambda^{2}}k_{-\overline{\beta}}k_{\frac{1}{\lambda}}^{\ast}$

∎

######  Proposition 4.6.

$P(A)\leq 1+\frac{c_{0}|\beta|}{\sqrt{1-|\beta|^{2}}}$

###### Proof.

Observe that

$\displaystyle\|T_{g}^{-1}T_{\frac{1}{\lambda-\overline{z}}}T_{g}\|$
$\displaystyle\leq\|T_{\frac{1}{\lambda-\overline{z}}}\|+\left\|\frac{\beta}{\lambda^{2}}k_{-\overline{\beta}}k_{\frac{1}{\lambda}}^{\ast}\right\|$
$\displaystyle=\frac{1}{|\lambda|-1}+\frac{|\beta|}{|\lambda|^{2}}\cdot\|k_{-\overline{\beta}}\|\cdot\|k_{\frac{1}{\lambda}}\|$
$\displaystyle=\frac{1}{|\lambda|-1}+\frac{|\beta|}{|\lambda|^{2}}\cdot\frac{1}{\sqrt{1-|\beta|^{2}}\sqrt{1-|\frac{1}{\lambda}|^{2}}}$
$\displaystyle=\frac{1}{|\lambda|-1}+\frac{|\beta|}{|\lambda|\cdot\sqrt{1-|\beta|^{2}}\sqrt{|\lambda|^{2}-1}}$

Therefore,

$(|\lambda|-1)\|T_{g}^{-1}T_{\frac{1}{\lambda-\overline{z}}}T_{g}\|\leq
1+\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}\cdot\frac{|\lambda|-1}{|\lambda|\sqrt{|\lambda|^{2}-1}}$

Since $\sigma(A)=\overline{\mathbf{D}}$, it follows that

$P(A)\leq\sup_{|\lambda|>1}1+\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}\cdot\frac{|\lambda|-1}{|\lambda|\sqrt{|\lambda|^{2}-1}}$

Using single variable calculus, we maximize
$\frac{|\lambda|-1}{|\lambda|\sqrt{|\lambda|^{2}-1}}$ as a function of
$|\lambda|$. The optimal value occurs at $|\lambda|=\frac{1+\sqrt{5}}{2}$ and
hence

$P(A)\leq
1+\left(\frac{\sqrt{2}(\sqrt{5}-1)}{(1+\sqrt{5})^{\frac{3}{2}}}\right)\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$

∎

###  4.4 Lower Bounds for the Resolvent Condition

######  Proposition 4.7.

$P(A)\geq 1$

###### Proof.

Because $A$ is a contraction, the result is immediate from Proposition 1.1. ∎

######  Proposition 4.8.

$\frac{c_{0}|\beta|}{\sqrt{1-|\beta|^{2}}}\leq P(A)$

###### Proof.

First, we estimate the norm of the resolvent of $A$. Recall from Lemma 4.2
that

$(\lambda
I-A)^{-1}=T_{\frac{1}{\lambda-\overline{z}}}+\frac{\beta}{\lambda^{2}}k_{-\overline{\beta}}k_{\frac{1}{\lambda}}^{\ast}$

In this proof, we first estimate the norm of the resolvent and then use that
to estimate $P(A)$. Let $Q(\lambda)$ be given by

$Q(\lambda)=\frac{\langle(T_{\frac{1}{\lambda-\overline{z}}}+\frac{\beta}{\lambda^{2}}k_{-\overline{\beta}}k_{\frac{1}{\lambda}}^{\ast})\lambda^{2}k_{\frac{1}{\lambda}},\beta
k_{-\overline{\beta}}\rangle}{\left(\frac{|\lambda|^{2}}{\sqrt{1-\frac{1}{|\lambda|^{2}}}}\right)\left(\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}\right)}$

We have

$\displaystyle Q(\lambda)$
$\displaystyle=\frac{\sqrt{|\lambda|^{2}-1}\sqrt{1-|\beta|^{2}}}{|\lambda|^{3}|\beta|}\langle
T_{\frac{1}{\lambda-\overline{z}}}k_{\frac{1}{\lambda}},k_{-\overline{\beta}}\rangle\lambda^{2}\overline{\beta}+\frac{\left(1-\frac{1}{|\lambda|^{2}}\right)^{-1}}{\frac{|\lambda|^{2}}{\left(1-\frac{1}{|\lambda|^{2}}\right)^{\frac{1}{2}}}}\frac{|\beta|^{2}\cdot\frac{1}{1-|\beta|^{2}}}{\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}}$
$\displaystyle=\frac{\sqrt{|\lambda|^{2}-1}\sqrt{1-|\beta|^{2}}\lambda^{2}\overline{\beta}}{|\lambda|^{3}|\beta|}\frac{1}{\lambda-\frac{1}{\overline{\lambda}}}\frac{1}{1+\frac{\overline{\beta}}{\overline{\lambda}}}+\frac{1}{|\lambda|\sqrt{|\lambda|^{2}-1}}\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$
$\displaystyle=\frac{\sqrt{|\lambda|^{2}-1}\sqrt{1-|\beta|^{2}}}{|\lambda|^{3}|\beta|}\frac{\overline{\lambda}^{2}\lambda^{2}\overline{\beta}}{(|\lambda|^{2}-1)(\overline{\lambda+\beta})}+\frac{1}{|\lambda|\sqrt{|\lambda|^{2}-1}}\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$
$\displaystyle=\frac{\frac{\overline{\beta}}{|\beta|}|\lambda|\sqrt{1-|\beta|^{2}}}{\sqrt{|\lambda|^{2}-1}(\overline{\lambda}+\overline{\beta})}+\frac{1}{|\lambda|\sqrt{|\lambda|^{2}-1}}\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$
$\displaystyle=\frac{F_{1}(|\lambda|)}{\overline{\lambda}+\overline{\beta}}+F_{2}(|\lambda|)$

where we have defined

$F_{1}(|\lambda|)=\frac{\frac{\overline{\beta}}{|\beta|}|\lambda|\sqrt{1-|\beta|^{2}}}{\sqrt{|\lambda|^{2}-1}}$

and

$F_{2}(|\lambda|)=\frac{1}{|\lambda|\sqrt{|\lambda|^{2}-1}}\cdot\frac{|\beta|}{\sqrt{1-|\beta|^{2}}}$

By definition of $P(A)$, we know that

$|Q(\lambda)|\leq\frac{P(A)}{|\lambda|-1}$

So,

$|Q(\lambda)|^{2}\leq\frac{P(A)^{2}}{(|\lambda|-1)^{2}}$

Taking the average value yields

$\frac{1}{2\pi}\int_{0}^{2\pi}|Q(re^{it})|^{2}\,dt\leq\frac{P(A)^{2}}{(r-1)^{2}}$

Observe that

$\displaystyle\int_{0}^{2\pi}|Q(re^{it})|^{2}\,dt$
$\displaystyle=\int_{0}^{2\pi}\left|\frac{F_{1}(r)}{re^{-it}+\overline{\beta}}+F_{2}(r)\right|^{2}\,dt$
$\displaystyle=\int_{0}^{2\pi}\left|\frac{\frac{1}{r}F_{1}(r)e^{it}}{1+\left(\frac{\overline{\beta}}{r}\right)e^{it}}+F_{2}(r)\right|^{2}\,dt$

Since
$e^{it}\mapsto\frac{\frac{1}{r}F_{1}(r)e^{it}}{1+\left(\frac{\overline{\beta}}{r}\right)e^{it}}$
and $e^{it}\mapsto F_{2}(r)$ are orthogonal as elements of $H^{2}$, it follows
by the Pythagorean Theorem that

$\int_{0}^{2\pi}|Q(re^{it})|^{2}\,dt=\int_{0}^{2\pi}\left|\frac{\frac{1}{r}F_{1}(r)}{1+\frac{\overline{\beta}}{r}e^{it}}\right|^{2}+F_{2}(r)^{2}\,dt$

Since
$\left|\frac{\frac{1}{r}F_{1}(r)}{1+\frac{\overline{\beta}}{r}e^{it}}\right|^{2}\geq
0$, it follows that

$\int_{0}^{2\pi}\left|\frac{\frac{1}{r}F_{1}(r)}{1+\frac{\overline{\beta}}{r}e^{it}}\right|^{2}+F_{2}(r)^{2}\,dt\geq
2\pi\cdot F_{2}(r)^{2}$

Therefore, we have shown that

$F_{2}(r)^{2}\leq\frac{1}{2\pi}\int_{0}^{2\pi}|Q(re^{it})|^{2}\,dt\leq\frac{P(A)^{2}}{(r-1)^{2}}$

and hence

$(r-1)F_{2}(r)\leq P(A)$

for all $r>1$. Through single-variable calculus, we find that the maximum of
$(r-1)F_{2}(r)$ is equal to
$c_{0}=\frac{\sqrt{2}(\sqrt{5}-1)}{(1+\sqrt{5})^{\frac{3}{2}}}$ and occurs
when $r=\frac{1+\sqrt{5}}{2}$. Hence,

$\frac{c_{0}|\beta|}{\sqrt{1-|\beta|^{2}}}\leq P(A)$

∎

##  5 Proof of Theorem 3.2

Here, we prove results inspired by El-Fallah and Ransford [10].

######  Proposition 5.1.

Let $A$ be a bounded operator with the property that $\sigma(A)=[-1,1]$.

$P_{\sigma(A)}(A)\leq P_{\\{-1,1\\}}(A)$

$M(A)\leq eP_{\\{-1,1\\}}(A)^{2}\leq 2eP_{\sigma(A)}(A)^{2}$

###### Proof.

By definition,
$\operatorname{dist}(\lambda,\sigma(A))=\operatorname{dist}(\lambda,[-1,1])$.
Observe that

$\operatorname{dist}(\lambda,A)=\begin{cases}\sqrt{(1-|\text{Re}\lambda|)^{2}+(\text{Im}\lambda)^{2}}&|\text{Re}\lambda|<1\\\
\operatorname{dist}(\lambda,\\{-1,1\\})&|\text{Re}\lambda|\geq 1\end{cases}$

For $\lambda$ such that $|\text{Re}\lambda|<1$, we have
$\operatorname{dist}(\lambda,\\{-1,1\\})=|\text{Im}\lambda|$. Obviously,
$\sqrt{(1-|\text{Re}\lambda|)^{2}+(\text{Im}\lambda)^{2}}\geq\sqrt{(\text{Im}\lambda)^{2}}=|\text{Im}\lambda|$,
so

$\operatorname{dist}(\lambda,\sigma(A))\leq\operatorname{dist}(\lambda,\\{-1,1\\})$

It follows that

$\displaystyle P_{\sigma(A)}(A)$
$\displaystyle=\sup_{|\lambda|>1}\operatorname{dist}(\lambda,\sigma(A))\|(\lambda
I-A)^{-1}\|$
$\displaystyle\leq\sup_{|\lambda|>1}\operatorname{dist}(\lambda,\\{-1,1\\})\|(\lambda
I-A)^{-1}\|$ $\displaystyle=P_{\\{-1,1\\}}(A)$

Now, we prove the first equality of the second result. Recall the result of
El-Fallah and Ransford, which states that for a finite set $S$,

$M(A)\leq\frac{e}{2}C\\#S$

where $P_{S}(A)\leq C$. It follows that

$M(A)\leq\frac{e}{2}P_{S}(A)^{2}\\#S$

Set $S=\\{-1,1\\}$ to get

$M(A)\leq eP_{\\{-1,1\\}}(A)^{2}$

Finally, we prove the second equality of the second result. Write
$\lambda=x+iy$. Our proof has two cases: $|x|\geq 1$ and $|x|<1$. If $|x|\geq
1$, then the point of the spectrum of $A$ that is closest to $\lambda$ is
either of $\\{-1,1\\}$. Therefore, in the case $|x|\geq 1$,

$\operatorname{dist}(\lambda,\\{-1,1\\})=\operatorname{dist}(\lambda,[-1,1])=\sqrt{(|x|-1)^{2}+y^{2}}$

Now, consider the case when $|x|<1$ and $|\lambda|>1$. Without loss of
generality, assume that $x\geq 0$. All such $\lambda$ must satisfy $1-x\leq
y$, hence $(1-x)^{2}\leq y^{2}$, hence $(1-x)^{2}+y^{2}\leq 2y^{2}$, and hence

$\operatorname{dist}(\lambda,\\{-1,1\\})=\sqrt{(1-x)^{2}+y^{2}}\leq\sqrt{2}y=\sqrt{2}\operatorname{dist}(\lambda,[-1,1])$

Therefore,

$\operatorname{dist}(\lambda,\\{-1,1\\})^{2}\leq
2\operatorname{dist}(\lambda,[-1,1])^{2}$

hence

$\sup_{|\lambda|>1}\operatorname{dist}(\lambda,\\{-1,1\\})^{2}\|(\lambda
I-A)^{-1}\|^{2}\leq\sup_{|\lambda|>1}2\operatorname{dist}(\lambda,[-1,1])^{2}\|(\lambda
I-A)^{-1}\|^{2}$

which means that

$P_{\\{-1,1\\}}(A)^{2}\leq 2P_{\sigma(A)}^{2}$

We have shown that

$M(A)\leq eP_{\\{-1,1\\}}(A)^{2}\leq 2eP_{\sigma(A)}(A)^{2}$

and since $P(A)=P_{\sigma(A)}(A)$, it follows immediately that

$\sqrt{\frac{M(A)}{2e}}\leq P(A)$

∎

##  6 Proof of Theorem 3.3

###  6.1 Estimates for the Power Bound

######  Proposition 6.1.

$M(A)\leq\frac{1+|\beta|}{1-|\beta|}$

###### Proof.

By submultiplicativity of the operator norm,

$\|T_{\frac{z+\overline{z}}{2}}^{n}\|=\|T_{\frac{z+\overline{z}}{2}}\|^{n}\leq\left\|\frac{z+\overline{z}}{2}\right\|_{\infty}^{n}=1$

Because $T_{\frac{z+\overline{z}}{2}}$ is a contraction, we can apply
Proposition 4.1 and obtain

$M(A)\leq\|T_{g}^{-1}\|\cdot\|T_{g}\|=\frac{1+|\beta|}{1-|\beta|}$

∎

We now reproduce the result of Proposition 6.1 more explicitly through
considering the powers of the commutator. The following lemma will be utilized
in the computation.

######  Lemma 6.1.

The expansion for $[(T_{z}+T_{z}^{*})^{n},T_{z}]$ has $2(n+1)$ terms.

###### Proof.

Consider a polynomial $p$ with $n$ terms such that the product of any two of
these terms equals the product of the same pair of terms in the reverse order.
By the Stars and Bars Theorem of combinatorics, $p^{k}$ is expressible
(through simplification) to a polynomial with $\binom{n+k-1}{k-1}$ terms.
Since every word in $T_{z}$ and $T_{z}^{\ast}$ can be reduced to a product of
the form $T_{z}^{k}\cdot T_{\overline{z}}^{l}$ or $T_{z}^{l}\cdot
T_{\overline{z}}^{k}$, the expansion of $(T_{z}+T_{z}^{\ast})^{n}$ has twice
as many terms as it would have if $T_{z}$ and $T_{z}^{\ast}$ commuted. Hence,
$(T_{z}+T_{z}^{\ast})^{n}$ has $n+1$ terms, which implies
$[(T_{z}+T_{z}^{*})^{n},T_{z}]$ has $2(n+1)$ terms. ∎

######  Proposition 6.2.

$M(A)\leq\frac{1+|\beta|}{1-|\beta|}$

###### Proof.

$\displaystyle M(A)$ $\displaystyle=\sup_{n\geq
0}\|T_{g}^{-1}T_{f}^{n}T_{g}\|$ $\displaystyle=\sup_{n\geq
0}\|T_{\frac{z+\overline{z}}{2}}^{n}+T_{g}^{-1}[T_{\frac{z+\overline{z}}{2}}^{n},T_{g}]\|$
$\displaystyle\leq\sup_{n\geq
0}\frac{\|T_{z+\overline{z}}^{n}\|+|\beta|\cdot\|T_{g}^{-1}\|\cdot\|[(T_{z}+T_{\overline{z}})^{n},T_{z}]\|}{2^{n}}$
$\displaystyle=1+\frac{|\beta|}{1-|\beta|}\sup_{n\geq
0}\frac{\|[(T_{z}+T_{\overline{z}})^{n},T_{z}]\|}{2^{n}}$ $\displaystyle\leq
1+\frac{|\beta|}{1-|\beta|}\sup_{n\geq 0}\frac{2(n+1)}{2^{n}}$
$\displaystyle=1+\frac{2|\beta|}{1-|\beta|}$

The penultimate step applies Lemma 6.1 and the final step follows from single-
variable calculus. ∎

######  Proposition 6.3.

$M(A)\geq\frac{\left|\beta-\overline{\beta}(1-|\beta|^{2})\right|}{2\sqrt{1-|\beta|^{2}}}$

###### Proof.

Observe that $A=T_{\frac{z+\overline{z}}{2}}+T_{1+\beta
z}^{-1}[T_{\frac{z+\overline{z}}{2}},T_{1+\beta z}]$. Clearly,

$[T_{\frac{z+\overline{z}}{2}},T_{g}]=\frac{\beta}{2}[T_{\overline{z}},T_{z}]=\frac{\beta}{2}(I-T_{z}T_{z}^{\ast})=\frac{\beta}{2}e_{0}e_{0}^{\ast}$

By substitution,

$A=T_{\frac{z+\overline{z}}{2}}+\frac{\beta}{2}k_{-\overline{\beta}}e_{0}^{\ast}$

Hence,

$\displaystyle\|A\|$
$\displaystyle\geq\frac{|\langle(T_{\frac{z+\overline{z}}{2}}+\frac{\beta}{2}k_{-\overline{\beta}}e_{0}^{\ast})e_{0},k_{-\overline{\beta}}\rangle|}{\|k_{-\overline{\beta}}\|\cdot\|e_{0}\|}$
$\displaystyle=\frac{|\langle
T_{\frac{z}{2}}e_{0}+\frac{\beta}{2}k_{-\overline{\beta}}\|e_{0}\|^{2},k_{-\overline{\beta}}\rangle|}{\|k_{-\overline{\beta}}\|}$
$\displaystyle=\frac{|\langle
T_{\frac{z}{2}}e_{0}+\frac{\beta}{2}k_{-\overline{\beta}},k_{-\overline{\beta}}\rangle|}{\|k_{-\overline{\beta}}\|}$
$\displaystyle=\frac{|\langle
1,T_{\frac{z}{2}}^{\ast}k_{-\overline{\beta}}\rangle+\frac{\beta}{2}\langle
k_{-\overline{\beta}},k_{-\overline{\beta}}\rangle|}{\|k_{-\overline{\beta}}\|}$
$\displaystyle=\frac{|\langle
1,\frac{-\beta}{2}k_{-\overline{\beta}}\rangle+\frac{\beta}{2}\|k_{-\overline{\beta}}\|^{2}|}{\|k_{-\overline{\beta}}\|}$
$\displaystyle=\frac{|\frac{-\overline{\beta}}{2}+\frac{\beta}{2}\|k_{-\overline{\beta}}\|^{2}|}{\|k_{-\overline{\beta}}\|}$
$\displaystyle=\frac{1}{2}\left|\beta\|k_{-\overline{\beta}}\|-\frac{\overline{\beta}}{\|k_{-\overline{\beta}}\|}\right|$
$\displaystyle=\frac{1}{2}\left|\beta\left(\frac{1}{\sqrt{1-|\beta|^{2}}}\right)-\overline{\beta}\sqrt{1-|\beta|^{2}}\right|$
$\displaystyle=\frac{|\beta-\overline{\beta}(1-|\beta|^{2})|}{2\sqrt{1-|\beta|^{2}}}$

Since $\|A\|\leq M(A)$,

$M(A)\geq\frac{|\beta-\overline{\beta}(1-|\beta|^{2})|}{2\sqrt{1-|\beta|^{2}}}$

∎

######  Proposition 6.4.

$M(A)\geq\frac{\sqrt{1-|\beta|^{2}}}{8}\cdot\left|\frac{2-\overline{\beta}^{3}}{1-|\beta|^{2}}-2\overline{\beta}-\overline{\beta}^{3}\right|$

###### Proof.

Note that

$\|A^{3}\|\geq\frac{\sqrt{1-|\beta|^{2}}}{8}\cdot\left|\Big{\langle}\left((T_{z}+T_{\overline{z}})^{3}+T_{g}^{-1}[(T_{z}+T_{\overline{z}})^{3}]\right)1,\widehat{k_{-\overline{\beta}}}\Big{\rangle}\right|$

Observe that $[(T_{z}+T_{\overline{z}})^{3},T_{z}]1=z^{2}+2$. Furthermore,
expansion gives us $(T_{z}+T_{\overline{z}})^{3}1=z^{3}+2z$. So,

$\|A^{3}\|\geq\frac{\sqrt{1-|\beta|^{2}}}{8}\cdot\left|\Big{\langle}z^{3}+2z+\frac{z^{2}+2}{1+\beta
z},\widehat{k_{-\overline{\beta}}}\Big{\rangle}\right|$

and hence

$M(A)\geq\frac{\sqrt{1-|\beta|^{2}}}{8}\cdot\left|\frac{2+\overline{\beta}^{2}}{1-|\beta|^{2}}-2\overline{\beta}-\overline{\beta}^{3}\right|$

∎

###  6.2 Estimates for the Resolvent Condition

######  Proposition 6.5.

$1\leq P(A)$

###### Proof.

Since $A$ is a contraction, we apply Proposition 1.1 and obtain this result
immediately. ∎

######  Proposition 6.6.

$\sqrt{\frac{M(A)}{2e}}\leq P(A)$

###### Proof.

Since $f(z)=\frac{z+\overline{z}}{2}$ is real, $T_{f}$ is self-adjoint. This
implies that $\sigma(T_{f})\in\mathbf{R}$. The Hartman-Wintner Theorem, stated
on page 163 of [7], indicates that $\sigma(T_{f})=[-1,1]$. Because $T_{g}$ is
an invertible operator, $\sigma(T_{g}^{-1}T_{f}T_{g})=[-1,1]$. Now, applying
Proposition 5.1, the proof is complete. ∎

######  Proposition 6.7.

$P(A)\leq\frac{1+|\beta|}{1-|\beta|}$

###### Proof.

$\displaystyle\|(\lambda I-A)^{-1}\|$ $\displaystyle=\|(T_{g}^{-1}(\lambda
I-T_{f})T_{g})^{-1}\|$ $\displaystyle=\|T_{g}^{-1}(\lambda
I-T_{f})^{-1}T_{g}\|$
$\displaystyle\leq\|T_{g}^{-1}\|\cdot\|T_{g}\|\cdot\|(\lambda I-T_{f})^{-1}\|$
$\displaystyle=\frac{|1|+|\beta|}{|1|-|\beta|}\cdot\|(\lambda I-T_{f})^{-1}\|$

It follows from Brown and Halmos’ characterization of normal Toeplitz
operators in [7] that $\lambda I-T_{f}$ is normal. The inverse of a normal
operator is also normal, so $(\lambda I-T_{f})^{-1}$ is normal. Now, observe
that

$\|(\lambda I-T_{f})^{-1}\|=\rho((\lambda
I-T_{f})^{-1})=\frac{1}{\operatorname{dist}(\lambda,\sigma(T_{f}))}=\frac{1}{\operatorname{dist}(\lambda,\sigma(A))}$

By substitution,

$P(A)\leq\sup_{|\lambda|>1}\frac{1+|\beta|}{1-|\beta|}(\operatorname{dist}(\lambda,\sigma(A)))\cdot\frac{1}{\operatorname{dist}(\lambda,\sigma(A))}=\frac{1+|\beta|}{1-|\beta|}$

∎

##  7 Stability Analysis of Methods for the Numerical Solution of the Cauchy
Problem for Linear Differential Equations

Several classes of nonlinear dynamical systems, such as time delays [12],
arise frequently in the natural sciences and engineering yet are difficult to
model. Therefore, the stability analysis of numerical approximations for the
differential equations used to model these systems is critical. Proving such
stability results requires demonstrating the power-boundedness of an operator
that results from numerical approximations [3, 6, 18]. As a result, much
recent work has gone into algorithmic approaches for accurate computation of
the Kreiss constant (e.g. [1], [14]) and into sharpening the analytic
conditions required for power-boundedness (e.g. [2], [4]). Proofs of the
numerical stability of approximation methods for particular examples can be
found in [6] and [18] for linear initial value problems and [16] for the
method of lines approach, which is a spectral method (as opposed to a finite
difference method) that reduces partial differential equations in space and
time to a system of ordinary differential equations in time.

Numerical methods for solving linear initial value problems produce equations
that can typically be reduced to the following recursive numerical process:

$u_{n}=Bu_{n-1}+b_{n}$ (3)

where the $b_{n}\in\mathbf{C}^{s}$, $B\in\mathcal{L}(\mathbf{C}^{s})$, and the
$u_{n}\in\mathbf{C}^{s}$ are computed recursively starting from an initial
guess $u_{0}$. We are concerned with determining whether the propagated error
$v_{n}=\tilde{u}_{n}-u_{n}$, resulting from a perturbed initial vector, is
finite. Symbolically, suppose we start with $\tilde{u_{0}}$ instead of
$u_{0}$. By applying 3, we have

$v_{n}=\tilde{u}_{n}-u_{n}=(B\tilde{u}_{n-1}+b_{n})-(Bu_{n-1}+b_{n})=Bv_{n-1}$

so $v_{n}=B^{n}v_{0}$. Thus, for arbitrary $v_{0}\in\mathbf{C}^{s}$, the
sharpest possible bound is $|v_{n}|\leq\|B^{n}\|\cdot|v_{0}|$. Clearly, the
propagated error, and therefore the stability of the numerical approximation,
depends on the uniform boundedness of $\|B^{n}\|$ for all $n$, that is,
$M(B)<+\infty$.

In this paper, we analyze composites of Toeplitz operators. In the numerical
solution of intial value problems, Toeplitz operators arise naturally; for an
example, see Section 3 of [18]. Furthermore, composites of the forward and
backward shift operators are common in differential equation-based models in
science and engineering. Given that the operator that results from numerically
solving a differential equation depends on the particular differential
equation, the numerical method being used, the step size(s) of the
discretization, and the guess $u_{0}$ for the initial state, it is not useful
to contrive specific examples of numerical methods in which a particular
operator with known stability appears (albeit such “backtracking” seems
feasible). Rather, our results provide a framework for using reproducing
kernel spaces to prove the stability of existing numerical methods for new
classes of differential equations as well as for the development of novel
approximation schemes.

Acknowledgments

The author would like to thank Dr. Edward Timko for introducing him to this
problem and mentoring him during his visit at the Georgia Institute of
Technology School of Mathematics.

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