# Implementations of Derivations on the Quantum Annulus

Slawomir Klimek  Department of Mathematical Sciences, Indiana University-
Purdue University Indianapolis, 402 N. Blackford St., Indianapolis, IN 46202,
U.S.A<EMAIL_ADDRESS>,  Matt McBride  Department of Mathematics and
Statistics, Mississippi State University, 175 President’s Cir., Mississippi
State, MS 39762, U.S.A<EMAIL_ADDRESS>and  Kaoru Sakai
Department of Mathematical Sciences, Indiana University-Purdue University
Indianapolis, 402 N. Blackford St., Indianapolis, IN 46202, U.S.A.
<EMAIL_ADDRESS>
###### Abstract.

We construct compact parametrix implementations of covariant derivations on
the quantum annulus.

##  1\. Introduction

The goal of this paper is to provide simple examples of Dirac type operators
on noncommutative compact manifolds. We study analogs of d-bar operators on
the quantum annulus using only inherent geometrical structures: rotations,
invariant states, covariant derivations and their implementations. In our
previous paper [3], we constructed similar d-bar type operators on the quantum
annulus using APS-type boundary conditions. The class of operators was
designed to mimic the classical Atiyah-Patodi-Singer theory and is different,
less geometrical than the class studied in this paper. The main outcome, as in
the past paper, is that we show that our quantum d-bar type operators have
compact parametrices, like elliptic differential operators on compact
manifolds.

Our another paper [5] on quantum annulus, following [4], contains a
description of unbounded derivations, covariant with respect to a natural
rotation, and their implementations in Hilbert spaces obtained from the GNS
construction with respect to invariant states. It turned out that no such
implementation in any GNS Hilbert space for a faithful, normal, invariant
state has compact parametrices for a large class of boundary conditions.
However, as demonstrated in [6], if we relax the concept of an implementation
by allowing operators to act between different Hilbert spaces, then there is
an interesting class of examples of quantum d-bar operators with compact
parametrices that can be constructed this way for the case of the quantum
disk. It is the purpose of this paper to extend those ideas to the quantum
annulus.

Spectral triples are a key tool in noncommutative geometry [1], as they allow
using analytical methods in studying quantum spaces. Since compact parametrix
property is a part of the spectral triples conditions, our papers [4] and [5]
demonstrate, using analytic techniques, that spectral triples, in general,
cannot be constructed on the quantum disk and the quantum annulus using
implementations of covariant derivations in GNS Hilbert spaces. Another,
topological reason was pointed out in [7], in the case of the quantum disk
$\mathcal{T}$. Namely, the pull-back map in K-Homology $K^{0}(C(S^{1}))\to
K^{0}(\mathcal{T})$ is an isomorphism and so the restriction map
$K^{0}(\mathcal{T})\to K^{0}(\mathcal{K})$ is a zero map. Consequently, any
spectral triple over the Toeplitz algebra, when restricted to the ideal of
compact operators $\mathcal{K}$ should be trivial in K-Homology. However, it
is easy to compute that implementations of covariant derivations pair
nontrivially with a minimal projection in $\mathcal{K}$, and hence they cannot
lead to spectral triples over $\mathcal{T}$. Similar arguments seem to also
apply to the quantum annulus.

Additionally, as pointed out in [2], there are fundamental reasons why APS
boundary conditions are not compatible with spectral triples even in classical
geometry for algebras of functions which are non-constant on the boundary, as
the corresponding domains of the Dirac-type operators are not preserved by the
representations of the algebra.

In [6], the authors claimed to construct an even spectral triple over the
quantum disk. Due to technicalities in the definition of an implementation of
a derivation, however, this was not true. We clarify the generalized concept
of implementations of unbounded derivations and when they lead to spectral
triples in the next section. In Section 3 we establish the notation and review
results from [5]. The main result, Theorem 4.3, is proved in Section 4. It
states that, for a class of exponential coefficients, the operator $D$ defined
in equation (3.6) as a suitable Hilbert spaces implementation of a covariant
derivation $\delta$ of (3.2), in the quantum annulus algebra $A$, see (3.1),
has a compact parametrix; in fact the inverse of $D$ is compact.

##  2\. Implementations of Unbounded Derivations

Let $A$ be a $C^{*}$-algebra, $\mathcal{A}\subseteq A$ a dense $*$-subalgebra
and $\delta:\mathcal{A}\mapsto A$ a derivation. Suppose that $H_{1}$ and
$H_{2}$ are Hilbert spaces carrying representations of $A$ and denoted
$\pi_{1}$ and $\pi_{2}$ respectively. The following is a natural concept of an
implementation of a derivation in $A$ between two Hilbert spaces, generalizing
the usual notion of an implementation of an unbounded derivation.

An implementation of $\delta$ between $H_{1}$ and $H_{2}$ consists of the
following:

  * •

A dense subspace dom$(D)\subseteq H_{1}$

  * •

An implementing operator $D:\textrm{dom}(D)\to H_{2}$

  * •

An intertwinner $i:\textrm{dom}(D)\to H_{2}$

such that

  1. (1)

$\forall a\in\mathcal{A}$, $\forall x\in\textrm{dom}(D)$

$\pi_{1}(a)x\in\textrm{dom}(D)$

  2. (2)

$\forall a\in\mathcal{A}$, $\forall x\in\textrm{dom}(D)$

$i(\pi_{1}(a)x)=\pi_{2}(a)i(x)$

  3. (3)

$\forall a\in\mathcal{A}$, $\forall x\in\textrm{dom}(D)$

$D\pi_{1}(a)x-\pi_{2}(a)Dx=\pi_{2}(\delta(a))i(x)$

A special case of the above definition is when $H_{1}=H_{2}=H$,
$\pi_{1}=\pi_{2}=\pi$ and $i$ is the identity map. Then the second condition
above is obviously satisfied, while the third condition can be written as

$(D\pi(a)-\pi(a)D)x=\pi(\delta(a))x.$

This coincides with the usual concept of an unbounded implementation of a
derivation as a commutator.

Recall that a closed operator $D$ is called a Fredholm operator if there are
bounded operators $Q_{1}$ and $Q_{2}$ such that $Q_{1}D-I$ and $DQ_{2}-I$ are
compact. The operators $Q_{1}$ and $Q_{2}$ are called left and right
parametrices respectively. We say that a Fredholm operator $D$ has compact
parametrices if at least one (and consequently both) of the parametrices
$Q_{1}$ and $Q_{2}$ is compact. More on general properties of operators with
compact parametrices can be found in the appendix of [4]. We also say that an
implementation $(\textrm{dom}(D),D,i)$ has compact parametrices if the closure
of the operator $D$ has compact parametrices.

Under additional conditions an implementation of a derivation can lead to an
even spectral triple over $A$. Namely, defining $H=H_{1}\bigoplus H_{2}$, with
grading $\Gamma\big{|}_{H_{1}}=1$ and $\Gamma\big{|}_{H_{2}}=-1$ and a
representation $\pi:A\to B(H)$ of $A$ in $H$ given by the formula:

$\pi(a)=(\pi_{1}(a),\pi_{2}(a)),$

and also defining a generally unbounded operator ${\mathcal{D}}$ in $H$ by:

${\mathcal{D}}=\left[\begin{array}[]{cc}0&D\\\ D^{*}&0\end{array}\right],$

we see that $\pi(a)$ are even and ${\mathcal{D}}$ is odd with respect to
grading $\Gamma$. If ${\mathcal{D}}$ is a self-adjoint operator with compact
parametrix and additionally the intertwinner $i$ is bounded, then the
conditions in the definition of a derivation implementation imply that
$\pi(a)$ preserve the domain of ${\mathcal{D}}$ for all $a\in\mathcal{A}$ and
the commutator $[{\mathcal{D}},\pi(a)]$ is bounded as can be seen by a simple
calculation. Consequently, under those additional conditions, we obtain an
even spectral triple over $A$.

A very natural class of implementations of derivations can be obtained from
GNS representations in the following way. Suppose $\tau_{1}$, $\tau_{2}$ are
faithful states on $A$. Let $H_{1}$, $H_{2}$ be the corresponding GNS Hilbert
spaces, obtained by completing $A$ with respect to the inner products

$(a,b)_{i}=\tau_{i}(a^{*}b),\ \ i=0,1.$

Because we assume that the states are faithful, $A$ sits densely in $H_{1}$,
$H_{2}$. More precisely, there are injective continuous linear maps
$\phi_{1}:A\to H_{1}$, $\phi_{1}:A\to H_{1}$ with dense ranges embedding $A$
into $H_{1}$, $H_{2}$.

The Hilbert spaces $H_{1}$, $H_{2}$ carry natural representations of $A$ given
by left multiplication; for $a,b\in A$ we have

$\pi_{i}(a)\phi_{i}(b)=\phi_{i}(ab).$

Suppose as before that we have a dense $*$-subalgebra $\mathcal{A}\subseteq A$
and a derivation $\delta:\mathcal{A}\mapsto A$. Then we have the following
implementation of $\delta$ between $H_{1}$ and $H_{2}$:

  * •

dom$(D):=\phi_{1}(\mathcal{A})\subseteq H_{1}$. If $x\in\textrm{dom}(D)$ then
we write $x=\phi_{1}(b)$ for some $b\in\mathcal{A}$, notation we use in the
formulas below.

  * •

$D(x):=\phi_{2}(\delta(b))$,

  * •

$i(x):=\phi_{2}(b)$

It is a matter of straightforward calculations to verify that indeed the three
conditions of the definition are satisfied and the above defines an
implementation of $\delta$ between $H_{1}$ and $H_{2}$.

##  3\. Quantum Annulus Preliminaries

We review the notation and basic concepts from [5] below and, in a number of
places, we use the results contained in that paper.

###  3.1. The Quantum Annulus

Let $\\{E_{l}\\}_{l\in{\mathbb{Z}}}$ be the canonical basis for
$\ell^{2}({\mathbb{Z}})$ and $V$ be the bilateral shift defined by

$VE_{l}=E_{l+1}\,.$

Notice that $V$ is a unitary. Let $\mathbb{L}$ be the diagonal label operator
defined by

$\mathbb{L}E_{l}=lE_{l}\,.$

It follows from the functional calculus that given a function
$a:{\mathbb{Z}}\to{\mathbb{C}}$, we have

$a(\mathbb{L})E_{l}=a(l)E_{l}\,.$

These are precisely the operators which are diagonal with respect to
$\\{E_{l}\\}$. The operators $(\mathbb{L},V)$ serve as noncommutative polar
coordinates, and they satisfy the following commutation relation:

$\mathbb{L}V=V(\mathbb{L}+I)\,.$

Let $c({\mathbb{Z}})$ be the set of $a(l)$, as above, which are convergent as
$l\to\pm\infty$ and let $c_{00}^{++}({\mathbb{Z}})$ be the set of all
eventually constant functions, i.e. functions $a(l)$ such that there exists a
$l_{0}$ where $a(l)$ is a constant for $l\geq l_{0}$ and also is a possibly
different constant for $l\leq-l_{0}$.

Let $A$ be the $C^{*}$-algebra generated by $V$ and $a(\mathbb{L})$, that is:

$A=C^{*}(V,a(\mathbb{L}):a(l)\in c({\mathbb{Z}}))$ (3.1)

This algebra is called the quantum annulus. The smallest reasonable domain of
derivations in $A$ is the following dense $*$-subalgebra of A:

$\mathcal{A}=\left\\{a=\sum_{n\in{\mathbb{Z}}}V^{n}a_{n}(\mathbb{L}):\
a_{n}(l)\in c_{00}^{++}({\mathbb{Z}}),\ \textrm{finite sums}\right\\}\,.$

###  3.2. Derivations in the Quantum Annulus

Let $\rho_{\theta}:A\to A$, $0\leq\theta<2\pi$, be a one parameter group of
automorphisms of A defined by:

$\rho_{\theta}(a)=e^{i\theta\mathbb{L}}ae^{-i\theta\mathbb{L}}.$

Since $\rho_{\theta}(a(\mathbb{L}))=a(\mathbb{L})$,
$\rho_{\theta}(V)=e^{i\theta}V$ and consequently
$\rho_{\theta}(V^{-1})=e^{-i\theta}V^{-1}$, the automorphisms $\rho_{\theta}$
are well defined on $A$ and they preserve $\mathcal{A}$. By Proposition 3.2 in
[5], any densely-defined derivation $\delta:\mathcal{A}\to A$, covariant with
respect to $\rho_{\theta}$ that is

$\rho_{\theta}(\delta(a))=e^{i\theta}\delta(\rho_{\theta}(a))\,,$

is of the following form:

$\delta(a)=[U\beta(\mathbb{L}),a]\,$ (3.2)

where $\\{\beta(l+1)-\beta(l)\\}\in c({\mathbb{Z}})$. We use notation:

$\lim_{l\to\pm\infty}(\beta(l+1)-\beta(l)):=\beta_{\pm\infty},$

and below we only consider covariant derivations with $\beta_{\pm\infty}\neq
0$. It follows that there are constants $c_{1}$ and $c_{2}$ so that

$c_{1}(|l|+1)\leq|\beta(l)|\leq c_{2}(|l|+1)\,.$ (3.3)

###  3.3. Covariant Implementations on the Quantum Annulus

Here we consider covariant implementations of derivations (3.2). We begin by
introducing the following family of states $\tau_{w}:A\to{\mathbb{C}}$ on $A$,
defined by

$\tau_{w}(a)=\textrm{tr}(w(\mathbb{L})a)\,,$

where $w(l)>0$ for all $l\in{\mathbb{Z}}$ and

$\sum_{l\in{\mathbb{Z}}}{w(l)}=1\,.$

As a result of Proposition 4.3 in [5], $\tau_{w}$ are precisely the
$\rho_{\theta}$-invariant, normal, faithful states on A. Let $H_{w}$ be the
Hilbert space obtained by Gelfand-Naimark-Segal (GNS) construction on $A$
using state $\tau_{w}$. Since the state is faithful, $H_{w}$ is the completion
of $A$ with respect to the inner product given by

$\langle a,b\rangle_{w}=\tau_{w}(a^{*}b)\,.$

A simple calculation leads to the following precise description:

$H_{w}=\left\\{f=\sum_{n\in{\mathbb{Z}}}V^{n}f_{n}(\mathbb{L}):\|f\|_{w}^{2}=\sum_{n\in{\mathbb{Z}}}\sum_{l\in{\mathbb{Z}}}w(l)|f_{n}(l)|^{2}<\infty\right\\}\,.$
(3.4)

With this identification we naturally have $A\subseteq H_{w}$ so the inclusion
maps $\phi_{w}:A\to H_{w}$ are the identity maps. Notice also that
$\mathcal{A}$ is dense in $H_{w}$. The GNS representation map $\pi_{w}:A\to
B(H_{w})$ is given by left-hand multiplication:

$\pi_{w}(a)f=af\,.$

Define a one parameter group of unitary operators $V_{\theta}^{w}:H_{w}\to
H_{w}$ via the formula:

$V_{\theta}^{w}f=\sum_{n\in{\mathbb{Z}}}V^{n}e^{in\theta}f_{n}(\mathbb{L})\,.$

An immediate calculation shows that:

$\pi_{w}(\rho_{\theta}(a))=V_{\theta}^{w}\pi_{w}(a)(V_{\theta}^{w})^{-1}\,,$

and therefore the operators $V_{\theta}^{w}$ are implementing the one
parameter group of automorphisms $\rho_{\theta}$.

Consider an additional weight, $w^{\prime}(l)$, possibly different from
$w(l)$, satisfying the same conditions. Proceeding like in the previous
section, we set

$\textrm{dom}(D):=\mathcal{A}\subset H_{w}\,,$

and choose for an implementing operator

$i:\textrm{dom}(D)=\mathcal{A}\to\mathcal{A}\subset H_{w^{\prime}}\,$

to be the identity operator $a\mapsto a$. Clearly, the first two properties of
an implementation are satisfied. We say that an operator
$D:H_{w}\supseteq\mathcal{A}\to H_{w^{\prime}}$ defines a covariant
implementation of the covariant derivation (3.2) if for every
$a\in\mathcal{A}$, and for every $f\in\mathcal{A}$ considered as an element of
both $H_{w}$ and $H_{w^{\prime}}$, we have:

$D\pi_{w}(a)f-\pi_{w^{\prime}}(a)Df=\pi_{w^{\prime}}(\delta(a))f\,,$

and, additionally, $D$ satisfies:

$V_{\theta}^{w^{\prime}}D(V_{\theta}^{w})^{-1}f=e^{i\theta}Df\,.$

Proceeding as in Proposition 5.2 in [5] shows the following result.

######  Proposition 3.1.

There exists a sequence $\\{\alpha(l)\\}$ satisfying

$\sum_{l\in{\mathbb{Z}}}|\beta(l)-\alpha(l)|^{2}w^{\prime}(l)<\infty$ (3.5)

such that any covariant implementation $D:H_{w}\supseteq\mathcal{A}\to
H_{w^{\prime}}$ is of the form:

$Df=V\beta(\mathbb{L})f-fV\alpha(\mathbb{L})\,.$ (3.6)

Conversely, for any $\\{\alpha(l)\\}$ satisfying (3.5), the formula (3.6)
defines a covariant implementation $D:H_{w}\supseteq\mathcal{A}\to
H_{w^{\prime}}$ of the derivation (3.2).

We assume below that for every $l$ we have:

$\alpha(l),\beta(l)\neq 0\,.$

It is convenient, like in [4] and [5], to write

$\alpha(l)=\beta(l)\frac{\mu(l+1)}{\mu(l)}$

for some sequence $\\{\mu(l)\\}$ such that $\mu(0)=1$.

###  3.4. Fourier Decomposition

To further analyze the operator $D$ of formula (3.6), we can decompose it into
a Fourier series and study its Fourier components which are operators acting
between weighted $\ell^{2}$-spaces, defined as follows:

$\ell^{2}_{w}=\left\\{\\{f(l)\\}_{l\in{\mathbb{Z}}}:\sum_{l\in{\mathbb{Z}}}|f(l)|^{2}w(l)<\infty\right\\}\,.$

We have the following decomposition proposition:

######  Proposition 3.2.

Let $f\in\textrm{dom}(D)$. Then

$Df=\sum_{n\in{\mathbb{Z}}}V^{n+1}(D_{n}f_{n})(\mathbb{L})\,,$

where $D_{n}:\ell_{w}^{2}\supseteq
c_{00}^{++}({\mathbb{Z}})\to\ell_{w^{\prime}}^{2}$ and is given by the
following formula:

$(D_{n}h)(l)=\beta(l+n)h(l)-\beta(l)\frac{\mu(l+1)}{\mu(l)}h(l+1)$

for some $h\in c_{00}^{++}({\mathbb{Z}})$.

###### Proof.

The proof follows by writing $f\in\textrm{dom}(D)$ as its Fourier series,
applying $D$ to it and using the commutation relation
$\mathbb{L}V=V(\mathbb{L}+I)$. ∎

In what follows, the purpose is to choose the parameters such that $D$ has a
compact parametrix.

###  3.5. Parametrices

Next we study a formal candidate for a parametrix of $D$. The closure of $D$,
defined as above on $c_{00}^{++}({\mathbb{Z}})$, will be denoted by $\bar{D}$
while its closure defined on the space $c_{00}({\mathbb{Z}})$ of eventually
zero functions will be denoted by $\bar{D}_{00}$. Also notice that $D$
preserves $c_{00}({\mathbb{Z}})$. We have the following simple observation.

######  Proposition 3.3.

If $\\{\alpha(l)\\}$ satisfies (3.5) then $\bar{D}=\bar{D}_{00}$.

###### Proof.

Notice that $c_{00}({\mathbb{Z}})\subset\textrm{dom}(D)$ is a co-dimension 2
subspace. Thus, it is enough to verify that 1 and the characteristic function
of ${\mathbb{Z}}_{\geq 0}$ are in the domain of $\bar{D}$. Approximating 1 by
characteristic functions $\chi_{N}$ of sets $-N\leq l\leq N$, $\chi_{N}\in
c_{00}({\mathbb{Z}})$, we see that $D(\chi_{N})$ converges in $\ell_{w}^{2}$
to $D(1)$, which is in $\ell_{w}^{2}$ by (3.5), implying that 1 is in the
closure of $D$. The characteristic function of ${\mathbb{Z}}_{\geq 0}$ is in
the domain of $\bar{D}$ by the same argument. ∎

Let $Q_{n}$ be given by the following formula:

$(Q_{n}g)(l)=\left\\{\begin{aligned}
&\sum_{j=l}^{\infty}\frac{\prod_{k=l}^{l+n-1}\beta(k)}{\prod_{k=j}^{j+n}\beta(k)}\cdot\frac{\mu(j)}{\mu(l)}g(j)&&n\geq
0\\\
&\sum_{j=l}^{\infty}\frac{\prod_{k=j+n+1}^{j-1}\beta(k)}{\prod_{k=l+n}^{l-1}\beta(k)}\cdot\frac{\mu(j)}{\mu(l)}g(j)&&n<0.\end{aligned}\right.$

This expression was obtained by inverting $D_{n}$ using techniques similar to
the calculations in Proposition 4.12 in [3]. Notice that we have:

$Q_{n}:c_{00}({\mathbb{Z}})\to\textrm{dom}(D),$

since the sums in the definitions of $Q_{n}$ are finite for $g\in
c_{00}({\mathbb{Z}})$ and the outcomes are eventually constant. Relations
between $D_{n}$ and $Q_{n}$ are explained in the following statements.

######  Proposition 3.4.

For every $f\in c_{00}({\mathbb{Z}})$ we have:

$D_{n}Q_{n}f=f\ \textrm{ and }\ Q_{n}D_{n}f=f\,.$

###### Proof.

The formulas follow from straightforward calculations. ∎

From this proposition we can formally define the inverse for $D$ to be
$Q=\sum_{n\in{\mathbb{Z}}}Q_{n}$. For this to be well-defined, the series
needs to converge. In fact, once we verify that $Q$ is bounded, the previous
two propositions imply that $Q$ is the inverse of $\bar{D}$.

##  4\. Results

For the remainder of this section we assume that $\beta(k)=k+\frac{1}{2}$.
Moreover we only consider the special choices of the weights $\\{w(l)\\}$,
$\\{w^{\prime}(l)\\}$ and the choice for $\\{\mu(l)\\}$ namely:

$w(l)=e^{-a|l|},\quad w^{\prime}(l)=e^{-b|l|},\quad\textrm{and
}\mu(l)=e^{-(\gamma l)/2},\textrm{ for }a,b,\gamma>0\,.$

It should be noted that with these specific choices
$\alpha(l)=e^{-\gamma/2}\beta(l)$ and the conditions in Proposition 3.1 are
trivially satisfied. By a simple perturbative argument the results below are
valid for a much larger class of coefficients, see the remark at the end of
the next subsection.

###  4.1. Compactness of Parametrices

We show that $Q_{n}$ are Hilbert-Schmidt operators for every $n$ and verify
that their respective Hilbert-Schmidt norms go to zero as $n$ goes to
infinity, implying that $Q$ is a compact operator. To prove this we need a few
helper lemmas. We postpone proofs of those lemmas until the next subsection.

######  Lemma 4.1.

For $0\leq l\leq n$, define the following product:

$q_{n}(l)=\frac{\left(\frac{1}{2}\right)^{2}\left(\frac{3}{2}\right)^{2}\cdots\left(n-\frac{1}{2}\right)^{2}}{\left(l-\frac{1}{2}\right)^{2}\left(l-\frac{3}{2}\right)^{2}\cdots\left(l-n+\frac{1}{2}\right)^{2}}\,.$

Then we have the identity:

$q_{n}(l)=\frac{((2n-2l+1)\cdots(2n-3)(2n-1))^{2}}{(1\cdot 3\cdot
5\cdots(2l-1))^{2}}\textrm{ for every natural number }n\,.$

Moreover, $q_{n}(l)$ satisfies the following three estimates:

$(1)\ q_{n}(l)\geq 1,\quad(2)\ q_{n}(l)\leq 2l\left(\begin{array}[]{c}2n\\\
2l\end{array}\right),\quad(3)\
\frac{q_{n}(j)}{q_{n}(l)}\leq\left(\begin{array}[]{c}2n\\\
2l\end{array}\right)\textrm{ for }0\leq j\leq l\,.$

######  Lemma 4.2.

For a nonnegative integer $j$, define the following sum:

$J_{n}(j)=\sum_{k\geq
0}\frac{\left(k+j+\frac{1}{2}\right)^{2}\cdots\left(k+j+n-\frac{1}{2}\right)^{2}e^{-\gamma
k}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\,.$

Then for $n\geq 0$

$J_{n}(j)\leq\frac{2n+1}{\left(1-e^{-\frac{\gamma}{2}}\right)^{2n+1}}\,.$

The following is the main technical result of the paper.

######  Theorem 4.3.

Suppose that $\gamma>a>b$ and that

$\textrm{exp}\left(-\frac{(a-b)}{2}\right)+\textrm{exp}\left(-\frac{\gamma+a}{2}\right)<1\,.$

Then $Q_{n}:\ell_{w^{\prime}}^{2}\to\ell_{w}^{2}$ is a Hilbert-Schmidt
operator for every $n\in{\mathbb{Z}}$ and $\|Q_{n}\|_{\textrm{HS}}\to 0$ as
$n\to\pm\infty$. Consequently, $Q=\sum_{n}Q_{n}$ is the inverse of $\bar{D}$
and is a compact operator.

###### Proof.

Notice that formula for $Q_{n}$ shows that it is an integral operator.
Therefore, by direct calculation we can compute the Hilbert-Schmidt norms:

$\|Q_{n}\|_{\textrm{HS}}^{2}=\left\\{\begin{aligned}
&\sum_{j=l}^{\infty}\frac{\prod_{k=l}^{l+n-1}|\beta(k)|^{2}}{\prod_{k=j}^{j+n}|\beta(k)|^{2}}\cdot\frac{|\mu(j)|^{2}}{|\mu(l)|^{2}}\cdot\frac{w(l)}{w^{\prime}(j)}&&n\geq
0\\\
&\sum_{j=l}^{\infty}\frac{\prod_{k=j+n+1}^{j-1}|\beta(k)|^{2}}{\prod_{k=l+n}^{l-1}|\beta(k)|^{2}}\cdot\frac{|\mu(j)|^{2}}{|\mu(l)|^{2}}\cdot\frac{w(l)}{w^{\prime}(j)}&&n<0\end{aligned}\right.$

Since the choice of $\beta$’s are linear, the following ratios:

$\left|\frac{\beta(l)\cdots\beta(l+n-1)}{\beta(l+n)\cdots\beta(l-1)}\right|^{2}\textrm{
and
}\left|\frac{\beta(j+n+1)\cdots\beta(j-1)}{\beta(l+n)\cdots\beta(l-1)}\right|^{2}$

are ratios of polynomials. Thus by our choice of exponential $\mu$’s, $w$’s
and $w^{\prime}$’s, $\|Q_{n}\|_{\textrm{HS}}$ exists if and only if

$\sum_{j\geq
l}\left|\frac{\mu(j)}{\mu(l)}\right|^{2}\cdot\frac{w(l)}{w^{\prime}(j)}<\infty\,.$

However, this easily follows if $\gamma>a>b$. Thus the Hilbert-Schmidt norm of
$Q_{n}$ exists for all $n\in{\mathbb{Z}}$. It remains to show that those norms
go to zero as $n\to\pm\infty$. We only need to study the case for $n\geq 0$ as
if $n<0$ then by doing a change of variables of $j\mapsto-l$, $l\mapsto-j$ and
$n\mapsto-n-1$ we are back in the $n\geq 0$ case. Thus, we only need to
estimate the sum:

$\|Q_{n}\|_{\textrm{HS}}^{2}=\sum_{j\geq
l}\frac{\left(l+\frac{1}{2}\right)^{2}\cdots\left(l+n-\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\cdot\frac{e^{-\gamma(j-l)-a|l|+b|j|}}{\left(j+n+\frac{1}{2}\right)^{2}}\,.$

The sum is over all $j\geq l$. It splits into sums over four main regions
which we will further subdivide as illustrated in the picture below:

We have

$\|Q_{n}\|_{\textrm{HS}}^{2}\leq S_{A}+S_{B}+S_{C}+S_{D},$

as for simplicity of estimations we let the regions overlap. Here the regions
are: region A: $j\geq l\geq 0$, region B: $j\geq-l\geq 0$, $l\leq 0$, region
C: $-l\geq j\geq 0$, $l\leq 0$, and region D: $l\geq j\geq 0$, which we will
handle separately. In our estimates, we double count the boundaries in some
places for convenience of different estimates.

Regions A and B: Notice the following observation: if $-j\leq l\leq j$ for
$j\geq 0$, then for any $r>0$ we have that $(l+r)^{2}\leq(j+r)^{2}$. Using
this fact, we can estimate:

$\frac{\left(l+\frac{1}{2}\right)^{2}\cdots\left(l+n-\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\leq
1.$

It follows that we have

$S_{A}\leq\sum_{j\geq l\geq
0}\frac{e^{-\gamma(j-l)-al+bj}}{\left(j+n+\frac{1}{2}\right)^{2}}\to 0\textrm{
as }n\to\infty\,.$

Changing $l\to-l$ we get exactly the same estimate for $S_{B}$, and so
$S_{B}\to 0$ as $n\to\infty$.

Region C: $0\leq j\leq-l$.

First we map $l\mapsto-l$, then we split this region into two sub-regions: C1:
$0\leq j\leq l\leq n$ and the complement of C1: $0\leq j\leq l$, $l\geq n$. In
the second region we make a change of variables of $l^{\prime}=l-n$, then
replace $l^{\prime}$ with $l$ and we further get two additional sub-regions:
C2: $0\leq j\leq l$ and C3: $0\leq l\leq j\leq l+n$.

First in sub-region C1: since $l\leq n$, using the first inequality from Lemma
4.1, we have

$\displaystyle S_{C1}$ $\displaystyle=\sum_{0\leq j\leq l\leq
n}\frac{\left(l-\frac{1}{2}\right)^{2}\cdots\left(l-n+\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\cdot\frac{e^{-\gamma(j+l)-al+bj}}{\left(j+n+\frac{1}{2}\right)^{2}}$
$\displaystyle\leq\sum_{0\leq j\leq l\leq
n}\frac{\left(l-\frac{1}{2}\right)^{2}\cdots\left(l-n+\frac{1}{2}\right)^{2}}{\left(\frac{1}{2}\right)^{2}\cdots\left(n-\frac{1}{2}\right)^{2}}\cdot\frac{e^{-\gamma(j+l)-al+bj}}{\left(j+n+\frac{1}{2}\right)^{2}}$
$\displaystyle\leq\sum_{0\leq j\leq l\leq
n}\frac{e^{-\gamma(j+l)-al+bj}}{\left(j+n+\frac{1}{2}\right)^{2}}\leq\frac{1}{\left(n+\frac{1}{2}\right)^{2}}\sum_{0\leq
j\leq
l<\infty}e^{(-\gamma+b)j-(\gamma+a)l}<\frac{\textrm{const}}{\left(n+\frac{1}{2}\right)^{2}}$

which clearly goes to zero as $n\to\infty$.

In the case of C2, we get

$S_{C2}=\sum_{0\leq j\leq
l}\frac{\left(l+\frac{1}{2}\right)^{2}\cdots\left(l+n-\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\cdot\frac{e^{-(\gamma+a)n+(-\gamma+b)j-(\gamma+a)l}}{\left(j+n+\frac{1}{2}\right)^{2}}\,.$

Letting $l=k+j$, the sum becomes

$S_{C2}=\sum_{j,k\geq
0}\frac{\left(k+j+\frac{1}{2}\right)^{2}\cdots\left(k+j+n-\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\cdot\frac{e^{-(\gamma+a)n+(-\gamma+b)j-(\gamma+a)(k+j)}}{\left(j+n+\frac{1}{2}\right)^{2}}\,.$

Implementing Lemma 4.2 we have

$S_{C2}\leq\frac{(2n+1)e^{-(\gamma+a)n}}{\left(1-e^{-\frac{\gamma+a}{2}}\right)^{2n+1}}\to
0\textrm{ as }n\to\infty\,.$

In sub-region C3 we have:

$S_{C3}=\sum_{0\leq l\leq j\leq
l+n}\frac{\left(l+\frac{1}{2}\right)^{2}\cdots\left(l+n-\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\cdot\frac{e^{-(\gamma-b)j-(\gamma+a)(l+n)}}{\left(j+n+\frac{1}{2}\right)^{2}}$

Notice that in this region we again have that

$\frac{\left(l+\frac{1}{2}\right)^{2}\cdots\left(l+n-\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\leq
1.$

Thus, by overestimating, we have that

$S_{C3}\leq\sum_{0\leq l\leq
j\leq\infty}\frac{e^{-(\gamma-b)j-(\gamma+a)(l+n)}}{\left(j+n+\frac{1}{2}\right)^{2}}\leq\frac{1}{\left(n+\frac{1}{2}\right)^{2}}\sum_{0\leq
l\leq j\leq\infty}e^{-(\gamma-b)j-(\gamma+a)(l+n)}$

which goes to zero as $n\to\infty$.

Region D: $l\leq j\leq 0$.

First we map $j\mapsto-j$ and $l\mapsto-l$ and the sum becomes

$S_{D}=\sum_{0\leq j\leq
l}\frac{\left(l-\frac{1}{2}\right)^{2}\cdots\left(l-n+\frac{1}{2}\right)^{2}}{\left(j-\frac{1}{2}\right)^{2}\cdots\left(j-n+\frac{1}{2}\right)^{2}}\cdot\frac{e^{\gamma
j-\gamma l-a|l|+b|j|}}{\left(j-n-\frac{1}{2}\right)^{2}}\,.$

Like with region C, we split this into three sub-regions: D1: $0\leq j\leq
n\leq l$, D2: $n\leq j\leq l$, and D3: $0\leq j\leq l\leq n$.

In sub-region D1, we map $l\mapsto l+n$ which yields

$S_{D1}=\sum_{0\leq j\leq n,l\geq
0}\frac{\left(l+\frac{1}{2}\right)^{2}\cdots\left(l+n-\frac{1}{2}\right)^{2}}{\left(j-\frac{1}{2}\right)^{2}\cdots\left(j-n+\frac{1}{2}\right)^{2}}\cdot\frac{e^{-(\gamma+a)n}e^{(\gamma+b)j-(\gamma+a)l}}{\left(j-n-\frac{1}{2}\right)^{2}}\,.$

Multiplying and dividing the sum by
$(\frac{1}{2})^{2}\cdots(n-\frac{1}{2})^{2}$ and using Lemma 4.2 we get

$S_{D1}\leq\frac{(2n+1)e^{-(\gamma+a)n}}{\left(1-e^{-\frac{\gamma+a}{2}}\right)^{2n+1}}\sum_{0\leq
j\leq
n}\frac{\left(\frac{1}{2}\right)^{2}\cdots\left(n-\frac{1}{2}\right)^{2}}{\left(j-\frac{1}{2}\right)^{2}\cdots\left(j-n+\frac{1}{2}\right)^{2}}\cdot\frac{e^{(\gamma+b)j}}{\left(j-n-\frac{1}{2}\right)^{2}}\,.$

The second inequality in Lemma 4.1 implies that

$\displaystyle S_{D1}$
$\displaystyle\leq\frac{(2n+1)e^{-(\gamma+a)n}}{\left(1-e^{-\frac{\gamma+a}{2}}\right)^{2n+1}}\sum_{0\leq
j\leq n}\left(\begin{array}[]{c}2n\\\
2j\end{array}\right)\frac{2je^{(\gamma+b)j}}{\left(j-n-\frac{1}{2}\right)^{2}}$
$\displaystyle\leq\frac{(2n+1)e^{-(\gamma+a)n}}{\left(1-e^{-\frac{\gamma+a}{2}}\right)^{2n+1}}\sum_{0\leq
j\leq n}\left(\begin{array}[]{c}2n\\\
j\end{array}\right)e^{\left(\frac{\gamma+b}{2}\right)j}$ $\displaystyle\leq
n(2n+1)\left(\frac{e^{-\frac{\gamma+a}{2}}+e^{-\frac{(a-b)}{2}}}{1-e^{-\frac{\gamma+a}{2}}}\right)^{2n}\,.$

The conditions on $a$, $b$ and $\gamma$ imply that the right hand side of the
above inequality goes to $0$ as $n\to\infty$ and thus $S_{D1}\to 0$ as
$n\to\infty$.

In sub-region D2, we map $l\mapsto l+n$ and $j\mapsto j+n$ to get

$S_{D2}=\sum_{0\leq j\leq
l}\frac{\left(l+\frac{1}{2}\right)^{2}\cdots\left(l+n-\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\cdot\frac{e^{-(\gamma+a)(l+n)+(\gamma+b)(j+n)}}{\left(j-\frac{1}{2}\right)^{2}}\,.$

Writing $l=k+j$ and using Lemma 4.2 we obtain:

$\displaystyle S_{D2}$ $\displaystyle=e^{-(a-b)n}\sum_{0\leq
j,k}\frac{\left(j+k+\frac{1}{2}\right)^{2}\cdots\left(j+k+n-\frac{1}{2}\right)^{2}}{\left(j+\frac{1}{2}\right)^{2}\cdots\left(j+n-\frac{1}{2}\right)^{2}}\cdot\frac{e^{-(a-b)j-(\gamma+a)k}}{\left(j-\frac{1}{2}\right)^{2}}$
$\displaystyle\leq\frac{(2n+1)e^{-(a-b)n}}{\left(1-e^{-\frac{\gamma+a}{2}}\right)^{2n+1}}\sum_{0\leq
j}\frac{e^{-(a-b)j}}{\left(j-\frac{1}{2}\right)^{2}}<\infty\,.$

Like in sub-region D1, the conditions on $a$, $b$ and $\gamma$ imply the last
term in the above inequality goes to zero as $n\to\infty$ and hence $S_{D2}$
goes to zero as $n\to\infty$.

Finally in the sub-region D3, by multiplying and dividing by
$(\frac{1}{2})^{2}\cdots(n-\frac{1}{2})^{2}$, we have that:

$S_{D3}=\sum_{0\leq j\leq l\leq
n}\frac{q_{n}(j)}{q_{n}(l)}\cdot\frac{e^{-(\gamma+a)l+(\gamma+b)j}}{\left(j-n-\frac{1}{2}\right)^{2}}\,.$

Observe in this region we have that $n+\frac{1}{2}-j>n+\frac{1}{3}-j$. Using
this observation and the third inequality in Lemma 4.1 we have

$\displaystyle S_{D3}$ $\displaystyle\leq\sum_{0\leq j\leq
l}\left(\begin{array}[]{c}2l\\\
2j\end{array}\right)\cdot\frac{e^{-(\frac{\gamma+a}{2})2l+(\frac{\gamma+b}{2})2j}}{\left(\frac{1}{2}\cdot
2j-n-\frac{1}{3}\right)^{2}}\leq\sum_{0\leq j\leq l\leq
2n}\left(\begin{array}[]{c}l\\\
j\end{array}\right)\cdot\frac{e^{-(\frac{\gamma+a}{2})l+(\frac{\gamma+b}{2})j}}{\left(\frac{j}{2}-n-\frac{1}{3}\right)^{2}}$
$\displaystyle=\sum_{l\geq 0}\left(\sum_{j=0}^{l}\left(\begin{array}[]{c}l\\\
j\end{array}\right)\frac{e^{(\frac{\gamma+b}{2})j}}{\left(\frac{j}{2}-n-\frac{1}{3}\right)^{2}}\right)e^{-(\frac{\gamma+a}{2})l}=\sum_{l\geq
0}\frac{1}{\left(\frac{l}{2}-n-\frac{1}{3}\right)^{2}}\left(e^{-(\frac{\gamma+a}{2})}+e^{-(\frac{a-b}{2})}\right)^{l}$

where the last sum is finite. Thus by the Lebesgue Dominated Convergence
Theorem, $S_{D2}\to 0$ as $n\to\infty$. This completes the proof. ∎

Remark. Since a bounded perturbation of an operator with compact parametrices
also has compact parametrices, see Appendix of [4], it follows for example
that, if $\beta(l)=\beta_{\infty}l+\tilde{\beta}(l)$ and
$\alpha(l)=e^{-\gamma/2}\beta_{\infty}l+\tilde{\alpha}(l)$, where
$\beta_{\infty}$ is a nonzero constant, $\tilde{\alpha}(l)$ and
$\tilde{\beta}(l)$ are bounded, then the corresponding operator $D$ has
compact parametrices. This observation substantially increases the class of
covariant implementations with compact parametrices.

###  4.2. Proofs of Lemmas

###### Proof.

(of Lemma 4.1) Notice that

$q_{n}(1)=\frac{\left(\frac{1}{2}\right)^{2}\left(\frac{3}{2}\right)^{2}\cdots\left(n-\frac{3}{2}\right)^{2}\left(n-\frac{1}{2}\right)^{2}}{\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{2}\left(\frac{3}{2}\right)^{2}\cdots\left(n-\frac{5}{2}\right)^{2}\left(n-\frac{3}{2}\right)^{2}}=\frac{(2n-1)^{2}}{1^{2}}\,.$

Similarly

$q_{n}(2)=\frac{(2n-3)^{2}(2n-1)^{2}}{1^{2}\cdot 3^{2}}\,.$

It follows by induction that

$q_{n}(l)=q_{n}(l-1)\frac{(2n-2l+1)^{2}}{(2l-1)^{2}}=\frac{((2n-2l+1)\cdots(2n-3)(2n-1))^{2}}{(1\cdot
3\cdot 5\cdots(2l-1))^{2}}\,.$ (4.1)

For the first inequality, notice that from the inductive formula (4.1) we see
that $q_{n}(l)$ as a function of $l$, $0\leq l\leq n$, first increases and
then decreases, so the minimum of it occurs at the endpoints. This means that
$q_{n}(l)\geq q_{n}(0)=q_{n}(n)=1$, yielding the first inequality.

To prove the second inequality notice that

$\displaystyle q_{n}(l)$
$\displaystyle=\frac{((2n-2l+1)\cdots(2n-3)(2n-1))^{2}}{(1\cdot 3\cdot
5\cdots(2l-1))^{2}}$
$\displaystyle\leq\frac{(2n-2l+1)(2n-2l+2)\cdots(2n-1)(2n)}{1\cdot 2\cdot
3\cdots(2l-2)(2l-1)}=\frac{(2n)!}{(2n-1)!(2n-2l)!}$
$\displaystyle=\frac{2l(2n)!}{(2l)!(2n-2l)!}=2l\left(\begin{array}[]{c}2n\\\
2l\end{array}\right)\,.$

To prove the final inequality we estimate as follows:

$\displaystyle\frac{q_{n}(j)}{q_{n}(l)}$
$\displaystyle=\frac{\left(l-\frac{1}{2}\right)^{2}\cdots\left(l-n+\frac{1}{2}\right)^{2}}{\left(j-\frac{1}{2}\right)^{2}\cdots\left(j-n+\frac{1}{2}\right)^{2}}=\frac{(2j+1)^{2}\cdots(2l-1)^{2}}{(2n-2l+1)^{2}\cdots(2n-2j+1)^{2}}$
$\displaystyle\leq\frac{(2j+1)(2j+2)\cdots(2l-1)(2l)}{(2n-2l)(2n-2l+1)\cdots(2n-2j+1)}=\frac{(2l)!}{(2j)!}\cdot\frac{(2n-2l-1)!}{(2n-2j-1)!}$
$\displaystyle=\frac{\left(\begin{array}[]{c}2l\\\
2j\end{array}\right)}{\left(\begin{array}[]{c}2n-2j-1\\\
2n-2l-1\end{array}\right)}\leq\left(\begin{array}[]{c}2l\\\
2j\end{array}\right)\,.$

∎

To prove Lemma 4.2, we need the following additional step.

######  Lemma 4.4.

For a nonnegative integer $j$, define the following sum:

$I_{m}(j)=\sum_{k\geq
0}\frac{(k+2j+1)\cdots(k+2j+m)e^{-\frac{\gamma}{2}k}}{m!}\textrm{ for
}m>0\textrm{ and }I_{0}=\left(1-e^{-\frac{\gamma}{2}}\right)^{-1}\,.$

Then, for $m\geq 0$, we have:

$I_{m}(j)\leq\frac{1}{\left(1-e^{-\frac{\gamma}{2}}\right)^{m+1}}\cdot\frac{(2j+1)\cdots(2j+m)}{m!}\,.$

###### Proof.

Notice $I_{m}(j)$ satisfies the following reduction formula:

$\displaystyle I_{m}(j)$
$\displaystyle=\frac{(2j+1)\cdots(2j+m)}{m!}+e^{-\frac{\gamma}{2}}I_{m}(j)+\sum_{k\geq
1}\frac{(k+2j+1)\cdots(k+2j+m-1)e^{-\frac{\gamma}{2}k}}{(m-1)!}$
$\displaystyle=\frac{(2j)(2j+1)\cdots(2j+m-1)}{m!}+e^{-\frac{\gamma}{2}}I_{m}(j)+I_{m-1}(j)\,.$

This implies that $I_{m}(j)$ satisfies the following recurrence relation:

$\left(1-e^{-\frac{\gamma}{2}}\right)I_{m}(j)-I_{m-1}(j)=\left(\begin{array}[]{c}2j+m-1\\\
2j-1\end{array}\right)\,.$

This can be solved recursively which yields

$I_{m}(j)=\frac{1}{\left(1-e^{-\frac{\gamma}{2}}\right)^{m+1}}\sum_{r=0}^{m}\left(\begin{array}[]{c}2j+r-1\\\
r\end{array}\right)\left(1-e^{-\frac{\gamma}{2}}\right)^{r}\,.$

Since $1-e^{-\gamma/2}\leq 1$ we get

$I_{m}(j)\leq\frac{1}{\left(1-e^{-\frac{\gamma}{2}}\right)^{m+1}}\sum_{r=0}^{m}\left(\begin{array}[]{c}2j+r-1\\\
r\end{array}\right)\,.$

By parallel summation we have

$\sum_{r=0}^{m}\left(\begin{array}[]{c}2j+r-1\\\
r\end{array}\right)=\left(\begin{array}[]{c}2j+m\\\
m\end{array}\right)=\frac{(2j+1)\cdots(2j+m)}{m!}$

and thus the result follows. ∎

###### Proof.

(of Lemma 4.2) Observe the following inequality

$\displaystyle J_{n}(j)$ $\displaystyle=\sum_{k\geq
0}\frac{(2k+2j+1)^{2}\cdots(2k+2j+2n-1)^{2}}{(2j+1)^{2}\cdots(2j+2n-1)^{2}}e^{-\frac{\gamma}{2}\cdot
2k}$ $\displaystyle\leq\sum_{k\geq
0}\frac{(2k+2j+1)(2k+2j+2)\cdots(2k+2j+2n)}{(2j+1)^{2}\cdots(2j+2n-1)^{2}}e^{-\frac{\gamma}{2}\cdot
2k}.$

Overestimating by adding odd $2k+1$ terms we get:

$\displaystyle J_{n}(j)$
$\displaystyle\leq\frac{(2n)!}{(2j+1)^{2}\cdots(2j+2n-1)^{2}}\sum_{k\geq
0}\frac{(k+2j+1)\cdots(k+2j+2n)}{(2n)!}e^{-\frac{\gamma}{2}k}$
$\displaystyle=\frac{(2n)!}{(2j+1)^{2}\cdots(2j+2n-1)^{2}}I_{2n}(j),$

where $I_{2n}(j)$ is defined in Lemma 4.4. By implementing Lemma 4.4 we arrive
at

$\displaystyle J_{n}(j)$
$\displaystyle\leq\frac{(2n)!}{(2j+1)^{2}\cdots(2j+2n-1)^{2}}\cdot\frac{(2j+1)\cdots(2j+2n)}{(2n)!}\cdot\frac{1}{\left(1-e^{-\frac{\gamma}{2}}\right)^{2n+1}}$
$\displaystyle=\frac{(2j+2)(2j+4)\cdots(2j+2n)}{(2j+1)(2j+3)\cdots(2j+2n-1)}\cdot\frac{1}{\left(1-e^{-\frac{\gamma}{2}}\right)^{2n+1}}$
$\displaystyle\leq\frac{2j+2n}{2j+1}\cdot\frac{1}{\left(1-e^{-\frac{\gamma}{2}}\right)^{2n+1}}\leq\frac{2n+1}{\left(1-e^{-\frac{\gamma}{2}}\right)^{2n+1}}\,.$

∎

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