# On the class of almost subadditive weight functions

Gerhard Schindl  G. Schindl: Fakultät für Mathematik, Universität Wien, Oskar-
Morgenstern-Platz 1, A-1090 Wien, Austria<EMAIL_ADDRESS>
###### Abstract.

We answer a question from A. V. Abanin and P. T. Tien about so-called almost
subadditive weight functions in the sense of Braun-Meise-Taylor. Using recent
knowledge of a growth index for functions, crucially appearing in the
ultraholomorphic setting, we are able to show the existence of weights such
that there does not exist an equivalent almost subadditive function and hence
almost subadditive weights are a proper subclass.

###### Key words and phrases:

Weight functions, weight sequences, almost subadditivity, growth index

###### 2020 Mathematics Subject Classification:

26A12, 26A48

G. Schindl is supported by FWF-Project 10.55776/P33417

##  1\. Introduction

In [1] it has been shown that almost subadditive weight functions are
sufficient to describe the theory of ultradistributions in the sense of Braun-
Meise-Taylor; see [4]. Thus the authors are dealing with non-quasianalytic
weight functions $\omega$; see Section 2.2 for precise definitions and
conditions. All relevant information about growth and regularity properties
for weight functions and weight sequences can be found in Section 2.

In their closing discussion in [1, Rem. 1.5] the authors have asked the
following:

Let $\omega$ be a given non-quasianalytic weight function (in the sense of
Braun-Meise-Taylor), does there always exist a weight $\sigma$ such that

  * $(*)$

$\sigma$ is equivalent to $\omega$ which means that
$0<\liminf_{t\rightarrow+\infty}\frac{\omega(t)}{\sigma(t)}\leq\limsup_{t\rightarrow+\infty}\frac{\omega(t)}{\sigma(t)}<+\infty$,
and such that

  * $(*)$

$\sigma$ is almost subadditive:

$\forall\;q>1\;\exists\;C_{q}\geq 1\;\forall\;s,t\geq 0:\;\;\;\sigma(s+t)\leq
q(\sigma(s)+\sigma(t))+C_{q}.$

The notion of almost subadditivity is more recent and more general than the
classical (standard) assumption subadditivity appearing in the Beurling-Björck
framework; see e.g. [2]. On the other hand we mention that in the literature
different ”subadditivity-like” conditions for (associated) weight functions
have been studied and used; see e.g. [18] and the citations there.

The aim of this article is to answer this question and to prove that in
general there does not exist an equivalent almost subadditive weight. This
fact shows that the class of almost subadditive weight functions is a proper
subclass and hence the main result [1, Thm. 1.4] means that this particular
subclass of weight functions is sufficient to describe the theory of
ultradistributions in the sense of Braun-Meise-Taylor. Our statements can be
found in Section 3. In this work, since we are mainly focusing on functions
and their growth properties, it is not required to introduce the associated
ultradifferentiable function spaces explicitly; we refer to [4], [3], [9],
[13].

In order to treat this problem, crucially we involve recent knowledge about
the growth index $\gamma(\omega)$ from [6]; see Section 2.3 for precise
definitions and explanations. This value becomes relevant when proving
extension results in the ultraholomorphic setting; see [7], [8]. We establish
a connection between the notion of almost subadditivity and $\gamma(\omega)$
(see Lemma 3.1 and Remark 3.2). Involving this information, intrinsic
properties of $\gamma(\omega)$, in particular using the fact that this growth
index is preserved under equivalence of weights, we show the main result
Theorem 3.3.

Note that using this method we are able to give more information: On the one
hand, we are working within a more general setting of weight functions; see
Definition 2.1. Thus our basic assumptions on $\omega$ are weaker than being a
weight in the sense of Braun-Meise-Taylor, but also weaker than the standard
requirements assumed in [2] and even slightly weaker than in [7], [8]. This is
possible by the general setting studied in [6, Sect. 2.3 & 2.4]. On the other
hand, instead of almost subadditivity, we treat an more general condition (see
(3.1)) and obtain that the above problem is not depending on the notion of
(non-)quasianalyticity. So the main result is valid for quasianalytic weights
as well; see Remark 3.5 for more detailed comments.

However, Theorem 3.3 provides abstract information and the aim in Theorem 3.6
and Theorem 3.7 is to construct explicitly weights with the desired property.
There we are focusing on so-called associated weight functions $\omega_{M}$
with $M\in\mathbb{R}_{>0}^{\mathbb{N}}$ being a weight sequence; see Section
2.4. The advantage in this case is that the desired growth properties for
$\omega_{M}$ can be expressed in terms of $M$ and so one is able to proceed by
constructing an appropriate weight sequence $M$. Such explicit constructions
of (counter-)examples can be useful for further applications in different
contexts. Here, the growth index $\gamma(M)$ originally introduced by V.
Thilliez in [19] is becoming relevant; see again [6] for more details and its
relation to $\gamma(\omega_{M})$.

##  2\. Weights, growth conditions and the index $\gamma(\omega)$

###  2.1. General notation

We use the notation $\mathbb{R}_{>0}:=(0,+\infty)$,
$\mathbb{N}:=\\{0,1,2,\dots\\}$ and $\mathbb{N}_{>0}:=\\{1,2,\dots\\}$.

######  Definition 2.1.

A function $\omega:[0,+\infty)\rightarrow[0,+\infty)$ is called a weight
function if

  * $(*)$

$\omega$ is non-decreasing and

  * $(*)$

$\lim_{t\rightarrow+\infty}\omega(t)=+\infty$.

These are the basic assumptions resp. the notion for being a weight which is
used in [6, Sect. 2.3 & 2.4]. These requirements are sufficient to introduce
and work with the crucial index $\gamma(\omega)$; see Section 2.3 for details.

###  2.2. Relevant growth conditions

We list now conditions for weight functions
$\omega:[0,+\infty)\rightarrow[0,+\infty)$; they are known and standard in the
ultradifferentiable setting involving weight functions in the sense of Braun-
Meise-Taylor; see [4]. We use the abbreviations appearing in [16].

  * $(\omega_{1})$

$\omega(2t)=O(\omega(t))$ as $t\rightarrow+\infty$.

  * $(\omega_{2})$

$\omega(t)=O(t)$ as $t\rightarrow+\infty$.

  * $(\omega_{3})$

$\log(t)=o(\omega(t))$ as $t\rightarrow+\infty$.

  * $(\omega_{4})$

$\varphi_{\omega}:t\mapsto\omega(e^{t})$ is a convex function (on
$\mathbb{R}$).

  * $(\omega_{5})$

$\omega(t)=o(t)$ as $t\rightarrow+\infty$.

  * $(\omega_{\text{nq}})$

$\int_{1}^{\infty}\frac{\omega(t)}{t^{2}}dt<+\infty.$

  * $(\omega_{\text{snq}})$

$\exists\;C>0\;\forall\;y>0:\int_{1}^{\infty}\frac{\omega(yt)}{t^{2}}dt\leq
C\omega(y)+C$.

In the literature, when being defined on $\mathbb{R}^{d}$, usually $\omega$ is
radially extended, i.e. $\omega(x):=\omega(|x|)$ for any $x\in\mathbb{R}^{d}$.

We call a weight function $\omega$ normalized if $\omega(t)=0$ for all $0\leq
t\leq 1$, non-quasianalytic if $(\omega_{\operatorname{nq}})$ is valid and
quasianalytic if $(\omega_{\operatorname{nq}})$ is violated.

######  Definition 2.2.

A function $\omega:[0,+\infty)\rightarrow[0,+\infty)$ is called a BMT-weight
function (BMT standing for Braun-Meise-Taylor), if $\omega$ is continuous,
non-decreasing, $\omega(0)=0$, and such that $\omega$ satisfies
$(\omega_{1})$, $(\omega_{3})$ and $(\omega_{4})$.

If $\omega$ is in addition normalized, then it is called a normalized BMT-
weight function; if in addition $(\omega_{\operatorname{nq}})$ is valid, then
$\omega$ is called a non-quasianalytic BMT-weight function and it is called
quasianalytic, if $(\omega_{\operatorname{nq}})$ fails.

In [1] it has been assumed that $\omega$ is continuous and non-decreasing.
Moreover, the following abbreviations have been used: $(\alpha)$ is
$(\omega_{1})$, $(\beta)$ is $(\omega_{\operatorname{nq}})$, $(\gamma)$
denotes $(\omega_{3})$ and $(\delta)$ is $(\omega_{4})$. Summarizing, in [1]
non-quasianalytic BMT-weight functions (not necessarily satisfying
$\omega(0)=0$, but this failure is not effecting the results) have been
considered and this set is denoted by $\Omega$ there. Note that in the proof
of the main technical statement [1, Prop. 1.3] non-quasianalyticity has been
applied crucially.

Let $\sigma,\tau:[0,+\infty)\rightarrow[0,+\infty)$ be arbitrary, then we
write $\sigma\hypertarget{ompreceq}{\preceq}\tau$ if

$\tau(t)=O(\sigma(t))\;\text{as}\;t\rightarrow+\infty,$

and $\sigma$ and $\tau$ are called equivalent, written
$\sigma\hypertarget{sim}{\sim}\tau$, if

$\sigma\hyperlink{ompreceq}{\preceq}\tau\;\text{and}\;\tau\hyperlink{ompreceq}{\preceq}\sigma.$

This relation is crucial in the ultradifferentiable weight function setting
(in the sense of Braun-Meise-Taylor), see [13, Lemma 5.16, Cor. 5.17], and it
is precisely the same notion of equivalence used in [1]; see [1, Sect. 1, p.
297 & Rem. 1.5]. Moreover, it also has been considered in [6]; see [6, Sect.
2.1 & Rem. 2.6 $(5)$]. It is immediate that all properties listed above,
except the convexity condition $(\omega_{4})$, are preserved under equivalence
and that normalization can be assumed w.l.o.g. by switching to an equivalent
weight.

According to [1, Def. 1.1] we say that a weight function
$\omega:[0,+\infty)\rightarrow[0,+\infty)$ is almost subadditive if

(2.1) $\forall\;q>1\;\exists\;C_{q}\geq 1\;\forall\;s,t\geq
0:\;\;\;\omega(s+t)\leq q(\omega(s)+\omega(t))+C_{q}.$

This should be compared with the more original condition subadditivity in the
Beurling-Björck framework (see [2]):

  * $(\omega_{\text{sub}})$

$\omega(s+t)\leq\omega(s)+\omega(t)$ for all $s,t\geq 0$.

Obviously $(\omega_{\operatorname{sub}})$ implies (2.1) with the uniform
choice $C_{q}=0$ for all $q>1$. Note: If the inequality in (2.1) is valid for
some $q_{0}>1$, then for all $q\geq q_{0}$ as well when choosing
$C_{q}=C_{q_{0}}$. So, in order to verify almost subadditivity, one has to
study $q\rightarrow 1$. Note that by definition neither
$(\omega_{\operatorname{sub}})$ nor (2.1) is preserved under equivalence of
weight functions in general.

###  2.3. The growth index $\gamma(\omega)$

We briefly recall the definition of the growth index $\gamma(\omega)$ and some
consequences; see [6, Sect. 2.3] and [7, Sect. 4.2] and the references
therein. For given $\gamma>0$ we say that a weight function $\omega$ has
property $(P_{\omega,\gamma})$ if

$\exists\;K>1:\;\;\;\limsup_{t\rightarrow+\infty}\frac{\omega(K^{\gamma}t)}{\omega(t)}<K.$

If $(P_{\omega,\gamma})$ holds for some $K>1$, then
$(P_{\omega,\gamma^{\prime}})$ is satisfied for all
$\gamma^{\prime}\leq\gamma$ with the same $K$ since $\omega$ is non-
decreasing. Moreover we can restrict to $\gamma>0$, because for $\gamma\leq 0$
condition $(P_{\omega,\gamma})$ is satisfied for all $\omega$ (again since
$\omega$ is non-decreasing and $K>1$). Then we put

$\gamma(\omega):=\sup\\{\gamma>0:(P_{\omega,\gamma})\;\;\text{is
satisfied}\\},$

and if none condition $(P_{\omega,\gamma})$ holds true then set
$\gamma(\omega):=0$. We recall now some connections between $\gamma(\omega)$
and subadditivity-like conditions for $\omega$.

######  Remark 2.3.

Let $\omega$ be a weight function.

  * $(i)$

By [6, Cor. 2.14] (see also [7, Lemma 4.2]) we get $\gamma(\omega)>0$ if and
only if $(\omega_{1})$ holds true. In particular, for each BMT-weight function
$\omega$ we have $\gamma(\omega)>0$.

  * $(ii)$

By [6, Thm. 2.11, Cor. 2.13] (see also [7, Lemma 4.3]) we have that
$\gamma(\omega)>1$ if and only if $\omega$ has
$(\omega_{\operatorname{snq}})$.

  * $(iii)$

Moreover, by (the proof of) [6, Thm. 2.11, Cor. 2.13] we know that
$(\omega_{\operatorname{snq}})$ is equivalent to the fact that
$\omega\hyperlink{sim}{\sim}\kappa_{\omega}$, where

(2.2) $\kappa_{\omega}(y):=\int_{1}^{+\infty}\frac{\omega(yt)}{t^{2}}dt.$

If $\omega$ is also continuous, then by (the proof of) [12, Prop. 1.3] we have
that $\kappa_{\omega}$ is concave, continuous and
$\kappa_{\omega}(0)=\omega(0)\geq 0$; i.e. $\kappa_{\omega}(0)=0$ if and only
if $\omega(0)=0$. Hence it follows that also $(\omega_{\operatorname{sub}})$
holds true for $\kappa_{\omega}$; see e.g. [16, Lemma 3.8.1 $(1)$].

(If continuity for $\omega$ fails, then the proof of [6, Thm. 2.11
$(i)\Rightarrow(ii)$] yields that we get the same properties for
$y\mapsto\widetilde{\kappa}_{\omega}(y):=\int_{1}^{+\infty}\frac{\kappa_{\omega}(yt)}{t^{2}}dt$,
and so $\omega$ in (2.2) is replaced by $\kappa_{\omega}$. Consequently, in
any case one has that $\omega$ is equivalent to a weight satisfying
$(\omega_{\operatorname{sub}})$.)

  * $(iv)$

By definition of $\gamma(\omega)$ we immediately get (see e.g. [7, $(4.3)$])

(2.3) $\forall\;s>0:\;\;\;\gamma(\omega^{1/s})=s\gamma(\omega),$

with $\omega^{1/s}(t):=\omega(t^{1/s})$, $t\geq 0$ and $s>0$. Note that
$\omega^{1/s}$ is again a weight function; if $\omega$ is a (normalized) BMT-
weight function, then each $\omega^{1/s}$, too.

  * $(v)$

The value of this growth index is preserved under equivalence of weight
functions; i.e. if $\sigma,\tau$ are weights such that
$\sigma\hyperlink{sim}{\sim}\tau$, then $\gamma(\sigma)=\gamma(\tau)$: This
follows from the characterization [6, Thm. 2.11], see also [6, Rem. 2.12].

Next we summarize some relations between $\gamma(\omega)$ and the notion of
(non-)quasianalyticity.

######  Remark 2.4.

We have the following:

  * $(i)$

Recall that by [6, Lemma 2.10, Rem. 2.15 $(i)\Rightarrow(iv)$] we know that
each quasianalytic weight function $\omega$ has to satisfy
$1\leq\frac{1}{\gamma(\omega)}(=:\alpha(\omega))$; i.e. $\gamma(\omega)\leq
1$.

  * $(ii)$

Any non-quasianalytic weight function $\omega$ satisfies $(\omega_{5})$ (and
hence $(\omega_{2})$) since
$\int_{a}^{+\infty}\frac{\omega(t)}{t^{2}}dt\geq\omega(a)\int_{a}^{+\infty}\frac{1}{t^{2}}dt=\frac{\omega(a)}{a}$
for all $a\geq 1$.

###  2.4. Weight sequences and associated weight functions

Let $M\in\mathbb{R}_{>0}^{\mathbb{N}}$ be given and consider the corresponding
sequence of quotients $\mu=(\mu_{j})_{j\in\mathbb{N}}$ by
$\mu_{j}:=\frac{M_{j}}{M_{j-1}}$, $j\geq 1$, $\mu_{0}:=1$. $M$ is called
normalized if $1=M_{0}\leq M_{1}$ and $M$ is called log-convex if

$\forall\;j\in\mathbb{N}_{>0}:\;M_{j}^{2}\leq M_{j-1}M_{j+1},$

equivalently if $(\mu_{j})_{j}$ is non-decreasing. Consider the following set
of sequences

$\hypertarget{LCset}{\mathcal{LC}}:=\\{M\in\mathbb{R}_{>0}^{\mathbb{N}}:\;M\;\text{is
normalized, log-convex},\;\lim_{j\rightarrow+\infty}(M_{j})^{1/j}=+\infty\\}.$

We see that $M\in\hyperlink{LCset}{\mathcal{LC}}$ if and only if
$1=\mu_{0}\leq\mu_{1}\leq\dots$, $\lim_{j\rightarrow+\infty}\mu_{j}=+\infty$
(see e.g. [13, p. 104]) and there is a one-to-one correspondence between $M$
and $\mu=(\mu_{j})_{j}$ by taking $M_{j}:=\prod_{k=0}^{j}\mu_{k}$.

$M$ has condition moderate growth, denoted by $(\text{mg})$, if

$\exists\;C\geq 1\;\forall\;j,k\in\mathbb{N}:\;M_{j+k}\leq
C^{j+k+1}M_{j}M_{k}.$

In [9] this is denoted by $(M.2)$ and also known under the name stability
under ultradifferential operators. $M$ is called non-quasianalytic, denoted by
$(\text{nq})$ and by condition $(M.3)^{\prime}$ in [9], if

$\sum_{j\geq 1}\frac{1}{\mu_{j}}<+\infty.$

We call two sequences $M,N\in\mathbb{R}_{>0}^{\mathbb{N}}$ equivalent if

$0<\inf_{j\in\mathbb{N}_{>0}}\left(\frac{M_{j}}{N_{j}}\right)^{1/j}\leq\sup_{j\in\mathbb{N}_{>0}}\left(\frac{M_{j}}{N_{j}}\right)^{1/j}<+\infty.$

Let $M\in\mathbb{R}_{>0}^{\mathbb{N}}$ (with $M_{0}=1$), then the associated
(weight) function
$\omega_{M}:\mathbb{R}\rightarrow\mathbb{R}\cup\\{+\infty\\}$ is defined by

$\omega_{M}(t):=\sup_{j\in\mathbb{N}}\log\left(\frac{|t|^{j}}{M_{j}}\right)\;\;\;\text{for}\;t\in\mathbb{R},\;t\neq
0,\hskip 30.0pt\omega_{M}(0):=0.$

For an abstract introduction of the associated function we refer to [10,
Chapitre I], see also [9, Def. 3.1] and the more recent work [15].

######  Remark 2.5.

Let $M\in\hyperlink{LCset}{\mathcal{LC}}$ be given, then we summarize:

  * $(*)$

By [17, Lemma 2.8] and [7, Lemma 2.4], see also the references mentioned in
the proofs there, $\omega_{M}$ satisfies all requirements to be a normalized
BMT-weight function except $(\omega_{1})$.

  * $(*)$

By [18, Thm. 3.1] $\omega_{M}$ is a normalized BMT-weight function, and so
$\omega_{M}$ has in addition $(\omega_{1})$, if and only if $M$ satisfies

$\exists\;L\in\mathbb{N}_{>0}:\;\;\;\liminf_{j\rightarrow+\infty}\frac{(M_{Lj})^{\frac{1}{Lj}}}{(M_{j})^{\frac{1}{j}}}>1.$

Moreover, by [18, Prop. 3.4] it follows that if $M$ has in addition
$(\operatorname{mg})$, then $\omega_{M}$ has $(\omega_{1})$ if and only if

(2.4) $\exists\;Q\in\mathbb{N}_{\geq
2}:\;\;\;\liminf_{j\rightarrow+\infty}\frac{\mu_{Q_{j}}}{\mu_{j}}>1.$

(2.4) appeared crucially in [3] and there it also has been shown that (2.4)
implies $(\omega_{1})$ for $\omega_{M}$ even without having
$(\operatorname{mg})$; see [3, Lemma 12, $(2)\Rightarrow(4)$].

Note that in [11, Sect. 2.6 $(2)$], see also [4, Rem. 8.9], it has been
mentioned that H.-J. Petzsche has already been able to prove the statements
[18, Thm. 3.1, Prop. 3.4] in a private communication but it seems that a proof
never has been published before.

  * $(*)$

By [9, Lemma 4.1] we have that $\omega_{M}$ satisfies
$(\omega_{\operatorname{nq}})$ if and only if $M$ has $(\operatorname{nq})$.

  * $(*)$

Finally we mention, see e.g. [14, Lemma 2.2] and the citations there, that
$(\operatorname{mg})$ is equivalent to have
$\sup_{j\in\mathbb{N}}\frac{\mu_{2j}}{\mu_{j}}<+\infty$.

##  3\. Main results

We formulate and prove now the main results of this article. In order to
proceed we crucially involve the information about the growth index
$\gamma(\omega)$. Section 3.1 deals with an abstract statement and in Sections
3.2 and 3.3 we are investigating in more detail the (non-)quasianalytic weight
sequence case.

###  3.1. General result for weight functions

The next technical and crucial result illustrates the intimate connection
between the notion of almost subadditivity and the value of $\gamma(\omega)$.
Indeed, in this context we study a more general growth relation.

######  Lemma 3.1.

Let $\omega$ be a weight function.

  * $(i)$

Assume that $\omega$ satisfies

(3.1) $\exists\;q_{0}\geq\frac{1}{2}\;\forall\;q>q_{0}\;\exists\;C_{q}\geq
1\;\forall\;s,t\geq 0:\;\;\;\omega(s+t)\leq q(\omega(s)+\omega(t))+C_{q}.$

If $q_{0}>\frac{1}{2}$, then
$\gamma(\omega)\geq\frac{\log(2)}{\log(2)+\log(q_{0})}=\frac{1}{1+\log(q_{0})/\log(2)}>0$
and if $q_{0}=\frac{1}{2}$, then $\gamma(\omega)=+\infty$.

  * $(ii)$

If $\omega$ is almost subadditive, i.e. if (2.1) holds true, then
$\gamma(\omega)\geq 1$.

  * $(iii)$

Each almost subadditive and quasianalytic weight function $\omega$ has to
satisfy $\gamma(\omega)=1$.

Concerning $(i)$ note that $\frac{1}{1+\log(q_{0})/\log(2)}>0$ because
$1+\log(q_{0})/\log(2)>0\Leftrightarrow\log(q_{0})>-\log(2)\Leftrightarrow
q_{0}>\frac{1}{2}$. Thus, in particular, (3.1) yields $\gamma(\omega)>0$ and
hence $(\omega_{1})$; see $(i)$ in Remark 2.3 but also the proof below.

Proof. $(i)$ First, (3.1) yields

$\exists\;q_{0}\geq\frac{1}{2}\;\forall\;q>q_{0}\;\exists\;C_{q}\geq
1\;\forall\;t\geq 0:\;\;\;\omega(2t)\leq 2q\omega(t)+C_{q},$

which precisely means

(3.2)
$\exists\;q_{0}\geq\frac{1}{2}\;\forall\;q>q_{0}:\;\;\;\limsup_{t\rightarrow+\infty}\frac{\omega(2t)}{\omega(t)}\leq
2q;$

in particular $(\omega_{1})$. Assume $q_{0}>\frac{1}{2}$. Let $\epsilon>0$ be
small and set $\gamma:=\frac{\log(2)}{\log(2q_{0}+\epsilon)}$,
$K:=2q_{0}+\epsilon>1$ and so $\gamma=\frac{\log(2)}{\log(K)}$, equivalently
$2=K^{\gamma}$. Then, when taking $q$ with $q_{0}<q<q_{0}+\frac{\epsilon}{2}$
in (3.2), and so $2q<2q_{0}+\epsilon=K$, this property verifies
$(P_{\omega,\gamma})$ with the aforementioned choice of $K$. One has
$0<\gamma<\frac{\log(2)}{\log(2q_{0})}$ and when we let $\epsilon\rightarrow
0$ then we get $\gamma\rightarrow\frac{\log(2)}{\log(2q_{0})}$ (and
$q\rightarrow q_{0}$). Hence $(P_{\omega,\gamma})$ holds true for all
$0<\gamma<\frac{\log(2)}{\log(2)+\log(q_{0})}$ and by definition this gives
$\gamma(\omega)\geq\frac{1}{1+\log(q_{0})/\log(2)}$.

However, in general it is not clear that (3.2) implies
$(P_{\omega,\gamma_{0}})$ with $\gamma_{0}:=\frac{1}{1+\log(q_{0})/\log(2)}$
and also $\gamma(\omega)>\frac{1}{1+\log(q_{0})/\log(2)}$ might be violated.

If $q_{0}=\frac{1}{2}$, then take $\gamma:=\frac{\log(2)}{\log(1+\epsilon)}$,
$K:=1+\epsilon$ and when $\epsilon\rightarrow 0$ by the arguments before we
have $(P_{\omega,\gamma})$ for any $\gamma>0$ and so $\gamma(\omega)=+\infty$.

$(ii)$ (2.1) is precisely (3.1) with $q_{0}=1$.

$(iii)$ If $\omega$ is quasianalytic, then the previous part $(ii)$ and $(i)$
in Remark 2.4 yield the conclusion. ∎

######  Remark 3.2.

We comment on Lemma 3.1:

  * $(*)$

The value $\frac{1}{2}$ for the choice of $q_{0}$ in (3.1) is critical: If
$0<q_{0}<\frac{1}{2}$ is allowed, then take $q$ with $q_{0}<q<\frac{1}{2}$ and
we get (see (3.2)) that

$\limsup_{t\rightarrow+\infty}\frac{\omega(2t)}{\omega(t)}\leq 2q<1.$

But this clearly contradicts the standard assumption that $\omega$ is non-
decreasing.

  * $(*)$

As $q_{0}\rightarrow+\infty$ the lower bound $\frac{1}{1+\log(q_{0})/\log(2)}$
tends to $0$.

  * $(*)$

We give now a partial converse to $(i)$ in Lemma 3.1:

If $\gamma(\omega)>0$, equivalently $(\omega_{1})$ (see $(i)$ in Remark 2.3),
then since $\omega$ is non-decreasing we estimate for all $s,t\geq 0$ as
follows

(3.3) $\omega(s+t)\leq\omega(2\max\\{s,t\\})\leq L\omega(\max\\{s,t\\})+L\leq
L(\omega(t)+\omega(s))+L.$

Consequently, $\gamma(\omega)>0$ yields (3.1) for $q_{0}=L$, with $L$ being
the constant such that $\omega(2t)\leq L\omega(t)+L$ for all $t\geq 0$; see
$(\omega_{1})$. Note that $L\geq 1$ because $L<1$ yields a contradiction to
the fact that $\omega$ is a weight function (so
$\lim_{t\rightarrow+\infty}\omega(t)=+\infty$ and $\omega$ is non-decreasing).

Similarly, it is not clear that (3.2) implies (3.1): The analogous estimate as
in (3.3) gives that (3.2) for $q_{0}$ yields (3.1) for $2q_{0}$.

Using this information we are able to show the following main statement.

######  Theorem 3.3.

For any weight function $\omega$ with $\gamma(\omega)<+\infty$ we get:

  * $(*)$

There exist uncountable infinite many pair-wise different (i.e. non-
equivalent) weight functions $\tau_{a}:=\omega^{1/a}$,
$a\in(0,\gamma(\omega)^{-1})=:\mathcal{I}_{\omega}$ (with
$\mathcal{I}_{\omega}=(0,+\infty)$ if $\gamma(\omega)=0$), such that each
$\tau_{a}$ has the following property:

There does not exist a weight function $\sigma$ being equivalent to $\tau_{a}$
and such that $\sigma$ is almost subadditive (see (2.1)). In particular, each
$\tau_{a}$ itself is violating (2.1).

  * $(*)$

If $\omega$ is a (normalized) BMT-weight function, then each $\tau_{a}$, too.

Roughly speaking this result means that ”many” BMT-weight functions are not
equivalent to an almost subadditive function in the sense of [1]; to each
weight with finite $\gamma(\omega)$ we can assign an uncountable infinite
family of pair-wise different equivalence classes of weights such that in each
of these classes no almost subadditive weight exists.

Proof. Take any $a>0$ with $\gamma(\omega)<\frac{1}{a}$, then set
$\tau_{a}:=\omega^{1/a}$ and by (2.3) we get

(3.4) $\gamma(\tau_{a})=\gamma(\omega^{1/a})=a\gamma(\omega)<1.$

Note: If $\gamma(\omega)=0$, then any choice for $a>0$ in (3.4) is admissible
and $\gamma(\tau_{a})=0$ for all $a>0$. This verifies the interpretation
$\mathcal{I}_{\omega}=(0,+\infty)$ if $\gamma(\omega)=0$. (On the other hand,
in order to choose $a$ the assumption $\gamma(\omega)<+\infty$ is
indispensable. In the case $\gamma(\omega)=+\infty$, which is excluded by
assumption, we put $\mathcal{I}_{\omega}:=\emptyset$.)

Consider now for such $a>0$ (fixed) the function $\tau_{a}=\omega^{1/a}$. By
$(ii)$ in Lemma 3.1 and (3.4) we have that $\omega_{a}$ is violating (2.1).
Since $\gamma(\cdot)$ is preserved under equivalence of weight functions, see
$(v)$ in Remark 2.3, we have that each weight $\sigma$ with
$\sigma\hyperlink{sim}{\sim}\tau_{a}$ satisfies $\gamma(\sigma)<1$ and hence
also cannot have (2.1).

Recall that if $\omega$ is a (normalized) BMT-weight, then each
$\tau_{a}=\omega^{1/a}$, too.

Finally, let $a_{1}>a_{2}>0$ satisfy
$\gamma(\omega)<\frac{1}{a_{1}}<\frac{1}{a_{2}}$ and let these choices be
arbitrary but from now on fixed. We prove that $\omega^{1/a_{1}}$ and
$\omega^{1/a_{2}}$ are not equivalent:

On the one hand, since $\omega$ is assumed to be non-decreasing, we clearly
have
$\omega^{1/a_{1}}(t)\leq\omega^{1/a_{2}}(t)\Leftrightarrow\omega(t^{1/a_{1}})\leq\omega(t^{1/a_{2}})$
for any $0<a_{2}<a_{1}$ and all $t\geq 1$. But, on the other hand, if the
weights are equivalent, i.e. if
$\omega^{1/a_{1}}\hyperlink{sim}{\sim}\omega^{1/a_{2}}$, then

$\exists\;C\geq 1\;\forall\;t\geq 0:\;\;\;\omega(t^{1/a_{2}})\leq
C\omega(t^{1/a_{1}})+C,$

and so

(3.5) $\exists\;C\geq 1\;\forall\;t\geq 0:\;\;\;\omega(t^{a_{1}/a_{2}})\leq
C\omega(t)+C.$

We show that this is equivalent to

(3.6) $\exists\;C\geq 1\;\forall\;t\geq 0:\;\;\;\omega(t^{2})\leq
C\omega(t)+C.$

(3.5)$\Rightarrow$(3.6): If $a_{1}\geq 2a_{2}$, then the claim follows because
$\omega$ is non-decreasing. If $2a_{2}>a_{1}>a_{2}$, then an $n$-times
iteration with $n\in\mathbb{N}_{>0}$ (minimal) such that
$(a_{1}/a_{2})^{n}\geq 2$ yields the conclusion.

(3.6)$\Rightarrow$(3.5): Similarly, if $2a_{2}\geq a_{1}>a_{2}$, then the
implication is clear. If $a_{1}>2a_{2}$, then an $n$-times iteration with
$n\in\mathbb{N}_{>0}$ (minimal) such that $a_{1}/a_{2}\leq 2^{n}$ yields the
conclusion.

Finally, we verify that (3.6) implies $\gamma(\omega)=+\infty$ for any weight
function: Fix some $K>1$ arbitrary and take any $\gamma>0$. (As we are going
to see the choice for $K$ can be taken uniformly for all $\gamma>0$.) Then we
can find $t_{\gamma,K}:=K^{\gamma}>0$ such that
$\omega(K^{\gamma}t)\leq\omega(t^{2})$ for all $t\geq t_{\gamma,K}$ since
$\omega$ is non-decreasing and so $\omega(K^{\gamma}t)\leq C\omega(t)+C$ for
all $t\geq t_{\gamma,K}$. Hence it suffices to choose $K:=C-\epsilon$ with $C$
denoting the constant from (3.6) and $\epsilon>0$ small and fixed.

But this is a contradiction to the basic assumption $\gamma(\omega)<+\infty$.
(This step of the proof should be compared with [8, $(A.1)$, p. 2171] and the
proof of [8, Lemma A.1, p. 2173]; formally there we have dealt with more
specific weight functions but the additional assumptions on the weights are
superfluous.) ∎

######  Example 3.4.

Consider the Gevrey weights $\omega_{s}(t):=t^{1/s}$, $s>0$. Each $\omega_{s}$
is a BMT-weight function; if $s>1$ then $\omega_{s}$ is non-quasianalytic and
for all $0<s\leq 1$ the weight is quasianalytic. It is straight-forward to see
that $\gamma(\omega_{s})=s$. Fix $s>0$ and by Theorem 3.3 for each
$a\in(0,s^{-1})$, and so $as<1,$ the weight
$\tau_{a}=\omega_{s}(t^{1/a})=t^{1/(sa)}$ has the desired property.

Note that for each $s\geq 1$ we have that $(\omega_{\operatorname{sub}})$ is
valid and for each $s>1$ even $(\omega_{\operatorname{snq}})$; see $(iii)$ in
Remark 2.3.

Summarizing, for any $s\geq 1$ the weight $\omega_{s}$ is (almost) subadditive
but, if $0<s<1$, then there does not exist a weight function $\tau$ such that
$\tau$ is almost subadditive and $\tau\hyperlink{sim}{\sim}\omega_{s}$.
Compare this also with $(iii)$ in Lemma 3.1. However, since for each such
small $s$ the weight is quasianalytic, formally this example does not solve
the question from [1, Rem. 1.5].

$\sigma_{s}(t):=\max\\{0,(\log(t))^{s}\\}$, $s>1$, is another prominent
standard example for a (normalized) BMT-weight function. However, here
$\gamma(\sigma_{s})=+\infty$ for any $s>1$ and hence Theorem 3.3 cannot be
applied.

######  Remark 3.5.

The (standard) assumptions on the weight $\omega$ in Theorem 3.3 are more
general than in [1] and, in particular, this statement applies to
quasianalytic weights as well since by $(i)$ in Remark 2.4 in this case
$\gamma(\omega)\leq 1$ is valid.

Moreover, if instead of almost subadditivity we consider the more general
condition (3.1), then Theorem 3.3 transfers to this situation with the
analogous proof; only the interval $\mathcal{I}_{\omega}$ changes. If
$q_{0}>\frac{1}{2}$, then we get
$\mathcal{I}_{\omega}=(0,\frac{1}{1+\log(q_{0})/\log(2)}\gamma(\omega)^{-1})$
(again with $\mathcal{I}_{\omega}=(0,+\infty)$ provided that
$\gamma(\omega)=0$): choose $a>0$ with
$\left(1+\frac{\log(q_{0})}{\log(2)}\right)\gamma(\omega)<\frac{1}{a}$ and so
by (2.3)

$\gamma(\tau_{a})=\gamma(\omega^{1/a})=a\gamma(\omega)<\frac{\log(2)}{\log(2)+\log(q_{0})}.$

The rest follows as before. If $q_{0}=\frac{1}{2}$, then we have
$\mathcal{I}_{\omega}=(0,+\infty)$ and any choice for $a>0$ is valid.

###  3.2. The non-quasianalytic weight sequence case

From now on we focus on the case $\omega=\omega_{M}$ for a specific sequence
$M\in\mathbb{R}_{>0}^{\mathbb{N}}$. The proofs are involved and use technical
preparations and therefore we split it into several steps. However, the ideas
and constructions can be useful for further applications.

######  Theorem 3.6.

There exist uncountable infinite many pair-wise different (i.e. non-
equivalent) weight sequences $M\in\hyperlink{LCset}{\mathcal{LC}}$ and
normalized and non-quasianalytic BMT-weight functions $\omega=\omega_{M}$ such
that each of these functions has the following property:

There does not exist a weight function $\tau$ such that $\tau$ is equivalent
to $\omega_{M}$ and $\tau$ is almost subadditive.

Note that $\omega_{s}$ in Example 3.4 corresponds, up to equivalence, to
$\omega_{G^{s}}$ with $G^{s}:=(j!^{s})_{j\in\mathbb{N}}$. But for $0<s\leq 1$
the weights $\omega_{s}$, $\omega_{G^{s}}$ (and the sequences $G^{s}$) are
quasianalytic and hence not an appropriate choice.

Proof. Step I \- Comments on sufficient growth and regularity properties.

By Lemma 3.1 in order to conclude it suffices to construct a normalized BMT-
weight function $\omega$ which is

  * $(*)$

non-quasianalytic and

  * $(*)$

such that $0<\gamma(\omega)<1$.

In order to achieve this goal the idea is now to consider
$\omega\equiv\omega_{M}$ with $M\in\hyperlink{LCset}{\mathcal{LC}}$ satisfying
explicit growth and regularity requirements and to define $M$ explicitly.

Step II \- Sufficient growth and regularity properties for the weight
sequence.

We want to find $M\in\hyperlink{LCset}{\mathcal{LC}}$ such that

  * $(a)$

$M$ satisfies (2.4),

  * $(b)$

$M$ satisfies $(\operatorname{nq})$,

  * $(c)$

$M$ satisfies $(\operatorname{mg})$,

  * $(d)$

$M$ satisfies

(3.7) $\exists\;0<\beta<1\;\forall\;\ell\in\mathbb{N}_{\geq
2}:\;\;\;\liminf_{j\rightarrow+\infty}\frac{\mu_{\ell
j}}{\mu_{j}}\leq\ell^{\beta}.$

Because for such $M$ we get the following properties; for the precise
definition of the growth index $\gamma(M)$, its comparison with
$\gamma(\omega_{M})$ and more details we refer to [6, Sect. 3-5]:

  * $(*)$

By $(a)$ and $(b)$ the function $\omega_{M}$ is a non-quasianalytic (and
normalized) BMT-weight (see Remark 2.5).

  * $(*)$

By $(c)$ and [6, Cor. 4.6 $(iii)$] we get $\gamma(M)=\gamma(\omega_{M})$.
Note: Without having $(\operatorname{mg})$ we only know
$\gamma(M)\leq\gamma(\omega_{M})$ and the difference can be large in general;
see [6, Cor. 4.6 $(i)$, Sect. 5].

  * $(*)$

By $(d)$ and [6, Thm. 3.11 $(v)\Leftrightarrow(viii)$] we see that
$\gamma(M)\leq\beta<1$. Recall that $M\in\hyperlink{LCset}{\mathcal{LC}}$
implies $\gamma(M)\geq 0$.

  * $(*)$

Note that in [6] a weight sequence denotes a sequence satisfying all
requirements from the class $\mathcal{LC}$ except necessarily $M_{0}\leq
M_{1}$ (see [6, Sect. 3.1]). Moreover, concerning the quotients an index-shift
appears; i.e. $m_{p}$ in [6] is corresponding to $\mu_{p+1}$. However, this
does not effect the value of the index and hence the conclusion; see [6, Rem.
3.8] and the comments after [6, Cor. 3.12].

Altogether, $\omega_{M}$ is a (normalized) non-quasianalytic BMT-weight
function satisfying $\gamma(\omega_{M})=\gamma(M)<1$ and by Step I this
function provides a convenient example. Thus it remains to construct
explicitly $M\in\hyperlink{LCset}{\mathcal{LC}}$ such that $(a)-(d)$ hold
true.

Step III \- Definition of the sequence $M$.

We introduce $M$ in terms of its quotient sequence
$\mu=(\mu_{k})_{k\in\mathbb{N}}$ (with $\mu_{0}:=1$) and so set
$M_{j}:=\prod_{k=0}^{j}\mu_{k}$, $j\in\mathbb{N}$. For this, first fix an
arbitrary real constant $A>2$. Second, we consider three auxiliary sequences
(of positive integers) $(a_{j})_{j\in\mathbb{N}_{>0}}$,
$(b_{j})_{j\in\mathbb{N}_{>0}}$ and $(d_{j})_{j\in\mathbb{N}_{>0}}$ with
$a_{j}<b_{j}<a_{j+1}$ for all $j\in\mathbb{N}_{>0}$ and being defined as
follows:

(3.8) $a_{1}:=1,\hskip 20.0ptb_{j}:=2^{d_{j}}a_{j},\hskip
20.0pta_{j+1}:=2^{j}b_{j},\hskip 20.0ptj\in\mathbb{N}_{>0}.$

For the sequence $(d_{j})_{j\in\mathbb{N}_{>0}}$ we assume that it is strictly
increasing with $d_{1}\geq 1$ and satisfying the following growth restriction:

(3.9)
$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\left(\frac{2}{A}\right)^{d_{j}}\leq\frac{1}{2^{\frac{j}{2}+1}}.$

This choice for $d_{j}$ is possible since $A>2$.

For notational reasons we split each interval $(a_{j},b_{j}]$,
$j\in\mathbb{N}_{>0}$, into $d_{j}$-many subintervals
$I_{a_{j},i}:=(2^{i}a_{j},2^{i+1}a_{j}]$, $i\in\mathbb{N}$ and $0\leq i\leq
d_{j}-1$. Similarly, each interval $(b_{j},a_{j+1}]$, $j\in\mathbb{N}_{>0}$,
is split into $j$-many subintervals $I_{b_{j},i}:=(2^{i}b_{j},2^{i+1}b_{j}]$,
$i\in\mathbb{N}$ and $0\leq i\leq j-1$. To be formally precise, we only
consider all integers belonging to these intervals.

We define $\mu$ on $a_{1}$ and on the intervals $(a_{j},b_{j}]$,
$j\in\mathbb{N}_{>0}$, as follows:

(3.10) $\mu_{a_{1}}:=2(=2a_{1}),\hskip
30.0pt\mu_{k}:=A\mu_{2^{i}a_{j}}=A^{i+1}\mu_{a_{j}},\;\;\;k\in
I_{a_{j},i},\;\;\;0\leq i\leq d_{j}-1,\;j\in\mathbb{N}_{>0}.$

This means that iteratively on $I_{a_{j},i}$, $i\geq 1$, the sequence of
quotients is given by multiplying the already defined quotients on
$I_{a_{j},i-1}$ with the constant $A$ and $\mu$ is constant on each
$I_{a_{j},i}$. And the values on $I_{a_{j},0}$ are obtained by multiplying
$\mu_{a_{j}}$ with $A$. In particular, we get

(3.11) $\forall\;j\in\mathbb{N}_{>0}:\;\;\;\mu_{b_{j}}=A^{d_{j}}\mu_{a_{j}}.$

Next, on each $(b_{j},a_{j+1}]$, $j\in\mathbb{N}_{>0}$, we introduce $\mu$ as
follows:

(3.12) $\mu_{k}:=\sqrt{2}^{\frac{1}{2^{i}b_{j}}}\mu_{k-1},\;\;\;2^{i}b_{j}\leq
k-1<k\leq 2^{i+1}b_{j},\hskip 20.0pt0\leq i\leq j-1,\;j\in\mathbb{N}_{>0}.$

Note: When considering in (3.12) the choices $i=j-1$ and
$k=2^{i+1}b_{j}=2^{j}b_{j}=a_{j+1}$ this gives the value of $\mu_{a_{j+1}}$
and so the definition of $M$ is complete.

Since
$\frac{\mu_{2k}}{\mu_{k}}=\frac{\mu_{2k}}{\mu_{2k-1}}\frac{\mu_{2k-1}}{\mu_{2k-2}}\cdots\frac{\mu_{k+1}}{\mu_{k}}$
for all $k\in\mathbb{N}_{>0}$ gives a product of $k$-many factors, (3.12)
yields

(3.13) $\forall\;j\in\mathbb{N}_{>0}\;\forall\;0\leq i\leq
j-1:\;\;\;\frac{\mu_{2^{i+1}b_{j}}}{\mu_{2^{i}b_{j}}}=\sqrt{2},$

and which implies by iteration

(3.14) $\forall\;j\in\mathbb{N}_{>0}\;\forall\;0\leq i\leq
j-1:\;\;\;\frac{\mu_{2^{i+1}b_{j}}}{\mu_{b_{j}}}=2^{\frac{i+1}{2}}.$

In particular, the choice $i=j-1$ gives

(3.15)
$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\mu_{a_{j+1}}=2^{\frac{j}{2}}\mu_{b_{j}}.$

Let us now show by induction the following consequence of (3.9) which is
needed in the estimates below:

(3.16)
$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\frac{a_{j}}{\mu_{a_{j}}}\leq\frac{1}{2^{j}}.$

The case $j=1$ follows by (3.10). Moreover, combining (3.8), (3.11) (3.15)
with the induction hypothesis and finally with (3.9) gives for all
$j\in\mathbb{N}_{>0}$

$\frac{a_{j+1}}{\mu_{a_{j+1}}}=\frac{2^{j}b_{j}}{2^{\frac{j}{2}}\mu_{b_{j}}}=\frac{2^{j}2^{d_{j}}a_{j}}{2^{\frac{j}{2}}A^{d_{j}}\mu_{a_{j}}}=\frac{a_{j}}{\mu_{a_{j}}}2^{\frac{j}{2}}\left(\frac{2}{A}\right)^{d_{j}}\leq\frac{1}{2^{\frac{j}{2}}}\left(\frac{2}{A}\right)^{d_{j}}\leq\frac{1}{2^{j+1}},$

as desired.

Step IV \- Verification of the desired properties for $M$. We divide this step
into several claims.

Claim a: $M\in\hyperlink{LCset}{\mathcal{LC}}$. This is obvious since
$M_{0}=\mu_{0}=1$ and $\frac{M_{1}}{M_{0}}=\mu_{1}=\mu_{a_{1}}=2$ ensure
normalization. Clearly $j\mapsto\mu_{j}$ is by construction non-decreasing and
satisfying $\lim_{j\rightarrow+\infty}\mu_{j}=+\infty$.

Claim b: $M$ satisfies $(\operatorname{nq})$.

On the one hand, by recalling definition (3.10) and by (3.16) we have for all
$j\in\mathbb{N}_{>0}$:

$\displaystyle\sum_{k=a_{j}+1}^{b_{j}}\frac{1}{\mu_{k}}$
$\displaystyle=\sum_{i=0}^{d_{j}-1}\sum_{k\in
I_{a_{j},i}}\frac{1}{\mu_{k}}=\sum_{i=0}^{d_{j}-1}\frac{2^{i}a_{j}}{A^{i+1}\mu_{a_{j}}}=\frac{a_{j}}{A\mu_{a_{j}}}\sum_{i=0}^{d_{j}-1}\left(\frac{2}{A}\right)^{i}\leq\frac{a_{j}}{A\mu_{a_{j}}}\sum_{i=0}^{+\infty}\left(\frac{2}{A}\right)^{i}$
$\displaystyle=\frac{a_{j}}{A\mu_{a_{j}}}\frac{A}{A-2}\leq\frac{1}{2^{j}}\frac{1}{A-2}.$

On the other hand, by (3.8), (3.12) and by combining (3.9) and (3.16) we get
for all $j\in\mathbb{N}_{>0}$:

$\sum_{k=b_{j}+1}^{a_{j+1}}\frac{1}{\mu_{k}}=\sum_{i=0}^{j-1}\sum_{k\in
I_{b_{j},i}}\frac{1}{\mu_{k}}\leq\frac{a_{j+1}-b_{j}}{\mu_{b_{j}}}=\frac{b_{j}(2^{j}-1)}{A^{d_{j}}\mu_{a_{j}}}=\left(\frac{2}{A}\right)^{d_{j}}\frac{a_{j}(2^{j}-1)}{\mu_{a_{j}}}\leq\frac{2^{j}-1}{2^{j+\frac{j}{2}+1}}\leq\frac{1}{2^{\frac{j}{2}+1}}.$

Summarizing, we obtain

$\displaystyle\sum_{k\geq 1}\frac{1}{\mu_{k}}$ $\displaystyle=\sum_{k\geq
a_{1}}\frac{1}{\mu_{k}}=\frac{1}{2}+\sum_{k>a_{1}}\frac{1}{\mu_{k}}\leq\frac{1}{2}+\frac{1}{A-2}\sum_{j\geq
1}\frac{1}{2^{j}}+\frac{1}{2}\sum_{j\geq 1}\frac{1}{2^{\frac{j}{2}}}$
$\displaystyle=\frac{A}{2(A-2)}+\frac{1}{2}\frac{1}{\sqrt{2}-1}<+\infty,$

and the claim is shown.

Claim c: $M$ satisfies $(\operatorname{mg})$.

Recall that in order to conclude we have to show
$\sup_{k\in\mathbb{N}}\frac{\mu_{2k}}{\mu_{k}}<+\infty$ and according to the
definition of $M$ we have to distinguish between several cases (also required
in Claim d below):

  * $(i)$

For $k=1$ we get
$\frac{\mu_{2k}}{\mu_{k}}=\frac{\mu_{2a_{1}}}{\mu_{a_{1}}}=A$, see (3.8) and
(3.10).

  * $(ii)$

When $k\in I_{a_{j},i}$ for $j\in\mathbb{N}_{>0}$ and $0\leq i\leq d_{j}-2$
arbitrary, then we get $2k\in I_{a_{j},i+1}$ and so
$\frac{\mu_{2k}}{\mu_{k}}=A$, see again (3.10).

  * $(iii)$

When $k\in I_{a_{j},d_{j}-1}$, $j\in\mathbb{N}_{>0}$ arbitrary, then $2k\in
I_{b_{j},0}$ and there exists $2\leq\ell\leq b_{j}$ such that $b_{j}+\ell=2k$.
This gives

$\frac{\mu_{2k}}{\mu_{k}}=\frac{\mu_{2k}}{\mu_{b_{j}}}\frac{\mu_{b_{j}}}{\mu_{k}}=\sqrt{2}^{\frac{\ell}{b_{j}}},$

see (3.10) and (3.12), and this equality implies the estimate
$\frac{\mu_{2k}}{\mu_{k}}\leq\sqrt{2}$.

  * $(iv)$

When $k\in I_{b_{j},j-1}$, $j\in\mathbb{N}_{>0}$ arbitrary, then $2k\in
I_{a_{j+1},0}$ and there exists $0\leq\ell\leq 2^{j-1}b_{j}-1$ such that
$k+\ell=a_{j+1}$. This implies

$\frac{\mu_{2k}}{\mu_{k}}=\frac{\mu_{a_{j+1}}}{\mu_{k}}\frac{\mu_{2k}}{\mu_{a_{j+1}}}=\sqrt{2}^{\frac{\ell}{2^{j-1}b_{j}}}A,$

see (3.10) and (3.12), and this equality yields the estimate
$\frac{\mu_{2k}}{\mu_{k}}\leq\sqrt{2}A$.

  * $(v)$

Finally, let $k\in I_{b_{j},i}$ with $0\leq i\leq j-2$ and
$j\in\mathbb{N}_{>0}$ arbitrary. In this case $2k\in I_{b_{j},i+1}$ and there
exists $0\leq\ell\leq 2^{i}b_{j}-1$ such that $k+\ell=2^{i+1}b_{j}$ and there
exists $2\leq\ell^{\prime}\leq 2^{i+1}b_{j}$ such that
$2^{i+1}b_{j}+\ell^{\prime}=2k$ (and $\ell,\ell^{\prime}$ satisfy the relation
$k+\ell+\ell^{\prime}=2k$). Thus we get

$\frac{\mu_{2k}}{\mu_{k}}=\frac{\mu_{2^{i+1}b_{j}}}{\mu_{k}}\frac{\mu_{2k}}{\mu_{2^{i+1}b_{j}}}=\sqrt{2}^{\frac{\ell}{2^{i}b_{j}}}\sqrt{2}^{\frac{\ell^{\prime}}{2^{i+1}b_{j}}},$

see (3.12). This equality implies the estimate

$\frac{\mu_{2k}}{\mu_{k}}\leq\sqrt{2}^{\frac{\ell}{2^{i}b_{j}}}\sqrt{2}^{\frac{\ell^{\prime}}{2^{i}b_{j}}}=\sqrt{2}^{\frac{\ell+\ell^{\prime}}{2^{i}b_{j}}}=\sqrt{2}^{\frac{k}{2^{i}b_{j}}}\leq\sqrt{2}^{\frac{2^{i+1}b_{j}}{2^{i}b_{j}}}=2.$

Summarizing, we have verified $\frac{\mu_{2k}}{\mu_{k}}\leq\sqrt{2}A$ for all
$k\in\mathbb{N}_{>0}$ and we are done. (The case $k=0$ is trivial and gives
$\frac{\mu_{2k}}{\mu_{k}}=1$).

Claim d: $M$ satisfies (2.4).

We verify (2.4) for the choice $Q=4$ and use the equalities and explanations
from $(i)-(v)$ in Claim c above. (In fact even $Q=2$ is sufficient to treat
all cases except $(iii)$.)

For all $k$ as considered in $(ii)$ in Claim c we get with $Q=2$ the same
equality and for $k=1$ in case $(i)$, too. The equality in $(iv)$ in Claim c
yields $\frac{\mu_{2k}}{\mu_{k}}=\sqrt{2}^{\frac{\ell}{2^{j-1}b_{j}}}A\geq A$
and analogous estimates as in $(v)$ in Claim c give

$\frac{\mu_{2k}}{\mu_{k}}\geq\sqrt{2}^{\frac{\ell}{2^{i+1}b_{j}}}\sqrt{2}^{\frac{\ell^{\prime}}{2^{i+1}b_{j}}}=\sqrt{2}^{\frac{\ell+\ell^{\prime}}{2^{i+1}b_{j}}}=\sqrt{2}^{\frac{k}{2^{i+1}b_{j}}}\geq\sqrt{2}^{\frac{2^{i}b_{j}}{2^{i+1}b_{j}}}=2^{1/4}.$

Finally, let $k$ be as in case $(iii)$ in Claim c. Then $4k\in I_{b_{j},1}$
and so

$\frac{\mu_{4k}}{\mu_{k}}=\frac{\mu_{4k}}{\mu_{b_{j}}}\geq\frac{\mu_{2b_{j}}}{\mu_{b_{j}}}=\sqrt{2},$

see (3.13).

Summarizing, even $\inf_{j\rightarrow+\infty}\frac{\mu_{4j}}{\mu_{j}}\geq
2^{1/4}>1$ is verified and thus the claim holds true.

Claim e: $M$ satisfies (3.7).

Let $\ell\in\mathbb{N}_{\geq 2}$ be given, arbitrary but from now on fixed.
Consider $n\in\mathbb{N}_{>0}$ such that $2^{n}\leq\ell<2^{n+1}$ is valid and
distinguish: When $\ell=2^{n}$ for some $n\in\mathbb{N}_{>0}$, in particular
when $\ell=2$, then (3.14) implies for all $j\geq n$

$\frac{\mu_{\ell
b_{j}}}{\mu_{b_{j}}}=\frac{\mu_{2^{n}b_{j}}}{\mu_{b_{j}}}=2^{n/2}=\sqrt{\ell},$

which verifies the $\liminf$-condition in (3.7) with $\beta=\frac{1}{2}$.

When $2^{n}<\ell<2^{n+1}$, then for all $j\geq n+1$ property (3.14) (and the
fact that $k\mapsto\mu_{k}$ is non-decreasing) yields

$\frac{\mu_{\ell
b_{j}}}{\mu_{b_{j}}}\leq\frac{\mu_{2^{n+1}b_{j}}}{\mu_{b_{j}}}=2^{\frac{n+1}{2}}=\sqrt{2}2^{\frac{n}{2}}\leq\sqrt{2\ell}\leq\ell^{\beta},$

where we have put $\beta:=\frac{1}{2}+\frac{1}{2}\frac{\log(2)}{\log(3)}$ and
therefore $\frac{1}{2}<\beta<1$. Note that with this choice for $\beta$ the
last estimate above is equivalent to $2\leq\ell^{\log(2)/\log(3)}$ and so to
$\log(3)\leq\log(\ell)$ which is obviously true for all $\ell$ under
consideration since we have $\ell\in\mathbb{N}_{\geq 3}$.

Summarizing, (3.7) is verified with
$\beta:=\frac{1}{2}+\frac{1}{2}\frac{\log(2)}{\log(3)}$.

Step V \- There exist uncountable infinite many pair-wise different weight
sequences/functions satisfying the required properties.

Given a parameter $A>2$, then in view of (3.9) for all $A^{\prime}>A$ we can
take the same auxiliary sequence $(d_{j})_{j\in\mathbb{N}_{>0}}$ and hence
also the sequences $(a_{j})_{j\in\mathbb{N}_{>0}}$ and
$(b_{j})_{j\in\mathbb{N}_{>0}}$ remain unchanged; see (3.8). So, when fixing
$A_{0}>2$ and varying $A$ in $[A_{0},+\infty)$, then in the construction all
claims stay valid and it is easy to compare the quotient sequences since the
positions where the growth behavior of this sequence is changing remain
invariant.

Consider now $A^{\prime}>A>2$ arbitrary but from now on fixed, then we write
$M^{A}$ and $M^{A^{\prime}}$ for the corresponding sequences and similarly
$\mu^{A}$ and $\mu^{A^{\prime}}$ for the corresponding sequences of quotients.
By (3.11) and (3.15) one has

$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\frac{\mu^{A^{\prime}}_{a_{j+1}}}{\mu^{A}_{a_{j+1}}}=\frac{2^{\frac{j}{2}}(A^{\prime})^{d_{j}}\mu^{A^{\prime}}_{a_{j}}}{2^{\frac{j}{2}}A^{d_{j}}\mu^{A}_{a_{j}}}=\left(\frac{A^{\prime}}{A}\right)^{d_{j}}\frac{\mu^{A^{\prime}}_{a_{j}}}{\mu^{A}_{a_{j}}}.$

Iterating this procedure we immediately get

$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\frac{\mu^{A^{\prime}}_{a_{j+1}}}{\mu^{A}_{a_{j+1}}}=\left(\frac{A^{\prime}}{A}\right)^{d_{j}+d_{j-1}+\dots+d_{1}}\frac{\mu^{A^{\prime}}_{a_{1}}}{\mu^{A}_{a_{1}}}=\left(\frac{A^{\prime}}{A}\right)^{d_{j}+d_{j-1}+\dots+d_{1}},$

hence
$\limsup_{k\rightarrow+\infty}\frac{\mu^{A^{\prime}}_{k}}{\mu^{A}_{k}}=+\infty$
is verified.

Next, we claim that this property implies that $M^{A}$ and $M^{A^{\prime}}$
are not equivalent: Both sequences satisfy $(\operatorname{mg})$ (see Claim c)
and since $M^{A^{\prime}}\in\hyperlink{LCset}{\mathcal{LC}}$ condition
$(\operatorname{mg})$ is equivalent to
$\sup_{k\in\mathbb{N}_{>0}}\frac{\mu^{A^{\prime}}_{k}}{(M^{A^{\prime}}_{k})^{1/k}}<+\infty$,
see e.g. again [14, Lemma 2.2]. Moreover, by log-convexity and normalization
we clearly have $(M^{A}_{k})^{1/k}\leq\mu^{A}_{k}$ for all
$k\in\mathbb{N}_{>0}$ and hence

(3.17)
$\sup_{j\in\mathbb{N}_{>0}}\left(\frac{M^{A^{\prime}}_{j}}{M^{A}_{j}}\right)^{1/j}=+\infty.$

Thus the sequences are not equivalent.

Now recall that $M\in\hyperlink{LCset}{\mathcal{LC}}$ has
$(\operatorname{mg})$ if and only if $M$ is equivalent to some/all auxiliary
sequences $\widetilde{M}^{c}$ with $\widetilde{M}^{c}_{j}:=(M_{cj})^{1/c}$,
$c\in\mathbb{N}_{>0}$; see [18, Lemma 2.2].

Consequently, since $M^{A}$ has $(\operatorname{mg})$, we see that the
estimate

(3.18) $\exists\;c\in\mathbb{N}_{>0}\;\exists\;B\geq
1\;\forall\;j\in\mathbb{N}:\;\;\;M^{A^{\prime}}_{j}\leq B(M^{A}_{cj})^{1/c}$

has to be violated: Otherwise $M^{A^{\prime}}_{j}\leq B(M^{A}_{cj})^{1/c}\leq
BC^{j}M^{A}_{j}$ for some $B,C\geq 1$ and all $j\in\mathbb{N}$ and this
contradicts (3.17). (3.18) is precisely [5, $(6.4)$] for $N\equiv
M^{A^{\prime}}$ and $M\equiv M^{A}$. Therefore, the first part in [5, Lemma
6.5] implies that $\omega_{M^{A}}(t)=O(\omega_{M^{A^{\prime}}}(t))$ as
$t\rightarrow+\infty$ fails and so $\omega_{M^{A}}$ and
$\omega_{M^{A^{\prime}}}$ are not equivalent. ∎

###  3.3. The quasianalytic weight sequence case

The aim is to transfer Theorem 3.6 also to quasianalytic BMT-weight functions;
so we illustrate independently that the question whether there does always
exist an equivalent almost subadditive BMT-weight function is not depending on
the notion of (non-)quasianalyticity.

First we comment in more detail on the growth of $j\mapsto\frac{\mu_{j}}{j}$
in the example of Theorem 3.6. Given $M\in\hyperlink{LCset}{\mathcal{LC}}$,
then $\omega_{M}$ satisfies $(\omega_{5})$ if and only if
$\lim_{j\rightarrow+\infty}(M_{j}/j!)^{1/j}=+\infty$, see e.g. [17, Lemma 2.8]
and [7, Lemma 2.4] and the references mentioned in the proofs there.

Moreover, $\lim_{j\rightarrow+\infty}(M_{j}/j!)^{1/j}=+\infty$ if and only if
$\lim_{j\rightarrow+\infty}\frac{\mu_{j}}{j}=+\infty$; see e.g. [13, p. 104]
and recall that by Stirling’s formula $j\mapsto j!^{1/j}$ and $j\mapsto j$
grow comparable up to some positive constant.

For $M\in\hyperlink{LCset}{\mathcal{LC}}$ it is known that non-
quasianalyticity implies
$\lim_{j\rightarrow+\infty}\frac{\mu_{j}}{j}=+\infty$; see e.g. [17, Prop.
4.4]. The growth of $j\mapsto\frac{\mu_{j}}{j}$ is crucial to see if the real-
analytic functions are (strictly) contained in the corresponding
ultradifferentiable class; more precisely
$\lim_{j\rightarrow+\infty}\frac{\mu_{j}}{j}=+\infty$ if and only if the real-
analytic functions are contained in the function classes under consideration,
see e.g. again [13, Prop. 2.12, p. 104].

If $0<s<1$, then for $\omega_{s}$ in Example 3.4 corresponding to
$\omega_{G^{s}}$ with $G^{s}:=(j!^{s})_{j\in\mathbb{N}}$ we get that
$\lim_{j\rightarrow+\infty}(j!^{s}/j!)^{1/j}=0$. If $s=1$, then the limit is
equal to $1$.

Let now, more generally, the parameter $A>1$ be arbitrary (but fixed) and set
$a_{j+1}:=2^{c_{j}}b_{j}$ for all $j\in\mathbb{N}_{>0}$; i.e. $j$ in the
exponent is replaced by a fourth auxiliary sequence of integers
$(c_{j})_{\in\mathbb{N}_{>0}}$. Note that the conclusion of Claim e in Step IV
in Theorem 3.6 stays valid when replacing $j$ by any arbitrary non-decreasing
$(c_{j})_{j\in\mathbb{N}_{>0}}$ satisfying $c_{j}\geq j$.

  * $(*)$

On each interval $I_{a_{j},i}$ the mapping $k\mapsto\frac{\mu_{k}}{k}$ is
clearly non-increasing.

  * $(*)$

If $k=2^{i}a_{j}$ for some $j\in\mathbb{N}_{>0}$ and $0\leq i\leq d_{j}-1$,
i.e. positions where a jump occurs, then
$\frac{\mu_{k}}{k}\leq\frac{\mu_{k+1}}{k+1}\Leftrightarrow 1+\frac{1}{k}\leq
A$. This estimate is valid if $A\geq 2$ for all $k\in\mathbb{N}_{>0}$ (as in
Theorem 3.6) and, if $A>1$, for all $k$ sufficiently large depending on chosen
$A$.

Moreover,
$\frac{\mu_{2^{i}a_{j}}}{2^{i}a_{j}}\leq\frac{\mu_{2^{i+1}a_{j}}}{2^{i+1}a_{j}}\Leftrightarrow
2\leq A$, for all $j\in\mathbb{N}_{>0}$ and $0\leq i\leq d_{j}-1$.

  * $(*)$

By (3.8) and (3.11) we see

$\frac{\mu_{b_{j}}}{b_{j}}=\left(\frac{A}{2}\right)^{d_{j}}\frac{\mu_{a_{j}}}{a_{j}},$

and so $\frac{\mu_{b_{j}}}{b_{j}}<\frac{\mu_{a_{j}}}{a_{j}}$ if $A<2$ and
$\frac{\mu_{b_{j}}}{b_{j}}\geq\frac{\mu_{a_{j}}}{a_{j}}$ if $A\geq 2$.

  * $(*)$

In view of (3.12) we have for all $2^{i}b_{j}\leq k<k+1\leq 2^{i+1}b_{j}$,
$j\in\mathbb{N}_{>0}$ and $0\leq i\leq c_{j}-1$, that
$\frac{\mu_{k}}{k}\geq\frac{\mu_{k+1}}{k+1}\Leftrightarrow
1+\frac{1}{k}\geq\sqrt{2}^{\frac{1}{2^{i}b_{j}}}\Leftrightarrow\left(1+\frac{1}{k}\right)^{2^{i+1}b_{j}}\geq
2$. Finally note that $k\leq 2^{i+1}b_{j}$ and $k\geq b_{1}\geq 2a_{1}=2$ and
so on each interval $I_{b_{j},i}$ the mapping $k\mapsto\frac{\mu_{k}}{k}$ is
non-increasing.

  * $(*)$

If $k=2^{i+1}b_{j}$, $j\in\mathbb{N}_{>0}$ and $0\leq i\leq c_{j}-2$, then
$\frac{\mu_{k}}{k}\geq\frac{\mu_{k+1}}{k+1}\Leftrightarrow
1+\frac{1}{k}\geq\sqrt{2}^{\frac{1}{2^{i+1}b_{j}}}$ holds as shown before. If
$k=2^{i+1}b_{j}$, $j\in\mathbb{N}_{>0}$ and $i=c_{j}-1$ and so $k=a_{j+1}$,
then $\frac{\mu_{k}}{k}\leq\frac{\mu_{k+1}}{k+1}\Leftrightarrow
1+\frac{1}{k}\leq A$ which holds for $A>1$ for all $k$ sufficiently large.

  * $(*)$

By (3.15) we get

$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\frac{\mu_{a_{j+1}}}{a_{j+1}}=\frac{2^{\frac{c_{j}}{2}}}{2^{c_{j}}}\frac{\mu_{b_{j}}}{b_{j}}=\frac{1}{2^{\frac{c_{j}}{2}}}\frac{\mu_{b_{j}}}{b_{j}}<\frac{\mu_{b_{j}}}{b_{j}}.$

Now we formulate and prove the second main statement concerning weights
$\omega_{M}$.

######  Theorem 3.7.

There exist (normalized) quasianalytic BMT-weight functions
$\omega=\omega_{M}$ which are not equivalent to an almost sub-additive weight
function.

In addition, we can achieve different situations concerning the growth of
$k\mapsto\frac{\mu_{k}}{k}$:

  * $(*)$

Case $(a)$: $\lim_{k\rightarrow+\infty}\frac{\mu_{k}}{k}=0$.

  * $(*)$

Case $(b)$: $\lim_{k\rightarrow+\infty}\frac{\mu_{k}}{k}=+\infty$.

  * $(*)$

Case $(c)$: $\liminf_{k\rightarrow+\infty}\frac{\mu_{k}}{k}=0$ and
$\limsup_{k\rightarrow+\infty}\frac{\mu_{k}}{k}=+\infty$.

Proof. For all cases we use the analogous constructions and definitions as in
Theorem 3.6 but apply some technical modifications/generalizations and involve
the comments on monotonicity before.

Case a: We take the sequences $(a_{j})_{j\in\mathbb{N}_{>0}}$,
$(b_{j})_{j\in\mathbb{N}_{>0}}$ and $(d_{j})_{j\in\mathbb{N}_{>0}}$ as
considered in (3.8) but then let us fix some $1<A_{0}<2$ and write
$q:=\frac{2}{A_{0}}$. For $(d_{j})_{j\in\mathbb{N}_{>0}}$ we assume that the
sequence is (strictly) increasing with $d_{1}\geq 1$. We also set
$\mu_{a_{1}}=\mu_{1}:=1$ as ”initial condition” instead of $\mu_{a_{1}}=2$
(which is required to prove (3.16) for $j=1$ but in the recent setting this
property fails in any case as we are going to see). Next let us show that

(3.19) $\forall\;j\in\mathbb{N}_{>0}\;\forall\;1<A\leq
A_{0}:\;\;\;\frac{a_{j}}{\mu_{a_{j}}}\frac{(2/A)^{d_{j}}-1}{(2/A)-1}\geq 1.$

First, since $\sum_{i=0}^{k}t^{i}=\frac{t^{k+1}-1}{t-1}$ for any $t>1$ we see
that $t\mapsto\frac{t^{k+1}-1}{t-1}$ is non-decreasing on $(1,+\infty)$ for
any fixed $k\in\mathbb{N}$ and hence in (3.19) is suffices to consider
$q=\frac{2}{A_{0}}$. Then we use induction on $j$: The case $j=1$ gives
$\frac{a_{1}}{\mu_{a_{1}}}\frac{q^{d_{1}}-1}{q-1}=\frac{q^{d_{1}}-1}{q-1}\geq
1$ which is clear. For the induction step we have

$\frac{a_{j+1}}{\mu_{a_{j+1}}}=\frac{2^{j}b_{j}}{2^{\frac{j}{2}}\mu_{b_{j}}}=\frac{2^{j}2^{d_{j}}a_{j}}{2^{\frac{j}{2}}A^{d_{j}}\mu_{a_{j}}}=2^{\frac{j}{2}}\left(\frac{2}{A}\right)^{d_{j}}\frac{a_{j}}{\mu_{a_{j}}}\geq
2^{\frac{j}{2}}q^{d_{j}}\frac{a_{j}}{\mu_{a_{j}}}\geq
2^{\frac{j}{2}}q^{d_{j}}\frac{q-1}{q^{d_{j}}-1}\geq\frac{q-1}{q^{d_{j+1}}-1},$

where the last desired estimate is equivalent to having
$2^{\frac{j}{2}}(q^{d_{j+1}}-1)\geq 1-\frac{1}{q^{d_{j}}}$ and this clearly
holds true. (However, note that for parameters $A\leq 2$ the choice (3.9) and
hence the estimate (3.16) is impossible.)

Then fix a real parameter $A$ with $1<A\leq A_{0}$ and follow the proof of
Step IV. All parts except Claim b follow by a word-by-word repetition of the
given arguments there. Concerning Claim a note that normalization is still
valid by $\mu_{1}=\mu_{a_{1}}=1$ and in Claim d we have
$\liminf_{j\rightarrow+\infty}\frac{\mu_{4j}}{\mu_{j}}\geq\min\\{2^{1/4},A\\}>1$;
so the value $2^{1/4}$ has to be replaced by $A$ if $A<2^{1/4}$ which is
excluded by the choice $A>2$ in the proof before.

However, $(\operatorname{nq})$ fails: By (3.19) we estimate as follows for all
$j\in\mathbb{N}_{>0}$:

$\displaystyle\sum_{k=a_{j}+1}^{b_{j}}\frac{1}{\mu_{k}}$
$\displaystyle=\sum_{i=0}^{d_{j}-1}\sum_{k\in
I_{a_{j},i}}\frac{1}{\mu_{k}}=\sum_{i=0}^{d_{j}-1}\frac{2^{i}a_{j}}{A^{i+1}\mu_{a_{j}}}=\frac{a_{j}}{A\mu_{a_{j}}}\sum_{i=0}^{d_{j}-1}\left(\frac{2}{A}\right)^{i}=\frac{a_{j}}{A\mu_{a_{j}}}\frac{(2/A)^{d_{j}}-1}{(2/A)-1}\geq\frac{1}{A}.$

Since all appearing summands in the series under consideration are non-
negative the previous estimate immediately verifies $\sum_{k\geq
1}\frac{1}{\mu_{k}}=+\infty$ as desired.

Step V follows again by varying the parameter $A\in(1,A_{0}]$; note that the
choice of the numbers $d_{j}$ is not depending on given $A$ and so a direct
comparison of the quotients is again possible. Finally, we get

$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\frac{\mu_{a_{j+1}}}{a_{j+1}}=\frac{1}{2^{\frac{j}{2}}}\frac{\mu_{b_{j}}}{b_{j}}=\left(\frac{A}{2}\right)^{d_{j}}\frac{1}{2^{\frac{j}{2}}}\frac{\mu_{a_{j}}}{a_{j}},$

and so by taking into account the comments just before the statement in this
section it follows that $\lim_{k\rightarrow+\infty}\frac{\mu_{k}}{k}=0$ (for
any parameter $A\in(1,A_{0}]$).

Case b: First let us fix a value $A>2$. Second, we introduce four auxiliary
sequences (of positive integers) $(a_{j})_{j\in\mathbb{N}_{>0}}$,
$(b_{j})_{j\in\mathbb{N}_{>0}}$, $(c_{j})_{j\in\mathbb{N}_{>0}}$ and
$(d_{j})_{j\in\mathbb{N}_{>0}}$ with $a_{j}<b_{j}<a_{j+1}$ for all
$j\in\mathbb{N}_{>0}$ and being defined as follows:

(3.20) $a_{1}:=1,\hskip 20.0ptb_{j}:=2^{d_{j}}a_{j},\hskip
20.0pta_{j+1}:=2^{c_{j}}b_{j},\hskip 20.0ptj\in\mathbb{N}_{>0}.$

Thus $2^{j}$ in (3.8) is replaced by $2^{c_{j}}$ and so $(b_{j},a_{j+1}]$,
$j\in\mathbb{N}_{>0}$, is now split into $c_{j}$-many subintervals
$I_{b_{j},i}:=(2^{i}b_{j},2^{i+1}b_{j}]$, $j\in\mathbb{N}_{>0}$,
$i\in\mathbb{N}$ with $0\leq i\leq c_{j}-1$.

The definition of $\mu$ is then analogous according to (3.10) and (3.12) and
again we consider the ”initial condition” $\mu_{a_{1}}=\mu_{1}:=1$. Hence

(3.21)
$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\mu_{b_{j}}=A^{d_{j}}\mu_{a_{j}},\hskip
30.0pt\mu_{a_{j+1}}=2^{\frac{c_{j}}{2}}\mu_{b_{j}},$

meaning that (3.11) remains unchanged and in (3.15) the power $j$ is replaced
by $c_{j}$. However, the crucial difference compared with the construction in
Theorem 3.6 is now that we choose $(d_{j})_{j\in\mathbb{N}_{>0}}$,
$(c_{j})_{j\in\mathbb{N}_{>0}}$ iteratively as follows: Both sequences are
required to be strictly increasing with

(3.22) $d_{1}\geq 1,\hskip 30.0ptc_{j}\geq
j,\;\;\;\forall\;j\in\mathbb{N}_{>0},$

such that

(3.23)
$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\frac{\mu_{b_{j}}}{b_{j}}\geq\sqrt{2}^{c_{j}-1}\log(j+1),$

and

(3.24)
$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\sqrt{2}^{c_{j}-1}\geq\frac{\mu_{b_{j}}}{b_{j}}\frac{1}{j\log(j+1)}+1.$

These choices are possible: Recall that $A>2$ and
$\frac{\mu_{b_{j}}}{b_{j}}=\frac{\mu_{a_{j}}}{a_{j}}\left(\frac{A}{2}\right)^{d_{j}}$
for all $j\in\mathbb{N}_{>0}$. Hence, when given $\mu_{a_{j}}$ and $a_{j}$,
then the difference
$\frac{\mu_{b_{j}}}{b_{j}}\frac{1}{\log(j+1)}-\frac{\mu_{b_{j}}}{b_{j}}\frac{1}{j\log(j+1)}-1$
becomes as large as desired when $d_{j}$ is chosen sufficiently large. Since
$c_{j}$ is only required to define $\mu_{a_{j+1}}$, $a_{j+1}$, and so not
depending on $\mu_{b_{j}}$, $b_{j}$, both (3.23) and (3.24) can be guaranteed.

All properties from Step IV except Claim b dealing with $(\operatorname{nq})$
follow analogously as shown in the claims there; for Claim e note that
$c_{j}\geq j$ for all $j$. Concerning the failure of $(\operatorname{nq})$ by
definition, (3.24) (and recall (3.14)) we estimate as follows for all
$j\in\mathbb{N}_{>0}$:

$\displaystyle\sum_{k=b_{j}+1}^{a_{j+1}}\frac{1}{\mu_{k}}$
$\displaystyle=\sum_{i=0}^{c_{j}-1}\sum_{k\in
I_{b_{j},i}}\frac{1}{\mu_{k}}\geq\sum_{i=0}^{c_{j}-1}\frac{2^{i+1}b_{j}-2^{i}b_{j}}{\mu_{2^{i+1}b_{j}}}=\sum_{i=0}^{c_{j}-1}\frac{2^{i}b_{j}}{2^{\frac{i+1}{2}}\mu_{b_{j}}}$
$\displaystyle=\frac{1}{\sqrt{2}}\frac{b_{j}}{\mu_{b_{j}}}\sum_{i=0}^{c_{j}-1}2^{\frac{i}{2}}=\frac{1}{\sqrt{2}}\frac{b_{j}}{\mu_{b_{j}}}\frac{\sqrt{2}^{c_{j}}-1}{\sqrt{2}-1}\geq\frac{1}{\sqrt{2}}\frac{b_{j}}{\mu_{b_{j}}}\frac{\sqrt{2}^{c_{j}}-\sqrt{2}}{\sqrt{2}-1}$
$\displaystyle=\frac{b_{j}}{\mu_{b_{j}}}\frac{\sqrt{2}^{c_{j}-1}-1}{\sqrt{2}-1}\geq\frac{1}{\sqrt{2}-1}\frac{1}{j\log(j+1)}.$

Since all appearing summands in the series under consideration are non-
negative the previous estimate immediately verifies $\sum_{k\geq
1}\frac{1}{\mu_{k}}=+\infty$ as desired.

Let us see that here we get
$\lim_{k\rightarrow+\infty}\frac{\mu_{k}}{k}=+\infty$ as well: By definition
$\frac{\mu_{b_{j}}}{b_{j}}=\frac{\mu_{a_{j}}}{a_{j}}\left(\frac{A}{2}\right)^{d_{j}}\rightarrow+\infty$
as $j\rightarrow+\infty$, whereas (3.20), (3.21) and (3.23) yield

$\frac{\mu_{a_{j+1}}}{a_{j+1}}=\frac{2^{\frac{c_{j}}{2}}\mu_{b_{j}}}{2^{c_{j}}b_{j}}=\frac{1}{2^{\frac{c_{j}}{2}}}\frac{\mu_{b_{j}}}{b_{j}}\geq\frac{1}{\sqrt{2}}\log(j+1)\rightarrow+\infty$

as $j\rightarrow+\infty$. By the comments made on the beginning of this
section we are done.

Case c: Here we take the same construction and definition as in Case b but for
the sequences $(d_{j})_{j\in\mathbb{N}_{>0}}$ and
$(c_{j})_{j\in\mathbb{N}_{>0}}$, we assume the following: Again we require
(3.22) but instead of (3.23) and (3.24) iteratively we choose
$(d_{j})_{j\in\mathbb{N}_{>0}}$ and $(c_{j})_{j\in\mathbb{N}_{>0}}$ to grow
fast enough to ensure

(3.25)
$\forall\;j\in\mathbb{N}_{>0}:\;\;\;\frac{\mu_{b_{j}}}{b_{j}}=\frac{\mu_{a_{j}}}{a_{j}}\left(\frac{A}{2}\right)^{d_{j}}>j,$

and

(3.26) $\forall\;j\in\mathbb{N}_{>0}:\;\;\;\sqrt{2}^{c_{j}}\geq
j\frac{\mu_{b_{j}}}{b_{j}}.$

(3.25) implies $\limsup_{k\rightarrow+\infty}\frac{\mu_{k}}{k}=+\infty$ and
(3.26) yields, by taking into account (3.20) and (3.21), that

$\frac{\mu_{a_{j+1}}}{a_{j+1}}=\frac{1}{2^{\frac{c_{j}}{2}}}\frac{\mu_{b_{j}}}{b_{j}}\leq\frac{1}{j},$

and so $\liminf_{k\rightarrow+\infty}\frac{\mu_{k}}{k}=0$ follows.

Now let us see that (3.26) already implies (3.24) for all $j$ large and this
is sufficient to verify the quasianalyticity as in Case b: We have
$\sqrt{2}^{c_{j}-1}-1\geq\frac{j}{\sqrt{2}}\frac{\mu_{b_{j}}}{b_{j}}-1$ and
clearly, in view of (3.25),
$\frac{j}{\sqrt{2}}\frac{\mu_{b_{j}}}{b_{j}}-1\geq\frac{\mu_{b_{j}}}{b_{j}}\frac{1}{j\log(j+1)}\Leftrightarrow\frac{\mu_{b_{j}}}{b_{j}}\left(\frac{j}{\sqrt{2}}-\frac{1}{j\log(j+1)}\right)\geq
1$ for all $j$ large enough. ∎

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