# The Borel map in the mixed Beurling setting

David Nicolas Nenning, Armin Rainer, and Gerhard Schindl  Fakultät für
Mathematik, Universität Wien, Oskar-Morgenstern-Platz 1, A-1090 Wien, Austria.
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

The Borel map takes a smooth function to its infinite jet of derivatives (at
zero). We study the restriction of this map to ultradifferentiable classes of
Beurling type in a very general setting which encompasses the classical
Denjoy–Carleman and Braun–Meise–Taylor classes. More precisely, we
characterize when the Borel image of one class covers the sequence space of
another class in terms of the two weights that define the classes. We present
two independent solutions to this problem, one by reduction to the Roumieu
case and the other by dualization of the involved Fréchet spaces, a
Phragmén–Lindelöf theorem, and Hörmander’s solution of the
$\overline{\partial}$-problem.

###### Key words and phrases:

Ultradifferentiable function classes, Beurling type, Borel map, extension
results, mixed setting, controlled loss of regularity

###### 2020 Mathematics Subject Classification:

26E10, 46A13, 46E10, 46E25

AR was supported by FWF-Project P 32905-N, DNN and GS by FWF-Project P 33417-N

##  1\. Introduction

The Borel map
$j^{\infty}_{0}:C^{\infty}(\mathbb{R})\to\mathbb{C}^{\mathbb{N}}$ at $0$ is
defined by $j^{\infty}_{0}f:=(f^{(n)}(0))_{n\in\mathbb{N}}$. We will study the
restriction of $j^{\infty}_{0}$ to ultradifferentiable classes in a general
setting which allows us to treat the classical Denjoy–Carleman and
Braun–Meise–Taylor classes at the same time. Our classes are defined in terms
of one-parameter families $\mathfrak{M}=(M^{(x)})_{x>0}$,
$\mathfrak{N}=(N^{(x)})_{x>0}$, etc., of weight sequences; we call them weight
matrices. In this article, we are mostly interested in classes of Beurling
type

$\mathcal{E}^{(\mathfrak{N})}(\mathbb{R}):=\Big{\\{}f\in
C^{\infty}(\mathbb{R}):\;\forall j,k,l\in\mathbb{N}_{\geq
1}:\sup_{x\in[-j,j]}\sup_{p\in\mathbb{N}}\frac{|f^{(p)}(x)|}{(\frac{1}{l})^{p}N^{(\frac{1}{k})}_{p}}<\infty\Big{\\}},$

but we shall have to use also results on the Roumieu counterpart

$\mathcal{E}^{\\{\mathfrak{N}\\}}(\mathbb{R}):=\Big{\\{}f\in
C^{\infty}(\mathbb{R}):\;\forall j\in\mathbb{N}_{\geq 1}\;\exists
k,l\in\mathbb{N}_{\geq
1}:\sup_{x\in[-j,j]}\sup_{p\in\mathbb{N}}\frac{|f^{(p)}(x)|}{l^{p}N^{(k)}_{p}}<\infty\Big{\\}}.$

By definition, the image
$j^{\infty}_{0}\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$ is contained in the
sequence space

$\Lambda^{(\mathfrak{N})}:=\Big{\\{}a=(a_{p})\in\mathbb{C}^{\mathbb{N}}:\;\forall
k,l\in\mathbb{N}_{\geq
1}:\sup_{p\in\mathbb{N}}\frac{|a_{p}|}{(\frac{1}{l})^{p}N^{(\frac{1}{k})}_{p}}<\infty\Big{\\}};$

and likewise
$j^{\infty}_{0}\mathcal{E}^{\\{\mathfrak{N}\\}}(\mathbb{R})\subseteq\Lambda^{\\{\mathfrak{N}\\}}$,
where $\Lambda^{\\{\mathfrak{N}\\}}$ is defined analogously. The goal of this
paper is to find necessary and sufficient conditions for

$\Lambda^{(\mathfrak{M})}\subseteq
j^{\infty}_{0}\mathcal{E}^{(\mathfrak{N})}(\mathbb{R}),$ (1.1)

in terms of $\mathfrak{M}$ and $\mathfrak{N}$.

The Roumieu case is well understood, see our recent article [24]: under some
mild assumptions on $\mathfrak{M}$ and $\mathfrak{N}$, we have
$\Lambda^{\\{\mathfrak{M}\\}}\subseteq
j^{\infty}_{0}\mathcal{E}^{\\{\mathfrak{N}\\}}(\mathbb{R})$ if and only if

$\;\forall x>0\;\exists y>0:~{}M^{(x)}\prec_{SV}N^{(y)},$

where $M^{(x)}\prec_{SV}N^{(y)}$ means

$\;\exists s\in\mathbb{N}:~{}\sup_{j\geq 1}\sup_{0\leq
i<j}\Big{(}\frac{M^{(x)}_{j}}{s^{j}N^{(y)}_{i}}\Big{)}^{\frac{1}{j-i}}\frac{1}{j}\sum_{k=j}^{\infty}\frac{N^{(y)}_{k-1}}{N^{(y)}_{k}}<\infty,$

a condition introduced by Schmets and Valdivia in [37].

The characterization of (1.1) is considerably more difficult (partly, because
the image of an intersection is, in general, smaller than the intersection of
the images). We solve this problem in two independent ways:

  1. 1

The first method reduces the Beurling to the Roumieu problem, and uses the
solution of the latter. This is a well-known approach which has been used, in
various disguises, in several settings; see e.g. [9], [37], [17], and [26].
The additional parameter (i.e., $x$ in the weight matrix) makes this delicate
reduction quite involved. As a result we prove in Theorem 3.1 that (again
under mild assumptions) (1.1) is equivalent to

$\;\forall y>0\;\exists x>0:~{}M^{(x)}\prec_{SV}N^{(y)}.$

  2. 2

The second approach is based on dualization of (1.1) and identification of the
strong duals $(\Lambda^{(\mathfrak{M})})^{\prime}$ and
$\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})^{\prime}$ with suitable spaces of
entire functions. This strategy has been implemented by [6] (for
Braun–Meise–Taylor classes, following [8] and [10]). In fact, our analysis is
based on the abstract functional-analytic result [6, Corollary 2.3] (which we
restate in Proposition 5.9). It translates the problem to a question about
bounded sets in the mentioned spaces of entire functions, where a
Phragmén–Lindelöf theorem and Hörmander’s solution of the
$\overline{\partial}$-problem can be brought to bear. We find in Theorem 5.1
that (under other mild assumptions) (1.1) is equivalent to

$\;\forall y>0\;\exists x>0:~{}M^{(x)}\prec_{L}N^{(y)}.$

The condition $M^{(x)}\prec_{L}N^{(y)}$ means

$\;\exists C>0\;\forall s\geq
0:~{}\frac{s}{\pi}\int_{-\infty}^{\infty}\frac{\omega_{N^{(y)}}(t)}{t^{2}+s^{2}}\,dt\leq\omega_{M^{(x)}}(Cs)+C,$

where $\omega_{M}(t):=\sup_{k\in\mathbb{N}}\log(\frac{t^{k}}{M_{k}})$ is the
pre-weight function associated with a weight sequence $M$. It appears as
(2.14’) in Langenbruch’s paper [20] and it is closely related to the condition
appearing in [6], but with a little twist; see Remark 5.2 and Section 6.

Let us briefly describe the structure of the paper. In Section 2, we gather
all relevant notation and conditions concerning weight sequences, functions,
and matrices and we introduce the corresponding ultradifferentiable function
and sequence spaces. The solution by reduction (1) is obtained in Section 3.
In Section 4, we identify the duals $(\Lambda^{(\mathfrak{M})})^{\prime}$ and
$\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})^{\prime}$ with certain weighted
spaces of entire functions. This allows us to carry out the solution by
dualization (2) in Section 5. In the final Section 6, we show that our
theorems specialize to the known results for Denjoy–Carleman and
Braun–Meise–Taylor classes; see Theorem 6.2, 6.3, and Theorem 6.4. In the
short appendix, we prove a technical statement needed in Section 4, namely
that the entire functions are dense in an auxiliary function space. Since the
inclusion of the entire functions is continuous, also the polynomials are
dense.

##  2\. Ultradifferentiable classes and weights

Ultradifferentiable classes are weighted classes of smooth functions.

###  2.1. Weight sequences

We call a sequence of positive real numbers $M=(M_{k})$ a _weight sequence_ ,
if $M_{0}=1$ and $M_{k}=\mu_{1}\cdots\mu_{k}$, $k\geq 1$, for an increasing
sequence $0<\mu_{1}\leq\mu_{2}\leq\cdots$ tending to $\infty$. We call a
weight sequence _normalized_ if $\mu_{1}\geq 1$ and put $\mu_{0}:=1$. Let us
also set $m_{k}:=\frac{M_{k}}{k!}$.

That $\mu_{k}$ is increasing means that $M_{k}$ is log-convex. Here are some
easy consequences of the definition: $M_{j}M_{k}\leq M_{j+k}$,
$(M_{k})^{1/k}\leq\mu_{k}$, and $(M_{k})^{1/k}\to\infty$ if and only if
$\mu_{k}\to\infty$ (cf. [27, Lemma 2.3]).

For a weight sequence $M$, we define the _Denjoy–Carleman class of Beurling
type_

$\mathcal{E}^{(M)}(\mathbb{R}):=\Big{\\{}f\in C^{\infty}(\mathbb{R}):\;\forall
K\subset\subset\mathbb{R}~{}\;\forall r>0:\|f\|^{M}_{K,r}:=\sup_{x\in
K,k\in\mathbb{N}}\frac{|f^{(k)}(x)|}{r^{k}M_{k}}<\infty\Big{\\}}.$

It is endowed with the natural projective topology and thus has the structure
of a Fréchet space. If the universal quantifier in front of $r$ is replaced by
an existential quantifier one gets the Denjoy–Carleman class
$\mathcal{E}^{\\{M\\}}(\mathbb{R})$ of _Roumieu type_.

It is immediate that the restriction of the Borel map $j^{\infty}_{0}$ to
$\mathcal{E}^{(M)}(\mathbb{R})$ takes values in the corresponding sequence
space

$\Lambda^{(M)}:=\Big{\\{}\lambda=(\lambda_{k})_{k}\in\mathbb{C}^{\mathbb{N}}:\;\forall
r>0:\|\lambda\|^{M}_{r}:=\sup_{k\in\mathbb{N}}\frac{|\lambda_{k}|}{r^{k}M_{k}}<\infty\Big{\\}},$

which again is endowed with its natural Fréchet topology. By the Denjoy-
Carleman theorem, $j^{\infty}_{0}|_{\mathcal{E}^{(M)}(\mathbb{R})}$ is
injective if and only if

$\sum_{k\geq 1}\frac{1}{\mu_{k}}=\infty;$

see e.g. [19, Theorem 4.2], [14, Theorem 1.3.8], or [27, Theorem 3.6]. In that
case, the class (and the weight sequence) is called _quasianalytic_ , and
_non-quasianalytic_ otherwise.

We say that $M$ has _moderate growth_ , if

$\;\exists C>0~{}\;\forall j,k\in\mathbb{N}:\quad M_{j+k}\leq
C^{j+k}M_{j}M_{k},$

and $M$ is _derivation closedness_ , if

$\;\exists C>0~{}\;\forall j\in\mathbb{N}:\quad M_{j+1}\leq C^{j+1}M_{j}.$

All these conditions are frequently used in the theory of ultradifferentiable
classes; in [19], non-quasianalyticity is denoted by $(M.3)^{\prime}$,
derivation closedness by $(M.2)^{\prime}$ and moderate growth by $(M.2)$.

Given two weight sequences $M$ and $N$, we write $M\leq N$ if $M_{k}\leq
N_{k}$ for all $k$, and $M\preccurlyeq N$ if
$\sup_{k>0}\big{(}\frac{M_{k}}{N_{k}}\big{)}^{1/k}<\infty$. We say that $M$
and $N$ are _equivalent_ if $M\preccurlyeq N$ and $N\preccurlyeq M$. Note that
both moderate growth and derivation closedness are preserved under
equivalence. Two weight sequences are equivalent if and only if they generate
the same class. In fact,

$\mathcal{E}^{(M)}(\mathbb{R})\subseteq\mathcal{E}^{(N)}(\mathbb{R})\quad\Longleftrightarrow\quad
M\preccurlyeq
N\quad\Longleftrightarrow\quad\Lambda^{(M)}\subseteq\Lambda^{(N)}.$

We shall also need the relation $M\lhd N$, defined by
$\lim_{k\to\infty}\big{(}\frac{M_{k}}{N_{k}}\big{)}^{1/k}=0$, which is
equivalent to
$\mathcal{E}^{\\{M\\}}(\mathbb{R})\subseteq\mathcal{E}^{(N)}(\mathbb{R})$ as
well as $\Lambda^{\\{M\\}}\subseteq\Lambda^{(N)}$. All this can be found in
[28, Proposition 2.12].

###  2.2. Weight functions

The second approach to ultradifferentiable classes is based on weight
functions, i.e., increasing continuous functions
$\omega:[0,\infty)\rightarrow[0,\infty)$ satisfying some additional properties
which will be specified shortly. Originally, $\omega$ was used, by Beurling
[1] and Björck [2], to impose growth restrictions at infinity on the Fourier
transform of the functions in question. In the modern approach due to Braun,
Meise, and Taylor [7], the derivatives of the functions are controlled by the
_Young conjugate_ of $y\mapsto\varphi_{\omega}(y):=\omega(e^{y})$, that is

$\varphi_{\omega}^{*}(x):=\sup\\{xy-\varphi_{\omega}(y):~{}y\geq
0\\},\;\;\;x\geq 0.$

Assuming that $\log(t)=o(\omega(t))$ as $t\rightarrow\infty$, which ensures
that $\varphi_{\omega}^{*}(x)$ is finite for all $x>0$, one defines the
_Braun–Meise–Taylor class of Beurling type_

$\mathcal{E}^{(\omega)}(\mathbb{R}):=\Big{\\{}f\in
C^{\infty}(\mathbb{R}):\;\forall K\subset\subset\mathbb{R}\;\forall
r>0:\|f\|^{\omega}_{K,r}:=\sup_{x\in
K,k\in\mathbb{N}}\frac{|f^{(k)}(x)|}{e^{\varphi_{\omega}^{*}(rk)/r}}<\infty\Big{\\}}$

and endows it with the natural Fréchet topology. Similarly, we have the
Fréchet space

$\Lambda^{(\omega)}:=\Big{\\{}\lambda=(\lambda_{k})_{k}\in\mathbb{C}^{\mathbb{N}}:\;\forall
r>0:\|\lambda\|^{\omega}_{r}:=\sup_{k\in\mathbb{N}}\frac{|\lambda_{k}|}{e^{\varphi_{\omega}^{*}(rk)/r}}<\infty\Big{\\}}$

and the map
$j^{\infty}_{0}|_{\mathcal{E}^{(\omega)}(\mathbb{R})}:\mathcal{E}^{(\omega)}(\mathbb{R})\to\Lambda^{(\omega)}$.
Again there is a Roumieu version of these classes of functions and sequences,
where $r$ is subjected to an existential quantifier; we refer to [24] for
details.

Let us now make precise the relevant regularity properties for $\omega$. We
say that an increasing continuous function
$\omega:[0,\infty)\rightarrow[0,\infty)$ with $\omega(0)=0$ is a _pre-weight
function_ , if $\log(t)=o(\omega(t))$ as $t\rightarrow\infty$ (in particular,
$\omega(t)\to\infty$), and $\varphi_{\omega}$ is convex. We call a pre-weight
$\omega$ a _weight function_ if it also fulfills

$\omega(2t)=O(\omega(t))\text{ as }t\rightarrow\infty.$ ($\omega_{1}$)

The map $j^{\infty}_{0}|_{\mathcal{E}^{(\omega)}(\mathbb{R})}$ is injective if
and only if

$\int_{0}^{\infty}\frac{\omega(t)}{1+t^{2}}\,dt=\infty;$

see e.g. [7], [35, Section 4], or [27, Theorem 11.17]. Then the class and the
weight function $\omega$ are called _quasianalytic_ , and _non-quasianalytic_
otherwise. It is straightforward to see that non-quasianalyticity implies
$\omega(t)=o(t)$ as $t\rightarrow\infty$.

Two pre-weight functions are called _equivalent_ , written $\omega\sim\sigma$,
if $\omega(t)=O(\sigma(t))$ and $\sigma(t)=O(\omega(t))$ as $t\to\infty$. This
is precisely the case if they generate the same classes. Indeed,

$\mathcal{E}^{(\omega)}(\mathbb{R})\subseteq\mathcal{E}^{(\sigma)}(\mathbb{R})\Longleftrightarrow\Lambda^{(\omega)}\subseteq\Lambda^{(\sigma)}\Longleftrightarrow\sigma(t)=O(\omega(t))\text{
as }t\to\infty,$

see [28, Corollary 5.17]. For every pre-weight function there is an equivalent
pre-weight function which vanishes on $[0,1]$.

######  Remark 2.1.

We will frequently consider the radially symmetric extension $\mathbb{C}\ni
z\mapsto\omega(|z|)$ of a pre-weight function $\omega$. By abuse of notation,
we will still write $\omega(z)$ instead of $\omega(|z|)$.

###  2.3. The associated weight function

Let $M$ be a weight sequence. Then

$\omega_{M}(t):=\sup_{k\in\mathbb{N}}\log\Big{(}\frac{t^{k}}{M_{k}}\Big{)},$

is a pre-weight function; cf. [21, Chapitre I] and [19, Section 3.1]. See [33,
Theorem 3.1] for necessary and sufficient conditions for $\omega_{M}$ being a
weight function. For $\lambda>0$, we set
$\mu_{M}(\lambda):=|\\{p\in\mathbb{N}_{\geq 1}:~{}\mu_{p}\leq\lambda\\}|$.
Then we have the following integral representation of $\omega_{M}$, cf. e.g.
[19, (3.11)] and references therein,

$\omega_{M}(t)=\int_{0}^{t}\frac{\mu_{M}(\lambda)}{\lambda}\,d\lambda.$ (2.1)

If $M$ is normalized, then $\omega_{M}|_{[0,1]}=0$. And $\omega_{M}$ is non-
quasianalytic if and only if $M$ is non-quasianalytic; see [19, Lemma 4.1].
Note that a weight sequence $M$ can be recovered from $\omega_{M}$ by

$M_{k}=\sup_{t>0}\frac{t^{k}}{\exp(\omega_{M}(t))},\quad k\in\mathbb{N}.$
(2.2)

In general, $\mathcal{E}^{(M)}(\mathbb{R})$ and
$\mathcal{E}^{(\omega_{M})}(\mathbb{R})$ may differ, unless $M$ has moderate
growth; see [5] and [28, Section 5].

###  2.4. Weight matrices

In [28] and [34], Denjoy–Carleman and Braun–Meise–Taylor classes were
understood as special cases of ultradifferentiable classes defined by weight
matrices. A _weight matrix_ is a one-parameter family of weight sequences
$\mathfrak{M}=(M^{(x)})_{x>0}$ such that $M^{(x)}\leq M^{(y)}$ if $x\leq y$
and, for all $x>0$,

$(m^{(x)}_{j})^{1/j}\to\infty\text{ as }j\to\infty.$ (2.3)

We call $\mathfrak{M}$ _normalized_ if all $M^{(x)}\in\mathfrak{M}$ are
normalized.

We define the classes of _Beurling type_

$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R}):=\Big{\\{}f\in
C^{\infty}(\mathbb{R}):\;\forall K\subset\subset\mathbb{R}~{}\;\forall
r,x>0:~{}\|f\|_{K,r}^{M^{(x)}}<\infty\Big{\\}},$

and

$\Lambda^{(\mathfrak{M})}:=\Big{\\{}\lambda=(\lambda_{k})_{k}\in\mathbb{C}^{\mathbb{N}}:\;\forall
r,x>0:~{}\|\lambda\|_{r}^{M^{(x)}}<\infty\Big{\\}},$

and endow both spaces with their natural Fréchet topology. Note that, by our
assumption (2.3), each class $\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$
contains all real analytic functions on $\mathbb{R}$ (cf. [28, Section 4.1]).

If _all_ $M^{(x)}\in\mathfrak{M}$ are non-quasianalytic, we call
$\mathfrak{M}$ _non-quasianalytic_. Non-quasianalyticity of $\mathfrak{M}$ is
equivalent to the existence of bump functions in
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$; see [35, Proposition 4.7] or [27,
Theorem 11.16].

Any weight sequence $M$ induces a weight matrix $\mathfrak{M}=(M^{(x)})_{x>0}$
with $M^{(x)}=M$ for all $x>0$. Then, obviously,
$\mathcal{E}^{(M)}(\mathbb{R})=\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$ and
$\Lambda^{(M)}(\mathbb{R})=\Lambda^{(\mathfrak{M})}(\mathbb{R})$.

###  2.5. Weight matrices associated with pre-weight functions

To a pre-weight function $\omega$ (vanishing on $[0,1]$) such that
$\omega(t)=o(t)$ as $t\to\infty$, we assign the normalized weight matrix
$\Omega=(W^{(x)})_{x>0}$ defined by

$W^{(x)}_{k}:=\exp\Big{(}\frac{1}{x}\varphi_{\omega}^{*}(xk)\Big{)}.$ (2.4)

If $\omega$ is actually a weight function, then

$\mathcal{E}^{(\omega)}(\mathbb{R})\cong\mathcal{E}^{(\Omega)}(\mathbb{R})\quad\text{
and }\quad\Lambda^{(\omega)}\cong\Lambda^{(\Omega)}$ (2.5)

as locally convex spaces; see [28] and [34]. Let us remark that here
$\omega(t)=o(t)$ as $t\to\infty$ is assumed so that $\Omega$ satisfies our
standard assumption (2.3).

Let us collect some useful properties of $\Omega$.

######  Lemma 2.2.

The weight matrix $\Omega=(W^{(x)})_{x>0}$ satisfies:

  1. 1

$\vartheta^{(x)}\leq\vartheta^{(y)}$ if $x\leq y$, where
$\vartheta^{(x)}_{k}:=\frac{W^{(x)}_{k}}{W^{(x)}_{k-1}}$.

  2. 2

$W^{(x)}_{j+k}\leq W^{(2x)}_{j}W^{(2x)}_{k}$ for all $x>0$ and
$j,k\in\mathbb{N}$.

  3. 3

$\omega\sim\omega_{W^{(x)}}$ for each $x>0$. More precisely,

$\;\forall x>0\;\exists D_{x}>0:x\omega_{W^{(x)}}\leq\omega\leq
2x\omega_{W^{(x)}}+D_{x}.$ (2.6)

  4. 4

$(w^{(x)}_{k})^{1/k}\to\infty$ for all $x>0$ if and only if $\omega(t)=o(t)$
as $t\to\infty$.

  5. 5

$\omega$ is non-quasianalytic if and only if each $W^{(x)}$ is non-
quasianalytic, i.e., if and only if $\Omega$ is non-quasianalytic.

  6. 6

If $\omega$ is a weight function, then

$\;\forall h\geq 1\;\exists A\geq 1\;\forall x>0\;\exists D\geq 1\;\forall
j\in\mathbb{N}:h^{j}W^{(x)}_{j}\leq DW^{(Ax)}_{j},$ (2.7)

which is crucial to have (2.5).

###### Proof.

Cf. [28, Section 5] and [29, Section 2.5]. For (3) see also [34, Theorem
4.0.3, Lemma 5.1.3] and [16, Lemma 2.5]. ∎

###  2.6. Order relations of weight matrices

For two weight matrices $\mathfrak{M}$ and $\mathfrak{N}$, we write
$\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ if for all $y$ there exists $x$ such
that $M^{(x)}\preccurlyeq N^{(y)}$. By [28, Proposition 4.6(1)],

$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})\subseteq\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})\quad\Longleftrightarrow\quad\Lambda^{(\mathfrak{M})}\subseteq\Lambda^{(\mathfrak{N})}\quad\Longleftrightarrow\quad\mathfrak{M}(\preccurlyeq)\mathfrak{N}.$

If $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ and
$\mathfrak{N}(\preccurlyeq)\mathfrak{M}$ hold simultaneously, then we say that
$\mathfrak{M}$ and $\mathfrak{N}$ are _equivalent_. This is the case if and
only if
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})=\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$
as well as $\Lambda^{(\mathfrak{M})}=\Lambda^{(\mathfrak{N})}$ (as sets and,
in turn, also as locally convex vector spaces).

######  Remark 2.3.

Typically, for each notion of Beurling type there is a related version of
Roumieu type. Since in this paper we are principally concerned with the
Beurling case, we will only mention the former without emphasizing every time
that it is the Beurling version.

If $M$ is a weight sequence and $\mathfrak{N}$ a weight matrix, then $M\lhd
N^{(x)}$ for all $x>0$ if and only if
$\mathcal{E}^{\\{M\\}}(\mathbb{R})\subseteq\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$;
see [28, Proposition 4.6(2)].

###  2.7. Moderate growth and derivation closedness

For weight sequences $M,N$, consider

$mg(M,N):=\sup_{j+k\geq
1}\Big{(}\frac{M_{j+k}}{N_{j}N_{k}}\Big{)}^{\frac{1}{j+k}}\in(0,\infty]$

and

$dc(M,N):=\sup_{j\in\mathbb{N}}\left(\frac{M_{j+1}}{N_{j}}\right)^{\frac{1}{j+1}}\in(0,\infty].$

A weight matrix $\mathfrak{M}=(M^{(x)})_{x>0}$ is said to have _moderate
growth_ if

$\;\forall y>0\;\exists x>0:~{}mg(M^{(x)},M^{(y)})<\infty,$
($\mathfrak{M}_{(mg)}$)

and to be _derivation closed_ if

$\;\forall y>0\;\exists x>0:~{}dc(M^{(x)},M^{(y)})<\infty.$
($\mathfrak{M}_{(dc)}$)

Note that moderate growth, derivation closedness, and non-quasianalyticity are
preserved under equivalence.

Derivation closedness allows for absorption of log-terms in associated weight
functions:

######  Lemma 2.4.

Let $M^{(k)}$, for $1\leq k\leq l+1$, be weight sequences such that
$dc(M^{(k)},M^{(k+1)})<\infty$ for all $1\leq k\leq l$. Then there exists
$C>0$ such that

$\omega_{M^{(l+1)}}(t)+\log(1+t^{l})\leq\omega_{M^{(1)}}(Ct)+C,\quad t\geq 0.$

###### Proof.

An iterated application of [4, Lemma 2] yields the result. ∎

###  2.8. Absorbing exponential growth

Inspired by (2.7) (cf. [28, Section 4.1]), we say that a weight matrix
$\mathfrak{M}$ _absorbs exponential growth_ if

$\;\forall y,h>0\;\exists x,A>0\;\forall k\in\mathbb{N}:\quad
h^{k}M^{(x)}_{k}\leq AM^{(y)}_{k}.$ ($\mathfrak{M}_{(L)}$)

The weight matrix $\Omega$ associated with a weight function always has this
property, by Lemma 2.2.

The following lemma states that for any weight matrix $\mathfrak{M}$ we may
find an equivalent weight matrix with the property $(\mathfrak{M}_{(L)})$. For
the sake of completeness, we mention that an analogous statement holds true in
the Roumieu setting as well.

######  Lemma 2.5.

Let $\mathfrak{M}$ be a (normalized) weight matrix. Then there exists an
equivalent (normalized) weight matrix $\mathfrak{N}$ that satisfies
$(\mathfrak{M}_{(L)})$. Actually, we can choose $\mathfrak{N}$ such that for
all $k\in\mathbb{N}_{\geq 1}$ there exist $A_{k}$ and $B_{k}$ such that for
all $j\in\mathbb{N}$

$A_{k}\Big{(}\frac{1}{2^{k}}\Big{)}^{j}M^{(\frac{1}{k})}_{j}\leq
N_{j}^{(\frac{1}{k})}\leq
B_{k}\Big{(}\frac{1}{2^{k}}\Big{)}^{j}M^{(\frac{1}{k})}_{j}.$ (2.8)

Consequently, for all $t\geq 0$,

$\omega_{M^{(1/k)}}(2^{k}t)-\log(B_{k})\leq\omega_{N^{(1/k)}}(t)\leq\omega_{M^{(1/k)}}(2^{k}t)-\log(A_{k}).$
(2.9)

###### Proof.

We will construct normalized weight sequences $N^{(\frac{1}{k})}$, indexed by
$k\in\mathbb{N}_{\geq 1}$, satisfying (2.8) and $N^{(\frac{1}{k+1})}\leq
N^{(\frac{1}{k})}$ for all $k$. If we set $N^{(x)}:=N^{(\frac{1}{k})}$ for
$\frac{1}{k+1}<x\leq\frac{1}{k}$, then (2.8) implies that $\mathfrak{M}$ and
$\mathfrak{N}$ are equivalent. Moreover, (2.9) follows from (2.8) and the
definition of the associated weight function. To see that $\mathfrak{N}$
fulfills $(\mathfrak{M}_{(L)})$, fix $y$ and $h$ and choose $k,n\in\mathbb{N}$
such that $h\leq 2^{k}$ and $\frac{1}{n}\leq y$. By (2.8),

$h^{j}N^{(\frac{1}{k+n})}_{j}\leq
B_{k+n}\Big{(}\frac{1}{2^{n}}\Big{)}^{j}M^{(\frac{1}{k+n})}_{j}\leq
B_{k+n}\Big{(}\frac{1}{2^{n}}\Big{)}^{j}M^{(\frac{1}{n})}_{j}\leq\frac{B_{k+n}}{A_{n}}N^{(\frac{1}{n})}_{j}\leq\frac{B_{k+n}}{A_{n}}N^{(y)}_{j},$

for all $j$.

Let us now construct the sequences $N^{(\frac{1}{k})}$. In the following, we
work with $\mu^{(x)}_{j}=\frac{M^{(x)}_{j}}{M^{(x)}_{j-1}}$ and
$\nu^{(x)}_{j}=\frac{N^{(x)}_{j}}{N^{(x)}_{j-1}}$. Choose $j_{0}\in\mathbb{N}$
minimal such that $\mu^{(1)}_{j}\geq 2$ for all $j\geq j_{0}$. Set

$\nu^{(1)}_{j}:=1\text{ for }j\leq
j_{0},\quad\nu^{(1)}_{j}:=\frac{1}{2}\mu^{(1)}_{j}\text{ for }j>j_{0},$

and $N^{(1)}_{j}:=\nu^{(1)}_{0}\nu^{(1)}_{1}\cdots\nu^{(1)}_{j}$, for
$j\in\mathbb{N}$. Thus $N^{(1)}$ is clearly log-convex and satisfies (2.8) for
$k=1$.

Now assume we have found sequences $N^{(\frac{1}{l})}$ such that (2.8) and
$N^{(\frac{1}{l})}\leq N^{(\frac{1}{l-1})}$ is satisfied for $l\leq k$. Then
we construct $N^{(\frac{1}{k+1})}$ as follows. Choose $j_{0}$ such that
$\mu_{j}^{(\frac{1}{k+1})}\geq 2^{k+1}$ for all $j\geq j_{0}$. By (2.8) and
the pointwise order of $\mathfrak{M}$, for $j\geq j_{0}$,

$\displaystyle\Big{(}\frac{1}{2^{k+1}}\Big{)}^{j-j_{0}}\mu_{j_{0}+1}^{(\frac{1}{k+1})}\cdots\mu_{j}^{(\frac{1}{k+1})}$
$\displaystyle=\Big{(}\frac{1}{2^{k+1}}\Big{)}^{j-j_{0}}\frac{M^{(\frac{1}{k+1})}_{j}}{M^{(\frac{1}{k+1})}_{j_{0}}}$
$\displaystyle\leq\frac{2^{(k+1)j_{0}}}{M^{(\frac{1}{k+1})}_{j_{0}}}\Big{(}\frac{1}{2^{k}}\Big{)}^{j}M^{(\frac{1}{k})}_{j}$
$\displaystyle\leq\frac{2^{(k+1)j_{0}}}{A_{k}M^{(\frac{1}{k+1})}_{j_{0}}}N^{(\frac{1}{k})}_{j}$
$\displaystyle=\frac{2^{(k+1)j_{0}}N_{j_{0}}^{(\frac{1}{k})}}{A_{k}M^{(\frac{1}{k+1})}_{j_{0}}}\nu^{(\frac{1}{k})}_{j_{0}+1}\cdots\nu_{j}^{(\frac{1}{k})}=:B_{k}\nu^{(\frac{1}{k})}_{j_{0}+1}\cdots\nu_{j}^{(\frac{1}{k})}.$

Since $\mu_{j}^{(\frac{1}{k+1})}\to\infty$ as $j\to\infty$, there exists
$j_{1}>j_{0}$ such that

$\Big{(}\frac{1}{2^{k+1}}\Big{)}^{j_{1}-j_{0}}\mu_{j_{0}+1}^{(\frac{1}{k+1})}\cdots\mu_{j_{1}}^{(\frac{1}{k+1})}\geq
B_{k}.$

Now set

$\nu^{(\frac{1}{k+1})}_{j}=1\text{ for }j\leq
j_{1},\quad\nu^{(\frac{1}{k+1})}_{j}=\frac{1}{2^{k+1}}\mu_{j}^{(\frac{1}{k+1})}\text{
for }j>j_{1}.$

Then (2.8) is immediate. Combining the last two estimates, we also get
$N^{(\frac{1}{k+1})}\leq N^{(\frac{1}{k})}$. This ends the proof. ∎

######  Corollary 2.6.

For any weight matrix $\mathfrak{M}$ there is an equivalent weight matrix
$\mathfrak{N}$ such that
$\\{\|\cdot\|^{N^{(1/k)}}_{[-k,k],1}:k\in\mathbb{N}_{\geq 1}\\}$ (resp.
$\\{\|\cdot\|^{N^{(1/k)}}_{1}:k\in\mathbb{N}_{\geq 1}\\}$) is a fundamental
system of seminorms for $\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$ (resp.
$\Lambda^{(\mathfrak{M})}$).

###  2.9. Strong $(\omega_{1})$ condition

Let us write $M\prec_{s\omega_{1}}N$ if and only if

$\;\exists C>0\;\forall t\geq 0:\quad\omega_{M}(2t)\leq\omega_{N}(t)+C,$
($s\omega_{1}$)

and say that $M$ and $N$ satisfy the _strong $(\omega_{1})$ condition_.

Then another immediate consequence of Lemma 2.5 is the following.

######  Corollary 2.7.

Up to equivalence, we can assume that a weight matrix $\mathfrak{M}$ satisfies

$\;\forall x>0\;\exists y>0:~{}M^{(x)}\prec_{s\omega_{1}}M^{(y)}.$ (2.10)

######  Remark 2.8.

In analogy to ($\omega_{1}$), one is led to the following condition:

$\;\forall x>0\;\exists
y>0:\quad\omega_{M^{(x)}}(2t)=O(\omega_{M^{(y)}}(t))\text{ as }t\to\infty;$
(2.11)

see [35] and [18]. But (2.10) is stronger than (2.11). Cf. the results from
[18, Section 3] and the citations therein as well as Remark 5.2.

##  3\. Reduction to the Roumieu case

The goal of this section is to prove the following theorem.

######  Theorem 3.1.

Let $\mathfrak{M},\mathfrak{N}$ be weight matrices that are ordered with
respect to their quotient sequences, i.e., $\mu^{(x)}\leq\mu^{(y)}$ and
$\nu^{(x)}\leq\nu^{(y)}$ if $x\leq y$. Then

$\Lambda^{(\mathfrak{M})}\subseteq
j^{\infty}_{0}\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})\quad\Longleftrightarrow\quad\forall
y>0\;\exists x>0:~{}M^{(x)}\prec_{SV}N^{(y)}.$ (SV)

We shall see in Lemma 3.3 that both sides of the equivalence (SV) imply
$\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ and non-quasianalyticity of
$\mathfrak{N}$. Recall that $M\prec_{SV}N$ means

$\;\exists C,s\in\mathbb{N}_{\geq 1}:~{}\sup_{j\geq 1}\sup_{0\leq
i<j}\Big{(}\frac{M_{j}}{s^{j}N_{i}}\Big{)}^{\frac{1}{j-i}}\frac{1}{j}\sum_{k=j}^{\infty}\frac{N_{k-1}}{N_{k}}\leq
C.$ (3.1)

We will deduce Theorem 3.1 from the following result for Denjoy–Carleman
classes of Roumieu type. It is due to [37] under slightly stronger conditions;
the version stated here is a special case of [17, Theorem 3.2].

######  Theorem 3.2.

Let $M\preccurlyeq N$ be weight sequences with
$\liminf_{p\rightarrow\infty}\left(m_{p}\right)^{1/p}>0$. Then
$\Lambda^{\\{M\\}}\subseteq j^{\infty}_{0}\mathcal{D}^{\\{N\\}}([-1,1])$ if
and only if $M\prec_{SV}N$.

Here $\mathcal{D}^{(N)}([-1,1])$ (resp. $\mathcal{D}^{\\{N\\}}([-1,1])$)
denotes the space of $\mathcal{E}^{(N)}$ (resp. $\mathcal{E}^{\\{N\\}}$)
functions supported in $[-1,1]$.

###  3.1. Auxiliary results

We show first that both sides of the equivalence (SV) imply
$\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ and non-quasianalyticity of
$\mathfrak{N}$. Similar results hold in the Roumieu case; cf. [24].

######  Lemma 3.3.

Let $\mathfrak{M}$ and $\mathfrak{N}$ be weight matrices. Both sides of the
equivalence (SV) imply $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ and non-
quasianalyticity of $\mathfrak{N}$.

###### Proof.

By [36, Lemma 3.2], $M^{(x)}\prec_{SV}N^{(y)}$ implies $M^{(x)}\preccurlyeq
N^{(y)}$ so that $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ is clearly a
consequence of the right-hand side of (SV).

To see that it also follows from the left-hand side, suppose that
$\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ is violated which means that there is
$y>0$ such that $(M^{(x)}_{k}/N^{(y)}_{k})^{1/k}$ is unbounded for all $x>0$.
Thus, for all $j\in\mathbb{N}_{\geq 1}$ we find $k_{j}\geq j$ such that

$\Big{(}\frac{M^{(1/j)}_{k_{j}}}{N^{(y)}_{k_{j}}}\Big{)}^{1/k_{j}}\geq j.$

Consider the sequence $a=(a_{\ell})$ with
$a_{k_{j}}=(\frac{1}{j})^{k_{j}}M_{k_{j}}^{(1/j)}$ and $a_{\ell}=0$ otherwise.
Then $a\in\Lambda^{(\mathfrak{M})}$, because for given $h,z>0$ and $j$ so
large that $\frac{1}{j}\leq\min\\{h,z\\}$, we have
$|a_{k_{j}}|=(\tfrac{1}{j})^{k_{j}}M_{k_{j}}^{(1/j)}\leq
h^{k_{j}}M^{(z)}_{k_{j}}$. On the other hand, we claim that $a\notin
j^{\infty}_{0}\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$. Indeed, if there is
$f\in\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$ with $j^{\infty}_{0}f=a$, then
$N^{(y)}_{k_{j}}\leq a_{k_{j}}=f^{(k_{j})}(0)\leq
A_{h,z}h^{k_{j}}N^{(z)}_{k_{j}}$ for all $h,z>0$ and $j$; a contradiction for
$z=y$ and $h=1/2$.

To infer non-quasianalyticity of $\mathfrak{N}$, we distinguish two cases. If
$\mathfrak{M}$ is non-quasianalytic so is $\mathfrak{N}$, since we already
know that $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$. If $\mathfrak{M}$ is
quasianalytic, then the assertion follows either from [30, Theorem 6], which
shows that no (proper) quasianalytic class is contained in the image of the
Borel map of any other quasianalytic class, or from the observation that
$M^{(x)}\prec_{SV}N^{(y)}$ cannot hold if $N^{(y)}$ is quasianalytic (since
then (3.1) is infinite). ∎

We restate [9, Lemme 16] which is crucial for the reduction.

######  Lemma 3.4.

Let $(\alpha_{j})$ be a sequence of nonnegative real numbers such that
$\sum_{j=1}^{\infty}\alpha_{j}<\infty$. Let $(\beta_{j})$ and $(\gamma_{j})$
be sequences of positive real numbers such that
$\lim_{j\rightarrow\infty}\beta_{j}=0=\lim_{j\rightarrow\infty}\gamma_{j}$,
and assume that $(\gamma_{j})$ is decreasing. Then there exists an increasing
sequence $(\theta_{j})$ tending to $\infty$ such that

  1. 1

$\theta_{j}\gamma_{j}$ is decreasing,

  2. 2

$\theta_{j}\beta_{j}\to 0$,

  3. 3

$\sum_{k=j}^{\infty}\theta_{k}\alpha_{k}\leq
8\theta_{j}\sum_{k=j}^{\infty}\alpha_{k}$ for all $j\geq 1$.

###  3.2. Scheme of proof

The direction $\Rightarrow$ in (SV) follows from a rather direct
generalization of the proof of [17, Theorem 4.7] which we sketch in Section
3.4.

The more delicate part is the converse implication. Our aim is to reduce its
proof to the Roumieu case, i.e., Theorem 3.2. More specifically, we show that
for any given $\lambda\in\Lambda^{(\mathfrak{M})}$ we find weight sequences
$R,S$ such that

  1. i

$\lambda\in\Lambda^{\\{R\\}}$,

  2. ii

$R\prec_{SV}S$,

  3. iii

$\mathcal{E}^{\\{S\\}}(\mathbb{R})\subseteq\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$.

Then Theorem 3.2 (together with Lemma 3.3) gives the desired conclusion.

###  3.3. Proof of Theorem 3.1($\Leftarrow$)

We construct the sequences $R,S$ in several steps.

######  Step (I).

Up to equivalence, we may assume that $\mathfrak{M}$ and $\mathfrak{N}$
satisfy the following conditions:

  1. a

For all $\alpha\in\mathbb{N}_{\geq 1}$,

$N^{(\frac{1}{\alpha})}_{j}\geq 2^{j}N_{j}^{(\frac{1}{\alpha+1})},\quad\text{
for large enough }j.$ (3.2)

  2. b

For all $y>0$ we have $M^{(y)}\prec_{SV}N^{(y)}$ with $C=s=1$ in (3.1).

  3. c

For all $y>0$ we have $M^{(y)}\leq N^{(y)}$.

###### Proof.

(a) follows from Lemma 2.5.

(b), (c) Fix $y>0$. By assumption, there is $x=x(y)$ such that
$M^{(x)}\prec_{SV}N^{(y)}$ and thus $M^{(x)}\preccurlyeq N^{(y)}$. We may
assume that $y\mapsto x(y)$ is increasing and $x(y)\leq y$ (by the order of
the weight matrices). Then $(M^{(x(y))})_{y>0}$ is equivalent to
$\mathfrak{M}$. Finally, there exists an increasing function $r$ with $r(0)=0$
such that the family $\mathfrak{M}^{\prime}$ with
$M^{\prime(y)}_{j}:=r(y)^{j}M^{(x(y))}_{j}$ satisfies the additional
assumption of (b) and (c). This matrix is not normalized, but we can use an
analogous technique as in the proof of Lemma 2.5 to force this as well.

Note that all constructions yield matrices that are still ordered with respect
to their quotients. ∎

We assume from now on that $\mathfrak{M}$ and $\mathfrak{N}$ satisfy (a),(b),
and (c). Fix $\lambda\in\Lambda^{(\mathfrak{M})}$.

######  Step (II).

There exist a decreasing 0-sequence $(\varepsilon_{j})$ and a strictly
increasing sequence of positive integers $(a_{\alpha})$ such that

$|\lambda_{j}|\leq\varepsilon_{1}\cdots\varepsilon_{j}M^{(\frac{1}{\alpha+1})}_{j},\quad\text{
if }~{}a_{\alpha}\leq j<a_{\alpha+1}.$ (3.3)

###### Proof.

By definition of $\Lambda^{(\mathfrak{M})}$, the sequence
$\varepsilon^{(\alpha)}:=(\varepsilon^{(\alpha)}_{j})$ defined by

$\varepsilon^{(\alpha)}_{j}:=\sup_{k\geq
j}\Big{(}\frac{|\lambda_{k}|}{M^{(\frac{1}{\alpha+1})}_{k}}\Big{)}^{1/k}$

is decreasing and tending to $0$ for each $\alpha$. By the order of
$\mathfrak{M}$, we also have
$\varepsilon^{(\alpha)}\leq\varepsilon^{(\alpha+1)}$. We define sequences
$(a_{\alpha})$ and $(a^{\prime}_{\alpha})$ of positive integers as follows:

  * •

Set $a_{1}:=1$.

  * •

For given $a_{\alpha}$, we choose $a^{\prime}_{\alpha}$ and in turn
$a_{\alpha+1}$ such that

$\varepsilon^{(\alpha+1)}_{a_{\alpha+1}}<\varepsilon^{(\alpha)}_{a^{\prime}_{\alpha}}\leq\frac{1}{1+\alpha}\varepsilon^{(\alpha)}_{a_{\alpha}}.$

It is clear that the sequences $(a_{\alpha})$ and $(a^{\prime}_{\alpha})$ are
strictly increasing and interlacing. Finally, define
$\varepsilon=(\varepsilon_{j})$ by

$\varepsilon_{j}:=\varepsilon^{(\alpha)}_{j}\;\;\;\text{for}\;a_{\alpha}\leq
j\leq a^{\prime}_{\alpha},\hskip
20.0pt\varepsilon_{j}:=\varepsilon^{(\alpha)}_{a^{\prime}_{\alpha}}\;\;\;\text{for}\;a^{\prime}_{\alpha}<j<a_{\alpha+1}.$

Then $\varepsilon$ is decreasing, tending to $0$, and, by construction

$\Big{(}\frac{|\lambda_{j}|}{M^{(\frac{1}{\alpha+1})}_{j}}\Big{)}^{1/j}\leq\varepsilon_{j},\quad\text{
if }~{}a_{\alpha}\leq j<a_{\alpha+1},$

which gives (3.3). ∎

######  Step (III).

There exist an increasing sequence $(\underline{\mu}_{j})$ with
$\underline{\mu}_{j}/j\to\infty$, and strictly increasing sequences of
integers $(b_{\alpha})$ and $(C_{\alpha})$ such that
$\underline{M}_{j}:=\underline{\mu}_{0}\underline{\mu}_{1}\cdots\underline{\mu}_{j}$
satisfies $\underline{M}\leq C_{\alpha}M^{(1/\alpha)}$, for all $\alpha$, and

$\displaystyle M^{(1/\alpha)}_{j}\leq\underline{M}_{j},\quad\text{ for all
}\alpha\text{ and }j\leq b_{\alpha}.$ (3.4)

###### Proof.

Let $(a_{\alpha})$ be the sequence from Step (II). We define sequences of
positive integers $(b_{\alpha})$ and $(b^{\prime}_{\alpha})$ as follows:

  * •

Set $b^{\prime}_{1}:=1$.

  * •

For given $b^{\prime}_{\alpha}$, we choose $b_{\alpha}$ such that

$b_{\alpha}>\max\\{a_{\alpha},b^{\prime}_{\alpha}\\}\quad\text{ and
}\quad\mu^{(\frac{1}{\alpha+1})}_{j}\geq\alpha j,\quad\text{ for }j\geq
b_{\alpha}.$ (3.5)

  * •

For given $b_{\alpha}$, we choose $b^{\prime}_{\alpha+1}>b_{\alpha}$ minimal
to ensure

$\mu^{(\frac{1}{\alpha+1})}_{b^{\prime}_{\alpha+1}}>\mu^{(\frac{1}{\alpha})}_{b_{\alpha}}.$

Note that $(b_{\alpha})$ and $(b^{\prime}_{\alpha})$ are strictly increasing,
interlacing, and
$\mu^{(\frac{1}{\alpha+1})}_{j}\leq\mu^{(\frac{1}{\alpha})}_{b_{\alpha}}$ for
all $j\leq b^{\prime}_{\alpha+1}-1$. Finally, set

$\underline{\mu}_{j}:=\mu^{(\frac{1}{\alpha})}_{j},\quad\text{ for
}b^{\prime}_{\alpha}\leq j\leq
b_{\alpha},\qquad\underline{\mu}_{j}:=\mu^{(\frac{1}{\alpha})}_{b_{\alpha}},\quad\text{
for }b_{\alpha}<j<b^{\prime}_{\alpha+1},$ (3.6)

and $\underline{\mu}_{0}:=1$. By construction, $\underline{\mu}_{j}$ is
increasing, $\underline{\mu}_{j}/j\to\infty$, and (3.4) holds. For fixed
$\alpha$, one has $\underline{\mu}_{j}\leq\mu_{j}^{(1/\alpha)}$ for all $j\geq
b_{\alpha}^{\prime}$ which yields $\underline{M}\leq C_{\alpha}M^{(1/\alpha)}$
for some positive constant $C_{\alpha}$. Clearly, we may assume that
$C_{\alpha}$ are integers, strictly increasing in $\alpha$. ∎

######  Step (IV).

There exist an increasing sequence $(\underline{\nu}_{j})$ tending to
$\infty$, strictly increasing sequences of positive integers $(c_{\alpha})$
and $(d_{\alpha})$, and an increasing sequence $(D_{\alpha})$ tending to
$\infty$ such that
$\underline{N}_{j}:=\underline{\nu}_{0}\underline{\nu}_{1}\cdots\underline{\nu}_{j}$
satisfies $\underline{N}\leq D_{\alpha}N^{(1/\alpha)}$, for all $\alpha$,

$\displaystyle N^{(1/\alpha)}_{j}\leq\underline{N}_{j},\quad\text{ for all
}\alpha\text{ and }j\leq d_{\alpha},$ (3.7)

and there is a constant $D\geq 1$ such that, for all $\alpha$ and
$c_{\alpha}\leq j<c_{\alpha+1}$,

$\displaystyle\sum_{k\geq j}\frac{1}{{\underline{\nu}}_{k}}\leq 2\sum_{k\geq
j}\frac{1}{\nu_{k}^{(\frac{1}{\alpha+2})}},$ (3.8) $\displaystyle
C_{\alpha+3}N^{(\frac{1}{\alpha+3})}_{i}\leq
D2^{j-i}\underline{N}_{i},\quad\text{ for all }~{}0\leq i<j,$ (3.9)

where $C_{\alpha}$ are the constants from Step (III).

###### Proof.

We define sequences $(c_{\alpha})$ and $(d_{\alpha})$ of positive integers as
follows:

  * •

Set $c_{1}:=1$ and $d_{0}:=0$.

  * •

For given $c_{\alpha}$, we choose $d_{\alpha}\geq C_{\alpha+4}+d_{\alpha-1}$
such that

$\displaystyle\sum_{k>d_{\alpha}}\frac{1}{\nu^{(\frac{1}{\alpha+1})}_{k}}\leq\frac{1}{2}\sum_{k>c_{\alpha}}\frac{1}{\nu^{(\frac{1}{\alpha})}_{k}},$
(3.10) $\displaystyle N^{(\frac{1}{\alpha+2})}_{j}\geq
2^{j}N^{(\frac{1}{\alpha+3})}_{j},\quad\text{ for }j\geq d_{\alpha}.$ (3.11)

  * •

For given $d_{\alpha}$, we choose $c_{\alpha+1}>d_{\alpha}$ minimal such that

$\nu^{(\frac{1}{\alpha+1})}_{c_{\alpha+1}}>\nu^{(\frac{1}{\alpha})}_{d_{\alpha}}.$
(3.12)

Then $(c_{\alpha})$ and $(d_{\alpha})$ are strictly increasing and
interlacing. Set

$\underline{\nu}_{j}:=\nu^{(\frac{1}{\alpha})}_{j},\quad c_{\alpha}\leq j\leq
d_{\alpha},\qquad\underline{\nu}_{j}:=\nu^{(\frac{1}{\alpha})}_{d_{\alpha}},\quad
d_{\alpha}<j<c_{\alpha+1},$ (3.13)

and $\underline{\nu}_{0}:=1$. Completely analogous to Step (III), we may
conclude that $\underline{N}\leq D_{\alpha}N^{(1/\alpha)}$ and (3.7).

Let us show (3.9). It clearly suffices to show the claim for $\alpha\geq 3$
(the finitely many remaining values can be controlled by possibly enlarging
$D$). So let $\alpha\geq 3$, $c_{\alpha}\leq j<c_{\alpha+1}$, and $0\leq i<j$.
Our construction yields $j\geq c_{\alpha}>d_{\alpha-1}\geq C_{\alpha+3}$, and
therefore

$C_{\alpha+3}N^{(\frac{1}{\alpha+3})}_{0}=C_{\alpha+3}\leq
2^{d_{\alpha-1}}\leq 2^{j}=2^{j}{\underline{N}}_{0}$

which finishes the case $i=0$. So let $1\leq i<j$. There is $\beta\leq\alpha$
such that $c_{\beta}\leq i<c_{\beta+1}$. If $\beta\geq 2$, then
$d_{\beta-1}\leq i\leq d_{\beta+1}$. By (3.7) and (3.11),

$\frac{\underline{N}_{i}}{N^{(\frac{1}{\alpha+3})}_{i}}\geq\frac{N^{(\frac{1}{\beta+1})}_{i}}{N^{(\frac{1}{\alpha+3})}_{i}}\geq\frac{N^{(\frac{1}{\beta+1})}_{i}}{N^{(\frac{1}{\beta+2})}_{i}}\geq
2^{i}.$

Since $2^{j}\geq d_{\alpha-1}\geq C_{\alpha+3}$, (3.9) follows. It remains to
consider $1\leq i\leq d_{1}$, in which case $N^{(\frac{1}{\alpha+3})}_{i}\leq
N^{(1)}_{i}=\underline{N}_{i}$ is clear and $2^{j-i}\geq
2^{d_{\alpha-1}-d_{1}}\geq C_{\alpha+3}$. Thus (3.9) is proved.

Let us now prove (3.8). First assume $c_{\alpha}\leq j\leq d_{\alpha}$. Then

$\sum_{k\geq
j}\frac{1}{\underline{\nu}_{k}}=\sum_{k=j}^{d_{\alpha}}\frac{1}{\nu^{(\frac{1}{\alpha})}_{k}}+\sum_{i\geq
1}\Big{(}\frac{c_{\alpha+i}-d_{\alpha+i-1}-1}{\nu^{(\frac{1}{\alpha+i-1})}_{d_{\alpha+i-1}}}+\sum_{k=c_{\alpha+i}}^{d_{\alpha+i}}\frac{1}{\nu^{(\frac{1}{\alpha+i})}_{k}}\Big{)}.$

By the minimal choice of $c_{\alpha+i}$ (see (3.12)),

$\frac{c_{\alpha+i}-d_{\alpha+i-1}-1}{\nu^{(\frac{1}{\alpha+i-1})}_{d_{\alpha+i-1}}}\leq\sum_{k=d_{\alpha+i-1}+1}^{c_{\alpha+i}-1}\frac{1}{\nu_{k}^{(\frac{1}{\alpha+i})}},$
(3.14)

whence

$\displaystyle\sum_{k\geq j}\frac{1}{\underline{\nu}_{k}}$
$\displaystyle\leq\sum_{k=j}^{d_{\alpha}}\frac{1}{\nu^{(\frac{1}{\alpha+1})}_{k}}+\sum_{i\geq
1}\sum_{k=d_{\alpha+i-1}+1}^{d_{\alpha+i}}\frac{1}{\nu_{k}^{(\frac{1}{\alpha+i})}}$
$\displaystyle=\sum_{k=j}^{d_{\alpha+1}}\frac{1}{\nu^{(\frac{1}{\alpha+1})}_{k}}+\sum_{i\geq
2}\sum_{k=d_{\alpha+i-1}+1}^{d_{\alpha+i}}\frac{1}{\nu_{k}^{(\frac{1}{\alpha+i})}}.$

Using (3.10), we find

$\displaystyle\sum_{k\geq
d_{\alpha+i-1}+1}\frac{1}{\nu_{k}^{(\frac{1}{\alpha+i})}}\leq\frac{1}{2^{i-1}}\sum_{k\geq
d_{\alpha}+1}\frac{1}{\nu_{k}^{(\frac{1}{\alpha+1})}}$

from which it is easy to conclude

$\displaystyle\sum_{k\geq j}\frac{1}{\underline{\nu}_{k}}$ $\displaystyle\leq
2\sum_{k\geq j}\frac{1}{\nu^{(\frac{1}{\alpha+1})}_{k}},$ (3.15)

in particular, (3.8). If $d_{\alpha}<j<c_{\alpha+1}$, then, using (3.15) for
$j=c_{\alpha+1}$, we find

$\sum_{k\geq
j}\frac{1}{\underline{\nu}_{k}}=\sum_{k=j}^{c_{\alpha+1}-1}\frac{1}{\nu^{(\frac{1}{\alpha})}_{d_{\alpha}}}+\sum_{k\geq
c_{\alpha+1}}\frac{1}{\underline{\nu}_{k}}\leq\sum_{k=j}^{c_{\alpha+1}-1}\frac{1}{\nu^{(\frac{1}{\alpha+2})}_{k}}+2\sum_{k\geq
c_{\alpha+1}}\frac{1}{\nu^{(\frac{1}{\alpha+2})}_{k}}\leq 2\sum_{k\geq
j}\frac{1}{\nu^{(\frac{1}{\alpha+2})}_{k}}.$

Thus (3.8) is proved. ∎

######  Step (V).

There exist weight sequences $R,S$ such that $(r_{j})^{1/j}\to\infty$ and

  1. i

$\lambda\in\Lambda^{\\{R\\}}$,

  2. ii

$R\prec_{SV}S$,

  3. iii

$\mathcal{E}^{\\{S\\}}(\mathbb{R})\subseteq\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$.

###### Proof.

For the construction of $R$, we apply Lemma 3.4 to

$\alpha_{j}:=0,\quad\beta_{j}:=\max\Big{\\{}\varepsilon_{j},\frac{j}{(\underline{M}_{j})^{1/j}}\Big{\\}},\quad\gamma_{j}:=\frac{1}{\underline{\mu}_{j}}.$

This yields an increasing sequence $(\theta_{j})$ tending to $\infty$ such
that $\theta_{j}\gamma_{j}$ is decreasing and $\theta_{j}\beta_{j}\to 0$. We
can assume $\theta_{0}=1$. Since $\theta_{j}\gamma_{j}\leq\theta_{j}\beta_{j}$
(as $(\underline{M}_{j})^{1/j}\leq\underline{\mu}_{j}$), also
$\theta_{j}\gamma_{j}\to 0$. Then

$R_{j}:=\prod_{i=0}^{j}\frac{{\underline{\mu}}_{j}}{\theta_{j}}=\frac{\underline{M}_{j}}{\theta_{0}\theta_{1}\cdots\theta_{j}}$

is a weight sequence (not necessarily normalized). We have
$(r_{j})^{1/j}\to\infty$, since

$\frac{j}{(R_{j})^{1/j}}=\frac{j(\theta_{1}\cdots\theta_{j})^{1/j}}{(\underline{M}_{j})^{1/j}}\leq\frac{j\theta_{j}}{(\underline{M}_{j})^{1/j}}\leq\theta_{j}\beta_{j}.$

By (3.3), (3.4), and (3.5),

$|\lambda_{j}|\leq\varepsilon_{1}\cdots\varepsilon_{j}M_{j}^{(\frac{1}{\alpha+1})}\leq\varepsilon_{1}\theta_{1}\cdots\varepsilon_{j}\theta_{j}R_{j}.$

Since $\varepsilon_{j}\theta_{j}\leq\beta_{j}\theta_{j}\rightarrow 0$, we get
$\lambda\in\Lambda^{\\{R\\}}$. This finishes the proof of (i).

To obtain $S$ we apply Lemma 3.4 to

$\alpha^{\prime}_{j}=\gamma^{\prime}_{j}:=\frac{1}{\underline{\nu}_{j}},\quad\beta^{\prime}_{j}:=\max\Big{\\{}\frac{1}{\sqrt{\theta_{\lfloor
j/2\rfloor}}},\frac{1}{\underline{\nu}_{j}}\Big{\\}},$

where $\lfloor j/2\rfloor$ denotes the integer part of $j/2$. We obtain an
increasing sequence $(\theta^{\prime}_{j})$ tending to $\infty$ such that
$\theta^{\prime}_{j}\gamma^{\prime}_{j}$ is decreasing,
$\theta^{\prime}_{j}\beta^{\prime}_{j}\to 0$, and

$\sum_{k=j}^{\infty}\frac{\theta^{\prime}_{k}}{\underline{\nu}_{k}}\leq
8\theta^{\prime}_{j}\sum_{k=j}^{\infty}\frac{1}{\underline{\nu}_{k}},\quad\text{
for all }j.$ (3.16)

Let $\theta^{\prime}_{0}:=1$. Then

$S_{j}:=A^{j}\prod_{i=0}^{j}\frac{{\underline{\nu}}_{j}}{\theta_{j}^{\prime}}=A^{j}\frac{\underline{N}_{j}}{\theta_{0}^{\prime}\theta_{1}^{\prime}\cdots\theta_{j}^{\prime}}$

is a weight sequence. Here $A$ is a constant chosen such that
$A\geq\max\\{1,\frac{\theta^{\prime}_{1}}{{\underline{\nu}}_{1}}\\}$ and

$\displaystyle\frac{\theta^{\prime}_{i}}{\theta_{i}}\leq A,$ (3.17)
$\displaystyle\frac{\theta^{\prime}_{j}}{(\theta_{i+1}\cdots\theta_{j})^{\frac{1}{j-i}}}\leq
A,\quad\text{ if }0\leq i<j.$ (3.18)

That (3.17) and (3.18) are possible is seen as follows. It is easy to see that
the choice of $\beta^{\prime}_{j}$ enables (3.17). For $0\leq i<\lfloor
j/2\rfloor$, we have

$\displaystyle(\theta_{i+1}\cdots\theta_{j})^{\frac{1}{j-i}}\geq(\theta_{\lfloor
j/2\rfloor}\cdots\theta_{j})^{\frac{1}{j-i}}\geq(\theta_{\lfloor
j/2\rfloor})^{\frac{j-\lfloor j/2\rfloor}{j-i}}\geq\sqrt{\theta_{\lfloor
j/2\rfloor}},$

since $\theta_{j}$ is increasing. If $\lfloor j/2\rfloor\leq i\leq j-1$, then
$(\theta_{i+1}\cdots\theta_{j})^{1/(j-i)}\geq\theta_{i+1}\geq\theta_{\lfloor
j/2\rfloor}$. The choice of $\beta^{\prime}_{j}$ shows that the left-hand side
of (3.18) is bounded.

Let us now show that $R\prec_{SV}S$. Fix $0\leq i<j$. There is $\alpha$ such
that $c_{\alpha}\leq j<c_{\alpha+1}$. We have (with
$\sigma_{k}=S_{k}/S_{k-1}$)

$\displaystyle\sum_{k=j}^{\infty}\frac{1}{\sigma_{k}}$
$\displaystyle\stackrel{{\scriptstyle\eqref{Beurling-
equ2}}}{{\leq}}\frac{8}{A}\theta^{\prime}_{j}\sum_{k=j}^{\infty}\frac{1}{\underline{\nu}_{k}}\stackrel{{\scriptstyle\eqref{eq:IV2}}}{{\leq}}\frac{16}{A}\theta^{\prime}_{j}\sum_{k=j}^{\infty}\frac{1}{\nu^{(\frac{1}{\alpha+2})}_{k}}\leq\frac{16}{A}\theta^{\prime}_{j}\sum_{k=j}^{\infty}\frac{1}{\nu^{(\frac{1}{\alpha+3})}_{k}}$
$\displaystyle\stackrel{{\scriptstyle{\rm
I}_{b}}}{{\leq}}\frac{16}{A}j\theta^{\prime}_{j}\Big{(}\frac{N^{(\frac{1}{\alpha+3})}_{i}}{M^{(\frac{1}{\alpha+3})}_{j}}\Big{)}^{\frac{1}{j-i}}\stackrel{{\scriptstyle{\rm
III}}}{{\leq}}\frac{16}{A}j\theta^{\prime}_{j}\Big{(}\frac{C_{\alpha+3}N^{(\frac{1}{\alpha+3})}_{i}}{\underline{M}_{j}}\Big{)}^{\frac{1}{j-i}}$
$\displaystyle\stackrel{{\scriptstyle\eqref{eq:IV3}}}{{\leq}}\frac{32D}{A}j\theta^{\prime}_{j}\Big{(}\frac{\underline{N}_{i}}{\underline{M}_{j}}\Big{)}^{\frac{1}{j-i}}=\frac{32D}{A}j\theta^{\prime}_{j}\Big{(}\frac{\theta^{\prime}_{1}\cdots\theta^{\prime}_{i}S_{i}}{A^{i}\theta_{1}\cdots\theta_{j}R_{j}}\Big{)}^{\frac{1}{j-i}}$
$\displaystyle\stackrel{{\scriptstyle\eqref{eq:th'th1}}}{{\leq}}\frac{32D}{A}j\theta^{\prime}_{j}\Big{(}\frac{S_{i}}{\theta_{i+1}\cdots\theta_{j}R_{j}}\Big{)}^{\frac{1}{j-i}}\stackrel{{\scriptstyle\eqref{eq:th'th2}}}{{\leq}}32Dj\Big{(}\frac{S_{i}}{R_{j}}\Big{)}^{\frac{1}{j-i}},$

which finishes the proof of (ii).

For (iii) observe that, by Step (IV),

$\displaystyle\frac{S_{j}}{N^{(1/\alpha)}_{j}}=\frac{A^{j}}{\theta_{1}^{\prime}\cdots\theta_{j}^{\prime}}\frac{\underline{N}_{j}}{N^{(1/\alpha)}_{j}}\leq
D_{\alpha}\frac{A^{j}}{\theta_{1}^{\prime}\cdots\theta_{j}^{\prime}}.$

We conclude that $S\lhd N^{(1/\alpha)}$ for all $\alpha$, since
$\theta_{j}^{\prime}\to\infty$. ∎

######  Step (VI).

There exists $f\in\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$ such that
$j^{\infty}_{0}f=\lambda$.

###### Proof.

By Step (V) and [36, Lemma 3.2], Theorem 3.2 can be applied to $R$ and $S$.
Thus there exists $f\in\mathcal{E}^{\\{S\\}}(\mathbb{R})$ with
$j^{\infty}_{0}f=\lambda$. By ${\rm V}_{iii}$, we know that
$f\in\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$. ∎

###  3.4. Proof of Theorem 3.1($\Rightarrow$)

Since $\Lambda^{(\mathfrak{M})}\subseteq
j^{\infty}_{0}\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$ implies that
$\mathfrak{N}$ is non-quasianalytic, by Lemma 3.3, and so there exist
$\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$-cutoff functions, e.g. by [27,
Corollary 3.2 and Theorem 11.16], we have $\Lambda^{(\mathfrak{M})}\subseteq
j^{\infty}_{0}\mathcal{D}^{(\mathfrak{N})}([-1,1])$.

Now we follow the ideas of [37, Proposition 4.3 and Theorem 4.4]. Let
$E_{m,k}$ be $\mathcal{D}^{(\mathfrak{N})}([-1,1])$ endowed with the norm
$\|f\|_{m,k}:=\|f\|_{[-1,1],\frac{1}{m}}^{N^{(\frac{1}{k})}}$ and let
$F_{m,k}$ be its completion. As in [37, Proposition 4.3], one sees that, for
all $m,k$, there exists a continuous linear right inverse
$T_{m,k}:\Lambda^{(\mathfrak{M})}\rightarrow F_{m,k}$ of
$j^{\infty}_{0}|_{F_{m,k}}$. Then for every $m\in\mathbb{N}_{\geq 1}$ we can
find $s\in\mathbb{N}_{\geq 1}$ and $C>0$ such that

$\|T_{m,1}(a)\|_{m,1}\leq C\|a\|_{s},\quad a\in\Lambda^{(\mathfrak{M})},$

where $\|\cdot\|_{s}:=\|\cdot\|^{M^{(\frac{1}{s})}}_{\frac{1}{s}}$. The proof
of [37, Theorem 4.4], applied to the sequences $M^{(\frac{1}{s})}$ and
$N^{(\frac{1}{m})}$, yields $M^{(\frac{1}{s})}\prec_{SV}N^{(\frac{1}{m})}$.
This ends the proof of Theorem 3.1.

##  4\. The duals of $\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$ and
$\Lambda^{(\mathfrak{M})}$

In this section, we identify the duals of
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$ and $\Lambda^{(\mathfrak{M})}$ with
weighted spaces of entire functions. This will be of crucial importance in the
proof of the second main result, i.e., Theorem 5.1.

###  4.1. Weighted spaces of entire functions

Let $g:\mathbb{C}\rightarrow[0,\infty)$ be a continuous function with
$\lim_{|z|\rightarrow\infty}g(z)=\infty$. We define the Banach space

$A_{g}:=\Big{\\{}f\in\mathcal{H}(\mathbb{C}):~{}\|f\|_{A_{g}}:=\sup_{z\in\mathbb{C}}\frac{|f(z)|}{e^{g(z)}}<\infty\Big{\\}}.$

Given an increasing sequence of continuous functions $\mathcal{G}=(g_{k})_{k}$
of the mentioned type, we define

$\mathcal{A}_{\mathcal{G}}:=\bigcup_{k\in\mathbb{N}}A_{g_{k}},$

and endow it with the natural inductive limit topology. Sometimes we need to
work with $L^{2}$ weights. To this end, we set

$A_{g}^{2}:=\Big{\\{}f\in\mathcal{H}(\mathbb{C}):~{}\|f\|_{A_{g}^{2}}:=\Big{(}\int_{\mathbb{C}}|f(z)|^{2}e^{-g(z)}\,d\lambda(z)\Big{)}^{1/2}<\infty\Big{\\}},$

where $\lambda$ denotes the Lebesgue measure in $\mathbb{C}$, and define the
corresponding inductive limit $\mathcal{A}^{2}_{\mathcal{G}}$ analogously.

######  Lemma 4.1.

Let $g:\mathbb{C}\rightarrow[0,\infty)$ be a continuous function with
$\lim_{|z|\rightarrow\infty}g(z)=\infty$. Then

$\|f\|_{A^{2}_{2g+\log(1+|z|^{4})}}\leq 3\pi\|f\|_{A_{g}},\quad f\in A_{g},$

in particular, $A_{g}\hookrightarrow A^{2}_{2g+\log(1+|z|^{4})}$.

Let $h:\mathbb{C}\rightarrow[0,\infty)$ be another continuous function and
assume there exists $K>0$ such that

$g(z+u)\leq h(z)+K,\quad z,u\in\mathbb{C},~{}|u|\leq 1.$ (4.1)

Then

$\|f\|_{A_{\frac{h}{2}}}\leq e^{K}\|f\|_{A^{2}_{g}},\quad f\in A^{2}_{g},$

in particular, $A^{2}_{g}\hookrightarrow A_{\frac{h}{2}}$.

###### Proof.

For $f\in A_{g}$,

$\displaystyle\|f\|_{A_{2g+\log(1+|z|^{4})}^{2}}^{2}$
$\displaystyle=\int_{\mathbb{C}}|f(z)|^{2}e^{-2g(z)-\log(1+|z|^{4})}\,d\lambda(z)\leq\|f\|_{A_{g}}^{2}\int_{\mathbb{C}}\frac{d\lambda(z)}{1+|z|^{4}}$

implies the first statement. For the second claim, we observe that an entire
function $f$ fulfills $f(z)=\frac{1}{\pi}\int_{|u|\leq 1}f(z+u)\,d\lambda(u)$
for each $z\in\mathbb{C}$, which follows from Cauchy’s integral formula and
switching to polar coordinates. Thus,

$f(z)^{2}=\frac{1}{\pi}\int_{|u|\leq
1}f(z+u)^{2}e^{g(z+u)-g(z+u)}\,d\lambda(u),$

and therefore

$\displaystyle|f(z)|^{2}\leq\frac{1}{\pi}e^{h(z)+K}\int_{\mathbb{C}}|f(z+u)|^{2}e^{-g(z+u)}\,d\lambda(u)$

which gives the desired result. ∎

######  Remark 4.2.

The proof shows that $\log(1+|z|^{4})$ can be replaced by any function $\rho$
such that $e^{-\rho}\in L^{1}(\mathbb{C})$; of course, the constant has to be
adjusted accordingly.

Let us now show that, under some mild constraints on the family $\mathcal{G}$,
the corresponding inductive limit is regular.

######  Proposition 4.3.

Let $\mathcal{G}=(g_{k})_{k}$ be an increasing family of continuous functions
$g_{k}:\mathbb{C}\to[0,\infty)$ tending to infinity as $|z|\to\infty$ such
that for all $k$

$\lim_{|z|\rightarrow\infty}g_{k+1}(z)-g_{k}(z)=\infty.$ (4.2)

Then $\mathcal{A}_{\mathcal{G}}$ is regular, complete, ultrabornological,
reflexive, and webbed.

###### Proof.

We will show that the connecting mappings are compact. Then the statements
follow from [23, Satz 25.19, 25.20, 24.23, and Bemerkung 24.36].

Take $p:=g_{k}$ and $q:=g_{k+1}$. We show that the inclusion
$A_{p}\hookrightarrow A_{q}$ is compact. Let $(f_{j})$ be a bounded sequence
in $A_{p}$, i.e., there exists $D>0$ such that

$|f_{j}(z)|\leq De^{p(z)},\quad z\in\mathbb{C},\text{ for all }j.$

Then this family of entire functions is locally uniformly bounded. Thus, by
Montel’s theorem, there exists a subsequence $(f_{j_{k}})$ that converges
uniformly on compact subsets to $f\in\mathcal{H}(\mathbb{C})$. Clearly,
$|f(z)|\leq De^{p(z)}$ for all $z\in\mathbb{C}$. Let us show that
$(f_{j_{k}})$ converges to $f$ in $A_{q}$. Let $\varepsilon>0$. Choose $R>0$
such that

$e^{p(z)-q(z)}<\frac{\varepsilon}{2D},\quad|z|>R,$

where we use (4.2), and let $k_{0}$ be such that

$|f_{j_{k}}(z)-f(z)|<\varepsilon,\quad k\geq k_{0},~{}|z|\leq R.$

It follows that $\|f_{j_{k}}-f\|_{A_{q}}<\varepsilon$ for $k\geq k_{0}$. ∎

###  4.2. The spaces $\mathcal{A}_{\Omega_{\mathfrak{M}}^{+}}$ and
$\mathcal{A}_{\Omega_{\mathfrak{M}}}$

Given a weight matrix $\mathfrak{M}=(M^{(x)})_{x>0}$, we consider the
sequences of functions

$\displaystyle\Omega_{\mathfrak{M}}^{+}$ $\displaystyle:=\big{(}z\mapsto
k|\operatorname{Im}z|+\omega_{M^{(1/k)}}(kz)\big{)}_{k},$
$\displaystyle\Omega_{\mathfrak{M}}$
$\displaystyle:=\big{(}z\mapsto\omega_{M^{(1/k)}}(kz)\big{)}_{k},$

and the associated spaces $\mathcal{A}_{\Omega_{\mathfrak{M}}^{+}}$ and
$\mathcal{A}_{\Omega_{\mathfrak{M}}}$.

If the weight matrix is clear from the context, we also write
$\omega^{(k)}(z):=\omega_{M^{(1/k)}}(z)$. Note that
$\omega^{(k)}\leq\omega^{(l)}$ if $k\leq l$ by the definition of associated
weight functions. Let us now see that Proposition 4.3 is applicable to
$\mathcal{A}_{\Omega_{\mathfrak{M}}^{+}}$ and
$\mathcal{A}_{\Omega_{\mathfrak{M}}}$.

Let $l>k>0$. Then (2.1) implies for all $t\geq 0$

$\omega^{(l)}(lt)-\omega^{(k)}(kt)\geq\omega^{(k)}(lt)-\omega^{(k)}(kt)=\int_{kt}^{lt}\frac{\mu_{M^{(1/k)}}(\lambda)}{\lambda}\,d\lambda\geq\mu_{M^{(1/k)}}(kt)\log(l/k).$

So for all $l>k>0$ we get that $\omega^{(l)}(lz)-\omega^{(k)}(kz)\to\infty$ as
$|z|\to\infty$ and thus we are able to infer the following corollary from
Proposition 4.3.

######  Corollary 4.4.

$\mathcal{A}_{\Omega_{\mathfrak{M}}^{+}}$ and
$\mathcal{A}_{\Omega_{\mathfrak{M}}}$ are regular, complete,
ultrabornological, reflexive, and webbed.

In what follows, unless mentioned otherwise, we assume that all weight
sequences and matrices are normalized.

###  4.3. The dual of $\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$

Let us recall a result of [38]. For this we need the Fourier transform of a
distribution $T\in\mathcal{E}(\mathbb{R})^{\prime}$,

$\widehat{T}(z):=T(x\mapsto e^{ixz}).$

For a weight sequence $M$, set

$\lambda_{M}(t):=\sum_{j\geq 0}\frac{t^{j}}{M_{j}},\quad t\geq 0.$

One immediately infers (cf. [19, Proposition 4.5$(a)\Rightarrow(b)$])

$e^{\omega_{M}(t)}\leq\lambda_{M}(t)\leq 2e^{\omega_{M}(2t)},\quad t\geq 0.$
(4.3)

######  Theorem 4.5 ([38, Theorem 2.8]).

Let $M$ be a weight sequence. Then, for

$\mathfrak{E}^{(M)}(\mathbb{R}):=\big{\\{}f\in
C^{\infty}(\mathbb{R}):\;\forall K\subset\subset\mathbb{R},\;\forall
m\in\mathbb{N},\;\forall r>0:\|f\|^{M}_{K,m,r}<\infty\big{\\}},$

where

$\|f\|^{M}_{K,m,r}:=\sup_{j\in\mathbb{N},\,0\leq k\leq m,\,x\in
K}\frac{|f^{(j+k)}(x)|}{r^{j}M_{j}},$ (4.4)

endowed with its natural Fréchet topology, we have

$\mathfrak{E}^{(M)}(\mathbb{R})^{\prime}\cong\mathcal{A}_{\Gamma_{M}},$

where

$\Gamma_{M}:=\big{(}z\mapsto
k\log(1+|z|)+\log(\lambda_{M}(k|z|))+k|\operatorname{Im}z|\big{)}_{k},$

and the isomorphism (of locally convex spaces) is realized by the Fourier
transform.

For a weight matrix $\mathfrak{M}$, we set

$\mathfrak{E}^{(\mathfrak{M})}(\mathbb{R}):=\bigcap_{x>0}\mathfrak{E}^{(M^{(x)})}(\mathbb{R}).$

Note that $\mathfrak{E}^{(M)}(\mathbb{R})=\mathcal{E}^{(M)}(\mathbb{R})$
(resp.,
$\mathfrak{E}^{(\mathfrak{M})}(\mathbb{R})=\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$),
if $M$ (resp., $\mathfrak{M}$) is derivation closed.

######  Proposition 4.6.

Let $\mathfrak{M}$ be derivation closed. Then, as locally convex spaces,

$\mathcal{A}_{\Gamma_{\mathfrak{M}}}\cong\mathcal{A}_{\Omega_{\mathfrak{M}}^{+}},$

where

$\Gamma_{\mathfrak{M}}:=\big{(}z\mapsto
k\log(1+|z|)+\log(\lambda_{M^{(1/k)}}(k|z|))+k|\operatorname{Im}z|\big{)}_{k}.$

###### Proof.

Using Lemma 2.4 together with (4.3), one easily checks that the respective
inductive systems are equivalent and thus the topologies coincide. ∎

We are ready to identify the dual of
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$.

######  Theorem 4.7.

Let $\mathfrak{M}$ be derivation closed. Then
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})^{\prime}\cong\mathcal{A}_{\Omega_{\mathfrak{M}}^{+}}$.

###### Proof.

We have
$\mathfrak{E}^{(\mathfrak{M})}(\mathbb{R})^{\prime}\cong\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})^{\prime}$,
since $\mathfrak{M}$ is derivation closed. First we show

$\bigcup_{k\in\mathbb{N}_{\geq
1}}\mathfrak{E}^{(M^{(1/k)})}(\mathbb{R})^{\prime}\cong\mathfrak{E}^{(\mathfrak{M})}(\mathbb{R})^{\prime},$

where the union on the left carries the locally convex inductive limit
topology and the isomorphism is given by the restriction map which we denote
by $R$. Observe that, for $k\leq l$, we have a continuous inclusion
$\mathfrak{E}^{(M^{(1/k)})}(\mathbb{R})^{\prime}\hookrightarrow\mathfrak{E}^{(M^{(1/l)})}(\mathbb{R})^{\prime}$,
and the locally convex inductive limit exists.

The map $R$ is surjective, since we can extend each continuous functional on
$\mathfrak{E}^{(\mathfrak{M})}(\mathbb{R})$ to some
$\mathfrak{E}^{(M^{(1/k)})}(\mathbb{R})$, by the Hahn–Banach theorem. By Lemma
A.1, this extension is unique and thus $R$ is also injective.

Let us now show continuity in both directions. Continuity of $R$ follows from
continuity of its restriction to any fixed
$\mathfrak{E}^{(M^{(1/k)})}(\mathbb{R})^{\prime}$ which is clear. Since
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$ is a Fréchet–Schwartz space (see
the proof of Proposition 5.10), the dual
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})^{\prime}$ is bornological (cf. [23,
Satz 24.23]). Thus it suffices to show that $R^{-1}$ maps bounded sets to
bounded sets. Now

$U_{n}:=\big{\\{}f\in\mathcal{E}^{(\mathfrak{M})}(\mathbb{R}):p_{n}(f):=\|f\|^{M^{(1/n)}}_{[-n,n],1/n}\leq\tfrac{1}{n}\big{\\}},\quad
n\in\mathbb{N}_{\geq 1},$

is a fundamental system of $0$-neighborhoods in
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})$, and the polars $U_{n}^{\circ}$
form a fundamental system of bounded sets in
$\mathcal{E}^{(\mathfrak{M})}(\mathbb{R})^{\prime}$ (cf. [23, Lemma 25.5]). If
$T\in U_{n}^{\circ}$, then

$|T(f)|\leq np_{n}(f),\quad f\in\mathcal{E}^{(\mathfrak{M})}(\mathbb{R}),$

and, by the Hahn–Banach theorem, $T$ extends to
$\mathfrak{E}^{(M^{(1/n)})}(\mathbb{R})$ and satisfies this estimate for all
$f\in\mathfrak{E}^{(M^{(1/n)})}(\mathbb{R})$. So $R^{-1}(U_{n}^{\circ})$ is
bounded in $\mathfrak{E}^{(M^{(1/n)})}(\mathbb{R})^{\prime}$, therefore
$R^{-1}$ is continuous.

By Theorem 4.5,
$\mathfrak{E}^{(M^{(1/k)})}(\mathbb{R})^{\prime}\cong\mathcal{A}_{\Gamma_{M^{(1/k)}}}$
so that Proposition 4.6 yields the desired result. ∎

###  4.4. The dual of $\Lambda^{(\mathfrak{M})}$

######  Theorem 4.8.

Let $\mathfrak{M}$ be a weight matrix. Then
$(\Lambda^{(\mathfrak{M})})^{\prime}\cong A_{\Omega_{\mathfrak{M}}}$, and this
isomorphism is realized by

$\widetilde{S}:(\Lambda^{(\mathfrak{M})})^{\prime}\rightarrow
A_{\Omega_{\mathfrak{M}}},\quad
T\mapsto\widetilde{S}(T):=\Big{(}z\mapsto\sum_{j\geq 0}T(e_{j})z^{j}\Big{)},$

with $e_{j}$ denoting the $j$-th unit vector.

Since $z\mapsto iz$ is an automorphism of $\mathbb{C}$, the map
$S(T):=(z\mapsto\sum_{j\geq 0}T(e_{j})i^{j}z^{j})$ realizes the isomorphism
$(\Lambda^{(\mathfrak{M})})^{\prime}\cong A_{\Omega_{\mathfrak{M}}}$, too.

###### Proof.

First observe that for any sequence $b=(b_{j})_{j}$ satisfying

$\exists A,B,k>0\;\forall j\in\mathbb{N}:\quad|b_{j}|\leq
AB^{j}\frac{1}{M_{j}^{(1/k)}},$ (4.5)

the map

$T_{b}(a):=\sum_{j\geq 0}a_{j}b_{j},\quad
a=(a_{j})_{j}\in\Lambda^{(\mathfrak{M})},$ (4.6)

is an element of $(\Lambda^{(\mathfrak{M})})^{\prime}$. Actually, every
$T\in(\Lambda^{(\mathfrak{M})})^{\prime}$ has the form (4.6). Indeed,

$T(a_{1}e_{1}+\cdots+a_{n}e_{n})=a_{1}T(e_{1})+\cdots+a_{n}T(e_{n})$

and since $a_{1}e_{1}+\cdots+a_{n}e_{n}\to a$ in $\Lambda^{(\mathfrak{M})}$ as
$n\to\infty$, the statement follows with $b_{j}=T(e_{j})$.

If $b$ satisfies (4.5), then

$f_{b}(z):=\widetilde{S}(T_{b})(z)=\sum_{j\geq 0}b_{j}z^{j}$

defines an element in $A_{\Omega_{\mathfrak{M}}}$. Indeed,

$\displaystyle|f_{b}(z)|$ $\displaystyle\leq A\sum_{j\geq
0}\frac{(B|z|)^{j}}{M_{j}^{(1/k)}}\leq
A\sup_{k\in\mathbb{N}}\frac{(2B|z|)^{k}}{M_{k}^{(1/k)}}\sum_{j\geq
0}\frac{1}{2^{j}}=2Ae^{\omega^{(k)}(2Bz)}.$

Conversely, if $f\in A_{\Omega_{\mathfrak{M}}}$, then, by the Cauchy estimates
and (2.2),

$\displaystyle\frac{|f^{(j)}(0)|}{j!}$ $\displaystyle\leq
A\inf_{r>0}\frac{e^{\omega^{(k)}(kr)}}{r^{j}}=Ak^{j}\frac{1}{M_{j}^{(1/k)}}.$

So $\widetilde{S}:(\Lambda^{(\mathfrak{M})})^{\prime}\rightarrow
A_{\Omega_{\mathfrak{M}}}$ is a linear isomorphism.

Next we show continuity of $\widetilde{S}^{-1}$. To this end, it is enough to
show that $\widetilde{S}^{-1}|_{A_{k}}$ is continuous, where
$A_{k}:=A_{\omega^{(k)}(kz)}$. A typical $0$-neighborhood in
$(\Lambda^{(\mathfrak{M})})^{\prime}$ is of the form $U=\\{T:T(C)\leq r\\}$
for some bounded set $C\subseteq\Lambda^{(\mathfrak{M})}$ and $r>0$. Let $D>0$
be such that $|a_{j}|\leq D\frac{1}{(2k)^{j}}M_{j}^{(\frac{1}{k})}$ for all
$j\in\mathbb{N}$ and all $a=(a_{j})_{j}\in C$. Then $\widetilde{S}^{-1}$ maps
the $A_{k}$-ball of radius $\frac{r}{2D}$ into $U$.

For the continuity of $\widetilde{S}$ we observe that
$(\Lambda^{(\mathfrak{M})})^{\prime}$ is ultrabornological, since
$\Lambda^{(\mathfrak{M})}$ is a Fréchet–Schwartz space (cf. [23, Satz 24.23]).
Indeed, the Fréchet space $\Lambda^{(\mathfrak{M})}$ is nuclear (by the
Grothendieck-Pietsch criterion, cf. [23, 28.15]) and hence a Schwartz space
(cf. [23, Corollary 28.5]). On the other hand, $A_{\Omega_{\mathfrak{M}}}$ is
webbed. So the assertion follows from the open mapping theorem (cf. [23, Satz
24.30]). ∎

##  5\. Proof by dualization

This section builds on the techniques developed in [6] for Braun–Meise–Taylor
classes. Let us first introduce some notation. For a (normalized) non-
quasianalytic pre-weight function $\omega$, we set

$\displaystyle\begin{split}P_{\omega}(x+iy)&:=\frac{|y|}{\pi}\int_{-\infty}^{\infty}\frac{\omega(t)}{(t-x)^{2}+y^{2}}\,dt,\quad
x,y\in\mathbb{R},~{}y\neq 0\\\ P_{\omega}(x)&:=\omega(x),\quad
x\in\mathbb{R},\end{split}$ (5.1)

the _harmonic extension_ of $\omega$ to the open upper and lower half plane
(and subharmonic extension to $\mathbb{C}$). A detailed exposition of its main
features is presented in Section 5.2. $P_{\omega}$ is closely related to the
concave weight function (cf. [6, Definition 3.1$(b)$])

$\kappa_{\omega}(r):=\int_{1}^{\infty}\frac{\omega(rt)}{t^{2}}dt=r\int_{r}^{\infty}\frac{\omega(t)}{t^{2}}dt.$
(5.2)

In fact,

$\frac{1}{\pi}\kappa_{\omega}(r)\leq
P_{\omega}(ir)\leq\frac{4}{\pi}\kappa_{\omega}(r),\quad r>0,$ (5.3)

by [6, Lemma 3.3]. It was proved in [6] that, for a non-quasianalytic weight
function $\omega$ and another weight function $\sigma$, the inclusion

$\Lambda^{(\sigma)}\subseteq
j^{\infty}_{0}\mathcal{E}^{(\omega)}(\mathbb{R}),$

holds if and only if

$\kappa_{\omega}(r)=O(\sigma(r))\quad\text{ as }r\rightarrow\infty.$ (5.4)

Note that (5.4) is also equivalent to $\Lambda^{\\{\sigma\\}}\subseteq
j^{\infty}_{0}\mathcal{E}^{\\{\omega\\}}(\mathbb{R})$; cf. Theorem 6.4, [6],
and [24] as well as [31], [32], and [26] for the more general mixed Whitney
extension problem.

When $M$ is a non-quasianalytic weight sequence, we also write $P_{M}$ instead
of $P_{\omega_{M}}$ and $\kappa_{M}$ instead of $\kappa_{\omega_{M}}$. The
crucial mixed condition for weight sequences $M,N$ in this section is
$M\prec_{L}N$ defined by

$\;\exists C>0\;\forall s\geq 0:~{}P_{N}(is)\leq\omega_{M}(Cs)+C.$ (5.5)

This condition appears in [20, (2.14’)].

Let us now formulate the main result of this section.

######  Theorem 5.1.

Let $\mathfrak{M},\mathfrak{N}$ be weight matrices, $\mathfrak{N}$ derivation
closed. Then

$\Lambda^{(\mathfrak{M})}\subseteq
j^{\infty}_{0}\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})\quad\Longleftrightarrow\quad\;\forall
y>0\;\exists x>0:~{}M^{(x)}\prec_{L}N^{(y)}.$ (L)

######  Remark 5.2.

Similarly to Remark 2.8, note that in (5.5) on the right-hand side the
constant $C$ appears in the argument of $\omega_{M}$ (not in front). This
subtle difference will become important later on; it stems from the fact that
we aim for results for classes defined by (a family of) weight sequences
instead of (associated) weight functions. Even though (5.3) implies
$\kappa_{N}\sim P_{N}$, generally we cannot replace $P_{N}$ by $\kappa_{N}$ in
(5.5).

###  5.1. Auxiliary results

We saw in Lemma 3.3 that the left-hand side of (L) entails
$\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ and non-quasianalyticity of
$\mathfrak{N}$. This is also true for the right-hand side.

######  Lemma 5.3.

Let $\mathfrak{M}$ and $\mathfrak{N}$ be weight matrices. The right-hand side
of (L) entails $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ and non-
quasianalyticity of $\mathfrak{N}$.

###### Proof.

Non-quasianalyticity is immediate, since $\omega_{N^{(y)}}$ is non-
quasianalytic if and only if so is $N^{(y)}$; see [19, Lemma 4.1].

Fix $y>0$. There is $x>0$ such that $M^{(x)}\prec_{L}N^{(y)}$, i.e., there is
$C>0$ such that

$\omega_{N^{(y)}}(s)\leq P_{N^{(y)}}(is)\leq\omega_{M^{(x)}}(Cs)+C,\quad s>0,$

where the first inequality will be justified later in (5.6). By (2.2), we
conclude

$\displaystyle N^{(y)}_{k}$
$\displaystyle=\sup_{t>0}\frac{t^{k}}{e^{\omega_{N^{(y)}}(t)}}\geq\sup_{t>0}\frac{t^{k}}{e^{\omega_{M^{(x)}}(Ct)+C}}=e^{-C}C^{-k}M^{(x)}_{k}.$

This shows $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$. ∎

###  5.2. Properties of $P_{\omega}$, the (sub-)harmonic extension of $\omega$

In this section we assume, without further mentioning, that $\omega$ has the
following properties:

  * •

$\omega:[0,\infty)\rightarrow[0,\infty)$ is increasing and continuous.

  * •

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

  * •

$\varphi_{\omega}=\omega\circ\exp$ is convex.

  * •

$\int_{0}^{\infty}\frac{\omega(t)}{1+t^{2}}\,dt<\infty$.

So $\omega$ may be any non-quasianalytic pre-weight function, in particular,
$\omega_{M}$ for a non-quasianalytic weight sequence $M$. Recall our
convention $\omega(z):=\omega(|z|)$ for $z\in\mathbb{C}$.

The harmonic extension $P_{\omega}$, defined in (5.1), will play a crucial
role as a weight for a weighted space of entire functions. Let us list some
obvious properties:

  1. 1

$P_{\omega}(z)\geq 0$ for all $z\in\mathbb{C}$,

  2. 2

$P_{\omega}$ is symmetric relative to the real and imaginary axis,

  3. 3

$\sigma\leq\omega$ implies $P_{\sigma}\leq P_{\omega}$,

  4. 4

$P_{\sigma+\omega}=P_{\sigma}+P_{\omega}$,

  5. 5

$P_{t\mapsto\omega(nt)}(z)=P_{\omega}(nz)$.

######  Remark 5.4.

For an increasing sequence of functions $\omega_{j}$ converging to $\omega$
uniformly on compact subsets of $\mathbb{R}$, we get directly from the
definition that $P_{\omega_{j}}\rightarrow P_{\omega}$ uniformly on compact
subsets of $\mathbb{C}$.

The following proposition is well-known.

######  Proposition 5.5.

$P_{\omega}$ is continuous on $\mathbb{C}$, harmonic in the open upper and
lower half plane, and subharmonic on $\mathbb{C}$.

###### Proof.

That $P_{\omega}$ is continuous on $\mathbb{C}$ and harmonic in the open upper
and lower half plane is clear. For the subharmonicity, note first that
$\omega$ is subharmonic on $\mathbb{C}$; indeed,
$\omega(z)=\varphi_{\omega}(\log|z|)$ and $\varphi_{\omega}$ is increasing and
convex.

Next we show that

$P_{\omega}(z)\geq\omega(z),\quad z\in\mathbb{C}.$ (5.6)

In fact, $\xi+i\eta\mapsto P_{\omega}(e^{\xi+i\eta})$ is harmonic on the
horizontal strip $\\{0<\eta<\pi\\}$, convex in $\xi$, and thus concave and
symmetric relative to $\frac{\pi}{2}$ in $\eta$ (cf. the arguments in [8, p.
198]). So for any fixed $\xi$ the map $\eta\mapsto P_{\omega}(e^{\xi+i\eta})$
takes its minimum at $\eta=0$ (and $\eta=\pi$). Since $P_{\omega}$ extends
$\omega$, this proves (5.6).

Now, for $x\in\mathbb{R}$ and $\delta>0$,

$P_{\omega}(x)=\omega(x)\leq\frac{1}{2\pi}\int_{0}^{2\pi}\omega(x+\delta
e^{i\theta})\,d\theta\leq\frac{1}{2\pi}\int_{0}^{2\pi}P_{\omega}(x+\delta
e^{i\theta})\,d\theta,$

which implies that $P_{\omega}$ is subharmonic on $\mathbb{C}$. ∎

###  5.3. Consequences of properties of weight sequences for $P_{M}$

If $M,N$ are weight sequences such that $M\prec_{s\omega_{1}}N$, then one
easily infers the existence of $K\geq 1$ such that

$\omega_{M}(z+w)\leq\omega_{N}(z)+\omega_{N}(w)+K,\quad z,w\in\mathbb{C}.$
(5.7)

Moreover, if $M,N$ are non-quasianalytic, there is a constant $C\geq 1$ such
that

$P_{M}(z)\leq P_{N}(z)+C,\quad z\in\mathbb{C}.$ (5.8)

######  Lemma 5.6.

Let $M$ and $N$ be weight sequences such that $M$ is non-quasianalytic and
$M\prec_{s\omega_{1}}N$. Then for all $\varepsilon>0$ there exists $K>0$ such
that

$P_{M}(x+iy)\leq\omega_{N}(x)+\varepsilon y+K,\quad x+iy\in\mathbb{C}.$

###### Proof.

Cf. [7, Lemma 2.2] with the obvious changes. ∎

Having this we prove the following mixed version of [22, Lemma 1.9].

######  Lemma 5.7.

Let $M^{(i)}$, $1\leq i\leq 3$, be non-quasianalytic weight sequences with
$M^{(1)}\prec_{s\omega_{1}}M^{(2)}\prec_{s\omega_{1}}M^{(3)}$. Then there
exists $A>0$ such that

$P_{M^{(1)}}(z+w)\leq P_{M^{(3)}}(z)+A,\quad z,w\in\mathbb{C},~{}|w|\leq 1.$

###### Proof.

First observe that $P_{M}$ has the following alternative form

$P_{M}(x+iy)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\omega_{M}(|y|t+x)}{t^{2}+1}\,dt,\quad(y\neq
0).$ (5.9)

Now take $w=u+iv\in\mathbb{C}$ with $|w|\leq 1$ and $z=x+iy\in\mathbb{C}$ with
$y>1$. Then $\operatorname{Im}(z+w)=y+v>0$ and, by (5.7),

$\displaystyle P_{M^{(1)}}(z+w)$
$\displaystyle=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\omega_{M^{(1)}}((y+v)t+x+u)}{t^{2}+1}\,dt$
$\displaystyle\leq\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\omega_{M^{(2)}}(yt+x)+\omega_{M^{(2)}}(vt+u)+K}{t^{2}+1}\,dt$
$\displaystyle\leq
P_{M^{(2)}}(z)+K\underbrace{\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\omega_{M^{(2)}}(|t|+1)+1}{t^{2}+1}\,dt}_{=:B>1},$

since $K\geq 1$ and
$\omega_{M^{(2)}}(vt+u)=\omega_{M^{(2)}}(|vt+u|)\leq\omega_{M^{(2)}}(|v||t|+|u|)\leq\omega_{M^{(2)}}(|t|+1)$.
By (5.8), the choice $A=BK+C^{\prime}$ establishes the claim for $y>1$, and by
symmetry for $y<-1$. If $|y|\leq 1$, then Lemma 5.6 and (5.7) yield constants
$K_{i}\geq 1$ such that

$P_{M^{(1)}}(z+w)\leq\omega_{M^{(2)}}(x+u)+K_{1}\leq\omega_{M^{(3)}}(x)+K_{2},$

and (5.6) finishes the proof. ∎

The effect of derivation closedness on $P_{M}$ is captured in the next lemma.

######  Lemma 5.8.

Let $l\in\mathbb{N}$ and let $M^{(k)}$, for $1\leq k\leq l+1$, be weight
sequences such that $dc(M^{(k)},M^{(k+1)})<\infty$ for all $k$. Then there
exists $C>0$ such that

$P_{M^{(l+1)}}(z)+\log(1+|z|^{l})\leq P_{M^{(1)}}(Cz)+C,\quad z\in\mathbb{C}.$

###### Proof.

By Lemma 2.4, we have, for some $C>0$,

$\sigma(t):=\omega_{M^{(l+1)}}(t)+\log(1+t^{l})\leq\omega_{M^{(1)}}(Ct)+C,\quad
t\geq 0.$

It is easy to see that $\sigma$ is a pre-weight function. By monotonicity and
additivity of $P_{\omega}$ in $\omega$, and (5.6) applied to $\log(1+t^{l})$,
we infer

$P_{M^{(l+1)}}(z)+\log(1+|z|^{l})\leq P_{\sigma}(z)\leq P_{M^{(1)}}(Cz)+C$

and are done. ∎

###  5.4. Scheme of the proof of Theorem 5.1

The key is the following proposition.

######  Proposition 5.9 ([6, Corollary 2.3]).

Let $E,F,G$ be Fréchet–Schwartz spaces and let $T\in L(E,F)$ and $R\in L(G,F)$
have dense range. Assume that $F^{\prime}$ endowed with the initial topology
with respect to $T^{t}:F^{\prime}\rightarrow E^{\prime}$ is bornological. Then
the following conditions are equivalent:

  1. 1

$R(G)\subseteq T(E)$.

  2. 2

If $B\subseteq F^{\prime}$ is such that $T^{t}(B)$ is bounded in $E^{\prime}$,
then $R^{t}(B)$ is bounded in $G^{\prime}$.

${E}$${F}$${G}$${E^{\prime}}$${F^{\prime}}$${G^{\prime}}$$\scriptstyle{T}$$\scriptstyle{R}$$\scriptstyle{T^{t}}$$\scriptstyle{R^{t}}$

Let $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ be weight matrices,
$\mathfrak{N}$ derivation closed and non-quasianalytic. We will apply
Proposition 5.9 to

${\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})}$${\Lambda^{(\mathfrak{N})}}$${\Lambda^{(\mathfrak{M})}}$$\scriptstyle{j^{\infty}_{0}}$incl
(5.10)

By Theorem 4.7 and Theorem 4.8, we have the following commuting diagram

${\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})^{\prime}}$${(\Lambda^{(\mathfrak{N})})^{\prime}}$${(\Lambda^{(\mathfrak{M})})^{\prime}}$${\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}}$${\mathcal{A}_{\Omega_{\mathfrak{N}}}}$${A_{\Omega_{\mathfrak{M}}}}$$\scriptstyle{\mathcal{F}}$$\scriptstyle{(j^{\infty}_{0})^{t}}$$\scriptstyle{S}$$\scriptstyle{\text{incl}^{t}}$$\scriptstyle{S}$inclincl
(5.11)

where the vertical arrows are isomorphisms. This will lead to

######  Proposition 5.10.

Let $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ be weight matrices,
$\mathfrak{N}$ derivation closed and non-quasianalytic. Then the following
conditions are equivalent:

  1. 1

$\Lambda^{(\mathfrak{M})}\subseteq
j^{\infty}_{0}\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$.

  2. 2

If $B\subseteq\mathcal{A}_{\Omega_{\mathfrak{N}}}$ is bounded in
$\mathcal{A}_{\Omega^{+}_{\mathfrak{N}}}$, then $B$ is bounded in
$\mathcal{A}_{\Omega_{\mathfrak{M}}}$.

We will prove Proposition 5.10 in Section 5.5. In Section 5.6 we will make the
connection between condition (2) and the right-hand side of (L), and thus
complete the proof of Theorem 5.1.

###  5.5. Proof of Proposition 5.10

The proof is based on the following Phragmén–Lindelöf theorem; cf. [3, Theorem
6.5.4].

######  Theorem 5.11.

Let $f$ be holomorphic in the upper half plane and continuous up to the
boundary. Assume that the zeros of $f$ have no finite limit point, and

$\liminf_{r\rightarrow\infty}\frac{\sup_{|z|=r}\log|f(z)|}{r}<\infty,\quad\int_{-\infty}^{\infty}\frac{\max(0,\log|f(t)|)}{1+t^{2}}\,dt<\infty.$
(5.12)

Then (writing $z=x+iy$)

$\displaystyle\log|f(z)|\leq\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\log|f(t)|}{(t-x)^{2}+y^{2}}\,dt+\frac{2y}{\pi}\lim_{r\rightarrow\infty}\frac{1}{r}\int_{0}^{\pi}\log|f(re^{i\theta})|\sin(\theta)\,d\theta.$

Every function $f\not\equiv 0$ in $\mathcal{A}_{\Omega_{\mathfrak{N}}}$, or
$\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}$, satisfies the assumptions of
Theorem 5.11. Indeed, since $f$ is entire, its zeros cannot have any finite
limit point unless $f\equiv 0$. Since all $N\in\mathfrak{N}$ are non-
quasianalytic and so $\omega_{N}(t)=o(t)$, see Section 2.3, also the
conditions (5.12) are clear.

A direct application of this result yields the following corollary.

######  Corollary 5.12.

Let $N$ be a non-quasianalytic weight sequence. Let
$f\in\mathcal{H}(\mathbb{C})$ be such that, for some positive integer $k$,

$\displaystyle\log|f(z)|=o(|z|)\quad\text{as }|z|\rightarrow\infty\quad\text{
and }\quad\log|f(x)|\leq\omega_{N}(kx),~{}x\in\mathbb{R}.$

Then

$|f(z)|\leq e^{P_{N}(kz)},\quad z\in\mathbb{C}.$

###### Proof of Proposition 5.10.

We have to verify the assumptions of Proposition 5.9 with the choices of
(5.10).

We saw in the proof of Theorem 4.8 that $\Lambda^{(\mathfrak{N})}$ and
$\Lambda^{(\mathfrak{M})}$ are Fréchet–Schwartz spaces. For
$\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})$, this is a consequence of the
compactness of the inclusions
$\mathcal{E}^{N^{(\frac{1}{k+1})},\frac{1}{k+1}}(K)\hookrightarrow\mathcal{E}^{N^{(\frac{1}{k})},\frac{1}{k}}(K)$
for compact intervals $K\subseteq\mathbb{R}$ (cf. [11, §22, Satz 3.1]); here
$\mathcal{E}^{M,a}(K)$ denotes the normed space of all functions $f\in
C^{\infty}(K)$ such that $\|f\|^{M}_{K,a}<\infty$.

Both maps in (5.10) have dense range, since the finite sequences are dense in
$\Lambda^{(\mathfrak{N})}$.

Next we prove that $(\Lambda^{(\mathfrak{N})})^{\prime}$ endowed with the
initial topology with respect to
$(j^{\infty}_{0})^{t}:(\Lambda^{(\mathfrak{N})})^{\prime}\to\mathcal{E}^{(\mathfrak{N})}(\mathbb{R})^{\prime}$
is bornological. By (5.11), this amounts to showing that

$\mathcal{A}_{\Omega_{\mathfrak{N}}}\text{ endowed with the trace topology of
}\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}\text{ is bornological}.$ (5.13)

To prove (5.13), we set

$\omega^{(k)}(z):=\omega_{N^{(\frac{1}{k})}}(z),\quad
A_{k}:=A_{k|\operatorname{Im}z|+\omega^{(k)}(kz)}.$

For every $k\in\mathbb{N}_{\geq 1}$, there exists $l>k$ such that
$A_{k}\hookrightarrow A_{l}$ is compact; cf. Section 4.2. Thus, by [6,
Proposition 2.6$(2)\Leftrightarrow(3)$], (5.13) holds if and only if

$\bigcup_{k\in\mathbb{N}_{\geq
1}}\overline{Y}_{k}^{A_{k}}=\overline{\mathcal{A}_{\Omega_{\mathfrak{N}}}}^{\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}},\quad\text{
where }Y_{k}:=\mathcal{A}_{\Omega_{\mathfrak{N}}}\cap A_{k}.$ (5.14)

The inclusion $\bigcup_{k\in\mathbb{N}_{\geq
1}}\overline{Y}_{k}^{A_{k}}\subseteq\overline{\mathcal{A}_{\Omega_{\mathfrak{N}}}}^{\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}}$
is clear and we are left to prove the converse. To this end, we will show the
two inclusions

$\overline{\mathcal{A}_{\Omega_{\mathfrak{N}}}}^{\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}}\subseteq\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}\subseteq\bigcup_{k\in\mathbb{N}_{\geq
1}}\overline{Y_{k}}^{A_{k}},$ (5.15)

where we put $P^{(k)}:=P_{\omega^{(k)}}$ and
$\mathcal{P}_{\mathfrak{N}}:=\big{(}z\mapsto P^{(k)}(kz)\big{)}_{k}$.

Let us start with the first inclusion in (5.15). By (5.6) and Lemma 5.6 (and
Corollary 2.7), for each $k\in\mathbb{N}_{\geq 1}$ there exist
$l\in\mathbb{N}_{\geq 1}$ and $A>0$ such that

$\omega^{(k)}(z)\leq
P^{(k)}(z)\leq\omega^{(l)}(z)+|\operatorname{Im}z|+A,\quad z\in\mathbb{C}.$
(5.16)

This shows
$\mathcal{A}_{\Omega_{\mathfrak{N}}}\subseteq\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}\subseteq\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}$.
Thus it suffices to show that $\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}$ is
closed in $\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}$, i.e.,
$\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}\cap A_{k}$ is closed in $A_{k}$ for
all $k$ (cf. [11, §25, Satz 1.2]). So let
$f\in\overline{\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}\cap A_{k}}^{A_{k}}$.
Then there is a sequence $f_{j}\in\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}\cap
A_{k}$ converging to $f$ in $A_{k}$. Clearly, there exists $C>0$ such that

$|f_{j}(z)|\leq Ce^{k|\operatorname{Im}z|+\omega^{(k)}(kz)},\quad
z\in\mathbb{C},~{}j\in\mathbb{N}.$

Since $f_{j}\in\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}$, there exist
$C_{j}>0$ and $k_{j}$ such that

$|f_{j}(z)|\leq C_{j}e^{P^{(k_{j})}(k_{j}z)},\quad z\in\mathbb{C},$

consequently, $\log|f_{j}(z)|=o(|z|)$ as $|z|\to\infty$. Now Corollary 5.12
implies that all $f_{j}$ are contained and uniformly bounded in some step of
$\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}$. This shows that
$f\in\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}$, and we are done.

It remains to prove the second inclusion in (5.15). To this end, we use [39,
Theorem 1] which states the following: _Let $(\varphi_{j})_{j}$ be an
increasing sequence of subharmonic functions on $\mathbb{C}$ converging to
some subharmonic function $\varphi$, and assume that $e^{-\varphi_{1}}$ is
locally integrable on $\mathbb{C}$. Then any function in $A^{2}_{\varphi}$ can
be approximated in $L^{2}_{\varphi(z)+\log(1+|z|^{2})}$ by a sequence in
$\bigcup_{k\in\mathbb{N}_{\geq 1}}A_{\varphi_{k}(z)+\log(1+|z|^{2})}^{2}$._

Let $f\in\mathcal{A}_{\mathcal{P}_{\mathfrak{N}}}$. By Lemma 4.1, there exists
$k\in\mathbb{N}_{\geq 1}$ such that $f\in A^{2}_{\varphi}$, where
$\varphi(z):=2P^{(k)}(kz)+\log(1+|z|^{4})$. For this $k$, we introduce the
function $\omega_{j}$ by

$\displaystyle\omega_{j}(t):=2\omega^{(k)}(kt),~{}|t|\leq
j,\quad\omega_{j}(t):=a_{j}\log|t|+b_{j},~{}|t|\geq j,$

where $a_{j},b_{j}\in\mathbb{R}$ are chosen such that $\omega_{j}$ is
continuous, increasing, and $t\mapsto\omega_{j}(e^{t})$ is convex. Then
$\big{(}\varphi_{j}(z):=P_{\omega_{j}}(z)+\log(1+|z|^{4})\big{)}_{j}$ is an
increasing sequence of subharmonic functions converging to $\varphi$; cf.
Remark 5.4. Thus there is a sequence $(f_{j})_{j}$ such that $f_{j}\in
A^{2}_{\varphi_{j}(z)+\log(1+|z|^{2})}$ and $f_{j}\to f$ in
$A^{2}_{\varphi(z)+\log(1+|z|^{2})}$. By (5.16) and Lemma 2.4, there exist
$s\in\mathbb{N}_{\geq 1}$ and $K\geq 1$ such that

$\displaystyle\varphi(z)+\log(1+|z|^{2})$ $\displaystyle\leq
2\omega^{(l)}(kz)+2k|\operatorname{Im}z|+\log(1+|z|^{4})+\log(1+|z|^{2})+2A$
$\displaystyle\leq 2\omega^{(s)}(sz)+2s|\operatorname{Im}z|+K$

for all $z\in\mathbb{C}$ so that
$A^{2}_{\varphi(z)+\log(1+|z|^{2})}\hookrightarrow
A^{2}_{2\omega^{(s)}(sz)+2s|\operatorname{Im}z|}$. By Lemma 5.7 and (5.16),
there exist $t\in\mathbb{N}_{\geq 1}$ and $L\geq 1$ such that

$2\omega^{(s)}(s(z+u))+2s|\operatorname{Im}(z+u)|\leq
2\omega^{(t)}(tz)+2t|\operatorname{Im}z|+L,\quad z,u\in\mathbb{C},~{}|u|\leq
1.$

Then Lemma 4.1 implies $A^{2}_{\varphi(z)+\log(1+|z|^{2})}\hookrightarrow
A_{\omega^{(t)}(tz)+t|\operatorname{Im}z|}=A_{t}$. Thus $f_{j}\to f$ in
$A_{t}$. Since $P_{\omega_{j}}(z)=O(\log|z|)$ as $|z|\rightarrow\infty$, all
$f_{j}$ are actually polynomials and hence contained in $Y_{t}$. So also the
second inclusion in (5.15) is proved. ∎

###  5.6. Proof of Theorem 5.1

Let $\mathfrak{M}(\preccurlyeq)\mathfrak{N}$ be weight matrices,
$\mathfrak{N}$ derivation closed and non-quasianalytic; cf. Lemma 5.3. We
write $\omega^{(k)}(z):=\omega_{N^{(\frac{1}{k})}}(z)$.

We need the following lemma.

######  Lemma 5.13.

Let $a_{j}\geq 1$ be a sequence tending to $\infty$ and $k_{0}$ a positive
integer. There exist a sequence of polynomials $(p_{j})_{j}$ and
$k\in\mathbb{N}_{\geq k_{0}}$ such that
$p_{j}(ia_{j})=e^{P^{(k_{0})}(ia_{j})}$ and

$\displaystyle|p_{j}(z)|\leq Ce^{P^{(k)}(Dz)},\quad z\in\mathbb{C},~{}j\geq
1,$ (5.17)

with uniform constants $C,D>0$.

###### Proof.

We follow closely the arguments in the proof of [22, Proposition 2.3]. Let
$k_{1}\leq k_{2}$ be chosen such that
$N^{(\frac{1}{k_{0}})}\prec_{s\omega_{1}}N^{(\frac{1}{k_{1}})}\prec_{s\omega_{1}}N^{(\frac{1}{k_{2}})}$;
cf. Corollary 2.7. We can find positive numbers $A_{j}$, $B_{j}$, and $R_{j}$
such that

$\omega_{j}(t):=\begin{cases}\omega^{(k_{2})}(t),&|t|\leq R_{j},\\\
A_{j}\log|t|+B_{j},&|t|>R_{j},\end{cases}$

is continuous, increasing, $t\mapsto\omega_{j}(e^{t})$ is convex,
$\omega_{j}\leq\omega^{(k_{2})}$, and

$\sup_{|z-ia_{j}|\leq
1}|P^{(k_{2})}(z)-P_{\omega_{j}}(z)|\leq\frac{1}{j},\quad\text{ for all }j;$
(5.18)

cf. Remark 5.4. Let $\varphi:\mathbb{C}\to[0,1]$ be a $C^{\infty}$-function
with support contained in the unit disc and $\varphi(z)=1$ for
$|z|\leq\frac{1}{2}$. As in [22], we set

$u_{j}(z):=\Big{(}1-\frac{z}{ia_{j}}\Big{)}^{-1}e^{P^{(k_{0})}(ia_{j})}\,\overline{\partial}\varphi(z-ia_{j}).$

By Lemma 5.7, there is $A>0$ such that, for all $j$,

$P^{(k_{0})}(ia_{j})\leq P^{(k_{2})}(z)+A,\quad|z-ia_{j}|\leq 1.$ (5.19)

Thus, there exists $M\geq 1$ such that for all $j$ we have

$\displaystyle\int_{\mathbb{C}}|u_{j}(z)|^{2}e^{-2P_{\omega_{j}}(z)-\log(1+|z|^{2})}d\lambda(z)$
$\displaystyle=\int_{|z-ia_{j}|\leq
1}|u_{j}(z)|^{2}e^{-2P_{\omega_{j}}(z)-\log(1+|z|^{2})}d\lambda(z)\leq M.$

Since $\overline{\partial}u_{j}=0$, we infer from [13, Theorem 4.4.2] the
existence of $v_{j}\in C^{\infty}(\mathbb{C})$ with
$\overline{\partial}v_{j}=u_{j}$ such that

$\int_{\mathbb{C}}|v_{j}(z)|^{2}e^{-2P_{\omega_{j}}(z)-3\log(1+|z|^{2})}\,d\lambda(z)\leq
M.$

Then

$p_{j}(z):=\varphi(z-ia_{j})e^{P^{(k_{0})}(ia_{j})}-\Big{(}1-\frac{z}{ia_{j}}\Big{)}v_{j}(z)$

is entire and $p_{j}(ia_{j})=e^{P^{(k_{0})}(ia_{j})}$.

We claim that there exists $M^{\prime}>0$ such that, for all $j$,

$\int_{\mathbb{C}}|p_{j}(z)|^{2}e^{-2P_{\omega_{j}}(z)-4\log(1+|z|^{2})}d\lambda(z)\leq
M^{\prime}.$ (5.20)

Indeed, by (5.19) and (5.18),

$\displaystyle\int_{\mathbb{C}}|\varphi(z-ia_{j})|^{2}e^{2P^{(k_{0})}(ia_{j})}e^{-2P_{\omega_{j}}(z)-4\log(1+|z|^{2})}d\lambda(z)$
$\displaystyle\leq$ $\displaystyle e^{2A}\int_{|z-ia_{j}|\leq
1}e^{2(P^{(k_{2})}(z)-P_{\omega_{j}}(z))-4\log(1+|z|^{2})}d\lambda(z)$
$\displaystyle\leq$ $\displaystyle e^{2A}\int_{|z-ia_{j}|\leq
1}e^{\frac{2}{j}-4\log(1+|z|^{2})}d\lambda(z),$

which is bounded in $j$. And, since $|1-\frac{z}{ia_{j}}|^{2}\leq
2(1+|z|^{2})$,

$\displaystyle\int_{\mathbb{C}}\Big{|}1-\frac{z}{ia_{j}}\Big{|}^{2}|v_{j}(z)|^{2}e^{-2P_{\omega_{j}}(z)-4\log(1+|z|^{2})}d\lambda(z)$
$\displaystyle\leq$
$\displaystyle~{}2\int_{\mathbb{C}}|v_{j}(z)|^{2}e^{-2P_{\omega_{j}}(z)-3\log(1+|z|^{2})}d\lambda(z)\leq
2M.$

This yields (5.20). Since
$2P_{\omega_{j}}+4\log(1+|z|^{2})=O(\log(1+|z|^{2}))$ as $|z|\to\infty$, we
infer that $p_{j}$ is actually a polynomial.

Let us show (5.17). Recall that $P_{\omega_{j}}\leq P^{(k_{2})}$. By Lemma
5.8, we find $k_{3}\in\mathbb{N}_{\geq 1}$ and $K>0$ such that

$P^{(k_{2})}(z)+\log(1+|z|^{2})\leq P^{(k_{3})}(k_{3}z)+K,\quad
z\in\mathbb{C}.$

Together with (5.20) this yields that $(p_{j})_{j}$ is bounded in
$A^{2}_{2P^{(k_{3})}(k_{3}z)}$. Take integers $k_{4}\leq k_{5}$ such that
$N^{(\frac{1}{k_{3}})}\prec_{s\omega_{1}}N^{(\frac{1}{k_{4}})}\prec_{s\omega_{1}}N^{(\frac{1}{k_{5}})}$.
By Lemma 5.7, there is $K_{1}>0$ such that

$2P^{(k_{3})}(k_{3}(z+w))\leq 2P^{(k_{5})}(k_{5}z)+K_{1},\quad
z,w\in\mathbb{C},~{}|w|\leq 1.$

Combining this with Lemma 4.1, we find that $(p_{j})_{j}$ is bounded in
$A_{P^{(k_{5})}(k_{5}z)}$, which shows (5.17) and thus finishes the proof. ∎

###### Proof of Theorem 5.1.

By Proposition 5.10, we need to show that the following conditions are
equivalent:

  1. 1

$\forall y>0\;\exists x>0:~{}M^{(x)}\prec_{L}N^{(y)}$.

  2. 2

If $B\subseteq\mathcal{A}_{\Omega_{\mathfrak{N}}}$ is bounded in
$\mathcal{A}_{\Omega^{+}_{\mathfrak{N}}}$, then $B$ is bounded in
$\mathcal{A}_{\Omega_{\mathfrak{M}}}$.

(1) $\Rightarrow$ (2) Let $B\subseteq\mathcal{A}_{\Omega_{\mathfrak{N}}}$ be
bounded in $\mathcal{A}_{\Omega_{\mathfrak{N}}}^{+}$. So there exist $C>0$ and
$k\in\mathbb{N}_{\geq 1}$ such that

$|f(z)|\leq Ce^{k|\operatorname{Im}z|+\omega^{(k)}(kz)},\quad
z\in\mathbb{C},~{}f\in B,$

where we again use the notation
$\omega^{(k)}(z):=\omega_{N^{(\frac{1}{k})}}(z)$. Since
$B\subseteq\mathcal{A}_{\Omega_{\mathfrak{N}}}$, we have

$|f(z)|\leq C_{f}e^{\omega^{(k_{f})}(k_{f}z)},\quad z\in\mathbb{C},$

which yields $\log|f(z)|=o(|z|)$ as $|z|\to\infty$. Then Corollary 5.12
implies

$|f(z)|\leq Ce^{P^{(k)}(kz)},\quad z\in\mathbb{C},~{}f\in B,$

where $P^{(k)}:=P_{\omega^{(k)}}$. By (1) (and the arguments after (5.6)),
there exist $l\in\mathbb{N}_{\geq 1}$ and $K>0$ such that

$P^{(k)}(kz)\leq P^{(k)}(ik|z|)\leq\omega_{M^{(\frac{1}{l})}}(lz)+K,\quad
z\in\mathbb{C}.$

This shows that $B$ is bounded in $\mathcal{A}_{\Omega_{\mathfrak{M}}}$.

(2) $\Rightarrow$ (1) We argue by contradiction. Suppose that there exist
$k_{0}\in\mathbb{N}_{\geq 1}$ and a sequence of real numbers $a_{j}\geq 1$
tending to infinity such that for all $j$

$P^{(k_{0})}(ia_{j})\geq\omega_{M^{(\frac{1}{j})}}(ja_{j})+j.$ (5.21)

By Lemma 5.13, there is a sequence of polynomials $(p_{j})_{j}$ and
$k\in\mathbb{N}_{\geq k_{0}}$ such that
$p_{j}(ia_{j})=e^{P^{(k_{0})}(ia_{j})}$. This gives the desired contradiction:
The sequence $(p_{j})_{j}$ is contained in
$\mathcal{A}_{\Omega_{\mathfrak{N}}}$, since $\log(|z|)=o(\omega^{(k)}(z))$ as
$|z|\to\infty$. By (5.17) and Lemma 5.6 (in view of Corollary 2.7),
$(p_{j})_{j}$ is bounded in $\mathcal{A}_{\Omega_{\mathfrak{N}}^{+}}$. But, by
(5.21), for every fixed $l\in\mathbb{N}_{\geq 1}$ and $j\geq l$, we have

$p_{j}(ia_{j})=\exp\big{(}P^{(k_{0})}(ia_{j})\big{)}\geq
e^{j}\exp\big{(}\omega_{M^{(\frac{1}{j})}}(ja_{j})\big{)}\geq
e^{j}\exp\big{(}\omega_{M^{(\frac{1}{l})}}(la_{j})\big{)}.$

Thus $(p_{j})_{j}$ is unbounded in every step of the inductive limit defining
$\mathcal{A}_{\Omega_{\mathfrak{M}}}$ and hence in
$\mathcal{A}_{\Omega_{\mathfrak{M}}}$, the limit being regular due to
Corollary 4.4. ∎

###  5.7. Theorem 5.1 without derivation closedness

If we do not require derivation closedness in Theorem 5.1 for $\mathfrak{N}$,
we still can infer some information on the image of the Borel map, but for a
(in general) smaller class. Let us be more precise. For a weight matrix
$\mathfrak{N}=(N^{(\frac{1}{k})})_{k\in\mathbb{N}_{\geq 1}}$, we may consider
the matrix
$\mathfrak{N}_{(dc)}=(N_{(dc)}^{(\frac{1}{k})})_{k\in\mathbb{N}_{\geq 1}}$
consisting of “shifted” sequences:

$(N_{(dc)}^{(\frac{1}{k})})_{j}:=N_{j-k}^{(\frac{1}{k})}\text{ for }j\geq
k,\quad(N_{(dc)}^{(\frac{1}{k})})_{j}:=1\text{ for }j<k.$

Then $\mathfrak{N}_{(dc)}$ is easily seen to be derivation closed, and
$N_{(dc)}^{(\frac{1}{k})}\leq N^{(\frac{1}{k})}$ for all $k$. (For a single
weight sequence $N$, we may still perform this construction with
$N^{(\frac{1}{k})}:=N$ for all $k$ which leads to a derivation closed matrix
$\mathfrak{N}_{(dc)}$ such that
$\mathcal{E}^{(\mathfrak{N}_{(dc)})}\subseteq\mathcal{E}^{(N)}$.)

We get the following version of Theorem 5.1.

######  Theorem 5.14.

Let $\mathfrak{M},\mathfrak{N}$ be weight matrices. Then

$\displaystyle\Lambda^{(\mathfrak{M})}\subseteq
j^{\infty}_{0}\mathcal{E}^{(\mathfrak{N}_{(dc)})}(\mathbb{R})$
$\displaystyle\Longleftrightarrow~{}\;\forall y>0\;\forall
n\in\mathbb{N}\;\exists x,C>0\;\forall t\geq
0:~{}P_{N^{(y)}}(it)+\log(1+t^{n})\leq\omega_{M^{(x)}}(Ct)+C.$

###### Proof.

Observe that

$\omega_{N^{(\frac{1}{n})}}(t)+\log(1+t^{n})-C\leq\omega_{N_{(dc)}^{(\frac{1}{n})}}(t)\leq\omega_{N^{(\frac{1}{n})}}(t)+\log(1+t^{n})+C.$

These inequalities transfer also to the respective harmonic extensions. And
this immediately yields the result via an application of Theorem 5.1. ∎

##  6\. Comparison and conclusions

Let us apply our results to Denjoy–Carleman and Braun–Meise–Taylor classes and
compare them with the known classical extension results.

###  6.1. Denjoy–Carleman classes

Taking $\mathfrak{M}=(M)$ and $\mathfrak{N}=(N)$ in Theorem 3.1, we recover
the Beurling result of [37] (see also [17]).

######  Remark 6.1.

It might be irritating that $\mathfrak{N}=(N)$ clearly fails (3.2), which was
used in the proof of Theorem 3.1 in a crucial way, but Lemma 2.5 associates
with $N$ an equivalent weight matrix with the desired properties (which
consists of infinitely many different weight sequences that are however all
equivalent to $N$).

Let us now discuss Theorem 5.1 in this special setting. Let $M,N$ be weight
sequences, and $N$ in addition derivation closed, such that $(m_{k})^{1/k}$
and $(n_{k})^{1/k}$ tend to $\infty$. Let $\mathfrak{M}=(M^{(x)})_{x>0}$,
$\mathfrak{N}=(N^{(y)})_{y>0}$ be the weight matrices equivalent to $M$, $N$,
respectively, provided by Lemma 2.5. Thus, for each $k\in\mathbb{N}_{\geq 1}$
there are constants $A_{k},B_{k}>0$ such that, for all $t\geq 0$,

$\displaystyle\omega_{M}(2^{k}t)-\log(B_{k})\leq\omega_{M^{(\frac{1}{k})}}(t)\leq\omega_{M}(2^{k}t)-\log(A_{k}),$
$\displaystyle\omega_{N}(2^{k}t)-\log(B_{k})\leq\omega_{N^{(\frac{1}{k})}}(t)\leq\omega_{N}(2^{k}t)-\log(A_{k}),$

and consequently,

$P_{N}(2^{k}t)-\log(B_{k})\leq P_{N^{(\frac{1}{k})}}(t)\leq
P_{N}(2^{k}t)-\log(A_{k}).$

This now shows that the right-hand side of (L) reduces to

$\;\exists C>0\;\forall t\geq 0:~{}P_{N}(it)\leq\omega_{M}(Ct)+C,$

i.e., $M\prec_{L}N$.

In this case, Theorem 5.1 specializes to a version of [20, Theorem 2.3] (see
also the remarks before Corollary 2.4 in said paper). Incorporating the
Roumieu case (see [37] and [17]) and the implications of Theorem 3.1, we
conclude

######  Theorem 6.2.

Let $M,N$ weight sequences, $N$ derivation closed, with
$(m_{k})^{1/k}\to\infty$ and $(n_{k})^{1/k}\to\infty$. Then the following are
equivalent:

  1. 1

$\Lambda^{(M)}\subseteq j^{\infty}_{0}\mathcal{E}^{(N)}(\mathbb{R})$.

  2. 2

$\Lambda^{\\{M\\}}\subseteq j^{\infty}_{0}\mathcal{E}^{\\{N\\}}(\mathbb{R})$.

  3. 3

$M\prec_{L}N$.

  4. 4

$M\prec_{SV}N$.

If $M$ has moderate growth, then the conditions are also equivalent to

  1. 5

There is $C>0$ such that $\kappa_{N}(s)=O(\omega_{M}(Cs))$ as $s\to\infty$.

In fact, that (3) implies (5) follows from (5.3). And, for (5) $\Rightarrow$
(3) note that moderate growth of $M$ is equivalent to

$\exists H\geq 1\;\forall t\geq 0:\quad 2\omega_{M}(t)\leq\omega_{M}(Ht)+H,$

see [19, Proposition 3.6], which allows to “move constant factors in front of
$\omega_{M}$ to its argument”.

Finally, we want to make the connection to the condition
$M\prec_{\gamma_{1}}N$ defined by

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

Note that $M\prec_{\gamma_{1}}M$ is the condition $(\gamma_{1})$ in [25] and
$(M.3)$ in [19]. If $M$ is a weight sequence, then $M\prec_{\gamma_{1}}M$ and
$M\prec_{SV}M$ are equivalent (see [37, Theorem 3.6] and [17, Theorem 5.2]),
but in the mixed setting they fall apart, in general. For weight sequences
$M,N$ such that $M\leq CN$ for some $C\geq 1$ we have that
$M\prec_{\gamma_{1}}N$ implies $M\prec_{SV}N$. If additionally $M$ has
moderate growth, then $M\prec_{\gamma_{1}}N$ if and only if $M\prec_{SV}N$
since these conditions persist if $M$ (or $N$) is replaced by an equivalent
weight sequence and since $M$ has moderate growth if and only if $\mu_{k}\leq
C_{1}(M_{k})^{1/k}$ (see e.g. [29, Lemma 2.2]). Thus, under this additional
requirement on $M$, if $M_{k}\leq CN_{k}$, then also $\mu_{k}\leq
C_{2}\nu_{k}$. Invoking [31, Lemma 5.7], we see that, under these
circumstances, $M\prec_{\gamma_{1}}N$ implies $\kappa_{N}(s)=O(\omega_{M}(s))$
as $s\to\infty$ as well.

So we have the following supplement.

######  Supplement 6.3.

In the setting of Theorem 6.2, if $M$ has moderate growth and $M\leq CN$, then
the conditions (1)–(5) are further equivalent to each of the following
conditions:

  1. 6

$M\prec_{\gamma_{1}}N$.

  2. 7

$\kappa_{N}(s)=O(\omega_{M}(s))$ as $s\to\infty$.

Clearly, (5) and (7) are equivalent if $\omega_{M}$ is a weight function, but
in general it is just a pre-weight function.

###  6.2. Braun–Meise–Taylor classes

Let $\Sigma=(S^{(x)})_{x>0}$ and $\Omega=(W^{(x)})_{x>0}$ be the matrices
associated with the weight functions $\sigma$ and $\omega$, respectively. By
Lemma 2.2, the basic assumptions in Theorem 5.1 hold for the choices
$\mathfrak{M}=\Sigma$, $\mathfrak{N}=\Omega$, provided that $\omega(t)=o(t)$
and $\sigma(t)=o(t)$ as $t\to\infty$. By (2.5),

$\Lambda^{(\sigma)}\cong\Lambda^{(\Sigma)},\quad\mathcal{E}^{(\Omega)}(\mathbb{R})\cong\mathcal{E}^{(\omega)}(\mathbb{R}).$

In this case, the right-hand side of (L), i.e., for all $y>0$ there is $x>0$
with $S^{(x)}\prec_{L}W^{(y)}$ which amounts to

$P_{W^{(y)}}(is)\leq\omega_{S^{(x)}}(Cs)+C,\quad s\geq 0,$ (6.1)

is equivalent to (5.4), i.e., $\kappa_{\omega}(t)=O(\sigma(t))$ as
$t\to\infty$. Indeed, by Lemma 2.2, we have
$\omega\sim\omega_{W^{(x)}},\sigma\sim\omega_{S^{(x)}}$ and thus, by
definition, $\kappa_{\omega}\sim\kappa_{W^{(x)}}$ and $P_{\omega}\sim
P_{W^{(x)}}$ for all $x>0$, whence one implication follows from (5.3).
Conversely, let $y>0$ be given. Then, for all $x>0$,

$\displaystyle
P_{W^{(y)}}(is)\stackrel{{\scriptstyle\eqref{heirswitch}}}{{\leq}}\frac{4}{\pi}\kappa_{W^{(y)}}(s)\stackrel{{\scriptstyle\eqref{goodequivalenceclassic}}}{{\leq}}\frac{4}{y\pi}\kappa_{\omega}(s)$
$\displaystyle\leq\frac{4C}{y\pi}(\sigma(s)+1)$
$\displaystyle\stackrel{{\scriptstyle\eqref{goodequivalenceclassic}}}{{\leq}}\frac{4C}{y\pi}(2x\omega_{S^{(x)}}(s)+D_{x}+1),$

and (6.1) follows if we put $x:=\frac{y\pi}{8C}$.

In this case, Theorem 5.1 specializes to [6, Theorem 3.6]. Incorporating also
the Roumieu part [6, Theorem 3.7] (see also [24, Section 5]) and the
implications of Theorem 3.1, we find

######  Theorem 6.4.

Let $\omega,\sigma$ be weight functions satisfying $\omega(t)=o(t)$,
$\sigma(t)=o(t)$ as $t\to\infty$ and let $\Omega=(W^{(x)})_{x>0}$,
$\Sigma=(S^{(x)})_{x>0}$ be the associated weight matrices. Then the following
conditions are equivalent:

  1. 1

$\Lambda^{(\sigma)}\subseteq
j^{\infty}_{0}\mathcal{E}^{(\omega)}(\mathbb{R})$.

  2. 2

$\Lambda^{\\{\sigma\\}}\subseteq
j^{\infty}_{0}\mathcal{E}^{\\{\omega\\}}(\mathbb{R})$.

  3. 3

$\kappa_{\omega}(t)=O(\sigma(t))$ as $t\to\infty$.

  4. 4

For all $y>0$ there is $x>0$ such that $S^{(x)}\prec_{SV}W^{(y)}$.

  5. 5

For all $y>0$ there is $x>0$ such that $S^{(x)}\prec_{L}W^{(y)}$.

  6. 6

There are $x,y>0$ such that $\kappa_{W^{(y)}}(t)=O(\omega_{S^{(x)}}(t))$ as
$t\to\infty$.

Note that any of the six conditions implies that $\omega$ is non-
quasianalytic; cf. Lemma 5.3.

##  Appendix A Density of entire functions

The following lemma is probably well-known, but we include a proof for the
convenience of the reader. The proof closely follows the arguments in [15,
Proposition 3.2] and [12, Proposition 3.2].

######  Lemma A.1.

Let $M$ be a weight sequence with $(m_{j})^{1/j}\rightarrow\infty$. Let
$f\in\mathfrak{E}^{(M)}(\mathbb{R})$, and $I_{k}:=[-k,k]$. Then there exists a
sequence of entire functions $f_{j}$ such that $\|f-f_{j}\|^{M}_{I_{k},m,r}\to
0$ for all $m\in\mathbb{N}$ and $r>0$; see (4.4) for the definition of the
seminorm.

###### Proof.

Let $\chi\in C^{\infty}(\mathbb{R})$ be a function with compact support and
$0\leq\chi\leq 1$. Suppose that $\chi$ is $1$ on $[-k-1,k+1]$ and $0$ outside
$[-k-2,k+2]$. Set $E_{j}(z):=\sqrt{\frac{j}{\pi}}e^{-jz^{2}}$ and
$f_{j}:=E_{j}\ast\chi f$. Then $f_{j}$ is entire. It is easily seen by
induction on $p$ that

$f_{j}^{(p)}(x)=E_{j}\ast\chi
f^{(p)}(x)+\sum_{\nu=1}^{p}E_{j}^{(p-\nu)}\ast\chi^{\prime}f^{(\nu-1)}(x).$

This yields

$|(f-f_{j})^{(p)}(x)|\leq|f^{(p)}(x)-E_{j}\ast\chi
f^{(p)}(x)|+\sum_{\nu=1}^{p}|E_{j}^{(p-\nu)}\ast\chi^{\prime}f^{(\nu-1)}(x)|.$

For $|x|\leq k$, we find (see [12, (3.5)])

$\displaystyle|f^{(p)}$ $\displaystyle(x)-E_{j}\ast\chi
f^{(p)}(x)|\leq\frac{C_{0}}{\sqrt{j}}\Big{(}\sup_{|\xi|\leq
k+1}|f^{(p+1)}(\xi)|+2\sup_{|\xi|\leq k+2}|f^{(p)}(\xi)|\Big{)},$

for some absolute constant $C_{0}$. Moreover, (see [12, (3.6) and (3.7)])

$\displaystyle|E_{j}^{(p-\nu)}\ast\chi^{\prime}f^{(\nu-1)}(x)|$
$\displaystyle\leq D\sup_{|\xi|\leq k+2}|f^{(\nu-1)}(\xi)|\sup_{|y|\geq
1}|E_{j}^{(p-\nu)}(y)|$ $\displaystyle\leq D\sup_{|\xi|\leq
k+2}|f^{(\nu-1)}(\xi)|(C_{1}(p-\nu))^{p-\nu}\sqrt{\frac{j}{\pi}}e^{-C_{2}j}$

for some constant $D$ depending on $\chi$ and absolute constants
$C_{1},C_{2}>0$. Since $(m_{p})^{1/p}\rightarrow\infty$, there is a constant
$C_{3}$ such that $(C_{1}(p+l))^{p+l}\leq C_{3}(\tfrac{r}{2})^{p}M_{p}$ for
all $p\in\mathbb{N}$ and $0\leq l\leq m$. Altogether, we find, for $|x|\leq
k$, $p\in\mathbb{N}$, and $0\leq l\leq m$,

$\displaystyle|(f-f_{j})^{(p+l)}(x)|\leq\frac{3C_{0}}{\sqrt{j}}\|f\|^{M}_{I_{k+2},m+1,r}r^{p}M_{p}$
$\displaystyle\quad+D\sqrt{\frac{j}{\pi}}e^{-C_{2}j}\|f\|^{M}_{I_{k+2},m,r/2}\Big{(}C_{3}\sum_{\nu=1}^{p}(\tfrac{r}{2})^{p-1}M_{\nu-1}M_{p-\nu}+C_{4}\sum_{\nu=p+1}^{p+l}(\tfrac{r}{2})^{p}M_{p}\Big{)}$
$\displaystyle\quad\leq\|f\|^{M}_{I_{k+2},m+1,r}r^{p}M_{p}\Big{(}\frac{3C_{0}}{\sqrt{j}}+DC_{5}\sqrt{\frac{j}{\pi}}e^{-C_{2}j}\Big{)}$

for absolute constants $C_{4},C_{5}$. This implies
$\|f-f_{j}\|^{M}_{I_{k},m,r}\to 0$ as $j\to\infty$. ∎

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