# Filling Gaps in Hardy Fields

Matthias Aschenbrenner  Kurt Gödel Research Center for Mathematical Logic
Universität Wien
1090 Wien
Austria<EMAIL_ADDRESS>,  Lou van den Dries  Department
of Mathematics
University of Illinois at Urbana-Champaign
Urbana, IL 61801
U.S.A<EMAIL_ADDRESS>and  Joris van der Hoeven  CNRS, LIX (UMR 7161)
Campus de l’École Polytechnique
91120 Palaiseau
France<EMAIL_ADDRESS>
(Date: July 2023)

###### Abstract.

We show how to fill “countable” gaps in Hardy fields. We use this to prove
that any two maximal Hardy fields are back-and-forth equivalent.

###### Contents

  1. 1 Preliminaries on Hausdorff Fields
  2. 2 Analytic Preliminaries
  3. 3 Pseudoconvergence in Hardy Fields
  4. 4 The Remaining Case
  5. 5 Constructing Overhardian Germs
  6. 6 Filling Wide Gaps
  7. 7 The Number of Maximal Hardy Fields
  8. 8 The $H$-couple of a Maximal Hardy Field
  9. 9 Filling Gaps of Type (b)
  10. 10 Isomorphism of Maximal Hardy Fields

## Introduction

By a “Hardy field” we mean in this paper a Hardy field at $+\infty$: a
subfield $H$ of the ring of germs at $+\infty$ of real valued differentiable
functions on intervals $(a,+\infty)$ ($a\in{\mathbb{R}}$) such that $H$ is
closed under differentiation. For basics on Hardy fields, see [26]. Each Hardy
field is an ordered differential field, the (total) ordering given by
$f\leqslant g$ iff $f(t)\leqslant g(t)$ eventually (that is, for all
sufficiently large $t$). Among functions whose germs at $+\infty$ live in
Hardy fields are all one-variable rational functions with real coefficients,
the real exponential and logarithm functions (more generally, Hardy’s
logarithmico-exponential functions [18]), Euler’s $\Gamma$-function and
Riemann’s $\zeta$-function [27], and many other “regularly growing” functions
arising in mathematical practice. As a case in point, by [15] every o-minimal
expansion of the ordered field of real numbers gives rise to a Hardy field (of
germs of definable functions). Our main result is as follows:

######  Theorem A.

Let $H$ be a Hardy field, and let $A$, $B$ be countable subsets of $H$ such
that $A<B$. Then $A<f<B$ for some $f$ in a Hardy field extension of $H$.

Some of the gaps $A<B$ in this theorem correspond to pseudo-cauchy sequences
(pc-sequences in our abbreviated terminology). The relevant pc-sequences have
length $\omega$, and we can handle them using results from our book [ADH] and
the manuscript [7] in an essential way, and various glueing techniques. This
is done in Sections 3 and 4. This dependence on [ADH] and [7] makes this the
deepest part of the present paper, but most of our work here deals with other
gaps.

Sjödin [28] deals with the case $B=\emptyset$ for $\mathcal{C}^{\infty}$-Hardy
fields (whose elements are germs of $\mathcal{C}^{\infty}$-functions). This
provides an important clue for other kinds of gaps: Sjödin’s construction of a
suitable $f$ can be varied in several ways, and that gives us a handle on the
relevant remaining cases. In Section 5 we treat $B=\emptyset$, basically as in
[28], and organized so that it helps in Section 6 where we deal with “wide”
gaps. For the remaining gaps we use results about asymptotic couples from [6]
and an elaboration of the “reverse engineering” in [28]; see Sections 8 and 9.
(Sections 1 and 2 contain mainly analytic preliminaries, and Section 7 applies
material in Sections 5 and 6 to show that there are $2^{\mathfrak{c}}$ many
maximal Hardy fields where $\mathfrak{c}=2^{\aleph_{0}}$ is the cardinality of
the continuum. Here and below, “maximal” means “maximal under inclusion”.)

Most of [7] concerns differentially algebraic extensions of Hardy fields. The
present paper complements this with a “good enough” overview of differentially
transcendental Hardy field extensions $H\langle y\rangle$ of a Liouville
closed Hardy field $H\supseteq{\mathbb{R}}$.

An equivalent formulation of Theorem A is that every maximal Hardy field is
$\eta_{1}$. The property $\eta_{1}$ (Hausdorff [20]) is defined at the end of
the introduction. The main result of [7] is that all maximal Hardy fields, as
ordered differential fields, are $\upomega$-free newtonian Liouville closed
$H$-fields, and thus by [ADH, 15.0.2, 16.6.3] elementarily equivalent to
${\mathbb{T}}$, the ordered differential field of transseries. (On
${\mathbb{T}}$, see [ADH, Appendix A] or [4].) Combining this fact with
Theorem A and a result from [5] we shall derive in Section 10:

######  Corollary B.

Assuming $\operatorname{CH}$ (the Continuum Hypothesis), every maximal Hardy
field is isomorphic as an ordered differential field to the ordered field
$\text{{\bf No}}(\omega_{1})$ of surreal numbers of countable length equipped
with the derivation ${}_{\operatorname{BM}}$ of [9].

Thus with $\operatorname{CH}$, all maximal Hardy field are isomorphic as
ordered differential fields. Without $\operatorname{CH}$, the proof yields a
nonempty back-and-forth system between any maximal Hardy field and the ordered
differential field $\text{{\bf No}}(\omega_{1})$. (See [ADH, B.5] for “back-
and-forth system”.) Then by Karp [21], cf. [8, Theorem 3], any maximal Hardy
field and the ordered differential field $\text{{\bf No}}(\omega_{1})$ are
$\infty\omega$-equivalent. This is a strengthening of [7, Corollary 1].

Key ingredients for proving Theorem A include Lemma 3.4, the construction of a
partition of unity in Section 4, the reduction to Case (b) stated in Lemma
8.11, and the elaborated reverse engineering in Section 9 that culminates in a
diagonal argument. (The idea behind the original reverse engineering from [28]
is sketched in the remarks that follow the statement of Theorem 5.12.)

Theorem A answers a question of Ehrlich [16] and establishes Conjecture B from
[4]. (For Conjecture A, see [7, Theorem A].)

In this paper our Hardy fields are not assumed to be
$\mathcal{C}^{\infty}$-Hardy fields, and we do not know if maximal
$\mathcal{C}^{\infty}$-Hardy fields are necessarily maximal Hardy fields (even
under $\operatorname{CH}$). So the question arises if our main results go
through for maximal $\mathcal{C}^{\infty}$-Hardy fields. This is indeed the
case, and it is not hard to refine some of our proofs to that effect. The same
question arises for the still more special $\mathcal{C}^{\omega}$-Hardy fields
(analytic Hardy fields). Our main results also go through in that setting, but
this is more delicate. We shall treat these refinements in a follow-up paper.

### Notations and conventions

We let $i$, $j$, $k$, $l$, $m$, $n$ range over
${\mathbb{N}}=\\{0,1,2,\dots\\}$. As in [ADH] the convention is that the
ordering of an ordered set, ordered abelian group, or ordered field is a total
ordering. Let $S$ be an ordered set. For any element $b$ in an ordered set
extending $S$ we set

$S^{<b}\ :=\ \\{s\in S:\,s<b\\},\qquad S^{>b}\ :=\ \\{s\in S:\,s>b\\}.$

We have the usual notion of a set $P\subseteq S$ being cofinal in $S$
(respectively, coinitial in $S$). In addition, sets $P,Q\subseteq S$ are said
to be cofinal if for every $p\in P$ there exists $q\in Q$ with $p\leqslant q$
and for every $q\in Q$ there exists $p\in P$ with $q\leqslant p$; replacing
here $\leqslant$ by $\geqslant$, we obtain the notion of $P$ and $Q$ being
coinitial. Thus $P$ and $P^{\downarrow}=\\{s\in S:\,\text{$s\leqslant p$ for
some $p\in P$}\\}$ are cofinal, hence $P$, $Q$ are cofinal iff
$P^{\downarrow}=Q^{\downarrow}$. Likewise, $P$ and $P^{\uparrow}=\\{s\in
S:\,\text{$s\geqslant p$ for some $p\in P$}\\}$ are coinitial, and $P$, $Q$
are coinitial iff $P^{\uparrow}=Q^{\uparrow}$. We let $\operatorname{cf}(S)$
and $\operatorname{ci}(S)$ denote the cofinality and coinitiality of $S$; see
[ADH, 2.1]. We say that $S$ is $\eta_{1}$ if for all countable $P,Q\subseteq
S$ with $P<Q$ there exists an $s\in S$ with $P<s<Q$; in particular, such $S$
is uncountable (cf. Lemma 7.2 below), has no least element, no largest
element, and is dense in the sense that for all $p,q\in S$ with $p<q$ there
exists $s\in S$ with $p<s<q$. We say that an ordered abelian group (ordered
field) is $\eta_{1}$ if its underlying ordered set is $\eta_{1}$. For basic
facts about various $\eta_{1}$-structures, see [25, Kapitel IV].

Let $(a_{\rho})$ be a well-indexed sequence. Its length is the (infinite
limit) ordinal that is the order type of its well-ordered set of indices
$\rho$ (cf. [ADH, p. 73]). Note that if $(a_{\rho})$ has countable length,
then its length has cofinality $\omega$, and thus $(a_{\rho})$ has a cofinal
subsequence $(a_{\rho_{n}})$ of length $\omega$.

Let $(\Gamma,\psi)$ be an asymptotic couple. As in [ADH, 6.5] we set
$\Gamma_{\infty}:=\Gamma\cup\\{\infty\\}$, and adopt the convention that
$\psi(0)=\psi(\infty)=\infty>\Gamma$. For $\alpha\in\Gamma_{\infty}$ we use
$\alpha^{\dagger}$ as an alternative notation for $\psi(\alpha)$ and define
$\alpha^{\langle n\rangle}\in\Gamma_{\infty}$ by recursion on $n$ by
$\alpha^{\langle 0\rangle}:=\alpha$ and $\alpha^{\langle
n+1\rangle}:=(\alpha^{\langle n\rangle})^{\dagger}$. We simplify terminology
by calling an $H$-field closed (“$H$-closed” in [4, 7]) if it is
$\upomega$-free, newtonian, and Liouville closed.

As in [7], $\mathcal{C}$ is the ring of germs at $+\infty$ of continuous
functions ${[a,+\infty)\to{\mathbb{R}}}$, ${a\in{\mathbb{R}}}$, and
$\mathcal{C}^{r}$ for $r\in{\mathbb{N}}\cup\\{\infty\\}$ its subring of germs
of $r$ times continuously differentiable functions
$[a,+\infty)\to{\mathbb{R}}$, $a\in{\mathbb{R}}$. Thus
$\mathcal{C}^{<\infty}:=\bigcap_{n}\mathcal{C}^{n}$ is a differential ring
with the obvious derivation, and has $\mathcal{C}^{\infty}$ as a differential
subring.

Given a Hardy field $H\supseteq{\mathbb{R}}$ we let $\operatorname{Li}(H)$ be
the Hardy-Liouville closure of $H$, that is, the smallest real closed Hardy
field extension of $H$ that contains with any $f$ also $\exp(f)$, and contains
any $g\in\mathcal{C}^{1}$ whenever it contains $g^{\prime}$; see [7, Section
5.3].

##  1\. Preliminaries on Hausdorff Fields

This section contains basic facts about Hausdorff fields. After a subsection
on pc-sequences of length $\omega$ in an ordered field we construct
pseudolimits of such pc-sequences in the setting of Hausdorff fields, and show
how to extend the value group of a Hausdorff field.

### Ordered fields

Let $K$ be an ordered field. We view ${\mathbb{Q}}$ as a subfield of $K$ in
the natural way, and consider $K$ also as a valued field with respect to the
standard valuation given by the valuation ring

$\mathcal{O}\ =\ \big{\\{}a\in K:\,\text{$|a|\leqslant n$ for some
$n$}\big{\\}},$

the smallest convex subring of $K$; see [ADH, p. 175].

######  Lemma 1.1 (Alling [1, 2]).

The following two conditions on $K$ are equivalent:

  1. (i)

$K$ is $\eta_{1}$;

  2. (ii)

the residue field of $K$ is isomorphic to ${\mathbb{R}}$, every pc-sequence of
length $\omega$ in $K$ has a pseudolimit in $K$, and the value group of $K$ is
$\eta_{1}$.

This is well-known, see [24, 1.4] or [25, p. 160]. For a maximal Hardy field
$H$ we have ${{\mathbb{R}}\subseteq H}$, and so the residue field of $H$ is
indeed isomorphic to ${\mathbb{R}}$. Thus in order to show that $H$ is
$\eta_{1}$ it remains to show that all pc-sequences in $H$ of length $\omega$
have a pseudolimit in $H$ and that the value group of $H$ is $\eta_{1}$. The
former will be taken care of in Sections 3, 4, and the latter will be handled
in Sections 5–9.

We continue with generalities on pc-sequences of length $\omega$ in our
ordered field $K$.

Let $(a_{n})$ be a pc-sequence in $K$ of length $\omega$. When does $(a_{n})$
have a pseudolimit in $K$? We indicate below a reduction of this question to
something that turns out to be more manageable. First, $(a_{n})$ and any
infinite subsequence have the same pseudolimits in $K$, and so by passing to
such a subsequence we can arrange that $(a_{n})$ is either strictly increasing
or strictly decreasing. Replacing $(a_{n})$ by $(-a_{n})$, the strictly
decreasing case reduces to the strictly increasing case. Replacing $(a_{n})$
by $(a+a_{n})$ for a suitable $a\in K$, the strictly increasing case reduces
to the strictly increasing case where in addition all terms are positive.
Next, assume $(a_{n})$ is strictly increasing and all terms are positive.
Dropping some initial terms, if necessary, we arrange in addition that
$a_{n}-a_{n-1}\succ a_{n+1}-a_{n}$ for all $n\geqslant 1$. Then we define
$b_{n}$ by $b_{0}:=a_{0}$ and $b_{n}:=a_{n}-a_{n-1}$ for $n\geqslant 1$, so
that $b_{n}>0$, $b_{n}\succ b_{n+1}$, and $a_{n}=b_{0}+\cdots+b_{n}$, for all
$n$.

Reversing this last step, starting with a sequence $(b_{n})$ in $K$ such that
$b_{n}>0$ and ${b_{n}\succ b_{n+1}}$ for all $n$, we obtain a strictly
increasing pc-sequence $(a_{n})$ of positive terms $a_{n}$ by
$a_{n}=b_{0}+\cdots+b_{n}$. This leads to:

######  Lemma 1.2.

The following are equivalent for $K$:

  1. (i)

all pc-sequences in $K$ of length $\omega$ have a pseudolimit in $K$;

  2. (ii)

for every sequence $(b_{n})$ in $K$ with $b_{n}>0$ and $b_{n}\succ b_{n+1}$
for all $n$, the pc-sequence $(a_{n})$ with $a_{n}=b_{0}+\cdots+b_{n}$ for all
$n$ has a pseudolimit in $K$.

### Hausdorff fields

As in [7] we define a Hausdorff field to be a subfield of $\mathcal{C}$, that
is, a subring of $\mathcal{C}$ that happens to be a field. Let $H$ be a
Hausdorff field. Then

$\big{\\{}f\in H:\,\text{$f(t)>0$, eventually}\big{\\}}$

is the strictly positive cone for a (total) ordering on $K$ that makes $H$ an
ordered field, and below we consider $H$ as an ordered field in this way. This
yields the convex subring

$\mathcal{O}\ :=\ \big{\\{}f\in H:\,\text{$|f|\leqslant n$ for some
$n$}\big{\\}},$

which is a valuation ring of $H$, and we consider $H$ accordingly as a valued
field as well. Restricting the relations $\preccurlyeq$, $\prec$, $\sim$ on
$\mathcal{C}$ to $H$ gives exactly the asymptotic relations $\preccurlyeq$,
$\prec$, $\sim$ on $H$ that it comes equipped with as a valued field.

### Extending Hausdorff fields with pseudolimits

Let $H$ be a Hausdorff field, and let a sequence

$f_{0}\succ f_{1}\succ f_{2}\succ\cdots$

in $H^{>}$ be given. Then $(f_{0}+\cdots+f_{n})$ is a pc-sequence in $H$. We
shall construct a pseudolimit of this pc-sequence in some Hausdorff field
extension of $H$ (possibly $H$ itself). To conform with some later parts we
let $t$ range over real numbers $\geqslant 1$ in this subsection. We take for
each $n$ a continuous function ${\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}$
that represents the germ $f_{n}$, to be denoted also by $f_{n}$, such that
$f_{n}(t)\geqslant 0$ and $f_{n+1}(t)\leqslant f_{n}(t)/2$ for all $t$. Now
the sequence $(f_{0}+\cdots+f_{n})$ of partial sums converges pointwise to a
function $f=\sum_{n=0}^{\infty}f_{n}\colon{\mathbb{R}}^{\geqslant
1}\to{\mathbb{R}}$, with the convergence being uniform on each compact subset
of ${\mathbb{R}}^{\geqslant 1}$, so $f$ is continuous. We claim that for all
$n$,

$f-(f_{0}+\cdots+f_{n})\ \prec\ f_{n}\ \text{ in $\mathcal{C}$.}$

Let $\varepsilon>0$, and take $t_{n}\in{\mathbb{R}}^{\geqslant 1}$ with
$f_{n+1}(t)\leqslant\varepsilon f_{n}(t)$ for all $t\geqslant t_{n}$. Then for
such $t$,

$\displaystyle f(t)-\big{(}f_{0}(t)+\cdots+f_{n}(t)\big{)}\ $ $\displaystyle=\
f_{n+1}(t)+f_{n+2}(t)+f_{n+3}(t)+\cdots$ $\displaystyle\leqslant\
f_{n+1}(t)+f_{n+1}(t)/2+f_{n+1}(t)/4+\cdots$ $\displaystyle=\ 2f_{n+1}(t)\
\leqslant\ 2\varepsilon f_{n}(t),$

which proves the claim. As usual we denote the germ of $f$ at $+\infty$ also
by $f$, so that $f\in\mathcal{C}$. In [7, Section 5.1] we defined for
$g,h\in\mathcal{C}$ that $g\leqslant h$ means $g(t)\leqslant h(t)$,
eventually, and $g<h$ means $g\leqslant h$ and $g\neq h$. Here we define for
$g,h\in\mathcal{C}$,

$g<_{\operatorname{e}}h\quad:\Longleftrightarrow\quad g(t)<h(t),\text{
eventually,}$

so $g<_{\operatorname{e}}h\Rightarrow g<h$, and if $g,h\in H$, then
$g<_{\operatorname{e}}h\Leftrightarrow g<h$.

######  Lemma 1.3.

Suppose $(f_{0}+\cdots+f_{n})$ has no pseudolimit in $H$. Let $g\in H$ be such
that $g>f_{0}+\cdots+f_{n}$ in $\mathcal{C}$, for all $n$. Then for all $n$ we
have

$f_{0}+\cdots+f_{n}<_{\operatorname{e}}f<_{\operatorname{e}}g\quad\text{ in
$\mathcal{C}$}.$

###### Proof.

As $g$ is not a pseudolimit of $(f_{0}+\cdots+f_{n})$, we have
$v\big{(}g-(f_{0}+\cdots+f_{n})\big{)}<v(f_{n+1})$ for some $n$. For such $n$
we have, eventually,
$g(t)-\big{(}f_{0}(t)+\cdots+f_{n}(t)\big{)}>2f_{n+1}(t)$, and thus,
eventually, $g(t)>f_{0}(t)+\cdots+f_{n}(t)+2f_{n+1}(t)\geqslant f(t)$. ∎

In view of [7, Lemma 5.1.17] this yields:

######  Corollary 1.4.

If $H$ is real closed and $(f_{0}+\cdots+f_{n})$ has no pseudolimit in $H$,
then $f$ generates over $H$ an immediate Hausdorff field extension $H(f)$ of
$H$ such that $f_{0}+\cdots+f_{n}\leadsto f$.

Even if $H$ is not real closed, $(f_{0}+\cdots+f_{n})$ pseudoconverges in some
Hausdorff field extension of $H$, since we can pass to the real closure of $H$
by [7, Proposition 5.1.4].

### Extending the value group of a Hausdorff field

This is closely connected to filling additive gaps in Hausdorff fields: see
Remark 1.7 and Lemma 1.11 below. For now, $H$ is just an ordered field and
$v\colon H^{\times}\to\Gamma$ is its standard valuation.

######  Lemma 1.5.

Let $A\subseteq H$. Then $A+A$, $2A$ are cofinal. Also, $A$, $2A$ are cofinal
iff $A$, $\frac{1}{2}A$ are cofinal. Likewise with “coinitial” in place of
“cofinal”.

###### Proof.

From $2A\subseteq A+A$ and $a+b\leqslant 2\max(a,b)$ for $a,b\in A$ it follows
that $A+A$ and $2A$ are cofinal. The rest is clear. ∎

######  Corollary 1.6.

Let $A,B\subseteq H^{>}$ be such that $A<B$ and there is no $h\in H$ with
$A<h<B$. Then the following are equivalent:

  1. (i)

$A$, $A+A$ are cofinal;

  2. (ii)

$A$, $2A$ are cofinal;

  3. (iii)

$B$, $B+B$ are coinitial;

  4. (iv)

$B$, $\frac{1}{2}B$ are coinitial.

###### Proof.

The equivalence of (i) and (ii) follows from Lemma 1.5; likewise with (iii)
and (iv), The equivalence of (ii) and (iv) is a consequence of
$B^{\uparrow}=H^{>}\setminus A^{\downarrow}$. ∎

An additive gap in $H$ is a pair $A$, $B$ of subsets of $H^{>}$ with $A<B$
such that there is no $h\in H$ with $A<h<B$, and one of the equivalent
conditions (i)–(iv) in Corollary 1.6 holds.

######  Remark 1.7.

As in [ADH], a cut in an ordered set $S$ is a downward closed subset of $S$.
Call a cut $A$ in the ordered set $H^{>}$ additive if $A$, $B:=H^{>}\setminus
A$ is an additive gap in $H$. Then $A\mapsto A\cup(-A)\cup\\{0\\}$ defines an
inclusion-preserving bijection

$\big{\\{}\text{additive cuts in $H^{>}$}\big{\\}}\to\\{\text{convex subgroups
of $H$}\\},$

with inverse $D\mapsto D^{>}$. (In some places additive cuts in $H^{>}$ are
therefore called “group cuts” in $H$; cf. [23].) Note: $D\mapsto v(D^{>})$ is
an inclusion-preserving bijection

$\\{\text{convex subgroups of $H$}\\}\to\\{\text{upward closed subsets of
$\Gamma$}\\},$

with inverse $P\mapsto v^{-1}(P)\cup\\{0\\}$.

In 1.8–1.10 below we assume that $H$ is real closed. We have multiplicative
versions of Lemma 1.5 and Corollary 1.6, obtained in the same way:

######  Lemma 1.8.

Let $A\subseteq H^{>}$. Then $A\cdot A$ and
$\operatorname{sq}(A):=\\{a^{2}:\,a\in A\\}$ are cofinal. Moreover, $A$ and
$\operatorname{sq}(A)$ are cofinal iff $A$ and $\sqrt{A}:={\\{b\in
H^{>}:\,b^{2}\in A\\}}$ are cofinal. Likewise with “coinitial” in place of
“cofinal”.

######  Corollary 1.9.

Let $A,B\subseteq H^{>}$ be such that $A<B$ and there is no $h\in H$ with
$A<h<B$. Then the following are equivalent:

  1. (i)

$A$, $A\cdot A$ are cofinal;

  2. (ii)

$A$, $\operatorname{sq}(A)$ are cofinal;

  3. (iii)

$B$, $B\cdot B$ are coinitial;

  4. (iv)

$B$, $\sqrt{B}$ are coinitial.

######  Lemma 1.10.

Let $A\subseteq H^{>{\mathbb{Q}}}$. If $A$, $\operatorname{sq}(A)$ are
cofinal, then so are $A$, $2A$, and if $A$, $\sqrt{A}$ are coinitial, then so
are $A$, $\frac{1}{2}A$.

###### Proof.

Let $a\in A$. For the first part, use $2a<a^{2}$; for the second, use
$\sqrt{a}<\frac{a}{2}$. ∎

Now suppose $H$ is a Hausdorff field, turned into an ordered field as
described earlier in this section. The following is Lemma 5.1.18 in [7]:

######  Lemma 1.11.

Suppose $\Gamma=v(H^{\times})$ is divisible. Let $P$ be a nonempty upward
closed subset of $\Gamma$, and let $f\in\mathcal{C}$ be such that $a<f$ for
all $a\in H^{>}$ with $va\in P$, and $f<b$ for all $b\in H^{>}$ with $vb<P$.
Then $f$ generates a Hausdorff field $H(f)$ with $P>vf>\Gamma\setminus P$.

We now specialize $H$ even further: in the rest of this subsection we assume
that $H$ is a Liouville closed Hardy field and $H\supseteq{\mathbb{R}}$.

######  Lemma 1.12.

Let $A\subseteq H^{>{\mathbb{R}}}$. Then:

  1. (i)

if $A$ and $\exp(A)$ are cofinal, then so are $A$ and $\operatorname{sq}(A)$;

Next, assume also that $\operatorname{e}^{x}\in A$, and that $A$ and
$\operatorname{sq}(A)$ are cofinal. Then:

  1. (ii)

$A$ and $A^{\prime}:=\\{a^{\prime}:\,a\in A\\}$ are cofinal;

  2. (iii)

$A$ and $\int\\!A:=\\{b\in H:\,b^{\prime}\in A\\}$ are cofinal, and
$\int\\!A\subseteq H^{>{\mathbb{R}}}$.

###### Proof.

Item (i) follows from $a^{2}\leqslant\exp a$ for $a\in A$.

Next, assume $\operatorname{e}^{x}\in A$, and $A$, $\operatorname{sq}(A)$ are
cofinal. Then $\operatorname{e}^{nx}\in A^{\downarrow}$ if $n\geqslant 1$. Now
for (ii), let $a\in A$. Then $1/a\prec 1$, so
$-a^{\prime}/a^{2}=(1/a)^{\prime}\prec 1$, and thus $0<a^{\prime}<a^{2}$. This
yields
$(A^{\prime})^{\downarrow}\subseteq\operatorname{sq}(A)^{\downarrow}=A^{\downarrow}$.
Suppose in addition $a\geqslant\operatorname{e}^{x}$, so
$a^{\dagger}\succcurlyeq 1$. If $a^{\dagger}\succ 1$, then $a<a^{\prime}$, and
if $a^{\dagger}\asymp 1$, then [ADH, 9.1.11] yields $n\geqslant 1$ with
$a\leqslant\operatorname{e}^{nx}$, and taking $b\in A$ with
$b\geqslant\operatorname{e}^{(n+1)x}\succ\operatorname{e}^{nx}$ we get
$b^{\prime}>(\operatorname{e}^{nx})^{\prime}\geqslant\operatorname{e}^{nx}\geqslant
a$. Thus $A^{\downarrow}\subseteq(A^{\prime})^{\downarrow}$.

As to (iii), let $a\in A$, $b\in H$, and $b^{\prime}=a$. Then
$b>{\mathbb{R}}$, even $b\succ x$. Moreover, $0<b^{\prime}=a<b^{2}$, so
$\sqrt{a}<b$. Thus
$A^{\downarrow}=(\sqrt{A})^{\downarrow}\subseteq(\int\\!A)^{\downarrow}$.
Next, assume also $a=b^{\prime}\succ\operatorname{e}^{x}$. Then
$b\succ\operatorname{e}^{x}$, since $H$ is asymptotic, so
$a/b=b^{\dagger}\succcurlyeq 1$, hence $b\preccurlyeq a\prec a^{2}$ and thus
$b<a^{2}$. This yields
$(\int\\!A)^{\downarrow}\subseteq\operatorname{sq}(A)^{\downarrow}=A^{\downarrow}$.
∎

######  Lemma 1.13.

Let $B\subseteq H$, $B>\operatorname{e}^{x}$, and assume $B$, $\sqrt{B}$ are
coinitial. Then:

  1. (i)

$B$ and $B^{\prime}:=\\{b^{\prime}:b\in B\\}$ are coinitial;

  2. (ii)

$B$ and $\int B:=\\{a\in H:\,a^{\prime}\in B\\}$ are coinitial;

  3. (iii)

$B^{-1}$ and ${-\int B^{-1}}:=\big{\\{}{-g}:\,g\in H^{\prec 1},\,g^{\prime}\in
B^{-1}\big{\\}}$ are cofinal.

###### Proof.

Since $B$, $\sqrt{B}$ are coinitial, so are $B$, $\frac{1}{2}B$, by Lemma
1.10. Thus $B$, ${\mathbb{R}}^{>}B$ are coinitial. Let $b\in B$. Then
$b\succ\sqrt{b}>\operatorname{e}^{x}$, so $\beta:=vb<0$ gives
$\beta^{\dagger}\leqslant 0$, hence $\beta^{\prime}\leqslant\beta$, and thus
$b^{\prime}\geqslant b$. Also $\beta^{\dagger}=o(\beta)$ by [ADH, 9.2.10], so
$\beta<\textstyle\frac{1}{2}\beta+\beta^{\dagger}=(\frac{1}{2}\beta)^{\prime}$
and thus $b\succ(\sqrt{b})^{\prime}\succcurlyeq d^{\prime}$ for some $d\in B$.
This proves (i).

For (ii), let $a\in H$ and $a^{\prime}=b\in B$. Then $a>{\mathbb{R}}$, and
also $a\succ\operatorname{e}^{x}$, since $a\preccurlyeq\operatorname{e}^{x}$
gives $b=a^{\prime}\preccurlyeq\operatorname{e}^{x}$, a contradiction. Hence
$\alpha^{\dagger}\leqslant 0$ for $\alpha:=va$, so
$\alpha\geqslant\alpha^{\prime}=\beta:=vb$, which gives $a\preccurlyeq b$, and
thus $a\leqslant b^{2}$. Since $b\in B$ was arbitrary and $B$ and
$\operatorname{sq}(B)$ are coinitial by Lemma 1.8, this shows that every
element of $B$ is $\geqslant a$ for some $a\in\int B$. With $a$ and $b$ as
above we also have $\alpha<\beta/2$, so $a>\sqrt{b}$. This proves (ii).

As to (iii), let $b\in B$, $\beta:=vb$, and $g\in H^{\prec 1}$ with
$g^{\prime}=b^{-1}$, so $g<0$, and for $\gamma:=vg$ we have
$-\beta=\gamma+\gamma^{\dagger}$. We have $\sqrt{b}>\operatorname{e}^{x}$, so
$b^{-1}<\operatorname{e}^{-2x}$. Claim: $\gamma^{\dagger}\leqslant 0$. If this
claim does not hold, then $0<\gamma^{\dagger}<v(x^{-2})$, so
$g\succ\operatorname{e}^{-x}$, and $g^{\dagger}\succ
x^{-2}\succ\operatorname{e}^{-x}$, and thus
$b^{-1}=g^{\prime}\succ\operatorname{e}^{-2x}$, a contradiction. Now
$\gamma^{\dagger}\leqslant 0$ gives $-\beta\leqslant\gamma$, hence
$b^{-1}=|b^{-1}|\succcurlyeq|g|=-g$, and thus $-g\leqslant d$ for some $d\in
B^{-1}$. From $\gamma^{\dagger}=(-\gamma)^{\dagger}\leqslant 0$ we get
$\gamma^{\dagger}=o(\gamma)$ by [ADH, 9.2.10], hence $-2\beta>\gamma$, and
thus $b^{-2}<-g$. It remains to use that $B^{-1}$, $\operatorname{sq}(B^{-1})$
are cofinal. ∎

##  2\. Analytic Preliminaries

In this section $a$, $b$, $c$, $s$, $t$ range over ${\mathbb{R}}$.

### Constructing smooth functions

We prove here some facts about smooth functions needed later. Let
$\rho\colon{\mathbb{R}}\to{\mathbb{R}}$ be the $\mathcal{C}^{\infty}$-function
of [14, (8.12), Exercise 2(a)]. It is defined by

$\rho(t)\ :=\ \exp\\!\left(-\frac{1}{(1+t)^{2}}-\frac{1}{(1-t)^{2}}\right)\
\text{if $-1<t<1$,}\quad\rho(t):=0\ \text{if $t\leqslant-1$ or $t\geqslant
1$.}$

Thus $\rho(t)>0$ for $-1<t<1$, $\rho$ is even, and
$\rho(0)=\operatorname{e}^{-2}$. (See Figure 1.)

$-1$$1$$\rho$$t$ Figure 1. Sketch of $\rho$

For any subset $I$ of ${\mathbb{R}}$ and
$r\in{\mathbb{N}}\cup\\{\infty,\omega\\}$ we define $\mathcal{C}^{r}(I)$ to be
the set of $f\colon I\to{\mathbb{R}}$ for which $f=g|_{I}$ for some
$\mathcal{C}^{r}$-function $g\colon U\to{\mathbb{R}}$ with $U$ an open
neighborhood of $I$ in ${\mathbb{R}}$; instead of “$f\in\mathcal{C}^{r}(I)$”
we also write “$f\colon I\to{\mathbb{R}}$ is a $\mathcal{C}^{r}$-function” or
“$f\colon I\to{\mathbb{R}}$ is of class $\mathcal{C}^{r}$”. We use this mainly
for sets $I=[a,b]$ with $a<b$ and sets $I=[a,\infty)$. As in [7, Section 5.2]
we denote $\mathcal{C}^{r}[a,+\infty)$ by $\mathcal{C}_{a}^{r}$, and
$\mathcal{C}_{a}:=\mathcal{C}_{a}^{0}$.

######  Lemma 2.1.

There is a $\mathcal{C}^{\infty}$-function
$\alpha\colon{\mathbb{R}}\to{\mathbb{R}}$ such that $\alpha=0$ on
$[-\infty,0]$, $\alpha$ is strictly increasing on $[0,1]$, and $\alpha=1$ on
$[1,+\infty)$.

###### Proof.

One can take $\alpha(t):=c^{-1}\int_{-\infty}^{t}\rho(2s-1)ds$ where
$c:=\int_{-\infty}^{\infty}\rho(2s-1)ds$. ∎

######  Lemma 2.2.

Let $\theta\colon[a,\infty)\to{\mathbb{R}}^{>}$ be continuous. Then there
exists a decreasing $\mathcal{C}^{\infty}$-function
$\zeta\colon[a,\infty)\to{\mathbb{R}}^{>}$ such that $\theta(t)>\zeta(t)$ and
$\zeta^{\prime}(t)>-1$ for all $t\geqslant a$.

###### Proof.

Replacing $\theta$ by the function $t\mapsto\min_{a\leqslant s\leqslant
t}\min\\!\big{(}\theta(s),1\big{)}\colon[a,\infty)\to{\mathbb{R}}^{>}$ we
arrange that $\theta$ is decreasing and $0\leqslant\theta\leqslant 1$ on
$[a,\infty)$. Next we follow Exercise 2 of [14, (8.12)], taking the
convolution with $\rho$; in other words, we extend $\theta$ to all of
${\mathbb{R}}$ by setting $\theta(t)=0$ for $t<a$, and then define
$f\colon{\mathbb{R}}\to{\mathbb{R}}$ by

$f(t):=\int_{-\infty}^{\infty}\theta(s)\rho(t-s)\,ds=\int_{-\infty}^{\infty}\theta(t-s)\rho(s)\,ds.$

Instead of $-\infty$, $\infty$ we can take in the left integral any real
bounds $c\leqslant t-1$, $d\geqslant t+1$, and in the right integral any real
bounds $c\leqslant-1$, $d\geqslant 1$. As in that exercise one shows that $f$
is of class $\mathcal{C}^{\infty}$ (in fact,
$f^{(p)}(t)=\int_{-\infty}^{\infty}\theta(s)\rho^{(p)}(t-s)ds$ for all
$p\in{\mathbb{N}}$ and all $t$) and decreasing on $[a+1,\infty)$. For
$t\geqslant a+1$ we have

$0<f(t)\ =\ \int_{-1}^{1}\theta(t-s)\rho(s)\,ds\ \leqslant\
2\operatorname{e}^{-2}\theta(t-1)\ <\ \theta(t-1).$

Using $\rho^{\prime}(s)\geqslant 0$ for $-1\leqslant s\leqslant 0$ and
$\rho^{\prime}(s)\leqslant 0$ for $0\leqslant s\leqslant 1$, we obtain for all
$t$,

$f^{\prime}(t)\ =\ \int_{-1}^{1}\theta(t-s)\rho^{\prime}(s)\,ds\ \geqslant\
\int_{0}^{1}\theta(t-s)\rho^{\prime}(s)\,ds\ \geqslant\
\int_{0}^{1}\rho^{\prime}(s)\,ds\ =\ -\operatorname{e}^{-2}\ >\ -1.$

Thus $\zeta\colon[a,\infty)\to{\mathbb{R}}^{>}$ defined by $\zeta(t):=f(t+1)$
has the desired property. ∎

######  Lemma 2.3.

Let $a<b$ and $\phi,\zeta\in\mathcal{C}^{\infty}[a,b]$ be such that
$\phi(a)=\zeta(a)$ and $\phi<\zeta$ on $(a,b]$, and let real numbers $c_{n}$
be given with $\phi(b)<c_{0}<\zeta(b)$. Then there exists a function
$\theta\in\mathcal{C}^{\infty}[a,b]$ such that $\theta^{(n)}(a)=\phi^{(n)}(a)$
for all $n$, $\phi<\theta<\zeta$ on $(a,b]$, and $\theta^{(n)}(b)=c_{n}$ for
all $n$. (See Figure 2.)

$a$$b$$\phi$$\zeta$$\theta$$t$ Figure 2. Sketch of $\phi$, $\theta$, $\zeta$
in Lemma 2.3

###### Proof.

By subtracting $\phi$ throughout we arrange $\phi=0$. A result due to E. Borel
[14, Exercise 4(a), p. 192] yields a function
$\beta\in\mathcal{C}^{\infty}[a,b]$ such that $\beta^{(n)}(b)=c_{n}$ for all
$n$. Take $\delta\in(0,b-a)$ with $\delta<\beta<\zeta-\delta$ on
$[b-\delta,b]$, and then $\alpha\in\mathcal{C}^{\infty}[a,b]$ such that

  * •

$\alpha=0$ on $[a,b-\delta]$,

  * •

$\alpha$ is strictly increasing on $[b-\delta,b-\frac{1}{2}\delta]$, and

  * •

$\alpha=1$ on $[b-\frac{1}{2}\delta,b]$.

Take $\varepsilon\in(0,\frac{1}{2})$ such that $\varepsilon\zeta<\delta$ on
$[b-\delta,b]$, and take $\gamma\in\mathcal{C}^{\infty}[a,b]$ such that

  * •

$\gamma^{(n)}(a)=0$ for all $n$,

  * •

$0<\gamma<\varepsilon$ on $(a,b-\frac{1}{2}\delta)$, and

  * •

$\gamma=0$ on $[b-\frac{1}{2}\delta,b]$.

Then the function $\theta:=\alpha\beta+\gamma\zeta$ has the desired
properties. ∎

######  Lemma 2.4.

Let $a<b$, $f,g\in\mathcal{C}[a,b]$, and $f<g$ on $[a,b]$. Then there are
$a_{0}<a_{1}<\cdots<a_{n}$ with $a_{0}=a,\ a_{n}=b$, and a function
$\phi\colon[a,b]\to{\mathbb{R}}$ such that

  1. (i)

$f<\phi<g$ on $[a,b]$,

  2. (ii)

$\phi(a)=\frac{1}{2}\big{(}f(a)+g(a)\big{)}$ and
$\phi(b)=\frac{1}{2}\big{(}f(b)+g(b)\big{)}$, and

  3. (iii)

for $i=0,\dots,n-1$, the restriction of $\phi$ to $[a_{i},a_{i+1}]$ is the
restriction of an affine function ${\mathbb{R}}\to{\mathbb{R}}$.

###### Proof.

Let $\varepsilon:=\frac{1}{2}\min\big{\\{}g(t)-f(t):t\in[a,b]\big{\\}}$, so
$\varepsilon>0$. Choose $n\geqslant 1$ such that for all $s,t\in[a,b]$ with
$\lvert s-t\rvert\leqslant\delta:=\frac{b-a}{n}$ we have $\lvert
f(s)-f(t)\rvert,\lvert g(s)-g(t)\rvert<\varepsilon$. For $i=0,\dots,n$ set
$a_{i}:=a+i\delta$, and for $i=0,\dots,n-1$ take affine
$\phi_{i}\colon{\mathbb{R}}\to{\mathbb{R}}$ with
$\phi_{i}(a_{i})=\frac{1}{2}\big{(}f(a_{i})+g(a_{i})\big{)}$ and
$\phi_{i}(a_{i+1})=\frac{1}{2}\big{(}f(a_{i+1})+g(a_{i+1})\big{)}$. It
suffices to show that $f<\phi_{i}<g$ on $[a_{i},a_{i+1}]$ for $i=0,\dots,n-1$.
For such $i$ and $s,t\in[a_{i},a_{i+1}]$,

$f(t)<\varepsilon+f(s)\leqslant\textstyle\frac{1}{2}\big{(}f(s)+g(s)\big{)}\leqslant-\varepsilon+g(s)<g(t),$

in particular, $f(t)<\phi_{i}(a_{i}),\phi_{i}(a_{i+1})<g(t)$. Since
$\phi_{i}(a_{i})\leqslant\phi_{i}(t)\leqslant\phi_{i}(a_{i+1})$ or
$\phi_{i}(a_{i+1})\leqslant\phi_{i}(t)\leqslant\phi_{i}(a_{i})$, we are done.
∎

######  Lemma 2.5.

Let $f,g\in\mathcal{C}_{a}$ be such that $f<g$ on $[a,+\infty)$. Then there
exists a function $y\in\mathcal{C}^{\infty}_{a}$ such that $f<y<g$ on
$[a,+\infty)$.

###### Proof.

Lemma 2.4 yields a piecewise affine intermediary $\phi$, more precisely, a
strictly increasing sequence $(a_{n})$ in ${\mathbb{R}}$ with $a_{0}=a$ and
$a_{n}\to+\infty$ as $n\to\infty$ and a $\phi\in\mathcal{C}_{a}$ such that for
each $n$ the restriction of $\phi$ to $[a_{n},a_{n+1}]$ is the restriction of
an affine function ${\mathbb{R}}\to{\mathbb{R}}$, and such that $f<\phi<g$ on
$[a,+\infty)$. This reduces the problem of constructing $y$ to proving the
next lemma. ∎

######  Lemma 2.6.

Let $a<b<c$ and $\phi,\theta\in\mathcal{C}^{\infty}[a,c]$ be such that
$\phi(b)=\theta(b)$, and let $0<\varepsilon<b-a,c-b$. Then there exists
$y\in\mathcal{C}^{\infty}[a,c]$ such that

$\displaystyle y(t)\ $ $\displaystyle=\ \phi(t)\ \text{ for $a\leqslant
t\leqslant b-\varepsilon$,}$ $\displaystyle\lvert y(t)-\phi(t)\rvert$
$\displaystyle\ <\ \varepsilon\ \text{ for $b-\varepsilon\leqslant t\leqslant
b$,}$ $\displaystyle y(t)\ $ $\displaystyle=\ \theta(t)\ \text{ for
$b+\varepsilon\leqslant t\leqslant c$,}$ $\displaystyle\lvert
y(t)-\theta(t)\rvert$ $\displaystyle\ <\ \varepsilon\ \text{ for $b\leqslant
t\leqslant b+\varepsilon$.}$

###### Proof.

Take $0<\delta<\varepsilon$ such that $|\phi-\theta|\leqslant\varepsilon/2$ on
$[b-\delta,b+\delta]$. Next, take $\beta\in\mathcal{C}^{\infty}[a,c]$ such
that

  * •

$\beta=0$ on $[a,b-\delta]$,

  * •

$0\leqslant\beta\leqslant 1$ on $[b-\delta,b+\delta]$, and

  * •

$\beta=1$ on $[b+\delta,c]$.

Then $y:=(1-\beta)\phi+\beta\theta$ has the desired property. ∎

######  Lemma 2.7.

For each $n$, let $f_{n},g_{n}\in\mathcal{C}$ be such that $f_{n}\leqslant
f_{n+1}$, $g_{n+1}\leqslant g_{n}$, and $f_{n}<_{\operatorname{e}}g_{n}$. Then
there exists $\phi\in\mathcal{C}^{\infty}$ such that
$f_{n}<_{\operatorname{e}}\phi<_{\operatorname{e}}g_{n}$ for each $n$.

###### Proof.

Take for each $n$ representatives of $f_{n}$ and $g_{n}$ in $\mathcal{C}_{0}$,
denoted also by $f_{n}$ and $g_{n}$, such that $f_{n}<g_{n}$ on $[0,\infty)$.
Next, take a strictly increasing sequence $(a_{n})$ of real numbers $\geqslant
0$ with $a_{n}\to\infty$ as $n\to\infty$, such that $f_{n}\leqslant f_{n+1}$
and $g_{n}\geqslant g_{n+1}$ on $[a_{n},\infty)$, and take continuous
functions $\alpha_{n},\beta_{n}\colon[0,\infty)\to[0,1]$ with
$\alpha_{n}(a_{n})=1$, $\alpha_{n}(a_{n+1})=0$ and $\alpha_{n}+\beta_{n}=1$ on
$[a_{n},a_{n+1}]$. Let $f,g\colon[0,\infty)\to{\mathbb{R}}$ be given by

$\displaystyle f$ $\displaystyle=f_{0}\text{ on }[0,a_{0}],\quad
f=\alpha_{n}f_{n}+\beta_{n}f_{n+1}\text{ on }[a_{n},a_{n+1}],$ $\displaystyle
g$ $\displaystyle=g_{0}\text{ on }[0,a_{0}],\quad
g=\alpha_{n}g_{n}+\beta_{n}g_{n+1}\text{ on }[a_{n},a_{n+1}],$

so $f$, $g$ are continuous, $f_{n}\leqslant f$ and $g\leqslant g_{n}$ on
$[a_{n},\infty)$, and $f<g$ on $[0,\infty)$. (See Figure 3.) Now Lemma 2.5
gives $\phi\in\mathcal{C}_{0}^{\infty}$ such that $f<\phi<g$ on $[0,\infty)$,
and then its germ at $+\infty$, denoted also by $\phi$, satisfies
$f_{n}<_{\operatorname{e}}\phi<_{\operatorname{e}}g_{n}$ for all $n$. ∎

$a_{n}$$a_{n+1}$$\beta_{n}$$\alpha_{n}$$t$

$a_{n}$$a_{n+1}$$\phi$$g_{n+1}$$g_{n}$$f_{n+1}$$f_{n}$$t$

Figure 3. Constructing $\phi$ in the proof of Lemma 2.7

######  Corollary 2.8.

Let $H$ be a Hausdorff field and $A$, $B$ nonempty countable subsets of $H$
with $A<B$. Then there exists $\phi\in\mathcal{C}^{\infty}$ such that
$A<_{\operatorname{e}}\phi<_{\operatorname{e}}B$.

###### Proof.

Take an increasing and cofinal sequence $(f_{n})$ in $A$ and a decreasing
coinitial sequence $(g_{n})$ in $B$, and apply the previous lemma. ∎

We shall also use the following variant of Lemma 2.3:

######  Lemma 2.9.

Let $a<b$, $f,g\in\mathcal{C}[a,b]$, and $c_{n},d_{n}\in{\mathbb{R}}$ for
$n=0,1,2,\dots$ be such that $f(a)<c_{0}<g(a)$, $f<g$ on $[a,b]$, and
$f(b)<d_{0}<g(b)$. Then there exists $y\in\mathcal{C}^{\infty}[a,b]$ with
$f<y<g$ on $[a,b]$ and $y^{(n)}(a)=c_{n}$, $y^{(n)}(b)=d_{n}$ for all $n$.

###### Proof.

Take $\varepsilon>0$ such that $f(a)+\varepsilon<c_{0}$, $f+\varepsilon<g$ on
$[a,b]$, and $f(b)+\varepsilon<d_{0}$. Lemma 2.5 gives
$\phi\in\mathcal{C}^{\infty}[a,b]$ with $f<\phi<f+\varepsilon$ on $[a,b]$, and
so replacing $f$ by $\phi$ and then subtracting $\phi$ throughout (replacing
$g$ by $g-\phi$ and $c_{n}$, $d_{n}$ by $c_{n}-\phi^{(n)}(a)$,
$d_{n}-\phi^{(n)}(b)$, respectively) we arrange $f=0$.

Borel’s result gives $\alpha,\beta\in\mathcal{C}^{\infty}[a,b]$ with
$\alpha^{(n)}(a)=c_{n}$ and $\beta^{(n)}(b)=d_{n}$ for all $n$. Take a real
number $M>0$ such that $|\alpha|,|\beta|\leqslant M$ on $[a,b]$. Take “small”
real numbers $\eta_{1},\eta_{2}>0$ such that $a+2\eta_{1}<b-2\eta_{1}$,
$2M\eta_{1}<\eta_{2}$, and $2M\eta_{1}+\eta_{2}<g$ on $[a,b]$. Take
$\gamma,\delta\in\mathcal{C}^{\infty}[a,b]$ such that

  * •

$\gamma=1$ on $[a,a+\eta_{1}]$,

  * •

$\gamma$ is decreasing on $[a+\eta_{1},a+2\eta_{1}]$,

  * •

$\gamma=\eta_{1}$ on $[a+2\eta_{1},b-2\eta_{1}]$,

  * •

$\gamma$ is decreasing on $[b-2\eta_{1},b-\eta_{1}]$, and

  * •

$\gamma=0$ on $[b-\eta_{1},b]$,

and $\delta$ behaves similarly in the opposite direction:

  * •

$\delta=0$ on $[a,a+\eta_{1}]$,

  * •

$\delta$ is increasing on $[a+\eta_{1},a+2\eta_{1}]$,

  * •

$\delta=\eta_{1}$ on $[a+2\eta_{1},b-2\eta_{1}]$,

  * •

$\delta$ is increasing on $[b-2\eta_{1},b-\eta_{1}]$, and

  * •

$\delta=1$ on $[b-\eta_{1},b]$.

Finally, take $\theta\in\mathcal{C}^{\infty}[a,b]$ such that

  * •

$\theta=0$ on $[a,a+\eta_{1}]$,

  * •

$\theta$ is increasing on $[a+\eta_{1},a+2\eta_{1}]$,

  * •

$\theta=\eta_{2}$ on $[a+2\eta_{1},b-2\eta_{1}]$,

  * •

$\theta$ is decreasing on $[b-2\eta_{1},b-\eta_{1}]$, and

  * •

$\theta=0$ on $[b-\eta_{1},b]$.

(See Figure 4.) Then $y:=\gamma\alpha+\delta\beta+\theta$ has the desired
property, provided $\eta_{1}$, $\eta_{2}$ are sufficiently small. ∎

$a$$a+\eta_{1}$$\hskip
22.50003pta+2\eta_{1}$$\hskip-12.50002ptb-2\eta_{1}$$\hskip
5.0ptb-\eta_{1}$$b$$\eta_{1}$$\eta_{2}$$1$$t$$\gamma$$\delta$$\theta$ Figure
4. The functions $\gamma$, $\delta$, $\theta$

### Constructing infinite sums

The next lemma follows from [14, (8.6.4)]:

######  Lemma 2.10.

Let $(f_{n})$ be a sequence of functions in $\mathcal{C}_{a}^{1}$ such that
$f_{n}(a)\to c$ as ${n\to\infty}$, for some $c\in{\mathbb{R}}$. Suppose also
that $(f_{n}^{\prime})$ converges to $g\in\mathcal{C}_{a}$, uniformly on
$[a,b]$ for every $b>a$. Then $(f_{n})$ converges to a function
$f\in\mathcal{C}_{a}^{1}$, uniformly on $[a,b]$ for every $b>a$, with
$f^{\prime}=g$.

We use this for infinite series where the $f_{n}$ are the partial sums, and
with higher derivatives where the assumptions allow us to apply the lemma
inductively. We shall also need a slight twist, where instead of the
derivation $\der\colon\mathcal{C}_{a}^{1}\to\mathcal{C}_{a}$ we use
${\derdelta\colon\mathcal{C}_{a}^{1}\to\mathcal{C}_{a}}$, with
$\derdelta:=\phi^{-1}\der$, $\phi\in(\mathcal{C}_{a})^{\times}$:

######  Lemma 2.11.

Let $(f_{n})$ be a sequence of functions in $\mathcal{C}_{a}^{1}$ such that
$f_{n}(a)\to c$ as ${n\to\infty}$, for some $c$. Suppose also that $(\derdelta
f_{n})$ converges to $g\in\mathcal{C}_{a}$, uniformly on $[a,b]$ for every
$b>a$. Then $(f_{n})$ converges to a function $f\in\mathcal{C}_{a}^{1}$,
uniformly on $[a,b]$ for every $b>a$, with $\derdelta f=g$.

This follows from the previous lemma in view of $\der=\phi\derdelta$.
Induction on $m$ yields:

######  Corollary 2.12.

Let $m\geqslant 1$, $\phi\in(\mathcal{C}_{a}^{m-1})^{\times}$, and
$\derdelta:=\phi^{-1}\der\colon\mathcal{C}_{a}^{1}\to\mathcal{C}_{a}$. Then
maps $\mathcal{C}_{a}^{j}$ into $\mathcal{C}_{a}^{j-1}$ for $j=1,\dots,m$. Let
$(f_{n})$ be a sequence of functions in $\mathcal{C}_{a}^{m}$ such that for
$k=0,\dots,m$ the series $\sum_{n=0}^{\infty}{}^{k}f_{n}$ converges to
$g_{k}\in\mathcal{C}_{a}$, uniformly on $[a,b]$ for every $b>a$. Then for
$k=0,\dots,m$ we have $g_{k}\in\mathcal{C}_{a}^{m-k}$ and
$f:=g_{0}\in\mathcal{C}_{a}^{m}$ satisfies ${}^{k}f=g_{k}$.

This corollary and the following results on infinite sums will be used in the
next two sections. Let $a\in{\mathbb{R}}$, and for $i=0,1,2,\dots$, let a
continuous function $f_{i}\colon[a,\infty)\to{\mathbb{R}}$ be given, and set
$M_{i}^{n}:=\max\limits_{a\leqslant t\leqslant a+n}|f_{i}(t)|$, so

$0\ \leqslant\ M_{i}^{0}\ \leqslant\ M_{i}^{1}\ \leqslant\ M_{i}^{2}\
\leqslant\ \cdots.$

Suppose the real numbers $\varepsilon_{i}>0$ are such that
$\sum_{i}\varepsilon_{i}M^{n}_{i}<\infty$ for every $n$. Then
$\sum_{i}\varepsilon_{i}f_{i}$ converges uniformly on each set $[a,a+n]$, and
so this sum defines a continuous function on $[a,\infty)$. We can certainly
take real numbers $\varepsilon_{i}>0$ such that
$\sum_{i}\varepsilon_{i}M_{i}^{i}<\infty$, and then we do indeed have for
every $n$ that $\sum_{i}\varepsilon_{i}M^{n}_{i}<\infty$, since

$\sum_{i}\varepsilon_{i}M^{n}_{i}\ =\
\sum_{i=0}^{n}\varepsilon_{i}M^{n}_{i}+\sum_{i>n}\varepsilon_{i}M^{n}_{i}\
\leqslant\
\sum_{i=0}^{n}\varepsilon_{i}M^{n}_{i}+\sum_{i>n}\varepsilon_{i}M^{i}_{i}.$

Thus there exist $\varepsilon_{i}$ as in the hypothesis of the next lemma. In
the rest of this subsection we assume that for every $i$ we have
$f_{i}\geqslant 0$ on $[a,\infty)$ and $f_{i}\prec f_{i+1}$ in $\mathcal{C}$.

######  Lemma 2.13.

Let the reals $\varepsilon_{i}>0$ be such that $\sum_{i}\varepsilon_{i}f_{i}$
converges to a function $f\colon[a,\infty)\to{\mathbb{R}}$, uniformly on each
compact subset of $[a,\infty)$. Then $f\succ f_{n}$ (in $\mathcal{C}$) for all
$n$. If all $f_{i}$ are increasing, then so is
$f=\sum_{i}\varepsilon_{i}f_{i}$.

###### Proof.

Note that $\sum_{i}\varepsilon_{i}f_{i}\geqslant\varepsilon_{n+1}f_{n+1}\succ
f_{n}$. ∎

######  Lemma 2.14.

Let for each $n$ a continuous function
$g_{n}\colon[a,\infty)\to{\mathbb{R}}^{>}$ be given such that $f_{i}\prec
g_{n}$, for all $i$ and $n$. Then there exist reals $\varepsilon_{i}>0$ for
which $\sum_{i}\varepsilon_{i}f_{i}$ converges to a function
$f\colon[a,\infty)\to{\mathbb{R}}$, uniformly on each compact subset of
$[a,\infty)$, such that $f\leqslant g_{n}$ in $\mathcal{C}$ for all $n$.

###### Proof.

For the moment we just consider one continuous function
$g\colon[a,\infty)\to{\mathbb{R}}^{>}$ with $f_{i}\prec g$ for all $i$. Then
we pick the $\varepsilon_{i}>0$ so small that
$\sum_{i}\varepsilon_{i}M_{i}^{i}<\infty$ and $\varepsilon_{i}f_{i}\leqslant
g/2^{i+1}$. This results in $f:=\sum_{i}\varepsilon_{i}f_{i}\leqslant g$. Let
a second continuous function $h\colon[a,\infty)\to{\mathbb{R}}^{>}$ be given
with $f_{i}\prec h$ for all $i$. Take $b\geqslant a$ such that
$\varepsilon_{0}f_{0}\leqslant h/2$ on $[b,\infty)$, and next decrease, if
necessary, the $\varepsilon_{i}$ with $i\geqslant 1$ so that
$\varepsilon_{i}f_{i}\leqslant h/2^{i+1}$ on $[b,\infty)$ for the new values
of $\varepsilon_{i}$. This results in $f\leqslant h$ on $[b,\infty)$ for the
new $f$; note that we did not change $\varepsilon_{0}$. Starting with
$g=g_{0}$ we apply this procedure successively to $g_{1},g_{2},\dots$ in the
role of $h$: we recursively pick $b_{1},b_{2},\dots\geqslant a$, decreasing
only the $\varepsilon_{i}$ for $i\geqslant n$ when dealing with $g_{n}$,
$n\geqslant 1$. Then at the end we have not only $f\leqslant g_{0}$ on
$[a,\infty)$, but also $f\leqslant g_{n}$ on $[b_{n},\infty)$, for all
$n\geqslant 1$ simultaneously. ∎

Note that if in Lemma 2.14 we have $g_{0}\succ g_{1}\succ g_{2}\succ\cdots$,
then $f\prec g_{n}$ for all $n$. Lemmas 2.13 and 2.14 are more precise
versions of results of du Bois-Reymond [10] and Hadamard [17, §19],
respectively; cf. [19, Chapter II].

Assume next that the $f_{i}$ are of class $\mathcal{C}^{\infty}$. Then we set

$M^{n}_{i}\ :=\ \max_{j\leqslant n,\ a\leqslant t\leqslant
a+n}\big{|}f_{i}^{(j)}(t)\big{|}.$

Again, $0\leqslant M_{i}^{0}\leqslant M_{i}^{1}\leqslant
M_{i}^{2}\leqslant\cdots$. Taking the $\varepsilon_{i}>0$ such that
$\sum_{i}\varepsilon_{i}M_{i}^{i}<\infty$, we have
$\sum_{i}\varepsilon_{i}M^{n}_{i}<\infty$ for every $n$, as before. Hence
$\sum_{i}\varepsilon_{i}f_{i}$ converges, say to the continuous function
$f\colon[a,\infty)\to{\mathbb{R}}^{\geqslant}$, uniformly on each $[a,a+n]$.
Also $\sum_{i}\varepsilon_{i}f_{i}^{(j)}$ converges for every $j$ to a
continuous function $f^{(j)}\colon[a,\infty)\to{\mathbb{R}}$, uniformly on
each $[a,a+n]$. An easy induction on $j$ shows that $f$ is in fact of class
$\mathcal{C}^{j}$ with $f^{(j)}$ as its $j$th derivative, as suggested by the
notation. Thus $f$ is of class $\mathcal{C}^{\infty}$.

### Useful inequalities in constructing Hardy fields

The lemmas in this subsection will be used in Sections 6 and 9.

######  Lemma 2.15.

Let $F,G\in\mathcal{C}^{1}_{a}$ satisfy $F^{\prime}(t)\leqslant G^{\prime}(t)$
for all $t\geqslant a$. Then there is a real constant $c$ such that $F<G+c$ on
$[a,\infty)$.

###### Proof.

The function $F-G$ is continuous and decreasing, hence on $[a,\infty)$ we have
$F-G\leqslant F(a)-G(a)<c:=F(a)-G(a)+1$. ∎

Here is a useful multiplicative version:

######  Lemma 2.16.

Let $F,G\colon[a,+\infty)\to{\mathbb{R}}^{>}$ of class $\mathcal{C}^{1}$ be
such that $F^{\dagger}\leqslant G^{\dagger}$ on $[a,\infty)$. Then there is
$c\in{\mathbb{R}}^{>}$ such that $F<cG$ on $[a,\infty)$.

###### Proof.

We have $F^{\dagger}=(\log F)^{\prime}$ and $G^{\dagger}=(\log G)^{\prime}$,
so Lemma 2.15 yields $d\in{\mathbb{R}}$ with $\log F<d+\log G$ on
$[a,\infty)$, and thus $F<c\,G$ on $[a,\infty)$ for
$c:=\operatorname{e}^{d}\in{\mathbb{R}}^{>}$. ∎

######  Lemma 2.17.

Suppose $f\in\mathcal{C}$ lies in a Hardy field. Then the germ $f(x+1)$
satisfies:

  1. (i)

$f(x)-x>_{\operatorname{e}}{\mathbb{R}}\ \Longrightarrow\
f(x)+1<_{\operatorname{e}}f(x+1)$;

  2. (ii)

$0<f(x)\succ\operatorname{e}^{x}\ \Longrightarrow\
f(x)<_{\operatorname{e}}f(x+1)/2$;

  3. (iii)

$0<f(x)\prec\operatorname{e}^{-x}\ \Longrightarrow\
f(x)-f(x+1)>_{\operatorname{e}}f(x)/2$.

###### Proof.

For (i), assume $f=x+g$ with $g\in\mathcal{C}$ and $g>{\mathbb{R}}$. Then $g$
lies in a Hardy field, so $g$ is eventually strictly increasing, hence
$g=g(x)<_{\operatorname{e}}g(x+1)$, and thus

$f(x)+1\ =\ x+1+g\ <_{\operatorname{e}}\ x+1+g(x+1)\ =\ f(x+1).$

Next, assume $0<f\succ\operatorname{e}^{x}$. Then $(\log
f)-x>_{\operatorname{e}}{\mathbb{R}}$, so $\big{(}\log
f(x)\big{)}+1<_{\operatorname{e}}\log f(x+1)$ by (i), and thus
$\operatorname{e}f(x)<_{\operatorname{e}}\ f(x+1)$, hence
$f(x)<_{\operatorname{e}}f(x+1)/\operatorname{e}<_{\operatorname{e}}f(x+1)/2$.
As to (iii), assume $0<f(x)\prec\operatorname{e}^{-x}$. Applying (ii) to
$f^{-1}$ gives $f(x)>_{\operatorname{e}}2f(x+1)$, so
$0<_{\operatorname{e}}f(x+1)<_{\operatorname{e}}f(x)/2$, and thus
$f(x)-f(x+1)>_{\operatorname{e}}f(x)/2$. ∎

##  3\. Pseudoconvergence in Hardy Fields

Let $H$ be a Hardy field. Let a sequence

$f_{0}\succ f_{1}\succ f_{2}\succ\dots$

in $H^{>}$ be given. Then we have the pc-sequence $(F_{i})$ in $H$, with
$F_{i}:=f_{0}+\cdots+f_{i}$. Our aim in this section and the next is to show:

######  Theorem 3.1.

$(F_{i})$ pseudoconverges in some Hardy field extension of $H$.

In view of Lemma 1.2 this has the following consequence:

######  Corollary 3.2.

Every pc-sequence of countable length in a maximal Hardy field has a
pseudolimit in that Hardy field.

Towards the proof of Theorem 3.1 we first recall from [7, Sections 5.1, 5.3]
the following. Let $\ell\in\mathcal{C}^{<\infty}$ be such that
$\ell>{\mathbb{R}}$ and $\ell^{\prime}\in H$. Then $\ell$ lies in a Hardy
field extension of $H$, $\phi:=\ell^{\prime}\in H^{>}$ is active in $H$, and
the compositional inverse $\ell^{\operatorname{inv}}>{\mathbb{R}}$ of $\ell$
yields an isomorphism $f\mapsto
f^{\circ}:=f\circ\ell^{\operatorname{inv}}\colon(\mathcal{C}^{<\infty})^{\phi}\to\mathcal{C}^{<\infty}$
of differential rings that maps $H$ onto the Hardy field
$H^{\circ}:=H\circ\ell^{\operatorname{inv}}$; moreover, $f_{1}\prec
f_{2}\Leftrightarrow f_{1}\circ\ell\prec f_{2}\circ\ell$, for all
$f_{1},f_{2}\in\mathcal{C}^{<\infty}$. Thus $(F_{i}^{\circ})$ is a pc-sequence
in $H^{\circ}$, and we have:

######  Lemma 3.3.

$(F_{i})$ pseudoconverges in some Hardy field extension of $H$ if and only if
$(F_{i}^{\circ})$ pseudoconverges in some Hardy field extension of
$H^{\circ}$.

We can also use [7, Theorem 6.7.22] to pass to an extension and arrange that
$H\supseteq{\mathbb{R}}$ and $H$ is closed. Then the following lemma is
relevant.

######  Lemma 3.4.

Let $H\supseteq{\mathbb{R}}$ be closed. Suppose $(F_{i})$ has no pseudolimit
in $H$, and let any element $F\in\mathcal{C}^{<\infty}$ be given. Then the
following are equivalent:

  1. (i)

$F$ generates a Hardy field $H\langle F\rangle$ over $H$ with $F_{i}\leadsto
F$;

  2. (ii)

for all $k$, $m$ with $k<m$ and active $\phi\in H^{>}$ we have

${}^{k}\\!\left(\frac{F-F_{m}}{f_{m}}\right)\preccurlyeq 1\text{ in
}\mathcal{C}^{<\infty}$

where $\derdelta:=\phi^{-1}\der$ is construed as a derivation of
$\mathcal{C}^{<\infty}$.

###### Proof.

Assume (i). Then for all $k$, $m$ and active $\phi\in H^{>}$ we have
$(F-F_{m})/f_{m}\prec 1$, and thus
${}^{k}\big{(}\frac{F-F_{m}}{f_{m}}\big{)}\prec 1$. This proves (i)
$\Rightarrow$ (ii). For (ii) $\Rightarrow$ (i), assume (ii). For $k=0$ we get
$F-F_{m}\preccurlyeq f_{m}$ for all $m\geqslant 1$. Let $P\in
H\\{Y\\}^{\neq}$. Now $(F_{m})$ is of $\operatorname{d}$-transcendental type
over $H$ by [ADH, 11.4, 14.0.2], so we have $m_{0}\geqslant\text{order}(P)$ in
${\mathbb{N}}^{\geqslant 1}$ such that $\operatorname{ndeg}_{\preccurlyeq
f_{m+1}}P_{+F_{m}}=0$ for all $m\geqslant m_{0}$, by [ADH, 11.4.11, 11.4.12].
Using $P_{+F_{m+1}}=(P_{+F_{m}})_{+f_{m+1}}$ and [ADH, 11.2.7] we obtain
$\operatorname{ndeg}_{\preccurlyeq f_{m+1}}P_{+F_{m+1}}=0$ for all $m\geqslant
m_{0}$. Thus for $m_{1}:=m_{0}+1$ and $Q:=P_{+F_{m_{1}},\times f_{m_{1}}}$ we
have an active $\phi_{0}\in H^{>}$ with $\operatorname{ddeg}Q^{\phi}=0$ for
all active $\phi\preccurlyeq\phi_{0}$ in $H^{>}$. This gives $h\in H^{\times}$
such that, with ${\boldsymbol{j}}$ ranging over ${\mathbb{N}}^{m_{0}}$ and
$Q^{\phi_{0}}_{{\boldsymbol{j}}}:=(Q^{\phi_{0}})_{{\boldsymbol{j}}}$,

$Q^{\phi_{0}}(Y)\ =\ h+\sum_{|{\boldsymbol{j}}|\neq
0}Q^{\phi_{0}}_{{\boldsymbol{j}}}Y^{{\boldsymbol{j}}},\qquad
Q^{\phi_{0}}_{{\boldsymbol{j}}}\prec h\text{ for }|{\boldsymbol{j}}|\neq 0.$

Thus with $G:=(F-F_{m_{1}})/f_{m_{1}}$ we have $G\preccurlyeq 1$ and
$F=F_{m_{1}}+f_{m_{1}}G$, so

$P(F)\ =\ Q^{\phi_{0}}(G)\ =\ h+\sum_{|{\boldsymbol{j}}|\neq
0}Q^{\phi_{0}}_{{\boldsymbol{j}}}G^{{\boldsymbol{j}}}$

where the factors $G^{{\boldsymbol{j}}}$ are evaluated in
$\mathcal{C}^{<\infty}$ using the derivation $\derdelta=\phi_{0}^{-1}\der$,
and so $G^{{\boldsymbol{j}}}\preccurlyeq 1$ for $|{\boldsymbol{j}}|\neq 0$, by
(ii). Hence $P(F)\,\sim\,h$. This yields (i). ∎

######  Corollary 3.5.

In Lemma 3.4 we can replace (ii) by any of the two variants below:

  1. (ii)∗

for all $m>k$ and active $\phi_{0}\in H$ there is an active
$\phi\preccurlyeq\phi_{0}$ in $H^{>}$ such that

${}^{k}\left(\frac{F-F_{m}}{f_{m}}\right)\preccurlyeq 1\text{ in
}\mathcal{C}^{<\infty}$

where $\derdelta:=\phi^{-1}\der$ is construed as a derivation of
$\mathcal{C}^{<\infty}$.

  2. (ii)∗∗

for all $m_{0}\geqslant 1$ and active $\phi_{0}\in H$ there is an active
$\phi\preccurlyeq\phi_{0}$ in $H^{>}$ and an $m\geqslant m_{0}$ such that for
$k=1,\dots,m_{0}$,

${}^{k}\left(\frac{F-F_{m}}{f_{m}}\right)\preccurlyeq 1\text{ in
$\mathcal{C}^{<\infty}$,}\ \text{with $\derdelta:=\phi^{-1}\der$.}$

###### Proof.

For (ii)∗∗ $\Rightarrow$ (i), assume (ii)∗∗ and take $m_{0},Q,\phi_{0}$ as in
the proof of (ii) $\Rightarrow$ (i), and set $Q_{m}:=P_{+F_{m+1},\times
f_{m+1}}$. For any $m\geqslant m_{0}$ and active $\phi\preccurlyeq\phi_{0}$ in
$H^{>}$ we have $\operatorname{ddeg}Q^{\phi}=0$, so
$\operatorname{ddeg}Q_{m}^{\phi}=0$ by [ADH, 6.6.12]. Now (ii) gives active
$\phi\preccurlyeq\phi_{0}$ in $H^{>}$ and $m\geqslant m_{0}$ such that for
$k=1,\dots,m_{0}$ we have
${}^{k}\big{(}\frac{F-F_{m}}{f_{m}}\big{)}\preccurlyeq 1$ in
$\mathcal{C}^{<\infty}$, with $\derdelta:=\phi^{-1}\der$. In view of
$\operatorname{ddeg}Q_{m}^{\phi}=0$, the last part of the proof of the lemma
with $\phi_{0}$, $Q$ replaced by $\phi$, $Q_{m}$, and $G$ replaced by
$G_{m}:=(F-F_{m+1})/f_{m+1}$, but ${\boldsymbol{j}}$ still ranging over
${\mathbb{N}}^{m_{0}}$, goes through, and yields the desired conclusion. ∎

Rather than Lemma 3.4 we shall use in what follows the implications (ii)∗
$\Rightarrow$ (i) and (ii)∗∗ $\Rightarrow$ (i) that are implicit in the proof
of that lemma, as we saw.

### Expressing the powers k in terms of

To facilitate the use of Lemma 3.4 and its variants we shall express k in
terms of . Let $R$ be any differential ring with derivation . Then $f\in R$
gives rise to a derivation $\derdelta:=f\der$ on the underlying ring of $R$.
For $k\geqslant 1$, $0\leqslant j\leqslant k$, we define
$G^{k}_{j}(Y)\in\mathbb{Q}\\{Y\\}\subseteq R\\{Y\\}$ by recursion:

  * •

$G^{k}_{0}=0$,

  * •

$G^{k}_{k}=Y^{k}$,

  * •

$G^{k+1}_{j}=Y\left(\der(G^{k}_{j})+G^{k}_{j-1}\right)$ for $1\leqslant
j\leqslant k$.

(See also [ADH, 5.7].) For the additive operators and on the underlying ring
$R$ this recursion gives:

${}^{k}\ =\ \sum_{j=1}^{k}G^{k}_{j}(f){}^{j}\qquad(k\geqslant 1).$

For $1\leqslant j\leqslant k$ the differential polynomial $G^{k}_{j}(Y)$ is
homogeneous of degree $k$ and of order $\leqslant k$, so we have a
differential polynomial $R^{k}_{j}(Z)\in{\mathbb{Q}}\\{Z\\}$ of order $<k$ and
depending only on $j$ and $k$ such that
$G^{k}_{j}(f)=f^{k}R^{k}_{j}(f^{\dagger})$ for all $f\in R^{\times}$. (See
also [7, Section 5.3].) For $g\in R$, $\phi\in R^{\times}$,
$\derdelta=\phi^{-1}\der$, this gives

(3.1) ${}^{k}(g)\ =\ \phi^{-k}\sum_{j=1}^{k}R^{k}_{j}(-\phi^{\dagger})g^{(j)}\
\text{ with }g^{(j)}:={}^{j}(g)\qquad(k\geqslant 1).$

Given $a\in{\mathbb{R}}$, the identity (3.1) also holds for
$g\in\mathcal{C}^{k}_{a}$ and $\phi\in(\mathcal{C}^{k}_{a})^{\times}$, where k
and the j for $j\leqslant k$ are construed in the obvious way as maps
$\mathcal{C}^{k}_{a}\to\mathcal{C}_{a}$.

For use in the next section we add the following observation:

######  Lemma 3.6.

Let $g\in H$ be active and $g\preccurlyeq h\in H$, and suppose
$f\in\mathcal{C}^{<\infty}$ satisfies $(g^{-1}\der)^{k}(f)\prec 1$ for
$k=0,\dots,m$. Then also $(h^{-1}\der)^{k}(f)\prec 1$ for $k=0,\dots,m$.

###### Proof.

Set $u:=g/h\in H^{\preccurlyeq 1}$, ${}_{g}:=g^{-1}\der$ and
${}_{h}:=h^{-1}\der$. Then ${}_{h}=u{}_{g}$, as derivations on $H$ and on
$\mathcal{C}^{<\infty}$. For $k\geqslant 1$ we have by an earlier identity

(3.2) ${}_{h}^{k}(f)\ =\ \sum_{j=1}^{k}G^{k}_{j}(u){}_{g}^{j}(f)$

where each $G^{k}_{j}(u)$ is evaluated according to the small derivation g on
the asymptotic field $H$, and thus $G^{k}_{j}(u)\preccurlyeq 1$. This gives
the desired result. ∎

######  Remark 3.7.

For later use we note that the identity (3.2) also holds for ${1\leqslant
k\leqslant m}$, $f,g\in\mathcal{C}_{a}^{m}$,
$h\in(\mathcal{C}_{a}^{m})^{\times}$, with $a\in{\mathbb{R}}$ and $u:=g/h$ (an
element of $\mathcal{C}_{a}^{m}$), and where ${}_{g}:=g^{-1}\der$,
${}_{h}:=h^{-1}\der$ are taken as derivations
$\mathcal{C}_{a}^{j}\to\mathcal{C}_{a}^{j-1}$, for $j=1,\dots,m$, and each
$G^{k}_{j}$ is evaluated according to g.

### Bump functions

In this subsection $t$ ranges over ${\mathbb{R}}$. From Lemma 2.1 we obtain an
increasing $\mathcal{C}^{\infty}$-function
$\alpha\colon{\mathbb{R}}\to{\mathbb{R}}$ with $\alpha(t)=0$ for $t\leqslant
0$ and $\alpha(t)=1$ for $t\geqslant 1$, and below we fix such an $\alpha$.
(Cf. Figure 5.)

$1$$1$$0$$\alpha$$t$ Figure 5. The bump function $\alpha$

For each $n$ we take a real constant $C_{n}$ such that $1\leqslant
C_{0}\leqslant C_{1}\leqslant C_{2}\leqslant\cdots$ and

(3.3) $|\alpha^{(n)}(t)|\ \leqslant\ C_{n}\ \text{ for all $n$ and $t$.}$

For reals $a<b$ we define the increasing $\mathcal{C}^{\infty}$-function
$\alpha_{a,b}\colon{\mathbb{R}}\to{\mathbb{R}}$ by

(3.4) $\alpha_{a,b}(t)\ :=\ \alpha\\!\left(\frac{t-a}{b-a}\right),$

so $\alpha_{a,b}(t)=0$ for $t\leqslant a$ and $\alpha_{a,b}(t)=1$ for
$t\geqslant b$. Also,

(3.5) $\big{|}\alpha_{a,b}^{(m)}(t)\big{|}\ \leqslant\
\frac{C_{m}}{(b-a)^{m}}\ \text{ for all $m$ and $t$}.$

### Constructing $F^{*}$

We go back to our Hardy field $H$ (not necessarily $\upomega$-free or
newtonian) and its elements $f_{n}$ and $F_{n}:=f_{0}+\cdots+f_{n}$, and in
the rest of this section $t$ ranges over ${\mathbb{R}}^{\geqslant 1}$. First,
we take for each $n$ a continuous function ${{\mathbb{R}}^{\geqslant
1}\to{\mathbb{R}}}$ that represents the germ $f_{n}$, to be denoted also by
$f_{n}$, such that $f_{n}(t)>0$ and $f_{n+1}(t)\leqslant f_{n}(t)/2$ for all
$t$: first choose the function $f_{0}$, then $f_{1}$, next $f_{2}$, and so on.

For each $n$ we fix an $a_{n}\in{\mathbb{R}}^{\geqslant 1}$ such that
$f_{0},\dots,f_{n}$ are of class $\mathcal{C}^{n}$ on $[a_{n},+\infty)$. Next,
let $c_{0}<c_{1}<c_{2}<\cdots$ be real numbers $\geqslant 1$ with
$c_{n}\to\infty$ as $n\to\infty$. We define
$\alpha_{n}\colon{\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}$ by
$\alpha_{n}(t):=\alpha_{c_{n},c_{n+1}}(t)$, so $\alpha_{n}$ is an increasing
$\mathcal{C}^{\infty}$-function with $\alpha_{n}(t)=0$ for $t\leqslant c_{n}$
and $\alpha_{n}(t)=1$ for $t\geqslant c_{n+1}$, and we set

$f_{n}^{*}\ :=\ \alpha_{n}f_{n}\ :\,{\mathbb{R}}^{\geqslant
1}\to{\mathbb{R}}^{\geqslant 0},$

so $f_{n}^{*}(t)=0$ for $t\leqslant c_{n}$ and $f_{n}^{*}(t)=f_{n}(t)$ for
$t\geqslant c_{n+1}$. Thus $f_{n}$ and $f_{n}^{*}$ have the same germ at
$+\infty$, and we still have $f^{*}_{n+1}(t)\leqslant f_{n}^{*}(t)/2$ for all
$n$ and $t$. As we saw in the subsection on Hausdorff fields in Section 1,
this yields a continuous function

$F^{*}\ :=\ \sum_{n=0}^{\infty}f_{n}^{*}\ :\,{\mathbb{R}}^{\geqslant
1}\to{\mathbb{R}}$

such that $F^{*}-F_{n}\prec f_{n}$ (in $\mathcal{C}$) for all $n$.

######  Lemma 3.8.

Assume $c_{n}>a_{0},\dots,a_{n}$ for all $n$. Then for all $n$, $f_{n}^{*}$ is
of class $\mathcal{C}^{n}$, and $F^{*}$ is of class $\mathcal{C}^{n}$ on
$[c_{n},+\infty)$. So the germ of $F^{*}$ at $+\infty$ belongs to
$\mathcal{C}^{<\infty}$.

###### Proof.

We have $f_{n}^{*}=0$ on $[1,c_{n}]$, and $f_{n}$ is of class
$\mathcal{C}^{n}$ on $[a_{n},+\infty)$, so $f_{n}^{*}$ is of class
$\mathcal{C}^{n}$. For $t\leqslant c_{n+1}$ we have
$F^{*}(t)=f_{0}^{*}(t)+\cdots+f_{n}^{*}(t)$, so $F^{*}$ is of class
$\mathcal{C}^{n}$ on $[a_{n},c_{n+1}]$. Likewise, $F^{*}$ is of class
$\mathcal{C}^{n+1}$ on $[a_{n+1},c_{n+2}]$. Continuing this way we obtain that
$F^{*}$ is of class $\mathcal{C}^{n}$ on $[c_{n},+\infty)$. ∎

We consider the $a_{n}$ as fixed, with the $c_{n}>a_{0},\dots,a_{n}$ to be
chosen as needed later. We set $\varepsilon_{m}:=f_{m+1}/f_{m}$, so
$0<\varepsilon_{m}(t)\leqslant 1/2$ for all $t$ and $\varepsilon_{m}\prec 1$
in $H$. For any $n>m$ we also set $\varepsilon_{n,m}:=f_{n}/f_{m}$, so
$\varepsilon_{n,m}$ is of class $\mathcal{C}^{n}$ on $[a_{n},+\infty)$ and
$0<\varepsilon_{n,m}(t)\leqslant 2\varepsilon_{m}(t)/2^{n-m}$ for all $t$.
Then for $n>m$ we have $\varepsilon_{n,m}\preccurlyeq\varepsilon_{m}\prec 1$
in $H$, so $\varepsilon_{n,m}^{(k)}\prec x^{-1}$ for all $k\geqslant 1$: use
[ADH, 9.1.9(iv), 9.1.10], first passing from $H$ to a Hardy field extension
containing $x$ if necessary.

### Proof in the fluent case

This case of Theorem 3.1 is as follows:

######  Proposition 3.9.

Suppose $\varepsilon\in H^{\prec 1}$ is such that
$f_{i+1}/f_{i}\prec\varepsilon$ for all $i$. Then $(F_{i})$ pseudoconverges in
some Hardy field extension of $H$.

###### Proof.

By passing to a suitable extension we first arrange that
$H\supseteq{\mathbb{R}}$ is closed. Then $\ell:=-\log|\varepsilon|\in
H^{>{\mathbb{R}}}$, $|\varepsilon|=\operatorname{e}^{-\ell}$, so
$\lvert\varepsilon\rvert\circ\ell^{\operatorname{inv}}=\operatorname{e}^{-x}$,
and thus
${(f_{i+1}/f_{i})\circ\ell^{\operatorname{inv}}}\prec\operatorname{e}^{-x}$
for all $i$. Replacing $H$ by $H\circ\ell^{\operatorname{inv}}$ and renaming
we can arrange in view of Lemma 3.3 that
$f_{i+1}/f_{i}\prec\operatorname{e}^{-x}$ for all $i$, and this is what we
assume below. Note that then $(f_{n}/f_{m})^{(k)}\prec\operatorname{e}^{-x}$
for all $n>m$ and all $k$. We also assume that $(F_{i})$ does not
pseudoconverge in $H$.

As in the subsection on constructing $F^{*}$ we choose for each germ $f_{n}$ a
continuous representative ${\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}$, also to
be denoted by $f_{n}$, and real numbers $a_{0},a_{1},a_{2},\dots$,
$c_{0},c_{1},c_{2},\dots$ with the properties listed there, and with
$c_{n}>a_{0},\dots,a_{n}$ for all $n$: the $a_{n}$ are fixed and the $c_{n}$
are adjustable. As in that subsection this yields an
$F^{*}=\sum_{n=0}^{\infty}f_{n}^{*}\in\mathcal{C}^{<\infty}$ with
$f_{n}^{*}=\alpha_{n}f_{n}$ for all $n$, and we introduce the functions
$\varepsilon_{m}=f_{m+1}/f_{m}$ and $\varepsilon_{n,m}=f_{n}/f_{m}$ for $n>m$.
For each $n$ we take $b_{n}\geqslant a_{0},\dots,a_{n}$ such that for all $k$,
$m$ with $0\leqslant k\leqslant m<n$,

$t\geqslant b_{n}\ \Longrightarrow\ \big{|}\varepsilon_{n,m}^{(k)}(t)\big{|}\
\leqslant\ \frac{\operatorname{e}^{-t}}{2^{n-m}}.$

Next, with the $C_{n}$ from (3.3), take $c_{n}>b_{0},\dots,b_{n}$ such that
$c_{n+1}-c_{n}\geqslant C_{n}$ (so $c_{n}\to\infty$). For $m\leqslant n$ we
have $C_{m}\leqslant C_{n}$, so $|\alpha_{n}^{(m)}(t)|\leqslant 1$ for all
$t$, by (3.5).

Let $\phi\in H^{>}$ be active and $\phi\prec 1$, so $\phi\succ x^{-2}$ and
$\phi^{\dagger}\preccurlyeq x^{-1}$. This gives a derivation
$\derdelta:=\phi^{-1}\der$ on $\mathcal{C}^{<\infty}$. Now we use (ii)∗
$\Rightarrow$ (i) from Corollary 3.5. It tells us that for $F^{*}$ to generate
a Hardy field over $H$ with $F_{i}\leadsto F^{*}$, it is enough to establish
that the present assumptions on $\phi$ imply:

Claim : for all $m>k$ we have
${}^{k}\\!\big{(}\frac{F^{*}-F_{m}}{f_{m}}\big{)}\prec 1$ in
$\mathcal{C}^{<\infty}$.

Let $m\geqslant 1$ be given and represent the germ $\phi$ by a
$\mathcal{C}^{m}$-function ${\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}^{>}$, to
be denoted also by $\phi$. For $1\leqslant j<k$, the coefficient of $Y^{k}$ in
the homogeneous differential polynomial $G^{k}_{j}$ of degree $k$ is $0$, so
$G^{k}_{j}(1)=R^{k}_{j}(0)=0$. Also $R^{k}_{k}=1$ for $k\geqslant 1$. Hence we
can take a real number $c_{m}^{*}\geqslant c_{m}$ such that for all
$t\geqslant c_{m}^{*}$,

$\phi(t)\geqslant\
t^{-2},\quad\big{|}R^{k}_{j}(-\phi^{\dagger})(t)\big{|}\leqslant 1\ \text{
whenever $1\leqslant j\leqslant k\leqslant m$.}$

Then (3.1) yields

$\left|{}^{k}\\!\left(\frac{f_{n}^{*}}{f_{m}}\right)(t)\right|\ \leqslant\
t^{2k}\sum_{j=1}^{k}\left|\left(\frac{f_{n}^{*}}{f_{m}}\right)^{(j)}(t)\right|\qquad(1\leqslant
k\leqslant m<n,\ t\geqslant c_{m}^{*}).$

Here it is relevant that the $f_{n}^{*}/f_{m}$ are of class $\mathcal{C}^{m}$
on $[c_{m},+\infty)$ for the derivatives to exist. Next, for $1\leqslant
j\leqslant m<n$ and $t\geqslant c_{m}^{*}$,

$\displaystyle\left|\left(\frac{f_{n}^{*}}{f_{m}}\right)^{(j)}(t)\right|\ $
$\displaystyle\leqslant\
\sum_{i=0}^{j}\binom{j}{i}\left|\alpha_{n}^{(j-i)}(t)\cdot\varepsilon_{n,m}^{(i)}(t)\right|$
$\displaystyle\leqslant\
\sum_{i=0}^{j}\binom{j}{i}\frac{\operatorname{e}^{-t}}{2^{n-m}}\ =\
2^{j}\frac{\operatorname{e}^{-t}}{2^{n-m}}.$

Combining this with the previous inequality we get

$\left|{}^{k}\\!\left(\frac{f_{n}^{*}}{f_{m}}\right)(t)\right|\ \leqslant\
2^{k+1}t^{2k}\frac{\operatorname{e}^{-t}}{2^{n-m}}\qquad(1\leqslant k\leqslant
m<n,\ t\geqslant c_{m}^{*}).$

Now $F_{m}^{*}:=f_{0}^{*}+\cdots+f_{m}^{*}$ is of class $\mathcal{C}^{m}$ on
$[c_{m}^{*},\infty)$, so by Lemma 3.8 the function

$\frac{F^{*}-F_{m}^{*}}{f_{m}}\ =\
\sum_{n=m+1}^{\infty}\frac{f_{n}^{*}}{f_{m}}$

is of class $\mathcal{C}^{m}$ on $[c_{m}^{*},\infty)$. Using also Corollary
2.12 we have for $t\geqslant c_{m}^{*}$,

$\left|{}^{k}\\!\left(\frac{F^{*}-F_{m}^{*}}{f_{m}}\right)(t)\right|\
\leqslant\ 2^{k+1}t^{2k}\operatorname{e}^{-t}\qquad(1\leqslant k\leqslant m).$

Hence ${}^{k}\\!\left(\frac{F^{*}-F_{m}^{*}}{f_{m}}\right)\prec 1\text{ in
}\mathcal{C}^{<\infty}$ for $1\leqslant k\leqslant m$. As $F_{m}^{*}$ and
$F_{m}$ are equal as germs in $\mathcal{C}^{<\infty}$, this proves the claim
when $k\geqslant 1$. For $k=0$, use that $F^{*}-F_{n}\prec f_{n}$ for all $n$.
∎

######  Corollary 3.10.

If $H^{>{\mathbb{R}}}$ has uncountable coinitiality, then $(F_{i})$
pseudoconverges in some Hardy field extension of $H$.

Thus to prove Theorem 3.1 it would be enough to show that in every maximal
Hardy field its set of positive infinite elements has uncountable
coinitiality. However, we were not able to prove the latter directly, and so
couldn’t exploit this remark. Instead we refine in the next section the
previous constructions in the remaining case where $H^{>{\mathbb{R}}}$ has
countable coinitiality.

### Remarks on $H^{>{\mathbb{R}}}$ having countable coninitiality

We show that the property of $H^{>{\mathbb{R}}}$ having countable coinitiality
is fairly robust; this is not used later but has independent interest. More
generally, in this subsection $K$ is a pre-$H$-field with
$\Gamma:=v(K^{\times})\neq\\{0\\}$. Note: $K^{>\mathcal{O}}=K^{>C}$ if $K$ is
an $H$-field. If $K$ is ungrounded, then
$\operatorname{ci}(K^{>\mathcal{O}})=\operatorname{cf}(\Gamma^{<})\geqslant\omega$,
and $\operatorname{cf}(\Gamma^{<})=\omega$ iff $K$ has a logarithmic sequence
(as defined in [ADH, 11.5]) of countable length. First we refine [7, Lemma
1.4.20]:

######  Lemma 3.11.

Suppose $K$ is not $\uplambda$-free, and $L$ is a Liouville closed
$\operatorname{d}$-algebraic $H$-field extension of $K$. Then $L$ is
$\upomega$-free with a logarithmic sequence of length $\omega$, and
$\Gamma^{<}$ is not cofinal in $\Gamma_{L}^{<}$.

###### Proof.

Suppose first that $K$ is grounded. Let $K_{\upomega}$ be the $\upomega$-free
pre-$H$-field extension of $K$ introduced before [ADH, 11.7.16] (with $K$ in
place of $F$ there), identified with a pre-$H$-subfield of $L$ containing $K$
as in the proof of [7, Lemma 1.4.18]. The sequence $(f_{n})$ constructed
before [ADH, 11.7.16] is a logarithmic sequence in $K_{\upomega}$ with
$\Gamma^{<}<v(f_{n})<0$ for all $n\geqslant 1$. By [7, Theorem 1.4.1], $L$ is
$\upomega$-free and $\Gamma_{K_{\upomega}}^{<}$ is cofinal in
$\Gamma_{L}^{<}$, so $(f_{n})$ remains a logarithmic sequence in $L$, and
$\Gamma^{<}$ is not cofinal in $\Gamma_{L}^{<}$. If $K$ is not grounded we
reduce to the grounded case by following the proofs of [7, Lemmas
1.4.18–1.4.20]. ∎

Next, let $\mathbf{K}=(K,I,\Lambda,\Omega)$ be a pre-$\Uplambda\Upomega$-field
with Newton-Liouville closure
$\mathbf{K}^{\operatorname{nl}}=(K^{\operatorname{nl}},\dots)$; see [ADH,
16.4]. Recall that $K^{\operatorname{nl}}$ is differentially algebraic over
$K$. The following proposition is analogous to the characterizations of
rational asymptotic integration and of $\uplambda$-freeness in [7,
Propositions 1.4.8, 1.4.12]:

######  Proposition 3.12.

The following are equivalent:

  1. (i)

$K$ is $\upomega$-free;

  2. (ii)

$\Gamma^{<}$ is cofinal in $\Gamma_{L}^{<}$ for every
$\operatorname{d}$-algebraic $H$-field extension $L$ of $K$;

  3. (iii)

$\Gamma^{<}$ is cofinal in $\Gamma_{K^{\operatorname{nl}}}^{<}$.

Moreover, if $K$ is not $\upomega$-free, then $K^{\operatorname{nl}}$ has a
logarithmic sequence of length $\omega$.

###### Proof.

The implication (i) $\Rightarrow$ (ii) holds by [7, Theorem 1.4.1], and (ii)
$\Rightarrow$ (iii) is clear. For the rest, note that if $K$ is not
$\uplambda$-free, then $K^{\operatorname{nl}}$ has a logarithmic sequence of
length $\omega$ and $\Gamma^{<}$ is not cofinal in
$\Gamma_{K^{\operatorname{nl}}}^{<}$, by Lemma 3.11. Suppose now that $K$ is
$\uplambda$-free but not $\upomega$-free. Then [ADH, 11.8.30] gives
$\upomega\in K$ with
$\omega\big{(}\Uplambda(K)\big{)}<\upomega<\sigma\big{(}\Upgamma(K)\big{)}$.
By the proof of [ADH, 16.4.6], either $\Omega=\omega(K)^{\downarrow}$ or
$\Omega=K\setminus\sigma\big{(}\Upgamma(K)\big{)}{}^{\uparrow}$. If
$\Omega=\omega(K)^{\downarrow}$, then the proof of [ADH, 16.4.6] yields a
$\upgamma\in K^{\operatorname{nl}}$ such that $\upgamma>0$,
$\sigma(\upgamma)=\upomega$, and the pre-$H$-subfield
$K_{\upgamma}:=K\langle\upgamma\rangle$ of $K^{\operatorname{nl}}$ has a gap.
Replacing $\mathbf{K}$ by the pre-$\Uplambda\Upomega$-subfield
$(K_{\upgamma},\dots)$ of $\mathbf{K}^{\operatorname{nl}}$ we reduce to the
case that $K$ is not $\uplambda$-free. If
$\Omega=K\setminus\sigma\big{(}\Upgamma(K)\big{)}{}^{\uparrow}$, then the
proof of [ADH, 16.4.6] yields $\uplambda\in K^{\operatorname{nl}}$ such that
$\omega(\uplambda)=\upomega$ and the pre-$H$-subfield
$K_{\uplambda}:=K\langle\uplambda\rangle$ of $K^{\operatorname{nl}}$ is not
$\uplambda$-free, so we can argue as before, with $K_{\uplambda}$ in place of
$K_{\upgamma}$. ∎

Now assume $H\supseteq{\mathbb{R}}$. Let $M$ be a maximal Hardy field
extension of $H$ and $H^{\operatorname{da}}$ the $\operatorname{d}$-closure of
$H$ in $M$, with canonical $\Uplambda\Upomega$-expansions $\mathbf{H}$,
$\mathbf{H}^{\operatorname{da}}$, respectively; see [7, Sections 7.1, 7.2].
Then $\mathbf{H}^{\operatorname{da}}$ is a Newton-Liouville closure of
$\mathbf{H}$, by [7, Corollary 7.2.17], and so the $H$-field
$H^{\operatorname{da}}$ is closed. If $M^{*}$ is also a maximal Hardy field
extending $H$, then there is an $H$-field embedding $H^{\operatorname{da}}\to
M^{*}$ over $H$ whose image is the $\operatorname{d}$-closure of $H$ in
$M^{*}$, by [ADH, 16.4.9]. By Proposition 3.12:

######  Corollary 3.13.

If $H$ is $\upomega$-free, then $H^{>{\mathbb{R}}}$ is coinitial in
$(H^{\operatorname{da}})^{>{\mathbb{R}}}$. If $H$ is not $\upomega$-free, then
$H^{\operatorname{da}}$ has a logarithmic sequence of length $\omega$. In
particular, if $H^{>{\mathbb{R}}}$ has countable coinitiality, then so does
$(H^{\operatorname{da}})^{>{\mathbb{R}}}$.

##  4\. The Remaining Case

We keep the assumptions on $H$ and $(f_{i})$ from the beginning of Section 3,
and let $t$ range over ${\mathbb{R}}^{\geqslant 1}$. For use in the “remaining
case” we first derive bounds like those of clause (ii) in Lemma 3.4 for
$\phi=1$.

### Useful bounds

We adopt the conventions and notations in the subsection on constructing
$F^{*}$ from the previous section; in particular, the $a_{n}$ are fixed, and
the $c_{n}$ will be adjusted so as to get the desired bounds on certain
derivatives of the functions $f_{n}^{*}/f_{m}$ with $n>m$. For each $n$ we
take $b_{n}\geqslant a_{0},\dots,a_{n}$ such that for all $k$, $m$ with
$1\leqslant k\leqslant m<n$,

$t\geqslant
b_{n}\quad\Longrightarrow\quad\left|\varepsilon_{n,m}^{(k)}(t)\right|\leqslant\
\frac{t^{-1}}{2^{n-m}}.$

Next, with the $C_{n}$ from (3.3), we take for each $n$ a
$c_{n}>b_{0},\dots,b_{n}$ with $c_{n+1}-c_{n}\geqslant C_{n}$ (so
$c_{n}\to\infty$ as $n\to\infty$). Then $|\alpha_{n}^{(m)}(t)|\leqslant 1$ for
all $t$ whenever $m\leqslant n$, using that $C_{m}\leqslant C_{n}$ for such
$m$, $n$. Note also that for $m\leqslant n$ the function $f_{n}^{*}/f_{m}$ is
of class $\mathcal{C}^{n}$ on its entire domain $[1,\infty)$ in view of
$f_{n}^{*}(t)=0$ for $t\leqslant c_{n}$.

######  Lemma 4.1.

For all $k$, $m$ with $k\leqslant m$ we have

${}^{k}\\!\left(\frac{F^{*}-F_{m}}{f_{m}}\right)\prec 1\quad\text{ in
$\mathcal{C}^{<\infty}$.}$

###### Proof.

Let $1\leqslant k\leqslant m<n$. From
$f_{n}^{*}/f_{m}=\alpha_{n}\varepsilon_{n,m}$ we get for $t\geqslant c_{n}$,

$\displaystyle\left|\left(\frac{f_{n}^{*}}{f_{m}}\right)^{(k)}(t)\right|\ $
$\displaystyle\leqslant\
\sum_{j=0}^{k}\binom{k}{j}\left|\alpha_{n}^{(k-j)}(t)\cdot\varepsilon_{n,m}^{(j)}(t)\right|$
$\displaystyle\leqslant\
|\alpha_{n}^{(k)}(t)|\frac{2\varepsilon_{m}(t)}{2^{n-m}}+\sum_{j=1}^{k}\binom{k}{j}\frac{t^{-1}}{2^{n-m}}\
\leqslant\ \frac{2\varepsilon_{m}(t)}{2^{n-m}}+\frac{2^{k}t^{-1}}{2^{n-m}}.$

This also holds for $t<c_{n}$, since $(f_{n}^{*}/f_{m})(t)=0$ for such $t$.
Now fix $m\geqslant 1$ and set $F_{m}^{*}:=f_{0}^{*}+\cdots+f_{m}^{*}$. By
Corollary 2.12 the function

$\frac{F^{*}-F_{m}^{*}}{f_{m}}\ =\
\sum_{n=m+1}^{\infty}\frac{f_{n}^{*}}{f_{m}}$

is of class $\mathcal{C}^{m+1}$ on its entire domain $[1,\infty)$, and for all
$t$,

$\left|{}^{k}\\!\left(\frac{F^{*}-F_{m}^{*}}{f_{m}}\right)(t)\right|\
\leqslant\ 2\varepsilon_{m}(t)+2^{k}t^{-1}\qquad(k=1,\dots,m).$

Hence ${}^{k}\big{(}\frac{F^{*}-F_{m}^{*}}{f_{m}}\big{)}\prec 1$ in
$\mathcal{C}^{<\infty}$ for $k=1,\dots,m$. As $F_{m}^{*}$ and $F_{m}$ are
equal as germs in $\mathcal{C}^{<\infty}$, this gives the desired result when
$k\geqslant 1$. For $k=0$, use that $F^{*}-F_{n}\prec f_{n}$ for all $n$. ∎

For later use we record the following consequence:

######  Corollary 4.2.

Let $\phi\in H^{>}$ be active, and $\derdelta:=\phi^{-1}\der$, as a derivation
on $\mathcal{C}^{<\infty}$. Then there exists
$F_{\phi}\in\mathcal{C}^{<\infty}$ such that for all $k$, $m$ with $k\leqslant
m$,

${}^{k}\left(\frac{F_{\phi}-F_{m}}{f_{m}}\right)\prec 1\quad\text{ in
$\mathcal{C}^{<\infty}$.}$

###### Proof.

Take an $\ell$ in a Hardy field extension of $H$ with $\ell^{\prime}=\phi$;
note that $\ell>{\mathbb{R}}$. The lemma above applied to the sequence
$(f_{i}\circ\ell^{\operatorname{inv}})$ in $H\circ\ell^{\operatorname{inv}}$
yields $F^{*}\in\mathcal{C}^{<\infty}$ with
${{}^{k}\big{(}\frac{F^{*}-F_{m}\circ\ell^{\operatorname{inv}}}{f_{m}\circ\ell^{\operatorname{inv}}}\big{)}\prec
1}$ in $\mathcal{C}^{<\infty}$ for all $m\geqslant k$. For
$F_{\phi}:=F^{*}\circ\ell\in\mathcal{C}^{<\infty}$ this gives the desired
result. ∎

In view of Lemma 3.4, the problem is that $F_{\phi}$ depends on $\phi$. The
idea, to be carried out in the next subsections, is to show that for suitable
$\phi_{n}$ and a kind of partition of unity $(\beta_{n})$ the infinite sum
$\sum_{n}\beta_{n}F_{\phi_{n}}$ has the desired properties. In the previous
section we proved Theorem 3.1 in the so-called fluent case, which includes the
case that $H^{>{\mathbb{R}}}$ has uncountable coinitiality. The remaining case
where $H^{>{\mathbb{R}}}$ has countable coinitiality will lead to the suitable
$\phi_{n}$ and the partition of unity $(\beta_{n})$ that we alluded to. The
$a_{n}$ and $b_{n}$ below are still real numbers but have little to do with
the earlier $a_{n}$ and $b_{n}$; reusing these symbols with another meaning
simply reflects the limitations of the alphabet.

### Towards constructing a good partition of unity

Until further notice the Hardy field $H\supseteq{\mathbb{R}}$ is Liouville
closed and $H^{>{\mathbb{R}}}$ has countable coinitiality. It follows that
there is a sequence $(\phi_{n})$ of active elements in $H^{>}$ such that
$\big{(}v(\phi_{n})\big{)}$ is strictly increasing and cofinal in $\Psi_{H}$.
Below we fix such a sequence $(\phi_{n})$, and set
${}_{n}:=\phi_{n}^{-1}\der$, a derivation on $\mathcal{C}^{<\infty}$. Then
Corollary 4.2 provides for each $n$ a $\Phi_{n}\in\mathcal{C}^{<\infty}$ such
that for all $k$, $m$ with $k\leqslant m$,

${}_{n}^{k}\\!\left(\frac{\Phi_{n}-F_{m}}{f_{m}}\right)\ \prec\ 1\quad\text{
in $\mathcal{C}^{<\infty}$,}$

and thus by Lemma 3.6, for all $k\leqslant m$ and all $i\leqslant n$,

${}_{i}^{k}\\!\left(\frac{\Phi_{n}-F_{m}}{f_{m}}\right)\ \prec\ 1\quad\text{
in $\mathcal{C}^{<\infty}$.}$

We represent the germs $\phi_{n}$, $f_{n}$, and $\Phi_{n}$ by
$\mathcal{C}^{n}$-functions ${\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}^{>}$,
denoted also by $\phi_{n}$, $f_{n}$, and $\Phi_{n}$. These functions
$\phi_{n}$, $f_{n}$ and $\Phi_{n}$ are fixed in the rest of this section, and
the notion of “admissible sequence” defined below is relative to these given
sequences $(\phi_{n})$, $(f_{n})$, $(\Phi_{n})$. Suppose the real numbers
$a_{n}\geqslant 1$ are such that:

  1. (I)

for each $n$, $f_{0},\dots,f_{n}$ and $\phi_{0},\dots,\phi_{n}$ are of class
$\mathcal{C}^{n}$ on $[a_{n},+\infty)$;

  2. (II)

for all $i$, $k$, $m$, $n$ with $k\leqslant m\leqslant n$, $i\leqslant n$, and
all $t\geqslant a_{n}$ we have

$\left|{}_{i}^{k}\\!\left(\frac{\Phi_{n}-F_{m}}{f_{m}}\right)(t)\right|\
\leqslant\ 1.$

Note that (II) makes sense in view of (I), and that (I) and (II) remain valid
upon increasing all $a_{n}$. We have $\phi_{n}/\phi_{i}\prec 1$ in $H$ for
$i<n$ and $\phi_{n}/\phi_{i}=1$ for $i=n$, and thus
${}_{n}^{k}(\phi_{n}/\phi_{i})\prec 1$ in $H$ for $i\leqslant n$ and
$k\geqslant 1$. Note also that ${}_{n}^{k}(\phi_{n}/\phi_{i})(t)$ is defined
for $i,k\leqslant n$ and $t\geqslant a_{n}$, since $\phi_{n}/\phi_{i}$ is of
class $\mathcal{C}^{n}$ on $[a_{n},+\infty)$ for $i\leqslant n$. Thus by
taking the $a_{n}$ large enough we can arrange in addition to (I) and (II):

  1. (III)

for all $n$ and $i,k\leqslant n$ and all $t\geqslant a_{n}$ we have

$|{}_{n}^{k}(\phi_{n}/\phi_{i})(t)|\ \leqslant\ 1.$

An admissible sequence is a sequence
$\big{(}(a_{n},b_{n},\beta_{n})\big{)}_{n\geqslant 0}$ of triples
$(a_{n},b_{n},\beta_{n})$ such that:

  1. (i)

$(a_{n})$ is a strictly increasing sequence of real numbers $\geqslant 1$ with
$a_{n}\to\infty$ as $n\to\infty$ for which (I), (II), (III) hold;

and such that for all $n$:

  1. (ii)

$b_{n}$ is a real number with $a_{n}<b_{n}<a_{n+1}$;

  2. (iii)

$\beta_{n}$ is a function ${\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}$ of class
$\mathcal{C}^{n}$;

  3. (iv)

$\beta_{n}(t)=0$ if $t\leqslant a_{n}$, $\beta_{n}$ is increasing on
$[a_{n},b_{n}]$, $\beta_{n}(t)=1$ if $b_{n}\leqslant t\leqslant a_{n+1}$,
$\beta_{n}$ is decreasing on $[a_{n+1},b_{n+1}]$, and $\beta_{n}(t)=0$ for
$t\geqslant b_{n+1}$;

  4. (v)

$\beta_{n}+\beta_{n+1}=1$ on $[a_{n+1},b_{n+1}]$.

(See Figure 6.)

$a_{n}$$b_{n}$$a_{n+1}$$b_{n+1}$$a_{n+2}$$b_{n+2}$$t$$\beta_{n}$$\beta_{n+1}$
Figure 6. The functions $\beta_{n}$, $\beta_{n+1}$

In the rest of this subsection $\big{(}(a_{n},b_{n},\beta_{n})\big{)}$ denotes
an admissible sequence. Note that
$\operatorname{supp}\beta_{n}\subseteq[a_{n},b_{n+1}]$ by (iv), and that (v)
expresses the “partition of unity” requirement. The series
$\sum_{n}\beta_{n}\Phi_{n}$ converges pointwise on ${\mathbb{R}}^{\geqslant
1}$ to a continuous function $\Phi$ such that on each segment
$[b_{n},a_{n+2}]$ we have

$\beta_{n}+\beta_{n+1}\ =\ 1\ \text{ and }\ \Phi\ =\
\beta_{n}\Phi_{n}+\beta_{n+1}\Phi_{n+1},$

so $\Phi$ is of class $\mathcal{C}^{n}$ on $[b_{n},a_{n+2})$. Likewise, $\Phi$
is of class $\mathcal{C}^{n+1}$ on the set $(b_{n+1},a_{n+3}]$, which overlaps
the previous set. Continuing this way we see that $\Phi$ is of class
$\mathcal{C}^{n}$ on $[b_{n},+\infty)$, and thus the germ of $\Phi$ lies in
$\mathcal{C}^{<\infty}$.

######  Lemma 4.3.

Suppose for all $m$ and $i,k\leqslant m$ there is a positive constant
$C=C(i,k,m)$ such that for all $n\geqslant m$,

$\left|{}_{i}^{k}\\!\left(\frac{\beta_{n}\Phi_{n}+\beta_{n+1}\Phi_{n+1}-F_{m}}{f_{m}}\right)\right|\
\leqslant\ C\ \text{ on }\ [a_{n+1},b_{n+1}].$

Then for all $m$ and $i,k\leqslant m$ we have

${}_{i}^{k}\\!\left(\frac{\Phi-F_{m}}{f_{m}}\right)\ \preccurlyeq\
1\quad\text{ in $\mathcal{C}^{<\infty}$.}$

###### Proof.

Let $i,k\leqslant m$, and take a $C\geqslant 1$ as in the hypothesis. Then for
all $n\geqslant m$ we have $|{}_{i}^{k}\big{(}\frac{\Phi-
F_{m}}{f_{m}}\big{)}|\leqslant C$ on $[a_{n+1},b_{n+1}]$, and also by (II)
above,

$\left|{}_{i}^{k}\\!\left(\frac{\Phi-F_{m}}{f_{m}}\right)\right|\leqslant
1\text{ on $[b_{n},a_{n+1}]$,}\qquad\left|{}_{i}^{k}\\!\left(\frac{\Phi-
F_{m}}{f_{m}}\right)\right|\leqslant 1\text{ on $[b_{n+1},a_{n+2}]$,}$

and thus $|{}_{i}^{k}\big{(}\frac{\Phi-F_{m}}{f_{m}}\big{)}|\leqslant C\text{
on }[b_{n},a_{n+2}]$. Taking the union over all $n\geqslant m$ we obtain

$\left|{}_{i}^{k}\\!\left(\frac{\Phi-F_{m}}{f_{m}}\right)\right|\leqslant
C\text{ on }[b_{m},+\infty),$

which gives the desired result. ∎

We didn’t use (III) yet, but we need it for a further reduction:

######  Lemma 4.4.

Suppose for all $m$ and $k\leqslant m$ there is a positive constant $C(k,m)$
such that for all $n\geqslant m$,

$\left|{}_{n}^{k}\left(\frac{\beta_{n}\Phi_{n}+\beta_{n+1}\Phi_{n+1}-F_{m}}{f_{m}}\right)\right|\
\leqslant\ C(k,m)\ \text{ on }\ [a_{n+1},b_{n+1}].$

Then for all $m$ and $i,k\leqslant m$ we have

${}_{i}^{k}\left(\frac{\Phi-F_{m}}{f_{m}}\right)\ \preccurlyeq\ 1\quad\text{
in $\mathcal{C}^{<\infty}$.}$

###### Proof.

Let $C(m):=\max_{k\leqslant m}C(k,m)$ where the $C(k,m)$ are as in the
hypothesis. Let $i,k\leqslant m\leqslant n$, and let $F,g,h$ be the
restrictions of $\frac{\beta_{n}\Phi_{n}+\beta_{n+1}\Phi_{n+1}-F_{m}}{f_{m}}$,
$\phi_{n}$, $\phi_{i}$ to $[a_{n+1},+\infty)$, respectively; these functions
are of class $\mathcal{C}^{n}$. For $j=1,\dots,m$ we denote the derivations

$f\mapsto g^{-1}f^{\prime}\
:\,\mathcal{C}^{j}_{a_{n+1}}\to\mathcal{C}^{j-1}_{a_{n+1}},\quad f\mapsto
h^{-1}f^{\prime}\ :\,\mathcal{C}^{j}_{a_{n+1}}\to\mathcal{C}^{j-1}_{a_{n+1}}$

by g and h, suppressing for convenience the dependence on $j$. Let
$u:=g/h\in\mathcal{C}^{m}_{a_{n+1}}$. Then Remark 3.7 gives for
$f\in\mathcal{C}^{m}_{a_{n+1}}$:

${}_{h}^{k}(f)\ =\ \sum_{j=0}^{k}G^{k}_{j}(u){}_{g}^{j}(f)\quad\text{ (with
$G^{0}_{0}:=1$ to handle the case $k=0$)},$

where $G^{k}_{j}$ is evaluated according to the derivation g. By our
hypothesis,

$|{}_{g}^{j}(F)|\leqslant C(m)\ \text{ on }[a_{n+1},b_{n+1}],\ j=0,\dots,k.$

Now (III) provides a positive constant $B(m)$ depending on $m$ but not on $n$,
such that $|G^{k}_{j}(u)|\leqslant B(m)$ on $[a_{n+1},b_{n+1}]$ for
$j=0,\dots,k$. Hence

$|{}_{h}^{k}(F)|\ \leqslant\ (k+1)B(m)C(m)\ \text{ on }[a_{n+1},b_{n+1}],$

and so the hypothesis of the previous lemma is satisfied. ∎

######  Corollary 4.5.

If $H$ is closed, $(F_{i})$ has no pseudolimit in $H$, and the hypothesis of
Lemma 3.4 is satisfied, then $\Phi$ generates a Hardy field over $H$ and
$F_{i}\leadsto\Phi$.

###### Proof.

Use the conclusion of Lemma 4.4, and the implication (ii)∗∗ $\Rightarrow$ (i)
(with $\Phi$ in the role of $F$) from Corollary 3.5. ∎

In order to make a further reduction, note that on $[a_{n+1},b_{n+1}]$ we have
$\beta_{n}=1-\beta_{n+1}$, and so, on $[a_{n+1},b_{n+1}]$,

($\ast$)
$\displaystyle\frac{\beta_{n}\Phi_{n}+\beta_{n+1}\Phi_{n+1}-F_{m}}{f_{m}}\ $
$\displaystyle=\
\beta_{n+1}\cdot\left(\frac{\Phi_{n+1}-\Phi_{n}}{f_{m}}\right)+\frac{\Phi_{n}-F_{m}}{f_{m}}$
$\displaystyle=\
\beta_{n+1}\cdot\left(\frac{\Phi_{n+1}-F_{m}}{f_{m}}-\frac{\Phi_{n}-F_{m}}{f_{m}}\right)+\frac{\Phi_{n}-F_{m}}{f_{m}}.$

This leads to a further simplification:

######  Lemma 4.6.

If for all $k$ there is a constant $B(k)>0$ such that for all $n\geqslant k$,
$|{}_{n}^{k}(\beta_{n+1})|\leqslant B(k)$ on $[a_{n+1},b_{n+1}]$, then the
hypothesis of Lemma 4.4 is satisfied.

###### Proof.

To simplify notation, set $G_{n,m}:=\frac{\Phi_{n}-F_{m}}{f_{m}}$, and let
$k\leqslant m\leqslant n$. Then by ($\ast$ ‣ 4),

$\displaystyle{}_{n}^{k}\left(\frac{\beta_{n}\Phi_{n}+\beta_{n+1}\Phi_{n+1}-F_{m}}{f_{m}}\right)\
$ $\displaystyle=\
\sum_{j=0}^{k}\binom{k}{j}{}_{n}^{j}(\beta_{n+1}){}_{n}^{k-j}(G_{n+1,m}-G_{n,m})$
$\displaystyle\qquad+{}_{n}^{k}(G_{n,m})$

on $[a_{n+1},b_{n+1}]$. Suppose $B\in{\mathbb{R}}^{>}$ and
$|{}_{n}^{j}(\beta_{n+1})|\leqslant B$ on $[a_{n+1},b_{n+1}]$ for
$j=0,\dots,k$. Then the above identity and (II) gives that on
$[a_{n+1},b_{n+1}]$,

$\left|{}_{n}^{k}\left(\frac{\beta_{n}\Phi_{n}+\beta_{n+1}\Phi_{n+1}-F_{m}}{f_{m}}\right)\right|\
\leqslant\ \left[\sum_{j=0}^{k}\binom{k}{j}\cdot B\cdot 2\right]+1\ =\
2^{k+1}B+1,$

which gives the desired result. ∎

### Using composition

In this subsection we explore how we might arrange that our admissible
sequence $\big{(}(a_{n},b_{n},\beta_{n})\big{)}$ satisfies the hypothesis of
Lemma 4.6, and thus of Lemma 4.4. In the next subsection we then construct
such a sequence.

By (iv) there is no problem for $k=0$, since $0\leqslant\beta_{n+1}\leqslant
1$. Assume ${1\leqslant k\leqslant n}$ and set $\phi:=\phi_{n}$, so
$\derdelta:=\phi^{-1}\der={}_{n}$, and set $a:=a_{n+1}$, $b:=b_{n+1}$, $\beta\
:=\ \beta_{n+1}$. We wish to bound $|{}^{k}(\beta)|$ on $[a,b]$ by a positive
constant that may depend on $k$ but not on ${n\geqslant k}$. To achieve this
goal we introduce the strictly increasing bijection
$g\colon{\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}^{\geqslant 0}$ given by
$g(r)=\int_{1}^{r}\phi(t)\,dt$, so $g\in\mathcal{C}_{1}^{n+1}$,
$g^{\prime}=\phi$, and $g$ has as compositional inverse the strictly
increasing bijection $g^{\operatorname{inv}}\colon{\mathbb{R}}^{\geqslant
0}\to{\mathbb{R}}^{\geqslant 1}$ of class $\mathcal{C}^{n+1}$. Induction on
$j\leqslant n+1$ gives ${}^{j}\beta=(\beta\circ
g^{\operatorname{inv}})^{(j)}\circ g$ on ${\mathbb{R}}^{\geqslant 1}$. For
$j=k$ this identity gives for any $B\in{\mathbb{R}}^{>}$ the equivalence

(4.1) $|{}^{k}(\beta)|\leqslant B\text{ on
$[a,b]$}\quad\Longleftrightarrow\quad|(\beta\circ
g^{\operatorname{inv}})^{(k)}|\ \leqslant\ B\text{ on
$\big{[}g(a),g(b)\big{]}$.}$

We shall arrange below that $b$ is given in terms of $a$ by $g(b)=g(a)+1$ and
that on $\big{[}g(a),g(b)\big{]}$ the function $\beta\circ
g^{\operatorname{inv}}$ equals $\alpha_{g(a),g(b)}$ with $\alpha$ the bump
function from Section 3. In the next subsection we show that then for
sufficiently fast growing $(a_{n})$ all our constraints are satisfied.

### The construction

In view of the dependence of the function $g$ on $n$ in the story above we
restore here indices, defining the strictly increasing bijection

$g_{n}\ :\,{\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}^{\geqslant 0},\qquad
g_{n}(r)\ :=\ \int_{1}^{r}\phi_{n}(t)\,dt,$

so $g_{n}\in\mathcal{C}^{n+1}_{1}$, $g_{n}^{\prime}=\phi_{n}$, and $g_{n}$ has
as compositional inverse the strictly increasing bijection
$g_{n}^{\operatorname{inv}}\colon{\mathbb{R}}^{\geqslant
0}\to{\mathbb{R}}^{\geqslant 1}$ of class $\mathcal{C}^{n+1}$.

Next we take a strictly increasing sequence $(a_{n})$ of real numbers
$\geqslant 1$ such that $a_{n}\to\infty$ and (I), (II), (III) hold, and such
that for every $n$ we have $b_{n+1}<a_{n+2}$ where the real number $b_{n+1}$
is defined by $g_{n}(b_{n+1})=g_{n}(a_{n+1})+1$. (It will be clear that there
are such sequences.) With $b_{0}$ any real number satisfying
$a_{0}<b_{0}<a_{1}$, we now have $a_{n}<b_{n}<a_{n+1}$ for all $n$. We define
for each $n$ the $\mathcal{C}^{\infty}$-function

$\alpha_{n}:=\alpha_{g_{n}(a_{n+1}),g_{n}(b_{n+1})}\colon{\mathbb{R}}\to{\mathbb{R}}.$

The bump function $\alpha$ came with constants $C_{k}>0$ such that
$|\alpha^{(k)}|\leqslant C_{k}$ on ${\mathbb{R}}$, thus
$|\alpha_{n}^{(k)}|\leqslant C_{k}$ on ${\mathbb{R}}$ for all $k$, $n$ in view
of $g_{n}(b_{n+1})-g_{n}(a_{n+1})=1$ and (3.5). Since
$\alpha_{n}\big{(}g_{n}(b_{n+1})\big{)}=1=1-\alpha_{n+1}\big{(}g_{n+1}(b_{n+1})\big{)}$
we can define $\beta_{n+1}\colon{\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}$ by

$\beta_{n+1}(t)\ =\ \begin{cases}\ \alpha_{n}\big{(}g_{n}(t)\big{)}&\text{ for
$t\leqslant b_{n+1}$,}\\\ \ 1-\alpha_{n+1}\big{(}g_{n+1}(t)\big{)}&\text{ for
$t\geqslant b_{n+1}$.}\end{cases}$

Then $\beta_{n+1}$ is continuous. We also take a continuous function
$\beta_{0}\colon{\mathbb{R}}^{\geqslant 1}\to{\mathbb{R}}$ such that (iii),
(iv), (v) hold for $n=0$. We now have constructed a sequence
$\big{(}(a_{n},b_{n},\beta_{n})\big{)}$ that satisfies conditions (i), (ii),
(iv), and (v) (and (iii) for $n=0$). In fact, it fulfills all our wishes:

######  Proposition 4.7.

The sequence $\big{(}(a_{n},b_{n},\beta_{n})\big{)}$ is admissible, and for
all $k$ and $n\geqslant k$ we have $|{}_{n}^{k}(\beta_{n+1})|\leqslant C_{k}$
on $[a_{n+1},b_{n+1}]$.

###### Proof.

Clearly $\beta_{n+1}$ is of class $\mathcal{C}^{n+1}$ on
${\mathbb{R}}^{\geqslant 1}\setminus\\{b_{n+1}\\}$. Now $\alpha_{n}\circ
g_{n}=1$ on $[b_{n+1},\infty)$, so ${(\alpha_{n}\circ
g_{n})^{(j)}(b_{n+1})}=0$ for $j=1,\dots,n+1$. Moreover, $\alpha_{n+1}\circ
g_{n+1}=0$ on $[1,a_{n+2}]$, so $\beta_{n+1}$ is $\mathcal{C}^{n+1}$ on all of
${\mathbb{R}}^{\geqslant 1}$. Therefore condition (iii) is satisfied, and so
$\big{(}(a_{n},b_{n},\beta_{n})\big{)}$ is admissible. The bound
$|{}_{n}^{k}(\beta_{n+1}|\leqslant C_{k}$ on $[a_{n+1},b_{n+1}]$ for
$n\geqslant k$ is clear from $\beta_{n+1}\circ
g_{n}^{\operatorname{inv}}=\alpha_{n}$ on
$\big{[}g_{n}(a_{n+1}),g_{n}(b_{n+1})\big{]}$ and the equivalence (4.1). ∎

### Finishing the proof of Theorem 3.1

As already mentioned we can use [7, Theorem 6.7.22] to pass to an extension
and arrange that $H\supseteq{\mathbb{R}}$ is closed. If $H^{>{\mathbb{R}}}$
has uncountable coinitiality, we are done by Corollary 3.10. Suppose
$H^{>{\mathbb{R}}}$ has countable coinitiality. Then we have an admissible
sequence as in Proposition 4.7, and so by Lemma 4.6 and Corollary 4.5, if
$(F_{i})$ has no pseudolimit in $H$, then $\Phi$ is a pseudolimit of $(F_{i})$
in a Hardy field extension of $H$. This concludes the proof.

######  Corollary 4.8.

Suppose $H$ is a maximal Hardy field. Then
$\operatorname{ci}(H^{>{\mathbb{R}}})>\omega$.

###### Proof.

If $\operatorname{ci}(H^{>{\mathbb{R}}})=\omega$, then $H$ being
$\uplambda$-free yields a divergent pc-sequence $(\uplambda_{\rho})$ in $H$
whose well-ordered index set has cofinality $\omega$, contradicting Corollary
3.2. ∎

##  5\. Constructing Overhardian Germs

Our goal in this section is the following:

######  Theorem 5.1.

If $H\supseteq{\mathbb{R}}$ is a Liouville closed Hardy field and
$\phi\in\mathcal{C}$, $\phi>_{\operatorname{e}}H$, then some
$y\in\mathcal{C}^{\infty}$ with $y>_{\operatorname{e}}\phi$ generates a Hardy
field $H\langle y\rangle$ over $H$.

This is Sjödin’s main result in [28], except that he considers only
$\mathcal{C}^{\infty}$-Hardy fields. Our construction of $y$ follows that of
Sjödin, with the material organized so that much of it will also be useful in
the next section where we fill more general gaps.

Boshernitzan [12, Theorems 1.1, 1.2] (see also [7, proof of Corollary 5.4.23])
showed that $y$ in Theorem 5.1 can be taken in $\mathcal{C}^{\omega}$, using a
result of Kneser [22] on solutions $E\in\mathcal{C}^{\omega}$ to the
functional equation ${\exp}\circ E=E\circ(x+1)$. (Our follow-up paper will
have a different argument that yields a $y\in\mathcal{C}^{\omega}$ in Theorem
5.1.)

We state here an easy consequence of Theorem 5.1:

######  Corollary 5.2.

If $H$ is a maximal Hardy field, then $\operatorname{cf}(H)>\omega$, and thus

$\operatorname{ci}(H)\ =\ \operatorname{cf}(H^{<a})\ =\
\operatorname{ci}(H^{>a})\ >\ \omega\ \text{ for all $a\in H$.}$

###### Proof.

If $H$ is a maximal Hardy field with a strictly increasing cofinal sequence
$(h_{n})$ in $H$, then Lemma 2.13 yields a $\phi\in\mathcal{C}$ such that
$h_{n}<_{\operatorname{e}}\phi$ for all $n$, contradicting Theorem 5.1. For
any Hausdorff field $F$ and $a\in F$ we have
$\operatorname{cf}(F)=\operatorname{ci}(F)=\operatorname{cf}(F^{<a})=\operatorname{ci}(F^{>a})$
(use fractional linear transformations). ∎

This corollary yields Theorem A in the case where $A$ or $B$ is finite.

### Lemmas on logarithmic derivatives

Let $f\in\mathcal{C}_{a}^{1}$. Note that if $f(t)>0$ and $f^{\prime}(t)>0$ for
all $t\geqslant a$, then $f$ is strictly increasing, and thus $f(t)\geqslant
f(a)>0$ for all $t\geqslant a$. It is convenient to replace here $f^{\prime}$
by $f^{\dagger}$, noting that if $f(t)>0$ for all $t\geqslant a$, then
$f^{\dagger}(t)$ is defined for all $t\geqslant a$. Thus if $f(t)>0$ and
$f^{\dagger}(t)>0$ for all $t\geqslant a$, then $f$ is strictly increasing,
and thus $f(t)\geqslant f(a)>0$ for all $t\geqslant a$.

######  Lemma 5.3.

Let $f\in\mathcal{C}_{a}^{2}$. Assume that
$f(t),f^{\dagger}(t),f^{\dagger\dagger}(t)>0$ for all $t\geqslant a$. Then
$f(t)\to+\infty$ as $t\to+\infty$.

###### Proof.

Applying the remark preceding the lemma to $f^{\dagger}$ in the role of $f$
gives $f^{\dagger}(t)\geqslant f^{\dagger}(a)$ for $t\geqslant a$, so
$f^{\prime}(t)=f(t)f^{\dagger}(t)\geqslant f(a)f^{\dagger}(a)$ for $t\geqslant
a$, where we apply that same remark also to $f$. Hence for $t\geqslant a$,

$f(t)\ =\ f(a)+\int_{a}^{t}f^{\prime}(s)\,ds\ \geqslant\
f(a)+(t-a)f(a)f^{\dagger}(a),$

which gives the desired conclusion. ∎

######  Lemma 5.4.

Let $f\in\mathcal{C}_{a}^{3}$, and suppose that for all $t\geqslant a$,

$f(t)>0,\quad f^{\dagger}(t)>0,\quad f^{\dagger\dagger}(t)>0,\ \text{ and
}f^{\dagger}(t)\ >\ f^{\dagger\dagger}(t)\ >\ f^{\dagger\dagger\dagger}(t).$

Then $f(t)/f^{\dagger}(t)\to+\infty$ as $t\to+\infty$.

###### Proof.

We have $(f/f^{\dagger})^{\dagger}=f^{\dagger}-f^{\dagger\dagger}$, so
$(f/f^{\dagger})^{\dagger}(t)>0$ for $t\geqslant a$. Also

$\displaystyle{(f/f^{\dagger})^{\dagger}}^{\prime}\ $ $\displaystyle=\
(f^{\dagger}-f^{\dagger\dagger})^{\prime}\ =\
{f^{\dagger}}^{\prime}-{f^{\dagger\dagger}}^{\prime}$ $\displaystyle=\
f^{\dagger\dagger}f^{\dagger}-f^{\dagger\dagger\dagger}f^{\dagger\dagger}\ =\
f^{\dagger\dagger}(f^{\dagger}-f^{\dagger\dagger\dagger})$

and thus $(f/f^{\dagger})^{\dagger\dagger}(t)>0$ for all $t\geqslant a$.
Applying Lemma 5.3 to $f/f^{\dagger}$ in the role of $f$ now gives the desired
result. ∎

######  Lemma 5.5.

Let $f\in\mathcal{C}_{a}^{4}$, and suppose that for all $t\geqslant a$,

$\displaystyle f(t)>0,\quad f^{\dagger}(t)>0,\quad
f^{\dagger\dagger}(t)>0,\quad f^{\dagger\dagger\dagger}(t)>0,\text{ and}$
$\displaystyle f^{\dagger}(t)\ >\ f^{\dagger\dagger}(t)\ >\
f^{\dagger\dagger\dagger}(t)\ >\ f^{\dagger\dagger\dagger\dagger}(t).$

Then $f^{\dagger}(t)\to+\infty$ as $t\to+\infty$, and for every $n$,
$f(t)>f^{\dagger}(t)^{n}$, eventually.

###### Proof.

Applying Lemma 5.3 to $f^{\dagger}$ in the role of $f$ gives
$f^{\dagger}(t)\to+\infty$ as $t\to+\infty$. Applying Lemma 5.4 to
$f^{\dagger}$ in the role of $f$ gives
$f^{\dagger}(t)/f^{\dagger\dagger}(t)\to+\infty$ as ${t\to+\infty}$. Let
$n\geqslant 1$ and take $a_{n}\geqslant a$ such that
$f^{\dagger}(t)/n>f^{\dagger\dagger}(t)$ for all $t\geqslant a_{n}$. So the
assumptions of Lemma 5.4 are satisfied for $a_{n}$ and the restriction of
$f^{1/n}$ to $[a_{n},+\infty)$ in the role of $a$ and $f$, hence
$f(t)^{1/n}/\big{(}f^{\dagger}(t)/n\big{)}\to+\infty$ as $t\to+\infty$, and
thus $f(t)/f^{\dagger}(t)^{n}\to+\infty$ as $t\to+\infty$. ∎

### Hardian and overhardian germs

Let $y\in\mathcal{C}^{<\infty}$. Following the terminology of [28] we say that
$y$ is hardian if $y$ generates a Hardy field ${\mathbb{Q}}\langle y\rangle$.

######  Lemma 5.6.

If $y$ is hardian and $y>_{\operatorname{e}}0$,
$y^{\dagger}>_{\operatorname{e}}0$, then
$y>_{\operatorname{e}}(y^{\dagger})^{n}$ for all $n$.

###### Proof.

Suppose $y$ is hardian and $y>_{\operatorname{e}}0$,
$y^{\dagger}>_{\operatorname{e}}0$. The case $y\prec 1$ is impossible, since
it would give $y^{\dagger}<_{\operatorname{e}}0$. If $y\asymp 1$, then
$y^{\dagger}\prec 1$, and we are done. If $y\succ 1$ and $y^{\dagger}\prec 1$,
we are done. If $y\succ 1$ and $y^{\dagger}\succcurlyeq 1$, then
$v(y^{\dagger})=o(vy)$ by [ADH, 9.2.10], which gives the desired conclusion. ∎

We set $y^{\langle 0\rangle}:=y$, and inductively, if $y^{\langle
i\rangle}\in\mathcal{C}^{<\infty}$ is defined and $y^{\langle
i\rangle}\in(\mathcal{C}^{<\infty})^{\times}$ (so either $y^{\langle
i\rangle}<_{\operatorname{e}}0$ or $y^{\langle
i\rangle}>_{\operatorname{e}}0$), then $y^{\langle i+1\rangle}:=(y^{\langle
i\rangle})^{\dagger}$, and otherwise $y^{\langle i+1\rangle}$ is not defined.
As in [28] we call $y$ overhardian if for all $i$,

$y^{\langle i\rangle}\text{ is defined},\quad y^{\langle
i\rangle}>_{\operatorname{e}}0,\text{ and }y^{\langle
i\rangle}>_{\operatorname{e}}y^{\langle i+1\rangle}.$

If $y$ is overhardian, then so is $y^{\dagger}$. By Lemma 5.5:

######  Corollary 5.7.

If $y$ is overhardian, then for all $i$, $n$ we have

$y^{\langle i\rangle}>_{\operatorname{e}}{\mathbb{R}},\quad y^{\langle
i\rangle}>_{\operatorname{e}}(y^{\langle i+1\rangle})^{n}.$

Next we recall from [ADH, 4.3] that a differential polynomial $P(Y)\in
K\\{Y\\}$ over a differential field $K$ has a unique logarithmic decomposition

$P(Y)\ =\
\sum_{{\boldsymbol{i}}}P_{\langle{\boldsymbol{i}}\rangle}Y^{\langle{\boldsymbol{i}}\rangle}\qquad(P_{\langle{\boldsymbol{i}}\rangle}\in
K).$

If $K$ is a Hardy field and $y^{\langle i\rangle}$ is defined for all $i$,
then we can substitute $y$ for the indeterminate $Y$ to get
$P(y)=\sum_{{\boldsymbol{i}}}P_{\langle
i\rangle}y^{\langle{\boldsymbol{i}}\rangle}$ in $\mathcal{C}^{<\infty}$, where
of course

$y^{\langle{\boldsymbol{i}}\rangle}:=(y^{\langle
0\rangle})^{i_{0}}\cdots(y^{\langle r\rangle})^{i_{r}}\quad\text{for
${\boldsymbol{i}}=(i_{0},\dots,i_{r})\in{\mathbb{N}}^{1+r}$.}$

Such a substitution is in particular possible if $y$ is overhardian. Thus for
overhardian $y$ and $P\in{\mathbb{R}}\\{Y\\}^{\neq}$ we obtain
$P(y)\in(\mathcal{C}^{<\infty})^{\times}$ from Corollary 5.7. Therefore:

######  Corollary 5.8.

If $y$ is overhardian, then $y$ is hardian.

######  Lemma 5.9.

If $y$ is overhardian, then $\log y\prec y^{\dagger}$.

###### Proof.

More generally, let $y$ be hardian, $y>_{\operatorname{e}}{\mathbb{R}}$,
$y^{\dagger}>_{e}{\mathbb{R}}$, and
$y^{\dagger\dagger}>_{\operatorname{e}}{\mathbb{R}}$; we claim that then $\log
y\prec y^{\dagger}$. To prove this, take a Liouville closed Hardy field
$H\supseteq{\mathbb{R}}$ with $y\in H$. Applying [ADH, 9.2.18] to $\alpha=vy$
in the asymptotic couple $(\Gamma,\psi)$ of $H$ gives $\log y\asymp
y^{\dagger}/y^{\dagger\dagger}\prec y^{\dagger}$. ∎

Given a Hardy field $H$, we say that a germ $y\in\mathcal{C}$ is $H$-hardian
if $y$ is contained in a Hardy field extension of $H$; see also [7, Section
5.3].

######  Corollary 5.10.

Suppose $H\supseteq{\mathbb{R}}$ is a Liouville closed Hardy field and
$y>_{\operatorname{e}}H$. Then the following are equivalent:

  1. (i)

$y$ is overhardian;

  2. (ii)

$y$ is $H$-hardian;

  3. (iii)

$y$ is hardian.

###### Proof.

From $\exp(H)\subseteq H$ and $y>_{\operatorname{e}}H$ we obtain $\log
y>_{\operatorname{e}}H$. If $y$ is overhardian, this gives by induction on $n$
and Lemma 5.9 that $y^{\langle n\rangle}>_{\operatorname{e}}H$ for all $n$,
and so ${P(y)<_{\operatorname{e}}H}$ or $P(y)>_{\operatorname{e}}H$ for all
$P(Y)\in H\\{Y\\}\setminus H$, hence $y$ is $H$-hardian. This proves (i)
$\Rightarrow$ (ii), and (ii) $\Rightarrow$ (iii) is trivial. To show (iii)
$\Rightarrow$ (i), assume (iii). From $\log y>_{\operatorname{e}}H$ and
$\exp(x^{2})\in H$ we obtain $\log y\succ\exp(x^{2})$. Working in a Hardy
field containing $y$, $\log y$, $x$, and $\exp(x^{2})$, we have $(\log
y)^{\prime}\succ\exp(x^{2})^{\prime}$, so

$v(y^{\dagger})\ <\ v\big{(}x\exp(x^{2})\big{)}\ <\
v\big{(}\\!\exp(x^{2})\big{)}\ <\ 0,$

hence $v(y^{\dagger\dagger})\leqslant
v\big{(}\\!\exp(x^{2})^{\dagger}\big{)}=v(x)<0$, and thus
$y^{\dagger\dagger}>_{\operatorname{e}}{\mathbb{R}}$. Hence
$y^{\dagger}>_{\operatorname{e}}\log y>_{\operatorname{e}}H$ by the proof of
Lemma 5.9. Since $y^{\dagger}$ is hardian, we can iterate this argument, which
by induction shows that all $y^{\langle n\rangle}$ are defined and
$>_{\operatorname{e}}H$. This yields (i). ∎

Corollary 5.10 combines [28, Theorems 3 and 4]; the implication (iii)
$\Rightarrow$ (ii) also follows from [11, Theorem 12.23].

######  Corollary 5.11.

If $y$ is overhardian, then so is $\log y$. Moreover,

$\text{$y$ is overhardian}\ \Longleftrightarrow\ \text{$y$ is hardian and
$y>_{\operatorname{e}}\exp_{n}(x)$ for all $n$.}$

###### Proof.

Suppose $y$ is overhardian. Then $y$ is hardian and hence so is $\log y$.
Moreover, $\log y>_{e}{\mathbb{R}}$, and $\log y\asymp
y^{\dagger}/y^{\dagger\dagger}$ by the proof of Lemma 5.9. Hence

$(\log y)^{\dagger}\ \sim\
\big{(}y^{\dagger}/y^{\dagger\dagger}\big{)}^{\dagger}\ =\
y^{\dagger\dagger}-y^{\dagger\dagger\dagger}\ \sim\ y^{\dagger\dagger}\ =\
y^{\langle 2\rangle}.$

Now an easy induction shows that all $(\log y)^{\langle n\rangle}$ are
defined, and that for $n\geqslant 1$ we have $(\log y)^{\langle n\rangle}\sim
y^{\langle n+1\rangle}$. This proves the first claim of the corollary. Also
$x\prec y$, since $1\prec y\preccurlyeq x$ would give $y^{\dagger}\preccurlyeq
x^{\dagger}=1/x$, contradicting $y^{\dagger}>_{\operatorname{e}}{\mathbb{R}}$.
Applying this to $\log_{n}y$ (which we now know to be overhardian), gives
$\log_{n}y>_{\operatorname{e}}x$, and thus $y>_{\operatorname{e}}\exp_{n}(x)$,
proving the direction $\Rightarrow$ of the equivalence.

For the converse, assume $y$ is hardian and $y>_{\operatorname{e}}\exp_{n}(x)$
for all $n$. Then $H:=\operatorname{Li}\\!\big{(}{\mathbb{R}}(x)\big{)}$, the
Liouville closure of ${\mathbb{R}}(x)$ as a Hardy field, embeds as an
$H$-field over ${\mathbb{R}}(x)$ into the Liouville closed $H$-field extension
${\mathbb{T}}$ of ${\mathbb{R}}(x)$. Since the sequence
$\big{(}\\!\exp_{n}(x)\big{)}$ is cofinal in ${\mathbb{T}}$, this is also the
case in $H$, so $y>_{\operatorname{e}}H$, and hence $y$ is overhardian by
Corollary 5.10. ∎

### Constructing overhardian germs

Our goal is the following:

######  Theorem 5.12.

For any $\phi\in\mathcal{C}$ there is an overhardian
$y\in\mathcal{C}^{\infty}$ such that ${y^{\langle
m\rangle}>_{\operatorname{e}}\phi}$ for all $m$.

Note that Theorem 5.1 follows from Corollary 5.10 and Theorem 5.12. To get an
idea of how to construct a $y$ as in Theorem 5.12, consider an overhardian $y$
represented by a function in $\mathcal{C}_{a}^{\infty}$, to be denoted also by
$y$. Then we have a strictly increasing sequence $(a_{m})$ of real numbers
$\geqslant a$ tending to $+\infty$ such that $y^{\langle m\rangle}(t)$ is
defined for $t\geqslant a_{m}$, for every $m$, and thus

$y^{\langle m-1\rangle}(t)\ =\ y^{\langle
m-1\rangle}(a_{m})\cdot\exp\int_{a_{m}}^{t}y^{\langle
m\rangle}(s)\,ds\quad\text{for $m\geqslant 1$, $t\geqslant a_{m}$.}$

It follows that $y$ is determined as a function on $[a_{0},+\infty)$ by the
family of restrictions $\big{(}y^{\langle
m\rangle}|_{[a_{m},a_{m+1}]}\big{)}$: $y$ on $[a_{0},a_{1}]$ and $y^{\langle
1\rangle}$ on $[a_{1},a_{2}]$ determine $y$ on $[a_{0},a_{2}]$; likewise,
$y^{\langle 1\rangle}$ on $[a_{1},a_{2}]$ and $y^{\langle 2\rangle}$ on
$[a_{2},a_{3}]$ determine $y^{\langle 1\rangle}$ on $[a_{1},a_{3}]$, and thus
$y$ on $[a_{0},a_{3}]$, and so on. We use this as a clue to reverse engineer
overhardian elements.

We start with $a\in{\mathbb{R}}$ and a strictly increasing sequence $(a_{m})$
in ${\mathbb{R}}^{\geqslant a}$ tending to $+\infty$ and for each $m\geqslant
1$ a continuous function $y_{m-1,m}\colon[a_{m-1},a_{m}]\to{\mathbb{R}}$. Let
${m\geqslant 1}$. We define the continuous function
$y_{k,m}\colon[a_{k},a_{m}]\to{\mathbb{R}}$ for $0\leqslant k<m$ by downward
recursion on $k$: $y_{m-1,m}$ is already given to us, and for $1\leqslant
k<m$,

$y_{k-1,m}(t)\ :=\ \begin{cases}y_{k-1,k}(t)&\text{ for $a_{k-1}\leqslant
t\leqslant a_{k}$,}\\\
y_{k-1,k}(a_{k})\cdot\exp\displaystyle\int_{a_{k}}^{t}y_{k,m}(s)\,ds&\text{
for $a_{k}\leqslant t\leqslant a_{m}$.}\end{cases}$

(See Figure 7.)

$a_{k-1}$$a_{k}$$a_{m}$$y_{k-1,m}$$y_{k,m}$$t$ Figure 7. Passing from
$y_{k,m}$ to $y_{k-1,m}$

Downward induction on $k$ gives $y_{k,m}=y_{k,m+1}$ on $[a_{k},a_{m}]$ for
${k<m}$. This fact gives for each $k$ a continuous function
$y_{k}\colon[a_{k},+\infty)\to{\mathbb{R}}$ such that $y_{k}=y_{k,m}$ on
$[a_{k},a_{m}]$, for all $m>k$. Thus for $k\geqslant 1$ we have

$y_{k-1}(t)\ =\
y_{k-1}(a_{k})\cdot\exp\int_{a_{k}}^{t}y_{k}(s)\,ds\quad\text{for $t\geqslant
a_{k}$.}$

In the next lemma we use the notation $E(t)^{(r)}_{t=a}$ where the expression
$E(t)$ defines a function $t\mapsto E(t)$ in $\mathcal{C}^{r}(I)$, where
$I=[b,c]$ ($b<c$ in ${\mathbb{R}}$) and $a\in I$. With $f$ this function,
$E(t)^{(r)}_{t=a}:=f^{(r)}(a)$.

######  Lemma 5.13.

Assume the following holds for all $k\geqslant 1$:

  1. (i)

$y_{k-1,k}\in\mathcal{C}^{\infty}[a_{k-1},a_{k}]$ and $y_{k-1,k}(t)>0$ for
$a_{k-1}\leqslant t\leqslant a_{k}$;

  2. (ii)

$y_{k-1,k}^{(r)}(a_{k})=y_{k-1,k}(a_{k})\cdot\big{(}\\!\exp\int_{a_{k}}^{t}y_{k,k+1}(s)\,ds\big{)}^{(r)}_{t=a_{k}}$
for all $r\in{\mathbb{N}}^{\geqslant 1}$.

Then for all $k$ we have $y_{k}\in\mathcal{C}_{a_{k}}^{\infty}$, $y_{k}(t)>0$
for $t\geqslant a_{k}$, and $y_{k}^{\dagger}=y_{k+1}$ on $[a_{k+1},+\infty)$.
Thus $y_{0}$ is overhardian if $y_{k}>_{\operatorname{e}}y_{k+1}$ for all $k$.

###### Proof.

Downward induction on $k$ shows that $y_{k,m}$ for $k<m$ has the corresponding
properties. Note: $y_{k}^{\langle m\rangle}$ is defined in
$\mathcal{C}^{<\infty}$ and equals $y_{k+m}$ in $\mathcal{C}^{<\infty}$ for
all $k$, $m$. ∎

Towards proving Theorem 5.12 we may assume $\phi\in\mathcal{C}$ to be
represented by a continuous function
$\phi\colon[a,+\infty)\to{\mathbb{R}}^{>}$, so $\phi$ denotes the function and
its germ.

######  Lemma 5.14.

There exists an increasing $\mathcal{C}^{\infty}$-function
$f\colon[a,+\infty)\to{\mathbb{R}}$ such that $\phi(t)<f(t)$ and
$f(t)>f^{\dagger}(t)$ for all $t\geqslant a$.

###### Proof.

Lemma 2.2 yields a decreasing $\mathcal{C}^{\infty}$-function
$\zeta\colon[a,+\infty)\to{\mathbb{R}}^{>}$ with $1/\phi(t)>\zeta(t)$ and
$\zeta^{\prime}(t)>-1$ for all $t\geqslant a$. Then $f:=1/\zeta$ works. ∎

Replacing $\phi$ by $f$ and renaming, we arrange that
$\phi\colon[a,+\infty)\to{\mathbb{R}}^{>}$ is increasing of class
$\mathcal{C}^{\infty}$ and $\phi(t)>\phi^{\dagger}(t)$ for all $t\geqslant a$.
With these assumptions:

######  Lemma 5.15.

Suppose for all $k\geqslant 1$ we have $y_{k-1,k}(t)>\phi(t)$ for
$a_{k-1}<t\leqslant a_{k}$. Then for all $k$ we have $y_{k}(t)>\phi(t)$ for
$t>a_{k}$.

###### Proof.

Let $1\leqslant k<m$, and assume as an inductive assumption that
$y_{k,m}(t)>\phi(t)$ for $a_{k}<t\leqslant a_{m}$. Our job is to show that
then $y_{k-1,m}(t)>\phi(t)$ for $a_{k-1}<t\leqslant a_{m}$, and this amounts
to showing for $a_{k}\leqslant t\leqslant a_{m}$ that

$y_{k-1,k}(a_{k})\cdot\exp\int_{a_{k}}^{t}y_{k,m}(s)\,ds\ >\ \phi(t).$

This holds for $t=a_{k}$, and for $a_{k}<t\leqslant a_{m}$ we have

$\displaystyle y_{k-1,k}(a_{k})\cdot\exp\int_{a_{k}}^{t}y_{k,m}(s)\,ds\ $
$\displaystyle>\ \phi(a_{k})\cdot\exp\int_{a_{k}}^{t}\phi(s)\,ds$
$\displaystyle>\ \phi(a_{k})\cdot\exp\int_{a_{k}}^{t}\phi^{\dagger}(s)\,ds$
$\displaystyle=\
\phi(a_{k})\cdot\exp\\!\big{(}\log\phi(t)-\log\phi(a_{k})\big{)}\ =\ \phi(t),$

which gives the desired result. ∎

For $b\geqslant a$ we define the $\mathcal{C}^{\infty}$-function
$\phi_{b}\colon[a,+\infty)\to{\mathbb{R}}^{>}$ by

(5.1) $\phi_{b}(t)\ =\ \phi(b)\cdot\exp\int_{b}^{t}\phi(s)\,ds,$

so $\phi(t)<\phi_{b}(t)$ for $t>b$, using again that
$\phi(s)>\phi^{\dagger}(s)$ for $s>b$.

######  Lemma 5.16.

Suppose that for all $k\geqslant 1$ we have
$\phi<y_{k-1,k}\leqslant\phi_{a_{k-1}}$ on $(a_{k-1},a_{k}]$. Then for $k+1<m$
we have $y_{k,m}>y_{k+1,m}$ on $[a_{k+1},a_{m}]$.

###### Proof.

For $m=k+2$ and $a_{k+1}\leqslant t\leqslant a_{m}$ we have

$\displaystyle y_{k,m}(t)\ $ $\displaystyle=\
y_{k,k+1}(a_{k+1})\cdot\exp\int_{a_{k+1}}^{t}y_{k+1,m}(s)\,ds$
$\displaystyle>\ \phi(a_{k+1})\exp\int_{a_{k+1}}^{t}\phi(s)\,ds\ =\
\phi_{a_{k+1}}(t)\ \geqslant\ y_{k+1,m}(t).$

Let $1\leqslant k<k+1<m$ and assume inductively that $y_{k,m}(t)>y_{k+1,m}(t)$
whenever $a_{k+1}\leqslant t\leqslant a_{m}$. Then for $a_{k}\leqslant
t\leqslant a_{k+1}$ the special case above yields

$y_{k-1,m}(t)\ =\ y_{k-1,k+1}(t)\ >\ y_{k,k+1}(t)\ =\ y_{k,m}(t),$

and for $a_{k+1}\leqslant t\leqslant a_{m}$ the inductive assumption gives

$\displaystyle y_{k-1,m}(t)\ $ $\displaystyle=\
y_{k-1,k}(a_{k})\cdot\exp\int_{a_{k}}^{t}y_{k,m}(s)\,ds$ $\displaystyle=\
y_{k-1,k}(a_{k})\cdot\exp\int_{a_{k}}^{a_{k+1}}y_{k,m}(s)\,ds\cdot\exp\int_{a_{k+1}}^{t}y_{k,m}(s)\,ds$
$\displaystyle=\ y_{k-1,m}(a_{k+1})\cdot\exp\int_{a_{k+1}}^{t}y_{k,m}(s)\,ds$
$\displaystyle>\ y_{k,m}(a_{k+1})\exp\int_{a_{k+1}}^{t}y_{k+1,m}(s)\,ds\ =\
y_{k,m}(t),$

which concludes the induction. ∎

######  Corollary 5.17.

Suppose that for all $k\geqslant 1$ we have

  1. (i)

$y_{k-1,k}\in\mathcal{C}^{\infty}[a_{k-1},a_{k}]$;

  2. (ii)

$\phi<y_{k-1,k}\leqslant\phi_{a_{k-1}}$ on $(a_{k-1},a_{k}]$;

  3. (iii)

$y_{k-1,k}^{(r)}(a_{k-1})\ =\ \phi^{(r)}(a_{k-1})$ for all $r\in{\mathbb{N}}$;

  4. (iv)

$y_{k-1,k}^{(r)}(a_{k})=y_{k-1,k}(a_{k})\cdot\big{(}\\!\exp\int_{a_{k}}^{t}\phi(s)\,ds\big{)}^{(r)}_{t=a_{k}}$
for all $r\in{\mathbb{N}}^{\geqslant 1}$.

Then $y:=y_{0}\in\mathcal{C}_{a_{0}}^{\infty}$, $y$ is overhardian, and
$y^{\langle k\rangle}>_{\operatorname{e}}\phi$ for all $k$.

###### Proof.

This follows from Lemmas 5.13, 5.15, and 5.16. ∎

The phrase “$y$ is overhardian” in the corollary above is short for “the germ
of $y$ at $+\infty$ is overhardian”. Given the strictly increasing sequence
$(a_{k})$ of real numbers $\geqslant a$ tending to $+\infty$ and the
increasing $\mathcal{C}^{\infty}$-function
$\phi\colon[a,+\infty)\to{\mathbb{R}}^{>}$ such that
$\phi(t)>\phi^{\dagger}(t)$ for all $t\geqslant a$, it follows from Lemma 2.3
that there exist functions $y_{k-1,k}$ for $k\geqslant 1$ satisfying
conditions (i)–(iv) of Corollary 5.17, where each value $y_{k-1,k}(a_{k})$ can
be chosen arbitrarily in the interval
$\big{(}\phi(a_{k}),\phi_{a_{k-1}}(a_{k})\big{)}$. Now the conclusion of that
corollary yields Theorem 5.12, which in turn gives us Theorem 5.1.

##  6\. Filling Wide Gaps

We now adapt the material from the previous section to filling a wide gap. To
describe this situation, let $H\supseteq{\mathbb{R}}$ be a Liouville closed
Hardy field. By a wide gap in $H$ we mean a pair $A$, $B$ of nonempty subsets
of $H^{>{\mathbb{R}}}$ such that $A<B$, there is no $h\in H$ with $A<h<B$, and
$A$ and $\exp A$ are cofinal; note that then $A$ and $\log A$ are cofinal,
that $B$, $\exp B$, $\log B$ are coinitial, and that for any
$\phi\in\mathcal{C}$ with $A<_{\operatorname{e}}\phi<_{\operatorname{e}}B$ we
have $A<_{\operatorname{e}}\log\phi,\exp\phi<_{\operatorname{e}}B$. Moreover,
if $A,B$ is a wide gap in $H$, then it is an additive gap in $H$, and $A$,
$\operatorname{sq}(A)$ are cofinal, and $B$, $2B$, $\sqrt{B}$ are coinitial,
by Corollary 1.9 and Lemmas 1.5, 1.10, and 1.12(i). Let us also record the
following, although we shall not explicitly use it:

######  Lemma 6.1.

Let $H\supseteq{\mathbb{R}}$ be a Liouville closed Hardy field and $A$, $B$
nonempty subsets of $H^{>{\mathbb{R}}}$ such that $A<B$ and there is no $h\in
H$ with $A<h<B$. Suppose there exists $\phi\in\mathcal{C}$ such that
$A<_{\operatorname{e}}\phi$ and
$\operatorname{e}^{\phi}<_{\operatorname{e}}B$. Then $A$, $B$ is a wide gap in
$H$.

###### Proof.

For $\phi$ as above and $h\in A$ we have $h<_{\operatorname{e}}\phi$, so
$\operatorname{e}^{h}<_{\operatorname{e}}\operatorname{e}^{\phi}<_{\operatorname{e}}B$,
and thus $\operatorname{e}^{h}\leqslant f$ for some $f\in A$. ∎

Here is the main result of this section:

######  Theorem 6.2.

If $H\supseteq{\mathbb{R}}$ is a Liouville closed Hardy field and $A$, $B$ is
a wide gap in $H$ with $\operatorname{cf}(A)=\operatorname{ci}(B)=\omega$,
then some $y\in\mathcal{C}^{<\infty}$ with
$A<_{\operatorname{e}}y<_{\operatorname{e}}B$ is $H$-hardian.

Wide gaps as in Theorem 6.2 do actually occur, as we show in the next
subsection. Towards proving Theorem 6.2 and some variants we begin with a
result that is mainly an exercise in valuation theory:

######  Lemma 6.3.

Let $H\supseteq{\mathbb{R}}$ be a Liouville closed Hardy field, let $A$, $B$
be a wide gap in $H$, and let $y\in\mathcal{C}^{<\infty}$ be overhardian with
$A<_{\operatorname{e}}y<_{\operatorname{e}}B$. Then $y$ is $H$-hardian and
$\operatorname{d}$-transcendental over $H$.

###### Proof.

It will be convenient to work with the $y^{\langle n\rangle}$. Note that Lemma
5.9 and the cofinality of $A$ and $\exp(A)$ give $A<_{\operatorname{e}}\log
y<_{\operatorname{e}}y^{\dagger}<_{\operatorname{e}}y<_{\operatorname{e}}B$.
Using this inductively we obtain $A<_{\operatorname{e}}y^{\langle
i\rangle}<_{\operatorname{e}}B$ for all $i$. We prove by induction on $n$ the
claim that $y,y^{\prime},\dots,y^{(n)}$ generate a Hausdorff field extension
$H_{n}\ :=\ H(y,y^{\prime},\dots,y^{(n)})$ of $H$. For $n=0$ this claim
follows by applying Lemma 1.11 to

$P\ :=\ \big{\\{}vh:\,\text{$h\in H^{>}$, $h\preccurlyeq g$ for some $g\in
A$}\big{\\}}.$

Assume the claim holds for a certain $n$. It is easy to check that then
$y^{\langle 0\rangle},\dots,y^{\langle n\rangle}$ lie in $H_{n}$, that $H_{n}\
=\ H\big{(}y^{\langle 0\rangle},\dots,y^{\langle n\rangle}\big{)}$, and that
$H_{n}$ has value group

$v(H_{n}^{\times})\ =\ v(H^{\times})\oplus{\mathbb{Z}}vy^{\langle
0\rangle}\oplus\cdots\oplus{\mathbb{Z}}vy^{\langle n\rangle}$

with $vB<vy^{\langle i\rangle}<vA$ for all $i\leqslant n$ and $vy^{\langle
i+1\rangle}=o(vy^{\langle i\rangle})$ for all $i<n$. Note that $vA<0$. Let
$\Delta$ be the smallest convex subgroup of $v(H^{\times})$ that includes
$vA$. Then $vA$ is coinitial in $\Delta$, and $\Delta+{\mathbb{Z}}vy^{\langle
0\rangle}+\cdots+{\mathbb{Z}}vy^{\langle n\rangle}$ is a convex subgroup of
$v(H_{n}^{\times})$ with $vB<\Delta+{\mathbb{Z}}vy^{\langle
0\rangle}+\cdots+{\mathbb{Z}}vy^{\langle n\rangle}$. Hence the real closure
$H_{n}^{\operatorname{rc}}$ of $H_{n}$, taken as a Hausdorff field extension
of $H_{n}$, has value group

$v\big{(}H_{n}^{\operatorname{rc},\times}\big{)}\ =\
v(H^{\times})\oplus{\mathbb{Q}}vy^{\langle
0\rangle}\oplus\cdots\oplus{\mathbb{Q}}vy^{\langle n\rangle},$

and $\Delta$ as well as $\Delta+{\mathbb{Q}}vy^{\langle
0\rangle}+\cdots+{\mathbb{Q}}vy^{\langle n\rangle}$ are convex subgroups of
$v(H_{n}^{\operatorname{rc},\times})$ with $vB<\Delta+{\mathbb{Q}}vy^{\langle
0\rangle}+\cdots+{\mathbb{Q}}vy^{\langle n\rangle}$. (See Figure 8.) In view
of $A<_{\operatorname{e}}(y^{\langle
n+1\rangle})^{i}<_{\operatorname{e}}y^{\langle n\rangle}$ for all $i\geqslant
1$ it now follows from Lemma 1.11 (with $H_{n}^{\operatorname{rc}}$ in the
role of $H$) that $y^{(n+1)}$ generates a Hausdorff field over
$H_{n}^{\operatorname{rc}}$. ∎

$v(H_{n}^{\times})$$0$$y^{\langle 2\rangle}$$y^{\langle 1\rangle}$$y^{\langle
0\rangle}$$\cdots$$vB$$\Delta^{<}$$(\Delta+{\mathbb{Z}}vy^{\langle
0\rangle}+{\mathbb{Z}}vy^{\langle 1\rangle}+\cdots+{\mathbb{Z}}vy^{\langle
n\rangle})^{<}$ Figure 8. Value group of $H_{n}$

Let us now consider the slightly different situation where
$H\supseteq{\mathbb{R}}$ is a Liouville closed Hardy field and
$y\in\mathcal{C}^{<\infty}$ is overhardian with $y>_{\operatorname{e}}H$. Then
the proof of the lemma above goes through for $A:=H^{>{\mathbb{R}}}$ and
$B=\emptyset$, although this pair $A$, $B$ is not a wide gap. The proof not
only shows in this situation that $y$ generates a Hardy field $H\langle
y\rangle$, but also that $v(H^{\times})$ is a convex subgroup of
$v\big{(}H\langle y\rangle^{\times}\big{)}$, and that $v\big{(}H\langle
y\rangle^{\times}\big{)}=v(H^{\times})\oplus\bigoplus_{i}{\mathbb{Z}}vy^{\langle
i\rangle}$, and so $y$ is $\operatorname{d}$-transcendental over $H$. (See
also [ADH, 16.6.10].)

### Constructing “countable” wide gaps

The Liouville closed Hardy field
$\operatorname{Li}({\mathbb{R}})=\operatorname{Li}\\!\big{(}{\mathbb{R}}(x)\big{)}$
is $\operatorname{d}$-algebraic over ${\mathbb{R}}$. Hence by [3, Theorem 3.4]
the sequence $\big{(}\\!\exp_{n}(x)\big{)}$ is cofinal in
$\operatorname{Li}({\mathbb{R}})$, so
$\operatorname{cf}\\!\big{(}\\!\operatorname{Li}({\mathbb{R}})\big{)}=\omega$.
More generally, let $H\supseteq{\mathbb{R}}$ be any Liouville closed Hardy
field with $\operatorname{cf}(H)=\omega$. Then [7, remarks after Lemma 5.4.17]
yields a $\phi\in\mathcal{C}$ with $\phi>_{\operatorname{e}}H$ and so Theorem
5.1 gives an $H$-hardian $y\in\mathcal{C}^{<\infty}$ such that
$y>_{\operatorname{e}}H$. We now consider the Hardy-Liouville closure
$\operatorname{Li}\\!\big{(}H\langle y\rangle\big{)}$ of $H\langle y\rangle$.
We have a wide gap $A$, $B$ in $\operatorname{Li}\\!\big{(}H\langle
y\rangle\big{)}$ given by

$\displaystyle A\ $ $\displaystyle:=\
\big{\\{}f\in\operatorname{Li}\\!\big{(}H\langle
y\rangle\big{)}:\,\text{${\mathbb{R}}<f<h$ for some $h\in H$}\big{\\}},$
$\displaystyle B\ $ $\displaystyle:=\
\big{\\{}g\in\operatorname{Li}\\!\big{(}H\langle
y\rangle\big{)}:\,g>H\big{\\}}.$

Note that $\operatorname{cf}(H)=\omega$ gives $\operatorname{cf}(A)=\omega$.
Moreover:

######  Lemma 6.4.

$B=\big{\\{}g\in\operatorname{Li}\\!\big{(}H\langle
y\rangle\big{)}:\,\text{$g>\log_{n}y$ for some $n$}\big{\\}}$.

###### Proof.

By the remarks preceding this subsection $y$ is
$\operatorname{d}$-transcendental over $H$ and $\\{y^{n}:\,n=0,1,2,\dots\\}$
is cofinal in $H\langle y\rangle$. Now $\operatorname{Li}\\!\big{(}H\langle
y\rangle\big{)}$ is $\operatorname{d}$-algebraic over $H\langle y\rangle$, so
$\big{\\{}\\!\exp_{n}(y):\,n=0,1,2,\dots\big{\\}}$ is cofinal in
$\operatorname{Li}\\!\big{(}H\langle y\rangle\big{)}$ by [3, Theorem 3.4]
applied to $K=H\langle y\rangle$. In particular,
$\operatorname{Li}\\!\big{(}H\langle y\rangle\big{)}$ has
$\operatorname{d}$-transcendence degree $1$ over $H$. Towards a contradiction,
suppose $g\in B$ and $g<\log_{n}y$ for all $n$. With $g$ instead of $y$ we
conclude that $\big{\\{}\\!\exp_{n}(g):\,n=0,1,2,\dots\big{\\}}$ is cofinal in
$\operatorname{Li}\\!\big{(}H\langle g\rangle\big{)}$ and
$\operatorname{Li}\\!\big{(}H\langle g\rangle\big{)}$ has
$\operatorname{d}$-transcendence degree $1$ over $H$. Hence
$y>\operatorname{Li}\\!\big{(}H\langle g\rangle\big{)}$, and so with
$\operatorname{Li}\\!\big{(}H\langle g\rangle\big{)}$ in the role of $H$ we
conclude that $\operatorname{Li}\\!\big{(}H\langle
y\rangle\big{)}=\operatorname{Li}\\!\big{(}\operatorname{Li}\\!\big{(}H\langle
g\rangle\big{)}\langle y\rangle\big{)}$ has $\operatorname{d}$-transcendence
degree $1$ over $\operatorname{Li}\\!\big{(}H\langle g\rangle\big{)}$ and thus
$\operatorname{d}$-transcendence degree $2$ over $H$, a contradiction. ∎

Thus $A$, $B$ is a wide gap in $\operatorname{Li}\\!\big{(}H\langle
y\rangle\big{)}$ with $\operatorname{cf}(A)=\operatorname{ci}(B)=\omega$. See
also Figure 9.

$y$$\log y$$\log_{2}y$$\log_{3}y$$\cdots$${\mathbb{R}}$$A$$B$ Figure 9. The
countable wide gap $A$, $B$

### Upper bounds

Assume that $a\geqslant 1$, $\phi\colon[a,+\infty)\to{\mathbb{R}}^{>}$ is
$\mathcal{C}^{\infty}$ and increasing, and ${\phi>\phi^{\dagger}}$ on
$[a,+\infty)$. Let $(a_{m})$ be a strictly increasing sequence of real numbers
${\geqslant a}$ tending to $+\infty$, and let for each $m\geqslant 1$ a
continuous function

$y_{m-1,m}\colon\ [a_{m-1},a_{m}]\to{\mathbb{R}}$

be given. As in the previous section this gives rise to functions $y_{k,m}$
for $k<m$ and functions $y_{m}$ and $y:=y_{0}$. Finally, assume that
$y_{k-1,k}\leqslant\phi_{a_{k-1}}$ on $(a_{k-1},a_{k}]$, for all $k\geqslant
1$. (See (5.1) for the definition of $\phi_{b}$ for $b\geqslant a$.) Our goal
is to find an upper bound for $y$ on $[a_{0},a_{n}]$ for $n\geqslant 1$ that
depends only on $\phi$ and $n$, not on the sequence $(a_{m})$ or the functions
$y_{m-1,m}$.

For $n\geqslant 1$ and $a_{n-1}\leqslant t\leqslant a_{n}$, we have

$\displaystyle y_{n-1,n}(t)\ $ $\displaystyle\leqslant\ \phi_{a_{n-1}}(t)\ =\
\phi(a_{n-1})\exp\int_{a_{n-1}}^{t}\phi(s)\,ds$ $\displaystyle\leqslant\
\phi(a_{n-1})\exp\\!\big{(}(t-a_{n-1})\phi(t)\big{)}\ =\
\frac{\phi(a_{n-1})}{\exp\\!\big{(}a_{n-1}\phi(t)\big{)}}\exp\\!\big{(}t\phi(t)\big{)}$
$\displaystyle\leqslant\ \exp\\!\big{(}t\phi(t)\big{)}.$

Let $1\leqslant k<n$. Then
$y_{k-1,n}(t)\leqslant\exp\\!\big{(}t\phi(t))\big{)}$ for $a_{k-1}\leqslant
t\leqslant a_{k}$. We assume inductively that for $a_{k}\leqslant t\leqslant
a_{n}$ we have
$y_{k,n}(t)\leqslant\exp_{n-k}\\!\big{(}t\phi(t)+(n-k)t\big{)}$. Then for
$a_{k}\leqslant t\leqslant a_{n}$,

$\displaystyle y_{k-1,n}(t)\ $ $\displaystyle=\
y_{k-1,k}(a_{k})\exp\int_{a_{k}}^{t}y_{k,n}(s)\,ds$ $\displaystyle\leqslant\
y_{k-1,k}(a_{k})\exp\\!\big{[}(t-a_{k})\exp_{n-k}\big{(}t\phi(t)+(n-k)t\big{)}\big{]}$
$\displaystyle\leqslant\frac{y_{k-1,k}(a_{k})}{\exp\\!\big{[}a_{k}\exp_{n-k}\\!\big{(}t\phi(t)+(n-k)t\big{)}\big{]}}\exp\\!\big{[}t\exp_{n-k}\\!\big{(}t\phi(t)+(n-k)t\big{)}\big{]}$
$\displaystyle\leqslant\ \exp_{n-(k-1)}\\!\big{(}t\phi(t)+(n-k+1)t\big{)},$

where we use that for $t\geqslant a_{k}$ we have the inequalities

$\displaystyle y_{k-1,k}(a_{k})\ $ $\displaystyle\leqslant\
\exp\\!\big{(}a_{k}\phi(a_{k})\big{)}\ \leqslant\
\exp\\!\big{[}a_{k}\exp_{n-k}\\!\big{(}t\phi(t)+(n-k)t\big{)}\big{]},$
$\displaystyle t\exp_{n-k}\\!\big{(}t\phi(t)+(n-k)t\big{)}\ $
$\displaystyle\leqslant\ \exp_{n-k}\\!\big{(}t\phi(t)+(n-k+1)t\big{)},$

the latter being a special case of the easily verified fact that

$t\exp_{n}\\!\big{(}t\phi(t)+nt\big{)}\ \leqslant\
\exp_{n}\\!\big{(}t\phi(t)+(n+1)t\big{)}\qquad(n\geqslant 1,\ t\geqslant 1).$

We have now proved by downward induction on $k$ that for all $k<n$,

$y_{k,n}(t)\ \leqslant\ \exp_{n-k}\\!\big{(}t\phi(t)+(n-k)t\big{)}\qquad\text{
for $a_{k}\leqslant t\leqslant a_{n}$.}$

For $y:=y_{0}$ this yields

$y(t)\ \leqslant\ \exp_{n}\\!\big{(}t\phi(t)+nt\big{)}\ \text{ for $n\geqslant
1$ and $a_{0}\leqslant t\leqslant a_{n}$.}$

To simplify notation, let $\phi_{n}\colon[a,+\infty)\to{\mathbb{R}}$ be the
function given by

$\phi_{n}(t)\ :=\ \exp_{n}\\!\big{(}t\phi(t)+nt\big{)},$

so that that $\phi<\phi_{1}<\phi_{2}<\phi_{3}<\cdots$ on $[a,+\infty)$, and
the bound above takes the form that for all $n\geqslant 1$ we have
$y\leqslant\phi_{n}$ on $[a_{0},a_{n}]$.

### Back to wide gaps

In the rest of this section $H$ is a Liouville closed Hardy field with
${\mathbb{R}}\subseteq H$, and $A$, $B$ is a wide gap in $H$. We say that
$\phi\in\mathcal{C}$ lies between $A$ and $B$ if
$A<_{\operatorname{e}}\phi<_{\operatorname{e}}B$. By an intermediary for $A$,
$B$ we mean a $\phi\in\mathcal{C}^{\infty}$ lying between $A$ and $B$ such
that $0<_{\operatorname{e}}\phi^{\dagger}<_{\operatorname{e}}\phi$; note that
the condition $0<_{\operatorname{e}}\phi^{\dagger}$ implies that $\phi$ is
eventually strictly increasing.

######  Proposition 6.5.

Suppose $x\in A$, $\operatorname{ci}(B)=\omega$, and $\phi$ is an intermediary
for $A$, $B$. Then there exists an overhardian $y\in\mathcal{C}^{\infty}$ such
that $\phi<_{\operatorname{e}}y^{\langle n\rangle}<_{\operatorname{e}}B$ for
all $n$, in particular $A<_{\operatorname{e}}y<_{\operatorname{e}}B$, and so
$y$ is $H$-hardian, by Lemma 6.3.

###### Proof.

Take a strictly increasing $\mathcal{C}^{\infty}$-function
$\phi\colon[a,+\infty)\to{\mathbb{R}}^{>}$ representing the germ $\phi$ such
that $a\geqslant 1$ and $0<\phi^{\dagger}<\phi$ on $[a,+\infty)$. For
$n\geqslant 1$, let $\phi_{n}\colon[a,+\infty)\to{\mathbb{R}}$ be the function
from the previous subsection given by

$\phi_{n}(t)\ =\ \exp_{n}\\!\big{(}t\phi(t)+nt\big{)}.$

From $x\in A$ and $B$ and $\log B$ being cofinal we obtain
$\phi_{n}<_{\operatorname{e}}\phi_{n+1}<_{\operatorname{e}}B$.

Take a strictly decreasing sequence $g_{1}>g_{2}>g_{3}>\cdots$ in $B$,
coinitial in $B$. Let $g_{n}$ also denote a continuous function
$[a,+\infty)\to{\mathbb{R}}$ representing the germ $g_{n}$. Choose a strictly
increasing sequence $b_{1}<b_{2}<b_{3}<\cdots$ of real numbers $\geqslant a$
tending to ${+\infty}$ such that $\phi_{n}<g_{n}$ and $g_{n+1}<g_{n}$, on
$[b_{n},+\infty)$. Next, set $a_{n}:=b_{n+1}$, and choose functions
$y_{k-1,k}\in\mathcal{C}^{\infty}[a_{k-1},a_{k}]$ for $k\geqslant 1$ such that
conditions (i)–(iv) of Corollary 5.17 are satisfied. (The discussion following
that corollary indicates how to construct such functions, using Lemma 2.3.)
This yields an overhardian $y:=y_{0}\in\mathcal{C}_{a_{0}}^{\infty}$ as in
that corollary, with $y^{\langle k\rangle}>_{\operatorname{e}}\phi$ for all
$k$.

Let $n\geqslant 1$. The upper bound from the previous subsection gives
$y\leqslant\phi_{n}$ on $[a_{0},a_{n}]$, so $y<g_{n}$ on $[b_{n},b_{n+1}]$.
With $n+1$ instead of $n$ this gives $y<g_{n+1}$ on $[b_{n+1},b_{n+2}]$, and
as $g_{n+1}<g_{n}$ on $[b_{n},+\infty)$, we get $y<g_{n}$ on
$[b_{n},b_{n+2}]$. Continuing this way we get $y<g_{n}$ on $[b_{n},b_{n+3}]$,
and so on, and thus $y<g_{n}$ on $[b_{n},+\infty)$. Since this holds for all
$n\geqslant 1$, this yields $y<_{\operatorname{e}}B$. ∎

######  Lemma 6.6.

Suppose $x\in A$, and some element of $\mathcal{C}$ lies between $A$ and $B$.
Then there exists an intermediary for $A$, $B$.

###### Proof.

Let $f\colon[a,+\infty)\to{\mathbb{R}}^{>}$ be a continuous function whose
germ at $+\infty$ lies between $A$ and $B$. Lemma 2.5 gives a
$\mathcal{C}^{\infty}$-function $f^{*}\colon[a,+\infty)\to{\mathbb{R}}^{>}$
such that $f<f^{*}<f+1$ on $[a,+\infty)$. Then $f^{*}<_{\operatorname{e}}B$,
and so replacing $f$ by $f^{*}$ we have arranged that
$f\in\mathcal{C}_{a}^{\infty}$. Defining $F(t):=1+\int_{a}^{t}f(s)\,ds$ we
obtain a strictly increasing $F\in\mathcal{C}_{a}^{\infty}$ with
$F^{\prime}=f$. By Lemma 2.15 we have $\int
A<_{\operatorname{e}}F<_{\operatorname{e}}\int B$, and so
$A<_{\operatorname{e}}F<_{\operatorname{e}}B$ by Lemma 1.12(iii) and Lemma
1.13(ii). Thus we can replace $f$ by $F$ and arrange in this way that $f$ is
also strictly increasing and $f\geqslant 1$. Next, consider the strictly
decreasing $\mathcal{C}^{\infty}$-function $\theta\colon[a,+\infty)\to(0,1]$
given by

$\theta(t)\ :=\ \int_{t}^{t+1}f(s)^{-1}\,ds\ =\
\int_{0}^{1}f^{-1}(s+t)\,ds,\qquad f^{-1}(s):=f(s)^{-1}\text{ for $s\geqslant
a$.}$

Claim : $\theta^{\prime}>-1$ on $[a,+\infty)$, and
$B^{-1}<_{\operatorname{e}}\theta<_{\operatorname{e}}A^{-1}$.

That $\theta^{\prime}>-1$ on $[a,+\infty)$ is clear from

$\theta^{\prime}(t)\ =\ \int_{0}^{1}(f^{-1})^{\prime}(s+t)\,ds\ =\
f(t+1)^{-1}-f(t)^{-1}.$

Also $\theta(t)<f(t)^{-1}$ for $t\geqslant a$, so
$\theta<_{\operatorname{e}}f^{-1}<_{\operatorname{e}}A^{-1}$.

To establish the claim it remains to show that
$B^{-1}<_{\operatorname{e}}\theta$, and this is where we shall need Lemma
1.13(iii). Let $g\in\int B^{-1}$, so $g\in H^{\prec 1}$ and
$g^{\prime}=h^{-1}$ with ${h\in B}$. We have $h\succ\operatorname{e}^{x}$, so
after increasing $a$ if necessary we can assume that the germ $h$ is
represented by a continuous function $h\colon[a,+\infty)\to{\mathbb{R}}$ with
$h(t)>\operatorname{e}^{t}$ and thus ${0<h(t)^{-1}<\operatorname{e}^{-t}}$,
for all $t\geqslant a$. This yields a $\mathcal{C}^{1}$-function

$t\mapsto\int_{+\infty}^{t}h(s)^{-1}\,ds\ :=\
-\int_{t}^{+\infty}h(s)^{-1}\,ds\ :\,[a,+\infty)\to{\mathbb{R}}$

with derivative $h^{-1}$ and tending to $0$ as $t\to+\infty$, so this function
represents the germ $g$, and will be denoted below by $g$. Thus for
$t\geqslant a$,

$g(t+1)-g(t)\ =\ \int_{t}^{t+1}h(s)^{-1}\,ds.$

Moreover, $h(s)^{-1}<f(s)^{-1}$ for all sufficiently large $s\geqslant a$, and
thus

(6.1) $g(t+1)-g(t)\ <\ \theta(t)\quad\text{ for all sufficiently large
$t\geqslant a$.}$

From $h^{-1}\prec\operatorname{e}^{-x}$ we get
$0<-g\prec\operatorname{e}^{-x}$, and so Lemma 2.17(iii) applied to $-g$ and
combined with (6.1) gives $-g(t)/2<\theta(t)$ for all sufficiently large
$t\geqslant a$. In view of Lemma 1.13(iii) and coinitiality of $B$, $2B$ this
yields $B^{-1}<_{\operatorname{e}}\theta$, as claimed.

From the claim it follows that the germ of
$\phi:=\theta^{-1}\colon[a,+\infty)\to{\mathbb{R}}$ is an intermediary for
$A$, $B$. ∎

######  Corollary 6.7.

Suppose $x\in A$ and $\operatorname{cf}(A)=\operatorname{ci}(B)=\omega$. Then
there exists an overhardian $y\in\mathcal{C}^{\infty}$ with
$A<_{\operatorname{e}}y<_{\operatorname{e}}B$, thus generating a Hardy field
over $H$.

###### Proof.

Using $\operatorname{cf}(A)=\operatorname{ci}(B)=\omega$, Lemmas 2.13 and 2.14
give an element of $\mathcal{C}$ that lies between $A$ and $B$. Then Lemma 6.6