Let $M \subseteq \H^{>0}$ be a pure scale on standard power domains. In this post, I gave the base step of the construction of a qaa field $(\F,L,T)$ as claimed here. The goal of this post is to finish this construction.
Step 0.5: apply a $\log$-shift to the qaa field $(\F_0,L_0,T_0)$, that is, set $$\F’_1 := \F_0 \circ \log,$$ let $L’_1:= \langle x \rangle^\times$ be the multiplicative $\RR$-vector space generated by $x$, and define $T’_1:\F’_1 \into L’_1$ by $$T’_1(f \circ \log):= (T_0 f) \circ \log, \quad\text{for } f \in \F_0.$$
Since $\blog$ maps every standard power domain into every standard power domain (in the sense of germs at $\infty$ of domains in the right half-plane), it follows that $\left(\F’_1, L’_1, T’_1\right)$ is a strong qaa field.
Remark
All germs in $\F’_1$ are polynomially bounded. Thus, if $m \in L_0$ is small (which means that $m = \exp^{-r}$ for some $r > 0$) and $a \in \F’_1$, then the holomorphic extension $\aa\mm$ is bounded on some standard power domain.
To iterate the construction, I now need to replace the real coefficients in the definition of (strong) asymptotic expansion by coefficients in $\C$ that have slower than exponential growth. In the setting of my general $M$, this means working with the following set of germs: set $$\C_M := \set{f \in \C:\ \frac1m \prec f \prec m \text{ for all large } m \in M}.$$ In other words, the set $\C_M$ consists of all germs whose comparability class is slower than that of any $m \in M$.
Examples
- $\RR \subseteq \C_{L} \subseteq \cdots \subseteq \C_{L_1} \subseteq \C_{L_0}$.
- $L’_1 \subseteq \C_{L_0}$.
- $\F’_1 \subseteq \C_{L_0}$.
Exercise
Show that, for $a \in \F’_1$, $n \in L_0$ and any standard power domain $\Omega$, we have $\aa \prec_\Omega \nn$.
Definition
- A series $F \in \Gs{\C_M}{M}$ is an $M$-generalized power series if there exist $k \in \NN$, small $m_0, \dots, m_k \in M$, a generalized power series $G \in \As{\C_M}{X_0^*, \dots, X_k^*}$ with natural support and a (not necessarily small) $m \in M$ such that $$F = m G(m_0, \dots, m_k).$$
- Let $f \in \C$ and $F = \sum a_m m \in \Gs{\C_M}{M}$ have $M$-natural support. The germ $f$ has strong asymptotic expansion $F$ if there exists a standard power domain $\Omega$ such that
- $f$ has holomorphic extension $\ff$ on $\Omega$;
- each $a_m$ has holomorphic extension $\aa_m$ on $\Omega$ such that $\aa_m \prec_\Omega \nn$ for every $n \in M$. In particular, for every $n \in M$ the truncation $F_n$ has a holomorphic extension $\FF_n$ on $\Omega$; and
- the condition $\ff – \FF_n \prec_\Omega \nn$ holds for each $n \in M$.
As before, we get:
Lemma
- The set $\C(M)^*_{\st}$ of all $f \in \C$ that have a strong asymptotic expansion in $\Gs{\C_M}{M}$ is an $\RR$-algebra.
- Every $f \in \C(M)^*_{\st}$ has exactly one asymptotic expansion $\tau_M(f)$ in $\Gs{\C_M}{M}$, and the map $\tau_M:\C(M)^*_{\st} \into \Gs{\C_M}{M}$ is an injective $\RR$-algebra homomorphism.
I am now ready for
Step 1: Set $$\A_1 := \set{f \in \C(L_0)^*_{\st}:\ f \text{ is bounded and } \tau_mf \in \As{\F’_1}{L_0}}$$ and, for $f \in \A_1$ with $\tau f = \sum a_m m$, set $$T_1 f:= \sum \left(T’_1 a_m\right) m \quad \in \Gs{R}{L_1}.$$
Proposition
The triple $(\A_1, L_1, T_1)$ is a strong qaa algebra.
(Show proof)Let now $f \in \A_1$ be such that $$T_1 f = \sum_{m \in L} a_m m \quad\text{and}\quad \tau_1 f = \sum_{r \ge 0} f_r \exp^{-r},$$ and let $n \in L$; we show there exists $g \in \A_1$ such that $T_1 g = (T_1 f)_n$. Considering $n$ as a function $n:\{-1,0\} \into \RR$ such that $n = \exp^{-n(-1)} x^{-n(0)}$, set $r:= n(-1)$ and $n’:= x^{-n(0)} \in L_1’$, so that $n = n’\exp^{-r}$ and
\begin{equation*}
\left(T_1f\right)_n = \sum_{m(-1) > n(-1)} a_m m + \left(T’_1 f_r\right)_{n’} \exp^{-r},
\end{equation*}
and let $\Omega$ be a strong asymptotic expansion domain of $f$. Note that each $f_s \exp^{-s}$ has a bounded holomorphic extension on $\Omega$. Since $$\sigma^{-1}\left(\sum_{m(-1) > n(-1)} a_m m\right) = \sum_{s \lt r} f_s \exp^{-s}$$ has finite support in $\Gs{\F’_1}{L_0}$, it follows that $$g_1:= \sum_{s \lt r} f_s \exp^{-s}$$ belongs to $\A_1$ and satisfies $\tau_1 g_1 = g_1$ and $T_1 g_1 = \sum_{m(-1) > n(-1)} a_m m$. On the other hand, by the inductive hypothesis, there exists $h \in \F’_1$ such that $T’_1 h = \left(T’_1 f_r\right)_{n’}$. Hence $h \exp^{-r} \in \A_1$ and, by definition of $T_1$, we obtain $T_1(h \exp^{-r}) = \left(T’_1 f_r\right)_{n’} \exp^{-r}$. Therefore, we can take $g:= g_1+h \exp ^{-r}$.
Finally, after shrinking $\Omega$ if necessary, we may assume that $\Omega$ is also a strong asymptotic expansion domain of $g$; we now claim that $\ff-\gg = o(\nn)$ in $\Omega$, which then proves the proposition. By the inductive hypothesis, we have $\ff_r – \hh = o(\nn’)$ in $\Omega$; therefore,
\begin{equation} \label{asym_1}
\ff_r \bexp^{-r} – \hh \bexp^{-r} = o(\nn) \quad\text{ in } \Omega.
\end{equation}
On the other hand, let $r’:= \min\set{s \in \RR:\ s > r \text{ and } f_r \ne 0}$. Then, by hypothesis, we have
\begin{equation} \label{asym_2}
\ff – \gg_1 – \ff_r \bexp^{-r} = o\left(\bexp^{-\frac{r+r’}2}\right) \quad\text{ in } \Omega.
\end{equation}
Since $\bexp^{-\frac{r+r’}2} = o(\nn)$ in $\Omega$, the proposition follows. $\qed$
Finally, let $\F_1$ be the field of fractions of $\A_1$ and extend $T_1$ accordingly; then $(\F_1,L_1,T_1)$, and hence $(\F_1, L, T_1)$, is a strong qaa field.
Remarks
- Ilyashenko’s class $\A$ of almost regular germs is contained in $\F_1$.
- We have $\F_0 \subseteq \F_1$ and $T_0 = T_1\rest{\F_0}$.
Continuing this iteration, we obtain qaa fields $(\F_i,L,T_i)$ such that $\F_i \subseteq \F_{i+1}$ and $T_i = T_{i+1} \rest{\F_i}$, for $i \in \NN$. So we set $$\F:= \bigcup_{i \in \NN} \F_i$$ and let $T$ be the common extension of all $T_i$ to $\F$. Then $(\F,L,T)$ is a qaa field, and in addition we have:
Proposition
$\F$ is closed under differentiation; in particular, $\F$ is a Hardy field.