\chapter{Operators with Support}

\noindent
Throughout $C$ is an abelian group (additively written), $\fM$ is a monomial set, and $\fm$ and $\fn$ range over $\fM$. 
In the first section we define {\em strong\/} operators on the Hahn space 
$C[[\fM]]$ and prove Lemma \ref{sumExt} to the effect that certain maps from $C\fM$ to $C[[\fM]]$ can be extended to strong operators on $C[[\fM]]$. This is useful in constructing strong derivations, see for example the proof of Lemma \ref{extder}. 

In the second section $C$ is a commutative ring and $\fM$ a monomial group. This allows us to introduce the more restricted
notions of {\em supported\/} operators and {\em small\/} operators on the Hahn ring $C[[\fM]]$. Corollary~\ref{cnPn} says how to use small operators in order to construct supported operators; this will be needed in the proof of Lemma \ref{OperatorLemma}. 

\section{Strong Operators} 
\noindent
Consider the Hahn space $C[[\fM]]$. 
An {\bf operator on $C[[\fM]]$} is by definition a map  
\[P\ :\ C[[\fM]]\rightarrow C[[\fM]].\]
An operator $P$ on $C[[\fM]]$ is said to be 
\textbf{additive} if  $P(f+g) = P(f)+P(g)$ for all $f,g\in C[[\fM]]$. If $P$ and $Q$ are additive operators on $C[[\fM]]$, then so are $P+Q$ and $-P$ defined by $(P+Q)(f):=P(f)+Q(f)$ and 
$(-P)(f):= -P(f)$. The set of additive operators on $C[[\fM]]$ with the above addition and multiplication given by composition is a ring with the null operator $O$ as its zero element,
the identity operator $I$ on $C[[\fM]]$ as its identity element, and
$P+(-P)=O$ for $P$ in this ring.  
 We call an operator $P$ on $C[[\fM]]$ \textbf{strongly additive}, or \textbf{strong} for short, if for every summable family $(f_i)$ in $C[[\fM]]$  the family $(P(f_i))$ is summable and $\sum P(f_i) = P(\sum f_i)$. Strong operators on $C[[\fM]]$ are additive, and the null operator $O$ and the identity operator 
$I$ are 
strong. The following is a routine consequence of the definition of ``strong''. 

\begin{lemma}
If $P,Q$ are strong operators on $C[[\fM]]$, then so are $-P$, $P+Q$, $PQ$.
\end{lemma}

\noindent
Thus the set of strong operators on $C[[\fM]]$ is a subring of the ring 
of additive operators on $C[[\fM]]$.


Let $(P_i)_{i\in I}$ be a family of strong operators on $C[[\fM]]$. Then we say that {\bf $\sum_{i\in I} P_i$ exists } if 
 for all $f\in C[[\fM]]$ the family $(P_i(f))_{i\in I}$ 
 is summable.
 


If $\sum_{i\in I} P_i$ exists, then the operator $f\mapsto \sum_i P_i(f)$ 
on $C[[\fM]]$ is additive, and is denoted by $\sum_{i\in I} P_i$ (or by $\sum P_i$ if $I$ is clear from the context). We do not claim that if 
$\sum_{i\in I} P_i$ exists, then $\sum_{i\in I} P_i$ is strong.

\begin{lemma} Let $(P_i)_{i\in I}$ be a family of strong operators on $C[[\fM]]$. 
\begin{enumerate}

    \item[\rm{(i)}] If $I$ is finite, then $\sum_{i\in I} P_i$ exists, and
   $\sum P_i$
    is the usual finite sum of additive operators. 
    \item[\rm{(ii)}] If $I$ and $J$ are disjoint sets, and $(P_j)_{j\in J}$ is also a family of strong operators on $C[[\fM]]$, and both $\sum_{i\in I} P_i$, and $\sum_{j\in J} P_j$ exist, then $\sum_{k\in I\cup J} P_k$ exists and equals $\sum_i P_i + \sum_j P_j$. 
    \item[\rm{(iii)}] If $\sigma : I \rightarrow J$ is a bijection and $\sum_{i\in I} P_i$ exists, then $\sum_{j\in J} P_{\sigma\inv(j)}$ exists and equals  $\sum_{i\in I} P_i$.
\end{enumerate}
\end{lemma}

\begin{lemma}\label{strpr}
Let $(P_i)_{i\in I}$ be a family of strong operators on $C[[\fM]]$ such that $\sum_i P_i$ exists, and let $Q$ be a strong operator on $C[[\fM]]$. Then 
$\sum_i P_iQ$ and $\sum_i QP_i$ exist, and are equal to
 $(\sum_i P_i)Q$ and $Q(\sum_iP_i)$, respectively.
\end{lemma}

\noindent
The next lemma modifies a result by 
van der Hoeven \cite[Proposition 3.5]{H} and is sometimes useful in
showing that an operator is strong. 


\begin{lemma}\label{sumExt} Let $C$ be a commutative ring with $1\ne 0$. Let
 $\phi: C\fM\rightarrow C[[\fM]]$ be a map such that: \begin{enumerate} 

\item[\rm{(i)}] $\phi((a+b)\fm) = \phi(a\fm) + \phi(b\fm)$ for all $a,b\in C$ and $\fm\in \fM$;
\item[\rm{(ii)}] $\supp(\phi(c\fm))\subseteq \supp(\phi(\fm)) \cup \{\fm\}$ for all $c\in C$ and $\fm\in \fM$;
\item[\rm{(iii)}] for every well-based $\fG\subseteq \fM$, 
the family $(\phi(\fg))_{\fg\in \fG}$ is summable.
\end{enumerate}
Then $\phi$ extends uniquely to a strong operator on $C[[\fM]]$.
\end{lemma}

\begin{proof}
Let $f = \sum f_\fm \fm\in C[[\fM]]$. We first show that the family $(\phi(f_\fm \fm))_{\fm\in \supp(f)}$ is summable. Indeed, by (ii),
\[\bigcup_{\fm\in \supp(f)}\supp(\phi(f_m\fm))\ \subseteq\  \left(\bigcup_{\fm\in\supp(f)}\supp \phi(\fm)\right)\cup \supp(f).\]
Since the right-hand side is well-based by (iii), so is the left-hand side.


Let any $\fn$ be given. Then by (ii) again,
\begin{align*} \{\fm\ \in \supp(f):\ 
\fn \in \supp(\phi(f_\fm\fm))\} \ &\subseteq\ \{\fm\in \supp(f):\ 
\fn \in \supp(\phi(\fm)) \cup \{\fm\}\}\\
     &\subseteq\  \{\fm\in \supp(f):\ 
\fn \in \supp(\phi(\fm))\} \cup \{\fn\}.
\end{align*}
The right-hand side is finite by (iii), and so is the left-hand side. Thus 
$\phi$ extends to an operator $\psi$ on
$C[[\fM]]$ by $\psi(f) := \sum_{\fm\in \supp(f)} \phi(f_\fm\fm)= \sum_{\fm} \phi(f_\fm \fm)$.
Now let $(f_i)_{i\in I}$ be a summable family in $C[[\fM]]$, and set $\fG:=\bigcup_{i\in I}\supp(f_i)$. We claim that $\big(\phi(f_{i,\fg}\fg)\big)_{(i,\fg)\in I\times \fG}$ is summable. First, 
\[\bigcup_{(i,\fg)\in I\times \fG} \supp(\phi(f_{i,\fg}\fg))\ \subseteq\  \bigcup_{\fg\in \fG}\supp(\phi(\fg)) \cup \fG,\]
so the left-hand side is well-based.
Secondly, let $\fn$ be given. Then
\[\{\fg\in \fG:\fn\in \supp(\phi(f_{i,\fg}\fg))\}\subseteq  \{\fg\in\fG:\fn\in\supp(\phi(\fg))\}\cup \{\fn\},\]
so the left-hand side is finite.
For $i\in I$ and $\fg\in \fG$ we have
\[\fn\in \supp(\phi(f_{i,\fg}\fg))\ \Rightarrow\ f_{i,\fg}\ne 0\ \Rightarrow\ 
\fg\in \supp f_i\]
and that  
for a fixed $\fg\in \fG$ the set $\{i\in I: \fg\in\supp(f_i)\}$ is finite.
 Thus the set
\[\{(i,\fg)\in I\times\fG:\  \fn\in \supp(\phi(f_{i,\fg}\fg))\}\]
is finite, and
our claim that $\big(\phi(f_{i,\fg}\fg)\big)_{(i,\fg)\in I\times\fG}$ is summable
has been established. 
It follows that 
$(\psi(f_i))_{i\in I}$ is summable and
\[\sum_{i\in I}\psi(f_i)\ =\ \sum_{i\in I} \sum_{\fg\in\fG}\phi(f_{i,\fg}\fg)\ =\ \sum_{(i,\fg)\in I\times\fG}\phi(f_{i,\fg}\fg)\ =\ \sum_{\fg\in\fG}\sum_{i\in I}\phi(f_{i,\fg}\fg)\]
\[=\ \sum_{\fg\in\fG}\phi\left(\big(\sum_{i\in I}f_{i,\fg}\big)\fg\right)\ =\ \psi\left(\sum_{\fg\in\fG} \big(\sum_{i\in I}f_{i,\fg}\big)\fg\right)\ =\ \psi\left(\sum_{i\in I}f_i\right),\]
showing that $\psi$ is strong.
\end{proof}


\subsection*{Strongly additive maps}
For use in the last chapter we now consider a second monomial set $\fN$ and define a map
$P: C[[\fM]]\to C[[\fN]]$ to be {\bf strongly additive} if for every summable family $(f_i)$ in $C[[\fM]]$  the family $(P(f_i))$ is summable in $C[[\fN]]$ and $\sum P(f_i) = P(\sum f_i)$. Thus the trivial map sending every $f\in C[[\fM]]$ to $0\in C[[\fN]]$ is strongly additive. 

Suppose $P$ as above is strongly additive. Then $P$ is additive: $P(f+g)=P(f)+P(g)$ for $f,g\in C[[\fM]]$. Moreover, $-P$ is strongly additive, and if $Q: C[[\fM]]\to C[[\fN]]$ is also a strongly additive map, then $P+Q$ is strongly additive. Given also a monomial set $\fG$ and a strongly additive map $Q:C[[\fG]]\to C[[\fM]]$, the composed map
$P\circ Q: C[[\fG]] \to C[[\fN]]$ is strongly additive. 


\section{Supported Operators and Small Operators}
\noindent
In this section $C$ and $\fM$ are equipped with product operations that 
make $C$ a commutative ring and $\fM$ a 
monomial group.
To say that an operator $P$ on $C[[\fM]]$ {\bf has support $\fG$\/} will mean that $P$ is additive, $\fG$ is well-based, and $\supp P(g) \subseteq \fG\supp(g)$ for all $g\in C[[\fM]]$.
An operator on $C[[\fM]]$ with support $\fG$ for some $\fG$
is said to be a {\bf supported operator\/}.

Let $f\in C[[\fM]]$. Then the operator
$g\mapsto fg$ on $C[[\fM]]$ has support $\supp f$. In particular, it is additive, and so
we have for each additive operator $P$ on $C[[\fM]]$ the additive operator $fP$ on $C[[\fM]]$ given by $(fP)(g):= f\cdot P(g)$.
In particular, the null operator $O$ has support $\emptyset$, and the identity operator $I$ has support $\{1\}$. Here is a key fact about supported operators:

\begin{lemma}\label{suppstrong} Every supported operator on $C[[\fM]]$ is strong.
\end{lemma}
\begin{proof} Suppose the operator $P$ on $C[[\fM]]$ has support $\fG$, and let $(f_i)\in C[[\fM]]^I$ be a summable family. Then
\[\bigcup_i \supp P(f_i)\ \subseteq\ \fG\cdot \bigcup_i \supp f_i\]
and the latter is a well-based subset of $\fM$. Let
$\fm \in \fM$ be given. Then there are only finitely many pairs $(\fg,\fh)\in \fG\times \bigcup_i \supp(f_i)$ with $\fm=\fg\fh$. This yields a finite $I(\fm)\subseteq I$ such that $i\in I(\fm)$
whenever $i\in I$, $(\fg,\fh)\in \fG\times \supp(f_i)$, and
$\fm=\fg\fh$. In particular, for every $i\in I$
with $\fm\in \supp P(f_i)$ we have $i\in I(\fm)$. 
Thus $\big(P(f_i)\big)$ is summable. Also $f = g+h$ with $f:=\sum_i f_i$, $g:=\sum_{i\in I(\fm)}f_i$ and $h:= \sum_{i\notin I(\fm)}f_i$, so
$P(f)=P(g)+P(h)$ with $\fm\notin \supp P(h)$. Hence
$$P(f)_{\fm}\ =\ P(g)_{\fm}\ =\ 
\sum_{i\in I(\fm)}P(f_i)_{\fm}\ =\ \sum_i P(f_i)_{\fm}\ =\ \big(\sum_i P(f_i)\big)_{\fm},$$
and thus $P(f)=\sum_i P(f_i)$. 
\end{proof}

\noindent
The following two lemmas are rather obvious:

\begin{lemma}\label{suppobv1} If $c\in C$ and the operators $P,Q$ on $C[[\fM]]$ have support
$\fG$ and $\fH$, respectively, then $cP$, $P+Q$, $PQ$ have support
$\fG$, $\fG\cup \fH$, $\fG\fH$, respectively.
\end{lemma}

\noindent
Thus the set of supported operators on $C[[\fM]]$ is a subring
of the ring of strong operators on $C[[\fM]]$. This subring also has a useful infinitary
property:

\begin{lemma}\label{suppobv2} Suppose $\fG\subseteq \fM$ is well-based, $(P_i)$ is a family of operators on $C[[\fM]]$, all having support 
$\fG$, and $\sum_i P_i$ exists. Then
$\sum_i P_i$ has support $\fG$. If in addition $I = \bigcup_{j\in J} I_j$ with pairwise disjoint $I_j$, then $\sum_{i\in I_j}P_i$ exists  
for every $j\in J$, and $\sum_{j\in J}\big(\sum_{i\in I_j}P_i\big)$ exists and equals 
$\sum_{i \in I }P_i$. 
\end{lemma}
 
\noindent
The next result requires a bit more effort:

\begin{lemma}\label{suppnonobv} Let $\fG, \fH\subseteq \fM$ be well-based, let $(P_i)_{i\in I}$ be a family of operators on $C[[\fM]]$, all having support $\fG$, and let $(Q_j)_{j\in J}$ be a family of operators on $C[[\fM]]$, all having support $\fH$. Assume also that $\sum_i P_i$ and $\sum_j Q_j$ exist.
Then $\sum_{i,j} P_iQ_j$ exists and equals $(\sum_iP_i)(\sum_j Q_j)$.
\end{lemma}
\begin{proof} Let $f\in C[[\fM]]$. Then $\supp P_iQ_j(f)\subseteq \fG\fH\supp f$, so $\bigcup_{i,j} \supp P_iQ_j(f)$ is well-based. Let $\fm\in \fM$ be given. By Lemma~\ref{fGfN} the set 
\[T\ :=\ \{(\fg, \fh, \fn)\in \fG\times \fH \times \supp f:\ \fg\fh\fn=\fm\}\]
is finite. As $\sum_jQ_j(f)$ exists, this yields a finite $J(\fm)\subseteq J$ such that $j\in J(\fm)$ whenever $j\in J$, $(\fg, \fh, \fn)\in T$, and 
$\fh\fn\in \supp Q_j(f)$. Now $P_iQ_j(f)=P_i(Q_j(f))$, so $j\in J(\fm)$ whenever $\fm\in \supp P_iQ_j(f)$. For each $j\in J(\fm)$ there are only
finitely many $i$ with $\fm\in \supp P_i(Q_j(f))$, so there are only
finitely many $(i,j)\in I\times J$ with $\fm\in \supp P_iQ_j(f)$. 
We have now shown that  $\sum_{i,j} P_iQ_j$ exists. That it equals $(\sum_iP_i)(\sum_j Q_j)$ is now an easy consequence of Lemmas~\ref{suppobv2} 
and  ~\ref{strpr}.
\end{proof}  


\noindent 
An operator $P$ on $C[[\fM]]$ is said to be \textbf{small}
\footnote{Small operators were defined in \cite[p. 66]{DMM} without requiring additivity, but this invalidates the assertion there about the inverse of $I-P$. Fortunately, this assertion is only used later in that paper for additive $P$, for which it is correct.}, if $P$ has support
$\fG$ for some $\fG\prec 1$. Note that if
$c\in C$, and $P$ and $Q$ are small operators on $C[[\fM]]$, then $cP$, $P+Q$, and $PQ$ are small operators on $C[[\fM]]$. Small operators are particularly useful in constructing inverses of operators:

\begin{corollary}\label{cnPn} Let $P$ be a small operator. Then for any sequence $(c_n)$ in $C$ the sum $\sum_{n} c_nP^n$ exists and is supported; such a sum 
is small if $c_0=0$. Moreover, the operator
$I-P$ is bijective with inverse $\sum_n P^n$.  
\end{corollary}
\begin{proof}
Let $P$ have support $\fG\prec 1$. Then $P^n$ has support $\fG^n$.
Let $(c_n)$ be a sequence in $C$. Then for $g\in C[[\fM]]$, 
\[\bigcup_{n} \supp c_nP^n(f)\ \subseteq\ \big(\bigcup_{n} \fG^n\big)\cdot \supp f\ =\ \fG^*\cdot \supp f,\]
the latter is well-based, and for every $\fm\in \fM$ there are only finitely many $n$ with $\fm\in \fG^n\supp f$ by \ref{fGfN}
and \ref{fGstar}, so $\sum_nc_n P^n$ exists. Note that all $c_nP^n$ have support $\fG^*$. The rest is clear from Lemmas~\ref{suppstrong}, \ref{suppobv1}, and ~\ref{suppobv2}. 
\end{proof}
{"config": "arxiv", "file": "1806.05309/supportedOperators.tex"}