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\begin{document}
%\baselineskip=23pt

\title{On functions whose graph is a Hamel basis}

\author{Krzysztof P\l otka}
\address{Department of Mathematics, University of Scranton,
Scranton, PA 18510, USA and Institute of Mathematics,
Gda{\'n}sk University, Wita Stwosza 57, 80-952 Gda{\'n}sk, Poland}
\email{plotkak2@scranton.edu}
\thanks{Most of this work was done when the author was working on his Ph.D. at West Virginia University. The author wishes to thank his advisor, Prof. K. Ciesielski, for many helpful conversations.}
\subjclass[2000]{Primary 15A03 54C40; Secondary 26A21, 54C30}
\date{09/11/2001}

%\dedicatory{This paper is dedicated to our authors.}

\keywords{Hamel basis, additive and Hamel functions.}

\begin{abstract}
     We  say  that  a  function  $h  \colon  \real  \to \real$ is a {\it Hamel
function\/} ($h \in {\rm HF}$) if $h$, considered as a subset of $\real^2$, is
a Hamel basis for $\real^2$. We prove that
{\emph  {every  function  from  $\real$  into  $\real$  can  be represented
as a pointwise sum of two Hamel functions}.}
The latter is equivalent to the statement: for all $f_1,f_2 \in \real^\real$
there is a $g\in\real^\real$ such that $g+f_1,g+f_2\in\ham$. We show that this
fails for infinitely many functions.
\end{abstract}

\maketitle

\section{Introduction} \label{S:Intro}

     The  terminology  is standard and follows \cite{cie}. The symbols $\real$
and  $\rational$  stand  for  the  sets  of all real and all rational numbers,
respectively.  A  basis  of  $\real^n$  as  a linear space over $\rational$ is
called {\it Hamel basis\/}. For $Y\subset\real^n$, the symbol $\lin(Y)$ stands
for  the  smallest linear subspace of $\real^n$ over $\rational$ that contains
$Y$. The zero element of $\real^n$ is denoted by $0$. The cardinality of a set
$X$  we denote by $|X|$. In particular, $\cont$ stands for $|\real|$. Given
a  cardinal  $\kappa$, we let $\cf(\kappa)$ denote the cofinality of $\kappa$.
We say that a cardinal $\kappa$ is regular if $\cf(\kappa)=\kappa$.
For any set $X$, the symbol $[X]^{<\kappa}$ denotes the set $\{Z\sq X \colon |Z|<\kappa\}$.
For $A,B \sq \real^n$, $A+B$ stands for $\{a+b\colon a\in A ,b\in B \}$.

We consider only real-valued functions. No distinction is made between
a function and its graph. For any two partial real functions $f,g$ we write $f+g$,
$f-g$ for the sum and difference functions defined on $\dom(f) \cap \dom(g)$.
The class of all functions from a set $X$ into a set $Y$ is denoted by $Y^X$.
We write $f|A$ for the restriction of $f \in Y^X$ to the set $A \sq X$.
For $B\sq\real^n$ its characteristic function is denoted by $\charf{B}$.
%If $f, g \in Y^X$, we denote the set $\{ x \in X : f(x)=g(x) \}$ by $[f=g]$.
For any function $g \in \real^X $ and any family of functions
$F \sq \real^X$ we define $g+F =\{g+f \colon f\in F\}$.
For any planar set $P$, we denote its $x$-projection by $\dom(P)$.

     The  cardinal  function  $\A({\rm  F})$,  for  ${\rm  F}  \varsubsetneq
\real^X$,  is  defined as the smallest cardinality of a family $G \sq \real^X$
for  which  there  is  no $g \in \real^X$ such that $g+G \sq {\rm F}$. It was
investigated   for   many  different  classes  of  real  functions,  see  e.g.
\cite{cie-nat,cie-rec,natkaniec}. Recall here that $\A({\rm  F})\ge 3$ is equivalent
to ${\rm  F}-{\rm  F}=\real^X$ (see \cite[Proposition 1]{kplotka}.)


One of the very important concepts in Real Analysis is {\it
additivity\/}. It dates back to the early 19th century when the
following functional equation was considered for the first time
$$
f(x+y)=f(x)+f(y) \mbox{ for all } x,y \in \real.
$$
     An  obvious  solution  to  this equation is a linear function, that is, a
function defined by $f(x)=ax$ for all $x\in\real$, where $a$ is some constant.
The fact that the linear functions are the only
continuous  solutions,  was first proved by A.  L. Cauchy \cite{cauchy}. Because of this, the above  equation  is known as Cauchy's Functional Equation. The problem of the
existence  of  a  discontinuous  solutions  of the Cauchy equation was  solved  by  G.  Hamel  in  1905 \cite{hamel} who constructed  a discontinuous function which satisfies the desired equation. To
construct  an  example  of  such  a  function  observe  first  that  $f \in
\real^\real$   satisfies   the  Cauchy  equation  if and only if  it  is  linear
over $\rational$,  i.e.,  for  all  $p,q \in \rational$ and $x,y \in \real$ we have
$f(px+qy)=pf(x)+qf(y)$.  So  to  define  an  additive function it is enough to
define  it  on  a Hamel basis. Thus, if $H \sq \real$ is a Hamel basis and $f$
identically  equals 1 on $H$ then clearly $f$ is not continuous. The family
of  all solutions of Cauchy's Functional Equation is called the family of {\it
additive functions\/} and we denote it by $\ad$. For the class of additive functions defined on $\real^n$ we use the symbol $\ad(\real^n)$.

It is obvious that not all functions are additive. But one could wonder how ``badly'' the
additive  condition can be violated. In particular, does there exist a function $f$ for which the
condition $f(x+y)= f(x)+f(y)$ fails for all $x$ and $y$? It turns out that the answer is positive.
We  give  two examples of families of such functions. In Section~\ref{chap2_1}
we  define  and  discuss  a  class  of  functions  whose  graph  is a linearly
independent set over $\rational$. Then, in Section~\ref{chap2_2}, we investigate
a proper subfamily of this class: functions whose graph is a Hamel basis.
In this section we state and prove the main result of this paper
(Theorem~\ref{p3-thm1}) which says that every real function is the pointwise
sum of two Hamel functions.

\section{Functions with linearly independent graphs}\label{chap2_1}

\defi{independent}{We say that a function $f\colon \real^n \to \real$
is {\it linearly independent over $\rational$}
(shortly: {\it linearly independent})
if $f$ is linearly independent subset of the space
$\la \real^{n+1};\rational;+;\cdot \ra$.}


The symbol $\lind(\real^n)$ stands for the family of all linearly independent
functions. In the case when $n=1$, we simply write $\lind$.
An easy example shows that the family $\lind(\real^n)$ is non-empty for all $n \ge
1$.

\ex{linex}{Every injection from $\real^n$ into a linearly independent set $H\sq \real$ is linearly independent over $\rational$.}
\proof Let $f\colon \real^n \to H$ be an injection. Assume that for some $p_1,\dots ,p_n\in\rational$ and pairwise different $x_1,\dots ,x_n\in\real^n$ we have
$
\sum_1^n p_i\la x_i,f(x_i) \ra=0.
$
Since $f(x_1),\dots ,f(x_n)\in H$ are all different and $H$ is linearly independent over $\rational$, we conclude that $p_1=\dots =p_n=0$. \qed

As mentioned in the introductory part of
this paper, the linearly independent functions lack the additive
property. Thus, $\ad(\real^n)\cap\lind(\real^n)=\e$.

Below we give some basic properties of the class $\lind(\real^n)$. Note that
Fact~\ref{p3-fact1}~(i) has its counterpart in the case of continuous and
Sierpi{\'n}ski-Zygmund functions (for the definition see \cite{kplotka}.)

\fact{p3-fact1}{

\noindent
\begin{description}

\item[(i)] $\lind(\real^n) + \ad(\real^n) = \lind(\real^n)$.

\item[(ii)] If $f\in\lind(\real^n)$ then $|f[\real^n]|=\cont$.

\item[(iii)] If $f\colon \real^n \to \real$ is continuous on a non-empty
open set then $f\notin\lind(\real^n)$.

\item[(iv)] There exists an $f\in\lind(\real^n)$ which is the union of countably many partial continuous functions.

\item[(v)] $\A(\lind(\real^n))=\cont$.

\end{description} }

\proof {\bf (i)} Let $f\in\lind(\real^n)$ and $g\in\ad(\real^n)$. Fix $x_1,\dots ,
x_k\in \real^n$ and $q_1,\dots q_k\in \rational$. Now suppose that
$
\sum_1^k q_i\la x_i,f(x_i)+g(x_i) \ra =0.
$
Thus, in particular, $\sum_1^k q_i x_i=0$. Since $g$ is additive we have
$\sum_1^k q_ig(x_i)=0$. Consequently, $\sum_1^k q_i\la x_i,f(x_i)\ra =0$.
The linear independence of $f$ implies that $q_1=\dots =q_k=0$. So
$f+g\in\lind(\real^n)$.

{\bf (ii)} Notice that it suffices to prove part (ii) for $n=1$. Assume, by
the way of contradiction, that $f\in\lind$ and $|f[\real]|=\kappa <\cont$. We
claim that there exist $x_1,x_2 \in\real $ with the following properties:
$$x_1\not=x_2,\; f(x_1)=f(x_2),\mbox{ and } f(-x_1)=f(-x_2).$$
To see the
claim choose $y_0\in\real$ such that
$|f^{-1}(y_0)\cap(0,\infty)|\ge \kappa^+$. Such
an element exists because $(0,\infty) \sq \bigcup_{y\in\real} f^{-1}(y)$ and $|f[\real]|=\kappa <\cont$.
Since $y_0$ satisfies the condition $|f[-f^{-1}(y_0)]|\le
\kappa < \kappa^+\le |-f^{-1}(y_0)|$, there exist different $x_1,x_2\in f^{-1}(y_0)\cap
(0,\infty)$ satisfying the equality $f(-x_1)=f(-x_2)$. Note that $x_1$ and $x_2$
are the required points. Next observe that

\centerline {$
\la x_1,f(x_1)\ra + \la -x_1,f(-x_1)\ra = \la x_2,f(x_2)\ra + \la
-x_2,f(-x_2)\ra.
$}
\vskip .1in
This leads to a contradiction with $f\in\lind$.

{\bf (iii)} Like in part (ii), it is enough to prove the case $n=1$.
Let $(a-h,a+h) \sq \real$ be a non-empty open interval such that
$f|(a-h,a+h)$ is continuous. Consider a function $g\colon [0,h)\to \real$
defined by $g(x)=f(a-x)+f(a+x)$. Obviously, $g$ is also continuous.
If $g(x)=g(0)=2f(a)$ for all $x\in[0,h)$ then $f$ is not linearly independent. Hence we may suppose that there
exist two different $x_1,x_2\in (0,h)$ such that $g(x_1)=2f(a)+p_1$ and $g(x_2)=2f(a)+p_2$ for some non-zero rationals $p_1,p_2$.
Then we have
\begin{equation}\label{(iii)}
p_2\la 2a,g(x_1)\ra - p_1\la 2a,g(x_2)\ra\in\lin(\la 2a,2f(a)\ra)=\lin(\la
a,f(a)\ra).
\end{equation}

Now, recall the definition of $g$ and note $\la a-x_i,f(a-x_i)\ra+\la
a+x_i,f(a+x_i)\ra=\la 2a,g(x_i)\ra$ for $i=1,2$. Based on (\ref{(iii)}), we
see that $f$ is not linearly independent.

{\bf (iv)} Let  us  first recall that $\real^n$ can be decomposed into $(n+1)$
0-dimensional  spaces $E_0,\dots ,E_n$. For every perfect set $Q\sq \real$ and
0-dimensional  space $E$ there exists an embedding $h_Q^E\colon E \to Q$. (See
e.g.,~\cite{hurewicz}.) It is also known that there exists a perfect set $P\sq
\real$    which    is    linearly    independent    over   $\rational$.   (See
e.g.,~\cite{kuczma}.)  Now,  if  $P=P_0\cup  P_1  \cup  \dots  \cup  P_n$ is a
partition  of  $P$  into  $(n+1)$  perfect  sets then, by Example~\ref{linex},
$h_{P_i}^{E_i}\colon  E_i  \to  P_i \; (i=0,\dots ,n)$ is  a  linearly  independent  subset  of $\real^{n+1}$.  It  is  easy  to  see that $h=\bigcup_0^n h_{P_i}^{E_i} \colon
\real^n  \to  P$  is  one-to-one.  So,  again  by  Example~\ref{linex}, $h$ is
linearly independent. Obviously, $h$ is the union of countably many partial continuous functions.

{\bf (v)} We start with showing that $\A(\lind(\real^n))\ge \cont$. Let
$\real^n=\{ x_\xi \colon \xi<\cont\}$. Fix an $F\sq \real^{\real^n}$ of
cardinality less than continuum. We will define, by induction, a function
$h\colon \real^n \to \real$ such that for every $f\in F$, $h+f$ is one-to-one
and $(h+f)[\real^n]$ is linearly independent. Then,  by Example~\ref{linex}, $h+F\sq \lind(\real^n)$.

Let $\alpha<\cont$. Assume that $h$ is defined on $\{x_\xi\colon \xi<
\alpha\}$, for all $f\in F$ the function $h+f$ is one-to-one, and
$(h+f)[\{x_\xi\colon
\xi< \alpha\}]$ is linearly independent. We will define $h(x_\alpha)$. Choose
$$
h(x_\alpha)\in \real\setminus\lin \left (\bigcup_{f\in
F}\left ( (h+f)[\{x_\xi\colon \xi<
\alpha\}] \cup \{f(x_\alpha)\} \right ) \right ).
$$
This choice is possible since $$\mbox{ $\left |\bigcup_{f\in
F} \left ( (h+f)[\{x_\xi\colon \xi< \alpha\}] \cup \{f(x_\alpha)\} \right ) \right
|\le (\alpha + 1)|F|<\cont$}.$$
It is
easy to see that all the required properties of $h$ are preserved. This ends
the proof of $\A(\lind(\real^n))\ge \cont$.

To see the opposite inequality consider $F$ consisting of all constant
functions. Then for any function $h \colon \real^n \to \real$ there is an
$f\in F$ such that $h(0)+f(0)=0$. Therefore
$h+f\notin\lind(\real^n)$.\qed

\section{Hamel functions}\label{chap2_2}

In this section we confine ourselves to a proper subclass of linearly
independent functions. More precisely, we consider the class of {\it Hamel
functions\/}. We say that a function $f\colon \real^n \to \real$ is a
Hamel function ($f\in \ham(\real^n)$ or $f\in\ham$ for $n=1$) if $f$, considered as a subset of
$\real^{n+1}$, is a Hamel basis for $\real^{n+1}$. Clearly, $\ham(\real^n)\sq
\lind(\real^n)$. A little more challenging argument, comparing with the case
of linearly independent functions, proves the existence of a Hamel function.
We do not present it here since this observation is a corollary of Theorem~\ref{p3-thm1}.

Fact~\ref{p3-fact1} states some basic properties of the class
$\lind(\real^n)$. It is interesting whether the same statements are true for
$\ham(\real^n)$. Since $\ham(\real^n)\sq\lind(\real^n)$, the properties (ii)
and (iii) hold trivially. A short additional argument shows that (i) is also
true. So we can state

\fact{p3-fact2}{
\noindent
\begin{description}

\item[(i)] $\ham(\real^n) + \ad(\real^n) = \ham(\real^n)$.

\item[(ii)] If $f\in\ham(\real^n)$ then $|f[\real^n]|=\cont$.

\item[(iii)] If $f\colon \real^n \to \real$ is continuous on a non-empty
open set then $f\notin\ham(\real^n)$.
\end{description}}

However, it remains an open problem whether Fact~\ref{p3-fact1} (iv) still holds when
$\lind(\real^n)$ is replaced by $\ham(\real^n)$.

\pr{prob3}{Does there exist an $h\in\ham(\real^n)$ which is the union of
countably many partial continuous functions?}

But it turns out that the statement of the last part of Fact~\ref{p3-fact1} is
false for the class $\ham(\real^n)$.

\fact{p3-fact3}{$\A(\ham(\real^n)) \le \omega$ for every $n \ge 1$.}
\proof For each $q \in \rational$ and each open ball $B$ with rational center
and radius (rational ball), let us define a function $f_q^B \colon \real^n \to
\real$ by $f_q^B=q\charf{B}$. We claim that for every function $f\colon
\real^n \to \real$ there exist a $q \in \rational$ and a rational ball $B$
such that $f+f_q^B \notin \ham(\real^n)$. To see this, first note that we may
assume that $f=f+f_0^B \in \ham(\real^n)$. Thus, $\la 0,1\ra \in
\lin(f)$. Consequently, there exist $x_1, \dots, x_k \in \real^n$ and $p_1,
\dots, p_k \in \rational$ satisfying
$
\sum_{i=1}^k p_i \la x_i,f(x_i)\ra=\la 0,1\ra.
$

Without loss of generality we may assume that $p_1\not= 0$. Now let
$q=\frac{-1}{p_1}$ and $B$ be a rational ball containing $x_1$ but not $x_2,
\dots, x_k$. It follows easily that $f+f_q^B$ is not linearly independent over
$\rational$. Indeed,
$$\sum_{i=1}^k p_i \la x_i,f(x_i)+f_q^B(x_i)\ra =
\sum_{i=1}^k p_i \la x_i,f(x_i)\ra +\sum_{i=1}^k p_i \la {0},f_q^B(x_i)\ra=
\la 0,1\ra +p_1 \la 0,q\ra=0.$$
\vskip -0.2 in
\qed
%\begin{eqnarray*}
%\sum_{i=1}^k p_i \la x_i,f(x_i)+f_q^B(x_i)\ra & = &
%\sum_{i=1}^k p_i \la x_i,f(x_i)\ra +\sum_{i=1}^k p_i \la {\bf0},f_q^B(x_i)\ra
% =  \\
%\la 0,1\ra +p_1 \la 0,q\ra= \la 0,1\ra-\la {\bf0},1\ra & = &
%0. \\
%\end{eqnarray*}

Notice here that $\A(\lind)=\cont$ (Fact~\ref{p3-fact1} (v)) implies, in particular, that every function
from $\real^\real$ can be written as the algebraic sum of two linearly
independent functions. In other words $\lind + \lind =\real^\real$. Since we
only found the upper bound for $\A(\ham)$, it would be very interesting to
determine whether $\ham + \ham =\real^\real$. The answer to the latter is given in the
main result of this paper - Theorem~\ref{p3-thm1}.

\thm{p3-thm1}{Every real function $f \colon \real^n \to \real$ can be represented as
a sum of two Hamel functions. In other words, $\real^{\real^n}=\ham(\real^n)+\ham(\real^n)$.}

Theorem~\ref{p3-thm1} and Fact~\ref{p3-fact3} give us the bounds for $\A(\ham(\real^n))$. Namely, $3\le \A(\ham(\real^n))\le \omega$. It is not known whether the techniques used in the proofs of these two results could also be used to determine $\A(\ham(
\real^n))$ exactly. We state the next open problem.

\pr{prob4}{$\A(\ham(\real^n))=\omega$?}

Before proving the theorem we introduce some definitions and show auxiliary
results. For $f \colon \real^n \to \real$, $x \in \real^n$, and $1\le k<\omega$ let
$$
\lc(f,k,x)=
\left \{\sum_1^k p_i f(x_i) \colon p_j \in \rational,\ x_j \in \real^n \ (j=1,
\dots, k),\
\sum_1^k  p_i x_i~=~x \right \}.
$$
When $x=0$ we write $\lc(f,k)$.
We also use $\lc(f)$ to denote $\bigcup_{1\le k<\omega} \lc(f,k)$. Observe
that $\lc(f)$ is a linear subspace of $\real$ over $\rational$, i.e.,
$\lc(f)=\lin(\lc(f))$. This is so because $\lc(f)$ is linearly isomorphic to $\lin(f)\cap (\{
0\}\times\real)$.

The sets $\lc(f)$ will play an important role in the proof of
Theorem~\ref{p3-thm1}. Hence, we will investigate properties of these sets.

\property{pr1}{$\lc(f,k)\sq \lc(f,3)+\lc(f,k-1)$  for every
$f\in\real^{\real^n}$ and
$3\leq k<\omega$.}

\proof
Let $y \in \lc(f,k)$. So $y = \sum_1^k p_i f(x_i)$ for some $x_1,
\dots , x_k \in \real^n$ and $ p_1, \dots , p_k \in \rational$
satisfying $\sum_1^k p_i x_i=0$. Define $x'=p_1 x_1 + p_2
x_2$, $q=1$, and $r=-1$. Observe that
$$p_1 x_1 + p_2 x_2 + r x'=q x' + p_3 x_3 +
\dots + p_n x_k =0.$$

Hence, $p_1 f(x_1) + p_2 f(x_2) + r f(x') \in \lc(f,3)$
and $q f(x') + p_3 f(x_3) + \dots + p_n f(x_k) \in
\lc(f,k-1)$. Since $y=p_1 f(x_1) + p_2 f(x_2) + r f(x')+
q f(x') + p_3 f(x_3) + \dots + p_n f(x_k)$ we conclude that
$y \in \lc(f,3)+\lc(f,k-1)$. \qed

Notice that Property~\ref{pr1} implies that

\begin{equation}\label{star}
\mbox{if $|\lc(f)|=\cont$ then $m_0=\min \{k\ge 1 \colon
|\lc(f,k)|=\cont\} \le 3$.}
\end{equation}

Next we show another property which is important for the proof of
Theorem~\ref{p3-thm1}. Note that if $\cont$ is regular (i.e., $\cf(\cont)=\cont$), then the set $Z$
from part (a) can be taken as a singleton.

\property{pr2}{Assume that $|\lc(f)|=\cont$. Then at least one of the
following two
cases hold.
\begin{description}

\item[(a)] There exists a set $Z \in [\real^n]^{< \cont}$ such that
$\left|\bigcup_{z \in Z} \lc(f,2,z)\right|=\cont$.

\item[(b)] For all $X\in[\real^n]^{<\cont} , Y \in [\real]^{< \cont}$
there exist $q_1, q_2, q_3 \in \rational \setminus \{ 0\} $ and pairwise
linearly
independent $x_1,x_2,x_3 \in \real^n $ such that $\sum_1^3 q_i f(x_i) \notin Y$,
$\sum_1^3 q_i x_i=0$, and $\lin(x_1,x_2,x_3) \cap \lin(X)=\{0\}$.
\end{description}
}

\proof
Notice first that if $|\lc(f,2)| = \cont$ then case (a) holds with $Z=\{ 0\}$.
Hence, using (\ref{star}), we may assume that
\begin{equation}\label{starstar}
|\lc(f,2)| < \cont \mbox{ and } |\lc(f,3)| = \cont .
\end{equation}
Based on the above assumption and the definition of the set $\lc(f,3)$, we
conclude that
there exist continuum many triples $\la x_1, x_2, x_3\ra\in (\real^n)^3$ and
$\la p_1, p_2, p_3\ra \in (\rational\setminus\{ 0\})^3$ such that
$\sum_1^3p_ix_i=0$ and
$\sum_1^3p_if(x_i)$ are all different. Thus, an easy cardinal argument implies
the existence of a sequence $ \la \la x_1^\xi ,x_2^\xi ,x_3^\xi
\ra \in (\real^n)^3 \colon \xi < \cont \ra$ and some non-zero rationals $q_1,q_2,q_3$
with the property that $q_1x_1^\xi + q_2x_2^\xi + q_3 x_3^\xi=0$
for every $\xi < \cont $, and all
$q_1f(x_1^\xi)+q_2f(x_2^\xi)+q_3f(x_3^\xi)$ are different.

Notice that, if $\dim(\lin(x_1^\xi,x_2^\xi ,x_3^\xi))=1$ for some $\xi$ then
$\lin(x_1^\xi,x_2^\xi ,x_3^\xi)=\lin(x_i^\xi)$ for some $i\in\{1,2,3\}$.
Say i=1. So there is an $s \in \rational$
such that $s q_1x_1^\xi + q_2x_2^\xi=0$. Combining this with the equality
$q_1 x_1^\xi+q_2x_2^\xi +q_3 x_3^\xi =0$ we obtain that
$s q_1x_1^\xi + q_2x_2^\xi= (1-s)q_1x_1^\xi+ q_3 x_3^\xi=0$.
Consequently,
\begin{eqnarray*}
[sq_1 f(x_1^\xi) + q_2f(x_2^\xi)],[(1-s)q_1f(x_1^\xi)+ q_3
f(x_3^\xi)] & \in & \lc(f,2) \\
\mbox{and \qquad \qquad} \\
q_1f(x_1^\xi)+q_2f(x_2^\xi)+q_3f(x_3^\xi) & = & \\
sq_1f(x_1^\xi) + q_2f(x_2^\xi) + (1-s)q_1f(x_1^\xi)+ q_3 f(x_3^\xi) & \in &
\lc(f,2)+\lc(f,2).\\
\end{eqnarray*}
\vskip -0.2in
So, if $\dim(\lin(x_1^\xi,x_2^\xi ,x_3^\xi))=1$ for
continuum many $\xi$
then $|\lc(f,2)|=\cont$. This contradicts
(\ref{starstar}).
Thus, we may assume that $\dim(\lin(x_1^\xi,x_2^\xi ,x_3^\xi))=2$ for all
$\xi<\cont$.

Now choose $X\in[\real^n]^{<\cont}$ and $Y \in [\real]^{< \cont}$. Notice that
\begin{itemize}
\item[($\bullet$)] if $\lin(x_1^\xi,x_2^\xi,x_3^\xi) \cap \lin(X)\not=\{0\}$
and $Z=\lin(X)$ then
$$
q_1f(x_1^\xi)+q_2f(x_2^\xi)+q_3f(x_3^\xi) \in \bigcup_{z\in Z}
\lc(f,2,z)+ \bigcup_{z\in Z} \lc(f,2,z).
$$
\end{itemize}

Indeed, if $\lin(x_1^\xi,x_2^\xi,x_3^\xi)$ $\cap \lin(X)\not=\{0\}$ then
there exist $a,b,c
\in \rational$ such that $ax_1^\xi + bx_2^\xi + cx_3^\xi \in
\lin(X)\setminus\{ 0\}$.  At
least one of the numbers $a,b,c$ is not equal to zero because $ax_1^\xi + bx_2^\xi +
cx_3^\xi\not= 0$.
Without loss of generality we may suppose that $c\not= 0$ and consequently
$c=q_3$
(multiply the above equation by $\frac{q_3}{c}$.) Then, by subtracting
$ax_1^\xi + bx_2^\xi
+ q_3x_3^\xi$ from $q_1x_1^\xi + q_2x_2^\xi + q_3 x_3^\xi=0$, we obtain that
$(q_1-a)x_1^\xi + (q_2-b)x_2^\xi \in \lin(X)\setminus\{ 0\}$. So at least one of
$(q_1-a),(q_2-b)$ is not 0. We may assume that $(q_2-b)\not=0$.
(If $(q_1-b)\not=0$ then the
following argument works analogously.) Now multiply $(q_1-a)x_1^\xi +
(q_2-b)x_2^\xi$ by
$\frac{q_2}{q_2-b}$. We get that $rq_1x_1^\xi + q_2x_2^\xi \in\lin(X)$
and consequently
$(1-r)q_1 x_1^\xi + q_3x_3^\xi=[q_1x_1^\xi + q_2x_2^\xi +
q_3 x_3^\xi]- [rq_1x_1^\xi + q_2x_2^\xi]=-[rq_1x_1^\xi + q_2x_2^\xi] \in
\lin(X)$, for some
$r\in\rational$. Hence
$$[rq_1f(x_1^\xi) + q_2f(x_2^\xi)],[(1-r)q_1f(x_1^\xi)+ q_3 f(x_3^\xi)]\in
\bigcup_{z\in Z} \lc(f,2,z).$$

Now the claim $(\bullet)$ follows from
\begin{eqnarray*}
q_1f(x_1^\xi)+q_2f(x_2^\xi)+q_3f(x_3^\xi) \!\!\!
& = & \!\!\!
rq_1f(x_1^\xi) + q_2f(x_2^\xi) + (1-r)q_1f(x_1^\xi)+ q_3 f(x_3^\xi)\\
& \in & \!\!\!\bigcup_{z\in Z} \lc(f,2,z)+ \bigcup_{z\in Z} \lc(f,2,z).
\end{eqnarray*}
From $(\bullet)$ we see that if $\lin(x_1^\xi,x_2^\xi,x_3^\xi) \cap
\lin(X)\not=\{0\}$
holds for $\cont$-many $\xi$ then the set $Z$ satisfies the condition
$|\bigcup_{z \in Z}
\lc(f,2,z)|=\cont$. Obviously $Z \in[\real^n]^{<\cont}$. Thus, case (a) holds.

Summarizing the above discussion, we just need to consider a situation when
$\dim(\lin(x_1^\xi,x_2^\xi ,x_3^\xi))=2$ and
$\lin(x_1^\xi,x_2^\xi,x_3^\xi) \cap \lin(X)=\{0\}$ for all $\xi$.
Recall that $q_1x_1^\xi + q_2x_2^\xi + q_3 x_3^\xi=0$, where $q_1,q_2,q_3 \in
\rational \setminus\{0\}$. If two of $x_1^\xi,x_2^\xi,x_3^\xi$ were
dependent over
$\rational$ then we would have $\dim(\lin(x_1^\xi,x_2^\xi ,x_3^\xi))\le 1$.
Thus, $x_1^\xi,x_2^\xi,x_3^\xi$ are pairwise independent. Now it is
easy to see that case (b) holds. \qed

\lem{p3-lem2}{Let $X \in [\real^n]^{< \cont}$, $x \notin X$, and $y \in
\real$. Suppose also that $h,g \colon X \to \real$ are functions linearly
independent over $\rational$. Then there exist extensions $h',\; g'$ of $h$
and $g$ onto $X \cup \{x\}$ such that $h'$ and $g'$ are linearly independent
over $\rational$ and $h'(x)+g'(x)=y$.}
\proof Choose $h'(x)\in\real\setminus\lin(h[X]\cup g[X]\cup\{y\})$. This
choice is possible since $|\lin(h[X]\cup g[X]\cup\{y\})|<\cont$. Then
define $g'(x)=y-h'(x)$. It is easy to see that $h'=h\cup \{\la
x,h'(x)\ra\}$ and $g'=g\cup \{\la x,g'(x)\ra\}$ are the desired extensions.
\qed


\noindent
{\sc Proof of Theorem~\ref{p3-thm1}.}
Let us start with fixing a function $f \colon \real^n \to \real$ and enumerations
$\{x_\xi\colon \xi <\cont\}$,
$\{v_\xi\colon \xi <\cont\}$ of $\real^n$ and $\{ 0\} \times \real \sq \real^{n+1}$, respectively.
We will construct functions $h,g\colon \real^n \to \real$ which are
linearly independent over $\rational$ and satisfy the property
that $h+g=f$ and $\{ 0\} \times \real \sq \lin(h)\cap \lin(g)$.

First, let us argue that this is enough to prove the theorem. What we have to
show is that $\lin(h)=\lin(g)=\real^{n+1}$. To see $\lin(h)=\real^{n+1}$ note that
$$
\forall x\in \real^n \; \forall z\in \real \; \la x,z \ra=
\la x,h(x) \ra + \la 0,z-h(x) \ra \in \lin(h)+\lin(h)=\lin(h).
$$
By the same argument $\lin(g)=\real^{n+1}$.

To construct the desired functions $h$ and $g$, we consider three cases. In
the first case we assume that $|\lc(f)|<\cont$. If the latter fails, that is
$|\lc(f)|=\cont$, then either part (a) (Case 2) or part (b) (Case 3) of
Property~\ref{pr2} holds.
\medskip

\noindent
{\bf Case 1: $|\lc(f)| <\cont$.}

Let $\kappa <\cont$ denote the cardinality of the basis of $\lc(f)$ over
$\rational$. There exist $c\in \lc(f)$ and a linearly independent
set $A\sq\real^n$ such that $|A|=\kappa$ and $f(-a)+f(a) \equiv c=const$
for all $a \in A$. Such a set can be found since $|\lc(f)|<\cont$ and
$f(x)+f(-x)\in\lc(f)$ for every $x\in\real^n$. Put $B=(-A)\cup A$.

First, we will construct functions $h,g\colon B\to \real$ linearly
independent over $\rational$ for which $h+g \sq f$ and
\begin{equation}\label{circ}
\{0\}\times \lc(f) \sq \lin(h)\cap (\{0\}\times \real)=\lin(g)\cap
(\{0\}\times \real).
\end{equation}
To accomplish this let us fix enumerations $\{a_\xi\colon \xi < \kappa\}$ of
$A$ and $\{m_\xi\colon \xi < \kappa \}$ of a linear basis of $\lc(f)$ over
$\rational$. We may assume that $m_0 = c$ if $c\not= 0$.
The construction of $h$ and $g$ is by induction. At every step
$\alpha < \kappa$ we will define $h$ and $g$ on $\{ -a_\alpha , a_\alpha\}$,
assuring that
\begin{description}
\item[(a)] $h|A_\alpha, g|A_\alpha$ are linearly independent and
$(h+g)|A_\alpha \sq f$,
\item[(b)] $\la 0 , m_\alpha \ra \in \lin(h|A_\alpha)\cap (\{0\}\times
\real)=\lin(g|A_\alpha)\cap (\{0\}\times \real)$,
\end{description}
where $A_\alpha=\{ ia_\xi \colon \xi \le \alpha, i=-1,1 \}$.

For $\alpha=0$ and $x=\pm a_0$ put $h(x)=\frac{1}{4}m_0$ and
$g(x)=f(x)-h(x)$. Observe that
$g(-a_0)+g(a_0)=[f(-a_0)+f(a_0)]-[h(-a_0)+h(a_0)]\in
\{-\frac{1}{2}m_0,\frac{1}{2}m_0 \}$. This holds because $f(-a_0)+f(a_0)=c$
and $m_0=c$ if $c\not= 0$. Thus $\la 0,c \ra ,\la 0,m_0 \ra \in \lin(h|A_0)
\cap \lin(g|A_0)$. It is easily seen that $h|A_0$ and $g|A_0$ satisfy (a) and
(b).


Now suppose that $h$ and $g$ are defined on $A_{<\alpha}=\bigcup_{\xi < \alpha
}A_\xi$, $\alpha < \kappa$, and they satisfy the conditions (a) and (b) for
all $\xi <\alpha$. We will extend $h$ and $g$ onto $A_\alpha$ preserving the
desired properties.

We may assume that $\la 0,m_\alpha\ra \not\in
\lin(h|A_{<\alpha})\cup\lin(g|A_{<\alpha})$. (Otherwise we could extend $h$
and $g$ using Lemma~\ref{p3-lem2} preserving the condition (a).) Put
$h(x)=\frac{1}{2}m_\alpha$ and $g(x)=f(x)-h(x)$ for $x \in \{-a_\alpha
,a_\alpha\}$. We claim that (a) and (b) are satisfied. Obviously,
$(h+g)|A_\alpha \sq f$. To see the linear independence of $h|A_\alpha$ and
$g|A_\alpha$ first note that, based on the inductive assumption,
$h|A_{<\alpha}$ and $g|A_{<\alpha}$ are linearly independent.
Next suppose that
$$
p\la -a_\alpha, h(-a_\alpha) \ra + q\la a_\alpha, h(a_\alpha) \ra =v
\mbox{ for some } p,q \in \rational \mbox{ and } v \in
\lin(h|A_{<\alpha}).
$$
Since $a_\alpha \notin \lin(A_{<\alpha})$ we conclude that $p=q$. Therefore
we have
$$ p\la -a_\alpha, h(-a_\alpha) \ra + q\la a_\alpha, h(a_\alpha) \ra =p\la
0, h(-a_\alpha)+ h(a_\alpha) \ra =p\la 0,m_\alpha \ra=v.$$ But we assumed
that
$\la 0,m_\alpha\ra \not\in \lin(h|A_{<\alpha})\cup\lin(g|A_{<\alpha})$, so
$p=0$ and $v=0$. This shows linear independence of $h|A_{\alpha}$.
Very similar argument works for $g|A_{\alpha}$: just notice that
$g(-a_\alpha)+ g(a_\alpha)=[f(-a_\alpha)+ f(a_\alpha)]-[h(-a_\alpha)+
h(a_\alpha)]= c - m_\alpha$ and recall that $\la 0,c\ra \in \lin(g|A_0) \sq
\lin(g|A_{<\alpha})$.

Now we show that (b) is also satisfied. From what has already been proved, we
conclude that
$\la 0,m_\alpha\ra \in \lin(h|A_\alpha)\cup\lin(g|A_\alpha)$.

Thus, what remains to prove is the equality part of (b). (The following
argument is also needed in the case when Lemma~\ref{p3-lem2} was used to
define $h$ and $g$ on $\{-a_\alpha,a_\alpha\}$.)
It follows from the fact that
$\la 0,y\ra \in \lin(h|A_\alpha)$ provided there exist
$p_i \in \rational$ and $a_i \in A_\alpha$, $i \le n$ such that
\begin{eqnarray*}
\la 0,y\ra
& = & \sum_1^m p_i[\la -a_i,h(-a_i)\ra + \la a_i,h(a_i)\ra]\\
& = & \sum_1^m p_i\la 0,h(-a_i)+h(a_i)\ra\\
& = & \sum_1^m p_i\la 0,f(-a_i)+f(a_i)\ra -p_i\la 0,g(-a_i)+g(a_i)\ra\\
& = & \sum_1^m p_i\la 0,c\ra -\sum_1^m p_i[\la -a_i,g(-a_i)\ra + \la a_i,g(a_i)\ra]\\
& \in & \lin(g|A_\alpha).
\end{eqnarray*}
This completes the inductive definition of $h$ and $g$. Note that
$(\ref{circ})$ implies that any extensions $h',g'$
of $h$ and $g$, with  $h'+g'\sq f$, satisfy also
\begin{equation}\label{circprime}
\{0\}\times \lc(f) \sq \lin(h')\cap (\{0\}\times
\real)=\lin(g')\cap (\{0\}\times \real).
\end{equation}
To see this choose $\la 0,y\ra \in \lin(h')\cap (\{0\}\times \real)$. So
for some $p_i \in \rational$ and $x_i \in \real^n$ we have
$\la 0,y\ra =\sum_1^mp_i \la x_i, h'(x_i) \ra =\sum_1^mp_i \la x_i, f(x_i)
\ra -\sum_1^mp_i \la x_i, g'(x_i) \ra \in \lin(g')\cap (\{0\}\times
\real)$. The latter holds because
$\sum_1^mp_ix_i=0$ and consequently $\sum_1^mp_if(x_i)\in \lc(f)$. This
ends the proof of $(\ref{circprime})$.

Next we extend $h$ and $g$ onto $\real^n=\{x_\xi\colon \xi <\cont\}$,
preserving the linear independence and the property that at a step $\xi$
of the inductive definition we assure that $x_\xi\in
\dom(h_\xi)=\dom(g_\xi)$ and $v_\xi \in \lin(h_\xi) \cap \lin(g_\xi)$,
where $h_\xi$ and $g_\xi$ denote the extensions obtained in the step $\xi$.

Let $\beta < \cont$. Assume that $v_\beta \notin 
\lin(\bigcup_{\xi <\beta }h_\xi) \cup \lin(\bigcup_{\xi <\beta
}g_\xi)$. Choose an $a \in \real \setminus \lin(\dom(\bigcup_{\xi <\beta
}h_\xi))$ and define $h(x)$ by $\la 0, h(x)\ra
=\frac{1}{2}v_\beta$ for $x \in \{ -a,a\}$. Put also $g(x)=f(x)-h(x)$. Since
$f(-a)+f(a)\in\lc(f)$, $(\ref{circ})$ implies that $v_\beta \in
\lin(h)\cap\lin(g)$. What remains to show is that $h$ and $g$ are still linearly
independent. But this follows from $(\ref{circprime})$ and almost the same
argument which is used to show linear independence of $h|A_\alpha$ and
$g|A_\alpha$ in the previous part of the proof. (Replace $a_\alpha$,
$h|A_{<\alpha}$, and $g|A_{<\alpha}$ by $a$, $\bigcup_{\xi <\beta }h_\xi$, and
$\bigcup_{\xi <\beta }g_\xi$, respectively.)

To finish the step $\beta$ of the inductive definition we need to make sure
that $h$ and $g$ are defined at $x_\beta$. If $x_\beta \notin \dom(h)=\dom(g)$
then we can use Lemma~\ref{p3-lem2} to define these functions at $x_\beta$,
preserving all the required properties. This ends the construction in Case 1.

\noindent
{\bf Case 2:} Property~\ref{pr2} (a) holds.

Let $Z\in [\real^n]^{<\cont}$ be a set satisfying $ \left | \bigcup_{z \in Z}
\lc(f,2,z) \right |=\cont$. We start with defining functions $h,g\colon Z \to
\real$ which are linearly independent over $\rational$ and whose sum is
contained in $f$ (i.e., $h+g\sq f$.) It can be easily done by using
Lemma~\ref{p3-lem2}.

 We will extend $h$ and $g$ onto $\real^n$ by induction. Let $\beta < \cont$.
Assume that $h$ and $g$ are linearly independent, $h+g \sq f$,
$\{ x_\xi \colon \xi < \beta \} \sq D_\beta =\dom(h)=\dom(g)$,
$\{ v_\xi \colon \xi < \beta \} \sq \lin(h)\cap \lin(g)$, and
$v_\beta \notin \lin(h)$.
The property of the set $Z$ implies the existence of a $z \in Z$ satisfying
$|\lc(f,2,z)| > \max(|h|,\omega)=\max(|g|,\omega)$. Thus, an easy cardinal
argument
shows that we can find
$z_1,z_2 \in \real^n\setminus \lin(D_\beta )$ and $p_1,p_2 \in \rational
\setminus \{ 0\}$ which satisfy

\begin{equation}\label{circcirc}
\qquad p_1z_1 + p_2z_2 = z \mbox{ and } \la z,p_1 f(z_1) + p_2f(z_2)\ra \notin \lin(g
\cup \{ \la 0, h(z) \ra, v_\beta\}).
\end{equation}
Define the values of $h$ at $z_1$ and $z_2$ so that
$$p_1\la z_1, h(z_1)\ra + p_2\la z_2, h(z_2)\ra=\la z, p_1h(z_1) + p_2h(z_2)\ra
=v_\beta +\la z, h(z) \ra.$$
Observe that
$v_\beta =[v_\beta +\la z, h(z) \ra] - \la z, h(z) \ra \in \lin(h)$.

Now we argue that $h$ and $g$ are linearly independent. To see linear
independence of $h$ suppose that for some $q,r \in \rational$ (not both
equal $0$) we have
$$q\la z_1, h(z_1)\ra + r\la z_2, h(z_2)\ra=\la qz_1+rz_2, qh(z_1) +
rh(z_2)\ra \in \lin(h|D_\beta).$$
Since $z_1,z_2\notin \lin(D_\beta)$ and $p_1z_1+p_2z_2=z\in Z \sq \lin(D_\beta)$ we
conclude that $\la q , r\ra $ and $\la p_1 , p_2\ra $ are linearly dependent. So
we may assume that $\la q , r\ra =\la p_1 , p_2\ra $.
Consequently,
$v_\beta +\la z, h(z) \ra =\la z, p_1h(z_1) + p_2h(z_2)\ra \in \lin(h|D_\beta)$.
This contradicts the assumption $v_\beta \notin \lin(h|D_\beta)$.
Hence, $h$ is linearly independent.

Based on the above argument, we see that linear independence of $g$
will follow from $\la z, p_1g(z_1) + p_2g(z_2)\ra \notin \lin(g|D_\beta)$.
But this holds since $(\ref{circcirc})$ implies
\begin{eqnarray*}
\la z, p_1g(z_1) + p_2 g(z_2)\ra & = & \\
\la z, p_1f(z_1)+p_2f(z_2)-[p_1h(z_1) + p_2h(z_2)]\ra & = & \\
\la z, p_1f(z_1)+p_2f(z_2)\ra -\la 0, h(z) \ra - v_\beta & \notin &
\lin(g|D_\beta). \\
\end{eqnarray*}
To assure that $v_\beta \in \lin(g)$ we repeat the same procedure as
above for the function $g$. Finally, if $x_\beta \notin \dom(h)=\dom(g)$ then
we use Lemma~\ref{p3-lem2} to define the functions at $x_\beta$. This ends the construction in Case 2.

\noindent
{\bf Case 3:} Property~\ref{pr2} (b) holds.

The inductive construction of functions $h$ and $g$ is somewhat similar to the
one from the previous case. So assume that $\beta < \cont$ and the
construction has been carried out for all $\xi < \beta $. If $v_\beta \notin
\lin(h)$ then let $X=\dom(h)=\dom(g)$ and $Y \in [\real]^{<\cont}$ be such a
set that $\lin(g\cup\{v_\beta\}) \sq \real^n\times Y$. By Property~\ref{pr2} (b), there
exist $p_1, p_2, p_3 \in \rational \setminus \{ 0\}$ and pairwise independent
$x_1,x_2,x_3 \in \real^n$ such that $\sum_1^3 p_i x_i=0$, $\lin(x_1,x_2,x_3)
\cap \lin(X)=\{0\}$, and $\sum_1^3 p_i f(x_i) \notin Y$.


We extend $h$ and $g$ onto $\{x_1,x_2,x_3\}$.
Choose $h(x_1),h(x_2),h(x_3) \in \real$ in such a way that
$$\sum_1^3 p_i \la x_i,h(x_i)\ra=
\left\langle 0, \sum_1^3 p_i h(x_i) \right\rangle = v_\beta.
$$
Then put $g(x_i)=f(x_i) - h(x_i) \mbox{ for } i \le 3$.
Obviously $v_\beta \in \lin(h)$ and $h+g\sq f$.
We claim that the linear independence of $h$ and $g$ is also preserved.

To show this claim note first that, if
$\sum_1^3 p_i^{\prime} x_i \in \lin(X)$ for some
$p_1^{\prime},p_2^{\prime},p_3^{\prime} \in \rational$ then
$\sum_1^3 p_i^{\prime} x_i=0$. Pairwise independence of
$x_1,x_2,x_3$ implies that $\sum_1^3 p_i^{\prime} x_i=0$ holds only
for triples $\la p_1^{\prime},p_2^{\prime},p_3^{\prime} \ra \in \lin(\la
p_1,p_2,p_3\ra)$.
Thus, our claim holds if $\sum_1^3 p_i \la x_i,h(x_i)\ra\notin\lin(h|X)$ and
$\sum_1^3 p_i \la x_i,g(x_i)\ra\notin\lin(g|X)$. But these two conditions
follow from

\vskip 0.15in
\noindent
$\bullet\sum_1^3 p_i \la x_i,h(x_i)\ra=v_\beta\notin\lin(h|X)$ and

\vskip 0.10in

\noindent
$\bullet \sum_1^3 p_i \la x_i,g(x_i)\ra=
\sum_1^3 p_i \la x_i,f(x_i) - h(x_i)\ra=
\la 0,\sum_1^3 p_i f(x_i) \ra -v_\beta \notin \lin(g|X)$

\vskip 0.05in

\noindent
\quad $\;\;\;$ (``$\notin$'' part holds because
$\lin((g|X) \cup\{v_\beta\}) \sq \real^n\times Y$ and $\sum_1^3 p_i f(x_i)
\notin Y$.)

\vskip 0.15in
To assure that $v_\beta \in \lin(g)$ we repeat the same steps as
above for the function $g$ and then, if necessary, define
$h$ and $g$ at $x_\beta$ using Lemma~\ref{p3-lem2}.
This ends the construction in Case 3.
\qed

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\end{document}