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\title{Uniformly approachable maps%
\thanks{Work partially supported by the
NATO Collaborative Research Grant CRG~950347
%research projects 60\% and 40\% of the Italian Ministero dell'
%Universit\`{a} e della Ricerca Scientifica e Tecnologica.
%\\ {\it Key words:}  
\endgraf AMS classification
numbers: Primary  54C30, 41A30; \endgraf
Secondary 54E35, 54B30, 26A15. \endgraf
Key words and phrases: 
{\it metric spaces, uniformly continuous maps,  
uniform approachability, complex analytic functions.} }}  
\author{
Krzysztof Ciesielski\\ 
{\footnotesize
Department of Mathematics, West Virginia University,} \\ 
{\footnotesize  Morgantown, WV 26506-6310} \\ 
{\footnotesize  KCies@wvnvms.wvnet.edu}
\and Dikran Dikranjan \\ 
{\footnotesize Dipartimento di Matematica e Informatica,
Universit\`{a} di Udine} \\ 
{\footnotesize Via Zanon 6, 33100 Udine, Italy} \\ 
{\footnotesize  dikranja@dimi.uniud.it}
}
\date{}
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\section{Preliminaries}\label{Preliminaries}

Throughout the paper we will use the standard definitions
and notation (\cite{ABR}, \cite{E}). 

Let $X$ and $Y$ be metric spaces. The goal is to study some 
intermediate classes of functions between the class $C_{uc}(X,Y)$ of all uniformly
continuous mappings (briefly, $UC$) from $X$ into $Y$
and the class $C(X,Y)$ of all continuous functions
$f\colon X\to Y$. These classes, defined below, have been 
intensively studied in \cite{BD} and \cite{BDP}
mainly in the case when $Y=\R$. 
(Compare also \cite{B,DP}.) In this paper we will study
them for general $Y$. In particular, we will consider the case
when $X=Y$ is the complex plane $\C$. 

Let $X$ and $Y$ be metric spaces and let $f\colon X\to Y$.   
\begin{enumerate}
\item For $K,M \subseteq X$ we say that 
$g\colon X\to Y$ is a {\em $(K,M)$-approximation} 
of $f$ if $g$ is a $UC$ map such that $g[M]\subseteq f[M]$ and 
$g(x)=f(x)$ for each $x\in K$.

\item The function $f$ is {\it uniformly approachable} (briefly, $UA$) 
if $f$ has a $(K,M)$-approximation
for every compact $K \subseteq X$ and every $M\subseteq X$.

\item The function $f$ is {\it weakly uniformly approachable} 
(briefly, $WUA$) if $f$ has a $(\{x\},M)$-approximation
for every $x\in X$ and every $M\subseteq X$.  
\end{enumerate}

The class of all uniformly approachable 
(weekly uniformly approachable, respectively) functions
$f\colon X\to Y$ will be denoted by
$C_{ua}(X,Y)$ ($C_{wua}(X,Y)$, respectively.) 
Notice that
\begin{equation}\label{eq1}
C_{uc}(X,Y)\subseteq C_{ua}(X,Y)\subseteq 
C_{wua}(X,Y)\subseteq C(X,Y).
\end{equation}
(See \cite{DP}. Compare also \cite[Section 12]{BD}.)

Indeed, every $UC$ function $f$ is $UA$, since $g=f$ is a
$(K,M)$-approximation of $f$ for every $M,K\subseteq X$.
Every $UA$ function is $WUA$, since $\{x\}$ is compact for every $x\in X$.
To see that every $WUA$ map $f$ is continuous, it is enough
to show that 
$f[\ov{M}]\subseteq\ov{f[M]}$ for every $M\subseteq X$,
where $\ov{M}$ stands for the closure of $M$. So, take
$x\in\ov{M}$ and an $(\{x\},M)$-approximation $g$ of $f$.
Then, $f(x)=g(x)\in g[\ov{M}]\subseteq\ov{g[M]}\subseteq\ov{f[M]}$.

We will discuss possible equations in (\ref{eq1}) for different spaces
$X$ and $Y$ in Section \ref{sec:Eq}.

For a topological space $X$ we say that a function $g\in C(X,\R)$
is a {\em truncation} of $f\in C(X,\R)$ if 
$g$ is constant on every connected component of 
$\{x\in X\colon f(x)\neq g(x)\}$.
In what follows we will use also the following fact
which can be proved by straightforward transfinite induction
diagonal argument.

\begin{proposition}\label{prop:magic} {\rm \cite[Thm 8.1]{BD}}
Let $X$ be a separable topological space. 
Then there is a set $M \subseteq X$
such that for every $f,g\in C(X,\R)$, if $g[M]\subseteq f[M]$
and $f^{-1}(y)$ is at most countable for every $y\in\R$
then $g$ is a truncation of $f$.
\end{proposition}

The set $M$ from Proposition \ref{prop:magic} will be called
a {\em magic set (for $X$).}
The following fact is an easy but useful corollary
of Proposition \ref{prop:magic}. It is 
a version of \cite[Cor 8.3]{BD}.

\begin{corollary}\label{cor:magic} 
Let $X$ be a separable metric space. 
If $f\in C(X,\R)$ is a one-to-one
function without non-constant $UC$ truncation
then $f$ is not $WUA$.
\end{corollary}




\section{Basic facts}\label{BasicFacts}

In this section we collect some basic properties 
of the classes of $UA$- and $WUA$-maps.

\begin{theorem}\label{thm:composition}  
Composition of $UA$ ($WUA$, respectively) 
maps is a $UA$ ($WUA$, respectively) map.
\end{theorem}

\begin{proof} 
Let $f\colon X\to Y$ and $f'\colon Y\to Z$ be $UA$ maps.  
To show that $f'\circ f\colon X\to Z$ is $UA$ take a compact subset 
$K$ of $X$ and a set $M\sq X$. 
Then $f[K]$ is a compact subset of $Y$.  
Choose a $(K,M)$-approximation $g\colon X\to Y$ of $f$ 
and a $(f[K],f[M])$-approximation $g'\colon Y\to Z$ of $f'$. 
It is enough to show that $g'\circ g$ is a 
$(K,M)$-approximation of $f'\circ f$. 
Obviously $g'\circ g$ coincides with  $f'\circ f$ on $K$. 
Also $g[M]\sq f[M]$ and $g'[f[M]]\sq f'[f[M]]$ implies that
$(g'\circ g)[M]\sq  (f'\circ f)[M]$. 
The proof is finished when we notice that $g'\circ g$ is
$UC$ as a composition of two $UC$ functions.

The argument for $WUA$ functions is identical, if we replace
$K$ with $\{x\}$. 
\end{proof}

\begin{proposition}\label{prop:restr}
Let $F$ be a subset of a metric space $X$.
If $f\colon X\to\R$ is $UA$ ($WUA$, respectively)
then so is its restriction $f|_F\colon F\to\R$.
\end{proposition}

\begin{proof} 
Let $f\in C_{ua}$ and $M,K\sq F$, $K$ being compact.
So, there exists $(K,M)$-approximation $g\colon X\to\R$ of $f$.
It is easy to see that $g|_F$ is an 
$(K,M)$-approximation of $f|_F$.
\end{proof}

In \cite{BD} it was shown that every continuous function
$f\colon\R\to\R$ is $UA$. In fact, one can show
the following stronger property.

\begin{theorem}\label{thm:R-R}
If $X$ is a uniform space then every continuous function
$f\colon\R\to X$ is $UA$.
\end{theorem}

\begin{proof}
Let $K$ be a compact subset of $\R$ and  $M\sq \R$. 
Choose an interval $[a,b]$ containing $K$
such that either $a\in M$ or $(-\infty,a]\cap M=\emptyset$
and  that either $b\in M$ or  $[b,\infty)\cap M=\emptyset$.
Let $r\colon\R\to[a,b]$ be the retraction, i.e.,
$r(x)=x$ for $x\in[a,b]$, $r(x)=a$ for $x<a$
and $r(x)=b$ for $x>b$. Clearly $r$ is $UC$. 
Define $g$ by $g=f|_{[a,b]}\circ r$. 
Then $g$ is $UC$ as a composition of two $UC$ functions, 
and $g$ coincides with $f$ on $K\sq [a,b]$.
It is also easy to see that $g[M]\sq f[M]$.
\end{proof}

In what follows we will often use the following criteria
which is similar to that of  \cite[Thm. 6.3]{BD}.  

\begin{proposition}\label{prop:crit}
Let $(X,\rho)$ be a separable metric space and 
let $f\colon X\to \R$.
If there exists an uncountable $Y\sq\R$ such that 
$f^{-1}(y)$ is non-empty and connected for every $y\in Y$
and that $\rho(f^{-1}(x),f^{-1}(y))=0$ 
for every $x,y\in Y$
then $f$ is not $WUA$.
\end{proposition} 

\begin{proof}
Replacing $X$ with $f^{-1}(Y)$, if necessary, we can assume that
$X=f^{-1}(Y)$. 
Now, let $D$ be a countable dense subset of $X$ and 
put $M=f^{-1}(f[D])$. 
Then $M$ is dense in $X$, since $D\sq M$.
Moreover, $M$ is a proper subset of $X$, since $Y=f[X]$ is uncountable
unlike $f[M]=f[f^{-1}(f[D])]=f[D]$.  
Pick some $x\in X\setminus M$. 
Then, $x\not\in M=f^{-1}(f[D])=f^{-1}(f[f^{-1}(f[D])])=f^{-1}(f[M])$,
i.e., $f(x)\not\in f[M]$.
Our aim is to show that $f$ has no $(\{x\},M)$-approximation. 
By way of contradiction assume that there exists an $UC$ function
$g\colon X\to \R$ such that $g[M]\sq f[M]$.
Then $g[M]\sq f[M]=f[D]$ is countable, hence
zero-dimensional. (See \cite[Chapter 7, \S2, Corollary]{E}.) 
Since $C_d=f^{-1}(f(d))\sq M$ is connected for every $d\in D$ and
$g[C_d]\sq g[M]\sq f[M]=f[D]$, 
we conclude that $g$ is constant on each $C_d$.
Since our hypothesis yields 
that $\rho(C_d, C_{d'})=0$ for every $d,d'\in D$, the
uniform continuity of $g$ implies that all these constants coincide.
Therefore $g|_M$ is constantly equal to some $b\in f[M]$. 
But $f(x)\not\in f[M]$. So, $g(x)=b\ne f(x)$. 
Therefore $g$ is not an $(\{x\},M)$-approximation of $f$. 
\end{proof}

As an immediate corollary we obtain the following.
(See also \cite[Lemma 5.4]{BD}.)

\begin{corollary}\label{cor:crit} 
The function $h\colon\R^2\to\R$ given by formula $h(x,y)=x^2-y^2$ is not $WUA$.
\end{corollary} 

\begin{proof}
By Proposition \ref{prop:crit} function 
$h|_{[0,\infty)\times[0,\infty)}$ is not $WUA$. So, $h$ is not $WUA$
by Proposition \ref{prop:restr}.
\end{proof}

One of the most interesting properties of
the class of $UA$ functions is the fact that every perfect map%
\footnote{ A map $f\colon X\to\R^n$ is perfect if $f^{-1}(r)$ is compact
for every $r\in\R^n$.}
from $\R^n$ into $\R$ is $UA$ \cite[Thm 5.2]{BD}. 
However, this theorem does not generalize for all functions
from $\R^2$ into $\R^2$, as shown by the next example. 
(Compare with Example \ref{Ex:zSq}.)

\begin{em-example}\label{ex:perfect}
The functions $f,g\colon\R^2\to\R$ given by the
formulas $f(x,y)=x^2$ and $g(x,y)=y^2$ are $UA$, while
the perfect map $H=(f,g)\colon\R^2\to\R^2$ is not $WUA$.
\end{em-example}

\begin{proof}
The function $f$ is $UA$ by Theorem \ref{thm:composition}
since it is a composite of a $UC$ function
$f_1\colon\R^2\to\R$, $f_1(x,y)=x$, and a $UA$ function
$f_2\colon\R\to\R$, $f_2(x)=x^2$. (See Theorem \ref{thm:R-R}.)
Similarly we show that $g$ is $UA$.
To see that $H$ is is not $WUA$ notice that the function 
$h\colon\R^2\to\R$ from Corollary \ref{cor:crit} 
is a composite
of $H$ and a $UC$ function $h\colon\R^2\to\R$, $h(x,y)=x-y$.
Thus, by Theorem \ref{thm:composition}, $H$ cannot be $WUA$. 
\end{proof}

Example \ref{ex:perfect} shows also that a map
$H=(f,g)\colon X\to Y\times Z$ does not have to be
$WUA$ even if $f\colon X\to Y$ and $g\colon X\to Z$
are both $UA$. 

\medskip

Now, if $X$ is a metric space then 
$C(X,\R)$ is a linear topological space
and $C_{uc}(X,\R)$ is its linear subspace. 
However, the classes $C_{ua}(X,\R)$ and $C_{wua}(X,\R)$
are not closed under addition, as shown by 
Example \ref{ex:perfect} and Corollary \ref{cor:crit}.
In fact, the next example shows that the sum of $WUA$ function and $UC$ function
does not have to be $WUA$. 
To formulate it easier, we need the following notation and lemma.

We will use the symbol $C$ to denote the unit circle on the complex
plane:
\[
C=\{z\in\C\colon |z|=1\}
\]
with the standard distance and we will write
$\R_c$ to denote the real line with the metric of
$C\setminus\{-1\}$, i.e., given by formula
\[
\rho(x,y)=\left|e^{2 {\bf i} \arctan x}-e^{2 {\bf i} \arctan y}\right|.
\]

\begin{lemma}\label{lem:SinFunc}
\begin{description}
\item[(i)]  Let $Y$ be a metric space. 
            A function $f\in C(\R_c,Y)$ is $UC$ if and only if 
            both limits $\lim_{x\to -\infty}f(x)$ and 
            $\lim_{x\to\infty}f(x)$ exist and are equal.
\item[(ii)] A function $f\in C(\R_c,\R)$ is $WUA$ provided\ 
            ${\rm int}(f[(-\infty,x)]\cap f[(x,\infty)])\neq\emptyset$
            for every $x\in\R_c$.
\end{description}
\end{lemma}

\begin{proof}
Since the spaces $\R_c$ and $C\setminus\{-1\}$ are isometric, 
we can assume that $f\colon C\setminus\{-1\}\to Y$.

(i) Now, if both limits $\lim_{x\to -\infty}f(x)$ and 
$\lim_{x\to\infty}f(x)$ exist and are equal to $b\in\R$, 
then we can extend $f$ to a continuous function $F$ on $C$
by putting $F(-1)=b$. Clearly $F$ is $UC$, 
as a continuous function on the compact set $C$. So, $f$ is $UC$.

Conversely, if $f\colon C\setminus\{-1\}\to Y$ is $UC$, then it can be 
extended uniquely to $UC$ function $F\colon C\to Y$
and then clearly both limits exist and are equal to $F(-1)$. 

(ii) Choose $x\in\R_c$ and $M\sq \R_c$. Clearly we can assume that
$M\not\sq\{x\}$. 

If $M$ is not dense in $\R_c$ then we can
find $a<b$ in $\R_c$ and $m\in M$
such that $(a,b)\cap M=\emptyset$ and either $a<b<x<m$ or $m<x<a<b$.
Assume that $a<b<x<m$. Define $g|_{[b,m]}=f|_{[b,m]}$, 
$g(x)=f(m)$ for $x\in(-\infty,a)\cup(m,\infty)$ and extend it
to $(a,b)$ in a continuous way. Then, $g$ is $UC$ by (i)
and it is easy to see that $g$ is an $(\{x\},M)$-approximation of $f$.
The case $m<x<a<b$ is handled in a similar way. 

So, assume that $M$ is dense in $\R_c$. 
Let $m\in M\cap f^{-1}({\rm int}(f[(-\infty,x)]\cap f[(x,\infty)]))$
and put $y=f(m)\in f[(-\infty,x)]\cap f[(x,\infty)]$.
Choose $a\in (-\infty,x)\cap f^{-1}(y)$ and 
$b\in (x,\infty)\cap f^{-1}(y)$. Then, $a<x<b$ and $f(a)=f(b)=y=f(m)$.
Define $g|_{[a,b]}=f|_{[a,b]}$ and 
$g(x)=f(m)$ for $x\in(-\infty,a)\cup(b,\infty)$. 
It is easy to see that $g$ is a $UC$ function which is
an $(\{x\},M)$-approximation of $f$. 
\end{proof}

Notice that the assumption in condition (ii) of
Lemma \ref{lem:SinFunc} is not necessary for its conclusion
as shown by the $UC$ function
$f\colon\R_c\to\R$, $f(x)=x/(1+x^2)$.
Notice also, that Lemma \ref{lem:SinFunc}(i) implies that
$C_{uc}(\R_c,\R)$ is a subring of $C(\R_c,\R)$,
i.e., that $C_{uc}(\R_c,\R)$ is closed under pointwise
product of functions. 


     \begin{figure}[htb]
     \[
       \vbox to 1.8in{\hrule width 4.75in height 0pt depth 0pt \vfill
         \special{illustration kcgraph.eps scaled 1100}}
     \]
    \caption{Graph of function $f$ from 
    Lemma \protect\ref{ex:SinFunc}.\label{Fig1} }
    \end{figure}


\begin{em-example}\label{ex:SinFunc}
Let $f,h\colon\R_c\to\R$ be given by formulas
\[
h(x)=\frac{12 \sin \frac{\pi}{2} x}{x^2+1}
\]
and $f(x)= -h(x) + \arctan x$. (See Figure \ref{Fig1}.)
Then $h$ is UC, $f$ is WUA while $f+h$ is not WUA. 
\end{em-example}

\begin{proof}
The function $h$ is $UC$ by Lemma \ref{lem:SinFunc}(i). 
The increasing function $f+h=\arctan$ is not $WUA$
by Corollary \ref{cor:magic} and the other implication of 
Lemma \ref{lem:SinFunc}(i).
To finish the proof, it is enough to
show that $f$ satisfies 
the assumption of Lemma \ref{lem:SinFunc}(ii).
So, let $x\in\R_c$ and choose an integer number 
$k$ such that $x\in [4k,4(k+1)]$.
Notice that
\[
f(4k-1)=
%\arctan(4k-1) - \frac{12 \sin \frac{\pi}{2} (4k-1)}{(4k-1)^2+1}=
\arctan(4k-1) + \frac{12}{(4k-1)^2+1}>\arctan(4k-1)>-\frac{\pi}{2}
\]
and
\[
f(4k+5)=
%\arctan(4k+5) - \frac{12 \sin \frac{\pi}{2} (4k+5)}{(4k+5)^2+1}=
\arctan(4k+5) - \frac{12}{(4k+5)^2+1}<\arctan(4k+5)<\frac{\pi}{2}.
\]
Thus, it is enough to show 
the inequality $f(4k-1)>f(4k+5)$, since then we have 
$(-\pi/2,f(4k-1))\sq f[(-\infty,x)]$, 
$(f(4k+5),\pi/2)\sq f[(x,\infty)]$, and
$(-\pi/2,f(4k-1))\cap (f(4k+5),\pi/2)\neq\emptyset$.
So, we can reduce our task to the proof of
\begin{equation}\label{eq1:sin}
\arctan(4k+5) - \arctan(4k-1)<
\frac{12}{(4k+5)^2+1} + \frac{12}{(4k-1)^2+1}.
\end{equation}
Now, if $4k+5\leq 0$, then $\arctan$ is concave on the interval
$[4k-1,4k+5]$ and using differential
approximation at the point $4k+5$
we obtain
\[
\arctan(4k+5) - \arctan(4k-1)\leq 6 \arctan^\prime(4k+5)=
\frac{6}{(4k+5)^2+1}.
\]
So, (\ref{eq1:sin}) holds.
The case when $4k-1\geq 0$ is handled by the fact that $f$ is an odd function.
Now, if $4k-1<0<4k+5$ then 
$k=0$ or $k=-1$, i.e., $\{4k-1,4k+5\}\cap\{-1,1\}\neq\emptyset$,
So, using the fact that the derivative of
$\arctan$ is at most $1$ we obtain
\[
\arctan(4k+5) - \arctan(4k-1)<6=
\frac{12}{(\pm 1)^2+1}<\frac{12}{(4k+5)^2+1} + \frac{12}{(4k-1)^2+1}.
\]\end{proof}

\begin{problem}\label{problem:UcUa}
If $X$ is a metric space, $f\in C_{ua}(X,\R)$ and $g\in C_{uc}(X,\R)$
is $f+g$ $UA$?
\end{problem}

\begin{problem}\label{problem:UcWua}
If $f\in C_{wua}(\R^2,\R)$ and $g\in C_{uc}(\R^2,\R)$ is $f+g$ $WUA$?
\end{problem}


\section{Equations between $C_{uc}$, $C_{ua}$, 
$C_{wua}$ and $C$.
}\label{sec:Eq}

In this section we will discuss all the possible 
spectra of equations and sharp inclusions
in the formula 
\begin{equation}\label{eq2}
%C_{uc}(X,Y)\buildrel (a)\over \subseteq C_{ua}(X,Y)\buildrel (b)\over
%\subseteq C_{wua}(X,Y)\buildrel (c)\over \subseteq C(X,Y).
C_{uc}(X,Y)\subseteq C_{ua}(X,Y)
\subseteq C_{wua}(X,Y)\subseteq C(X,Y).
\end{equation}

An important contribution for this discussion comes from the following
fact.

\begin{proposition}\label{prop:ats} {\rm \cite[Thm. 12.1]{BD}}
For any metric space $X$ the equation
$C_{uc}(X,\R)=C_{ua}(X,\R)$ implies that
$C_{uc}(X,\R)=C(X,\R)$. 
\end{proposition}

In particular, the space $Y$ in the following cases (ii)-(iv)
cannot be equal to $\R$. On the other hand, we have 
$Y=\R$ in all remaining cases. 

\begin{description}
\item[(i)] $C_{uc}(X,Y)=C_{ua}(X,Y)=C_{wua}(X,Y)=C(X,Y)$.
Any compact space $X$ will do the job. For example $X=[0,1]$, $Y=\R$.
\end{description}

\begin{description}
\item[(ii)] $C_{uc}(X,Y)=C_{ua}(X,Y)=C_{wua}(X,Y)\neq C(X,Y)$.
Let $M$ be a magic set for $\R$ from Proposition \ref{prop:magic}.
Then $X=Y=M$ considered with the usual metric have the 
above property. (See \cite[Example 13]{B}.)
Spaces $X$ and $Y$ can be also chosen to be
complete metric spaces \cite[Cor. 18]{B}.
(Notice that in \cite{B} term $UA$ is used for what we call here $WUA$.)
\end{description}

\begin{description}
\item[(iii)] $C_{uc}(X,Y)=C_{ua}(X,Y)\neq C_{wua}(X,Y)=C(X,Y)$.
\end{description}

We do not know such an example. So, the following problem remains open. 

\begin{problem}\label{problem:Equations}
Do there exist metric spaces $X$ and $Y$ with the property that
$C_{uc}(X,Y)=C_{ua}(X,Y)\neq C_{wua}(X,Y)=C(X,Y)$?
\end{problem}

\begin{description}
\item[(iv)] $C_{uc}(X,Y)=C_{ua}(X,Y)\neq C_{wua}(X,Y)\neq C(X,Y)$.
%Notice first, that as in (2) $Y$ cannot be equal to $\R$. 
Now, let $f
%h=f_{c_0}
\colon\R_c\to\R$
be as in Example \ref{ex:SinFunc} and let $M\sq\R_c$ be a magic set. 
We claim, that the above chain of equations and 
inequalities holds for $X=M$ and $Y=f[M]$. 

First notice that both $M$  and the complement of $M$ are dense in $\R_c$.
Otherwise, if $(a,b)$ is a non-empty interval contained in either $M$ or
$\R_c\setminus M$ and if $g\in C(\R_c,\R)$ is such that 
$g|_{\R_c\setminus(a,b)}=\arctan |_{\R_c\setminus(a,b)}$, 
$g[(a,b)]=\arctan [(a,b)]$, and $g\ne \arctan$, then $g$ is an
$(\{a\},M)$-approximation of $\arctan$ which is not a 
truncation of $\arctan$.

Next we will show that 
\begin{equation}\label{eq4:u_ext}
\mbox{ any }h\in C_{wua}(M,f[M]) 
\mbox{ can be uniquely extended to }C(\R_c,\R).
\end{equation}

This immediately implies $C_{wua}(M,f[M])\ne C(M,f[M])$ since there exists a
function $h\in C(M,f[M])$ with $h[M]$ having precisely two elements and, by
(\ref{eq4:u_ext}), such $h$ is not  $WUA$.

To see (\ref{eq4:u_ext}) let $h\colon M\to \R$
be a function without continuous extension into $\R_c$. 
Then there exist $x\in \R_c\setminus M$, $\varepsilon>0$
and two sequences $\{x_n\}_{n\in\N}$ and  
$\{y_n\}_{n\in\N}$ in $M$
converging to $x$ such that
$|h(x_n)-h(y_n)|\geq \varepsilon$ for all $n\in \N$.
Let $a<b$ be such that $x_n, y_n\in [a,b]$ for all $n\in \N$. 
Since $f|_{[a,b]}$ is $UC$, there exists $\delta>0$ such that 
$|f(x)-f(y)|<\varepsilon$ for every $x,y\in [a,b]$ with $|x-y|<\delta$. 
It is easy to see that this implies essentially the same property for 
any truncation $g$ of $f$, i.e., that 
\begin{equation}\label{eq5:**}
|g(x)-g(y)|<\varepsilon
\end{equation}
for every truncation $g$ of $f$ and any
$x,y\in [a,b]$ with $|x-y|<\delta$. 

Now, choose $k\in \N$ such that $|x_k-y_k|<\delta$ and 
put $ M_0=\{y_k\}$. We claim that 
if $g\colon M\to h[M]$ is an $(\{x_k\},M_0)$-approximation 
of $h$ then $g$ is not $UC$. 
Indeed, if $g$ is an $(\{x_k\},M_0)$-approximation of $h$ then
$g(x_k)=h(x_k)$ and $\{g(y_k)\}=g[M]\sq h[M]=\{h(y_k)\}$. So 

\begin{equation}\label{eq6:***}
                       |g(x_k)-g(y_k)|=|h(x_k)-h(y_k)|\geq\varepsilon.
\end{equation}

On the other hand, if $g$ were $UC$, then it could be extended
to a $UC$ function $\hat g:\R_c\to \R$ which must be a truncation of 
$f$, since $\hat g[M]=g[M]\subseteq f[M]$.
But this and (\ref{eq6:***}) contradict (\ref{eq5:**}). 
This finishes the proof of (\ref{eq4:u_ext}).

Next we will show that for any $h\in C_{wua}(M,f[M])$
\begin{eqnarray}
h\in C_{ua}(M,f[M]) & \mbox{ if and only if } & \hat{h} 
\mbox{ is a truncation of $f$ with} \\ \label{eq:o}
& & 
\lim_{x\to\infty}\hat h(x)=\lim_{x\to-\infty}\hat h(x), \nonumber
\end{eqnarray}
%\begin{equation}\label{eq:o}
%h\in C_{ua}(M,f[M]) \mbox{ if and only if } \hat{h} 
%\mbox{ is a truncation of $f$ with }
%\lim_{x\to\infty}\hat h(x)=\lim_{x\to-\infty}\hat h(x),
%\end{equation}
where $\hat h\colon \R_c\to \R$ is the unique extension of $h$, 
existing by (\ref{eq4:u_ext}). 
Notice that by Lemma \ref{lem:SinFunc}(i) and (\ref{eq4:u_ext})
this will immediately imply $C_{uc}(M,f[M])=C_{ua}(M,f[M])$.

The implication ``$\Leftarrow$" follows immediately from 
Lemma \ref{lem:SinFunc}(ii). 
To prove the other implication take $h\in C_{ua}(M,f[M])$. 
Then $\hat{h}$ is a truncation of $f$ by the definition of $M$. 
It is also easy to see that the limits under question exist. 
%\neq C(X,Y)$ choose  $y_0,y_1\in h[M]$ such that $y_0<y_1<\pi/2$
%and let $a\in\R\setminus M$ be such that 
%Let $f\in C(X,Y)$ be such that $f(x)=y_0$ for $x\in M\cap(-\infty,x_0)$
%and $f(x)=y_1$ for $x\in M\cap(x_0,\infty)$.
%We will show that $f\not\in C_{wua}(X,Y)$. So, take $x\in M\cap(-\infty,x_0)$,  
%$m_0\in M\cap(x_0,\infty)$ and e $M_0=\{m_0\}$.
%It is enough to argue But if $g\in C_{uc}(X,Y)$ is such that
%$g(x_0)=f(x_0)=y_0<h(x_0)$, then , has all values less then 
Put $s_0=\lim_{x\to-\infty}\hat h(x)$ and $s_1=\lim_{x\to\infty}\hat h(x)$,
and  by way of contradiction assume that $s_0\ne s_1$. 
Let $\varepsilon=|s_0-s_1|/6$ and choose $a_0,a_1\in M$ such that 
$h(a_i)=\hat{h}(a_i)\in (s_i-\varepsilon, s_i+\varepsilon)$ for
$i=0,1$, $f[(-\infty,a_0]]\sq (-\pi/2-\varepsilon,-\pi/2+\varepsilon)$ and
$f[(a_1, \infty)]\sq (\pi/2-\varepsilon, \pi/2+\varepsilon)$.
We claim  that there is no $(\{a_0, a_1\},M)$-approximation 
$g\colon M\to f[M]$ of $h$ which is $UC$. 
To see it assume by way of contradiction that there exists such $g$. 
Then $g$ is $UC$ so
that $g$ has a $UC$ extension $\hat{g}\colon \R_c\to\R$ which must be a 
truncation of $f$ by the choice of $M$. Now 

\begin{equation}\label{eq:oo}
             |\hat g(a_0)-\hat g(a_1)|=|h(a_0)-h(a_1)|>4\varepsilon.
\end{equation}

Moreover, if $b_0=\lim_{x\to-\infty}\hat g(x)$ then either $b_0=h(a_0)$,
if $h(a_0)\not \in (-\pi/2-\varepsilon,-\pi/2+\varepsilon)$, or otherwise
$b_0\in (-\pi/2-\varepsilon,-\pi/2+\varepsilon)$. 
In any case, $|b_0-h(a_0)|\leq 2\varepsilon$. 
Similarly, we argue that $|b_1-h(a_1)|\leq 2\varepsilon$,
where  $b_1=\lim_{x\to\infty}\hat g(x)$.
Combining this with (\ref{eq:oo}) we obtain that
$b_0\ne b_1$. So, by Lemma \ref{lem:SinFunc}(i),  $\hat g$ is not $UC$.  This
contradiction finishes the proof of (\ref{eq:o}). 

To finish the proof it is enough to show that 

\begin{equation}\label{eq:ooo}
       f|_M\in C_{wua}(M,f[M])\setminus C_{ua}(M,f[M]).
\end{equation}

But in Example 2.9 we proved that $f$ is $WUA$, so Proposition \ref{prop:restr}
implies that  $f|_M\in C_{wua}(M,f[M])$. 
The fact that  $f|_M\not \in C_{ua}(M,f[M])$ 
follows immediately from (\ref{eq:o}). 
\end{description}

\begin{description}
\item[(v)] $C_{uc}(X,Y)\neq C_{ua}(X,Y)=C_{wua}(X,Y)=C(X,Y)$
holds for $X=Y=\R$ by Theorem \ref{thm:R-R}.
\end{description}

\begin{description}
\item[(vi)] $C_{uc}(X,Y)\neq C_{ua}(X,Y)=C_{wua}(X,Y)\neq C(X,Y)$
holds for $X$ being the Hedgehog with
continuum many spikes \cite[Example 4.1.3]{E} and $Y=\R$.
For the proof see \cite[Theorem 5.9]{BDP}.
(Actually it suffices to take just $b$ spikes, where $b$ is the smallest
cardinality of an unbounded family in $\omega^\omega$.)
\end{description}

\begin{problem}\label{problem:Separable}
Does there exist a separable metric space $X$ with the property that
$C_{uc}(X,\R)\neq C_{ua}(X,\R)=C_{wua}(X,\R)\neq C(X,\R)$?
\end{problem}


\begin{description}
\item[(vii)] $C_{uc}(X,Y)\neq C_{ua}(X,Y)\neq C_{wua}(X,Y) = C(X,Y)$
holds for $X=\R\setminus\{0\}$ and $Y=\R$.  
(See \cite[Corollary 7.2 and Lemma 3.1]{BD}.)
\end{description}

\begin{problem}\label{problem:Connected}
Does there exist a connected metric space $X$ with the property that
$C_{uc}(X,\R)\neq C_{ua}(X,\R)\neq C_{wua}(X,\R)=C(X,\R)$?
\end{problem}

\begin{description}
\item[(viii)] $C_{uc}(X,Y)\neq C_{ua}(X,Y)\neq C_{wua}(X,Y) \neq C(X,Y)$.
Holds for $Y=\R$ and $X=\R_c$. The relation 
$C_{wua}(\R_c,\R) \neq C(\R_c,\R)$ is witnessed by the identity function,
which is not $WUA$ by Corollary \ref{cor:magic} and Lemma
\ref{lem:SinFunc}(i).  
This and Proposition \ref{prop:ats}
imply $C_{uc}(\R_c,\R)\neq C_{ua}(\R_c,\R)$.
The relation $C_{ua}(\R_c,\R)\neq C_{wua}(\R_c,\R)$
is justified by the function $f$ from Example \ref{ex:SinFunc}.
It is shown there that $f\in C_{wua}(\R_c,\R)$.
The fact that $f\not\in C_{ua}(\R_c,\R)$ follows
from (\ref{eq:ooo}) and Proposition \ref{prop:restr}.
\end{description}

\section{Complex polynomials and harmonic functions}\label{harmonic}

In this section we will show that most complex
analytic functions $f\colon\C\to\C$ are not $WUA$. 
We start with the following easy example.

\begin{em-example}\label{Ex:zSq} 
The function $f\colon\C\to\C$ given by $f(z)=z^2$ is not $WUA$.
\end{em-example}

\begin{proof} 
If $f$ were $WUA$ than the real part of $f$, $\re f=\pr_x\circ f$, 
would be $WUA$, since $\pr_x\colon\R^2\to\R$, $\pr_x(x,y)=x$, is $UC$.
But $\re f(x+{\bf i}y)=x^2-y^2$ is not $WUA$. 
(See Corollary \ref{cor:crit}.)
\end{proof}

Notice that $f(z)=z^2$ is perfect. So, even the simplest
analytic perfect mappings do not have to be $WUA$. 

Clearly all linear functions from $\R^2$ into $\R^2$ are $UC$.
Hence all mappings $\C\to \C$, $z\mapsto a z + b$ $(a, b \in \C)$ are $UC$.
In fact, it seems that these are the only 
complex analytic functions that are $WUA$. 
We do not have a proof of this general statement.
However, the examples below should explain 
the reasons that stands behind this conjecture. 

In all the examples that follow we will prove 
that the complex-valued function $f(z)$ in question
is not $WUA$ by showing that its imaginary part 
$h(x,y)=\im f(x+\iu y)$ is not $WUA$. This is enough 
by Theorem \ref{thm:composition}.   
To prove that $h$ is not $WUA$ we will 
apply Proposition \ref{prop:restr}  
to an appropriate restriction $h_A$ of the function $h$ 
to a subspace $A$ of $\R^2$. 
The verification that $h_A$ is not $WUA$ will be done by
applying Proposition \ref{prop:crit}.

\begin{example}\label{two_functions}
\begin{description}
\item[(1)] The for $n>1$ the $n$-th power function $\C\to \C$, 
$z\mapsto z^n$, is not $WUA$.
\item[(2)] 
The exponential function  $\C\to \C$, $z\mapsto e^z$,  is not $WUA$.
\end{description}
\end{example}

\begin{proof} 
In case (1) 
we have $h(x,y)=(x^2+y^2)^{n/2}\sin [n\arctan(y/x)]$. 
Take as $A$ those points of the upper half-plain which have 
the argument in the interval $(0, \pi/n)$.
Then for every $c>0$ the level curve  
$L_c= h^{-1}(c)$ is connected and has two
asymptotes: 
the rays $\mbox{Arg}(z)=0$ and $\mbox{Arg}(z)=\pi/n$. 
%(See Figure 2.)
Thus the distance between $L_c$
and $L_{c'}$ is zero for $c, c'>0$. 
The application of Proposition \ref{prop:crit}
finishes the proof.

In case (2) we have $h(x,y)=e^x\sin y$. 
Take $A=\{z\in \C: \re z> 0\ \&\ 0< \im z< \pi/2\}$.
Then for $c>1$ the level curve 
$L_c= h^{-1}(c)=\{(\ln[c/\sin y],y)\colon 0<y<\pi/2\}$ 
is connected as a graph of continuous function
$x(y)=\ln[c/\sin y]$. 
%(See Figure 3.)
Moreover, all curves $\{L_c\}_{c>1}$
have as common asymptote the line 
$\im z= 0$. Thus the distance between $L_c$
and $L_{c'}$ is zero for $c, c'>1$. 
Now Proposition \ref{prop:crit} can be applied.
\end{proof}

In both cases above we used the fact that
the function $h(x,y)=\im f(x+\iu y)$ has
many level curves whose connected components are 
of distance zero. (This was done by noticing that they 
have the same asymptote.)
The same argument works also
for other elementary entire functions like
$\sin z$, $\cos z$ and 
complex polynomial functions of degree greater than $1$. 
In fact, by Liouville's theorem for positive harmonic functions
\cite[Theorem 3.1]{ABR},
%the Great Theorem of Picard easily implies that 
for every entire analytic function $f$ the level curves of $h=\im f$ 
are unbounded and it seems conceivable that
for any non-linear $f$ the function $h$ always has
many level curves whose connected components are 
of distance zero. This leads us to the following conjecture.

\begin{conjecture}\label{conj:analituc}
If $f\colon\C\to\C$ is entire analytic function then
$f$ is $WUA$ if and only if $f$ is a linear. 
\end{conjecture}

We will finish this discussion with one more example.

\begin{example}\label{third_function}
The inverse function 
$f\colon\C\setminus\{0\}\to\C$, $f(z)= 1/z$, is not $WUA$.
\end{example}

\begin{proof} 
Consider $A$ to be the first quadrant 
$\{(x,y)\colon x>0\ \&\ y>0\}$. 
Since $h(x,y)=\im f(x+\iu y)=-y/\sqrt{x^2+y^2}$ it is easy to see
that for $-1<c<0$ the level curve $L_c= h^{-1}(c)$ is 
the ray in $A$ of the line 
$y=-c(1-c^2)^{-1/2}x$. The distance between $L_c$
and $L_{c'}$ is clearly zero for $c,c'>0$.
So, Proposition \ref{prop:crit} can be applied.
\end{proof}

It is not difficult to extend the argument of 
Example \ref{third_function}
to all rational functions with essential poles. 
This suggests that Conjecture \ref{conj:analituc}
can be true also for all meromorphic functions. 

\medskip

We also do not know the answer for the following
natural question. (Compare with Example
\ref{ex:SinFunc}.)

\begin{question}
If $f\in C_{ua}(\C,\C)$ 
($f\in C_{wua}(\C,\C)$, respectively) and $g:\C\to \C$ is linear, 
is it always true that $f+g$ is
$UA$ ($WUA$, respectively)? \end{question}

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Rend. Ist. Matematico Univ.  di Trieste {\bf 25} (1993), 23--56.

\bibitem[BDP]{BDP} A. Berarducci,  D. Dikranjan and Jan Pelant, 
{\it Totally continuous functions},  Preprint. 

\bibitem[B]{B} M. Burke,  {\it Characterizing uniform continuity with
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\bibitem[DP]{DP} D. Dikranjan and J. Pelant, {\it The impact of
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\bibitem[E]{E} R. Engelking, {\it General Topology, Revised and
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\end{thebibliography}

\end{sloppypar}
\end{document}

