% Analytic functions are I-density continuous
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\markright{Analytic functions are \I-density continuous}

\title{Analytic functions are \idens\ continuous} 
\author{Krzysztof Ciesielski \and Lee Larson}
\date{}

\begin{document} \maketitle

Let \densitytop\ stand for the density topology on the real 
line, \reals. A
function $f\RtoR$ is {\em density continuous} at the point $x$ 
if it is
continuous at $x$ when \densitytop\ is used on both the domain 
and the range.
The class of all everywhere density continuous functions is written as
\denscont. It is known that all locally convex functions are 
density
continuous, and it follows quite easily from this that all 
analytic functions are
in \denscont. But, there are \cinfinity\ functions which 
are not in \denscont\ 
\cite{CL:SpDensCont}.

W. Wilczy\'nski \cite{Wilczynski:GenDense} introduced 
the \idens\ topology on
\reals, which has many properties in common with the 
density topology, except
that it is based upon category instead of measure. 
(For its definition see
\cite{Wilczynski:GenDense} or \cite{PWW:CatAn}.) 
The \idens\ topology is denoted
here by \itopology. The \idens\ continuous functions, 
\idenscont, are those
functions $f\RtoR$ which are continuous when the 
domain and range are both given
the topology \itopology.

It is natural to ask if the known properties of the 
density continuous functions
can be proved in the case of the \idens\ continuous 
functions. It turns out
that some properties can and some cannot be proved. 
Theorem \ref{analytic},
given below, establishes that analytic functions are 
\idens\ continuous, but the
proof is necessarily different from the case of the 
density continuous functions
because we also exhibit in Example \ref{cor:Cinfinityconvex}, 
a convex and
\cinfinity\  function which is not \idens\ continuous.

The notation used here is fairly standard. The set of 
subsets of \reals\ with
the Baire property is written as \BaireSets. \I\ 
stands for the ideal of first
category subsets of \reals. \cinfinity\ is the set of 
all functions $f\RtoR$
which are infinitely differentiable at every point and 
\analytic\ stands for the
collection of all real analytic functions. A set $E$ 
is a {\em right interval
set} at a point $a\in\reals$, if $E=\UoverN [a_n,b_n]$ 
or $E=\UoverN (a_n,b_n)$
where $a_n\to a$ and $a_n>b_{n+1}>a_{n+1}$ for all 
$n\in\nats$. The definition
of a left interval set at $a$ is similar. The set
 $E$ is an interval set at $a$,
if it is the union of a right and left interval 
set at $a$. Any interval set at
$0$ is just called an interval set. 

An open set $S$ is said to be {\em regular}, 
if $S=\interior{\closure{S}}$. In
particular, it can be shown that for any 
$B\in\BaireSets$, there is a unique
regular open set, $\tilde{B}$ such that 
$B\triangle\tilde{B}\in\I$. This
observation is important below because 
it often enables us to replace an
arbitrary $B\in\itopology$ by $\tilde{B}$ 
without losing any generality in a
proof. 

We begin by stating 
several known results which are needed below. The
first is essentially the same as
\cite[Theorem 2]{Wilczynski:CatAn}.

\lemma{intervalratio}{
Let $\{c_n\}_{n\in\smallnats}$ be a decreasing sequence of positive
numbers converging to zero
and, for each $n\in\nats$, let $(a_n,b_n)$ be an open 
interval centered at 
$c_n$.
If 
\[
\lim_{n\to\infty}\frac{c_{n+1}}{c_n} = 0\,\,\,\,\,
\mbox{ and }\,\,\,\,\,
\lim_{n\to\infty}\frac{b_n - a_n}{c_n} = 0,
\]
then $0$ is an \idisp\ point of
\[
\bigcup_{n\in\smallnats} [a_n,b_n].
\]
}

\theorem{list}{Let $B$ be a regular open set. The following statements
are equivalent:
\begin{description}
\item[(i)] $0$ is an \idisp\ point of $B$.
\item[(ii)] For every increasing sequence
$\{t_k\}$ of positive numbers diverging to infinity there
exists a subsequence $\{t_{k_i}\}$ such that
\begin{equation}\label{tndef}
\limsup_{i\to\infty} t_{k_i}B\cap(-1,1)\in\I.
\end{equation}
\item[(iii)] For every increasing sequence
$\{t_k\}$ of positive numbers diverging to infinity and 
every nonempty interval $(a,b)\subset(-1,1)$ there
exists a nonempty subinterval $(c,d)\subset (a,b)$ and a
 subsequence
$\{t_{k_i}\}$ such that for every $i\in\nats$  
\[
(c,d)\cap t_{k_{i}} B = \emptyset. 
\]
\end{description}
}

Proof.
The fact that (i) and (ii) are equivalent is known 
\cite[Theorem 1]{PWW:CatAn}.

Assume that (ii) is true, but that there exists an 
interval $(a,b)\subset(-1,1)$
for which (iii) fails. Then every subinterval 
$(c,d)\subset(a,b)$ has the
property that $\{ k:(c,d)\cap t_k B = \emptyset\}$ is finite.
From this it is apparent that
$\limsup_i t_{k_i}B$ is a dense \Gdelta\ subset of 
$(a,b)\subset(-1,1)$ for 
every subsequence $\{t_{k_i}\}$ of $\{t_k\}$. 
This contradicts (\ref{tndef}), so (iii) must be true.

Finally, suppose that (iii) is true. 
Let $d_n$ be a countable dense subset of
$(-1,1)$ and suppose $I_n$ is a sequential 
representation of the set
$\{ (d_n,d_m): n,m\in\nats,\ d_n<d_m\}$.
Applying (iii), there must exist an interval 
$J_1\subset I_1$ and a subsequence
$\{t_{k_m^1}\}$ of $\{t_k\}$ so that 
$t_{k_m^1}B\cap J_1=\emptyset$ for all $m$.
Proceeding inductively, for each $i\in\nats$ 
there must exist an interval
$J_{i+1}\subset I_{i+1}$ and a subsequence 
$t_{k_m^{i+1}}$ of $t_{k_m^i}$ such
that $t_{k_m^{i+1}}B\cap J_{i+1}=\emptyset$ 
for each $m$.
Since $\{ d_n: n\in\nats\}$ is dense in 
$(-1,1)$ it is clear that
$\limsup_i t_{k_i^i}B\cap(-1,1)\in\I$, and (ii) follows.

\bigskip

The following theorem is a consequence of 
\cite[Corollary 1]{AversaWilczynski:Homeomorphisms}.

\theorem{lipschitz}{If $f\RtoR$ is monotone and 
satisfies the Lipschitz
condition 
\[
0<\alpha|b-a|<|f(b)-f(a)|<\beta|b-a|<\infty
\] 
for all distinct $a$ and $b$ in some
interval $I$, then $f$ is \idens\ continuous on $I$.}

The first order of business is to prove that 
$\analytic\subset\idenscont$. The
following two technical lemmas are needed for the proof.

\lemma{limitshrinking}{ Let 
$f,h\colon [0,+\infty)\to [0,+\infty)$ be
homeomorphisms such that
\[
\lim_{x \rightarrow 0^+} \frac{h^{-1}(x)}{f^{-1}(x)} = 1.
\]
Then for every $0<c<c^\prime <d^\prime <d$ 
there exists $\varepsilon_0 >0$
such that for every $\varepsilon\in (0,\varepsilon_0 )$,
\[
f\left((\varepsilon c^\prime, \varepsilon d^\prime) \right) \subset
h\left((\varepsilon c, \varepsilon d) \right).
\]
}
 
Proof. Since $c/c^\prime < 1$ and $d/d^\prime > 1$ we can find
$\delta_0 > 0$ such that for every $x\in (0,\delta_0)$
\begin{equation} \label{inequality1}
\frac{c}{c^\prime} < \frac{h^{-1}(x)}{f^{-1}(x)} < \frac{d}{d^\prime}.
\end{equation}
Using the continuity of
$\inv{f}$ at $0$ we can find $\varepsilon_0 >0$ such that
$f( (0,\varepsilon_0 d) )\subset (0,\delta_0)$.
 
Now let $\varepsilon\in (0,\varepsilon_0)$ and
$x\in f((\varepsilon c^\prime, \varepsilon d^\prime))\subset
f( (0,\varepsilon_0 d) )\subset (0,\delta_0)$. 
So, (\ref{inequality1}) holds
and $f^{-1}(x)\in (\varepsilon c^\prime, \varepsilon d^\prime)$;
i.e.,
\[
\varepsilon c^\prime < f^{-1}(x) < \varepsilon d^\prime.
\]
Multiplying the above inequality by (\ref{inequality1}), 
we obtain
\[
\varepsilon c < h^{-1}(x) < \varepsilon d,
\]
which implies 
$x\in h\left((\varepsilon c, \varepsilon d) \right)$.

\bigskip

\lemma{ratio}{If $f,h:[0,\infty)\to[0,\infty)$ are 
homeomorphisms satisfying
\begin{equation}\label{theRatio}
\lim_{x\to0^+}\frac{\inv{h}(x)}{\inv{f}(x)}=1,
\end{equation}
then $h$ is right \idens\ continuous at $0$ 
iff $f$ is right \idens\ continuous
at $0$.}

Proof. Without loss of generality we may assume 
that both functions are
increasing, as the decreasing case is essentially the same.
 
So assume that $h$ is right \idens\ continuous 
at $0$. It will be shown that $f$
is right \idens\ continuous at $0$. This will 
finish the proof, as the converse
implication follows by exchanging $f$ with $h$.
 
Let us choose $B\in\BaireSets$,  $0\not\in B$, 
which has $0$ as an \I-dispersion
point.  We will use Theorem \ref{list} to prove 
that $0$ is a  right
\I-dispersion point of $\inv{f}(B)$. 

First, notice that since $f$ and $h$ are both 
homeomorphisms, we may assume that
$B$ is a regular open set. Choose a divergent 
increasing sequence of positive
real numbers $\{t_k\}_{k\in\smallnats}$ and a 
nonempty interval $(a,b) \subset
(0,1)$. Since $0$ is a right \I-dispersion 
point  of $h^{-1}(B)$, there exists a
nonempty interval $(c,d) \subset (a,b)$ and a subsequence
$\{t_{k_p}\}_{p\in\smallnats}$ of 
$\{t_k\}_{k\in\smallnats}$ such that for every
$p\in\nats$  
\[
(c,d)\cap t_{k_p} h^{-1}(B) = \emptyset.
\]
But this last condition is equivalent to
\[
h\left( \left(\frac{1}{t_{k_p}} c,
\frac{1}{t_{k_p}} d \right) \right)
\cap B = \emptyset.
\]
Now let $0<c<c^\prime<d^\prime<d$. Then, 
by Lemma \ref{limitshrinking},
\[
f\left(\frac{1}{t_{k_p}} c^\prime,\frac{1}{t_{k_p}} 
d^\prime \right) \subset
h\left(\frac{1}{t_{k_p}} c,\frac{1}{t_{k_p}} d \right)
\]
for almost all $p\in\nats$. This implies that for 
almost all $p\in\nats$
\[
f\left(\left(\frac{1}{t_{k_p}} c^\prime,\frac{1}{t_{k_p}} 
d^\prime\right)\right)
\cap B = \emptyset,
\]
or
\[
(c^\prime,d^\prime)\cap t_{k_p} f^{-1}(B) = \emptyset.
\]
This finishes the proof of Theorem \ref{ratio}.

\bigskip

The following theorem, which is interesting in its own right, 
is also needed in
what follows. Its analogue for ordinary density 
continuity is also known to be
true \cite{CL:SpDensCont}.


\theorem{xtoalpha}{For any $\alpha\in\reals$, the 
function $f(x)=x^\alpha$ is
\idens\ continuous on its domain.}

Proof.
If $x\ne 0$ and $f(x)$ exists, then it is clear 
that on a neighborhood of $x$,
$f$ satisfies the conditions of 
Theorem \ref{lipschitz}, so $f$ is \idens\
continuous at $x$.

Suppose $x=0$ and $\alpha>0$. It suffices 
to show $f$ is right \idens\
continuous at $0$. Let $B\in\BaireSets$ 
such that $0$ is an \idisp\ point of
$B$. It must be shown that $0$ is a right 
\idisp\ point of $\inv{f}(B)$.

To do this, first note that $f$ is a 
homeomorphism on $(0,\infty)$, so
$\inv{f}(S)\in\I$ whenever $S\in\I$ and 
there is no generality lost with the
assumption that $B$ is a regular open set. 
Choose any nonempty interval
$(a,b)\subset(0,1)$ and an increasing sequence
$\{s_k\}_{k\in\smallnats}$
of positive numbers diverging to infinity. 
Let $(a',b')=f((a,b))$ and define the
increasing sequence
\[
t_k=\frac{1}{f(1/s_k)}\to\infty.
\]
Using Theorem \ref{list}, there exists an 
interval $(c',d')\subset(a',b')$ and
a subsequence $\{t_{k_i}\}$ of $\{t_k\}$ such that
\[
(c',d')\cap t_{k_i}B=\emptyset\ \ 
\mbox{ for all } i\in\nats.
\]
Suppose that $(c,d)=\inv{f}((c',d'))$. 
Then a straightforward calculation shows
\begin{eqnarray*}
\emptyset & = & \inv{f}\left( (c',d')\cap t_{k_i}B \right)\\
&=& (c,d)\cap\inv{f}\left(\frac{1}{f(1/s_{k_i})}B\right)\\
&=& (c,d)\cap\left( s_{k_i}^{-\alpha}B \right)^{-1/\alpha}\\
&=& (c,d)\cap s_{k_i}(B)^{-1/\alpha}\\
&=& (c,d)\cap s_{k_i}\inv{f}(B).
\end{eqnarray*}
From Theorem \ref{list}, we see that $0$ is a 
right \idisp\ point of
$\inv{f}(B)$, and the theorem follows.

\theorem{analytic}{$\analytic\subset\idenscont$.}

Proof. Let $h\in\cal A$. It is enough to prove that
$h$ is \idens\ continuous at $0$. 
We prove that $h$ is right \idens\ 
continuous at $0$. 
The left-hand argument is similar.
 
Let $h(x)=\sum_{n=0}^{\infty} a_n x^n$.
We can assume that $a_0=0$. Since the \idens\ 
topology is closed under
homothetic transformations of its open sets, 
we can also assume that for
$i=\min\{n\colon a_n \neq 0\}$ we have $a_i = 1$. 
Now let $f(x)=x^i$.  Because
$h$ is analytic, 
$h^{-1}$ exists on some right neighborhood of $0$.
Let us assume that $h^{-1}$ is positive on this 
neighborhood, the other
case being similar. Then
\begin{eqnarray*}
1=\lim_{x\rightarrow 0^+} \frac{h(x)}{x^i} 
&=&
\lim_{x\rightarrow 0^+} 
\frac{h(h^{-1}(x))}{(h^{-1}(x))^i}\\
&=&
\lim_{x\rightarrow 0^+} \left( 
\frac{x^\frac{1}{i}}{h^{-1}(x)}\right)^i\\
&=&
\left( \lim_{x\rightarrow 0^+} 
\frac{f^{-1}(x)}{h^{-1}(x)}\right)^i.
\end{eqnarray*}
Hence,
\[
\lim_{x\rightarrow 0^+} \frac{h^{-1}(x)}{f^{-1}(x)}=1
\]
and, by Theorems \ref{xtoalpha} and \ref{ratio}, $h$ is
\idens\ continuous at $0$.

\bigskip

After seeing that $\analytic\subset\idenscont$, 
it is natural to ask whether
the same can be claimed for \cinfinity. 
This turns out not to be true. The
lemma and theorem given below are used 
to establish this fact.

\bigskip

\lemma{lem:convergence}{ Let $f\in\cal C^{\infty}$ 
be such that for every
$n\geq 0$
\[
f^{(n)}(0)=0 \,\,\, \mbox{ and } \,\,\, 
f^{(n)}( (0,\varepsilon_n)) \subset
(0,\infty), \,\,\, \mbox{ for some } \,\, 
\varepsilon_n > 0.
\]
Then
\[
\lim_{x \rightarrow 0^{+}} \frac{f(ax)}{f(x)} = 0,
\]
for every $a\in (0,1)$.
}
 
Proof. Let $a\in (0,1)$ and $n\in\nats$. 
Moreover, let us choose
$\varepsilon > 0$ such that 
$0<\varepsilon < \varepsilon_k$ for every
$k \leq n+1$. In particular, $f^{(n)}$ is 
increasing on $(0, \varepsilon)$, and
so
\[
\left| \frac{f^{(n)}(a\xi)}{f^{(n)}(\xi)} 
\right| < 1 \,\,\,\, \mbox{ for every
}
\,\, \xi\in (0,\varepsilon).
\]
Now let $x\in (0,\varepsilon)$ and let 
$g(x)=f(ax)$. Using Cauchy's
Theorem
$n$-times we can find $\xi\in (0,x)$ such that
\[
\left| \frac{f(ax)}{f(x)} \right| =
\left| \frac{g(x)}{f(x)} \right| =
\left| \frac{g^{(n)}(\xi)}{f^{(n)}(\xi)} \right| =
\left| a^n \right| \left| 
\frac{f^{(n)}(a\xi)}{f^{(n)}(\xi)} \right| < a^n.
\]
Thus,
\[ 
\lim_{x \rightarrow 0^{+}} \frac{f(ax)}{f(x)} = 0.
\]

\bigskip

\theorem{th:Cinfinityex}{ Let $f\in\cal C^{\infty}$ 
be such that for every
$n\geq 0$
\[
f^{(n)}(0)=0 \,\,\, \mbox{ and } \,\,\, 
f^{(n)}( (0,\varepsilon_n)) \subset
(0,\infty) \,\,\, \mbox{ for some } \,\, \varepsilon_n > 0.
\]
Then $f$ is not \I-density
continuous.
}

Proof.  
We start with a proof that $f$ is not right
\idens\ continuous at $0$.
Let $D_n=\{\frac{i}{2^n}\colon i=1,2,\ldots,2^n\}$ 
for $n\in\nats$.
First notice that if a sequence $\{n_k\}_{k\in\smallnats}$
is such that
\begin{equation}\label{cond1}
n_{k+1} > 2^k n_k \,\,\, \mbox{ for every } \, k\in\nats,
\end{equation}
then
\[
\min \frac{1}{n_k} D_k = \frac{1}{n_k}\frac{1}{2^k} > 
\frac{1}{n_{k+1}} =
\max \frac{1}{n_{k+1}} D_{k+1}.
\]
This means that if $\{s_i\}_{i>1}$ is a 
decreasing ordering of
$D=\bigcup_{k\in\smallnats} \frac{1}{n_k} D_k$, then
\[
\frac{1}{n_k} D_k = \{s_i\colon 2^k \leq i < 2^{k+1} \}.
\]
We also define a sequence $\{n_k\}_{k\in\smallnats}$ 
by induction on $k$ such
that it will satisfy condition (\ref{cond1}) 
and for every $k>0$
\begin{equation}\label{indcond}
\frac{f(s_i)}{f(s_{i-1})} \leq \frac{1}{k} \,\,\, 
\mbox{ for } \,\,
2^k \leq i<2^{k+1}.
\end{equation}
Put $n_1=1$ and assume that $n_{k-1}$ has already 
been choosen for some
$k>1$. Choose $n_k > 2^{k-1} n_{k-1}$ such that
\[
\frac{f(\frac{2^k -1}{2^k}x)}{f(x)} < \frac{1}{k}, \,\,\,
\mbox{ for all } \,\, x\in (0,\frac{1}{n_k}).
\]
Such a choice is possible by Lemma \ref{lem:convergence}. 
Then, the above
condition obviously implies condition (\ref{indcond}) 
for $2^k < i<2^{k+1}$.
Increasing $n_k$, if necessary, we can also obtain 
condition (\ref{indcond})
for $i=2^k$. This finishes the construction of $D$.
 
Now let $\{(a_n,b_n)\}_{n\in\smallnats}$ be a 
sequence of pairwise disjoint intervals
such that every interval $(a_n,b_n)$ is centered at 
$c_n=f(s_n)$ and that
\[
\lim_{n\to\infty}\frac{b_n - a_n}{c_n} = 0.
\]
By (\ref{indcond}),
\[
\lim_{n\to\infty}\frac{c_{n+1}}{c_n} = 0
\]
so, by Lemma \ref{intervalratio}, $0$ is an 
\I-dispersion point of the interval set
\[
E=\bigcup_{n\in\smallnats} (a_n,b_n).
\]
On the other hand, we notice that
for every subsequence 
$\{n_{k_i}\}_{i\in\smallnats}$ of
$\{n_k\}_{k\in\smallnats}$, the set
\[
\bigcup_{i\in\smallnats} n_{k_i} f^{-1}(E) 
\supset \bigcup_{i\in\smallnats} D_{k_i}
\]
is dense and open in $[0,1]$. So, $0$ is 
not a right \I-dispersion
point of
$f^{-1}(E)$ and $f$ is not \idens\ continuous at $0$.
 
\bigskip

\example{cor:Cinfinityconvex}{ There exists a 
convex \cinfinity\
function that is not \idens\ continuous. 
}
 
Proof. Define $g\colon (-\infty, 0.5)\to\reals$ by
\[
g(x)= \left\{
\begin{array}{lll}
e^{-x^{-2}} & x\in(0,1/2)     \\
0           & x\in(-\infty,0] \\
\end{array}\right.
\]
Examining the second derivative of $g$ 
it is easy to see that $g$ is
convex on $(-\infty,1/2)$.
It is well-known that $f\in\cinfinity$ 
and that $f^{(n)}(0)=0$ for all $n$.
Repeated differentiation of $f$ makes 
it apparent that for each $n$ there is
an $\e_n>0$ such that $f^{(n)}(x)>0$ 
whenever $0<x<\e_n$. Now an application of
Theorem \ref{th:Cinfinityex} finishes the argument.

\bigskip

It is also not difficult to see that the 
function described in Theorem
\ref{th:Cinfinityex} does not preserve 
\idens\ points. In particular, the
function $g$ from Example \ref{cor:Cinfinityconvex} 
does not preserve \idens\ points.

%\markboth{Bibliography}{Bibliography}
%\bibliographystyle{plain}
%\bibliography{analytic}


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\end{thebibliography}


\end{document}
