% Analytic functions are I-density continuous % LaTeX document Version sent 9/2/94 \documentstyle[12pt]{article} \pagestyle{myheadings} %%%%%%%%%%% MACROS %%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%% HYPHENATIONS \hyphenation{ap-prox-im-ate-ly} \hyphenation{dif-fer-ent-iab-le} \hyphenation{den-sity} \hyphenation{con-tin-uous} \hyphenation{func-tion} \hyphenation{dis-per-sion} \hyphenation{di-ver-gent} %%%%%%%%%%%%%%%%%%%%% SUBSETS OF THE REALS %sets with the Baire property \newcommand{\BaireSets}{\mbox{$\cal B$}} %Lebesgue measurable sets \newcommand{\Lebesgue}{\mbox{$\cal L$}} % Use blackboard bold for the normal real subsets \newfont{\amssymten}{msbm10 scaled\magstep1} \newfont{\amssymnine}{msbm9} % natural numbers \newcommand{\nats}{\mbox{{\amssymten N}}} \newcommand{\smallnats}{\mbox{{\amssymnine N}}} % rational numbers \newcommand{\rationals}{\mbox{{\amssymten Q}}} \newcommand{\smallrationals}{\mbox{{\amssymnine Q}}} % real numbers \newcommand{\reals}{\mbox{{\amssymten R}}} \newcommand{\smallreals}{\mbox{{\amssymnine R}}} %%%%%%%%%%%%%%%%%%%%% IDEALS %the ideal of first category sets \newcommand{\I}{\mbox{$\cal I$}} %the ideal of measure zero sets \newcommand{\N}{\mbox{$\cal N$}} %%%%%%%%%%%%%%%%%%% TOPOLOGIES \newcommand{\T}{\mbox{$\cal T$}} %ordinary topology on R \newcommand{\ordinarytop}{\mbox{$\T_{\cal O}$}} %ordinary density topology on R \newcommand{\densitytop}{\mbox{$\T_{\cal N}$}} %I-density topology \newcommand{\itopology}{\mbox{$\T_{\cal I}$}} %%%%%%%%%%%%%%%%%%%%% FUNCTION SPACES \newcommand{\C}{\mbox{$\cal C$}} %continuous functions \newcommand{\ordcont}{\mbox{$\C_{\cal OO}$}} %density continuous functions \newcommand{\denscont}{\mbox{$\C_{\cal NN}$}} %I-dens cont funct \newcommand{\idenscont}{\mbox{$\C_{\cal II}$}} % analytic functions \newcommand{\analytic}{\mbox{$\cal A$}} % C-infinity functions \newcommand{\cinfinity}{\mbox{$C^\infty$}} %%%%%%%%%%%%%%%%%%%%%% FUNCTION OPERATIONS %inverse of a function \newcommand{\inv}[1]{\mbox{$#1^{-1}$}} %reals to reals \newcommand{\RtoR}{\mbox{$\colon\reals\to\reals$}} %oscillation \newcommand{\osc}{\omega} %restriction \newcommand{\fP}{\mbox{$f|_P$}} %support \newcommand{\supp}[1]{\mbox{supp$#1$}} %%%%%%%%%%%%%%%%%%%%%% SET OPERATIONS %interior of a set \newcommand{\interior}[1]{\mbox{$\mbox{{\rm int}}\left(#1\right)$}} %closure of a set \newcommand{\closure}[1]{\mbox{$\mbox{\rm cl}\left(#1\right)$}} %complement of a set \newcommand{\complement}[1]{\mbox{$#1^c$}} % big union \newcommand{\Cup}{\bigcup} % big intersection \newcommand{\Cap}{\bigcap} % union over the natural numbers \newcommand{\UoverN}{\Cup_{n\in\smallnats}} % characteristic function \newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}} % distance between sets \newcommand{\dist}[2]{\mbox{dist$(#1,#2)$}} %measure of a set \newcommand{\measure}[1]{\mbox{$\mbox{m}(#1)$}} \newcommand{\innermeasure}[1]{\mbox{$\mbox{m}_i(#1)$}} \newcommand{\outermeasure}[1]{\mbox{$\mbox{m}_o(#1)$}} %density points \newcommand{\idensitypts}[1]{\mbox{$\Phi_{\cal I}(#1)$}} \newcommand{\deepidensitypts}[1]{\mbox{$\Phi_{\cal D}(#1)$}} \newcommand{\jdensitypts}[2]{\mbox{$\Phi_{\cal #1}(#2)$}} \newcommand{\densitypts}[1]{\mbox{$\Phi_{\cal N}(#1)$}} %%%%%%%%%%%%%%%%%%%%%% THEOREM-LIKE THINGS %Note the capitals to differentiate them from the commands. \newtheorem{Theorem}{Theorem} \newtheorem{Observation}[Theorem]{Observation} \newtheorem{Lemma}[Theorem]{Lemma} \newtheorem{Corollary}[Theorem]{Corollary} \newtheorem{Proposition}[Theorem]{Proposition} \newtheorem{Example}[Theorem]{Example} \newtheorem{Problem}[Theorem]{Problem} %define shorter commands for these environments \newcommand{\theorem}[2]{\begin{Theorem}\Label{#1}#2\end{Theorem}} \newcommand{\lemma}[2]{\begin{Lemma}\Label{#1}#2\end{Lemma}} \newcommand{\corollary}[2]{\begin{Corollary}\Label{#1}#2\end{Corollary}} \newcommand{\example}[2]{\begin{Example}\Label{#1}#2\end{Example}} \newcommand{\proposition}[2]{\begin{Proposition}\Label{#1}#2 \end{Proposition}} \newcommand{\observation}[2]{\begin{Observation}\Label{#1}#2 \end{Observation}} \newcommand{\problem}[2]{\begin{Problem}\Label{#1}#2\end{Problem}} %%%%%%%%%%%%%%%%%%%%%% LABEL MACRO %Macro to include theorem labels in the margins. \newcommand{\marginlabels}[1]{\def\margintags{#1}} \marginlabels{n} \def\nolabels{n} \newcommand{\Label}[1] {\if\margintags\nolabels \label{#1} \else \label{#1}\marginpar{{\tiny #1}} \fi} %%%%%%%%%%%%%%%%%%%%% MISCELLANEOUS \newcommand{\e}{\mbox{$\varepsilon$}} \renewcommand{\d}{\mbox{$\delta$}} \newcommand{\iapp}{\I-approximately} \newcommand{\idens}{\I-den\-sity} \newcommand{\idisp}{\I-dispersion} \hyphenation{-appro-xi-ma-te-ly} \hyphenation{-dif-fer-ent-iable} \newcommand{\sumi}[2]{\mbox{$\sum_{i=1}^{#1} #2$}} \newcommand{\Gdelta}{\mbox{$\mbox{\bf G}_\delta$}} \newcommand{\Y}{\mbox{$\Psi$}} \newcommand{\Lazarow}{{\L}azarow} %%%% End Macros \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 $\{ (d_n,d_m): n,m\in\nats,\ d_n0$ 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 $00$. 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