ThereAreContinuousFunctionsInTheInterval01WhichAreNotMonotonicAtAnySubinterval 2005-01-31 03:22:23 2007-09-22 23:05:19 Result Baire category theorem van der Waerden function % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them \usepackage{graphicx} % define commands here \newtheorem{prop}{Proposition} Let $f$ be a real-valued continuous function defined on the unit interval $[0,1]$. It seems intuitively clear that $f$ should be monotonic on some subinterval $I$ of $[0,1]$. Most of the concrete examples seem to support this. A counterexample is termed \emph{nowhere monotonic}, meaning that the function is not monotonic in any subinterval of $[0,1]$. Surprisingly, nowhere monotonic functions do exist: \begin{prop} There exists a real-valued continuous function defined on $[0,1]$ that is nowhere monotonic. \end{prop} A sketch of the proof goes as follows: \begin{enumerate} \item Let $X$ be the set of all continuous real-valued functions on [0,1]. Then $X$ is a complete metric space given the sup norm. Clearly $X$ is non-empty. \item Given any subinterval $I$ of $[0,1]$, the subset $P(I)\subseteq X$ consisting of all non-decreasing functions, the subset $Q(I)\subseteq X$ consisting of all non-increasing functions, and hence their union $M(I)$, are closed. \item Furthermore, $M(I)$ is nowhere dense. \item Let $S$ be the set of all rational intervals in $[0,1]$ (a rational interval is an interval whose endpoints are rational numbers). Then $S$ is countably infinite. Take the union $M$ of all $M(I)$, where $I$ ranges over $S$. \item If $f$ is monotone on some interval $J$ in $[0,1]$, then $f$ is monotone on some rational interval $I\subseteq J$. If the theorem is false, then every continuous function is monotone on some rational interval, which means $M=X$. \item However, $M$ is a countable union of nowhere dense sets and $X$ is a non-empty complete metric space. By Baire Category Theorem, this can not happen. Therefore, $M\subset X$ strictly and there exists a continuous nowhere monotone real-valued function defined on $[0,1]$. \end{enumerate} \subsubsection*{Example : van der Waerden function} The above shows the existence of such a function. Here is an actual example of a nowhere monotonic continuous function, called the {\bf van der Waerden function}. This function, which we designate by $f$, is given by a series \begin{displaymath} f(x) = \sum_{k=0}^{\infty} f_k(x) \end{displaymath} where the functions $f_k$ are defined by \begin{displaymath} f_0(x) = \begin{cases} x, & \text{if}\;\; 0 \leq x \leq \frac{1}{2} \\ -x +1,\quad & \text{if}\;\; \frac{1}{2} \leq x \leq 1 \end{cases} \quad\quad \text{and} \quad\quad f_k(x) = \frac{1}{2^k}f_0(2^kx) \end{displaymath} where each $f_k(x)$ is defined on $[0,2^{-k}]$. Since $f_k$ agrees on the endpoints, we can extend the its domain to the entire unit interval by periodic extension (so that the graph of $f_k(x)$ has the shape of a sawtooth). \begin{figure} \begin{centering} \includegraphics[scale=0.6]{graf1} \includegraphics[scale=0.6]{graf2} \includegraphics[scale=0.6]{graf4} \caption{The graphics of $f_0$ (left), $f_1$ (middle) and $f_2$ (right).} \end{centering} \end{figure} $\quad\;$ \begin{figure} \begin{centering} \includegraphics[scale=0.7]{graf3} \caption{The graphic of the van der Waerden function (the dashed lines are the graphics of each $f_k$).} \end{centering} \end{figure} - It is easy to check that each $f_k$ is continuous. Using the Weierstrass M-test we can also see that the series converges uniformly, and therefore conclude that $f$ itself is a continuous function (it is the uniform limit of continuous functions). - We now prove that $f$ is nowhere monotonic: The set $\{\frac{L}{2^k} : k \in \mathbb{N}, \;0< L < 2^k\}$ is dense in $[0,1]$. Given any interval $I\subset [0,1]$ we can then find a point of the form $\frac{L}{2^k}$ in its interior. It is easily seen that $f_j(\frac{L}{2^k})=0$ for $j \geq k$. For any integer $j>k$, consider the points $a_j :=\frac{2^{j-k}L-1}{2^j}$ and $b_j := \frac{2^{j-k}L+1}{2^j}$. The points $a_j$ (resp. $b_j$) are just the points on the left (resp. on the right) of $\frac{L}{2^k}$ when we divide the unit interval in segments of size $\frac{1}{2^j}$. A direct calculation would show that \begin{displaymath} \begin{cases} f_s(a_j)=0, & s\geq j \\ f_s(a_j)=\frac{1}{2^j}, & j> s \geq k \\ f_s(a_j)= f_s(\frac{L}{2^k}) \pm \frac{1}{2^j},\;\;\; & k > s \geq 0 \end{cases} \end{displaymath} and similarly for $b_j$. Evaluating $f$ in the points $a_j$ and $b_j$ we obtain \begin{eqnarray*} f(a_j) & = & \sum_{s =0}^{\infty}f_s(a_j) \\ & = & \sum_{s =0}^{j-1}f_s(a_j) \\ & = & \sum_{s =0}^{k-1}f_s(a_j) + \sum_{s =k}^{j-1}f_s(a_j) \\ & = & \sum_{s =0}^{k-1}f_s(\frac{L}{2^k}) + \sum_{s =0}^{k-1}(\pm \frac{1}{2^j}) + \frac{j-k}{2^j} \\ & = & f(\frac{L}{2^k}) + \sum_{s =0}^{k-1}(\pm \frac{1}{2^j}) + \frac{j-k}{2^j} \end{eqnarray*} and similarly for $f(b_j)$. The least value we can obtain is $\displaystyle f(a_j) = f(\frac{L}{2^k}) - \frac{k}{2^j} + \frac{j-k}{2^j}= f(\frac{L}{2^k})+ \frac{j-2k}{2^j}$, and even in this extreme case we can still choose $j$ large enough so that $a_j \in I$ and $j > 2k$. For this appropriate $j$ we see that $f(a_j)> f(\frac{L}{2^k})$, and similarly $f(b_j) > f(\frac{L}{2^k})$. Recall that $a_j < \frac{L}{2^k} < b_j$. We conclude that $f$ is not monotonic in $I$, and hence it is nowhere monotonic. \textbf{Remark}. The van der Waerden function turns out to be nowhere differentiable as well.