Exercise 6.2.12

Question

[Cantor Function]
  Review the construction of the Cantor set $C \subseteq[0,1]$ from Section 3.1. This exercise makes use of results and notation from this discussion.
  \begin{enumerate}[label=(\alph*)] 
  \item Define $f_{0}(x)=x$ for all $x \in[0,1]$. Now, let
    $$
    f_{1}(x)= \begin{cases}(3 / 2) x & 	ext { for } 0 \leq x \leq 1 / 3 \ 1 / 2 & 	ext { for } 1 / 3<x<2 / 3 \ (3 / 2) x-1 / 2 & 	ext { for } 2 / 3 \leq x \leq 1\end{cases}
    $$
    Sketch $f_{0}$ and $f_{1}$ over $[0,1]$ and observe that $f_{1}$ is continuous, increasing, and constant on the middle third $(1 / 3,2 / 3)=[0,1] \backslash C_{1}$.
  \item Construct $f_{2}$ by imitating this process of flattening out the middle third of each nonconstant segment of $f_{1}$. Specifically, let
    $$
    f_{2}(x)= \begin{cases}(1 / 2) f_{1}(3 x) & 	ext { for } 0 \leq x \leq 1 / 3 \ f_{1}(x) & 	ext { for } 1 / 3<x<2 / 3 \ (1 / 2) f_{1}(3 x-2)+1 / 2 & 	ext { for } 2 / 3 \leq x \leq 1\end{cases}
    $$
    If we continue this process, show that the resulting sequence $\left(f_{n}ight)$ converges uniformly on $[0,1]$.
  \item Let $f=\lim f_{n}$. Prove that $f$ is a continuous, increasing function on $[0,1]$ with $f(0)=0$ and $f(1)=1$ that satisfies $f^{\prime}(x)=0$ for all $x$ in the open set $[0,1] \backslash C$. Recall that the ``length'' of the Cantor set $C$ is 0 . Somehow, $f$ manages to increase from 0 to 1 while remaining constant on a set of ``length 1.''

\end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $f_0$ in red, $f_1$ in blue:

\begin{figure}[h]
    \centering
    \begin{tikzpicture}[scale=3]
      \draw[red](0,0) -- (1, 1);
      \draw[blue] (0, 0)   -- (1/3, 1/2) -- (2/3, 1/2) -- (1, 1);
    \end{tikzpicture}
  \end{figure}
  \item Define
   $$f_{n+1}(x) = \begin{cases}
       (1 / 2) f_n(3 x) & \text { for } 0 \leq x \leq 1 / 3 \
       f_n(x) & \text { for } 1 / 3<x<2 / 3 \
       (1 / 2) f_n(3 x-2)+1 / 2 & \text { for } 2 / 3 \leq x \leq 1
   \end{cases}$$
We will aim to use the Cauchy Criterion to show $f_n$ uniformly converges. I first prove through induction that $\left|f_n(x) - f_{n-1}(x)\right| < 1/2^n$. The base case of $n=1$ is trivial by the definitions of $f_1$ and $f_0$.

Now first assume $\left|f_n(x) - f_{n-1}(x)\right| < 1/2^n$; our goal is to show $\left|f_{n+1}(x) - f_{n}(x)\right| < 1/2^{n+1}$. Note this is obviously true over $(1/3,2/3)$ since $f_{n+1}(x) = f_n(x)$ in this interval. Consider $x \in [0, 1/3]$, then
$$\left|f_{n+1}(x) - f_n(x)\right| = \left|\frac{1}{2}f_n(3x) - \frac{1}{2}f_{n-1}(3x)\right| = \frac{1}{2} \left|f_n(3x) - f_{n-1}(3x)\right| < \frac{1}{2} \frac{1}{2^n} = \frac{1}{2^{n+1}}$$
The case for $[2/3, 1]$ is similar.

Given this, it should be clear that for $m \geq n$, $\left|f_m(x) - f_n(x)\right| < 1/2^n$. Since we can make $1/2^n$ arbitrarily small by increasing $n$, we can readily conclude that $f_n$ satisfies Theorem 6.2.5 (Cauchy Criterion for Uniform Convergence), and hence $f_n$ converges uniformly.

\item $f_n(0) = 0$ and $f_n(1) = 1$ can easily be proven with induction, and imply that each $f_n$ is continuous.  Since each $f_n$ is continuous, by the Continuous Limit Theorem $f$ is continuous. One of the sub-results from Exercise 6.2.10 was that all $f_n$ increasing implies $f$ is increasing, which can be applied here to show $f$ is increasing. Since $f_n(0)$ and $f_n(1)$ are constant with respect to $n$, $f(0) = 0$ and $f(1) = 1$.

Observe that for any $n$, $f_{n+1}(x) = f_n(x)$ for $x \in [0,1]\backslash C_n$. To prove this formally, we can use induction. By definition $f_{n+1} = f_n(x)$ over $(1/3, 2/3)$. Focusing on $[0, 1/3]$, we have
\begin{align*}
   &\frac{1}{2} f_n(3x) = f_n(x) \text{ over } [0,1/3] \backslash C_n\
   \Longleftrightarrow &\frac{1}{2} f_n(3x) = \frac{1}{2}f_{n-1}(3x) \text{ over } [0, 1/3] \backslash \left(C_{n-1} / 3\right)\
   \Longleftrightarrow & f_n(x) = f_{n-1}(x) \text{ over } [0,1] \backslash C_{n-1}\
\end{align*}
which is the inductive hypothesis. The case over $[2/3, 1]$ is similar.

Moreover, since $C_m \subseteq C_n$ for $m \geq n$, we can make the stronger statement $f(x) = f_{m}(x) = f_n(x)$ for $x \in [0,1]\backslash C_n$ and $m \geq n$.

Finally, we can show by induction that $f_n(x)$ is constant over $[0,1]\backslash C_n$, implying that $f$ is constant and $f'(x) = 0$ over $[0,1] \backslash C$.
   \end{enumerate}

Exercise 6.2.12

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Exercise 6.2.12

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