Cantor Function (Abbott's Definition)

Question

In Exercise 6.2.13 page 162 Abbott define Cantor Function.

He defines it as follows:

$f_0(x)=x $ for all $ x \in [0,1]$,

$f_1(x)=(3/2)x$, for $ x \in [0,1/3]$,

$f_1(x)=1/2$, for $x \in (1/3,2/3)$,

$f_1(x)=(3/2)x-1/2$, for $x \in [2/3,1]$.

Then defines

$f_n(x)=1/2f_{n-1}(3x)$, for $ x \in [0,1/3]$,

$f_n(x)=f_{n-1}(x)$, for $x \in (1/3,2/3)$,

$f_n(x)=1/2f_{n-1}(3x-2)+1/2$, for $x \in [2/3,1]$.

I have proved that for every $x \in [0,1] \ (f_n(x))$ is increasing and $|f_1(x)-f_n(x)|<1/2$ for every n and for every x in our domain.

Then in order to prove uniform convergence I have to prove $|f_m(x)-f_n(x)|<1/2^m$ for $m<n$ AND THIS IS THE POINT THAT I HAVE STUCK

Please any help. THANKS!!

Answer 1

That wont work to prove its convergent, you need it to go to $0$, not just being smaller than $1/2$. We want to prove $||f_m - f_n||$ is a cauchy sequence under the uniform bound, thus it converges to some $f$ and it is continuous. Here is my attemp at it:

Proof:

First, we prove that $$||f_n - f_{n-1}||_{\infty} = \max_{x \in [0,1]} |f_n(x) - f_{n-1}(x)| < \dfrac{1}{2^n} $$

We use induction (I'll leave the base case for you), suppose its valid for $k = 1, 2, \dots , n$, that is $$ \, \max_{x \in [0, 1]} |f_{n}(x) - f_{n-1}(x)| \le \frac 1 2 \max_{x \in [0, 1]} |f_{n-1}(x) - f_{n-2}(x)|$$

Now if $x \in [0,\frac 1 3]$

$$\max_{x \in [0, \frac 1 3 ]} |f_{n+1}(x) - f_{n}(x)| = \frac 1 2 \max_{x \in [0, \frac 1 3]} |f_{n}(3x) - f_{n-1}(3x)| $$ $$ \leq \frac 1 4 \max_{x \in [0, \frac 1 3]} |f_{n-1}(3x) - f_{n-2}(3x)| = \frac 1 2 \max_{x \in [0, \frac 1 3]} |f_{n}(x) - f_{n-1}(x)| $$

Edit I'll leave the other two cases to you, the second one should be straightforward and the third one has a similar line of thought ($ x \in [1/3, 2/3]$ and $x \in [2/3,1]$)

From this we deduce that $$||f_{n+1} - f_{n}||_{\infty} \leq \dfrac{1}{2^n}$$

Now, take WLOG, suppose $m > n$, then using the triangle inequality $$||f_m - f_n||_{\infty} \leq ||f_m - f_{m-1}||_{\infty} + ||f_{m-1} - f_{m-2}||_{\infty} + \dots + ||f_{n+1} - f_n||_{\infty}$$ $$ \leq \sum_{k = n}^{m-1} \dfrac{1}{2^k} $$

Because this last serie is convergent, we have that $$\lim_{n,m \to \infty} \sum_{k = n}^{m-1} \dfrac{1}{2^k} = 0$$

Thus $f_n$ is a Cauchy sequence under the $|| \cdot ||_{\infty}$ norm, and this implies that it converges uniformly to some (continuous) $f$.

Answer 2

Disclaimer:

1. I feel that the proof is somehow the same as the mostly upvoted one. However, the jargons I adopted are completely different. In other words, if you have only studied real analysis from Abbott's Understanding Analysis, then you will most likely understand my elaboration.
2. Please feel free to correct my mistakes. I would be more than grateful.

Notice that $ f_n(x)= \begin{cases} \frac{1}{2}f_{n-1}(3x),& \text{if } x\in [0,\frac{1}{3}]\\ f_{n-1}(x), & \text{if }x\in(\frac{1}{3},\frac{2}{3})\\ \frac{1}{2}f_{n-1}(3x-2)+\frac{1}{2},&\text{ otherwise} \end{cases} $

Notice that,

1. $x$ will be continuously mapped to different points. Yet it is always confined within unit interval. Since I am dumb, it took me 2 hours to realize this.
2. The above definition is derived implicitly from Abbott's definition. Again, it took me a while to realize recursive definition is the best. This may not happen if you have a solid understanding.

Now, $\forall m,n\in\mathbb{N}\exists z\in[0,1]$

\begin{align*} |f_m(x)-f_n(x)|\leq\frac{|f_{m-1}(z)-f_{n-1}(z)|}{2} \end{align*}

There are 3 cases, Case: $x\in[0,\frac{1}{3}]$

\begin{align*} |f_m(x)-f_n(x)|&=\frac{|f_{m-1}(3x)-f_{n-1}(3x)|}{2}\\ &\leq\frac{|f_{m-1}(z)-f_{n-1}(z)|}{2} \end{align*}

Case: $x\in(\frac{1}{3},\frac{2}{3})$

\begin{align*} |f_m(x)-f_n(x)|&=|f_{m-1}(x)-f_{n-1}(x)|\\ &=|f_1(x)-f_1(x)|\\ &=0\\ &\leq\frac{|f_{m-1}(z)-f_{n-1}(z)|}{2} \end{align*}

Case: Otherwise

\begin{align*} |f_m(x)-f_n(x)|&=|\frac{1}{2}f_{m-1}(3x-2)+\frac{1}{2}-\frac{1}{2}f_{n-1}(3x-2)-\frac{1}{2}|\\ &=|\frac{1}{2}f_{m-1}(3x-2)-\frac{1}{2}f_{n-1}(3x-2)|\\ &\leq\frac{|f_{m-1}(z)-f_{n-1}(z)|}{2} \end{align*}

Hence, $\forall\epsilon>0\exists n_0-1>\frac{1}{\epsilon}$.

\begin{align*} \forall\epsilon>0\forall m\geq n\geq n_0 |f_m(x)-f_n(x)|&\leq\frac{|f_{m-1}(y)-f_{n-1}(y)|}{2},y\in[0,1]\text{, $y=3x$}\\ &\leq\frac{|f_{m-(n-1)}(z)-f_1(z)|}{2^{n-1}},z\in[0,1]\text{, z is mapped}\\ &\leq\frac{1}{2^{n-1}}\\ &\leq\frac{1}{n_0-1}\\ &<\epsilon \end{align*}

By cauchy criterion for uniform convergence, cantor function is uniformly convergent. This completes the proof.