Cantor Function (Abbott's Definition) 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!!
 A: 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) $f$.
A: Disclaimer:

*

*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.

*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,

*

*$x$ 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.

*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.
