I know two definitions of the Cantor function $c: [0,1] \to [0,1]$.
$$ c(x) = \begin{cases} \sum_{n=1}^{\infty} \frac{a_{n}}{2^{n}}, \; x \in C\\ \sup_{y \leq x, \; y \in C} c(y), \; x \notin C \end{cases} $$ where $\{a_{n}\}$ is the ternary expansion of $x$.
I also know the recursive definition:
$$ c(x) = \lim_{n \to \infty} c_{n}(x) $$ where $$ c_{n+1}(x) = \begin{cases} \frac{1}{2} c_{n}(3x), & x\in [0, \frac{1}{3}); \\ \frac{1}{2}, & x\in[\frac{1}{3}, \frac{2}{3}];\\ \frac{1}{2}(1 + c_{n}(3x-2)), & x\in (\frac{2}{3}, 1]. \end{cases} $$ and $$ c_{0}(x) = x $$
I have seen and understood the proof that the recursive definition converges uniformly to some continuous function. This function is the Cantor function of course, but I cannot see how this is. Is there some easy way to demonstrate that this recursive definition of the function is the equivalent the function above it?
Consider the metric space of continuous non decreasing functions $f$ on $[0,1]\to\Bbb R$ so that $f(0)=0$, $f(1)=1$ with the maximum norm. Then this space is complete (as the space of continuous functions is complete and by pointwise limit the conditions on $f(0)$ and $f(1),$ as well as inequalities, carry on to limits).
The map $\Gamma$ where $$\Gamma(f)(x)=\begin{cases}\frac12 f(3x),&\text{ if }x\in\left[0,\frac13\right)\\\frac12,&\text{ if } x\in\left[\frac13,\frac23\right]\\\frac12(1+f(3x-2)),&\text{ if }x\in\left(\frac23,1\right]\end{cases}$$ (it is easy to see that this satisfies the above conditions) is a strict contraction, as $$ \|\Gamma f-\Gamma g\|_\infty \le \frac12\|f-g\|_\infty$$
Thus, by the Banach fixed point theorem, the function $c$ from the second definition is in fact the unique fixed point of $\Gamma$. Thus, we get $$ \Gamma(c) = c $$
So, if $a_0$ is the first ternary digit of $x$, then this shows us: $$ c(x) = \frac12\left(\frac{a_0}2+c(3x-a_0)\right) $$ if $a_0\neq 1$ or $c(x)\frac12$ if so. By iteration on $c(3x-a_i)$ this shows us that if $I$ is the first time that the terniary digit $a_I$ of $x$ is $1$ we have $$ c(x) = \sum_{i<I}\frac{a_i}{2^{i+1}} + \frac1{2^I}$$ and if this never happens (i.e. $x\in C$) we get by taking limits $$ c(x) = \sum_{i\in\Bbb N} \frac{a_i}{2^{i+1}}$$
Note that as $c$ is continous and non-decreasing we get if $x\not \in C$ that $\sup\limits_{y<x,y\in C} c(y) = c(x)$.
Thus we have proven that $c$ has the form of the first definition.