Editing Cantor function (section)

=== Iterative construction ===
[[File:Cantor function sequence.png|250px|right]]

Below we define a sequence <math>(f_n)_n</math> of functions on the unit interval that converges to the Cantor function.

Let <math>f_0(x) = x</math>.

Then, for every integer <math>n \geq 0</math>, the next function <math>f_{n + 1}(x)</math> will be defined in terms of <math>f_n(x)</math> as follows:<math display="block">f_{n + 1}(x) = \begin{cases} \displaystyle \frac{1}{2} f_n(3 x) &\text{if } 0 \leq x \leq \frac{1}{3} \\ \displaystyle \frac{1}{2} &\text{if } \frac{1}{3} \leq x \leq \frac{2}{3} \\ \displaystyle \frac{1}{2} + \frac{1}{2} f_n(3 x - 2) &\text{if } \frac{2}{3} \leq x \leq 1 \end{cases}</math>The three definitions are compatible at the end-points <math>\tfrac{1}{3}</math> and <math>\tfrac{2}{3}</math>, because <math>f_n(0) = 0</math> and <math>f_n(1) = 1</math> for every <math>n</math>, by induction.  One may check that <math>(f_n)_n</math> converges pointwise to the Cantor function defined above.  Furthermore, the convergence is uniform.  Indeed, separating into three cases, according to the definition of <math>f_{n + 1}</math>, one sees that

:<math>\max_{x \in [0, 1]} |f_{n+1}(x) - f_n(x)| \le \frac 1 2 \, \max_{x \in [0, 1]} |f_{n}(x) - f_{n-1}(x)|, \quad n \ge 1.</math>

If <math>f</math> denotes the limit function, it follows that, for every <math>n \geq 0</math>,

:<math>\max_{x \in [0, 1]} |f(x) - f_n(x)| \le 2^{-n+1} \, \max_{x \in [0, 1]} |f_1(x) - f_0(x)|.</math>