Cantor function: Difference between revisions

Revision as of 13:58, 12 May 2025

Revision as of 14:15, 12 May 2025

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

Below we define a sequence {''f''<submath>''n''(f_n)_n</submath>} of functions on the unit interval that converges to the Cantor function.

Let ''f''<sub>0</submath>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

Then, for every integer {{nowrap|''n'' &ge; 0}}, the next function ''f''_''n''+1(''x'') will be defined in terms of ''f''_''n''(''x'') as follows:

Let ''f''_''n''+1(''x'')&nbsp;= {{nowrap|1/2 &times; ''f''_''n''(3''x'')}},&nbsp; when {{nowrap|0 ≤ ''x'' ≤ 1/3&thinsp;}};

Let ''f''_''n''+1(''x'')&nbsp;= 1/2,&nbsp; when {{nowrap|1/3 ≤ ''x'' ≤ 2/3&thinsp;}};

Let ''f''_''n''+1(''x'')&nbsp;= {{nowrap|1/2 + 1/2 &times; ''f''_''n''(3&thinsp;''x'' &minus; 2)}},&nbsp; when {{nowrap|2/3 ≤ ''x'' ≤ 1}}.

The three definitions are compatible at the end-points 1/3 and 2/3, because ''f''_''n''(0)&nbsp;= 0 and ''f''_''n''(1)&nbsp;= 1 for every&nbsp;''n'', by induction. One may check that ''f''_''n'' converges pointwise to the Cantor function defined above. Furthermore, the convergence is uniform. Indeed, separating into three cases, according to the definition of ''f''_''n''+1, 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''&nbsp;&ge;&nbsp; \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>