Advanced Calculus May 23, 2011 Cantor-Lebesgue function (uniformly continuous, not Lipschitz)
Construction of Cantor set. Let I = [0, 1]. The first stage of the construction is to subdivide [0, 1]
into thirds and remove the interior of the middle third; that is remove the open interval (13,23). Each successive step of the construction is essentially the same. Thus, at the second stage, we subdivide each of the remaining two intervals [0,13] and [23,1] into thirds and remove the interiors, (19,29) and (79,89), of the middle thirds. We continue the construction for each of the remaining intervals. The sets removed in the first three successive stages are indicated below by darkened intervals:
stage 1
1 3
2
0 3 1
stage 2
1 3
2 3 1
9 2 9
7 9
8
0 9 1
stage 3
For each k = 1, 2, . . . , let Ckdenote the union of the interval left at the kth stage. The set C =
∞
\
k=1
Ck is called the Cantor set. Note that each Ck consists of 2k closed interval, each of length 3−k, and that C contains the endpoints of all these intervals. Any point of C belongs to an interval in Ck for every k and is therefore a limit point of the endpoints of the intervals, e.g. 23 = 2 33kk−1 =
k→∞lim
2 3k−1+1
3k and note that 2 3k3−1k+1 is an endpoint of an interval in Ck. Let Dk = I \ Ck. Then Dk
consists of the 2k − 1 open intervals Ijk (ordered from left to right) removed in the first k stages of construction of the Cantor set. Let fk be the continuous function on [0, 1] defined by fk(x) =
0 if x = 0
1 if x = 1
j2−k if x ∈ Ijk, j = 1, . . . , 2k− 1, linear on each interval of Ck
Note that
(1) each fk is monotone increasing,
(2) {Ijk}2j=1k−1 ⊆ {Iml }2m=1l−1 whenever k ≤ l, we have fk+1 = fk on Ijk, j = 1, . . . , 2k− 1, (3) |fk− fk+1| < 2−k for all k.
Hence, P(fk− fk+1) converges uniformly on [0, 1], and therefore, {fk} converges uniformly on [0, 1].
Let f = lim
k→∞fk. Then f (0) = 0, f (1) = 1, f is monotone increasing and (uniformly) continuous on [0, 1], and f is constant on every interval removed in constructing C. (Alternative argument: By (3) and the Azel`a-Ascoli Theorem, the sequence {fk} is a Cauchy sequence and it converges uniformly to a continuous function f.) This f is called the Cantor-Lebesgue function.
Note that
(1) 23 = 2 33kk−1, 2 3k3−1k+1 ∈ Ck for all k,
(2) f (2 33kk−1) = f (23) = 12, and f (2 3k3−1k+1) = 12 + 21k. Thus, we have
f(2 3k−1+1 3k )−f (2
3) 2 3k−1+1
3k −2 3
=
1 2+1
2k−1 2 2 3k−1+1
3k −2 3k−1 3k
= 32kk → ∞ as k goes to ∞ which implies that f is Not Lipschitz on [0, 1].
Advanced Calculus May 23, 2011
0 1
3
2 3 1
9 2 9
7 9
8
9 1
1 2
1 4 3 4
1
f1 (dotted)
f2
