Absolute continuity vs total singularity {Part II{ Absolute continuity vs total singularity {Part II{ E. Arthur (Robbie) Robinson The George Washington University February 16, 2012 Absolute continuity vs total singularity {Part II{ Outline 1 Introduction 2 A \converse" to Villani's theorem 3 An easy TS example 4 Raphael Salem's examples 5 The Hartman-Kreshner approach 6 Minkowski's ?(x) and Conway's (x) 7 Spectral measures 8 References Absolute continuity vs total singularity {Part II{ Introduction 1 Introduction 2 A \converse" to Villani's theorem 3 An easy TS example 4 Raphael Salem's examples 5 The Hartman-Kreshner approach 6 Minkowski's ?(x) and Conway's (x) 7 Spectral measures 8 References Absolute continuity vs total singularity {Part II{ Introduction Review In the first lecture I proved the following theorem: Lemma (Soichi Kakeya, 1924 ) If f : [0; 1] ! R is continuous, strictly monotone and satisfies jf 0(x)j ≥ a > 0 a.e., then f −1(x) is absolutely continuous, and j(f −1)0(x)j ≤ 1=a a.e.. The proof is based on the following result. Lemma (A. Villani, 1985) −1 Let ' :[a; b] ! R be continuous, strictly monotone. Then ' (x) is absolutely continuous if and only if j'0(x)j > 0 a.e.. Villani says he believes this lemma is known (clearly it was known to Kakeya) but he could not find it in the literature. Absolute continuity vs total singularity {Part II{ Introduction Today Definition We call an continuous increasing function f :[a; b] ! R singular (continuous) if f 0(x) = 0 λ-a.e.. We call such an f totally singular (TS) if it is strictly increasing. The best known example of a singular continuous function is the Cantor function (we review it below). However, it is not strictly increasing, so it is not totally singular. Examples of totally singular functions abound in the literature. The best known example is probably Minkowski's \question mark function" ?(x) (we discuss it later). However, these examples are not so widely known. Absolute continuity vs total singularity {Part II{ A \converse" to Villani's theorem 1 Introduction 2 A \converse" to Villani's theorem 3 An easy TS example 4 Raphael Salem's examples 5 The Hartman-Kreshner approach 6 Minkowski's ?(x) and Conway's (x) 7 Spectral measures 8 References Absolute continuity vs total singularity {Part II{ A \converse" to Villani's theorem Converse The following result seems to be well known (many authors state it as a fact and it is often attributed to Lebesgue). Theorem (Lebesgue) If f :[a; b] ! R is totally singular (i.e., continuous and strictly increasing with f 0(x) = 0 λ-a.e.), then so is f −1. The proof uses a lemma from Natanson, related to one from the first lecture. Lemma (Natanson) 0 Let ' :[c; d] ! R and let ' (x) ≥ η > 0 for x 2 E ⊆ [c; d]. Then λ∗('(E)) ≥ ηλ∗(E); where λ∗ denotes Lebesgue outer measure. Absolute continuity vs total singularity {Part II{ A \converse" to Villani's theorem Converse The following result seems to be well known (many authors state it as a fact and it is often attributed to Lebesgue). Theorem (Lebesgue) If f :[a; b] ! R is totally singular (i.e., continuous and strictly increasing with f 0(x) = 0 λ-a.e.), then so is f −1. The proof uses a lemma from Natanson, related to one from the first lecture. Lemma (Natanson) 0 Let ' :[c; d] ! R and let ' (x) ≥ η > 0 for x 2 E ⊆ [c; d]. Then λ∗('(E)) ≥ ηλ∗(E); where λ∗ denotes Lebesgue outer measure. Absolute continuity vs total singularity {Part II{ A \converse" to Villani's theorem Proof of converse lemma −1 Let ' = f and note that ' :[f(a); f(b)] ! R is continuous and strictly increasing. Suppose λ(F ) > 0 where F = fy : '0(y) > 0g. There exists γ > η > 0 and E ⊆ F (measurable) so that λ(E) > η and η ≤ '0(x) ≤ γ for x 2 E. Note that '(E) = f −1(E) is measurable. By Natanson's lemma, λ('(E)) = λ∗('(E)) ≥ ηλ∗(E) = ηλ(E) > η2 > 0: But for (λ-a.e.) x 2 '(E), 1 1 f 0(x) = ≥ > 0: '0(f(x)) γ Absolute continuity vs total singularity {Part II{ An easy TS example 1 Introduction 2 A \converse" to Villani's theorem 3 An easy TS example 4 Raphael Salem's examples 5 The Hartman-Kreshner approach 6 Minkowski's ?(x) and Conway's (x) 7 Spectral measures 8 References Absolute continuity vs total singularity {Part II{ An easy TS example Definition of the Cantor function For x 2 [0; 1] write x = :d1d2d3 ::: in base 2 (so di 2 f0; 1g) and x = :c1c2c3 ::: in base 3 (so di 2 f0; 1; 2g). The Cantor set C ⊆ [0; 1] is the set of all x = :c1c2c3 ::: so that ci 6= 18i. Define the Cantor function g(:c1c2c3c3 ::: ) = :d1d2d3 ::: as follows: If ci 6= 18i then i0 = 1, else let i0 be the first i so that ci = 1. Then for i < i0, di = 0 if ci = 0 and di = 1 if ci = 2. And for all i ≥ i0 put di = 0. Absolute continuity vs total singularity {Part II{ An easy TS example Graph of Cantor function (c 1883) The Cantor functiong : [0; 1] ! [0; 1] is an increasing continuous surjection with g0(x) = 0 a.e. (singular). It is not strictly increasing (not 1:1) so not TS. (Source: http://mathworld.wolfram.com/CantorFunction.html) Absolute continuity vs total singularity {Part II{ An easy TS example Brown's example This is related to an example I recall from grad school. It is easier to state than mine, but harder to visualize. Let g be the Cantor function. For an interval I = [a; b] ⊆ [0; 1] define gI (x) = 0 if x < a, gI (x) = 1 if x > b, and x − a g (x) = g if x 2 [a; b]. I b − a Note that I is continuous, increasing and singular. Let I1;I2;I3;::: be an enumeration of the dyadic intervals in [0; 1]. Define 1 X 1 f(x) = g (x): 2k Ik k=1 Absolute continuity vs total singularity {Part II{ An easy TS example Brown's example. The function f is TS because it has the following properties: continuous; (uniform limit of continuous functions), strictly increasing; (any interval I contains a dyadic interval −k Ik = [ak; bk], and f(bk) − f(ak) > 2 ), and has f 0(x) = 0; (follows from next lemma). Lemma (Riesz-Sz. Nagy) P If f(x) = k fk(x) for λ-a.e. x 2 [0; 1], where each fk(x) is 0 P 0 increasing, then f (x) = k fk(x). Absolute continuity vs total singularity {Part II{ An easy TS example The \jimmied" Cantor function This is the example I remember from grad school. Start with f1(x) = (1=2)g(x). 2 Take f2(x) = (1=2 )g[1=3;2=3](x). This fills in the biggest \hole" of length 1=3 in C, but creates new holes of lengths 1=9; 1=27;::: . These are in addition to the pre-existing holes of the same sizes. Make f3(x) to fill in all the holes of length 1=9 (they are finite in number) with a total \rise" of 1=23, ... k−1 Make fk(x) to fill in all the holes of length 1=3 (they are finite in number) with a total \rise" of 1=2k, ... P Define f(x) = k fk(x). Absolute continuity vs total singularity {Part II{ An easy TS example Picture 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 Figure: The jimmied Cantor function Absolute continuity vs total singularity {Part II{ Raphael Salem's examples 1 Introduction 2 A \converse" to Villani's theorem 3 An easy TS example 4 Raphael Salem's examples 5 The Hartman-Kreshner approach 6 Minkowski's ?(x) and Conway's (x) 7 Spectral measures 8 References Absolute continuity vs total singularity {Part II{ Raphael Salem's examples The induction step Suppose g :[a; b] ! [c; d] has g(a) = c, g(b) = d, and is linear in between. 1 1 1 1 Let e = 2 a + 2 b and note that g(e) = 2 c + 2 d. For µ > 0, µ 6= 2, define a new function Bµg :[a; b] ! [c; d] by (Bµg)(a) = c, (Bµg)(b) = d, 1 1 (B g)(e) = c + 1 − d; µ µ µ and linear in between. Note that g and (Bµg) are both strictly increasing and continuous. Absolute continuity vs total singularity {Part II{ Raphael Salem's examples Picture d (1/6)c+(5/6)d c a (1/2)(a+b) b Figure: The functions g and Bµg in the case µ = 3. Absolute continuity vs total singularity {Part II{ Raphael Salem's examples Definition Fix some µ > 0, µ 6= 2. Define f0 : [0; 1] ! [0; 1] by f0(x) = x. Now suppose (by induction) that we have a strictly increasing, continuous fn that satisfies: fn(0) = 0, fn(1) = 1, and k−1 k fn is linear on each interval of the form In;k = [ 2n ; 2n ]. Construct fn+1 by replacing (fn)jIn;k with Bµ(fn)jIn;k , for k = 1; 2;:::; 2n. Define f(x) = lim fn(x): n!1 Absolute continuity vs total singularity {Part II{ Raphael Salem's examples The Salem function 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 Figure: The Salem function f(x) for µ = 4.