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 |f 0(x)| ≥ a > 0 a.e., then f −1(x) is absolutely continuous, and |(f −1)0(x)| ≤ 1/a a.e..

The proof is based on the following result.

Lemma (A. Villani, 1985) −1 Let ϕ :[a, b] → R be continuous, strictly monotone. Then ϕ (x) is absolutely continuous if and only if |ϕ0(x)| > 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 ϕ :[c, d] → R and let ϕ (x) ≥ η > 0 for x ∈ E ⊆ [c, d]. Then

λ∗(ϕ(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 ϕ :[c, d] → R and let ϕ (x) ≥ η > 0 for x ∈ E ⊆ [c, d]. Then

λ∗(ϕ(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 ϕ = f and note that ϕ :[f(a), f(b)] → R is continuous and strictly increasing. Suppose λ(F ) > 0 where F = {y : ϕ0(y) > 0}. There exists γ > η > 0 and E ⊆ F (measurable) so that λ(E) > η and η ≤ ϕ0(x) ≤ γ for x ∈ E. Note that ϕ(E) = f −1(E) is measurable. By Natanson’s lemma,

λ(ϕ(E)) = λ∗(ϕ(E)) ≥ ηλ∗(E) = ηλ(E) > η2 > 0.

But for (λ-a.e.) x ∈ ϕ(E), 1 1 f 0(x) = ≥ > 0. ϕ0(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 ∈ [0, 1] write x = .d1d2d3 ... in base 2 (so di ∈ {0, 1}) and x = .c1c2c3 ... in base 3 (so di ∈ {0, 1, 2}).

The Cantor set C ⊆ [0, 1] is the set of all x = .c1c2c3 ... so that ci 6= 1∀i.

Deﬁne the Cantor function g(.c1c2c3c3 ... ) = .d1d2d3 ... as follows:

If ci 6= 1∀i then i0 = ∞,

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 ∈ [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

∞ 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 ∈ [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

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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

(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)|In,k with Bµ(fn)|In,k , for k = 1, 2,..., 2n. Deﬁne f(x) = lim fn(x). n→∞ Absolute continuity vs total singularity –Part II– Raphael Salem’s examples The Salem function

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Figure: The Salem function f(x) for µ = 4. Absolute continuity vs total singularity –Part II– Raphael Salem’s examples The Salem function

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Figure: f −1(x) for µ = 4 or f(x) for µ = 4/3. Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Koch snowflake (1904) Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Properties

Put µ = µ and let µ satisfy 1 + 1 = 1. 0 1 µ0 µ1 f is continuous since fn → f uniformly: indeed 1 1 1 n |fn+1(x) − fn(x)| ≤ max , 2 µ 0 µ 1 .

f is strictly increasing: the vertices of each fn lie on the graph of f. Thus f is constant on no interval. f 0(x) = 0 a.e: The proof will follow below. Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Dynamical interpretation

Fix µ > 0. Deﬁne Tµ : [0, 1] → [0, 1] by ( µ x if 0 ≤ x < 1 , 0 µ0 Tµ(x) = µ (x − 1 ) if 1 ≤ x ≤ 1. 1 µ0 µ0

Also deﬁne D (x) = χh i(x). µ 1 ,1 µ0 Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Interval map

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Figure: The binary generalized L¨urothfunction, µ = 3. Absolute continuity vs total singularity –Part II– Raphael Salem’s examples µ-representations and µ-expansions

For x ∈ [0, 1], deﬁne Rµ(x) = d = .d1d2d3 · · · ∈ {0, 1}N by

k−1 dk = Dµ(Tµ (x)).

For d = .d1d2d3 · · · ∈ {0, 1}N deﬁne

d1 d2 d3 Eµ(d) = + + + ... µd1 µd2 µd1 µd3 µd2 µd1 Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Properties

Rµ(x) exists for all x ∈ [0, 1], and Eµ(d) converges for all d ∈ {0, 1}N (comparison to geometric series). These are “f-expansions”a la R´enyi.They satisfy Kakeya’s 0 criterion Tµ(x) > 1, so Eµ(Rµ(x)) = x (“valid”). The case µ = 2 is the usual base 2. For µ 6= 2 these are “binary” (cases of) “Generalized L¨urothSeries” (GLS) Dajani and Kraaikamp.

For all µ > 0, Tµ preserves Lebesgue measure and is ergodic.

Let µ 6= 2. The Salem function is given by f(x) = Eµ(R2(x)). Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Properties

For all µ > 0, Tµ preserves Lebesgue measure and is ergodic.

Let µ 6= 2. The Salem function is given by f(x) = Eµ(R2(x)). Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Proof that f 0(x) = 0 a.e.

For x ∈ [0, 1] with d = .d1d2d3 ··· = R2(x), and for p ∈ N, let xp = E2(.d1d2 . . . dp). 0 0 Note that if x < x with xp = xp, then 1 f(x0) − f(x) < . µd1 µd2 . . . µdp

Now a.e. x ∈ [0, 1] is normal, meaning p d + d + ··· + d = + ϕ(p), 1 2 p 2 where ϕ(p)/p → 0 as p → ∞. (This is Borel’s normal number theorem or the Birkhoﬀ ergodic theorem.) Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Proof that f 0(x) = 0 a.e., cont.. . .

Denote the normal numbers N, so that λ(N) = 0, and ﬁx x ∈ N. p+1 d Let hp = p+1/2 where p+1 = (−1) p+1 . Thus 1 |f(x + hp) − f(x)| < µd1 µd2 . . . µdp st (x to x + hp ﬂips the (p + 1) digit).

We also have 1 p+1 = 2 . hp Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Proof that f 0(x) = 0 a.e., cont.. . .

Since x ∈ N, 1 1 1 1 = · < . p/2+ϕ(p) p/2−ϕ(p) p/2−|ϕ(p)| µd µd . . . µd (µ0µ1) 1 2 p µ0 µ1 We have 2 p 2p+1 |f(x + h ) − f(x)| < · 2(µ µ )|ϕ(p)|. p 1/2 0 1 (µ0µ1) Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Proof that f 0(x) = 0 a.e., cont.. . .

We have 2 2 p < 1, so that = K−p for K > 1. 1/2 1/2 1 1 (µ0µ1) (µ0µ1) and |ϕ(p)| |ϕ(p)| (µ0µ1) = K2 for K2 > 1.

Thus

0 |f(x + hp) − f(x)| −p |ϕ(p)| f (x) = lim = 2K1 K2 = 0, p→∞ |hp|

for x ∈ N such that f 0(x) exists, since |ϕ(p)|/p → 0. Absolute continuity vs total singularity –Part II– Raphael Salem’s examples Additional results by Salem

f satisﬁes a H¨oldercondition

|f(x) − f(y)| ≤ 2γ−1|x − y|log γ/ log 2,

where γ = max{1/µ0, 1/µ1}. The Fourier coeﬃcients of the Lebesgue-Steltjes measure df on [0, 1] are given by

Z 1 ∞ inx Y 1 1 2πin/2k cn := e df(x) = + e µ0 µ1 0 k=1

cn 6→ 0 (not a Rajchman measure) since

2 2 2 2 2 |c2m | > (1 − 2/µ0) cos (π/4) cos (π/8) cos (π/16) .... Absolute continuity vs total singularity –Part II– The Hartman-Kreshner approach

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– The Hartman-Kreshner approach Good binary trees

n n Consider a sequence Pn = {Ik : n = 1, 2,... 2 } of partitions of n n n n n n [0, 1), with Ik = [ak−1, ak ), and 0 = a0 < a1 < ··· < a2n = 1.

Call {Pn} a good binary tree if:

each interval in Pn−1 is cut into two intervals n−1 n n Ik = I2k−1 ∪ I2k in Pn. n ||Pn|| = max{λ(In): i = 1, 2,..., 2 } → 0 as n → ∞.

For the last condition, it suﬃces to show ∪nPn is dense in [0, 1]. Absolute continuity vs total singularity –Part II– The Hartman-Kreshner approach The Hartmen-Kreshner function

Let {Pn} be a good binary tree, and let {Qn} denote the partition of [0, 1) into 2n dyadic intervals of length 2−n (a good binary tree). n n n Deﬁne fn(i/2 ) = ai for i = 0, 1,... 2 , and extend to fn : [0, 1] → [0, 1] by linear interpolation.

Deﬁne f(x) = limn→∞ fn(x).

Clearly f(x) is continuous and strictly increasing.

1 Note: f : Z 2 → ∪nPn is a strictly increasing bijection. 1 Here, Z 2 = ∪nQn. Absolute continuity vs total singularity –Part II– The Hartman-Kreshner approach Expansions using good binary trees

Let {Pn} be a good binary tree. For x ∈ [0, 1), deﬁne

R{Pn}(x) = .d1d2d3 ... by ( 0 if x ∈ In−1 ∩ In , and d = k 2k−1 n n−1 n 1 if x ∈ Ik ∩ I2k.

Given d = .d1d2d3 ... let rn(d) = dndn−1 . . . d1.02 ∈ N (base 2 integer), and let n E{P }(d) = lim a . n n→∞ rn(d)

Then E{Pn}(R{Pn}(x)) = x (“valid”). The Hartman-Kreshner function is given by

y = f(x) = E{Pn}(R{Qn}(x)),

where {Qn} denotes the dyadic binary tree. Absolute continuity vs total singularity –Part II– The Hartman-Kreshner approach Some definitions. . .

Let {Pn} be a good binary tree, and let {Qn} be the dyadic binary tree. n Let I (y) denote the interval of Pn that contains y. Write

n n n n n I (y) = A (y) ∪ B (y) where B (y) < A (y) ∈ Pn+1.

n Note that (for d = EPn (y)) dn = 0 if y ∈ A (y) and dn = 1 if y ∈ Bn(y). Let λ(Bn(y)) − λ(An(y)) (y) = (−1)dn(y) . n λ(In(y)) Absolute continuity vs total singularity –Part II– The Hartman-Kreshner approach The Hartman-Kreshner theorem

Theorem (Hartman-Kreshner) The strictly increasing continuous function y = f(x) is totally singular, mixed, or absolutely continuous according whether the set of points y for which the inﬁnite product

∞ Y (1 + n(y)) k=1 diverges, is of measure one, of measure between zero and one, or of measure zero. Absolute continuity vs total singularity –Part II– The Hartman-Kreshner approach A corollary

Corollary If λ(B (y)) lim inf n < 1 n→∞ λ(An(y)) for λ-a.e. y, then y = f(x) is totally singular.

Proof. n n n λ(A ) − λ(B ) 1 λ(B ) |n(y)| = > 1 − . λ(An) + λ(Bn) 2 λ(An) This implies lim sup |n(y)| > 0 n→∞ for λ-a.e. y, which implies the divergence of the inﬁnite product in the theorem. Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x)

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– Minkowski’s ?(x) and Conway’s (x) Farey fractons

p r Let q < s be two fractions in lowest terms. Deﬁne the Farey sum p r p + r ⊕ = . q s q + s Then p+r q+s is in lowest terms, and p p+r r q < q+s < s . Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) The Farey tree

0 1 Start with P1 = 1 < 1 .

Make Pn+1 out of Pn by dividing each interval at the Farey sum of its two endpoints.

n 0 0 1 0+1 1 1 o P2 = 1 < 1 ⊕ 1 = 1+1 = 2 < 1 . 0 1 1 2 1 P3 = 1 < 3 < 2 < 3 < 1 . 0 1 1 2 1 3 2 3 1 P4 = 1 < 4 < 3 < 5 < 2 < 5 < 3 < 4 < 1 . ...

Theorem

Q = ∪nPn.

p r If q < s are adjacent in Pn then rq − ps = 1. Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) The function

Deﬁne f(x) using the Hartman-Kreshner recipe, using the Farey tree Pn, and the dyadic tree Qn. This function f(x) is J. H. Conway’s box function (x). Either f(x) or f −1(x) is the Minkowski “question mark” function ?(x) (depending on whose account one accepts). Theorem Both f(x) and f −1(x) are totally singular.

Comment: f : Z[1/2] → Q is an order preserving bijection. Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) Conway’s box

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Figure: The Box function (x). Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) Minkowski’s question mark (1897)

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Figure: The function ?(x). Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) Proof

By the corollary, if f is not TS then (since 0 < λ(Bn)/λ(An) ≤ 1), lim inf λ(Bn)/λ(An) = 1 on a set of positive measure. This implies ∃ N and a set P , with λ(P ) > 0 so that

λ(Bn(y)) > 1/2 for y ∈ P and n ≥ N. λ(An(y))

N N Now consider how an interval I (y) = Ik = [p/q, r/s) ∈ Pn is cut into four intervals in Pn+2:

N+2 N+2 N+2 N+2 I4k ,I4k+1,I4k+2,I4k+3, and each of these is cut

N+2 N+2 N+2 I4k+i = B4k+i ∪ A4k+i. Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) Proof

N+2 The endpoints are of the I4k+i are p 2p + r p + r p + 2r q , , , , . q 2q + s q + s q + 2s s

N+2 N+2 A calculation shows λ(B4k )/λ(A4k ) = q/(2q + s) < 1/2. N+2 Thus y 6∈ I4k . N+2 N+2 N+2 A similar calculation shows y 6∈ I4k+3. Thus y ∈ I4k+1 ∪ I4k+2 N+` N+` In fact, we have y ∈ I2`k+2`−1 ∪ I2`k+2`−1+1 for all ` ≥ 2. It follows that #(P ) < ∞, so λ(P ) = 0. Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) Proof of Hartmen-Kreshner

We consider only the TS case. We will show

−1 0 1 (f ) (x) = Q∞ . k=1(1 + n(x)) The result will then follow from the Lebesgue lemma. n n Write In(y) = (yk , yk+1) (the endpoints are a set of measure 0, which we ignore. Deﬁne

−1 n −1 n f (yk+1) − f (yk ) ∆n(y) := n n . yk+1 − yk Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) Proof of Hartmen-Kreshner

−1 0 −1 0 Then ∆n(y) → (f ) (y) whenever (f ) (y) exists (i.e., a.e.), and y not an endpoint. Suppose In+1(y) = An(y). Then

n n n λ(A ) − λ(B ) = n(y)λ(I (y)),

or 1 λ(An) = λ(In)(1 + (y)), 2 n or 1 λ(In+1) = λ(In)(1 + (y)). 2 n The same formula holds if In+1(y) = Bn(y). Absolute continuity vs total singularity –Part II– Minkowski’s ?(x) and Conway’s (x) Proof of Hartmen-Kreshner

Since λ(I0) = 1,

n n 1 Y λ(In(y)) = (1 + (y)). 2 n k=0 The result follows since 1n 1 ∆ (y) = . n 2 λ(In(y)) Absolute continuity vs total singularity –Part II– Spectral measures

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– Spectral measures Ergodic theory

Let T :(X, µ) → (X, µ) be an invertible ergodic measure preserving transformation, and the the induced unitary 2 (UT f)(x) = f(T x) on L (X, µ) 2 n For h(x) ∈ L (X, µ), ||h|| = 1, deﬁne cn = hUT h, hi.

Since cn is positive deﬁnite ∃ probability measure σh on [0, 1] so that Z 1 2πin e dσh = cn. 0

There is an (almost unique) hmax (“maximal spectral type”) so

that σT := σh << σhmax for all h. Absolute continuity vs total singularity –Part II– Spectral measures Ergodic theory

Apply the Lebesgue decomposition σT = σac + σsc + σd. Deﬁne 0 σT = σac + σsc. Assume T is weakly mixing (iﬀ σd = δ0. since T ergodic)

If σac = 0, we say T has singular continuous spectrum.

If σsc = 0, we say T has Lebesgue spectrum. 0 T is mixing iﬀ cn → 0 as n → ±∞ (σT is a Rajchman measure). 0 If T has Lebesgue spectrum, cn → 0 (σT is a Rajchman by Riemann-Lebesgue lemma): Lebesgue spectrum → mixing. Mixing 6→ Lebesgue spectrum (∃ SC Rajchman measures: rank 1 mixing). Absolute continuity vs total singularity –Part II– Spectral measures Ergodic theory

Examples: SC, non-Rajchman: Chacon, weakly mixing substitutions, Morse ( span eigenvalues). SC, Rajchman: Rank 1 mixing. AC: Bernoulli shifts, Rudin-Shapiro ( span eigenvalues).

Question: Is σh(I) > 0 for any interval I (True or false)? Answer: Yes! (thanks to Karl Petersen) sp(UT ) = T (an ergodic theory result), and any open set I ⊆ sp(UT ) = T = 2π[0, 1) has σT (I) > 0 (see Rudin, Functional Analysis). Absolute continuity vs total singularity –Part II– Spectral measures Morse sequence shift T

Figure: The distribution for σsc for the Morse sequence: 0 → 01, 1 → 10 (Baake & Grimm, 2008) Absolute continuity vs total singularity –Part II– Spectral measures Challenge

Draw a graph of the distribution of a TS Rajchman measure. Preferably it should be the spectral measure for a mixing T with singular continuous spectrum (e.g. rank 1 mixing). How does the graph of a non-Rajchman distribution look compared to how the graph of a Rajchman distribution? What does a AC one look like? Absolute continuity vs total singularity –Part II– References

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– References

Arlen Brown, “An Elementary Example of a Continuous Singular Function”, Monthly 76, (1969), 295-297. John H. Conway. “Contorted Fractions”, On Numbers and Games, (1976). Philip Hartman and Richard Kershner, “The Structure of Monotone Functions”, American J. of Math., 59, (1937), 809-822. Hermann Minkowski, “Zur Geometrie der Zahlen.” (circa 1897) In Gesammelte Abhandlungen, Chelsea, 44-52, (1991). Raphael Salem, “On Some Singular Monotone Functions which Are Strictly Increasing.” Trans. Amer. Math. Soc. 53, 427-439, (1943).