Absolute continuity vs total singularity Part IIrobinson/Documents/Total Singularity.pdf ·...

$: Absolute continuity vs total singularity Part IIrobinson/Documents/Total Singularity.pdf · Absolute continuity vs total singularity {Part II{Introduction 1 Introduction 2 A \converse"$
Absolute continuity vs total singularity –Part II–

Absolute continuity vs totalsingularity–Part II–

E. Arthur (Robbie) Robinson

The George Washington University

February 16, 2012


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


Introduction

1 Introduction






7 Spectral measures

8 References


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

The proof is based on the following result.

Lemma (A. Villani, 1985)

Let ϕ : [a, b]→ R be continuous, strictly monotone. Then ϕ−1(x)is absolutely continuous if and only if |ϕ′(x)| > 0 a.e..

Villani says he believes this lemma is known (clearly it was knownto Kakeya) but he could not find it in the literature.


Introduction

Today

Definition

We call an continuous increasing function f : [a, b]→ R singular(continuous) if f ′(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 theCantor function (we review it below). However, it is not strictlyincreasing, so it is not totally singular.

Examples of totally singular functions abound in the literature.The best known example is probably Minkowski’s “question markfunction” ?(x) (we discuss it later).

However, these examples are not so widely known.


A “converse” to Villani’s theorem

1 Introduction






7 Spectral measures

8 References



Converse

The following result seems to be well known (many authors state itas a fact and it is often attributed to Lebesgue).

Theorem (Lebesgue)

If f : [a, b]→ R is totally singular (i.e., continuous and strictlyincreasing with f ′(x) = 0 λ-a.e.), then so is f−1.

The proof uses a lemma from Natanson, related to one from thefirst lecture.

Lemma (Natanson)

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.



Proof of converse lemma

Let ϕ = f−1 and note that ϕ : [f(a), f(b)]→ R is continuous andstrictly increasing.

Suppose λ(F ) > 0 where F = {y : ϕ′(y) > 0}. There existsγ > η > 0 and E ⊆ F (measurable) so that λ(E) > η andη ≤ ϕ′(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),

f ′(x) =1

ϕ′(f(x))≥ 1

γ> 0.


An easy TS example

1 Introduction






7 Spectral measures

8 References


An easy TS example

Definition of the Cantor function

For x ∈ [0, 1] write x = .d1d2d3 . . . in base 2 (so di ∈ {0, 1}) andx = .c1c2c3 . . . in base 3 (so di ∈ {0, 1, 2}).

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

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

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 alli ≥ i0 put di = 0.


An easy TS example

Graph of Cantor function (c 1883)

The Cantor functiong : [0, 1]→ [0, 1] is an increasing continuoussurjection with g′(x) = 0 a.e. (singular). It is not strictlyincreasing (not 1:1) so not TS.

(Source: http://mathworld.wolfram.com/CantorFunction.html)


An easy TS example

Brown’s example

This is related to an example I recall from grad school. It is easierto 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

gI(x) = g

(x− ab− a

)if x ∈ [a, b].

Note that I is continuous, increasing and singular.

Let I1, I2, I3, . . . be an enumeration of the dyadic intervals in[0, 1]. Define

f(x) =

∞∑k=1

1

2kgIk(x).


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 intervalIk = [ak, bk], and f(bk)− f(ak) > 2−k), and

has f ′(x) = 0; (follows from next lemma).

Lemma (Riesz-Sz. Nagy)

If f(x) =∑

k fk(x) for λ-a.e. x ∈ [0, 1], where each fk(x) isincreasing, then f ′(x) =

∑k f′k(x).


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).

Take f2(x) = (1/22)g[1/3,2/3](x).

This fills in the biggest “hole” of length 1/3 in C, but creates newholes of lengths 1/9, 1/27, . . . . These are in addition to thepre-existing holes of the same sizes.

Make f3(x) to fill in all the holes of length 1/9 (they are finitein number) with a total “rise” of 1/23,

. . .

Make fk(x) to fill in all the holes of length 1/3k−1 (they arefinite in number) with a total “rise” of 1/2k,

. . .

Define f(x) =∑

k fk(x).


An easy TS example

Picture

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

Figure: The jimmied Cantor function


Raphael Salem’s examples

1 Introduction






7 Spectral measures

8 References



The induction step

Suppose g : [a, b]→ [c, d] has g(a) = c, g(b) = d, and is linear inbetween.

Let e =(12

)a+

(12

)b and note that g(e) =

(12

)c+

(12

)d.

For µ > 0, µ 6= 2, define a new function Bµg : [a, b]→ [c, d] by(Bµg)(a) = c, (Bµg)(b) = d,

(Bµg)(e) =

(1

µ

)c+

(1− 1

µ

)d,

and linear in between.

Note that g and (Bµg) are both strictly increasing and continuous.



Picture

a b

c

d

(1/2)(a+b)

(1/6)c+(5/6)d

Figure: The functions g and Bµg in the case µ = 3.



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

fn is linear on each interval of the form In,k = [k−12n ,k2n ].

Construct fn+1 by replacing (fn)|In,k with Bµ(fn)|In,k , fork = 1, 2, . . . , 2n.

Definef(x) = lim

n→∞fn(x).



The Salem function

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: The Salem function f(x) for µ = 4.



The Salem function

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: f−1(x) for µ = 4 or f(x) for µ = 4/3.



Koch snowflake (1904)



Properties

Put µ0 = µ and let µ1 satisfy 1µ0

+ 1µ1

= 1.

f is continuous since fn → f uniformly: indeed

|fn+1(x)− fn(x)| ≤ 1

2

(max

{1

µ0

,1

µ1

})n.

f is strictly increasing: the vertices of each fn lie on thegraph of f . Thus f is constant on no interval.

f ′(x) = 0 a.e: The proof will follow below.



Dynamical interpretation

Fix µ > 0. Define Tµ : [0, 1]→ [0, 1] by

Tµ(x) =

{µ0x if 0 ≤ x < 1

µ0,

µ1(x− 1µ0

) if 1µ0≤ x ≤ 1.

Also define Dµ(x) = χ[1µ0,1](x).



Interval map

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: The binary generalized Luroth function, µ = 3.



µ-representations and µ-expansions

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

dk = Dµ(T k−1µ (x)).

For d = .d1d2d3 · · · ∈ {0, 1}N define

Eµ(d) =d1µd1

+d2

µd2µd1+

d3µd3µd2µd1

+ . . .



Properties

Rµ(x) exists for all x ∈ [0, 1], and Eµ(d) converges for alld ∈ {0, 1}N (comparison to geometric series).

These are “f -expansions”a la Renyi. They satisfy Kakeya’scriterion T ′µ(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 Luroth Series” (GLS) Dajaniand 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)).



Proof that f ′(x) = 0 a.e.

For x ∈ [0, 1] with d = .d1d2d3 · · · = R2(x), and for p ∈ N, letxp = E2(.d1d2 . . . dp).

Note that if x < x′ with xp = x′p, then

f(x′)− f(x) <1

µd1µd2 . . . µdp.

Now a.e. x ∈ [0, 1] is normal, meaning

d1 + d2 + · · ·+ dp =p

2+ ϕ(p),

where ϕ(p)/p→ 0 as p→∞. (This is Borel’s normal numbertheorem or the Birkhoff ergodic theorem.)



Proof that f ′(x) = 0 a.e., cont.. . .

Denote the normal numbers N , so that λ(N) = 0, and fix x ∈ N .

Let hp = εp+1/2p+1 where εp+1 = (−1)dp+1 .

Thus

|f(x+ hp)− f(x)| < 1

µd1µd2 . . . µdp

(x to x+ hp flips the (p+ 1)st digit).

We also have ∣∣∣∣ 1

hp

∣∣∣∣ = 2p+1.




Since x ∈ N ,

1

µd1µd2 . . . µdp=

1

µp/2+ϕ(p)0

· 1

µp/2−ϕ(p)1

<1

(µ0µ1)p/2−|ϕ(p)|.

We have

2p+1 |f(x+ hp)− f(x)| <(

2

(µ0µ1)1/2

)p· 2(µ0µ1)

|ϕ(p)|.




We have

2

(µ0µ1)1/2< 1, so that

(2

(µ0µ1)1/2

)p= K−p1 for K1 > 1.

and(µ0µ1)

|ϕ(p)| = K|ϕ(p)|2 for K2 > 1.

Thus

f ′(x) = limp→∞

|f(x+ hp)− f(x)||hp|

= 2K−p1 K|ϕ(p)|2 = 0,

for x ∈ N such that f ′(x) exists, since |ϕ(p)|/p→ 0.



Additional results by Salem

f satisfies a Holder condition

|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 dfon [0, 1] are given by

cn :=

∫ 1

0einx df(x) =

∞∏k=1

(1

µ0+

1

µ1e2πin/2

k

)cn 6→ 0 (not a Rajchman measure) since

|c2m |2 > (1− 2/µ0)2 cos2(π/4) cos2(π/8) cos2(π/16) . . . .


The Hartman-Kreshner approach

1 Introduction






7 Spectral measures

8 References



Good binary trees

Consider a sequence Pn = {Ink : n = 1, 2, . . . 2n} of partitions of[0, 1), with Ink = [ank−1, a

nk), and 0 = an0 < an1 < · · · < an2n = 1.

Call {Pn} a good binary tree if:

each interval in Pn−1 is cut into two intervalsIn−1k = In2k−1 ∪ In2k in Pn.

||Pn|| = max{λ(In) : i = 1, 2, . . . , 2n} → 0 as n→∞.

For the last condition, it suffices to show ∪nPn is dense in [0, 1].



The Hartmen-Kreshner function

Let {Pn} be a good binary tree, and let {Qn} denote the partitionof [0, 1) into 2n dyadic intervals of length 2−n (a good binary tree).

Define fn(i/2n) = ani for i = 0, 1, . . . 2n, and extend tofn : [0, 1]→ [0, 1] by linear interpolation.

Define f(x) = limn→∞ fn(x).

Clearly f(x) is continuous and strictly increasing.

Note: f : Z[12

]→ ∪nPn is a strictly increasing bijection.

Here, Z[12

]= ∪nQn.



Expansions using good binary trees

Let {Pn} be a good binary tree. For x ∈ [0, 1), defineR{Pn}(x) = .d1d2d3 . . . by

dn =

{0 if x ∈ In−1k ∩ In2k−1, and

1 if x ∈ In−1k ∩ In2k.

Given d = .d1d2d3 . . . let rn(d) = dndn−1 . . . d1.02 ∈ N (base 2integer), and let

E{Pn}(d) = limn→∞

anrn(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.



Some definitions. . .

Let {Pn} be a good binary tree, and let {Qn} be the dyadic binarytree.

Let In(y) denote the interval of Pn that contains y. Write

In(y) = An(y) ∪Bn(y) where Bn(y) < An(y) ∈ Pn+1.

Note that (for d = EPn(y)) dn = 0 if y ∈ An(y) and dn = 1 ify ∈ Bn(y).

Let

εn(y) = (−1)dn(y)λ(Bn(y))− λ(An(y))

λ(In(y)).



The Hartman-Kreshner theorem

Theorem (Hartman-Kreshner)

The strictly increasing continuous function y = f(x) is totallysingular, mixed, or absolutely continuous according whether the setof points y for which the infinite product

∞∏k=1

(1 + εn(y))

diverges, is of measure one, of measure between zero and one, orof measure zero.



A corollary

Corollary

If

lim infn→∞

λ(Bn(y))

λ(An(y))< 1

for λ-a.e. y, then y = f(x) is totally singular.

Proof.

|εn(y)| =∣∣∣∣λ(An)− λ(Bn)

λ(An) + λ(Bn)

∣∣∣∣ > 1

2

∣∣∣∣1− λ(Bn)

λ(An)

∣∣∣∣ .This implies

lim supn→∞

|εn(y)| > 0

for λ-a.e. y, which implies the divergence of the infinite product inthe theorem.


Minkowski’s ?(x) and Conway’s �(x)

1 Introduction






7 Spectral measures

8 References



Farey fractons

Let pq <

rs be two fractions in lowest terms. Define the Farey sum

p

q⊕ r

s=p+ r

q + s.

Thenp+rq+s is in lowest terms, and

pq <

p+rq+s <

rs .



The Farey tree

Start with P1 ={01 <

11

}.

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

P2 ={

01 <

01 ⊕

11 = 0+1

1+1 = 12 <

11

}.

P3 ={01 <

13 <

12 <

23 <

11

}.

P4 ={01 <

14 <

13 <

25 <

12 <

35 <

23 <

34 <

11

}.

. . .

Theorem

Q = ∪nPn.

If pq <

rs are adjacent in Pn then rq − ps = 1.



The function

Define f(x) using the Hartman-Kreshner recipe, using the Fareytree Pn, and the dyadic tree Qn.

This function f(x) is J. H. Conway’s box function �(x). Eitherf(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.



Conway’s box

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: The Box function �(x).



Minkowski’s question mark (1897)

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: The function ?(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))

λ(An(y))> 1/2 for y ∈ P and n ≥ N.

Now consider how an interval IN (y) = INk = [p/q, r/s) ∈ Pn iscut into four intervals in Pn+2:

IN+24k , IN+2

4k+1, IN+24k+2, I

N+24k+3,

and each of these is cut

IN+24k+i = BN+2

4k+i ∪AN+24k+i.



Proof

The endpoints are of the IN+24k+i are

p

q,2p+ r

2q + s,p+ r

q + s,p+ 2r

q + 2s,q

s.

A calculation shows λ(BN+24k )/λ(AN+2

4k ) = q/(2q + s) < 1/2.

Thus y 6∈ IN+24k .

A similar calculation shows y 6∈ IN+24k+3. Thus y ∈ IN+2

4k+1 ∪ IN+24k+2

In fact, we have y ∈ IN+`2`k+2`−1 ∪ IN+`

2`k+2`−1+1for all ` ≥ 2.

It follows that #(P ) <∞, so λ(P ) = 0.



Proof of Hartmen-Kreshner

We consider only the TS case. We will show

(f−1)′(x) =1∏∞

k=1(1 + εn(x)).

The result will then follow from the Lebesgue lemma.

Write In(y) = (ynk , ynk+1) (the endpoints are a set of measure 0,

which we ignore. Define

∆n(y) :=f−1(ynk+1)− f−1(ynk )

ynk+1 − ynk.




Then ∆n(y)→ (f−1)′(y) whenever (f−1)′(y) exists (i.e., a.e.),and y not an endpoint.

Suppose In+1(y) = An(y). Then

λ(An)− λ(Bn) = εn(y)λ(In(y)),

or

λ(An) =1

2λ(In)(1 + εn(y)),

or

λ(In+1) =1

2λ(In)(1 + εn(y)).

The same formula holds if In+1(y) = Bn(y).




Since λ(I0) = 1,

λ(In(y)) =

(1

2

)n n∏k=0

(1 + εn(y)).

The result follows since

∆n(y) =

(1

2

)n 1

λ(In(y)).


Spectral measures

1 Introduction






7 Spectral measures

8 References


Spectral measures

Ergodic theory

Let T : (X,µ)→ (X,µ) be an invertible ergodic measurepreserving transformation, and the the induced unitary(UT f)(x) = f(Tx) on L2(X,µ)

For h(x) ∈ L2(X,µ), ||h|| = 1, define cn = 〈UnT h, h〉.

Since cn is positive definite ∃ probability measure σh on [0, 1] sothat ∫ 1

0e2πin dσh = cn.

There is an (almost unique) hmax (“maximal spectral type”) sothat σT := σh << σhmax for all h.


Spectral measures

Ergodic theory

Apply the Lebesgue decomposition σT = σac + σsc + σd. Defineσ0T = 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.

T is mixing iff cn → 0 as n→ ±∞ (σ0T is a Rajchmanmeasure).

If T has Lebesgue spectrum, cn → 0 (σ0T is a Rajchman byRiemann-Lebesgue lemma): Lebesgue spectrum → mixing.

Mixing 6→ Lebesgue spectrum (∃ SC Rajchman measures:rank 1 mixing).


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 ergodictheory result), and any open set I ⊆ sp(UT ) = T = 2π[0, 1) hasσT (I) > 0 (see Rudin, Functional Analysis).


Spectral measures

Morse sequence shift T

Figure: The distribution for σsc for the Morse sequence:0→ 01, 1→ 10 (Baake & Grimm, 2008)


Spectral measures

Challenge

Draw a graph of the distribution of a TS Rajchman measure.

Preferably it should be the spectral measure for a mixing Twith singular continuous spectrum (e.g. rank 1 mixing).

How does the graph of a non-Rajchman distribution lookcompared to how the graph of a Rajchman distribution?What does a AC one look like?


References

1 Introduction






7 Spectral measures

8 References


References

Arlen Brown, “An Elementary Example of a ContinuousSingular Function”, Monthly 76, (1969), 295-297.

John H. Conway. “Contorted Fractions”, On Numbers andGames, (1976).

Philip Hartman and Richard Kershner, “The Structure ofMonotone Functions”, American J. of Math., 59, (1937),809-822.

Hermann Minkowski, “Zur Geometrie der Zahlen.” (circa1897) In Gesammelte Abhandlungen, Chelsea, 44-52, (1991).

Raphael Salem, “On Some Singular Monotone Functionswhich Are Strictly Increasing.” Trans. Amer. Math. Soc. 53,427-439, (1943).

Absolute continuity vs total singularity Part IIrobinson/Documents/Total Singularity.pdf ·...

Documents

Transcript of Absolute continuity vs total singularity Part IIrobinson/Documents/Total Singularity.pdf ·...