Number theory and maps of the intervalhome.gwu.edu/~robinson/Documents/NTMI.pdfNumber theory and maps of the interval Number theory and maps of the interval E. Arthur (Robbie) Robinson

Number theory and maps of the interval

Number theory and maps of theinterval

E. Arthur (Robbie) Robinson

The George Washington University

May 22, 2012

Outline

The New Math

Figure: Mullen-Hall Elementary School, Falmouth, Massuahusetts.

In 1963, during the Cold War, my third grade math teacher wastrained in the “New Math” and taught us base 2 and base 5.

Some examples

Continued fractions

Any x ∈ (0, 1]\Q is given by its unique continued fractionexpansion:

x =1

d1 +1

d2 +1

d3 + · · ·

.

Here d = .d1d2d3 · · · = [d1, d2, . . . ] is an arbitrary infinite sequenceof positive integers. (We call D = N the digit set.)

Some examples

Continued fractions

x =1

d1 +1

d2 +1

d3 + · · ·

.

Some examples

Continued fractions

x =1

d1 +1

d2 +1

d3 + · · ·

.

Some examples

Example 1

√5 + 1

2(≈ 1.6180 . . . ) = 1 +

1

1 +1

1 + · · ·

.

2 = 1 +1

1,

3

2= 1 +

1

1 +1

1

,

The next few “convergents”: 53 ≈ 1.667,85 ≈ 1.6,

138 ≈ 1.625. . .

Some examples

Example 1

√5 + 1

2(≈ 1.6180 . . . ) = 1 +

1

1 +1

1 + · · ·

.

2 = 1 +1

1,

3

2= 1 +

1

1 +1

1

,

138 ≈ 1.625. . .

Some examples

Example 1

√5 + 1

2(≈ 1.6180 . . . ) = 1 +

1

1 +1

1 + · · ·

.

2 = 1 +1

1,

3

2= 1 +

1

1 +1

1

,

138 ≈ 1.625. . .

Some examples

Example 1

√5 + 1

2(≈ 1.6180 . . . ) = 1 +

1

1 +1

1 + · · ·

.

2 = 1 +1

1,

3

2= 1 +

1

1 +1

1

,

138 ≈ 1.625. . .

Some examples

Example 1

√5 + 1

2(≈ 1.6180 . . . ) = 1 +

1

1 +1

1 + · · ·

.

2 = 1 +1

1,

3

2= 1 +

1

1 +1

1

,

138 ≈ 1.625. . .

Some examples

Example 1

√5 + 1

2(≈ 1.6180 . . . ) = 1 +

1

1 +1

1 + · · ·

.

2 = 1 +1

1,

3

2= 1 +

1

1 +1

1

,

138 ≈ 1.625. . .

Some examples

Example 2

π = 3 +1

7 +1

15 +1

1 +1

292 + · · ·

.

22

7= 3 +

1

7≈ 3.1428, (Arhimedes, c 240 BC)

333

106= 3 +

1

7 +1

15

≈ 3.141509,

One more gives 355/113 ≈ 3.1415929 (Zu Chongzhi, c 480).

Some examples

Example 2

π = 3 +1

7 +1

15 +1

1 +1

292 + · · ·

.

22

7= 3 +

1

7≈ 3.1428, (Arhimedes, c 240 BC)

333

106= 3 +

1

7 +1

15

≈ 3.141509,

Some examples

Example 2

π = 3 +1

7 +1

15 +1

1 +1

292 + · · ·

.

22

7= 3 +

1

7≈ 3.1428, (Arhimedes, c 240 BC)

333

106= 3 +

1

7 +1

15

≈ 3.141509,

Some examples

Example 2

π = 3 +1

7 +1

15 +1

1 +1

292 + · · ·

.

22

7= 3 +

1

7≈ 3.1428, (Arhimedes, c 240 BC)

333

106= 3 +

1

7 +1

15

≈ 3.141509,

Some examples

Example 2

π = 3 +1

7 +1

15 +1

1 +1

292 + · · ·

.

22

7= 3 +

1

7≈ 3.1428, (Arhimedes, c 240 BC)

333

106= 3 +

1

7 +1

15

≈ 3.141509,

Some examples

Example 3

e = 2 +1

1 +1

2 +1

1 +1

[4, 1, 1, 6, 1, 1, 8, . . . ]

.

3 = 2+1

1;

19

7= 2+

1

1 +1

2 +1

1 +1

1

≈ 2.71429; 1264465

≈ 2.71827957

Some examples

Example 3

e = 2 +1

1 +1

2 +1

1 +1

[4, 1, 1, 6, 1, 1, 8, . . . ]

.

3 = 2+1

1;

19

7= 2+

1

1 +1

2 +1

1 +1

1

≈ 2.71429; 1264465

≈ 2.71827957

Some examples

Example 3

e = 2 +1

1 +1

2 +1

1 +1

[4, 1, 1, 6, 1, 1, 8, . . . ]

.

3 = 2+1

1;

19

7= 2+

1

1 +1

2 +1

1 +1

1

≈ 2.71429; 1264465

≈ 2.71827957

Some examples

Example 3

e = 2 +1

1 +1

2 +1

1 +1

[4, 1, 1, 6, 1, 1, 8, . . . ]

.

3 = 2+1

1;

19

7= 2+

1

1 +1

2 +1

1 +1

1

≈ 2.71429; 1264465

≈ 2.71827957

Some examples

Continued fraction coefficients

Here is the “greedy algorithm” for continued fraction coefficients.

Start with d1 = b1/xc and x1 = 1/x− d1.Then

x =1

d1 + x1.

Next d2 = b1/x1c and x2 = 1/x1 − d2.Then

x =1

d1 + x1=

1

d1 +1

d2 + x2

.

Continue . . .

Some examples

x =1

d1 + x1.

x =1

d1 + x1=

1

d1 +1

d2 + x2

.

Continue . . .

Some examples

x =1

d1 + x1.

x =1

d1 + x1=

1

d1 +1

d2 + x2

.

Continue . . .

Some examples

x =1

d1 + x1.

x =1

d1 + x1=

1

d1 +1

d2 + x2

.

Continue . . .

Some examples

x =1

d1 + x1.

x =1

d1 + x1=

1

d1 +1

d2 + x2

.

Continue . . .

Some examples

x =1

d1 + x1.

x =1

d1 + x1=

1

d1 +1

d2 + x2

.

Continue . . .

Some examples

Base-r expansions

Let r ∈ N, r > 1. Any x ∈ [0, 1) is given by a base-r radixexpansion:

x =

∞∑n=1

dnrn,

where d = .d1d2d3 . . . is a sequence from D = {0, 1, . . . , r − 1}.

Here is an alternative way to write this (due to Kakeya, 1924)

x =d1 +

d2 +d3 + . . .

rr

r.

Some examples

Base-r expansions

x =

∞∑n=1

dnrn,

x =d1 +

d2 +d3 + . . .

rr

r.

Some examples

Base-r expansions

x =

∞∑n=1

dnrn,

x =d1 +

d2 +d3 + . . .

rr

r.

Some examples

Base-r expansions

x =

∞∑n=1

dnrn,

x =d1 +

d2 +d3 + . . .

rr

r.

Some examples

Example 4

In base r = 2:

1/3 = .0101010101010101010101 . . .

=0 +

1 +0 + . . .

22

2.

π = 11 + .0010010000111111011010101000100 . . .

=0 +

0 +1 + . . .

22

2=

1 +1 +

1 + . . .

211

26

23

Some examples

Example 4

In base r = 2:

1/3 = .0101010101010101010101 . . .

=0 +

1 +0 + . . .

22

2.

π = 11 + .0010010000111111011010101000100 . . .

=0 +

0 +1 + . . .

22

2=

1 +1 +

1 + . . .

211

26

23

Some examples

Example 4

In base r = 2:

1/3 = .0101010101010101010101 . . .

=0 +

1 +0 + . . .

22

2.

π = 11 + .0010010000111111011010101000100 . . .

=0 +

0 +1 + . . .

22

2=

1 +1 +

1 + . . .

211

26

23

Some examples

Base-r coefficients

Here is the “greedy algorithm” for base r

Start with d1 = brxc and x1 = rx− d1.Then

x =d1 + x1

r.

Next d2 = brx1c and x2 = rx1 − d2.Then

x =d1 + x1

r=d1 +

d2 + x2

rr

.

Continue . . .

Some examples

Base-r coefficients

x =d1 + x1

r.

x =d1 + x1

r=d1 +

d2 + x2

rr

.

Continue . . .

Some examples

Base-r coefficients

x =d1 + x1

r.

x =d1 + x1

r=d1 +

d2 + x2

rr

.

Continue . . .

Some examples

Base-r coefficients

x =d1 + x1

r.

x =d1 + x1

r=d1 +

d2 + x2

rr

.

Continue . . .

Some examples

Base-r coefficients

x =d1 + x1

r.

x =d1 + x1

r=d1 +

d2 + x2

rr

.

Continue . . .

Some examples

Base-r coefficients

x =d1 + x1

r.

x =d1 + x1

r=d1 +

d2 + x2

rr

.

Continue . . .

Some examples

An interesting fact

In base r = 3 the digit set is D = {0, 1, 2}. Consider the set of Knumbers x ∈ [0, 1] so that the expansion

d = .d1d2d3 . . .

in base r = 3, has dn 6= 1 for all n.

The set K is the standard (middle thirds) Cantor set.

Some examples

An interesting fact

In base r = 3 the digit set is D = {0, 1, 2}. Consider the set of Knumbers x ∈ [0, 1] so that the expansion

d = .d1d2d3 . . .

in base r = 3, has dn 6= 1 for all n.

The set K is the standard (middle thirds) Cantor set.

Some examples

Base-r (continued – 2)

Let N(d : w) the number of digits d in word w ∈ D∗ = ∪n≥1Dn.For x ∈ [0, 1) let d = .d1d2d3 . . . and let dn = d1d2 . . . dn ∈ Dn.For a.e. x we can recover r, the base, by

1/r = limn→∞

1

nN(d : dn).

Let C(n,d) be the number of words of length n in d. Then fora.e. x

r = limn→∞

1

nlogC(n,d).

The generalized algorithm

A generalization

Kakeya (1924) (and independently Bissinger, (1944) and Everett(1946)) identified a common algorithm that generalizes bothcontinued fraction and base-r.

It is based on a monotonic (increasing or decreasing) functionf : R→ [0, 1] (satisfying a few additional hypotheses), togetherwith its inverse f−1(x).

For continued fractions: f(x) = 1/x, so that f−1(x) = 1/x.

For base-r: f(x) = x/r, so that f−1(x) = rx.

A generalization

The f-expansion

Given a sequence d = .d1d2d3 . . . with the digits dj in some digitset D ⊆ Z, we define the f -expansion

ε(d) = f(d1 + f(d2 + f(d3 . . . ))),

provided it converges.

That is, εn(d)→ ε(d), where

εn(d) = f(d1 + f(d2 + f(d3 + · · ·+ f(dn)))).

The f-expansion

Given a sequence d = .d1d2d3 . . . with the digits dj in some digitset D ⊆ Z, we define the f -expansion

ε(d) = f(d1 + f(d2 + f(d3 . . . ))),

provided it converges.

That is, εn(d)→ ε(d), where

εn(d) = f(d1 + f(d2 + f(d3 + · · ·+ f(dn)))).

The proper digits

The proper digits are the ones obtained by the greedy algorithm.

For x ∈ [0, 1) define x1 = x.Then for each n ≥ 1,dn = bf−1(xn)c.xn+1 = f

−1(xn)− dn.

We call the infinite sequence ρ(x) = .d1d2d3 . . . (if it exists) thef -representation of x.

Note that dj ∈ D = bf−1[0, 1)c. These are the proper digits.

The proper digits

−1(xn)− dn.

The proper digits

−1(xn)− dn.

The proper digits

−1(xn)− dn.

The proper digits

−1(xn)− dn.

The proper digits

−1(xn)− dn.

Validity

If for a.e. x ∈ [0, 1), the number x may then be recovered by theformula

x = f(d1 + f(d2 + f(d3 . . . ))),

we say f -expansions are valid.

In effect,x = ε(ρ(x)),

for a.e. x.

Validity

x = f(d1 + f(d2 + f(d3 . . . ))),

for a.e. x.

Validity

x = f(d1 + f(d2 + f(d3 . . . ))),

for a.e. x.

A dynamical interpretation

Rényi (1957) observed that the greedy algorithm can beimplemented by iterating a map of the interval:

T (x) = f−1(x) mod 1.

The proper digits are obtained by dn = ξ(Tn−1(x)), where

ξ(x) = bf−1(x)c. This is “symbolic dynamics”.

T (x) = f−1(x) mod 1.

0 1 2 3 4 50.0

0.2

0.4

0.6

0.8

1.0

Figure: Continued fractions: f(x) = 1/x

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 Gauss map T (x) = f−1(x) mod 1 = 1/x mod 1.

0.0 0.5 1.0 1.5 2.0 2.5 3.00.0

0.2

0.4

0.6

0.8

1.0

Figure: Base 3: f(x) = x/3 (r = 3).

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 “period tripling” map T (x) = f−1(x) mod 1 = 3x mod 1.

Basic hypotheses

The following hypotheses on T are essentially those of Bill Parry(1957).

D ⊆ Z, an interval of integers, |D| > 2.For d ∈ D, ∆(d) = [cd, cd+1) ⊆ I, the unit interval.cd increasing, with inf cd = 0 and sup cd = 1.

I = ∪d∈D∆(d).

Let T : I → I be:strictly monotone and continuous on each ∆(d) (all in samedirection),

onto I.

Call T Type A/Type B if all T |∆(d) decreasing/increasing,

The function f

Define ξ : I → D by ξ(x) = d if x ∈ ∆(d).Think of ξ = {∆(d) : d ∈ D} as a finite or countable partition.Define T ∗(x) = T (x) + ξ(x). Then T ∗ : I → R is strictlymonotone.

Define f : R→ I by f(x) = (T ∗)−1(x), extended to R to be(non-strictly) monotone and continuous (i.e., constant on thegaps).

Proper digits dn = ξ(Tn−1x).

Valid if x = f(d1 + f(d2 + f(d3 . . . ))).

The function f

Valid if x = f(d1 + f(d2 + f(d3 . . . ))).

The function f

Valid if x = f(d1 + f(d2 + f(d3 . . . ))).

Kakeya’s Theorem

Theorem (Kakeya, 1924)

Let T be Type A or B and satisfy the basic hypotheses. Suppose|T ′(x)| > 1 almost everywhere. Then f -expansions are valid.

Bissinger (1944), Everett (1946) have similar results, with relatedhypotheses.

Bissinger (1944), Everett (1946), Rényi (1957), and Parry(1964)were apparently unaware of Kakeya’s result.

Parry’s Theorem

A mapping T : I → I is called topologically transitive if there is anx ∈ I so that {Tnx : n = 0, 1, 2, . . . } is dense in I.

Theorem (Parry, 1964)

Let T be Type A or B and satisfy the basic hypotheses. Supposethat T is topologically transitive. Then f -expansions are valid.

Note that both Kakeya’s and Parry’s theorems apply both tocontinued fractions and to radix expansions.

Comment: Both theorems essentially show that ξ is a generatingpartition.

Examples

Kakeya’s negative base expansions

1 2 3 40.0

0.2

0.4

0.6

0.8

1.0

Figure: Kakeya (1924), base −3: f(x) = 1− x/3.

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: Reverse period tripling T (x) = 4− 3x mod 1.

Examples

Rényi’s β-expansions

0 1 2 3 40.0

0.2

0.4

0.6

0.8

1.0

Figure: Rényi (1957), β-expansions: f(x) = x/β, β 6∈ N (hereβ = 1+

√5

2 ).

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 β-transformation : T (x) = f−1(x) mod 1 = βx mod 1.

Examples

Basic properties of β-expansions

1 Proper digit sequences (and their shifts) lexicographically <“carry sequence”: 1 = .d1d2d3 . . . (a “coded” subshift).(Parry, 1960; Gelfond, 1959).

2 Convergence (highly non unique) holds for all sequences.d1d2d3 · · · ∈ DN.

3 This β = (1/2)(1 +√

5) ≈ 1.6180 is “finite type”: nodjdj+1 = 11. Here 1 = .110000 · · · = .1010101010 . . .

4 Any x ∈ Q(β) has eventually periodic digits. This β is “Pisotunit”: β2 − b− 1 = 0, |β′| < 1 (see Schmidt, 1980, Ito-Rao,2005).

5 The base β = 3/2 = 1.5 is much harder (!)

Examples

Parry’s βx+ α

0.5 1.0 1.5 2.00.0

0.2

0.4

0.6

0.8

1.0

Figure: βx+ α expansions: f(x) = (x− α)/β.

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: Map: T (x) = βx+ α mod 1 (α =√

2− 1, β = 2√

3).

Not always topologically transitive.

Examples

Rényi-Bolyai

2 4 6 8

0.20.40.60.81.0

Figure: f(x) = −1 +√

1 + x

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: T (x) = x2 + 2x mod 1.

Examples

Rényi-Bolyai. . .

1 +√

5

2=

√1 +

√1 +√

1 + . . .

Examples

Irrational Rotation

0.5 1.0 1.5 2.00.0

0.2

0.4

0.6

0.8

1.0

Figure: f(x) = x− α for α ≤ x < α+ 1.

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: Irrational rotation map: T (x) = x+ α mod 1.

Examples

“Sturmian” expansions

For the irrational rotation map

T (x) = x+ α mod 1,

put

ξ(x) =

{0 if x ∈ [0, 1− α)1 if x ∈ [1− α, 1),

and define f as before.

The proper digit sequences ρ(x) = .d1d2d3 . . . in this case areSturmian sequences.

Parry’s Theorem implies these Sturmian expansions are valid.

Examples

Sturmian sequences

A Sturmian sequence is any d = .d1d2d3 · · · ∈ {0, 1}N such thatthere are exactly n+ 1 distinct subsequences djdj+1 . . . dj+n−1 oflength n.

Morse and Hedlund (1940) showed that any sequence with fewersubsequences must be periodic.

Theorem (Morse, Hedlund: 1940)

A sequence d ∈ {0, 1}N is Sturmian if and only if it is given by

dn = bα(n+ 1) + xc − bαn+ xc,

ordn = dα(n+ 1) + xc − bαn+ xe,

for some α ∈ [0, 1)\Q (the “slope”) and x ∈ [0, 1).

Examples

Sturmian sequences

A Sturmian sequence is any d = .d1d2d3 · · · ∈ {0, 1}N such thatthere are exactly n+ 1 distinct subsequences djdj+1 . . . dj+n−1 oflength n.

Morse and Hedlund (1940) showed that any sequence with fewersubsequences must be periodic.

Theorem (Morse, Hedlund: 1940)

A sequence d ∈ {0, 1}N is Sturmian if and only if it is given by

dn = bα(n+ 1) + xc − bαn+ xc,

ordn = dα(n+ 1) + xc − bαn+ xe,

for some α ∈ [0, 1)\Q (the “slope”) and x ∈ [0, 1).

Examples

An example of convergence

Let α =√

5−12 . Let x = .322. Then

d = .0110110101101011011010110110101101011011010110101 . . .

Here are the first 20 convergents:0, 12(7− 3

√5), 12(7− 3

√5), 7− 3

√5, 7− 3

√5, 7−

3√

5, 7− 3√

5, 7−3√

5, 7− 3√

5

All belong to N + αN. Convergence is SLOW.

Examples

Graph of convergence

The first 1000 has only a few values:12(7−3

√5), 7−3

√5, 12(61−27

√5), 92−41

√5, 12(1027−459

√5) . . .

0 200 400 600 800 1000

0.25

0.30

0.35

0.40

Figure: List Plot of first 1000 convergents

Examples

Interval Exchange

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: A 3-interval exchange

Almost always topologically transitive (Keane, 1975) & thusvalid.

Sometimes not ergodic.

Examples

Homeomorphisms of the circle

View a homeomorphism as a map T : [0, 1)→ [0, 1) with onediscontinuity.

Assume the rotation number α is irrational. Poincare (1885)proved T is semi-conjugate to a rotation by α.

1 If T is conjugate to rotation, f -expansions valid. (In this casethere is a Lebesgue-equivalent invariant measure).

2 If not, there is a wandering interval: J = [a, b) ⊆ [0, 1) so thatTn(J), n ∈ Z are pairwise disjoint. Thus T is not topologicallyconjugate, and (one can show) f - expansions are not valid.

Examples

Infinite interval exchanges

Consider an “abstract” invertible ergodic measure preservingtransformation τ on a Lebesgue probability space (Y, ν), andsuppose hν(τ)

Examples

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

Figure: von Neumann adding machine T is a ∞-interval exchange

Kakeya’s Theorem

Proof. . .

The proof requires two lemmas:

Lemma (1)

If f : [a, b]→ R is strictly monotone and satisfies |f ′(x)| ≥ a > 0a.e., then f−1(x) is absolutely continuous, and |(f−1)′(x)| ≤ 1/aa.e..

We will come back to this. And,

Kakeya’s Theorem

Proof. . .

Lemma (1)

Kakeya’s Theorem

Proof. . .

Lemma (1)

Kakeya’s Theorem

Fundamental interval lemma

Define the fundamental interval

∆(d1d2 . . . dn) = ∆(d1) ∩ T−1∆(d2) ∩ · · · ∩ T−n+1∆(dn).

Lemma (2)

The endpoints of ∆(d1 . . . dn) are

f(d1 + f(d2 + · · ·+ f(dn)))

andf(d1 + f(d2 + · · ·+ f(dn + 1))).

We will prove this, then proceed to the proof of the theorem.

Kakeya’s Theorem

Fundamental interval lemma

Define the fundamental interval

∆(d1d2 . . . dn) = ∆(d1) ∩ T−1∆(d2) ∩ · · · ∩ T−n+1∆(dn).

Lemma (2)

The endpoints of ∆(d1 . . . dn) are

f(d1 + f(d2 + · · ·+ f(dn)))

andf(d1 + f(d2 + · · ·+ f(dn + 1))).

We will prove this, then proceed to the proof of the theorem.

Kakeya’s Theorem

Proof of Lemma (2)

The inverse (T |∆(d))−1 of the dth branch of T is the restriction off(d+ x) from x ∈ [0, 1) to a subinterval.

Note that f(d+x) is constant on the two complementary intervals.

Thus ∆(d) has endpoints f(d) and f(d+ 1).

Kakeya’s Theorem

Proof of Lemma (2). . .

Now

∆(d1d2 . . . dn) = ∆(d1) ∩ T−1∆(d2) ∩ · · · ∩ T−n+1∆(dn)= ∆(d1) ∩ T−1(∆(d2) ∩ · · · ∩ T−n+1∆(dn))= ∆(d1) ∩ T−1∆(d2d3 . . . dn).

Kakeya’s Theorem

Assume by induction that ∆(d2 . . . dn) has endpoints

y0 = f(d2 + f(d2 + · · ·+ f(dn)))

andy1 = f(d2 + f(d3 + · · ·+ f(dn + 1))).

Note thatT−1{y} = ∪d∈D{f(d+ y)},

soT−1{y} ∩∆(d1) = f(d1 + y).

The lemma follows by induction.

Proof of Kakeya’s Theorem

Proof:

Let ρ(x) = .d1d2d3 . . . , and Dn := |∆(d1d2 . . . dn)|.

We will show Dn → 0.

Define fd : [0, 1]→ [0, 1], d ∈ D, fd(y) = f(d+ y).

Then, since |T ′(x)| > 1 > 0 a.e.,

|f ′d(y)| < 1 almost everywhere, andfd is absolutely continuous.

Proof:

Let ρ(x) = .d1d2d3 . . . , and Dn := |∆(d1d2 . . . dn)|.

We will show Dn → 0.

Define fd : [0, 1]→ [0, 1], d ∈ D, fd(y) = f(d+ y).

Then, since |T ′(x)| > 1 > 0 a.e.,

|f ′d(y)| < 1 almost everywhere, andfd is absolutely continuous.

Proof. . .

Fix n. For y ∈ [0, 1] define F0(y) = y.

For k = 1, 2, . . . , n, put

Fk(y) = fdn−(k−1)(Fk−1(y)). (1)

Thus

Fk(y) = f(dn−(k−1) + f(dn−(k−2) + · · ·+ f(dn + y) . . . ).

Proof. . .

Taking k = n gives

Fn(y) = f(d1 + f(d2 + · · ·+ f(dn + y) . . . ).

Thus by Lemma (2), ∆(d1d2 . . . dn) has endpoints Fn(0) andFn(1). ”

It follows that

Dn = |Fn(1)− Fn(0)|.

Proof. . .

For k = 0, 1, . . . , n, let En,k = |Fk(1)−Fk(0)|, so that Dn = En,n.

We have

En,k = |Fk(1)− Fk(0)| =

∣∣∣∣∣∫ Fk−1(1)Fk−1(0)

|f ′dn−(k−1)(y)| dy

∣∣∣∣∣ (2)< |Fk−1(1)− Fk−1(0)| = En,k−1. (3)

Here (2) and (3) follow from the FTC and |f ′d(x)| < 1 a.e., usingLemma (1).

Proof. . .

Now we have

En,0 = 1, En,k−1 > En,k, En,n = Dn.

All En,k lie in [Dn, 1].

It follows that two consecutive ones are ≤ (1−Dn)/n apart.

Thus there is 1 ≤ kn ≤ n so that

Cn := En,kn−1 − En,kn ≤1−Dnn

. (4)

So Cn → 0.

Proof. . .

Now we have

En,0 = 1, En,k−1 > En,k, En,n = Dn.

All En,k lie in [Dn, 1].

It follows that two consecutive ones are ≤ (1−Dn)/n apart.

Thus there is 1 ≤ kn ≤ n so that

Cn := En,kn−1 − En,kn ≤1−Dnn

. (4)

So Cn → 0.

Proof. . .

Using (2), (3) and (4),∣∣∣∣∣∫ Fkn−1(1)Fkn−1(0)

|f ′dn−(kn−1)(y)| dy

∣∣∣∣∣ = |Fkn−1(1)−Fkn−1(0)|+Cn. (5)

Proof. . .

There are now two cases:

1 There exists d ∈ D so that dnk−1 = d for infinitely many k. Inthis case, by passing to a further subsequence, we assume thisholds for all k.

2 Alternatively, dnk−1 is unbounded. In this case, by passing toa further subsequence, we assume dnk−1 → ±∞.

Proof. . .

Consider (Fn,kn−1(0), Fn,kn−1(1)) ∈ [0, 1]2.

By passing a subsequence, assume

(Fkn−1(0), Fkn−1(1))→ (a, b) ∈ [0, 1]2.

In case (1), apply the Dominated Convergence Theorem to (5),with dkn−1 = d.

Using Cn → 0, we have∫ ba|f ′d(y)| dy = b− a.

If a 6= b then |f ′d(y)| < 1 a.e implies b− a < b− a. Thus a = b.

Proof. . .

In case (2) we have |f ′dnk−1(y)| → 0 uniformly.

This is because limx→±∞ f(x) ∈ {0, 1}.

The Dominated Convergence Theorem gives∫ ba 0 dy = b− a,

which again implies a = b.

To finish, it follows that

limDn ≤ limEn,kn = lim |Fkn−1(1)− Fkn−1(0)| = |b− a| = 0.

Proof. . .

In case (2) we have |f ′dnk−1(y)| → 0 uniformly.

This is because limx→±∞ f(x) ∈ {0, 1}.

The Dominated Convergence Theorem gives∫ ba 0 dy = b− a,

which again implies a = b.

To finish, it follows that

limDn ≤ limEn,kn = lim |Fkn−1(1)− Fkn−1(0)| = |b− a| = 0.

Lemmas on absolute continuity

Lemma(1) and Banach’s lemma

Lemma (1)

If ϕ : [a, b]→ R is continuous, strictly monotone and satisfies|ϕ′(x)| ≥ a > 0 a.e., then ϕ−1(x) is absolutely continuous, and|(ϕ−1)′(x)| ≤ 1/a a.e..

Let I be an interval and let λ denote Lebesgue measure. Afunction f : I → R is said to satisfy Lusin’s property-N ifλ(f(E)) = 0 whenever E ⊆ I has λ(E) = 0.

Lemma (Banach)

A continuous function f : I → R of bounded variation is absolutelycontinuous if and only if it satisfies Lusin’s property-N.

Proof of Banach’s lemma

We only prove property-N implies absolute continuity (the converseis easier, but not needed).

For simplicity, we assume f is strictly increasing.

The case of strictly decreasing is the same.

The case of bounded variation is harder, but not needed here.

Proof of Banach’s lemma. . .

Let I denote a finite set of disjoint open intervals in Ii ⊆ I.Let |I| = ∪Ii∈IIi.

Let J = f(I) = {f(Ii)} be the collection of their images (here weuse that f is strictly monotone).

Recall that ϕ is absolutely continuous if for all � > 0 there existsδ > 0 so that λ(|I|) < δ implies λ(|f(I)|) < �.

Suppose f satisfies property-N but is not absolutely continuous.

Let I denote a finite set of disjoint open intervals in Ii ⊆ I.Let |I| = ∪Ii∈IIi.

Let J = f(I) = {f(Ii)} be the collection of their images (here weuse that f is strictly monotone).

Recall that ϕ is absolutely continuous if for all � > 0 there existsδ > 0 so that λ(|I|) < δ implies λ(|f(I)|) < �.

Suppose f satisfies property-N but is not absolutely continuous.

Since ϕ is not absolutely continuous, there exists �0 > 0 and

I1, I2, I3, . . .

withλ(|In|) ≥ �0 (6)

for all n = 1, 2, 3, . . . , and

∞∑n=1

λ(|f(In)|)

Let Nn(y) = χ|f(In)|(y) (the characteristic function of |f(In)|).

Then by (6) ∫ϕ(I)

Nn(y) ≥ �0 > 0 (8)

for all n.

LetB = lim sup |In|.

By (7), the Borel-Cantelli lemma implies λ(B) = 0.

LetA = lim sup |f(In)|.

Note that A = f(B).

Then λ(A) = 0 since f satisfies property-N.

Now, y ∈ A iff Nn(y) 6= 0 infinitely often.

It follows that limn→∞Nn(y) = 0 for y ∈ f(I)\A (that is, a.e. y).

By the Dominated Convergence Theorem

limn→∞

∫f(I)

Nn(y)dy = 0,

contradicting (8).

Proof of Lemma (1).

Lemma (1’)

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

Proof of Lemma (1’):

Let

[c, d] = ϕ([a, b]),

λ=Lebeague measure on [a, b], and

λ′=Lebeague measure on [c, d].

Note that

|ϕ′(x)| = d(λ′ ◦ ϕ)dλ

(x).

We haveλ′ ◦ ϕ = |ϕ′| · λ+ σ where σ ⊥ λ.

By the Banach lemma, it suffices to show ϕ−1 has property-N(i.e., λ ◦ ϕ−1 0 implies λ′(ϕ(A)) > 0.

But

(λ′ ◦ ϕ)(A) =∫A|ϕ′(x)| dλ(x) + σ(A) > 0,

since |ϕ′(x)| > 0 a.e.

Lemma (1”)

Let ϕ : [a, b]→ R be continuous, strictly monotone and satisfying|ϕ′(x)| > a > 0 a.e., then |ϕ−1(y)| < 1/a a.e..

Proof of Lemma (1”):For a continuous monotonic function ϕ, the inverse functiontheorem

(ϕ−1)′(ϕ(x)) =1

ϕ′(x),

holds whenever ϕ′(x) exists and is nonzero [?].

Note that ϕ′(x) exists λ a.e.. Let E = {x : ϕ′(x) ≤ a}

By assumption, λ(E) = 0.

Now ϕ(E) = {y : ϕ′(ϕ−1)′(y))) ≤ a}

Let F be the set of y so that (ϕ−1)′(y) does not exist.

Since ϕ−1 is monotone, λ(F ) = 0.

It suffices to show λ(ϕ(E) ∪ F ) = λ(ϕ(E)) = 0, because thenλ((ϕ(E) ∪ F )c) = b− a, and the inverse function theorem implies

(ϕ−1)′(y) =1

ϕ′(ϕ−1(y))< 1/a for λ a.e. y.

This follows from the next lemma.

Lemma

Let ϕ : [a, b]→ R and let E = {x : ϕ′(x) ≤ a}. Then

λ∗(ϕ(E)) ≤ aλ∗(E),

where λ∗ denotes Lebesgue outer measure.

Comment: If λ∗(E) = 0 then E is measurable, and this impliesλ(E) = 0. This is what happens in our case.

References

Adler, Roy L., f -expansions revisited. Recent advances intopological dynamics, Springer Lecture Notes in Math., 318,(1973). 1–5.

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References (cont.)

Parry, W., On the β-expansions of real numbers. Acta Math.Acad. Sci. Hungar. 11 (1960) 401–416.

Parry, W. Representations for real numbers. Acta Math.Acad. Sci. Hungar. 15 (1964) 95–105.

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Some examplesThe generalized algorithmExamplesKakeya's TheoremProof of Kakeya's TheoremLemmas on absolute continuity