Proof of the transversality condition

To verify (1.4.15) we fix an i= ((i0, κ0)(i1, κ1)· · ·)∈ Σ and m ∈ Qⁱ. Using that a_bⁱ

oim+1

=a_bⁱ

ri m

=ai0 by definition we have

fi(ai₀) = a_bⁱ

ri m+1+

oⁱ_m−1

X

l=rⁱm+1

(a_bⁱ

l+1−a_bⁱ

l)λ^♯i(p

i l)−♯i(pⁱ

ri m

)+(a_bⁱ

oi

m+1−a_bⁱ

oi m

)λ^♯i(p

i oi

m

)−♯i(pⁱ

ri m

)

and

d^m_i =

oⁱ_m

X

l=r_mⁱ

(a_bⁱ

l+1−a_bⁱ

l)λ^♯i(pil)

λ^m_(i0,2) = λ^♯i(p

i ri

m)

λ^m_(i0,2) fi(ai0)−a_bⁱ

ri m

! . Which completes the proof of (1.4.15). Therefore

|d^m_i | ≤ λ^♯i(pirim)

λ^m_(i0,2) max{aN−1−ai0, ai0 −a₀}.

Now let us suppose that 0 ∈ Qⁱ then r₀ⁱ = 0. Moreover bⁱ₀ contains only (i0,1). Then by |bⁱ₀|=k₀ⁱ we have

d^m_i =λ^k_(iⁱ⁰_0,1) fi^′(ai0)−ai0

, where i^′ = (bⁱ₁· · ·bⁱ_oi

0). By definition, bⁱ₁ does not contain elements from {(i0,1),(i0,2)}. Then by (1.4.3) and λ(i,2) < λ(i,1) we have

|fi^′(ai0)−ai0| ≥min

f_(i0+1,1)(a₀)−ai0, ai0 −f_(i0−1,1)(aN−1) which completes the proof.

We prove in Lemma 1.4.2 below that for every k ≥ 2 the IFS Ψk satisfies transversality on a certain parameter domain Rε. Using this, in Proposi-tion 1.4.4, we verify that the transversality holds on a domain which ap-proximates the parameter domain that appears in Theorem 1.1.2. First we introduce the corresponding notation. Let us denote the attractor of Ψk by Ω^λ_k and the natural projection from Σk := Uk^N onto Ω^λ_k by π_k^λ. Denote the elements of Σk by i^′ = (i₀i₁· · ·).

Lemma 1.4.2. Let0< εi < λ_(i,1) for everyi= 0, . . . , N−1. Then for every k ≥2 and every i^′ = (i₀i₁· · ·),j^′ = (j₀j₁· · ·)∈Σk such that i₀ 6=j₀ ∈ U^k,

π^eλ_k(i^′) =π_k^eλ(j^′) =⇒

∂

∂λ(i,2)

π_k^λ(i^′)−π_k^λ(j^′)

λ=eλ

>0, (1.4.18) for some i and for every

λe₂ ∈Rε =Y

i∈N





εi,min









λ_(i,1), 1

1 + r

λmaxαi

1 + ^α_εⁱ

i













, (1.4.19)

if it exists, where λmax= maxi=0,...,N−1

λ(i,1) and

αi = max{aN−1 −ai, ai−a0} min

f_(i+1,1)(a₀)−ai, ai−f_(i−1,1)(aN−1) . To prove Lemma 1.4.2 we need the following Sublemma:

Sublemma 1.4.3. Let i, j finite length word of symbols such that

i=

k1

z }| {

(i,1)· · ·(i,1)(l1, κ1) j =

k₂

z }| {

(i,2)· · ·(i,2)(l2, κ2)

where l1, l2 6=i. If fi([a0, aN−1])∩fj([a0, aN−1])6=∅ then λ^k_(i,2)²

λ^k_(i,1)¹ ≤αi.

Proof. Since for every (i,2) ∈ J, λ(i,2) < λ(i,1), we have that fi([a0, aN−1])∩ fj([a0, aN−1])6= ∅ implies

λ^k_(i,1)¹ λ_(l1,κ₁)a₀+λ^k_(i,1)¹ al₁(1−λ_(l1,κ₁)) +ai(1−λ^k_(i,1)¹ )≤

λ^k_(i,2)² λ_(l2,κ2)aN−1+λ^k_(i,2)² al2(1−λ_(l2,κ2)) +ai(1−λ^k_(i,2)² ), λ^k_(i,2)² λ(l2,κ2)a0+λ^k_(i,2)² al₂(1−λ(l2,κ2)) +ai(1−λ^k_(i,2)² )≤

λ^k_(i,1)¹ λ(l1,κ₁)aN−1+λ^k_(i,1)¹ al₁(1−λ(l1,κ₁)) +ai(1−λ^k_(i,1)¹ ).

Using the fact that F satisfies (1.4.3), we have l1, l2 > i or l1, l2 < i. One can finish the proof by some obvious algebraic manipulations.

Proof of Lemma 1.4.2. Let 0 < εi < λ(i,1) and suppose that εi < λ(i,2) for every i ∈ N. Let i^′,j^′ ∈ Σk such that i₀ 6=j₀ and π_k^λ(i^′) = π_k^λ(j^′). Divide i₀ and j₀ into blocks such that i₀ = (bⁱ₀⁰· · ·bⁱ_l⁰) and j₀ = (b^j₀⁰· · ·b^jq⁰). By defi-nition, a block consists of such pairs which share the same first component.

If u is the common first element in the case of the block bⁱ₀⁰ and v for b^j₀⁰ then applying (1.4.3) we obtain that u=v. That is the first elements of all of the pairs that are contained either in bⁱ₀⁰ or in b^j₀⁰ are the same. First let us assume that both of i₀ and j

0 begin with (i,2). Then by the definition of U^k (see (1.4.16)), bⁱ₀⁰, b^j₀⁰ contain only (i,2). Since S satisfies (1.4.3) we have that |bⁱ₀⁰|=|b^j₀⁰|=n. This implies that

0 =π_k^λ(i^′)−π_k^λ(j^′) =λⁿ_(i,2)

π_k^λ(i^′∗)−π_k^λ(j^′∗)

where the first element of i^′∗ is (bⁱ₁⁰· · ·bⁱ_l⁰)∈Σk and the first element of j^′∗ is (b^j₁⁰· · ·b^jq⁰)∈Σk. Since λ(i,2) > εi, without loss of generality we can assume that i₀ = (i,1) and b^j₀⁰ contains only (i,2) for an i ∈ N. Let us write i,j for the elements of Σ = (I ∪ J)^N that correspond to i^′,j^′ respectively. Then π_k^λ(i^′)≡πλ,a(i) and π_k^λ(j^′)≡πλ,a(j).

If♯(i,2)i(kⁱ₀)≥♯(i,2)j(k₀^j) then by (1.4.3),πλ,a(i)6=πλ,a(j) therefore without loss of generality we assume that ♯(i,2)i(kⁱ₀)< ♯(i,2)j(k^j₀). Then

πλ,a(i)−πλ,a(j) = λ^♯_(i,2)^(i,2)i(ki0)(πλ,a(i^∗)−πλ,a(j^∗)), where

i^∗ = (

♯_(i,1)i(kⁱ₀)

z }| {

(i,1)· · ·(i,1)bⁱ₁· · ·) andj^∗ = (

♯_(i,2)j(k₀^j)−♯_(i,2)i(k₀ⁱ)

z }| {

(i,2)· · ·(i,2) b^j₁b^j₂· · ·).

Since λ(i,2) > εi >0 it is enough to prove that

f(λ) = 0 =⇒ kgradf(λ)k>0, (1.4.20) where f(λ) =πλ,a(i^∗)−πλ,a(j^∗). Let m = minQ^j∗ then by (1.4.11) we have

f(λ) =d⁰_i^∗



1 + X

k∈Q^i∗\{0}

d^k_i^∗

d⁰_i^∗λ^k_(i,2)− X

k∈Q^j∗

d^k_j^∗ d⁰_i^∗λ^k_(i,2)



=

d⁰_i^∗



1 + X

k∈Q^i∗\{0}

d^k_i^∗

d⁰_i^∗λ^k_(i,2) − X

k∈Q^j∗

d^k_j^∗λ^m_(i,2) d⁰_i^∗λ(i,2)

λ^k_(i,2)^−m+1



.

Now we give upper bound for the absolute value of the coefficients. It is easy to see by Lemma 1.4.1 and Sublemma 1.4.3 that

^dd^ki0_i^∗∗

≤λ_maxαi for every k ∈Qⁱ_i^∗\ {0}

^d

m j∗λ^m_(i,2) d⁰_i∗λ_(i,2)

≤ ^α_εi²ⁱ and

d^k_j∗λ^m_(i,2) d⁰_i∗λ_(i,2)

≤λ_max^α_ε²ⁱ

i for every k ∈Q^j_i^∗\ {m}.

Therefore absolute value of the coefficient of λ_(i,2) is at most λ_maxαi + ^α_ε²ⁱ

i

and the absolute value of the coefficient of λ^k_(i,2) for k ≥ 2 is at most λmaxαi+λmax

α²_i

εi. If f(eλ) = 0 then

∂f

∂λ(i,2)

(λ) =d⁰_i^∗



 X

k∈Q^i∗\{0}

d^k_i^∗

d⁰_i^∗kλ^k_(i,2)⁻¹ − X

k∈Q^j∗

d^k_j^∗λ^m_(i,2) d⁰_i^∗λ(i,2)

(k−m+ 1)λ^k_(i,2)^−m

− X

k∈Q^j∗

(m−1)d^k_j^∗λ^m_(i,2)⁻²

d⁰_i^∗ λ^k_(i,2)^−m+1



,

and by Lemma 1.2.3 we obtain that for λ(i,2) ∈



ε_i, ¹

1+

r λmaxαi

1+^αi_εi



 the following inequality holds:

X

k∈Q^i∗\{0}

d^k_i^∗

d⁰_i^∗kλ^k_(i,2)⁻¹ − X

k∈Q^j∗

d^k_j^∗λ^m_(i,2)

d⁰_i^∗λ_(i,2)(k−m+ 1)λ^k_(i,2)^−m <0. (1.4.21)

On the other hand, (1.4.15) yields that for suitable i^′, j^′ we have

d^m_j^∗ d⁰_i^∗ =

λ

♯j(pj rj

m )

λ^m_(i,2)

fj^′(ai)−ai

λ^ki

∗ 0

(i,1) fi^′(ai)−ai

.

Let i^′₀ and j₀^′ be the first element of the first component of i^′, j^′. Then by (1.4.3), i^′₀, j₀^′ > i or i^′₀, j₀^′ < i which implies that ^d

m j∗

d⁰_i∗ > 0. Therefore by Lemma 1.4.1 we have for λ_(i,2) < _1+λ¹

maxαi that X

k∈Q^j∗

(m−1)d^k_j^∗λ^m_(i,2)⁻²

d⁰_i^∗ λ^k_(i,2)^−m+1 = (m−1)d^m_j^∗ d⁰_i^∗λ^m_(i,2)⁻¹



1 + X

k∈Q^j∗\{m}

d^k_j^∗ d^m_j^∗λ^k_(i,2)^−m



≥

(m−1)d^m_j^∗

d⁰_i^∗λ^m_(i,2)⁻¹ 1− X∞ k=1

λmaxαiλ^k_(i,2)

!

≥0. (1.4.22) Observe that ¹

1+

r λmaxαi

1+^αi_εi < _1+λmax¹ _α

i holds for every 0 < εi <1. Using this (1.4.21) and (1.4.22) we have

f(eλ) = 0 =⇒ ∂f

∂λ(i,2)

(eλ)<0 which was to be proved.

Proposition 1.4.4. For every k ≥2, the IFS Ψk satisfies the transversality condition on

λ₂ ∈ T^N(ξ) = Y

i∈N

(ξ,min

λ_(i,1), 2

(1 +√

2)(α_i²λmax+ 2)

−ξ) (1.4.23) where ξ >0 is arbitrary small and

αi = max{aN−1−ai, ai−a0}

min{fi+1(a0)−ai, ai−fi−1(aN−1)} for i∈ N. Proof. Let

gi(x) = 1

1 +q

λmaxαi 1 + ^α_xⁱ.

We can extend gi onto [0,∞) as gi(0) = 0, which is a fixed point of gi. It is easy to see by simple calculations that gi is strictly monotone increasing and has a unique positive fixed point ε^∗_i.

Hence, we can cover the rectangle Q

i∈N(0,min

λ(i,1), ε^∗_i ) by countable many rectangles in the type Rε, see (1.4.19).

It follows from Lemma 1.4.2 that for every k ≥ 2 and i^′,j^′ ∈ Σk with i₀ 6=j₀ the function π_k^λ(i^′)−π_k^λ(j^′) satisfies (1.2.3) on the rectangle

Q

i∈N(0,min

λ_(i,1), ε^∗_i ).

Now we are going to prove that 2 (√

2 + 1)(α²_iλmax+ 2) ≤ε^∗_i. (1.4.24) To verify this, observe that

ε^∗_i = 2

p(α_i²λmax+ 2)²+ 4(αiλmax−1) +α²_iλmax+ 2.

If the second term under the square root is non-positive, that is if αiλmax≤1 then clearly (1.4.24) holds. Otherwise, αiλmax >1. Then αi > 1. A simple calculation yields: 4(αiλ_max−1)≤(α_i²λ_max+ 2)² which follows that (1.4.24) holds. To complete the proof we apply Lemma 1.2.2 for the rectangle on the right hand side of (1.4.23) with ξ = 0.

1.4.3 Hausdorff dimension

Before we prove the theorems we have to introduce a sequence of functions.

For everyk ≥2 we introduce the functionhλ,k(s) which is defined as the sum of the s-powers of the contraction ratios of the IFS Ψk. That is

hλ,k(s) =

NX−1 i=0

λ^s_(i,1)+

k−2

X

l=0

X

i∈N

λ^s_(i,2)

!l

X

i∈N NX−1 j=0,j6=i

λ^s_(i,2)λ^s_(j,1). (1.4.25)

Letsk(λ) be the unique solution ofhλ,k(s) = 1. Therefore dimHΩ^λ_k ≤ min{1, sk(λ)}, where Ω^λ_k is the attractor of Ψk.

Since the sequence sk(λ) is monotone increasing and bounded, it is con-vergent. It is easy to see by some algebraic manipulation that the limit of sk(λ) is the unique solution of

NX−1 i=0

λ^s_(i,1)+X

i∈N

λ^s_(i,2) 1−λ^s_(i,1)

= 1.

This equation corresponds to (1.1.5).

Moreover, we need to introduce a sequence of subsets of Σ^∗. Let

C¹ =I ={(0,1), . . . ,(N −1,1)} (1.4.26) and by induction let

Ck+1 =

N[−1 j=0

[

i∈Ck

{(j,1)i} ∪ [

j∈N

[

i∈Ck

(i0,κ₀)6=(j,1)

{(j,2)i}. (1.4.27)

Then we can look at the elements ofC^k either as certain sequences of lengthk of symbols from I ∪ J or juxtapositions of at most k elements of U^k. Lemma 1.4.5. Let esk(λ) be the unique solution of

X

i∈Ck

λ^s_i = 1, and let es(λ) = sup_kesk(λ) then

dimHΩλ,a≤min{1,es(λ)}. Moreover,

H^es(λ)(Ωλ,a)≤(aN−1−a₀)^es(λ). Note thatesk(λ) is bounded since C^k⊂(I ∪ J)^k. Proof. Using that for every i∈ N

f(i,1)◦f(i,2) ≡f(i,2)◦f(i,1),

and 0< λ(i,2) < λ(i,1) <1 we have that the set of closed intervals {fi([a0, aN−1])}_i∈Ck

gives a cover of Ωλ,a with diameter at most λ^k_max. Then H_λ^es(λ)k_max(Ωλ,a)≤X

i∈Ck

|fi([a0, aN−1])|^es(λ)= (aN−1−a0)^s(λ)e X

i∈Ck

f_i^′(0)^es(λ) ≤ (aN−1−a0)^es(λ)X

i∈Ck

f_i^′(0)^esk(λ)

| {z }

1

= (aN−1−a0)^s(λ)e .

This proves the upper bound of the dimension and the measure claim of the Lemma.

Proof of Theorem 1.1.2. Letξ > 0. By the definition ofC^k we have that for every k≥1

C^k ⊂ [k l=1

Uk^l. (1.4.28)

As it was mentioned above, every i ∈ C^k can be decomposed as a juxtapo-sition i = j₁· · ·j_r, where each j_l is in U^k and 1 ≤ r ≤ k. By using this fact and Proposition 1.4.4 we have that the system Ψek = {fi}_i∈Ck satisfies transversality on T^N(ξ). By Theorem 1.2.1 we have

dimHΩe^λ_k = min{1,esk(λ)} for L-a.e. λ₂ ∈ T^N(ξ), (1.4.29) where Ωe^λ_k denotes the attractor of {fi}i∈Ck. Using (1.4.28)

dimH Ωe^λ_k ≤dimHΩ^λ_k.

Moreover by Proposition 1.4.4 and Theorem 1.2.1 we have dimHΩ^λ_k = min{1, sk(λ)} for L-a.e. λ₂ ∈ T^N(ξ).

Since Ωe^λ_k,Ω^λ_k ⊆Ωλ,a for every k ≥2 by Lemma 1.4.5 we have min{1,esk(λ)} ≤min{1, sk(λ)} ≤min{1,es(λ)}.

Since sk(λ) is strictly monotone increasing limk→∞sk(λ) = sup_ksk(λ). This implies that min{1, s(λ)}= min{1,es(λ)}, moreover

dimHΩλ,a= min{1, s(λ)}.

To complete the proof of the last assertion of Theorem 1.1.2 first observe that whenever s(λ) > 1 then there exists a k ≥ 2 such that sk(λ) > 1.

Therefore, by Theorem 1.2.1 and Proposition 1.4.4, L(Ωλ,a) ≥ L Ω^λ_k

>0 for a.e. λ₂ ∈ T^N(ξ)∩ {λ₂ :s(λ)>1}. Since ξ was arbitrary, this completes the proof.

1.4.4 Example

To visualize the behavior of the vector of contracting ratios we consider an easy example, where the functions of F are uniformly distributed with uniform contracting ratio, that is

F ={fi(x) =λx+i(1−λ)}^Ni=0⁻¹,

0.05 0.10 0.15 0.20Λ

0.1 0.2 0.3 0.4

Γ0,Γ4

Λ 2

J1+ 2N IΛ Α₀²+2M

0.05 0.10 0.15 0.20Λ

0.1 0.2 0.3 0.4

Γ₁,Γ3

Λ 2

J1+ 2N IΛ Α₁²+2M

0.05 0.10 0.15 0.20Λ

0.1 0.2 0.3 0.4

Γ2

Λ 2

J1+ 2N IΛ Α₂²+2M

Figure 1.1: Transversality region for N = 5 fixed points

where 0< λ < _N¹. Let us add to the system the following N functions:

G={gi(x) =γix+i(1−γi)}^Ni=0⁻¹.

Note that the fixed point of both fi and gi is i, i = 0, . . . , N −1. It is easy to see that for every i= 1, . . . , N −2

αi =αN−1−i = max{N −1−i, i}

min{1−(i+ 1)λ,1−(N −i)λ} and α₀ =αN−1 = N −1 1−λ, whereαi is as in Theorem 1.1.2. To satisfy the assumptions of Theorem 1.1.2 it is enough to require that

0< γi <min

λ, 2

(1 +√

2)(α²_iλ+ 2)

(1.4.30) holds for i= 0, . . . , N −1. For example, when N = 5 then we can choose γi

from the appropriate shaded region of Figure 1.1. In general, first we observe that

αi ≤α1 =αN−2 = N −2 1−(N −1)λ,

holds for every i = 0, . . . , N −1. So by (1.4.30) the assumptions of Theo-rem 1.1.2 hold if we assume that

0< γi <min









λ, 2

(1 +√

2)

N−2 1−(N−1)λ

2

λ+ 2









, 0≤i≤N −1.

(1.4.31) We know that 0 < λ must be smaller than 1/N. By (1.4.31) we obtain that whenever λ < 0.4764/N holds then the assumptions of Theorem 1.1.2 are satisfied for γi < λ.

In document Bal´azsB´ar´any DimensionTheoryofNon-conformalAttractorsandOverlappingSelf-similarSets (Pldal 37-46)