[SOLVED] Dynamical -Midterm

Description

Problem 1.

Let                                                         √

x <

T(x) =

1≥ 4

2                       x    x

Note then that,
	 x2 −1                     T      (x) =	x < 14

     x

In addition, note that for x , we have T and for x, we have

1. T. Thus, for any y in the image of T, we will need to consider two pre-images under the inverse function T ⁻¹(x).

Now let C be the set of intervals in . This is a sufficient semi-ring for the domain. Let us fix some interval I with endpoints a,b ∈C with b ≥ a (whether it is open or closed will not change its Lebesgue measure). We will assume the interval is open for now. Then,

λ(I) = b − a

Now let us consider T ⁻¹(I). We have two inverse images to consider,

T1−1(I) = (a2,b2)

and,

T₂⁻¹(I) = (−a² + a + ,−b² + b + )

Note that the inverse of T preserves the interval structure, so T ⁻¹(I) is a measurable set.

Now let us check λ(T ⁻¹(I)). We have,

λ(T ⁻¹(I)) = λ(T₁⁻¹(I)) + λ(T₂⁻¹(I))

= b² − a² + (−b² + b − (−a² + a + ))

= b² − a² − b² + b +  a² − a −

= b − a

= λ(I)

Hence, by Theorem 3.4.1, T is measure preserving.

Problem 2.

We have that T is continuous and measure preserving and that f is continuous and f(T(x)) ≥ f(x).

Now suppose T is recurrent. Then for every measurable set A of positive measure, there is a null set N ⊂ A such that for all x ∈ A\N there is an integer n = n(x) > 0 with Tⁿ(X) ∈ A.

Fix x₀ ∈R² and suppose f(T(x₀)) > f(x₀). Since f,T are continuous, for points x₁ near x₀, we would expect to see f(T(x₁)) > f(x₁) as well.

Problem 3.

• Let C_m,n be the set of points x ∈ X for which m and n are consecutive visit times to A. Suppose C_m,n is not measurable. Then C_m,n 6∈ S. But since T is measurable and T ⁻¹(T^m+1(X)) = T^m(X) ∈ S. Furthermore, since A measurable, we must have A∩T^m(X) is measurable as well. The same applies to T−1(Tⁿ⁺¹(X)) = Tⁿ(x). Hence, A ∩ Tⁿ(X) is measurable as well. Then we have that

A ∩ Tⁿ(X) ∪ A ∩ T^m(X)

must be measurable. Finally, we have,

n−m                                   !

A ∩ T^m(X) ∪ A ∩ Tⁿ(X) \ ^[ T^m+i(X) ∩ A

i=1

is measurable since we are taking a finite union of measurable sets and then set minusing this finite union from another measurable set. However, note that this is exactly C_m,n, and so C_m,n is measurable.

• Now let us take C_m,n and define the following set:

n−m−1                                   !

Cm,n ∩ [ Tm+i(X) ∩ B

i=1

That is, we have taken the intersection of C_m,n with the union of the set of points in X such that Tⁱ(X) ∈ B for some i with m < i < n. Note that, by the same arguments as above T^m+i(X) ∩ B is measurable for every i. Furthermore, we are taking a finite union of measurable sets, so ^Sn_i=1^−m−1 T^m+i(X) ∩ B is measurable as well. And lastly,

C_m,n is measurable by (a), so

n−m−1                                   !

Dm,n = Cm,n ∩ [ Tm+i(X) ∩ B

i=1

is measurable.

(c) We have that x ∈ E if and only if x ∈ C_m,n implies x ∈ D_m,n for all integers m and n with 0 ≤ m < n. That is,

Problem 4.

Let us show that T is continuous. That is, we want to show that for each x , we have that for all  > 0, there exists δ > 0 such that d(T(x),T(y)) <  whenever y and d(x,y) < δ. First let us fix x  and  > 0. Let m be the length of the initial constant sequence in x. Let δ = ₂¹_m. Then for any y  with d(x,y) < δ, we have that the first position where x 6= y is some k such that

k                   1

1/2 < δ =

2m

=⇒ 2^m2^−k <

=⇒ 2m−k <

=⇒ 1/2^k−m <

That is, after removing the first m elements of x and y (call these new elements x⁰, y⁰), we still have that d(x⁰,y⁰) < . Now note that T(x) will remove exactly the first m elements. Since k ≥ m and x_i = y_i for all 0 ≤ i ≤ k − 1, we thus also have that T(y) removes exactly the first m elements of y. Hence, by what we have just shown, for any fixed x and

 > 0, we have that for all y. Thus, T is continuous.