Dynamical -Midterm -Solved

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

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.

(a)      Let Cm,n be the set of points x ∈ X for which m and n are consecutive visit times to A. Suppose Cm,n is not measurable. Then Cm,n 6∈ S. But since T is measurable and T −1(Tm+1(X)) = Tm(X) ∈ S. Furthermore, since A measurable, we must have A∩Tm(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 ∩ Tm(X)

must be measurable. Finally, we have,

                                                                                                                         n−m                                   !

A ∩ Tm(X) ∪ A ∩ Tn(X) \ [ Tm+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 Cm,n, and so Cm,n is measurable.

(b)     Now let us take Cm,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 Cm,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 Tm+i(X) ∩ B is measurable for every i. Furthermore, we are taking a finite union of measurable sets, so Sni=1−m−1 Tm+i(X) ∩ B is measurable as well. And lastly,

Cm,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 ∈ Cm,n implies x ∈ Dm,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 = 21m. 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

=⇒ 2m2−k < 

=⇒ 2m−k < 

=⇒ 1/2k−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 xi = yi 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.