13. Examples of measure-preserving tranformations: rotations of a torus, the doubling map

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1 3. Examples of measure-preserving tranformations: rotations of a torus, the doubling map 3. Rotations of a torus, the doubling map In this lecture we give two methods by which one can show that a given dynamical system preserves a given measure. We shall illustrate these two methods by proving that (i) a rotation of a torus, and (ii) the doubling map preserve Lebesgue measure. Let us first recall how these examples are defined. 3.. Rotations on tori Take X = R k /Z k, the k-dimensional torus. Recall that Lebesgue measure µ is defined by first defining the measure of a k-dimensional cube [a, b ] [a k, b k ] to be k k µ [a j, b j ] = (b j a j ) j= and then extending this to the Borel σ-algebra by using the Kolmogorov Extension Theorem. Fix α = (α,..., α k ) R k and define T : X X by j= T (x,..., x k ) = (x + α,..., x k + α k ) mod. (In multiplicative notation this becomes: T (e 2πiθ,..., e 2πiθ k ) = (e 2πi(θ +α ),..., e 2πi(θ k+α k ) ). This is the rotation of the k-dimensional torus R k /Z k by the vector (α,..., α k ). In dimension k = we get a rotation of a circle defined by 3..2 The doubling map T : R/Z R/Z : x x + α mod. Let X = R/Z denote the circle. The doubling map is defined to be T : R/Z R/Z : x 2x mod.

2 3.2 Using the Kolmogorov Extension Theorem Recall the Kolmogorov Extension Theorem: Theorem 3. (Kolmogorov Extension Theorem) Let A be an algebra of subsets of X. Suppose that µ : A R + { } satisfies: (i) µ( ) = ; (ii) there exists finitely or countably many sets X n A such that X = n X n and µ(x n ) < ; (iii) if E n A, n, are pairwise disjoint and if n= E n A then ( ) µ E n = µ(e n ). n= Then there is a unique measure µ : B(A) R + { } which is an extension of µ : A R + { }. That is, if something looks like a measure on an algebra A, then it extends uniquely to a measure defined on the σ-algebra B(A) generated by A. Corollary 3.2 Let A be an algebra of subsets of X. Suppose that µ and µ 2 are two measures on B(A) such that µ (E) = µ 2 (E) for all E A. Then µ = µ 2 on B(A). To show that a dynamical system T preserves a probability measure µ we have to show that T µ = µ. By the above corollary, we see that it is sufficient to check that T µ = µ on an algebra that generates the σ-algebra. Recall that the collection of all finite unions of sub-intervals forms an algebra of subsets of both [, ] and R/Z that generates the Borel σ-algebra. Similarly, the collection of all finite unions of k-dimensional sub-cubes of R k /Z k forms an algebra of subsets of the k-dimensional torus R k /Z k that generates the Borel σ-algebra. Thus to show that for a dynamical system T defined on R/Z preserves a measure µ we need only check that for all subintervals (a, b). n= T µ(a, b) = µt (a, b) = µ(a, b) 2

3 3.2. Rotations of a circle We claim that the rotation T (x) = x + α mod preserves Lebesgue measure µ. First note that Hence T (a, b) = {x T (x) (a, b)} = (a α, b α). T µ(a, b) = µt (a, b) = µ(a α, b α) = (b α) (a α) = b a = µ(a, b). Hence T µ = µ on the algebra of finite unions of subintervals. As this algebra generates the Borel σ-algebra, by uniqueness in the Kolmogorov Extension Theorem we see that T µ = µ; i.e. Lebesgue measure is T -invariant The doubling map We claim that the doubling map T (x) = 2x mod preserves Lebesgue measure µ. First note that ( a T (a, b) = {x T (x) (a, b)} = 2, b ) ( a +, b + ) Hence T µ(a, b) = µt (a, b) ( a = µ 2, b ) ( a +, b + ) = b 2 a (b + ) (a + ) = b a = µ(a, b). Hence T µ = µ on the algebra of finite unions of subintervals. As this semialgebra generates the Borel σ-algebra, by uniqueness in the Kolmogorov Extension Theorem we see that T µ = µ; i.e. Lebesgue measure is T -invariant. 3.3 Using Fourier series Let B denote the Borel σ-algebra on R/Z and let µ be Lebesgue measure. Given a Lebesgue integrable function f L (R/Z, B, µ), we can associate to f the Fourier series a 2 + (a n cos 2πnx + bn sin 2πnx), n= 3

4 where a n = 2 f(x) cos 2πnx dµ, b n = 2 f(x) sin 2πnx dµ. (Notice that we are not claiming that the series converges we are just formally associating the Fourier series to f.) We shall find it more convenient to work with a complex form of the Fourier series: c n e 2πinx, (3.) where c n = n= f(x)e 2πinx dµ. (In particular, c = f dµ.) We are still not making any assumption as to (i) whether the series (3.) converges at all, or (ii) whether, if the series does converge, it converges to f(x). In general, answering these questions relies on the class of function to which f belongs. The weakest class of function is f L (X, B, µ). In this case, we only know that the coefficients c n as n. Although this condition is clearly necessary for (3.) to converge, it is not sufficient, and there exist examples of functions f L (X, B, µ) for which (3.) does not converge to f(x). Lemma 3.3 (Riemann-Lebesgue Lemma) If f L (R/Z, B, µ) then c n as n, i.e.: lim n ± f(x)e 2πinx dµ =. It is of great interest and practical importance to know when and in what sense the Fourier series converges to the original function f. For convenience, we shall write the nth partial sum of (3.) as s n (x) = l= n c l e 2πilx and the average of the first n partial sums as σ n (x) = n (s (x) + s (x) + + s n (x)). We define L 2 (X, B, µ) to be the set of all functions f : X R such that f 2 dµ <. Notice that L 2 L. 4

5 Theorem 3.4 (i) If f L 2 (R/Z, B, µ) then s n converges to f in L 2, i.e., s n f 2 dµ, as n. (ii) If f C(R/Z) then σ n converges uniformly to f as n, i.e., σ n f, as n. In summary: Class of Property of Fourier Fourier series converges function coefficients to function L c n Not in general L 2 partial sums s n converge Yes, s n f (convergence in L 2 sense) continuous averages σ n of partial Yes, σ n f sums converge (uniform convergence) 3.3. Rotations of a circle Let T (x) = x + α mod be a circle rotation. We now give an alternative method of proving that µ is T -invariant using Fourier series. Recall Lemma 2.3: µ is T -invariant if and only if f T dµ = f dµ for all f C(X, R). Heuristically, the argument is as follows. First note that { e 2πinx, if n dµ =, if n =. If f C(X, R) has Fourier series n Z c ne 2πinx then f T has Fourier series n Z c ne 2πinα e 2πinx. The underlying idea is the following: f T dµ = c n e 2πinα e 2πinx dµ n Z = c n e 2πinα n Z = c = f dµ. e 2πinx dµ Notice that the above involves saying the the integral of an infinite sum is the infinite sum of the integrals. This is not necessarily the case, so to make this argument rigorous we need to use Theorem 4.4(ii) to justify this step. 5

6 Let f C(X, R). Then f has a Fourier series c n e 2πinx. n Z Let s n (x) denote the nth partial sum: Then s n (x) = s n (T x) = l= n l= n c l e 2πilx. c l e 2πilα e 2πilx and this is the nth partial sum for the Fourier series of f T. As e 2πilx dµ = unless l =, it follows that s n dµ = c = s n T dµ. Consider σ n (x) = n (s + + s n )(x). Then σ n (x) f(x) uniformly. Moreover, σ n (T x) f(t x) uniformly. Hence f dµ = lim n σ n dµ = c = lim n σ n T dµ = f T dµ and Lemma 3.3 implies that Lebesgue measure is invariant. 3.4 The doubling map Define T : X X by T (x) = 2x mod. Heuristically, the argument is as follows: If f has Fourier series n c ne 2πinx then f T has Fourier series n c ne 2πi2nx. Hence f T dµ = c n e 2πi2nx dµ n = c n n = c = f dµ. e 2πi2nx dµ Again, this needs to be made rigorous, and the argument is similar to that above. 6

7 3.4. Higher dimensional Fourier series Let X = R k /Z k be the k-dimensional torus and let µ denote Lebesgue measure on X. Let f L (X, B, µ) be an integrable function defined on the torus. For each n = (n,..., n k ) Z k define c n = f(x)e 2πi n,x dµ where n, x = n x + + n k x k. Then we can associate to f the Fourier series: n Z k c n e 2πi n,x, where n = (n,..., n k ), x = (x,..., x k ). Essentially the same convergence results hold as in the case k =, provided that we write s n (x) = l = n l k = n c l e 2πi l,x. Exercise 3. For an integer k 2 define T : R/Z R/Z by T (x) = kx mod. Show that T preserves Lebesgue measure. Exercise 3.2 Let β > denote the golden ratio (so that β 2 = β + ). Define T : [, ] [, ] by T (x) = βx mod. Show that T does not preserve Lebesgue measure. Define the measure µ by µ(b) = B k(x) dx where k(x) = β + β 3 β β + on [, /β) β 3 on [/β, ). By using the Kolmogorov Extension Theorem, show that T preserves µ. Exercise 3.3 Define the logistic map T : [, ] [, ] by T (x) = 4x( x). Define the measure µ by µ(b) = dx. π x( x) (i) Check that µ is a probability measure. B (ii) By using the Kolmogorov Extension Theorem, show that T preserves µ. 7