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An introduction to ergodic theory - Walters P.

Walters P. An introduction to ergodic theory - London, 1982. - 251 p.
Download (direct link): anintroduction1982.djvu
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(ii) A measure-preserving transformation always has Я = 1 as an eigenvalue and any non-zero constant function is a corresponding eigenfunction.
Definition 1.8. We say that a measure-preserving transformation T of a probability space (Х,Ув,т) has continuous spectrum if 1 is the only eigenvalue of T and the only eigenfunctions are the constants.
Observe that T has continuous spectrum iff Я = 1 is the only eigenvalue and T is ergodic.
We shall need the following result from spectral theory to prove the next Theorem. The proof can be found in Halmos [2].
Theorem 1.25 (Spectral Theorem for Unitary Operators). Suppose U is a unitary operator on a complex Hilbert space У?. Then for each f e Ж there exists a unique finite Borel measure jij- on К such that
(uv,/) = JK^/W v«eZ-
If T is an invertible measure-preserving transformation then UT is unitary, and if T has continuous spectrum and (/, 1) = 0 then nf has no atoms {i.e. each point of К has zero цmeasure).
Theorem 1.26. If T is an invertible measure-preserving transformation of a probability space {X, &l, m) then T is weak-mixiiig iff T has continuous spectrum.
Proof. Suppose T is weak-mixing and let UTf = Я/, f e L2{m). If Я # 1 then integration give^ (/,!) = 0 and by the weak-mixing property we have
§1.7 Mixing
49
and hence
Since |A| = 1 this gives (/, f) = 0 and therefore / = 0 a.e. If Я = 1 then f = constant a.e. by the ergodicity of T. (This part of the proof did not use the spectral theorem.)
Now suppose T has continuous spectrum. We show that if/ e L2(m) then
-"l WrfJ) -(/,i)(i, /)|2-o. n i=0
If / is constant a.e. this is true. Hence all we need to show is that (/, 1) = 0 implies
j-z1 i(^*r/,/)i2-o.
n i = 0
By the spectral theorem it suffices to show that if is a continuous (non-atomic) measure on К then
We have
■0.
1 n_1 -I
n i=o
JЯ' dM/(X)2 = i "l (Jягс^(Я) • Ja-*-d/z/Я) = ~ E f Ja1 ^/(Я) • Jt■ «fo/т)
i = 0 n — 1
1 n-1
= -E [T {ft)'d(pif x 1) (by Fubini’s Theorem) " 1=0 w
= JJ (~ E (W}<*(AV x ^/)(Я, t).
KxlE ' i_0 '
If (Я,т) is not in the diagonal of К x К then
‘1 - (Ят)"'
i = 0
1 - (Ят)
• 0
as л -> oo. Since /.if has no atoms the diagonal has measure 0 for jiI x j.if and therefore (l/и) (Ят)‘ -» 0 a.e. {jj.f x nf). The modulus of the integrand is bounded by 1, so we can apply the bounded convergence theorem to obtain the result. □
We now investigate the mixing properties of the examples mentioned in §1.1.
50
1 Measure-Preserving Transformations
Examples
(1) Clearly the identity transformation I of (X,3d,m) Is ei'^od.c iff all the elements of M have measure 0 or 1 itT I is strong-mixin-..
(2) A rotation T{z) = a: of the unit circle К is never weak-mixing. This follows because because if f(z) = z then f(Tz) = f(az) = af(z) and we can apply the easy part of Theorem 1.26.
(3) Theorem 1.27. No rotation Tx = ax on a compact group is weak-mixing.
Proof. We know that if T is ergodic then the group G is abelian, and if у is any character of G we have y(Tx) = y(a)y(x), which shows that T is not weak-mixing by the easy part of Theorem 1.26. □
(4) Theorem 1.28. For an endomorphism of a compact group strong-mixing, weak-mixing and ergodicity are all equivalent. (The condition for ergodicity was given in Theorem 1.10.)
Proof. We shall give the proof when G is abelian. It suffices to show that if the endomorphism A G -* G is ergodic then A is strong-mixing. If.7, 6s G then (UAy,S) = 0 eventually unless у = S = 1. So always (UAy,S) —*■ (у, 1)(1,й). Fix SeG. The collection
Ж6 = {/e L\m):{U“Af,S) -> (/, 1)(1,<5)}
is a subspace of L2(m) which is closed. (To check Жд is closed, suppose fk e .‘/C and fk~* f e L2(m). For 3 = 1 it is clear that = L2(m), so suppose (1, (>) = 0. Then
\{U”Af.SI\ ^ I(U*Af,6) - (UnAfk,6)\ + 5)|
< II/ - fk\\i IHI2 + \(1&А,д)\ (by the Schwarz inequality)
= \\f-M2 + \(UAfk,d)\.
If e > 0 is given choose к so that l|/ — fk||2 < e/2 and then choose N(c) so that n > N(c) implies \(V'Afk,b)\ < e/2.) Since Жь contains G it is equal to L2(m). Fix JeL2(m) and consider J&f} = (ff e L2(m):(UAf,g) -> (f, l)(l,g)j. Then is a closed subspace of L2(m), contains G by the above, and so equals L2(m). Hence A is strong-mixing. □
(5) Theorem 1.29. For an affine transformation T = a - A on a compact metric abehun group the following are equivalent:
(i) T is strong-mixing.
(ii) T is weak-mixing.
(iii) A is ergodic.
§1.7 Mixing
51
Proof. We shall give the proof in the case when G is connected. Let Bx = x_1/t(x) and recall that T is ergodic iff
(a) у ° Ak = у, к > 0, implies у ° A = у, and
(b) [a,BG] = G.
If A is ergodic then BG = G since the endomorphism В of 0 is one-to-one. ChoosebeGsothatB(b) = a. Define<f>\G-* Gby^(x) = bx.ThenфТ = Аф and ф preserves Haar measure in. By (4) above A is strong-mixing and so if C, De3d we have m{T~nCn Б) = т(ф(Т~"Сn D) *= т{ф T '"C r\ фО) = hi(/4_"0C П фО) ”»т[фС)т(ф) = m(C)m{D). Therefore Г is strong-mixing.
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