Proof of Weak-Mixing Equivalences

Ergodicity, Recurrence and Mixing

2.8 Proof of Weak-Mixing Equivalences

Exercise 2.7.12.WriteT^(k) for the k-fold Cartesian productT × · · · ×T. Prove⁽²⁹⁾ thatT^(k) is ergodic for allk2 if and only ifT⁽²⁾ is ergodic.

Exercise 2.7.13.Let T be an ergodic endomorphism of T^d. The following exponential error rate for the mixing property⁽³⁰⁾,

f₁, U_Tⁿf₂ −

f₁

f₂

S(f₁)S(f₂)θⁿ

for some θ < 1 depending on T and for a pair of constants S(f1), S(f2) depending onf1, f2∈C^∞(T^d), is known to hold.

(a) Prove an exponential rate of mixing for the map Tn : T → T defined byTn(x) =nx (mod 1).

(b) Prove an exponential rate of mixing for the automorphism ofT² defined byT : (^xy)→ y

x+y

.

(c) Could an exponential rate of mixing hold for all continuous functions?

2.8 Proof of Weak-Mixing Equivalences 55

1

n|Jk∩[0, n)|k1 n

n−1

i=0

ai−→0

asn→ ∞for eachk1, so eachJ_k has zero density. We will construct the setJ by taking a union of segments of each setJ_k. Since each of the sets J_k has zero density, we may inductively choose numbers 0< 1< 2<· · · with the property that

1

n|J_k∩[0, n)| 1

k (2.33)

forn_k and anyk1. Define the setJ by J =

∞ k=0

Jk∩[k, k+1) .

We claim two properties for the setJ, namely

• an−→

n /∈J0 asn→ ∞;

• J has density zero.

For the first claim, note thatJk∩[k,∞)⊆J by (2.32), so if J nk

thenn /∈Jk, and soan ¹_k. This shows thatan−→

n /∈J0 as claimed.

For the second claim, notice that ifn∈[_k, _k+1) then again by (2.32)J∩ [0, n)⊆Jk∩[0, n) and so

1

n|J ∩[0, n)| 1 k by (2.33), showing thatJ has density zero.

(2) =⇒ (1): The sequence (an) is bounded, so there is some R > 0 withanRfor alln1. For eachk1 chooseNk so that

J nNk =⇒ an< 1 k and so that

nNk =⇒ 1

n|J∩[0, n)| 1 k. Then fornkNk,

1 n

n−1 i=0

ai = 1 n

⎛

⎜⎝

Nk−1 i=0

ai+

Nki<ni∈J,

ai+

i /∈J, Nki<n

ai

⎞

⎟⎠

< 1 n

RNk+R|J∩[0, n)|+n1 k

2R+ 1

k ,

showing (1).

(3) ⇐⇒ (1): This is clear from the characterization (2) of property (1).

Proof of Theorem 2.36. Properties (1), (6) and (7) are equivalent by Lemma2.41applied withan=|μ(A∩T⁻ⁿB)−μ(A)μ(B)|.

(6) =⇒ (3): Given sets A1, B1, A2, B2 ∈ B, property (6) gives sets J1

andJ2 of density zero with μ

A1∩T⁻ⁿB1

−→

n /∈J₁μ(A1)μ(B1) and

μ

A2∩T⁻ⁿB2

−→

n /∈J₂μ(A2)μ(B2).

LetJ=J1∪J2; this still has density zero and lim

Jn→∞(μ×μ)

(A1×A2)∩(T×T)⁻ⁿ(B1×B2)

−(μ×μ)(A1×A2)·(μ×μ)(B1×B2)

= lim

Jn→∞μ(A1∩T⁻ⁿB1)·μ(A2∩T⁻ⁿB2)

−μ(A1)μ(A2)μ(B1)μ(B2)

= 0,

soT×T is weak-mixing since the measurable rectangles generateB×B. (3) =⇒ (1): IfT×T is weak-mixing, then property (1) holds in particular for subsets ofX×X of the form A×X and B×X, which shows that (1) holds forT, soT is weak-mixing.

(1) =⇒ (4): Let (Y,BY, ν, S) be an ergodic system and assume thatT is weak-mixing. For measurable setsA1, B1∈BandA2, B2∈BY,

1 N

N−1 n=0

(μ×ν)

A₁×A₂∩(T×S)⁻ⁿ(B₁×B₂)

= 1 N

N−1 n=0

μ(A1∩T⁻ⁿB1)ν(A2∩S⁻ⁿB2)

= 1 N

N−1 n=0

μ(A1)μ(B1)ν(A2∩S⁻ⁿB2)

+1 N

N−1 n=0

μ(A1∩T⁻ⁿB1)−μ(A1)μ(B1)&

ν(A2∩S⁻ⁿB2). (2.34)

2.8 Proof of Weak-Mixing Equivalences 57

By the characterization in (2.31) and ergodicity ofS, the expression on the right in (2.34) converges to

μ(A1)μ(B1)ν(A2)ν(B2).

The second term in (2.34) is dominated by 1

N

N−1 n=0

μ(A1∩T⁻ⁿB1)−μ(A1)μ(B1)

which converges to 0 sinceT is weak-mixing. It follows that 1

N

N−1 n=0

(μ×ν)

A₁×A₂∩(T×S)⁻ⁿ(B₁×B₂)

−→μ(A₁)μ(B₁)ν(A₂)ν(B₂)

soT×S is ergodic by the characterization in (2.31).

(4) =⇒ (2): Let (Y,BY, ν, S) be the ergodic system defined by the identity map on the singletonY ={y}. ThenT×Sis isomorphic toT, so (4) shows thatT is ergodic. Invoking (4) again now shows thatT×T is ergodic, proving (2).

(2) =⇒ (7): We must show that 1

N

N−1 n=0

μ(A∩T⁻ⁿB)−μ(A)μ(B)²−→0

as N → ∞, for everyA, B ∈ B. Letμ² denote the product measure μ×μ on (X×X,B⊗B). By the ergodicity ofT×T,

1 N

N−1 n=0

μ

A∩T⁻ⁿB

= 1

N

N−1 n=0

μ²

(A×X)∩(T×T)⁻ⁿ(B×X)

−→μ²(A×X)·μ²(B×X) =μ(A)μ(B) and

1 N

N−1 n=0

μ

A∩T⁻ⁿB2

= 1

N

N−1 n=0

μ²

(A×A)∩(T×T)⁻ⁿ(B×B)

−→μ²(A×A)·μ²(B×B) =μ(A)²μ(B)². It follows that

1 N

N−1 n=0

μ

A∩T⁻ⁿB

−μ(A)μ(B)&2

= 1 N

N−1 n=0

μ

A∩T⁻ⁿB2

+μ(A)²μ(B)²

−2μ(A)μ(B)1 N

N−1 n=0

μ

A∩T⁻ⁿB

→2μ(A)²μ(B)²−2μ(A)²μ(B)²= 0, so (7) holds.

(2) =⇒ (5): Suppose thatf is a measurable eigenfunction forT, so UTf =λf

for someλ∈S¹. Define a measurable function onX×X by g(x₁, x₂) =f(x₁)f(x₂);

then

UT×Tg(x, y) =g(T x, T y) =λλg(x, y) =g(x, y)

so by ergodicity of T×T, g (and hencef) must be constant almost every- where.

All that remains is to prove that (5) =⇒ (2), and this is considerably more difficult. There are several different proofs, each of which uses a non- trivial result from functional analysis⁽³¹⁾. Assume thatT×T is not ergodic, so there is a non-constant function f ∈ L²_μ2(X×X) that is almost every- where invariant underT×T. We would like to have the additional symmetry property f(x, y) =f(y, x) for all (x, y)∈X ×X. To obtain this additional property, consider the functions

(x, y)→f(x, y) +f(y, x) and

(x, y)→i(f(x, y)−f(y, x)).

Notice that if both of these functions are constant, thenf must be constant. It follows that one of them must be non-constant. So without loss of generality we may assume thatf satisfies f(x, y) = f(y, x). We may further suppose (by subtracting

fdμ²) that

fdμ² = 0. It follows that the operator F onL²_μ defined by

(F(g)) (x) =

X

f(x, y)g(y) dμ(y)

2.8 Proof of Weak-Mixing Equivalences 59

is a non-trivial self-adjoint compact⁽³²⁾ operator, and so by Theorem B.3 has at least one non-zero eigenvalue λ whose corresponding eigenspace Vλ

is finite-dimensional. We claim that the finite-dimensional spaceVλ⊆L²_μ is invariant underT. To see this, assume thatF(g) =λg. Then

λg(T x) =

X

f(T x, y)g(y) dμ(y)

=

X

f(T x, T y)g(T y) dμ(y) (sinceμisT-invariant)

=

X

f(x, y)g(T y) dμ(y),

sincef isT×T-invariant, soF(g◦T) =λ(g◦T) and thusg◦T ∈V_λ. It follows that U_T restricted to V_λ is a non-trivial linear map of a finite-dimensional linear space, and therefore has a non-trivial eigenvector. Since

fdμ² = 0,

any such eigenvector is non-constant.

2.8.1 Continuous Spectrum and Weak-Mixing

A more conventional proof of the difficult step in Theorem2.36, which may be taken to be (5) =⇒ (1), proceeds via the Spectral theorem (Theorem B.4) in the following form.

Alternative proof of (5) =⇒ (1) in Theorem 2.36.Definition2.35is clearly equivalent to the property that

Nlim→∞

1 N

N−1 n=0

|U_Tⁿf, g − f,1 · 1, g|= 0

for anyf, g∈L²_μ, and by polarization this is in turn equivalent to

Nlim→∞

1 N

N−1 n=0

|U_Tⁿf, f − f,1 · 1, f|= 0

for any f ∈ L²_μ (see Exercise 2.7.8 and page 441). By subtracting

Xfdμ fromf, it is therefore enough to show that iff ∈L²_μ has

Xfdμ= 0, then 1

N

N−1 n=0

|U_Tⁿf, f|²−→0

asN → ∞. By (B.1), it is enough to show that for the non-atomic measureμf

onS¹,

1 N

N−1 n=0

S¹

zⁿdμ_f(z)

²−→0 (2.35)

asN → ∞. Sincezⁿ=z⁻ⁿforz∈S¹the product in (2.35) may be expanded to give

1 N

N−1 n=0

S¹

zⁿdμ_f(z) ² = 1

N

N−1 n=0

S¹

zⁿdμ_f(z)·

S¹

w⁻ⁿdμ_f(w)

= 1 N

N−1 n=0

S¹×S¹

(z/w)ⁿdμ²_f(z, w) (by Fubini)

=

S¹×S¹

1 N

N−1 n=0

(z/w)ⁿ

dμ²_f(z, w).

The measureμ_f is non-atomic so the diagonal set{(z, z)|z∈S¹} ⊆S¹×S¹ has zeroμ²_f-measure. Forz=w,

1 N

N−1 n=0

(z/w)ⁿ= 1 N

1−(z/w)^N 1−(z/w)

−→0

as N → ∞, so the convergence (2.35) holds by the dominated convergence

theorem (Theorem A.18).

Exercises for Sect. 2.8

Exercise 2.8.1.Is the hypothesis that the sequence (a_n) be bounded neces- sary in Lemma2.41?

Exercise 2.8.2.Give an alternative proof of (1) =⇒ (5) in Theorem2.36 by proving the following statements:

(1) Any factor of a weak-mixing transformation is weak-mixing.

(2) A complex-valued eigenfunctionf ofUT has constant modulus.

(3) Iff is an eigenfunction ofUT, thenx→arg (f(x)/|f(x)|) is a factor map from (X,B, μ, T) to (T,BT, m_T, Rα) for someα.

Exercise 2.8.3.Show the following converse to Exercise2.5.6: if a measure- preserving system (Y,BY, ν, S) is not totally ergodic then there exists a measure-preserving system (X,B, μ, T) and a K > 1 with the property that (Y,BY, ν, S) is measurably isomorphic to the system

(X^(K),B^(K), μ^(K), T^(K)) constructed in Exercise2.5.6.