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Mixing hierarchy for a selected probability-preserving map

Statement

For one fixed probability-preserving map $T$ , strong mixing implies weak mixing, and weak mixing implies ergodicity, under the exact definitions in Ergodicity and mixing for a selected probability map. No converse is asserted.

Facts & Assumptions

Given: One probability-preserving system $(X, A, μ, T)$ and the definitions of strong mixing, weak mixing, and ergodicity for this same map.

[F1]

Ergodicity and mixing for a selected probability map: strong mixing means $d_{n} (E, F) \to 0$ for each measurable pair; weak mixing means the Cesàro averages of $∣ d_{n} (E, F) ∣$ tend to zero; the map preserves the probability; ergodicity means each strictly invariant set $E$ with $T^{- 1} E = E$ has probability zero or one.

[F2]

Mixing implies weak mixing, which implies ergodicity: in a probability-preserving system, strong mixing implies weak mixing and weak mixing implies ergodicity. Its complete proof uses the Cesàro estimate and tests weak mixing on a strictly invariant set.

Proof

Given: [F1], [F2], and the system in the statement.

1.1F1F2given

Apply the first implication of [F2] to the fixed system. Strong mixing therefore implies weak mixing with the absolute value inside the Cesàro average.

1.2F1given

Let $E \in A$ satisfy the strict invariance condition $T^{- 1} E = E$ . Then $T^{- n} E = E$ for every $n \geq 0$ , so $d_{n} (E, E) = μ (E) - μ (E)^{2}$ for every $n$ . Weak mixing makes the Cesàro mean of $∣ d_{n} (E, E) ∣$ tend to zero, so its constant value $μ (E) (1 - μ (E))$ is zero. As $0 \leq μ (E) \leq 1$ , $μ (E)$ is zero or one. This is ergodicity under the exact convention in [F1]. Thus weak mixing implies ergodicity for this same map, without asserting a converse or a property of every Hamiltonian flow.

2.1step 1.1step 1.2∎

Combining steps 1.1 and 1.2 proves strong mixing implies weak mixing implies ergodicity.

Remarks

The implications concern only the selected probability-preserving map. Hamiltonian volume preservation by itself does not supply either mixing hypothesis.

Mixing hierarchy for a selected probability-preserving map

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources