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Ergodic averages converge in Lp on finite-measure spaces

Statement

Let $μ (X) < \infty$ , let $T$ preserve $μ$ , let $1 \leq p < \infty$ , and let $f \in L^{p} (μ)$ be real or complex valued. If $f^{*}$ is its Birkhoff limit, then $f^{*} \in L^{p} (μ)$ and

$∥ A_{n} f - f^{*} ∥_{p} ⟶ 0.$

No $L^{\infty}$ convergence is asserted.

Facts & Assumptions

Given: The finite measure space, $T$ , $p$ , $f$ , and $f^{*}$ in the Statement.

[F1]

Ergodic averages are $L^{p}$ contractions (Ergodic averages are measurable, representative independent, and Lp contractive) and converge a.e. by Birkhoff (Birkhoff pointwise ergodic theorem).

[F2]

On finite measure spaces, a.e. convergence implies convergence in measure, and convergence in measure plus uniform integrability gives $L^{1}$ convergence (On a finite measure space, almost-everywhere convergence implies convergence in measure, Vitali convergence theorem on finite and sigma-finite measure spaces).

[F3]

Markov's inequality, dominated convergence, and Fatou's lemma have their usual integral forms (Chebyshev-Markov inequality for the integral, Dominated convergence, Fatou's lemma).

[F4]

A bounded function on a finite measure space belongs to every finite $L^{p}$ (Finite-measure $L^{r}$ includes into $L^{p}$ for $p < r$ ).

Proof

technique · cases $p=1$ and $1<p<\infty$

1.1assume-case poneF1F3

Assume $p = 1$ . Put $E_{n, M} = {∣ A_{n} f ∣ > M}$ . Contractivity and Markov give $μ (E_{n, M}) \leq ∥ f ∥_{1} / M$ . For $K > 0$ , split $f = f_{K} + r_{K}$ by radial clipping, so $∣ f_{K} ∣ \leq K$ and $∣ r_{K} ∣ = (∣ f ∣ - K)_{+}$ . Pointwise, $∣ A_{n} f ∣ \leq A_{n} ∣ f ∣ \leq K + A_{n} ∣ r_{K} ∣ .$ Consequently $\int_{E_{n, M}} ∣ A_{n} f ∣ d μ \leq \frac{K ∥ f ∥_{1}}{M} + ∥ r_{K} ∥_{1},$ where invariance gives $\int A_{n} ∣ r_{K} ∣ = ∥ r_{K} ∥_{1}$ .

1.2assume-case pgreatF3F4

Now assume $1 < p < \infty$ . Let $f_{m} = f 1_{{∣ f ∣ \leq m}}$ . Then $f_{m}$ is bounded and belongs to $L^{p}$ by [F4], while dominated convergence applied to $∣ f - f_{m} ∣^{p}$ gives $∥ f - f_{m} ∥_{p} \to 0$ .

2.1F3step 1.1

As $K \to \infty$ , $∣ r_{K} ∣ ↓ 0$ and is dominated by $∣ f ∣$ , so its integral tends to zero. Choose $K$ and then $M$ in step 1.1; the bound is uniform in $n$ and proves uniform integrability of $(A_{n} f)$ . For complex $f$ , it also proves uniform integrability of the real and imaginary parts because each component modulus is bounded by $∣ A_{n} f ∣$ .

2.2F1F3F4step 1.2

Let $f_{m}^{*}$ be the Birkhoff limit of $f_{m}$ . For fixed $m$ , both $∣ A_{n} f_{m} ∣$ and $∣ f_{m}^{*} ∣$ are bounded by $m$ . Their pointwise difference tends to zero a.e.; dominated convergence on the finite measure space therefore gives $∥ A_{n} f_{m} - f_{m}^{*} ∥_{p} \to 0$ .

3.1F1F2step 2.1cases: p-one

By [F1] the averages converge a.e., hence by [F2] in measure. Vitali applied to the real and imaginary parts gives $L^{1}$ convergence to their corresponding components of $f^{*}$ . The complex triangle inequality combines the two component conclusions.

4.1F1F3step 1.2step 2.2cases-exhaustive∎

Since $A_{n} (f - f_{m}) \to f^{*} - f_{m}^{*}$ a.e., Fatou and contractivity imply $∥ f^{*} - f_{m}^{*} ∥_{p} \leq {lim inf}_{n} ∥ A_{n} (f - f_{m}) ∥_{p} \leq ∥ f - f_{m} ∥_{p} .$ Thus $f^{*} \in L^{p}$ , and ${lim sup}_{n} ∥ A_{n} f - f^{*} ∥_{p} \leq 2 ∥ f - f_{m} ∥_{p} .$ Letting $m \to \infty$ proves the claim. The two cases exhaust $1 \leq p < \infty$ ; the proof never supplies uniform-norm convergence.

Depends on

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Sources

Omri Sarig, Lecture Notes on Ergodic Theory (2023) (standard reference, not scraped)
Charles Walkden, Ergodic Theory lecture notes (standard reference, not scraped)