Dynamics from iterated averaging

TL;DR

This paper proves that the closure of the semigroup generated by conditional expectations on a Lebesgue space includes all measure-preserving automorphisms, linking ergodic theory with a water tank equilibration puzzle.

Contribution

It establishes a connection between iterated averaging operators and measure-preserving transformations, providing a new perspective on their closure properties.

Findings

The strong operator closure contains all measure-preserving automorphisms.

The result is based on solving a water tank equilibration puzzle.

Provides a novel link between ergodic theory and a physical analogy.

Abstract

We prove that for a standard Lebesgue space $X$, the strong operator closure of the semigroup generated by conditional expectations on $L^\infty(X)$ contains the group of measure-preserving automorphisms. This is based on a solution to the following puzzle: given $n$ full water tanks, each containing one unit of water, and $n$ empty ones, how much water can be transferred from the full tanks to the empty ones by repeatedly equilibrating the water levels between pairs of tanks?

Equations

$$\frac{n}{4^n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=\sqrt{\frac{n}{\pi }}+O(n^{-1/2})$$

$$\mathopen{}\mathclose{{\left\|f_{i}\circ T-\mathbb{E}_{m}\cdots\mathbb{E}_{1}f_{i}}}\right\|_{\infty}<\varepsilon\qquad\forall i=1,\dots,n,$$

Full text

Acknowledgements. We thank Paolo Perrone for raising the question which led to this work, Omer Tamuz and puzzling.SE users Pranay and Nitrodon for discussion and Jakub Czartowski for further comments including pointers to the literature on quantum thermodynamics. Nicolás Rivera was supported by ANID FONDECYT grant No 11251301 and ANID SIA grant No 85220033.

1. Introduction

Consider $n$ full water tanks, each containing one unit of water, and $n$ empty ones. The tanks are all identical and at the exact same elevation. Suppose that you have a hose at your disposal, using which you can transfer water from one tank to another. But without a pump, all that you can do is to equilibrate the water levels between two tanks. Then under these conditions, how much water can you transfer from the full tanks to the empty ones?

This problem has a surprising answer: for large $n$, it is possible to transfer almost all water from the full tanks to the empty ones.

Theorem 1.1**.**

For this problem, there is a strategy which ends up with only

[TABLE]

units of water in the originally full tanks, and this is optimal.

Remark 1.2**.**

We present the proof in Section 2, but already point out now that most of this result is not due to us. First of all, the strategy is a straightforward discretization of the well-known countercurrent heat exchanger method for transferring heat between two fluids (LABEL:heat). Moreover, the strategy and its asymptotics also appear in recent work by Czartowski, de Oliveira Junior and Korzekwa in the context of quantum thermodynamics [COK, Lemma 1], but without the exact expression or an optimality proof. Finally, the optimality proof is largely due to user Nitrodon on puzzling.stackexchange.com [nitrodon]. It applies to any initial configuration of water levels.

Curiously, the fraction of water which remains in the originally full tanks is precisely the probability for $2n$ tosses of a fair coin to yield exactly $n$ heads, namely $4^{-n}\binom{2n}{n}$. The fact that we can transfer all water as $n\to\infty$ may seem surprising: equilibrating the water between two tanks is a kind of averaging operation, and averaging should result in a general tendency towards equalization. Instead, the theorem asserts that there is a strategy which amounts to a permutation of the full tanks with the empty ones—at least to a very good approximation as $n\gg 1$.

An equilibration of two tanks can equivalently be described as taking the conditional expectation of the function which assigns to each tank its water level with respect to the $\sigma$-algebra which identifies these two tanks while distinguishes all others. Thus a closely related question to the water tank puzzle is: which functions can be obtained by repeatedly taking conditional expectations of a function on a probability space, say a standard Lebesgue space? Here the answer is similarly surprising: such repeated averaging can approximate any dynamics, by which we mean the action of a measure-preserving automorphism on the function. Moreover, this works for any finite number of functions at a time.

Theorem 1.3**.**

Let $(X,\Sigma,\mu)$ be a standard Lebesgue probability space and $T:X\to X$ a measure-preserving automorphism. Then for all $f_{1},\dots,f_{n}\in L^{\infty}(X)$ and $\varepsilon>0$, there is a finite sequence of sub-$\sigma$-algebras $\mathcal{F}_{1},\ldots,\mathcal{F}_{m}\subseteq\Sigma$ such that

[TABLE]

where each $\mathbb{E}_{j}:L^{\infty}(X)\to L^{\infty}(X)$ is the conditional expectation operator with respect to $\mathcal{F}_{j}$.

In other words, if one equips the space of bounded operators $L^{\infty}(X)\to L^{\infty}(X)$ with the strong operator topology, then the closure of the semigroup generated by conditional expectations contains the group of measure-preserving automorphisms. Note that it does not matter whether we take $L^{\infty}(X)$ to consist of the real-valued or the complex-valued bounded measurable functions.

In the following two sections, we prove these results and give some additional context.

Remark 1.4**.**

For us, the equilibration puzzle arose from the following question asked by Paolo Perrone. Given $f,g\in L^{\infty}(X)$ on a standard probability space $X$, write $f\geq g$ if $g$ can be written as a conditional expectation of $f$ with respect to some sub-$\sigma$-algebra. Then is the transitive closure of this relation an interesting new preorder?

Our answer is negative: In order for it to be interesting, one will also at least want to take the topological closure in $L^{\infty}$. But then Theorem 1.3 forces $f\geq f\circ T\geq f$ for any measure-preserving automorphism $T:X\to X$, and thus one is forced to identify all functions related by such automorphisms. The resulting equivalence classes are precisely the compactly supported distributions on $\mathbb{R}$, and the preorder reduces to the usual Choquet order on distributions [winkler, Theorem 1.3.6]. We discovered the water tank puzzle and its solution in the process of proving Theorem 1.3.

2. Optimal equilibration strategies and proof of Theorem 1.1