# Oscillation of Ergodic Averages and Other Stochastic Processes

S. Mondal  Department of Mathematics, The Ohio State University
Columbus, OH 43210, USA
<EMAIL_ADDRESS>,  J. Rosenblatt  Department of Mathematics
University of Illinois at Urbana-Champaign
Urbana, IL 61801<EMAIL_ADDRESS>and  M. Wierdl  Department of
Mathematical Sciences
University of Memphis
Memphis, TN 38152<EMAIL_ADDRESS>
###### Abstract.

For an ergodic map $T$ and a non-constant, real-valued $f\in L^{1}$, the
ergodic averages $A_{N}f(x)=\frac{1}{N}\sum\limits_{n=1}^{N}f(T^{n}x)$
converge a.e., but the convergence is never monotone. Depending on particular
properties of the function $f$, the averages $A_{N}f(x)$ may or may not
actually oscillate around the mean value infinitely often a.e.

We will prove that a.e. oscillation around the mean is the generic behavior.
That is, for a fixed ergodic $T$, the generic non-constant $f\in L^{1}$ has
the averages $A_{N}f(x)$ oscillating around the mean infinitely often for
almost every $x$.

We also consider oscillation for other stochastic processes like subsequences
of the ergodic averages, convolution operators, weighted averages, uniform
distribution and martingales. We will show that in general, in these settings
oscillation around the limit infinitely often persists as the generic
behavior.

###### Contents

  1. 1 Introduction
  2. 2 Organization of the paper
  3. 3 Summary of results
  4. 4 Preliminaries
    1. 4.1 Notation
    2. 4.2 Complete recurrence property
  5. 5 Oscillation and non-monotonicity of the classical ergodic averages
    1. 5.1 Proof of the oscillation and non-monotonicity results
    2. 5.2 Irrational rotation case
    3. 5.3 Some quantitative results
  6. 6 Oscillation of convolution operators
  7. 7 Martingales
  8. 8 Uniformly distributed sequences
  9. 9 General weighted averages in ergodic theory

##  1\. Introduction

The aim of this paper is to analyze the non-monotonicity and the oscillation
of a variety of convergent stochastic process. Suppose that
$(X,\mathcal{B},\mu)$ is a standard probability space, $T_{N}:L^{1}\to L^{1}$
is a bounded, linear operator and $f\in L^{1}$ is a real valued function such
that $(T_{N}f)$ converges to $\int f$ for almost every $x$. We are interested
in the following question.

######  Question 1.1.

[Non-monotonicity] For any function $f\in L^{1}$, is the convergence of
$T_{N}f(x)$ non-monotone in the following sense:
There cannot exist any set $M\subset X$ with $\mu(M)>0$ such that whenever
$x\in M$, $T_{N}f(x)\geq T_{N+1}f(x)$ eventually, or $T_{N}f(x)\leq
T_{N+1}f(x)$ eventually?

When this question has an affirmative answer, we ask a more pertinent question
about the behavior of convergence. Namely,

######  Question 1.2.

[Oscillation] Can we say that for almost every $x\in X$, $T_{N}f(x)$
_oscillates_ around its limit infinitely often, that is, is $(T_{N}f(x)-\int
f)>0$ for infinitely many $N$, and $(T_{N}f(x)-\int f)<0$ for infinitely many
(different) $N$?

The study of non-monotonicity and oscillation of convergent stochastic
processes is a topic of intrinsic interest in probability theory. For a
symmetric random walk, non-monotonicity was first investigated by Erdős-Rényi
[10]. They proved that the length of the longest run of heads in $n$ tosses of
a fair coin will approximately be equal to $\log n$. The question of
oscillation for such stochastic processes was first studied by Chung-Hunt [8],
and later by Chung-Erdős [7] and Csáki-Erdős-Révész [9]. They found bounds on
the length of the longest excursion 111There is a subtle difference between
oscillation and excursion. The former one refers to strict inequality whereas
the later one does not require the inequality to be strict. in a symmetric
random variable.

While our primary focus is on various ergodic averages, we will also study
Lebesgue differentiation, dyadic martingales, uniform distribution and
weighted averages. It turns out that for these stochastic processes, there are
functions $f$ for which $T_{N}f$ fails to oscillate on a set of positive
measure of $x$, nonetheless, for the generic function $f$, the stochastic
process $T_{N}f$ oscillate around the limit infinitely often a.e.

##  2\. Organization of the paper

The rest of the paper is organized as follows:

In Section 3, we will state the main results. In Section 4 we discuss some
preliminary concepts that will be needed for the rest of the paper. In Section
5, we will prove the results related to non-monotonicity and oscillation of
ergodic averages including 3.1, 3.5, 3.8 and 3.9. We deal with the oscillation
of convolution operators in Section 6. In this section, we prove 3.11. 3.12 is
proved in Section 7. In Section 8 and Section 9, we discuss the non-
monotonicity and oscillation of uniformly distributed sequences and weighted
ergodic averages, respectively.

##  3\. Summary of results

In this section we state a variety of results about non-monotonicity and
oscillation. The proofs for these results are given in the upcoming sections
as stated in Section 2.

As a basic model case, take the classical ergodic averages. We have an
ergodic, invertible, measure preserving mapping $T$ on a standard probability
space $(X,\mathcal{B},\mu,T)$. For a function $f\in
L^{1}=L^{1}(X,\mathcal{B},\mu)$, by the Pointwise Ergodic Theorem,
$\mathbb{A}_{[N]}f(T^{n}x)$ converges to $\int f$ for a.e. $x$. (See Eq. 4.1
for the meaning of the averaging notation).

In the special case, if we consider the irrational rotation
$({\mathbb{T}},{\Sigma},\lambda,R_{\alpha})$ and a continuous function $f$,
one can see that for almost every $x$, the convergence of
$\mathbb{A}_{[N]}f(x+n\alpha)$ to $\int f$ is never monotone. Indeed, compare
$\mathbb{A}_{[N]}f(x+n\alpha)$ with $\mathbb{A}_{[N+1]}f(x+n\alpha)$, and use
the uniform distribution of $(n\alpha)$ mod 1 in ${\mathbb{T}}$. See Section 5
for details.

Our first proposition shows that this is, in fact, true for any ergodic
transformation $T$ and any integrable function $f$.

######  Proposition 3.1.

Suppose $f\in L^{1}$ is real-valued and not a constant function. Then for a.e.
$x$, the averages $\mathbb{A}_{[N]}f(T^{n}x)$ are non-monotone in the sense of
1.1.

We can also give a somewhat more detailed result about the gap that occurs
between two consecutive averages.

######  Proposition 3.2.

Suppose $f\in L^{1}$ is real-valued and not a constant function. Then there
exists $d>0$ such that for all $T$ ergodic, we have for a.e. $x$,

$\displaystyle\mathbb{A}_{[N+1]}f(T^{n}x)$
$\displaystyle>\frac{d}{N}+\mathbb{A}_{[N]}f(T^{n}x)\text{ infinitely often,
and }$ $\displaystyle\mathbb{A}_{[N+1]}f(T^{n}x)$
$\displaystyle<-\frac{d}{N}+\mathbb{A}_{[N]}f(T^{n}x)\text{ infinitely
often.}$

Different aspects of oscillation for the classical ergodic averages have
already been studied by several authors. For instance, Bishop [6] showed that
one can give universal estimates for the probability that there are many
fluctuations in the ergodic averages of $L^{1}$ functions. In fact, he proved
an exact analogy of Doob’s upcrossing inequality for martingales which was
later improved by Kalikow and Weiss [15]. The most relevant result related to
our paper is the following theorem of Halász [12, Theorem 4].

######  Theorem 3.3.

[Halász] For any $f\in L^{1}$, $\big{(}\mathbb{A}_{[N]}f(T^{n}x)-\int
f\big{)}$ almost everywhere changes sign infinitely often in the weaker sense
that it cannot be eventually positive or negative.

Halász’s theorem suggests that for every $f\in L^{1}$,
$\mathbb{A}_{[N]}f(T^{n}x)$ may oscillate around the mean infinitely often for
almost every $x$. In the following example, we see that it may not be the
case.

######  Example 3.4.

Take our probability space to be $[0,1)$ with the usual Lebesgue measure
$\lambda$. Take any map $T$ and let $f$ be the coboundary with the transfer
function $F=\mathbbm{1}_{[0,1/2]}-\mathbbm{1}_{[1/2,1)}$, that is, $f=F-F\circ
T$. Then the possible values of $F(Tx)-F\circ T^{N+1}(x)$ for $x\in
T^{-1}[0,1/2]$ are $1-1=0$ and $1-(-1)=2$. So for these $x$, we get
$F(Tx)-F\circ T^{N+1}(x)$ changing value infinitely often, but the sign is
never strictly negative. At the same time, the possible values of
$F(Tx)-F\circ T^{N+1}(x)$ for $x\in[1/2,1)$ are $-1-1=-2$ and $-1-(-1)=0$. So,
for these $x$, we get $F(Tx)-F\circ T^{N+1}(x)$ again changing value
infinitely often, but the sign is never strictly positive. So, we do not have
the averages $\mathbb{A}_{[N]}f(T^{n}x)$ oscillating around the mean in either
case.

In the next theorem, we will characterize those functions for which
$\mathbb{A}_{[N]}f(T^{n}x)$ fails to oscillate in a set of positive measure.

We say that $F$ has an upper barrier if there is a set $P$ of positive measure
on which $F$ is a constant $p$ a.e. and such that $F(y)\leq p$ for a.e. $y\in
X$. We say that $F$ has a lower barrier if there is a set $Q$ of positive
measure on which $F$ is a constant $q$ a.e. and such that $F(y)\geq q$ for
a.e. $y\in X$.

######  Theorem 3.5.

Assume that $F$ has either an upper barrier or a lower barrier, and
$f=F-F\circ T$. Then there exists a set $E$ of positive measure such that
$\mathbb{A}_{[N]}f(T^{n}x)$ fails to oscillate on $E$, that is, for every
$x\in E$, either $\mathbb{A}_{[N]}f(T^{n}x)\geq 0$ eventually or
$\mathbb{A}_{[N]}f(T^{n}x)\leq 0$ eventually.

Conversely, assume that for some $f\in L^{1}$, $\mathbb{A}_{[N]}f(T^{n}x)$
fails to oscillate on a set of positive measure. Then $f$ can be expressed as
$F-F\circ T$ for some measurable function $F$ which has either an upper or a
lower barrier.

Next, for a given increasing sequence $(N_{i})_{i}$ of positive integers, we
initiate the study of similar type of questions for the subsequence of
averages $\mathbb{A}_{[N_{i}]}f(T^{n}x)$. More precisely, we ask the
following:

######  Question 3.6.

For any real valued function $f\in L^{1}$, is the convergence of
$\mathbb{A}_{[N_{i}]}f(T^{n}x)$ non-monotone for almost every $x$ in the sense
of 1.1?

Similarly,

######  Question 3.7.

For almost every $x$, do the averages $\mathbb{A}_{[N_{i}]}f(T^{n}x)$
oscillate almost everywhere around its mean $\int fd\mu$, that is, does
$\big{(}\mathbb{A}_{[N_{i}]}f(T^{n}x)-\int f\big{)}_{i}$ change sign
infinitely often?

One would expect that the non-monotonicity and the oscillation of the averages
$\mathbb{A}_{[N]}f(T^{n}x)$ is not something that persists when one takes a
subsequence $\mathbb{A}_{[N_{i}]}f(T^{n}x)$. In 5.6, we will construct a
system $(X,\mathcal{B},\mu,T)$ and a subsequence $(N_{i})$ such that for every
$x$,
$\Big{(}\mathbb{A}_{[N_{i+1}]}f(T^{n}x)-\mathbb{A}_{[N_{i}]}f(T^{n}x)\Big{)}_{i}$
will be eventually positive or negative depending only on $x$ (not on $i$),
thereby answering 3.6 negatively. On the other hand, the following theorem
shows that for any increasing sequence of positive integers $(N_{i})$ and for
the generic function $f$, the averages $\mathbb{A}_{[N_{i}]}f(T^{n}x)$ exhibit
an oscillating behavior around the mean infinitely often a.e.

######  Theorem 3.8.

If $T$ is ergodic, and $(N_{i})$ is an unbounded sequence, then there is a
dense $G_{\delta}$ set $\mathcal{O}$ in $L^{1}$ such that for every
$f\in\mathcal{O}$ and for almost every $x\in X$,
$\mathbb{A}_{[N_{i}]}f(T^{n}x)$ oscillates infinitely often around its mean.
That is, for a.e. $x$ we have $\mathbb{A}_{[N_{i}]}f(T^{n}x)>\int f$ for
infinitely many $i$ and $\mathbb{A}_{[N_{i}]}f(T^{n}x)<\int f$ for infinitely
many (different) $i$.

A salient feature of the proof of Theorem 3.8 is that it uses the _strong
sweeping out property_ of lacunary sequences.

The next natural question would be to describe those functions $f$ for which
the averages $\mathbb{A}_{[N_{i}]}f(T^{n}x)$ exhibit non-monotonicity or
oscillating behavior (around the mean) for a set of positive measure. Our
results in this regard are far from complete. It turns out that these
questions are related to an intrinsic dynamical property of the sequence
$(N_{i})$ which we will call as _complete recurrence property_. See Subsection
4.2 for the definition of the complete recurrence property.

In the special case, when $(N_{i})$ is a syndetic sequence (i.e. a sequence
which has bounded gaps between the consecutive terms), for a ‘typical’
$\alpha$ we can say something definitive about the non-monotonicity of the
averages with respect to the rotation by $\alpha$.

######  Theorem 3.9.

Let $(N_{i})$ be a syndetic sequence. Then there exists a countable subset $E$
of ${\mathbb{T}}$ such that the following holds. If
$\alpha\in{\mathbb{T}}\setminus E$ and $f$ is a non-constant, integrable
function, then for a.e. $x$, the averages $\mathbb{A}_{[N_{i}]}f(x+n\alpha)$
are non-monotone in the sense of 1.1.

The next topic of our discussion will be the oscillation of convolution
operators.

######  Definition 3.10.

An _approximate identity_ is a sequence $(\phi_{n})\in L^{1}({\mathbb{R}})$
such that for all $f\in L^{1}({\mathbb{R}})$, we have
$\lim_{n\to\infty}\|\phi_{n}\star f-f\|_{1}=0$. A sequence $(\phi_{n})$ is
_normalized_ if $\|\phi_{n}\|_{1}=1$ for all $n\geq 1$. A sequence is
_eventually local_ if the functions all have bounded support and for all
$\delta>0$, there exists $N\geq 1$, such that for all $n\geq N$, we have
$\phi_{n}(x)=0$ a.e. on ${\mathbb{R}}\setminus[-\delta,\delta].$ A _proper_
approximate identity is a normalized, eventually local, approximate identity
consisting of positive functions.

Suppose that $(\phi_{n})$ is a proper approximate identity on
$L^{1}(\mathbb{R})$. While the convolutions $\phi_{n}\star f$ converge in
$L^{1}$-norm to $f$ for all $f\in L^{1}(\mathbb{R})$, these convolutions may
or may not converge almost everywhere. This is a topic for which there is a
great deal of literature of all types; see Rosenblatt [20] for some background
and theorems about these issues.

In this paper, we are more concerned with what could be considered the other
side of the issue, the rate of this convergence. Let $CB(\mathbb{R})$ denote
the continuous, bounded functions on $\mathbb{R}$ in the uniform norm
topology. The main result concerning this topic is the following:

######  Theorem 3.11.

Let $B=L^{p}(\mathbb{R})$ for some fixed $p,1\leq p\leq\infty$, or
$B=CB(\mathbb{R})$. Suppose that $(\phi_{n})$ is a proper approximate identity
and $(\epsilon_{n})$ is a sequence of positive numbers with
$\lim\limits_{n\to\infty}\epsilon_{n}=0$. Then there is a dense $G_{\delta}$
subset $\mathcal{A}$ of $B$ such that for all $f\in\mathcal{A}$, we have both
$\limsup\limits_{n\to\infty}\frac{\phi_{n}\star
f(x)-f(x)}{\epsilon_{n}}=\infty$ a.e. and
$\liminf\limits_{n\to\infty}\frac{\phi_{n}\star
f(x)-f(x)}{\epsilon_{n}}=-\infty$ a.e.

The final topic of this paper is the oscillation of convergent martingales.
The classical dyadic martingale is probably the best example to consider
because of its connections to Lebesgue differentiation, and through that to
dynamical systems.

The dyadic martingale is constructed as follows. Take $\mathcal{D}_{n}$ to be
the dyadic $\sigma$-algebra whose atoms are the intervals
$D_{n,j}=[\frac{j}{2^{n}},\frac{j+1}{2^{n}}],j=0,\dots,2^{n}-1$. The finite
$\sigma$-algebras $\mathcal{D}_{n}$ are increasing, with the measure-theoretic
completion of $\bigcup\limits_{n\in{\mathbb{N}}}\mathcal{D}_{n}$ being the
Lebesgue $\sigma$-algebra in $[0,1]$.

Now for $G\in L^{1}[0,1]$, as $n\to\infty$, we have the conditional
expectations $E(G|\mathcal{D}_{n})$ converge to $G$ in $L^{1}$-norm and a.e.
with respect to Lebesgue measure $\lambda$. We claim that this convergence is
generically oscillating a.e.

######  Theorem 3.12.

For a dense $G_{\delta}$ set $\mathcal{O}$ of functions in $L^{1}[0,1]$, if
$G\in\mathcal{O}$, for a.e. $x\in[0,1]$, $E(G|\mathcal{D}_{n})(x)>G(x)$ for
infinitely many $n\geq 1$, and $E(G|\mathcal{D}_{n})(x)<G(x)$ for infinitely
many (other) $n\geq 1$.

######  Remark 3.13.

It is clear that we have to consider oscillation of this strong type on only a
subset $\mathcal{O}$ of $L^{1}[0,1]$. Indeed, if $G$ were measurable with
respect to $\mathcal{D}_{m}$, then $E(G|\mathcal{D}_{n})=G$ for all $n\geq m$.
So for such functions, there is no oscillation. If we vary $m$, then such
functions are dense in $L^{1}[0,1]$ in $L^{1}$-norm.

##  4\. Preliminaries

###  4.1. Notation

Let $(X,\mathcal{B},\mu,T)$ be a measure preserving system and $f\in
L^{1}(X)$. For a finite set of $S$ of positive integers and a sequence
$(a_{n})$ of positive integers, we use the following averaging notation:

$\mathbb{A}_{S}f(T^{a_{n}}x):=\dfrac{1}{|\\#S|}\sum_{n\in S}f(T^{a_{n}}x).$
(4.1)

Intervals in the averaging notation means intervals of integers.
In particular,

$\mathbb{A}_{[N]}f(T^{n}x):=\dfrac{f(Tx)+f(T^{2}x)+\dots+f(T^{N}x)}{N}.$ (4.2)

When we are concerned with the function space and not their functional value,
we will use $\mathbb{A}_{S}^{T}$ where $\mathbb{A}_{S}^{T}:X\to X$ is defined
by $\mathbb{A}_{S}^{T}(x)=\mathbb{A}_{S}f(T^{n}x).$

${\mathbb{T}}$ stands for the one dimensional torus, that is,
${\mathbb{R}}/{\mathbb{Z}}$ and $\lambda$ is reserved for denoting the
Lebesgue measure.

Sometimes we assume that $T$ is totally ergodic to avoid the following
obstacle regarding oscillation.

Suppose that $T$ is not totally ergodic. Then there is a non-zero $f$ such
that $f\circ T=\lambda f$ with $\lambda$ a root of unity. Say $\lambda^{N}=1$.
Then $\sum\limits_{n\in[N]}\lambda^{n}=0$. So the same thing occurs with any
multiple of $N$, and hence, for example,
$\mathbb{A}_{[N]}f(T^{n}x)=\mathbb{A}_{[2N]}f(T^{n}x)=0$ for almost every $x$.

On the other hand, if we assume total ergodicity then two averages cannot be
equal.

######  Proposition 4.1.

If $T$ is ergodic and for (almost) every $x\in X$,
$\mathbb{A}_{[M]}f(T^{n}x)=$ $\mathbb{A}_{[N]}f(T^{n}x)$ for some $M>N$ and a
non-zero $f\in L^{1}$, then $T$ is not totally ergodic.

###### Proof.

Since $\mathbb{A}_{[M]}^{T}f=\mathbb{A}_{[M]}^{T}f$, we have $f\circ
T^{M}=\frac{M}{N}\sum\limits_{[N]}f\circ T^{n}-\sum\limits_{[M-1]}f\circ
T^{n}$. So let $\mathcal{S}$ be the linear span of $\\{f\circ T,\dots,f\circ
T^{M-1}\\}$. We have $f\circ T^{M}\in\mathcal{S}$. That is, $\mathcal{S}\circ
T\subset\mathcal{S}$. So, because $\mathcal{S}$ is finite-dimensional,
$\mathcal{S}$ is invariant under $T$.

Since $T$ is ergodic, it follows that there is a non-zero $F\in\mathcal{S}$
and $\lambda\in\mathbb{T},\lambda\not=1,$ such that $F\circ T=\lambda F$. We
claim that in fact $\lambda$ is a root of unity and so $T$ is not totally
ergodic. To see this, we take $F=\sum\limits_{m\in[M]}c_{m}f\circ T^{m}$ for
some $c_{1},\dots,c_{M}\in\mathbb{C}$. So we have
$\mathbb{A}_{[M]}^{T}F=\sum\limits_{m\in[M]}c_{m}\mathbb{A}_{[M]}^{T}(f\circ
T^{m})=\sum\limits_{m\in[M]}c_{m}\mathbb{A}_{[M]}^{T}f\circ
T^{m}=\sum\limits_{m\in[M]}c_{m}\mathbb{A}_{[N]}^{T}f\circ
T^{m}=\sum\limits_{m\in[M]}c_{m}\mathbb{A}_{[N]}^{T}(f\circ
T^{m})=\mathbb{A}_{[N]}^{T}F$. Because $F$ is not zero and $F\circ T=\lambda
F$, we have
$\frac{1}{M}\sum\limits_{m\in[M]}\lambda^{m}=\frac{1}{N}\sum\limits_{[N]}\lambda^{n}$.
Using the geometric sum formula, it follows that
$\frac{1}{M}(\lambda^{M}-1)=\frac{1}{N}(\lambda^{N}-1)$. That is,
$M-N=M\lambda^{N}-N\lambda^{M}=(M-N\lambda^{M-N})\lambda^{N}$. Since
$|\lambda|=1$, it follows that $M-N=|M-N\lambda^{M-N}|$. That is, in
$\mathbb{C}$ the distance from $N$ to $M$ is the distance of $N\lambda^{M-N}$
to $M$. But this can only happen if $\lambda^{M-N}=1$. That is, $\lambda$ is a
root of unity. ∎

######  Remark 4.2.

The argument in Proposition 4.1 about powers of $\lambda$ cannot be made too
general. When
$\frac{1}{M}\sum\limits_{m\in[M]}\lambda^{m}=\frac{1}{N}\sum\limits_{[N]}\lambda^{n}$,
we know right away that $\lambda$ is algebraic. But that is not enough to show
that $\lambda$ is a root of unity. Indeed, $\gamma=(3+4i)/5$ is modulus one
and not a root of unity, but it is algebraic since $5\gamma^{2}-6\gamma+5=0$.
∎

######  Corollary 4.3.

If $T$ is totally ergodic, and $f$ is a non-constant $L^{1}$ function, then
$\mathbb{A}_{[M]}^{T}f$ is also non-constant for all $M\geq 1$.

###### Proof.

Suppose, to the contrary, there is a constant $c$ such that
$\mathbb{A}_{[M]}^{T}f(x)=c$ for almost every $x\in X$. Applying this equality
on $T^{M}x$, we get that $\mathbb{A}_{[M]}^{T}f(T^{M}x)=c$ for almost every
$x$. Adding these two equalities, and dividing both side by $2$, we have
$\mathbb{A}_{[2M]}^{T}f(x)=c$ for almost every $x$. Thus we have
$\mathbb{A}_{[M]}^{T}f(x)=\mathbb{A}_{[2M]}^{T}f(x)=c$ for almost every $x$.
But this is a contradiction to Proposition 4.1. ∎

###  4.2. Complete recurrence property

In this subsection, we will introduce the notion of complete recurrence
property. This has some consequences related to the non-monotonicity and
oscillation of the subsequence ergodic averages (see 5.1 and 5.13). It is also
a topic of independent interest.

######  Definition 4.4 (Complete recurrence property).

A sequence $(N_{i})$ of positive integers is said to satisfy the complete
recurrence property for the system $(X,\mathcal{B},\mu,T)$ if every set $E$ of
positive measure and every positive integer $L$ satisfy the following
condition:

$\displaystyle\mu\Big{(}\cup_{i\geq L}T^{-N_{i}}E\Big{)}=1.$ (4.3)

In particular, if a sequence $(N_{i})$ satisfies Equation 4.3 for the
irrational rotation $({\mathbb{T}},\Sigma,\lambda,R_{\alpha})$, then we say
that $(N_{i})$ satisfies the complete recurrence property for $\alpha$. We say
that a sequence $(N_{i})$ satisfies the complete recurrence property if it
satisfies the same for _every_ ergodic system.

An equivalent formulation of the complete recurrence property is the
following:
For every positive integer $L$ and every set $E$ and $F$ of positive measures,
there exists $i\geq L$ such that

$\mu\big{(}E\cap T^{-N_{i}}F\big{)}>0.$ (4.4)

######  Remark 4.5.

  1. (1)

Recall that a set $R\subset{\mathbb{N}}$ is said to be a _set of recurrence_
if for any measure preserving system $(X,\mathcal{B},\mu,T)$ and for all
$A\in\mathcal{B}$ with $\mu(A)>0$ if we have $\mu(A\cap T^{r}A)>0$ for some
$r\in R$. This suggests us the name _complete recurrence_. Sets of recurrence
have been studied extensively in ergodic theory. For a beautiful account on
the set of recurrence, the reader is suggested to look at the book of
Furstenberg [11].

  2. (2)

While the set of recurrence is defined for any measure preserving system, to
define complete recurrence property, we must restrict ourselves to ergodic
systems.

One can check the following fact.

Fact: Let $(X,\mathcal{B},\mu,T)$ be an ergodic system and $(N_{i})$ be an
increasing sequence of positive numbers. Suppose that for every $f\in
L^{\infty}$

$\frac{1}{K}\sum_{i\in[K]}f(T^{N_{i}}x)\to\int f\text{ in $L^{2}$ norm as
}K\to\infty.$ (4.5)

Then $(N_{i})$ satisfies the complete recurrence property for
$(X,\mathcal{B},\mu,T)$.

A particular class of sequences which satisfies the above hypothesis is the
class of _ergodic sequences_.

######  Definition 4.6.

(Ergodic sequence) Let $T$ be an ergodic transformation on the probability
space $(X,\mathcal{B},\mu)$. A sequence $(N_{i})$ of positive increasing
integers is said to be an ergodic sequence for the system
$(X,\mathcal{B},\mu,T)$ if for every $f\in L^{2}$,
$\displaystyle\frac{1}{K}\sum_{i\in[K]}f(T^{N_{i}}x)\to\int f\text{ in $L^{2}$
norm as }K\to\infty.$ If the sequence $(N_{i})$ is ergodic for every ergodic
system, then it is said to be an ergodic sequence.

For a detailed discussion on ergodic sequence, one can look at the survey
article of Rosenblatt-Wierdl [24, Chapter II]. Here we will list some examples
of ergodic sequence (and hence sequence of complete recurrence).

######  Example 4.7.

Let $c>1$ be a non-integer real number and $p_{n}$ denote the $n$-th prime
number.

  1. (a)

Then from Boshernitzan-Kolensik-Quas-Wierdl [4, Theorem A ], it follows that
$([n^{c}])$ is an ergodic sequence.

  2. (b)

From Bergelson-Kolesnik-Madritsch-Son-Tichy [2, Theorem 3.1], it follows that
$([p_{n}^{c}])$ is an ergodic sequence.

######  Example 4.8.

Let $p_{i}$ denotes the $i$-th prime number. It is well known that both
$(i^{2})$ and $(p_{i})$ both are ergodic sequences for totally ergodic
systems.

The following example shows that the complete recurrence property is indeed a
special property.

######  Example 4.9.

Let $I=[a,b]\subset{\mathbb{T}}$ be an interval with $\lambda(I)<\frac{1}{4}$.
Let $\alpha$ be a fixed irrational number. Define
$S=\\{n\in{\mathbb{N}}:I\cap(I-n\alpha)\neq\emptyset\\}$. We claim that $S$ is
a set of recurrence but it does not satisfy the complete recurrence property.
Let $(X,\mathcal{B},\mu,T)$ be any measure preserving system. Let
$(\tilde{X},\tilde{\mathcal{B}},\tilde{\mu},\tilde{T})$ be the product system
of $(X,\mathcal{B},\mu,T)$ and $({\mathbb{T}},\beta,\lambda,R_{\alpha}).$ For
an arbitrary $A\in\mathcal{B}$, let us consider the set $\tilde{A}=A\times I$.
By the Poincare recurrence theorem, there exists $n\in{\mathbb{N}}$ such that
$\tilde{\mu}\Big{(}\tilde{A}\cap\tilde{T}^{-n}\tilde{A}\Big{)}>0.$ Such $n$
must belong to $S$ and it also satisfies that $\mu(A\cap T^{-n}A)>0$. Thus $S$
is a set of recurrence. On the other hand, there exists an interval $J$ with
$\lambda(J)\leq\frac{1}{4}$ such that $\\{n\alpha:n\in S\\}\subset J$. Hence,
for any interval $E$ with $\lambda(E)<\frac{3}{4}$, we must have
$\lambda(J+E)<1.$ So, $\lambda\Big{(}\cup_{n\in S}(E-n\alpha)\Big{)}<1$.
Hence, $S$ cannot satisfy the complete recurrence property. ∎

##  5\. Oscillation and non-monotonicity of the classical ergodic averages

###  5.1. Proof of the oscillation and non-monotonicity results

We begin by proving 3.1.

First observe that $\mathbb{A}_{[N+1]}f(T^{n}x)\leq\mathbb{A}_{[N]}f(T^{n}x)$
implies that
$\frac{1}{N+1}f(T^{N+1}x)\leq(\frac{1}{N}-\frac{1}{N+1})\sum\limits_{[N]}f(T^{n}x)$.
This means that
$f(T^{N+1}x)\leq\frac{1}{N}\sum\limits_{[N]}f(T^{n}x)=\mathbb{A}_{[N]}f(T^{n}x)$.
Indeed, $\mathbb{A}_{[N+1]}f(T^{n}x)\leq\mathbb{A}_{[N]}f(T^{n}x)$ if and only
if $f(T^{N+1}x)\leq\mathbb{A}_{[N]}f(T^{n}x)$. Similarly,
$\mathbb{A}_{[N+1]}f(T^{n}x)\geq\mathbb{A}_{[N]}f(T^{n}x)$ if and only if
$f(T^{N+1}x)\geq\mathbb{A}_{[N]}f(T^{n}x)$. Now if either of these
inequalities hold for all large enough values of $N$, then we can compare
$f(T^{N+1}x)$ to $\mathbb{A}_{[N]}f(T^{n}x)$, which is going to zero for a.e.
$x$, as $N\to\infty$.

###### Proof of 3.1.

Without loss of generality, we can assume that $f$ is mean-zero, but not
constant. Hence, there is some $\delta>0$, and there are sets $A$ and $B$ of
positive measure, such that $f\geq\delta$ on $A$ and $f\leq-\delta$ on $B$.

By ergodicity, for a.e. $x$, there is some $N_{0}$, depending on $x$, such
that $|\mathbb{A}_{[N]}f(T^{n}x)|\leq\delta/2$ for all $N\geq N_{0}$. In
addition, the ergodicity of $T$ implies that for a.e. $x$, we have
$T^{N+1}x\in A$ for infinitely many $N$, and $T^{N+1}x\in B$ for infinitely
many $N$.

Therefore, for a.e. $x$, we cannot have for all $N\geq N_{0}$,
$f(T^{N+1}x)\leq\mathbb{A}_{[N]}f(T^{n}x)$ because using recurrence into $A$
this would imply $\delta\leq\delta/2$. But at the same time, for a.e. $x$, we
cannot have for all $N\geq N_{0}$, $f(T^{N+1}x)\geq\mathbb{A}_{[N]}f(T^{n}x)$
because using recurrence into $B$ this would imply $-\delta\geq-\delta/2$.

Using both of these computations, we see that for a.e. $x$, for infinitely
many $N$, we do not have
$\mathbb{A}_{[N+1]}f(T^{n}x)\leq\mathbb{A}_{[N]}f(T^{n}x)$ and for infinitely
many $N$ we do not have
$\mathbb{A}_{[N+1]}f(T^{n}x)\geq\mathbb{A}_{[N]}f(T^{n}x)$. Hence, for a.e.
$x$, the ergodic averages are not monotonically decreasing, and they are not
monotonically increasing. ∎

Now, we will prove 3.2.

###### Proof of 3.2.

Without loss of generality, we can assume that $f$ is mean-zero, but not
constant. Hence, there is some $\delta>0$, and there are sets $A$ and $B$ of
positive measure, such that $f\geq\delta$ on $A$ and $f\leq-\delta$ on $B$.
Let $d=\frac{\delta}{6}$.

If we have
$\mathbb{A}_{[N+1]}f(T^{n}x)\leq\frac{d}{N}+\mathbb{A}_{[N]}f(T^{n}x)$ for all
sufficiently large $N$, then using the computation above, we have
$f(T^{N+1}x)\leq\frac{(N+1)d}{N}+\mathbb{A}_{[N]}f(T^{n}x)$ for all
sufficiently large $N$. But for a.e. $x$, $T^{N+1}x\in A$ for infinitely many
$N$. Also, for a.e. $x$, $\mathbb{A}_{[N]}f(T^{n}x)\to\int f\,d\mu=0$ as
$N\to\infty$. Hence, for a.e. $x$, we have for infinitely many $N$,
$\delta\leq 2d+\mathbb{A}_{[N]}f(T^{n}x)\leq
2d+\frac{\delta}{2}=\frac{\delta}{3}+\frac{\delta}{2}$. This is impossible. So
we must have for a.e. $x$,
$\mathbb{A}_{[N+1]}f(T^{n}x)>\frac{d}{N}+\mathbb{A}_{[N]}f(T^{n}x)$ for
infinitely many $N$.

Similarly, if we have
$\mathbb{A}_{[N+1]}f(T^{n}x)\geq-\frac{d}{N}+\mathbb{A}_{[N]}f(T^{n}x)$ for
alll sufficiently large $N$, then using the computation above, we have
$f(T^{N+1}x)\geq-\frac{(N+1)d}{N}+\mathbb{A}_{[N]}f(T^{n}x)$ for all
sufficiently large $N$. But for a.e. $x$, $T^{N+1}x\in B$ for infinitely many
$N$. Also, for a.e. $\mathbb{A}_{[N]}f(T^{n}x)\to\int f\,d\mu=0$ as
$N\to\infty$. Hence, for a.e. $x$, we have for infinitely many $N$,
$-\delta\geq-2d+\mathbb{A}_{[N]}f(T^{n}x)\geq-2d-\frac{\delta}{2}=-\frac{\delta}{3}-\frac{\delta}{2}$.
This is impossible. So we must have for a.e. $x$,
$\mathbb{A}_{[N+1]}f(T^{n}x)<-\frac{d}{N}+\mathbb{A}_{[N]}f(T^{n}x)$ for
infinitely many $N$. ∎

We need the following results to prove 3.5. The definitions of upper barrier
and lower barrier are given right before the statement of 3.5.

######  Proposition 5.1.

Suppose that $(N_{i})$ satisfies the complete recurrence property for the
system $(X,\mathcal{B},\mu,T)$. Let $F$ be a measurable function and
$f=F-F\circ T$. Assume that $F$ has no upper barrier and no lower barrier.
Then for a.e. $x$, the averages $\mathbb{A}_{[N_{i}]}f(T^{n}x)<0$ infinitely
often and $\mathbb{A}_{[N_{i}]}f(T^{n}x)>0$ infinitely often.

###### Proof.

Since $F$ does not have an upper barrier, $F\circ T$ also does not have an
upper barrier. Hence, for a.e. $x$, there exists a set $B=B(x)$ of positive
measure such that $F(Ty)>F(Tx)$ for all $y\in B$. Since $(N_{i})$ is a
satisfies the complete recurrence property, for every positive integer $L$, we
have
$\displaystyle\mu\Big{(}\cup_{i\geq L}T^{-N_{i}}B\Big{)}=1.$ Thus, for
infinitely many $i$, $T^{N_{i}}x$ falls into $B$. For such an $i$,
$\mathbb{A}_{[N_{i}]}f(T^{n}x)=\frac{1}{N_{i}}\Big{(}F(Tx)-F(T^{N_{i}+1}x)\Big{)}<0$.
Similarly, for infinitely many $j$, $\mathbb{A}_{[N_{j}]}f(T^{n}x)>0$. ∎

Since ${\mathbb{N}}$ (with respect to the natural order) satisfies the
complete recurrence property, we have the following corollary.

######  Corollary 5.2.

Suppose that $F$ is a measurable function and $f=F-F\circ T$. Assume that $F$
has no upper barrier and no lower barrier. Then for a.e. $x$, the averages
$\mathbb{A}_{[N]}f(T^{n}x)<0$ infinitely often and
$\mathbb{A}_{[N]}f(T^{n}x)>0$ infinitely often.

######  Remark 5.3.

One cannot expect that the conclusion of 5.2 will hold for _every_ $x\in X$.
In fact, to the contrast of 5.2, it was proved by Peres [18] that if $X$ is a
compact space, $\mu$ is a Borel probability measure on $X$, $T:X\rightarrow X$
a measure preserving continuous transformation and
$g:X\rightarrow{\mathbb{R}}$ a continuous function, then for some $y\in X$

$\frac{1}{N}\sum_{n\in[N]}g(T^{n}y)\geq\int gd\mu\text{ for all }N\geq 1.$
(5.1)

The collection of those $y$ which satisfy Equation 5.1 is called _Heavy set_
for $g$. See also [5] for further results on Heavy set.

Now, we are ready to prove Theorem 3.5. The converse part of this theorem is
based on 3.3.

###### Proof of Theorem 3.5.

Assume that $F$ an upper barrier (the lower barrier case is similar). Then by
definition, there is a set $E$ of positive measure on which $F$ is a constant
$p$ a.e. and $F(x)\leq p$ for a.e. $x\in X$. Now consider the function
$f=\mathbbm{1}_{E}-\mathbbm{1}_{E}\circ T$. Observe that
$\displaystyle\mathbb{A}_{[N]}f(T^{n}x)\geq 0$ for a.e. $x$.

Conversely, assume that there exists a set $E\in\mathcal{B}$ such that
$\mu(E)>0$ and for all $x\in E$, $\mathbb{A}_{[N]}f(T^{n}x)\geq\int f$
eventually (The $`\leq^{\prime}$-case is similar). Without loss of generality,
let us assume that $\int f=0$. This implies that
$\mathbb{A}_{[0,N)}f(T^{n}x)\geq 0$ eventually on $T^{-1}E.$ Our goal is to
show that $f=F-F\circ T$ for some measurable function $F$. Consider the
following function:

$F(x)=\inf_{N\in{\mathbb{N}}}\sum_{n\in[0,N)}f(T^{n}x).$ (5.2)

Observe that $F$ is bounded below on $T^{-1}E$, otherwise we cannot have
$\mathbb{A}_{[0,N)}f(T^{n}x)\geq 0$ eventually on $T^{-1}E$. Since, the set
$G:=\\{x:F(x)\text{ is bounded below }\\}$ is invariant under $T$, and $T$ is
ergodic, $\mu(G)=1$. So, $F$ is a measurable function which is finite a.e. We
have

$\displaystyle F(Tx)$
$\displaystyle=\inf_{N\in{\mathbb{N}}}\sum_{n\in[N]}f(T^{n}x)$
$\displaystyle=\Big{(}\inf_{N\in{\mathbb{N}}}\sum_{n\in[0,N]}f(T^{n}x)\Big{)}-f(x)$
$\displaystyle\geq\Big{(}\inf_{N\in{\mathbb{N}}}\sum_{n\in[0,N)}f(T^{n}x)\Big{)}-f(x)$
$\displaystyle=F(x)-f(x)$

Define $K(x)=f(x)+F(Tx)-F(x)$. By the above discussion, we have $K(x)\geq 0$
for almost every $x$ and by the ergodic theorem

$\dfrac{1}{N}\sum_{n\in[0,N)}K(T^{n}x)\to\int_{X}K(x)d\mu.$ (5.3)

(even if $\int_{X}K(x)d\mu=\infty$, it follows easily from the finite case).
The left hand side is

$\dfrac{1}{N}\sum_{n\in[0,N)}f(T^{n}x)+\dfrac{1}{N}\sum_{n\in[0,N)}\Big{(}F(T^{N}x)-F(x)\Big{)}.$
(5.4)

The first term tends to $\int f=0$ a.e. by the ergodic theorem. Since $F(x)$
is finite a.e., there is a level set $\\{x:|F(x)|\leq M\\}$ with positive
measure and hence $T^{N}x$ falls into this set for a infinitely many $N$. For
such a $N$, $\big{(}F(T^{N}x)-F(x)\big{)}$ is bounded and so the second term
also tends to $0$. We conclude that $\int_{X}K(x)d\mu=0$. Since, $K(x)$ is
non-negative

$K(x)=0\text{ a.e.},\text{ i. e. }f(x)=F(x)-F(Tx).$ (5.5)

This proves that $f=F-F\circ T$. By assumption, we have
$\mathbb{A}_{[N]}f(T^{n}x)\geq 0$ eventually on a set of positive measure $E$.
Invoke, 5.2 to conclude that $F$ must have either an upper barrier or a lower
barrier, finishing the proof of the theorem. ∎

Now, we will show that if $T$ is totally ergodic then the averages of an
indicator functions will always oscillate.

######  Theorem 5.4.

Let $(X,\mathcal{B},\mu,T)$ be a totally ergodic system. Then for every
$E\in\mathcal{B}$ , $\displaystyle\mathbb{A}_{[N]}\mathbbm{1}_{E}(T^{n}x)$
oscillates infinitely often around $\mu(E)$ a.e.

###### Proof.

Suppose, to the contrary,
$\displaystyle\mathbb{A}_{[N]}\mathbbm{1}_{E}(T^{n}x)$ fails to oscillate in a
set of positive measure $F$. Since $T$ is ergodic, by 3.3 for almost every $x$
we can find two subsequences $N_{i}=N_{i}(x)$ and $N_{j}=N_{j}(x)$ such that
$\mathbb{A}_{[N_{i}]}\mathbbm{1}_{E}(T^{n}x)\geq\mu(E)$ and
$\mathbb{A}_{[N_{j}]}\mathbbm{1}_{E}(T^{n}x)\leq\mu(E)$. Hence, the failure of
oscillation imply that for all $x\in F$, we can find a subsequence
$N_{i}=N_{i}(x)$ such that
$\mathbb{A}_{[N_{i}]}\mathbbm{1}_{E}(T^{n}x)=\mu(E).$ Since
$\mathbb{A}_{[N]}\mathbbm{1}_{E}(T^{n}x)$ is always a rational number, the
equality can hold only if $\mu(E)$ is a rational number. So, let us assume
that $\mu(E)$ is a rational number. Now, we need the following lemma of Halász
[12, Theorem 2].

######  Lemma 5.5.

Let $(X,\mathcal{B},\mu,T)$ be a totally ergodic system. If for a measurable
set $E$,
$\displaystyle\Big{(}\sum_{n\in[N]}\mathbbm{1}_{E}(T^{n}x)-N\mu(E)\Big{)}_{N\in{\mathbb{N}}}$
is bounded below on a set of positive measure, then $e^{2\pi i\mu(E)}$ is an
eigenvalue of $T$, that is , there exists a function $g(x)$ not identically
$0$ a.e. such that $g(Tx)=e^{2\pi i\mu(E)}g(x)$.

Since $T$ is totally ergodic and $\mu(E)$ is a rational number, $e^{2\pi
i\mu(E)}$ cannot be an eigenvalue of $T$. Hence, for a.e. $x$, we have

$\inf_{N\in{\mathbb{N}}}\Big{(}\displaystyle\sum_{n\in[N]}\mathbbm{1}_{E}(T^{n}x)-N\mu(E)\Big{)}=-\infty.$
(5.6)

This gives us that for almost every $x$,
$\displaystyle\mathbb{A}_{[N]}\mathbbm{1}_{E}(T^{n}x)-\mu(E)<0$ infinitely
often. A similar argument can show that for almost every $x$,
$\displaystyle\mathbb{A}_{[N]}\mathbbm{1}_{E}(T^{n}x)-\mu(E)>0$ infinitely
often. This is a contradiction! This finishes the proof. ∎

The questions of non-monotonicity and oscillation for the subsequence of
ergodic averages are much more subtle than that of the classical ergodic
averages. As the following example shows, one can construct a totally ergodic
system $(X,\mathcal{B},\mu,T)$ (which is going to be an irrational rotation on
the usual one dimensional torus for us), a function $f\in L^{1}(X)$ and a
sequence $(N_{i})$ of positive integers such that for almost every $x$, the
averages $\displaystyle\mathbb{A}_{[N_{i}]}f(T^{n}x)$ are eventually, strictly
monotone.

######  Proposition 5.6.

Let $\alpha$ be an irrational number and let the sequence $(N_{i})$ be
strictly increasing and satisfy $0<(N_{i}+1)\alpha<(N_{i-1}+1)\alpha\mod 1$
for every $i$.

Then there is a function $f\in L^{1}(\mathbb{T},\lambda)$ such that for almost
every $x\in{\mathbb{T}}$ we have

$\mathbb{A}_{[N_{i-1}]}f(x+n\alpha)-\mathbb{A}_{[N_{i}]}f(x+n\alpha)>0\text{
for }i\geq i(x).$ (5.7)

In particular, the sequence
$\big{(}\mathbb{A}_{[N_{i}]}f(x+n\alpha)\big{)}_{i}$ of averages are almost
everywhere strictly monotone.

###### Proof.

In fact, we identify a certain set $F_{+}$ of coboundaries the elements of
which will all satisfy the claim of the proposition, that is, 5.7 for
$\lambda$-a.e. $x$.

Let us denote by $G_{+}$ those functions $g\in L^{1}$ which are piecewise
increasing, that is, there is a partition of $\mathbb{T}=[0,1)$ into finitely
many intervals222We assume that these intervals are subintervals of the
interval $[0,1)$. so that on each interval $g$ is strictly increasing. Let
$F_{+}$ be the set of coboundaries with transfer functions from $G_{+}$,

$F_{+}=\\{f:f(x)=g(x+\alpha)-g(x),g\in G_{+}\\}.$ (5.8)

Let $f\in F_{+}$ and let $g\in G_{+}$ be the corresponding transfer function,
so $f(x)=g(x+\alpha)-g(x)$. Let $x$ be in the interior of one of the intervals
$I$ where $g$ is strictly increasing. Then there is an index $i=i(x)$ so that
$x+(N_{i-1}+1)\alpha\in I$ for all $i\geq i(x)$. By definition, we have

$\displaystyle\mathbb{A}_{[N_{i-1}]}f(x+n\alpha)-\mathbb{A}_{[N_{i}]}f(x+n\alpha)$
(5.9)
$\displaystyle=\frac{1}{N_{i-1}}\Big{(}g(x+(N_{i-1}+1)\alpha)-g(x)\Big{)}-\frac{1}{N_{i}}\Big{(}g(x+(N_{i}+1)\alpha)-g(x)\Big{)}$
(5.10)

If $i\geq i(x)$ then
$g\big{(}x+(N_{i-1}+1)\alpha\big{)}>g\big{(}x+(N_{i}+1)\alpha\big{)}$ since we
assumed that $(N_{i-1}+1)\alpha>(N_{i}+1)\alpha$ for every $i$, hence
$x+(N_{i-1}+1)\alpha>x+(N_{i}+1)\alpha$ for $i\geq i(x)$, and $g$ is strictly
increasing on the interval $I\subset[0,1)$. But then
$g\big{(}x+(N_{i-1}+1\big{)}\alpha)-g(x)>g\big{(}x+(N_{i}+1\big{)}\alpha)-g(x)$
as well. Since $\frac{1}{N_{i-1}}>\frac{1}{N_{i}}$, we have

$\frac{1}{N_{i-1}}\Big{(}g\big{(}x+(N_{i-1}+1)\alpha\big{)}-g(x)\Big{)}>\frac{1}{N_{i}}\Big{(}g\big{(}x+(N_{i}+1)\alpha\big{)}-g(x)\Big{)}\text{
for }i\geq i(x)$ (5.11)

which then implies

$\mathbb{A}_{[N_{i-1}]}f(x+n\alpha)-\mathbb{A}_{[N_{i}]}f(x+n\alpha)>0\text{
for }i\geq i(x).$ (5.12)

∎

######  Remark 5.7.

$G_{+}$ is fairly large, namely it is dense in $L^{1}$, since every continuous
function can be approximated in $L^{1}$-norm by functions from $G_{+}$. It
follows that $F_{+}$ is dense in the orthocomplement of the constant
functions, that is, in the set of zero mean $L^{1}$ functions.

Next, we will prove 3.8. We need a couple of basic facts about the _strong
sweeping out property_.

######  Definition 5.8 (Strong sweeping out property).

Let $(X,\mathcal{B},\mu,T)$ be a measure preserving system. Let $(M_{n})$ be a
sequence of linear operators from $L^{1}$ to $L^{1}$. We say that the _strong
sweeping out_ property holds for the operators $M_{n}$ if for every
$\epsilon>0$ there is a set $E\in\mathcal{B}$ with $\mu(E)<\epsilon$ and

$\limsup_{N\to\infty}M_{N}\mathbbm{1}_{E}(x)=1\text{ a.e. and
}\liminf_{N\to\infty}M_{N}\mathbbm{1}_{E}(x)=0\text{ a.e.}$

A sequence $(a_{n})$ is said to satisfy the strong sweeping property, if for
every aperiodic system $(X,\mathcal{B},\mu,T)$ and for every $\epsilon>0$,
there exists a set $E$ with $\mu(E)<\epsilon$ such that

$\limsup\mathbb{A}_{[N]}\mathbbm{1}_{E}(T^{a_{n}}x)=1\text{ a.e. and
}\liminf\mathbb{A}_{[N]}\mathbbm{1}_{E}(T^{a_{n}}x)=0\text{ a.e.}$

######  Lemma 5.9.

Suppose that $(a_{n})$ is a lacunary sequence, that means, there exists
$\rho>1$ such that $\frac{a_{n+1}}{a_{n}}\geq\rho$ for all large $n$. Then
$(a_{n})$ satisfies the strong sweeping out property.

###### Proof.

This is proved in Corollary 1.2 of [1]. See Jones [13] or Mondal-Roy-Wierdl
[17] for a more transparent proof. ∎

###### Proof of 3.8.

Define

$\displaystyle S_{M}^{+}=\Big{\\{}f\in
L^{1}:\mu\\{x:\mathbb{A}_{[N_{i}]}f(T^{n}x)>0\text{ for some }i\geq
M\\}>1-\frac{1}{M}\Big{\\}}$ and $\displaystyle S_{M}^{-}=\Big{\\{}f\in
L^{1}:\mu\\{x:\mathbb{A}_{[N_{i}]}f(T^{n}x)<0\text{ for some }i\geq
M\\}>1-\frac{1}{M}\Big{\\}}.$

We will prove that $S_{M}^{+}$ and $S_{M}^{-}$ are both open and dense sets.
If we can show this then
$\displaystyle{\mathcal{O}}=\cap_{M}S_{M}^{+}\cap\mathcal{S}_{M^{-}}$ will
satisfy the desired properties of the theorem. Notice that $f\in S_{M}^{+}$
iff $-f\in S_{M}^{-}$. Hence, we will work only with $S_{M}^{+}$. The proof of
the theorem follows from 5.10 and 5.11.

######  Proposition 5.10.

$S_{M}^{+}$ is open in $L^{1}$.

###### Proof.

Let $f\in S_{M}^{+}$. Choose $\rho>0$ satisfying $\mu(D)>\rho>1-\frac{1}{M}$,
where

$D=\\{x:\mathbb{A}_{[N_{i}]}f(T^{n}x)>0\text{ for some }i\geq M\\}.$ (5.13)

Define

$D_{L}:=\\{x:\mathbb{A}_{[N_{i}]}f(T^{n}x)>0\text{ for some }i\text{ with
}L\geq i\geq M\\}.$ (5.14)

Observe that $D_{L}\uparrow D$. Hence, there exists $L_{0}$ such that
$\mu(D_{L_{0}})>\rho.$ Similarly, define

$D_{L_{0}}^{\delta}:=\\{x:\mathbb{A}_{[N_{i}]}f(T^{n}x)>\delta\text{ for some
}L_{0}\geq i\geq M\\}.$ (5.15)

Observe that as $\delta\downarrow 0$, $D_{L_{0}}^{\delta}\uparrow D_{L_{0}}.$
Hence, there exists $\delta_{0}>0$ such that
$\mu(D_{L_{0}}^{\delta_{0}})>\rho$. Choose
$\epsilon=\frac{1}{4}\big{(}\rho-(1-\frac{1}{M})\big{)}\frac{\delta_{0}}{(M-{L_{0}})}$.
We claim that whenever $\|f-g\|_{1}<\epsilon$, $g\in S_{M}^{+}$. Let $h=g-f$.
By Markov’s inequality, for every $i\in{\mathbb{N}}$ we have

$\mu\\{x:\mathbb{A}_{[N_{i}]}h(T^{n}x)<-\frac{\delta_{0}}{2}\\}<\frac{2\|h\|_{1}}{\delta_{0}}.$
(5.16)

Hence,

$\mu\Big{\\{}x:\mathbb{A}_{[N_{i}]}h(T^{n}x)<-\frac{\delta_{0}}{2}\text{ for
some }i\text{ with }L_{0}\geq i\geq
M\Big{\\}}<\frac{2\|h\|_{1}(M-{L_{0}})}{\delta_{0}}$ (5.17)

Since $\|h\|_{1}<\epsilon$, we must have

$\mu\Big{\\{}x:\mathbb{A}_{[N_{i}]}h(T^{n}x)<-\frac{\delta_{0}}{2}\text{ for
some }i\text{ with }L_{0}\geq i\geq
M\Big{\\}}<\frac{1}{2}\big{(}\rho-(1-\frac{1}{M})\big{)}$

This implies that

$\mu\Big{\\{}x:\mathbb{A}_{[N_{i}]}g(T^{n}x)>\frac{\delta_{0}}{2}\text{ for
some }L_{0}\geq i\geq
M\Big{\\}}>\rho-\frac{1}{2}\big{(}\rho-(1-\frac{1}{M})\big{)}.$ (5.18)

By our choice of $\rho$,
$\rho-\frac{1}{2}\big{(}\rho-(1-\frac{1}{M})\big{)}>\big{(}1-\frac{1}{M}\big{)}$.
Thus, we have

$\mu\Big{\\{}x:\mathbb{A}_{[N_{i}]}g(T^{n}x)>0\text{ for some }i\geq
M\Big{\\}}>\big{(}\rho-(1-\frac{1}{M})\big{)}$ (5.19)

This completes the proof of the proposition. ∎

######  Proposition 5.11.

$S_{M}^{+}$ is dense in $L^{1}$.

###### Proof.

It will be sufficient to show that $S_{M}^{+}$ is dense in $L^{1}_{0}$, where
$L^{1}_{0}$ denotes the space of $L^{1}$ functions with mean $0$. Since $T$ is
ergodic, coboundaries are dense in $L^{1}_{0}$. Furthermore, $L^{\infty}$ is
dense in $L^{1}$. Hence, the proof reduces to showing that $S_{M}^{+}$ is
dense in $\mathcal{F}$ where

$\mathcal{F}:=\Big{\\{}g:g=h\circ T-h\text{ for some }h\in
L^{\infty}\Big{\\}}.$ (5.20)

Let $\epsilon>0$, $M\in\mathbb{N}$ and $g=(h\circ T-h)\in\mathcal{F}$ be
arbitrary. Assume that $0<\epsilon<\frac{1}{M}$ and $\|h\|_{\infty}=K$.

######  Lemma 5.12.

Let $E,\epsilon,K,M,(N_{i})$ be as in the proposition. Then there exists a set
$E$ such that $\mu(E)<\frac{\epsilon}{6K}$ and a set $F$ with $\mu(F)=1$ such
that for every $x\in F$, there exist $i\geq M$ such that $T^{N_{i}+1}x\in E$.

###### Proof.

By deleting some terms if needed, we can assume that $(N_{i})$ satisfies
$\frac{N_{i+1}}{N_{i}}>2$. Then by 5.9, there exists a set $E$ with
$\mu(E)<\frac{\epsilon}{6K}$ such that

$\limsup_{i}\dfrac{\mathbbm{1}_{E}(T^{N_{1}+1}x)+\mathbbm{1}_{E}(T^{N_{2}+1}x)+\dots\mathbbm{1}_{E}(T^{N_{i}+1}x)}{i}=1$
(5.21)

holds for almost every $x\in X$. This gives our desired conclusion. ∎

Now, we want to construct a function $f$ such that $\|f-g\|_{1}<\epsilon$ and
$f\in S_{M}^{+}$.
Define $f=\tilde{h}\circ T-\tilde{h}$ where $\tilde{h}=h+3K\mathbbm{1}_{E}$.
Observe that

$\displaystyle\int|h-\tilde{h}|$ $\displaystyle=\int_{E}|h-\tilde{h}|\leq
3K\frac{\epsilon}{6K}=\frac{\epsilon}{2}.$ (5.22)
$\displaystyle\text{Similarly,}\int|h\circ T-\tilde{h}\circ T|$
$\displaystyle=\int_{{T^{-1}E}}|h-\tilde{h}|\leq
3K\frac{\epsilon}{6K}=\frac{\epsilon}{2}.\text{ Hence }\|f-g\|_{1}$
$\displaystyle<\epsilon.$ (5.23)

To check that $f\in S_{M}^{+}$, let $x\in(E^{c}\cap F)$. Then

$\displaystyle\mathbb{A}_{[N_{i}]}f(T^{n}x)$
$\displaystyle=\dfrac{\tilde{h}(T^{N_{i}+1}x)-\tilde{h}(x)}{N_{i}}$ (5.24)
$\displaystyle=\dfrac{\Big{(}h(T^{N_{i}+1}x)-h(x)\Big{)}}{N_{i}}+3K\dfrac{\mathbbm{1}_{E}(T^{N_{i}+1}x)-\mathbbm{1}_{E}(x)}{N_{i}}$
(5.25)

Since $x\in E^{c}$, we get

$\displaystyle\mathbb{A}_{[N_{i}]}f(T^{n}x)=\dfrac{\Big{(}h(T^{N_{i}+1}x)-h(x)\Big{)}+3K\mathbbm{1}_{E}(T^{N_{i}+1}x)}{N_{i}}$
(5.26)

By the above lemma, there exists $i_{0}>M$ such that $T^{N_{i_{0}}+1}x\in E$
i.e. $\mathbbm{1}_{E}(T^{N_{i_{0}}+1}x)=1$. We want to show that for this
$i_{0}$, the expression in 5.39 is positive. Since $\|h\|_{\infty}=K,$ we must
have $\Big{(}h(T^{N_{i_{0}}+1}x)-h(x)\Big{)}\leq 2K$. This implies that the
expression in 5.39 is at least $\dfrac{K}{N_{i}}$ and hence positive. Thus, we
have

$\mu\Big{\\{}x:\mathbb{A}_{[N_{i}]}f(T^{n}x)>0\text{ for some }i\geq
M\Big{\\}}\geq\mu(E^{c})\geq 1-\epsilon\geq 1-\frac{1}{M}.$

Hence we conclude that $f\in S_{M}^{+}$, thereby finishing the proof of the
theorem. ∎

∎

The following theorem provides a sufficient condition for non-monotonicity of
the averages in terms of the complete recurrence property.

######  Theorem 5.13.

Let $T$ be totally ergodic transformation and $(N_{i})$ be an increasing
sequence of positive integers which admits a subsequence $(N_{i_{j}})$ such
that

  1. (1)

$(N_{i_{j}+1}-N_{i_{j}})=K$ for some fixed constant $K$.

  2. (2)

$(N_{i_{j}})$ satisfies the complete recurrence property.

Then for any non-constant, integrable $f$ , $\mathbb{A}_{[N_{i}]}f(T^{n}x)$ is
non-monotone a.e. in the sense of 1.1.

###### Proof.

Without loss of generality, let us assume that $f\in L^{1}_{0}$, that is,
$\int f=0$.
One can check that

$\displaystyle\mathbb{A}_{[N_{i_{j}+1}]}f(T^{n}x)-\mathbb{A}_{[N_{i_{j}}]}f(T^{n}x)$
$\displaystyle=\Big{(}\mathbb{A}_{[N_{i_{j}}+1,N_{i_{j}+1}]}f(T^{n}x)-\mathbb{A}_{[N_{i_{j}}]}f(T^{n}x)\Big{)}\dfrac{(N_{i_{j}+1}-N_{i_{j}})}{N_{i_{j}+1}}.$

Since $\mathbb{A}_{[N_{i_{j}}]}f(T^{n}x)\to 0$ for almost every $x$, it will
be sufficient to show that there exists $\delta>0$ such that for a.e. $x$,
$\mathbb{A}_{[N_{i_{j}}+1,N_{i_{j}+1}]}f(T^{n}x)>\delta$ for infinitely many
$j$ and $\mathbb{A}_{[N_{i_{l}}+1,N_{i_{l}+1}]}f(T^{n}x)<-\delta$ for
infinitely many $l$.
Define

$g(x)=\mathbb{A}_{[K]}f(T^{n}x).$ (5.27)

Since $(N_{i_{j}+1}-N_{i_{j}})=K$ for every $j$,

$\mathbb{A}_{[N_{i_{j}}+1,N_{i_{j}+1}]}f(T^{n}x)=g(T^{N_{i_{j}}}x).$ (5.28)

This suggests that we should analyze the function $g$.
Since $f$ is non-constant and $T$ is totally ergodic, by 4.3, $g$ is also a
non-constant function. Since $f\in L^{1}_{0}$, we must have $g\in L^{1}_{0}$.
Hence, there exists $\delta>0$ such that both $\mu(S_{0}^{+})$ and
$\mu(S_{0}^{-})$ are positive, where

$\displaystyle S_{0}^{+}=\\{x:g(x)>\delta\\},\text{ and }\
S_{0}^{-}=\\{x:g(x)<-\delta\\}.$

Since $(N_{i_{j}})$ satisfies the complete recurrence property, for a.e. $x$,
there are infinitely many $j$ and $l$ such that $g(T^{N_{i_{j}}}x)>\delta$ and
$g(T^{N_{i_{l}}}x)<-\delta$. This finishes the proof. ∎

###  5.2. Irrational rotation case

Let us prove some corollaries of 5.13.

######  Corollary 5.14.

If $(X,\mathcal{B},\mu,T)$ is a totally ergodic transformation and $(N_{i})$
is a sequence of positive integers which contains an infinite arithmetic
progression, then for any non-constant, integrable $f$ ,
$\mathbb{A}_{[N_{i}]}f(T^{n}x)$ is non-monotone a.e. in the sense of 1.1.

The proof is immediate from Theorem 5.13.

For the next corollaries, we need the following lemmas.

######  Lemma 5.15.

Let $E\subset{\mathbb{T}}$ such that $\lambda(E)>0$ and $(x_{j})$ is a
sequence which is dense in ${\mathbb{T}}$, then
$\lambda\Big{(}\cup_{j\in{\mathbb{N}}}(x_{j}+E)\Big{)}=1$.

###### Proof.

Denote $F=\cup_{j}(E+x_{j})$. If we had $\lambda(F)<1$, then
$\lambda(F^{c})>0$. Let $I$ be an interval in which the density of $E$ is
greater than $1/2$, that is, $\lambda(E\cap I)>1/2\lambda(I)$. Let $J$ be an
interval in which the density of $F^{c}$ is greater than $1/2$. By the
Lebesgue density theorem we can assume that $I$ and $J$ have the same length.
Because of the density of $x_{j}$, there is a subsequence $j_{k}$ so that
$\lim_{k}(I+x_{j_{k}})=J$. Let $G=\lim_{k}(E\cap I)+x_{j_{k}}$. Then
$\lambda(G\cap J)>1/2\lambda(J)$ and also $\lambda(F^{c}\cap
J)>1/2\lambda(J)$. But $\lambda(F^{c}\cap G\cap J)=0$, an impossibility. ∎

######  Corollary 5.16.

Let $p_{i}$ denote the $i$-th prime. There exists a set $F\subset{\mathbb{T}}$
with $\lambda(F)=1$ which satisfies the following:
For each $\alpha\in F$ and non-constant $f\in L^{1}$,
$\mathbb{A}_{[p_{i}]}f(x+n\alpha)$ is non-monotone a.e. in the sense of 1.1

###### Proof.

By a famous theorem of Zhang [25], we know that there exists a constant $k$
and a subsequence $(p_{i_{j}})$ of $(p_{i})$ such that
$p_{i_{j}+1}-p_{i_{j}}=k$. By Weyl’s theorem, there exists a set $F$ with
$\lambda(F)=1$ such that $(p_{i_{j}}\alpha)_{j}$ mod 1 is dense in
${\mathbb{T}}$. Now, the corollary follows from 5.15 and 5.13. ∎

######  Corollary 5.17.

Let $(N_{i})$ be an increasing sequence such that
$\max_{i}(N_{i+1}-N_{i})<+\infty$. Then there exists a set $F$ with
$\lambda(F)=1$ which satisfies the following:
For each $\alpha\in F$ and non-constant $f\in L^{1}$,
$\mathbb{A}_{[N_{i}]}f(x+n\alpha)$ is non-monotone a.e. in the sense of 1.1

The proof is similar to the previous corollary.

Next, we will prove 3.9 which strengthens 5.17. We require the following lemma
for proving the theorem.

######  Lemma 5.18.

If a sequence $(N_{i})$ of positive integers has positive lower density, then
for all but countably many $\alpha$, $(N_{i}\alpha)_{i}$ (mod 1) is dense in
$\mathbb{T}$.

###### Proof.

This was (independently) proved by Rosenblatt [21] and Boshernitzan [3,
Proposition 8.1]. ∎

###### Proof of 3.9.

Let $M$ be the maximum gap of $(N_{i})$. For each $j\in[M]$, construct the set
$S_{j}$ as follows:
If for any $i\in{\mathbb{N}}$, $N_{2i}-N_{2i-1}=j$, then $N_{2i-1}\in S_{j}$.

Clearly,
$S=\displaystyle\cdot\bigcup_{j\in[M]}\Big{(}S_{j}\cup(S_{j}+j)\Big{)}$. For
$N_{l}\in S_{j}$ and $\alpha\in{\mathbb{T}}$, we have

$\displaystyle\mathbb{A}_{[N_{l+1}]}f(x+n\alpha)-\mathbb{A}_{[N_{l}]}f(x+n\alpha)$
$\displaystyle=\Big{(}\mathbb{A}_{[N_{l}+1,N_{l+1}]}f(x+n\alpha)-\mathbb{A}_{[N_{l}]}f(x+n\alpha)\Big{)}\dfrac{(N_{l+1}-N_{l})}{N_{l+1}}$
(5.29)
$\displaystyle=\Big{(}\mathbb{A}_{[j]}f\big{(}x+({N_{l}+n})\alpha\big{)}-\mathbb{A}_{[N_{l}]}f(x+n\alpha)\Big{)}\dfrac{j}{N_{l+1}}.$
(5.30)

For a $M$-tuple of positive integers $\mathsf{k}=(k_{1},k_{2},\dots,k_{M})$
and $j\in[M]$, we consider

$\displaystyle\text{ the shifted sets }S^{\mathsf{k}}_{j}=S_{j}+k_{j}\text{
and their union }S^{\mathsf{k}}=\cup_{j\in[M]}S^{\mathsf{k}}_{j}.$ (5.31)

Clearly, for every $\mathsf{k}$, $S^{\mathsf{k}}$ is a set of bounded gaps and
hence, it has positive lower density. But this implies, by 5.18, that
$E^{\mathsf{k}}$ is countable, where

$\displaystyle E^{\mathsf{k}}$
$\displaystyle=E(S^{\mathsf{k}})=\big{\\{}\alpha\in{\mathbb{T}}:\\{\alpha
s:s\in S^{\mathsf{k}}\\}\text{ is not dense}\in{\mathbb{T}}\big{\\}}.$ (5.32)

Hence,
$E=\displaystyle\bigcup_{\mathsf{k}\in{{\mathbb{N}}}^{M}}E^{\mathsf{k}}$ is
also countable. Now, we claim that if $\alpha\in{\mathbb{T}}\setminus E$ and
$f$ is a non-constant, integrable function, then for a.e. $x$, the averages
$\mathbb{A}_{[N_{i}]}f(x+n\alpha)$ are non-monotone.
For every $j\in[M]$, consider the following function

$g_{j}(x)=\mathbb{A}_{[j]}f(x+n\alpha).$

We know that for almost every $x$, $\mathbb{A}_{[N_{l}]}f(x+n\alpha)\to 0$ as
$l\to\infty$. We will show that there is a $\delta>0$ such that for a.e. $x$,
there are infinitely many $N_{l}\in S_{j}$ for some $j\in[M]$ satisfying
$g_{j}(x+N_{l}\alpha)>\delta$ and infinitely many (different) $N_{l}\in S_{w}$
for some $w\in[M]$ satisfying $g_{w}(x+N_{l}\alpha)<-\delta$. If we can show
this, then the expression in 5.30 will be $>0$ infinitely often and $<0$
infinitely often which will finish the proof of the claim. We will show here
the ‘$>$’ part only (other part is similar).

By using the total ergodicty of irrational rotation and 4.3, we can choose
$\delta>0$ small enough so that for every $j\in[M]$, $\lambda(B_{j,0}^{+})$ is
positive where

$B_{j,0}^{+}=\\{x:g_{j}(x)>\delta\\}.$ (5.33)

Again, by using the ergodicity of irrational rotation, we can find
$k_{1},k_{2},\dots,k_{M}$ such that $\lambda(B)>0$ where

$B=\displaystyle\bigcap_{j\in[M]}\Big{(}k_{j}\alpha+B^{+}_{j,0}\Big{)}.$
(5.34)

Fix $\mathsf{k}=(k_{1},k_{2},\dots,k_{M})$. Let $(s^{\mathsf{k}}_{n})_{n}$ be
the sequence obtained by arranging the terms of $S^{\mathsf{k}}$ in an
increasing order. (See 5.31 for the definition of $S^{\mathsf{k}}$).

Since $\alpha\in{\mathbb{T}}\setminus E$, $(\alpha s^{\mathsf{k}}_{n})_{n}$ is
dense in ${\mathbb{T}}$. Hence, by 5.15, we have

$\displaystyle\lambda\Big{(}\bigcup_{n\in{\mathbb{N}}}(-\alpha
s^{\mathsf{k}}_{n}+B\Big{)}=1.$

Hence, for a.e. $x\in{\mathbb{T}}$ and every large $L\in{\mathbb{N}}$, there
exists $n\geq L$ such that $x\in(-\alpha s^{\mathsf{k}}_{n}+B)$ for some $n$.
Let $s^{\mathsf{k}}_{n}\in S^{\mathsf{k}}_{j}$ for some $j\in[M]$, so
$s^{\mathsf{k}}_{n}=s_{j}+k_{j}$ for some $s_{j}\in S_{j}$. This implies that
$x\in-(s_{j}+k_{j})\alpha+B\subset-(s_{j}+k_{j})\alpha+(k_{j}\alpha+B^{+}_{j,0})=-s_{j}\alpha+B^{+}_{j,0}$.
Hence, we get $g_{j}(x+s_{j}\alpha)>\delta$, which is our desired conclusion.
∎

######  Remark 5.19.

  1. (1)

In view of 5.13 and 5.15, for an easier proof one would try to find a
subsequence $(N_{i_{j}})$ of $(N_{i})$ such that $(N_{i_{j}})$ has positive
lower density and for some fixed constant $K$, it satisfies
$N_{i_{j}+1}-N_{i_{j}}=K$ for every $j\in{\mathbb{N}}$. However, in general,
such subsequence may not exist. For example, take a large block of 2-gap
numbers, a large block of 3-gap numbers, another large block of 2-gap numbers,
another large block of 3-gap numbers, and so on. While it is always possible
to find a subsequence $(N_{i_{j}})$ of $(N_{i})$ that has positive upper
density and that satisfies $N_{i_{j}+1}-N_{i_{j}}=K$ for every
$j\in{\mathbb{N}}$ for some fixed constant $K$, 5.15 does not hold for a
sequence of positive upper density (see Boshernitzan [3] for a counter
example).

  2. (2)

The conclusion of Theorem 1.9 may not hold for all $\alpha$. Indeed, take
$(N_{i})$ to be the increasing sequence which is obtained by arranging those
$n\in\mathbb{N}$ which satisfies $n\alpha\in[\frac{1}{4},\frac{1}{2}]$ (mod
$1$) and

$f(x):=\mathbbm{1}_{E}(x)-\mathbbm{1}_{E}(x+\alpha)\text{ where
}E=[\frac{1}{2},1].$ (5.35)

With the above choice of $N_{i}$ and $\alpha$, one can check that for any
$x\in[\frac{1}{4},\frac{1}{2}]$, we have
$\mathbb{A}_{[0,N_{i+1})}f(x+n\alpha)-\mathbb{A}_{[0,N_{i})}f(x+n\alpha)=\frac{1}{N_{i}}-\frac{1}{N_{i+1}}\geq
0$. Hence, there is monotonicity for a set of positive measure.

######  Problem 5.20.

Let $(X,\mathcal{B},\mu,T)$ be an ergodic system and $(N_{i})$ be an
increasing sequence of positive integer. Can we characterize those functions
which oscillate around its mean infinitely often?

###  5.3. Some quantitative results

In 3.2 we have seen a quantitative result between consecutive averages. Now we
will investigate when the gap term $\frac{d}{N}$ can be made larger. Take
$w(N)$ that increases to $\infty$ as $N\to\infty$. We want to see for a mean-
zero $f\in L^{1}$ what can be said about having for a.e. $x$,
$\mathbb{A}_{[N+1]}f(T^{n}x)>\frac{w(N)}{N}+\mathbb{A}_{[N]}f(T^{n}x)$ for
infinitely many $N$. This condition clearly forces $w(N)=o(N)$ because a.e.
$\mathbb{A}_{[N]}f(T^{n}x)\to 0$ as $N\to\infty$.

By our computation above, this inequality basically means is
$f(T^{N+1}x)>2w(N)$ for infinitely many $N$. So we cannot have a gap like this
for $f\in L^{\infty}$. But if $f\in L^{1}\setminus L^{\infty}$, then it is
possible to have this occur. Similarly, at the same time,
$\mathbb{A}_{[N+1]}f(T^{n}x)<-\frac{w(N)}{N}+\mathbb{A}_{[N]}f(T^{n}x)$
infinitely often occurs if $f(T^{N+1}x)\leq-2w(N)$ for infinitely many $N$.
For unbounded $f\in L^{1}$ this is possible.

The level $w(N)$ for which this can occur depends on $f$ and $T$. For example,
suppose we have $w(N)=o(N)$. A related question is when can we have
$\sum\limits_{N\in{\mathbb{N}}}\mu\\{x:f(x)>w(N)\\}=\infty$? It is likely that
this occurs for a dense $G_{\delta}$ of functions in $L^{1}$. But then also
suppose we can take a map $T$ such that $f\circ T^{n}$ are IID. By the Borel-
Cantelli Lemma, this will allow us to conclude that for a.e. $x$,
$f(T^{N+1}x)>w(N)$ occurs for infinitely many $N$. Finding such a $T$ may not
work for $f$ on $X$, but we can certainly get many instances of the above by
changing our measure space $X$ to $\prod\limits_{n\in{\mathbb{Z}}}$ and
letting $T$ be the shift map. Then we take any $f\in L^{1}$ as above and
replace it by $f_{0}=f\circ\pi_{0}$ where $\pi_{0}$ is the projection of
$\prod\limits_{n\in{\mathbb{Z}}}$ onto the coordinate with $n=0$.

The conclusion above is that when $w(N)$ increases to $\infty$ but $w(N)/N\to
0$ as $N\to\infty$, then there are many functions $f$ and ergodic maps $T$
such that for a.e. $x$, both
$\mathbb{A}_{[N+1]}f(T^{n}x)>\frac{w(N)}{N}+\mathbb{A}_{[N]}f(T^{n}x)$ and
$\mathbb{A}_{[N+1]}f(T^{n}x)<-\frac{w(N)}{N}+\mathbb{A}_{[N]}f(T^{n}x)$.
Indeed, this is the typical case. We take here $\mathcal{M}$ to be the
invertible, measure-preserving maps of $(X,\mathcal{B},\mu)$ given the usual
weak topology.

######  Proposition 5.21.

Suppose $w(N)$ is increasing to $\infty$ and $w(N)/N\to 0$ as $N\to\infty$.
Then there is a dense $G_{\delta}$ set $\mathcal{G}\subset L^{1}$ such that
for all $f\in\mathcal{G}$, there is a dense $G_{\delta}$ set
$M(f)\subset\mathcal{M}$ of ergodic maps $T$ such that for a.e. $x$, we have
both $\mathbb{A}_{[N+1]}f(T^{n}x)>\frac{w(N)}{N}+\mathbb{A}_{[N]}f(T^{n}x)$
infinitely often and
$\mathbb{A}_{[N+1]}f(T^{n}x)<-\frac{w(N)}{N}+\mathbb{A}_{[N]}f(T^{n}x)$
infinitely often.

###### Proof.

We know that $f$ is integrable on a probability space if and only if
$\sum\limits_{N\in{\mathbb{N}}}\mu\\{|f|\geq N\\}<\infty$.

It is not hard to see this is best possible since if $\rho_{N}\to 0$, then for
a dense $G_{\delta}$ set of functions in $L^{1}$, we have
$\sum\limits_{N\in{\mathbb{N}}}\mu\\{|f|>\rho_{N}N\\}=\infty$. So if $w_{N}>0$
and $w_{N}/N\to 0$, taking this as $\rho_{N}$, we see that for a dense
$G_{\delta}$ set of functions
$\sum\limits_{N\in{\mathbb{N}}}\mu\\{|f|>w_{N}\\}=\infty$.

A little more work will show even more, as stated in the proof: for a dense
$G_{\delta}$ set of functions both
$\sum\limits_{N\in{\mathbb{N}}}\mu\\{f>2w_{N}\\}=\infty$ and
$\sum\limits_{N\in{\mathbb{N}}}\mu\\{f<-2w_{N}\\}=\infty$.

We take such a class $\mathcal{F}$ as above in $L^{1}$. We will use the
argument of Rosenblatt-Roychowdhury [22, Theorem 4.3]. However, here we will
use $B_{N}=\\{f>2w_{N}\\}$ and $B_{N}=\\{f<-2w_{N}\\}$ and use $T$ to denote
the system, not $\tau$.

By Rosenblatt-Roychowdhury [22, Theorem 4.3] , for all the $f\in\mathcal{F}$,
we have a dense $G_{\delta}$ set of maps $T$ such that for a.e. $x$,
$T^{N}x\in B_{N}$ infinitely often, for both types of $B_{N}$.

Now we use the analysis before 5.21 to both the following inequalities:
$\mathbb{A}_{[N+1]}f(T^{n}x)>w_{N}/N+\mathbb{A}_{[N]}f(T^{n}x)$ and
$\mathbb{A}_{[N+1]}f(T^{n}x)<-w_{N}/N+\mathbb{A}_{[N]}f(T^{n}x)$. Because we
have used $2w_{N}$ in defining the $B_{N}$, we have a little more room in the
inequalities. So this gives us the divergent series results as stated in 5.21.
∎

######  Remark 5.22.

It is worth noting that if $p>1$, then $L^{p}$ is first category in $L^{1}$,
and indeed the complement $L^{1}\backslash L^{p}$ is a dense $G_{\delta}$. So
if we ask for generic behavior for
$\sum\limits_{n\in{\mathbb{N}}}\mu\\{|f|>w_{n}\\}=\infty$, and
$w_{n}=n^{1/p}$, we can use this.

However, for more general $w_{n}$ with $w_{n}/n\to 0$, something more is
needed. Probably, in this case there is always an increasing
$\phi:[0,\infty)\to[0,\infty)$ such that $\phi(n)\geq w_{n}$ and yet still
$\phi(n)/n\to 0$. Then one argues that the $f\in L^{1}$ such that
$\phi^{-1}(|f|)$ are not integrable is a generic set. So generically we would
have $\sum\limits_{n\in{\mathbb{N}}}\mu\\{|f|\geq
w_{n}\\}\geq\sum\limits_{n\in{\mathbb{N}}}\mu\\{|f|\geq\phi(n)\\}=\sum\limits_{n\in{\mathbb{N}}}\mu\\{\phi^{-1}(|f|)\geq
n\\}=\infty$.

We use $\phi$ and consider an Orlicz space of functions where $\phi^{-1}(|f|)$
is integrable and show that it is first category in $L^{1}$, with a dense
$G_{\delta}$ complement, just as is done in the special cases where
$\phi(n)=n^{1/p}$.

∎

Now, we will deduce some quantitative results about non-monotonicity of
subsequence of averages by using some strong sweeping out results of moving
averages.

As we determined earlier,

$\mathbb{A}_{[0,N_{i+1})}f(T^{n}x)-\mathbb{A}_{[0,N_{i})}f(T^{n}x)=\Big{(}\mathbb{A}_{[N_{i},N_{i+1})}f(T^{n}x)-\mathbb{A}_{[0,N_{i})}f(T^{n}x)\Big{)}\dfrac{N_{i+1}-N_{i}}{N_{i+1}}$
(5.36)

If we let $l_{i}=N_{i+1}-N_{i}$, then we get

$\mathbb{A}_{[N_{i},N_{i+1})}f(T^{n}x)=\frac{1}{l_{i}}\sum_{j\in[0,l_{i}-1]}f(T^{N_{i}+j}x).$
(5.37)

We introduce a sequence of operators $M_{i}:L^{1}\to L^{1}$ as follows

$M_{i}f(x)=\frac{1}{l_{i}}\sum_{j\in[0,l_{i}-1]}f(T^{N_{i}+j}x).$ (5.38)

This operators are called the _moving averages_ with parameters
$(N_{i},l_{i})$.

######  Proposition 5.23.

Assume that $T$ is ergodic, and $(M_{i})$ satisfies the strong sweeping out
property. Then there is a residual set $\beta\subset\mathcal{B}$ with the
following property:
For any $E\in\beta$ there exists a constant $d=d(E,N_{i})>0$ such that for
a.e. $x$,
$\mathbb{A}_{[0,N_{i+1})}\mathbbm{1}_{E}(T^{n}x)>\frac{d}{N_{i+1}}+\mathbb{A}_{[0,N_{i})}\mathbbm{1}_{E}(T^{n}x)$
infinitely often and for a.e. $x$,
$\mathbb{A}_{[0,N_{i+1})}\mathbbm{1}_{E}(T^{n}x)<-\frac{d}{N_{i+1}}+\mathbb{A}_{[0,N_{i})}\mathbbm{1}_{E}(T^{n}x)$
infinitely often. The constant $d$ can be taken to be any number smaller than
$\big{(}\min\\{l_{i}\\}\big{)}.\big{(}\min\\{\mu(E),1-\mu(E)\\}\big{)}.$

###### Proof.

Let $d$ be as mentioned in the proposition. Since $T$ is ergodic,

$\mathbb{A}_{[0,N_{i})}\mathbbm{1}_{E}(T^{n}x)\to\mu(E)\text{ for a.e. }x.$
(5.39)

Suppose $(M_{i})$ satisfies the strong sweeping out. Then, by using
[14][Theorem 1.2] and observing that
$\mathbb{A}_{[N_{i},N_{i+1})}f(T^{n}x)=M_{i}(x)$, we can find a residual set
$\beta\subset\mathcal{B}$ such that for every $E\in\beta$,

$\limsup_{i}\mathbb{A}_{[N_{i},N_{i+1})}\mathbbm{1}_{E}(T^{n}x)=1\text{ for
a.e. }x$ (5.40)

and

$\liminf_{i}\mathbb{A}_{[N_{i},N_{i+1})}\mathbbm{1}_{E}(T^{n}x)=0\text{ for
a.e. }x.$ (5.41)

Let $E\in\beta$. Let $\tilde{X}$ be the common set of full measure where Eq.
5.39, Eq. 5.40 and Eq. 5.41 hold. Hence, for $x\in\tilde{X}$, there are two
subsequences $(N_{i_{j}})=(N_{i_{j}})(x)$ and $(N_{i_{k}})=(N_{i_{k}})(x)$
such that

$\displaystyle\lim_{j}\mathbb{A}_{[N_{i_{j}}+l_{i_{j}})}\mathbbm{1}_{E}(T^{n}x)=1$
(5.42)

and

$\displaystyle\lim_{k}\mathbb{A}_{[N_{i_{k}},N_{i_{k}}+l_{i_{k}})}\mathbbm{1}_{E}(T^{n}x)=0.$
(5.43)

Thus we have

$\displaystyle\lim_{j}\mathbb{A}_{[N_{i_{j}},N_{i_{j}}+l_{i_{j}})}\mathbbm{1}_{E}(T^{n}x)-\mathbb{A}_{N_{i_{j}}}\mathbbm{1}_{E}(T^{n}x)\to(1-\mu(E))$
(5.44) and
$\displaystyle\lim_{k}\mathbb{A}_{[N_{i_{k}},N_{i_{k}}+l_{i_{k}})}\mathbbm{1}_{E}(T^{n}x)-\mathbb{A}_{N_{i_{k}}}\mathbbm{1}_{E}(T^{n}x)\to(0-\mu(E)).$
(5.45)

By Equation 5.36, we have

$\displaystyle\mathbb{A}_{N_{i_{j}+1}}\mathbbm{1}_{E}(T^{n}x)-\mathbb{A}_{N_{i_{j}}}\mathbbm{1}_{E}(T^{n}x)$
$\displaystyle=\Big{(}\mathbb{A}_{[N_{i_{j}},N_{i_{j}}+l_{i_{j}})}\mathbbm{1}_{E}(T^{n}x)-\mathbb{A}_{N_{i_{j}}}\mathbbm{1}_{E}(T^{n}x)\Big{)}\dfrac{l_{i_{j}}}{N_{i_{j}+1}}$

By using Equation 5.44, we get that

$\displaystyle\Big{(}\mathbb{A}_{N_{i_{j}+1}}\mathbbm{1}_{E}(T^{n}x)-\mathbb{A}_{N_{i_{j}}}\mathbbm{1}_{E}(T^{n}x)\Big{)}$
$\displaystyle\to(1-\mu(E))\dfrac{l_{i_{j}}}{N_{i_{j}+1}}\text{ as
}j\to\infty.$ Thus for large $j$, we have
$\displaystyle\Big{(}\mathbb{A}_{N_{i_{j}+1}}\mathbbm{1}_{E}(T^{n}x)-\mathbb{A}_{N_{i_{j}}}\mathbbm{1}_{E}(T^{n}x)\Big{)}$
$\displaystyle\geq\dfrac{d}{N_{i_{j}+1}}.$ Similarly, using Equation 5.36 and
Equation 5.45 we get,
$\displaystyle\Big{(}\mathbb{A}_{N_{i_{k}+1}}\mathbbm{1}_{E}(T^{n}x)-\mathbb{A}_{N_{i_{k}}}\mathbbm{1}_{E}(T^{n}x)\Big{)}$
$\displaystyle\leq-\dfrac{d}{N_{i_{k}+1}}.$

This finishes the proof. ∎

We say a sequence $(N_{i})$ is sublacunary if it satisfies the following

$\lim_{i\to\infty}\dfrac{N_{i+1}}{N_{i}}=1.$

######  Corollary 5.24.

Assume that $T$ is an ergodic transformation on a non-atomic probability space
$(X,\mathcal{B},\mu)$ and $(N_{i})$ is sublacunary. Then there is a residual
set $\beta\subset\mathcal{B}$ with the following property:
For any $E\in\beta$ there exists a constant $d=d(E,N_{i})>0$ such that for
a.e. $x$,
$\mathbb{A}_{[0,N_{i+1})}\mathbbm{1}_{E}(T^{n}x)>\frac{d}{N_{i+1}}+\mathbb{A}_{[0,N_{i})}\mathbbm{1}_{E}(T^{n}x)$
infinitely often and for a.e. $x$,
$\mathbb{A}_{[0,N_{i+1})}\mathbbm{1}_{E}(T^{n}x)<-\frac{d}{N_{i+1}}+\mathbb{A}_{[0,N_{i})}\mathbbm{1}_{E}(T^{n}x)$
infinitely often.

###### Proof.

This result follows from Rosenblatt-Wierdl [23][Theorem 2.5] and 5.23. ∎

##  6\. Oscillation of convolution operators

Given $\phi\in L^{1}(\mathbb{R})$ and a suitable function $f$, the convolution
of $\phi$ and $f$ will be denoted by $\phi\star
f(x)=\int\limits_{-\infty}^{\infty}\phi(y)f(x-y)d\lambda(y)$. The first basic
general principle is this one.

######  Theorem 6.1.

Let $B=L^{p}(\mathbb{R})$ for some fixed $1\leq p\leq\infty$, or
$B=CB(\mathbb{R})$. Assume that $(\phi_{n})$ is a proper approximate identity.
Let $(\epsilon_{n})$ be a sequence of positive numbers with
$\lim\limits_{n\to\infty}\epsilon_{n}=0$. Then there is a dense $G_{\delta}$
subset $\mathcal{A}$ of $B$ such that for all $f\in\mathcal{A}$,
$\limsup\limits_{n\to\infty}\frac{|\phi_{n}\star
f(x)-f(x)|}{\epsilon_{n}}=\infty$ a.e.

###### Proof.

We use Theorem 1.1 in del Junco and Rosenblatt [14]. Because our approximate
identity is normalized and asymptotically local, we can use this theorem,
which is stated for probability spaces, in the context of convolutions on
$\mathbb{R}$. Indeed, what we show here is that for any interval $I=(a,b)$,
there is a dense $G_{\delta}$ subset $\mathcal{A}_{I}$ in $B$ such that for
$f\in\mathcal{A}_{I}$, $\sup\limits_{n\geq 1}\frac{|\phi_{n}\star
f(x)-f(x)|}{\epsilon_{n}}=\infty$ a.e. on $I$. Intersecting these sets over
all rational numbers $a$ and $b$ gives us a set $\mathcal{A}$ as in the
statement of the theorem.

First, consider the case that $B=L^{1}(\mathbb{R})$. To use Theorem 1.1 in del
Junco-Rosenblatt [14], we need to show that for any $K$ and
$\epsilon<\frac{1}{2}$, there exists $f\in L^{1}(\mathbb{R})$ with
$\|f\|_{1}\leq 1$ such that $\lambda\\{x\in I:\sup\limits_{n\geq
1}\frac{|\phi_{n}\star f-f|}{\epsilon_{n}}\geq K\\}\geq\lambda(I)-\epsilon$.
For now also assume that $I=(0,1)$. Then, taking $M$ to be a whole number, let
$f(x)=\frac{1}{\lambda(I)}\frac{\sin(Mx)}{|\sin(Mx)|}\mathbbm{1}_{I}(x)$
except when $\sin(Mx)=0$, and then let $f(x)=1$. We see that $\|f\|_{1}\leq
1$. Fix $N\geq 1$, and choose $M$ sufficiently large such that the rapid
oscillation of $f$ causes $\|\phi_{n}\star f\|_{\infty}\leq\epsilon$ for
$1\leq n\leq N$. So for a fixed $n,1\leq n\leq N$, if
$\epsilon_{n}\leq\frac{1}{2K}$, this gives $\frac{|\phi_{n}\star
f-f|}{\epsilon_{n}}\geq K$ a.e. on $I$ because $|\phi_{n}\star f-f|\geq
1-\epsilon\geq\frac{1}{2}$. So certainly we can construct $f,\|f\|_{1}\leq 1$,
with $\lambda\big{\\{}x\in I:\sup\limits_{n\geq 1}\frac{|\phi_{n}\star
f-f|}{\epsilon_{n}}\geq K\big{\\}}\geq\lambda(I)-\epsilon$. By translating and
dilating the construction of $f$, we can achieve the same result for any
interval $I$, although the choice of $M$ will need to take into account the
size of $I$.

The same argument works in any $L^{p}(\mathbb{R})$ with some small changes. If
$1\leq p<\infty$, we take
$f(x)=\frac{1}{\lambda(I)^{1/p}}\frac{\sin(Mx)}{|\sin(Mx)|}\mathbbm{1}_{I}(x)$,
and proceed as above. For $L^{\infty}(\mathbb{R})$ the normalizing factor is
taken to be $1$, not $\frac{1}{\lambda(I)^{1/p}}$. A little more care in the
construction of $f$ is needed when $B=CB(\mathbb{R})$, since we must make sure
that the function is continuous. In this case, again letting $I=(0,1)$ at the
outset, take $g\in CB(\mathbb{R})$ to be a piecewise linear function such that
$g=1$ on $(0,\frac{1}{2}-\epsilon)$, $g=-1$ on $(\frac{1}{2}+\epsilon,1)$, and
$g=0$ otherwise. The let $f(x)=g(Mx)$ for all $x\in\mathbb{R}$. As we decrease
$\epsilon$ and increase $M$, the increased oscillation in $f$ on $I$ will
allow us to have simultaneously $|\phi_{n}\star f|\leq\epsilon$ for all
$n,1\leq n\leq N$, and $|f|=1$ on some $J\subset I$ with $\lambda(J)\geq
1-\epsilon$. Again, since $\lim\limits_{n\to\infty}\epsilon_{n}=0$, this gives
the desired estimate on the function on $(0,1)$. By translating and dilating
the construction of $f$, we can achieve the the same result for any interval
$I$. ∎

######  Remark 6.2.

a) Theorem 6.1 is really meant to address the case the $(\phi_{n}\star f)$
converges to $f$ a.e. for all $f\in B$. If not, having a ratio divergence does
not have the same value. But even in this situation, the fact that the
divergence could occur for different functions because of different very
sparse subsequences of $(\phi_{n})$ suggests that even in this case the
theorem tells us something worthwhile.

b) One can possibly think that dividing by a sequence of constants
$\epsilon_{n}$ that tend to $0$ is too strict, but rather one should look for
a sequence of positive functions $(\epsilon_{n}(x))$ that converges $0$ a.e.
to gauge the rate of convergence of the convolutions. However, if
$\lim\limits_{n\to\infty}\epsilon_{n}(x)=0$ a.e., then for any $\epsilon>0$
and any bounded interval $I$, there is $J\subset I$ such that
$\lambda(I\backslash J)<\epsilon$ and
$\lim\limits_{n\to\infty}\epsilon_{n}(x)=0$ uniformly on $J$. Using this the
argument in Theorem 6.1 shows that there is a dense $G_{\delta}$ of functions
$f$ in $B$ such that $\limsup\limits_{n\to\infty}\frac{|\phi_{n}\star
f(x)-f(x)|}{\epsilon_{n}(x)}=\infty$ a.e.

c) We could also develop other types of measures of the rate of convergence of
$(\phi_{n}\star f)$ as was done in the ergodic theory context in del Junco and
Rosenblatt [14] using the principle Theorem 1.1 formulated there. For example,
even in the best situation, where $\phi_{n}=n1_{[-1/n,0]},n\geq 1$, as long as
a sequence of positive numbers $(w_{n})$ has
$\sum\limits_{n\in{\mathbb{N}}}w_{n}=\infty$, the series
$\sum\limits_{n\in{\mathbb{N}}}w_{n}(\phi_{n}\star f-f)$ will diverge almost
everywhere for a dense $G_{\delta}$ set of functions in $B$. The proof of this
and related variations on this theme are reasonably straightforward
adaptations of the ergodic theoretic results that are given in del Junco and
Rosenblatt [14].

d) It is clear that there are non-trivial functions, even continuous
functions, such that we do not have any such divergence. Having a
characterization of this, or at least some simple constructions, would be
worthwhile.

With a little more attention to the construction, we can improve Theorem 6.1
to prove 3.11.

###### Proof of 3.11.

The mechanics of the proof are the same as for the proof of Theorem 6.1. We
illustrate this first for $B=L^{1}(\mathbb{R})$. What we need to show is that
for a fixed interval interval $I$, we have for all $K>0$ and $\epsilon>0$,
there exists some function $f\in L^{1}(\mathbb{R})$ with $\|f\|_{1}\leq 1$
such that $\lambda\\{x\in I:\sup\limits_{n\geq 1}\frac{\phi_{n}\star
f(x)-f(x)}{\epsilon_{n}}\geq K\\}\geq\lambda(I)-\epsilon$. There is no harm in
assuming that $I=(0,1)$. Let $(r_{m})$ be the sequence of Rademacher
functions. Let $E_{m}=\\{x\in I:r_{m}(x)=-1\\}$. We will be choosing a finite
subsequence $r_{m_{1}},\dots,r_{m_{L}}$. The mutually independence of the
Rademacher functions shows that
$\lambda(\bigcup\limits_{l=1}^{L}E_{m_{l}})\geq 1-(\frac{1}{2})^{L}$. Take a
set of positive numbers $\\{a_{1},\dots,a_{L}\\}$. Let
$f=\sum\limits_{l\in[L]}a_{l}r_{m_{l}}$. We will want to have $\|f\|_{1}\leq
1$, so for now we just fix $a_{l}>0$ so that $\sum\limits_{l\in[L]}a_{l}\leq
1$. We consider some finite subsequence $(\phi_{n_{l}}:l=1,\dots,L)$ and a
fixed $j=1,\dots,L$. We now estimate how large $\frac{\phi_{n_{j}}\star
f-f}{\epsilon_{n_{j}}}$ can be made by using the fact that

$\displaystyle\frac{\phi_{n_{j}}\star f-f}{\epsilon_{n_{j}}}$
$\displaystyle\geq\frac{\phi_{n_{j}}\star(a_{j}r_{m_{j}})-a_{j}r_{m_{j}}}{\epsilon_{n_{j}}}$
$\displaystyle-\sum\limits_{l\in[j-1]}\frac{\phi_{n_{j}}\star(a_{l}r_{m_{l}})-a_{l}r_{m_{l}}}{\epsilon_{n_{j}}}-\sum\limits_{l\in[j+1,L]}\frac{\phi_{n_{j}}\star(a_{l}r_{m_{l}})-a_{l}r_{m_{l}}}{\epsilon_{n_{j}}}.$

We make the term
$\frac{\phi_{n_{j}}\star(a_{j}r_{m_{j}})-a_{j}r_{m_{j}}}{\epsilon_{n_{j}}}$
large on $E_{m_{j}}$ just as in the proof of Theorem 6.1 except that we also
have to compensate for the scaling by $a_{j}$. That is, we can fix $n_{j}$
large and then choose $m_{j}$ so that
$\phi_{n_{j}}\star(a_{j}r_{m_{j}})-a_{j}r_{m_{j}}\geq\frac{a_{j}}{2}$ on
$E_{m_{j}}$. So if we have originally chosen $n_{j}$ so that
$\frac{a_{j}}{2\epsilon_{n_{j}}}\geq 2K$, we get
$\frac{\phi_{n_{j}}\star(a_{j}r_{m_{j}})-a_{j}r_{m_{j}}}{\epsilon_{n_{j}}}\geq
2K$ on $E_{n_{j}}$. This makes it possible to choose $n_{j}$ to give us this
estimate even thought $a_{j}$ may be very small.

Our construction of $(a_{l}:l=1,\dots,L),(n_{l}:l=1,\dots,L)$ and
$(m_{l}:l=1,\dots,L)$ will be inductive. We begin with any fixed
$a_{1},0<a_{1}<1$, and then choose $n_{1}$ and $m_{1}$ so that
$\frac{\phi_{n_{1}}\star(a_{1}r_{m_{1}})-a_{1}r_{m_{1}}}{\epsilon_{n_{1}}}\geq
2K$ on $E_{n_{1}}$. Now suppose that $j=1,\dots,L$ is fixed and we have chosen
$a_{l},n_{l}$ and $m_{l}$ for $l=1,\dots,j-1$. We then choose $n_{j}$ large
enough for the convolutions $\phi_{n_{j}}\star(a_{l}r_{m_{l}})$ to be equal to
$a_{l}r_{m_{l}}$ everywhere except for a set of small measure made up of
intervals around the discontinuities of $r_{m_{l}}$ for all $l=1,\dots,j-1$.
So we can arrange for
$\sum\limits_{l\in[j-1]}\frac{\phi_{n_{j}}\star(a_{l}r_{m_{l}})-a_{l}r_{m_{l}}}{\epsilon_{n_{j}}}$
to be $0$ except for a set $F_{j}$ of measure as small as we like in $I$. We
can also have chosen $n_{j}$ and $m_{j}$ as discussed above so that we have
$\frac{\phi_{n_{j}}\star(a_{j}r_{m_{j}})-a_{j}r_{m_{j}}}{\epsilon_{n_{j}}}\geq
2K$ on $E_{n_{j}}$. So this means that we can arrange that
$\frac{\phi_{n_{j}}\star(a_{j}r_{m_{j}})-a_{j}r_{m_{j}}}{\epsilon_{n_{j}}}-\sum\limits_{l\in[j-1]}\frac{\phi_{n_{j}}\star(a_{l}r_{m_{l}})-a_{l}r_{m_{l}}}{\epsilon_{n_{j}}}\geq
2K$ on $\widetilde{E_{n_{j}}}=E_{n_{j}}\backslash F_{j}$. But now the sum
$\sum\limits_{l\in[j+1,L]}\frac{\phi_{n_{j}}\star(a_{l}r_{m_{l}})-a_{l}r_{m_{l}}}{\epsilon_{n_{j}}}$
comes from terms chosen later in the inductive construction, and it could
affect that estimate that we have made already. However, we do have the
freedom to place restrictions on the values of $a_{l},l=j+1,\dots,L$ making
them sufficiently small, without disturbing the necessary fact that
$\sum\limits_{l\in[L]}a_{l}\leq 1$, so we can arrange that
$\sum\limits_{l\in[j+1,L]}\frac{\phi_{n_{j}}\star(a_{l}r_{m_{l}})-a_{l}r_{m_{l}}}{\epsilon_{n_{j}}}$
is uniformly less than $\frac{K}{2}$ on $I$. In summary, on
$\widetilde{E_{m_{j}}}$, we have $\frac{\phi_{n_{j}}\star
f-f}{\epsilon_{n_{j}}}\geq\frac{3K}{2}$. In this inductive step, we also
assume of course that we have chosen $a_{j}$ so that $a_{j}>0$ and
$\sum\limits_{l\in[j]}a_{l}<1$. This ends the inductive step. We notice that
at the $j$th stage of the induction we are revising the restrictions on
$(a_{l}:l=j+1,\dots,L)$ that were imposed form earlier steps in the induction,
to make the future choices of $a_{l}$ even smaller than before. In any case,
we can choose the sets $F_{n_{j}}$ of sufficiently small measure so that the
function $f$ has $\|f\|_{1}\leq 1$ and $\lambda\\{x\in I:\sup\limits_{n\geq
1}\frac{\phi_{n}\star f(x)-f(x)}{\epsilon_{n}}\geq
K\\}\geq\lambda(I)-\epsilon$.

We now know that there is a dense $G_{\delta}$ set $\mathcal{A}^{+}$ in
$L^{1}(\mathbb{R})$ such that for all $f\in\mathcal{B}^{+}$ we have
$\limsup\limits_{n\to\infty}\frac{\phi_{n}\star f-f}{\epsilon_{n}}=\infty$
almost everywhere. By working with the sets $\\{x\in I:r_{m}(x)=1\\}$, we
could similarly arrange to have this companion lemma: for all $K>0$ and
$\epsilon>0$, there exists some function $f\in L^{1}(\mathbb{R})$ with
$\|f\|_{1}\leq 1$ such that $\lambda\\{x\in I:\inf\limits_{n\geq
1}\frac{\phi_{n}\star
f(x)f(x)}{\epsilon_{n}}\leq-K\\}\geq\lambda(I)-\epsilon$. Then we can conclude
that there is a dense $G_{\delta}$ set $\mathcal{A}^{-}$ in
$L^{1}(\mathbb{R})$ such that for all $f\in\mathcal{A}^{-}$ we have
$\liminf\limits_{n\to\infty}\frac{\phi_{n}\star f-f}{\epsilon_{n}}=-\infty$
almost everywhere. The intersection
$\mathcal{A}=\mathcal{A}^{+}\cap\mathcal{A}^{-}$ gives the desired dense
$G_{\delta}$ set.

The case where $B=L^{p}(\mathbb{R})$ is handled similarly except for the
normalization inherent in the choice of $(a_{l}:l=1,\dots,L)$. The case of
$CB(\mathbb{R})$ requires more care as before, because the functions we
construct for our lemmas must be continuous. But there is not much difficulty
in handling this, in the same way that it was dealt with in Theorem 6.1, at
the expense of having a somewhat large set $F_{n_{j}}$ to deal with, albeit
one that can be chosen inductively to have measure as small as we like at any
stage of the construction. ∎

######  Remark 6.3.

  1. (1)

There may be value in having the two separate Baire category arguments. But
again, we can use the first one to get the result. Once we know that for a
generic set $\mathcal{C}$, we have for $f\in\mathcal{C}$,
$\limsup\limits_{n\to\infty}\frac{\phi_{n}\star
f(x)-f(x)}{\epsilon_{n}}=\infty$ almost everywhere, then we would have a
generic set $\mathcal{A}=-\mathcal{C}$ such that for $f\in\mathcal{A}$, we
have
$\limsup\limits_{n\to\infty}\frac{\phi_{n}\star(-f(x)-(-f)(x)}{\epsilon_{n}}=\infty$
almost everywhere. But then because in general we have
$\limsup\limits_{n\to\infty}-\rho_{n}=-\liminf\limits_{n\to\infty}\rho_{n}$,
we would have for $f\in\mathcal{A}$,
$\liminf\limits_{n\to\infty}\frac{\phi_{n}\star
f(x)-f(x)}{\epsilon_{n}}=-\infty$ almost everywhere. So taking
$\mathcal{A}\cap\mathcal{C}$, we would get a generic set on which we have the
oscillation around the limit that we want to demonstrate.

  2. (2)

In the case that we are considering $B=CB(\mathbb{R})$, it seems reasonable to
ask to improve Theorem 6.1 by requiring _everywhere_ divergence of the ratios.
This is possible, but not without some technical difficulties. In our upcoming
work, we will explore this topic.

##  7\. Martingales

In this section, we will prove 3.12.

###### Proof of 3.12.

We can prove the generic behavior occurs above in the following fashion.
First, fix $l\geq 1$ and $N_{0}\geq 1$. Then let $\mathcal{O}_{l,N_{0}}$
consisting of all $G\in L^{1}[0,1]$ such that
$\lambda\\{x:E(G|\mathcal{D}_{n})(x)>G(x)\,\text{for some}\,\,n,N_{0}\leq
n\\}>1-\frac{1}{l}$. By using a similar argument as in 3.8, one can check that
$\mathcal{O}_{l,N_{0}}$ is open in the $L^{1}$-norm topology. Suppose we show
that it is dense in that topology too. Then
$\mathcal{O}^{+}=\bigcap\limits_{l=1}^{\infty}\bigcap\limits_{N_{0}=1}^{\infty}\mathcal{O}_{l,N_{0}}$
is a dense $G_{\delta}$ set. Clearly, any $G\in\mathcal{O}^{+}$ has the
property that for a.e. $x$, there are infinitely many $n$ with
$E(G|\mathcal{D}_{n})(x)>G(x)$. But now let
$\mathcal{O}^{-}=\\{G:-G\in\mathcal{O}^{+}\\}$. Then this is also a dense
$G_{\delta}$ set and $G\in\mathcal{O}^{-}$ has the property that for a.e. $x$,
there is some $n$ with $E(G|\mathcal{D}_{n})(x)<G(x)$. Hence, any function $G$
in the dense $G_{\delta}$ set $\mathcal{O}=\mathcal{O}^{+}\cap\mathcal{O}^{-}$
will have the oscillation property that we claimed.

So what we want to prove is that each $\mathcal{O}_{l,N_{0}}$ is a dense set
in the $L^{1}$-norm topology. So fix a function $G\in L^{1}[0,1]$. Then, for
any fixed $N_{0}$, we want to construct a function $F\in L^{1}[0,1]$ such that
$\|G-F\|_{1}$ is small and such that the conditional expectations
$E(F|\mathcal{D}_{n})$ oscillate around the limit $F$ on most of $[0,1]$, once
we take a large number $N$ and consider all $n,N_{0}\leq n\leq N$.

Here is what we do. First, for any $\epsilon>0$, we can find $M$ sufficiently
large and $F_{0}$ which is measurable with respect to $\mathcal{D}_{M}$ and
such that $\|G-F_{0}\|_{1}<\epsilon$. Now let $\delta>0$. Then we consider the
Rademacher functions $r_{k}$ with $k\geq 0$. Let
$\Sigma_{N}=\sum\limits_{k\in[0,N]}r_{k}$. We take $F=F_{0}+\delta\Sigma_{N}$.
For $n\geq M$, we have $E(F|\mathcal{D}_{n})=F_{0}+\delta
E(\Sigma_{N}|\mathcal{D}_{n})$. Now fix in addition $N_{0}$ such that $M\leq
N_{0}$ without loss of generality. For $N$ sufficiently large, we will show
that $E(\Sigma_{N}|\mathcal{D}_{n})$ oscillates around its limit $\Sigma_{N}$
on most of $[0,1]$, with $N_{0}\leq n\leq N$. Since $M\leq N_{0}\leq n$, we
have $E(F|\mathcal{D}_{n})=F_{0}+\delta E(\Sigma_{N}|\mathcal{D}_{n})$. So we
have also $E(F|\mathcal{D}_{n})$ oscillating in the same manner around its
limit $F$. Since below we will be fixing $N$, even though it is large, we can
now choose $\delta$ sufficiently small that $\|G-F\|_{1}\leq 2\epsilon$. This
gives us the approximation of $G$ by suitably oscillating conditional
expectations that we need to prove the Baire category theorem - that the
dyadic martingale is oscillating generically.

Let us now look more closely at what the Rademacher functions do for us here.
We have $r_{0}=\mathbbm{1}_{[0,1/2]}-\mathbbm{1}_{[1/2,1]}$, and
$r_{k}(x)=r_{0}(\\{2^{k}x\\})=r_{0}(2^{k}x\mod 1)$ for all $x\in[0,1]$, and
$k\geq 0$. So $r_{k}$ is measurable with respect to $\mathcal{D}_{k+1}$. Also,
for $1\leq n\leq k$, we have $E(r_{k}|\mathcal{D}_{n})=0$. The Rademacher
functions are IID mean-zero functions taking the values of $\pm 1$. So the
functions $n\to\Sigma_{n}=\sum\limits_{k\in[0,n]}r_{k}$ are a model for the
classical symmetric random walk on the integers.

So now consider the conditional expectations $E(\Sigma_{N}|\mathcal{D}_{n})$.
For $n\geq N+1$, we would have all $r_{k},0\leq k\leq N$ measurable with
respect to $\mathcal{D}_{n}$, and hence
$E(\Sigma_{N}|\mathcal{D}_{n})=\Sigma_{N}$. Our goal is to show that for all
$N_{0}$ and $\eta>0$, there is $N\geq N_{0}$ large enough such that for all
$x\in[0,1]$ except for a set of measure smaller than $\eta$, we have

$E(\Sigma_{N}|\mathcal{D}_{n})(x)>\Sigma_{N}(x)\,\text{for
some}\,\,n,N_{0}\leq n\leq N$ (7.1)

and

$E(\Sigma_{N}|\mathcal{D}_{n})(x)<\Sigma_{N}(x)\,\text{for some
(other)}\,n,N_{0}\leq n\leq N.$ (7.2)

By the definition of $r_{k}$ and $\mathcal{D}_{n}$, we actually have
$E(\Sigma_{N}|\mathcal{D}_{n})=\sum\limits_{k\in[0,n-1]}r_{k}$. So we are
asking for oscillation of $\sum\limits_{k\in[0,n-1]}r_{k}$ around
$\sum\limits_{k\in[0,N]}r_{k}$ using values of $n,N_{0}\leq n\leq N$.
Subtracting
$E(\Sigma_{N}|\mathcal{D}_{n})(x)=\sum\limits_{k\in[0,n-1]}r_{k}(x)$ from both
sides of the inequalities in Equation (7.1) and Equation (7.2) we are asking
for oscillation of $\sum\limits_{k\in[n,N]}r_{k}$ around $0$. This is like the
oscillation of the symmetric random walk around $0$. Here what we need to use
is that the terms $r_{k}$ are IID mean-zero functions taking the values of
$\pm 1$, and so for large enough $N$, even though we will be restricting
$n,N_{0}\leq n\leq N$, we have $\sum\limits_{k\in[n,N]}r_{k}$ oscillating
around $0$ on $[0,1]$ except for a set of measure less than $\eta$.

∎

######  Remark 7.1.

There is clearly a index, aka“time reversal”, occurring here. If we were
dealing with a classical random walk, we would be looking at where
$\sum\limits_{k\in[1,n]}r_{k}$ oscillates around $0$ a.e., but because we are
looking at conditional expectations, which are strongly related to Lebesgue
differentiation, we are looking instead at where
$\sum\limits_{k\in[n,N]}r_{k}$ oscillates on most of the measure space as
$N\geq N_{0}$ gets large, if we take $n$ with $N_{0}\leq n\leq N$.

######  Problem 7.2.

  1. (1)

Can we describe completely the functions for which we do not have oscillation
of the dyadic martingales in the limit, on some set of positive measure?

  2. (2)

Is this class actually much different from the class of functions where
$E(G|\mathcal{D}_{n})$ is not monotone i.e. a.e.
$E(G|\mathcal{D}_{n})>E(G|\mathcal{D}_{n+1})$ infinitely often, and a.e.
$E(G|\mathcal{D}_{n})<E(G|\mathcal{D}_{n+1})$?

  3. (3)

When is $E_{n}G$ monotone without being eventually constant?

##  8\. Uniformly distributed sequences

Next, we consider the non-monotonicity and oscillation of averages of
uniformly distributed sequences. We generally can take a sequence
$\mathsf{x}:=(x_{n})$ in $[0,1]$ that is uniformly distributed. That is, for
every continuous $f:[0,1]\to\mathbb{C}$, the averages
$\mathbb{A}_{[N]}f(\mathsf{x})=\frac{1}{N}\sum\limits_{n\in[N]}f(x_{n})\to\int_{0}^{1}f(t)\,dt$
as $N\to\infty$. The classical example closest to ergodic averages is where we
take $\theta\in\mathbb{R}$ irrational and let $x_{n}$ be given by the
fractional parts $\\{n\theta\\}$ for all $n\geq 1$. As usual, if we let
$Tx=\\{\theta+x\\}$ for $x\in[0,1]$, then $T$ is ergodic and
$\mathbb{A}_{[N]}f(T^{n}x)=\frac{1}{N}\sum\limits_{n\in[N]}f(T^{n}x)=\frac{1}{N}\sum\limits_{n\in[N]}f(\\{n\theta+x\\})$.
For this average, we could take $f\in L^{1}([0,1],\lambda)$ where $\lambda$ is
the normalized Lebesgue measure on $[0,1]$. However, when we restrict
$\mathbb{A}_{[N]}f(T^{n}x)$ to $x=0$, then we have to also restrict $f$, for
example by assuming it is continuous, in order that we would have convergence
of the averages
$\mathbb{A}_{[N]}f(\mathsf{x})=\frac{1}{N}\sum\limits_{n\in[N]}f(\\{n\theta\\})$.
For simplicity, let $I=\int_{0}^{1}f(t)\,dt$.

As in ergodic dynamical systems, we have some basic inequalities. Suppose
$\mathbb{A}_{N+1}f(\mathsf{x})\leq\mathbb{A}_{N}f(\mathsf{x})$. Then

$\frac{1}{N+1}f(x_{N+1})\leq\Big{(}\frac{1}{N}-\frac{1}{N+1}\Big{)}\sum\limits_{n\in[N]}f(x_{n})=\frac{1}{N(N+1)}\sum\limits_{n\in[N]}f(x_{n}).$

This means that
$f(x_{N+1})\leq\frac{1}{N}\sum\limits_{n\in[N]}f(x_{n})=\mathbb{A}_{[N]}f$.
Similarly, if
$\mathbb{A}_{[N+1]}f(\mathsf{x})\geq\mathbb{A}_{[N]}f(\mathsf{x})$ then
$f(x_{N+1})\geq\mathbb{A}_{[N]}f(\mathsf{x})$.

These basic inequalities give the following analogue to 3.1

######  Proposition 8.1.

Suppose a continuous $f:[0,1]\to\mathbb{R}$ is not a constant function. Then
for infinitely many $N$ we have
$\mathbb{A}_{[N+1]}f(\mathsf{x})>\mathbb{A}_{[N]}f(\mathsf{x})$, and for
infinitely many $N$ we have
$\mathbb{A}_{[N+1]}f(\mathsf{x})<\mathbb{A}_{[N+1]}f(\mathsf{x})$. Hence, the
averages $\mathbb{A}_{[N]}f(\mathsf{x})$ are not eventually monotonically
increasing or eventually monotonically decreasing.

###### Proof.

Since $f$ is not a constant, there must be some $\delta>0$ such that
$f\geq\delta+I$ on some non-empty open interval $I$, and $f\leq-\delta+I$ on
some non-empty open interval $J$.

Assume $\mathbb{A}_{[N+1]}f(\mathsf{x})\leq\mathbb{A}_{[N]}f(\mathsf{x})$
holds for all large enough values of $N$. This means
$f(x_{N+1})\leq\mathbb{A}_{[N]}f(\mathsf{x})$ for all large enough $N$. But
$\mathbb{A}_{[N]}f(\mathsf{x})\to I$ as $N\to\infty$. Hence, for all large
enough $N$, $\mathbb{A}_{[N]}f(\mathsf{x})\leq\delta/2+I$. However, we have
$x_{N+1}\in I$ for infinitely many $N$. So $\delta+I\leq f(x_{N+1})$ for
infinitely many $N$. If we put these inequalities together, we get values of
$N$ such that $\delta+I\leq f(x_{N+1})\leq\delta/2+I$. Then
$\delta\leq\delta/2$, which is not possible.

Similarly, assume
$\mathbb{A}_{[N+1]}f(\mathsf{x})\geq\mathbb{A}_{[N]}f(\mathsf{x})$ holds for
all large enough values of $N$. This means that
$f(x_{N+1})\geq\mathbb{A}_{[N]}f(\mathsf{x})$ for all large enough $N$. But
$\mathbb{A}_{[N]}f(\mathsf{x})\to I$ as $N\to\infty$. Hence, for all large
enough $N$, $\mathbb{A}_{[N]}f(\mathsf{x})\geq-\delta/2+I$. However, we have
$x_{N+1}\in J$ for infinitely many $N$. So $-\delta+I\geq f(x_{N+1})$ for
infinitely many $N$. If we put these inequalities together, we get values of
$N$ such that $-\delta+I\geq f(x_{N+1})\geq-\delta/2+I$. Then
$-\delta\geq-\delta/2$, which is not possible. ∎

As in 3.2, we can prove this more detailed result.

######  Proposition 8.2.

Suppose $f:[0,1]\to\mathbb{R}$ is not a constant function. Then there exists
$d>0$ such that we have
$\mathbb{A}_{[N+1]}f(\mathsf{x})>\frac{d}{N}+\mathbb{A}_{[N]}f(\mathsf{x})$
infinitely often, and
$\mathbb{A}_{[N+1]}f(\mathsf{x})<-\frac{d}{N}+\mathbb{A}_{[N]}f(\mathsf{x})$
infinitely often.

8.1 answers one possible question about how $\mathbb{A}_{[N]}f(\mathsf{x})$
behaves as $N\to\infty$. But it does not address whether or not there are
infinitely many sign changes for $\mathbb{A}_{[N]}f(\mathsf{x})-I$. In fact,
as we will see, infinitely many sign changes may or may not occur, depending
on the function $f$ and the context of the dynamics. It is not clear how to
characterize when it does occur in general.

We consider the special case of $\mathsf{x}=\\{x_{n}\\}_{n}=\\{n\theta\\}_{n}$
for an irrational $\theta\in\mathbb{R}$. In addition, we take the special case
of a continuous coboundary $f$. This is a continuous function
$f(t)=F(t)-F(\\{t+\theta\\})$ for some continuous $F:[0,1]\to\mathbb{R}$. The
function $F$ here is called the transfer function. Then

$\mathbb{A}_{[N]}f(\mathsf{x})=\frac{1}{N}\sum\limits_{n\in[N]}f(\\{n\theta\\})=\frac{1}{N}\sum\limits_{n\in[N]}\Bigl{(}F(\\{n\theta\\})-F(\\{(n+1)\theta\\})\Bigr{)}=\frac{1}{N}\Bigl{(}F(\\{\theta\\})-F(\\{(N+1)\theta\\})\Bigr{)}$

because the last sum here telescopes. While this converges to $0$ as
$N\to\infty$, whether or not there are changes of sign depend on whether or
not $F(\\{\theta\\})-F(\\{(N+1)\theta\\})$ changes sign infinitely often.

In the same way we can argue for oscillation in the ergodic theory case, with
suitable hypotheses on the transfer function we can argue for oscillation in
the case of averages of uniformly distributed sequences.

######  Proposition 8.3.

Suppose a continuous, real-valued function $f$ is a coboundary with the
transfer function $F$ that is continuous, real-valued, and not a constant.

1) Suppose that $F(\\{\theta\\})$ is the maximum value of $F$. Then the
averages $\mathbb{A}_{[N]}f(\mathsf{x})$ converge to $0$ as $N\to\infty$, but
$\mathbb{A}_{[N]}f(\mathsf{x})\geq 0$ for all $N\geq 1$

2) Suppose that $F(\\{\theta\\})$ is not a maximum value for $F$. Then the
averages $\mathbb{A}_{[N]}f(\mathsf{x})$ converge to $0$ as $N\to\infty$ and
$\mathbb{A}_{[N]}f(\mathsf{x})$ is infinitely often less than $0$.

3) Suppose that $F(\\{\theta\\})$ is the minimum value of $F$. Then the
averages $\mathbb{A}_{[N]}f(\mathsf{x})$ converge to $0$ as $N\to\infty$, but
$\mathbb{A}_{[N]}f(\mathsf{x})\leq 0$ for all $N\geq 1$

4) Suppose that $F(\\{\theta\\})$ is not a minimum value for $F$. Then the
averages $\mathbb{A}_{[N]}f(\mathsf{x})$ converge to $0$ as $N\to\infty$ and
$\mathbb{A}_{[N]}f(\mathsf{x})$ is infinitely often more than $0$.

5) Suppose that $F(\\{\theta\\})$ is not a maximum value nor a minimum value
for $F$. Then the averages $\mathbb{A}_{[N]}f(\mathsf{x})$ converge to $0$ as
$N\to\infty$ and $\mathbb{A}_{[N]}f(\mathsf{x})$ is both infinitely often less
than $0$, and it is infinitely often more than $0$.

######  Remark 8.4.

We have taken advantage of $f$ being a coboundary to get a simpler expression
for $\mathbb{A}_{[N]}f(\mathsf{x})$. This has allowed us to obtain 8.3. But
then the following question is left unresolved.

Suppose $f$ is continuous and $I=0$. Under what necessary and sufficient
conditions do we have $\mathbb{A}_{[N]}f(\mathsf{x})$ infinitely often greater
than $0$, and infinitely often less than $0$? In the light of 3.5 , we
conjecture the following:

######  Conjecture 8.5.

If for a continuous function $f$, $\mathbb{A}_{[N]}f(\mathsf{x})$ fails to
oscillate in a set of positive measure, then $f$ can be expressed as
$f(t)=F(t)-F(\\{t+\theta\\})$ with $F$ satisfying either property (1) or (3)
from 8.3.

On the other hand, as in 3.8, we can easily prove this result.

######  Proposition 8.6.

For a given irrational $\theta$, there is a dense $G_{\delta}$ set
$\mathcal{O}$ in the real-valued functions in $C[0,1]$, with the uniform norm,
such that for all $f\in\mathcal{O}$, we have
$\mathbb{A}_{[N]}f(\mathsf{x})>\int_{0}^{1}f(t)\,dt$ infinitely often, and we
have $\mathbb{A}_{[N]}f(\mathsf{x})<\int_{0}^{1}f(t)\,dt$

######  Remark 8.7.

As remarked before, we only need to get one of these inequalities to hold
generically, and then we can argue that they both do, at least generically of
course. ∎

##  9\. General weighted averages in ergodic theory

Here we take a dissipative sequence $\mu_{N}$ of probability measures on
$\mathbb{Z}$. Our operators
$\mu_{N}^{T}f=\sum\limits_{n\in\mathbb{Z}}\mu_{N}(n)f\circ T^{n}$. Now we ask
when these always oscillate. Because we are not assuming there is pointwise
a.e. convergence, or even perhaps $L^{1}$-norm convergence, there is no clear
limit about which we could consider oscillation. However, in the case that
there is norm convergence, oscillation around the limit will be worthwhile to
consider.

######  Conjecture 9.1.

For a dissipative sequence $(\mu_{N})$, the generic function $f\in L^{1}$ has
for a.e. $x$, the averages $\mu_{N}^{T}f(x)$ oscillation infinitely often.

The oscillation behavior of the classical averages clearly depended on the
fact that these are Cesàro averages. It may be worthwhile to ask when similar
behavior holds for other methods of averaging like the ones above.

######  Example 9.2.

Consider an ergodic map $T$ that has $-1$ as an eigenvalue. Let $f\in L^{1}$
be mean-zero eigenvalue for $-1$, so that $f+f\circ T=0$. Hence,
$\mathbb{A}^{\frac{1}{2}}f(Tx):=\frac{1}{2}\Big{(}f(Tx)+f(T^{2}x\Big{)}=0$.
But this does not remain so, indeed for even $N$,
$\mathbb{A}^{\frac{1}{2}}_{[N]}f(T^{n}x)=0$ but for odd N
$\mathbb{A}^{\frac{1}{2}}_{[N]}f(T^{n}x)=-\frac{1}{N}f$. So the averages
converge to $0$, but not monotone, and they do not switch sign infinitely
often.

However, if we use a different averaging method we can get a very different
result. Take
$\mathbb{A}^{b}_{[N]}f=\frac{1}{2^{N}}\sum\limits_{n\in[N]}{N\choose n}f\circ
T^{n}$ for all $N\geq 1$. These averages do converge in $L^{1}$-norm to $\int
f\,d\lambda$ for all $f\in L^{1}$, although they do not converge a.e. in
general. See [19]. And we certainly can have monotonicity since we have
$\mathbb{A}^{b}_{[N]}f=0$ for all $N$ when $f\circ T=-f$ as above. Of course,
we do not have oscillation of the sign. ∎

This example makes it clear that whether, and when, we have monotonicity of
the averages depends on the method of averaging that we use. Moreover, for
which $f$ and $T$ we have oscillation of the values about the mean will also
depend on the averaging method used.

On the other hand, we can get some similar results to what happens using the
classical ergodic averages, at least when using more general Cesàro averages.
This is in line with what was done in Section 8, where general uniformly
distributed sequences were used instead of just the classical sequence
$(\\{n\theta\\}:n\geq 1)$ with $\theta\in\mathbb{R}$ irrational.

######  Proposition 9.3.

Consider an increasing sequence $(m_{n}:n\geq 1)$ in $\mathbb{Z}^{+}$. Assume
that for all the $f\in L^{2}$, averages
$\mathbb{A}_{[N]}f(T^{m_{n}}x):=\frac{1}{N}\sum\limits_{n\in[N]}f(T^{m_{n}}x)$
converge to $\int f\,d\lambda$ a.e. Then for a non-constant $f\in L^{2}$, for
a.e. $x$, the averages
$\mathbb{A}_{[N+1]}f(T^{m_{n}}x)>\mathbb{A}_{[N]}f(T^{m_{n}}x)$ infinitely
often, and the averages
$\mathbb{A}_{[N+1]}f(T^{m_{n}}x)<\mathbb{A}_{[N]}f(T^{m_{n}}x)$ infinitely
often.

## Acknowledgement

The authors would like to thank Terence Adams for his valuable input
throughout the paper.

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