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 \date{\dateline{Sep 12, 2013}{Jun 21, 2014}{Jul 3, 2014}\\
 \small Mathematics Subject Classifications: 05D10, 37A15}


\title{\bf A Characteristic Factor for the\\ 3-Term IP Roth Theorem in
  $\Z^\N_3$}

\author{Randall McCutcheon \qquad Alistair Windsor\\
\small Department of Mathematical Sciences\\[-0.8ex]
\small The University of Memphis\\[-0.8ex] 
\small Tennessee, U.S.A.\\
\small\tt rmcctchn@memphis.edu \qquad awindsor@memphis.edu\\
}


\begin{document}


\maketitle

\begin{abstract}
  Let $\Omega = \bigoplus_{i=1}^\infty \Z_3$ and $e_i = (0, \dots, 0 ,
  1, 0, \dots)$ where the $1$ occurs in the $i$-th coordinate. Let $\F
  =\{ \alpha \subset \N : \varnothing \neq \alpha, \alpha \text{ is
    finite} \}$. There is a natural inclusion of $\F$ into $\Omega$
  where $\alpha \in \F$ is mapped to $e_\alpha = \sum_{i \in \alpha}
  e_i$. We give a new proof that if $E \subset \Omega$ with $d^*(E)
  >0$ then there exist $\omega \in \Omega$ and $\alpha \in \F$ such
  that
  \begin{align*}
        \{ \omega, \omega+ e_\alpha, \omega + 2 e_\alpha \} \subset E.
  \end{align*}
  Our proof establishes that for the ergodic reformulation of the
  problem there is a characteristic factor that is a one step compact
  extension of the Kronecker factor.
\end{abstract}

\section{Introduction}

Let $\Omega = \bigoplus_{i=1}^\infty \Z_3$. $\Omega$ is an abelian
group and hence amenable. Let $e_i = (0, \dots, 0 , 1, 0, \dots)$
where the $1$ occurs in the $i$-th coordinate. For a set $S$ let
$$\F(S) = \{ \gamma \subset S : \gamma \text{ is non-empty and finite}
\}.$$ We will denote $\F(\N)$ by simply $\F$ and endow it with the
discrete topology.  There is a natural inclusion of $\F$ into $\Omega$
where $\alpha \in \F$ is mapped to $e_\alpha = \sum_{i \in \alpha}
e_i$.

The upper Banach density of a set $E \subset \Omega$, denoted
$d^*(E)$, is defined as
\begin{align*}
  d^*(E) = \sup_{ (\Phi_n) \text{ F\o lner}} \limsup_{n \rightarrow
    \infty} \frac{|E \cap \Phi_n|}{|\Phi_n|}
\end{align*}
where the supremum is taken over the set of F\o lner sequences,
i.e. over the set of sequences of finite sets $(\Phi_n)_{n=1}^\infty$
in $\Omega$ such that for all $\omega \in \Omega$
\begin{equation*}
  \lim_{n\rightarrow \infty} \frac{| (\omega+\Phi_n) \triangle \Phi_n |}{|\Phi_n|}=0.
\end{equation*}

We give a new proof of the following theorem:
\begin{thm}\label{thm:Main}
  Let $E \subset \Omega$ with $d^*(E) >0$. There exists $\omega \in
  \Omega$ and $\alpha \in \F$ such that
  \begin{equation}
    \label{eq:1}
    \{ \omega, \omega+ e_\alpha, \omega + 2 e_\alpha \} \subset E.
  \end{equation}
\end{thm}


One can derive Theorem \ref{thm:Main} from Furstenberg's correspondence
principle and the following recurrence theorem.

\begin{thm}\label{thm:Recurrence} 
  Let $(T_\omega)_{\omega \in \Omega}$ be a measure-preserving action
  of $\Omega$ on a probability space $(X, \A, \mu)$. If $A \in \A$
  with $\mu(A)>0$ then there exists $\alpha \in \F$ such that
  \begin{align*}
    \mu(A \cap T_{e_\alpha} A \cap T_{2 e_\alpha} A ) >0. 
  \end{align*}
\end{thm}

Theorem \ref{thm:Recurrence} is not new; it follows from the
Furstenberg-Katznelson IP-Szemer\'edi Theorem \cite{FK85}. However,
our proof identifies a characteristic factor that is a 1-step compact
extension of the Kronecker factor of $T_\omega$. Identifying a
characteristic factor is suggestive of a first step in obtaining
a decent quantitative result. 

\section{Ultrafilter Preliminaries}

We will be working with the Stone-\v{C}ech compactification of $\F$,
$\beta \F$. Since $\F$ is discrete we may identify points of $\beta\F$
with ultrafilters on $\F$.  An ultrafilter $\p$ on $\F$ is a subset $\p
\subset \mathscr{P}(\F)$ that satisfies the following axioms
\begin{enumerate}
\item $\emptyset \not\in \p$,
\item If $A \subset B$ and $A \in \p$ then $B \in \p$.
\item If $A, B \in \p$ then $A \cap B \in \p$.  
\item if $A \subset \F$ then either $A \in \p$ or $A^c \in \p$. 
\end{enumerate}
We identify $\alpha \in \F$ with the
principal ultrafilter $\p_\alpha = \{ A \subset \F : \alpha \in A \}$.
We can endow $\beta \F$ with the Stone topology, that is for $A \subset \F$,
we define $\overline{A} = \{ \p \in \beta \F : A \in \p \}$, and the
set $\{\overline{A} : A \subset \F \}$ is a basis for the closed sets
of $\beta\F$. Indeed, from the ultrafilter property $\overline{A}^c =
\overline{A^c}$ and so this is also a basis for the open sets.


For $ \alpha , \beta \in \F $ we write $ \alpha< \beta $ if $\max
\alpha < \min \beta$. When $\alpha < \beta$ we define $\alpha \ast
\beta = \alpha \cup \beta$ and we leave $\alpha \ast \beta$ undefined
otherwise. This makes $( \F, \ast)$ into  an adequate partial semigroup in
the sense of \cite{BBH94} (see also \cite{HM01}]). Briefly this means that $\ast$
maps a subset of $\F \times \F$ to $\F$, is associative for all triples where
defined, and for any $\alpha_1, \dots, \alpha_n \in \F$ there exists
$\beta \in \F$ such that $\alpha_i \ast \beta$ is defined for all $1
\leq i \leq n$. Notice that if $\alpha < \beta$ then 
\begin{equation*}
  e_{\alpha \ast \beta} = e_{\alpha} + e_{\beta}.
\end{equation*}



In the case of a semi-group $\F$ the operation extends to an
operation on $\beta \F$ that makes $\beta \F$ a semi-group. In our
case however $\F$ is only a partial semi-group and $\ast$ does not
extend to all of $\beta \F\times \beta\F$. We extend the operation $\ast$ to a
partial semi-group operation on $\beta \F$ using the same definition
as for semi-groups
\begin{align*}
  A \in \p \ast \q  \iff \{ p: \{q: p \ast q \in A\} \in \q \} \in \p
\end{align*}
where $ p \ast q \in A $ means both that $p \ast q$ is defined and
$p \ast q \in A$. This extends the existing operation $\ast$ in the
following sense: if $\alpha, \beta \in \F$ then $\p_\alpha \ast
\p_\beta$ is defined if $\alpha \ast \beta$ is defined,  and in this
case $\p_\alpha
\ast \p_\beta = \p_{\alpha \ast \beta}$. If we let $\F_n
=\F\bigl(\{n+1, n+2, \dots\}\bigr)$ then we can define $\delta \F =
\cap_{n=1}^\infty \overline{\F_n} \subset \beta \F$. If we wish $\p
\ast \q$ to be defined for all $\p \in \beta \F$ then it turns out
that we must have $\q \in \delta \F$. 

Given an ultrafilter $\p$ on $\F$ we say that a sequence
$(x_\alpha)_{\alpha \in \F}$ in a Banach space $X$ $\p$-converges to
$L$ in norm if for every $\epsilon >0$ we have $\{\alpha \in \F :
\|x_\alpha - L\|< \epsilon \} \in \p$. Since the limit, if it exists,
is unique we write $\plim x_\alpha = L$.  It can be shown that every
pre-compact sequence $\p$-converges. Given an ultrafilter $\p$ on $\F$
we say that a sequence $(x_\alpha)_{\alpha \in \F}$ in a Hilbert space
$H$ $\p$-converges to $L$ weakly if for every $x \in H$ we have $\plim
\langle x_\alpha, x\rangle = \langle L, x \rangle$.

It may be shown that $( \delta\F, \ast)$ is a compact Hausdorff right
topological semigroup. An idempotent ultrafilter $\p$ is one that
satisfies $\p \ast \p = \p$. For $E \subset \F$ write 
\begin{align*}
  \overline{d}(E) = \limsup_{n\rightarrow \infty} \frac{|E \cap
    \F(\{1, \dots, n\}) |}{2^n} .
\end{align*}
An ultrafilter $\p \in \delta \F$ is called essential if for every $A
\in \p$ we have $\bar{d}(A \cap \F_n) >0$. By \cite[Proposition
2.1]{BM12} there exists an essential idempotent ultrafilter in $\delta
\F$.

\section{Factors and Joinings}

Crucial to our proof will be the following theorem. It is a
combination of \cite[Theorem 3.3]{BM12} and \cite[Theorem
4.3]{BM12}. In \cite {BM12} Theorem 4.3 is derived from
\cite[Corollary 1]{BKMP88}; in our case, where we need only consider
$\bigoplus_{n=1}^\infty \Z_3$, one may derive the appropriate version
of \cite[Theorem 4.3]{BM12} from \cite[Proposition 2.7] {BM83}, which
is a direct consequence of a result of Woodall in
\cite{w77}. This theorem is the only place where the fact that $\p$ is
an essential ultrafilter is used. 
%The arguments in this paper will use only
%that the ultrafilter $\p$ is idempotent and $\p \in \delta \F$ %except
%insofar as they refer to Theorem \ref{thm:Projection}.

\begin{thm}\label{thm:Projection}\cite[Corollary 4.6]{Bergelson}
  Let $(T_\omega)_{\omega \in \Omega}$ be a measure-preserving action
  of $\Omega$ on a probability space $(X, \A, \mu)$.  The action
  $T$ extends to a unitary action on $L^2(X, \A,\mu)$. Let $\p$ be an
  essential idempotent ultrafilter in $\delta\F$. Define an operator
  $P$ on $L^2(X,\A,\mu)$ by
\begin{align*}
  \plim_{\alpha \in \F} T_{e_\alpha} f = Pf \text{ weakly}
\end{align*}
for $f \in L^2(X,\A,\mu)$. The operator $P$ is the orthogonal projection onto
the Kronecker factor
\begin{align*}
  K = \bigl\{f \in L^2(X,\A,\mu) : \{T_ \omega f: \omega \in \Omega\}
  \text{ is norm precompact} \bigr\}.
\end{align*}
\end{thm}

Now we observe that $K = L^2(X, \K, \mu)$ for some
sub-$\sigma$-algebra $\K \subset \A$ with $\K$ being $T_\omega$
invariant (see \cite[Lemma 3.1]{FK91} or \cite[Theorem 2.7]{McC00}).
Let $(\mu_x)_{x \in X}$ be the disintegration of $\mu$ over the
Kronecker algebra $\K$, so that
\begin{align*}
  \int f \, d \mu_x = E\bigl(f|\K \bigr)(x) = P f (x) \text{  a.e.,}
\end{align*}
where $E\bigl(f|\K \bigr)$ is the conditional expectation of $f$ over
$\K$ and $P$ is the orthogonal projection onto $K$. To see that these
are equal we observe that if $f \in L^2(X, \A,\mu)$ with
$E\bigl(f|\K\bigr) =0$ then $\int f \, g \, d \mu = 0$ for all $g \in
K$. 

Using the measures $\mu_x$ on $X$ we may define a family of norms
on $X$ indexed by $x \in X$ as follows
\begin{align*}
  \|f\|_x = \bigl(\int |f|^2 d\mu_x\bigr)^\frac{1}{2}.
\end{align*}
We will use $\|\cdot\|$ without any subscript to denote the
appropriate $L^2$ norm. 

\begin{defn}
  A function $f \in L^2(X,\A,\mu)$ is called \emph{almost periodic
    over $K$}, or AP over $K$ for short, if for all $\epsilon >0$
  there exist $g_1, \dots, g_k \in L^2(X,\A,\mu)$ such that for
  almost every $x \in X$ and every $\omega \in \Omega$ there exists $1
  \leq i=i(x,\omega) \leq k$ such that
  \begin{equation*}
    \| T_\omega f - g_i \|_x \leq \epsilon. 
  \end{equation*}
\end{defn}

One may show that the bounded AP over $K$ functions are dense in the
AP over $K$ functions, the constant functions are AP over $K$, sums and
products of bounded AP over $K$ functions are again bounded AP over
$K$ functions. Therefore, the closure of the set of AP over $K$
functions is of the form $L^2(X, \B, \mu)$ for some $\sigma$-algebra
$\B \subset \A$ with $\B$ being $T_\omega$ invariant (see \cite[Lemma
3.1]{FK91} or \cite[Theorem 2.7]{McC00}). One may also observe that
any bounded $f \in K$ is AP over $K$ and hence any $f \in K$ is
$\B$-measurable, so that $\K \subset \B$.  We observe that if $E( f |
\B)=0$ then $E(f | \K) = P f =0$.

Consider the relative product measure $\tilde{\mu}$ on $X \times X$ defined by 
\begin{align*}
  \tilde{\mu}(A) = \int (\mu_x \times \mu_x )(A) \,   d\mu(x)
\end{align*}
so that 
\begin{align*}
  \int f(x) g(y) \, d \tilde{\mu}(x,y) = \int P(f)(x) P(g)(x) \, d \mu(x).
\end{align*}
This measure is invariant under $\tilde{T}_\omega = T_\omega \times
T_\omega$ for every $\omega \in \Omega$. If $\p$ is an essential
idempotent ultrafilter in $\delta\F$ then by Theorem
\ref{thm:Projection}
\begin{align*}
  \plim_\alpha \tilde{T}_{e_\alpha} h = \plim_\alpha \tilde{T}_{2
    e_\alpha} h = Q h,
\end{align*}
where $Q$ is the orthogonal projection onto the Kronecker factor of
$\tilde{T}$.  

% $\{f \in L^2(X\times X, \tilde\mu) : \{\tilde{T}_ \omega
% f: \omega \in \Omega\} \text{ is precompact} \bigr\}$

Given $H \in L^2(X \times X, \A \otimes \A, \tilde\mu)$, where $\A
\otimes \A$ denotes the (completion of the) sigma algebra generated by
the rectangles $\{ A_1 \times A_2: A_1, A_2 \in \A \}$, and $\phi \in
L^\infty(X, \A,\mu)$ we may define
\begin{equation*}
  H \star \phi (x) = \int H(x,s) \phi(s) \, d \mu_x(s).
\end{equation*}

\begin{lem}\label{lem:One}
  If $Q H = H$ then for all $\phi \in L^\infty(X,\A,\mu)$ $H \star \phi$ is
  $\B$-measurable.
\end{lem}

\begin{proof}
  Since $Q$ is the projection onto the Kronecker factor of $(X\times
  X, \tilde{T}_\omega, \tilde{\mu} )$ we have that $H$ is expressible
  as a countable sum of eigenfunctions for $\tilde{T}_\omega$. Without
  loss of generality we may assume that $H$ is an eigenfunction, i.e.\
  that
  \begin{align*}
    \bigl( \tilde{T}_\omega H\bigr) (t,s) &= H\bigl(T_\omega t,
    T_\omega s) = \lambda(\omega)\, H(t,s)
  \end{align*}
  where $\lambda : \Omega \rightarrow S^1$ is a character. Notice that 
  \begin{equation}\label{eq:TomIntertwine}
  \begin{aligned}
    \bigl( T_\omega (H \star \phi) \bigr) (x) &= (H \star \phi)
    (T_\omega x)\\
    & =\int H(T_\omega x, s) \phi(s) \, d \mu_{T_\omega x} (s)\\
    & =\int H(T_\omega x, T_\omega s) \phi(T_\omega s) \, d \mu_{x} (s)\\
    &= \lambda(\omega) \int H(x,s) T_\omega \phi(s) \, d \mu_x (s)\\
    & = \lambda(\omega)\, \bigl( H \star T_\omega \phi\bigr) (x).
  \end{aligned}
\end{equation}
Now for almost every $y \in X$ we have $\phi \mapsto H \star \phi$ is
a compact operator on $L^2(X, \A, \mu_y)$ and
$\mathrm{range}(\lambda)$ is finite so \eqref{eq:TomIntertwine} shows
that $\{ T_\omega (H \star \phi) : \omega \in \Omega\}$ is precompact
in $L^2(X, \A, \mu_y)$. Hence for almost every $y \in X$ and every
$\epsilon >0$ there exists $M(y,\epsilon)$ such that $\big\{ T_\omega
(H \star \phi) : \omega \in \mathrm{span} \{ e_1, \dots,
e_{M(y,\epsilon)}\} \big\}$ is $\epsilon$-dense in $\{ T_\omega (H
\star \phi) : \omega \in \Omega\}$ in $L^2(X, \A, \mu_y)$. Let
$\epsilon >0$ be arbitrary. Choose $M_n$ sufficiently large that $M_n
> M(y,\frac{1}{n})$ except for $y \in E_n$ where $E_n \in \K$ and
$\mu(E_n) \leq \epsilon 2^{-n}$. Define 
\begin{equation*}
   f_\epsilon(x) =
  \begin{cases}
    0 & \text{if $x \in \bigcup_{n=1}^\infty E_n$,}\\
      H\star \phi(x)& \text{otherwise.}
  \end{cases}
\end{equation*}
It can be easily shown that $\| f_\epsilon - H \star \phi\| < \|H\|_\infty \, 
\|\phi\|_\infty \, \epsilon$. Thus it suffices to show that
$f_\epsilon$ is almost periodic over $K$. From the construction it is
easy to observe that $\{0\} \cup \bigl\{T_\omega (H \star \phi) : \omega
\in \mathrm{span } \{e_1, \dots, e_{M_n} \} \bigr\}$ is
$\frac{1}{n}$-dense in $\{T_\omega f_\epsilon : \omega \in
\Omega\}$ in $L^2(\mu_y)$ for almost every $y\in X$. Thus $f_\epsilon$
is almost periodic over $K$ as required. Since $H \star \phi$ lies in
the closure of the functions that are almost periodic over $K$,
$H\star \phi$ is $\B$-measurable.
\end{proof}

\section{Projection Results}

We now give applications of our results on joinings to the projections
of products. 

\begin{lem}\label{lem:FirstVanishing}
  Let $\p$ be an essential idempotent in $\delta\F$ and let $f \in
  L^\infty(X,\A,\mu)$.  If $E(f|\B)=0$ then 
  \begin{align*}
    \plim_\alpha \|P(f \, T_{e_\alpha} f) \| = \plim_\alpha
    \|P(f \, T_{2e_\alpha} f) \| = 0. 
  \end{align*}
\end{lem}

\begin{proof}
  Let$f,g \in L^\infty(X,\A,\mu)$. One has
  \begin{equation}
  \begin{aligned}
    \plim_{\alpha} \| P &\bigl( g (T_{e_\alpha} f) \bigr) \|^2\\
    &= \plim_\alpha \int \Bigl( P \bigl( g (T_{e_\alpha} f) \bigr)(x)
    \Bigr)^2 \, d
    \mu(x)\\
    & = \plim_\alpha \int \Bigl( \int g(t) \, (T_{e_\alpha} f) (t) \, d
    \mu_x(t) \Bigr)^2 \, d
    \mu(x)\\
    & =
    \begin{aligned}[t]
      \plim_\alpha \int &\Bigl( \int g(t) \, (T_{e_\alpha} f) (t) \, d
      \mu_x(t) \Bigr) \\
      &\qquad\Bigl( \int g(s)\, (T_{e_\alpha} f)
      (s) \, d \mu_x(s) \Bigr) \, d \mu(x)
    \end{aligned}\\
    &=\plim_\alpha \int (g \otimes g ) \, \tilde{T}_{e_\alpha} (f
    \otimes f )\, d \tilde{\mu}\\
    &= \int (g \otimes g ) \, Q  (f
    \otimes f)\, d \tilde{\mu},
  \end{aligned}\label{eq:2} 
\end{equation}
and the same equality holds for $ \plim_{\alpha} \| P \bigl( f (T_{2
  e_\alpha} f) \bigr) \|^2$. Clearly $Q(f \otimes f) (t,s)= Q(f
\otimes f) (s,t)$ and $Q(f \otimes f)$ is (essentially) bounded. We
thus have that $ Q (f \otimes f )$ is a positive definite
symmetric kernel in the sense of \cite[Section 3.6]{FK91}. By
\cite[equation (3.6)]{FK91} $Q (f \otimes f) (t,s) = \sum_{k}
\lambda_k(t) \phi_k(t) \phi_k(s)$ where $\lambda_k$ is
$\K$-measurable (so that $\lambda_k(s)= \lambda_k (t)$ for
$\tilde{\mu}$-almost every $(t,s)$), and for almost every $y \in X$
$\{\phi_k\}$ is orthonormal in $L^2(\mu_y)$. One may then check that
$Q (f \otimes f) \star \phi_k = \lambda_k \phi_k$, so by Lemma
\ref{lem:One} $\lambda_k \phi_k$ is $\B$-measurable.

Applying \eqref{eq:2} with $g=f$, we get 
  \begin{align*}
    \plim_{\alpha} \| P\bigl( f \, (T_{e_\alpha} f) \bigr) \|^2&=\int (f
    \otimes f) \, Q (f \otimes f) \, d\tilde{\mu} \\
    &=
    \int f(t) \, f(s)\bigl( \sum _k \lambda_k(t) \, \phi_k(t) \,
    \phi_k(s) \bigr) \,  d \tilde\mu
    \\
    &= \sum_k \int f(t) \, f(s) \, \lambda_k(t) \,\phi_k(t)\,
    \phi_k(s) \, d \tilde\mu\\
    &= \sum_k \int P(f  \, \lambda_k \, \phi_k) P(f \, \phi_k)\, 
    d\mu .
  \end{align*}
  However, since $\lambda_k \phi_k $ is 
  $\B$-measurable we have 
  \begin{equation*}
    E(f \, \lambda_k \, \phi_k |
    \B) =\lambda_k \, \phi_k \, E(f | \B) =0
  \end{equation*}
  and consequently 
  $P(f \, \lambda_k \, \phi_k)=0 $.
\end{proof}

\begin{lem}\label{lem:Three}
  Let $\p$ be an essential idempotent on $\F$. If $f,g \in
  L^\infty(X,\A, \mu)$ with either $E(f|\B) =0$ or $E(g|\B)=0$
  then $\plim_\alpha T_{e_\alpha} f \, T_{2 e_\alpha} g = 0$ weakly. 
\end{lem}

As is common in proofs of this type, a version of the Van der Corput
lemma is crucial. This version appears as \cite[Theorem 3.5]{BM12}. 

\begin{lem}[Van der Corput Lemma]
  Let $(x_\alpha)_{\alpha \in \F}$ be a bounded $\F$-sequence of
  vectors in a Hilbert space and let $\p \in \delta\F$ be an
  idempotent. If $\plim_\alpha \plim_\beta \langle x_{\alpha \ast
    \beta}, x_\beta \rangle =0$ then $\plim_\alpha x_\alpha = 0$
  weakly. 
\end{lem}

We now use this Van der Corput Lemma to prove Lemma \ref{lem:Three}.

\begin{proof}[Proof of Lemma \ref{lem:Three}] Let $x_\alpha =
  T_{e_\alpha} f T_{2 e_\alpha} g$. Then 
  \begin{align*}
    \plim_\beta \plim_\alpha &\langle x_{\alpha \ast
      \beta}, x_\alpha \rangle \\
    &= \plim_\beta \plim_\alpha\int T_{e_\alpha} T_{e_\beta} f \, T_{2 e_\alpha} T_{2 e_\beta} g \,
    T_{e_\alpha}f \, T_{2 e_\alpha} g \, d \mu\\
    &=\plim_\beta \plim_\alpha\int T_{e_\beta} f \, T_{ e_\alpha} T_{2 e_\beta} g \,
    f \, T_{e_\alpha} g \, d \mu\\
    &=\plim_\beta \plim_\alpha
    \int \bigl(f\, T_{e_\beta} f \bigr)\, T_{ e_\alpha}
    \bigl( g\, T_{2 e_\beta} g \bigr) \, d \mu\\
    &= \plim_\beta \int P\bigl(f\, T_{e_\beta} f \bigr)\,
    P\bigl( g\, T_{2 e_\beta} g \bigr) \, d \mu\\
    &\leq \plim_\beta \|P(f T_{e_\beta} f) \| \, \|P(g T_{2 e_\beta} g)\| =0
  \end{align*}
  since either $ \plim_\beta\|P(f T_{e_\beta} f) \|=0$ or $
  \plim_\beta \|P(g T_{2 e_\beta}
  g)\| =0$ by Lemma \ref{lem:FirstVanishing}. Hence by the conclusion of the Van
  der Corput lemma we obtain that 
  \begin{equation*}
    \plim_\alpha T_{e_\alpha} f \, T_{2  e_\alpha} g = 0 \text{ weakly,}
\end{equation*}

as required. 
\end{proof}




% \begin{lem}
%   Suppose that $Q H = H$ and $\phi \in L^\infty(X, \B, \mu)$. Then
%   $H \star \phi$ is AP over $K$, i.e. for all $\epsilon > 0$ there
%   exists $D \in \B$, with $\mu(D) < \epsilon$, and $g_1, \dots, g_k
%   \in L^2(X, \B,\mu)$ such that for all $ y \in X \setminus D$ and for
%   all $\omega \in \Omega$
%   \begin{align*}
%     \| T_\omega H \star \phi - g_i \|_y < \epsilon
%   \end{align*}
% \end{lem}

% \begin{proof}
%   Let $\epsilon > 0$ be arbitrary. Since $\Omega$ is countable we may
%   fix an ordering $\Omega = \{ w_1, w_2, w_3, \dots \}$. Since $Q H
%   =H$ we have that $H$ is in the Kronecker factor of $\tilde T$ and
%   consequently the closure of the $\tilde T_\omega$ orbit of $H$ is
%   compact. Thus there exists $H_1, \dots, H_k\in L^2(X \times X, \A
%   \times \A , \tilde \mu)$ that form an
%   $\sqrt{\frac{\epsilon}{2}}$-dense net in $\{ \tilde T_\omega H :
%   \omega \in \Omega \}$ in the norm topology on $ L^2(X \times X, \A
%   \times \A , \tilde \mu)$. Each $H_j$ is a compact operator on
%   $L^2(\mu_y)$ almost everywhere so there exists $M=M(y,j)$ such that
%   $\{ H_j \star T_{\omega_i} \phi : 1 \leq i \leq M \}$ is
%   $\frac{\epsilon}{2}$-dense in $\{ H_j \star T_\omega \phi : \omega
%   \in \Omega\}$. Fix $M$ such that $M(y,j) < M$ off a set of $D \in
%   \B$, $\mu(D)<\epsilon$. Let $g_1, \dots, g_k$ be an enumeration of
%   $\{ H_j \star T_{\omega_i} \phi : 1 \leq j \leq k_1, 1\leq i \leq M
%   \}$. Now for $\psi \in X \setminus D$, $\omega \in \Omega$, $1 \leq
%   j \leq k$ one has
%   \begin{align*}
%     \| T_\omega H \star \phi - H_j \star T_\omega \phi \|_2 &= \int \bigl| H
%     \star \phi(T_\omega x) - H_j \star T_\omega \phi (x) \bigr|^2 \, d \mu(x)\\
%   &=
%   \begin{aligned}[t]
%     \int \Bigl| &\int H(T_\omega x, x') \phi(x') \, d \mu_{T_\omega x}(x')\\
%     &- \int H_j(x, x') \phi(T_\omega x') \, d \mu_x (x') \Bigr|^2 \, d \mu(x)
%   \end{aligned}\\
%   &=
%   \begin{aligned}[t]
%     \int \Bigl| &\int \bigl(  H(T_\omega x, T_\omega x') \phi(T_w x')\\
%     &- H_j(x,x') \phi(T_\omega x') \bigr) d\mu_x(x') \Bigr|^2 d\mu(x) 
%   \end{aligned}\\
%   &\leq \| \tilde T_\omega H - H_j \|^2_{L^2(\tilde \mu)} \| \phi
%   \|_{\infty}^2.
%   \end{align*}
%   For some $j$ we have $\| \tilde T_\omega H - H_j \|_{L^2(\tilde
%     \mu)}< \frac{\epsilon}{2}$. Fix this $j$ and pick $i =
%   i(\omega,y)$ with $\| H_j \star T_\omega \phi - g_i \|_{L^2(\mu_y)} <
%   \frac{\epsilon}{2}$ then $\| T_\omega H \star \phi - g_i \|_y <
%   \epsilon$. 
%  \end{proof}

\section{Proof of Theorem \ref{thm:Recurrence}}

Our goal is to show that for $A \in \A$ with $\mu(A)>0$, 
\begin{equation*}
\plim_{\alpha} \mu(A \cap T_{e_\alpha} A \cap T_{2 e_\alpha} A) >0. 
\end{equation*}
It is equivalent that for  $A \in \A$ with $\mu(A)>0$ the
characteristic function $f = \mathbbm{1}_A$ satistfies
\begin{equation*}
  \plim_{\alpha} \int f \; T_{e_\alpha} f \; T_{2 e_\alpha} f \; d \mu >
  0.  
\end{equation*}
To show this we will decompose $f$ into 
\begin{align*}
  f_1 &= E(f | \B),\\
  f_2 &= f - E(f|\B).
\end{align*}
Clearly $E(f_2 | \B)=0$. Expanding, we get
\begin{align*}
  \int f \; &T_{e_\alpha} f \; T_{2 e_\alpha} f \; d \mu\\
 &= \int f \;
  T_{e_\alpha}\bigl(  f_1 + f_2 \bigr) \; T_{2 e_\alpha} \bigl(  f_1 +
  f_2 \bigr) \; d \mu,\\
 &=
 \begin{aligned}[t]
   \int f \; &T_{e_\alpha} f_1 \; T_{2 e_\alpha} f_1  \; d
   \mu+\int f \; T_{e_\alpha} f_1 \; T_{2 e_\alpha} f_2  \; d
   \mu\\
   &+\int f \; T_{e_\alpha} f_2 \; T_{2 e_\alpha} f_1  \; d
   \mu+\int f \; T_{e_\alpha} f_2 \; T_{2 e_\alpha} f_2  \; d \mu.
 \end{aligned}
\end{align*}
From Lemma \ref{lem:Three} we have 
\begin{align*}
 \plim_{\alpha}\int f \; T_{e_\alpha} f_1 \; T_{2 e_\alpha} f_2  \; d
   \mu &= 0,\\
   \plim_{\alpha}\int f \; T_{e_\alpha} f_2 \; T_{2 e_\alpha} f_1  \; d
   \mu&=0, \hbox{ and}\\
    \plim_{\alpha}\int f \; T_{e_\alpha} f_2 \; T_{2 e_\alpha} f_2  \; d \mu &= 0.
\end{align*}
so the only term which contributes is 
\begin{align*}
   \plim_{\alpha}\int f \; T_{e_\alpha} f_1 \; T_{2 e_\alpha} f_1  \; d \mu.
\end{align*}

 \begin{prop}
   Let $f = \mathbbm{1}_A$ for $A \in \A$ with $\mu(A)>0$ and let $ f_1 =
   E( f| \mathscr{B})$. Then
   \begin{equation*}
     \plim_{\alpha} \int f \; T_{e_\alpha} f_1 \; T_{2 e_\alpha} f_1 \; d \mu >
     0.  
   \end{equation*}
 \end{prop}

 \begin{proof}
   By the decomposition of measures $f_1(x) >0$ for $\mu$-a.e. $x \in
   A$. So for some $ a >0$, if we let
   \begin{equation*}
     A' = \{ x \in X : f_1(x)^2 > a\}
   \end{equation*}
   then there exist $b >0$ and a set $B_1 \in \K$ with $\mu(B_1) =
   5 \xi >0$ such that for all $y \in B_1$ we have  $\mu_y(A') > b$. 

   Note that
   $\int f \, f_1 \, f_1 \,d \mu_y > a b$ for all $y \in B_1$.

   Let $\epsilon = \frac{a}{36}$. Now $f_1$ is $\B$-measurable and
   thus is in the closure of the AP over $K$ functions.  Hence we may
   choose an almost periodic over $K$ function $\phi_1 $ such that
   $\|f_1 - \phi_1\|< \epsilon \sqrt{\xi}$. This means that $\|f_1
   - \phi_1\|_y < \epsilon$ for every $y \in X \setminus C_1$, where
   $\mu(C_1) < \xi$. Since $\|f_1 - \phi_1\|_y$ is a $\K$-measurable
   function of $y$ we have that $C_1$ is $\K$-measurable. We let $B_2
   = B_1 \setminus C_1$. We have $\mu(B_2) > 4 \xi$ and $\|f_1 -
   \phi_1\|_y < \epsilon$ for $y \in B_2$.


   Since $\phi_1$ is AP over $K$ there exist $g_1, \dots, g_M \in
   L^2(X, \mu)$ such that for a.e. $y \in X$ and all $\omega \in
   \Omega$ one has $i =i(\omega, y)$ such that $1 \leq i \leq M$ and
   \begin{equation*}
     \|T_\omega \phi_1 - g_i\|_y < \epsilon. 
   \end{equation*}

   We claim that $\plim_\alpha \int f \; T_{e_\alpha} f_1 \;
   T_{2e_\alpha} f_1\; d \mu \geq \frac{a b \xi}{4 M^2}.$

   Let $E_1 \in \p$ be arbitrary. It suffices to find a single $\alpha
   \in E_1$ for which
   \begin{equation*}
     \int f  T_{e_\alpha} f_1 \; T_{2e_\alpha} f_1 \; 
     d \mu > \frac{a b \xi}{4 M^2}.
   \end{equation*}
   Let $N = M^2 +1$. Since $B_2 \in \K$ we have $\plim_\alpha
   T_{e_\alpha} \mathbbm{1}_{B_2} = \mathbbm{1}_{B_2}$ weakly. However
   since there is no loss of norm we must have $\plim_\alpha
   T_{e_\alpha} \mathbbm{1}_{B_2} = \mathbbm{1}_{B_2}$ in
   norm. Writing this in terms of measure we immediately obtain
   $\plim_\alpha \mu(B_2 \triangle T_{e_\alpha} B_2 ) =0$, and hence
   there exists $E_2 \in \p$ such that for all $\alpha \in E_2$,
   \begin{align} \label{eq:ExclusionEstimate1}
     \mu(B_2\triangle T_{e_\alpha} B_2 ) &< \frac{\xi}{ 2^N}.\\
     \intertext{The same holds true under $T_{2 e_\alpha}$, hence
       there exists $E_3 \in \p$ such that for all $\alpha \in
       E_3$,} \label{eq:ExclusionEstimate2} \mu(B_2 \triangle
     T_{2e_\alpha} B_2) &< \frac{\xi}{2^N}.
   \end{align}
   Now let $E= E_1 \cap E_2 \cap E_3 \subset E_1$. Since $\plim_\alpha
   T_{e_\alpha} h = \plim_\alpha T_{e_\alpha} h = h$ in norm for all
   $h \in L^2(X, \mathscr{K}, \mu)$, we have
   \begin{align*}
     \plim_\alpha \| T_{e_\alpha} h - T_{2 e_\alpha} h \| =0.
   \end{align*}
   Taking $h(y) = \|f_1 - T_{2 e_\alpha} f_1 \|_y$ we obtain that for
   all $\alpha$,
   \begin{align*}
     \plim_\beta \int \bigl| \|f_1 - T_{2 e_\alpha}
     f_1\|_{T_{e_\beta}y} - \|f_1 - T_{2 e_\alpha}
     f_1\|_{T_{2e_\beta}y} \bigr|^2 d \mu(y) = 0.
   \end{align*}
   Let
   \begin{equation*}
     A_\alpha = \bigl\{ \beta \in E:  \int \bigl| \|f_1 - T_{2 e_\alpha} f_1\|_{T_{e_\beta}y}
     - \|f_1 - T_{2 e_\alpha} f_1\|_{T_{2e_\beta}y} \bigr|^2 d \mu(y) <
     \frac{\epsilon^2 \xi}{2 M^2} \bigr\}.
   \end{equation*}
   We have $A_\alpha \in \p$ for all $\alpha \in \F$.  We will choose
   $\alpha_1 < \alpha_2 < \dots < \alpha_N$ inductively such that for
   all $\alpha,\beta \in FU(\alpha_1, \dots, \alpha_N)$, where
   $FU(\alpha_1, \dots, \alpha_N)$ denotes all finite unions of sets
   from $\{\alpha_1, \dots, \alpha_N\}$, with $\alpha < \beta$ one has
   $\beta \in A_{\alpha}$ i.e.
   \begin{align}\label{eq:FiberComparison}
     \int \bigl| \|f_1 - T_{2 e_\alpha} f_1\|_{T_{e_\beta}y} - \|f_1 -
     T_{2 e_\alpha} f_1\|_{T_{2e_\beta}y} \bigr|^2 \, d \mu(y) <
     \frac{\epsilon^2 \xi}{2  M^2}.
   \end{align}
   We let $\alpha_1 \in E_2$ be arbitrary. We have $A_{\alpha_1} \in
   \p$. Since $\p$ is idempotent we have $A_{\alpha_1} \in \p \ast \p$,
   so that $\{ \gamma \in \F : \gamma^{-1} A_{\alpha_1} \in \p \} \in
   \p$. Intersecting this set with $A_{\alpha_1} \in \p$ and $\F_{\max
     \alpha_1} \in \p$ we get $\{ \gamma \in \F_{\max \alpha_1} \cap
   A_{\alpha_1} : \gamma^{-1} A_{\alpha_1} \in \p \} \in \p$ and
   consequently $\{ \gamma \in \F_{\max \alpha_1} \cap A_{\alpha_1} :
   \gamma^{-1} A_{\alpha_1} \in \p \} \neq \varnothing$. We choose
   $\alpha_2 \in \{ \gamma \in \F_{\max \alpha_1} \cap A_{\alpha_1} :
   \gamma^{-1} A_{\alpha_1} \in \p \} $ so that $\alpha_1 < \alpha_2$,
   $\alpha_2 \in A_{\alpha_1}$, and $\alpha_2^{-1} A_{\alpha_1} \in
   \p$. Our inductive hypothesis is that for all $\alpha, \beta \in
   FU(\alpha_1, \dots, \alpha_n)$ with $\alpha < \beta$ we have $\beta
   \in A_\alpha$ and $\beta^{-1} A_\alpha \in \p$.  For conciseness we
   will write $F_n = FU(\alpha_1, \dots, \alpha_n)$. By the inductive
   hypothesis
   \begin{equation*}
     \bigcap_{\alpha \in F_n}
     A_\alpha \cap\bigcap_{\substack{\alpha,\beta \in F_n\\ \alpha < \beta}} \beta^{-1} A_\alpha \in \p.
   \end{equation*}
   Since $\p$ is an idempotent we have
   \begin{equation*}
     \bigl\{\gamma \in \F : \gamma^{-1} \bigl(\bigcap_{\alpha \in F_n}
     A_\alpha \cap
     \bigcap_{\substack{\alpha,\beta \in F_n \\ \alpha < \beta}} \beta^{-1} A_\alpha \in \p \bigr)
     \bigr\} \in \p
   \end{equation*}
   intersecting we then have
   \begin{equation*}
     \begin{aligned}
       \Bigl\{\gamma \in \F_{\max \alpha_n}\cap\bigcap_{\alpha \in
         F_n}& A_\alpha \cap \bigcap_{\substack{\alpha,\beta \in F_n
           \\ \alpha <
           \beta}} \beta^{-1} A_\alpha\\
       &: \gamma^{-1} \bigl(\bigcap_{\alpha \in F_n} A_\alpha \cap
       \bigcap_{\substack{\alpha,\beta \in F_n\\ \alpha < \beta}}
       \beta^{-1} A_\alpha \in \p \bigr) \Bigr\} \in \p.
     \end{aligned}
   \end{equation*}
   We choose $\alpha_{n+1}$ from this set. The reader should verify
   that all the conditions for the induction to continue are
   satisfied. We define $B_3 \in \K$ as
   \begin{equation*}
     B_3 = B_2 \cap  \bigcap_{\alpha \in F_N} \bigl(
     T_{e_\alpha}^{-1}  B_2 \cap T_{2 e_\alpha}^{-1} B_2 
     \bigr).
   \end{equation*}
   Using \eqref{eq:ExclusionEstimate1} and
   \eqref{eq:ExclusionEstimate2} we have that $\mu(B_3) > \xi$. Also
   for all $y \in B_3$, $\alpha \in F_N$, $T_{e_\alpha} y \in B_2$ and
   $T_{2 e_\alpha} y \in B_2$. 

   Since $N = M^2 + 1$ for all $y \in B_3$ there exists
   $\ell=\ell(y)$, $m=m(y)$ with $1 \leq \ell < m \leq N$ such that
   \begin{align}
     i(e_{\alpha_\ell \cup \dots \cup \alpha_N},y) &= i( e_{\alpha_m
       \cup
       \dots \cup \alpha_N} ,y)\\
     \intertext{and} i( 2 e_{\alpha_\ell \cup \dots \cup \alpha_N},y)
     &= i(2 e_{\alpha_m \cup \dots \cup \alpha_N} ,y).
   \end{align}
   We may divide $B_3$ into at most $M^2$ cells on which both $\ell$
   and $m$ are constant. At least one of these cells must have measure
   at least $\mu(B_3)/M^2$. Let $B_4 \subset B_3$ be such a cell. Now
   $\mu(B_4) > \frac{\xi}{M^2}$. For conciseness we write $\beta_j =
   \alpha_j \cup \dots \cup \alpha_N$.

   For $y \in B_4$ we have
   \begin{equation}\label{eq:APEstimate1}
     \begin{aligned}
       \| T_{e_{\beta_m}} \phi_1 &- T_{e_{\beta_\ell}} \phi_1 \|_y \\
       &< \| T_{e_{\beta_m}} \phi_1 - g_{i(e_{\beta_m},y)}\|_y +
       \|g_{i(e_{\beta_\ell},y)}- T_{e_{\beta_\ell}} \phi_1 \|_y < 2
       \epsilon.
     \end{aligned}
   \end{equation}

   Similarly,
   \begin{align}\label{eq:APEstimate2}
     \| T_{2 e_{\beta_m}} \phi_1 - T_{2e_{\beta_\ell}} \phi_1 \|_y < 2
     \epsilon.
   \end{align}
   Since $T_\omega$ is measure-preserving, by \eqref{eq:APEstimate1}
   and \eqref{eq:APEstimate2} we have
   \begin{align}
     \label{eq:4}
     \begin{aligned}
       \| \phi_1 - T_{e_{\alpha_\ell \cup \dots \cup \alpha_{m-1} }}
       \phi_1 \|_{T_{e_{\beta_m}} y} &< 2 \epsilon\hbox{ and}\\
       \| \phi_1 - T_{2e_{\alpha_\ell \cup \dots \cup \alpha_{m-1} }}
       \phi_1 \|_{T_{2e_{\beta_m}} y} &< 2 \epsilon.
     \end{aligned}
   \end{align}
   Since $\beta_m, \beta_\ell \in F_N$ we have $T_{e_{\beta_m}} y, T_{e_{\beta_\ell}}y \in B_2$ and
   consequently 
   \begin{align*}
     \| \phi_1 - f_1\| _{T_{e_{\beta_m}} y} < \epsilon\text{ and }\|
     \phi_1 - f_1\| _{T_{e_{\beta_\ell}} y} < \epsilon.
   \end{align*}
   Now  $\|
   T_{e_{\alpha_\ell \cup \dots \cup \alpha_{m-1}}} f_1 -
   T_{e_{\alpha_\ell \cup \dots \cup \alpha_{m-1}}} \phi_1
   \|_{T_{e_{\beta_m}}y} = \| f_1 - \phi_1 \|_{T_{e_{\beta_\ell}}y} <
   \epsilon$.   Write $\alpha = \alpha_\ell \cup \dots \cup \alpha_{m-1}$ and $\beta
   = \beta_m$.  Now by the triangle inequality
   \begin{align}
     \label{eq:T1Estimate}
     \| f_1 - T_{e_{\alpha}} f_1 \|_{T_{e_{\beta}}y} < 4 \epsilon.
   \end{align}
   Similarly we can  conclude that 
   \begin{align}\label{eq:WrongFiberEstimate}
     \| f_1 - T_{2e_{\alpha}} f_1\|_{T_{2e_{\beta}}y} < 4 \epsilon.
   \end{align}
  Since $\alpha < \beta$, by
   \eqref{eq:FiberComparison} we have
   \begin{align}
     \label{eq:LocalizedFiberComparison}
     \bigl| \| f_1 - T_{2 e_\alpha} f_1 \|_{T_{e_\beta} y} - \| f_1 -
     T_{2 e_{\alpha}} f_1 \|_{ T_{2 e_\beta} y} \bigr| < \epsilon
   \end{align}
   for all $y \in X \setminus C_2$, where $\mu(C_2) < \frac{ \xi}{2
     M^2}$. Let $B_5 =B_4 \setminus C_2$. Clearly  $\mu (B_5) >
   \frac{\xi}{2 M^2}$. For all $y \in B_5$ we can combine
   \eqref{eq:LocalizedFiberComparison} with
   \eqref{eq:WrongFiberEstimate}  to get 
   \begin{align}
     \|f_1 - T_{2 e_\alpha} f_1\|_{T_{e_\beta}y} < 5 \epsilon .\label{eq:T2Estimate}
   \end{align}
   Now
   \begin{align*}
      \int f &\, T_{e_\alpha} f_1 T_{2 e_\alpha} f_1 \, d \mu_y \\
      &= \int  f \, \bigl(f_1+ (T_{e_\alpha}f_1- f_1)\bigr)
      \bigl(f_1+(T_{2 e_\alpha} f_1-f_1) \bigr) \, d \mu_y \\
      &= 
      \begin{aligned}[t]
        \int f &\, f_1\,  f_1\, d\mu_y + \int f \, (T_{e_\alpha}f_1-
      f_1)\, f_1 \, d\mu_y +\int f \,f_1\, (T_{2e_\alpha}f_1-f_1) \, d\mu_y \\
      &+ \int f \, (T_{e_\alpha}f_1-
      f_1)\,  (T_{2e_\alpha}f_1-
      f_1) \, d\mu_y.
      \end{aligned}
   \end{align*}
   Using $|f| \leq 1$ and $|f_1|\leq 1$ together with
   \eqref{eq:T1Estimate} and \eqref{eq:T2Estimate} we obtain
   \begin{equation*}
     \int f \, T_{e_\alpha} f_1 T_{2 e_\alpha} f_1 \, d \mu_{T_{e_{\beta}}y} > (a -
     18 \epsilon) b > \frac{a b }{2}
   \end{equation*}
   for all $ y \in B_5$.  Since $\mu(B_5) > \frac{\xi}{2 M^2}$ we have
   \begin{equation*}
     \int f \, T_{e_\alpha} f_1 \, T_{2 e_\alpha} f_1 \, d \mu > \frac{a
       b \xi}{4 M^2}
   \end{equation*}
   as claimed.
 \end{proof}

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\end{document} 




\begin{defn}
  A function $f \in L^2(X, \A, \mu)$ is called almost periodic over
  $K$ if for every $\epsilon >0$ there exists functions $g_1, \dots
  ,g_k \in L^2(X, \A,\mu)$ such that for all $\delta >0$ there exists
  a $D^*$-set $C \subset \F$ such that for all $\alpha \in C$ there
  exists $E(\alpha) \in \K$ with $\mu\bigl(E(\alpha)\bigr) <\delta$
  having the property that for all $x \in (D \cup E(\alpha))^c$ there
  exists $0 \leq i=i(x,\alpha) \leq k$ such that
  \begin{align*}
    \| T_{e_\alpha} f - g_i \|_x < \epsilon.
  \end{align*}
\end{defn}


One may show that the closure of the set of $(T_{e_\alpha})$-almost
periodic over $K$ functions has the form $L^2(X, \B_1, \mu)$ for some
$\sigma$-algebra $\B_1$ containing $\K$. Similarly, one may show that
the closure of the set of $(T_{2e_\alpha})$-almost periodic over $K$
functions has the form $L^2(X, \B_1, \mu)$ for some $\sigma$-algebra
$\B_1$ containing $\K$.

% Old Lemma 2
\begin{lem}
  Let $\p$ be an essential idempotent in $\delta\F$ and let $f \in
  L^\infty(X,\A,\mu)$ 
  \begin{enumerate}
  \item If $E(f|\B_1)=0$ then $\plim_\alpha \|P(f \, T_{e_\alpha} f) \| =
    0. $
 \item If $E(f|\B_1)=0$ then $\plim_\alpha \|P(f \, T_{2e_\alpha} f) \| =
    0. $
  \end{enumerate}
\end{lem}

%%Old Lemma 5
\begin{lem}
  For $f$ in the Kronecker subspace of $T_\omega$ (or
  $\tilde{T}_\omega$) we have 
  \begin{enumerate}
  \item $\plim_\alpha T_{e_\alpha} f = f$, and 
  \item $\plim_\alpha T_{2 e_\alpha} f = f$.
  \end{enumerate}
\end{lem}

\begin{proof}
  As a consequence of the spectral theorem the Kronecker subspace of
  the unitary action of $\Omega$ is spanned by eigenfunctions, namely
  by functions $h$ for which there is a character $\chi: \Omega
  \rightarrow S^1$ where $S^1 =\{z \in \mathbb{C}
: |z|=1\}$ such that $T_\omega h =
  \chi(\omega) h$ for all $\omega \in \Omega$.  By linearity it suffices to
  show that $\plim_\alpha U_{e_\alpha} h = h$ for all eigenfunctions $h$. Since
  \begin{equation*}
    \plim_\alpha T_{e_\alpha} h = \plim_\alpha \chi\bigl(e_\alpha \bigr) h
  \end{equation*}
  we see that this reduces to the statement
  \begin{equation*}
    \plim_\alpha \chi\bigl( e_\alpha \bigr) = 1
  \end{equation*}
  for all characters $\chi$.
  
  Since $S^1$ is compact we have that $ \plim_\alpha \chi\bigl(
  e_\alpha \bigr)$ exists. Let $z = \plim_\alpha \chi\bigl( e_\alpha
  \bigr)$. Now
  \begin{align*}
    z &= \plim_\alpha \plim_\beta \chi\bigl( e_{\alpha+ \beta}
    \bigr)\\
    &=\plim_\alpha \plim_\beta \chi\bigl( e_{\alpha}e_{\beta} \bigr)\\
    &=\plim_\alpha \plim_\beta \chi\bigl( e_{\alpha}\bigr)
    \chi\bigl(e_{\beta} \bigr)\\
    &=\plim_\alpha \chi\bigl( e_{\alpha}\bigr)  \plim_\beta \chi\bigl(e_{\beta} \bigr)\\
    &= z^2. 
  \end{align*}
  Thus $z=1$. An identical argument works for $2 e_\alpha$. 
\end{proof}