# Mean ergodic theorems in $L^{r}(\mu)$ and $H^{r}(\mathbb{T})$, $0<r<1$

el Houcein el Abdalaoui  Laboratoire Mathématique Raphaël Salem, Université de
Rouen-Normandie, Rouen, France<EMAIL_ADDRESS>and
Michael Lin  Department of Mathematics, Ben-Gurion University of the Negev,
Beer-Sheva, Israel<EMAIL_ADDRESS>
###### Abstract.

Let $T$ be the Koopman operator of a measure preserving transformation
$\theta$ of a probability space $(X,\Sigma,\mu)$. We study the convergence
properties of the averages $M_{n}f:=\frac{1}{n}\sum_{k=0}^{n-1}T^{k}f$ when
$f\in L^{r}(\mu)$, $0<r<1$. We prove that if $\int|M_{n}f|^{r}d\mu\to 0$, then
$f\in\overline{(I-T)L^{r}}$, and show that the converse fails whenever
$\theta$ is ergodic aperiodic. When $\theta$ is invertible ergodic aperiodic,
we show that for $0<r<1$ there exists $f_{r}\in(I-T)L^{r}$ for which
$M_{n}f_{r}$ does not converge a.e. (although $\int|M_{n}f|^{r}d\mu\to 0$). We
further establish that for $1\leq p<\frac{1}{r},$ there is a dense
$G_{\delta}$ subset ${\mathcal{F}}\subset L^{p}(X,\mu)$ such that
$\limsup_{n}\frac{|T^{n}h|}{n^{r}}=\infty$ a.e. for any $h\in{\mathcal{F}}$.

When $T$ is induced by an irrational rotation of $\mathbb{T}$, the Hardy
spaces $H^{r}(\mathbb{T})$ are $T$-invariant. For $0<r<1$, we prove that
$H^{r}(\mathbb{T})=\\{constants\\}\oplus\overline{(I-T)H^{r}(\mathbb{T})}$,
and $\int|M_{n}f|^{r}d\mu\to 0$ for $f\in\overline{(I-T)H^{r}(\mathbb{T})}$.
However, there exists $f\in(I-T)H^{r}(\mathbb{T})$ such that $M_{n}f$ does not
converge a.e.

###### Key words and phrases:

ergodic theorem, measure preserving transformations, non-integrable functions,
$L^{r}(\mu)$ with $0<r<1$, circle rotations, Hardy spaces $H^{r}(\mathbb{T})$

###### 2010 Mathematics Subject Classification:

Primary: 37A30; secondary : 42B30, 37E10

##  1\. Introduction

Let $\theta$ be a measure preserving transformation of a probability space
$(X,\Sigma,\mu)$ and for $f$ measurable put $Tf(x)=f(\theta x)$. For $1\leq
p<\infty$, the linear operator $T$ is an isometry of $L^{p}(\mu)$, and the
mean ergodic theorem yields the $L^{p}$-convergence of the averages
$M_{n}f:=\frac{1}{n}\sum_{k=0}^{n-1}T^{k}f$.

When $0<r<1$, the space $L^{r}(\mu):=\\{f:\int|f|^{r}d\mu<\infty\\}$ is a
vector space, which is a complete metric space under the (invariant) metric
$d(f,g)=\int|f-g|^{r}d\mu$ [20, p. 37], and $T$ is an isometry of this metric
space. Note that an isometry in a complete metric space is continuous. Dalibor
Volný asked one of us about the ergodic properties of $T$ in $L^{r}$.

When $T$ is induced by an irrational rotation of $\mathbb{T}$, the Hardy
spaces $H^{r}(\mathbb{T})$ are $T$-invariant subspaces of $L^{r}(\mathbb{T})$.
For $0<r<1$, we study the ergodic properties of $T$ in $H^{r}(\mathbb{T})$.

##  2\. Ergodic theorems in $L^{r}(\mu)$, $0<r<1$

In this section $0<r<1$ is fixed, and $f_{n}\to f$ for $\\{f_{n}\\}\subset
L^{r}$ means convergence in the metric, i.e. $\int|f_{n}-f|^{r}d\mu\to 0$.

######  Lemma 2.1.

Let $T$ be induced on $L^{r}(\mu)$ by a measure preserving transformation
$\theta$, and let $f\in L^{r}$. If $\frac{1}{n}\sum_{k=0}^{n-1}T^{k}f\to g$,
then $Tg=g$. Hence $g$ is constant when $\theta$ is ergodic.

###### Proof.

Clearly $d(Tg,M_{n}f)\leq d(Tg,TM_{n}f)+d(TM_{n}f,M_{n}f)$. Since $T$ is an
isometry,
${d(Tg,TM_{n}f)=d(g,M_{n}f)\to 0}$. We also have

$d(TM_{n}f,M_{n}f)=\int|TM_{n}f-M_{n}f|^{r}d\mu=\int\Big{|}\frac{T^{n}f}{n}-\frac{f}{n}\Big{|}^{r}d\mu\leq
2\int\Big{|}\frac{f}{n}\Big{|}^{r}d\mu\to 0,$

so $M_{n}f\to Tg$, hence $Tg=g$. ∎

######  Corollary 2.2.

Let $0<r<1$ and $T$ as above. If $f\in L^{r}$ satisfies
$\frac{1}{n}\sum_{k=0}^{n-1}T^{k}f\to g$, then
$\frac{1}{n}\sum_{k=0}^{n-1}T^{k}(f-g)\to 0$.

The next theorem is a partial analogue of the result in Banach spaces proved
in [23]. Yosida used the Hahn-Banach separation theorem, but in $L^{r}$ there
are no non-zero continuous functionals [20, p. 37], so a different proof is
needed.

######  Theorem 2.3.

Let $T$ be induced on $L^{r}(\mu)$ by a measure preserving transformation, and
let $f\in L^{r}$. If $f\in(I-T)L^{r}$, then
$\frac{1}{n}\sum_{k=0}^{n-1}T^{k}f\to 0$. If
$\frac{1}{n}\sum_{k=0}^{n-1}T^{k}f\to 0$, then $f\in\overline{(I-T)L^{r}}$.

###### Proof.

Let $f=h-Th$. Then, as before,

$\int|M_{n}f|^{r}d\mu=\int\Big{|}M_{n}h-TM_{n}h\Big{|}^{r}d\mu=\int\Big{|}\frac{f}{n}-\frac{T^{n}h}{n}\Big{|}^{r}d\mu\to
0.$

Assume now that $M_{n}f\to 0$. Put

$g_{n}:=f-M_{n}f=\frac{1}{n}\sum_{k=0}^{n-1}(f-T^{k}f)=\frac{1}{n}\sum_{k=1}^{n-1}(I-T)\sum_{j=0}^{k-1}T^{j}f.$

Hence $g_{n}\in(I-T)L^{r}$, and $d(f,g_{n})=\int|M_{n}f|^{r}d\mu\to 0$, so
$f\in\overline{(I-T)L^{r}}$. ∎

We will show below that there is no complete analogue of Yosida’s result.

######  Theorem 2.4.

Let $T$ be induced on $L^{r}$, $0<r<1$, by an ergodic probability preserving
transformation $\theta$, and let $f\in L^{r}$. Then $f\in(I-T)L^{r}$ if and
only if

(1) $\sup_{n\geq 1}\int\big{|}\sum_{k=0}^{n-1}T^{k}f\big{|}^{r}d\mu<\infty.$

###### Proof.

If $f=(I-T)g$ with $g\in L^{r}$, then $\sum_{k=0}^{n-1}T^{k}f=g-T^{n}g$, so

$\sup_{n\geq 1}\int\Big{|}\sum_{k=0}^{n-1}T^{k}f\Big{|}^{r}d\mu\leq
2\int|g|^{r}d\mu<\infty.$

Assume now that (1) holds. Then

$\sup_{N\geq
1}\frac{1}{N}\sum_{n=1}^{N}\int\Big{|}\sum_{k=0}^{n-1}T^{k}f\Big{|}^{r}d\mu<\infty.$

We can therefore apply Sato’s [21, Corollary, p. 287] and obtain that
$f\in(I-T)L^{r}$. ∎

Remark. Theorem 2.4 is valid for $1<r<\infty$ as a special case of Browder
[4], and for $r=1$ by Lin and Sine [18]. One of the novelties in [1] and [21]
is in the results for $0<r<1$.

When $\theta$ is ergodic and $f\in L^{1}(\mu)$, the ergodic theorems yield
that $M_{n}f\to\int fd\mu$ in $L^{1}$-norm and a.e., so the limit does not
depend on the ergodic transformation. When $f\in L^{r}$, $0<r<1$, and $M_{n}f$
converges in the $L^{r}$ metric, the limit is still a constant, by Lemma 2.1,
but we show that the constant may depend on $\theta$.

######  Proposition 2.5.

There exist two ergodic measure preserving transformations on the unit circle,
with induced operators $T$ and $S$, and a function $f\in L^{r}$,
$0<r<\frac{1}{3}$, such that, in the $L^{r}$ metric,
$\frac{1}{n}\sum_{k=0}^{n-1}T^{k}f\to\mathbf{1}$ and
$\frac{1}{n}\sum_{k=0}^{n-1}S^{k}f\to\mathbf{0}$,

###### Proof.

The two transformations $T$ and $S$ are given by irrational rotations
$\alpha,\beta$ which are independent over the rationals. Buchzolich [5]
constructed a measurable function $f$ (necessarily not integrable) such that
$\frac{1}{n}\sum_{k=0}^{n-1}T^{k}f\to 1$ and
$\frac{1}{n}\sum_{k=0}^{n-1}S^{k}f\to 0$ a.e.

Claim: It is possible to choose the parameters in [5] so that $f\in L^{r}$,
for $0<r<1/3$.

Let ${\varepsilon}_{0}\in(\frac{1}{2},1)$, and for $j\geq 1$ define
inductively

$\frac{1}{{\varepsilon}_{j}}=\Big{[}\frac{1}{\sqrt{\prod_{k=0}^{j-1}{\varepsilon}_{k}}}\Big{]}+1.$

Then ${\varepsilon}_{1}=\frac{1}{2}<{\varepsilon}_{0}$. For any $j$ we have
${\varepsilon}_{j}<1$ and $1/{\varepsilon}_{j}$ an integer. By definition
${\varepsilon}_{j}^{2}<\prod_{k=0}^{j-1}{\varepsilon}_{k}$. Since
${\varepsilon}_{j}<1$, $\
{\varepsilon}_{j+1}<{\varepsilon}_{j}<{\varepsilon}_{0}$ for $j\geq 1$. By
induction, $\prod_{k=0}^{j}{\varepsilon}_{k}\leq{\varepsilon}_{0}^{j+1}$,
hence
${\varepsilon}_{j}^{2}<\prod_{k=0}^{j-1}{\varepsilon}_{k}\leq{\varepsilon}_{0}^{j}$.
Thus, for $0<r<1/3$, we have

(2) $\sum_{j\geq
0}{\varepsilon}_{j}^{1-3r}\leq\sum_{j=0}^{\infty}{\varepsilon}_{0}^{j(1-3r)/2}=\sum_{j=0}^{\infty}({\varepsilon}_{0}^{(1-3r)/2})^{j}<+\infty.$

The sequence ${\varepsilon}_{j}$ satisfies all the assumptions in [5, p. 250].

The function $f$, constructed in [5], is given by
$f=\sum_{j=0}^{+\infty}(-1)^{j}f_{j}$, with the definition of $f_{j}$
dependent on the fixed irrationals $\alpha$ and $\beta$; the support of
$f_{j}$, denoted by $E^{j}$, satisfies $\mu(E^{j})<{\varepsilon}_{j}.$
Furthermore, the construction yields

(i) $\int f_{j}d\mu=1$ for $j\geq 0$.

(ii) $\displaystyle\sup_{x\in
X}|f_{2j}(x)|\leq{\varepsilon}_{2j}^{-1}\sup_{x\in X}|f_{2j-1}(x)|,$

(iii) $\displaystyle\sup_{x\in
X}|f_{2j+1}(x)|\leq{\varepsilon}_{2j+1}^{-1}\sup_{x\in X}|f_{2j}(x)|.$

By the construction $|f_{0}|\leq 1$, so with $f_{-1}\equiv 1$ we have

(3) $\sup_{x\in X}|f_{k}(x)|\leq{\varepsilon}_{k}^{-1}\sup_{x\in
X}|f_{k-1}(x)|=1/\prod_{j=0}^{k}{\varepsilon}_{j}\quad\rm{for}\quad k\geq 0.$

Applying the triangle inequality in $L^{r}$ and the estimate
${\varepsilon}_{j}^{2}<\prod_{k=0}^{j-1}{\varepsilon}_{k}$, we get

$\int_{X}|f|^{r}d\mu\leq\sum_{j\geq
0}\int|f_{j}|^{r}d\mu\leq\sum_{j=0}^{\infty}\mu(E^{j})\sup_{x\in
X}|f_{j}(x)|^{r}\leq\sum_{j=0}^{\infty}{\varepsilon}_{j}\Big{(}\frac{1}{\prod_{k=0}^{j}{\varepsilon}_{k}}\Big{)}^{r}\leq\sum_{j=0}^{\infty}{\varepsilon}_{j}^{1-3r}.$

Therefore, by (2), $f\in L^{r}$ when $0<r<\frac{1}{3}$, which proves the
claim.

We now prove the claimed convergence in $L^{r}$ metric, for fixed
$0<r<\frac{1}{3}$. We denote $M_{n}(T):=\frac{1}{n}\sum_{k=0}^{n-1}T^{k}$ and
$M_{n}(S):=\frac{1}{n}\sum_{k=0}^{n-1}S^{k}$. We use the choice of
${\varepsilon}_{j}$ made above, and the corresponding $f$, which is in $L^{r}$
by the claim.

We first prove that $M_{n}(T)f\to\bf 1$ in $L^{r}$. Let ${\varepsilon}>0$ and
fix $J_{0}$ such that
$\sum_{j=2J_{0}}^{\infty}{\varepsilon}_{2j+1}^{1-3r}<{\varepsilon}/2$.

In the construction in [5], it is shown that there exist sets $\hat{X}_{2j+1}$
with $\mu(\hat{X}_{j})>1-2{\varepsilon}_{2j+1}$, such that
$|M_{n}(T)(f_{2j+2}-f_{2j+1})(x)|<{\varepsilon}_{2j+1}$ for every
$x\in\hat{X}_{2j+1}$, $n>0$ and $j\geq 0$. Put $\hat{X}=\cap_{j\geq
2J_{0}}\hat{X}_{2j+1}$. Then $\mu(\hat{X})>1-{\varepsilon}/2$. Put
$g_{J_{0}}:=f_{0}+\sum_{j=0}^{2J_{0}-1}(f_{2j}-f_{2j+1})$. Recall that
$f=\sum_{j=0}^{\infty}(-1)^{j}f_{j}=\sum_{j=0}^{\infty}(f_{2j}-f_{2j+1}$.
Since

$|M_{N}(T)f(x)-1|\leq|M_{N}(T)g_{J_{0}}(x)-1|+\sum_{j=2J_{0}}^{\infty}|M_{N}(T)(f_{2j+2}-f_{2j+1})(x)|,$

the estimates for $x\in\hat{X}$ yield that on $\hat{X}$

$|M_{N}(T)f-1|\leq|M_{N}(T)g_{J_{0}}-1|+\sum_{j=2J_{0}}^{\infty}|M_{N}(T)(f_{2j+2}-f_{2j+1})|$
$\leq|M_{N}(T)g_{J_{0}}-1|+\sum_{j=2J_{0}}^{\infty}{\varepsilon}_{2j+1}\leq|M_{N}(T)g_{J_{0}}-1|+{\varepsilon}/2.$

By the ergodic theorem $|M_{N}(T)g_{J_{0}}-1|\to 0$ almost surely, since by
the construction in [5] $g_{J_{0}}$ is bounded with integral 1. Hence for
$x\in\hat{X}$ and $N\geq N_{1}(x)$ we have on $\hat{X}$

$|M_{N}(T)f(x)-1|<{\varepsilon}.$

Whence $|M_{N}(T)f-1|$ converge almost surely to $0$ (as shown in [5]). By the
triangle inequality in $L^{r}$,

(4)
$\int|M_{N}(T)f-1|^{r}d\mu\leq\int|M_{N}(T)g_{J_{0}}-1|^{r}d\mu+\sum_{j=2J_{0}}^{\infty}\int|M_{N}(T)(f_{2j+2}-f_{2j+1})|^{r}d\mu.$

Since $g_{J_{0}}$ is bounded, by the dominated convergence theorem
$\int|M_{N}(T)g_{J_{0}}-1|^{r}d\mu\to 0$ as $N\to\infty$. We estimate the
series in (4). For each $j$, let $E^{2j+2,2j+1}$ be the support of
$f_{2j+2}-f_{2j+1}$. It is shown in [5, p. 251] that for some sequence
$(N_{j})$, we have

(i) $\mu(\bigcup_{0}^{N_{2j+2}}(T^{-k}E^{2j+2,2j+1}))<2{\varepsilon}_{2j+2}.$

(ii)
$\big{|}\sum_{k=0}^{M}(f_{2j+2}-f_{2j+1})(T^{k}x)\big{|}<2{\varepsilon}_{2j+2}^{-1}\sup_{y\in
X}|f_{2j+1}(y)|$, for all $M\in\mathbb{N}^{*}$ and $x\in X$, strengthening
(3).

Applying (ii), (3) and (i) we see that

$\int|M_{N}(T)(f_{2j+2}-f_{2j+1})|^{r}d\mu=\int_{\bigcup_{k=0}^{N_{2j+2}}(T^{-k}E^{2j+2,2j+1})}|M_{N}(T)(f_{2j+2}-f_{2j+1})|^{r}d\mu\leq$
$\frac{1}{N^{r}}2^{r}{\varepsilon}_{2j+2}^{-r}\Big{(}\frac{1}{\prod_{k=0}^{2j+1}{\varepsilon}_{k}}\Big{)}^{r}\mu(\bigcup_{k=0}^{N_{2j+2}}(T^{-k}E^{2j+2,2j+1}))\leq\frac{1}{N^{r}}2^{r+1}{\varepsilon}_{2j+2}{\varepsilon}_{2j+2}^{-r}\Big{(}\frac{1}{\prod_{k=0}^{2j+1}{\varepsilon}_{k}}\Big{)}^{r}$

Hence, since
${\varepsilon}_{2j+2}^{2}\leq\prod_{k=0}^{2j+1}{\varepsilon}_{k}$, by (2)

$\sum_{j=2J_{0}}^{\infty}\int|M_{N}(T)(f_{2j+2}-f_{2j+1})|^{r}d\mu\leq\frac{2^{r+1}}{N^{r}}\sum_{j=2J_{0}}^{+\infty}{\varepsilon}_{2j+2}^{1-3r}\leq
C\frac{2^{r+1}}{N^{r}}.$

Letting $N\to\infty$, (4) yields

$\int|M_{N}(T)f-1|^{r}d\mu\to 0.$

This complete the proof of the first part of the proposition.

The proof for $S$ is similar, as noted in [5] (we have not listed all the
properties of the $f_{j}$ obtained in [5]). For the limit of $M_{n}(S)f$ to be
0 and not 1, we define in this case
$g_{J_{0}}=\sum_{j=0}^{2J_{0}-1}(f_{2j}-f_{2j+1})$, so now $\int g_{J_{0}}=0$.
∎

Remark. By [5] we have also a.e. convergence $M_{n}(T)f\to 1$ and
$M_{n}(S)f\to 0$.

######  Proposition 2.6.

Let $T$ be induced by an irrational rotation of the circle, and let
$0<r<\frac{1}{3}$. Then ${\bf 1}\in\overline{(I-T)L^{r}}$.

###### Proof.

Let $T$ be induced by $\alpha$, and let $S$ be induced by some $\beta$ so that
$\alpha$ and $\beta$ are independent over the rationals. We need the following
lemma.

######  Lemma 2.7.

Fix $0<r<1$. Then $\overline{(I-T)L^{r}}=\overline{(I-S)L^{r}}$.

###### Proof.

By symmetry, it is enough to show
$\overline{(I-T)L^{r}}\supset\overline{(I-S)L^{r}}.$ Obviously, we have

$\overline{(I-T)L^{r}}\supset\overline{(I-T^{n})L^{r}}.$

By density of the orbits of $\alpha$, there exists a sequence $(n_{j})$ of
integers such that $\alpha^{n_{j}}\to\beta$. Hence for each $f\in
L^{1}(X,\mu)$ we have $\|T^{n_{j}}f-Sf\|_{1}\to 0$ (by approximation; easy for
$f$ continuous). By Hölder’s inequality
$\int|f|^{r}d\mu\leq(\int|f|d\mu)^{r}$, so for $f\in L^{1}$ we obtain
$\int|T^{n_{j}}f-Sf|^{r}d\mu\to 0$. But $L^{1}(X,\mu)$ is dense in
$L^{r}(X,\mu).$ Therefore, for each $f\in L^{r}$, we have
$\int|T^{n_{j}}f-Sf|^{r}d\mu\to 0$, which yields $(I-T^{n_{j}})f\to(I-S)f$ in
$L^{r}$. This implies that $(I-S)f\in\overline{(I-T)L^{r}}$ since, for each
$j$, $(I-T^{n_{j}})(f)\in\overline{(I-T)L^{r}}$. This completes the proof of
the Lemma. ∎

Proof of Proposition 2.6. Let $f$ be given by Proposition 2.5. Then $f\in
L^{r}$, and by Theorem 2.1, $f-{\bf 1}\in\overline{(I-T)L^{r}}$ and
$f\in\overline{(I-S)L^{r}}$. Whence, by Lemma 2.7, ${\bf
1}\in\overline{(I-T)L^{r}}$. ∎

By Proposition 2.6, for $0<r<1/3$ and any irrational rotation, ${\bf
1}\in\overline{(I-T)L^{r}}$, while its averages do not converge to zero (but
do converge). Kosloff and Volný suggested the next general proposition (with
an outline of its proof), in which the averages of a function in
$\overline{(I-T)L^{r}}$, $0<r<1$, do not converge.

######  Proposition 2.8.

Let $\theta$ be an ergodic aperiodic measure preserving tansformation of a
probability space $(X,\Sigma,\mu)$ and denote $Tf=f\circ\theta$. Then for
$0<r<1$ there exists $f\in\overline{(I-T)L^{r}}$ such that $M_{n}(T)f$ does
not converge in $L^{r}$ metric.

###### Proof.

Fix $0<r<1$ and fix $r<\alpha<1$. It is proved in [16, Theorem 1] that there
exists a measurable (real) $f$ such that
$\frac{1}{n^{1/\alpha}}\sum_{k=0}^{n-1}T^{k}f$ converges in distribution to a
symmetric $\alpha$-stable distribution with parameter $0<\sigma<1$. We show
that $f$ is the desired function.

By formula (5.13) in [9, Theorem 1, p. 576] (see also [24, Property 2.5, p.
63]), when $g$ has $\alpha$-stable symmetric distribution and $r<\alpha$, we
have $\int|g|^{r}d\mu<\infty$.

Claim: There exist $C$ and $K_{0}$, such that when $g$ has a symmetric
$\alpha$-stable distribution with parameter $\sigma<1$, for every $r<\alpha$
and $K>K_{0}$,

$\int_{\\{|g|>K\\}}|g|^{r}d\mu\leq C\sigma^{\alpha}K^{r-\alpha}.$

For the proof, fix $g$ as in the claim, with distribution
$F(x)=\mu\big{\\{}\omega:~{}~{}g(\omega)\leq x\big{\\}},~{}~{}x\in\mathbb{R},$
and following [19], put $q(x)=1-F(x)+F(-x),~{}~{}x\in\mathbb{R}.$ Then
$q(x)=\mu\big{\\{}\omega:~{}~{}|g(\omega)|>x\big{\\}}$ for $x\geq 0$. By
symmetry of $F$ we have $q(x)=2(1-F(x)).$ We further have, by [19, p.367, B.
3],

$\lim_{x\to+\infty}x^{\alpha}q(x)=\sigma^{\alpha},$

Whence,

$\lim_{x\to+\infty}x^{\alpha}\mu\Big{\\{}\omega~{}~{}:~{}~{}|g(\omega)|>x\Big{\\}}=2\sigma^{\alpha}.$

Now, write

$\int_{\\{|g|>K\\}}|g|^{r}d\mu=\int_{\\{|x|>K\\}}|x|^{r}dF(x)=-2\int_{\\{|x|>K\\}}|x|^{r}dq(x),$

and apply an integration by parts to get

$\int_{\\{|g|>K\\}}|g|^{r}d\mu=K^{r}q(K)+2r\int_{K}^{+\infty}x^{r-1}q(x)dx.$

Let $K_{0}>0$ such that for any $K>K_{0}$, we have

$|K^{\alpha}q(K)-\sigma^{\alpha}|<\sigma^{\alpha}.$

Then, for any $K>K_{0}$,

$K^{\alpha}q(K)\leq 2\sigma^{\alpha},$

and

$K^{r}q(K)\leq 2\sigma^{\alpha}K^{r-\alpha}.$

Therefore,

(5) $\displaystyle\int_{\\{|g|>K\\}}|g|^{r}d\mu$ $\displaystyle\leq
2\sigma^{\alpha}K^{r-\alpha}+2\sigma^{\alpha}r\int_{K}^{+\infty}x^{r-1}x^{-\alpha}dx$
(6) $\displaystyle\leq
2\sigma^{\alpha}K^{r-\alpha}+2\sigma^{\alpha}\frac{2r}{\alpha-r}K^{r-\alpha}.$

The claim thus follows by taking
$C=\max\big{\\{}2,\frac{2r}{\alpha-r}\big{\\}}.$

By the construction in [16], $f$ is of the form
$f=\sum_{j=1}^{\infty}(f_{j}-T^{d_{j}}f_{j})$, where
$f_{j}=g_{j}1_{\\{2^{j}<g_{j}<2^{j^{2}}\\}}$, with $g_{j}$ having a symmetric
$\alpha$-stable distribution with parameter $\sigma=j^{-1/\alpha}$. Using the
triangle inequality in $L^{r}$ and the claim, for $\log_{2}K_{0}<n<m$ we
obtain

$\int\Big{|}\sum_{j=n}^{m}(f_{j}-T^{d_{j}}f)\Big{|}^{r}d\mu\leq
2\sum_{j=n}^{m}\int_{\\{|g_{j}|>2^{j}\\}}|g_{j}|^{r}d\mu\leq
2C\sum_{j=n}^{m}\frac{1}{j}2^{j(r-\alpha)}.$

Since $\sum_{j=1}^{\infty}\frac{1}{j}2^{j(r-\alpha)}<\infty$, the partial sums
of $f$ are a Cauchy sequence in $L^{r}$ metric, so converge to $f$ in $L^{r}$,
and since the partial sums are in $(I-T)L^{r}$ by their definition, $f$ is in
$\overline{(I-T)L^{r}}$.

We denote below convergence in distribution by $\Rightarrow$. Let
$\xi\not\equiv 0$ be a random variable on $(X^{\prime},\mu^{\prime})$ with the
symmetric $\alpha$-stable distribution such that
$n^{-1/\alpha}\sum_{k=0}^{n-1}T^{k}\Rightarrow\xi$. Since $\xi$ has a
continuous distribution, using [3, Theorem 2.1], we obtain that

$\Big{|}n^{-1/\alpha}\sum_{k=0}^{n-1}T^{k}f\Big{|}^{r}\Rightarrow|\xi|^{r}.$

As noted above, $\int|\xi|^{r}d\mu^{\prime}<\infty$. By [3, Theorem 3.4] we
have

$0<a:=\int|\xi|^{r}d\mu^{\prime}\leq\liminf_{n\to\infty}\int\big{|}n^{-1/\alpha}\sum_{k=0}^{n-1}T^{k}f\big{|}^{r}d\mu.$

Denoting $p=1/\alpha$, the last inequality yields that for large $n$

$\int\big{|}n^{-1}\sum_{k=0}^{n-1}T^{k}f\big{|}^{r}d\mu=n^{r(p-1)}\int\big{|}n^{-1/\alpha}\sum_{k=0}^{n-1}T^{k}f\big{|}^{r}d\mu>\frac{1}{2}an^{r(p-1)}\to\infty.$

Hence $M_{n}(T)f$ does not converge in $L^{r}$. ∎

Remark. The averages $M_{n}(T)f$ of the above $f$ do not a.e. converge.
Indeed, if $M_{n}(T)f\to g$ a.e., then

$\frac{1}{n^{1/\alpha}}\sum_{k=0}^{n-1}T^{k}f=an^{1-\alpha^{-1}}M_{n}(T)f\to
0\quad\text{a.e.}$

contradicting the convergence in distribution to $\xi\not\equiv 0$.

######  Proposition 2.9.

Let $T$ be induced by an ergodic aperiodic measure preserving transformation
$\theta$ on a Lebesgue space $(X,\Sigma,\mu)$. Then for any $0<r<1$ there
exists $f_{r}\in(I-T)L^{r}(\mu)$ such that $M_{n}f_{r}$ does not converge a.e.

The proof will use the following proposition.

######  Proposition 2.10.

Let $T$ be induced by an ergodic aperiodic measure preserving transformation
$\theta$ on a Lebesgue space $(X,\Sigma,\mu)$. Then there exists $0\leq h\in
L^{1}(\mu)$ such that for every $0<r<1$ we have
$\limsup_{n}n^{-r}T^{n}h(x)=\infty$ a.e.

###### Proof.

We first note that $\|n^{-r}T^{n}g\|_{1}\to 0$ for $g\in L^{1}$ implies
$\liminf|n^{-r}T^{n}g|=0$.

We use Rokhlin’s lemma for non-invertible transformations [12], [2]111[2] does
not require the additional assumption that $\theta$ is onto, used in [12]. to
construct the function $h$. Let $\varepsilon_{n}={\rm e}^{-n}$ and
$k_{n}=2^{n}$. By Rokhlin’s lemma there exists $B_{n}\in\Sigma$ such that
$(\theta^{-j}B_{n})_{0\leq j<k_{n}}$ are disjoint and
$k_{n}\mu(B_{n})=\mu(\cup_{j=0}^{k_{n}-1}\,\theta^{-j}B_{n})>1-\varepsilon_{n}$.
Put $X_{n}:=\cup_{j=0}^{k_{n}-1}\,\theta^{-j}B_{n}$ and define

$h:=\sum_{n=1}^{\infty}n^{-2}\frac{1}{\mu(B_{n})}1_{B_{n}}.$

Then $\int h\,d\mu=\sum_{n=1}^{\infty}n^{-2}<\infty$, so $0\leq h\in
L^{1}(\mu)$.

Let $x\in X_{n}\setminus B_{n}$. Then there exists $1\leq j\leq k_{n}-1$ such
that $x\in\theta^{-j}B_{n}$. For $0<r<1$ we then have

$\frac{T^{j}h(x)}{j^{r}}\geq\frac{h(\theta^{j}x)}{k_{n}^{r}}\geq\frac{n^{-2}}{\mu(B_{n})}\cdot\frac{1}{k_{n}^{r}}=\frac{n^{-2}k_{n}^{1-r}}{k_{n}\mu(B_{n})}\geq\frac{k_{n}^{1-r}}{n^{2}}=\frac{2^{n(1-r)}}{n^{2}}.$

We now fix $r\in(0,1)$. Given $M>0$, for $n>N_{M}$ we have
$2^{n(1-r)}/n^{2}>M$, so

$\mu\Big{(}\Big{\\{}x:\sup_{j\geq
1}\frac{T^{j}h(x)}{j^{r}}>M\Big{\\}}\Big{)}\geq\mu(X_{n}\setminus
B_{n})=(k_{n}-1)\mu(B_{n})>1-\varepsilon_{n}-k_{n}^{-1}\qquad(n>n_{M}).$

Letting $n\to\infty$ we obtain $\sup_{j}j^{-r}T^{j}h>M$ a.e. Since this is for
any $M>0$, we obtain $\limsup_{j\to\infty}j^{-r}T^{j}h=\infty$ a.e. ∎

Proof of Proposition 2.9. Let $0\leq h\in L^{1}$ the function obtained in
Proposition 2.10. For $0<r<1$ put $g_{r}=h^{1/r}$ and $f_{r}=(I-T)g_{r}$. Then
$M_{n}(T)f_{r}=\frac{1}{n}(g_{r}-T^{n}g_{r})$. But
$\limsup(T^{n}g_{r}/n)^{r}=\limsup T^{n}h/n^{r}=\infty$ a.e., so
$M_{n}(T)f_{r}$ does not converge a.e. $\hfill\square$

Remark. When $T$ is induced by a probability preserving transformation on
$(X,\Sigma,\mu)$, Theorem 2.3 and Fatou’s Lemma yield that if
$f\in(I-T)L^{r}$, then $\liminf|M_{n}(T)f|=0$ a.e.

In Proposition 2.10 we exhibit a positive function $h$ in $L^{1}$ for which
for any $r\in(0,1)$ we have $\sup_{n}\frac{T^{n}h}{n^{r}}=\infty$. In
Proposition 2.12 below, we will establish that there is an abundance of such
functions in $L^{1}$, and in the more general setting, for a given $r$ there
are many functions in $L^{p}$, for any $1\leq p<1/r$, with
$\sup_{n}\frac{|T^{n}h|}{n^{r}}=\infty$. Note that when $g\in L^{p}$,
$1<p<\infty$, we have $n^{-1/p}T^{n}g\to 0$ a.e. by the pointwise ergodic
theorem.

We will use the following special case of a theorem by del Junco and
Rosenblatt [7]. Its proof is inspired by Garsia’s proof of Banach’s principle
[10].

######  Theorem 2.11.

[7, Theorem 1.1]. Fix $1\leq p\leq\infty$ and let $(T_{n})$ be a sequence of
linear maps on $L^{p}(X,\mu)$ which are continuous in measure. Assume that for
every $\epsilon>0$ and $K>0$, there exists $h\in L^{p}(X,\mu)$ with $\|h\|\leq
1$ such that $\mu\Big{\\{}x~{}~{}:~{}~{}\sup_{n\geq
1}|T_{n}h|>K\Big{\\}}>1-\epsilon$. Then there is a dense $G_{\delta}$ subset
${\mathcal{F}}\subset L^{p}(X,\mu)$ such that $\sup_{n}|T_{n}h|=\infty$ a.e.
for any $h\in{\mathcal{F}}$.

######  Proposition 2.12.

Let $T$ be induced by an ergodic aperiodic probability preserving
transformation $\theta$ on a Lebesgue space $(X,\Sigma,\mu)$, and fix
$r\in(0,1)$. Then for any $1\leq p<1/r$ there is a dense $G_{\delta}$ subset
${\mathcal{F}}\subset L^{p}(X,\mu)$, such that
$\limsup_{n}\frac{|T^{n}h|}{n^{r}}=\infty$ a.e. for any $h\in{\mathcal{F}}$.

###### Proof.

We will apply Theorem 2.11. For $f\in L^{p}$, put
$T_{n}f=\frac{T^{n}f}{n^{r}}$. Then $T_{n}$ is continous on $L^{p}$, and
therefore continuous in measure. Fix $\epsilon\in(0,1)$ and $K>0$, We use
Rokhlin’s lemma as in the proof of Proposition 2.10, and define $h$ as
follows:

$h:=\sum_{n=1}^{\infty}n^{-2}\frac{1}{\mu(B_{n})^{1/p}}1_{B_{n}}.$

Then $0\leq h\in L^{p}(\mu)$ with $\|h\|_{p}\leq\sum_{n=1}^{\infty}n^{-2}$. As
in the proof of Proposition 2.10, there is $n_{K}\in{\mathbb{N}}$ such that
for any $n>n_{K}$, we have

$\mu\Big{\\{}x~{}~{}:~{}~{}\sup_{n\geq 1}|T_{n}h|>K\Big{\\}}>1-\epsilon.$

Let $h^{\prime}=h/\|h\|_{p}$, and replace $K$ in the previous inequality by
$K\|h\|_{p}$. Then $\|h^{\prime}\|_{p}=1$, and

$\mu\Big{\\{}x~{}~{}:~{}~{}\sup_{n\geq
1}|T_{n}h^{\prime}|>K\Big{\\}}>1-\epsilon.$

We then deduce from Theorem 2.11 that there is a dense $G_{\delta}$ subset
${\mathcal{F}}\subset L^{p}(X,\mu)$ such that
$\sup_{n}\frac{|T^{n}h|}{n^{r}}=\infty$ a.e. for any $h\in{\mathcal{F}}$. ∎

Proposition 2.10 was inspired by its special case below, first proved by JP
Conze; its ”number theoretic” proof, which does not use Rokhlin’s lemma, is of
independent interest.

######  Proposition 2.13.

Let $T$ be induced by an irrational rotation of the unit circle. Then there
exists $0\leq h\in L^{1}$ such that for every $0<r<1$ we have
$\limsup_{n}n^{-r}T^{n}h(x)=\infty$ a.e.

###### Proof.

We represent the transformation by $\theta x=x+\alpha\mod 1$, with $\alpha$
irrational. Let $(q_{n})$ be the sequence of denominators in the continued
fractions representation of $\alpha$, and define

$h:=\sum_{n=1}^{\infty}n^{-2}q_{n}1_{[0,2/q_{n}]}.$

Then $\int_{0}^{1}h\,dx=2\sum_{n\geq 1}n^{-2}<\infty$, so $h$ is integrable.

For $x\in[0,1)$ there exists $k_{n}(x)$, $0\leq k_{n}(x)\leq q_{n}$, such that
$x+k_{n}(x)\alpha\mod 1\in[0,2/q_{n}]$. The sequence $k_{n}(x)$ converges to
$\infty$, and for $0<r<1$ we have

$\frac{1}{k_{n}(x)^{r}}h(\theta^{k_{n}(x)}x)\geq\frac{1}{q_{n}^{r}}h(\theta^{k_{n}(x)}x)\geq
n^{-2}q_{n}^{1-r}\to\infty,$

since $q_{n}$ increases to infinity at least exponentially [15, Theorem 12, p.
13].

Hence $\limsup_{n}n^{-r}T^{n}h(x)=\infty$ a.e. ∎

######  Proposition 2.14.

Let $T$ be induced by a probability preserving transformation on
$(X,\Sigma,\mu)$ and fix $0<r<1$. Given $\rho>1$, put $n_{j}:=[\rho^{j}]$.
Then for $f\in(I-T)L^{r}(\mu)$ we have $\lim_{j}M_{n_{j}}f=0$ a.e.

###### Proof.

We show that if $g\in L^{r}$, then $n_{j}^{-1}T^{n_{j}}g\to 0$ a.e. Since
$n_{j}+1>\rho^{j}$, we have

$\int\Big{|}\frac{T^{n_{j}}g}{n_{j}}\Big{|}^{r}d\mu=\Big{(}\frac{n_{j}+1}{n_{j}}\Big{)}^{r}\Big{(}\frac{1}{n_{j}+1}\Big{)}^{r}\int|g|^{r}d\mu\leq
2^{r}(\rho^{-r})^{j}\int|g|^{r}d\mu.$

By the triangle inequality in $L^{r}$, we have

$\int\Big{|}\sum_{j=1}^{\infty}\frac{T^{n_{j}}|g|}{n_{j}}\Big{|}^{r}d\mu\leq\sum_{j=1}^{\infty}\int\Big{(}\frac{T^{n_{j}}|g|}{n_{j}}\Big{)}^{r}d\mu\leq
2^{r}\int|g|^{r}d\mu\sum_{j=1}^{\infty}(\rho^{-r})^{j}<\infty.$

Hence the series $\displaystyle\sum_{j=1}^{\infty}\frac{T^{n_{j}}|g|}{n_{j}}$
converges a.e., so
$\displaystyle\frac{|T^{n_{j}}g|}{n_{j}}=\frac{T^{n_{j}}|g|}{n_{j}}\to 0$ a.e.

When $f=(I-T)g$ we have $\displaystyle M_{n_{j}}f=n_{j}^{-1}(g-T^{n_{j}}g)\to
0$ a.e. ∎

##  3\. Ergodic theorems in the Hardy spaces $H^{r}(\mathbb{T})$, $0<r<1$

In this section, the mean ergodic theorem for rotations is extended to
$H^{r}(\mathbb{T})$, $0<r<1$.

The Hardy space $H^{r}=H^{r}(\mathbb{D})$, $r>0$, is the linear space of all
holomorphic functions $f$ in the open unit disc $\mathbb{D}$ such that

$\sup_{R<1}\frac{1}{2\pi}\int_{0}^{2\pi}|f(Re^{it})|^{r}dt<+\infty.$

Since the left hand-side in the above is increasing with $R$ [8, Theorem 1.5],
we have the limit

$\|f\|_{H^{r}}^{r}:=\lim_{R\to
1^{-}}\frac{1}{2\pi}\int_{0}^{2\pi}|f(Re^{it})|^{r}dt<+\infty.$

When $f\in H^{r}$, it has radial limits $f(e^{it}):=\lim_{R\to
1^{-}}f(Re^{it})$ for a.e. $t\in[0,2\pi)$ (e.g. [25, Theorem VII.7.25]). By
Fatou’s lemma
$\frac{1}{2\pi}\int_{0}^{2\pi}|f(e^{it})|^{r}dt\leq\|f\|_{H^{r}}^{r}$, so
$f(e^{it})\in L^{r}(\mathbb{T})$. We denote by $H^{r}(\mathbb{T})$ the set of
all these radial limits. In fact,
$\frac{1}{2\pi}\int_{0}^{2\pi}|f(e^{it})|^{r}dt=\|f\|_{H^{r}}^{r}$ [25,
Theorem VII.7.32], [8, Theorem 2.6(i)], and where there is no risk of
confusion we write $\|f\|_{r}^{r}$ for both sides of the latter equality.
Moreover [8, Theorem 2.6], $f\in H^{r}$ satisfies

(7) $\lim_{R\to
1^{-}}\frac{1}{2\pi}\int_{0}^{2\pi}\big{|}f(Re^{it})-f(e^{it})\big{|}^{r}dt=0.$

For $R<1$, we denote by $\overline{\mathbb{C}_{R}}$ the circle of radius $R$.

For nice accounts on Hardy spaces, we refer to [25, Chap. VII, Vol I, pp.
271-287], [14, Chap. III], [8, Chap. 1].

It was observed in [25, Chapter VII, p. 284] that $H^{r}(\mathbb{T})$ can be
defined as the closure, in the $L^{r}$ metric, of the analytic trigonometric
polynomials, and when $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k}$ is in
$H^{r}(\mathbb{D})$, we have convergence of $\sum_{k=0}^{n}a_{k}e^{ikt}$ to
$f(e^{it})$ in $L^{r}(\mathbb{T})$ metric. Moreover, $H^{r}(\mathbb{T})$ is
complete, so is a closed subspace of $L^{r}(\mathbb{T})$. If $\theta$ is an
irrational rotation of the circle with induced operator $Tf=f\circ\theta$,
then $e^{ikt}$ is an eigenfunction of $T$ for any $k\in\mathbb{Z}$, hence
$H^{r}(\mathbb{T})$ is invariant under $T$.

In contrast to $L^{r}(\mathbb{T})$, here we can take advantage of the theorem
of Duren-Romberg-Shields (cf. [8, Theorem 7.5, p.115]), which describes the
dual space of $H^{r}$. According to that theorem, the dual space of $H^{r}$
can be identified with a certain space of Lipshitz functions
$\Lambda_{\alpha}$ for a certain $\alpha>0.$

We prove the following results.

######  Theorem 3.1.

Fix $0<r<1$ and let $T$ be induced by an ergodic irrational rotation $\theta$
of the circle. Then,

$H^{r}(\mathbb{T})=\mathcal{C}\oplus\overline{(I-T)H^{r}(\mathbb{T})},$

where $\mathcal{C}$ is the subspace of constant functions.

We further have the following ergodic theorem.

######  Theorem 3.2.

Fix $0<r<1$ and let $T$ be induced by an ergodic irrational rotation $\theta$
of the circle. Then, for each $f\in H^{r}(\mathbb{T})$ there is a constant
$a(f)\in\mathbb{C}$ such that

$\Big{\|}\frac{1}{N}\sum_{k=0}^{N-1}T^{k}f-a(f)\Big{\|}_{r}\to 0.$

Classically, the mean ergodic theorem for power-bounded operators in Banach
spaces is equivalent to the ergodic decomposition. But, here, since the Hahn-
Banach separation theorem fails, we need to give separate proofs for Theorem
3.2 and Theorem 3.1.

###### Proof of Theorem 3.1..

By Lemma 2.1 and the proof of Theorem 2.3, we proceed first by proving that
$\mathbf{1}\not\in\overline{(I-T)H^{r}}$. Assume, by contradiction, that
$\mathbf{1}\in\overline{(I-T)H^{r}}$. Then, there is a sequence of functions
$(g_{n})\subset H^{r}$, such that $g_{n}-Tg_{n}\to\mathbf{1}$ in $L^{r}$
metric.

By the Duren-Romberg-Shields’s theorem, elements $\psi$ in the dual of $H^{r}$
can be identified with certain analytic Lipschitz functions $g$ by the
relation

$\psi(f)=\lim_{R\to 1}\frac{1}{2\pi}\int_{0}^{2\pi}f(Re^{it})g(e^{-it})dt.$

Since $g\equiv 1$ is in that class, the linear form $\phi$ on $H^{r}$ given by

$\phi(f)=\lim_{R\to 1}\frac{1}{2\pi}\int_{0}^{2\pi}f(Re^{it})dt,$

is an element of the dual space, i.e. well-defined and continuous [8, p. 115].
Therefore,

$\phi(g_{n}-Tg_{n})\to\phi(\mathbf{1})=\mathbf{1}.$

But, for each $n\geq 1$,

$\displaystyle\phi(g_{n}-Tg_{n})$ $\displaystyle=\lim_{R\to
1}\Big{(}\frac{1}{2\pi}\int_{0}^{2\pi}g_{n}(Re^{it})-g_{n}(Re^{i\theta(t)})dt\Big{)}$
$\displaystyle=\lim_{R\to
1}\Big{(}\frac{1}{2\pi}\int_{0}^{2\pi}g_{n}(Re^{it})dt-\frac{1}{2\pi}\int_{0}^{2\pi}g_{n}(Re^{i\theta(t)})dt\Big{)}$
$\displaystyle=\lim_{R\to
1}\Big{(}\frac{1}{2\pi}\int_{0}^{2\pi}g_{n}(Re^{it})dt-\frac{1}{2\pi}\int_{0}^{2\pi}g_{n}(Re^{it})dt\Big{)}=0,$

The last equality since $\theta$ is a measure preserving transformation.
Whence

$\phi(g_{n}-Tg_{n})\to 0.$

This leads to a contradiction, so
$\mathbf{1}\not\in\overline{(I-T)H^{r}(\mathbb{T})}$.

Now, take $f\in H^{r}(\mathbb{T})$ and write, in the $L^{r}$ metric,

$f(e^{it})=\sum_{k=0}^{+\infty}a_{k}e^{ikt}.$

For each $K\in\mathbb{N}^{*}$ we have
$P_{K}(e^{it}):=\sum_{k=1}^{K}a_{k}e^{ikt}\in{(I-T)H^{r}(\mathbb{T})}$, since
for each $0\neq k\in\mathbb{Z}$, the function $t\mapsto e^{ikt}$ is an
eigenfunction with unimodular eigenvalue $\neq 1$. Moreover, by von Neumann’s
ergodic theorem, we have

$\lim_{N\to+\infty}\frac{1}{2\pi}\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N-1}P_{K}(e^{i\theta^{n}(t)})\Big{|}^{r}dt\leq\lim_{N\to+\infty}\Big{\|}\frac{1}{N}\sum_{n=1}^{N-1}P_{K}(e^{i\theta^{n}(t)})\Big{\|}_{2}^{r}=0.$

We thus get $f-a_{0}=L^{r}-\lim_{K}P_{K}\in\overline{(I-T)H^{r}(\mathbb{T})}.$
Hence $H^{r}(\mathbb{T})=\mathcal{C}+\overline{(I-T)H^{r}(\mathbb{T})}$; the
sum is a direct sum, since we have proved that
$\mathcal{C}\cap\overline{(I-T)H^{r}(\mathbb{T})}=\\{0\\}$. ∎

Remark. In Proposition 2.6 we proved for $T$ induced by an irrational rotation
that $\mathbf{1}\in\overline{(I-T)L^{r}(\mathbb{T})}$ when $0<r<\frac{1}{3}$;
the proof of Theorem 3.1 shows
$\mathbf{1}\not\in\overline{(I-T)H^{r}(\mathbb{T})}$.

For the the proof of Theorem 3.2, we need the following lemma; for its proof
we refer to [25, Chap. VII, Vol I pp. 284/5].

######  Lemma 3.3.

Let $r\in(0,1)$ and $f\in H^{r}$ such that $\big{\|}f\big{\|}_{r}\leq M$, for
some $M>0.$ Then, for $R<1$ and for $|z|\leq R,$ we have

$\big{|}f(z)\big{|}\leq\frac{M}{\big{(}1-R\big{)}^{\frac{1}{r}}}.$

We proceed now to the proof of Theorem 3.2.

###### Proof of Theorem 3.2..

Let $f(z)=\sum_{k=0}^{+\infty}a_{k}z^{k}$ and
$P_{K}(z)=\sum_{k=0}^{K}a_{k}z^{k}.$ Then, $\Big{\|}f-P_{K}\Big{\|}_{r}\to 0$
as $K\to\infty$. Fix $\epsilon>0$ and $K_{0}$ such that for each $K>K_{0}$, we
have

$\Big{\|}f-P_{K}\Big{\|}_{r}<\epsilon.$

Take $R<1$ sufficiently close to 1. Then, by Lemma 3.3, we have

$\big{|}f(z)-P_{K}(z)\big{|}<\frac{\epsilon}{\big{(}1-R\big{)}^{\frac{1}{r}}}\quad\rm{for}\quad|z|\leq
R.$

Therefore, the sequence $\Big{(}P_{K}(z)-f(z)\Big{)}_{K\geq 1}$ is uniformly
bounded in each compact subset of $|z|\leq R<1$. We thus apply Montel-
Stieltjes-Osgood theorem to extract a subsequence $K_{j}$ such that
$P_{K_{j}}(z)-f(z)$ converges uniformly to zero in each compact subset of
$|z|\leq R<1$. Whence, there is $j_{0}\in\mathbb{N}$ such that for each $j\geq
j_{0}$, for any $t\in[0,2\pi)$ we have

$\displaystyle\Big{|}P_{K_{j}}\big{(}Re^{i\theta^{n}(t)}\big{)}-f\big{(}Re^{i\theta^{n}(t)}\big{)}\Big{|}<\epsilon$

Hence, for any $N\geq 1,$ and for any $t\in[0,2\pi)$

(8)
$\displaystyle\Big{|}\frac{1}{N}\sum_{n=0}^{N-1}P_{K_{j}}\big{(}Re^{i\theta^{n}(t)}\big{)}-\frac{1}{N}\sum_{n=0}^{N-1}f\big{(}Re^{i\theta^{n}(t)}\big{)}\Big{|}<\epsilon$

But, by von Neumann’s mean ergodic theorem, we have

(9)
$\displaystyle\lim_{N\to+\infty}\Big{\|}\frac{1}{N}\sum_{n=0}^{N-1}P_{K_{j}}\big{(}e^{i\theta^{n}(t)}\big{)}-a_{0}\Big{\|}_{2}=0.$

Therefore, there is $N_{0}\in\mathbb{N}$, such that for any $N>N_{0}$, we have

(10)
$\displaystyle\int\Big{|}\frac{1}{N}\sum_{n=0}^{N-1}P_{K_{j}}\big{(}Re^{i\theta^{n}(t)}\big{)}-a_{0}\Big{|}^{r}dt$
$\displaystyle<\int\Big{|}\frac{1}{N}\sum_{n=0}^{N-1}P_{K_{j}}\big{(}e^{i\theta^{n}(t)}\big{)}-a_{0}\Big{|}^{r}dt$
(11)
$\displaystyle<\Big{\|}\frac{1}{N}\sum_{n=0}^{N-1}P_{K_{j}}\big{(}e^{i\theta^{n}(t)}\big{)}-a_{0}\Big{\|}_{2}^{r}<\epsilon.$

Combining (8) and (10) with the triangle inequality, we get, for any $N\geq
N_{0}$, for each $R<1$,

$\displaystyle\int\Big{|}\frac{1}{N}\sum_{n=0}^{N-1}f\big{(}Re^{i\theta^{n}(t)}\big{)}-a_{0}\Big{|}^{r}dt$
(12)
$\displaystyle\leq\int\Big{|}\frac{1}{N}\sum_{n=0}^{N-1}f\big{(}Re^{i\theta^{n}(t)}\big{)}-\frac{1}{N}\sum_{n=0}^{N-1}P_{K_{j}}\big{(}e^{i\theta^{n}(t)}\big{)}\Big{|}^{r}dt+\int\Big{|}\frac{1}{N}\sum_{n=0}^{N-1}P_{K_{j}}\big{(}e^{i\theta^{n}(t)}\big{)}-a_{0}\Big{|}^{r}dt$
$\displaystyle\leq\epsilon+\epsilon=2\epsilon.$

Whence, for any $N\geq N_{0}$,

$\int\Big{|}\frac{1}{N}\sum_{n=0}^{N-1}f\big{(}Re^{i\theta^{n}(t)}\big{)}-a_{0}\Big{|}^{r}dt<2\epsilon.$

This finishes the proof of the theorem. ∎

The following corollary is in contrast to Propositions 2.6 and 2.8.

######  Corollary 3.4.

Fix $0<r<1$. Let $T$ be induced by an ergodic irrational rotation $\theta$ of
the circle, and let $f\in H^{r}(\mathbb{T})$. Then
$f\in\overline{(I-T)H^{r}(\mathbb{T})}$ if and only if

$\int_{0}^{2\pi}\Big{|}\frac{1}{n}\sum_{k=0}^{n-1}(T^{k}f)(e^{it})\Big{|}^{r}dt\to
0.$

###### Proof.

Let $f\in\overline{(I-T)H^{r}(\mathbb{T})}$. Then the averages $M_{n}f$ are
also in $\overline{(I-T)H^{r}(\mathbb{T})}$. By the proof of Theorem 3.2
$M_{n}f\to a_{0}$ in $L^{r}$ metric, which yields
$a_{0}\in\overline{(I-T)H^{r}(\mathbb{T})}$, and by Theorem 3.1 $a_{0}=0$.

Assume now that $M_{n}f\to 0$ in $L^{r}$ norm. Since $H^{r}(\mathbb{T})$ is
invariant under $T$, the functions
$g_{n}:=f-M_{n}f=\frac{1}{n}\sum_{k=1}^{n-1}(I-T)\sum_{j=0}^{k-1}T^{j}f$,
defined in the proof of Theorem 2.3, are in $(I-T)H^{r}(\mathbb{T})$, and
$\|f-g_{n}\|_{r}\to 0$. ∎

######  Theorem 3.5.

Fix $0<r<1$. Let $T$ be induced on $H^{r}(\mathbb{T})$ by an ergodic
irrational rotation of the circle, and let $f\in H^{r}(\mathbb{T})$. Then
$f\in(I-T)H^{r}(\mathbb{T})$ if and only if

(13) $\sup_{N\geq
1}\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f(e^{it})\Big{|}^{r}dt<\infty.$

###### Proof.

Assume (13). By the construction, $f(e^{it})\in H^{r}(\mathbb{T})$ is a.e. the
boundary limit of $f(z)\in H^{r}$. By (13), there exists $M>0$ such that
$\|\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f\|_{r}\leq M$ for every
$N\geq 1$. By Lemma 3.3, for every $N\geq 1$ and $R<1$ we have

(14)
$\displaystyle\Big{|}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f(z)\Big{|}\leq\frac{M}{(1-R)^{1/r}}\qquad\forall\quad|z|\leq
R.$

Hence the sequence
$\big{\\{}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f(z)\big{\\}}\subset
H^{r}$ is uniformly bounded on any compact subset of $\mathbb{D}$. By the
Montel-Stieltjes-Osgood theorem, there is a subsequence $\\{N_{j}\\}$ such
that $g_{j}(z):=\frac{1}{N_{j}}\sum_{n=1}^{N_{j}}\sum_{k=0}^{n-1}T^{k}f(z)$
converges uniformly on compact subsets of $\mathbb{D}$, to a function $g$
continuous on $\mathbb{D}$. Since on any triangle in $\mathbb{D}$, $g$ is the
uniform limit of analytic functions, $g$ is analytic in $\mathbb{D}$ by
Morera’s theorem. Clearly $g_{j}\in H^{r}$, so for $R<1$, by Lebesgue’s
theorem and (13),

$\int_{0}^{2\pi}|g(Re^{it})|^{r}dt=\lim_{j\to\infty}\int_{0}^{2\pi}|g_{j}(Re^{it})|^{r}dt\leq\liminf_{j\to\infty}\int_{0}^{2\pi}|g_{j}(e^{it})|^{r}dt\leq
M^{r}.$

Hence $g\in H^{r}$, with $\|g\|_{r}\leq\liminf_{j}\|g_{j}\|_{r}\leq M$, so
$g(e^{it})\in H^{r}(\mathbb{T})$. We show that ${(I-T)g(e^{it})=f(e^{it})}$.

Since $g_{j}\to g$ converges uniformly on compact subsets of $\mathbb{D}$, it
follows that for $R<1$,

$\int_{0}^{2\pi}|g_{j}(Re^{it})-g(Re^{it})|^{r}dt\to_{j\to\infty}0.$

This with continuity of $T$ yields, with limits in
$L^{r}(\overline{\mathbb{C}_{R}})$ metric,

$(I-T)g(Re^{it})=\lim_{j}(I-T)g_{j}(Re^{it})=$
$\lim_{j}\frac{1}{N_{j}}\sum_{n=1}^{N_{j}}(I-T^{n})f(Re^{it})=f(Re^{it})-\lim_{j}M_{N_{j}}(T)f(Re^{it})$

Let $f(z)=\sum_{k=0}^{+\infty}a_{k}z^{k}$. Then, by (12), $M_{n}(T)f\to a_{0}$
in $L^{r}(\overline{\mathbb{C}_{R}})$ metric, so, for almost all $t$,

$[(I-T)g](Re^{it})=f(Re^{it})-a_{0}.$

Taking radial limits we obtain $(I-T)g=f-a_{0}$ a.e. on $\mathbb{T}$.

Claim: $a_{0}=0$.

Assume, by contradiction, that $a_{0}\neq 0$. Let $R<1$. Then, for almost all
$t$,

$(I-T)g(Re^{it})+a_{0}=f(Re^{it}).$

We can thus write

$\displaystyle\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f(Re^{it})\Big{|}^{r}dt$
$\displaystyle=\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N}\Big{(}(I-T^{n})g(Re^{it})-na_{0}\Big{)}\Big{|}^{r}dt$
$\displaystyle=\int_{0}^{2\pi}\Big{|}g(Re^{it})-\frac{1}{N}\sum_{n=1}^{N}T^{n}g(Re^{it})-\frac{(N+1)}{2}a_{0}\Big{|}^{r}dt$

Applying the triangle inequality, we get

$\frac{1}{2\pi}\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f(e^{it})\Big{|}^{r}dt\geq\frac{1}{2\pi}\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f(Re^{it})\Big{|}^{r}dt\geq$
$\frac{(N+1)^{r}}{2^{r}}|a_{0}|^{r}-\frac{1}{2\pi}\int_{0}^{2\pi}\Big{|}g(Re^{it})-\frac{1}{N}\sum_{n=1}^{N}T^{n}g(Re^{it})\Big{|}^{r}dt\geq$
$\frac{(N+1)^{r}}{2^{r}}|a_{0}|^{r}-\|g\|_{r}^{r}-\|M_{N}(T)g\|_{r}^{r}.$

By Theorem 3.2, $\\{\|M_{N}(T)g\|_{r}^{r}\\}$ is bounded; hence

$\liminf_{N}\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f(e^{it})\Big{|}^{r}dt=+\infty,$

which contradict (13). Hence $a_{0}=0$. We thus obtain $(I-T)g=f$ in
$H^{r}(\mathbb{T})$.

Assume now that $f\in(I-T)H^{r}(\mathbb{T})$, so $f=(I-T)g$. Then

$\frac{1}{2\pi}\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}T^{k}f(e^{it})\Big{|}^{r}dt=\frac{1}{2\pi}\int_{0}^{2\pi}\Big{|}\frac{1}{N}\sum_{n=1}^{N}\Big{(}(I-T^{n})g(e^{it})\Big{)}\Big{|}^{r}dt=$
$\frac{1}{2\pi}\int_{0}^{2\pi}\Big{|}g(e^{it})-M_{N}(T)g(e^{it})\Big{|}^{r}dt\leq\|g\|_{r}^{r}+\|M_{N}(T)g\|_{r}^{r}.$

Since $M_{N}(T)g$ converges in $H^{r}(\mathbb{T})$, the last term is bounded,
so (13) holds. ∎

Remark. When $f\in(I-T)H^{r}(\mathbb{T})$, computations as in the end of the
proof of Theorem 3.5 yield

(15) $\displaystyle\sup_{n\geq
1}\int_{0}^{2\pi}\Big{|}\sum_{k=0}^{n-1}T^{k}f(e^{it})\Big{|}^{r}dt<\infty,$

If we assume (15), then $f\in(I-T)L^{r}(\mathbb{T})$ by Theorem 2.4, but
neither [21] nor [1] (which also applies) yield an indication of the
”location” of the function $g\in L^{r}$ satisfying $(I-T)g=f$. The assumption
(15) yields that $M_{n}(T)f\to 0$ in $H^{r}(\mathbb{T})$, so $a_{0}=0$ by
Theorem 3.2, but we have not succeeded to deduce from that assumption that
$g\in H^{r}(\mathbb{T})$, nor that it implies (13).

######  Proposition 3.6.

Let $T$ be induced by an irrational rotation of the circle. Then there exists
$f$ such that $f\in(I-T)H^{r}(\mathbb{T})$ for each $0<r<1$, and $M_{n}(T)f$
does not converge a.e.

###### Proof.

We produce a function $g$ with $g\in H^{r}(\mathbb{T})$ for each $r<1$, such
that $\frac{T^{n}g}{n}$ does not converge a.e.

Let $g(z)=\frac{1}{1-z}$; we show that $g({\rm e}^{it})$ is in
$H^{r}(\mathbb{T})$, for each $r<1$. We first prove $g\in L^{r}(\mathbb{T})$:

$\int_{0}^{2\pi}\Big{|}\frac{1}{1-e^{it}}\Big{|}^{r}dt=2^{-r}\int_{0}^{2\pi}\Big{|}\frac{1}{\sin(\frac{t}{2})}\Big{|}^{r}dt=2^{1-r}\int_{0}^{\pi/2}|\frac{1}{\sin(t)}\Big{|}^{r}dt.$

But for each $t\in[0,\pi/2)$, $\sin(t)\geq\frac{2}{\pi}t$. Therefore

$\int_{0}^{2\pi}\Big{|}\frac{1}{1-e^{it}}\Big{|}^{r}dt=2^{1-r}\sum_{n=1}^{+\infty}\int_{\frac{\pi}{2(n+1)}}^{\frac{\pi}{2n}}\Big{|}\frac{1}{\sin(t)}\Big{|}^{r}dt\leq
2^{1-r}\sum_{n=1}^{+\infty}\int_{\frac{\pi}{2(n+1)}}^{\frac{\pi}{2n}}\frac{\pi^{r}}{2^{r}}\frac{1}{t^{r}}dt\leq$
$2^{1-2r}\pi^{r}\sum_{n=1}^{+\infty}\frac{1}{1-r}\Big{(}\Big{(}\frac{\pi}{2n}\Big{)}^{1-r}-\Big{(}\frac{\pi}{2(n+1)}\Big{)}^{1-r}\Big{)}\leq$
$\lim_{N\to\infty}2^{1-2r}\pi^{r}\frac{1}{1-r}\Big{(}\Big{(}\frac{\pi}{2}\Big{)}^{1-r}-\Big{(}\frac{\pi}{2(N+1)}\Big{)}^{1-r}\Big{)}\leq\frac{2^{1-3r}\pi}{1-r}.$

Thus $g\in L^{r}(\mathbb{T})$. We now show $g$ is in $H^{r}(\mathbb{T})$. Let
$P_{n}({\rm e}^{it})=\frac{1}{N}\sum_{n=1}^{N}\sum_{k=0}^{n-1}{\rm e}^{ikt}$.
Then for $0<t<2\pi$ we have

$P_{N}({\rm e}^{it})=(1-\frac{1}{N}\sum_{n=1}^{N}{\rm e}^{int})/(1-{\rm
e}^{it}).$

Hence $P_{N}({\rm e}^{it})\to g({\rm e}^{it})$ for $0<t<2\pi$, and $g({\rm
e}^{it})-P_{N}({\rm e}^{it})=(\frac{1}{N}\sum_{n=1}^{N}{\rm e}^{int})/(1-{\rm
e}^{it}).$ Thus, almost everywhere $|g({\rm e}^{it})-P_{N}({\rm
e}^{it}|^{r}\leq\frac{1}{|1-{\rm e}^{it}|^{r}}\in L^{1}(\mathbb{T})$. By
Lebesgue’s dominated convergence theorem, $\int_{0}^{2\pi}|g({\rm
e}^{it})-P_{N}({\rm e}^{it}|^{r}dt\to 0$. Thus $g$ is approximated in $L^{r}$
distance by analytic trigonometric polynomials, so $g\in H^{r}(\mathbb{T})$,

As in Proposition 2.13, we represent the transformation by $\theta
x=x+\alpha\mod 1$, with $\alpha$ irrational. Let $(q_{n})$ be the denominators
in the continued fractions representation of $\alpha$. As noted in the proof
of Proposition 2.13, for $x\in[0,1)$ there is $\ell_{n}(x)\in[0,q_{n}]$ such
that $x+\ell_{n}(x)\alpha\mod 1\in[0,2/q_{n}]$. Then, for any $x\in[0,1)$ we
have

$\Big{|}1-e^{2\pi i(\ell_{n}(x)\cdot\alpha+x)}\Big{|}\leq\frac{2}{q_{n}},$

and we obtain

$\frac{(T^{\ell_{n}(x)}g)({\rm e}^{2\pi
ix})}{\ell_{n}(x)}=\frac{1}{\ell_{n}(x)\Big{|}1-e^{i(\ell_{n}(x).\alpha+x)}\Big{|}}\geq\frac{q_{n}}{2\ell_{n}(x)}\geq\frac{1}{2},$

so $\limsup\Big{|}\frac{T^{n}g}{n}\Big{|}\geq\frac{1}{2}$ a.e. Since
$\int_{0}^{2\pi}\Big{|}\frac{T^{n}g}{n}\Big{|}^{r}dt=\int_{0}^{2\pi}\Big{|}\frac{g}{n}\Big{|}^{r}dt\to
0$, $\liminf\Big{|}\frac{T^{n}g}{n}\Big{|}=0$ a.e. by Fatou’s lemma. We
conclude that $\frac{T^{n}g}{n}$ does not converge a.e. To finish the proof,
we put $f=(I-T)g$ and the proof is complete. ∎

Remarks. 1\. By Corollary 3.4 and Fatou’s lemma, every
$f\in\overline{(I-T)H^{r}(\mathbb{T})}$ satisfies $\liminf|M_{n}f|=0$ a.e.

2\. Note that Proposition 2.14 applies to any $f\in(I-T)H^{r}(\mathbb{T})$.

3\. Here is an alternative proof that $1/(1-{\rm e}^{it})\in
H^{r}(\mathbb{T})$ for any $0<r<1$. We prove that $g(z)=\frac{1}{1-z}$ is in
$H^{r}(\mathbb{D})$, for each $r<1$. Indeed, for $0<R<1$, write

$\int_{0}^{2\pi}\Big{|}\frac{1}{1-Re^{it}}\Big{|}^{r}dt=\int_{0}^{2\pi}\Big{(}\Big{|}\frac{1}{1-Re^{it}}\Big{|}^{2}\Big{)}^{r/2}dt=\int_{-\pi}^{\pi}\Big{(}\frac{1}{1+2R\cos(s)+R^{2}}\Big{)}^{r/2}ds.$

The last term is obtained by change of variable $s=t-\pi$. Now, put
$u=\tan(s/2)$ to obtain

$\int_{0}^{2\pi}\Big{|}\frac{1}{1-Re^{it}}\Big{|}^{r}dt=\frac{1}{2}\int_{-\infty}^{+\infty}\Big{(}\frac{1}{1+2R\frac{1-u^{2}}{1+u^{2}}+R^{2}}\Big{)}^{r/2}\frac{du}{1+u^{2}}.$

But for each $u\in{\mathbb{R}}$,
$1+2R\frac{1-u^{2}}{1+u^{2}}+R^{2}=\Big{(}R+\frac{1-u^{2}}{1+u^{2}}\Big{)}^{2}+1-\Big{(}\frac{1-u^{2}}{1+u^{2}}\Big{)}^{2}\geq\frac{4u^{2}}{(1+u^{2})^{2}}$.
Therefore

$\sup_{0<R<1}\int_{0}^{2\pi}\Big{|}\frac{1}{1-Re^{it}}\Big{|}^{r}dt\leq\frac{1}{2}\int_{-\infty}^{\infty}\Big{(}\frac{(1+u^{2})^{2}}{4u^{2}}\Big{)}^{r/2}\frac{du}{1+u^{2}}=$
$\frac{1}{2^{r+1}}\int_{-\infty}^{\infty}\frac{1}{|u|^{r}}\frac{du}{(1+u^{2})^{1-r}}<+\infty.$

Hence $g\in H^{r}(\mathbb{D})$, for each $r<1$, so $g({\rm e}^{it})\in
H^{r}(\mathbb{T})$.

4\. Let $T=T_{\alpha}$ be induced by an irrational rotation of the circle, and
$g(z)=1/(1-z)$ for $z\in\mathbb{D}$. We showed above that $g(e^{it})\in
H^{r}(\mathbb{T})$. Since $g(z)=\sum_{n=0}^{\infty}z^{n}$ on $\mathbb{D}$, by
Theorem 3.2
$\int_{0}^{2\pi}\Big{|}\frac{1}{n}\sum_{k=0}^{n-1}T^{k}_{\alpha}g({\rm
e}^{it})-1\Big{|}^{r}dt\to 0$. Motivated by [22], for $\varepsilon>0$ we
define

$E_{n}:=\\{(\alpha,x):\Big{|}\frac{1}{n}\sum_{k=0}^{n-1}T^{k}_{\alpha}g({\rm
e}^{2\pi ix})-1\Big{|}>\varepsilon\\}\quad\text{and}\quad
E_{n,\alpha}:=\\{x:\Big{|}\frac{1}{n}\sum_{k=0}^{n-1}T^{k}_{\alpha}g({\rm
e}^{2\pi ix})-1\Big{|}>\varepsilon\\}.$

We denote by $\lambda$ Lebesgue’s measure on $[0,1)$. By Markov’s inequality,
for each irrational $\alpha$ in $[0,1]$ we obtain $\lambda(E_{n,\alpha})\to 0$
as $n\to\infty$. By Fubini’s theorem and Lebesgue’s theorem,

$\lambda\times\lambda(E_{n})=\int_{0}^{1}\lambda(E_{n,\alpha})d\lambda(\alpha)\to
0\quad\text{as}\quad n\to\infty,$

which means that $\displaystyle\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{1-{\rm
e}^{2\pi i(k\alpha+x)}}$ converges to 1 in $\lambda\times\lambda$ measure.
Then there exists a subsequence $(n_{j})$ such that
$\displaystyle\frac{1}{n_{j}}\sum_{k=0}^{n_{j}-1}\frac{1}{1-{\rm e}^{2\pi
i(k\alpha+x)}}$ converges to 1 $\ \lambda\times\lambda$ a.e. Hence the
limiting distribution in the Sinai-Ulcigrai Theorem [22, Theorem 1] is
$\delta_{1}$.

The proof of the next proposition shows that for $r<\frac{1}{2}$ there are
more functions $f$ satisfying the property in Proposition 3.6.

######  Proposition 3.7.

Let $T$ be induced by an irrational rotation of the circle. Then there exists
$f$ such that $f\in(I-T)H^{r}(\mathbb{T})$ for any $0<r<\frac{1}{2}$, and
$M_{n}(T)f$ does not converge a.e.

###### Proof.

Let $0\leq g\in L^{1}(\mathbb{T})$ such that $\limsup_{n}n^{-r}T^{n}g=\infty$
a.e. for any $r<1$, e.g. the function obtained in Proposition 2.13; we may
assume (by adding 1) that $g>0$ everywhere. Let

$g(R{\rm e}^{it}):=\sum_{n=-\infty}^{\infty}R^{|n|}\hat{g}(n){\rm
e}^{int},\qquad(0\leq R<1)$

where $\hat{g}(n)$ is the Fourier coefficient of $g$ at $n$, be the extension
of $g$ to the unit disc, and denote by $\tilde{g}$ its harmonic conjugate,
that is,

$\tilde{g}(Re^{it})=-i\sum_{n=-\infty}^{+\infty}\textrm{sgn}(n)R^{|n|}\hat{g}(n)e^{int},\qquad(0\leq
R<1)$

where $\textrm{sgn}(n)$ is the sign function. Then $\tilde{g}({\rm
e}^{it})=\lim_{R\to 1}\tilde{g}(Re^{it})$ is defined a.e. [14, Lemma 1.3, p.
64]. The function $G(z):=g(R{\rm e}^{it})+i\tilde{g}(R{\rm e}^{it})$ is
analytic on $\mathbb{D}$, and $\lim_{R\to 1}G(R{\rm e}^{it})=h({\rm
e}^{it}):=g({\rm e}^{it})+i\tilde{g}({\rm e}^{it})$ exists a.e. Since $g\in
L^{1}(\mathbb{T})$, by [14, Corollary 1.6, p. 66] (see the proof of [14,
Corollary 3.9, p. 87]), we have $h({\rm e}^{it}):=g({\rm
e}^{it})+i\tilde{g}({\rm e}^{it})\in H^{\beta}(\mathbb{{\mathbb{T}}})$ for any
$\beta<1$.

Fix $0<r<\frac{1}{2}$. Put $h_{r}({\rm e}^{it}):=h^{2}({\rm
e}^{it})=\lim_{R\to 1}[G(R{\rm e}^{it})]^{2}$. Since $G^{2}$ is analytic, and
$G\in H^{2r}(\mathbb{D})$, we have $G^{2}\in H^{r}(\mathbb{D})$. Then
$\lim_{R\to 1}G^{2}(R{\rm e}^{it})=h^{2}({\rm e}^{it})$ is in
$H^{r}(\mathbb{T})$. Since $\Re(h)\geq 1$, $\log h$ is well-defined, and
$h={\rm e}^{\frac{1}{2}2{\rm{Log}}h}=(h^{2})^{1/2}$. But
$\frac{T^{n}h_{r}}{n}$ does not converge a.e., since
$\big{(}\frac{T^{n}h_{r}}{n}\big{)}^{1/2}=\frac{T^{n}g+iT^{n}\tilde{g}}{n^{1/2}}$
does not converge. Consequently, when $f=(I-T)h^{2}$, $M_{n}(T)f$ does not
converge a.e. ∎

##  4\. Appendix by Guy Cohen: on rates of convergence

Proposition 2.10 raises the question of the rate of a.e. convergence to zero
of $\frac{1}{n}T^{n}g$ when $g\in L^{1}(\mu)$. When $g\in L^{p}$,
$1<p<\infty$, we have $n^{-1/p}T^{n}g\to 0$ a.e., so we have a rate of
$1/n^{1-1/p}$. Proposition 2.10 yields $h\in L^{1}$ which does not have this
rate for any $p>1$. The following proposition extends Proposition 2.10, with
similar use of Rokhlin’s lemma.

######  Proposition 4.1.

Let $T$ be induced by an ergodic aperiodic measure preserving transformation
$\theta$ on a Lebesgue space $(X,\Sigma,\mu)$, and let $(b_{n})$ be a sequence
tending to infinity, such that $b_{n}/n$ decreases to zero. Then there exists
$0\leq h\in L^{1}(\mu)$ such that $\limsup_{n\to\infty}b_{n}\cdot
n^{-1}T^{n}h=\infty$ a.e.

###### Proof.

Put $a_{n}=n/b_{n}$. Then $a_{n}$ increases to infinity and $a_{n}/n$ tends to
zero. For a fast increasing sequence $(\ell_{n})\subset\mathbb{N}$ (to be
determined later), let $k_{n}$ be the first index such that
$a_{k_{n}}\geq\ell_{n}$, so $k_{n}$ is increasing, and $k_{n}^{-1}\ell_{n}\leq
k_{n}^{-1}a_{k_{n}}\to 0$.

Let $\varepsilon_{n}$ decrease to zero. By Rokhlin’s lemma there is
$B_{n}\in\Sigma$ such that $(\theta^{-j}B_{n})_{0\leq j<k_{n}}$ are disjoint
and $\mu(X_{n})>1-\varepsilon_{n}$, where
$X_{n}=\cup_{j=0}^{k_{n}-1}\theta^{-j}B_{n}$. We define
$h=\sum_{n=1}^{\infty}n^{-2}\frac{1}{\mu(B_{n})}1_{B_{n}}$, so $0\leq h\in
L^{1}$. For $x\in X_{n}\setminus B_{n}$ there exists $1\leq j\leq k_{n}-1$
such that $x\in\theta^{-j}B_{n}$, and then

$\frac{T^{j}h(x)}{a_{j}}\geq\frac{h(\theta^{j}x)}{a_{k_{n}-1}}\geq\frac{n^{-2}}{\mu(B_{n})}\cdot\frac{1}{a_{k_{n}-1}}\geq\frac{n^{-2}}{\mu(B_{n})}\cdot\frac{1}{\ell_{n}}\geq\frac{k_{n}}{n^{2}\ell_{n}}.$

The sequence $(a_{n})$ is given (through $(b_{n})$ ), and the sequence
$(\ell_{n})$ can be chosen to grow fast enough so
$\frac{k_{n}}{n^{2}\ell_{n}}\to\infty$ (see below). Then for $M>0$ we have
$\frac{k_{n}}{n^{2}\ell_{n}}>M$ for $n>N_{M}$. We then have

$b_{j}\cdot\frac{T^{j}h(x)}{j}=\frac{T^{j}h(x)}{a_{j}}>M\qquad\qquad(n>N_{M},\
x\in X_{n}\setminus B_{n}).$

Hence

$\mu\Big{(}\Big{\\{}x:\sup_{j\geq
1}b_{j}\cdot\frac{T^{j}h(x)}{j}>M\Big{\\}}\Big{)}\geq\mu(X_{n}\setminus
B_{n})=(k_{n}-1)\mu(B_{n})>1-\varepsilon_{n}-k_{n}^{-1}\qquad(n>n_{M}).$

Letting $n\to\infty$ we obtain $\sup_{j}b_{j}\cdot j^{-1}T^{j}h>M$ a.e. Since
this is for any $M>0$, we obtain $\limsup_{j\to\infty}b_{j}\cdot
j^{-1}T^{j}h=\infty$ a.e.

We now show how to choose $(\ell_{n})$. We define an ”inverse” of $(a_{n})$ by
$c_{n}=\min\\{m:a_{m}\geq n\\}$. It is easy to show that $c_{n}$ is increasing
to infinity. When $c_{n}=m$, we have $a_{m-1}<n\leq a_{m}$, so

$\frac{c_{n}}{n}=\frac{m}{n}\geq\frac{m}{a_{m}}\to\infty.$

Fix $n$; then for $j>J_{n}$ we have $j^{-1}c_{j}>n^{3}$. Put
$\ell_{n}=J_{n}+1$. By definition $k_{n}=c_{\ell_{n}}$, so we have

$\frac{k_{n}}{n^{2}\ell_{n}}=\frac{c_{\ell_{n}}}{n^{2}\ell_{n}}>n,$

hence $\frac{k_{n}}{n^{2}\ell_{n}}\to\infty$. ∎

Remarks. 1\. The function $h$ depends in general on $(b_{n})$, since
$(\ell_{n})$ and $(k_{n})$ do.

2\. In Proposition 2.10, obtained when $b_{n}=n^{1-r}$, $0<r<1$, $h$ does not
depend on $r$.

3\. Proposition 4.1 does not apply to the sequence $b_{n}=n$. However, for any
unbouded $g\in L^{1}$, infinite returns of $\theta^{n}x$ to the set
$\\{y:|g(y)|>M\\}$ show that $\limsup_{n}|T^{n}g|\geq M$ a.e. Hence
$\limsup_{n}|T^{n}g|=\infty$ a.e. (this is the idea in [11, p. 391] for
proving $\limsup_{n}|T^{n}g|=\|g\|_{\infty}$ for $g\in L_{\infty}$).

######  Corollary 4.2.

Let $T$ be induced by an ergodic aperiodic measure preserving transformation
$\theta$ on a Lebesgue space $(X,\Sigma,\mu)$, and let $0<\delta_{n}\to 0$. If
there exists $(\gamma_{n})$ such that $\gamma_{n}\to 0$, $\ n\gamma_{n}$
increase to $\infty$, and $\delta_{n}\leq\gamma_{n}$ for every $n$, then there
exists an $L^{1}$-coboundary $f\in(I-T)L^{1}$ such that
$\limsup_{n\to\infty}\delta_{n}^{-1}|M_{n}f|\to\infty$ a.e.

###### Proof.

Let $b_{n}=1/\gamma_{n}$; then $(b_{n})$ satisfies the assumptions of
Proposition 4.1, and let $h$ be the function obtained in Proposition 4.1. Then
$f=(I-T)h$ satisfies

$\limsup_{n\to\infty}\delta_{n}^{-1}|M_{n}f|\geq\limsup_{n\to\infty}\gamma_{n}^{-1}|M_{n}f|=\limsup_{n}b_{n}n^{-1}|h-T^{n}h|=\infty\quad
a.e.$

∎

Remark. Krengel [17] and Kakutani-Petersen [13] proved that for any
$0<\delta_{n}\to 0$ (with no additional conditions) there exists a bounded $f$
such that $\limsup_{n\to\infty}\delta_{n}^{-1}|M_{n}f|\to\infty$ a.e. The
novelty in Corollary 4.2 is in obtaining $f$ a coboundary.

######  Proposition 4.3.

Let $T$ be induced by an ergodic aperiodic probability preserving
transformation $\theta$ on a Lebesgue space $(X,\Sigma,\mu)$. and let
$(b_{n})$ be a sequence tending to infinity, such that $b_{n}/n$ decreases to
zero. Then there exists a dense $G_{\delta}$ subset ${\mathcal{F}}\subset
L^{1}(X,\mu)$, such that $\limsup_{n}b_{n}\cdot n^{-1}|T^{n}h|=\infty$ a.e.
for any $h\in{\mathcal{F}}$.

The proof is similar to that of Proposition 2.12, using Proposition 4.1.

Acknowledgement. The first author is grateful to the Center for Advanced
Studies in Mathematics of Ben-Gurion University for its hospitality and
support. The second author is grateful to the University of Rouen-Normandie
for its hospitality and support.

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