# Low complexity of optimizing measures over an expanding circle map

Rui Gao  Rui Gao: College of Mathematics, Sichuan University, Chengdu 610064,
China<EMAIL_ADDRESS>and  Weixiao Shen  Weixiao Shen: Shanghai Center
for Mathematical Sciences, Jiangwan Campus, Fudan University, No 2005 Songhu
Road, Shanghai 200438, China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we prove that for real analytic expanding circle maps, all
optimizing measures of a real analytic potential function have zero entropy,
unless the potential is cohomologous to constant. We use the group structure
of the symbolic space to solve a transversality problem involved. We also
discuss applications to optimizing measures for generic smooth potentials and
to Lyapunov optimizing measures.

##  1\. Introduction

Given a continuous map $T:X\to X$ from a compact metric space into itself and
a continuous function $f:X\to\mathbb{R}$, the problem of ergodic optimization
looks for maximization/minimization of $\int f\>\\!\mathrm{d}\mu$, where $\mu$
runs over the collection $\mathcal{M}(T)$ of all $T$-invariant Borel
probability measures. Let

$\beta(T,f)=\beta(f)=\sup_{\mu\in\mathcal{M}(T)}\int
f\>\\!\mathrm{d}\mu,\quad\alpha(T,f)=\alpha(f)=\inf_{\mu\in\mathcal{M}(T)}\int
f\>\\!\mathrm{d}\mu.$ (1)

By Birkhoff’s Ergodic Theorem, these quantities also optimize the time average
of $f$ along orbits.

A measure $\mu\in\mathcal{M}(T)$ is called a maximizing measure if
$\beta(f)=\int f\>\\!\mathrm{d}\mu$ and an orbit
$\\{T^{n}(x)\\}_{n=0}^{\infty}$ is called a maximizing orbit if
$\beta(f)=\lim_{n}\frac{1}{n}\sum_{i=0}^{n-1}f(T^{i}(x)).$ Minimizing
measures/orbits are similarly defined. As $-\alpha(f)=\beta(-f)$, the problems
of maximization and minimization are equivalent. In this paper, we shall
mainly discuss the maximizing problem.

Numerical experiments by Hunt-Ott in [11, 12] indicate that typically
maximizing orbits are periodic and maximizing measures are supported on
periodic orbits or at least of zero entropy. The mathematical theory of
ergodic optimization has been developed since then, mainly in the case that
$T$ is hyperbolic. Yuan-Hunt [23] dealt with the case that $T$ is either an
Axiom A diffeomorphism or a non-invertible uniformly expanding map, and $f$ is
Lipschitz, and proved that only periodic orbits (measures) can be persistently
maximal in the Lipschitz topology. Building upon [23, 5, 19, 21], Contreras
[6] proved that for $T$ uniformly expanding, a generic $f$ has a unique
maximizing measure which is supported on a periodic orbit in the Lipschitz
topology. Recently, Huang et al [9] extended the results and proved that when
$T$ is either uniformly expanding or Axiom A, for a generic $f$ in the $C^{1}$
topology, the maximizing measure is unique and supported on a periodic orbit.
However, just as in many other circumstances in the state-of-the-art theory of
dynamical systems, the local perturbation technique is tied to the $C^{1}$ (or
coarser) topology.

The simplest uniformly expanding maps are probably the maps
$T:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$, $x\mapsto dx\mod 1$ with
$d\geq 2$. This case has received much attention. When $d=2$, for
$f_{\theta}(x)=\cos(2\pi(x-\theta))$, $\theta\in\mathbb{R}$, Bousch [2] proved
that the maximizing measure of $f_{\theta}$ is unique and is a Sturmian
measure which, in particular, has entropy zero. See [14, 15, 8] for similar
results in this direction.

We refer to [13, 1, 16] for development on other aspects of ergodic
optimization.

The goal of this paper is to prove the following theorem.

######  Main Theorem.

Let $T:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ be a real analytic
expanding map and let $f:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ be a real
analytic function. Then

  * •

either $f$ is real analytically cohomologous to constant, i.e. there exists a
real analytic function $g$ and a constant $c$ such that $f=g\circ T-g+c$;

  * •

or any optimizing measure of $f$ is of zero entropy.

In particular, the Main Theorem gives a complete solution to a problem raised
by O. Jenkinson, see [13, Problem 3.12], which was also mentioned in [1, page
1847].

Morris [19] proved that for generic Hölder or Lipschitz functions, there is a
unique maximizing measure and this measure has entropy zero. This result
played an important role in the work [6] cited above which proves the
outstanding TPO conjecture (TPO means “typically maximizing measures are
supported on a periodic orbit”) in the Lipschitz topology. Our Main Theorem
implies that Morris’ result remains true for $C^{r}$ functions for any
$r\in\\{0,1,2,\ldots,\infty\\}$, see Theorem 1.2. It is our hope that our Main
Theorem will throw some insight in proving the TPO conjecture in $C^{r}$
topology with $r\geq 2$ in the case that the underlying dynamics is an
expanding circle map.

One of the key ingredients in our proof is a transversality result. To reduce
the technicality, we shall only state a special version that is needed for the
proof of the Main Theorem here. See Theorem 2.1 for the general statement. Let
$T:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ and
$f:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ be as in the Main Theorem. We assume
that $T$ is orientation preserving and that $T$ fixes $0$. Let
$\widehat{T}:\mathbb{R}\to\mathbb{R}$ denote the unique lift of $T$ with
$\widehat{T}(0)=0$, via the natural projection
$\pi:\mathbb{R}\to\mathbb{R}/\mathbb{Z}$. Let $\tau=\widehat{T}^{-1}$ and
$\tau_{i}(x)=\tau(x+i)$ for each $i\in\mathbb{Z}$. Let
$\widehat{f}=f\circ\pi$. Let $\mathbb{N}=\\{1,2,\ldots\\}$. For each
$\mathbf{i}=(i_{n})_{n\geq
1}\in\Sigma_{d}:=\\{0,1,\ldots,d-1\\}^{\mathbb{N}},$ consider

$h_{\mathbf{i}}(x):=\sum_{n=1}^{\infty}\left(\widehat{f}\circ\tau_{i_{n}}\circ\tau_{i_{n-1}}\circ\cdots\circ\tau_{i_{1}}(x)-\widehat{f}\circ\tau_{i_{n}}\circ\tau_{i_{n-1}}\circ\cdots\circ\tau_{i_{1}}(0)\right),\quad
x\in\mathbb{R},$

which is well-defined and real analytic on $\mathbb{R}$.

######  Theorem 1.1 (Transversality).

Under the above circumstances, we have

  * •

either $f$ is real analytically cohomologous to constant and
$h_{\mathbf{i}}\equiv h_{\mathbf{j}}$ holds for all
$\mathbf{i},\mathbf{j}\in\Sigma_{d}$;

  * •

or $h_{\mathbf{i}}\not\equiv h_{\mathbf{j}}$ holds whenever
$\mathbf{i},\mathbf{j}\in\Sigma_{d}$ and $\mathbf{i}\not=\mathbf{j}$.

We prove this theorem using the odometer structure of the space $\Sigma_{d}$.
An important observation is that
$G_{0}:=\\{\mathbf{i}\in\Sigma_{d}:h_{\mathbf{i}}=h_{\mathbf{0}}\\}$ is a
closed subgroup of $\Sigma_{d}$, where $\mathbf{0}$ denotes the element of
$\Sigma_{d}$ whose entries are all $0$. When $d$ is prime, this immediately
implies that either $G_{0}=\\{\mathbf{0}\\}$ or $G_{0}=\Sigma_{d}$. The same
conclusion remains true for an arbitrary integer $d\geq 2$ which will be dealt
with by Fourier analysis. See Proposition 2.3. Note that this kind of
transversality problem appears naturally in the study of skew product over
circle expanding maps. In particular, a special case of this type of dichotomy
was obtained in [22, Theorem A] using a different method, which represents an
important step in the study of a dimension dichotomy for the graphs of
Weierstrass-type functions. The method here provides a much more general
result with simpler proofs.

Let us explain how Theorem 1.1 is used in the proof of the Main Theorem.
Assume that $f$ is not real analytically cohomologous to a constant. Then the
second alternative of Theorem 1.1 holds. By ergodic decomposition, we only
need to show that any ergodic maximizing measure $\mu$ has zero entropy. Let
$S$ denote the support of $\mu$, We may assume that $T(0)=0$ and $0\not\in S$
so that $S$ can be identified naturally with a compact subset of $(0,1)$. The
space of inverse orbits in $S$ is naturally identified with

$\mathcal{S}=\\{((i_{n})_{n\geq 1},x):i_{n}\in\\{0,1,\ldots,d-1\\},x\in
S,\tau_{i_{n}}\tau_{i_{n-1}}\cdots\tau_{i_{1}}(x)\in S,\forall n\\}.$

With the help of a sub-action, we deduce from Theorem 1.1 that if the section

$\mathcal{S}_{\mathbf{i}}=\\{x\in S:(\mathbf{i},x)\in\mathcal{S}\\}$

has a limit point $x_{0}$, then $\mathbf{i}$ belongs to a set which has at
most two elements and which is uniquely determined by $x_{0}$, see Proposition
3.2. (It is here that we need real analyticity of $f$ and $T$.) This enables
us to apply well-known results in dimension theory of dynamical systems (for
example [18]) to conclude that $\mu$ has entropy zero.

Now we state a few corollaries of the Main Theorem.

######  Theorem 1.2.

Let $T:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ be expanding and real
analytic. Let $r\in\\{0,1,2,\cdots,\infty\\}$. Then for generic $f\in
C^{r}(\mathbb{R}/\mathbb{Z})$, any optimizing measure of $f$ is of zero
entropy.

As mentioned before, Morris [19] proved a similar result for Hölder and
Lipschitz functions, using a result of Bressaud and Quas [5]. Our argument is
based on our Main Theorem and the fact that $C^{r}$ functions can be
approximated by real analytic ones. It is worthy noting that the $C^{0}$ case
is particularly special, as generically, the unique optimizing measure is
fully supported but has zero entropy, see [3, 17, 4].

The special case $f=\pm\log|T^{\prime}|$ is often referred to as Lyapunov
optimization, see [7]. Our result has the following consequences.

######  Theorem 1.3.

Let $T:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ be expanding and real
analytic. Then

  * •

either there exists a real analytic diffeomorphism
$\phi:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ such that $\phi\circ
T\circ\phi^{-1}(x)=\deg(T)\cdot x$;

  * •

or any Lyapunov optimizing measure of $T$ is of zero entropy.

For $r\in\\{1,2,3,\cdots,\infty\\}$, let $\mathcal{C}^{r}$ denote the
collection of $C^{r}$ maps from $\mathbb{R}/\mathbb{Z}$ to itself, endowed
with the $C^{r}$ topology. Let $\mathcal{E}^{r}$ be the open subset of
$\mathcal{C}^{r}$ consisting of $C^{r}$ expanding maps on
$\mathbb{R}/\mathbb{Z}$.

######  Theorem 1.4.

Let $r\in\\{1,2,3,\cdots,\infty\\}$. For generic $T\in\mathcal{E}^{r}$, any
Lyapunov optimizing measure of $T$ is of zero entropy.

In [10], it is proved that for generic $T\in\mathcal{E}^{2}$, a Lyapunov
optimizing measure is supported on a periodic orbit. See [7, 17] for earlier
results on $C^{1+\alpha}$ and $C^{1}$ cases respectively.

The rest of the paper is organized as follows. In § 2, we prove Theorem 1.1
and a more general transversality result; in § 3, we prove the Main Theorem;
in § 4, we prove Theorems 1.2, 1.3 and 1.4.

Acknowledgment. RG would like to thank Bing Gao and Zeng Lian for helpful
discussions. We would also like to thank Jairo Bochi for reading a first
version of the paper and several useful remarks. WS is supported by NSFC grant
No. 11731003.

##  2\. A transversality result

In this section, we state and prove a more general version of Theorem 1.1. The
proof exploits the structure of $\Sigma_{d}=\\{0,1,\ldots,d-1\\}^{\mathbb{N}}$
as a compact abelian group.

###  2.1. Statement of result

We start with the statement of the general transversality result. Let
$T:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ be a $C^{1}$ orientation-
preserving expanding map with $T(0)=0$ and let $d\geq 2$ be the degree. Let
$f:\mathbb{R}/\mathbb{Z}\to{\mathbb{C}}$ be a $C^{\alpha}$ function for some
$\alpha\in(0,1)$. As before, denote by $\widehat{T}:\mathbb{R}\to\mathbb{R}$
the lift of $T$ with $\widehat{T}(0)=0$, $\tau=\widehat{T}^{-1},$ and
$\tau_{i}(x)=\tau(x+i)$. Given
$\mathbf{i}=(i_{1},\cdots,i_{n},\cdots)\in\Sigma_{d}$, for any $n\geq 1$,
denote:

$\tau_{\mathbf{i},n}:=\tau_{i_{n}}\circ\cdots\circ\tau_{i_{2}}\circ\tau_{i_{1}}.$

Let
$\lambda_{0}=\sup_{n=1}^{\infty}\min_{x\in\mathbb{R}}|(\widehat{T}^{n})^{\prime}(x)|^{1/n}>1$.
111By [20, Theorem A.3], $\log\lambda_{0}=\alpha(T,\log|T^{\prime}|)$, where
$\alpha$ is as in (1). Note that for any $0<\lambda_{1}<\lambda_{0}$, there
exists $C\geq 1$ such that

$\sup_{x\in\mathbb{R},\mathbf{i}\in\Sigma_{d}}|\tau^{\prime}_{\mathbf{i},n}(x)|\leq
C\lambda_{1}^{-n},\quad\forall~{}n\geq 0.$

Fix a $\lambda\in\mathbb{C}$ such that $0<|\lambda|<\lambda_{0}^{\alpha}$.
Then, for each $\mathbf{i}\in\Sigma_{d}$,

$h_{\mathbf{i}}(x):=\sum_{n=1}^{\infty}\lambda^{n}\left(\widehat{f}\circ\tau_{\mathbf{i},n}(x)-\widehat{f}\circ\tau_{\mathbf{i},n}(0)\right),\quad\quad
x\in\mathbb{R},$

is a well-defined continuous function from $\mathbb{R}$ to $\mathbb{C}$, where
$\widehat{f}=f\circ\pi$ and $\pi:\mathbb{R}\to\mathbb{R}/\mathbb{Z}$ is the
natural projection. The following theorem is our main transversality result.

######  Theorem 2.1.

Under the circumstances above, the following are equivalent:

  1. (i)

there exist $\mathbf{i}\neq\mathbf{j}$ with $h_{\mathbf{i}}\equiv
h_{\mathbf{j}}$;

  2. (ii)

$h_{\mathbf{i}}=h_{\mathbf{0}}$ for any $\mathbf{i}$;

  3. (iii)

$h_{\mathbf{0}}$ is $1$-periodic;

  4. (iv)

$h_{\mathbf{0}}$ is $1$-periodic and $\widehat{f}=\lambda^{-1}\cdot
h_{\mathbf{0}}\circ\widehat{T}-h_{\mathbf{0}}+\widehat{f}(0)$;

  5. (v)

there exists a $C^{\alpha}$ function $\phi:\mathbb{R}/\mathbb{Z}\to\mathbb{C}$
such that $f=\lambda^{-1}\cdot\phi\circ T-\phi+const$.

###### Proof of Theorem 1.1 assuming Theorem 2.1.

Assume that the second alternative in the conclusion of Theorem 1.1 does not
hold. Then (i) and hence (ii)-(v) all hold in Theorem 2.1 (for $\lambda=1$).
By (iv), $f$ is real analytically cohomologous to constant since
$h_{\mathbf{0}}$ is real analytic. By (ii), $h_{\textbf{i}}=h_{\textbf{j}}$
for all $\mathbf{i},\mathbf{j}\in\Sigma_{d}$. Thus the first alternative in
the conclusion of Theorem 1.1 holds. ∎

###  2.2. Group structure of $\Sigma_{d}$

Let us recall the well-known structure of $\Sigma_{d}$ as a compact abelian
group. The topology on $\Sigma_{d}$ is the product topology of the discrete
topology on $\\{0,1,\ldots,d-1\\}$. The space $\Sigma_{d}$ becomes a compact
abelian group with addition defined as follows. For $\mathbf{i}=(i_{n})_{n\geq
1},\mathbf{j}=(j_{n})_{n\geq 1}\in\Sigma_{d}$,
$\mathbf{i}+\mathbf{j}\in\Sigma_{d}$ is the unique element
$\mathbf{k}=(k_{n})_{n\geq 1}\in\Sigma_{d}$ such that for each $n=1,2,\ldots$,

$k_{1}+k_{2}d+\cdots+k_{n}d^{n-1}=(i_{1}+j_{1})+(i_{2}+j_{2})d+\cdots+(i_{n}+j_{n})d^{n-1}\mod
d^{n}.$

Note that the semi-group $\mathbb{Z}_{+}=\\{0,1,2,\cdots\\}$ can be naturally
embedded into $\Sigma_{d}$ as a dense sub-semi-group in the following way:

$m=i_{1}+i_{2}d+\cdots+i_{n}d^{n-1}\mapsto(i_{1},\cdots,i_{n},0,0,\cdots)=:\iota(m).$

In particular, $\mathbf{0}=\iota(0)$ is the zero element of $\Sigma_{d}$. The
map $\mathbf{i}\mapsto\mathbf{i}+\iota(1)$ is usually called a ($d$-adic)
adding machine.

The following lemma is a simple observation which plays an important role in
our proof.

######  Lemma 2.2.

Given $\mathbf{i},\mathbf{j},\mathbf{k}\in\Sigma_{d}$,
$h_{\mathbf{i}}=h_{\mathbf{j}}$ if and only if
$h_{\mathbf{i}\mathbf{+}\mathbf{k}}=h_{\mathbf{j}+\mathbf{k}}$ . As a result,
$G_{0}:=\\{\mathbf{i}:h_{\mathbf{i}}=h_{\mathbf{0}}\\}$ is a closed subgroup
of $\Sigma_{d}$.

###### Proof.

Since $\Sigma_{d}$ is a group, we only need to prove the “only if” part. By
definition, for each positive integer $m$,

$\tau_{\mathbf{i},n}(x+m)=\tau_{\mathbf{i}+\iota(m),n}(x)\mod
1,\quad\forall~{}\mathbf{i},~{}\forall~{}n\geq 1~{},\forall~{}x\in\mathbb{R}.$

Since $\widehat{f}$ is of period $1$, it follows that for $m\geq 1$, we have:

$h_{\mathbf{i}}(x+m)=h_{\mathbf{i}+\iota(m)}(x).$

In particular,

$h_{\mathbf{i}}=h_{\mathbf{j}}\Rightarrow
h_{\mathbf{i}+\iota(m)}=h_{\mathbf{j}+\iota(m)}.$

Since $\iota(\mathbb{Z}_{+})$ is dense in $\Sigma_{d}$ and since
$\mathbf{k}\mapsto h_{\mathbf{k}}(x)$ is continuous for any fixed
$x\in\mathbb{R}$, it follows that $h_{\mathbf{i}}=h_{\mathbf{j}}$ implies that
$h_{\mathbf{i}+\mathbf{k}}=h_{\mathbf{j}+\mathbf{k}}$.

In particular, $G_{0}$ is a subgroup of $\Sigma_{d}$. Its closedness follows
from the fact that $\mathbf{i}\mapsto h_{\mathbf{i}}(x)$ is continuous for
each $x$. ∎

######  Proposition 2.3.

Under the circumstances of Theorem 2.1, (i) implies (iii).

The proof of this proposition will be given in the next subsection, using
Fourier analysis. Here we give a short proof in the case that $d$ is a prime.

###### Proof of Proposition 2.3 assuming that $d$ is a prime.

It suffices to prove that $\iota(1)\in
G_{0}=\\{\mathbf{i}\in\Sigma_{d}:h_{\mathbf{i}}=h_{\mathbf{0}}\\}$, because
$h_{\iota(1)}=h_{\mathbf{0}}$ means that $h_{\mathbf{0}}$ is $1$-periodic.
Since (i) holds, by Lemma 2.2, $G_{0}$ is a non-trivial closed subgroup of
$\Sigma_{d}$. For each $m\geq 1$,

$A_{m}:=\\{i_{1}+i_{2}d+\cdots+i_{m}d^{m-1}\mod d^{m}:(i_{n})_{n\geq 1}\in
G_{0}\\}$

is a subgroup of the cyclic group $\mathbb{Z}/(d^{m}\mathbb{Z})$. Since the
group $G_{0}$ also has the following property:

$0i_{1}i_{2}\cdots\in G_{0}\implies i_{1}i_{2}\cdots\in G_{0},$

$A_{m}$ is non-trivial for each $m$. Since $d$ is prime, there exists a unique
$k_{m}\in\\{0,1,\ldots,m-1\\}$ such that $A_{m}$ is generated by
$d^{k_{m}}\mod d^{m}$. In particular, $k_{1}=0$. Since $j\mod d^{m+1}\mapsto
j\mod d^{m}$ induces a surjective homomorphism from $A_{m+1}$ to $A_{m}$,
$d^{k_{m+1}}\mod d^{m}$ is also a generator of $A_{m}$, and thus
$k_{m+1}=k_{m}$. In conlcusion, $k_{m}=0$ for all $m\geq 1$. Since $G_{0}$ is
closed, it follows that $\iota(1)\in G_{0}$. ∎

###### Proof of Theorem 2.1.

(ii) $\implies$ (i). This is trivial.

(i) $\implies$ (iii). This is Proposition 2.3.

(iii) $\implies$ (iv). Let $\widehat{f}_{0}=\widehat{f}-\widehat{f}(0)$. By
definition,

$h_{\mathbf{0}}=\sum_{n=1}^{\infty}\lambda^{n}\cdot\widehat{f}_{0}\circ\tau^{n}.$

Then

$h_{\mathbf{0}}\circ\widehat{T}=\sum_{n=1}^{\infty}\lambda^{n}\cdot\widehat{f}_{0}\circ\tau^{n-1}=\lambda(\widehat{f}_{0}+h_{\mathbf{0}}).$

(iv) $\implies$ (v). This is trivial.

(v) $\implies$ (ii). Denote $\tau_{\mathbf{i},0}=\mathrm{id}$ and
$\widehat{\phi}=\phi\circ\pi$. There exists $c\in\mathbb{C}$ such that for any
$\mathbf{i}$ and any $n\geq 1$,

$\widehat{f}\circ\tau_{\mathbf{i},n}=\lambda^{-1}\widehat{\phi}\circ\widehat{T}\circ\tau_{\mathbf{i},n}-\widehat{\phi}\circ\tau_{\mathbf{i},n}+c=\lambda^{-1}\widehat{\phi}\circ\tau_{\mathbf{i},n-1}-\widehat{\phi}\circ\tau_{\mathbf{i},n}+c.$

It can be rewritten as

$\widehat{f}\circ\tau_{\mathbf{i},n}-\widehat{f}\circ\tau_{\mathbf{i},n}(0)=\lambda^{-1}(\widehat{\phi}\circ\tau_{\mathbf{i},n-1}-\widehat{\phi}\circ\tau_{\mathbf{i},n-1}(0))-(\widehat{\phi}\circ\tau_{\mathbf{i},n}-\widehat{\phi}\circ\tau_{\mathbf{i},n}(0)).$

It follows that

$h_{\mathbf{i}}=\widehat{\phi}-\widehat{\phi}(0)$

does not depend on $\mathbf{i}$.

∎

###  2.3. Fourier analysis

The goal of this subsection is to prove Proposition 2.3.

We first describe a procedure which reduces the problem to the case that $T$
is linear. It is well-known that $T$ is topologically conjugate to the linear
map $\mathbf{m}_{d}:x\mapsto dx\mod 1$ via a homeomorphism
$\theta:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ with $\theta(0)=0$; in
particular, $\theta\circ T=\mathbf{m}_{d}\circ\theta$. Let
$\Theta:\mathbb{R}\to\mathbb{R}$ be the lift of $\theta$ with $\Theta(0)=0$.
Then

$\Theta\circ\widehat{T}=d\cdot\Theta,$

which implies that

$\Theta\circ\tau_{\mathbf{i},n}\circ\Theta^{-1}(x)=\frac{x+i_{1}+i_{2}d+\cdots+i_{n}d^{n-1}}{d^{n}}=:\xi_{n}(\mathbf{i},x).$

Let $F=\widehat{f}\circ\Theta^{-1}$ and let

$H_{\mathbf{i}}(x):=h_{\mathbf{i}}\circ\Theta^{-1}(x)=\sum_{m=1}^{\infty}\lambda^{m}[F(\xi_{m}(\mathbf{i},x))-F(\xi_{m}(\mathbf{i},0))].$

By definition, we have the following.

  * •

$F$ is $1$-periodic.

  * •

$h_{\mathbf{i}}$ is $1$-periodic if and only if $H_{\mathbf{i}}$ is
$1$-periodic.

  * •

$h_{\mathbf{i}}\equiv h_{\mathbf{j}}$ if and only if $H_{\mathbf{i}}\equiv
H_{\mathbf{j}}$.

  * •

If $T=\mathbf{m}_{d}$, then $\Theta=\mathrm{id}$ and
$h_{\mathbf{i}}=H_{\mathbf{i}}$.

For each $s\in\mathbb{Z}$, consider

$\mathcal{H}^{s}:\Sigma_{d}\times[0,1)\to\mathbb{C},\quad(\mathbf{k},x)\mapsto
H_{\mathbf{k}}(x+s)-H_{\mathbf{k}}(x).$

To prove Proposition 2.3, we shall use Fourier analysis to show that
$\mathcal{H}^{s}$ is constant for each $s\in\mathbb{Z}$.

The space $\Sigma_{d}\times[0,1)$ carries a unique Borel probability measure
$\mu$ such that for any $m\geq 1$ and any Borel set $U\subset[0,1)$,

$\mu(\\{(\mathbf{k},x)\in\Sigma_{d}\times[0,1):\xi_{m}(\mathbf{k},x)\in
U\\})=|U|,$

where $|U|$ denote the standard Lebesgue measure of $U$. Moreover, there is a
$\mu$-preserving Borel measurable bijection
$m_{d}:\Sigma_{d}\times[0,1)\to\Sigma_{d}\times[0,1)$ defined as

$m_{d}(\mathbf{k},x)=(k_{0}\mathbf{k},dx-k_{0}),$

where $k_{0}=\lfloor dx\rfloor$ is the largest integer which is not greater
than $dx$, and $k_{0}\mathbf{k}=(k_{n})_{n\geq 0}$ for
$\mathbf{k}=(k_{n})_{n\geq 1}$.

Indeed, $\xi:(\mathbf{k},x)\mapsto(\xi_{m}(\mathbf{k},x))_{m\geq 0}$ provides
a natural identification between $\Sigma_{d}\times[0,1)$ and the space $X(d)$
of backward orbits of $\mathbf{m}_{d}$:

$X(d)=\\{(x_{m})_{m=0}^{\infty}:x_{m}\in\mathbb{R}/\mathbb{Z},\mathbf{m}_{d}(x_{m+1})=x_{m},\forall
m\geq 0\\}.$

The map $m_{d}$ corresponds to the homeomorphism $(x_{m})_{m\geq
0}\mapsto(\mathbf{m}_{d}x_{m})_{m\geq 0}$ and $\mu$ corresponds to the lift of
the $\mathbf{m}_{d}$-invariant Lebesgue measure on $\mathbb{R}/\mathbb{Z}$ to
the space $X(d)$. Let us note that the measure $\mu$ corresponds to the Haar
measure on the compact abelian group $X(d)$, although we do not need this fact
explicitly.

The following lemma describes symmetry of the functions $\mathcal{H}^{s}$.

######  Lemma 2.4.

For any $s\in\mathbb{Z}$,

$\mathcal{H}^{ds}\circ m_{d}=\lambda\cdot\mathcal{H}^{s}.$ (2)

Moreover, for any $\mathbf{i}\in G_{0}$, we have

$\mathcal{H}^{s}(\mathbf{i}+\mathbf{k},x)=\mathcal{H}^{s}(\mathbf{k},x),\quad\forall\,(\mathbf{k},x)\in\Sigma_{d}\times[0,1).$
(3)

###### Proof.

We first prove (2). Note that for each $y\in\mathbb{R}$, $m\geq 1$,
$k_{0}\in\\{0,1,\ldots,d-1\\}$ and $\mathbf{k}\in\Sigma_{d}$, we have

$\xi_{m}(k_{0}\mathbf{k},dy-k_{0})=\xi_{m-1}(\mathbf{k},y),$

where $\xi_{0}(\mathbf{k},y)=y$. So for $k_{0}=\lfloor dx\rfloor$,

$\displaystyle\mathcal{H}^{ds}\circ m_{d}(\mathbf{k},x)$
$\displaystyle=\mathcal{H}^{ds}(k_{0}\mathbf{k},dx-k_{0})$
$\displaystyle=H_{k_{0}\mathbf{k}}(d(x+s)-k_{0})-H_{k_{0}\mathbf{k}}(dx-
k_{0})$
$\displaystyle=\sum_{m=1}^{\infty}\lambda^{m}\cdot\left[F(\xi_{m}(k_{0}\mathbf{k},d(x+s)-k_{0}))-F(\xi_{m}(k_{0}\mathbf{k},dx-
k_{0}))\right]$
$\displaystyle=\sum_{m=1}^{\infty}\lambda^{m}\cdot\left[F(\xi_{m-1}(\mathbf{k},x+s))-F(\xi_{m-1}(\mathbf{k},x))\right]$
$\displaystyle=\lambda\cdot(F(x+s)-F(x))+\lambda\cdot\mathcal{H}^{s}(\mathbf{k},x)$
$\displaystyle=\lambda\cdot\mathcal{H}^{s}(\mathbf{k},x),$

where the last equality follows from the fact that $F$ is $1$-periodic and
$s\in\mathbb{Z}$.

Now let us prove (3). Given $\mathbf{i}\in G_{0}$, by Lemma 2.2,
$h_{\mathbf{i}+\mathbf{k}}\equiv h_{\mathbf{k}}$, and hence
$H_{\mathbf{i}+\mathbf{k}}\equiv H_{\mathbf{k}}$ for all
$\mathbf{k}\in\Sigma_{d}$. The equality follows. ∎

We shall also need the following two lemmas.

######  Lemma 2.5.

For $\mathbf{i}=(i_{n})_{n\geq 1}\in\Sigma_{d}\setminus\\{\mathbf{0}\\}$, let
$z_{m}=(i_{1}+i_{2}d+\cdots+i_{m}d^{m-1})/d^{m}$ for each $m\geq 1$. Then for
any integer $q\not=0$, there exists a positive integer $m_{*}$ such that for
any $m\geq m_{*}$,

$qz_{m}\not=0\mod 1.$

###### Proof.

Fix $q$ and let $Y=\\{\frac{k}{q}\mod 1:k\in\mathbb{Z}\\}$. Then $Y$ is a
$\mathbf{m}_{d}$-invariant finite set. Assuming that the conclusion fails,
i.e., $z_{m}\in Y$ for all $m\geq 1$, it remains to show that
$\mathbf{i}=\mathbf{0}$. Since $\mathbf{m}_{d}(z_{m+1})=z_{m}$ for all $m\geq
1$, it follows that each $z_{m}$ is a periodic point of $\mathbf{m}_{d}$ in
$Y$. Since each $z_{m}$ is eventually mapped to $0\mod 1$, it follows that
$z_{m}=0\mod 1$ for all $m\geq 1$, and hence $\mathbf{i}=\mathbf{0}$. ∎

######  Lemma 2.6.

If $\mathcal{H}^{1}$ is constant, then $h_{\mathbf{0}}$ (or equivalently
$H_{\mathbf{0}}$) is $1$-periodic.

###### Proof.

Let $c\in\mathbb{C}$ be such that $\mathcal{H}^{1}\equiv c$. By definition,
$H_{\mathbf{k}}(x+1)-H_{\mathbf{k}}(x)=c$ for any
$(\mathbf{k},x)\in\Sigma_{d}\times[0,1)$. Since
$H_{\mathbf{k}+\iota(m)}(y)=H_{\mathbf{k}}(y+m)$ for any $y\in\mathbb{R}$, any
integer $m\geq 0$ and any $\mathbf{k}\in\Sigma_{d}$, it follows that
$H_{\mathbf{k}}(y+1)-H_{\mathbf{k}}(y)=c$ for any $y\in\mathbb{R}$ and any
$\mathbf{k}\in\Sigma_{d}$. In particular, $\mathcal{H}^{s}\equiv sc$ for any
$s\in\mathbb{Z}$. By (2), it follows that

$dc=\lambda c.$

By the Mean Value Theorem, $\lambda_{0}\leq d$, so
$|\lambda|<\lambda_{0}^{\alpha}\leq d$. Thus $c=0$, which completes the proof.
∎

We are ready to complete the proof of Proposition 2.3.

###### Completion of proof of Proposition 2.3.

Assume that (i) holds. By Lemma 2.6, it suffices to show that
$\mathcal{H}^{s}$ is constant for any $s\in\mathbb{Z}$. To this end, let

$E_{m,q}(\mathbf{k},x)=e^{2\pi i\xi_{m}(\mathbf{k},x)q},\quad
m,q\in\mathbb{Z},m\geq 0.$

We shall show that for any integers $m\geq 0$, $q\not=0$ and $s$,

$c_{m,q}^{s}:=\int_{\Sigma_{d}\times[0,1)}\mathcal{H}^{s}\cdot\overline{E_{m,q}}\>\\!\mathrm{d}\mu=0.$
(4)

Before the proof of (4), let us show how it implies that $\mathcal{H}^{s}$ is
constant. Note that $\mathcal{H}^{s}$ is bounded and continuous on
$\Sigma_{d}\times[0,1)$. Let $\mathcal{B}$ denote the Borel $\sigma$-algebra
of $[0,1)$ and let $\mathcal{B}_{m}=\xi_{m}^{-1}(\mathcal{B})$, which is a
$\sigma$-algebra in $\Sigma_{d}\times[0,1)$. Then $\mathcal{B}_{m}$ is
monotone increasing to the Borel $\sigma$-algebra in $\Sigma_{d}\times[0,1)$.
By the Martingale Convergence theorem,
$\mathcal{H}^{s}_{m}:=\mathbb{E}[\mathcal{H}^{s}|\mathcal{B}_{m}]$ converges
$\mu$-a.e. to $\mathcal{H}^{s}$. Since $E_{m,q}$ is
$\mathcal{B}_{m}$-measurable,

$\int_{\Sigma_{d}\times[0,1)}\mathcal{H}^{s}_{m}\cdot\overline{E_{m,q}}\>\\!\mathrm{d}\mu=c_{m,q}^{s}=0$

for any $m\geq 0$ and $q\not=0$, where the last equality follows from (4).
Since $(\xi_{m})_{*}\mu$ is the standard Lebesgue measure on $[0,1)$, each
$\mathcal{H}^{s}_{m}$, which can be viewed as a function in $[0,1)$, must be
constant a.e.. Thus $\mathcal{H}^{s}$ is constant $\mu$-a.e.. Since
$\mathcal{H}^{s}$ is continuous, it is constant.

It remains to prove that (4) holds for all $m\geq 0$ and $q\not=0$. First let
us show

$c_{m+1,q}^{ds}=\lambda c_{m,q}^{s}.$ (5)

Indeed, since $(m_{d})_{*}\mu=\mu$ and $\xi_{m+1}\circ m_{d}=\xi_{m}$, we have

$\int\mathcal{H}^{ds}\circ
m_{d}\cdot\overline{E_{m,q}}\,\>\\!\mathrm{d}\mu=\int\mathcal{H}^{ds}\circ
m_{d}\cdot\overline{E_{m+1,q}\circ
m_{d}}\,\>\\!\mathrm{d}\mu=\int\mathcal{H}^{ds}\cdot\overline{E_{m+1,q}}\,\>\\!\mathrm{d}\mu=c_{m+1,q}^{ds}.$

Combining this with (2), we obtain (5).

Next, by Lemma 2.2, $G_{0}\not=\\{\mathbf{0}\\}$. Let
$\mathbf{i}=(i_{n})_{n\geq 1}\in G_{0}\setminus\\{\mathbf{0}\\}$. Then by (3),
for any $s\in\mathbb{Z}$, $m\geq 1$ and $q\in\mathbb{Z}$,

$c_{m,q}^{s}=e^{2\pi iz_{m}q}c_{m,q}^{s},$

where $z_{m}=(i_{1}+i_{2}d+\cdots+i_{m}d^{m-1})/d^{m}$. By Lemma 2.5, for each
$q\not=0$, there exists $m_{*}(q)$ such that for any $m\geq m_{*}(q)$,
$qz_{m}\not=0\mod 1$, so that $e^{2\pi iz_{m}q}\not=1$, which implies that
$c_{m,q}^{s}=0$ for all $s\in\mathbb{Z}$. By (5), it follows that
$c_{m,q}^{s}=0$ for all $m\geq 0$ and $0\neq q\in\mathbb{Z}$. The proof is
completed. ∎

##  3\. Proof of the Main Theorem

This section is devoted to the proof of the Main Theorem. The following is an
equivalent reformulation of the Main Theorem.

######  Main Theorem’.

Let $T:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ be a real analytic
expanding map and let $f:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ be a real
analytic function. Assume that $f$ is not real analytically cohomologous to
constant. Let $\mu$ be a maximizing measure of $(T,f)$. Then the measure-
theoretic entropy $h_{T}(\mu)=0$.

In §§ 3.1– 3.3, we shall prove the Main Theorem’ under the following technical
condition:

$(\ast):T\mbox{ is orientation-perserving and
}\text{Fix}(T)\setminus\text{supp}(\mu)\not=\emptyset,$

where $\text{Fix}(T)$ denotes the set of fixed points of $T$. In § 3.4, we
shall show how to remove this condition and complete the proof, using the
equivalence of the maximization problem between $(T,f)$ and $(T^{k},f+f\circ
T+\cdots+f\circ T^{k-1})$.

###  3.1. Strategy of the proof assuming ($\ast$)

Let $d$ be the degree of $T$. Without loss of generality, we may assume that
$0\in\mathbb{R}/\mathbb{Z}$ is a fixed point of $T$ which is not contained in
$\text{supp}(\mu)$. As before, let $\widehat{T}$ be the unique lift of $T$
with $\widehat{T}(0)=0$ and denote $\tau:=\widehat{T}^{-1}$. Moreover, denote
$\tau_{i}(x):=\tau(x+i)$ for $0\leq i<d$. Let $S=\text{supp}(\mu)$.
Identifying $\mathbb{R}/\mathbb{Z}$ with $[0,1)$ in the natural way, $S$ is a
non-empty compact subset of $(0,1)$ with $T(S)=S$.

For $\mathbf{i}=(i_{n})_{n\geq 1}\in\Sigma_{d}$, recall

$\tau_{\mathbf{i},n}=\tau_{i_{n}}\circ\cdots\circ\tau_{i_{2}}\circ\tau_{i_{1}},\quad\forall
n\geq 1.$

In order to show that $h_{T}(\mu)=0$, we shall analyze the inverse limit of
$T:S\to S$. So let

$\mathcal{S}=\\{(\mathbf{i},x)\in\Sigma_{d}\times S:\tau_{\mathbf{i},n}(x)\in
S,\forall n\geq 1\\}.$

Given $x\in S$, denote

$\mathcal{S}^{x}:=\\{\mathbf{i}\in\Sigma_{d}:(\mathbf{i},x)\in\mathcal{S}\\}.$

Note that $\mathcal{S}^{x}$ is a non-empty compact subset in $\Sigma_{d}$.
Given $\mathbf{i}\in\Sigma_{d}$, denote

$\mathcal{S}_{\mathbf{i}}:=\\{x\in S:(\mathbf{i},x)\in\mathcal{S}\\}=\\{x\in
S:\mathbf{i}\in\mathcal{S}^{x}\\}.$

Then $\mathcal{S}_{\mathbf{i}}$ is a compact subset of $S$ (possibly empty).

Let us apply Theorem 1.1, so that

$h_{\mathbf{i}}(x)=\sum_{n=1}^{\infty}(\widehat{f}\circ\tau_{\mathbf{i},n}(x)-\widehat{f}\circ\tau_{\mathbf{i},n}(0))),\quad
x\in\mathbb{R}.$

As we are assuming that $f$ is not real analytically cohomologous to constant,
the second alternative of Theorem 1.1 holds. So for distinct
$\mathbf{i},\mathbf{j}\in\Sigma_{d}$, the real analytic functions
$h_{\mathbf{i}}$ and $h_{\mathbf{j}}$ are not identical, and hence
$h^{\prime}_{\mathbf{i}}-h^{\prime}_{\mathbf{j}}$ has only isolated zeros.
This allows us to define, for each $x\in S$, two total orders $\prec^{x}_{+}$
and $\prec^{x}_{-}$ on $\mathcal{S}^{x}$ as follows.

######  Definition 3.1.

Given $x\in S$, define two total orders $\prec^{x}_{\pm}$ on $\mathcal{S}^{x}$
as follows. Given $\mathbf{i}\neq\mathbf{j}$ in $\mathcal{S}^{x}$,

  * •

$\mathbf{i}$ is strictly less than $\mathbf{j}$ with respect to
$\prec^{x}_{+}$ if there exists $\delta>0$ such that
$h^{\prime}_{\mathbf{i}}(y)<h^{\prime}_{\mathbf{j}}(y)$ holds for all
$y\in(x,x+\delta)$;

  * •

$\mathbf{i}$ is strictly less than $\mathbf{j}$ with respect to
$\prec^{x}_{-}$ if there exists $\delta>0$ such that
$h^{\prime}_{\mathbf{i}}(y)>h^{\prime}_{\mathbf{j}}(y)$ holds for all
$y\in(x-\delta,x)$.

######  Lemma 3.1.

Suppose that $f$ is not real analytically cohomologous to constant. Then for
any $x\in S$, $(\mathcal{S}^{x},\prec^{x}_{+})$ has a unique maximal element
$\kappa_{+}(x)\in\mathcal{S}^{x}$, and $(\mathcal{S}^{x},\prec^{x}_{-})$ has a
unique maximal element $\kappa_{-}(x)\in\mathcal{S}^{x}$, which define two
maps $\kappa_{\pm}:S\to\Sigma_{d}$.

###### Proof.

Uniqueness follows from the fact that both orders are total orders. For
existence, let us focus on “$\prec^{x}_{+}$”; the discussion on
“$\prec^{x}_{-}$” is totally similar and omitted. For each $m\geq 1$, since
$h_{\mathbf{i}}^{(m)}(x)$ is continuous in $\mathbf{i}\in\mathcal{S}^{x}$,
$\mathcal{I}_{m}$ defined inductively below is a decreasing sequence of non-
empty compact subsets of $\mathcal{S}^{x}$, where
$\mathcal{I}_{0}=\mathcal{S}^{x}$:

$\mathcal{I}_{m}=\left\\{\mathbf{i}\in\mathcal{I}_{m-1}:h_{\mathbf{i}}^{(m)}(x)=\max\\{h_{\mathbf{j}}^{(m)}(x):\mathbf{j}\in\mathcal{I}_{m-1}\\}\right\\}.$

Therefore $\bigcap_{m=1}^{\infty}\mathcal{I}_{m}$ contains at least one
element $\mathbf{i}$. Then for any $\mathbf{j}\in\mathcal{S}^{x}$,
$h_{\mathbf{i}}^{(m)}(x)\geq h_{\mathbf{j}}^{(m)}(x)$ for any $m\geq 1$, which
implies that $\mathbf{i}$ is a maximal element in $\mathcal{S}^{x}$ with
respect to $\prec^{x}_{+}$. ∎

We shall prove the following proposition in § 3.3.

######  Proposition 3.2.

Assume that $\mathcal{S}_{\mathbf{i}}$ has a limit point $x$. Then
$\mathbf{i}\in\\{\kappa_{+}(x),\kappa_{-}(x)\\}.$

###  3.2. Proof of the Main Theorem’ assuming ($\ast$)

Let $\mathcal{T}:\mathcal{S}\to\mathcal{S}$ be the inverse limit of $T:S\to
S$, i.e., for $x\in\tau_{i_{0}}([0,1))\cap S$,

$\mathcal{T}((i_{n})_{n\geq 1},x)=((i_{n-1})_{n\geq 1},\widehat{T}(x)-i_{0}).$

As $0\not\in S$, $T:S\to S$ is naturally topologically conjugate to a one-
sided subshift with $d$ symbols and $\mathcal{T}:\mathcal{S}\to\mathcal{S}$ is
topologically conjugate to a two-sided subshift with $d$ symbols.

We shall need the following well-known result.

######  Proposition 3.3.

If $\nu$ is an ergodic invariant probability Borel measure of the two-sided
full shift $\sigma:\\{0,1,\ldots,d-1\\}^{\mathbb{Z}}\circlearrowleft$ such
that $h_{\sigma}(\nu)>0$, then for any Borel subset $U$ of
$\\{0,1,\ldots,d-1\\}^{\mathbb{Z}}$ with $\nu(U)>0$, the following holds for
$\nu$-a.e. $(i_{n})_{n\in\mathbb{Z}}\in U$:

$\\{(j_{n})_{n\in\mathbb{Z}}\in U:j_{n}=i_{n},\,\forall n\geq 0\\},$
$\\{(j_{n})_{n\in\mathbb{Z}}\in U:j_{n}=i_{n},\,\forall n<0\\}$

are both uncountable.

###### Proof.

It is well-known that the two-sided full shift is topologically conjugate to a
linear horseshoe $F:\Lambda\to\Lambda$, with the sets in question
corresponding to (local) stable and unstable manifolds. If $\nu_{F}$ is the
$F$-invariant ergodic probability measure corresponding to $\nu$, then
$h_{F}(\nu_{F})>0$. It is well-known that (see e.g. [18]) for $\nu_{F}$-a.e.
$(y,x)\in\Lambda$, the conditional measures $\nu_{F}^{x}$ and $\nu_{F,y}$
along the local stable and unstable manifolds have positive local dimensions
and thus admit no atom. So if $\nu_{F}(W)>0$, then for $\nu_{F}$-a.e. $z\in
W$, the intersection of $W$ with the stable and unstable manifolds of $z$ must
be both uncountable. ∎

The following lemma deals with the measurability issue involved.

######  Lemma 3.4.

The set $B$ defined below is Borel:

$B:=\\{(\mathbf{i},x)\in\mathcal{S}:x~{}\text{is a limit point of
}\mathcal{S}_{\mathbf{i}}\\}.$

###### Proof.

Define $\phi:\Sigma_{d}\times S\to\mathbb{R}$ as
$\phi(\mathbf{i},x)=\operatorname{dist}(x,(\mathcal{S}_{\mathbf{i}}\cup\\{2\\})\setminus\\{x\\})$
(here use $\mathcal{S}_{\mathbf{i}}\cup\\{2\\}$ instead of
$\mathcal{S}_{\mathbf{i}}$ because $\mathcal{S}_{\mathbf{i}}\setminus\\{x\\}$
might be empty), where $\operatorname{dist}(x,\cdot)$ means the distance of
$x$ to a subset of the real line in the usual sense. Then $B=\phi^{-1}(0)$. It
suffices to show that $\phi$ is Borel. In fact, the following hold.

  * •

Given $\mathbf{i}\in\Sigma_{d}$, $x\mapsto\phi(\mathbf{i},x)$ is continuous.
This is easy to check.

  * •

Given $x\in S$, $\mathbf{i}\mapsto\phi(\mathbf{i},x)$ is Borel. To see this,
for each $n\geq 1$, let

$\varphi_{n}(\mathbf{i}):=\operatorname{dist}\left(x,(\mathcal{S}_{\mathbf{i}}\cup\\{2\\})\setminus(x-\tfrac{1}{n},x+\tfrac{1}{n})\right),\quad\mathbf{i}\in\Sigma_{d}.$

Then, by compactness of $\mathcal{S}$, $\varphi_{n}$ is lower semi-continuous
for each $n$, while
$\lim_{n\to\infty}\varphi_{n}(\mathbf{i})=\phi(\mathbf{i},x)$.

It follows that $\phi$ is Borel. ∎

Let us now complete the proof of the Main Theorem assuming ($\ast$).

###### Proof of the Main Theorem’ assuming ($\ast$).

Assume by contradiction that $h_{T}(\mu)>0$. By ergodic decomposition and
affinity of the entropy function, we may assume that $\mu$ is ergodic with
respect to $T$. As the inverse limit of $T:S\to S$, the map
$\mathcal{T}:\mathcal{S}\to\mathcal{S}$ has an ergodic invariant measure
$\widetilde{\mu}$ with positive entropy.

We shall show that $\widetilde{\mu}(\mathcal{S})=0$ to get a contradiction. To
this end, let $B$ be the set defined in Lemma 3.4. By the definition of $B$
and Proposition 3.2, $\mathbf{i}\in\\{\kappa_{\pm}(x)\\}$ for any
$(\mathbf{i},x)\in B$, so by Proposition 3.3, $\widetilde{\mu}(B)=0$. By the
definition of $B$ again, for any $(\mathbf{i},x)\in\mathcal{S}\setminus B$,
the set $\\{y\in S:(\mathbf{i},y)\in\mathcal{S}\setminus B\\}$ is countable,
so by Proposition 3.3 again, $\widetilde{\mu}(\mathcal{S}\setminus B)=0$. The
proof is done. ∎

###  3.3. Complexity of $\mathcal{S}$

This subsection is devoted to the proof of Proposition 3.2. Let
$g:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ be a sub-action for $(T,f)$, i.e.,
$g:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ is a continuous function such that

$f(x)\leq g\circ T(x)-g(x)+\beta(f)$

for all $x\in\mathbb{R}/\mathbb{Z}$. If we put

$S_{0}=\\{x\in\mathbb{R}/\mathbb{Z}:f(x)=g(T(x))-g(x)+\beta(f)\\},$ (6)

then a $T$-invariant measure is a maximizing measure $f$ if and only if it is
supported in $S_{0}$. In particular, $S\subset S_{0}$.

Sub-actions played an important role in the ergodic maximization problem. When
$T$ is expanding and $f$ is Lipschitz, it is well-known that there exists a
sub-action $g$ which is Lipschitz. This is often referred to as Mãné’s lemma.
However, in general we cannot expect higher regularity of $g$, even when we
assume $f$ and $T$ are both real analytic, see [3].

######  Lemma 3.5.

Let $(\mathbf{i},x)\in\mathcal{S}$. Then for any $y\in S$,

$g(y)-g(x)\geq h_{\mathbf{i}}(y)-h_{\mathbf{i}}(x).$

Moreover, equality holds if $(\mathbf{i},y)\in\mathcal{S}$.

###### Proof.

Denote $\tau_{\mathbf{i},0}=\mathrm{id}$. Since $g$ is a sub-action and
$x\in\mathcal{S}_{\mathbf{i}}$, for each $n\geq 0$,

$g(\tau_{\mathbf{i},n}(x))=f(\tau_{\mathbf{i},n+1}(x))+g(\tau_{\mathbf{i},n+1}(x))-\beta(f),$
$g(\tau_{\mathbf{i},n}(y))\geq
f(\tau_{\mathbf{i},n+1}(y))+g(\tau_{\mathbf{i},n+1}(y))-\beta(f).$

Moreover, equality holds in the last inequality if
$y\in\mathcal{S}_{\mathbf{i}}$. Therefore, for each $m\geq 1$,

$g(x)=\sum_{n=1}^{m}f(\tau_{\mathbf{i},n}(x))+g(\tau_{\mathbf{i},m}(x))-m\beta(f),$
$g(y)\geq\sum_{n=1}^{m}f(\tau_{\mathbf{i},n}(y))+g(\tau_{\mathbf{i},m}(y))-m\beta(f).$

Consequently,

$g(y)-g(x)\geq\sum_{n=1}^{m}\big{(}f(\tau_{\mathbf{i},n}(y))-f(\tau_{\mathbf{i},n}(x))\big{)}+\big{(}g(\tau_{\mathbf{i},m}(y))-g(\tau_{\mathbf{i},m}(x))\big{)}.$

Letting $m\to\infty$, we obtain the desired inequality.

If $y\in\mathcal{S}_{\mathbf{i}}$, then all the inequalities above become
equalities. ∎

###### Proof of Proposition 3.2.

We may assume that there exists a sequence $(x_{k})$ in
$\mathcal{S}_{\mathbf{i}}$ converging to $x$ from the right side; the other
situation is similar and omitted. We shall show that
$\mathbf{i}=\kappa_{+}(x)$. To this end, consider an arbitrary
$\mathbf{j}\in\mathcal{S}^{x}$. By Lemma 3.5, we have the following.

  * •

Since $x,x_{k}\in\mathcal{S}_{\mathbf{i}}$,

$g(x_{k})-g(x)=h_{\mathbf{i}}(x_{k})-h_{\mathbf{i}}(x)=\int_{x}^{x_{k}}h^{\prime}_{\mathbf{i}}(t)\>\\!\mathrm{d}t.$

  * •

Since $x\in\mathcal{S}_{\mathbf{j}}$,

$g(x_{k})-g(x)\geq
h_{\mathbf{j}}(x_{k})-h_{\mathbf{j}}(x)=\int_{x}^{x_{k}}h^{\prime}_{\mathbf{j}}(t)\>\\!\mathrm{d}t.$

Therefore,

$\int_{x}^{x_{k}}h^{\prime}_{\mathbf{j}}(t)\>\\!\mathrm{d}t\leq\int_{x}^{x_{k}}h^{\prime}_{\mathbf{i}}(t)\>\\!\mathrm{d}t.$

Since $x_{k}$ converges to $x$ from the right side, the inequality above
implies that $h^{\prime}_{\mathbf{j}}(t)\leq h^{\prime}_{\mathbf{i}}(t)$ holds
for a sequence of points converging to $x$ from the right side. It follows
that $\mathbf{i}\prec_{+}^{x}\mathbf{j}$ cannot hold. Therefore
$\mathbf{i}=\kappa_{+}(x)$. ∎

###  3.4. Iteration

We shall show that the technical condition ($\ast$) can be removed. Let
$T\in\mathcal{E}^{\omega}$ and let $f\in C^{\omega}(\mathbb{R}/\mathbb{Z})$.
For any positive integer $k$, let $\mathscr{S}_{k}f=f+f\circ T+\cdots+f\circ
T^{k-1}$.

The following elementary observations should be well-known. Let us include a
proof for completeness.

######  Lemma 3.6.

  1. (1)

There exists $g\in C^{\omega}(\mathbb{R}/\mathbb{Z})$ such that
$\mathscr{S}_{k}f=g\circ T^{k}-g+const$ for some $k\geq 1$ if and only if the
same holds for $k=1$.

  2. (2)

A maximizing measure of $(T,f)$ is also a maximizing measure of
$(T^{k},\mathscr{S}_{k}f)$ for any $k\geq 1$.

###### Proof.

(1) If $f=g\circ T-g+c$, then $\mathscr{S}_{k}f=g\circ T^{k}-g+kc.$ For the
other direction, suppose that $\mathscr{S}_{k}f=g\circ T^{k}-g+c$ for some
$g\in C^{\omega}(\mathbb{R}/\mathbb{Z})$. Then

$f\circ T^{k}-f=(\mathscr{S}_{k}f)\circ T-\mathscr{S}_{k}f=(g\circ
T^{k}-g)\circ T-(g\circ T^{k}-g),$

which can be rewritten as follows:

$(f+g-g\circ T)\circ T^{k}=f+g-g\circ T.$

Since $T^{k}:\mathbb{T}\to\mathbb{T}$ is topologically transitive and since
$f+g-g\circ T$ is continuous, the equality above implies that $f+g-g\circ
T=const$.

(2) If $\nu$ is a maximizing measure for $(T^{k},\mathscr{S}_{k}f)$, then
$\mu=\frac{1}{k}\sum_{j=0}^{k-1}T^{j}_{*}\nu$ is $T$-invariant, and

$\int f\>\\!\mathrm{d}\mu=\frac{1}{k}\sum_{j=0}^{k-1}\int f\circ
T^{j}\>\\!\mathrm{d}\nu=\int\frac{1}{k}\mathscr{S}_{k}f\>\\!\mathrm{d}\nu.$

This shows $\beta(T,f)\geq\beta(T^{k},\frac{1}{k}\mathscr{S}_{k}f)$. On the
other hand, if $\mu$ is a maximizing measure of $(T,f)$, then $\mu$ is
$T^{k}$-invariant and

$\beta(T,f)=\int
f\>\\!\mathrm{d}\mu=\int\frac{1}{k}\mathscr{S}_{k}f\>\\!\mathrm{d}\mu\leq\beta(T^{k},\frac{1}{k}\mathscr{S}_{k}f).$

Hence $\beta(T,f)=\beta(T^{k},\frac{1}{k}\mathscr{S}_{k}f)$ and a maximizing
measure for $(T,f)$ is also maximizing for $(T^{k},\mathscr{S}_{k}f)$.

∎

###### Completion of the proof of the Main Theorem’.

As $f$ is not analytically cohomologous to a constant, for $S_{0}$ defined by
(6), $S_{0}\subsetneq\mathbb{R}/\mathbb{Z}$, so $S=\text{supp}(\mu)$ is
nowhere dense in $\mathbb{R}/\mathbb{Z}.$ In particular, there is a periodic
point $p$ of $T$ such that $p\not\in S$. Let $k$ be an even positive integer
such that $T^{k}(p)=p$. By Lemma 3.6, $\mu$ is a maximizing measure for
$(T^{k},\mathscr{S}_{k}f)$ and $\mathscr{S}_{k}f$ is not analytically
cohomologous to a constant with respect to $T^{k}$. Thus by what we have
proved before, $h_{T^{k}}(\mu)=0$ and hence $h_{T}(\mu)=0$. ∎

##  4\. Proof of Theorems 1.2, 1.3 and 1.4

In this section, we prove Theorems 1.2, 1.3 and 1.4. The basic idea is to
approximate $C^{r}$ functions or maps with real analytic ones and then apply
the Main Theorem.

We shall need upper semi-continuity of the following function:

$\mathcal{H}(T,f):=\sup_{\mu\in\mathcal{M}_{max}(T,f)}h_{T}(\mu),\,\,(T,f)\in\mathcal{E}^{1}\times
C^{0}(\mathbb{R}/\mathbb{Z}).$ (7)

Here $\mathcal{M}_{max}(T,f)$ denote the collection of $(T,f)$ maximizing
measures. This result is essentially contained in [4, 17].

######  Proposition 4.1.

The function $\mathcal{H}$ is upper semi-continuous on $\mathcal{E}^{1}\times
C^{0}(\mathbb{R}/\mathbb{Z})$.

###### Proof.

It was observed in [4] that for any fixed $T\in\mathcal{E}^{1}$,
$f\mapsto\mathcal{H}(T,f)$ is upper semi-continuous on
$C^{0}(\mathbb{R}/\mathbb{Z})$. Indeed, if $\mu_{n}$ is a maximizing measure
for $(T,f_{n})$ with $f_{n}\to f$ in $C^{0}(\mathbb{R}/\mathbb{Z})$, then any
accumulation point $\mu$ of $\mu_{n}$ in the weak-* topology is a maximizing
measure for $(T,f)$. So the result is a consequence of upper semi-continuity
of the entropy map $\nu\mapsto h_{T}(\nu)$.

Now suppose that $(T_{n},g_{n})\to(T,f)$ in $\mathcal{E}^{1}\times
C^{0}(\mathbb{R}/\mathbb{Z})$. By [17, Lemma 2], for each $n$ sufficiently
large, there exists a homeomorphism
$h_{n}:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ such that
$T=h_{n}^{-1}\circ T_{n}\circ h_{n}$, and moreover,
$\lim_{n\to\infty}\max_{x\in\mathbb{R}/\mathbb{Z}}d_{\mathbb{R}/\mathbb{Z}}(h_{n}(x),x)=0$,
where $d_{\mathbb{R}/\mathbb{Z}}$ is the standard metric on
$\mathbb{R}/\mathbb{Z}$. Put $f_{n}:=g_{n}\circ h_{n}\in
C^{0}(\mathbb{R}/\mathbb{Z})$. Then
$\mathcal{H}(T,f_{n})=\mathcal{H}(T_{n},g_{n})$ holds and $f_{n}\to f$ in
$C^{0}(\mathbb{R}/\mathbb{Z})$. Thus

$\limsup_{n\to\infty}\mathcal{H}(T_{n},g_{n})=\limsup_{n\to\infty}\mathcal{H}(T,f_{n})\leq\mathcal{H}(T,f).$

∎

###### Proof of Theorem 1.2.

We only need to consider the maximizing case. Let
$T\in\mathcal{E}^{\omega}(\mathbb{R}/\mathbb{Z})$ and let
$r\in\\{0,1,2,\cdots,\infty\\}$. Since the inclusion map from
$C^{r}(\mathbb{R}/\mathbb{Z})$ to $C^{0}(\mathbb{R}/\mathbb{Z})$ is
continuous, according to Proposition 4.1, the function
$f\mapsto\mathcal{H}(T,f)$ defined on $C^{r}(\mathbb{R}/\mathbb{Z})$ is upper
semi-continuous. Therefore, $C_{ze}^{r}:=\\{f\in
C^{r}(\mathbb{R}/\mathbb{Z}):\mathcal{H}(T,f)=0\\}$ is a $G_{\delta}$ subset
of $C^{r}(\mathbb{R}/\mathbb{Z})$. On the other hand, by our Main Theorem,
$C_{ze}^{r}\supset C_{ncc}^{\omega}$, where $C_{ncc}^{\omega}$ denote the
collection of functions in $C^{\omega}(\mathbb{R}/\mathbb{Z})$ that are not
analytically cohomologous to constant. To complete the proof, it remains to
show that $C_{ncc}^{\omega}$ is dense in $C^{r}(\mathbb{R}/\mathbb{Z})$. To
this end, let $C_{wcb}^{r}=\\{f\in C^{r}(\mathbb{R}/\mathbb{Z}):\int
f\>\\!\mathrm{d}\mu~{}\text{is independent of }\mu\in\mathcal{M}(T)\\}$.
Clearly, $C_{wcb}^{r}$ is a closed subset of $C^{r}(\mathbb{R}/\mathbb{Z})$
with empty interior, and $C^{\omega}(\mathbb{R}/\mathbb{Z})\setminus
C_{wcb}^{r}\subset C_{ncc}^{\omega}$. Since
$C^{\omega}(\mathbb{R}/\mathbb{Z})$ is dense in
$C^{r}(\mathbb{R}/\mathbb{Z})$, it follows that $C_{ncc}^{\omega}$ dense in
$C^{r}(\mathbb{R}/\mathbb{Z})$, which completes the proof. ∎

We shall need the following well-known result for the proof of Theorems 1.3
and 1.4. See, for example, [7, Proposition 28] for a more comprehensive
version of this result under the $C^{1+\alpha}$ (and orientation-preserving)
setting.

######  Proposition 4.2.

For each $T\in\mathcal{E}^{r}$, $r\in\\{1,2,\ldots,\infty,\omega\\}$, the
following are equivalent:

  1. (i)

$\log|T^{\prime}|$ is $C^{0}$ cohomologous to constant with respect to $T$;

  2. (ii)

$T$ is $C^{r}$ conjugate to the linear map $x\mapsto\deg(T)\cdot x$.

###### Proof.

The implication (ii) $\implies$ (i) is trivial. Let us show (i) $\implies$
(ii). Since $\log|T^{\prime}|$ is $C^{0}$ cohomologous to constant, there
exist $\psi\in C^{0}(\mathbb{R}/\mathbb{Z})$ and
$c\in\mathbb{R}\setminus\\{0\\}$ such that the following hold:

  * •

$\psi>0$ and $\int_{\mathbb{R}/\mathbb{Z}}\psi(x)\>\\!\mathrm{d}x=1$;

  * •

$\psi\circ T\cdot T^{\prime}=c\cdot\psi$.

The first item above implies that there exists a $C^{1}$-diffeomorphism
$\phi:\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ with
$\phi^{\prime}=\psi$. Then the second item above can be rewritten as
$(\phi\circ T)^{\prime}=c\cdot\phi^{\prime}$. Integrating both sides over
$\mathbb{R}/\mathbb{Z}$ yields that $\deg(T)=c$. Therefore, $T$ is $C^{1}$
conjugate to the linear map $x\mapsto\deg(T)\cdot x$ via $\phi$.

It remains to show that $\phi$ is $C^{r}$ when $T\in\mathcal{E}^{r}$ for
$r\in\\{2,3,\ldots,\infty,\omega\\}$. In this situation, $T$ admits a unique
absolutely continuous invariant probability measure $\mu$, which has $C^{r-1}$
density. On the other hand, $\phi_{*}\mu$ is the unique absolutely continuous
invariant probability measure of $x\mapsto\deg(T)\cdot x$, which is exactly
the standard Lebesgue measure. Thus $\phi$ is also $C^{r}$. ∎

###### Proof of Theorem 1.3.

Let us only consider the maximizing case, i.e. $f=\log|T^{\prime}|$, as the
minimizing case is similar. By the Main Theorem, either $f$ is analytically
cohomologous to constant, or any Lyapunov maximizing measure has zero entropy.
If the first case happens, then by Proposition 4.2, $T$ is $C^{\omega}$
conjugate to $x\mapsto\deg(T)\cdot x$. ∎

###### Proof of Theorem 1.4.

We only need to consider the maximizing case. Let
$r\in\\{1,2,\cdots,\infty\\}$. Since the map $T\mapsto(T,\log|T^{\prime}|)$
from $\mathcal{E}^{r}$ to $\mathcal{E}^{1}\times C^{0}(\mathbb{R}/\mathbb{Z})$
is continuous, according to Proposition 4.1, the function
$T\mapsto\mathcal{H}(T,\log|T^{\prime}|)$ defined on $\mathcal{E}^{r}$ is
upper semi-continuous. Therefore,
$\mathcal{E}_{ze}^{r}:=\\{f\in\mathcal{E}^{r}:\mathcal{H}(T,\log|T^{\prime}|)=0\\}$
is a $G_{\delta}$ subset of $\mathcal{E}^{r}$. On the other hand, by Theorem
1.3, $\mathcal{E}_{ze}^{r}\supset\mathcal{E}_{ncc}^{\omega}$, where
$\mathcal{E}_{ncc}^{\omega}$ denote the collection of maps in
$\mathcal{E}^{\omega}$ that are not analytically conjugate to linear map. To
complete the proof, it remains to show that $\mathcal{E}_{ncc}^{\omega}$ is
dense $\mathcal{E}^{r}$. To this end, let
$\mathcal{E}_{wcb}^{r}=\\{T\in\mathcal{E}^{r}:\int\log|T^{\prime}|\>\\!\mathrm{d}\mu~{}\text{is
independent of }\mu\in\mathcal{M}(T)\\}$. Clearly, $\mathcal{E}_{wcb}^{r}$ is
a closed subset of $\mathcal{E}^{r}$ with empty interior and
$\mathcal{E}^{\omega}\setminus\mathcal{E}_{wcb}^{r}\subset\mathcal{E}_{ncc}^{\omega}$.
Since $\mathcal{E}^{\omega}$ is dense in $\mathcal{E}^{r}$, it follows that
$\mathcal{E}_{ncc}^{\omega}$ dense in $\mathcal{E}^{r}$, which completes the
proof. ∎

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