# A converse to the neo-classical inequality with an application to the
Mittag-Leffler function

Stefan Gerhold
TU Wien
<EMAIL_ADDRESS>Thomas Simon
Université de Lille
<EMAIL_ADDRESS>
###### Abstract

We prove two inequalities for the Mittag-Leffler function, namely that the
function $\log E_{\alpha}(x^{\alpha})$ is sub-additive for $0<\alpha<1,$ and
super-additive for $\alpha>1.$ These assertions follow from two new binomial
inequalities, one of which is a converse to the neo-classical inequality. The
proofs use a generalization of the binomial theorem due to Hara and Hino
(Bull. London Math. Soc. 2010). For $0<\alpha<2,$ we also show that
$E_{\alpha}(x^{\alpha})$ is log-concave resp. log-convex, using analytic as
well as probabilistic arguments.

MSC2020: 26D15, 33E12

##  1 Introduction and main results

The Mittag-Leffler function

$E_{\alpha}(x):=\sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(\alpha
k+1)},\quad\alpha>0,$

is a well-known special function with a large number of applications in pure
and applied mathematics; see [9, 13] for surveys. Clearly, we have
$E_{1}(x)=e^{x}.$ Somewhat surprisingly, the “identity”

$E_{\alpha}((x+y)^{\alpha})\stackrel{{\scriptstyle?}}{{=}}E_{\alpha}(x^{\alpha})E_{\alpha}(y^{\alpha})$
(1.1)

has been used in a few papers. As discussed in [6, 18], it is not correct for
$\alpha\neq 1$. In [18], a correct identity involving integrals of
$E_{\alpha}(x^{\alpha})$ is proven, which reduces to $e^{x+y}=e^{x}e^{y}$ as
$\alpha\to 1$. Besides this, it seems natural to ask whether the left and
right hand sides of (1.1) are comparable. This is indeed the case:

######  Theorem 1.1.

For $0<\alpha<1,$ we have

$E_{\alpha}((x+y)^{\alpha})\leq
E_{\alpha}(x^{\alpha})E_{\alpha}(y^{\alpha}),\quad x,y\geq 0,$ (1.2)

and for $\alpha>1$

$E_{\alpha}((x+y)^{\alpha})\geq
E_{\alpha}(x^{\alpha})E_{\alpha}(y^{\alpha}),\quad x,y\geq 0.$ (1.3)

These inequalities are strict for $x,y>0$.

The strictness assertion strengthens the observation made at the beginning of
Section 2 of [6], where it is argued that the validity of (1.1) _for all_
$x,y\geq 0$ implies $\alpha=1.$ According to Theorem 1.1, this equality
_never_ holds, except in the obvious cases ($\alpha=1,$ or $xy=0$). Although
apparently not made explicit in the literature, the lower estimate

$E_{\alpha}((x+y)^{\alpha})\geq\alpha
E_{\alpha}(x^{\alpha})E_{\alpha}(y^{\alpha}),\quad x,y\geq 0,\ 0<\alpha<1,$

complementing (1.2), follows from the calculation

$\displaystyle\begin{split}E_{\alpha}((x+y)^{\alpha})&=\sum_{k=0}^{\infty}\frac{(x+y)^{\alpha
k}}{\Gamma(\alpha k+1)}\\\ &\geq\sum_{k=0}^{\infty}\frac{\alpha}{\Gamma(\alpha
k+1)}\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}x^{\alpha j}y^{\alpha(k-j)}\\\
&=\alpha\sum_{k=0}^{\infty}\sum_{j=0}^{k}\frac{x^{\alpha
j}y^{\alpha(k-j)}}{\Gamma(\alpha j+1)\Gamma(\alpha(k-j)+1)}=\alpha
E_{\alpha}(x^{\alpha})E_{\alpha}(y^{\alpha}).\end{split}$ (1.4)

In the second line, we have used the following result.

######  Theorem 1.2 (Neo-classical inequality; Theorem 1.2 in [12]).

For $k\in\mathbb{N}$ and $0<\alpha<1,$ we have

$\alpha\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}x^{\alpha
j}y^{\alpha(k-j)}\leq(x+y)^{\alpha k},\quad x,y\geq 0.$ (1.5)

With a slightly weaker factor of $\alpha^{2}$ instead of $\alpha$, this result
was proven by Lyons in 1998, who also coined the term neo-classical
inequality, in a pioneering paper on rough path theory [15]. Later, it has
been applied by several other authors, see e.g. [2, 5, 8, 14]. Analogously to
(1.4), it is clear that Theorem 1.1 follows from the following new binomial
inequalities.

######  Theorem 1.3.

For $k\in\mathbb{N}$ and $0<\alpha<1,$ we have the converse neo-classical
inequality

$\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}x^{\alpha
j}y^{\alpha(k-j)}\geq(x+y)^{\alpha k},\quad x,y\geq 0,$ (1.6)

and for $\alpha>1$ we have

$\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}x^{\alpha
j}y^{\alpha(k-j)}\leq(x+y)^{\alpha k},\quad x,y\geq 0.$ (1.7)

These inequalities are strict for $x,y>0$.

The inequalities (1.5)–(1.7) look deceptively simple. For instance, it is not
obvious that the proof of (1.7) – at least with the approach used here – is
much harder when the integer $\lfloor\alpha\rfloor$ is even than when it is
odd. Lyons’s proof of (the weaker version of) (1.5) applies the maximum
principle for sub-parabolic functions in a non-trivial way. Hara and Hino [12]
use fractional calculus to derive an extension of the binomial theorem (see
Theorem 3.1 below), which immediately implies Theorem 1.2. This extended
binomal theorem will be our starting point when proving Theorem 1.3. We also
note that in a preliminary version of [7] (available at arXiv:1104.0577v2
[math.PR]), it was shown that a multinomial extension can be derived from
(1.5), by induction over the number of variables.

Our first theorem, Theorem 1.1, says that the function $\log
E_{\alpha}(x^{\alpha})$ is sub- resp. super-additive on
$\mathbb{R}^{+}=(0,\infty).$ In Section 2, we will show stronger statements
for $0<\alpha<2$: The function $E_{\alpha}(x^{\alpha})$ is log-concave for
$0<\alpha<1,$ and log-convex for $1<\alpha<2.$ Sections 3–6 are devoted to
proving Theorem 1.3. Some preliminaries and the plan of the proof are given in
Section 3. In that section we also state a conjecture concerning a converse
inequality to (1.7).

##  2 Proof of Theorem 1.1 for $0<\alpha<2$ and related statements

For brevity, we do not discuss strictness in this section. This would be
straightforward, and the strictness assertion in Theorem 1.1 will follow
anyways from Theorem 1.3. The following easy fact is well-known; see, e.g.,
[3].

######  Lemma 2.1.

Let $f:[0,\infty)\to\mathbb{R}$ be convex with $f(0)=0.$ Then $f$ is
superadditive, i.e.,

$f(x+y)\geq f(x)+f(y),\quad x,y\geq 0.$

Thus, the following theorem implies (1.2), and (1.3) for $1<\alpha<2$.

######  Theorem 2.2.

For $0<\alpha<1,$ the function $x\mapsto E_{\alpha}(x^{\alpha})$ is log-
concave on $\mathbb{R}^{+}$. For $1<\alpha<2,$ it is log-convex.

For $\alpha>2,$ it seems that $E_{\alpha}(x^{\alpha})$ is _not_ log-convex.
For instance, $E_{4}(x^{4})=\tfrac{1}{2}(\cos x+\cosh x),$ and

$\big{(}\log E_{4}(x^{4})\big{)}^{\prime\prime}=\frac{2\sin x\ \sinh x}{(\cos
x+\cosh x)^{2}}$

changes sign. To prove (1.3) for $\alpha>2,$ we thus rely on the binomial
inequality (1.7), which is proven later.

###### Proof of Theorem 2.2.

We start from the representation, given in (3.4) of [21],

$E_{\alpha}(x^{\alpha})=\frac{e^{x}}{\alpha}-\varphi_{\alpha}(x),\quad x>0,\
0<\alpha<1,$

where

$\varphi_{\alpha}(x):=\frac{\sin\alpha\pi}{\pi}\int_{0}^{\infty}\frac{t^{\alpha-1}e^{-xt}}{t^{2\alpha}-2\cos(\alpha\pi)t^{\alpha}+1}\,dt$

is a completely monotone function. (Recall that a smooth function $f$ on
$\mathbb{R}^{+}$ is completely monotone if

$(-1)^{n}f^{(n)}(x)\geq 0,\quad n\geq 0,\ x>0.)$

By well-known closure properties of completely monotone functions (see e.g.
Corollaries 1.6 and 1.7 in [20]),

$\frac{1}{E_{\alpha}(x^{\alpha})}=\alpha
e^{-x}\sum_{n=0}^{\infty}\big{(}\alpha e^{-x}\varphi_{\alpha}(x)\big{)}^{n}$

is completely monotone as well and hence log-convex. Thus,
$E_{\alpha}(x^{\alpha})$ is log-concave.

Now suppose that $1<\alpha<2.$ Setting $\beta=\alpha/2\in(1/2,1),$ we have

$E_{\alpha}(x^{\alpha})=\frac{E_{\beta}(x^{\beta})+E_{\beta}(-x^{\beta})}{2}=\frac{e^{x}}{\alpha}+\frac{E_{\beta}(-x^{\beta})-\varphi_{\beta}(x)}{2}\cdot$

Inserting the well-known representation

$E_{\beta}(-x^{\beta})=\frac{\sin\beta\pi}{\pi}\int_{0}^{\infty}\frac{t^{\beta-1}e^{-xt}}{t^{2\beta}+2\cos(\beta\pi)t^{\beta}+1}dt,$

which is e.g. a consequence of the Perron-Stieltjes inversion formula applied
to (3.7.7) in [9], and making some trigonometric simplifications, we get the
representation

$E_{\alpha}(x^{\alpha})=\frac{e^{x}}{\alpha}-\frac{\sin\alpha\pi}{\pi}\int_{0}^{\infty}\frac{t^{\alpha-1}e^{-xt}}{t^{2\alpha}-2\cos(\alpha\pi)t^{\alpha}+1}\,dt,$

which is also given in (3.6) of [21]. This implies

$\log(E_{\alpha}(x^{\alpha}))=x-\log\alpha+\log(\psi_{\alpha}(x))$

with

$\psi_{\alpha}(x):=1-\frac{\alpha\sin\alpha\pi}{\pi}\int_{0}^{\infty}\frac{t^{\alpha-1}e^{-x(1+t)}}{t^{2\alpha}-2\cos(\alpha\pi)t^{\alpha}+1}\,dt$

a completely monotone function. Thus, $E_{\alpha}(x^{\alpha})$ is log-convex.
∎

Observe that Theorem 2.2 extends to the boundary case $\alpha=2,$ where

$E_{2}(x^{2})=\cosh x$

is clearly log-convex. Let us also mention an alternative probabilistic
argument for (1.2) based on the $\alpha$-stable subordinator
$\\{Z_{t}^{(\alpha)},\;t\geq 0\\}$ with normalization
$\mathbb{E}\big{[}e^{-\lambda
Z_{t}^{(\alpha)}}\big{]}=e^{-t\lambda^{\alpha}}.$ It is indeed well-known –
see e.g. Exercise 50.7 in [19] – that

$E_{\alpha}(x^{\alpha})=\mathbb{E}[e^{R^{(\alpha)}_{x}}],$

where $R^{(\alpha)}_{x}:=\inf\\{t>0:Z^{(\alpha)}_{t}>x\\}.$ Now if
${\tilde{R}}^{(\alpha)}_{y}$ is an independent copy of $R^{(\alpha)}_{y},$ the
Markov property implies

$\displaystyle R^{(\alpha)}_{x}+{\tilde{R}}^{(\alpha)}_{y}$
$\displaystyle\stackrel{{\scriptstyle\mathrm{d}}}{{=}}\inf\big{\\{}t>\\!R^{(\alpha)}_{x}:Z_{t}^{(\alpha)}>Z_{R^{(\alpha)}_{x}}^{(\alpha)}+y\big{\\}}$
$\displaystyle\geq\inf\big{\\{}t>\\!R^{(\alpha)}_{x}:Z_{t}^{(\alpha)}\\!>x+y\big{\\}}$
$\displaystyle=\inf\big{\\{}t>0:Z_{t}^{(\alpha)}\\!>x+y\big{\\}}\stackrel{{\scriptstyle\mathrm{d}}}{{=}}R^{(\alpha)}_{x+y},$

where the inequality follows from the obvious fact that
$Z_{R^{(\alpha)}_{x}}^{(\alpha)}\geq x.$ This shows the desired inequality

$E_{\alpha}(x^{\alpha})E_{\alpha}(x^{\alpha})\geq E_{\alpha}((x+y)^{\alpha}).$

To conclude this section, we give some related results for the function
$E_{\alpha}(x).$

######  Proposition 2.3.

For $0<\alpha<1,$ the function $x\mapsto E_{\alpha}(x)$ is log-convex on
$\mathbb{R}$. For $\alpha>1,$ it is log-concave on $\mathbb{R}^{+}$.

###### Proof.

The logarithmic derivative of $E_{\alpha}(x)$ is the ratio of series

$\frac{E_{\alpha}^{\prime}(x)}{E_{\alpha}(x)}=\frac{{\displaystyle\sum_{n\geq
0}\frac{x^{n}}{\Gamma(\alpha+\alpha n)}}}{{\displaystyle\sum_{n\geq
0}\frac{\alpha\,x^{n}}{\Gamma(1+\alpha n)}}},$

and it is clear by log-convexity of the gamma function that the sequence

$n\mapsto\frac{\Gamma(1+\alpha n)}{\Gamma(\alpha+\alpha n)}$

is increasing for $\alpha<1$ and decreasing for $\alpha>1.$ By Biernacki and
Krzyż’s lemma (see [1]), this shows that

$x\mapsto\frac{E_{\alpha}^{\prime}(x)}{E_{\alpha}(x)}$

is non-decreasing on $\mathbb{R}^{+}$ for $\alpha<1$ and non-increasing on
$\mathbb{R}^{+}$ for $\alpha>1.$ Finally, the log-convexity of $E_{\alpha}(x)$
on the whole real line for $0<\alpha<1$ is a consequence of Hölder’s
inequality and the classic m.g.f. representation

$E_{\alpha}(x)=\mathbb{E}[e^{xR^{(\alpha)}_{1}}],$

where we have used the above notation and the easily established self-similar
identity
$R^{(\alpha)}_{x}\stackrel{{\scriptstyle\mathrm{d}}}{{=}}xR^{(\alpha)}_{1}.$ ∎

For $\alpha\geq 2,$ there is actually a stronger result. It has been shown by
Wiman [22] that the zeros of the Mittag-Leffler function are real and negative
for $\alpha\geq 2.$ As this function is of order $1/\alpha<1,$ the Hadamard
factorization theorem (Theorem XI.3.4 in [4]) implies that

$\frac{1}{E_{\alpha}(x)}=\prod_{n=1}^{\infty}\Big{(}1+\frac{x}{x_{n,\alpha}}\Big{)}^{-1},$

where

$0<x_{1,\alpha}\leq x_{2,\alpha}\leq\cdots$

are the absolute values of the zeros of $E_{\alpha}(x).$ We conclude that
$1/E_{\alpha}(x)$ is completely monotone, and thus log-convex, on
${\mathbb{R}}^{+}.$ An interesting open question is whether $1/E_{\alpha}(x)$
remains completely monotone on ${\mathbb{R}}^{+}$ for $1<\alpha<2.$
Unfortunately, the above argument fails because the large zeroes of
$E_{\alpha}(x)$ have non-trivial imaginary part (see Proposition 3.13 in [9]).

##  3 Proof of Theorem 1.3: preliminaries

By symmetry and scaling, it is clearly sufficient to prove

$\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}\lambda^{\alpha
j}>(1+\lambda)^{\alpha k},\quad\alpha\in(0,1),\ \lambda\in(0,1],\
k\in\mathbb{N},$ (3.1)

and

$\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}\lambda^{\alpha
j}<(1+\lambda)^{\alpha k},\quad\alpha>1,\ \lambda\in(0,1],\ k\in\mathbb{N}.$
(3.2)

As a sanity check, we verify these statements for $\lambda>0$ sufficiently
small:

$1+\binom{\alpha
k}{\alpha}\lambda^{\alpha}+\mathrm{O}(\lambda^{2\alpha})>1+\alpha
k\lambda+\mathrm{O}(\lambda^{2}),\quad\alpha\in(0,1),\ k\in\mathbb{N},$

and

$1+\mathrm{O}(\lambda^{\alpha})<1+\alpha
k\lambda+\mathrm{O}(\lambda^{2}),\quad\alpha>1,\ k\in\mathbb{N}.$

These inequalities are obviously correct for small $\lambda>0,$ the first one
by $\binom{\alpha k}{\alpha}>0.$ We now recall a remarkable generalization of
the binomial theorem, due to Hara and Hino [12] . Their proof, using
fractional Taylor series, builds on earlier work by Osler [17]. Following
[12], for $\alpha>0$ define

$\displaystyle K_{\alpha}$
$\displaystyle:=\\{\omega\in\mathbb{C}:\omega^{\alpha}=1\\}$
$\displaystyle=\\{e^{i\theta}:-\pi<\theta\leq\pi,\ e^{i\theta\alpha}=1\\}$
$\displaystyle=\Big{\\{}\exp\Big{(}\frac{2k\pi
i}{\alpha}\Big{)}:k\in\mathbb{Z},\
-\frac{\alpha}{2}<k\leq\frac{\alpha}{2}\Big{\\}}.$ (3.3)

For $t,\lambda\in(0,1]$ and $k\in\mathbb{N},$ define

$F(t,\lambda,k):=t^{\alpha-1}(1-t)^{\alpha
k}\bigg{(}\frac{1}{|t^{\alpha}-\lambda^{\alpha}e^{-i\alpha\pi}|^{2}}+\frac{\lambda^{\alpha
k}}{|e^{-i\alpha\pi}-(\lambda t)^{\alpha}|^{2}}\bigg{)}.$ (3.4)

######  Theorem 3.1 (Theorem 3.2 in [12]).

Let $\alpha>0,$ $\lambda\in(0,1]$ and $k\in\mathbb{N}_{0}.$ Then we have

$\alpha\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}\lambda^{\alpha
j}=\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha
k}-\frac{\alpha\lambda^{\alpha}\sin\alpha\pi}{\pi}\int_{0}^{1}F(t,\lambda,k)dt.$
(3.5)

In [12], the theorem was stated for $k\in\mathbb{N}$, but it is not hard to
see that the proof also works for $k=0.$ Clearly, the classical binomial
theorem is recovered from (3.5) by putting $\alpha=1.$ As noted in [12], for
$\alpha\in(0,1)$ we have $K_{\alpha}=\\{1\\},$ and thus Theorem 1.2 is an
immediate consequence of Theorem 3.1. Hara and Hino also mention that Theorem
3.1 implies

$\alpha\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}x^{\alpha
j}y^{\alpha(k-j)}\geq(x+y)^{\alpha k},\quad x,y\geq 0,\ \alpha\in(1,2],$

a partial converse to (1.7). It seems that this inequality does not hold for
$\alpha>2.$ We leave it to future research to find an appropriate inequality
comparing the binomial sum with $(x+y)^{\alpha k}$ for $\alpha>2.$ Different
methods than in the subsequent sections will be required, as a _lower_
estimate for $\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha k}$ is
needed. The following statement might be true:

######  Conjecture 3.2.

For $k\in\mathbb{N}$ and $\alpha>2,$ we have

$2^{\alpha-1}\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}x^{\alpha
j}y^{\alpha(k-j)}\geq(x+y)^{\alpha k},\quad x,y\geq 0.$

The factor $2^{\alpha-1}=\sup_{0<\lambda\leq
1}(1+\lambda)^{\alpha}/(1+\lambda^{\alpha})$ would be sharp for $k=1,$
$x=\lambda\in(0,1],$ $y=1.$ In the following sections, we will use arguments
based on Theorem 3.1 to prove (3.1) and (3.2), which imply Theorem 1.3, from
which Theorem 1.1 follows. The proof of (3.1) is presented in Section 4. It
profits from the fact that

$\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha k}=(1+\lambda)^{\alpha
k},\quad\alpha\in(0,1),$

and requires only obvious properties of the function $F$ defined in (3.4). At
the beginning of Section 5, we show that, for $2\leq\alpha\in\mathbb{N},$ the
inequality (3.2) immediately follows from the classical binomial theorem. We
then continue with the case where $\alpha>1$ is not an integer, and
$\lfloor\alpha\rfloor$ is odd. Then, the set $K_{\alpha}$ has
$\lfloor\alpha\rfloor$ elements, and the crude estimate

$\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha
k}\leq\lfloor\alpha\rfloor(1+\lambda)^{\alpha k}$

suffices to show (3.2), again using only simple properties of $F.$ The case
where $\lfloor\alpha\rfloor$ is even is more involved, and is handled in
Section 6. In this case, $|K_{\alpha}|=\lceil\alpha\rceil$, and the obvious
estimate

$\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha
k}\leq\lceil\alpha\rceil(1+\lambda)^{\alpha k}$ (3.6)

is too weak to lead anywhere. We first show (Lemma 6.1) that, for
$\lambda\in[\tfrac{1}{2},1],$ the right hand side of (3.6) can be strengthened
to $\alpha(1+\lambda)^{\alpha k},$ and that (3.2) easily follows from this for
these values of $\lambda.$ For smaller $\lambda>0,$ more precise estimates for
the sum $\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha k}$ and the
integral in (3.5) are needed, which are developed in the remainder of Section
6.

##  4 Proof of Theorem 1.3 for $0<\alpha<1$

As mentioned above, it suffices to prove (3.1). For $\alpha\in(0,1),$ we have
$K_{\alpha}=\\{1\\}.$ Since $\sin\alpha\pi>0,$ we see from (3.5) that the
desired inequality (3.1) is equivalent to

$\int_{0}^{1}F(t,\lambda,k)dt<G(\lambda,k),\quad\lambda\in(0,1],\
k\in\mathbb{N},$ (4.1)

where

$G(\lambda,k):=\frac{\pi(1-\alpha)}{\alpha\lambda^{\alpha}\sin\alpha\pi}(1+\lambda)^{\alpha
k}.$

Define

$\tilde{\delta}:=\inf_{k\in\mathbb{N}}\Big{(}(1+\lambda)^{\alpha
k}-1)\Big{)}^{\frac{1}{\alpha
k}}=(1+\lambda)\inf_{k\in\mathbb{N}}\Big{(}1-(1+\lambda)^{-\alpha
k}\Big{)}^{\frac{1}{\alpha k}}>0.$

This number is positive, as it is defined by the infimum of a sequence of
positive numbers converging to a positive limit. Moreover, we define

$\delta:=\tfrac{1}{2}\min\\{1,\tilde{\delta}\\}>0.$

By definition, $F(t,\lambda,k)$ decreases w.r.t. $k$, and so

$F(t,\lambda,k)\leq F(t,\lambda,0),\quad t,\lambda\in(0,1],\ k\in\mathbb{N}.$

Applying Theorem 3.1 with $k=0$ yields

$\int_{0}^{1}F(t,\lambda,0)dt=G(\lambda,0),\quad\lambda\in(0,1].$

By these two observations,

$\int_{0}^{1-\delta}F(t,\lambda,k)dt\leq\int_{0}^{1-\delta}F(t,\lambda,0)dt\leq
G(\lambda,0).$ (4.2)

Since $\lambda^{\alpha k}\leq 1,$ and $(1-t)^{\alpha k}$ decreases w.r.t. $t$,
it is clear from the definition of $F$ that

$\int_{1-\delta}^{1}F(t,\lambda,k)dt\leq\delta^{\alpha
k}\int_{0}^{1}F(t,\lambda,0)dt=\delta^{\alpha k}G(\lambda,0).$ (4.3)

By definition of $\delta$, we have

$1+\delta^{\alpha k}<(1+\lambda)^{\alpha k},\quad\lambda\in(0,1],\
k\in\mathbb{N}.$ (4.4)

Now note that

$\int_{0}^{1}F(t,\lambda,k)dt\leq(1+\delta^{\alpha
k})G(\lambda,0)<G(\lambda,k),$ (4.5)

where the first estimate follows from (4.2) and (4.3), and the second one from
(4.4). Thus, (4.1) is established.

##  5 Proof of Theorem 1.3 for $\alpha\in\mathbb{N}$ or $\lfloor\alpha\rfloor$
odd

If $2\leq\alpha\in\mathbb{N}$ is an integer, then the proof of (3.2) is very
easy, as we are dealing with a classical binomial sum with some summands
removed:

$\sum_{j=0}^{k}\binom{\alpha k}{\alpha j}\lambda^{\alpha j}<\sum_{j=0}^{\alpha
k}\binom{\alpha k}{j}\lambda^{j}=(1+\lambda)^{\alpha k}.$

In the remainder of this section, we prove (3.2) for
$1<\alpha\notin\mathbb{N}$ with $\lfloor\alpha\rfloor$ odd. Our approach is
similar to the preceding section. First, observe that

$\displaystyle|K_{\alpha}|=\Big{\lfloor}\frac{\alpha}{2}\Big{\rfloor}-\Big{\lfloor}{-\frac{\alpha}{2}}\Big{\rfloor}=\lfloor\alpha\rfloor,$

and that

$\displaystyle\alpha(1+\lambda)^{\alpha k}$ $\displaystyle-\sum_{\omega\in
K_{\alpha}}(1+\lambda\omega)^{\alpha k}$
$\displaystyle\geq\alpha(1+\lambda)^{\alpha
k}-\lfloor\alpha\rfloor(1+\lambda)^{\alpha k}>\alpha-\lfloor\alpha\rfloor.$

Therefore, the sequence

$A_{k}:=\bigg{(}\frac{\alpha(1+\lambda)^{\alpha k}-\sum_{\omega\in
K_{\alpha}}(1+\lambda\omega)^{\alpha
k}}{\alpha-\lfloor\alpha\rfloor}-1\bigg{)}^{\frac{1}{\alpha k}},\quad
k\in\mathbb{N},$

is well-defined and positive, and

$\displaystyle\lim_{k\to\infty}A_{k}$
$\displaystyle=(1+\lambda)\lim_{k\to\infty}(\alpha-1)^{\frac{1}{\alpha k}}$
$\displaystyle\qquad\qquad\times\bigg{(}\frac{1-\sum_{\omega\in
K_{\alpha}\setminus\\{1\\}}\frac{(1+\lambda\omega)^{\alpha
k}}{(\alpha-1)(1+\lambda)^{\alpha
k}}}{\alpha-\lfloor\alpha\rfloor}-\frac{1}{(\alpha-1)(1+\lambda)^{\alpha
k}}\bigg{)}^{\frac{1}{\alpha k}}$ $\displaystyle=1+\lambda>0.$

We can thus define the positive number

$\hat{\delta}:=\frac{1}{2}\min\Big{\\{}1,\inf_{k\in\mathbb{N}}A_{k}\Big{\\}}>0.$

Since odd $\lfloor\alpha\rfloor$ implies $\sin\alpha\pi<0$ for
$\alpha\notin\mathbb{N},$ it is clear from (3.5) that (3.2) is equivalent to

$\int_{0}^{1}F(t,\lambda,k)dt<\hat{G}(\lambda,k),\quad\lambda\in(0,1],\
k\in\mathbb{N},$ (5.1)

where

$\hat{G}(\lambda,k):=-\frac{\pi}{\alpha\lambda^{\alpha}\sin\alpha\pi}\Big{(}\alpha(1+\lambda)^{\alpha
k}-\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha k}\Big{)}.$

By the same argument that gave us the first inequality in (4.5), we find

$\int_{0}^{1}F(t,\lambda,k)dt\leq(1+\hat{\delta}^{\alpha
k})\hat{G}(\lambda,0),\qquad\lambda\in(0,1],\ k\in\mathbb{N}.$

Note that

$\hat{G}(\lambda,0)=-\frac{\pi}{\alpha\lambda^{\alpha}\sin\alpha\pi}(\alpha-\lfloor\alpha\rfloor).$

The proof of (3.2) with $\lfloor\alpha\rfloor$ odd will thus be finished if we
can show that

$(1+\hat{\delta}^{\alpha
k})\hat{G}(\lambda,0)<\hat{G}(\lambda,k),\quad\lambda\in(0,1],\
k\in\mathbb{N}.$

But this is equivalent to

$(1+\hat{\delta}^{\alpha
k})(\alpha-\lfloor\alpha\rfloor)<\Big{(}\alpha(1+\lambda)^{\alpha
k}-\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha k}\Big{)},$

which follows from the definition of $\hat{\delta}.$

##  6 Proof of Theorem 1.3 for $\lfloor\alpha\rfloor$ even

We now prove (3.2) in the case that

$2<\alpha\notin\mathbb{N},\quad\lfloor\alpha\rfloor=2m,\quad m\in\mathbb{N}.$
(6.1)

As $\sin\alpha\pi>0,$ it follows from (3.5) that (3.2) is equivalent to

$\int_{0}^{1}F(t,\lambda,k)dt>\frac{\pi}{\alpha\lambda^{\alpha}\sin\alpha\pi}\Big{(}\sum_{\omega\in
K_{\alpha}}(1+\lambda\omega)^{\alpha k}-\alpha(1+\lambda)^{\alpha
k}\Big{)},\\\ \lambda\in(0,1],\ k\in\mathbb{N}.$ (6.2)

Since $F(\cdot,\lambda,k)$ is positive on $(0,1),$ the following lemma
establishes this for $\lambda\in[\tfrac{1}{2},1].$

######  Lemma 6.1.

Let $\alpha>2$ be as in (6.1). Then we have

$\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha
k}\leq\alpha(1+\lambda)^{\alpha k},\quad\lambda\in[\tfrac{1}{2},1],\
k\in\mathbb{N}.$

###### Proof.

By (3.3), we have

$\displaystyle\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha k}$
$\displaystyle\leq\sum_{\omega\in K_{\alpha}}|1+\lambda\omega|^{\alpha k}$
$\displaystyle=\sum_{j=-m}^{m}\Big{(}1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}\Big{)}^{\alpha
k/2},\quad\lambda\in(0,1],\ k\in\mathbb{N}.$ (6.3)

We will show that

$\sum_{j=-m}^{m}\Big{(}1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}\Big{)}\leq\alpha(1+\lambda)^{2},\quad\lambda\in[\tfrac{1}{2},1].$
(6.4)

Then, (6.3) and (6.4) imply

$\displaystyle\sum_{j=-m}^{m}\Big{(}1+$ $\displaystyle
2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}\Big{)}^{\alpha
k/2}=(1+\lambda)^{\alpha
k}\sum_{j=-m}^{m}\bigg{(}\frac{1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}}{(1+\lambda)^{2}}\bigg{)}^{\alpha
k/2}$ $\displaystyle\qquad\qquad\leq(1+\lambda)^{\alpha
k}\sum_{j=-m}^{m}\frac{1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}}{(1+\lambda)^{2}}\leq\alpha(1+\lambda)^{\alpha
k},$

which proves the lemma. To prove (6.4), observe that (6.1) implies

$\displaystyle|K_{\alpha}|=\Big{\lfloor}\frac{\alpha}{2}\Big{\rfloor}-\Big{\lfloor}{-\frac{\alpha}{2}}\Big{\rfloor}=m-(-m-1)=2m+1=\lceil\alpha\rceil.$

Using the geometric series to evaluate the cosine sum (see, e.g., p. 102 in
[16]), we obtain

$\sum_{j=-m}^{m}\Big{(}1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}\Big{)}=\lceil\alpha\rceil(1+\lambda^{2})\\\
+2\lambda\bigg{(}1+2\frac{\cos\big{(}(m+1)\pi/\alpha\big{)}\sin(m\pi/\alpha)}{\sin(\pi/\alpha)}\bigg{)}.$
(6.5)

Here, $\cos\big{(}(m+1)\pi/\alpha\big{)}<0,$ and the sines are both positive.
By (6.5) and the elementary inequalities

$\displaystyle\cos x$ $\displaystyle\leq-1+\tfrac{1}{2}(x-\pi)^{2},\quad
x\in\mathbb{R},$ $\displaystyle\sin x$ $\displaystyle\leq x,\quad x\geq 0,$
$\displaystyle\sin x$ $\displaystyle\geq x-\tfrac{1}{6}x^{3},\quad x\geq 0,$

the following statement is sufficient for the validity of (6.4):

$(A+1)(1+\lambda^{2})+2\lambda\bigg{(}1+2\frac{\big{(}{-1}+\frac{1}{2}\big{(}\frac{(M+1)\pi}{A}-\pi\big{)}^{2}\big{)}\frac{M\pi}{A}}{\frac{\pi}{A}-\frac{1}{6}(\frac{\pi}{A})^{3}}\bigg{)}\leq
A(1+\lambda)^{2},\\\ \quad A>2,\ M>0,\ 2M<A<2M+1,\ \tfrac{1}{2}\leq\lambda\leq
1.$ (6.6)

This is a polynomial inequality in real variables with polynomial constraints,
which can be verified by cylindrical algebraic decomposition, using a computer
algebra system. For instance, using Mathematica’s Reduce command on (6.6),
with the first $\leq$ replaced by $>$, yields False. This shows that (6.6) is
correct, which finishes the proof. ∎

The following two lemmas will be required for some estimates of the function
$F(t,\lambda,k)$ for small $\lambda.$

######  Lemma 6.2.

For $\alpha$ as in (6.1), we have

$\int_{0}^{\infty}\frac{s^{\alpha}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds=\frac{\pi\sin\big{(}\frac{\lfloor\alpha\rfloor+1}{\alpha}\pi\big{)}}{\alpha\sin(\alpha\pi)\sin\big{(}\frac{\alpha+1}{\alpha}\pi\big{)}}$
(6.7)

and

$\int_{0}^{\infty}\frac{s^{\alpha-1}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds=\frac{\pi(\lceil\alpha\rceil-\alpha)}{\alpha\sin\alpha\pi}.$
(6.8)

###### Proof.

By substituting $s^{\alpha}=w,$

$\displaystyle\int_{0}^{\infty}\frac{s^{\alpha}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds$
$\displaystyle=\frac{1}{\alpha}\int_{0}^{\infty}\frac{w^{1/\alpha}}{w^{2}-2w\cos\alpha\pi+1}dw$
$\displaystyle=\frac{1}{\alpha}\int_{0}^{\infty}\frac{w^{1/\alpha}}{w^{2}+2w\cos(\pi(\alpha-\lfloor\alpha\rfloor-1))+1}dw,$

where we have used that $\lfloor\alpha\rfloor$ is even. The first formula now
follows from 3.242 on p. 322 of [10], with $m=1/(2\alpha),$ $n=\tfrac{1}{2}$
and $t=\pi(\alpha-\lfloor\alpha\rfloor-1).$ As for (6.8),

$\displaystyle\int_{0}^{\infty}\frac{s^{\alpha-1}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds$
$\displaystyle=\frac{1}{\alpha}\int_{0}^{\infty}\frac{1}{w^{2}-2w\cos\alpha\pi+1}dw$
$\displaystyle=\frac{1}{\alpha}\int_{0}^{\infty}\frac{1}{w^{2}+2w\cos(\pi(\alpha-\lfloor\alpha\rfloor-1))+1}dw.$

The identity (6.8) then follows from 11a) on p. 14 of [11], with

$\lambda=\pi(\alpha-\lfloor\alpha\rfloor-1)=\pi(\alpha-\lceil\alpha\rceil).\qed$

######  Lemma 6.3.

Again, suppose that $\alpha$ satisfies (6.1). For $k\in\mathbb{N},$ we have

$\int_{0}^{1/\lambda}\frac{s^{\alpha-1}(1-\lambda s)^{\alpha
k}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds=\int_{0}^{\infty}\frac{s^{\alpha-1}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds\\\
-\alpha
k\lambda\int_{0}^{\infty}\frac{s^{\alpha}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds+\mathrm{o}(\lambda),\quad\lambda\downarrow
0.$

###### Proof.

Fix some $\beta\in(\tfrac{1}{\alpha},\tfrac{1}{2}).$ We have

$\int_{\lambda^{-\beta}}^{\infty}\frac{s^{\alpha-1}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds=\mathrm{O}(\lambda^{\alpha\beta})=\mathrm{o}(\lambda),\quad\lambda\downarrow
0,$ (6.9)

and

$\int_{\lambda^{-\beta}}^{\infty}\frac{s^{\alpha}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds=\mathrm{O}(\lambda^{\beta(\alpha-1)})=\mathrm{o}(1),\quad\lambda\downarrow
0.$ (6.10)

These assertions easily follow from the fact that the integrand is of order
$\mathrm{O}(s^{-\alpha-1})$ resp. $\mathrm{O}(s^{-\alpha})$ at infinity.
Moreover, we have the uniform expansion

$\displaystyle(1-\lambda s)^{\alpha k}$ $\displaystyle=1-\alpha k\lambda
s+\mathrm{O}(\lambda^{2-2\beta})$ $\displaystyle=1-\alpha k\lambda
s+\mathrm{o}(\lambda),\quad 0\leq s\leq\lambda^{-\beta},\ \lambda\downarrow
0.$ (6.11)

Since $0\leq(1-\lambda s)^{\alpha k}\leq 1,$ (6.9) implies

$\displaystyle\int_{0}^{1/\lambda}\frac{s^{\alpha-1}(1-\lambda s)^{\alpha
k}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds$
$\displaystyle=\int_{0}^{\lambda^{-\beta}}\frac{s^{\alpha-1}(1-\lambda
s)^{\alpha k}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds+\mathrm{o}(\lambda).$

The statement now follows from (6.11), (6.10) and (6.9). ∎

We now continue the proof of (6.2). From the definition of $F$, it is clear
that

$\displaystyle\int_{0}^{1}F(t,\lambda,k)dt$
$\displaystyle\geq\int_{0}^{1}\frac{t^{\alpha-1}(1-t)^{\alpha
k}}{|t^{\alpha}-\lambda^{\alpha}e^{-i\alpha\pi}|^{2}}dt$
$\displaystyle=\lambda^{-\alpha}\int_{0}^{1/\lambda}\frac{s^{\alpha-1}(1-\lambda
s)^{\alpha k}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds.$ (6.12)

As noted above, from (3.3), we have

$\displaystyle\sum_{\omega\in K_{\alpha}}(1+\lambda\omega)^{\alpha k}$
$\displaystyle\leq\sum_{\omega\in K_{\alpha}}|1+\lambda\omega|^{\alpha k}$
$\displaystyle=1+2\sum_{j=1}^{m}\Big{(}1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}\Big{)}^{\alpha
k/2},\quad\lambda\in(0,1],\ k\in\mathbb{N}.$ (6.13)

Since Lemma 6.1 settles the case $\lambda\in[\tfrac{1}{2},1],$ we may assume
$\lambda\in(0,\tfrac{1}{2})$ in what follows. Using (6.12) and (6.13) in
(6.2), we see that it is sufficient to show

$2\sum_{j=1}^{m}\Big{(}1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}\Big{)}^{\alpha
k/2}+(1-\alpha)(1+\lambda)^{\alpha k}\\\
-\frac{\alpha\sin\alpha\pi}{\pi}\int_{0}^{1/\lambda}\frac{s^{\alpha-1}(1-\lambda
s)^{\alpha
k}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds<0,\quad\lambda\in(0,\tfrac{1}{2}),\
k\in\mathbb{N}.$ (6.14)

This will be proven in the following two lemmas.

######  Lemma 6.4.

For $\lambda\in(0,\tfrac{1}{2}),$ the left hand side of (6.14) decreases
w.r.t. $\lambda$.

###### Proof.

The derivative of this expression is

$\displaystyle 2\alpha k\sum_{j=1}^{m}$
$\displaystyle\Big{(}\cos\frac{2j\pi}{\alpha}+\lambda\Big{)}\Big{(}1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}\Big{)}^{\alpha
k/2-1}+\alpha k(1-\alpha)(1+\lambda)^{\alpha k-1}$
$\displaystyle\qquad+\frac{\alpha\sin\alpha\pi}{\pi}\int_{0}^{1/\lambda}\frac{s^{\alpha}(1-\lambda
s)^{\alpha k-1}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds$ $\displaystyle\leq
2\alpha k\sum_{1\leq
j\leq\alpha/3}\Big{(}\cos\frac{2j\pi}{\alpha}+\lambda\Big{)}\Big{(}1+2\lambda\cos\frac{2j\pi}{\alpha}+\lambda^{2}\Big{)}^{\alpha
k/2-1}$ $\displaystyle\qquad+\alpha k(1-\alpha)(1+\lambda)^{\alpha
k-1}+\frac{\alpha\sin\alpha\pi}{\pi}\int_{0}^{\infty}\frac{s^{\alpha}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds.$
(6.15)

Note that it is easy to show that $\cos(2j\pi/\alpha)+\lambda<0$ for
$j>\alpha/3,$ which is where we use our assumption that
$\lambda<\tfrac{1}{2}$. Thus, we are discarding only negative terms when
passing from $\sum_{j=1}^{m}$ to $\sum_{1\leq j\leq\alpha/3}$ in (6.15). By
Lemma 6.2, the last term in (6.15) satisfies

$\frac{\alpha\sin\alpha\pi}{\pi}\int_{0}^{\infty}\frac{s^{\alpha}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds=\frac{\sin\big{(}\frac{\lfloor\alpha\rfloor+1}{\alpha}\pi\big{)}}{\sin\big{(}\frac{\alpha+1}{\alpha}\pi\big{)}}<1,$
(6.16)

where the inequality follows from

$1<\frac{\lfloor\alpha\rfloor+1}{\alpha}<\frac{\alpha+1}{\alpha}<\frac{3}{2}.$
(6.17)

We can thus estimate (6.15) further by

$\displaystyle 2\alpha k\sum_{1\leq j\leq\alpha/3}(1+\lambda)$
$\displaystyle\big{(}(1+\lambda)^{2}\big{)}^{\alpha k/2-1}+\alpha
k(1-\alpha)(1+\lambda)^{\alpha k-1}+1$ $\displaystyle=2\alpha
k\Big{\lfloor}\frac{\alpha}{3}\Big{\rfloor}(1+\lambda)^{\alpha k-1}+\alpha
k(1-\alpha)(1+\lambda)^{\alpha k-1}+1$ $\displaystyle\leq\alpha
k(1+\lambda)^{\alpha
k-1}\Big{(}2\Big{\lfloor}\frac{\alpha}{3}\Big{\rfloor}+2-\alpha\Big{)}\leq 0.$

Indeed, it is easy to see that $2\lfloor\alpha/3\rfloor+2-\alpha\leq 0$ for
$\alpha$ as in (6.1). ∎

Presumably, the preceding lemma can be extended to $\lambda\in(0,1],$ but this
would require a much better estimate for $\sum_{\omega\in
K_{\alpha}}(1+\lambda\omega)^{\alpha k}$ than the one we have used. The proof
of (3.2) could then possibly be streamlined, because Lemma 6.1 would no longer
be required.

######  Lemma 6.5.

Let $\alpha$ be as in (6.1), and $k\in\mathbb{N}.$ Then (6.14) holds for
$\lambda>0$ sufficiently small.

###### Proof.

Since $m=(\lceil\alpha\rceil-1)/2,$ expanding the first terms of (6.14) gives

$\lceil\alpha\rceil-\alpha+\alpha
k\bigg{(}2\sum_{j=1}^{m}\cos\frac{2j\pi}{\alpha}+1-\alpha\bigg{)}\lambda+\mathrm{O}(\lambda^{2})\\\
-\frac{\alpha\sin\alpha\pi}{\pi}\int_{0}^{1/\lambda}\frac{s^{\alpha-1}(1-\lambda
s)^{\alpha k}}{s^{2\alpha}-2s^{\alpha}\cos\alpha\pi+1}ds.$

By Lemmas 6.2 and 6.3, this further equals

$\displaystyle\lceil\alpha\rceil-\alpha+\alpha$ $\displaystyle
k\bigg{(}2\sum_{j=1}^{m}\cos\frac{2j\pi}{\alpha}+1-\alpha\bigg{)}\lambda$
$\displaystyle\qquad\qquad\quad+\alpha-\lceil\alpha\rceil+\alpha
k\lambda\frac{\sin\big{(}\frac{\lfloor\alpha\rfloor+1}{\alpha}\pi\big{)}}{\sin\big{(}\frac{\alpha+1}{\alpha}\pi\big{)}}+\mathrm{o}(\lambda)$
$\displaystyle=\alpha
k\Bigg{(}2\sum_{j=1}^{m}\cos\frac{2j\pi}{\alpha}+1-\alpha+\frac{\sin\big{(}\frac{\lfloor\alpha\rfloor+1}{\alpha}\pi\big{)}}{\sin\big{(}\frac{\alpha+1}{\alpha}\pi\big{)}}\Bigg{)}\lambda+\mathrm{o}(\lambda),$
(6.18)

and we see that (6.14) becomes sharp as $\lambda\downarrow 0,$ as the left
hand side is $\mathrm{O}(\lambda).$ This is no surprise, since the inequality
(3.2) we are proving is obviously sharp for $\lambda\downarrow 0.$ It remains
to show that the coefficient of $\lambda$ in (6.18) is negative. Similarly to
(6.15), we have the bound

$\displaystyle 2\sum_{j=1}^{m}\cos\frac{2j\pi}{\alpha}\leq 2\sum_{1\leq
j\leq\alpha/3}\cos\frac{2j\pi}{\alpha}\leq
2\Big{\lfloor}\frac{\alpha}{3}\Big{\rfloor}.$

Thus, we wish to show that

$2\Big{\lfloor}\frac{\alpha}{3}\Big{\rfloor}+1-\alpha+\frac{\sin\big{(}\frac{\lfloor\alpha\rfloor+1}{\alpha}\pi\big{)}}{\sin\big{(}\frac{\alpha+1}{\alpha}\pi\big{)}}<0.$
(6.19)

As the sine quotient is $<1,$ by (6.16), this is clearly true for $\alpha>6.$
Now consider $\alpha\in(4,5).$ It is easy to verify that

$\sin\Big{(}\frac{4+1}{\alpha}\pi\Big{)}<-\frac{1}{\sqrt{2}}+\frac{6\pi(\alpha-4)}{16\sqrt{2}},\quad
4<\alpha<5,$

and that

$\sin\Big{(}\frac{\alpha+1}{\alpha}\pi\Big{)}>-\frac{1}{\sqrt{2}}+\frac{\pi(\alpha-4)}{20\sqrt{2}},\quad
4<\alpha<5.$

Using these estimates in (6.19) leads to a quadratic inequality, which is
straightforward to check. The proof of (6.19) for $\alpha\in(2,3)$ is similar.
∎

Clearly, Lemmas 6.4 and 6.5 establish (6.14). As argued above (6.14), we have
thus proven (6.2), hence (3.2), and the proof of Theorem 1.3 is complete.

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which implies that $\mathop{\mathrm{Fix}}R$ has a null Lebesgue measure. Since
the remainder
$\mathop{\mathrm{Cl}}(\mathop{\mathrm{Fix}}R)\smallsetminus\mathop{\mathrm{Fix}}R$
is countable, also $\mathop{\mathrm{Cl}}(\mathop{\mathrm{Fix}}R)$ has a null
Lebesgue measure.

As a consequence, $\mathop{\mathrm{Cl}}(\mathop{\mathrm{Fix}}R)$ is
topologically 0-dimensional. On the other hand, it is a perfect set, because
for every $a\in\mathchoice{\text{\eusmtwelve A}}{\text{\eusmtwelve
A}}{\text{\eusmeight A}}{\text{\eusmsix A}}$, we have
$\xi({\langle{a}\rangle}^{\mkern-3.0mu\scriptscriptstyle\text{F}})=\lim_{n\to\infty}\xi({\langle{a_{n}}\rangle}^{\mkern-3.0mu\scriptscriptstyle\text{F}})$
with $a_{n}\in\mathchoice{\text{\eusmtwelve A}}{\text{\eusmtwelve
A}}{\text{\eusmeight A}}{\text{\eusmsix A}}$ defined by
$(a_{n})_{i}=a_{i}+\delta_{i,n}$, for any $i,n\geq 1$. So,
$\mathop{\mathrm{Cl}}(\mathop{\mathrm{Fix}}R)$ is a Cantor set (see [23]). ∎

We observe that the proof of Proposition 6.6 is essentially based on the
parametrization of $\mathop{\mathrm{Fix}}\rho$ given by Lemma 6.5, which is
nothing else than an explicit reformulation in the case of fixed points of the
general construction provided in the proof of Proposition 6.4 of periodic
points of any period. So, we expect that similar results hold for periodic
points of any given period.

The idea is that, in order to generate all the periodic orbits of a given
period $n>1$, one could start from a certain finite set of “simplest” ones,
and than progressively enlarge this set by iterating in a suitable order the
following procedure. Take any such $n$-orbit $(b_{1},b_{2},\dots,b_{n})$,
insert some pairs of consecutive 0’s in any position of any $b_{i}$, and then
reconstruct all the orbit accordingly.

For example, starting from the 2-cycle $(1^{\infty},(01)^{\infty})$ of $\rho$,
corresponding to the 2-cycle $C_{1}$ of $R$ (see Proposition 6.3), one can
obtain in this way the Thue-Morse 2-cycle $(u,\widetilde{u})$ of $\rho$
described in the next remark, and hence a corresponding 2-cycle for $R$.

######  Remark 6.7.

Consider the well known Thue-Morse substitution rule on B

$\widehat{\tau}:\left\\{\vrule
width=0.0pt,height=12.0pt\right.\\!\\!\\!\begin{array}[]{ll}0\mapsto 01\\\
1\mapsto 10\end{array},$

and the digit-wise generated map $\widehat{\tau}:\mathchoice{\text{\eusmtwelve
B}}{\text{\eusmtwelve B}}{\text{\eusmeight B}}{\text{\eusmsix
B}}\to\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve B}}{\text{\eusmeight
B}}{\text{\eusmsix B}}$. The map $\widehat{\tau}$ has two fixed points, the
Thue-Morse sequence starting with $0$

$u=01101001100101101001011001101001\dots$

and its complementary sequence $\widetilde{u}$, the Thue-Morse sequence
starting with $1$, which was used in [42] to found the combinatorics of words.
Recalling the definition of $\,\overline{\tau}$ in (13) and noting that

$\overline{\tau}^{2}(0)=\widehat{\tau}^{2}(0)=0110\quad\text{and}\quad\overline{\tau}^{2}(1)=\widehat{\tau}^{2}(1)=1001\,,$

we can easily verify that $(u,\widetilde{u})$ is a 2-cycle of $\rho$. Then,
letting $t=\xi(u)$ be the Thue-Morse number, we have the corresponding 2-cycle
$(t,1-t)$ of $R$.

Concerning the periodic points of $R$ having higher order $n$, in analogy with
the description of the maximal fixed point $x^{0}$, we limit ourselves to
consider the periodic points whose binary expansion is the “simplest” infinite
binary word among those generated by the construction described in the proof
of Proposition 6.4. Namely, the binary word obtained by starting from the word
$w_{0}=0$ and choosing each $w_{k}$ to be the shortest word in
$(\rho^{n}_{w_{0}w_{1}\dots w_{k-1}})^{-1}(w_{k-1})$, that is the image of
$w_{k-1}$ under a suitable composition of $\sigma$’s and
$\widetilde{\sigma}$’s.

The case of $n$ even is not so interesting, since we always obtain
$0.1^{\infty}$, while the binary expression of the “simplest” periodic point
of odd order $n=2\ell+1\geq 3$ is given by

$\begin{array}[]{ll}x^{\ell}=0.(01)^{2^{\ell-1}}\\!\prod_{h=0}^{\infty}&\\!\\!\\!\\!\\!\big{(}1^{2^{(2\ell+1)h+\ell}}(01)^{2^{(2\ell+1)h+\ell-1}}1^{(2^{\ell}-1)\,2^{(2\ell+1)h+\ell}}\\\\[2.0pt]
&\\!\\!\\!\\!\\!\ \
(01)^{2^{(2\ell+1)h+2\ell}}1^{2^{(2\ell+1)h+2\ell}}(01)^{(2^{\ell}-1)\,2^{(2\ell+1)h+2\ell}}\big{)}\,.\end{array}$
(54)

In order to see that this point has order $n=2\ell+1$, we write it as
$0.w\prod_{h=0}^{\infty}(u_{h}v_{h})$, with

$\begin{array}[]{rl}w=(01)^{2^{\ell-1}},&\\!\\!u_{h}=1^{2^{(2\ell+1)h+\ell}}(01)^{2^{(2\ell+1)h+\ell-1}}1^{(2^{\ell}-1)\,2^{(2\ell+1)h+\ell}},\\\\[4.0pt]
&\\!\\!v_{h}=(01)^{2^{(2\ell+1)h+2\ell}}1^{2^{(2\ell+1)h+2\ell}}(01)^{(2^{\ell}-1)\,2^{(2\ell+1)h+2\ell}}.\end{array}$
(55)

Then, we first note that the word $w$ vanish to the empty word $\epsilon$ in
exactly $n$ iterates of $\rho$. Moreover, by a somewhat tedious but
straightforward calculation, repeatedly using the rules

$\rho:\left\\{\vrule
width=0.0pt,height=15.0pt\right.\\!\\!\\!\begin{array}[]{ll}(01)^{k}\mapsto
1^{k}\\\\[4.0pt] 1^{2k}\mapsto(01)^{k}\end{array},$

for every $k\geq 1$, one checks that

$\rho^{n}_{w}(u_{0})=\widetilde{\rho}\,^{2}\circ\rho^{2\ell-1}(u_{0})=w\,,$

while

$\rho^{n}_{wu_{0}v_{0}\dots
u_{h}}(v_{h})=\rho^{2\ell+1}(v_{h})=u_{h}\quad\text{and}\quad\rho^{n}_{wu_{0}v_{0}\dots
v_{h}}(u_{h+1})=\rho^{2\ell+1}(u_{h+1})=v_{h}$

for every $h\geq 0$. Putting all those equation together, one can conclude
that $w\prod_{h=0}^{\infty}(u_{h}v_{h})$ is a fixed point for $\rho^{n}$, and
hence that $0.w\prod_{h=0}^{\infty}(u_{h}v_{h})$ is a fixed point for $R^{n}$.

The rest of this section is devoted to analyse some chaotic properties of the
map $R$.

######  Proposition 6.8.

The set of points with a dense $R$-orbit is uncountable and dense in $[0,1]$.

###### Proof.

To get the existence of uncountably many points with a dense $R$-orbit, it
suffices to take the sequence $(w_{n})_{n\geq 0}$ in Lemma 6.2 with the
property that it contains all the non-empty finite binary words. The density
of the points with a dense $R$-orbit follows immediately observing that we can
take any non-empty finite word as $w_{0}$. ∎

######  Proposition 6.9.

For every interval $I\subset[0,1]$, there exists $n\geq 1$ such that
$R^{n}(I)=[0,1]$.In particular, $R$ is topologically $($strongly$)$ mixing and
has sensitive dependence on initial conditions.

###### Proof.

Let $w$ be any non-empty finite binary word such that $\xi([\mkern
0.5muw\mkern 0.5mu])\subset I$, and set $n=n_{\epsilon}(w)$. Pick $y\in[0,1]$.
By the surjectivity of $\rho_{w}^{n}$, there is
$v\in\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve C}}{\text{\eusmeight
C}}{\text{\eusmsix C}}$ such that $\rho_{w}^{n}(v)=\beta(y)$, and hence
$\rho^{n}(wv)=\rho^{n}(w)\rho^{n}_{w}(v)=\beta(y)$ as well. Then, for
$x=\xi(wv)$ we have $x\in I$ and $R^{n}(x)=y$.

This implies sensitivity to initial conditions with sensitivity constant
$1/2$. Since every open set is a countable union of open intervals, it also
follows that, for every open sets $A,B\subset[0,1]$, there is $n\geq 1$ such
that $R^{m}(A)\cap B\neq\emptyset$ for every $m>n$, that is $R$ is
topologically mixing. ∎

######  Corollary 6.10.

The map $R$ is chaotic in the sense of Devaney $($see e.g. [13]$)$.

###### Proof.

It follows by Propositions 6.4, 6.8 and 6.9. ∎

Concerning distributional chaos, a concept introduced within the context of
topological dynamical systems in 1994 (see [37]), the map $R$ satisfies the
strongest form of distributional chaos, namely the distributional chaos of
type 1, DC1-chaos in short [1]. Actually, we prove that the DC1-chaos
exhibited by $R$ is uniform in the sense of [30], as specified in the next
proposition.

######  Proposition 6.11.

The map $R$ is uniformly DC1-chaotic, meaning that it admits an uncountable
uniformly DC1-scrambled set, that is an uncountable subset $S\subset[0,1]$
such that for every $\delta>0$ and some $\overline{\delta}>0$ the following
two conditions hold, where $\\#$ denotes the cardinality,

$\displaystyle\limsup_{n\to\infty}\displaystyle\frac{1}{n}\,\\#\\{k=0,\dots,n-1\text{
such that }|R^{k}(x)-R^{k}(y)|<\delta\\}=1\,,$ (56)
$\displaystyle\liminf_{n\to\infty}\displaystyle\frac{1}{n}\,\\#\\{k=0,\dots,n-1\text{
such that }|R^{k}(x)-R^{k}(y)|<\overline{\delta}\\}=0\,,$ (57)

for every $x\neq y$ in $S$.

###### Proof.

We will explicitly construct a set $S=\\{x_{\alpha}\\}_{\alpha\in A}$
satisfying the properties (56) and (57).As the set of indices $A$ we take the
image $A=\mu(\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve
B}}{\text{\eusmeight B}}{\text{\eusmsix
B}})\subset\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve
B}}{\text{\eusmeight B}}{\text{\eusmsix B}}$ of the map
$\mu:\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve B}}{\text{\eusmeight
B}}{\text{\eusmsix B}}\to\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve
B}}{\text{\eusmeight B}}{\text{\eusmsix B}}$ defined by

$\mu(b)_{n}=\left\\{\vrule
width=0.0pt,height=15.0pt\right.\\!\\!\\!\begin{array}[]{ll}b_{k}&\text{if
$n=p^{k}$ with $p$ prime and $k\geq 1$}\\\\[2.0pt]
0&\text{otherwise}\end{array},$

for every $b\in\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve
B}}{\text{\eusmeight B}}{\text{\eusmsix B}}$. We observe that $A$ is an
uncountable set, being $\mu$ injective, and that all the elements $\alpha\in
A$ share the same $n$-th component $\alpha_{n}=0$ for infinitely many $n\geq
1$. Furthermore, since each $\alpha=\mu(b)$ contains infinite copies of $b$ as
subsequences, any two different elements $\alpha,\alpha^{\prime}\in A$ cannot
be ultimately coinciding, that is for every $m\geq 1$ there is $n\geq m$ with
$\alpha_{n}\neq\alpha^{\prime}_{n}$.

Now, let $(e_{n})_{n\geq 1}$ be the increasing sequence of natural numbers
inductively defined by

$e_{1}=1\quad\text{and}\quad e_{n}=n^{2}\textstyle\sum_{k=1}^{n-1}e_{k}\
\text{for $n>1$}\,,$ (58)

and let $(v_{n})_{n\geq 1}$ and $(u_{n})_{n\geq 1}$ be the two sequences of
binary words inductively defined by

$\begin{array}[]{llll}v_{1}=1&\text{and}&v_{n}=\sigma^{2e_{n}}(v_{n-1})&\text{for
$n>1$}\,,\\\\[2.0pt]
u_{1}=001&\text{and}&u_{n}=00\,\sigma^{2e_{n}}(u_{n-1})&\text{for
$n>1$}\,.\end{array}$ (59)

Clearly, we have
$n_{\epsilon}(v_{n})=n_{\epsilon}(u_{n})=\sum_{k=1}^{n}2e_{k}$ for every
$n\geq 1$. Moreover, the equality $\sigma^{2}(1^{k})=1^{2k}$ implies that

$v_{n}=1^{2^{\sum_{k=2}^{n}e_{k}}},$ (60)

for every $n>1$.

To any index $\alpha\in A$ we associate the sequence of non-empty finite
binary words $(w^{\alpha}_{n})_{n\geq 0}$, where $w_{0}^{\alpha}$ can be
arbitrarily chosen, while for $n\geq 1$

$w^{\alpha}_{n}=\left\\{\vrule
width=0.0pt,height=12.0pt\right.\\!\\!\\!\begin{array}[]{ll}v_{n}&\text{if
$\alpha_{n}=0$}\\\\[2.0pt] u_{n}&\text{if $\alpha_{n}=1$}\end{array}.$

We set
$k_{n}=n_{\epsilon}(w^{\alpha}_{1})+\ldots+n_{\epsilon}(w^{\alpha}_{n-1})$ and
observe that $k_{n}$ is the same for all $\alpha\in A$.

Then, we apply Lemma 6.1 the sequences $(w^{\alpha}_{n})_{n\geq 0}$ and
$(k_{n})_{n\geq 1}$ and pick a single binary word $b_{\alpha}\in[\mkern
0.5muw^{\alpha}_{0}\mkern 0.5mu]\cap\mathchoice{\text{\eusmtwelve
C}}{\text{\eusmtwelve C}}{\text{\eusmeight C}}{\text{\eusmsix C}}$, out of the
uncountable resulting set $B$, such that $\rho^{k_{n}}(b_{\alpha})\in[\mkern
0.5muw_{n}^{\alpha}\mkern 0.5mu]$ for every $n\geq 1$. Finally, we put
$S=\\{x_{\alpha}=\xi(b_{\alpha})\\}_{\alpha\in A}$. Since all the words
$b_{\alpha}$ and $\rho^{k}(b_{\alpha})$ belong to C, by their construction in
the proof of Lemma 6.1, we have that all the points $x_{\alpha}$ are distinct
and that $R^{k}(x_{\alpha})=\xi(\rho^{k}(b_{\alpha}))$ for every $k\geq 1$ and
$\alpha\in A$.

To see that $S$ is uniformly DC1-scrambled, take any
$\alpha\neq\alpha^{\prime}$ in $A$. For any $\delta>0$, let $c(\delta)\geq 1$
be such that

$1/2^{2^{c(\delta)/2}}<\delta\,,$

in such a way that the interval $\xi([\mkern
0.5mu\rho^{h}(w^{\alpha}_{n})\mkern 0.5mu])$ has width

$\lambda(\xi([\mkern 0.5mu\rho^{h}(w^{\alpha}_{n})\mkern 0.5mu])<\delta\text{
\ for every
$h=1,\dots,n_{\epsilon}(w^{\alpha}_{n})-c(\delta)=\textstyle\sum_{k=1}^{n}2e_{k}-c(\delta)$}.$
(61)

By construction, there is a sequence $(n_{i})_{i\geq 1}$ such that
$\alpha_{n_{i}}=\alpha^{\prime}_{n_{i}}$ for every $i\geq 1$. By (61), this
implies that

$\frac{1}{k_{n_{i}+1}}\,\\#\\{k=1,\dots,k_{n_{i}+1}\text{ such that
}|R^{k}(x_{\alpha})-R^{k}(x_{\alpha^{\prime}})|<\delta\\}\geq\frac{\sum_{k=1}^{n_{i}}2e_{k}-c(\delta)}{k_{n_{i}+1}}\,.$
(62)

Thanks to equation (58), we have

$\textstyle\sum_{k=1}^{n_{i}}2e_{k}=\textstyle\sum_{k=1}^{n_{i}-1}2e_{k}+2e_{n_{i}}=2(n_{i}^{2}+1)\displaystyle\frac{e_{n_{i}}}{n_{i}^{2}}\,,$
(63)

while

$\begin{array}[]{rcl}k_{{n_{i}}+1}&\\!\\!\\!=\\!\\!\\!&\textstyle\sum_{k=1}^{n_{i}}n_{\epsilon}(w_{k}^{\alpha})=\textstyle\sum_{k=1}^{n_{i}}\textstyle\sum_{h=1}^{k}2e_{h}\\\\[6.0pt]
&\\!\\!\\!=\\!\\!\\!&\textstyle\sum_{k=1}^{n_{i}-1}\textstyle\sum_{h=1}^{k}2e_{h}+\sum_{h=1}^{n_{i}}2e_{h}\\\\[6.0pt]
&\\!\\!\\!\leq\\!\\!\\!&2(n_{i}-1)\textstyle\sum_{k=1}^{n_{i}-1}e_{k}+\sum_{k=1}^{n_{i}}2e_{k}\\\\[6.0pt]
&\\!\\!\\!=\\!\\!\\!&2(n_{i}-1)\frac{e_{n_{i}}}{n_{i}^{2}}+\textstyle\sum_{k=1}^{n_{i}}2e_{k}\\\\[4.0pt]
&\\!\\!\\!=\\!\\!\\!&2(n_{i}-1)\frac{e_{n_{i}}}{n_{i}^{2}}+2(n_{i}^{2}+1)\displaystyle\frac{e_{n_{i}}}{n_{i}^{2}}\,.\end{array}$
(64)

Therefore, the right hand side of (62) is bounded below by

$\frac{n_{i}^{2}+1}{n_{i}^{2}+n_{i}}-\frac{c(\delta)}{k_{n_{i}+1}}\,,$

which tends to 1 as $n_{i}$ diverges. This completes the proof of property
(56).

To prove (57), we consider a different sequence $(n_{i})_{i\geq 1}$ such that
$\alpha_{n_{i}}\neq\alpha^{\prime}_{n_{i}}$ for every $i\geq 1$, whose
existence is once again guaranteed by construction. Then, (59) and (60) imply
that $|R^{k_{n_{i}}}(x_{\alpha})-R^{k_{n_{i}}}(x_{\alpha^{\prime}})|>1/8$. So,
putting $\overline{\delta}=1/8$ and arguing as above, we get

$\frac{1}{k_{n_{i}+1}}\,\\#\\{k=1,\dots,k_{n_{i}+1}\text{ such that
}|R^{k}(x_{\alpha})-R^{k}(x_{\alpha^{\prime}})|<\overline{\delta}\\}\leq\frac{k_{n_{i}+1}-\sum_{k=1}^{n_{i}}2e_{k}}{k_{n_{i}+1}}\,.$

Recalling the equations (63) and (64), we see that the last quantity vanishes
as $n_{i}$ diverges, which gives property (57). ∎

We emphasize that in the above proof the family of starting words
$(w^{\alpha}_{0})_{\alpha\in A}$ is completely arbitrary. This means that the
DC1-scrambled sets for $R$ form a dense subset in the space of all subsets of
$[0,1]$ with the Hausdorff pseudo-distance. In particular, the DC1-scrambled
set $S$ can be chosen to be a dense subset of any interval $I\subset[0,1]$.
Moreover, by suitably modifying the sequences $(w^{\alpha}_{n})_{n\geq 1}$ in
the proof, it could be shown that actually $R$ is transitively DC1-chaotic,
that is all the points of the uncountable DC1-scrambled set $S$ have dense
orbits.

######  Remark 6.12.

Proposition 6.11 is interesting in view of the results of [39] and [40], in
which the author proves that there is a residual subset in the space of
bounded functions of Baire class $1$, and of Baire class $2$ as well, whose
elements are neither Devaney nor Li-Yorke chaotic$($see [20] and [26]$)$. It
follows that the map $R$, although generated by a very simple erasing
substitution rule, is “much more chaotic” than the topologically generic map
in its Baire class and also in the higher one.

In the final part of this section we address topological entropy.

We recall that for a map $f:X\to X$ with $X$ a metric space, the topological
entropy can be written as [14, 4]

$h(f)=\lim_{\varepsilon\to 0}\limsup_{n\to\infty}\frac{\log
r(n,\varepsilon)}{n}\,,$ (65)

where $r(n,\varepsilon)$ is the maximum cardinality of an
$\varepsilon$-separated set (that is a set whose points are at least
$\varepsilon$-apart from each other) in the metric

$d_{n}(x,y)=\max_{0\leq i\leq n}|f^{i}(x)-f^{i}(y)|\,.$ (66)

Although this definition is expressed in metric terms, it turns out to depend
only on the underlying topology — for a modern general reference on
topological entropy see [15].

For continuous interval maps, positive topological entropy is known to be
equivalent distributional chaos, and in particular to DC1-chaos. In our more
general context, the positivity of the entropy of the map $R$ should be proved
directly. In fact, in the next proposition we prove that $h(R)=\infty$. This
can be thought as a consequence of the fact that the substitution $\rho$ is
“efficient enough” in erasing the prefixes, so that all the points “forget”
quickly where they came from. Namely, Proposition 2.1 tells us that the
vanishing order of any finite binary word $w$ satisfies the inequality

$n_{\epsilon}(w)\leq 2\left\lfloor{\log_{2}{|w|}}\right\rfloor+2\,.$ (67)

######  Proposition 6.13.

The map $R$ has infinite topological entropy.

###### Proof.

For every $k,n\geq 1$ and every sequence of words
$w_{0},w_{1},\dots,w_{n}\in\\{0,1\\}^{k}$, we apply Lemma 6.1 to get a single
binary word $b_{w_{0},w_{1},\dots,w_{n}}\in[\mkern 0.5muw_{0}0\mkern
0.5mu]\cap\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve
C}}{\text{\eusmeight C}}{\text{\eusmsix C}}$ such that for every
$i=0,1,\dots,n$

$\rho^{i\ell}(b_{w_{0},w_{1},\dots,w_{n}})\in[\mkern 0.5muw_{i}0\mkern
0.5mu]\,,$

where $\ell=2\lfloor\log_{2}|w|\rfloor+2\geq n_{\epsilon}(w_{i})$ according to
Proposition 2.1. Then, we put

$S_{k,n}=\\{x_{w_{0},w_{1},\dots,w_{n}}=\xi(b_{w_{0},w_{1},\dots,w_{n}})\;|\;w_{0},w_{1},\dots,w_{n}\in\\{0,1\\}^{k}\\}\subset[0,1]\,.$
(68)

Since all the words $b_{w_{0},w_{1},\dots,w_{n}}$ and
$\rho^{k}(b_{w_{0},w_{1},\dots,w_{n}})$ belong to C, by their construction in
the proof of Lemma 6.1, we have that all the points
$x_{w_{0},w_{1},\dots,w_{n}}$ are distinct and that
$R^{k}(x_{w_{0},w_{1},\dots,w_{n}})=\xi(\rho^{k}(b_{w_{0},w_{1},\dots,w_{n}}))$
for every $k\geq 1$ and $w_{0},w_{1},\dots,w_{n}\in\\{0,1\\}^{k}$.

We note that, for every $k,n\geq 1$ the set $S_{k,n}$ has cardinality
$\\#S_{k,n}=2^{(n+1)k}$ and it is $\varepsilon$-separated in the metric
$d_{n}$ if $\varepsilon<1/2^{k+1}$. Indeed, given any two different points
$x=x_{w_{0},w_{1},\dots,w_{n}}$ and
$y=x_{w^{\prime}_{0},w^{\prime}_{1},\dots,w^{\prime}_{n}}$ in $S_{k,n}$, we
have $w_{i}\neq w_{i}^{\prime}$ for some $i=0,\dots,n$, and hence
$|R^{i}(x)-R^{i}(y)|\geq\varepsilon$, as it can be easily deduced from
$x\in\xi([\mkern 0.5muw_{i}0\mkern 0.5mu])$ and $y\in\xi([\mkern
0.5muw^{\prime}_{i}0\mkern 0.5mu])$.

Then, based on equation (65) and putting
$k=\lceil-\log_{2}2\varepsilon\rceil-1$ we get

$h(R)\geq\lim_{\varepsilon\to 0}\limsup_{n\to\infty}\frac{\log
2^{(n+1)k}}{n}=\lim_{\varepsilon\to 0}\lim_{n\to\infty}\frac{(n+1)\log
2^{(\lceil-\log_{2}2\varepsilon\rceil-1)}}{n}=\infty\,.$

∎

Recently some attention has been devoted to the study of finer properties of
topological entropy, in particular to the characterization of the points
around which the entropy concentrates. More precisely, one can introduce the
notion of relative entropy $h(f,K)$ for a map $f:X\to X$, with $X$ a metric
space, and a subset $K\subset X$, as the limit (65) where $r(n,\epsilon)$ is
defined by constraining the $\varepsilon$-separated sets to be contained in
$K$. Then, following [44], if $f$ has positive entropy, $x\in X$ is called an
entropy point (resp. a full entropy point) if $h(f,K)>0$ (resp. $h(f,K)=h(f)$)
for every closed neighborhood of $x$ in $X$.

######  Remark 6.14.

It is known that, in case of continuous maps with positive entropy on a
compact metric space, minimality implies that every point is a full entropy
point [44]. The system $([0,1],R)$ is not minimal (there are periodic points
of every period), nevertheless every point is a full entropy point, as an
immediate consequence of Proposition 6.9.

##  7 Arithmetic properties

We recall that a Liouville number is a real number $x$ such that, for every
$n$, there exists a sequence of rationals $p_{i}/q_{i}$
($p_{i},q_{i}\in\mathbb{Z},q_{i}\neq 0$) verifying

$\left|\,\vrule
height=12.0pt,width=0.0pt\smash{x-\frac{p_{i}}{q_{i}}}\,\right|<\frac{1}{q_{i}^{n}}\,.$
(69)

######  Proposition 7.1.

The map $R$ admits uncountably many fixed points that are Liouville numbers,
hence transcendental real numbers.

###### Proof.

For the existence of uncountably many Liouville fixed points, it suffices to
observe that, by construction, the fixed point
$0.{\langle{a}\rangle}^{\mkern-3.0mu\scriptscriptstyle\text{F}}$ is a
Liouville number as soon as the sequence $a\in A$ increases fast enough (see
definition (53) and Proposition 6.6). By a classical result [25], all the
fixed points which are Liouville numbers are transcendental. ∎

A similar result could be established analogously for periodic points of any
given period. However, this simple argument does not apply to the largest
fixed point, for which we have to provide a different proof of transcendence.

######  Proposition 7.2.

The largest fixed point $x^{0}$ of the map $R$ is transcendental.

###### Proof.

Let $p(n,x^{0})$ be the cardinality of the set of words of length $n$
appearing in the binary expansion of $x^{0}$. Then we have

$\limsup_{n\to\infty}\frac{p(n,x^{0})}{n}<8\,.$ (70)

Indeed, given any $n>3$, let $k$ be the least positive integer such that
$k\geq\log_{2}(n/3)$, hence $3\cdot 2^{k}\geq n$ and $3\cdot 2^{k-1}<n$. Then,
we split the binary expansion $b^{0}$ of $x^{0}$ as the product
$b^{0}=b_{0}^{\prime}b_{0}^{\prime\prime}$, where
$b_{0}^{\prime}=00101\prod_{h=0}^{k-1}1^{3\cdot 2^{h}}(01)^{3\cdot 2^{h}}$ and
$b_{0}^{\prime\prime}=\prod_{h=k}^{\infty}1^{3\cdot 2^{h}}(01)^{3\cdot
2^{h}}$. The length of $b_{0}^{\prime}$ is

$5+9\textstyle\sum_{h=0}^{k-1}2^{h}=5+9(2^{k}-1)=9\cdot 2^{k}-4<6n-4<6n\,.$
(71)

Therefore, there are less than $6n$ different subwords of $b^{0}$ having
length $n$ and starting in $b_{0}^{\prime}$. We are left to consider the
different subwords of length $n$ that are contained in $b_{0}^{\prime\prime}$.
Due to the inequality $3\cdot 2^{k}\geq n$, each of them consists of a
(possibly empty) sequence of 1’s followed/preceded by a (possibly empty or
truncated on the right/left) sequence of 01’s. So, there are at most $2n+3$
such subwords. Therefore, $p(n,x^{0})<8n+3$ for every $n\geq 1$, which gives
(70). Thus, by Theorem 3.1 in [7], $x^{0}$ is transcendental. ∎

Along the same lines we can prove the following.

######  Proposition 7.3.

The “simplest” periodic point $x^{\ell}$ of period $2\ell+1$ defined in (54)
is transcendental for each $\ell\geq 1$.

###### Proof.

We know that the general form for the binary expansion of $x^{\ell}$ is
$b^{\ell}=w\prod_{h=0}^{\infty}u_{h}v_{h}$ where $w,u_{h}$ and $v_{h}$ are the
words in (55). Proceeding as above, let $k$ be the least positive integer such
that $2^{(2\ell+1)k+\ell}\geq n$ and $2^{(2\ell+1)(k-1)+\ell}<n$. We then
split the binary expansion of $b^{\ell}$ as
$b_{\ell}^{\prime}b_{\ell}^{\prime\prime}$, where
$b_{\ell}^{\prime}=w\prod_{h=0}^{k-1}u_{h}v_{h}$ and
$b_{\ell}^{\prime\prime}=\prod_{h=k}^{\infty}u_{h}v_{h}$. Now, one shows that
the length of $b_{\ell}^{\prime}$ is bounded above by $12\cdot 2^{2\ell+1}n$,
which also gives a bound for the number of different subwords having length
$n$ and starting in $b_{\ell}^{\prime}$. Moreover, by the same argument used
above yields that $b_{\ell}^{\prime\prime}$ contributes with at most $2n+3$ to
$p(n,b^{\ell})$. ∎

##  8 Some open questions

Many natural questions concerning the map $R$ do not have yet an answer. We
limit ourselves to list a few of them.

Concerning basic properties of the map, we propose the following.

######  Question 8.1.

What is the Hausdorff dimension of the graph $G$ of $R$?

We note that the self-affine structure of $G$ discussed in Section 4 seems not
to fit into any of the known criteria for the computation of the Hausdorff
dimension.

From the ergodic point of view, the main question is probably the next one.

######  Question 8.2.

Does $R$ admit an invariant measure which is not purely atomic?

We recall that the singularity of $R$ prevents the existence of absolutely
continuous (with respect to Lebesgue) invariant measures.

Although the main interest in studying the dynamics of a map such as R comes
from a natural generalization of topological dynamics questions to a wider
class of functions, like Baire class 1 functions, it could be of interest also
in an ergodic context to model specific “non-equilibrium” situations in which
events that are impossible at one time become possible at another time.

Concerning points with a dense orbit, heuristics indicate that there is plenty
of them, as we have uncountability for two different reasons: in the procedure
described in the proof of Lemma 6.2 there are countably many steps each
involving countably many choices; in Proposition 6.8 there are uncountably
many possible choices for the sequence $(w_{n})_{n\geq 1}$.

Therefore, it seems reasonable to ask the following.

######  Question 8.3.

Is the set of points with a dense $R$-orbit of second category $($or even co-
meagre$)$? Does it have positive $($or even full$)$ Lebesgue measure?

About the arithmetic properties of $R$, in the light of the results of Section
7, and of the fact that the Thue-Morse number is known to be transcendental,
one could ask the question below.

######  Question 8.4.

Is every fixed point of $R$ transcendental? Is every periodic point of $R$ not
in $C_{0}$ or $C_{1}$ transcendental?

Finally, as said in the Introduction, the map $R$ is the model-case of a more
general class of interval maps generated by erasing block substitutions,
meaning (in the binary case) substitution rules
$s:\\{0,1\\}^{k}\to\\{0,1\\}^{*}$ such that at least one word $w$ of length
$k$ is mapped to $\epsilon$. Looking at these more general maps poses of
course new problems.

######  Question 8.5.

What of the properties we established for the map $R$ are true for interval
maps generated by which classes of $($binary$)$ erasing block substitutions?
What new phenomena arise in this extended context?

Let us list some speculation about this last question.

  1. 1.

While the map $\rho$ has an “almost-inverse” map $\sigma$ (up to insertions of
the word 00), this is not true for a general erasing substitution rule $s$ as
above. Indeed, in the $s$-preimages of a given infinite word
$v\in\\{0,1\\}^{\omega}$, there could be in principle infinitely many words
$\prod_{i=1}^{\infty}w_{i}$ with $w_{i}\in\\{0,1\\}^{k}$, such that
$s(w_{i})\neq\epsilon$ for every $i\geq 1$. This could make the topological
structure of the fibers much more complex than in our present case.

  2. 2.

Even assuming that there is only one word $w_{\epsilon}\in\\{0,1\\}^{k}$
mapped by $s$ to the empty word, if $w_{\epsilon}\neq 0^{k}$ we have to
consider separately three objects, that is the set D of dyadic rationals in
$[0,1]$ and the two sets

$\displaystyle S_{1}=\\{x\in[0,1]\;|\;x=0.v,\ v=s(w0^{\infty}),\
w\in\\{0,1\\}^{*}\\}\,,$ $\displaystyle
S_{2}=\\{x\in[0,1]\;|\;x=0.ww_{\epsilon}^{\infty},\ w\in\\{0,1\\}^{nk}\\}\,.$

It is not difficult to see that these sets coincide in the case when
$w_{\epsilon}=0^{k}$, but this is not true in general. Moreover, if
$s^{-1}(\epsilon)\subset\\{0,1\\}^{k}$ has more than one element, the interval
map $f_{s}:[0,1]\to[0,1]$ generated by the symbolic action of $s$ is in
general discontinuous also at some (actually uncountably many) irrationals.

  3. 3.

For suitable erasing substitution rules $s$, the map $f_{s}$ can have stronger
properties than $R$. For instance, $s$ can be chosen so that $f_{s}$ is a
Darboux function.

A first study of interval maps generated by general erasing block
substitutions is done in [12], but several problems are still open, in
particular concerning point 1 above.

## Acknowledgements

This work is partially supported by the research project PRIN
2017S35EHN$\textunderscore$004 “Regular and stochastic behaviour in dynamical
systems” of the Italian Ministry of Education and Research.

## List of symbols

$\mathbb{N}$ |  natural numbers (including 0)
---|---
$\mathbb{D}\subset\mathbb{Q}$ |  dyadic rationals and all rationals, respectively
$\varphi$ |  the golden ratio $(1+\sqrt{5})/2$
$\mathchoice{\text{\eusmtwelve Q}}{\text{\eusmtwelve Q}}{\text{\eusmeight Q}}{\text{\eusmsix Q}}=\mathbb{Q}\cap[0,1]$ |  rationals in $[0,1]$
$\mathchoice{\text{\eusmtwelve D}}{\text{\eusmtwelve D}}{\text{\eusmeight D}}{\text{\eusmsix D}}=\mathbb{D}\cap[0,1]$ |  dyadic rationals in $[0,1]$
$\mathchoice{\text{\eusmtwelve N}}{\text{\eusmtwelve N}}{\text{\eusmeight N}}{\text{\eusmsix N}}\subset[0,1]$ |  normal binary numbers in $[0,1]$
$\mathchoice{\text{\eusmtwelve A}}{\text{\eusmtwelve A}}{\text{\eusmeight A}}{\text{\eusmsix A}}=\mathbb{N}^{\omega}$ |  infinite sequences of natural numbers
$\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve B}}{\text{\eusmeight B}}{\text{\eusmsix B}}=\\{0,1\\}^{\omega}$ |  infinite binary sequences/words
$\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve C}}{\text{\eusmeight C}}{\text{\eusmsix C}}\subset\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve B}}{\text{\eusmeight B}}{\text{\eusmsix B}}$ |  infinite binary sequences/words which are not eventually $0$
$\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve C}}{\text{\eusmeight C}}{\text{\eusmsix C}}^{\prime}\subset\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve B}}{\text{\eusmeight B}}{\text{\eusmsix B}}$ |  infinite binary sequences/words which are not eventually $1$
$\mathchoice{\text{\eusmtwelve S}}{\text{\eusmtwelve S}}{\text{\eusmeight S}}{\text{\eusmsix S}}\subset\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve C}}{\text{\eusmeight C}}{\text{\eusmsix C}}$ |  infinite binary sequences/words with no pair of consecutive 0’s
$\eta:\mathchoice{\text{\eusmtwelve A}}{\text{\eusmtwelve A}}{\text{\eusmeight A}}{\text{\eusmsix A}}\to\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve C}}{\text{\eusmeight C}}{\text{\eusmsix C}}$ |  map $(a_{1},a_{2},\dots)\mapsto 0^{a_{1}}10^{a_{2}}1\dots$ (1)
$\xi:\mathchoice{\text{\eusmtwelve B}}{\text{\eusmtwelve B}}{\text{\eusmeight B}}{\text{\eusmsix B}}\to[0,1]$ |  map $(b_{1},b_{2},\dots)\mapsto x=0.b_{1}b_{2}\dots$ (3)
$\beta:(0,1]\to\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve C}}{\text{\eusmeight C}}{\text{\eusmsix C}}$ |  map $x\mapsto(b_{1},b_{2},\dots)$, the unique sequence in C such that $x=0.b_{1}b_{2}\dots$ (4)
$\beta^{\prime}:[0,1)\to\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve C}}{\text{\eusmeight C}}{\text{\eusmsix C}}^{\prime}$ |  map $x\mapsto(b_{1},b_{2},\dots)$, the unique sequence in $\mathchoice{\text{\eusmtwelve C}}{\text{\eusmtwelve C}}{\text{\eusmeight C}}{\text{\eusmsix C}}^{\prime}$ such that $x=0.b_{1}b_{2}\dots$ (4)
$|s|\kern 2.0pt,|w|$ |  length of a sequence $s$ or word $w$, respectively
$|s|_{1}\kern 2.0pt,|w|_{1}$ |  number of 1’s in a binary sequence or word, respectively
$\widetilde{s}\kern 2.0pt,\widetilde{w}$ |  binary sequence/word obtained by complementing $s/w$, respectively
$s\smash{\text{{\raise 0.21529pt\hbox{$\shortmid$}}\hbox to0.0pt{\hss\raise-2.36806pt\hbox{$\shortmid$}}}}_{n}\kern 2.0pt,w\smash{\text{{\raise 0.21529pt\hbox{$\shortmid$}}\hbox to0.0pt{\hss\raise-2.36806pt\hbox{$\shortmid$}}}}_{n}$ |  $n$-truncation of a sequence $s$ or a word $w$, respectively (5)
$[\mkern 0.5mus\mkern 0.5mu]\kern 2.0pt,[\mkern 0.5muw\mkern 0.5mu]$ |  set of all infinite sequences/words having prefix $s$/$w$, respectively (6)
$\\{0,1\\}^{*}$ |  possibly empty finite binary sequences/words
$\\{0,1\\}^{\infty}$ |  possibly empty finite or infinite binary sequences/words

$\rho:\\{0,1\\}^{\infty}\to\\{0,1\\}^{\infty}$ |  the main substitution map (7)
---|---
$\tau:\\{0,1\\}^{\infty}\to\\{0,1\\}^{\infty}$ |  an equivalent 2-block substitution map (defined on even/infinite length binary sequences/words) (11)
$\overline{\tau}:\\{0,1\\}^{\infty}\to\\{0,1\\}^{\infty}$ |  the substitution map inverse to $\tau$ (13)
$\sigma:\\{0,1\\}^{\infty}\to\\{0,1\\}^{\infty}$ |  the “simplest” substitution map inverse to $\rho$ (14)
$R:[0,1]\to[0,1]$ |  the real map corresponding to $\rho$ (16)
$S:[0,1]\to[0,1]$ |  the section of $R$ corresponding to $\sigma$ (22)

$\epsilon$ |  the empty sequence/word
---|---
$n_{\epsilon}$ |  vanishing order function of binary sequences/words under the action or $\rho$ (12)
$\widetilde{\rho}$ |  the substitution map complementary to $\rho$ (8)
$\rho_{k}$ |  $\rho$ or $\widetilde{\rho}$, depending on $k$ being an even or odd natural, respectively (9)
$\rho_{v}$ |  auxiliary substitution map, coinciding with $\rho_{|v|}$, such that $\rho(vw)=\rho(v)\rho_{v}(w)$ (10)
$\rho_{v}^{n}$ |  auxiliary substitution map such that $\rho^{n}(vw)=\rho^{n}(v)\rho^{n}_{v}(w)$ (48)
${\langle{a}\rangle}^{\mkern-3.0mu{b}}$ |  binary sequence/word obtained from $b$ by inserting $0^{2a_{k}}$ before each $k$-th 1 (2)
$\langle\mkern-4.5mu\langle{b}\rangle\mkern-4.5mu\rangle$ |  the auxiliary set $\\{{\langle{a}\rangle}^{\mkern-3.0mu{b}}\,|\,a\in\mathchoice{\text{\eusmtwelve A}}{\text{\eusmtwelve A}}{\text{\eusmeight A}}{\text{\eusmsix A}}\\}$ used to construct the fibers of $\rho$ and $R$ (15,21)
${\langle{a}\rangle}^{\mkern-3.0mu\scriptscriptstyle\text{F}}$ |  the element of $\mathop{\mathrm{Fix}}\rho$ generated from the “simples” one $b^{0}$ according to $a\in\mathchoice{\text{\eusmtwelve A}}{\text{\eusmtwelve A}}{\text{\eusmeight A}}{\text{\eusmsix A}}$ (53)
$b^{0}$ |  the “simplest” fixed point ${\langle{0^{\infty}}\rangle}^{\mkern-3.0mu\scriptscriptstyle\text{F}}$ of the map $\rho$ (51)
$x^{0}$ |  the largest fixed point $\xi(b^{0})$ of the map $R$ (52)
$x^{\ell}$ |  the “simplest” periodic point of the map $R$ having odd order $2\ell+1$ (54)
$C_{0}\kern 2.0pt,C_{1}$ |  the two rational $R$-cycles $\\{0,2/3\\}$ and $\\{1,1/3\\}$ respectively
$Q_{0}\kern 2.0pt,Q_{1}$ |  the subsets of Q attracted by $C_{0}$ and $C_{1}$ respectively (50)

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