Proof. First we replace k and l by two smaller numbers. Below by $\nu _p(n)$ for prime p and $n\in {\mathbb {Z}}^+$ we mean the largest integer satisfying $p^{\nu _p(n)}|n$ .

Lemma 3.2. There exist numbers $k'>l'\geqslant 1$ for which there are infinitely many primes p satisfying

$$ \begin{align*} \gcd(p-1, k)=k'\quad\text{and}\quad\gcd(p-1, l)=l'. \end{align*} $$

If l is odd, and $q|k$ and $q\nmid l$ for some prime q, then we can take $l'=1$ .

Proof. Since $k\nmid l$ , there exists a prime q for which $\nu _q(k)>\nu _q(l)$ (if the last sentence of the theorem statement is true, we can also assume $\nu _q(l)=1$ ). If $q=2$ , then we can set $k'=2l'=2^{\nu _2(l)+1}$ , and let p be any prime satisfying

$$ \begin{align*} \gcd(p-1, kl)=2^{\nu_2(l)+1}. \end{align*} $$

Otherwise we let $l'{\kern-1pt}={\kern-1pt}q^{\nu _q(l)}\cdot \gcd (l, 2)$ and $k'{\kern-1pt}={\kern-1pt}q^{\nu _q(k)+1}\cdot \gcd (k, 2)$ , so that ${k'/l'{\kern-1pt}\geqslant{\kern-1pt} q/2{\kern-1pt}>{\kern-2pt}1}$ . Then p just has to satisfy

$$ \begin{align*} \gcd(p-1, 2kl)=2q^{\nu_q(l)+1}. \end{align*} $$

In both cases there are infinitely many possible primes p by Dirichlet’s theorem about primes in arithmetic progressions.

We now recursively define the sequence $(n_t)$ by setting $n_0=1$ and letting ${n_{t+1}=n_tp_{t+1}}$ for a strictly increasing sequence $(p_t)$ of primes satisfying the condition in Lemma 3.2 and the following:

(5) $$ \begin{align} \frac{\varphi(n_t)}{n_t}>\frac{9}{10}, \end{align} $$

(6) $$ \begin{align} \sum\limits_{t=1}^{\infty}\frac{8kn_t}{\sqrt{p_{t+1}}}<\frac{1}{10}, \end{align} $$

(7) $$ \begin{align} p_{t+1}>30n_t^k. \end{align} $$

We also define a sequence $x_t\in \{0, 1, ?\}^{n_t}$ . We start with $x_0=?$ . Having defined $x_t$ , we define $x_{t+1}$ as follows. First concatenate $n_{t+1}/n_t=p_{t+1}$ copies of $x_t$ , creating $x_{t+1}'$ . If the resulting word has the symbol ‘?’ at position i for

(8) $$ \begin{align} i\in [0, n_t-1]\cup [n_{t+1}-n_t, n_{t+1}-1]\cup ([0, n_{t+1}-1]\setminus \widetilde{R^k_{n_{t+1}}}), \end{align} $$

we change it to a 0 or 1 arbitrarily, creating $x_{t+1}"$ . Then, for $i\in [0,n_{t+1}^{1/k}]$ , we fill the positions $i^k$ which still contain ‘?’ with 0s for even t and 1s for odd t, creating $x_{t+1}$ .

The pair $(n_t), (x_t)$ defined this way will be viable, since the first and last $n_{t-1}$ symbols of $x_t$ are not ‘?’. Also, we have

$$ \begin{align*} ?_t\leqslant |\widetilde{R^k_{n_t}}|=\prod_{i=1}^{t}|\widetilde{R^k_{p_i}}|=\prod_{i=1}^{t} \frac{p_i-1}{k'}=\frac{\varphi(n_t)}{(k')^t}, \end{align*} $$

where $?_t$ was defined in Definition 2.1. But $\rho _l(n_t)=(l')^t=o((k')^t)$ and $\varphi (n_t)<n_t$ , so the conditions of Theorem 2.2 are satisfied, and the limit (3) exists. Also, $X_x$ is regular since $?_t=o(n_t)$ .

We now estimate $?_t$ from below. Notice that by the Chinese remainder theorem for any $a\in \widetilde {R^k_{n_t}}$ there are exactly $(p-1)/k'$ elements of $\widetilde {R^k_{n_{t+1}}}$ congruent to a modulo $n_t$ . Hence, in $x_{t+1}"$ there are at least

$$ \begin{align*} \frac{p_{t+1}-1}{k'}?_t-2n_t \end{align*} $$

question marks. Taking into account the last step of the construction, we get

$$ \begin{align*} ?_{t+1}\geqslant \frac{p_{t+1}-1}{k'}?_t-2n_t -{n_{t+1}}^{1/k}, \end{align*} $$

so that

$$ \begin{align*} ?_{t+1}\cdot \frac{(k')^{t+1}}{\varphi(n_{t+1})}\geqslant ?_t\cdot \frac{(k')^{t}}{\varphi(n_{t})}-\frac{2k'n_t}{p_{t+1}-1}-\frac{(n_tp_{t+1})^{1/k}}{p_{t+1}-1}\geqslant ?_t\cdot \frac{(k')^{t}}{\varphi(n_{t})}-\frac{8k'n_t}{p_{t+1}^{(k-1)/k}}. \end{align*} $$

By induction and from (6) we obtain

(9) $$ \begin{align} ?_t\geqslant \frac{9}{10}\frac{\varphi(n_t)}{(k')^t}. \end{align} $$

We now estimate the number of kth powers whose positions contain question marks in $x_{t+1}'$ .

Let $C_t=\lfloor {n_{t+1}}^{1/k}\rfloor =an_t+r$ for $a\in \mathbb {Z}$ and $r\in [0, n_t-1]$ , where $a\geqslant 30$ by (7). Then

$$ \begin{align*} |\{i<C_t: x_{t+1}'(i^k)=?\}|\geqslant a|\{i<n_t: x_{t+1}'(i^k)=?\}|. \end{align*} $$

But we know that $\rho _k(n_t; n_t, j)=(k')^t$ for any $j\in \widetilde {R_{n_t}^k}$ , and all the question marks in $x_t$ are at positions coprime with $n_t$ , so the above is at least

$$ \begin{align*} a\cdot ?_t\cdot (k')^t\geqslant \frac{9}{10}\frac{C_t}{n_t}\cdot \frac{9}{10}\varphi(n_t)\geqslant \frac{7}{10} C_t \end{align*} $$

by (9) and condition (5). By construction we have

$$ \begin{align*} &|\{i<C_t: x_{t+1}'(i^k)\ne ?\text{ and }x_{t+1}"(i^k)=?\}|\\ & \quad\leqslant 2n_t+|\{i<C_t: p_{t+1}|i\}| \leqslant 2n_t+\frac{C_t}{p_{t+1}}+1\leqslant \frac{1}{10}C_t \end{align*} $$

for large t. Therefore, in the last step of the construction of $x_{t+1}$ , we fill in at least $6C_t/10$ question marks, and so

$$ \begin{align*} \frac{1}{C_t}\sum\limits_{m<C_t}G(\sigma^{m^k}x)=\frac{1}{C_t}\sum\limits_{m<C_t}(-1)^{x_{t+1}(m^k)}\geqslant \frac{1}{C_t}\bigg(\frac{6}{10}C_t-\frac{4}{10}C_t\bigg)= \frac{1}{5} \end{align*} $$

for even t, and

$$ \begin{align*} \frac{1}{C_t}\sum\limits_{m<C_t}G(\sigma^{m^k}x)\leqslant -\frac{1}{5} \end{align*} $$

for odd t, and so the limit (4) does not exist.

Now let us assume that the last sentence in the statement of Theorem 3.1 is true, so that $l'$ is odd and coprime with all $p-1$ for $p|n_t$ prime. Then the polynomial $m^l$ is a permutation modulo each $n_t$ , so the orbit $(\sigma ^{m^l}y)$ is equidistributed in $X_x$ by Theorem 5.7.

Proof. We follow the proof of [Reference Frączek, Kanigowski and Lemańczyk3, Theorem 6.8].

It suffices to show that for every continuous $F:X_x\to \mathbb {C}$ and every $\varepsilon>0$ there exists $N_\varepsilon $ so that for every $N, M>N_\varepsilon $ and every $r\in \mathbb {Z}$ , we have

(12) $$ \begin{align} \bigg\lvert \frac{1}{N}\sum\limits_{m<N}F(\sigma^{P(m)+r}x)-\frac{1}{M}\sum\limits_{m<M}F(\sigma^{P(m)+r}x)\bigg\rvert<\varepsilon. \end{align} $$

Indeed, then the sequence of averages in (11) converges for all $y=\sigma ^r x$ , uniformly in the choice of $r\in \mathbb {Z}$ , and these points form a dense set in $X_x$ . Hence, not only does the limit (11) exist for all $y\in X_x$ , but the convergence is uniform in $y\in X_x$ .

It is enough to consider only F depending on finitely many coordinates, since the set of such functions is dense in $C(X_x)$ with the supremum norm topology. So let us assume that whenever $y, y'\in X_x$ agree on $[-C, C]$ for some $C\in {\mathbb {Z}}^+$ , we have $F(y)=F(y')$ . We can also assume that $F:X_x\to [0, 1]$ .

Fix $\varepsilon>0$ and t such that the left-hand side of (10) is at most $\varepsilon n_t/(2C+1)$ . Then fix $r\in {\mathbb {Z}}$ and let

$$ \begin{align*} A_a=\{i\in [0, n_t-1]: X_t(P(i)+a+r)=?\}, \end{align*} $$

and consider the set

$$ \begin{align*} A=\bigcup_{a=-C}^{C}A_a. \end{align*} $$

Clearly $|A|\leqslant \varepsilon n_t$ , and if $i\notin A$ , then

$$ \begin{align*} F(\sigma^{P(i)+r}x)=F(\sigma^{P(i)+n_tj+r}x) \end{align*} $$

for any $j\in \mathbb {Z}$ . Therefore,

$$ \begin{align*} \bigg\lvert\! \sum\limits_{m<kn_t}F(\sigma^{P(m)+r}x)-k\sum\limits_{m<n_t}F(\sigma^{P(m)+r}x)\bigg\rvert\leqslant 2k|A|\leqslant 2k\varepsilon n_t \end{align*} $$

for any $k\in \mathbb {Z}^+$ .

Then, for any $N{\kern-1pt}>{\kern-1pt}(\varepsilon ^{-1}{\kern-1pt}+{\kern-1pt}1)n_t$ , we can write $N{\kern-1pt}={\kern-1pt}kn_t{\kern-1pt}+{\kern-1pt}b$ for $k{\kern-1pt}>{\kern-1pt}\varepsilon ^{-1}$ and ${b{\kern-1pt}\in{\kern-1pt} [0, n_t-1]}$ , and we have

$$ \begin{align*} \bigg\lvert \frac{1}{N}\sum\limits_{m<N}F(\sigma^{P(m)+r}x)-\frac{1}{n_t}\sum\limits_{m<n_t}F(\sigma^{P(m)+r}x)\bigg\rvert\leqslant \frac{b}{N}+\bigg(\frac{1}{N}-\frac{1}{kn_t}\bigg)\cdot N\\ +\,\bigg\lvert \frac{1}{kn_t}\sum\limits_{m<N}F(\sigma^{P(m)+r}x)-\frac{1}{n_t}\sum\limits_{m<n_t}F(\sigma^{P(m)+r}x)\bigg\rvert\leqslant 2\varepsilon+2\varepsilon=4\varepsilon. \end{align*} $$

Therefore, for $M, N>(\varepsilon ^{-1}+1)n_t$ the estimate (12) holds (with the constant $8\varepsilon $ in place of $\varepsilon $ ).

Proof. This is an application of the Weil bound for character sums:

Theorem 4.3. Fix a prime p, a monic polynomial $g\in \mathbb {F}_p[X]$ , an integer $d|p-1$ , and a multiplicative character $\chi $ of $\mathbb {F}_p$ of order d. Then, provided there is no polynomial ${h\in \overline {\mathbb {F}_p}[X]}$ such that $g=h^d$ , we have

$$ \begin{align*} \bigg\lvert\! \sum\limits_{x\in \mathbb{F}_p}\chi(g(x))\bigg\rvert\leqslant (\deg g-1)\sqrt{p}. \end{align*} $$

For a discussion and proof see, for example, [Reference Kowalski6, Theorem 3.1].

We will apply this result to all non-trivial characters of $\mathbb {F}_p$ of order dividing k, and to the polynomial $g(x)=x^l+a$ . Notice that $g'(x)=lx^{l-1}$ , and $a\ne 0$ and $p\nmid l$ since $p>l$ , so g and $g'$ are coprime. Therefore, g is not a power of any polynomial.

For any $i\in \mathbb {F}_p$ we have

$$ \begin{align*} |\{x\in \mathbb{F}_p: x^k=i\}|=1+\sum\limits_{\substack{\chi\ne 1 \\ \chi^k=1}}\chi(i). \end{align*} $$

Therefore, the number of pairs $(x, y)$ satisfying $x^k=y^l+a$ is

$$ \begin{align*} \sum\limits_{y\in\mathbb{F}_p}|\{x\in\mathbb{F}_p: x^k=g(y)\}|=\sum\limits_{y\in\mathbb{F}_p}\bigg(1+\sum\limits_{\substack{\chi\ne 1 \\ \chi^k=1}}\chi(g(y))\bigg). \end{align*} $$

This sum consists of the main term p and at most $k-1$ character sums, each bounded by $(l-1)\sqrt {p}$ by the Weil bound.

Proof. Let

$$ \begin{align*} A_p=\{a\in(({\mathbb{Z}}/n{\mathbb{Z}})^*)^k: a\ \pmod{p}\in (\mathbb{F}_p)^l\}. \end{align*} $$

Since n is squarefree and $k|p-1$ , we have $|(({\mathbb {Z}}/n{\mathbb {Z}})^*)^k|=\varphi (n)/k^{\omega (n)}$ . Since $l|p-1$ , and by the Chinese remainder theorem,

$$ \begin{align*} \frac{|A_p|}{\varphi(n)/k^{\omega(n)}}=\frac{(p-1)/l}{(p-1)/k}\leqslant \frac{1}{2}.\end{align*} $$

Let $\mathcal {P}$ be the family of all subsets of $\{p: p|n_t\}$ of size at least $\omega (n)-\ln \omega (n)$ . From the above estimate and by the Chinese remainder theorem, for any $P\in \mathcal {P}$ we have

$$ \begin{align*} \frac{k^{\omega(n)}}{\varphi(n)}\cdot \bigcap_{p\in P}A_p\leqslant \bigg(\frac{1}{2}\bigg)^{\omega(n)-\ln\omega(n)}, \end{align*} $$

and since $|\mathcal {P}|\leqslant \omega (n)^{\ln \omega (n)}$ , we have

$$ \begin{align*} \frac{k^{\omega(n)}}{\varphi(n)}\cdot \bigcup_{P\in\mathcal{P}}\bigcap_{p\in P}A_p\leqslant \omega(n)^{\ln\omega(n)}\cdot\bigg(\frac{1}{2}\bigg)^{\omega(n)-\ln\omega(n)}\\ =\bigg(\frac{1}{2}\bigg)^{\omega(n)-\ln\omega(n)-\ln(2)(\ln\omega(n))^2}<\bigg(\frac{1}{2}\bigg)^{\omega(n)/4}. \end{align*} $$

But by definition we have

$$ \begin{align*} A=(({\mathbb{Z}}/n{\mathbb{Z}})^*)^k\setminus\bigg(\bigcup_{P\in\mathcal{P}}\bigcap_{p\in P}A_p\bigg), \end{align*} $$

so we obtain (13).

Now assume that (14) does not hold, say for some $i\in \mathbb {Z}/n{\mathbb {Z}}$ . We first show that

(15) $$ \begin{align} |\{p|n: p\nmid i\}|>\ln\omega(n). \end{align} $$

Indeed, by our assumption the set $(A+i)\cap ({\mathbb {Z}}/n{\mathbb {Z}})^l$ is non-empty, so some $a\in A$ satisfies $a+i\in ({\mathbb {Z}}/n{\mathbb {Z}})^l$ . Then

$$ \begin{align*} \{p|n: p|i\}\subset \{p|n: a\ \pmod{p}\in(\mathbb{F}_p)^l\}, \end{align*} $$

and by the definition of the set A this set has less than $\omega (n)-\ln \omega (n)$ elements.

If $p|n$ and $p\nmid i$ , then by Lemma 4.2 we have

$$ \begin{align*} |\{x\in \mathbb{F}_p: x^l\in (\mathbb{F}_p)^k+a\}|&\leqslant \frac{1}{k}|\{(x, y)\in \mathbb{F}_p\times\mathbb{F}_p: x^l=y^k+a\}|+l\\& \leqslant \frac{p}{k}+2kl\sqrt{p}\leqslant \frac{2}{3}p. \end{align*} $$

Here the first inequality follows from the fact that the number of kth roots modulo p of ${x^l-a}$ is exactly 0 or k unless $x^l-a=0$ , which happens at most l times. The last inequality follows from $p>(12kl)^2$ .

Since $A\subset (\mathbb {Z}/n{\mathbb {Z}})^k$ , and by the Chinese remainder theorem, we have

$$ \begin{align*} |\{x\in {\mathbb{Z}}/n{\mathbb{Z}}: x^l\in A+i\}|\leqslant \prod_{p|n}|\{x\in \mathbb{F}_p: x^l\in (\mathbb{F}_p)^k+i\}|\leqslant \bigg(\frac{2}{3}\bigg)^{\ln\omega(n)}\cdot n, \end{align*} $$

where we bound each term in the product by $2p/3$ if $p\nmid i$ and by p otherwise.

Proof. We first recursively define a sequence $(n_t)$ so that $n_t|n_{t+1}$ , each $n_t$ is squarefree, and all prime divisors p of $n_t$ satisfy $l|p-1$ and $p>(12kl)^2$ . Additionally, we set $n_0=1$ and require

(18) $$ \begin{align} \sum\limits_{t=0}^{\infty}\bigg(2^{-\omega(n_{t+1})/4}+ \frac{2n_t+\sqrt[k]{n_{t+1}}}{\varphi(n_{t+1})/k^{\omega(n_{t+1})}}\bigg)<\frac{1}{10}, \end{align} $$

(19) $$ \begin{align} \frac{\varphi(n_t)}{n_t}>\frac{9}{10} \end{align} $$

(20) $$ \begin{align} n_{t+1}>10n_t^k. \end{align} $$

Let $A_t\subset ({\mathbb {Z}}/n_t{\mathbb {Z}})^*$ be the set given by Lemma 4.4 for $n_t$ . We treat $A_t$ as a subset of $[0, n_t-1]$ .

We also define a sequence $x_t\in \{0, 1, ?\}^{n_t}$ . We start with $x_0=?.$ Having defined $x_t$ , we define $x_{t+1}$ as follows. First we concatenate $n_{t+1}/n_t$ copies of $x_t$ , creating $x_{t+1}'$ . We then fill all the positions $i^k$ for $i\in [0,n_{t+1}^{1/k}]$ which still contain ‘?’ with 0s for even t and 1s for odd t, creating $x_{t+1}"$ . If the resulting word has the symbol ‘?’ at position i for

(21) $$ \begin{align} i\in [0, n_t-1]\cup [n_{t+1}-n_t, n_{t+1}-1]\cup ([0, n_{t+1}-1]\setminus A_{t+1}), \end{align} $$

we change it to a 0 or 1 arbitrarily, creating $x_{t+1}$ .

The pair $(n_t), (x_t)$ defined this way will be viable. Let us now show that condition (10) is satisfied. Also, all question marks in $x_t$ are at positions in $A_t\subset \widetilde {R^k_{n_t}}$ , so

$$ \begin{align*} ?_t\leqslant |\widetilde{R^k_{n_t}}|=o(n_t), \end{align*} $$

and so $X_x$ is regular.

Fix t and $a\in \mathbb {Z}$ . Notice that

(22) $$ \begin{align} \{i\in [0, n_t-1]: X_t(i^l+a)=?\}\subset\{i\in [0, n_t-1]: i^l+a\in A_t+n_t{\mathbb{Z}}\}, \end{align} $$

where we let $X_t\in \{0, 1, ?\}^{\mathbb {Z}}$ be the unique infinite $n_t$ -periodic sequence which agrees with $x_t$ on $[0, n_t-1]$ . By (14) the right-hand side of (22) has at most $n_t\cdot (2/3)^{\ln \omega (n_t)}=o(n_t)$ elements. Therefore, by Lemma 4.1, we obtain convergence of averages (16).

We now estimate $?_t$ from below. By the Chinese remainder theorem

$$ \begin{align*} \{i\in \widetilde{R^k_{n_{t+1}}}: x_{t+1}'(i)=?\}=?_t\cdot\prod_{\substack{p|n_{t+1}\\ p\nmid n_t}}\frac{p-1}{k}=?_t\cdot \frac{\varphi(n_{t+1}/n_t)}{k^{\omega(n_{t+1}/n_t)}}, \end{align*} $$

so from (13) we get

$$ \begin{align*} \{i\in A_{t+1}: x_{t+1}'(i)=?\}\geqslant ?_t\cdot \frac{\varphi(n_{t+1}/n_t)}{k^{\omega(n_{t+1}/n_t)}}-\frac{\varphi(n_{t+1})}{k^{\omega(n_{t+1})}}\cdot 2^{-\omega(n_{t+1})/4} \end{align*} $$

and

$$ \begin{align*} ?_{t+1}\geqslant ?_t\cdot \frac{\varphi(n_{t+1}/n_t)}{k^{\omega(n_{t+1}/n_t)}}-\frac{\varphi(n_{t+1})}{k^{\omega(n_{t+1})}}\cdot 2^{-\omega(n_{t+1})/4}-2n_t-\sqrt[k]{n_{t+1}}. \end{align*} $$

By induction we get

$$ \begin{align*} \frac{?_{t}}{\vphantom{\frac{1^1_1}{3^2_2}}|\widetilde{R^k_{n_{t}}}|}=?_{t}\cdot\frac{k^{\omega(n_t)}}{\varphi(n_t)}\geqslant 1-\sum\limits_{s=0}^{t-1}\bigg(2^{-\omega(n_{s+1})/4}+ \frac{2n_s+\sqrt[k]{n_{s+1}}}{\varphi(n_{s+1})/k^{\omega(n_{s+1})}}\bigg)\geqslant \frac{9}{10}, \end{align*} $$

where we use (18).

Now we estimate the number of question marks we fill in when going from $x_{t+1}'$ to $x_{t+1}"$ . First recall that all question marks in $x_t$ are at positions in $\widetilde {R^k_{n_t}}$ , and for all $a\in \widetilde {R^k_{n_t}}$ we have

$$ \begin{align*} |\{i<n_t: i^k\equiv a\ \pmod{n_t}\}|=k^{\omega(n_t)}, \end{align*} $$

so

$$ \begin{align*} |\{i<n_t: X_t(i^k)=?\}|=?_t\cdot k^{\omega(n_t)}\geqslant \tfrac{9}{10}\varphi(n_t)\geqslant \tfrac{8}{10}n_t \end{align*} $$

by (19).

Let $C_t=\lfloor {n_{t+1}}^{1/k}\rfloor =an_t+r$ for $a\in \mathbb {Z}$ and $r\in [0, n_t-1]$ , where $a\geqslant 10$ by (20). Then

$$ \begin{align*} |\{i<C_t: x_{t+1}'(i^k)=?\}|\geqslant a|\{i<n_t: x_{t+1}'(i^k)=?\}|\geqslant \tfrac{8}{10}an_t\geqslant \tfrac{7}{10}C_t. \end{align*} $$

Therefore, when constructing $x_{t+1}"$ we fill in at least $7C_t/10$ question marks, and so

$$ \begin{align*} \frac{1}{C_t}\sum\limits_{i<C_t}G(\sigma^{i^k}x)=\frac{1}{C_t}\sum\limits_{i<C_t}(-1)^{x_{t+1}(i^k)}\geqslant \frac{1}{C_t}\bigg(\frac{7}{10}C_t-\frac{3}{10}C_t\bigg)= \frac{2}{5} \end{align*} $$

for even t, and

$$ \begin{align*} \frac{1}{C_t}\sum\limits_{i<C_t}G(\sigma^{i^k}x)\leqslant -\frac{2}{5} \end{align*} $$

for odd t, and so the limit (4) does not exist.

Proof. Assume without loss of generality that $0\in \operatorname {\mathrm {Per}}_s(x)$ but $0\notin \operatorname {\mathrm {Per}}_{s'}(x)$ for any $s'<s$ , and assume that $x(0)=0$ .

First notice that for some N we have

$$ \begin{align*} \operatorname{\mathrm{Per}}^{(0)}_s(x|_{[0, N]})=\operatorname{\mathrm{Per}}^{(0)}_s(x). \end{align*} $$

Indeed, for any $i\in [0, s-1]\setminus \operatorname {\mathrm {Per}}^{(0)}_s(x)$ there exists a natural number $N_i$ such that $x(N_i)=1$ and $N_i\equiv i\ \pmod {s}$ . It is enough to take

$$ \begin{align*} N=\max\{N_i: i\in [0, s-1]\setminus \operatorname{\mathrm{Per}}^{(0)}_s(x)\}. \end{align*} $$

By minimality of $X_x$ , we can find a natural number M such that any word in $\mathcal {L}_{X_x}$ of length M contains $x|_{[0, N]}$ as a subword. We denote $x|_{[0, N]}=\nu $ .

We will now show that

$$ \begin{align*} d(y, \sigma^n y)\geqslant 2^{-M} \end{align*} $$

whenever $s\nmid n$ . This will clearly end the proof.

Pick n such that the above inequality does not hold. Then $y|_{[0, M]}=y|_{[n, n+M]}$ ; let us denote this word by $\omega $ . We know that $\omega |_{[k, k+N]}=\nu $ for some k, so

$$ \begin{align*} \operatorname{\mathrm{per}}^{(0)}_s(x)\leqslant \operatorname{\mathrm{per}}^{(0)}_s(\omega)\leqslant \operatorname{\mathrm{per}}^{(0)}_s(\nu)=\operatorname{\mathrm{per}}^{(0)}_s(x). \end{align*} $$

Here the first inequality follows from $\omega \in \mathcal {L}_{X_x}$ , the second from $\nu \subset \omega $ , and the equality follows from the definition of $\nu $ . In particular, both of the inequalities are actually equalities.

Since $\omega $ appears in $y|_{[0, n+M]}$ at positions starting with $0$ and n, we have

$$ \begin{align*} \operatorname{\mathrm{Per}}^{(0)}_s(y|_{[0, n+M]})\subset \operatorname{\mathrm{Per}}^{(0)}_s(\omega)\cap (n+\operatorname{\mathrm{Per}}^{(0)}_s(\omega)), \end{align*} $$

but

$$ \begin{align*} \operatorname{\mathrm{per}}^{(0)}_s(y|_{[0, n+M]})\geqslant \operatorname{\mathrm{per}}^{(0)}_s(x)=\operatorname{\mathrm{per}}^{(0)}_s(\omega) \end{align*} $$

since $y|_{[0, n+M]}\subset x$ . Hence,

$$ \begin{align*} \lvert \pi_s(\operatorname{\mathrm{Per}}_s^{(0)}(\omega))\rvert=\operatorname{\mathrm{per}}^{(0)}_s(\omega)\leqslant \lvert \pi_s(\operatorname{\mathrm{Per}}_s^{(0)}(\omega))\cap \pi_s(n+\operatorname{\mathrm{Per}}^{(0)}_s(\omega))\rvert. \end{align*} $$

The inequality must actually be an equality, and in particular we obtain

$$ \begin{align*} \pi_s(\operatorname{\mathrm{Per}}_s^{(0)}(\omega))=\pi_s(n+\operatorname{\mathrm{Per}}^{(0)}_s(\omega)). \end{align*} $$

We have $\nu \subset \omega $ , but $\operatorname {\mathrm {per}}^{(0)}_s(\nu )=\operatorname {\mathrm {per}}^{(0)}_s(\omega )$ , so

$$ \begin{align*} \pi_s(\operatorname{\mathrm{Per}}_s^{(0)}(\nu))=\pi_s(\operatorname{\mathrm{Per}}_s^{(0)}(\omega)-k)=\pi_s(n-k+\operatorname{\mathrm{Per}}^{(0)}_s(\omega))= \pi_s(n+\operatorname{\mathrm{Per}}^{(0)}_s(\nu)), \end{align*} $$

and so

$$ \begin{align*} \operatorname{\mathrm{Per}}_s^{(0)}(\nu)=n+\operatorname{\mathrm{Per}}_s^{(0)}(\nu), \end{align*} $$

and by the definition of $\nu $ we get

$$ \begin{align*} \operatorname{\mathrm{Per}}_s^{(0)}(x)=n+\operatorname{\mathrm{Per}}_s^{(0)}(x). \end{align*} $$

But then $x(an+bs)=x(0)=0$ for all $a, b\in {\mathbb {Z}}$ , and so $0\in \operatorname {\mathrm {Per}}_{\gcd (n, s)}(x)$ . We assumed that s is an essential period of 0, so we obtain $\gcd (n, s)=s$ , so $s|n$ .

Proof. Pick $i\in [0, s-1]$ . Since $\sigma ^i y$ is in the closure of the orbit $(\sigma ^{a_n}y)$ and $\sigma $ is a homeomorphism, we have

$$ \begin{align*} y\in \overline{\{\sigma^{a_n-i}y: n\in{\mathbb{Z}}^+\}}, \end{align*} $$

so by the previous lemma we have $s|a_n-i$ for some n.

Proof. Assume otherwise, and let $(n_t)$ be the sequence given by Corollary 5.3. Then $\Omega (n_{l+2})>l$ , so $n_{l+2}$ does not divide any element of $\mathbb {P}_l$ , and so we reach a contradiction.

Proof. It is enough to show that this orbit visits each set of the form

$$ \begin{align*}U=\{z\in X_x: z[-k, k]=\omega\}\end{align*} $$

for $\omega \in \mathcal {L}_{X_x}$ of length $2k+1$ . Fix such an $\omega $ , and let $x[a, a+2k]=\omega $ . Then for ${i\in [a, a+2k]}$ we let $p_i$ be the essential period of $x_i$ in x. This way the word $\omega $ appears in x starting at positions $a+np$ for $n\in {\mathbb {Z}}$ and $p=\operatorname {\mathrm {lcm}}\{p_i: i\in [a, a+2k]\}$ . Since the polynomial P is a permutation modulo each $p_i$ , it is also a permutation modulo p.

Now, since $y\in X_x$ , the word $\omega $ must appear in y periodically with period p, say starting at positions $a'+np$ for $n\in {\mathbb {Z}}$ . If $b>0$ is such that $P(b)\equiv a'+k\ \pmod {p}$ , then ${y[P(b)-k, P(b)+k]=\omega }$ , so $\sigma ^{P(b)}y\in U$ .

Proof. This follows directly from the definition of permutativity and Corollary 5.3.

Proof. Let $(x_t), (n_t)$ be a viable pair producing the Toeplitz word x, and such that each $n_t$ is a least common multiple of some essential periods of $X_x$ . This way P is a permutation modulo each $n_t$ . We proceed as in the proof of Lemma 4.1. It is enough to show that

$$ \begin{align*} \lim_{N\to\infty}\frac{1}{N}\sum\limits_{m<N}F(\sigma^{P(m)+r}x)=\int_{X_x}F\, d\mu \end{align*} $$

uniformly in $r\in {\mathbb {Z}}$ . We again assume that F only depends on coordinates from $-C$ to C for some $C\in {\mathbb {Z}}^+$ , and that $F:X_x\to [0, 1]$ . Fix t, and let

$$ \begin{align*} A_a=\{i\in [0, n_t-1]: X_t(i+a+r)=?\} \end{align*} $$

and

$$ \begin{align*} A=\bigcup_{a=-C}^{C}A_a. \end{align*} $$

Then $|A|\leqslant (2C+1)?_t$ , and for $i\notin A$ we have

$$ \begin{align*} F(\sigma^{P(i)+a+r}x)=F(\sigma^{P(i)+a+r \pmod{n_t}}x), \end{align*} $$

so since $P+r$ is a permutation modulo $n_t$ , we get

$$ \begin{align*} \bigg\lvert\! \sum\limits_{m<n_t}F(\sigma^{P(m)+r}x)-\sum\limits_{m<n_t}F(\sigma^{m}x)\bigg\rvert\leqslant 2\cdot (2C+1)?_t, \end{align*} $$

and as in the proof of Lemma 4.1 we obtain

$$ \begin{align*} \bigg\lvert \frac{1}{N}\sum\limits_{m<N}F(\sigma^{P(m)+r}x)- \frac{1}{n_t}\sum\limits_{m<n_t}F(\sigma^{m}x)\bigg\rvert\leqslant \frac{2\cdot (2C+1)?_t}{n_t}+\frac{n_t}{N}+\frac{n_t}{N-n_t} \end{align*} $$

for $N>n_t$ . Since

$$ \begin{align*} \lim\limits_{t\to\infty}\frac{1}{n_t}\sum\limits_{m<n_t}F(\sigma^{m}x)=\int_{X_x}F\, d\mu, \end{align*} $$

we obtain the desired convergence, and it is clearly uniform in $r\in {\mathbb {Z}}$ .

Proof. Let M be the absolute value of the leading coefficient of P. Since $d>1$ , by replacing $P(x)$ with $\pm P(x+k)$ for some large k we can assume that ${P(n{\kern-1pt}+{\kern-1pt}1){\kern-1pt}-{\kern-1pt}P(n){\kern-1pt}>{\kern-1pt}n}$ and $P(n)>Mn^d$ for $n\geqslant 0$ . Let $(n_t)$ be a strictly increasing sequence starting with $n_0=1$ satisfying $n_t|n_{t+1}$ such that P is permutative modulo each $n_t$ . By restricting to a subsequence, we can additionally assume that $n_{t+1}>(M+1)(10n_t)^{d}$ and

(24) $$ \begin{align} \sum\limits_{t=0}^{\infty}\frac{2n_t}{n_{t+1}}<\frac{1}{5}, \end{align} $$

and that all $m_t:=n_{t+1}/n_t$ are of the same parity. We now recursively define sequences of words $B_t^{(0)}, B_t^{(1)}\in \{0, 1\}^{n_t}$ . We start with $B_0=0$ and $B_0'=1$ .

Given $B_t^{(0)}$ and $B_t^{(1)}$ , we let

$$ \begin{align*} A_t=\{i\in [0, n_t-1]: B_t^{(0)}(i)\ne B_t^{(1)}(i)\}. \end{align*} $$

We let

$$ \begin{align*} B_{t+1}^{(0)}=B_t^{(\varepsilon_t(0))}B_t^{(\varepsilon_t(1))}\cdots B_t^{(\varepsilon_t(m_t-1))} \end{align*} $$

for a sequence $\varepsilon _t\in \{0, 1\}^{m_t}$ satisfying the following:

1. (1) $\varepsilon _t(0)=0$ and $\varepsilon _t(m_t-1)=1$ ;

2. (2) for $i>n_t$ satisfying $P(i)<n_{t+1}-n_t$ and $P(i)\in A_t+n_t{\mathbb {Z}}$ we have $B_{t+1}^{(0)}(i)=0$ if t is odd and 1 if t is even;

3. (3)

   $$ \begin{align*}|\{i\in [0, m_t-1]:\varepsilon_t(i)=0\}|=\begin{cases}m_t/2\quad&\text{if } m_t \text{ is even}\\ (m_t+1)/2\quad &\text{if } m_t \text{ is odd.}\end{cases}\end{align*} $$

Notice that the second condition can be satisfied, since for i in the specified range we have $P(i+1)-P(i)>n_t$ , and so the values of $\lfloor P(i)/n_t\rfloor \in [1, m_t-2]$ are all distinct, and so we can pick each of the $\varepsilon _t(\lfloor P(i)/n_t\rfloor )$ separately and without interfering with condition (1). This way we specify the value of $\varepsilon _t$ at no more than

$$ \begin{align*} 2+\sqrt[d]{n_{t+1}}<3\frac{n_{t+1}}{\sqrt[d]{n_{t+1}}}<3\frac{n_{t+1}}{10n_t}<\frac{m_t}{2} \end{align*} $$

places, so we can now fulfil condition (3).

We then define

$$ \begin{align*} B_{t+1}^{(1)}=B_t^{(\varepsilon_t'(0))}B_t^{(\varepsilon_t'(1))}\cdots B_t^{(\varepsilon_t'(m_t-1))}, \end{align*} $$

where $\varepsilon '(i)=1-\varepsilon (i)$ for $i\notin \{ 0, m_t-1\}$ , but $\varepsilon '(0)=0$ and $\varepsilon '(m_t-1)=1$ .

Letting $W_t=\{B_t^{(0)}, B_{t}^{(1)}\}$ , we construct $x\in \{0, 1\}^{{\mathbb {Z}}}$ as in §2.1; by Fact 2.3 the word x is then Toeplitz. If the $m_t$ are even, by induction we have

$$ \begin{align*} \operatorname{\mathrm{ap}}(B_t^{(\varepsilon)}, B_s^{(\varepsilon')})=\tfrac{1}{2} \end{align*} $$

for any $t<s$ and $\varepsilon , \varepsilon '\in \{0, 1\}$ . By Theorem 2.4 in this case $X_x$ is uniquely ergodic.

If the $m_t$ are odd, by a similar computation to Example 2.3 in [Reference Iwanik and Lacroix4] we obtain

$$ \begin{align*} \operatorname{\mathrm{ap}}(B^{(\varepsilon)}_t, B^{(\varepsilon')}_s)=\frac{1}{2}\bigg(1\pm \frac{1}{m_tm_{t+1}\cdots m_{s-1}}\bigg), \end{align*} $$

where the sign is positive if $\varepsilon =\varepsilon '$ and negative otherwise. Hence, in this case for any $t, \varepsilon , \varepsilon '$ we have

$$ \begin{align*} \lim\limits_{s\to\infty}\operatorname{\mathrm{ap}}(B_t^{(\varepsilon)}, B_s^{(\varepsilon')})=\tfrac{1}{2}, \end{align*} $$

so by Theorem 2.4 the system $X_x$ is uniquely ergodic.

All essential periods of $X_x$ are clearly divisors of some $n_t$ . By Theorem 5.5 we therefore see that $(\sigma ^{P(n)}y)$ is dense in $X_x$ for any $y\in X_x$ .

All that remains is to see that the limit (23) does not exist.

We first estimate $|A_t|$ inductively: we have $|A_0|=1$ , and

$$ \begin{align*} |A_{t+1}|\geqslant (m_t-2)|A_{t}|, \end{align*} $$

so

$$ \begin{align*} \frac{1}{n_t}|A_t|\geqslant \prod_{s=0}^{t-1}\bigg(1-\frac{2n_t}{n_{t+1}}\bigg)\geqslant \frac{4}{5} \end{align*} $$

by (24).

Let us now estimate the number of $i> n_t$ that satisfy the assumption of condition (2). Since P is permutative, among any $n_t$ consecutive values of i precisely $|A_t|$ satisfy $P(i)\in A_t+n_t{\mathbb {Z}}$ . Let $C_t=\max \{i\in {\mathbb {Z}}^+: P(i)<n_{t+1}-n_t\}$ . For large t we have ${C_t>\sqrt [d]{n_{t+1}/(M+1)}>10n_t}$ , so

$$ \begin{align*} \lvert \{i{\kern-1pt}>{\kern-1pt}n_t: P(i){\kern-1pt}<{\kern-1pt}n_{t+1}-n_t\text{ and }P(i){\kern-1pt}\in{\kern-1pt} A_t+n_t{\mathbb{Z}}\}\rvert>|A_t|\cdot \frac{(C_t-2n_t)}{n_t}{\kern-1pt}>{\kern-1pt}\frac{4}{5}\cdot \frac{4}{5}C_t{\kern-1pt}>{\kern-1pt}\frac{3}{5}C_t. \end{align*} $$

Therefore, for large even t we have

$$ \begin{align*} \frac{1}{C_t}\sum\limits_{m<C_t}G(\sigma^{P(m)}x)=\frac{1}{C_t}\sum\limits_{m<C_t}(-1)^{x_{t+1}(P(m))}\geqslant \frac{1}{C_t}\bigg(\frac{3}{5}C_t-\frac{2}{5}C_t\bigg)= \frac{1}{5}, \end{align*} $$

and for large odd t we have

$$ \begin{align*} \frac{1}{C_t}\sum\limits_{m<C_t}G(\sigma^{P(m)}x)\leqslant-\frac{1}{5}, \end{align*} $$

and so the limit (23) does not exist.