Proof. Assume that $x < ({2a-1})/({1-a})$ , otherwise, (33) is trivial. We prove by induction in k for each fixed n. In the basis case $k=0$ , inequality (33) is trivial.

We now prove the step of induction. Put

$$ \begin{align*} m= \max\{0 \le i \le \min(k, n): t_j(x)=t_j\text{ for all }j\le i\} \end{align*} $$

and

$$ \begin{align*} m'=\min(m, \sigma_a-1). \end{align*} $$

If $m=k$ , then (33) immediately follows from the definition of $d_n$ . From now on, we assume that $m < k$ and we claim that

(34) $$ \begin{align} T_a^{t_i}(0) \le T_a^{t_i}(x)=T_a^{t_i(x)}(x)<\frac{2a-1}{1-a}, \quad 0 \le i \le m'. \end{align} $$

We prove this by induction, assuming that $n \ge 1$ , otherwise, there is nothing to prove. In the basis case $i=0$ , this is true by the assumption. If we already established this for all $i \le j$ for some $j < m'$ , then $ 1 \le T_a^{t_j+1}(0) \le T_a^{t_j +1}(x)$ since $T_a$ increases on the interval $[0, ({2a-1})/({1-a}))$ . Then, $T_a^{\ell }(0) \le T_a^{\ell }(x)$ for all $t_j+1 \le \ell \le t_{j+1}$ since $T_a$ increases on $[1, {1}/({1-a})]$ . Moreover, $T_a^{t_{j+1}}(x) = T_a^{t_{j+1}(x)}(x) < ({2a-1})/({1-a})$ since $j +1 \le m' < \min (\sigma _a (x), \sigma _a)$ . This establishes (34).

Furthermore, by the same reasoning, it follows from (34) that $t_{m'+1}(x) \ge t_{m'+1}(0)$ and

(35) $$ \begin{align} T_a^{t_{m'+1}}(0) <1 \ \text{ and } \ T_a^{t_{m'+1}}(0) \le T_a^{t_{m'+1}}(x). \end{align} $$

Consider three cases. If $m'=n$ , then $n< \sigma _a$ , and by the definition of $d_n$ ,

$$ \begin{align*} t_{m'+1}(x) \ge t_{m'+1}(0) \ge (m'+1) d_n. \end{align*} $$

If $m'<n$ and $m'=m$ , then $m < \sigma _a$ and $m'=m<n$ , and $t_{m'+1}(x) = t_{m'+1}(0)$ is impossible by the definition of m. Hence, by the definition of $d_n$ ,

$$ \begin{align*} t_{m'+1}(x) \ge t_{m'+1}(0) +1 \ge (m'+1) d_n. \end{align*} $$

The remaining case $m'<n$ and $m'=\sigma _a-1<m$ is impossible, because otherwise, $m=n = \sigma _a$ and it follows from the definition of m and inequalities (35) that $({2a-1})/({1-a}) < T_a^{t_m}(x) <1$ . Thus, $\sigma _a(x)=m$ , which contradicts our earlier assumption that $m<k$ .

Thus, in all possible cases, we have $t_{m'+1}(x) \ge (m'+1) d_n$ , where $1 \le m' +1 \le k$ . Consequently,

$$ \begin{align*} t_k(x) = t_{m'+1}(x)+ t_{k-m'-1}(T_a^{t_{m'+1}(x)}(x)) \ge (m'+1) d_n + t_{k-m'-1}(T_a^{t_{m'+1}(x)}(x)). \end{align*} $$

Therefore, if we already proved (33) for all $k \le j$ so some $0 \le j < \sigma _a(x)$ , then

This finishes the proof of (33). In particular, (33) implies that $t_k \ge k d_0 = k t_1$ for all integer $1 \le k \le \sigma _a(x)$ . This in turn implies that $d_n \ge t_1$ for all integer $0 \le n \le \sigma _a $ , as needed.

Proof. (a) We argue by induction. In the basis case $k\kern1.2pt{=}\kern1.2pt 1$ , we have $D_1\kern1.2pt{=}\kern1.2pt\{({2a\kern1.2pt{-}\kern1.2pt1})/({1\kern1.2pt{-}\kern1.2pt a}), {1}/({1-a})\}$ , and hence, the set $\{x: \varkappa _a(x)\ge 1\} $ , which is $[0,{1}/({1-a})] \setminus I_a$ , is of the form stated. The claims concerning $\delta _0(x)$ and $T_a(x)$ are clearly satisfied.

Assume that the statements are proved for all $1 \le k \le n$ . Fix a $y \in D_n$ and consider three cases.

If $T_a^n(g_n(y)) \ge 1$ , then $\{x \in [g_n(y),y]: T_a^n(x) \not \in I_a \}=[g_n(y),y]$ , and this is a maximal interval contained in $\text {dom}(T^{n+1}_a)$ . We have $g_n(y) \in G_{n+1}$ by $G_n \cap \{x: \varkappa _a(x) \ge n+1\} \subset G_{n+1}$ and the fact that $\varkappa _a(y)\ge n+1$ , and also $y \in D_{n+1}$ by $D_n \subset D_{n+1}$ . The interval $(g_n(y),y)$ contains no points of $G_{n+1}$ since it contains no points of $G_n$ and $T_a^{n+1}(y) \neq 0$ for every $x\in (g_n(y),y)$ by $T_a^n(x)>T_a^n(g_n(y)) \ge 1$ . Hence, $g_{n+1}(y)=g_n(y)$ .

If $({2a-1})/({1-a}) < T_a^n(g_n(y)) < 1$ , then by $T_a^n(y) = {1}/({1-a})>1$ , there is a unique $z \in [g_n(y),y]$ such that $T_a^n(z)=1$ . Then, $z \in G_{n+1}$ by $T_a^{n+1}(z)=T_a(1)=0$ , and $ y \in D_{n+1}$ . By the same argument as in the previous case, $[z,y]$ is the maximal subinterval of $[g_n(y),y]$ contained in $\text {dom}(T^{n+1}_a)$ , and hence, $g_{n+1}(y)=z$ .

Lastly, if $0 \le T_a^n(g_n(y)) \le ({2a-1})/({1-a})$ , then by $T_a^n(y)>1$ , there exist unique $z_1, z_2 \in [g_n(y),y]$ such that $T_a^n(z_1) = ({2a-1})/({1-a})$ and $T_a^n(z_2)=1$ . Similarly to the previous cases, $[g_n(y),z_1]$ and $[z_2, y]$ are the maximal subintervals of $[g_n(y),y]$ contained in $\text {dom}(T^{n+1}_a)$ , and there are no other ones. It holds $z_1, y \in D_{n+1}$ and $g_n(y), z_2 \in D_{n+1}$ . We also have $g_{n+1}(z_1)=g_n(y)$ , $g_{n+1}(y)=z_2$ .

The above consideration of the three cases combined with the representation

$$ \begin{align*} \{x: \varkappa_a(x)\ge n+1\} = \{x: \varkappa_a(x)\ge n, T_a^n(x) \not \in I_a \} =\bigcup_{z \in D_n} \{x \in [g_n(z),z]: T_a^n(x) \not \in I_a \} \end{align*} $$

imply that the set $\{x: \varkappa _a(x)\ge n+1\}$ is a union of the intervals $\{[g_{n+1}(z),z]: z \in D_{n+1}\}$ and moreover, we have

$$ \begin{align*} \{g_{n+1}(z): z \in D_{n+1}\}=\{g_n(z): z \in D_n, \varkappa_a(z) \ge n+1\} \cup T_a^{-n}(1). \end{align*} $$

Hence, $G_{n+1} = \{g_{n+1}(z): z \in D_{n+1}\}$ by the assumption of induction and the fact that $T_a^{-n}(1)=T_a^{-(n+1)}(0)$ . The set $D_{n+1}$ is finite because so is $D_n$ . The intervals $\{[g_{n+1}(z),z]: z \in D_{n+1}\}$ are disjoint as subintervals of disjoint intervals $\{ [g_n(z),z]: z \in D_n\}$ .

Furthermore, in either of the three cases, the function $T_a^{n+1}$ is continuous and strictly increasing on $[g_{n+1}(y),y]$ as a composition of $T_a$ and $T_a^n$ , which have these properties on $[1, {1}/({1-a})]$ and $[g_n(y),y]$ , respectively. Then, $\delta _n$ is constant on $[g_{n+1}(y),y]$ . In the third case, $T_a^{n+1}$ and $\delta _n$ also have these respective properties on $[g_{n+1}(z_1), z_1]$ because on this interval $T_a^n$ does not exceed $({2a-1})/({1-a})$ and $T_a$ is continuous and strictly increasing on $[0, ({2a-1})/({1-a})]$ .

(b) The proof for $a=\tfrac 23$ is analogous. It suffices to replace throughout $[g_n(y), y]$ by $[g_n(y-), y)$ for all $y \in D_n \setminus \{3\}$ and use that $T_{2/3}^n(y-)=3$ instead of $T^n_a(y) = {1}/({1-a})$ .

Proof of Proposition 10

Let us show that the equation $R_a(\unicode{x3bb} )=1$ has exactly one positive solution $\unicode{x3bb} =\unicode{x3bb} _a$ . This is evident when $\varkappa _a$ is finite because in this case, $R_a$ is a continuous strictly decreasing function on $(0, \infty )$ which satisfies $R_a(p)>1$ by $\delta _0=1$ and $R_a\, (+\infty ) =0$ . Then, there is a unique solution $\unicode{x3bb} _a$ , which satisfies $\unicode{x3bb} _a> p$ . Moreover, this reasoning applies easily when $\varkappa _a = \infty $ and $q \le p$ . In this case, $R_a$ is finite on $(p, \infty )$ and satisfies $R_a(p)=R_a(p+)$ by the monotone convergence theorem, and hence, $1<R_a(p')< \infty $ for some $p'>p$ .

In the remaining case where $\varkappa _a=\infty $ and $p<1/2$ , put $r=q/p$ . We will use the following representation:

$$ \begin{align*} R_a(\unicode{x3bb}) = \sum_{k=0}^{\varkappa_a} \delta_k (p/\unicode{x3bb})^{k+1} (q/p)^{L_k}= \sum_{k=0}^{\sigma_a} (p/\unicode{x3bb})^{t_k+1} r^k. \end{align*} $$

We first assume that $\sigma _a = \infty $ . Denote $d=\sup _{n \ge 0} d_n$ and $\theta =\liminf _{n \to \infty } t_n/n $ . By the Cauchy–Hadamard formula, $R_a(\unicode{x3bb} )$ is finite for $\unicode{x3bb}>p$ when

$$ \begin{align*} r < \frac{1}{\limsup_{k \to \infty} (p/\unicode{x3bb})^{(t_k+1)/k}} =\frac{1}{(p/\unicode{x3bb})^\theta}, \end{align*} $$

that is, for $\unicode{x3bb}> p r^{1/\theta }$ . Notice that $\theta \ge d \ge t_1$ by Lemma 11.

If $d_n= {t_{n+1}}/({n+1})$ for all $n \ge 1$ , then $d \ge t_n/n$ for all $n \ge 1$ . Therefore, for $C=1/d$ , we have $n \ge C t_n$ for all $n \ge 1$ . Hence, by $r>1$ ,

$$ \begin{align*} R_a(pr^C) =\sum_{k=0}^\infty r^{-(t_k+1)C+k} \ge r^{-C} \sum_{k=0}^\infty r^0 =\infty. \end{align*} $$

If there exists an $n_0 \ge 1$ such that $d_{n_0}< {t_{n_0+1}}/({n_0+1})$ , then $d_{n_0} = ({t_{k_0} +1})/{k_0}$ for some $1 \le k_0 \le n_0$ , and hence, for $C=1/d_{n_0}$ ,

$$ \begin{align*} R_a(p r^C)> r^{-(t_{k_0}+1)C+k_0} = r^0=1. \end{align*} $$

In either case, we have $C \ge 1 /\theta $ . Therefore, $1<R_a(p')< \infty $ for some $p'>p r^C$ by the same argument as above for $p\ge 1/2$ , and there exists a unique solution $\unicode{x3bb} _a$ to $R_a(\unicode{x3bb} )=1$ , which satisfies $\unicode{x3bb} _a> p' > p r^C$ .

If $\sigma _a$ is finite, then we have $R_a(p r^C)>1$ for $C=1/d_{\sigma _a}$ and $\unicode{x3bb} _a> p r^C$ , exactly as above.

We now prove inequality (32). Fix an $x \in [0, 1/(1-a)]$ .

First, assume that $\sigma _a(x) = \infty $ . Fix a $k \ge 0$ and put $x'= T_a^k(x)$ if $T_a^k(x)<1$ and ${x'=T_a^{t_1(T_a^k(x))}}$ otherwise. It follows from Lemma 11 that $t_{\lceil Ck' \rceil }(x') \ge \lceil kC' \rceil /C -1\ge k'-1$ for all integer $k' \ge 1$ such that $1 \le \lceil Ck' \rceil \le \sigma _a(x')$ . Substituting $n-k=k'-1$ , we get

$$ \begin{align*} L_n(x) - L_k(x) = L_{n-k}(T_a^k(x)) \le L_{n-k}(x') \le L_{t_{\lceil Ck'\rceil}(x')}(x') = \lceil C(n-k+1)\rceil \end{align*} $$

for all integer $n \ge k$ since $\sigma _a(x')=\sigma _a(x)=\infty $ . Then,

$$ \begin{align*} L_n(x) - L_k(x) \le C(n-k+1) \end{align*} $$

because the left-hand side is integer. This proves inequality (32) when $\sigma _a(x) = \infty $ .

We now assume that $\sigma _a(x) <\infty $ , which is possible only when $a <\tfrac 23$ . We argue by reduction to the previous case. Assume that $\varkappa _a(x)\ge 1$ , otherwise, there is nothing to prove. Fix an integer $1 \le n \le \varkappa _a(x)$ . By Lemma 12(a), there exists a unique $y \in D_n$ such that $x \in [g_n(y), y]$ .

We first assume that $g_n(y) \neq y$ . Since the functions $z \mapsto L_k(z)$ for $0 \le k \le n$ are constant on $[g_n(y), y]$ by Lemma 12(a), inequality (32) follows instantly from the result in the previous case if we prove that there is an $x' \in [g_n(y), y]$ such that $\sigma _a(x') =\infty $ . We will use that there exist periodic orbits of $T_a$ starting arbitrarily close to the left of y. Consider two cases. If $T_a^m(y) = ({2a-1})/({1-a})$ for a unique $0 \le m \le n-1$ , then $T_a^m < ({2a-1})/({1-a})$ on $[g_n(y), y)$ by Lemma 12(a). By piecewise continuity of $T_a$ , the equation $z = T_a^{t_1(z)}(z)$ admits infinitely many solutions, which accumulate to $({2a-1})/({1-a})$ from the left. Pick any solution $z' \in [ T_a^m(g_n(y)), ({2a-1})/({1-a}))$ and then take the unique $x' \in [g_n(y), y]$ that solves $T_a^m(x')=z'$ . In the second case when ${y = {1}/({1-a})}$ , we argue similarly and take any $x' \ge g_n(y)$ that satisfies $x' = T_a^{t_1(x')+1}(x')$ .

We now assume that $g_n(y) =y$ . This is possible only when $G_n$ intersects $D_n$ , that is, when the orbit of zero hits $({2a-1})/({1-a})$ . Then, $T_a^{k_1}(x)=0$ and $T_a^{k_2}(x) = ({2a-1})/({1-a})$ for the unique $k_1$ , $k_2$ such that $0 \le k_1 < k_2 \le n-1$ . Thus, $x \in G_{k_2}$ and by Lemma 12(a), there exists a unique $y' \in D_{k_2}$ such that $g_{k_2}(y')=x$ . In other words, $\varkappa _a(z) = k_2$ for every $z \in (g_{k_2}(y'), y')$ . Clearly, $g_{k_2}(y') \neq y'$ and we can apply the result of the previous case for all k satisfying $0 \le k \le k_2$ . Therefore, since the functions $z \mapsto L_k(z)$ for $0 \le k \le k_2$ are constant on $[g_{k_2}(y'), y']$ ,

$$ \begin{align*} L_n(x) - L_k(x)= L_{k_2}(y') + 1 - L_k(y') \le C (k_2 - k+1) +1 \le C (n - k+1) + 1. \end{align*} $$

However, $L_n(x) - L_k(x)=0$ for all k satisfying $k_2 +1 \le k \le n$ . This finishes the proof of (32). Notice that in the last case, $\delta _n (x)=0$ for all $n \ge k_2+1$ , therefore, we actually proved that

(36) $$ \begin{align} \delta_n (x) L_n(x) \le C (n+1), \quad 0 \le n \le \varkappa_a(x). \end{align} $$

Finally, we prove the upper bound in (31). For $p\ge 1/2$ , we have $r \le 1$ and we use the $(1/a, 1/2)$ -expansion (11) to estimate

$$ \begin{align*} R_a (p/a) &= \sum_{k=0}^{\varkappa_a} \delta_k a^{k+1} r^{L_k} \le \sum_{k=0}^\infty (1- d_k(0)) a^{k+1} \\ &=\frac{a}{2(1-a)}+ \sum_{k=0}^\infty \bigg(\dfrac12- d_k(0) \bigg)a^{k+1} = \dfrac{a}{2(1-a)} \le 1. \end{align*} $$

Hence, $\unicode{x3bb} _a \le p/a$ , and both inequalities above turn into equalities only when $a = 2/3$ and $p=1/2$ . In this case, $R_{2/3}(3/4)=1$ and thus, $\unicode{x3bb} _{2/3}(1/2) = 3/4$ .

For $p<1/2$ , we use a similar bound based on (36):

$$ \begin{align*} R_a ( p r^C/a ) = \sum_{k=0}^{\varkappa_a} \delta_k a^{k+1} r^{L_k - C (k+1)} \le \sum_{k=0}^\infty (1- d_k(0)) a^{k+1} \le 1. \end{align*} $$

The second inequality turns into an equality only in the case where $a=2/3$ , when the first inequality is strict since the orbit of zero under $T_{2/3}$ is not periodic. Hence, $\unicode{x3bb} _a < p r^C/a$ .

Proof of Proposition 14

In the proof, we shall omit the subscript a in $T_a$ , $\varkappa _a$ and $\varkappa _a'$ .

For every function $f \in U$ , which is of the form , one has

Consider the first sum in the last line. We notice that for every $k<\varkappa $ , it is true that

and

If $\varkappa $ is finite, then $T^\varkappa (0)\in (({2a-1})/({1-a}),1)$ and therefore, $\delta _\varkappa =1$ and $(T^\varkappa (0)-1)/ a<0$ . Hence,

(44)

To rewrite the second sum, we notice that for every $k<\varkappa $ , it is true that

and

because in the case when $a =2/3$ , where $2/a=1/(1-a)$ , we have $T^k(0) \neq 1$ ; see §2.3. If $\varkappa $ is finite, then $T^\varkappa (0){\kern-1pt}\in{\kern-1pt} (({2a{\kern-1pt}-{\kern-1pt}1})/({1{\kern-1pt}-{\kern-1pt}a}),1)$ and therefore, $(T^\varkappa (0){\kern-1pt}+{\kern-1pt}1)/a{\kern-1pt}>{\kern-1pt} 1/(1{\kern-1pt}-{\kern-1pt}a)$ . Hence,

(45)

Putting (44) and (45) together, and using that $c_k=q\delta _k+p(1-\delta _k)$ , we conclude that

(46)

We now consider the two cases separately.

The orbit of zero is aperiodic. Comparing the coefficients at in the equalities and (46), we see that the restriction of P to the space U is equivalent to the linear operator A on $(\mathbb R^{\varkappa _a +1}, {\| \cdot \|}_1)$ .

Consider the eigenvalue problem $Au=\unicode{x3bb} u$ . By the definition of A,

$$ \begin{align*} p\sum_{k=0}^\varkappa \delta_ku_k=\unicode{x3bb} u_0 \quad\text{and}\quad c_{k-1}u_{k-1}=\unicode{x3bb} u_k, \quad 1 \le k < \varkappa +1. \end{align*} $$

These equations imply that

(47) $$ \begin{align} u_k= u_0 \frac{c_0 c_1 \cdot \cdots \cdot c_{k-1}}{\unicode{x3bb}^k},\quad 1 \le k < \varkappa +1. \end{align} $$

Since $\delta _0=1$ , we see that there is a non-trivial solution only if $\unicode{x3bb} $ satisfies the equation

(48) $$ \begin{align} \unicode{x3bb}= p + p\sum_{k=1}^\varkappa \delta_k \frac{c_0 c_1 \cdot \cdots \cdot c_{k-1}}{\unicode{x3bb}^k}. \end{align} $$

Recalling that $L_k = \sum _{i=0}^{k-1} \delta _i$ and $c_i=q\delta _i+p(1-\delta _i)$ , we get

(49) $$ \begin{align} c_0 c_1 \cdot \cdots \cdot c_{k-1}= p^k (q/p)^{L_k}. \end{align} $$

Hence, (48) is equivalent to (15), which has a unique positive solution $\unicode{x3bb} _a>p$ , as we showed in §3.

Notice that in the case when $\varkappa _a = \infty $ , the vectors u defined by (47) with $\unicode{x3bb} =\unicode{x3bb} _a$ satisfy $u \in \ell _1$ by $\unicode{x3bb} _a>p$ if $p \ge 1/2$ and by (32) if $p<1/2$ . Then, it follows that the vector v, which is proportional to such u, is the eigenvector of A corresponding to $\unicode{x3bb} _a$ .

We now consider the eigenvalue problem $A^* u^* =\unicode{x3bb} _a u^*$ . Invoking the definition of A, we get the following coordinate-wise equations:

(50)

Setting

where $c_0c_1\cdots c_{-1} = 1$ by convention, we get the equations

Therefore,

and using the definition of $\unicode{x3bb} _a$ and (49), we obtain

Consequently,

Notice that in the case of infinite $\varkappa $ , the vectors $u^*$ satisfy $u^* \in \ell _\infty $ by $\unicode{x3bb} _a>p$ if $p \ge 1/2$ and by (32) if $p<1/2$ . In the case of finite $\varkappa $ , the equation in the second line of (50) is satisfied since $\delta _\varkappa =1$ . It follows that the vector $v^*$ , which is proportional to such $u^*$ , is the eigenvector of $A^*$ corresponding to $\unicode{x3bb} _a$ .

Furthermore, we already know that there are no positive eigenvalues of A other than $\unicode{x3bb} = \unicode{x3bb} _a$ . Assume that $\unicode{x3bb} \in \mathbb C \setminus (0, \infty )$ is a different eigenvalue of A. By (48), we have

(51) $$ \begin{align} |\unicode{x3bb}|-p<|\unicode{x3bb}-p| =\bigg|p\sum_{k=1}^\varkappa \delta_k (p/\unicode{x3bb})^k (q/p)^{L_k}\bigg| \le p\sum_{k=1}^\varkappa \delta_k (p/|\unicode{x3bb}|)^k (q/p)^{L_k}. \end{align} $$

This implies that $R(|\unicode{x3bb} |)>1$ , and hence, $|\unicode{x3bb} |<\unicode{x3bb} _a$ . Thus, $\unicode{x3bb} _a$ is the leading eigenvalue of A. Denote $\varrho '=0$ if A has no other eigenvalues, otherwise, put

Assume that $\varkappa $ is finite. It is easy to see that the matrix $\mathcal A$ that represents A in the standard basis has non-negative entries and is irreducible, that is, for some $m \in \mathbb N$ , all entries of $\mathcal A+{\mathcal A}^2 + \cdots + {\mathcal A}^m$ are strictly positive. Then, $\unicode{x3bb} _a$ is a simple root of the characteristic polynomial of $\mathcal A$ and ${\mathcal A}^n/\unicode{x3bb} _a^n \to v v^*/ v^* v$ (with $v^*$ regarded as a row vector) in the operator norm as $n \to \infty $ by the classical Perron–Frobenius theorem; see [Reference Meyer19, p. 673 and (8.3.10)]. For the rate of this convergence, the spectral resolution theorem for $A^n$ implies the bound (39) for any $\gamma \in (\varrho '/\unicode{x3bb} _a, 1)$ ; see [Reference Meyer19, (7.9.9) and the last equation on p. 629].

From now on, we assume that the orbit of zero is infinite (and still aperiodic). Notice that $\varrho '< \unicode{x3bb} _a$ , otherwise, there is a convergent sequence of eigenvalues $\{\mu _n\}$ such that ${|\mu _n| \to |\unicode{x3bb} _a|}$ . If $p\ge 1/2$ , then R is analytic on the set $D=\{\unicode{x3bb} \in \mathbb C:|\unicode{x3bb} |>p \}$ , which contains $\mu _n$ for all n large enough by $\unicode{x3bb} _a>p$ . If $p< 1/2$ , then it follows from (32) that R is analytic on the set $D=\{\unicode{x3bb} \in \mathbb C:|\unicode{x3bb} |>p (q/p)^C \}$ , which contains $\mu _n$ for all n large enough since $\unicode{x3bb} _a>p (q/p)^C$ by (31). In either case, it must be that $R \equiv 1 $ on D by $R(\mu _n)\equiv 1$ , which is a contradiction.

We now show that the spectrum of A without the point $\unicode{x3bb} _a$ is contained in the closed centred ball of radius $\varrho =\max (\varrho ', p, p (q/p)^C)$ , where $\varrho < \unicode{x3bb} _a$ . Equivalently, the resolvent operator $(A-\unicode{x3bb} I)^{-1}$ is bounded for every $\unicode{x3bb} \neq \unicode{x3bb} _a$ with $|\unicode{x3bb} |>\varrho $ .

We first prove that the equation $(A-\unicode{x3bb} I)u=w$ has a unique solution for every $w\in \ell _1$ . Writing this equation coordinatewise, we have

(52) $$ \begin{align} \begin{array}{cl} &(p-\unicode{x3bb})u_0+p\sum_{k=1}^\infty\delta_ku_k=w_0,\\ &c_{k-1}u_{k-1}-\unicode{x3bb} u_k=w_k,\quad k\ge1. \end{array} \end{align} $$

It is easy to see that

(53) $$ \begin{align} u_k=\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}u_0- \sum_{j=1}^k w_j\frac{c_jc_{j+1}\cdots c_{k-1}}{\unicode{x3bb}^{k-j+1}}, \quad k \ge 1, \end{align} $$

is the unique solution to the equations in the second line of (52). Therefore,

(54) $$ \begin{align} \bigg(p-\unicode{x3bb}+p\sum_{k=1}^\infty\delta_k\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}\bigg)u_0 =w_0 + p\sum_{k=1}^\infty\delta_k \sum_{j=1}^k w_j\frac{c_jc_{j+1}\cdots c_{k-1}}{\unicode{x3bb}^{k-j+1}}, \end{align} $$

and thus, $u_0$ is defined uniquely since $\unicode{x3bb} $ is not an eigenvalue of A and therefore, the factor of $u_0$ on the left-hand side is non-zero by (48).

We now estimate the norm of u. Assume first that $p \ge 1/2$ . Then,

(55) $$ \begin{align} \nonumber \sum_{k=1}^\infty|u_k| &\le |u_0|\sum_{k=1}^\infty\frac{c_0c_1\cdots c_{k-1}}{|\unicode{x3bb}|^k} +\sum_{k=1}^\infty\sum_{j=1}^k|w_j| \frac{c_jc_{j+1}\cdots c_{k-1}}{|\unicode{x3bb}|^{k-j+1}}\\ \nonumber &\le |u_0|\sum_{k=1}^\infty\bigg(\frac{p}{|\unicode{x3bb}|}\bigg)^k +\sum_{j=1}^\infty|w_j|\sum_{k=j}^\infty\frac{p^{k-j}}{|\unicode{x3bb}|^{k-j+1}}\\ &\le \frac{p|u_0|}{|\unicode{x3bb}|-p}+\frac{1}{|\unicode{x3bb}|-p}\sum_{j=1}^\infty|w_j| \end{align} $$

and

(56) $$ \begin{align} |\unicode{x3bb} (R(\unicode{x3bb})-1) ||u_0| & \le |w_0|+p\sum_{k=1}^\infty\delta_k \sum_{j=1}^k|w_j|\frac{p^{k-j}}{|\unicode{x3bb}|^{k-j+1}} \le |w_0|+\frac{p}{|\unicode{x3bb}|-p}\sum_{j=1}^\infty|w_j|. \end{align} $$

Consequently, for every $\unicode{x3bb} \neq \unicode{x3bb} _a$ with $|\unicode{x3bb} |>\varrho $ , there exists a constant $C(\unicode{x3bb} )$ such that

(57) $$ \begin{align} { \|(A-\unicode{x3bb} I)^{-1} \|}_1\le C(\unicode{x3bb}) {\|w\|}_1. \end{align} $$

Let us estimate the norm of u in the case when $p < 1/2$ . Using the first line in (55) and recalling that $c_0c_1\cdots c_{j-1}=p^jr^{L_j}$ with $r=q/p$ , we obtain

$$ \begin{align*} \sum_{k=1}^\infty|u_k| &\le |u_0|\sum_{k=1}^\infty\frac{p^kr^{L_k}}{|\unicode{x3bb}|^k} +\sum_{k=1}^\infty\sum_{j=1}^k\frac{p^{k-j}r^{L_k-L_j}}{|\unicode{x3bb}|^{k-j+1}}|w_j|\\ &=|u_0|\sum_{k=1}^\infty\frac{p^kr^{L_k}}{|\unicode{x3bb}|^k} +\sum_{j=1}^\infty|w_j|\sum_{k=j}^\infty\frac{p^{k-j}r^{L_k-L_j}}{|\unicode{x3bb}|^{k-j+1}}. \end{align*} $$

Applying now (32), we get

$$ \begin{align*} \sum_{k=1}^\infty|u_k| &\le |u_0|r^{C+1}\sum_{k=1}^\infty\frac{p^k r^{Ck}}{|\unicode{x3bb}|^k} +\frac{r^{C+1}}{|\unicode{x3bb}|}\sum_{j=1}^\infty|w_j| \sum_{k=j}^\infty\bigg(\frac{pr^C}{|\unicode{x3bb}|}\bigg)^{k-j}\\ &\le |u_0|r^{C+1}\frac{pr^C}{|\unicode{x3bb}|-pr^C} +r^{C+1}{\|w\|}_1\frac{1}{|\unicode{x3bb}|-pr^C}. \end{align*} $$

Similar arguments lead to the following estimate:

$$ \begin{align*} |\unicode{x3bb}(R(\unicode{x3bb})-1)||u_0| &\le |w_0|+p\sum_{k=1}^\infty\delta_k\sum_{j=1}^k|w_j|\frac{p^{k-j}r^{L_k-L_j}}{|\unicode{x3bb}|^{k-j+1}}\\ &\le|w_0|+\frac{pr^{C+1}}{|\unicode{x3bb}|}\sum_{j=1}^\infty|w_j| \sum_{k=j}^\infty\bigg(\frac{pr^C}{|\unicode{x3bb}|}\bigg)^{k-j}\\ &\le |w_0|+pr^{C+1}{\|w\|}_1\frac{1}{|\unicode{x3bb}|-pr^C}. \end{align*} $$

We know from (31) that $\unicode{x3bb} _a>pr^C$ , and hence, (57) is also valid in the case $p<1/2$ . Consequently, for every $p \in (0,1)$ , the spectrum of A without the point $\unicode{x3bb} _a$ is contained in the closed centred ball of radius $\varrho <\unicode{x3bb} _a$ .

The property of the spectrum shown just above implies that the operator A is quasi-compact. If one shows additionally that A possesses certain additional positivity properties, then one can apply a Krein–Rutman type result, for example, [Reference Sasser23, Theorems 6 and 7]. In our particular case, we will arrive at the same conclusion by the direct calculations presented below, which are quite standard for the area of quasi-compact operators.

Define the linear operators $Qu=({(u,v^*)}/{(v,v^*)})v$ and $Bu=A(u-Qu)$ on $\ell _1$ , where $(v,v^*) \ge v_0 v_0^*>0$ by (38). Clearly, $Q^2=Q$ . Since v and $v^*$ are the eigenvectors corresponding to $\unicode{x3bb} _a$ , we have $AQu=\unicode{x3bb} _a Qu$ and

Thus, A and Q are commuting and $QB=BQ=0$ . These properties of the operators A, B and Q imply that

(58)

We claim that the spectral radius $\rho (B)$ of B satisfies $\rho (B)\le \varrho $ . Then, the estimate (39) for any $\gamma \in (\varrho /\unicode{x3bb} _a, 1)$ follows from the representation (58) combined with Gelfand’s formula $\rho (B)=\lim _{n \to \infty } \|B^n\|^{1/n}$ .

To prove the claim, we consider the closed linear subspace

$$ \begin{align*} Y=\{u\in\ell_1: (u,v^*)=0\}. \end{align*} $$

Since $Bu\in Y$ for any $u\in \ell _1$ , it suffices to consider the restriction of B to this subspace, which we denote by $B_Y$ . In fact, we have $\rho (B)=\rho (B_Y)$ , where the inequality ${\rho (B)\ge \rho (B_Y)}$ is immediate from Gelfand’s formula and the reverse one follows from the said formula by

$$ \begin{align*} \rho(B)= \lim_{n\to \infty} \sup_{u\in \ell_1: \|u\| \le 1} \| B^n u\|^{1/n} \le \lim_{n\to \infty} \sup_{v\in Y: \|v\| \le \| B\|} \| B^{n-1} v\|^{1/n} = \rho(B_Y). \end{align*} $$

We first show that the spectrum of $B_Y$ is a subset of the spectrum of A. Let $\unicode{x3bb} $ be any regular value of A, that is, $(A-\unicode{x3bb} I)u=w$ is solvable for every $w\in \ell _1$ . Assume now that $w\in Y$ and let $u_w$ denote the solution to $(A-\unicode{x3bb} I)u=w$ . Then,

Since $\unicode{x3bb} \neq \unicode{x3bb} _a$ , we have $(u_w,v^*)=0$ . In other words, $u_w\in Y$ and $Qu_w=0$ . Consequently,

$$ \begin{align*} (B_Y-\unicode{x3bb} I_Y)u_w=(A-\unicode{x3bb} I)u_w=w \end{align*} $$

and $\unicode{x3bb} $ is regular for $B_Y$ . Since the spectrum of A without point $\unicode{x3bb} _a$ is contained in the closed centred ball of radius $\varrho $ , it remains to show that $\unicode{x3bb} _a$ is regular for $B_Y$ . We need to show that the equation $(B_Y-\unicode{x3bb} _a I_Y)u=w$ for $w \in Y$ has a solution $u \in Y$ . Since $B_Y=A$ on Y, this is equivalent to showing that $(A- \unicode{x3bb} _a I)u=w$ has a solution $u \in Y$ . Therefore, we can take $\unicode{x3bb} =\unicode{x3bb} _a$ in (53) and notice that (54) is satisfied when $\unicode{x3bb} =\unicode{x3bb} _a$ for any $u_0$ because in this case, the right-hand side of (54) is $(w, v^*)$ . It remains to determine $u_0$ from the equation $(u,v^*)=0$ .

Alternatively, we can solve $(B_Y-\unicode{x3bb} _a I_Y)u=w$ as follows. Let $u^{(\unicode{x3bb} )}$ denote the solution to $(B_Y-\unicode{x3bb} I_Y)u=w$ for $\unicode{x3bb} $ such that $0<|\unicode{x3bb} -\unicode{x3bb} _a|<\unicode{x3bb} _a - \varrho $ . Then, by (54),

$$ \begin{align*} \unicode{x3bb}(R(\unicode{x3bb})-1)u_0^{(\unicode{x3bb})} =w_0+p\sum_{k=1}^\infty\delta_k \sum_{j=1}^k w_j\frac{c_jc_{j+1}\cdots c_{k-1}}{\unicode{x3bb}^{k-j+1}}. \end{align*} $$

Using the condition $(w,v^*)=0$ and the explicit form of the vector $v^*$ , one gets easily

Dividing both sides by $\unicode{x3bb} -\unicode{x3bb} _a$ and letting $\unicode{x3bb} \to \unicode{x3bb} _a$ , we obtain

Using now (53) with $\unicode{x3bb} =\unicode{x3bb} _a$ and $u_0=u_0^{(\unicode{x3bb} _a)}$ defined just above, we find the unique solution $u^{(\unicode{x3bb} _a )} $ to $(B_Y-\unicode{x3bb} _a I_Y)u=w$ . It follows from Proposition 10 that $u^{(\unicode{x3bb} _a )} \in \ell _1$ . Then, ${u^{(\unicode{x3bb} _a )} \in Y}$ since Y is closed in $\ell _1$ . This completes the proof in the aperiodic case.

The orbit of zero is eventually periodic. By definition, $k_0 \ge 0$ is the minimal integer such that $T^{k_0}(0)=T^{\varkappa '+1}(0)$ . The orbit of zero has $\varkappa '+1$ points and the space $(U, \| \cdot \|)$ is isometric to $(\mathbb R^{\varkappa '+1}, {\| \cdot \|}_1)$ . The isometry M between these spaces can be written as follows: for any function $f \in U$ , which is of the form for some $u \in \ell _1$ ,

(59) $$ \begin{align} (Mf)_k=u_k,\quad 0 \le k < k_0 \quad \text{and} \quad (Mf)_k= \sum_{m=0}^\infty u_{k+m(\varkappa'-k_0+1)},\quad k_0 \le k \le \varkappa'. \end{align} $$

Every $f \in U$ admits a unique representation with ${s_k = (Mf)_k}$ . Hence, by (46),

Since in the second sum the term with $k=\varkappa '+1$ is , we see that P is equivalent to the finite dimensional linear operator $\widehat {A}$ on $\mathbb R^{\varkappa '+1}$ given by

(60)

We first note that the operator $\widehat A$ is non-negative and irreducible. Moreover, the matrix that represents $\widehat A$ in the standard basis is primitive, that is, it does not have a cyclic structure, see [Reference Meyer19, p. 680]. Therefore, by the Perron–Frobenius theorem, $\widehat {A}$ has a positive simple eigenvalue that strictly exceeds the maximum $\varrho '$ of absolute values of all other eigenvalues.

Let us solve the equation $\widehat {A}s=\unicode{x3bb} s$ to determine this leading eigenvalue. It is immediate from the second line in (60) that

(61) $$ \begin{align} s_k=\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}s_0,\quad k=1, \ldots,k_0-1, \end{align} $$

and

(62) $$ \begin{align} s_k=\frac{c_{k_0}c_{k_0+1}\cdots c_{k-1}}{\unicode{x3bb}^{k-k_0}}s_{k_0}, \quad k=k_0+1,\ldots,\varkappa'. \end{align} $$

Assume first that $k_0 \neq 0$ . From the second line in (60) for $k=k_0$ , we get

$$ \begin{align*} \unicode{x3bb} s_{k_0}&=c_{k_0-1}s_{k_0-1}+c_{\varkappa'}s_{\varkappa'}=\frac{c_0c_1\cdots c_{k_0-1}}{\unicode{x3bb}^{k_0-1}}s_0 +\frac{c_{k_0}c_{k_0+1}\cdots c_{\varkappa'}}{\unicode{x3bb}^{\varkappa' - k_0}}s_{k_0}. \end{align*} $$

The positive solution to $\unicode{x3bb} ^{\varkappa '-k_0+1}=c_{k_0}c_{k_0+1}\cdots c_{\varkappa '}$ is not an eigenvalue of $\widehat A$ . Indeed, in this case, $s_0 = 0$ , and plugging (61) and (62) into the first line of (60) gives an impossible identity

$$ \begin{align*} \sum_{k=k_0}^{\varkappa'}\delta_k\frac{c_{k_0}c_1\cdots c_{k-1}}{\unicode{x3bb}^{k-k_0}} =0. \end{align*} $$

Consequently,

(63) $$ \begin{align} s_{k_0}=\frac{c_0c_1\cdots c_{k_0-1}}{\unicode{x3bb}^{k_0}} \bigg(1-\frac{c_{k_0}c_{k_0+1}\cdots c_{\varkappa'}}{\unicode{x3bb}^{\varkappa'-k_0+1}}\bigg)^{-1} s_0. \end{align} $$

Plugging (61), (62) and (63) into the first line of (60), we conclude that the leading eigenvalue satisfies the equation

(64) $$ \begin{align} \unicode{x3bb}=p\sum_{k=0}^{k_0-1}\delta_k\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k} +p\bigg(1-\frac{c_{k_0}c_{k_0+1}\cdots c_{\varkappa'}}{\unicode{x3bb}^{\varkappa'-k_0+1}}\bigg)^{-1} \sum_{k=k_0}^{\varkappa'}\delta_k\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}. \end{align} $$

Let us show that this equation is equivalent to (15) for all possible $k_0$ , including $k_0=0$ . To this end, we notice that periodicity of the orbit implies that $\delta _k=\delta _{(\varkappa '-k_0+1)m+k}$ and $c_k=c_{(\varkappa '-k_0+1)m+k}$ for all $k=k_0,k_0+1,\ldots ,\varkappa '$ and all $m\ge 0$ . Then,

(65) $$ \begin{align} \nonumber p\sum_{k=k_0}^\infty\delta_k\bigg(\frac{p}{\unicode{x3bb}}\bigg)^k \bigg(\dfrac{q}{p}\bigg)^{L_k} &=p\sum_{k=k_0}^\infty\delta_k\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}\\ \nonumber &=p\sum_{m=0}^\infty\sum_{k=k_0}^{\varkappa'} \delta_{(\varkappa'-k_0+1)m+k}\frac{c_0c_1\cdots c_{(\varkappa'-k_0+1)m+k-1}}{\unicode{x3bb}^{(\varkappa'-k_0+1)m+k}}\\ \nonumber &=p\sum_{k=k_0}^{\varkappa'}\delta_k\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}\sum_{m=0}^\infty\bigg(\frac{c_{k_0}c_{k_0+1}\cdots c_{\varkappa'}}{\unicode{x3bb}^{\varkappa'-k_0+1}}\bigg)^m\\ &=\bigg(1-\frac{c_{k_0}c_{k_0+1}\cdots c_{\varkappa'}}{\unicode{x3bb}^{\varkappa'-k_0+1}}\bigg)^{-1} p\sum_{k=k_0}^{\varkappa'}\delta_k\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}, \end{align} $$

and thus, (15) and (64) are indeed equivalent.

We now consider the purely periodic case $k_0 = 0$ . Plugging (62) into the first line of (60), we get

$$ \begin{align*} \unicode{x3bb} s_0=p\sum_{k=0}^{\varkappa'}\delta_k \frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}s_0+ \frac{c_0c_1\cdots c_{\varkappa'}}{\unicode{x3bb}^{\varkappa'}}s_0. \end{align*} $$

Consequently, a non-trivial solution exists if and only if $\unicode{x3bb} $ satisfies the equation

(66) $$ \begin{align} \unicode{x3bb}=p\sum_{k=0}^{\varkappa'}\delta_k \frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}+ \frac{c_0c_1\cdots c_{\varkappa'}}{\unicode{x3bb}^{\varkappa'}}. \end{align} $$

However, by (65), we have that (15) is equivalent to

$$ \begin{align*} \unicode{x3bb} \bigg(1-\frac{c_0c_1\cdots c_{\varkappa'}}{\unicode{x3bb}^{\varkappa'+1}}\bigg) = p\sum_{k=0}^{\varkappa'}\delta_k\frac{c_0c_1\cdots c_{k-1}}{\unicode{x3bb}^k}. \end{align*} $$

We now easily see that (66) is equivalent to (15) in the purely periodic case. Therefore, $\unicode{x3bb} _a$ is the leading eigenvalue of $\widehat A$ for all possible $k_0$ .

The eigenspace corresponding to $\unicode{x3bb} _a$ has dimension one by the Perron–Frobenius theorem. It is spanned by the vector $MV$ . In the case $k_0\neq 0$ , this can be seen from (61), (62), (63) and (59). In the purely periodic case $k_0=0$ , this is readily seen from (62) and (59).

Lastly, we find the eigenvectors $ {\widehat A}^* u^* = \unicode{x3bb} _a u^*$ . Using the definition of $\widehat A$ , we get the coordinate-wise equations

We solve them exactly as we did in the aperiodic case for (50). This gives us the same answer $u_k^* = u_0^* v_k^* $ for $1 \le k \le \varkappa '$ because the first $\varkappa '$ equations are the same and we did not use the equation in the second line. It is not hard to check that this last equation is indeed satisfied for such $u^*$ . Then, $v^*$ , which is proportional to $u^*$ , is the eigenvector of ${\widehat A}^*$ corresponding to $\unicode{x3bb} _a$ . Therefore, (39) follows for any $\gamma \in (\varrho '/\unicode{x3bb} _a, 1)$ as in the aperiodic case with finite $\varkappa $ .

Proof. We argue by induction. In the basis case $k=2$ , we have $\{a \in [\tfrac 12, \tfrac 23]: \varkappa _a \ge k\}= [\tfrac 12, \tfrac 23]$ , where both functions $T_a^1(0)=1/a$ and $T_a^2(0)=({1-a})/{a^2}$ are strictly decreasing and continuous. Therefore, since the range of the second one is $[\tfrac 34,2]$ and the function $({2a-1})/({1-a})$ increases on $[\tfrac 12, \tfrac 23]$ from $0$ to $1$ , we have