2 Preliminaries

In this section, we give some preliminaries for locally compact Hausdorff spaces, which we will work on in this paper. Unless mentioned particularly, let $X, Y, Z$ denote topological spaces and $C(X,Y)$ the set of all continuous maps of X into Y. Define on ${C}(X, Y)$ the compact-open topology (see [Reference Munkres21, p. 285]), that is, the topology generated by the collection of all finite intersections of sets in

$$ \begin{align*} \{\mathcal{F}(K,U): K~\text{is a compact subset of}~X~\text{and}~U~\text{is open in}~Y\}, \end{align*} $$

where $\mathcal {F}(K,U):=\{f\in {C}(X, Y):f(K)\subseteq U\}$ . When Y is a metric space equipped with metric d, we have another topology on ${C}(X,Y)$ , viz. the topology of compact convergence generated by

$$ \begin{align*} \{\mathcal{B}_K(f, \epsilon): f\in {C}(X,Y), K~\text{is a compact subset of}~X, \text{and}~\epsilon>0\}, \end{align*} $$

where $\mathcal {B}_K(f,\epsilon ):=\{g\in {C}(X,Y): \sup \{d(f(x), g(x)):x\in K\}<\epsilon \}$ .

As defined in [Reference Arens1], a topological space X is said to be hemicompact if there exists a sequence $(K_j)_{j\in \mathbb {N}}$ of compact sets of X such that if K is any compact subset of X, then $K\subseteq \bigcup _{i=1}^{k}K_{m_i}$ for some finitely many $K_{m_1}, K_{m_2}, \ldots , K_{m_k}$ .

3 Continuity and periodic points

Let X be a locally compact Hausdorff space. In this section, we prove $(\mathcal {C}(X),\mathcal {J}_n)$ to be a discrete semi-dynamical system indeed by showing the continuity of $\mathcal {J}_n$ on $\mathcal {C}(X)$ .

Proof. First, we prove by induction that the iterated evaluation map $\mathcal {E}_n:X\times \mathcal {C}(X)\rightarrow X$ defined by

(3.5) $$ \begin{align} \mathcal{E}_n(x,f)=f^n(x) \end{align} $$

is continuous on $X\times \mathcal {C}(X)$ for each $n\in \mathbb {N}$ . The case $n=1$ follows by Lemma 2, where $Y=X$ and $\mathcal {E}_1=\mathcal {E}$ . Suppose that $\mathcal {E}_n$ is continuous for certain $n\ge 2$ . To prove $\mathcal {E}_{n+1}$ is continuous on $X\times \mathcal {C}(X)$ , consider the map $\mathcal {H}_n:X\times \mathcal {C}(X)\rightarrow X\times \mathcal {C}(X)$ defined by $\mathcal {H}_n{\kern-1pt}={\kern-1pt}(\mathcal {E}_n, p)$ , where $p:X{\kern-1pt}\times{\kern-1pt} \mathcal {C}(X){\kern-1pt}\rightarrow{\kern-1pt} \mathcal {C}(X)$ is the projection map defined by $p(x,f)=f.$ Since $\mathcal {E}_n$ and p are continuous, so is $\mathcal {H}_n$ . Now,

$$ \begin{align*} (\mathcal{E}\circ \mathcal{H}_n)(x,f)=\mathcal{E}(\mathcal{E}_n(x,f), p(x,f))=\mathcal{E}(f^n(x),f)=f^{n+1}(x)=\mathcal{E}_{n+1}(x,f) \end{align*} $$

for each $(x,f)\in X\times \mathcal {C}(X)$ , implying that $\mathcal {E}_{n+1}=\mathcal {E}\circ \mathcal {H}_n$ . Therefore, $\mathcal {E}_{n+1}$ is continuous, being the composition of continuous maps $\mathcal {E}$ and $\mathcal {H}_n$ . Hence, the continuity of $\mathcal {E}_n$ is proved for all $n\in \mathbb {N}$ .

In fact, by putting $Y=X$ , $Z=C(X)$ , and $h=\mathcal {E}_n$ in Lemma 3, we get $F:C(X)\to C(X)$ and $F(f)(x)=h(x,f)=\mathcal {E}_n(x,f)=\mathcal {J}_n(f)(x)$ for all $f\in \mathcal {C}(X)$ and $x\in X$ , implying that $F=\mathcal {J}_n$ . Thus, by Lemma 3, we see that $\mathcal {J}_n$ is continuous on $\mathcal {C}(X)$ for each $n\in \mathbb {N}$ .

Theorem 1 looks the same, but is more general than [Reference Veerapazham, Gopalakrishna and Zhang23, Theorem 2.1]. Unlike [Reference Veerapazham, Gopalakrishna and Zhang23], Theorem 1 does not assume X to be metrizable, which causes more difficulties in its proof. In the special case that X is a compact metric space with a metric d, the uniform topology induced by the metric $\rho $ gives the compact-open topology on $\mathcal {C}(X)$ because of Lemma 1.

As indicated in §1, $\mathcal {J}_n$ defines a discrete semi-dynamical system on the space $\mathcal {C}(X)$ of continuous functions with its iteration semigroup $\{\mathcal {J}_n^k: k\ge 0\}$ . Remark that $\mathcal {J}_n$ usually does not define a dynamical system because a homeomorphism f of a compact interval may have infinitely many iterative roots as indicated in [Reference Kuczma14].

3.1 Periodic points in general case

One of the most fundamental problems on dynamical systems or semi-dynamical systems concerns fixed points and periodic points. To emphasize the dependence on the space X, we use $\mathrm {Fix}(f;X)$ and $\mathrm {Per}(f;X)$ to denote the set of all fixed points and the set of all periodic points of f in X, respectively. Since $ \mathcal {J}_n^k f=f^{n^k} $ by equation (1.1), we see that $f\in \mathcal {C}(X)$ is a fixed point of $\mathcal {J}_n$ if and only if f satisfies the functional equation

(3.6) $$ \begin{align} f^n=f, \end{align} $$

and $f\in \mathcal {C}(X)$ is a k-periodic point of $\mathcal {J}_n$ if and only if f satisfies

(3.7) $$ \begin{align} f^{n^k}=f,\ f^{n^{k-1}}\ne f\quad \mbox{and}\quad f^{n^i}\ne f\quad \text{for all } i=2,\ldots,k-2 \quad\text{with } i\nmid (k-1). \end{align} $$

So, the fixed points and periodic points of $\mathcal {J}_n$ are related to solutions of the Babbage equation [Reference Kuczma14]

(3.8) $$ \begin{align} \phi^m=\mathrm{id}, \end{align} $$

where $m>0$ is a certain integer. The functional equations in equations (3.6) and (3.7) cannot be simply treated as the Babbage equation (3.8) because the range of f may not be the whole X. In what follows, we need to consider restriction of maps. For any $f\in \mathcal {C}(X)$ and $A\subseteq X$ , let $\bar {A}$ denote the closure of A, $R(f)$ the range of f, and $f|_A$ the restriction of f to A. As in [Reference Munkres21], a space X is said to be first countable if each point has a countable local basis, that is, for each $x\in X$ , there exists a countable collection $\mathcal {B}_x$ of neighborhoods of x such that each open set U containing x contains an element B of $\mathcal {B}_x$ .

Lemma 5. Let $n\in \mathbb {N}$ and suppose that X is first countable. Then $f\in \mathcal {C}(X)$ is a solution of the equation

(3.9) $$ \begin{align} \phi^n=\phi \end{align} $$

on X if and only if $R(f)$ is closed and there exists $g\in \mathcal {C}(X)$ such that $R(g)=R(f)$ , $g|_{{R(g)}}$ satisfies the Babbage equation

$$ \begin{align*} \phi^{n-1}=\mathrm{id} \end{align*} $$

on $R(g)$ , and $f|_{{R(g)}}=g|_{{R(f)}}$ .

Proof. If $f\in \mathcal {C}(X)$ is a solution of equation (3.9), then $f^{n-1}(f(x))=f(x)$ for all $x\in X$ , that is, $f^{n-1}(y)=y$ for all $y\in R(f)$ . Also, since X is first countable, for each $y\in \overline {R(f)}$ , there exists a sequence $(y_n)$ in $R(f)$ such that $y_k\to y$ , implying by continuity of f that $f(y_k)\to f(y)$ as $k \to \infty $ . However, $y_k=f^{n-1}(y_k)\to f^{n-1}(y)$ as $k \to \infty $ . Therefore, as X is Hausdorff, we have $f(y)=y$ . Thus, $y\in R(f)$ , and so $\overline {R(f)}=R(f)$ . Let $g:=f$ . Then f is a continuous extension of the solution $g|_{R(g)}$ on $R(g)$ to X such that $R(f)$ is closed, $R(g)= R(f)$ , and $f|_{{R(g)}}=g|_{{R(f)}}$ . The proof of the converse is trivial, since for every $g\in \mathcal {C}(X)$ , whose restriction $g|_{R(g)}$ satisfies $\phi ^{n-1}=\mathrm {id}$ on $R(f)$ , we have $f^n(x)=f^{n-1}(f(x))=g^{n-1}(f(x))=f(x)$ for all $x\in X$ .

Usually, a solution $\phi $ of equation (3.8) is referred to as an mth order unit iterative root if m is the smallest positive integer such that equation (3.8) is satisfied. Clearly, each unit iterative root is invertible. As in [Reference Kuczma14, p. 290], for $m=2$ , every solution of equation (3.8), whose inverse is itself, is called an involutory function. In general, if f is an mth order unit iterative root and $f^k=\mathrm {id}$ , then m divides k clearly. On a general set E, the mth order unit iterative roots are formulated below.

Lemma 6. [Reference Kuczma14, Theorem 15.1]

Let $\{m_0,\ldots ,m_r\}$ , where $1=m_0<m_1<\cdots < m_r=m$ , be the complete set of divisors of m and let $E=\bigcup _{i=0}^{r}\bigcup _{j=1}^{m_i}U_j^i$ be a decomposition of E into disjoint sets such that the sets $U_1^i, U_{m_1}^i,\ldots ,U_{m_i}^i$ have the same cardinality for each $1\le i \le r$ . For $1\le i\le r$ and $1\le j \le m_i-1$ , let $f_{ij}$ be an arbitrary one-to-one map of $U_j^i$ onto $U_{j+1}^i$ . Then the formula

$$ \begin{align*} f(x):=\begin{cases} x&\text{for } x\in U_1^0,\\ f_{ij}(x)&\text{for } x\in U_j^i, j=1,2,\ldots,m_i-1, i\ge 1,\\ f_{i1}^{-1}(\cdots(f_{i,m_i-1}^{-1}(x))\cdots)&\text{for } x\in U_{m_i}^i, i\ge 1 \end{cases} \end{align*} $$

defines the general solution of $\phi ^m=\mathrm { id}$ on E.

Having Lemma 6, we are ready to define

$$ \begin{align*} \mathcal{U}_E^m:=\{f\in \mathcal{C}(X): f|_E ~\text{is an}~m\text{th} ~\text{order unit iterative root on}~E~\text{and}~ R(f)= E\} \end{align*} $$

for any subset E of X and $m\in \mathbb {N}$ . Since $\mathrm {Fix}(\mathcal {J}_1;X)=\mathcal {C}(X)$ , that is, the problem of fixed points of $\mathcal {J}_1$ is trivial, we focus on $\mathcal {J}_n$ with $n\ge 2$ . We give the following results on fixed points and periodic points, whose proof is similar to that of [Reference Veerapazham, Gopalakrishna and Zhang23, Theorem 2.4].

Theorem 2. Let $n, k\ge 2$ be integers and suppose that X is first countable. Then: (i) $f\in \mathcal {C}(X)$ is a fixed point of $\mathcal {J}_n$ if and only if $f\in \mathcal {U}_E^m$ for a closed subset E of X and an integer $m\ge 1$ dividing $n-1$ exactly; (ii) $f\in \mathcal {C}(X)$ is a k-periodic point of $\mathcal {J}_n$ if and only if $f\in \mathcal {U}_E^m$ for a closed subset E of X and an integer $m>1$ satisfying that

(3.10) $$ \begin{align} m\mid(n^k-1)\quad\text{and}\quad m\nmid (n^j-1) \quad\text{for } 1\le j\le k-1. \end{align} $$

Remark that we were able to infer that E is compact in [Reference Veerapazham, Gopalakrishna and Zhang23] because E is indeed $R(f)$ and domain X was assumed to be compact, but in Theorem 2, we can only conclude that E is closed because of using Lemma 5.

Example 1. Consider $f(x)=|x|$ . Then $f\in \mathcal {C}(\mathbb {R})$ . Clearly, $f\in \mathcal {U}_{[0,\infty )}^1$ . By result (i) of Theorem 2, f is a fixed point of $\mathcal {J}_n$ for each $n\ge 2$ .

Example 2. The doubling map $f\in \mathcal {C}(S^1)$ defined by $f(e^{it})=e^{2it}$ is not a k-periodic point of $\mathcal {J}_n$ for $n,k\ge 2$ . In fact, if $f^{n^k}=f$ for some $n,k\ge 2$ , then $e^{it(2^{n^k}-2)}=0$ for all $t\in [0,2\pi )$ , which is a contradiction.

Example 3. The rotation map $R_{{2\pi }/({n^k-1})}:S^1\to S^1$ defined by $R_{{2\pi }/({n^k-1})}(e^{it})=e^{i(t+({2\pi }/({n^k-1})))}$ is a k-periodic point of $\mathcal {J}_n$ in $\mathcal {C}(S^1)$ for $n\ge 2$ and $k\ge 1$ .

3.2 Periodic points in $\mathcal {\mathcal {C}}(\mathbb {R})$

For more detailed results, in this section, we focus on the case $X=\mathbb {R}$ , the real line in the usual topology. For $a,b \in \mathbb {R}$ with $a<b$ , let $|a,b|$ denote either an open interval $(a,b)$ , a semi-closed interval $[a,b)$ or $(a,b]$ , or a closed interval $[a,b]$ , where one or both of the endpoints may be infinite. We have the following result, which can also be deduced from Lemma 6.

Lemma 7. [Reference McShane18, Reference Vincze24]

If $f\in \mathcal {C}(|a,b|)$ is a solution of equation (3.8), then either $f=\mathrm {id}$ or f is a decreasing involutory function on $|a,b|$ . More precisely, any solution of equation (3.8) on $|a,b|$ for general m is also a second-order unit iterative root. If m is odd, then the solution is uniquely the identity $\mathrm {id}$ ; if m is even, then the solution is either id or a decreasing involutory function on $|a,b|$ .

In the following theorem, the proof of which is similar to that of [Reference Veerapazham, Gopalakrishna and Zhang23, Theorem 3.2] on a compact interval, we refer to monotonically increasing (or decreasing) functions satisfying equation (3.9) as monotonically increasing (or decreasing) fixed point of $\mathcal {J}_n$ .

Theorem 3. Every monotonic fixed point f of $\mathcal {J}_2$ in $\mathcal {C}(\mathbb {R})$ is of the first form in equation (3.11). Every monotonic fixed point f of $\mathcal {J}_3$ in $\mathcal {C}(\mathbb {R})$ is of one of the following two forms:

(3.11) $$ \begin{align} f(x)= \begin{cases} a & \text{if } x\in (-\infty,a|, \\ x & \text{if } x\in |a,b|, \\ b & \text{if } x\in |b, \infty), \end{cases} \ \ \ f(x)= \begin{cases} b & \text{if } x\in (-\infty,a|, \\ g(x) & \text{if } x\in |a,b|, \\ a & \text{if } x\in |b, \infty), \end{cases} \end{align} $$

where $|a,b|$ is a closed interval in $\mathbb {R}$ satisfying that $a\le b$ and $g\in \mathcal {C}(|a,b|)$ is a decreasing involutory map.

The above theorem is only applicable to monotonic fixed points. There exist fixed points of $\mathcal {J}_2$ or $\mathcal {J}_3$ which are not monotonic. For example, as considered in Example 1, the function $f(x)=|x|$ on $\mathbb {R}$ is not a monotonic map on $\mathbb {R}$ but a fixed point of $\mathcal {J}_2$ .

The ‘monotone’ in Theorem 3 need not mean ‘strict monotone’. In fact, if $a\in \mathbb {R}$ in the first form of equation (3.11), then we have $f(x)=a$ for $x\in (-\infty , a]$ , implying that f is not strictly monotone. For strictly monotonic ones, we have the following result, the proof of which is similar to that of [Reference Veerapazham, Gopalakrishna and Zhang23, Corollary 3.3] on a compact interval.

Corollary 1. If f is a strictly monotonic fixed point of $\mathcal {J}_3$ in $\mathcal {C}(\mathbb {R})$ , then either $f=\mathrm {id}$ or f is a strictly decreasing involutory function on $\mathbb {R}$ .

Theorem 3 gives results only for $\mathcal {J}_2$ and $\mathcal {J}_3$ , but not for the generic $\mathcal {J}_n$ . In what follows, we show that those monotonic fixed points of $\mathcal {J}_2$ and $\mathcal {J}_3$ are important representatives for the generic $\mathcal {J}_n$ . Let $\mathcal {C}_{\mathrm {id}}(\mathbb {R})$ and $\mathcal {C}_{\mathrm {inv}}(\mathbb {R})$ consist of all continuous self-maps of $\mathbb {R}$ which are the identity and decreasing involutions on their range, respectively. By Theorem 3, monotonic fixed points of $\mathcal {J}_3$ are in both classes $\mathcal {C}_{\mathrm {id}}(\mathbb {R})$ and $\mathcal {C}_{\mathrm {inv}}(\mathbb {R})$ but monotonic fixed points of $\mathcal {J}_2$ are all in the same class $\mathcal {C}_{\mathrm {id}}(\mathbb {R})$ . The following theorem, the proof of which is similar to that of [Reference Veerapazham, Gopalakrishna and Zhang23, Theorem 3.4] on a compact interval, describes all fixed points of $\mathcal {J}_n$ for any $n\ge 2$ .

Theorem 4. The following statements are true for system $(\mathcal {C}(\mathbb {R}), \mathcal {J}_n)$ :

1. (i) $\mathrm {Fix}(\mathcal {J}_m;\mathcal {C}(\mathbb {R}))=\mathrm {Fix}(\mathcal {J}_n;\mathcal {C}(\mathbb {R}))$ if integers $m,n\ge 2$ satisfy $m\equiv n(\mathrm {mod} ~2)$ ;

2. (ii) $\mathrm {Fix}(\mathcal {J}_2;\mathcal {C}(\mathbb {R}))\subsetneq \mathrm {Fix}(\mathcal {J}_3;\mathcal {C}(\mathbb {R}))$ . More concretely,

   $\mathrm {Fix}(\mathcal {J}_2;\mathcal {C}(\mathbb {R}))=\mathcal {C}_{\mathrm {id}}(\mathbb {R})$ and $\mathrm {Fix}(\mathcal {J}_3;\mathcal {C}(\mathbb {R}))=\mathcal {C}_{\mathrm {id}}(\mathbb {R})\cup \mathcal {C}_{\mathrm {inv}}(\mathbb {R})$ .

Theorem 3 together with result (i) of Theorem 4 shows that Theorem 3 is indeed true for every integer $n\ge 2$ . More precisely, monotonic fixed points of $\mathcal {J}_n$ coincide with those of $\mathcal {J}_2$ and $\mathcal {J}_3$ accordingly as n is even and odd, respectively. So, Theorem 3 actually gives results for the representatives.

Although we can find many fixed points of ${\cal J}_n$ , there are no non-trivial periodic points as seen from the following result, the proof of which is similar to that of [Reference Veerapazham, Gopalakrishna and Zhang23, Theorem 3.5] on a compact interval.

Theorem 5. For each $n\in \mathbb {N}, ~{\cal J}_n$ does not have periodic points of period $k\ge 2$ in $\mathcal {C}(\mathbb {R})$ .

As observed earlier, every element of $\mathcal {C}(\mathbb {R})$ is a fixed point for $\mathcal {J}_1$ . So, $\mathcal {J}_1$ has a dense set of periodic points in $\mathcal {C}(\mathbb {R})$ . However, since $\mathcal {C}(\mathbb {R})$ is metrizable with metric D by result (i) of Lemma 4 and $\mathcal {J}_n$ is not the identity operator on $\mathcal {C}(\mathbb {R})$ , it follows that $\mathrm {Fix}(\mathcal {J}_n;\mathcal {C}(\mathbb {R}))$ is not dense in $\mathcal {C}(\mathbb {R})$ for $n\ge 2$ . This implies by Theorem 5 that $\mathrm {Per}(\mathcal {J}_n;\mathcal {C}(\mathbb {R}))$ is not dense in $\mathcal {C}(\mathbb {R})$ for $n\ge 2$ .

3.3 Periodic points in $\mathcal {\mathcal {C}}(S^1)$

As illustrated in Example 3, ${\cal J}_n$ may have periodic points of period $k\ge 2$ in $\mathcal {C}(S^1)$ but, in contrast, ${\cal J}_n$ only has fixed points, that is, trivial periodic points of period 1, in $\mathcal {C}(\mathbb {R})$ as shown in Theorem 5. In this section, we investigate all periodic points of $\mathcal {J}_n$ in $\mathcal {C}(S^1)$ .

Given $z_0, z_1, \ldots , z_{m-1}\in S^1$ with $m\ge 2$ , we write $ z_0\prec z_1\prec \cdots \prec z_{m-1} $ if there exist $t_1,t_2,\ldots ,t_{m-1}\in \mathbb {R}$ such that $0<t_1<t_2<\cdots <t_{m-1}<1 ~\text {and}~ z_j=z_0e^{2\pi i t_j}~\text {for}~1\le j \le m-1. $ In this case, we have

$$ \begin{align*} z_{j(\mathrm{mod~m})}\prec z_{j+1(\mathrm{mod~m})}\prec \cdots \prec z_{j+m-1(\mathrm{mod~m})} \quad\text{for all } j\in \mathbb{N}, \end{align*} $$

so that ‘ $\prec $ ’ is indeed a cyclic order on $S^1$ .

For any two distinct points $z_1, z_2\in S^1$ , define the arcs $(z_1, z_2)$ , $[z_1, z_2)$ , and $(z_1, z_2]$ by $(z_1,z_2):=\{z\in S^1:z_1\prec z\prec z_2\},$ $[z_1, z_2):=(z_1, z_2)\cup \{z_1\},~ \text {and}~(z_1, z_2]:=(z_1, z_2)\cup \{z_2\}.$ Then we have $ (z_1,z_2)=\{e^{2\pi it}\in S^1:t\in (t_1, t_2)\},$ where $t_1, t_2$ are unique real such that $z_1=e^{2\pi it_1}, z_2=e^{2\pi it_2}$ , and $0\le t_1<t_2<t_1+1<2$ .

It is known (cf. [Reference Block and Coppel4, Reference Wall25]) that for every homeomorphism $F:S^1\to S^1$ , there exists a homeomorphism $f:\mathbb {R}\to \mathbb {R}$ satisfying one of the Abel equations:

$$ \begin{align*} &f(t+1)=f(t)+1 \quad\text{if }f\text{ is strictly increasing;} \\ &f(t+1)=f(t)-1 \quad\text{if }f\text{ is strictly decreasing} \end{align*} $$

such that $ F(e^{2\pi it})=e^{2\pi if(t)} \ \text { for all } t\in \mathbb {R}. $ Every such f is called a lift of F. We say that F preserves (or reverses) orientation accordingly as f is strictly increasing (or decreasing) on $\mathbb {R}$ .

The following two lemmas together describe the general solution of equation (3.8) on $S^1$ , each of which can also be deduced from Lemma 6.

Lemma 8. [Reference Jarczyk12]

Let $\phi \in \mathcal {C}(S^1)$ be a solution of equation (3.8) and have a fixed point in $S^1$ . Then: (i) $\phi $ is the identity map if $\phi $ is orientation-preserving; or (ii) $\phi $ is an involution if $\phi $ is orientation-reversing.

Lemma 9. [Reference Jarczyk12]

All mth-order iterative roots of identity in $\mathcal {C}(S^1)$ having no fixed points in $S^1$ are given by

(3.12) $$ \begin{align} f(z)=\begin{cases} \phi_0(z)&\text{if } z\in [z_0,z_{m-1}),\\ (\phi_1\circ \phi_2\circ \cdots \circ \phi_{m-1})^{-1}(z)& \text{if } z\in [z_{m-1},z_0) \end{cases} \end{align} $$

with $\phi _j:=\phi _0|_{[z_{j(m-k)-1}, z_{j(m-k)})}~\text {for}~1\le j \le m-1$ , where k is an integer in $\{1,2,\ldots , m-1\}$ relatively prime to m and $z_0, z_1, \ldots , z_{m-1}$ are some points in $S^1$ such that $z_0\prec z_1\prec \cdots \prec z_{m-1}$ , and $\phi _0:[z_0, z_{m-1}) \to [z_k,z_{k-1})$ is any arbitrary homeomorphism such that $ \phi _0([z_{j-1},z_j))=[z_{j-1+k},z_{j+k})~\text {for}~ 1\le j\le m-1 $ with $z_j:=z_{j(\mathrm {mod~m})}$ for $j\ge m$ .

The following two theorems characterize fixed points and periodic points of $\mathcal {J}_n$ in $\mathcal {C}(S^1)$ .

Theorem 6. Let $n\in \mathbb {N}$ . Then $f\in \mathcal {C}(S^1)$ is a fixed point of $\mathcal {J}_n$ if and only if one of the following conditions is satisfied: (i) $f|_{R(f)}$ is the identity map; (ii) $f|_{R(f)}$ is an orientation-reversing involution; or (iii) f is of the form equation (3.12) for a divisor m of $n-1$ .

Proof. Let $n\in \mathbb {N}$ . To find all fixed points of $\mathcal {J}_n$ in $\mathcal {C}(S^1)$ , in view of Lemma 5, it suffices to find all $f\in \mathcal {C}(S^1)$ satisfying the equation $\phi ^{n-1}=\text {id}$ on $R(f)$ . So let $f\in \mathcal {C}(S^1)$ be such that $f^{n-1}=\text {id}$ on $R(f)$ . Since $S^1$ is connected and compact, $R(f)$ is either a singleton set, an arc, or the whole of $S^1$ .

Suppose $R(f)=S^1$ . If f has a fixed point, then by Lemma 8, f is either the identity map or an involution according as f is orientation-preserving or orientation-reversing. If f has no fixed points, then by Lemma 9, f is of the form in equation (3.12) for some divisor m of $n-1$ .

Now suppose that $R(f)\subsetneq S^1$ . Then $R(f)$ is either a singleton set or an arc in $S^1$ . If $R(f)$ is a singleton set, then f is a constant map on $S^1$ . If $R(f)$ is an arc, say $[z_1,z_2]$ for some $z_1, z_2 \in S^1$ , then consider a homeomorphism $h_f:R(f)\to [t_1, t_2]$ , where $t_1, t_2$ are unique reals satisfying the conditions $z_1{\kern-1pt}={\kern-1pt}e^{2\pi it_1}, z_2{\kern-1pt}={\kern-1pt}e^{2\pi it_2}$ and $0{\kern-1pt}\le{\kern-1pt} t_1{\kern-1pt}<{\kern-1pt}t_2{\kern-1pt}<{\kern-1pt}t_1{\kern-1pt}+{\kern-1pt}1<2$ . Define a map $\mathcal {H}_f:\mathcal {C}(R(f)) \to \mathcal {C}([t_1, t_2])$ by

$$ \begin{align*}\mathcal{H}_f(g):=h_f\circ g\circ h_f^{-1} \quad \text{for all } g\in \mathcal{C}(R(f)).\end{align*} $$

Then $\mathcal {H}_f$ is a bijective, bi-continuous map such that $\mathcal {H}_f\circ \mathcal {J}_n=\mathcal {J}_n\circ \mathcal {H}_f$ (that is, $(\mathcal {C}(R(f)), \mathcal {J}_n)$ is topologically conjugate to $(\mathcal {C}([t_1, t_2]), \mathcal {J}_n)$ ). Now since $f^{n-1}=\text {id}$ on $R(f)$ , we have $\mathcal {H}_f(f^{n-1})=\mathcal {H}_f(\text {id})=\text {id} $ on $[t_1, t_2]$ , that is, $(h_f\circ f\circ h_f^{-1})^{n-1}=\text {id}$ on $[t_1, t_2]$ . Therefore, by Lemma 7, $h_f\circ f\circ h_f^{-1}$ is either the identity map or a decreasing involutory map on $[t_1, t_2]$ . This implies that $f|_{R(f)}$ is either an identity map or an orientation-reversing involutory map.

Conversely, if $f\in \mathcal {C}(S^1)$ satisfies either of the conditions (i) or (ii), then $f^n=f$ on $S^1$ implying that f is a fixed point of $\mathcal {J}_n$ . If f satisfies condition (iii), then by Lemma 9, we have $f^m=\mathrm {id}$ on $S^1$ , and therefore $f^{n-1}=\mathrm {id}$ on $S^1$ as m divides $n-1$ . Therefore, f is a fixed point of $\mathcal {J}_n$ .

Theorem 7. Let $n, k\ge 2$ . Then $f\in \mathcal {C}(S^1)$ is a k-periodic point of $\mathcal {J}_n$ if and only if f is of the form in equation (3.12) for some $m>1$ such that $m\mid (n^k-1)$ and $m\nmid (n^j-1)$ for $1\le j\le k-1$ .

Proof. Let $f\in \mathcal {C}(S^1)$ be a k-periodic point of $\mathcal {J}_n$ . Then f satisfies

(3.13) $$ \begin{align} f^{n^k}=f \quad\text{and}\quad f^{n^j}\ne f \quad\text{for } 1\le j\le k-1 \end{align} $$

on $S^1$ and also by result (ii) of Theorem 2, $f\in \mathcal {U}_E^m$ for some compact subset E of K and $m>1$ satisfying equation (3.10). In fact, here $E=R(f)$ . We assert that $E=S^1$ . Note that E, being the image of connected and compact set $S^1$ under f, is either a singleton set, an arc, or $S^1$ . If E is a singleton, then f is a constant map on $S^1$ , and therefore $f^n=f$ , which is a contradiction to equation (3.13). If E is an arc, then $f|_E$ is either the identity map or an orientation-reversing involution. In any case, we arrive at a contradiction to equation (3.13). Therefore, $E=S^1$ so that $f\in \mathcal {U}_{S^1}^m$ . This implies by Lemma 9 that f is of the form in equation (3.12).

Conversely, if f is of the form in equation (3.12) for some $m>1$ such that $m\mid (n^k-1)$ and $m\nmid (n^j-1)$ for $1\le j\le k-1$ , then clearly $f\in \mathcal {U}_{S^1}^m$ with $m>1$ satisfying equation (3.10) so that by result (ii) of Theorem 2, f is a k-periodic point of $\mathcal {J}_n$ .

4 Stability in $\mathcal {J}_n$

In this section, we study the (Lyapunov) stability of fixed points of the iteration operator $\mathcal {J}_n$ . In view of Lemma 4, to investigate boundedness of orbits, in the following, we consider X to be a hemicompact metrizable space. Clearly, all the orbits of $\mathcal {J}_n$ are bounded because $D(f,g)=\sum _{j=1}^{\infty }\mu _j(f,g)\le \sum _{j=1}^{\infty }{1}/{2^j}=1$ for all $f, g\in \mathcal {C}(X).$ In particular, when X is a compact metric space with metric d, we can let $X=\bigcup _{j=1}^{\infty }K_j$ such that $K_1=X$ and $K_j=\emptyset $ for all $j\ge 2$ , implying that the metric D of $\mathcal {C}(X)$ is $D(f,g)=\mu (f,g)=\min \{\tfrac 12, \rho (f,g)\}$ , which is also equivalent to the uniform metric $\rho $ on it. Further, the boundedness of orbits of $\mathcal {J}_n$ in $\mathcal {C}(X)$ can also be proved as follows. Since X is compact, there exist $y\in X$ and $M>0$ such that $d(x,y)\le {M}/{2}$ , implying that $d(f^{n^k}(x), f^{n^l}(x)) \le d(f^{n^k}(x), y)+d(y, f^{n^l}(x)) \le M$ for all $x \in X$ and $k, l \in \mathbb {N}$ . So, we can work with metric $\rho $ of $\mathcal {C}(X)$ instead of D whenever X is compact.

As defined in [Reference Milnor19], the orbit $(f^n(x))_{n\in \mathbb {N}\cup \{0\}}$ of a discrete semi-dynamical system $(X,f)$ , where X is a metric space equipped with the metric d, is said to be (Lyapunov) stable if for every $\epsilon>0$ , there exists $\delta>0$ such that $ d(f^k(x),f^k(y))<\epsilon $ for all $k\in \mathbb {N}$ whenever $y\in X$ satisfies $d(x,y)<\delta $ . We say that a point $x\in X$ is (Lyapunov) stable for f if its orbit $(f^n(x))_{n\in \mathbb {N}\cup \{0\}}$ is (Lyapunov) stable. The following two theorems prove that most fixed points of $\mathcal {J}_n$ in $\mathcal {C}(X)$ are not stable for $X=\mathbb {R}$ and $X=S^1$ , respectively.

Proof. Let $f\in \mathcal {C}_{\mathrm {id}}(\mathbb {R})$ be a non-constant map. Then there exist $a, b\in \mathbb {R}$ with $a<b$ such that $[a,b]\subseteq R(f)$ and $f|_{[a,b]}=\mathrm {id}$ . For each $\eta>0$ , let $g_\eta :\mathbb {R}\to \mathbb {R}$ be the map defined by

$$ \begin{align*} g_\eta(x)=\begin{cases} f(x)&\text{if } x\in (-\infty,a]\cup[b,\infty),\\ a+(x-a)(1-\eta)&\text{if } x\in \bigg[a,\dfrac{a+b}{2}\bigg],\\[6pt] x(1+\eta)-b\eta&\text{if } x\in \bigg[\dfrac{a+b}{2}, b\bigg]. \end{cases} \end{align*} $$

Then $g_\eta \in \mathcal {C}(\mathbb {R})$ for each $\eta>0$ . Consider the metric D on $\mathcal {C}(\mathbb {R})$ defined as in equations (2.2)–(2.4) with the partition $\mathbb {R}=\bigcup _{j=1}^{\infty }K_j$ , where

$$ \begin{align*} K_j:=\begin{cases} [a,b]&\text{if } j=1,\\ \bigg[b+\dfrac{j-3}{2},b+\dfrac{j-1}{2}\bigg]&\text{if } j=3,5,\ldots,\\[6pt] \bigg[a-\dfrac{j}{2},a-\dfrac{j-2}{2}\bigg] &\text{if } j=2,4,\ldots. \end{cases} \end{align*} $$

Let $\epsilon =\min \{({b-a})/{8},\tfrac 12\}$ and for any $\delta>0$ , choose $\eta _\delta>0$ such that $\eta _\delta <\min \{{\delta _0}/({b-a}),\tfrac 12\}$ for some $0< \delta _0<\min \{\delta ,\tfrac 12\}$ .

Claim. $D(f,g_{\eta _\delta })<\delta $ and $D(f,g_{\eta _\delta }^{n^{k_0}})\ge \epsilon $ for some $k_0\in \mathbb {N}$ .

Consider any $x\in \mathbb {R}$ . If $x\in K_1$ with $x\le ({a+b})/{2}$ , then

$$ \begin{align*} |f(x)-g_{\eta_\delta}(x)|=|x-(a+(x-a)(1-\eta_\delta))|=(x-a)\eta_\delta<(x-a)\frac{\delta_0}{b-a}<\delta_0. \end{align*} $$

If $x\in K_1$ with $x>({a+b})/{2}$ , then

$$ \begin{align*} |f(x)-g_{\eta_\delta}(x)|=|x-( x(1+\eta_\delta)-b\eta_\delta)|=(b-x)\eta_\delta<(b-x)\frac{\delta_0}{b-a}<\delta_0. \end{align*} $$

If $x\in K_j$ with $j>1$ , then

$$ \begin{align*} |f(x)-g_{\eta_\delta}(x)|=|f(x)-f(x)|=0. \end{align*} $$

Therefore, $\rho _1(f,g_{\eta _\delta })\le \delta _0$ and $\rho _j(f,g_{\eta _\delta })=0$ for all $j>1$ , implying that $\mu _1(f,g_{\eta _\delta }) \le \delta _0$ and $\mu _j(f,g_{\eta _\delta })=0$ for all $j>1$ . Hence, $D(f,g_{\eta _\delta })\le \delta _0<\delta $ .

Now for any $x\in [a,({a+b})/{2}]$ , we have

$$ \begin{align*} &g_{\eta_\delta}(x)= x(1-\eta_\delta)+a\eta_\delta\ge a(1-\eta_\delta)+a\eta_\delta =a, \\ &g_{\eta_\delta}(x)\le\frac{a+b}{2} (1-\eta_\delta)+a\eta_\delta =\frac{a+b}{2}-\frac{b-a}{2}\eta_\delta<\frac{a+b}{2}, \end{align*} $$

implying that $g_{\eta _\delta }(x)\in [a, ({a+b})/{2}]$ . Therefore, $g_{\eta _\delta }([a,({a+b})/{2}])\subseteq [a,({a+b})/{2}]$ , and hence by induction,

(4.14) $$ \begin{align} g_{\eta_\delta}^k(x)=a+(x-a)(1-\eta_\delta)^k \end{align} $$

for every $x\in [a, ({a+b})/{2}]$ and $k\in \mathbb {N}$ . Let $y=({a+b})/{2}$ . Since $\eta _\delta \in (0,1)$ , equation (4.14) implies that $g_{\eta _\delta }^k(y)\to a$ as $k\to \infty $ . So there exists $N\in \mathbb {N}$ such that $|g_{\eta _\delta }^k(y)-a|<\epsilon $ for all $k\ge N$ . Choose $k_0\in \mathbb {N}$ so large that $n^{k_0}>N$ . Then $g_{\eta _\delta }^{n^{k_0}}(y)-a<({b-a})/{8}$ , that is, $-g_{\eta _\delta }^{n^{k_0}}(y)>-({7a+b})/{8}$ . Thus,

$$ \begin{align*} f^{n^{k_0}}(y)-g_{\eta_\delta}^{n^{k_0}}(y)=\frac{b-a}{2} (1-(1-\eta_\delta)^{n^{k_0}})>0, \end{align*} $$

and therefore

$$ \begin{align*} |f^{n^{k_0}}(y)-g_{\eta_\delta}^{n^{k_0}}(y)|=f^{n^{k_0}}(y)-g_{\eta_\delta}^{n^{k_0}}(y)> \frac{a+b}{2}-\frac{7a+b}{8}=\frac{3(b-a)}{8}>\frac{b-a}{8}, \end{align*} $$

implying that

$$ \begin{align*} \rho_1(f, g_{\eta_\delta}^{n^{k_0}})=\rho_1(f^{n^{k_0}},g_{\eta_\delta}^{n^{k_0}})\ge |f^{n^{k_0}}(y)-g_{\eta_\delta}^{n^{k_0}}(y)|> \frac{b-a}{8}. \end{align*} $$

This proves that

$$ \begin{align*} \mu_1(f, g_{\eta_\delta}^{n^{k_0}})=\min\bigg\{\frac{1}{2}, \rho_1(f, g_{\eta_\delta}^{n^{k_0}})\bigg\}\ge \min\bigg\{\frac{1}{2}, \frac{b-a}{8}\bigg\}=\epsilon. \end{align*} $$

Therefore, the claim holds, and hence f is not stable for $\mathcal {J}_n$ .

Proof. Let $n\in \mathbb {N}$ and $f\in \mathcal {C}_{\mathrm {id}}(S^1)$ be a constant map on $S^1$ . For given $\epsilon>0$ , choose $\delta =\epsilon $ . Then for every $k\in \mathbb {N}$ and $g\in \mathcal {C}(S^1)$ with $\rho (f,g)<\delta $ , we have

$$ \begin{align*} |f(x)- g^k(x)|=|f(g^{k-1}(x))-g(g^{k-1}(x))|\le \rho(f,g)<\epsilon \quad \text{for all } x\in S^1, \end{align*} $$

implying $\rho (f,g^k)<\epsilon $ , and hence in particular $\rho (f^{n^k},g^{n^k})<\epsilon $ . Therefore, f is stable.

Conversely, suppose that $f\in \mathcal {C}_{\mathrm {id}}(S^1)$ is a non-constant map on $S^1$ . Then $f|_{R(f)}=\mathrm {id}$ such that either $R(f)=S^1$ or $R(f)=[z_1,z_2]$ for some $z_1=e^{it_1}$ , $z_2=e^{it_2}\in S^1$ with $0\le t_1<t_2< 2\pi $ . For each $\eta>0$ , let $g_\eta :S^1\to S^1$ be the map defined by

$$ \begin{align*} g_\eta(e^{it})=\begin{cases} f(e^{it})&\text{if } t\in [0,2\pi)\setminus[t_1,t_2],\\ e^{i[t_1+(t-t_1)(1-\eta)]}&\text{if } t\in \bigg[t_1,\dfrac{t_1+t_2}{2}\bigg],\\[6pt] e^{i[t(1+\eta)- t_2\eta]}&\text{if } t\in \bigg[\dfrac{t_1+t_2}{2}, t_2\bigg]. \end{cases} \end{align*} $$

Then $g_\eta \in \mathcal {C}(S^1)$ for each $\eta>0$ . Let $\epsilon =({t_2-t_1})/{2\sqrt {2}\pi }$ and for any $\delta>0$ , choose $\eta _\delta>0$ such that $\eta _\delta <\min \{{2\delta _0}/({t_2-t_1}),1\}$ for some $0<\delta _0<\delta $ .

Claim. $\rho (f,g_{\eta _\delta })<\delta $ and $\rho (f,g_{\eta _\delta }^{n^{k_0}})\ge \epsilon $ for some $k_0\in \mathbb {N}$ .

Consider any $t\in [0,2\pi )$ . If $t\in [0,2\pi )\setminus [t_1,t_2]$ , then

$$ \begin{align*} |f(e^{it})-g_{\eta_\delta}(e^{it})|=|f(e^{it})-f(e^{it})|=0<\delta_0. \end{align*} $$

If $t\in (t_1, ({t_1+t_2})/{2})$ , then

$$ \begin{align*} |f(e^{it})-g_{\eta_\delta}(e^{it})|&=|e^{it}-e^{i[t_1+(t-t_1)(1-\eta_\delta)]}|\\ &=|1-e^{i\eta_\delta(t_1-t)}|\\ &\le 2\bigg|\sin\bigg(\dfrac{\eta_\delta(t_1-t)}{2}\bigg)\bigg|\\ &\le |\eta_\delta(t_1-t)|=(t-t_1)\eta_\delta<\dfrac{t_2-t_1}{2} \dfrac{2\delta_0}{t_2-t_1}=\delta_0. \end{align*} $$

If $t\in [({t_1+t_2})/{2},t_2)$ , then by a similar argument, we have $|f(e^{it})-g_{\eta _\delta }(e^{it})|<\delta _0$ . Therefore, $|f(e^{it})-g_{\eta _\delta }(e^{it})|< \delta _0$ for all $t\in [0,2\pi )$ and hence $\rho (f,g_{\eta _\delta })\le \delta _0<\delta $ .

Now for any $t\in [t_1,({t_1+t_2})/{2}]$ , we have

$$ \begin{align*} t_1=t_1(1-\eta_\delta)+t_1\eta_\delta &\le t(1-\eta_\delta)+t_1\eta_\delta\\ &=t_1+(t-t_1)(1-\eta_\delta)\\ &\le\frac{t_1+t_2}{2} (1-\eta_\delta)+t_1\eta_\delta\\ &=\frac{t_1+t_2}{2}-\frac{t_2-t_1}{2}\eta_\delta<\frac{t_1+t_2}{2}, \end{align*} $$

implying $g_{\eta _\delta }(e^{it})\in [z_1, w]$ , where $w=e^{i({t_1+t_2})/{2}}$ . Therefore, $g_{\eta _\delta }([z_1,w])\subseteq [z_1,w]$ . Hence, it can be shown by induction that $g_{\eta _\delta }^k(e^{it})=e^{i[t_1+(t-t_1)(1-\eta _\delta )^k]}$ for every $t\in [t_1, ({t_1+t_2})/{2}]$ and $k\in \mathbb {N}$ . Also, $\sin ({t}/{2})\ge {t}/{\pi }$ for all $t\in [0,\pi ],$ and therefore,

(4.15) $$ \begin{align} |e^{it}-1|=\sqrt{2}\sin \bigg(\dfrac{t}{2}\bigg)\ge\dfrac{\sqrt{2}t}{\pi} \quad \text{for all } t\in [0,\pi]. \end{align} $$

Now for each $k\in \mathbb {N}$ , we have

(4.16) $$ \begin{align} |f(w)-g_{\eta_\delta}^k(w)|&=|e^{i({t_1+t_2})/{2}}-e^{i[t_1+(({t_1+t_2})/{2}-t_1)(1-\eta_\delta)^k]}| \nonumber\\ &=|1-e^{-i[({t_2-t_1})/{2}(1-(1-\eta_\delta)^k)]}| \nonumber\\ &=|e^{i[({t_2-t_1})/{2}(1-(1-\eta_\delta)^k)]}-1| \end{align} $$

and

$$ \begin{align*} 0\le \frac{t_2-t_1}{2}[1-(1-\eta_\delta)^k]<\frac{t_2-t_1}{2}<\frac{t_2}{2}\le \frac{2\pi}{2}=\pi, \end{align*} $$

implying by equation (4.15) that

$$ \begin{align*} |e^{i[({t_2-t_1})/{2}(1-(1-\eta_\delta)^k)]}-1|&\ge \dfrac{\sqrt{2}}{\pi}\cdot\frac{t_2-t_1}{2}[1-(1-\eta_\delta)^k]\\ &=\dfrac{t_2-t_1}{\sqrt{2}\pi}[1-(1-\eta_\delta)^k], \end{align*} $$

for each $k\in \mathbb {N}$ . Then equation (4.16) implies that

(4.17) $$ \begin{align} |f(w)-g_{\eta_\delta}^k(w)|\ge \dfrac{t_2-t_1}{\sqrt{2}\pi}[1-(1-\eta_\delta)^k], \end{align} $$

for each $k\in \mathbb {N}$ . Since $1-(1-\eta _\delta )^k\rightarrow 1$ as $k\rightarrow \infty $ , there exists $N\in \mathbb {N}$ such that $(1-\eta _\delta )^k<\tfrac 12$ for all $k\ge N$ . Choose $k_0$ sufficiently large such that $n^{k_0}>N$ . Then $1-(1-\eta _\delta )^{n^{k_0}}> \tfrac 12$ , and therefore from equation (4.17), we have

$$ \begin{align*} |f(w)-g_{\eta_\delta}^{n^{k_0}}(w)|\ge \dfrac{t_2-t_1}{\sqrt{2}\pi}\cdot \frac{1}{2}=\dfrac{t_2-t_1}{2\sqrt{2}\pi}=\epsilon, \end{align*} $$

which implies that

$$ \begin{align*}\rho(f, g_{\eta_\delta}^{n^{k_0}})=\rho(f^{n^{k_0}},g_{\eta_\delta}^{n^{k_0}})\ge |f^{n^{k_0}}(w)-g_{\eta_\delta}^{n^{k_0}}(w)|\ge \epsilon.\end{align*} $$

This proves the claim, from which it follows that f is not stable, which is a contradiction. Hence, f is a constant map on $S^1$ .

As defined in [Reference Holmgren10], a discrete semi-dynamical system $(X,f)$ , where X is a metric space equipped with the metric d, is said to be topologically transitive if for every pair of open sets $U, V$ in X, there exist $x \in U$ and $n \in \mathbb {N}$ such that $f^n(x)\in V$ . Here, f is said to be sensitively dependent on initial conditions if there exists $\delta>0$ such that for every $x \in X$ and every $\epsilon>0$ , there exist $y \in X$ and $n \in \mathbb {N}$ such that $d(x,y)<\epsilon $ and $d(f^n(x), f^n(y))>\delta $ . As defined in [Reference Devaney8], f is said to be chaotic in Devaney’s sense if: (i) the set of periodic points of f is dense in X; (ii) f is topologically transitive; and (iii) f exhibits sensitive dependence on initial conditions. Since $\mathrm {Per}(\mathcal {J}_n;\mathcal {C}(\mathbb {R}))$ is not dense in $\mathcal {C}(\mathbb {R})$ , as seen in §3, $\mathcal {J}_n$ is not chaotic $\mathcal {C}(\mathbb {R})$ for $n\ge 2$ . Also, since $S^1$ is compact, by [Reference Veerapazham, Gopalakrishna and Zhang23, Theorem 5.1], $\mathcal {J}_n$ is not chaotic on $\mathcal {C}(S^1)$ for $n\ge 2$ . However, $\mathcal {J}_1$ is also not chaotic on $\mathcal {C}(\mathbb {R})$ and $\mathcal {C}(S^1)$ since it is not sensitively dependent on initial conditions. Thus, although we see that all orbits of $\mathcal {J}_n$ in $\mathcal {C}(\mathbb {R})$ and $\mathcal {C}(S^1)$ are bounded but most of the fixed points are unstable, which actually exhibits a complicated behavior of $\mathcal {J}_n$ , the complicated behavior is not chaotic. The following examples show how complicated an orbit of $\mathcal {J}_n$ can be.

Example 4. Let $f:\mathbb {R}\to \mathbb {R}$ be the map defined by

$$ \begin{align*} f(x)=\begin{cases} 0&\text{if } x\le 0,\\ x&\text{if }0\le x\le 1,\\ 1&\text{if } x\ge 0. \end{cases} \end{align*} $$

Let $n=2$ , $\epsilon =\tfrac 18$ , $\delta =0.13$ , $\delta _0=0.07$ , $\eta _\delta =0.04$ , and $g_2$ be the map $g_{\eta _\delta }$ as defined in Theorem 8 with $a=0$ and $b=1$ . Define the maps $g_1$ , $g_3:\mathbb {R}\to \mathbb {R}$ by

$$ \begin{align*} g_1(x)=\begin{cases} 0&\text{if } x\le 0,\\ 0.99x&\text{if } 0\le x\le 1,\\ 0.99&\text{if } x\ge 1, \end{cases}\quad \text{and} \quad g_3(x)=\begin{cases} 0&\text{if } x\le 0,\\ 1.1x&\text{if } 0\le x \le 0.5,\\ x+0.05&\text{if } 0.5\le x\le 0.55,\\ -2x+1.7&\text{if } 0.55\le x\le 0.6,\\ 1.25x-0.25&\text{if } 0.6\le x\le 1,\\ 1&\text{if } x\ge 1. \end{cases} \end{align*} $$

Then $f\in \mathrm {Fix}(\mathcal {J}_2; \mathcal {C}(\mathbb {R}))$ and $g_1, g_2, g_3 \in \mathcal {C}(\mathbb {R})$ . Consider the metric D on $\mathcal {C}(\mathbb {R})$ defined as in equations (2.2)–(2.4) with the partition $\mathbb {R}=\bigcup _{j=1}^{\infty }K_j$ , where

$$ \begin{align*} K_j:=\begin{cases} [0,1]&\text{if } j=1,\\ \bigg[-\dfrac{j-1}{2},-\dfrac{j-1}{2}+1\bigg]&\text{if } j=3,5,\ldots,\\[6pt] \bigg[\dfrac{j}{2},\dfrac{j}{2}+1\bigg] &\text{if } j=2,4,\ldots. \end{cases} \end{align*} $$

An easy computation shows that $D(g_j,f)<\delta $ for all $j=1,2,3$ . However, $g_1^{2^6}(0.4)=0.2102$ , $g_2^{2^4}(0.4)=0.2082$ , and $g_3^{2^3}(0.4)=0.5796$ , implying that $D(g_1^{2^6} f)>\epsilon $ , $D(g_2^{2^4} f)>\epsilon $ , and $D(g_3^{2^3}, f)>\epsilon $ . This illustrates the claim in the proof of Theorem 8 for f for the above-mentioned specific choices of $\delta $ , $\delta _0$ , and $\eta _\delta $ with $\epsilon =\tfrac 18$ . Indeed, f is not stable for $\mathcal {J}_2$ as seen from Theorem 8.

To illustrate the complexity of $\mathcal {J}_2$ , we investigate the asymptotic behavior of the orbits of $g_1, g_2$ , and $g_3$ . We have

$$ \begin{align*} g_1^k(x)=\begin{cases} 0&\text{if } x\le 0,\\ 0.99^kx&\text{if } 0\le x\le 1,\\ 0.99^k&\text{if } x\ge 1, \end{cases} \end{align*} $$

for all $k\in \mathbb {N}$ . Therefore, the sequence of maps $(g_1^k)_{k\in \mathbb {N}}$ and hence the orbit $(g_1^{2^k})_{k\in \mathbb {N}\cup \{0\}}$ of $g_1$ converges uniformly to the zero map on $\mathbb {R}$ . As seen in the proof of Theorem 8, $g_2^k(x)\to 0 ~\text {as}~k\to \infty $ , for each $x\in [0,0.5]$ . Also, for each $x\in [0.5,1)$ , there exists $k_x\in \mathbb {N}$ such that $g_2^{k_x}(x)\in [0,0.5]$ . Further,

$$ \begin{align*} g_2^k(x)=\begin{cases} 0&\text{if } x\le 0,\\ 1&\text{if } x\ge 1, \end{cases} \end{align*} $$

for all $k\in \mathbb {N}$ . Thus, the sequence of maps $(g_2^k)_{k\in \mathbb {N}}$ and hence the orbit $(g_2^{2^k})_{k\in \mathbb {N}\cup \{0\}}$ of $g_2$ converges pointwise to the discontinuous map $f_2:\mathbb {R}\to \mathbb {R}$ defined by

$$ \begin{align*} f_2(x)=\begin{cases} 0&\text{if } x<1,\\ 1&\text{if } x\ge 1. \end{cases} \end{align*} $$

The point $x=0.5$ is a $3$ -periodic point of $g_3$ and therefore, the orbit $(g_3^{2^k})_{k\in \mathbb {N}\cup \{0\}}$ of $g_3$ does not converge. In fact, by [Reference Li and Yorke16, Theorem 1], $g_3$ is chaotic in the sense of Li and Yorke. However, $g_3([0,1])\subseteq [0,1]$ , implying that $g_3$ is not topologically transitive and therefore is not chaotic in the sense of Devaney.

Thus, although all the orbits of $\mathcal {J}_2$ in $\mathcal {C}(\mathbb {R})$ are bounded, it is possible that an orbit may not converge or, even if it converges, the limit function may not be continuous. Furthermore, the choice of $\delta =0.13$ is not special. Indeed, we can construct maps in $\mathcal {C}(\mathbb {R})$ that are similar to those of $g_1$ , $g_2$ , and $g_3$ in each $\delta $ -neighborhood of f.

Example 5. Let f be the identity map on $S^1$ . Let $n=2$ , $\epsilon =\tfrac 18$ , $\delta =0.15$ , $\delta _0=0.07$ , $\eta _\delta =0.05$ , and $g_2$ be the map $g_{\eta _\delta }$ as defined in Theorem 9 with $t_1=0$ and $t_2={\pi }/{2}$ . Define the maps $g_1$ , $g_3:S^1\to S^1$ by $g_1(e^{it})=e^{0.9it}$ for all $t\in [0,2\pi ),$ and

$$ \begin{align*} g_3(e^{it})=e^{i(t+{2\pi}/{7})} \quad \text{for all } t\in [0,2\pi). \end{align*} $$

Then $f\in \mathrm {Fix}(\mathcal {J}_2; \mathcal {C}(S^1))$ and $g_1, g_2, g_3 \in \mathcal {C}(S^1)$ . A computation as in Theorem 9 shows that $\rho (g_2,f)<\delta $ . However,

$$ \begin{align*} |g_2^{2^4}(e^{i{\pi}/{4}})-e^{i{\pi}/{4}}|&=|1-e^{i{\pi}/{4}(1-0.95^{2^4})}|\\ &\ge \frac{1}{2\sqrt{2}}(1-0.95^{2^4})\quad(\text{using equation~}(4.15))\\ &\ge\frac{1}{4\sqrt{2}}>\epsilon, \end{align*} $$

implying that $\rho (g_2^{2^4}, f)>\epsilon $ . Indeed, f is not stable for $\mathcal {J}_2$ as seen from Theorem 9.

To illustrate the complexity of iteration operator $\mathcal {J}_2$ , we investigate the asymptotic behavior of the orbits of $g_1, g_2$ , and $g_3$ . We have $g_1^k(e^{it})=e^{0.9^kit} \text { for all} k\in \mathbb {N}~\text {and for all } t\in [0,2\pi ).$ Therefore, the sequence of maps $(g_1^k)_{k\in \mathbb {N}}$ and hence the orbit $(g_1^{2^k})_{k\in \mathbb {N}\cup \{0\}}$ of $g_1$ converge uniformly to the constant map $f_1:S^1 \to S^1$ defined by $f_2(e^{it})=1$ for all $t\in [0,2\pi )$ . As noted in the proof of Theorem 9, $g_2^k(e^{it})\to 1 ~\text {as} k\to \infty $ , for each $t\in [0,{\pi }/{4}]$ . Also, for each $t\in [{\pi }/{4},{\pi }/{2})$ , there exists $k_t\in \mathbb {N}$ such that $\arg (g_2^{k_t}(e^{it}))\in [0,{\pi }/{4}]$ . Moreover, $g_2^k(e^{it})=e^{it} \text { for all } k\in \mathbb {N}$ and $\text { for all } t\in [{\pi }/{2}, 2\pi )$ . Thus, the sequence of maps $(g_2^k)_{k\in \mathbb {N}}$ and hence the orbit $(g_2^{2^k})_{k\in \mathbb {N}\cup \{0\}}$ of $g_2$ converge pointwise to the discontinuous map $f_2:S^1\to S^1$ defined by

$$ \begin{align*} f_2(e^{it})=\begin{cases} 1&\text{if } 0\le t<\dfrac{\pi}{2},\\ e^{it}&\text{if } \dfrac{\pi}{2}\le t<2\pi. \end{cases} \end{align*} $$

The map $g_3$ is a $3$ -periodic point of $\mathcal {J}_2$ and therefore the orbit $(g_3^{2^k})_{k\in \mathbb {N}\cup \{0\}}$ of $g_3$ converges to the periodic orbit $\{g_3, g_3^2, g_3^{2^2}\}$ .

Thus, similar to $\mathcal {C}(\mathbb {R})$ , although all the orbits of $\mathcal {J}_2$ in $\mathcal {C}(S^1)$ are bounded, it is possible that an orbit may not converge or, even if it converges, the limit function may not be continuous.

5 Classification up to conjugacy

Since $\mathbb {R}$ is not homeomorphic to $S^1$ , it is natural to expect that the dynamics of $\mathcal {J}_n$ on $\mathcal {C}(\mathbb {R})$ is ‘different’ from that on $\mathcal {C}(S^1)$ . This is also evident from our discussion as $\mathcal {J}_n$ has periodic points of all periods in $\mathcal {C}(S^1)$ whereas it has no non-trivial periodic points in $\mathcal {C}(\mathbb {R})$ for $n\ge 2$ . This leads to the question: Is the dynamics of $\mathcal {J}_n$ on $\mathcal {C}(X)$ identical to that on $\mathcal {C}(Y)$ whenever X is homeomorphic to Y? We investigate this question in this section in locally compact spaces.

Let $(X,f)$ and $(Y, g)$ be two discrete semi-dynamical systems. As defined in [Reference Holmgren10], we say that $(X,f)$ is topologically conjugate (or simply conjugate) to $(Y,g)$ if there exists a homeomorphism $h:X\to Y$ such that $h\circ f=g\circ h$ . In this case, h is called a topological conjugacy.

Proof. Let h be a homeomorphism of X onto Y. Define a map $\mathcal {H}:\mathcal {C}(X) \to \mathcal {C}(Y)$ as

$$ \begin{align*} \mathcal{H}(f):=h\circ f\circ h^{-1} \quad \text{for all } f\in \mathcal{C}(X). \end{align*} $$

Then $\mathcal {H}$ is a well-defined bijective map. To prove the continuity of $\mathcal {H}$ , first we claim that the map $\mathcal {F}:Y\times \mathcal {C}(X)\to Y$ defined by $\mathcal {F}(y,f):=h\circ f\circ h^{-1}(y)$ is continuous. In fact, consider the map $\mathcal {G}:Y\times \mathcal {C}(X)\to X\times \mathcal {C}(X)$ defined by $\mathcal {G}:=(h^{-1}\circ p_1, p_2)$ , where the $p_i$ are the projection maps on $Y\times \mathcal {C}(X)$ given by $p_1(y,f):=y$ and $p_2(y,f):=f$ . Since $h^{-1}\circ p_1$ and $p_2$ are continuous, so is $\mathcal {G}$ . Also, by Lemma 2, the evaluation map $\mathcal {E}:X \times \mathcal {C}(X)\to X$ , defined by $\mathcal {E}(x,f):=f(x)$ , is continuous. Now,

$$ \begin{align*} h\circ \mathcal{E}\circ \mathcal{G}(y, f)&=h(\mathcal{E}(h^{-1}\circ p_1(y,f), p_2(y,f))) \\ &=h(\mathcal{E}(h^{-1}(y), f))\\ &=h(f(h^{-1}(y)))=\mathcal{F}(y,f) \end{align*} $$

for each $(y,f)\in Y\times \mathcal {C}(X)$ , implying that $\mathcal {F}=h\circ \mathcal {E}\circ \mathcal {G}$ . Therefore, being the composition of continuous maps h, $\mathcal {E}$ , and $\mathcal {G}$ , $\mathcal {F}$ is continuous, and the claim is proved.

The map $\mathcal {H}$ is actually the induced map of $\mathcal {F}$ . Therefore, since $\mathcal {F}$ is continuous, by Lemma 3, we get that $\mathcal {H}$ is continuous on $\mathcal {C}(X)$ . By a similar argument, it follows that $\mathcal {H}^{-1}$ is continuous on $\mathcal {C}(Y)$ . Hence, $\mathcal {H}$ is a homeomorphism of $\mathcal {C}(X)$ onto $\mathcal {C}(Y)$ .

Finally, for each $f\in \mathcal {C}(X)$ , we have

$$ \begin{align*} \mathcal{J}_n\circ \mathcal{H}(f)=\mathcal{J}_n(h\circ f\circ h^{-1})&=(h\circ f\circ h^{-1})^n\\ &=h\circ f^n\circ h^{-1}=h\circ \mathcal{J}_n(f)\circ h^{-1}=\mathcal{H}\circ \mathcal{J}_n(f), \end{align*} $$

implying that $\mathcal {J}_n\circ \mathcal {H}=\mathcal {H}\circ \mathcal {J}_n$ . Therefore, $(\mathcal {C}(X), \mathcal {J}_n)$ is conjugate to $(\mathcal {C}(Y), \mathcal {J}_n)$ .

In contrast to the above theorem, it is more interesting to see whether $(\mathcal {C}(X), \mathcal {J}_n)$ is still conjugate to $(\mathcal {C}(Y), \mathcal {J}_n)$ for some $n\in \mathbb {N}$ and some spaces X and Y which are not homeomorphic to each other, unless we can prove the converse of the above theorem. Concerning the converse, we need to consider the weaker converse first and ask: Does the notion $\mathcal {C}(X)$ is homeomorphic to $\mathcal {C}(Y)$ imply that X is homeomorphic to Y? The answer is no in general because of the following counter-example.

In fact, as shown in [Reference Narici and Beckenstein22, pp. 116–122], a seminorm on a real vector space X is a map $p:X\to \mathbb {R}$ satisfying the following conditions: (i) $p(x)\ge 0$ for all $x\in X$ ; (ii) $p(\unicode{x3bb} x)=|\unicode{x3bb} |p(x)$ for all $\unicode{x3bb} \in \mathbb {R}$ and $x\in X$ ; (iii) $p(x+y)\le p(x)+p(y)$ for all $x,y\in X$ . A real topological vector space X along with a family $\mathcal {P}$ of seminorms on it is said to be a locally convex space. As shown in [Reference Willard26, pp. 175–176], a topological space X is said to be completely metrizable if there exists a complete metric on it inducing its topology. As seen at the end of §3.2, $\mathcal {C}(\mathbb {R})$ in the the compact-open topology is metrizable with metric D defined as in equations (2.2)–(2.4), where d is the usual metric on $\mathbb {R}$ . Also, by [Reference Conway7, Proposition 1.12, p. 145], D is complete. Hence, $\mathcal {C}(\mathbb {R})$ is completely metrizable. Further, $\mathcal {C}(\mathbb {R})$ is locally convex because the family $\mathcal {P}=\{\rho _j:j\in \mathbb {N}\}$ of seminorms induces the compact-open topology on it. Moreover, by of [Reference Kundu and Garg15, Theorem 5.2, p. 694], $\mathcal {C}(\mathbb {R})$ is separable. However, ${C}([a,b], \mathbb {R})$ , the space of all continuous functions of $[a,b]$ into $\mathbb {R}$ in the uniform topology induced by the uniform metric $\rho $ , is also a completely metrizable space. Further, by using the Weierstrass polynomial approximation theorem, it follows that ${C}([a,b], \mathbb {R})$ is also separable. Moreover, it is locally convex, being a normed linear space. Thus, both ${C}([a,b], \mathbb {R})$ and $\mathcal {C}(\mathbb {R})$ are infinite dimensional separable completely metrizable locally convex linear spaces. Hence, by the Anderson–Kadec theorem (cf. [Reference Bessaga and Pełczyński3, Theorem 5.2, p. 189]), which states that every infinite dimensional separable Banach space is homeomorphic to the countable product $\mathbb {R}^\omega $ of $\mathbb {R}$ with itself in the product topology, it follows that they are homeomorphic as topological spaces. Moreover, by a known result (cf. [Reference Bessaga and Pełczyński3, Theorem 6.2 p. 190]), which states that every closed convex set of non-empty interior in an infinite dimensional Banach space is homeomorphic to the whole space, we get that $\mathcal {C}([a,b])$ is homeomorphic to ${C}([a,b],\mathbb {R})$ . Therefore, $\mathcal {C}([a,b])$ is homeomorphic to $\mathcal {C}(\mathbb {R})$ .

Another concern is the question: Even if $\mathcal {C}(X)$ is homeomorphic to $\mathcal {C}(Y)$ , is $(\mathcal {C}(X),\mathcal {J}_m)$ conjugate to $(\mathcal {C}(Y),\mathcal {J}_n)$ for different m and n? As observed from Theorem 4, the existence of decreasing involutory fixed points ensures that $\mathcal {J}_3$ is not conjugate to $\mathcal {J}_2$ on $\mathcal {C}(\mathbb {R})$ . More generally, we have the following theorem.

Proof. Suppose that $\mathcal {H}$ is a conjugacy of $(\mathcal {C}(Y),\mathcal {J}_m)$ and $(\mathcal {C}(X),\mathcal {J}_2)$ . Then $\mathcal {H}:\mathcal {C}(Y)\to \mathcal {C}(X)$ is a homeomorphism such that

$$ \begin{align*} \mathcal{H}(g^m)=\mathcal{H}(g)^2\quad \text{for all } g\in \mathcal{C}(Y). \end{align*} $$

Let $f_1=f_0$ and $f_2=\mathrm {id}$ . Then there exist $g_1\ne g_2 \in \mathcal {C}(Y)$ such that $f_1=\mathcal {H}(g_1)$ and $f_2=\mathcal {H}(g_2)$ . We have $\text {id}=f_1^2=\mathcal {H}(g_1)^2=\mathcal {H}(g_1^m)$ and $\text {id}=f_2^2=\mathcal {H}(g_2)^2=\mathcal {H}(g_2^m)$ , implying that $g_1^m=g_2^m$ , because $\mathcal {H}$ is one-to-one. Since m is odd, $f_1$ is a fixed point for $\mathcal {J}_m$ , implying that $g_1$ is a fixed point for $\mathcal {J}_2$ , that is, $g_1^2=g_1$ and therefore $g_1^m=g_1$ . Similarly, since $f_2$ is a fixed point for $\mathcal {J}_m$ , it follows that $g_2$ is a fixed point for $\mathcal {J}_2$ , that is, $g_2^2=g_2$ , and therefore $g_2^m=g_2$ . Hence, we have $g_1=g_2$ , which is a contradiction. So $(\mathcal {C}(Y), \mathcal {J}_m)$ is not conjugate to $(\mathcal {C}(X), \mathcal {J}_2)$ and the result follows.

6 Remarks and questions

The iteration operator $\mathcal {J}_1$ has no non-trivial periodic points in both $\mathcal {C}(\mathbb {R})$ and $\mathcal {C}(S^1)$ . Also, unlike $\mathcal {C}(\mathbb {R})$ , where there are no non-trivial periodic points as seen in Theorem 5, $\mathcal {J}_n$ has periodic points of all periods in $\mathcal {C}(S^1)$ for $n\ge 2$ (see Example 3). Further, we have the following result.

In fact, in the cases of $(\mathcal {C}(\mathbb {R}), \mathcal {J}_n)$ with $n\ge 1$ and $(\mathcal {C}(S^1), \mathcal {J}_1)$ , the result follows vacuously since there are no $3$ -periodic points. However, in the case of $(\mathcal {C}(S^1), \mathcal {J}_n)$ with $n\ge 2$ , there are k-periodic points for all $k\ge 1$ as seen from Example 3. Observe that this proposition makes a contrast with the Sharkovskii theorem on the underlying spaces $\mathbb {R}$ and $S^1$ because the Sharkovskii theorem holds valid on $\mathbb {R}$ but not on $S^1$ (see, for example, the rotation map $R_{{2\pi }/{3}}$ on $S^1$ defined as in Example 3).

Additionally, from Theorem 10, we can conclude that it suffices to discuss the dynamics of $\mathcal {J}_n$ on $\mathcal {C}(X)$ for X to be a representative of the equivalence class $[X]$ under the homeomorphism equivalence relation. In view of this observation, it follows that dynamics of $\mathcal {J}_n$ on $\mathcal {C}(X)$ is identical to that on:

1. (i) $\mathcal {C}(\mathbb {R})$ (respectively $\mathcal {C}([a,b])$ ) whenever X is, for example, a curve in $\mathbb {C}$ homeomorphic to $\mathbb {R}$ (respectively $[a,b]$ );

2. (ii) $\mathcal {C}(S^1)$ whenever X is, for example, a simple closed curve in $\mathbb {C}$ , or the real projective line $\mathbb {RP}^1$ .

Moreover, by the Hahn–Mazurkiewicz theorem [Reference Willard26], a Hausdorff topological space X is a continuous image of $[a,b]$ or $S^1$ if and only if it is a Peano space (that is, a compact, connected, locally connected, metric space). Therefore, the equivalence class $[[a,b]]$ of $[a,b]$ (respectively $[S^1]$ of $S^1$ ) also contains subspaces of some ‘complicated’ spaces in the Euclidean plane $\mathbb {R}^2$ like the space filling curves—Peano curve, Hilbert curve, Sierpiński curve (respectively like Sierpiński triangle).

Finally, we conclude the paper with some problems for future discussion. Since any non-constant map f in $\mathcal {C}_{\mathrm {id}}(\mathbb {R})$ (respectively $\mathcal {C}_{\mathrm {id}}(S^1)$ ) has a unique choice for $f|_{R(f)}$ , namely $\mathrm {id}$ , we were indeed able to use the definition of stability to prove the instability of such maps in Theorem 8 (respectively Theorem 9). However, if f lies in $\mathcal {C}_{\mathrm {inv}}(S^1)$ (consisting of all continuous self-maps of $S^1$ which are orientation-reversing involutions on their range), then $f|_{R(f)}$ has uncountably many choices and therefore the case-wise approach used in the proof of the above theorem is impractical. Thus, the problem of stability of fixed points of $\mathcal {J}_n$ , which are in $\mathcal {C}_{\mathrm {inv}}(S^1)$ , is highly non-trivial. For the same reason, the problem of stability of fixed points of $\mathcal {J}_n$ in $\mathcal {C}_{\mathrm {inv}}(\mathbb {R})$ is also highly non-trivial. Additionally, although $\mathcal {J}_n$ does not have a non-trivial periodic point in $\mathcal {C}(\mathbb {R})$ by Theorem 5, the problem of stability of periodic points of $\mathcal {J}_n$ in $\mathcal {C}(S^1)$ is also difficult. Further, we remind that we did not conclude the discussion in Theorem 11 for all pairs n and m, which is interesting to consider. We observed through the counter-example given in §5 that a weaker converse of Theorem 10 is not true in general. However, it is interesting to investigate if the actual converse of Theorem 10 is true, that is, does $(\mathcal {C}(X), \mathcal {J}_n)$ is conjugate to $(\mathcal {C}(Y), \mathcal {J}_n)$ imply that X is homeomorphic to Y?