Iteration changes discontinuity into smoothness (Ⅱ): oscillating case

1.   Introduction

The $n$-th iterate $f^n$ of a mapping $f:E\to E$ is defined by $f^n(x) = f(f^{n-1}(x))$ and $f^0(x) = x$ for all $x\in E$ inductively, where $E$ is a nonempty set and $n$ is a positive integer. The research on the iteration of mappings can be traced back more than one hundred years ago at least (^[1,2,7]). The iterative operation is much more complicated than the general algebraic operation, especially the iteration of nonlinear functions, so the research work is very difficult and tortuous. Iteration is a common phenomenon in nature, which has become the focus of many disciplines. Under such circumstances, dynamical system theory has developed rapidly.

It is often thought that iteration turns a bad function into a worse one. For example, the function and its iterate

It is easy to see that $f$ has exactly one discontinuity at $\frac{1}{2}$ (see Figure 1), but its iterate $f^{2}$ has exactly three discontinuities at $\frac{3}{10}$, $\frac{1}{2}$ and $\frac{3}{4}$ (see Figure 2). However, a discontinuous function may have a continuous second-order iterate as shown in ^[3] and ^[5], which shows that iteration can also convert a "bad" function to a "good" one. This encourages efforts to study of such a converting. In ^[3] and ^[5] all self-mappings on a compact interval with exactly one discontinuity and more than one but finitely many discontinuities of the same type were classified for such a converting respectively. In ^[4] all continuous self-mappings with exactly one nonsmooth point were classified for the converting to $C^1$ iterates. Recently, we investigated the $C^1$ smoothness iterate of the second-order for self-mappings with exactly one removable or jumping discontinuity, we obtained necessary and sufficient conditions for those self-mappings whose second-order iterates are $C^1$ smooth in ^[6]. For a continuation, we are also interested in $C^1$ smoothness iterate of the second-order for self-mappings with exactly one oscillatory discontinuity, the remaining case of discontinuity.

It is possible to find an example with exactly one oscillatory discontinuity which is $C^1$ smooth by its iteration. The function and its iterate

One can see that $f$ has exactly one oscillatory discontinuity at $\frac{1}{2}$ (see Figure 3), but its iterate $f^{2}$ is $C^1$ smooth on the whole interval $(0, 1)$(see Figure 4).

Let $I: = (0, 1)$ and $V_{o}(I, I)$ consist of all $C^1$ self-mappings on $I$ with exactly one oscillatory discontinuity. Each $f\in V_{o}(I, I)$ can be presented as

where $x_{0}\in (0, 1)$ is the unique oscillatory discontinuity, $f_{i}$ is $C^{1}$ smooth on $I_{i}$ for each $i\in\{1, 2\}$ and $c\in (0, 1)$ is a constant.

In this paper we continue the work of ^[6], investigating the second-order $C^1$ smoothness of mappings in $V_{o}(I, I)$. By the definition of oscillating discontinuities, either $\lim_{x\to x_0-0}f_1(x)$ or $\lim_{x\to x_0+0} f_2(x)$ does not exist but both $f_{1}$ and $f_{2}$ are bounded. Thus, each mapping $f$ in $V_{o}(I, I)$ has 3 possibilities:

In order to investigate $C^{1}$ smoothness of the second-order iterates of mappings in $V_{o}(I, I)$, we also need to consider three subclasses for $V_{o+}(I, I)$ and $V_{o-}(I, I)$ respectively, i.e., $V_{o+}(I, I) = V^{E}_{o+}(I, I)\cup V^{O}_{o+}(I, I)\cup V^{\infty}_{o+}(I, I)$ and $V_{o-}(I, I) = V^{E}_{o-}(I, I)\cup V^{O}_{o-}(I, I)\cup V^{\infty}_{o-}(I, I)$, where

Then $V_{o}(I, I) = V^{E}_{o+}(I, I)\cup V^{O}_{o+}(I, I)\cup V^{\infty}_{o+}(I, I)\cup V^{E}_{o-}(I, I)\cup V^{O}_{o-}(I, I)\cup V^{\infty}_{o-}(I, I)\cup V_{o*}(I, I)$.

In this paper, we discuss $C^{1}$ smoothness of the second-order iterates of mapping $f$ in $V_o(I, I)\backslash V^{\infty}_{o-}(I, I)\cup V^{\infty}_{o+}(I, I)$. We give necessary and sufficient conditions for $C^{1}$ smooth $f^{2}$. We obtain necessary conditions and remark the difficulties in finding sufficient conditions for those self-mappings in $V^{\infty}_{o-}(I, I)$ and $V^{\infty}_{o+}(I, I)$ to have a $C^{1}$ smooth iterate of the second-order respectively. Moreover, we give sufficient conditions for those self-mappings in $V(I, I)$ whose second-order iterates are not $C^{1}$ smooth, where $V(I, I)$ consists of all $C^1$ self-mappings on $I$ having only one discontinuity. Finally, we use examples to demonstrate our theorems.

For convenience, let $I_0: = \{x_0\}$, then $I = I_1\cup I_0\cup I_2$. For $i, j = 0, 1, 2$ we use the notations

where $U^{-}_{\alpha}$ and $U^{+}_{\alpha}$ denote a sufficiently small left-half and right-half neighborhood of $\alpha$ respectively. We use $D_-f$ and $D_+f$ to denote the left derivative of $f$ and the right derivative of $f$ respectively.

2.   Iteration for $V_o(I, I)$

In this section, we consider $C^{1}$ smoothness of the second-order iterates of $f\in V_{o}(I, I)$ to be defined as in (1.1). In order to discuss $C^{1}$ smoothness of the second-order iterates of mappings in $V_{o}(I, I)$, we need to consider constantization of a mapping near the boundary of the domain, as shown in the Fourth part of ^[3]. Assume that $h_{1}:H_{1}: = (c, d)\to H_{2}$ and $h_{2}:H_{2}\to H_{3}$ are continuous mapping, where $H_{i}$ s ($i = 1, 2, 3$) are all nonempty intervals. $h_{1}$ is said to be $constantized$ by $h_{2}$ near $c$ (or $d$) if there exist a closed interval $L\subseteq H_{2}$ and a vicinity (hollow neighborhood) $U\subseteq H_{1}$ of $c$ (or $d$) such that $h_{1}(U)\subseteq L$ and $h_{2}$ is identical to a constant on $L$. For convenience, let $\theta(h_1, h_2)$ denote the constant. Moreover, we also need to define two mappings $f_{10}:I_{10}: = I_1\cup I_0\to I$ and $f_{20}:I_{20}: = I_0\cup I_2\to I$ such that

where $x_0\in (0, 1)$ is the unique oscillatory discontinuity. For convenience, for $i\in\{1, 2, 10, 20\}$, $m, j\in\{1, 2\}$, $\tau\in\{E, O, \infty\}$, $\lambda\in\{-, +\}$, $l\in\{1, 2\}\backslash\{m\}$, $\mu\in\{O, \infty\}$ and $\xi_{j}\in f^{-1}(I_0)\cap I_{j}$ let

Theorem 2.1. Suppose that $f\in V_o(I, I)\backslash V^{\infty}_{o+}(I, I)\cup V^{\infty}_{o-}(I, I)$ with the unique oscillatory discontinuity $x_0\in (0, 1)$ and that $\xi_{j}\in f^{-1}(I_0)\cap I_{j}$ for $j = 1$ or $2$. Let $y_1: = \lim_{x\to x_0-0}f_1(x)$ and $y_{2}: = \lim_{x\to x_0+0} f_2(x)$. The following results hold:

(o-) In the case that $f\in V^{\tau}_{o-}(I, I)$ for $\tau\in\{E, O\}$, $f^2$ is $C^{1}$ smooth on $I$ if and only if there is $i\in\{1, 2, 20\}$ such that $f_1$ is constantized by $f_i$ near $x_0$ and for $j\in\{1, 2\}$ the following two conditions are both fulfilled:

(o-1) $f\in \complement^{\tau 1ij}_{o-}(I, I)$ if $y_{2}\in I_j$, either $f\in\hat{\complement}^{E1i}_{o-}(I, I)$ as $x_{0}\in\Delta^{+}_{0}$ or $f\in\tilde{\complement}^{E1i}_{o-}(I, I)$ as $x_{0}\in\Delta^{+}_{2}$ if $y_{2} = x_{0}$.

(o-2) $\xi_{j}\in\Delta_{00}\cup\Delta_{22}\cup\Delta_{20}\cup\Delta_{02}$ and $f\in \bar{\complement}^{\tau 2j}_{o-}(I, I)$ if $\xi_{j}\in\Delta_{22}\cup\Delta_{20}\cup\Delta_{02}$.

(o+) In the case that $f\in V^{\lambda}_{o+}(I, I)$ for $\lambda\in\{E, O\}$, $f^2$ is $C^{1}$ smooth on $I$ if and only if there is $k\in\{1, 2, 10\}$ such that $f_2$ is constantized by $f_k$ near $x_0$ and for $j\in\{1, 2\}$ the following two conditions are both fulfilled:

(o+1) $f\in \complement^{\lambda 2kj}_{o+}(I, I)$ if $y_{1}\in I_j$, either $f\in\hat{\complement}^{E2k}_{o+}(I, I)$ as $x_{0}\in\Delta^{-}_{0}$ or $f\in\tilde{\complement}^{E2k}_{o+}(I, I)$ as $x_{0}\in\Delta^{-}_{1}$ if $y_{1} = x_{0}$.

(o+2) $\xi_{j}\in\Delta_{00}\cup\Delta_{11}\cup\Delta_{10}\cup\Delta_{01}$ and $f\in \bar{\complement}^{\lambda 1j}_{o+}(I, I)$ if $\xi_{j}\in\Delta_{11}\cup\Delta_{10}\cup\Delta_{01}$.

(o*) In the case that $f\in V_{o*}(I, I)$, $f^2$ is $C^{1}$ smooth on $I$ if and only if both $f_1$ and $f_2$ are constantized by $f_j$ near $x_0$, $f(I_{1}\cup I_{2})\subseteq I_{j}$ holds and $f\in\complement^{j}_{0*}(I, I)$ for $j = 1$ or $2$.

Proof. Since $f\in V_o(I, I)\backslash V^{\infty}_{o-}(I, I)\cup V^{\infty}_{o+}(I, I)$ and $V_{o}(I, I) = V^{E}_{o-}(I, I)\cup V^{O}_{o-}(I, I)\cup V^{\infty}_{o-}(I, I)\cup V^{E}_{o+}(I, I)\cup V^{O}_{o+}(I, I)\cup V^{\infty}_{o+}(I, I)\cup V_{o*}(I, I)$. Then there are five cases to be discussed: $f\in V^{E}_{o-}(I, I)$, $f\in V^{O}_{o-}(I, I)$, $f\in V^{E}_{o+}(I, I)$, $f\in V^{O}_{o+}(I, I)$ and $f\in V_{o*}(I, I)$.

For (o-), i.e., $f\in V^{\tau}_{o-}(I, I)$ for $\tau\in\{E, O\}$, it implies that $y_2: = \lim_{x\to x_0+0}f_{2}(x)$ exists but $\lim_{x\to x_0-0}f_{1}(x)$ does not exist.

Sufficiency of (o-). Since we have assumed that there is $i\in\{1, 2, 20\}$ such that $f_1$ is constantized by $f_i$ near $x_0$. In the following, we only discuss the situation that $f_1$ is constantized by $f_1$ near $x_0$ since the other situations can be discussed similarly. Under condition (o-1), we prove that $f^2$ is $C^{1}$ smooth at $x_{0}$. In fact, if $y_{2}\in I_1$, by the definition that $f_1$ is constantized by $f_1$ near $x_0$ and the continuity of $f_{2}$ on $I_{2}$, there exist a sufficiently small left-half neighborhood $U^{-}_{x_{0}}$ of $x_{0}$ and a sufficiently small right-half neighborhood $U^{+}_{x_{0}}$ of $x_{0}$ such that

It follows from (2.1) that

Since we assumed that $f\in \complement^{\tau 111}_{o-}(I, I)$ for $\tau\in\{E, O\}$. Thus, we need to discuss in two situations: $f\in \complement^{E111}_{o-}(I, I)$ and $f\in \complement^{O111}_{o-}(I, I)$. In the first situation that $f\in \complement^{E111}_{o-}(I, I)$, by the definition of $\complement^{E111}_{o-}(I, I)$ we see that $f\in V^{E}_{o-}(I, I)$, it implies that $\tilde y_{2}: = \lim_{x\to x_0+0}f_2'(x)$ exists. Note that $\lim_{x\to x_0+0}f_2(x) = y_{2}$. From (2.1) we get that

By our assumption that $f\in \complement^{E111}_{o-}(I, I)$ and the definition of $\complement^{E111}_{o-}(I, I)$, we get from (2.2)–(2.5) that $\lim_{x\to x_0-0}f^{2}(x) = \lim_{x\to x_0+0}f^{2}(x) = f^{2}(x_{0})$ and $D_-f^{2}(x_{0}) = D_+f^{2}(x_{0}) = 0$. It implies that $f^2$ is $C^{1}$ smooth at $x_{0}$. In the second situation that $f\in \complement^{O111}_{o-}(I, I)$, by the definition of $\complement^{O111}_{o-}(I, I)$ we see that $f\in V^{O}_{o-}(I, I)$, it implies that $\lim_{x\to x_0+0}f_2'(x)$ does not exist but $f'_{2}$ is bounded. From (2.1) we get that

By our assumption that $f\in \complement^{O111}_{o-}(I, I)$ and the definition of $\complement^{O111}_{o-}(I, I)$, we get from (2.2) and (2.3) that $\lim_{x\to x_0-0}f^{2}(x) = \lim_{x\to x_0+0}f^{2}(x) = f^{2}(x_{0})$. Moreover, from (2.6) and (2.7) we obtain $D_-f^{2}(x_{0}) = D_+f^{2}(x_{0}) = 0$ since $\lim_{x\to x_0+0}f'_{1}(f_{2}(x)) = f'_{1}(y_{2}) = 0$ and $f'_{2}$ is bounded. It implies that $f^2$ is $C^{1}$ smooth at $x_{0}$. The proof of $y_{2}\in I_2$ is similar to the proof of $y_{2}\in I_1$. Next, we consider that $y_{2} = x_{0}$, we have

when $x_{0}\in\Delta^{+}_{0}$. It follows from (2.8) that

By our assumption that $f\in \hat{\complement}^{E11}_{o-}(I, I)$ and the definition of $\hat{\complement}^{E11}_{o-}(I, I)$, we get from (2.9)–(2.11) that $\lim_{x\to x_0-0}f^{2}(x) = \lim_{x\to x_0+0}f^{2}(x) = f^{2}(x_{0})$ and $D_-f^{2}(x_{0}) = D_+f^{2}(x_{0}) = 0$. It implies that $f^2$ is $C^{1}$ smooth at $x_{0}$. Moreover, we have

when $x_{0}\in\Delta^{+}_{2}$. By our assumption that $f\in \tilde{\complement}^{E11}_{o-}(I, I)$, we see that $f\in V^{E}_{o-}(I, I)$, it implies that $\tilde y_{2}: = \lim_{x\to x_0+0}f_2'(x)$ exists. Note that $f_{2}(x)\rightarrow x_{0}+0$ as $x\rightarrow x_{0}+0$. From (2.12) we get that

By the assumption that $f\in \tilde{\complement}^{E11}_{o-}(I, I)$ and the definition of $\tilde{\complement}^{E11}_{o-}(I, I)$, we get from (2.13)–(2.16) that $\lim_{x\to x_0-0}f^{2}(x) = \lim_{x\to x_0+0}f^{2}(x) = f^{2}(x_{0})$ and $D_-f^{2}(x_{0}) = D_+f^{2}(x_{0}) = 0$. It implies that $f^2$ is $C^{1}$ smooth at $x_{0}$. Similarly to the proof of the condition (ii) of the sufficiency in Theorem 3 in ^[6], one can prove that $f^2$ is $C^1$ smooth at $\xi_{j}$ under condition (o-2), where $\xi_{j}\in f^{-1}(I_0)\cap I_{j}$ for $j = 1$ or $2$. Condition (o-1) and condition (o-2) imply that $f^2$ is $C^1$ smooth on the whole domain $I$. Therefore, the proof of sufficiency of (o-) is completed.

Necessity of (o-). Since $f\in V^{\tau}_{o-}(I, I)$ for $\tau\in\{E, O\}$, by the definition of $V^{\tau}_{o-}(I, I)$, implying that $f\in V_{o-}(I, I)$. Then we obtain from the definition of $V_{o-}(I, I)$ that $\lim_{x\to x_0-0}f_{1}(x)$ does not exist but $y_{2}: = \lim_{x\to x_0+0}f_{2}(x)$ exists. From the location of $y_{2}$, we need to consider two possibilities: either $y_{2}\in I_{j}$ for $j = 1$ or $2$, or $y_{2} = x_{0}$. Assume that $f^2$ is $C^1$ smooth on $I$, it implies that $f^2$ is continuous on $I$. By (i) of Theorem 3 in reference [3], one sees that there is $i\in\{1, 2, 20\}$ such that $f_1$ is constantized by $f_i$ near $x_0$. In what follows, we only consider the situation that $f_1$ is constantized by $f_1$ near $x_0$ since the other situations can be considered similarly. Note that $f\in V^{\tau}_{o-}(I, I)$ for $\tau\in\{E, O\}$. Thus, we need to discuss in two situations: $f\in V^{E}_{o-}(I, I)$ and $f\in V^{O}_{o-}(I, I)$. If $y_{2}\in I_1$, one sees that (2.1) holds. It follows that (2.2) and (2.3) hold. In the first situation that $f\in V^{E}_{o-}(I, I)$, it implies from the definition of $V^{E}_{o-}(I, I)$ that $\tilde y_{2}: = \lim_{x\to x_0+0}f_2'(x)$ exists. It follows that (2.4) and (2.5) hold. Because we have assumed that $f^2$ is $C^1$ smooth on $I$, we have $\lim_{x\to x_0-0}f^{2}(x) = \lim_{x\to x_0+0}f^{2}(x) = f^{2}(x_{0})$ and $D_-f^{2}(x_{0}) = D_+f^{2}(x_{0})$. Then we get from (2.2)–(2.5) that $\theta(f_1, f_1) = f_{1}(y_{2}) = f(c)$ and $f'_{1}(y_{2})\tilde y_{2} = 0$. It implies from the definition of $\complement^{E111}_{o-}(I, I)$ that $f\in\complement^{E111}_{o-}(I, I)$. In the second situation that $f\in V^{O}_{o-}(I, I)$, it implies from the definition of $V^{O}_{o-}(I, I)$ that $\lim_{x\to x_0+0}f_2'(x)$ does not exist but $f'_{2}$ is bounded. From (2.1) we obtain (2.6) and (2.7). Since $f^2$ is $C^1$ smooth on $I$, we see that $D_+f^{2}(x_{0})$ exists. Note that $\lim_{x\to x_0+0}f'_{1}(f_2(x)) = f'_{1}(y_{2})$. We claim that

In fact, if $f'_{1}(y_2)\neq0$, It follows from (2.7) that $\lim_{x\to x_0+0}f'_2(x)$ exists since

which contradicts to our assumption that $\lim_{x\to x_0+0}f'_2(x)$ does not exist. Thus, the claim that (2.17) is proved. On the other hand, by the smoothness of $f^{2}$ on $I$ we see that $f^2$ is continuous on $I$, then we have $\lim_{x\to x_0-0}f^{2}(x) = \lim_{x\to x_0+0}f^{2}(x) = f^{2}(x_{0})$. It follows from (2.2) and (2.3) that $\theta(f_1, f_1) = f_{1}(y_{2}) = f(c)$. It implies from (2.17) and the definition of $\complement^{O111}_{o-}(I, I)$ that $f\in\complement^{O111}_{o-}(I, I)$. We use a similar discussion to the proof of the situation $y_{2}\in I_1$, One can get that $f\in\complement^{\tau112}_{o-}(I, I)$ for $\tau\in\{E, O\}$ when $y_{2}\in I_2$. Finally, if $y_{2} = x_{0}$, by the continuity of $f_{2}$ on $I_{2}$, we see that

In the first situation that $f\in V^{E}_{o-}(I, I)$, we claim that

By (2.18) we need to deny the case $x_{0}\in \Delta^{+}_{1}$. In fact, if $x_{0}\in \Delta^{+}_{1}$, from the definition of $\Delta^{+}_{1}$, we have $f^{2}(x) = f_{1}(f_{2}(x)),$ $\forall x\in U^{+}_{x_{0}}$. Note that $f\in V^{E}_{o-}(I, I)$ and $f_{2}(x)\rightarrow x_{0}-0$ as $x\rightarrow x_{0}+0$. It follows that $\lim_{x\to x_0+0}f^{2}(x) = \lim_{x\to x_0+0}f_{1}(f_{2}(x)) = \lim_{y\to x_0-0}f_{1}(y)$ does not exist, which implies that $f^2$ is not continuous at $x_{0}$, it follows that $f^2$ is not $C^1$ smooth on $I$, a contradiction to our assumption. This proves the claimed (2.19). By (2.19), we need to discuss the two cases $x_{0}\in\Delta^{+}_{0}$ and $x_{0}\in\Delta^{+}_{2}$. For the case $x_{0}\in\Delta^{+}_{0}$, we see that (2.8) holds. It follows from (2.8) that (2.9) and (2.10) hold. Since we have assumed that $f^2$ is $C^1$ smooth on $I$, we have $\lim_{x\to x_0-0}f^{2}(x) = \lim_{x\to x_0+0}f^{2}(x) = f^{2}(x_{0})$. Then we get from (2.9) and (2.10) that $\theta(f_1, f_1) = c = f(c)$. It implies from the definition of $\hat{\complement}^{E11}_{o-}(I, I)$ that $f\in\hat{\complement}^{E11}_{o-}(I, I)$. For the case $x_{0}\in\Delta^{+}_{2}$, we see that (2.12) holds. It follows from (2.12) that (2.13-2.16) hold. Since we have assumed that $f^2$ is $C^1$ smooth on $I$, we have $\lim_{x\to x_0-0}f^{2}(x) = \lim_{x\to x_0+0}f^{2}(x) = f^{2}(x_{0})$ and $D_-f^{2}(x_{0}) = D_+f^{2}(x_{0})$. Then we get from (2.13-2.16) that $\theta(f_1, f_1) = y_{2} = f(c)$ and $\tilde y_{2} = 0$. It implies from the definition of $\tilde{\complement}^{E11}_{o-}(I, I)$ that $f\in\tilde{\complement}^{E11}_{o-}(I, I)$. Thus, condition (o-1) holds.

Remark that condition (o-1) does not give the results of the second situation that $f\in V^{O}_{o-}(I, I)$ when $y_{2} = x_{0}$. In fact, by (2.18), if $x_{0}\in \Delta^{+}_{1}$, using a similar discussion to the proof of the first situation that $f\in V^{E}_{o-}(I, I)$, we can get that $\lim_{x\to x_0+0}f^{2}(x)$ does not exist, a contradiction to our assumption. If $x_{0}\in\Delta^{+}_{0}$, from the definition of $\Delta^{+}_{0}$, there exists a sufficiently small right-half neighborhood $U^{+}_{x_{0}}$ of $x_{0}$ such that that $f_{2}(x) = x_{0}$ for $\forall x\in U^{+}_{x_{0}}$. It implies that $\lim_{x\to x_0+0}f_2'(x) = 0$, which contradicts the fact that $f\in V^{O}_{o-}(I, I)$. If $x_{0}\in\Delta^{+}_{2}$, from (2.12) we have $D_+f^{2}(x_{0}) = \lim_{x\to x_0+0}f'_{2}(f_{2}(x))f'_{2}(x)$. Note that $\lim_{x\to x_0+0}f'_{2}(f_{2}(x)) = \lim_{y\to x_0+0}f'_{2}(y)$ and $f\in V^{O}_{o-}(I, I)$, i.e., $\lim_{x\to x_0+0}f_2'(x)$ does not exist but $f'_{2}$ is bounded. Thus, it is hard to determine the existence of the limit $D_+f^{2}(x_{0}) = \lim_{x\to x_0+0}f_2'(f_2(x))f_2'(x)$.

We use a similar discussion to the proof of the condition (ii) of the necessity in Theorem 3 in ^[6], one can prove that condition (o-2) holds. Thus, the proof of result (o-) is completed.

Result (o+) can be discussed totally in a similar way to result (o-). In what follows, we consider result (o*).

We first prove necessity of (o*). Suppose that $f^2$ is $C^1$ smooth on $I$, it follows that $f^2$ is continuous on $I$. By (iii) of Theorem 3 in reference [3], we see that both $f_1$ and $f_2$ are constantized by $f_j$ near $x_0$, $f(I_{1}\cup I_{2})\subseteq I_{j}$ holds and $f\in\complement^{j}_{0*}(I, I)$ for $j = 1$ or $2$. Hence, the proof of necessity is completed.

Next, we prove sufficiency of (o*), we assumed that both $f_1$ and $f_2$ are constantized by $f_1$ near $x_0$, $f(I_{1}\cup I_{2})\subseteq I_{1}$ holds and $f\in\complement^{1}_{0*}(I, I)$. By (iii) of Theorem 3 in reference [3], we see that $f^2$ is continuous on $I$. By the definition of constantization, there exist a sufficiently small left-half neighborhood $U^{-}_{x_{0}}$ of $x_{0}$ and a sufficiently small right-half neighborhood $U^{+}_{x_{0}}$ of $x_{0}$ such that

we get from (2.20) that

Thus, the derivative of $f^{2}$ is continuous at $x_{0}$. It follows that $f^2$ is $C^{1}$ on $I$ because $f_{1}$ and $f_{2}$ are $C^{1}$ on $I_{1}$ and $I_{2}$ respectively. Similarly, we can prove that $f^2$ is $C^{1}$ on $I$ if both $f_1$ and $f_2$ are constantized by $f_2$ near $x_0$, $f(I_{1}\cup I_{2})\subseteq I_{2}$ holds and $f\in\complement^{2}_{0*}(I, I)$. Therefore, the proof of sufficiency is completed and the theorem is proved.

Theorem 2.2. Suppose that $f\in V^{\infty}_{o-}(I, I)\cup V^{\infty}_{o+}(I, I)$ with the unique oscillatory discontinuity $x_0\in (0, 1)$ and that $\xi_{j}\in f^{-1}(I_0)\cap I_{j}$ for $j = 1$ or $2$. Let $y_1: = \lim_{x\to x_0-0}f_1(x)$ and $y_{2}: = \lim_{x\to x_0+0} f_2(x)$. The following results hold:

(o-$\infty$) In the case that $f\in V^{\infty}_{o-}(I, I)$, assume that $f^2$ is $C^{1}$ smooth on $I$, then $y_{2}\neq x_{0}$ and there exists $i\in\{1, 2, 20\}$ such that $f_1$ is constantized by $f_i$ near $x_0$ and for $j\in\{1, 2\}$ the following two conditions are both fulfilled:

(o-$\infty$1) $f\in \complement^{\infty 1ij}_{o-}(I, I)$ if $y_{2}\in I_j$;

(o-$\infty$2) $\xi_{j}\in\Delta_{00}\cup\Delta_{22}\cup\Delta_{20}\cup\Delta_{02}$ and $f\in \bar{\complement}^{\infty 2j}_{o-}(I, I)$ if $\xi_{j}\in\Delta_{22}\cup\Delta_{20}\cup\Delta_{02}$.

(o+$\infty$) In the case that $f\in V^{\infty}_{o+}(I, I)$, assume that $f^2$ is $C^{1}$ smooth on $I$, then $y_{1}\neq x_{0}$ and there is $k\in\{1, 2, 10\}$ such that $f_2$ is constantized by $f_k$ near $x_0$ and for $j\in\{1, 2\}$ the following two conditions are both fulfilled:

(o+$\infty$1) $f\in \complement^{\infty 2kj}_{o+}(I, I)$ if $y_{1}\in I_j$;

(o+$\infty$2) $\xi_{j}\in\Delta_{00}\cup\Delta_{11}\cup\Delta_{10}\cup\Delta_{01}$ and $f\in \bar{\complement}^{\infty 1j}_{o+}(I, I)$ if $\xi_{j}\in\Delta_{11}\cup\Delta_{10}\cup\Delta_{01}$.

Proof. For (o-$\infty$), i.e., $f\in V^{\infty}_{o-}(I, I)$, by the definition of $V^{\infty}_{o-}(I, I)$, implying that $f\in V_{o-}(I, I)$. It follows from the definition of $V_{o-}(I, I)$ that $\lim_{x\to x_0-0}f_{1}(x)$ does not exist but $y_{2}: = \lim_{x\to x_0+0}f_{2}(x)$ exists. Moreover, we have $\lim_{x\to x_0+0}f_2'(x) = \infty$. By the $C^1$ smoothness of $f^{2}$ on $I$, We claim that $y_{2}\neq x_{0}$. In fact, if $y_{2} = x_{0}$, by the continuity of $f_{2}$ on $I_{2}$, we see that (2.18) holds. From (2.18) we need to discuss in three situations: $x_{0}\in \Delta^{+}_{1}$, $x_{0}\in \Delta^{+}_{0}$ and $x_{0}\in \Delta^{+}_{2}$. In the first situation that $x_{0}\in \Delta^{+}_{1}$, from the definition of $\Delta^{+}_{1}$, we have $f^{2}(x) = f_{1}(f_{2}(x)),$ $\forall x\in U^{+}_{x_{0}}$. Note that $f_{2}(x)\rightarrow x_{0}-0$ as $x\rightarrow x_{0}+0$. It follows that $\lim_{x\to x_0+0}f^{2}(x) = \lim_{x\to x_0+0}f_{1}(f_{2}(x)) = \lim_{y\to x_0-0}f_{1}(y)$ does not exist, which implies that $f^2$ is not continuous at $x_{0}$, it follows that $f^2$ is not $C^1$ smooth at $x_{0}$, which contradicts to our assumption that $f^2$ is $C^{1}$ smooth on $I$. In the second situation that $x_{0}\in \Delta^{+}_{0}$, from the definition of $\Delta^{+}_{0}$, we have $f_{2}(x) = x_{0},$ $\forall x\in U^{+}_{x_{0}}$. It follows that $\lim_{x\to x_0+0}f'_{2}(x) = 0$, which contradicts to our assumption that $\lim_{x\to x_0+0}f_2'(x) = \infty$. Finally, in the third situation that $x_{0}\in \Delta^{+}_{2}$, from the definition of $\Delta^{+}_{2}$, we have $f^{2}(x) = f_{2}(f_{2}(x)),$ $\forall x\in U^{+}_{x_{0}}$. It follows that $D_+f^{2}(x_{0}) = \lim_{x\to x_0+0}f'_{2}(f_{2}(x))f'_{2}(x)$. Note that $\lim_{x\to x_0+0}f'_{2}(f_{2}(x)) = \lim_{y\to x_0+0}f'_{2}(y) = \infty$. Thus, we obtain $D_+f^{2}(x_{0}) = \infty$. It implies that $f^2$ is not $C^{1}$ smooth on $I$, a contradiction to our assumption. Therefore, the claim that $y_{2}\neq x_{0}$ is proved. We use a similar discussion to the proof of the situation $f\in V^{O}_{o-}(I, I)$ of the necessity in Theorem 2.1, one can get that there exists $i\in\{1, 2, 20\}$ such that $f_1$ is constantized by $f_i$ near $x_0$ and both condition (o-$\infty$1) and condition (o-$\infty$2) hold.

Case (o+$\infty$) can be discussed totally in a similar way to case (o-$\infty$). Therefore, the theorem is proved.

Remark that the above Theorem 2.2 does not give sufficient conditions of $f^{2}$ to be $C^{1}$ because it is hard to determine the existence of either $\lim_{x\to x_0-0}f_i'(f_1(x))f_1'(x)$ or $\lim_{x\to x_0+0}f_i'(f_2(x))f_2'(x)$ for $i = 1$ or $2$. In fact, if $f\in V^{\infty}_{o-}(I, I)$, it follows from the definition of $V^{\infty}_{o-}(I, I)$ that $\lim_{x\to x_0-0}f_{1}(x)$ does not exist but $y_{2}: = \lim_{x\to x_0+0}f_{2}(x)$ exists. Moreover, we have $\lim_{x\to x_0+0}f_2'(x) = \infty$. We assume that $f_1$ is constantized by $f_1$ near $x_0$ and $f\in \complement^{\infty 111}_{o-}(I, I)$. A similar discussion to the proof in Theorem 2.1, we can get that

By the definition of $\complement^{\infty 111}_{o-}(I, I)$, we have $\lim_{x\to x_0+0}f_1'(f_2(x)) = f'_{1}(y_{2}) = 0$. Note that $\lim_{x\to x_0+0}f_2'(x) = \infty$. It follows from (2.21) that $\lim_{x\to x_0+0}f'_{1}(f_2(x))f_2'(x)$ is of $0\cdot\infty$ type. Thus, it is hard to judge the existence of the right derivative $D_+f^{2}(x_{0})$. We similarly see difficulty in other cases.

The following theorem gives conditions for $f^{2}$ not to be $C^{1}$. For convenience, for $i = 1$ or $2$ we use the notation

Theorem 2.3. Let $f\in V(I, I)$ and $x_0\in (0, 1)$ be the unique discontinuity. Suppose that $y_1: = \lim_{x\to x_0-0}f_1(x)$ and $y_{2}: = \lim_{x\to x_0+0} f_2(x)$. Then $f^{2}$ is not $C^{1}$ on $I$ if for $i = 1$ or $2$ and $j\in \{1, 2\}\backslash\{i\}$ either

(ⅰ)$y_{i} = x_{0}$, $f(I_{i})\subseteq I_{i}\cup I_{0}$ in the case that $\lim f'_{i}(x) = \infty$, or

(ⅱ)$y_{i} = x_{0}$, $f(I_{i})\subseteq I_{j}\cup I_{0}$ in the case that $\tilde y_{i}: = \lim f'_i(x)$ exists and $\tilde y_{i}\neq 0$ but $\lim f'_{j}(x)$ does not exist, or

(ⅲ)$y_{i} = x_{0}$ in the case that $\lim f'_{i}(x) = \infty$ and $\lim f'_{j}(x) = \infty$, or

(ⅳ)$y_{i}\in I_{i}$ in the case that $\lim f'_{i}(x)$ does not exist and $f'_{i}(y_{i})\neq 0$, or

(ⅴ)$y_{i}\in I_{j}$ in the case that $\lim f'_{i}(x)$ does not exist and $f'_{j}(y_{i})\neq 0$.

Proof. we only prove the situation that $i = 1$ because the situation that $i = 2$ can be proved similarly.

For (i), we have $y_{1} = x_{0}$ and $f(I_{1})\subseteq I_{1}\cup I_{0}$. By $\lim_{x\to x_0-0}f'_1(x) = \infty$ and the continuity of $f_1$ on $I_{1}$, there exists a sufficiently small left-half neighborhood $U^{-}_{x_{0}}$ of $x_{0}$ such that $f_1(x) < x_0$ for all $x\in U^{-}_{x_{0}}\subset I_{1}$, which implies that $f^{2}(x) = f_{1}(f_{1}(x))$ for all $x\in U^{-}_{x_{0}}$. It follows that $D_-f^{2}(x_{0}) = \lim_{x\to x_0-0}f_1'(f_1(x))f_1'(x)$. Note that $f_{1}(x)\rightarrow x_{0}-0$ as $x\rightarrow x_{0}-0$. Then we have $\lim_{x\to x_0-0}f_1'(f_1(x)) = \lim_{y\to x_0-0}f_1'(y) = \infty$. Thus, $D_-f^{2}(x_{0}) = \lim_{x\to x_0-0}f_1'(f_1(x))f_1'(x) = \infty$. This implies that $f^2$ is not $C^1$ on $I$.

For (ii), we have $y_{1} = x_{0}$ and $f(I_{1})\subseteq I_{2}\cup I_{0}$. By $\tilde y_{1} = \lim_{x\to x_0-0}f'_1(x)$ exists and $\tilde y_{1}\neq 0$, there exists a sufficiently small left-half neighborhood $U^{-}_{x_{0}}$ of $x_{0}$ such that $f_1(x) > x_0$ for all $x\in U^{-}_{x_{0}}\subset I_{1}$, which implies $f^{2}(x) = f_{2}(f_{1}(x))$ for all $x\in U^{-}_{x_{0}}$. It follows that $D_-f^{2}(x_{0}) = \lim_{x\to x_0-0}f_2'(f_1(x))f_1'(x)$. Note that $f_{1}(x)\rightarrow x_{0}+0$ as $x\rightarrow x_{0}-0$ and $\lim_{x\to x_0+0}f_2'(x)$ does not exist. Then we have $\lim_{x\to x_0-0}f_2'(f_1(x)) = \lim_{y\to x_0+0}f_2'(y)$ does not exist. We claim that $D_-f^{2}(x_{0})$ does not exists. In fact, assume that $D_-f^{2}(x_{0}) = \lim_{x\to x_0-0}f_2'(f_1(x))f_1'(x)$ exists, then

exists since $\tilde y_{1} = \lim_{x\to x_0-0}f'_1(x)$ exists and $\tilde y_{1}\neq 0$. However, this contradicts the fact that $\lim_{x\to x_0+0}f'_2(x)$ does not exist. This implies that $f^2$ is not $C^1$ on $I$.

For (ⅲ), we have both $\lim_{x\to x_0-0}f'_1(x) = \infty$ and $\lim_{x\to x_0+0}f'_2(x) = \infty$. By $\lim_{x\to x_0-0}f'_1(x) = \infty$ and $y_{1} = x_{0}$, one sees that there exists a sufficiently small left-half neighborhood $U^{-}_{x_{0}}$ of $x_{0}$ such that either $f_1(x) < x_0$ or $f_1(x) > x_0$ for all $x\in U^{-}_{x_{0}}\subset I_{1}$. If $f_1(x) < x_0$ for all $x\in U^{-}_{x_{0}}$, one can prove that $f^2$ is not $C^1$ on $I$ with a similar discussion to the proof of case (ⅰ). If $f_1(x) > x_0$ for all $x\in U^{-}_{x_{0}}$, we have $f^{2}(x) = f_{2}(f_{1}(x))$ for all $x\in U^{-}_{x_{0}}$. It follows that $D_-f^{2}(x_{0}) = \lim_{x\to x_0-0}f_2'(f_1(x))f_1'(x)$. Note that $\lim_{x\to x_0-0}f'_1(x) = \infty$ and $\lim_{x\to x_0-0}f_2'(f_1(x)) = \lim_{y\to x_0+0}f_2'(y) = \infty$. It follows that $D_-f^{2}(x_{0}) = \lim_{x\to x_0-0}f_2'(f_1(x))f_1'(x) = \infty$. This implies that $f^2$ is not $C^1$ on $I$.

For (ⅳ), we have $y_{1}\in I_{1}$. By the continuity of $f_1$ on $I_1$, there exists a sufficiently small left-half neighborhood $U^{-}_{x_{0}}$ of $x_{0}$ such that $f_1(x) < x_0$ for all $x\in U^{-}_{x_{0}}\subset I_{1}$. Then we get that $f^{2}(x) = f_{1}(f_{1}(x))$ for all $x\in U^{-}_{x_{0}}$. It follows that $D_-f^{2}(x_{0}) = \lim_{x\to x_0-0}f_1'(f_1(x))f_1'(x)$. Note that $\lim_{x\to x_0-0}f_1'(f_1(x)) = f_1'(y_{1})\neq 0$. We claim that $D_-f^{2}(x_{0})$ does not exists. In fact, assume that $D_-f^{2}(x_{0})$ exists, then

exists. However, this contradicts the fact that $\lim_{x\to x_0-0}f'_1(x)$ does not exist. This implies that $f^2$ is not $C^1$ on $I$.

For (ⅴ), we have $y_{1}\in I_{2}$. By the continuity of $f_1$ on $I_1$, there exists a sufficiently small left-half neighborhood $U^{-}_{x_{0}}$ of $x_{0}$ such that $f_1(x) > x_0$ for all $x\in U^{-}_{x_{0}}\subset I_{1}$. Then we get that $f^{2}(x) = f_{2}(f_{1}(x))$ for all $x\in U^{-}_{x_{0}}$. It follows that $D_-f^{2}(x_{0}) = \lim_{x\to x_0-0}f_2'(f_1(x))f_1'(x)$. Note that $\lim_{x\to x_0-0}f_2'(f_1(x)) = f_2'(y_{1})\neq 0$. Similarly to the above (ⅳ), we can also get that $D_-f^{2}(x_{0})$ does not exists. It implies that $f^2$ is not $C^1$ on $I$. Therefore, this completes the proof.

3.   Examples

We demonstrate our theorems with some examples.

Example 3.1. Consider the mapping $F_1:(0, 1)\to (0, 1)$ (see Figure 5) defined by

which has a unique oscillating discontinuous point $x_0 = \frac{1}{2}$ since $y_{1} = \lim\limits_{x\to \frac{1}{2}-0}F_1(x) = \frac{1}{4}$ exists but $\lim\limits_{x\to \frac{1}{2}+0}F_1(x)$ does not exist, i.e., $F_{1}\in V_{o+}(I, I)$. Moreover, $\tilde y_{1} = \lim_{x\to \frac{1}{2}-0}F'_1(x) = 0$ exists. It implies that $F_{1}\in V^{E}_{o+}(I, I)$. Note that $I_{1} = (0, \frac{1}{2}), I_{2} = (\frac{1}{2}, 1)$ and $y_{1} = c = \frac{1}{4}\in I_{1}$. It is easy to check that $f_2$ is constantized by $f_1$ near $x_0$, $\theta(f_2, f_1) = f_{1}(y_{1}) = f_{1}(c) = \frac{1}{4}$ and $f'_{1}(y_{1})\tilde y_{1} = 0$, i.e., $F_{1}\in \complement^{E211}_{o+}(I, I)$, where

It implies that the assumption (o+1) of (o+) in Theorem 2.1 is satisfied. Furthermore, we can check that $\frac{2}{7}\in F_{1}^{-1}(I_0)\cap I_{1}$ and $\frac{2}{5}\in F_{1}^{-1}(I_0)\cap I_{1}$ since $\{\frac{2}{7}, \frac{2}{5}\}\subset I_{1}$ and $F_{1}(\frac{2}{7}) = F_{1}(\frac{2}{5}) = x_{0} = \frac{1}{2}$. Moreover, one can also check that $\frac{2}{7}\in \Delta_{11}, \frac{2}{5}\in \Delta_{11}$, $y_{1} = c = \frac{1}{4}$ and $f'_{1}(\frac{2}{7}) = f'_{1}(\frac{2}{5}) = 0$, i.e., $F_{1}\in \bar{\complement}^{E11}_{o+}(I, I)$. It implies that the assumption (o+2) of (o+) in Theorem 2.1 is satisfied. On the other hand, one can compute

which is $C^{1}$ smooth on $(0, 1)$ as shown in Figure 6.

Example 3.2. Consider the mapping $F_2:(0, 1)\to (0, 1)$ (see Figure 7) defined by

which has a unique oscillating discontinuous point $x_0 = \frac{1}{3}$ since neither $\lim_{x\to \frac{1}{3}-0}F_2(x)$ nor $\lim_{x\to \frac{1}{3}+0}F_2(x)$ exists, i.e., $F_{2}\in V_{o*}(I, I)$. Note that $I_{1} = (0, \frac{1}{3}), I_{2} = (\frac{1}{3}, 1), c = \frac{1}{18}\in I_{1}$. One can check that both $f_1$ and $f_2$ are constantized by $f_1$ near $x_0$, $F_{2}(I_{1}\cup I_{2})\subseteq I_{1}$ and $\theta(f_1, f_1) = \theta(f_2, f_1) = f_{1}(c) = \frac{1}{9}$, i.e., $F_{2}\in \complement^{1}_{o*}(I, I)$, where

It implies that the assumptions of (o*) in Theorem 2.1 are satisfied. Actually, one can compute

which is $C^{1}$ smooth on $(0, 1)$ as shown in Figure 8.

Example 3.3. Consider the mapping $F_3:(0, 1)\to (0, 1)$ (see Figure 9) defined by

which has a unique oscillating discontinuous point $x_0 = \frac{1}{2}$ since $y_{1} = \lim\limits_{x\to \frac{1}{2}-0}F_3(x) = \frac{3}{8}$ exists but $\lim\limits_{x\to \frac{1}{2}+0}F_3(x)$ does not exist, i.e., $F_{3}\in V_{o+}(I, I)$. Moreover, $\tilde y_{1} = \lim_{x\to \frac{1}{2}-0}F'_3(x) = \frac{3}{2}$ exists. It implies that $F_{3}\in V^{E}_{o+}(I, I)$. Note that $I_{1} = (0, \frac{1}{2}), I_{2} = (\frac{1}{2}, 1), c = \frac{1}{8}\in I_{1}$ and $y_{1} = \frac{3}{8}\in I_{1}$. It is easy to check that $f_2$ is constantized by $f_1$ near $x_0$, $\theta(f_2, f_1) = \frac{1}{4}\neq f_{1}(y_{1}) = \frac{17}{64}$, i.e., $F_{3}\not\in \complement^{E211}_{o+}(I, I)$, where

It implies that the assumptions of (o+) in Theorem 2.1 are not satisfied. Actually, one can compute

which is not $C^{1}$ smooth on $(0, 1)$ with nonsmooth point $\frac{1}{2}$ as shown in Figure 10.

Example 3.4. Consider the mapping $F_4:(0, 1)\to (0, 1)$ (see Figure 11) defined by

which has a unique jumping discontinuous point $x_0 = \frac{1}{2}$ since

Note that

and

Moreover, $y_{1} = \frac{1}{2} = x_{0}$, $\lim_{x\to \frac{1}{2}-0}f'_1(x) = \infty$ and $f_{1}(I_{1})\subseteq I_{1}$, i.e., the assumption (i) in Theorem 2.3 is satisfied. Actually, one can compute

which is not $C^{1}$ smooth on $(0, 1)$ with nonsmooth point $\frac{1}{2}$ as shown in Figure 12.

4.   Conclusions

Difference from ^[6], where the function has exactly a removable or a jumping discontinuity, in this paper, we show how a function with exactly one oscillating discontinuity may have a $C^{1}$ smooth iterate of second-order.

Acknowledgments

Thanks for the comments and suggestions given by anonymous colleagues and experts who have carefully examined this paper. This research was supported by the Scientific Research Fund of Sichuan Provincial Education Department of China (No. 18ZA0242), Key Project of Leshan Normal University of China (No. JG2018-1-03) and Key Project of Leshan Normal University of China (No. LZD014).

Conflict of interest

The authors declare no conflict of interest.