The $C^0$ solutions of the Feigenbaum-like functional equation

Abstract

By using the Schauder’s fixed point theorem, and constructing the special functional space and the construction operator, the existence, uniqueness, quasi-convexity (or quasi-concavity), symmetry and stability of the $C^0$ solutions of the Feigenbaum-like functional equations are discussed.

MSC:39B22, 39B12, 39A10.

1 Introduction

As early as 1978, Feigenbaum found the period-doubling bifurcation phenomenon by researching the iteration of a single-peak function class [1]. To reveal the mechanism of the Feigenbaum phenomenon, many years ago, the Feigenbaum functional equation had been researched extensively. McCarthy [2] obtained the general continuous exact bijective solutions. Epstein [3] gave a new proof of the existence of analytic, unimodal solutions by taking advantage of the normality properties of Herglotz functions and the Schauder-Tikhonov theorem. Eokmann and Wittwer [4] studied the Feigenbaum fixed point by using the computer. Thompson [5] investigated an essentially singular solution by expressing Feigenbaum’s equation as a singular Schroder functional equation whose solution was obtained using a scaling ansatz, and so on. Thus, some solutions in specific cases were found.

Specifically, in 1985, to give a feasible method, the second kind of the Feigenbaum functional equation,

$$\{\begin{array}{c}f(x)=\frac{1}{\lambda }f(f(\lambda x)),0<\lambda <1,\hfill \\ f(0)=1,\hfill \\ 0\le f(x)\le 1,x\in [0,1]\hfill \end{array}$$

a kind of the equivalent equation, was given by Yang and Zhang [6]. The continuous valley-unimodal solutions were shown by using the constructive method. Recently, there have been a lots of results about the polynomial-like iterative equation. In 1987, by using Schauder’s fixed point theorem to an operator defined by a linear combination of iterates of the unknown map** f, a result on the existence of continuous solutions of the polynomial-like iterative equation was given in [7]. Furthermore, the results were given for its differentiable solutions [8]. Then the convex solutions and concave solutions [9, 10], the analytic solutions [11–13], the symmetric solutions [14], the higher-dimensional solutions [15], and the results on the unit circle [16] were obtained. In order to understand the dynamics of a second order delay differential equation with a piecewise constant argument, the derived planar map**s and their invariant curves were studied [17]. Based on the iterative root theory for monotone functions, an algorithm for computing polygonal iterative roots of increasing polygonal functions was given [18]. Else, a problem about the Hyers-Ulam stability was raised first by Ulam in 1940 and solved for Cauchy equation by Hyers [19]. Later, many papers on the Hyers-Ulam stability have been published, especially, for the polynomial-like iterative equation [20–22].

In this paper, by using Schauder’s fixed point theorem, and constructing the special functional space and the construction operator, we consider the properties of the solutions of the Feigenbaum-like functional equation, which is a non-extended iterative equation,

$$\{\begin{array}{c}f(x)=\frac{1}{\lambda +1}{f}^2(\lambda x)+\frac{\lambda }{\lambda +1}g(x),0<\lambda <1,\hfill \\ a\le f(x)\le b,x\in I,\hfill \end{array}$$

(1.1)

where $g(x)$ is a given disturbance function, $f(x)$ is an unknown function, and $f^2(x)=f(f(x))$, $I=[a,b]$. It is clear that $a\le 0\le b$, since $\lambda x\in I$ for all $x\in I$. We give not only the existence of continuous solutions of (1.1) but also their uniqueness, stability (the continuous dependence and the Hyers-Ulam stability), quasi-convexity (or quasi-concavity), symmetry by applying fixed point theorems. Finally, we give an example to verify those conditions given in theorems.

2 Preliminary

In this section, we give several important definitions, lemmas and notions.

Let $C^0(I,\mathbf{R})=\{f:I\to \mathbf{R},f\text{ is continuous}\}$. Obviously, $C^0(I,\mathbf{R})$ is a Banach space with the norm ${\parallel \cdot \parallel }_{c^0}$, where the norm ${\parallel f\parallel }_{c^0}={max}_{x\in I}|f(x)|$ for $f\in C^0(I,\mathbf{R})$.

Let $C^0(I)=\{f\in C^0(I,\mathbf{R}):a\le f(x)\le b,a\le f(\lambda x)\le b,f\text{ is continuous}\}$. Then $C^0(I)$ is a complete metric space.

Let $X(I;M)=\{f\in C^0(I):|f(x)-f(y)|\le M|y-x|,\mathrm{\forall }x,y\in I\}$, where M is a positive constant.

Let $X(I;m,M)=\{f\in X(I;M):|f(x)-f(y)|\ge m|x_2-x_1|,\mathrm{\forall }x\in I,0<m\le M\}$, where m is a positive constant.

Let $f(\lambda x):=f^{(\lambda )}(x)$, $f^2(\lambda x):=f(f(\lambda x)):=f^{2(\lambda )}(x)$, $f^k(\lambda x):=f^{k(\lambda )}(x)$.

Definition 2.1 $f:I\to R$ is a quasi-convex (or quasi-concave) function [23] if for $\mathrm{\forall }x,y\in I$ and $\lambda \in [0,1]$, we have

$$f(\lambda x+(1-\lambda )y)\le max\{f(x),f(y)\}(\text{or}f(\lambda x+(1-\lambda )y)\ge min\{f(x),f(y)\}).$$

Let $X_{\sigma }(I;M)$ denote the families consisting of all quasi-convex functions or quasi-concave ones in $X(I;M)$, where $\sigma =\mathit{qcv}$ or $\sigma =\mathit{qcc}$.

The following Lemma 2.1 and Lemma 2.2 are useful, and the methods of their proofs are similar to ones in the paper [24].

Lemma 2.1 $X(I;M)$, $X_{\mathit{qcv}}(I;M)$, and $X_{\mathit{qcc}}(I;M)$ are compact convex subsets of $C^0(I,R)$.

Lemma 2.2 The composition $f\circ g$ is quasi-convex (or quasi-concave) if f is increasing and g is quasi-convex (or quasi-concave). In particular, for an increasing quasi-convex (or quasi-concave) function f, $f^k$ is also quasi-convex (or quasi-concave).

Lemma 2.3 If $f,h\in X(I;M)$, then

$${\parallel f^{2(\lambda )}-h^{2(\lambda )}\parallel }_{c^0}\le (M+1){\parallel f-h\parallel }_{c^0}.$$

(2.1)

Proof Note that

$$\begin{array}{rl}{\parallel f^{2(\lambda )}-h^{2(\lambda )}\parallel }_{c^0}& =\underset{x\in I}{max}|f^2(\lambda x)-h^2(\lambda x)|\\ \le \underset{x\in I}{max}|f(f(\lambda x))-f(h(\lambda x))|+\underset{x\in I}{max}|f(h(\lambda x))-h(h(\lambda x))|\\ \le M{\parallel f^{(\lambda )}-h^{(\lambda )}\parallel }_{c^0}+{\parallel f-h\parallel }_{c^0}.\end{array}$$

Let $y=\lambda x$. Then $y\in \lambda I$, and

$$\begin{array}{rl}{\parallel f^{(\lambda )}-h^{(\lambda )}\parallel }_{c^0}& =\underset{y\in \lambda I}{max}|f(y)-h(y)|\\ \le \underset{y\in I}{max}|f(y)-h(y)|\\ \le {\parallel f-h\parallel }_{c^0}.\end{array}$$

Then

$$\begin{array}{rl}{\parallel f^{2(\lambda )}-h^{2(\lambda )}\parallel }_{c^0}& \le M{\parallel f-h\parallel }_{c^0}+{\parallel f-h\parallel }_{c^0}\\ =(M+1){\parallel f-h\parallel }_{c^0}.\end{array}$$

Thus, (2.1) holds. □

Lemma 2.4 Suppose that $\phi \in X(I;m,M)$. If the positive constants m, M and λ satisfy

$$\frac{1}{\lambda }+1>\frac{M^2}{m},$$

(2.2)

then Lφ, defined by

$$L\phi (x)=(1+\frac{1}{\lambda })x-\frac{1}{\lambda }{\phi }^2(\lambda {\phi }^{-1}(x)),$$

(2.3)

is an orientation-preserving homeomorphism from I onto itself, and

$${(L\phi )}^{-1}\in X(I;\frac{1}{{\xi }_2},\frac{1}{{\xi }_1}),$$

(2.4)

where

$${\xi }_1=1+\frac{1}{\lambda }+\frac{M^2}{m},{\xi }_2=1+\frac{1}{\lambda }-\frac{M^2}{m}>0.$$

(2.5)

Proof Because $\phi \in X(I;m,M)$, by the paper [7], ${\phi }^{-1}\in X(I;\frac{1}{M},\frac{1}{m})$. Thus, for any $x_1\ne x_2\in I$, by (2.3) and (2.5)

$$\begin{array}{rl}|L\phi (x_2)-L\phi (x_1)|& =|(1+\frac{1}{\lambda })(x_2-x_1)+\frac{1}{\lambda }({\phi }^2(\lambda {\phi }^{-1}(x_2))-{\phi }^2(\lambda {\phi }^{-1}(x_1)))|\\ \ge (1+\frac{1}{\lambda })|x_2-x_1|-M^2|{\phi }^{-1}(x_2)-{\phi }^{-1}(x_1)|\\ \ge (1+\frac{1}{\lambda })|x_2-x_1|-\frac{M^2}{m}|x_2-x_1|\\ \ge {\xi }_2|x_2-x_1|>0.\end{array}$$

On the other hand,

$$\begin{array}{rl}|L\phi (x_2)-L\phi (x_1)|& =|(1+\frac{1}{\lambda })(x_2-x_1)+\frac{1}{\lambda }({\phi }^2(\lambda {\phi }^{-1}(x_2))-{\phi }^2(\lambda {\phi }^{-1}(x_1)))|\\ \le (1+\frac{1}{\lambda })|x_2-x_1|+\frac{M^2}{m}|x_2-x_1|\\ \le {\xi }_1|x_2-x_1|.\end{array}$$

Therefore, $L\phi \in X(I;{\xi }_1,{\xi }_2)$. This implies that Lφ is strictly increasing and invertible on I, and ${(L\phi )}^{-1}\in X(I;\frac{1}{{\xi }_2},\frac{1}{{\xi }_1})$. □

Lemma 2.5 Suppose that $g\in X(I;m_1,M_1)$ and ${\phi }_0\in X(I;m,M)$. If

$$1+\frac{1}{\lambda }>max\{\frac{M^2}{m},\frac{1}{2}(\frac{M_1}{M}+\frac{m_1}{m})\},$$

(2.6)

then

$${\phi }_k:={(L{\phi }_{k-1})}^{-1}\circ g$$

(2.7)

and

$$L{\phi }_{k-1}(x):=(1+\frac{1}{\lambda })x-\frac{1}{\lambda }{\phi }_{k-1}^2(\lambda {\phi }_{k-1}^{-1}(x))$$

(2.8)

are well defined, and ${\phi }_k\in X(I;m,M)$, $k=1,2,\dots$ .

Proof Let

$$L{\phi }_0:=(1+\frac{1}{\lambda })x-\frac{1}{\lambda }{\phi }_0^2(\lambda {\phi }_0^{-1}(x)).$$

(2.9)

From Lemma 2.4, $L{\phi }_0$ is well defined and is an orientation-preserving homeomorphism from I onto itself, and ${(L{\phi }_0)}^{-1}\in X(I;\frac{1}{{\xi }_2},\frac{1}{{\xi }_1})$. Then ${\phi }_1(x):={(L{\phi }_0)}^{-1}\circ g(x)$ is well defined and ${\phi }_1\in X(I;\frac{m_1}{{\xi }_2},\frac{M_1}{{\xi }_1})\subset X(I;m,M)$ by (2.6) and Lemma 2.4. If

$$L{\phi }_k:=(1+\frac{1}{\lambda })x-\frac{1}{\lambda }{\phi }_k^2(\lambda {\phi }_k^{-1}(x))$$

(2.10)

is well defined and is an orientation-preserving homeomorphism from I onto itself, then ${(L{\phi }_k)}^{-1}\in X(I;\frac{1}{{\xi }_2},\frac{1}{{\xi }_1})$. We similarly see that

$${\phi }_{k+1}(x):={(L{\phi }_k)}^{-1}\circ g(x)$$

(2.11)

is well defined, and

$${\phi }_{k+1}\in X(I;\frac{m_k}{{\xi }_2},\frac{M_k}{{\xi }_1})\subset X(I;m,M).$$

(2.12)

This implies that the results in Lemma 2.5 are also true for $k+1$, which completes the proof. □

3 Main results

In this section, we give several important theorems on the existence, uniqueness, quasi-convex (quasi-concave), symmetry and stability of equation (1.1).

Theorem 3.1 (Existence)

Given a positive constant $M_1$ and $g(x)\in X(I;M_1)$. If there exist constants M and λ such that

$$\frac{\lambda }{\lambda +1}(M^2+M_1)\le M,$$

(3.1)

then equation (1.1) has a solution f in $X(I;M)$.

Proof Define $T:X(I;M)\to C^0(I)$ by

$$Tf(x)=\frac{1}{\lambda +1}{f}^2(\lambda x)+\frac{\lambda }{\lambda +1}g(x),\mathrm{\forall }x\in I.$$

(3.2)

Because f, $f(\lambda x)$ and g are continuous for all $x\in I$, we obtain that Tf is continuous for all $x\in I$, and $Tf\in C^0(I)$. By (3.1), for any x, y in I,

$$\begin{array}{rl}|Tf(x)-Tf(y)|& =|\frac{1}{\lambda +1}{f}^2(\lambda x)+\frac{\lambda }{\lambda +1}g(x)-\frac{1}{\lambda +1}{f}^2(\lambda y)-\frac{\lambda }{\lambda +1}g(y)|\\ \le \frac{1}{\lambda +1}|f^2(\lambda x)-f^2(\lambda y)|+\frac{\lambda }{\lambda +1}|g(x)-g(y)|\\ \le \frac{M}{\lambda +1}|f(\lambda x)-f(\lambda y)|+\frac{\lambda M_1}{\lambda +1}|x-y|\\ \le \frac{\lambda M^2}{\lambda +1}|x-y|+\frac{\lambda M_1}{\lambda +1}|x-y|\\ =\frac{\lambda }{\lambda +1}(M^2+M_1)|x-y|\\ \le M|x-y|.\end{array}$$

Thus, $T_f\in X(I,M)$. Now we prove the continuity of T under the norm ${\parallel \cdot \parallel }_{c^0}$. For arbitrary $f_1,f_2\in X(I,M)$,

$$\begin{array}{rl}{\parallel T{f}_1-T{f}_2\parallel }_{c^0}& =\underset{x\in I}{max}|T{f}_1(x)-T{f}_2(x)|\\ =\underset{x\in I}{max}|\frac{1}{\lambda +1}{f}_1^2(\lambda x)-\frac{1}{\lambda +1}{f}_2^2(\lambda x)|\\ =\frac{1}{\lambda +1}\underset{x\in I}{max}|f_1^2(\lambda x)-f_2^2(\lambda x)|\\ =\frac{1}{\lambda +1}\parallel f_1^{2(\lambda )}-f_2^{2(\lambda )}\parallel \\ \le E{\parallel f_1-f_2\parallel }_{C_0},\end{array}$$

where Lemma 2.3 is used and

$$E=\frac{M+1}{\lambda +1}.$$

(3.3)

Thus, T is continuous under the norm ${\parallel \cdot \parallel }_{c^0}$. Summarizing all the above, we see that T is a continuous map** from the compact convex subset $X(I,M)$ of the Banach space $C^0(I,R)$ into itself. By Schauder’s fixed point theorem, we assert that there is a map** $f\in X(I,M)$ such that

$$f(x)=Tf(x)=\frac{1}{\lambda +1}{f}^2(\lambda x)+\frac{\lambda }{\lambda +1}g(x),\mathrm{\forall }x\in I.$$

(3.4)

This completes the proof. □

Theorem 3.2 (Uniqueness)

Suppose that (3.1) is satisfied and

$$M<\lambda .$$

(3.5)

For any function $g\in X(I,M_1)$, equation (1.1) has a unique solution $f\in X(I,M)$.

Proof The existence of equation (1.1) in $X(I;M)$ is given by Theorem 3.1. Note that $X(I;M)$ is a closed subset of $C^0(I)$. By (3.3) and (3.5), we see that $T:X(I;M)\to X(I;M)$ is a contraction map**. Therefore T has a unique fixed point $f(x)$ in $X(I;M)$, that is, equation (1.1) has a unique solution $f(x)$ in $X(I;M)$. □

Below, we discuss the quasi-convex (or quasi-concave) solutions of equation (1.1).

Definition 3.1 Suppose that Γ is a Lie group of all linear transformations on R. A map** $f:R\to R$, is said to be Γ-equivariant [25] if $f(\gamma x)=\gamma f(x)$, $\mathrm{\forall }\gamma \in \mathrm{\Gamma }$, $\mathrm{\forall }x\in R$.

This implies that $f^i$ is the Γ-equivariant. Let $G_{\mathrm{\Gamma }}(I)=\{g\in C^0(I)\mid g(\gamma x)=\gamma g(x),\mathrm{\forall }\gamma \in \mathrm{\Gamma },\mathrm{\forall }x\in I\}$ and $G_{\mathrm{\Gamma }}(I;M)=G_{\mathrm{\Gamma }}(I)\cap X(I;M)$.

Lemma 3.1 $G_{\mathrm{\Gamma }}(I)$ is a closed convex subset of $C^0(I)$, and $G_{\mathrm{\Gamma }}(I;M)$ is a compact convex subset of $C^0(I)$.

The methods of the proofs are similar to the paper [24].

Theorem 3.3 (Quasi-convexity (Quasi-concavity))

If $g\in X_{\sigma }(I;M_1)$, (3.1) and (3.5) are satisfied, then (1.1) has a solution $f\in X_{\sigma }(I;M)$.

Proof Define a map** $T:X_{\mathit{qcv}}(I;M)\to C^0(I)$ as in (3.2). Note that each $f\in X_{\mathit{qcv}}(I;M)$ is an increasing function. In fact, if $x<y$ in I, there exists $t_0\in (0,1)$ such that $x=t_{0}a+(1-t_0)y$, and by the quasi-convexity, we get

$$f(x)\le max(f(a),f(y))=f(y).$$

(3.6)

Thus, for $f\in X_{\mathit{qcv}}(I,M)$, $x,y\in I$, and $t\in [0,1]$, we get

$$\begin{array}{rl}Tf(tx+(1-t)y)& =\frac{1}{\lambda +1}{f}^2\left(\lambda (tx+(1-t)y)\right)+\frac{\lambda }{\lambda +1}g(tx+(1-t)y)\\ \le \frac{1}{\lambda +1}f(max\{f(\lambda x),f(\lambda y)\})+\frac{\lambda }{\lambda +1}max\{g(x),g(y)\}\\ \le \frac{1}{\lambda +1}max\{f^2(\lambda x),f^2(\lambda y)\}+\frac{\lambda }{\lambda +1}max\{g(x),g(y)\}\\ \le max\{Tf(x),Tf(y)\}.\end{array}$$

Thus, T maps $X_{\mathit{qcv}}(I,M)$ into itself. Similarly, we can prove that T is continuous. By Lemma 2.1, $X_{\mathit{qcv}}(I,M)$ is a compact convex subset of the Banach space $C^0(I)$. Then Schauder’s fixed point theorem guarantees the existence of a fixed point f of T in $X_{\mathit{qcv}}(I;M)$. In the same way, the proof of the quasi-concave solution of equation (1.1) is similarly obtained. □

Now, we study the symmetric solutions of (1.1).

Theorem 3.4 (Symmetry)

If (3.1) and (3.5) are satisfied, and $g\in G_{\mathrm{\Gamma }}(I;M_1)$, then equation (1.1) has a unique Γ-equivariant solution $f\in G_{\mathrm{\Gamma }}(I;M_2)$.

Proof By Lemma 3.1 and (3.2), we have

$$\begin{array}{rl}Tf(\gamma x)& =\frac{1}{\lambda +1}{f}^2(\gamma \lambda x)+\frac{\lambda }{\lambda +1}g(\gamma x)\\ =\gamma \frac{1}{\lambda +1}{f}^2(\lambda x)+\gamma \frac{\lambda }{\lambda +1}g(x)\\ =\gamma Tf(x).\end{array}$$

From Theorem 3.1, we can find that T is a contraction map** in $G_{\mathrm{\Gamma }}(I;M_2)$. Since $G_{\mathrm{\Gamma }}(I;M_2)$ is a compact convex subset of $C^0(I)$, by Banach’s fixed point theorem, we assert that there is a unique fixed point $f\in G_{\mathrm{\Gamma }}(I;M_2)$. □

In the following, we give the conditions to guarantee two kinds of stability: the continuous dependence and the Hyers-Ulam stability [21].

Theorem 3.5 (Continuous dependence)

If (3.1) and (3.5) are satisfied, the solutions of (1.1) in $X(I;M)$ is continuously dependent on the given function $g(x)$ in $X(I;M_1)$.

Proof For $g_1,g_2\in X(I;M_1)$, Theorem 3.1 implies that there are functions $f_1,f_2\in X(I;M_1)$ such that

$$f_1(x)=\frac{1}{\lambda +1}{f}_1^2(\lambda x)+\frac{\lambda }{\lambda +1}{g}_1(x),$$

(3.7)

$$f_2(x)=\frac{1}{\lambda +1}{f}_2^2(\lambda x)+\frac{\lambda }{\lambda +1}{g}_2(x).$$

(3.8)

Thus, by Lemma 2.3

$$\begin{array}{rl}{\parallel f_1-f_2\parallel }_{c^0}& =\underset{x\in I}{max}|f_1(x)-f_2(x)|\\ \le \frac{1}{\lambda +1}\underset{x\in I}{max}|f_1^2(\lambda x)-f_2^2(\lambda x)|+\frac{\lambda }{\lambda +1}\underset{x\in I}{max}|g_1(x)-g_2(x)|\\ =\frac{1}{\lambda +1}{\parallel f_1^{2(\lambda )}-f_2^{2(\lambda )}\parallel }_{c^0}+\frac{\lambda }{\lambda +1}{\parallel g_1-g_2\parallel }_{c^0}\\ \le \frac{M+1}{\lambda +1}{\parallel f_1-f_2\parallel }_{c^0}+\frac{\lambda }{\lambda +1}{\parallel g_1-g_2\parallel }_{c^0},\end{array}$$

which implies

$$(1-\frac{M+1}{\lambda +1}){\parallel f_1-f_2\parallel }_{c^0}\le \frac{\lambda }{\lambda +1}{\parallel g_1-g_2\parallel }_{c^0},$$

so

$${\parallel f_1-f_2\parallel }_{c^0}\le \frac{\lambda }{\lambda -M}{\parallel g_1-g_2\parallel }_{c^0}.$$

(3.9)

Inequality (3.5) yields that the solution of (1.1) in $X(I;M)$ is continuously dependent on the given function g in $X(I;M_1)$. □

Definition 3.2 The functional equation

$$E_1(\phi )=E_2(\phi )$$

(3.10)

has the Hyers-Ulam stability [26] if for any approximate solution ${\phi }_s$ such as $\parallel E_1({\phi }_s)=E_2({\phi }_s)\parallel \le \delta$, there exist a solution φ of equation (3.10) such as $\parallel \phi -{\phi }_s\parallel \le \epsilon$, where $\delta \ge 0$, $\epsilon >0$, and a constant ε dependent only on δ.

Theorem 3.6 (Hyers-Ulam stability)

Suppose that $g\in X(I;m_1,M_1)$, and ${\phi }_s\in X(I;m,M)$ satisfy

$$|g(x)-(1+\frac{1}{\lambda }){\phi }_s(x)+\frac{1}{\lambda }{\phi }_s^2(\lambda x)|\le \delta ,\mathrm{\forall }x\in I,$$

(3.11)

where $\delta >0$ is a positive constant. If (2.6), (3.1) and (3.5) are satisfied, there exists a unique continuous solution $\phi \in X(I;m,M)$ of (1.1) such that

$$|\phi (x)-{\phi }_s(x)|\le \zeta \delta ,\mathrm{\forall }x\in I,$$

(3.12)

where $\zeta =\frac{1}{{\xi }_1(1-\eta )}$, $\eta =\frac{M+1}{\lambda {\xi }_1}+\frac{M^2}{m_1}<1$.

Proof Construct a sequence $\{{\phi }_k\}$ of functions as follows. Take ${\phi }_0={\phi }_s$, and then define ${\phi }_k$ as in (2.7) and $L{\phi }_k$ as in (2.8). By Lemma 2.4, both ${\phi }_k$ and $L{\phi }_k$ are well defined for any integer $k\ge 1$. Lemma 2.4 and Lemma 2.5 also imply that ${\phi }_k$ or $L{\phi }_k$ is an orientation-preserving homeomorphism from I into itself with ${(L{\phi }_k)}^{-1}\in X(I;\frac{1}{{\xi }_2},\frac{1}{{\xi }_1})$, where ${\xi }_1$ and ${\xi }_2$ are given in (2.5).

Now we claim that

$$\parallel g-L{\phi }_{k-1}\circ {\phi }_{k-1}\parallel \le {\eta }^{k-1}\delta ,$$

(3.13)

$$\parallel {\phi }_k-{\phi }_{k-1}\parallel \le {\eta }^{k-1}\frac{\delta }{{\xi }_1}$$

(3.14)

for all $x\in I$ and $k=1,2,\dots$ .

The case $k=1$ is trivial. Assume that (3.13) and (3.14) hold for k. Since

$$\begin{array}{rl}\parallel g(x)-L{\phi }_k\circ {\phi }_k\parallel & \le \parallel L{\phi }_{k-1}\circ {\phi }_k-L{\phi }_k\circ {\phi }_k\parallel \\ \le \frac{1}{\lambda }\parallel -{\phi }_{k-1}^2(\lambda {\phi }_{k-1}^{-1}({\phi }_k))+{\phi }_k^2(\lambda {\phi }_k^{-1}({\phi }_k))\parallel \\ \le \frac{1}{\lambda }\parallel {\phi }_k^2\left(\lambda {\phi }_k^{-1}\right)-{\phi }_{k-1}^2\left(\lambda {\phi }_{k-1}^{-1}\right)\parallel \\ \le \frac{1}{\lambda }(\parallel {\phi }_k^2\left(\lambda {\phi }_k^{-1}\right)-{\phi }_k^2\left(\lambda {\phi }_{k-1}^{-1}\right)\parallel +\parallel {\phi }_k^2\left(\lambda {\phi }_{k-1}^{-1}\right)-{\phi }_{k-1}^2\left(\lambda {\phi }_{k-1}^{-1}\right)\parallel )\\ \le \frac{1}{\lambda }(\lambda M^2\parallel {\phi }_k^{-1}-{\phi }_{k-1}^{-1}\parallel +(M+1)\parallel {\phi }_k-{\phi }_{k-1}\parallel ),\end{array}$$

where Lemma 2.3 is applied. From the hypothesis of induction, it follows that

$$\begin{array}{rl}\parallel g(x)-L{\phi }_k\circ {\phi }_k\parallel & \le M^2\parallel {\phi }_k^{-1}-{\phi }_{k-1}^{-1}\parallel +\frac{(M+1)}{\lambda }\parallel {\phi }_k-{\phi }_{k-1}\parallel \\ \le {(\frac{M^2}{m_1}+\frac{M+1}{\lambda {\xi }_1})}^k\delta ={\eta }^k\delta .\end{array}$$

Moreover,

$$\begin{array}{rl}\parallel {\phi }_{k+1}-{\phi }_k\parallel & =\parallel {(L{\phi }_k)}^{-1}\circ g-{(L{\phi }_k)}^{-1}\circ L{\phi }_k\circ {\phi }_k\parallel \\ \le \frac{1}{{\xi }_1}\parallel g-L{\phi }_k\circ {\phi }_k\parallel \\ \le {\eta }^k\frac{\delta }{{\xi }_1}.\end{array}$$

Thus, (3.13) and (3.14) hold.

On the other hand, for any positive integers k and l with $k>l$, by (3.14)

$$\begin{array}{rl}\parallel {\phi }_k-{\phi }_l\parallel & \le \parallel {\phi }_k-{\phi }_{k-1}\parallel +\parallel {\phi }_{k-1}-{\phi }_{k-2}\parallel +\cdots +\parallel {\phi }_{l+1}-{\phi }_l\parallel \\ \le {\eta }^{k-1}\frac{\delta }{{\xi }_1}+{\eta }^{k-2}\frac{\delta }{{\xi }_1}+\cdots +{\eta }^l\frac{\delta }{{\xi }_1}\\ \le \frac{{\eta }^k-{\eta }^l}{1-\eta }\frac{\delta }{{\xi }_1}.\end{array}$$

(3.15)

Note that $\eta <1$, so from (3.15), it follows that

$$\parallel {\phi }_k-{\phi }_l\parallel \to 0\text{as }k,l\to +\mathrm{\infty }.$$

This implies that $\{{\phi }_k(x)\}$ is a Cauchy sequence. Hence, $\{{\phi }_k(x)\}$ uniformly converges in the Banach space $C^0(I)$. Let $\phi ={lim}_{k\to +\mathrm{\infty }}{\phi }_k(x)$. Clearly, $\phi \in X(I;m,M)$. From (3.13),

$$\parallel g-L\phi \circ \phi \parallel =\underset{k\to +\mathrm{\infty }}{lim}\parallel g-L{\phi }_k\circ {\phi }_k\parallel \le \underset{k\to +\mathrm{\infty }}{lim}{\eta }^k\delta =0,$$

(3.16)

i.e., φ is a solution of (1.1). Furthermore, from (3.14)

$$\begin{array}{rl}\parallel \phi -{\phi }_s\parallel & =\underset{k\to +\mathrm{\infty }}{lim}\parallel {\phi }_k-{\phi }_0\parallel \\ \le \underset{k\to +\mathrm{\infty }}{lim}(\parallel {\phi }_k-{\phi }_{k-1}\parallel +\parallel {\phi }_{k-1}-{\phi }_{k-2}\parallel +\cdots +\parallel {\phi }_1-{\phi }_0\parallel )\\ \le \underset{k\to +\mathrm{\infty }}{lim}({\eta }^{k-1}\frac{\delta }{{\xi }_1}+{\eta }^{k-2}\frac{\delta }{{\xi }_1}+\cdots +\frac{\delta }{{\xi }_1})\\ \le \frac{\frac{\delta }{{\xi }_1}}{1-\eta }=\frac{1}{{\xi }_1(1-\eta )}\delta .\end{array}$$

Thus, $\parallel \phi -{\phi }_s\parallel <\zeta \delta$. Then (3.12) holds.

Concerning the uniqueness, we assume that there is another continuous solution $\varphi \in X(I;m,M)$ ($\varphi \ne \phi$), such that

$$|\varphi (x)-{\phi }_s(x)|\le \epsilon ,$$

where $\epsilon >0$ only depends on δ. Then

$$\begin{array}{rl}\parallel \phi -\varphi \parallel & =\parallel {(L\phi )}^{-1}\circ g-{(L\varphi )}^{-1}\circ g\parallel \\ \le \parallel {(L\phi )}^{-1}-{(L\varphi )}^{-1}\parallel \\ \le \parallel {(L\phi )}^{-1}-{(L\phi )}^{-1}\circ (L\phi )\circ {(L\varphi )}^{-1}\parallel \\ \le \frac{1}{{\xi }_1}\parallel (L\varphi )\circ {(L\varphi )}^{-1}-(L\phi )\circ {(L\varphi )}^{-1}\parallel \\ \le \frac{1}{{\xi }_1}\parallel L\phi -L\varphi \parallel \\ \le \frac{1}{\lambda {\xi }_1}(\parallel {\phi }^2\left(\lambda {\phi }^{-1}\right)-{\varphi }^2\left(\lambda {\phi }^{-1}\right)\parallel +\parallel {\varphi }^2\left(\lambda {\phi }^{-1}\right)-{\varphi }^2\left(\lambda {\varphi }^{-1}\right)\parallel )\\ \le \frac{M+1}{\lambda {\xi }_1}\parallel \phi -\varphi \parallel +\frac{M^2}{{\xi }_1}\parallel {\phi }^{-1}-{\varphi }^{-1}\parallel \\ \le \frac{M+1}{\lambda {\xi }_1}\parallel \phi -\varphi \parallel +\frac{M^2}{{\xi }_1}\parallel {\phi }^{-1}-{\phi }^{-1}\circ \phi \circ {\varphi }^{-1}\parallel \\ \le \frac{M+1}{\lambda {\xi }_1}\parallel \phi -\varphi \parallel +\frac{M^2}{m{\xi }_1}\parallel \varphi \circ {\varphi }^{-1}-\phi \circ {\varphi }^{-1}\parallel \\ \le \frac{M+1}{\lambda {\xi }_1}\parallel \phi -\varphi \parallel +\frac{M^2}{m{\xi }_1}\parallel \varphi -\phi \parallel ,\end{array}$$

that is,

$$(1-\frac{M+1}{\lambda {\xi }_1}-\frac{M^2}{m{\xi }_1})\parallel \phi -\varphi \parallel \le 0.$$

(3.17)

The assumption of Theorem 3.6 yields $\parallel \phi -\varphi \parallel =0$, i.e., $\phi =\varphi$, which contradicts with the assumption. The proof is completed. □

4 Example

Example 1 Consider the equation

$$f(x)=\frac{2}{3}{f}^2\left(\frac{1}{2}x\right)+\frac{1}{3}g(x),$$

(4.1)

where $x\in [0,1]$ and $\lambda =\frac{1}{2}$. Let

$$g(x)=\{\begin{array}{cc}\frac{1}{8}x,\hfill & 0\le x\le \frac{1}{4},\hfill \\ \frac{1}{4}x-\frac{1}{32},\hfill & \frac{1}{4}<x\le \frac{1}{2},\hfill \\ \frac{1}{8}x+\frac{1}{32},\hfill & \frac{1}{2}<x\le \frac{3}{4},\hfill \\ \frac{1}{2}x-\frac{1}{4},\hfill & \frac{3}{4}<x\le 1.\hfill \end{array}$$

(4.2)

Then $g(x)$ is quasi-convex and nonconvex (see Figure 1). Note that, for $x\in [0,\frac{1}{4}]$ and $y\in [\frac{3}{4},1]$

$$\begin{array}{rl}|f(x)-f(y)|& =|\frac{1}{2}y-\frac{1}{4}-\frac{1}{8}x|=|\frac{1}{8}(y-x)+\frac{3}{8}(y-\frac{2}{3})|\\ \le \frac{1}{8}|y-x|+\frac{3}{8}|y-x|=\frac{1}{2}|y-x|.\end{array}$$

Similarly, we can show that for any $x,y\in [0,1]$, $|f(x)-f(y)|\le \frac{1}{2}|y-x|$. Thus, $M_1=\frac{1}{2}$.

Figure 1

The graph of $\mathit{g}\mathbf{(}\mathit{x}\mathbf{)}$ .

For $x\in [\frac{1}{4},\frac{1}{2}]$ and $y\in [\frac{3}{4},1]$, we have

$$|f(x)-f(y)|=|\frac{1}{2}y-\frac{1}{4}-\frac{1}{4}x+\frac{1}{32}|\ge \frac{1}{4}|y-x|-\frac{1}{4}|y-\frac{7}{8}|.$$

If $1\ge y\ge \frac{7}{8}$, then $|f(x)-f(y)|\ge \frac{1}{4}|y-x|\ge \frac{1}{8}|y-x|$; if $\frac{7}{8}\ge y\ge \frac{3}{4}$, then

$$|f(x)-f(y)|\ge \frac{1}{4}|y-x|-\frac{1}{32}\ge \frac{3}{16}|y-x|\ge \frac{1}{8}|y-x|,$$

since $\frac{1}{2}\le |y-x|\le \frac{3}{4}$. Similarly, we can show that for any $x,y\in [0,1]$, $|f(x)-f(y)|\ge \frac{1}{8}|y-x|$. Thus, $m_1=\frac{1}{8}$. Therefore, we can get a quasi-convex solution $f(x)$ of equation (4.1) by Theorem 3.3, which is continuously dependent on the given function $g(x)\in X(I;M_1)$ with $M=\frac{3-\sqrt{7}}{2}$ by Theorem 3.5. Moreover, equation (4.1) satisfies the Hyers-Ulam stability in $X(I;\frac{131-49\sqrt{7}}{152-48\sqrt{7}},\frac{3-\sqrt{7}}{2})$ by Theorem 3.6.