Fixed point theorems for cyclic self-maps involving weaker Meir-Keeler functions in complete metric spaces and applications

Abstract

We obtain fixed point theorems for cyclic self-maps on complete metric spaces involving Meir-Keeler and weaker Meir-Keeler functions, respectively. In this way, we extend several well-known fixed point theorems and, in particular, improve some very recent results on weaker Meir-Keeler functions. Fixed point results for well-posed property and for limit shadowing property are also deduced. Finally, an application to the study of existence and uniqueness of solutions for a class of nonlinear integral equations is presented.

MSC:47H10, 54H25, 54E50, 45G10.

1 Introduction

In their paper [1], Kirk, Srinavasan and Veeramani started the fixed point theory for cyclic self-maps on (complete) metric spaces. In particular, they obtained, among others, cyclic versions of the Banach contraction principle [2], of the Boyd and Wong fixed point theorem [3] and of the Caristi fixed point theorem [4]. From then, several authors have contributed to the study of fixed point theorems and best proximity points for cyclic contractions (see, e.g., [5–13]). Very recently, Chen [14] (see also [15]) introduced the notion of a weaker Meir-Keeler function and obtained some fixed point theorems for cyclic contractions involving weaker Meir-Keeler functions.

In this paper we obtain a fixed point theorem for cyclic self-maps on complete metric spaces involving Meir-Keeler functions and deduce a variant of it for weaker Meir-Keeler functions. In this way, we extend in several directions and improve, among others, the main fixed point theorem of Chen’s paper [[14], Theorem 3]. Some consequences are given after the main results. Fixed point results for well-posedness property and for limit shadowing property in complete metric spaces are also given. Finally, an application to the study of existence and uniqueness of solution for a class of nonlinear integral equations is presented.

We recall that a self-map f of a (non-empty) set X is called a cyclic map if there exists $m\in \mathbb{N}$ such that $X={\bigcup }_{i=1}^m{A}_i$, with $A_i$ non-empty and $f(A_i)\subseteq A_{i+1}$, $i=1,\dots ,m$, where $A_{m+1}=A_1$.

In this case, we say that $X={\bigcup }_{i=1}^m{A}_i$ is a cyclic representation of X with respect to f.

2 Fixed point results

In the sequel, the letters ℝ, ${\mathbb{R}}^{+}$ and ℕ will denote the set of real numbers, the set of non-negative real numbers and the set of positive integer numbers, respectively.

Meir and Keeler proved in [16] that if f is a self-map of a complete metric space $(X,d)$ satisfying the condition that for each $\epsilon >0$ there is $\delta >0$ such that, for any $x,y\in X$, with $\epsilon \le d(x,y)<\epsilon +\delta$, we have $d(fx,fy)<\epsilon$, then f has a unique fixed point $z\in X$ and $f^{n}x\to z$ for all $x\in X$.

This important result suggests the notion of a Meir-Keeler function:

A function $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is said to be a Meir-Keeler function if for each $\epsilon >0$, there exists $\delta >0$ such that for $t>0$ with $\epsilon \le t<\epsilon +\delta$, we have $\varphi (t)<\epsilon$.

Remark 1 It is obvious that if ϕ is a Meir-Keeler function, then $\varphi (t)<t$ for all $t>0$.

In [14], Chen introduced the following interesting generalization of the notion of a Meir-Keeler function.

Definition 1 [[14], Definition 3]

A function $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is called a weaker Meir-Keeler function if for each $\epsilon >0$, there exists $\delta >0$ such that for $t>0$ with $\epsilon \le t<\epsilon +\delta$, there exists $n_0\in \mathbb{N}$ such that ${\varphi }^{n_0}(t)<\epsilon$.

Now let $\varphi ,\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$. According to Chen [[14], Section 2], consider the following conditions for ϕ and φ, respectively.

• (${\varphi }_1$) $\varphi (t)=0\iff t=0$;

• (${\varphi }_2$) for all $t>0$, the sequence ${\{{\varphi }^n(t)\}}_{n\in \mathbb{N}}$ is decreasing;

• (${\varphi }_3$) for $t_n>0$,

  • (a) if ${lim}_{n\to \mathrm{\infty }}{t}_n=\gamma >0$, then ${lim}_{n\to \mathrm{\infty }}\varphi (t_n)<\gamma$, and

  • (b) if ${lim}_{n\to \mathrm{\infty }}{t}_n=0$, then ${lim}_{n\to \mathrm{\infty }}\varphi (t_n)=0$;

• (${\phi }_1$) φ is non-decreasing and continuous with $\phi (t)=0\iff t=0$;

• (${\phi }_2$) φ is subadditive, that is, for every $t_1,t_2\in {\mathbb{R}}^{+}$, $\phi (t_1+t_2)\le \phi (t_1)+\phi (t_2)$;

• (${\phi }_3$) for $t_n>0$, ${lim}_{n\to \mathrm{\infty }}{t}_n=0$ if and only if ${lim}_{n\to \mathrm{\infty }}\phi (t_n)=0$.

Definition 2 [[14], Definition 4]

Let $(X,d)$ be a metric space. A self-map f of X is called a cyclic weaker $(\varphi \circ \phi )$-contraction if there exist $m\in \mathbb{N}$, for which $X={\bigcup }_{i=1}^m{A}_i$ (each $A_i$ a non-empty closed set), and two functions $\varphi ,\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfying conditions (${\varphi }_i$), $i=1,2,3$, and (${\phi }_i$), $i=1,2,3$, respectively, with ϕ a weaker Meir-Keeler function such that

• (1) $X={\bigcup }_{i=1}^m{A}_i$ is a cyclic representation of X with respect to f;

• (2) for any $x\in A_i$, $y\in A_{i+1}$, $i=1,2,\dots ,m$,

  $$\phi (d(fx,fy))\le \varphi \left(\phi (d(x,y))\right),$$

where $A_{m+1}=A_1$.

By using the above concept, Chen established the following fixed point theorem.

Theorem 1 [[14], Theorem 3]

Let $(X,d)$ be a complete metric space. Then every cyclic weaker $(\varphi \circ \phi )$-contraction f of X has a unique fixed point z. Moreover, $z\in {\bigcap }_{i=1}^m{A}_i$, where $X={\bigcup }_{i=1}^m{A}_i$ is the cyclic representation of X with respect to f of Definition  2.

We shall establish fixed point theorems which improve in several directions the preceding theorem. To this end, we start by obtaining a fixed point theorem for cyclic contractions involving Meir-Keeler functions.

Theorem 2 Let f be a self-map of a complete metric space $(X,d)$, and let $X={\bigcup }_{i=1}^m{A}_i$ be a cyclic representation of X with respect to f, with $A_i$ non-empty and closed, $i=1,\dots ,m$. If $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a Meir-Keeler function such that for any $x\in A_i$, $y\in A_{i+1}$, $i=1,2,\dots ,m$,

$$d(fx,fy)\le \varphi (d(x,y)),$$

where $A_{m+1}=A_1$, then f has a unique fixed point z. Moreover, $z\in {\bigcap }_{i=1}^m{A}_i$.

Proof Let $x_0\in A_m$. For each $n\in \mathbb{N}\cup \{0\}$, put $x_n=f^n{x}_0$. Note that $x_{nm+i}\in A_i$ whenever $n\in \mathbb{N}\cup \{0\}$ and $i=1,2,\dots ,m$.

If $x_{n_0}=x_{n_0+1}$ for some $n_0$, then $x_n$ is a fixed point of f. So, we assume that $x_n\ne x_{n+1}$ for all $n\in \mathbb{N}\cup \{0\}$. By Remark 1 and the contraction condition, it follows that ${\{d(x_n,x_{n+1})\}}_{n\in \mathbb{N}}$ is a strictly decreasing sequence in ${\mathbb{R}}^{+}$, so there exists $r\in {\mathbb{R}}^{+}$ such that ${lim}_{n\to \mathrm{\infty }}d(x_n,x_{n+1})=r$. If $r>0$, there is $n_0\in \mathbb{N}$ such that $\varphi (d(x_n,x_{n+1}))<r$ for all $n\ge n_0$ by our assumption that ϕ is a Meir-Keeler function. Hence, $d(x_{n+1},x_{n+2})<r$ for all $n\ge n_0$, a contradiction. Therefore ${lim}_{n\to \mathrm{\infty }}d(x_n,x_{n+1})=0$.

Next we prove that ${\{x_n\}}_{n\in \mathbb{N}}$ is a Cauchy sequence in $(X,d)$. Choose an arbitrary $\epsilon >0$. Then, there is $\delta \in (0,\epsilon )$ such that for $t>0$ with $\epsilon \le t<\epsilon +\delta$, we have $\varphi (t)<\epsilon$. Let $k_0\in \mathbb{N}$ be such that $d(x_k,x_{k+1})<\delta /2$, $d(x_k,x_{k+m-1})<\epsilon /2$ and $d(x_k,x_{k+m+1})<\delta /2$ for all $k\ge k_0$.

Take any $k>k_0$. Then $k=nm+i$ for some $n\in \mathbb{N}$ and some $i\in \{1,2,\dots ,m\}$. By induction we shall show that $d(x_{nm+i},x_{(n+j)m+i+1})<\epsilon$ for all $j\in \mathbb{N}$.

Indeed, for $j=1$, we have

$$d(x_{nm+i},x_{nm+i+m+1})=d(x_k,x_{k+m+1})<\frac{\delta }{2}<\epsilon .$$

Now, assume that $d(x_{nm+i},x_{(n+j)m+i+1})<\epsilon$ for some $j\in \mathbb{N}$. Thus

$$\begin{array}{rcl}d(x_{nm+i-1},x_{(n+j+1)m+i})& \le & d(x_{nm+i-1},x_{nm+i})+d(x_{nm+i},x_{(n+j)m+i+1})\\ +d(x_{(n+j)m+i+1},x_{(n+j+1)m+i})\\ <& \frac{\delta }{2}+\epsilon +\frac{\delta }{2}=\delta +\epsilon .\end{array}$$

If $\epsilon \le d(x_{nm+i-1},x_{(n+j+1)m+i})$, then $\varphi (d(x_{nm+i-1},x_{(n+j+1)m+i}))<\epsilon$, and, by the contraction condition,

$$d(x_{nm+i},x_{(n+j+1)m+i+1})<\epsilon .$$

If $d(x_{nm+i-1},x_{(n+j+1)m+i})<\epsilon$, we deduce

$$\begin{array}{rcl}d(x_{nm+i},x_{(n+j+1)m+i+1})& \le & \varphi (d(x_{nm+i-1},x_{(n+j+1)m+i}))\\ <& d(x_{nm+i-1},x_{(n+j+1)m+i})<\epsilon .\end{array}$$

It immediately follows that ${\{x_n\}}_{n\in \mathbb{N}}$ is a Cauchy sequence in $(X,d)$. Hence, there exists $z\in X$ such that $x_n\to z$. Since each $A_i$ is closed, we deduce that $z\in {\bigcap }_{i=1}^m{A}_i$.

Moreover, $z=fz$. Indeed, let $i_0\in \{1,\dots ,m\}$ be such that $fz\in A_{i_0+1}$. Then

$$\begin{array}{rcl}d(z,fz)& \le & d(z,x_{nm+i_0})+d(x_{nm+i_0},fz)\le d(z,x_{nm+i_0})+\varphi (d(x_{nm+i_0-1},z))\\ <& d(z,x_{nm+i_0})+d(x_{nm+i_0-1},z),\end{array}$$

for all $n\in \mathbb{N}$. Since ${lim}_{n\to \mathrm{\infty }}d(z,x_{nm+i_0})={lim}_{n\to \mathrm{\infty }}d(z,x_{nm+i_0-1})=0$, it follows that $d(z,fz)=0$, i.e., $z=fz$.

Finally, let $u\in X$ with $u=fu$ and $u\ne z$. Since $z\in {\bigcap }_{i=1}^m{A}_i$, we have $d(fz,fu)\le \varphi (d(z,u))$, so $d(z,u)<d(z,u)$, a contradiction. Hence $u=z$, and thus z is the unique fixed point of f. □

Next we analyze some relations between Chen’s conditions (${\varphi }_i$), $i=1,2,3$.

Lemma 1 If $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfies (${\varphi }_3$)(a), then ϕ is a Meir-Keeler function that satisfies conditions (${\varphi }_2$) and (${\varphi }_3$)(b).

Proof Suppose that ϕ is not a Meir-Keeler function. Then there exists $\epsilon >0$ such that for each $n\in \mathbb{N}$ we can find a $t_n>0$ with $\epsilon \le t_n<\epsilon +1/n$ and $\varphi (t_n)\ge \epsilon$. Then ${lim}_{n\to \mathrm{\infty }}{t}_n=\epsilon >0$, but $\varphi (t_n)\ge \epsilon$ for all n, so condition (${\varphi }_3$)(a) is not satisfied. We conclude that condition (${\varphi }_3$)(a) implies that ϕ is a Meir-Keeler function. Hence, by Remark 1, $\varphi (t)<t$ for all $t>0$, so the sequence ${\{{\varphi }^n(t)\}}_{n\in \mathbb{N}}$ is (strictly) decreasing for all $t>0$, and thus condition (${\varphi }_2$) is satisfied. Finally, if ${lim}_{n\to \mathrm{\infty }}{t}_n=0$, with $t_n>0$, we deduce that ${lim}_{n\to \mathrm{\infty }}\varphi (t_n)=0$ because $\varphi (t_n)<t_n$ for all n, so condition (${\varphi }_3$)(b) also holds. □

Proposition 1 Let $\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ be a function satisfying conditions (${\phi }_1$) and (${\phi }_2$). If $(X,d)$ is a metric space, then the function $p:X\times X\to {\mathbb{R}}^{+}$, given by

$$p(x,y)=\phi (d(x,y)),$$

is a metric on X. If, in addition, $(X,d)$ is complete and φ satisfies condition (${\phi }_3$), then the metric space $(X,p)$ is complete.

Proof We first show that p is a metric on X. Let $x,y,z\in X$:

• Suppose $p(x,y)=0$. Then $\phi (d(x,y))=0$, so $d(x,y)=0$ by (${\phi }_1$). Hence $x=y$.

• Clearly, $p(x,y)=p(y,x)$.

• Since $d(x,y)\le d(x,z)+d(z,y)$, and φ is non-decreasing and subadditive, we deduce that $\phi (d(x,y))\le \phi (d(x,z))+\phi (d(z,y))$, i.e., $p(x,y)\le p(x,z)+p(z,y)$.

Finally, suppose that $(X,d)$ is complete with φ satisfying (${\phi }_i$), $i=1,2,3$. Let ${\{x_n\}}_{n\in \mathbb{N}}$ be a Cauchy sequence in $(X,p)$. If ${\{x_n\}}_{n\in \mathbb{N}}$ is not a Cauchy sequence in $(X,d)$, there exist $\epsilon >0$ and sequences ${\{n_k\}}_{k\in \mathbb{N}}$ and ${\{m_k\}}_{k\in \mathbb{N}}$ in ℕ such that $k<n_k<m_k<n_{k+1}$ and $d(x_{n_k},x_{m_k})\ge \epsilon$ for all $k\in \mathbb{N}$. By (${\phi }_3$), the sequence ${\{p(x_{n_k},x_{m_k})\}}_{k\in \mathbb{N}}$ does not converge to zero, which contradicts the fact that ${\{x_n\}}_{n\in \mathbb{N}}$ is a Cauchy sequence in $(X,p)$. Consequently, ${\{x_n\}}_{n\in \mathbb{N}}$ is a Cauchy sequence in $(X,d)$, so it converges in $(X,d)$ to some $x\in X$. From (${\phi }_3$) we deduce that ${\{x_n\}}_{n\in \mathbb{N}}$ converges to x in $(X,p)$. Therefore $(X,p)$ is a complete metric space. □

Remark 2 Note that the continuity of φ is not used in the preceding proposition.

Now we easily deduce the following improvement of Chen’s theorem.

Theorem 3 Let f be a self-map of a complete metric space $(X,d)$, and let $X={\bigcup }_{i=1}^m{A}_i$ be a cyclic representation of X with respect to f, with $A_i$ non-empty and closed, $i=1,\dots ,m$. If $\varphi ,\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfy conditions (${\varphi }_3$)(a) and (${\phi }_i$), $i=1,2,3$, respectively, and for any $x\in A_i$, $y\in A_{i+1}$, $i=1,2,\dots ,m$, it follows

$$\phi (d(fx,fy))\le \varphi \left(\phi (d(x,y))\right),$$

where $A_{m+1}=A_1$, then f has a unique fixed point z. Moreover, $z\in {\bigcap }_{i=1}^m{A}_i$.

Proof Define $p:X\times X\to {\mathbb{R}}^{+}$ by $p(x,y)=\phi (d(x,y))$ for all $x,y\in X$. By Proposition 1, $(X,p)$ is a complete metric space. Moreover, from the condition

$$\phi (d(fx,fy))\le \varphi \left(\phi (d(x,y))\right),$$

for all $x\in A_i$, $y\in A_{i+1}$, $i=1,\dots ,m$, it follows that

$$p(fx,fy)=\phi (d(fx,fy))\le \varphi \left(\phi (d(x,y))\right)=\varphi (p(x,y))$$

for all $x\in A_i$, $y\in A_{i+1}$, $i=1,\dots ,m$.

Finally, since by Lemma 1 ϕ is a Meir-Keeler function, we can apply Theorem 2, so there exists $z\in {\bigcap }_{i=1}^m{A}_i$, which is the unique fixed point of f. □

Note that the continuity of φ can be omitted in Theorem 3. Moreover, the condition that ϕ is a weaker Meir-Keeler function turns out to be irrelevant by virtue of Lemma 1. This fact suggests the question of obtaining a fixed point theorem for cyclic contractions involving explicitly weaker Meir-Keeler functions. In particular, it is natural to wonder if Theorem 2 remains valid when we replace ‘Meir-Keeler function’ by ‘weaker Meir-Keeler function’. In the sequel we answer this question. First we give an easy example which shows that it has a negative answer in general, but the answer is positive whenever the weaker Meir-Keeler function is non-decreasing as Theorem 5 below shows.

Example 1 Let $X=\{0,1\}$ and let d be the discrete metric on X, i.e., $d(0,0)=d(1,1)=0$ and $d(x,y)=1$ otherwise. Of course $(X,d)$ is a complete metric space. Define $f:X\to X$ by $f0=1$ and $f1=0$, and consider the function $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ defined by $\varphi (t)=t/2$ for all $t\in [0,1)$, $\varphi (1)=2$ and $\varphi (t)=1/2$ for all $t>1$. Clearly, ϕ is a weaker Meir-Keeler function (note, in particular, that ${\varphi }^2(1)=1/2<1$), but it is not a Meir-Keeler function because $\varphi (1)>1$. Finally, since $d(f0,f1)=1$ and $\varphi (d(0,1))=2$, we deduce that $d(fx,fy)\le \varphi (d(x,y))$ for all $x,y\in X$. However, f has no fixed point.

The function ϕ of the preceding example is not non-decreasing. This fact is not casual as Theorem 5 below shows.

Lemma 2 Let $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ be a non-decreasing weaker Meir-Keeler function. Then the following hold:

• (i) $\varphi (t)<t$ for all $t>0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\varphi }^n(t)=0$ for all $t>0$.

Proof (i) Suppose that there exists $t_0>0$ such that $t_0\le \varphi (t_0)$. Since ϕ is non-decreasing, we deduce that ${\{{\varphi }^n(t_0)\}}_{n\in \mathbb{N}\cup \{0\}}$ is a non-decreasing sequence in ${\mathbb{R}}^{+}$, so, in particular, $t_0\le {\varphi }^n(t_0)$ for all $n\in \mathbb{N}$. Finally, since ϕ is a weaker Meir-Keeler function, there exists $n_0\in \mathbb{N}$ such that ${\varphi }^{n_0}(t_0)<t_0$, which yields a contradiction.

(ii) Fix $t>0$. By (i) the sequence ${\{{\varphi }^n(t)\}}_{n\in \mathbb{N}}$ is (strictly) decreasing, so there exists $r\ge 0$ such that $r={lim}_{n\to \mathrm{\infty }}{\varphi }^n(t)$. If $r>0$, there is $\delta >0$ such that for $s>0$ with $r\le s<r+\delta$, there exists $n_s\in \mathbb{N}$ with ${\varphi }^{n_s}(s)<r$. Let $n_r\in \mathbb{N}$ be such that $r<{\varphi }^n(t)<r+\delta$ for all $n\ge n_r$. Putting $s={\varphi }^{n_r}(t)$, we deduce that ${\varphi }^{n_s}(s)<r$, i.e., ${\varphi }^{n_s+n_r}(t)<r$, a contradiction. We conclude that ${lim}_{n\to \mathrm{\infty }}{\varphi }^n(t)=0$. □

Remark 3 Observe that, as a partial converse of the above lemma, if $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfies ${lim}_{n\to \mathrm{\infty }}{\varphi }^n(t)=0$ for all $t>0$, then ϕ is a weaker Meir-Keeler function. Indeed, otherwise, there exist $\epsilon >0$ and a sequence ${\{t_n\}}_{n\in \mathbb{N}}$ with $t_n\ge \epsilon$ for all $n\in \mathbb{N}$, ${lim}_{n\to \mathrm{\infty }}{t}_n=\epsilon$ but ${\varphi }^k(t_n)\ge \epsilon$ for all $k,n\in \mathbb{N}$, a contradiction.

We also will use the following cyclic extension of the celebrated Matkowski fixed point theorem [[17], Theorem 1.2], where for a self-map f of a metric space $(X,d)$, we define

$$M_d(x,y)=max\{d(x,y),d(x,fx),d(y,fy),\frac{1}{2}[d(x,fy)+d(fx,y)]\}$$

for all $x,y\in X$.

Theorem 4 (cf. [[18], Corollary 2.14])

Let f be a self-map of a complete metric space $(X,d)$, and let $X={\bigcup }_{i=1}^m{A}_i$ be a cyclic representation of X with respect to f, with $A_i$ non-empty and closed, $i=1,\dots ,m$. If $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing function such that ${lim}_{n\to \mathrm{\infty }}{\varphi }^n(t)=0$ for all $t>0$, and for any $x\in A_i$, $y\in A_{i+1}$, $i=1,2,\dots ,m$,

$$d(fx,fy)\le \varphi (M_d(x,y)),$$

where $A_{m+1}=A_1$, then f has a unique fixed point z. Moreover, $z\in {\bigcap }_{i=1}^m{A}_i$.

Then from Lemma 2 and Theorem 4 we immediately deduce the following theorem.

Theorem 5 Let f be a self-map of a complete metric space $(X,d)$, and let $X={\bigcup }_{i=1}^m{A}_i$ be a cyclic representation of X with respect to f, with $A_i$ non-empty and closed, $i=1,\dots ,m$. If $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing weaker Meir-Keeler function such that for any $x\in A_i$, $y\in A_{i+1}$, $i=1,2,\dots ,m$,

$$d(fx,fy)\le \varphi (M_d(x,y)),$$

where $A_{m+1}=A_1$, then f has a unique fixed point z. Moreover, $z\in {\bigcap }_{i=1}^m{A}_i$.

Corollary Let f be a self-map of a complete metric space $(X,d)$, and let $X={\bigcup }_{i=1}^m{A}_i$ be a cyclic representation of X with respect to f, with $A_i$ non-empty and closed, $i=1,\dots ,m$. If $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing weaker Meir-Keeler function such that for any $x\in A_i$, $y\in A_{i+1}$, $i=1,2,\dots ,m$,

$$d(fx,fy)\le \varphi (d(x,y)),$$

where $A_{m+1}=A_1$, then f has a unique fixed point z. Moreover, $z\in {\bigcap }_{i=1}^m{A}_i$.

Proof Since ϕ is non-decreasing, we deduce that for each $x,y\in X$, $\varphi (d(x,y))\le \varphi (M_d(x,y))$, so $d(fx,fy)\le \varphi (M_d(x,y))$. Hence, by Theorem 5, f has a unique fixed point z and $z\in {\bigcap }_{i=1}^m{A}_i$. □

Theorem 5 can be generalized according to the style of Chen’s theorem as follows.

Theorem 6 Let f be a self-map of a complete metric space $(X,d)$, and let $X={\bigcup }_{i=1}^m{A}_i$ be a cyclic representation of X with respect to f, with $A_i$ non-empty and closed, $i=1,\dots ,m$. If $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing weaker Meir-Keeler function, $\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a function satisfying conditions (${\phi }_i$), $i=1,2,3$, and for any $x\in A_i$, $y\in A_{i+1}$, $i=1,2,\dots ,m$, it follows

$$\phi (d(fx,fy))\le \varphi \left(\phi (M_d(x,y))\right),$$

where $A_{m+1}=A_1$, then f has a unique fixed point z. Moreover, $z\in {\bigcap }_{i=1}^m{A}_i$.

Proof Construct the complete metric space $(X,p)$ of Proposition 1, and observe that from the well-known fact that for $a_i\in {\mathbb{R}}^{+}$, $i=1,\dots ,k$, one has $\varphi ({max}_i{a}_i)={max}_i\varphi (a_i)$, one has

$$M_p(x,y)=\phi (M_d(x,y))$$

for all $x,y\in X$. Therefore, for any $x\in A_i$, $y\in A_{i+1}$, $i=1,2,\dots ,m$, we deduce that

$$p(fx,fy)\le \varphi (M_p(x,y)).$$

Theorem 5 concludes the proof. □

We finish this section with two examples illustrating Theorem 5 and its corollary.

Example 2 Let $A=\{n\in \mathbb{N}:n\text{ is even}\}\cup \{0\}$, $B=\{n\in \mathbb{N}:n\text{ is odd}\}\cup \{0\}$, $X=A\cup B=\mathbb{N}$, and let d be the complete metric on X defined by $d(x,x)=0$ for all $x\in X$ and $d(x,y)=x+y$ otherwise. Since d induces the discrete topology on X, we deduce that A and B are closed subsets of $(X,d)$.

Let f be the self-map of X defined by $f0=0$ and $fx=x-1$ otherwise. It is clear that $X=A\cup B$ is a cyclic representation of X with respect to f.

Now we define the function $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ by $\varphi (0)=0$, and $\varphi (t)=n-1$ if $t\in (n-1,n]$, $n\in \mathbb{N}$. It is immediate to check that ϕ is a non-decreasing weaker Meir-Keeler function which is not a Meir-Keeler function.

Furthermore, we have:

• For $x=0$ and $y=1$, $d(fx,fy)=d(0,0)=0$.

• For $x=0$ and $y=n\in \mathbb{N}\mathrm{\setminus }\{1\}$,

  $$d(fx,fy)=d(0,n-1)=n-1=\varphi (n)=\varphi (d(x,y)).$$

• For $x=n\in A\mathrm{\setminus }\{0\}$ and $y=m\in B\mathrm{\setminus }\{0\}$,

  $$\begin{array}{rcl}d(fx,fy)& =& d(n-1,m-1)=n+m-2<n+m-1\\ =& \varphi (n+m)=\varphi (d(x,y)).\end{array}$$

Consequently, the conditions of the corollary of Theorem 5 are verified; in fact, $z=0\in A\cap B$ is the unique fixed point of f.

Example 3 Let $A=[0,1/2]\cup \{1\}$, $B=[1,2]$, $X=A\cup B$ and let d be the restriction to X of the Euclidean metric on ℝ. Obviously, $(X,d)$ is a complete metric space (in fact, it is compact), with A and B closed subsets of $(X,d)$.

Let f be the self-map of X defined by $fx=2-x$ if $x\in A$, and $fx=1$ if $x\in B$. It is clear that $X=A\cup B$ is a cyclic representation of X with respect to f.

Now we define the function $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ by $\varphi (t)=t/2$ if $t\in [0,1]$, and $\varphi (t)=1$ if $t>1$. (Notice that ϕ is a non-decreasing weaker Meir-Keeler function which is not a Meir-Keeler function.)

Furthermore, we have:

• For $x=1\in A$ and $y\in B$, $d(fx,fy)=d(1,1)=0$.

• For $x=1/2\in A$ and $y\in B$,

  $$d(fx,fy)=d(3/2,1)=1/2=\varphi (1)=\varphi (d(x,fx)).$$

• For $x\in A\mathrm{\setminus }\{1,1/2\}$ and $y\in B$,

  $$d(fx,fy)=d(2-x,1)=1-x\le 1=\varphi (2-2x)=\varphi (d(x,fx)).$$

Consequently, the conditions of Theorem 5 are verified; in fact, $z=1\in A\cap B$ is the unique fixed point of f.

Finally, observe that the corollary of Theorem 5 cannot be applied in this case because for $x=1/2\in A$ and $y=1\in B$, we have

$$d(fx,fy)=1/2>\varphi (1/2)=\varphi (d(x,y)).$$

3 Applications to well-posedness and limit shadowing property of a fixed point problem

The notion of well-posedness of a fixed point problem has evoked much interest to several mathematicians, for example, De Blasi and Myjak [19], Lahiri and Das [20], Popa [21, 22] and others.

Definition 3 [19]

Let f be a self-map of a metric space $(X,d)$. The fixed point problem of f is said to be well posed if:

• (i) f has a unique fixed point $z\in X$;

• (ii) for any sequence ${\{x_n\}}_{n\in \mathbb{N}}$ in X such that ${lim}_{n\to \mathrm{\infty }}d(f{x}_n,x_n)=0$, we have ${lim}_{n\to \mathrm{\infty }}d(x_n,z)=0$.

Definition 4 [22]

Let f be a self-map of a metric space $(X,d)$. The fixed point problem of f is said to have limit shadowing property in X if for any sequence ${\{x_n\}}_{n\in \mathbb{N}}$ in X satisfying ${lim}_{n\to \mathrm{\infty }}d(f{x}_n,x_n)=0$, it follows that there exists $z\in X$ such that ${lim}_{n\to \mathrm{\infty }}d(f^{n}z,x_n)=0$.

Concerning the well-posedness and limit shadowing of the fixed point problem for a self-map of a complete metric space satisfying the conditions of Theorem 5, we have the following results.

Theorem 7 Let $(X,d)$ be a complete metric space. If f is a self-map of X and $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing weaker Meir-Keeler function satisfying the conditions of Theorem  5, then the fixed point problem of f is well posed.

Proof Owing to Theorem 5, we know that f has a unique fixed point $z\in X$. Let $\{x_n\}$ be a sequence in X such that ${lim}_{n\to \mathrm{\infty }}d(x_n,f{x}_n)=0$. Then

$$\begin{array}{rcl}d(x_n,z)& \le & d(x_n,f{x}_n)+d(f{x}_n,fz)\\ \le & d(x_n,f{x}_n)\\ +\varphi (max\{d(x_n,z),d(x_n,x_{n+1}),d(z,fz),\frac{d(x_n,fz)+d(z,x_{n+1})}{2}\}).\end{array}$$

Passing to the limit as $n\to \mathrm{\infty }$ in the above inequality, it follows that ${lim}_{n\to \mathrm{\infty }}d(x_n,z)=0$. □

Theorem 8 Let $(X,d)$ be a complete metric space. If f is a self-map of X and $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing weaker Meir-Keeler function satisfying the conditions of Theorem  5, then f has the limit shadowing property.

Proof From Theorem 5, we know that f has a unique fixed point $z\in X$. Let ${\{x_n\}}_{n\in \mathbb{N}}$ be a sequence in X such that ${lim}_{n\to \mathrm{\infty }}d(x_n,f{x}_n)=0$. Then, as in the proof of the previous theorem,

$$\begin{array}{rcl}d(x_n,z)& \le & d(x_n,f{x}_n)\\ +\varphi (max\{d(x_n,z),d(x_n,x_{n+1}),d(z,fz),\frac{d(x_n,fz)+d(z,x_{n+1})}{2}\}).\end{array}$$

Passing to the limit as $n\to \mathrm{\infty }$ in the above inequality, it follows that ${lim}_{n\to \mathrm{\infty }}d(x_n,f^{n}z)=d(x_n,z)=0$. □

4 An application to integral equations

In this section we apply Theorem 5 to study the existence and uniqueness of solutions for a class of nonlinear integral equations.

We consider the nonlinear integral equation

$$u(t)={\int }_0^{T}G(t,s)K(s,u(s))ds\text{for all }t\in [0,T],$$

(1)

where $T>0$, $K:[0,T]\times {\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ and $G:[0,T]\times [0,T]\to {\mathbb{R}}^{+}$ are continuous functions, and $M:={max}_{(s,x)\in {[0,T]}^2}K(s,x)$.

We shall suppose that the following four conditions are satisfied:

• (I) ${\int }_0^{T}G(t,s)ds\le 1$ for all $t\in [0,T]$.

• (II) $K(s,\cdot )$ is a non-increasing function for any fixed $s\in [0,1]$, that is,

  $$x,y\in {\mathbb{R}}^{+},x\ge y⟹K(s,x)\le K(s,y).$$

• (III) There exists a Meir-Keeler function $\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ that is non-decreasing on $[0,2M]$ and such that

  $$|K(s,x)-K(s,y)|\le \psi (|x-y|)$$

  for all $s,x\in [0,T]$ and $y\in {\mathbb{R}}^{+}$ with $|x-y|\le 2M$.

• (IV) There exists a continuous function $\alpha :[0,T]\to [0,T]$ such that:

  For all $t\in [0,T]$, we have

  $$\alpha (t)\le {\int }_0^{T}G(t,s)K(s,T)ds$$

  and

  $$T\ge {\int }_0^{T}G(t,s)K(s,\alpha (s))ds.$$

Now denote by $C([0,T],{\mathbb{R}}^{+})$ the set of non-negative real continuous functions on $[0,T]$. We endow $C([0,T],{\mathbb{R}}^{+})$ with the supremum metric

$$d_{\mathrm{\infty }}(u,v)=\underset{t\in [0,T]}{max}|u(t)-v(t)|,\text{for all }u,v\in C([0,T],{\mathbb{R}}^{+}).$$

It is well known that $(C([0,T],{\mathbb{R}}^{+}),d_{\mathrm{\infty }})$ is a complete metric space.

Consider the self-map $f:C([0,T],{\mathbb{R}}^{+})\to C([0,T],{\mathbb{R}}^{+})$ defined by

$$fu(t)={\int }_0^{T}G(t,s)K(s,u(s))ds\text{for all }t\in [0,T].$$

Clearly, u is a solution of (1) if and only if u is a fixed point of f.

In order to prove the existence of a (unique) fixed point of f, we construct the closed subsets $A_1$ and $A_2$ of $C([0,T],{\mathbb{R}}^{+})$ as follows:

$$A_1=\{u\in C([0,T],{\mathbb{R}}^{+}):u(s)\le T\text{ for all }s\in [0,T]\},$$

and

$$A_2=\{u\in C([0,T],{\mathbb{R}}^{+}):u\ge \alpha \}.$$

We shall prove that

$$f(A_1)\subseteq A_2\text{and}f(A_2)\subseteq A_1.$$

(2)

Let $u\in A_1$, that is,

$$u(s)\le T\text{for all }s\in [0,T].$$

Since $G(t,s)\ge 0$ for all $t,s\in [0,T]$, we deduce from (II) and (IV) that

$${\int }_0^{T}G(t,s)K(s,u(s))ds\ge {\int }_0^{T}G(t,s)K(s,T)ds\ge \alpha (t)$$

for all $t\in [0,T]$. Then we have $fu\in A_2$.

Similarly, let $u\in A_2$, that is,

$$u(s)\ge \alpha (s)\text{for all }s\in [0,T].$$

Again, from (II) and (IV), we deduce that

$${\int }_0^{T}G(t,s)K(s,u(s))ds\le {\int }_0^{T}G(t,s)K(s,\alpha (s))ds\le T$$

for all $t\in [0,T]$. Then we have $fu\in A_1$. Thus, we have shown that (2) holds.

Hence, if $X:=A_1\cup A_2$, we have that X is closed in $C([0,T],{\mathbb{R}}^{+})$, so the metric space $(X,d_{\mathrm{\infty }})$ is complete.

Moreover, $X:=A_1\cup A_2$ is a cyclic representation of the restriction of f with respect to X, which will be also denoted by f.

Now construct the function $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ given by

$$\varphi (t)=\psi (t)\text{if }t\in [0,2M],$$

and

$$\varphi (t)=2M\text{if }t>2M.$$

Since ψ is a Meir-Keeler function that is non-decreasing on $[0,2M]$, it immediately follows that ϕ is a non-decreasing weaker Meir-Keeler function. Note also that ϕ is not continuous at $t=2M$ (in fact, it is not a Meir-Keeler function).

Finally we shall show that for each $u\in A_1$ and $v\in A_2$, one has $d_{\mathrm{\infty }}(fu,fv)\le \varphi (d_{\mathrm{\infty }}(u,v))$.

To this end, let $u\in A_1$ and $v\in A_2$. Since $u(s)\in [0,T]$ for each $s\in [0,T]$, we have that

$$\begin{array}{rcl}fu(t)& =& {\int }_0^{T}G(t,s)K(s,u(s))ds\\ \le & M{\int }_0^{T}G(t,s)ds\le M\end{array}$$

for all $t\in [0,T]$.

Similarly, since $v\ge \alpha$ and $\alpha (s)\in [0,T]$ for each $s\in [0,T]$, we deduce that

$$fv(t)\le {\int }_0^{T}G(t,s)K(s,\alpha (s))ds\le M$$

for all $t\in [0,T]$. Therefore

$$|fu(t)-f(v(t)|\le fu(t)+fv(t)\le 2M$$

for all $t\in [0,T]$. So,

$$d_{\mathrm{\infty }}(fu,fv)\le 2M.$$

If $d_{\mathrm{\infty }}(u,v)>2M$, we have $\varphi (d_{\mathrm{\infty }}(u,v))=2M$, so

$$d_{\mathrm{\infty }}(fu,fv)\le \varphi (d_{\mathrm{\infty }}(u,v)).$$

If $d_{\mathrm{\infty }}(u,v)\le 2M$, then $|u(s)-v(s)|\le 2M$ for all $s\in [0,T]$, so by (III), we deduce that for each $t\in [0,T]$,

$$\begin{array}{rcl}|fu(t)-f(v(t)|& \le & {\int }_0^{T}G(t,s)|K(s,u(s))-K(s,v(s))|ds\\ \le & {\int }_0^{T}G(t,s)\psi \left(\right|u(s)-v(s)\left|\right)ds\\ \le & \psi (d_{\mathrm{\infty }}(u,v)){\int }_0^{T}G(t,s)ds\\ \le & \psi (d_{\mathrm{\infty }}(u,v))\\ =& \varphi (d_{\mathrm{\infty }}(u,v)).\end{array}$$

Consequently, by the corollary of Theorem 5, f has a unique fixed point $u^{\ast }\in A_1\cap A_2$, that is, $u^{\ast }\in \mathcal{C}$ is the unique solution to (1) in $A_1\cup A_2$.

Remark 4 The first author studied in [[9], Section 3] a variant of the problem discussed above for the case that ψ is the non-decreasing Meir-Keeler function given by $\psi (t)={(ln(t^2+1))}^{1/2}$ for all $t\in {\mathbb{R}}^{+}$.

The next example illustrates the preceding development.

Example 4 Consider the integral equation

$$u(t)={\int }_0^{T}G(t,s)K(s,u(s))ds\text{for all }t\in [0,T],$$

where $T=1$, $G(t,s)=t$ for all $t,s\in [0,1]$, and

$$K(s,x)=\frac{coss}{1+x}$$

for all $s\in [0,1]$ and $x\ge 0$.

Hence, $M={max}_{(s,x)\in {[0,1]}^2}K(s,x)=K(0,0)=1$.

Furthermore, it is obvious that G satisfies condition (I), whereas K satisfies condition (II).

Now construct a Meir-Keeler function $\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ as

$$\psi (t)=t/(1+t)\text{if }t\in [0,2],$$

and

$$\psi (t)=0\text{if }t>2.$$

Note that ψ is non-decreasing on $[0,2]$ and not continuous at $t=2$.

Moreover, for each $s,x\in [0,1]$ and each $y\in {\mathbb{R}}^{+}$ with $|x-y|\le 2$, we have

$$|K(s,x)-K(s,y)|=coss|\frac{1}{1+x}-\frac{1}{1+y}|\le \frac{|x-y|}{1+|x-y|}=\psi (|x-y|),$$

so condition (III) is also satisfied.

Finally, define $\alpha :[0,1]\to [0,1]$ as $\alpha (t)=t/3$ for all $t\in [0,1]$. It is not hard to check that α verifies condition (IV), and consequently the integral equation has a unique solution $u^{\ast }$ in $A_1\cup A_2$, where $A_1=\{u\in C([0,1],{\mathbb{R}}^{+}):u(s)\le 1\text{ for all }s\in [0,1]\}$ and $A_2=\{u\in C([0,1],{\mathbb{R}}^{+}):u(s)\ge s/3\text{ for all }s\in [0,1]\}$. In fact $u^{\ast }\in A_1\cap A_2$, i.e., $t/3\le u^{\ast }(t)\le 1$ for all $t\in [0,1]$.

Note that, according to our constructions, for each pair $u,v\in C([0,1],{\mathbb{R}}^{+})$ with $u\le 1$ and $v\ge \alpha$, we have $d_{\mathrm{\infty }}(fu,fv)\le \varphi (d_{\mathrm{\infty }}(u,v))$, where ϕ is the non-decreasing weaker Meir-Keeler function defined as $\varphi (t)=t/(t+1)$ if $t\in [0,2]$ and $\varphi (t)=2$ if $t>2$.

In particular, we can deduce the following approximation to the value of $u^{\ast }(t)$ for each $t\in [0,1]$:

$$\begin{array}{rcl}|u^{\ast }(t)-\frac{sin1}{2}t|& =& |u^{\ast }(t)-{\int }_0^{1}t\frac{coss}{2}ds|=|f{u}^{\ast }(t)-{\int }_0^{1}G(t,s)K(s,1)ds|\\ \le & \varphi \left(d_{\mathrm{\infty }}(u^{\ast },1)\right)=\frac{{max}_{t\in [0,1]}(1-u^{\ast }(t))}{1+{max}_{t\in [0,1]}(1-u^{\ast }(t))}\\ =& \frac{1-{min}_{t\in [0,1]}{u}^{\ast }(t)}{2-{min}_{t\in [0,1]}{u}^{\ast }(t)}\\ \le & \frac{1}{2}.\end{array}$$

Note also that the contraction inequality $d_{\mathrm{\infty }}(fu,fv)\le \varphi (d_{\mathrm{\infty }}(u,v))$ does not follow when the weaker Meir-Keeler function ϕ is replaced by our initial Meir-Keeler function ψ: Take, for instance, the constant functions $u=0$ and $v=3$; then $u\le 1$, $v\ge \alpha$, and

$$\psi (d_{\mathrm{\infty }}(u,v))=\psi (3)=0<d_{\mathrm{\infty }}(fu,fv).$$

Remark 5 In Example 4 above, the inequality $|K(s,x)-K(s,y)|\le \psi (|x-y|)$ is not globally satisfied, i.e., there exist $s,x\in [0,1]$ and $y\in {\mathbb{R}}^{+}$ such that $|K(s,x)-K(s,y)|>\psi (|x-y|)$. In fact, this happens for all $x,y\in {\mathbb{R}}^{+}$ with $y>x+2$. However, it is clear that for each $s\in [0,1]$, and $x,y\in {\mathbb{R}}^{+}$, one has $|K(s,x)-K(s,y)|\le {\psi }_1(|x-y|)$ for all $s\in [0,1]$, and $x,y\in {\mathbb{R}}^{+}$, where ${\psi }_1(t)=t/(t+1)$ for all $t\in {\mathbb{R}}^{+}$.

We conclude the paper with an example where conditions (I)-(IV) also hold (in particular, (III) for the function ${\psi }_1$ defined above) but the inequality $|K(s,x)-K(s,y)|\le {\psi }_1(|x-y|)$ is not globally satisfied.

Example 5 We modify Example 4 as follows. Consider the integral equation

$$u(t)={\int }_0^{T}G(t,s)K(s,u(s))ds\text{for all }t\in [0,T],$$

where $T=2$, $G(t,s)=t/2$ for all $t,s\in [0,2]$, and

$$\begin{array}{c}K(s,x)=e^{-s}/(1+x)\text{if }s\in [0,2],x\in [0,1];\hfill \\ K(s,x)=e^{-s}/(1+x^{1/2})\text{if }s\in [0,2],x\in (1,4];\hfill \\ K(s,x)=e^{-s}/(4x-13)\text{if }s\in [0,2],x>4.\hfill \end{array}$$

Clearly K is continuous on $[0,2]\times {\mathbb{R}}^{+}$. Moreover, $M=1$, and G and K satisfy conditions (I) and (II), respectively.

Now, construct a Meir-Keeler function ${\psi }_1:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ as ${\psi }_1(t)=t/(1+t)$ for all $t\in {\mathbb{R}}^{+}$.

By discussing the different cases, it is routine to show that for each $s,x\in [0,2]$ and each $y\in {\mathbb{R}}^{+}$ with $|x-y|\le 2$, we have

$$|K(s,x)-K(s,y)|\le {\psi }_1(|x-y|),$$

so condition (III) is also satisfied.

Finally, define $\alpha :[0,2]\to [0,2]$ as $\alpha (t)=6t/35$ for all $t\in [0,2]$. Then, for each $t\in [0,2]$, we have

$${\int }_0^{2}G(t,s)K(s,2)ds=\frac{t}{2}{\int }_0^2\frac{e^{-s}}{1+\sqrt{2}}ds=t\frac{(1-e^{-2})}{2(1+\sqrt{2})}>\frac{6t/7}{5}=\alpha (t).$$

Now observe that $\alpha (s)<1$ for all $s\in [0,2]$, so $K(s,\alpha (s))=e^{-s}/(1+\alpha (s))$. Hence, for each $t\in [0,2]$,

$$\begin{array}{rcl}{\int }_0^{2}G(t,s)K(s,\alpha (s))ds& =& \frac{t}{2}{\int }_0^2\frac{e^{-s}}{1+(6s/35)}ds=\frac{t}{2}{\int }_0^2\frac{35e^{-s}}{35+6s}ds\\ \le & \frac{t}{2}{\int }_0^{2}ds=t\le 2.\end{array}$$

Therefore α verifies condition (IV), and consequently the integral equation has a unique solution $u^{\ast }$ in $A_1\cup A_2$, where $A_1=\{u\in C([0,1],{\mathbb{R}}^{+}):u(s)\le 2\text{ for all }s\in [0,2]\}$ and $A_2=\{u\in C([0,1],{\mathbb{R}}^{+}):u(s)\ge 6s/35\text{ for all }s\in [0,2]\}$. In fact $u^{\ast }\in A_1\cap A_2$, i.e., $6t/35\le u^{\ast }(t)\le 2$ for all $t\in [0,2]$.

It is interesting to observe that the Meir-Keeler function ${\psi }_1$ is continuous on ${\mathbb{R}}^{+}$ but condition (III) is not globally satisfied: Indeed, take $x=0$ and $y>14/3$. Then, for each $s\in [0,1]$, we obtain

$$K(s,x)-K(s,y)=e^{-s}(1-\frac{1}{4y-13})>e^{-s}\frac{y}{1+y}.$$

Hence, $K(0,0)-K(0,y)>{\psi }_1(y)$.