Fixed point theorems for cyclic selfmaps involving weaker MeirKeeler functions in complete metric spaces and applications
 Hemant Kumar Nashine^{1} and
 Salvador Romaguera^{2}Email author
https://doi.org/10.1186/168718122013224
© Nashine and Romaguera; licensee Springer. 2013
Received: 7 May 2013
Accepted: 5 August 2013
Published: 22 August 2013
Abstract
We obtain fixed point theorems for cyclic selfmaps on complete metric spaces involving MeirKeeler and weaker MeirKeeler functions, respectively. In this way, we extend several wellknown fixed point theorems and, in particular, improve some very recent results on weaker MeirKeeler functions. Fixed point results for wellposed 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.
Keywords
1 Introduction
In their paper [1], Kirk, Srinavasan and Veeramani started the fixed point theory for cyclic selfmaps 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 MeirKeeler function and obtained some fixed point theorems for cyclic contractions involving weaker MeirKeeler functions.
In this paper we obtain a fixed point theorem for cyclic selfmaps on complete metric spaces involving MeirKeeler functions and deduce a variant of it for weaker MeirKeeler 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 wellposedness 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 selfmap f of a (nonempty) set X is called a cyclic map if there exists $m\in \mathbb{N}$ such that $X={\bigcup}_{i=1}^{m}{A}_{i}$, with ${A}_{i}$ nonempty 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 nonnegative real numbers and the set of positive integer numbers, respectively.
Meir and Keeler proved in [16] that if f is a selfmap 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 MeirKeeler function:
A function $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is said to be a MeirKeeler 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 MeirKeeler function, then $\varphi (t)<t$ for all $t>0$.
In [14], Chen introduced the following interesting generalization of the notion of a MeirKeeler function.
Definition 1 [[14], Definition 3]
A function $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is called a weaker MeirKeeler 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 ${lim}_{n\to \mathrm{\infty}}\varphi ({t}_{n})=0$;
 (a)

(${\phi}_{1}$) φ is nondecreasing 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]
 (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 MeirKeeler functions.
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 MeirKeeler 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 $d({x}_{k},{x}_{k+1})<\delta /2$, $d({x}_{k},{x}_{k+m1})<\epsilon /2$ 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 $j\in \mathbb{N}$.
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 $z\in {\bigcap}_{i=1}^{m}{A}_{i}$.
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 MeirKeeler function that satisfies conditions (${\varphi}_{2}$) and (${\varphi}_{3}$)(b).
Proof Suppose that ϕ is not a MeirKeeler 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 MeirKeeler 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. □
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 nondecreasing 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.
where ${A}_{m+1}={A}_{1}$, then f has a unique fixed point z. Moreover, $z\in {\bigcap}_{i=1}^{m}{A}_{i}$.
for all $x\in {A}_{i}$, $y\in {A}_{i+1}$, $i=1,\dots ,m$.
Finally, since by Lemma 1 ϕ is a MeirKeeler 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 MeirKeeler 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 MeirKeeler functions. In particular, it is natural to wonder if Theorem 2 remains valid when we replace ‘MeirKeeler function’ by ‘weaker MeirKeeler 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 MeirKeeler function is nondecreasing 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 MeirKeeler function (note, in particular, that ${\varphi}^{2}(1)=1/2<1$), but it is not a MeirKeeler 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 nondecreasing. This fact is not casual as Theorem 5 below shows.
 (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 nondecreasing, we deduce that ${\{{\varphi}^{n}({t}_{0})\}}_{n\in \mathbb{N}\cup \{0\}}$ is a nondecreasing 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 MeirKeeler 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 MeirKeeler 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.
for all $x,y\in X$.
Theorem 4 (cf. [[18], Corollary 2.14])
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.
where ${A}_{m+1}={A}_{1}$, then f has a unique fixed point z. Moreover, $z\in {\bigcap}_{i=1}^{m}{A}_{i}$.
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 nondecreasing, 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.
where ${A}_{m+1}={A}_{1}$, then f has a unique fixed point z. Moreover, $z\in {\bigcap}_{i=1}^{m}{A}_{i}$.
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 selfmap of X defined by $f0=0$ and $fx=x1$ 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 $\varphi (t)=n1$ if $t\in (n1,n]$, $n\in \mathbb{N}$. It is immediate to check that ϕ is a nondecreasing weaker MeirKeeler function which is not a MeirKeeler function.
Furthermore, we have:

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

For $x=0$ and $y=n\in \mathbb{N}\mathrm{\setminus}\{1\}$,$d(fx,fy)=d(0,n1)=n1=\varphi (n)=\varphi (d(x,y)).$

For $x=n\in A\mathrm{\setminus}\{0\}$ and $y=m\in B\mathrm{\setminus}\{0\}$,$\begin{array}{rcl}d(fx,fy)& =& d(n1,m1)=n+m2<n+m1\\ =& \varphi (n+m)=\varphi (d(x,y)).\end{array}$
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 selfmap of X defined by $fx=2x$ 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 nondecreasing weaker MeirKeeler function which is not a MeirKeeler 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 $y\in B$,$d(fx,fy)=d(2x,1)=1x\le 1=\varphi (22x)=\varphi (d(x,fx)).$
Consequently, the conditions of Theorem 5 are verified; in fact, $z=1\in A\cap B$ is the unique fixed point of f.
3 Applications to wellposedness and limit shadowing property of a fixed point problem
The notion of wellposedness 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]
 (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 selfmap 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 wellposedness and limit shadowing of the fixed point problem for a selfmap 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 selfmap of X and $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a nondecreasing weaker MeirKeeler function satisfying the conditions of Theorem 5, then the fixed point problem of f is well posed.
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 selfmap of X and $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a nondecreasing weaker MeirKeeler function satisfying the conditions of Theorem 5, then f has the limit shadowing property.
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.
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 $M:={max}_{(s,x)\in {[0,T]}^{2}}K(s,x)$.
 (I)
${\int}_{0}^{T}G(t,s)\phantom{\rule{0.2em}{0ex}}ds\le 1$ for all $t\in [0,T]$.
 (II)$K(s,\cdot )$ is a nonincreasing function for any fixed $s\in [0,1]$, that is,$x,y\in {\mathbb{R}}^{+},\phantom{\rule{1em}{0ex}}x\ge y\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}K(s,x)\le K(s,y).$
 (III)There exists a MeirKeeler function $\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ that is nondecreasing on $[0,2M]$ and such that$K(s,x)K(s,y)\le \psi (xy)$
for all $s,x\in [0,T]$ and $y\in {\mathbb{R}}^{+}$ with $xy\le 2M$.
 (IV)
There exists a continuous function $\alpha :[0,T]\to [0,T]$ such that:
For all $t\in [0,T]$, we have$\alpha (t)\le {\int}_{0}^{T}G(t,s)K(s,T)\phantom{\rule{0.2em}{0ex}}ds$and$T\ge {\int}_{0}^{T}G(t,s)K(s,\alpha (s))\phantom{\rule{0.2em}{0ex}}ds.$
It is well known that $(C([0,T],{\mathbb{R}}^{+}),{d}_{\mathrm{\infty}})$ is a complete metric space.
Clearly, u is a solution of (1) if and only if u is a fixed point of f.
for all $t\in [0,T]$. Then we have $fu\in {A}_{2}$.
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.
Since ψ is a MeirKeeler function that is nondecreasing on $[0,2M]$, it immediately follows that ϕ is a nondecreasing weaker MeirKeeler function. Note also that ϕ is not continuous at $t=2M$ (in fact, it is not a MeirKeeler function).
Finally we shall show that for each $u\in {A}_{1}$ and $v\in {A}_{2}$, one has ${d}_{\mathrm{\infty}}(fu,fv)\le \varphi ({d}_{\mathrm{\infty}}(u,v))$.
for all $t\in [0,T]$.
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 nondecreasing MeirKeeler function given by $\psi (t)={(ln({t}^{2}+1))}^{1/2}$ for all $t\in {\mathbb{R}}^{+}$.
The next example illustrates the preceding development.
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).
Note that ψ is nondecreasing on $[0,2]$ and not continuous at $t=2$.
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 nondecreasing weaker MeirKeeler function defined as $\varphi (t)=t/(t+1)$ if $t\in [0,2]$ and $\varphi (t)=2$ if $t>2$.
Remark 5 In Example 4 above, the inequality $K(s,x)K(s,y)\le \psi (xy)$ is not globally satisfied, i.e., there exist $s,x\in [0,1]$ and $y\in {\mathbb{R}}^{+}$ such that $K(s,x)K(s,y)>\psi (xy)$. 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 $K(s,x)K(s,y)\le {\psi}_{1}(xy)$ 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 $K(s,x)K(s,y)\le {\psi}_{1}(xy)$ is not globally satisfied.
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 MeirKeeler function ${\psi}_{1}:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ as ${\psi}_{1}(t)=t/(1+t)$ for all $t\in {\mathbb{R}}^{+}$.
so condition (III) is also satisfied.
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 $t\in [0,2]$.
Hence, $K(0,0)K(0,y)>{\psi}_{1}(y)$.
Declarations
Acknowledgements
The second author thanks for the support of the Ministry of Economy and Competitiveness of Spain under grant MTM201237894C0201, and the Universitat Politècnica de València, grant PAID0612SP20120471.
Authors’ Affiliations
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