Open Access

Fixed point theorems for a class of generalized nonexpansive mappings

Fixed Point Theory and Applications20162016:82

https://doi.org/10.1186/s13663-016-0571-y

Received: 13 February 2016

Accepted: 20 July 2016

Published: 1 August 2016

Abstract

In this paper, we introduce a new class of generalized nonexpansive mappings. Some new fixed point theorems for these mappings are obtained.

Keywords

monotone mappingnonexpansive mappingfixed point \(L_{p}\)

MSC

47H10

1 Introduction and preliminaries

A nonexpansive mapping has a Lipschitz constant equal to 1. The fixed point theory for such mappings is very rich [15] and has many applications in nonlinear functional analysis [6].

We first commence some basic concepts about generalization of nonexpansive mappings as formulated by Suzuki et al. [7, 8].

Definition 1

[8]

Let C be a nonempty subset of a Banach space X. We say that a mapping \(T:C \rightarrow C\) satisfies condition \((C)\) on C if \(\frac{1}{2}\|x-T(x)\| \leq\|x-y\|\) implies \(\|T(x)-T(y)\| \leq\|x-y\|\), for \(x,y\in C\).

Of course, every nonexpansive mapping satisfies condition \((C)\) but the converse is not correct and you can find some counterexamples for it in [8]. So the class of mappings which has condition \((C)\) is broader than the class of nonexpansive mappings.

In [7], condition \((C)\) is generalized as follows.

Definition 2

[7]

Let C be a nonempty subset of a Banach space X and \(\lambda\in(0,1)\). We say that a mapping \(T:C \rightarrow X\) satisfies (\(C_{\lambda}\))-condition on C if \(\lambda\|x-T(x)\| \leq\|x-y\|\) implies \(\|T(x)-T(y)\| \leq\|x-y\|\), for \(x,y\in C\).

So if \(\lambda=\frac{1}{2}\), we will have condition \((C)\). There are examples that show the converse is false; see [7].

In [9], monotone nonexpansive mappings are defined in \(L_{1}[0,1]\).

We next review some notions in \(L_{p}[0,1]\). All of them can be found in [10].

Consider the Riesz Banach space \(L_{p}[0,1]\), where \(\int_{0}^{1}|f(x)|^{p} \,dx<+\infty\) and \(p\in(0,+\infty)\). Also, we have \(f=0\) when the set
$$\bigl\{ x\in[0,1]:f(x)=0\bigr\} , $$
has Lebesgue measure zero. In this case, we say \(f=0\) almost everywhere. An element of \(L_{p}[0,1]\) is therefore seen as a class of functions. The norm of any \(f\in L_{p}[0,1]\) is given by \(\|f\|_{p}=(\int_{0}^{1}|f(x)|^{p} \,dx)^{\frac{1}{p}}\). Throughout this paper, we will write \(L_{p}\) instead of \(L_{p}[a,b]\), \(a,b\in\mathbb{R}\) and \(\|\cdot\|\) instead of \(\|\cdot\|_{p}\).

In this paper, we redefine Definition 2 on a subset of Banach space \(L_{p}\) and those theorems which are proved in [9] generalize to a wider class of monotone (\(C_{\lambda}\))-condition with preserving their fixed point property.

2 Main results

Let C be a nonempty subset of \(L_{p}\) which is equipped with a vector order relation . A map \(T:C\rightarrow C\) is called monotone if for all \(f\preceq g\) we have \(T(f)\preceq T(g)\).

We generalize the (\(C_{\lambda}\))-condition as follows.

Definition 3

Let C be a nonempty subset of a Banach space \(L_{p}\). For \(\lambda\in(0,1)\), we say that a mapping T monotone (\(C_{\lambda}\))-condition on C if T is monotone and for all \(f\preceq g\), \(\lambda\|f-T(f)\| \leq\|g-f\|\) implies \(\|T(g)-T(f)\| \leq\|g-f\|\).

Note Definition 3 is a generalization of the monotone nonexpansive mapping which is defined in [9] as follows.

A map T is said to be monotone nonexpansive if T is monotone and for \(f\preceq g\), we have \(\|T(g)-T(f)\| \leq\|g-f\|\).

The next example is a direct generalization of monotone nonexpansive mapping.

Example 1

Let \(C=\{f\in L_{p}[0,3]: f(x)=a\}\), where \(a\in[0,3]\). For \(f,g\in C\), consider the partial order relation
$$f\preceq g \quad\mbox{iff} \quad f(x)\leq g(x). $$
Let \(T:C\rightarrow C\) be defined by
$$T(f)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l@{}} 1, & f=3, \\ 0, & f\neq3. \end{array}\displaystyle \right . $$
Then the mapping T satisfies the monotone (\(C_{\frac{1}{2}}\))-condition but it fails monotone nonexpansiveness. Indeed, whenever \(f\preceq g\), if \(0\leq f(x)\leq g(x)<3\), then \(\|T(f)-T(g)\|\leq\|f-g\|\). On the other hand, \(0\leq f(x)<3\) and \(g=3\), so if \(0\leq f(x)\leq2\) and \(g=3\), then we have again \(\|T(f)-T(g)\|\leq\|f-g\|\), but if \(2< f(x)< 3\) and \(g=3\), then \(\frac{1}{2}\|f\|\nleq\|f-3\|\). Thus, the mapping T satisfying monotone (\(C_{\frac{1}{2}}\))-condition on \([0,3]\).

Let \(f=2.9\) and \(g=3\). Then \(f\preceq g\) while \(\|T(f)-T(g)\|\nleq \|f-g\|\). Thus, T is not monotone nonexpansive.

The following lemmas will be crucial to prove the main result of this paper.

Lemma 1

Let C be convex and T monotone. Assume that for some \(f_{1}\in C\), \(f_{1}\preceq T(f_{1})\). Then the sequence \(f_{n}\) defined by
$$(\star)\qquad f_{n+1}=\lambda T(f_{n})+(1- \lambda)f_{n}, $$
\(\lambda\in(0,1)\), satisfies
$$f_{n}\preceq f_{n+1}\preceq T(f_{n})\preceq T(f_{n+1}). $$
for \(n\geq1\).

Proof

First, we prove that \(f_{n}\preceq T(f_{n})\). By assumption, we have \(f_{1}\preceq T(f_{1})\). Assume that \(f_{n}\preceq T(f_{n})\), for \(n\geq1\). Then we have
$$f_{n}=\lambda f_{n}+(1-\lambda)f_{n}\preceq \lambda T(f_{n})+(1-\lambda)f_{n}=f_{n+1} $$
i.e. \(f_{n}\preceq f_{n+1}\). Since T is monotone, \(T(f_{n})\preceq T(f_{n+1})\). We have
$$f_{n+1}=\lambda T(f_{n})+(1-\lambda)f_{n}\preceq \lambda T(f_{n})+ (1-\lambda)T(f_{n})=T(f_{n}). $$
Thus
$$f_{n}\preceq f_{n+1}\preceq T(f_{n})\preceq T(f_{n+1}), $$
for \(n\geq1\). The proof is closely modeled on Lemma 3.1 of [9]. □
Note that under the assumption of Lemma 1, if we assume \(T(f_{1})\preceq f_{1}\), then we have
$$T(f_{n+1})\preceq T(f_{n})\preceq f_{n+1}\preceq f_{n} $$
for any \(n\geq1\).

A sequence \(\{f_{n}\}\) in C is called an almost fixed point sequence for T, if \(\|f_{n}-T(f_{n})\|\rightarrow0\) (a.f.p.s. in short).

Lemma 2

Let \(T:C\rightarrow L_{p}\) be a monotone \((C_{\lambda })\)-condition mapping and \(f_{n}\) be a bounded a.f.p.s. for T. Then
$$\liminf_{n}\bigl\| f_{n}-T(f)\bigr\| \leq\liminf _{n_{k}}\|f_{n}-f\|, $$
for \(f\in C\) which \(f_{n}\preceq f\) and \(\liminf_{n}\|f_{n}-f\|>0\), for all \(n\geq1\).

Proof

Fix \(f\in C\) such that \(f_{n}\preceq f\). Since \(f_{n}\) is an a.f.p.s., for \(\epsilon=\frac{1}{2}\liminf_{n}\|f_{n}-f\|\), there is \(n_{0}\) such that \(\|f_{n}-T(f_{n})\|<\epsilon\), for all \(n\geq n_{0}\). This implies that
$$\lambda\bigl\Vert f_{n}-T(f_{n})\bigr\Vert \leq\bigl\Vert f_{n}-T(f_{n})\bigr\Vert < \epsilon< \Vert f_{n}-f\Vert , $$
for all \(n\geq n_{0}\). Since T satisfies the monotone (\(C_{\lambda}\))-condition, we have
$$ \bigl\Vert T(f_{n})-T(f)\bigr\Vert \leq \Vert f_{n}-f\Vert , $$
(1)
for all \(n\geq n_{0}\). So by the triangle inequality and (1), we have
$$\bigl\Vert f_{n}-T(f)\bigr\Vert \leq\bigl\Vert f_{n}-T(f_{n})\bigr\Vert +\bigl\Vert T(f_{n})-T(f) \bigr\Vert \leq \bigl\Vert f_{n}-T(f_{n})\bigr\Vert + \Vert f_{n}-f\Vert . $$
Thus \(\liminf_{n}\|f_{n}-T(f)\|\leq \liminf_{n}\|f_{n}-f\|\). The proof is closely modeled on Lemma 1 of [7]. □

Lemma 3

[11]

If \(\{f_{n}\}\) is a sequence of \(L_{p}\)-uniformly bounded functions on a measure space, and \(f_{n} \rightarrow f\) almost everywhere, then
$$\liminf_{n}\Vert f_{n}\Vert ^{p} = \liminf_{n}\Vert f_{n}-f\Vert ^{p}+ \Vert f\Vert ^{p}, $$
for all \(p\in(0,\infty)\).

In the following, let C be a nonempty, convex, and bounded set and \(T:C\rightarrow C\) be a monotone \((C_{\lambda})\)-condition, for some \(\lambda\in(0,1)\).

Theorem 1

Let \(f_{1}\in C\) such that \(f_{1}\preceq T(f_{1})\). Then \(f_{n}\) defined in () is an a.f.p.s.

Proof

Since \(f_{n+1}=\lambda T(f_{n})+(1-\lambda)f_{n}\), for \(n\geq1\), we have
$$\lambda\bigl\Vert f_{n}-T(f_{n})\bigr\Vert = \|f_{n}-f_{n+1}\|. $$
By Lemma 1, we have \(f_{n}\preceq f_{n+1}\). Therefore, monotone (\(C_{\lambda}\))-condition implies that \(\|T(f_{n})-T(f_{n+1})\|\leq\|f_{n}-f_{n+1}\|\). Now, we can apply Lemma 3 of [1] to conclude that \(\lim_{n}\|f_{n}-T(f_{n})\|=0\). □

Example 2

We show that T, which is defined in Example 1, has an a.f.p.s. It is easy to see that C is a nonempty, convex, and bounded subset of \(L_{p}\). Also, we proved T obeys the monotone \((C_{\frac{1}{2}})\)-condition. Moreover, \(0\preceq T(0)\). Thus, by Theorem 1, T has an a.f.p.s.

Now, we construct an a.f.p.s. according (). Let \(f_{1}=0\). So \(f_{n}=0\). Therefore
$$\bigl\Vert f_{n}-T(f_{n})\bigr\Vert =0. $$
Thus \(f_{n}\) is an a.f.p.s.

Theorem 2

Let C be compact. Assume there exists \(f_{1}\in C\) such that \(f_{1}\) and \(T(f_{1})\) are comparable. Then T has a fixed point.

Proof

Let \(f_{n}\) be a sequence which is defined in (). By Theorem 1, \(f_{n}\) is an a.f.p.s. Since C is compact, \(f_{n}\) has a convergent subsequence \(f_{n_{k}}\) to f. By triangle inequality, we get
$$\liminf_{n_{k}}\bigl\Vert T(f_{n_{k}})-T(f)\bigr\Vert \leq \lim_{n_{k}}\bigl\Vert T(f_{n_{k}})-f_{n_{k}} \bigr\Vert +\liminf_{n_{k}}\bigl\Vert f_{n_{k}}-T(f) \bigr\Vert . $$
Since \(f_{n}\) is an a.f.p.s., we have
$$ \liminf_{n_{k}}\bigl\Vert T(f_{n_{k}})-T(f) \bigr\Vert \leq \liminf_{n_{k}}\bigl\Vert f_{n_{k}}-T(f) \bigr\Vert . $$
(2)
Again, by triangle inequality, we have
$$\liminf_{n_{k}}\bigl\Vert f_{n_{k}}-T(f)\bigr\Vert \leq \lim_{n_{k}}\bigl\Vert f_{n_{k}}-T(f_{n_{k}}) \bigr\Vert +\liminf_{n_{k}}\bigl\Vert T(f)-T(f_{n_{k}}) \bigr\Vert . $$
Therefore,
$$ \liminf_{n_{k}}\bigl\Vert f_{n_{k}}-T(f) \bigr\Vert \leq \liminf_{n_{k}}\bigl\Vert T(f_{n_{k}})-T(f) \bigr\Vert . $$
(3)
From equations (2) and (3), we have
$$ \liminf_{n_{k}}\bigl\Vert f_{n_{k}}-T(f) \bigr\Vert =\liminf_{n_{k}}\bigl\Vert T(f_{n_{k}})-T(f) \bigr\Vert . $$
(4)
By using the partially order and convergent properties \(f_{n_{k}}\preceq f\). Lemma 1 implies \(f_{n_{k}}\preceq f_{n_{k}+1}\preceq f\). So \(\Vert f_{n_{k}+1}-f_{n_{k}}\Vert \leq \Vert f-f_{n_{k}}\Vert \). Since \(f_{n_{k}+1}-f_{n _{k}}=\lambda(f_{n_{k}}-T(f_{n_{k}}))\), we get
$$\lambda \bigl\Vert f_{n_{k}}-T(f_{n_{k}})\bigr\Vert = \Vert f_{n_{k}+1}-f_{n_{k}}\Vert . $$
Therefore
$$\lambda\bigl\Vert \bigl(f_{n_{k}}-T(f_{n_{k}})\bigr)\bigr\Vert \leq \Vert f-f_{n_{k}}\Vert . $$
Thus the monotone (\(C_{\lambda}\))-condition implies
$$ \bigl\Vert T(f_{n_{k}})-T(f)\bigr\Vert \leq \Vert f_{n_{k}}-f\Vert . $$
(5)
Since \(f_{n_{k}}\) is bounded, Lemma 3 implies
$$\liminf_{n_{k}}\bigl\Vert f_{n_{k}}-T(f)\bigr\Vert = \liminf_{n_{k}}\Vert f_{n_{k}}-f\Vert +\bigl\Vert f-T(f)\bigr\Vert . $$
From equation (4), we get
$$\liminf_{n_{k}}\Vert f_{n_{k}}-f\Vert +\bigl\Vert f-T(f)\bigr\Vert =\liminf_{n_{k}}\bigl\Vert T(f_{n_{k}})-T(f)\bigr\Vert . $$
From equation (5), we get
$$\liminf_{n_{k}}\Vert f_{n_{k}}-f\Vert +\bigl\Vert f-T(f)\bigr\Vert \leq \liminf_{n_{k}}\Vert f_{n_{k}}-f \Vert . $$
This implies that \(T(f)=f\). □

By Theorem 2, we can see that T in Example 1, has a fixed point.

The following example shows that monotone \((C_{\lambda})\)-condition is a direct generalization of \((C_{\lambda})\)-condition.

Example 3

Let \(C=co\{x,\sin(x)\}\), where \(x\in[-\frac{\pi }{2},\frac{\pi}{2}]\). Define a partial order on C as follows:
$$f\preceq g \quad \mbox{iff} \quad f(x)\leq g(x). $$
Let \(T:C\rightarrow C\) be
$$T(f)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sin(x) & f\neq x, \\ x& f=x. \end{array}\displaystyle \right . $$
Since C is convex hull of a compact set \(\{x,\sin(x)\}\), so it is a nonempty, convex and compact subset of \(L_{p}\). Put \(f=x\). Then f and \(T(f)\) are comparable. Also, T obeys the monotone \((C_{\lambda})\)-condition. Thus, by Theorem 2, T has a fixed point.

Note, for \(\lambda\in(0,1)\), T does not obey the \((C_{\lambda})\)-condition. Because, for \(f=x\) and \(g=\frac{x}{2}+\frac{1}{2}\sin(x)\), we have \(\lambda\|f-T(f)\|\leq\|f-g\|\), but \(\|T(g)-T(f)\|\nleq\|f-g\|\).

Theorem 3

Let C be a weakly compact subset of \(L_{2}\). Assume, there is \(f_{1}\in C\) such that \(f_{1}\preceq T(f_{1})\). Then T has a fixed point.

Proof

By Theorem 1, T has an a.f.p.s. \(f_{n}\). Since C is weakly compact, there is a weakly convergent subsequence \(f_{n_{k}}\) to some \(f\in C\). If \(\liminf_{n_{k}}\|f_{n_{k}}-f\|=0\), then \(f_{n_{k}}\) is convergent and we will have the same proof of Theorem 2. On the other hand, if \(\liminf_{n_{k}}\|f_{n_{k}}-f\|>0\), then by Lemma 2,
$$ \liminf_{n_{k}}\bigl\Vert f_{n_{k}}-T(f) \bigr\Vert \leq\liminf_{n_{k}}\Vert f_{n_{k}}-f\Vert . $$
(6)
We claim that \(f= T(f)\). Because if \(f\neq T(f)\), since \(L_{2}\) satisfies Opial condition, we have
$$\liminf_{n_{k}}\Vert f_{n_{k}}-f\Vert < \liminf _{n_{k}}\bigl\Vert f_{n_{k}}-T(f)\bigr\Vert , $$
which is a contradiction with inequality (6). □

This result is a generalization of the original existence theorem in [7, 9] form monotone nonexpansive to monotone \((C_{\lambda})\)-condition. Therefore this class is bigger and is used to answer the question asked by T Benavides [12]: Does X also satisfy the fixed point property for Suzuki-type mappings?

Declarations

Acknowledgements

The first author acknowledges Buein Zahra Technical University for supporting this research.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Buein Zahra Technical University
(2)
Department of Mathematics, Payame Noor University

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© Lael and Heidarpour 2016