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# 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:

## 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 mapping
• nonexpansive mapping
• fixed point
• $$L_{p}$$

• 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  and has many applications in nonlinear functional analysis .

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

### Definition 1



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 . So the class of mappings which has condition $$(C)$$ is broader than the class of nonexpansive mappings.

In , condition $$(C)$$ is generalized as follows.

### Definition 2



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 .

In , 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 .

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  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  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 . □
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 . □

### Lemma 3



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  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 : 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. 