Fixed point theorems for a class of generalized nonexpansive mappings

Abstract

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

1 Introduction and preliminaries

A nonexpansive mapping has a Lipschitz constant equal to 1. The fixed point theory for such mappings is very rich [1–5] 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?