数理解析研究所講究録
第 1685 巻 2010 年 177-189

177

Nonlinear Operators and Fixed Point Theorems
in Hilbert Spaces
台湾国立中山大学理学院応用数学系
高橋渉 (Wataru Takahashi)
Department of Applied Mathematics

National Sun Yat-sen University, Taiwan

Abstract. In this article, we first consider new classes of nonlinear mappings containing theclass of firmly nonexpansive mappings which can be deduced from an equilibrium problem ina Hilbert space. Further, we deal with fixed point theorems and ergodic theorems for these
nonlinear mappings.

1

lntroduction

Let $H$ be a real Hilbert space and let $C$ be a nonempty closed convex subset of $H$ . Then amapping $T:Carrow H$ is said to be nonexpansive if $\Vert Tx-Ty\Vert\leq\Vert x-y\Vert$ for all $x,$ $y\in C$ . We
know that if $C$ is a bounded closed convex subset of $H$ and $T$ : $Carrow C$ is nonexpansive, thenthe set $F(T)$ of fixed points of $T$ is nonempty. Further, from Baillon [1] we know the first
nonlinear ergodic theorem in a Hilbert space: Let $C$ be a nonempty bounded closed convex
subset of $H$ and let $T$ : $Carrow C$ be nonexpansive. Then, for any $x\in C$ ,$S_{n}x= \frac{1}{n}\sum_{k=0}^{n-1}T^{k_{X}}$converges weakly to an element $z\in F(T)$ . An important example of nonexpansive mappings
is said to be firmly
in a Hilbert space is a firmly nonexpansive mapping. A mapping$T$nonexpansive if
$||$

Tx–Ty

$\Vert^{2}\leq\langle x-y$

, Tx–Ty ,$\}$$\forall x,$$y\in C$;see, for instance, Goebel and Kirk [8]. It is also known that a firmly nonexpansive mapping
can be deduced from an equilibrium problem in a Hilbert space as follows: Let $C$ be a
be a bifunction satisfying the:nonempty closed convex subset of $H$ and let
following conditions:
$T$

$f$

$C\cross Carrow \mathbb{R}$(Al) $f(x, x)=0$ , $\forall x\in C$ ;$y\in C$ ;(A2) $f$ is monotone, i.e., $f(x, y)+f(y, x)\leq 0$ ,$y,$ $z\in C$ ;(A3) $\lim_{t\downarrow 0}f(tz+(1-t)x, y)\leq f(x, y)$ ,(A4) for each $x\in C,$ $y\mapsto f(x, y)$ is convex and lower semicontinuous.
$\forall x,$$\forall x,$

We know the following lemma; see, for instance, [2] and [7].

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Lemma 1.1. Let
be a nonempty closed convex subset of $H$ and let be a bifunction frominto
satisfying (A 1), $(A2),$
and $(A4)$ . Then, for any $r>0$ and $x\in H$ , there$C$$C\cross C$$f$$(\Lambda 3)$$\mathbb{R}$exists $z\in C$ such that$f(z, y)+ \frac{1}{r}\langle y-z,$ $z-x\}\geq

0$ ,$\forall y\in C$.

Further, if $T_{r}x= \{z\in C : f(z, y)+\frac{1}{r}\langle y-z, z-x\rangle\geq 0, \forall y\in C\}$ for all $x\in H$ , then thefollowing hold:(1)(2)$T_{r}$$T_{r}$

is single-valued;
is firmly nonexpansive, i. e.,$\Vert T_{r}x-T_{r}y\Vert^{2}\leq\langle T_{r}x-T_{r}y,$ $x-y\rangle$,$\forall x,$$y\in H$.

Recently, Kohsaka and Takahashi [12] introduced the following nonlinear mapping: Let $E$
be a smooth, strictly convex and reflexive Banach space, let be the duality mapping of $E$
and let $C$ be a nonempty closed convex subset of $E$ . Then, a mapping: $Carrow E$ is said tobe nonspreading if
$J$

$S$

$\phi(Sx, Sy)+\phi(Sy, Sx)\leq\phi(Sx, y)+\phi(Sy, x)$ ,$\forall x,$$y\in C$ ,where $\phi(x, y)=\Vert x\Vert^{2}-2\langle x,$
for all $x,$ $y\in E$ . They considered such a mapping tostudy the resolvents of a maximal monotone operator in the Banach space. In the case when
$E$ is a
Hilbert space, we know that $\phi(x, y)=\Vert x-y\Vert^{2}$ for all $x,$ $y\in E$ . So, a nonspreading
mapping in a Hilbert space is defined as follows:$Jy\}+\Vert y\Vert^{2}$$S$

2 $\Vert Sx-Sy\Vert^{2}\leq\Vert Sx-y\Vert^{2}+\Vert Sy-x\Vert^{2}$ ,$\forall x,$$y\in C$.

On the other hand, Takahashi [29] found another new nonlinear mapping called a hybrid
mapping which is also deduced from a firmly nonexpansive mapping.
In this paper, we first discuss new classes of nonlinear mappings containing the class of
firmly nonexpansive mappings which can be deduced from a firmly nonexpansive mapping ina Hilbert space. Further, we deal with fixed point theorems and ergodic theorems for these
nonlinear mappings.

2

Preliminaries

Throughout this paper, we denote by the set of positive integers and by
numbers. Let $H$ be a real Hilbert space with inner productand norm
In a Hilbert space, it is known that for all $x,$ $y\in H$ and,$\mathbb{N}$$\langle\cdot,$$\cdot\}$$\mathbb{R}$$\Vert\cdot\Vert$

the set of real
, respectively.

$\alpha\in \mathbb{R}$$\Vert\alpha x+(1-\alpha)y\Vert^{2}=\alpha\Vert x\Vert^{2}+(1-\alpha)\Vert y\Vert^{2}-\alpha(1-\alpha)||x-y||^{2}$

;

(2.1)

see, for instance, [28]. Further, in a Hilbert space, we have that for all $x,$ $y,$ $z,$ $w\in H$ ,$2\langle x-y,$ $z-w\rangle=\Vert x-w\Vert^{2}+\Vert y-z\Vert^{2}-||x-z\Vert^{2}-||y-w\Vert^{2}$

.

(2.2)

Let $C$ be a nonempty subset of $H$ and let $T$ be a mapping of $C$ into $H$ . We denote by $F(T)$the set of all fixed points of $T$ , that is, $F(T)=\{z\in C : Tz=z\}$ . We denote the strong
convergence and the weak convergence ofto $x\in H$ by $x_{n}arrow x$ and $x_{n}arrow x$ , respectively.
The following lemma is in [20].
$\{x_{n}\}$

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Lemma 2.1. Let $C$ be a nonempty closed convex subset of $H$ and let : $Carrow(-\infty, \infty]$ bea proper convex lower semicontinuous function such that $f(z_{m})arrow\infty$ as. Then1
there exists an element $z_{0}\in C$ such that$f$$z_{m}\Vertarrow\infty$$f(z_{0})= \min\{f(z):z\in C\}$ .

Let be the set of positive integers and let
be the Banach space of bounded sequenceswith supremum norm. Let be an element of(the dual space of). Then, we denote
by $\mu(f)$ the value of at $f=(x_{1}, x_{2}, x_{3}, \ldots)\in l^{\infty}$ . Sometimes, we denote by
the value$\mu(f)$ . A linear functionalon
is called a mean if $\mu(e)=\Vert\mu\Vert=1$ , where $e=(1,1,1, \ldots)$ .
A mean is called a Banach limit on
if $\mu_{n}(x_{n+1})=\mu_{n}(x_{n})$ . We know that there exists a
Banach limit on; see [20] for more details.
$\mathbb{N}$$l^{\infty}$$(l^{\infty})^{*}$$\mu$$l^{\infty}$$\mu$$\mu_{n}(x_{n})$$l^{\infty}$$\mu$$l^{\infty}$$\mu$$l^{\infty}$

3

Nonlinear Mappings

Let $H$ be a Hilbert space. Let $C$ be a nonempty closed convex subset of $H$ and let $T$ be amapping of $C$ into $H$ . Then, from [29], we have the following equality:$\Vert$Tx–Ty $\Vert^{2}=\Vert x-y-(Tx-Ty)\Vert^{2}-\Vert x-y\Vert^{2}+2\langle x-y$, Tx–Ty(3. 1)$\}$

for all $x,$ $y\in C$ . We have also from (2.2) that2$\langle x-y,$$Tx-Ty\}=\Vert x-Ty||^{2}+\Vert y-Tx\Vert^{2}-||x-Tx\Vert^{2}-\Vert y-Ty\Vert^{2}$ .

(3.2)

Further, we have that$\Vert x-y-(Tx-Ty)\Vert^{2}=\Vert x-Tx\Vert^{2}+\Vert y-Ty\Vert^{2}-2\langle x-Tx,$ $y-Ty\rangle$

.

(3.3)

If $T:Carrow H$ is firmly nonexpansive, then$\Vert Tx-Ty\Vert^{2}\leq\langle x-y$

, Tx–Ty ,$\}$$\forall x,$$y\in C$.

So, we have from (3.1) that for all $x,$ $y\in C$ ,$2\Vert Tx-Ty\Vert^{2}\leq 2\langle x-y,$ $Tx-Ty\rangle$$=\Vert Tx-Ty\Vert^{2}-\Vert x-y-(Tx-Ty)||^{2}+\Vert x-y\Vert^{2}$$\leq\Vert Tx-Ty\Vert^{2}+\Vert x-y\Vert^{2}$.

Then, we have $\Vert Tx-Ty\Vert^{2}\leq||x-y\Vert^{2}$ and hence $Tx-Ty\Vert\leq\Vert x-y\Vert$ . Such a mappingis1
nonexpansive. We know that $T:Carrow H$ is nonexpansive if and only if$2\Vert Tx-Ty\Vert^{2}\leq\Vert x-Ty\Vert^{2}+\Vert y-Ts\Vert^{2}-2(x-Tx,$ $y-Ty\rangle,$$\forall x,$$y\in C$;

see $[29|$ . Thus, we can get new classes of nonlinear operators which contain the class of firmly
nonexpansive mappings in a Hilbert space. For example, Kohsaka and Takahahi [12] obtained
a nonspreading mapping. Let $T:Carrow H$ be a firmly nonexpansive mapping. Then, we havethat for all $x,$ $y\in C$ ,$2\Vert Tx-Ty||^{2}\leq 2\langle x-y,$ $Tx-Ty\rangle$

.

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From (3.2), we obtain$2\Vert Tx-Ty\Vert^{2}\leq\Vert x-Ty\Vert^{2}+\Vert y-Tx\Vert^{2}-\Vert x-Tx\Vert^{2}-\Vert y-Ty\Vert^{2}$$\leq\Vert x-Ty\Vert^{2}+\Vert y-Tx\Vert^{2}$.

So, we have
$2\Vert Tx-Ty\Vert^{2}\leq\Vert x-Ty\Vert^{2}+\Vert y-Tx\Vert^{2}$.

This is a nonspreading mapping. From Iemoto and Takah $as$ hi [10], we know the following
lemma.
Lemma 3.1. Let $C$ be a nonempty closed convex subset of H. Then a mapping $S:Carrow H$
is nonspreading if and only if$\Vert Sx-Sy\Vert^{2}\leq\Vert x-y\Vert^{2}+2\langle x-Sx,$ $y-Sy\rangle$

for all$x,$ $y\inC$.

Further, from a firmly nonexpansive mapping, i.e.,$\Vert Tx-Ty\Vert^{2}\leq${Tx--Ty, $x-y$ },$\forall x,$$y\in C$ ,we have$\Vert Tx-Ty\Vert^{2}\leq 2\langle Tx-Ty,$ $x-y\rangle$,$y\in C$$\forall x,$.

Such a mapping $T:Carrow H$ is called -inverse strongly monotone. Takahashi [29] also defined
the following hybnid mapping, i.e.,$\frac{1}{2}$$3\Vert Tx-Ty\Vert^{2}\leq\Vert x-y\Vert^{2}+\Vert y-Tx\Vert^{2}+\Vert x-Ty\Vert^{2}$,$\forall x,$$y\in C$.

From Takahashi [29], we know the following lemma.
Lemma 3.2. Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subset of $H$ .
Then a mapping $T:Carrow H$ is hybntd if and only if$\Vert Tx-Ty\Vert^{2}\leq\Vert x-y||^{2}+\langle x-Tx$

,

$y$

–

$Ty$

$\}$

,

$\forall x,$$y\in C$.

So, a hybrid mapping $T$ : $Carrow H$ is different from a nonspreading mapping. Further, we
define a new nonlinear operator from a firmly nonexpansive mapping. We have that for any$x,$ $y\in

C$ ,$2\Vert Tx-Ty\Vert^{2}\leq 2\langle x-y$$\Leftrightarrow\Vert$

$Tx$

–

, Tx–Ty$\}$$Ty$ $\Vert^{2}+||Tx\Vert^{2}+\Vert Ty\Vert^{2}-2\langle Tx,$ $Ty\rangle\leq 2\langle x-y,$ $Tx-Ty\rangle$$\Rightarrow\Vert Tx-Ty\Vert^{2}-2\langle Tx,$ $Ty\rangle\leq 2\langle x-y,$ $Tx-Ty\rangle$$\Leftrightarrow\Vert$

Tx–Ty$\Vert^{2}\leq 2\langle Tx,$

$Ty\rangle+2\langle x-y$

, Tx–Ty .
$\}$So, we can define a new mapping called a metric mapping, i.e.,$\Vert Tx-Ty||^{2}\leq 2\langle Tx,$ $Ty\rangle+2\langle x-y,$ $Tx-Ty\rangle$,$\forall x,$$y\in C$.

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Let $T:Carrow H$ be a nonexpansive mapping and put $A=I-T$ . Then, we have from [28] thatis 1/2-inverse strongly monotone, i.e.,$A$$\frac{1}{2}\Vert Ax-Ay\Vert^{2}\leq\langle x-y,$,$Ax-Ay\rangle$$\forall x,$$y\in C$.

Let $T:Carrow H$ be a nonspreading mapping and put $A=I-T$ . Then, we have from Lemma
3.1 and (3.1) that for any $x,$ $y\in C$ ,$\Vert$$Ax$–$Ay$ $\Vert^{2}=\Vert x-y-(Ax-Ay)\Vert^{2}-\Vert x-y\Vert^{2}+2\langle x-y,$ $Ax-Ay\rangle$$=\Vert Tx-Ty\Vert^{2}-\Vert x-y\Vert^{2}+2\langle x-y$

, Ax–Ay$\}$$\leq\Vert x-y\Vert^{2}+2\langle x-Tx,$ $y-Ty\rangle-\Vert x-y\Vert^{2}+2\langle x-y,$ $Ax-Ay\rangle$$=2\langle Ax,$ $Ay\rangle+2\langle x-y,$ $Ax-Ay\rangle$.

This implies that $A$ is a metric mapping.

4

Generalized Fixed Point Theorem and its Applications

In this section, we obtain a generalized fixed point theorem in a Hilbert space. Before
proving the theorem, we can obtain the following lemma from Lemma 2.1.
Lemma 4.1. Let $C$ be a nonempty closed convex subset of a Hilbert space $H$ , letbounded sequence in $H$ and let be a Banach limit. If $g:Carrow \mathbb{R}$ is defined by$\{x_{n}\}$

be a

$\mu$

$g(z)=\mu_{n}\Vert x_{n}-z\Vert^{2}$,$\forall z\in C$,

then there exists a unique $z_{0}\in C$ such that$g(z_{0})= \min\{g(z):z\in C\}$ .

Using Lemma 4.1, we can prove the following generalized fixed point theorem [31].
Theorem 4.2. Let $H$ be a Hilbert space, let $C$ be a nonempty closed convex subset of $H$ and
let $T$ be a mapping of $C$ into itself. Suppose that there exists an element $x\in C$ such that$\{T^{n}x\}$ is bounded and$\mu_{n}\Vert T^{n}x-Ty\Vert^{2}\leq\mu_{n}\Vert T^{n}x-y\Vert^{2}$for some Banach limit . Then,
Proof. Using a Banach limit on

$T$

$\mu$

$\mu$

,

$\forall y\in C$

has a fixed point in $C$ .

$\iota\infty$, we can define $g:Carrow \mathbb{R}$ as follows:$g(z)=\mu_{n}\Vert T^{n}x-z\Vert^{2}$,$\forall z\in C$.

From Lemma 4.1, there exists a unique $z_{0}\in C$ such that$g(z_{0})= \min\{g(z):z\in C\}$.

So, we have$g(Tz_{0})=\mu_{n}\Vert T^{n}x-Tz_{0}\Vert^{2}\leq\mu_{n}\Vert T^{n}x-z_{0}\Vert^{2}=g(z_{0})$

.

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Since

$Tz_{0}$is in $C$ and $z_{0}\in C$ is a unique element such that$g(z_{0})= \min\{g(z):z\in C\}$ ,we have $Tz_{0}=z_{0}$ . This completes the proof.

口Using Theorem 4.2, we can obtain some fixed point theorems. The following is the wellknown fixed point theorem for nonexpansive mappings in a Hilbert space; see, for instance,[28].
Theorem 4.3. Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subset of $H$ .
Let $T:Carrow C$ be a nonexpansive mapping, i. e.,$\Vert Tx-Ty\Vert\leq\Vert x-y\Vert$,$\forall x,$$y\in C$.

Suppose that there exists an element $x\in C$ such that $\{T^{n}x\}$ is bounded. Then, $T$ has a fixed
point in $C$ .

Proof. Let$\mu$be a Banach limit on$\iota\infty$. For any$n\in \mathbb{N}$and $y\in C$ , we have$\Vert T^{n+1}x-Ty\Vert^{2}\leq\Vert T^{n}x-y\Vert^{2}$.

So, we have$\mu_{n}\Vert T^{n}x-Ty\Vert^{2}=\mu_{n}\Vert T^{n+1}x-Ty\Vert^{2}\leq\mu_{n}\Vert T^{n}x-y\Vert^{2}$for all $y\in C.$ By Theorem 4.2, $T$ has a fixed point in C.

口

The following is a fixed point theorem for nonspreading mappings in a Hilbert space.

Theorem 4.4 ([12]). Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subsetof H. Let $T:Carrow C$ be a nonspreading mapping, i. e.,$2\Vert Tx-Ty\Vert^{2}\leq\Vert Tx-y\Vert^{2}+\Vert Ty-x\Vert^{2}$,$\forall x,$$y\in C$.

Suppose that there exists an element $x\in C$ such that $\{T^{n}x\}$ is bounded. Then, $T$ has a fixed
point in $C$ .

Proof. Let$\mu$be a Banach limit on$l^{\infty}$. For any$n\in \mathbb{N}$and $y\in C$ , we have

$2\Vert T^{n+1}x-Ty\Vert^{2}\leq\Vert T^{n+1}x-y\Vert^{2}+\Vert T^{n}x-Ty\Vert^{2}$.

So, we have
$2\mu_{n}\Vert T^{n}x-Ty\Vert^{2}=2\mu_{n}\Vert T^{n+1}x-Ty||^{2}$$\leq\mu_{n}\Vert T^{n+1}x-y||^{2}+\mu_{n}\Vert T^{n}x-Ty\Vert^{2}$$=\mu_{n}\Vert T^{n}x-y||^{2}+\mu_{n}||T^{n}x-Ty\Vert^{2}$and hence$\mu_{n}\Vert T^{n}x-Ty\Vert^{2}\leq\mu_{n}\Vert T^{n}x-y\Vert^{2}$.

By Theorem 4.2, $T$ has a fixed point in C.
The following is a fixed point theorem for hybrid mappings in a Hilbert space.

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Theorem 4.5 ([29]). Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subsetof H. Let $T:Carrow C$ be a hybrid mapping, i. e.,$\Vert Tx-Ty\Vert^{2}\leq\Vert x-y\Vert^{2}+\langle x-Tx,$ $y-Ty\rangle$,$\forall x,$$y\in C$.

Suppose that there exists an element $x\in C$ such that $\{T^{n}x\}$ is bounded. Then,point in $C$ .

Proof. Let be a Banach limit on
is hybrid if and only if$\mu$$\iota\infty$$n\in \mathbb{N}$

has a fixed. We know from Lemma 3.2 that a mapping $T:Carrow C$$3\Vert Tx-Ty\Vert^{2}\leq\Vert x-y\Vert^{2}+\Vert Tx-y\Vert^{2}+\Vert Ty-x\Vert^{2}$

So, for any$T$,$\forall x,$$y\in C$.

and $y\in C$ , we have.

$3\Vert T^{n+1}x-Ty\Vert^{2}\leq\Vert T^{n}x-y\Vert^{2}+\Vert T^{n+1}x-y\Vert^{2}+\Vert T^{n}x-Ty\Vert^{2}$

So, we have
$3\mu_{n}\Vert T^{n}x-Ty\Vert^{2}=3\mu_{n}\Vert T^{n+1}x-Ty\Vert^{2}$$\leq 2\mu_{n}\Vert T^{n}x-y\Vert^{2}+\mu_{n}\Vert T^{n}x-Ty\Vert^{2}$and hence$\mu_{n}\Vert T^{n}x-Ty\Vert^{2}\leq\mu_{n}\Vert T^{n}x-y\Vert^{2}$.

By Theorem 4.2, $T$ has a fixed point in C.

$\square$We can also prove the following two fixed point theorems in a Hilbert space.
Theorem 4.6. Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subset of $H$ .
Let $T:Carrow C$ be a mapping such that
$2\Vert Tx-Ty\Vert^{2}\leq\Vert x-y\Vert^{2}+\Vert Tx-y\Vert^{2}$,$\forall x,$$y\in C$.

Suppose that there exists an element $x\in C$ such that $\{T^{n}x\}$ is bounded. Then, $T$ has a
fixedpoint in $C$ .

Proof. Let$\mu$be a Banach limit on$l^{\infty}$. For any$n\in \mathbb{N}$and $y\in C$ , we have

$2\Vert T^{n+1}x-Ty\Vert^{2}\leq\Vert T^{n}x-y\Vert^{2}+\Vert T^{n+1}x-y\Vert^{2}$.

So, we have
$2\mu_{n}\Vert T^{n}x-Ty\Vert^{2}=2\mu_{n}\Vert T^{n+1}x-Ty\Vert^{2}$$\leq 2\mu_{n}\Vert T^{n}x-y||^{2}$and hence$\mu_{n}\Vert T^{n}x-Ty\Vert^{2}\leq\mu_{n}\Vert T^{n}x-y\Vert^{2}$

By Theorem 4.2, $T$ has a fixed point in $C$ .

.
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Theorem 4.7. Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subset of $H$ .
Let $T:Carrow C$ be a mapping such that
$3\Vert Tx-Ty\Vert^{2}\leq 2\Vert Tx-y\Vert^{2}+\Vert Ty-x\Vert^{2}$,$\forall x,$$y\in C$.

Suppose that there exists an element $x\in C$ such that $\{T^{n}x\}$ is bounded. Then, $T$ has a fixed
point in $C$ .

Proof. Let$\mu$be a Banach limit on$l^{\infty}$. For any$n\in \mathbb{N}$and $y\in C$ , we have.

$3\Vert T^{n+1}x-Ty\Vert^{2}\leq 2\Vert T^{n+1}x-y\Vert^{2}+\Vert T^{n}x-Ty\Vert^{2}$

So, we have
$3\mu_{n}\Vert T^{n}x-Ty\Vert^{2}\leq 2\mu_{n}\Vert T^{n}x-y\Vert^{2}+\mu_{n}\Vert T^{n}x-Ty\Vert^{2}$and hence$\mu_{n}||T^{n}x-Ty\Vert^{2}\leq\mu_{n}||T^{n}x-y\Vert^{2}$

By Theorem 4.2,

$T$.

has a fixed point in C.

$\square$We also know the following theorem by Ray [15].
Theorem 4.8. Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subset of $H$ .
Then, the following are equivalent:(i) Every nonexpansive mapping of $C$ into itself has a fixed point in $C$ ;(ii) $C$ is bounded.

Using Ray’s theorem, we can prove the following theorem [29].

Theorem 4.9. Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subset of $H$ .
Then, the following are equivalent:(i) Every hybred mapping of $C$ into itself has a fixed point in $C$ ;(ii) $C$ is bounded.

Proof. From Theorem 4.5, we know that (ii) implies (i). Let us show that (i) implies (ii). We
know that every firmly nonexpansive mapping is a hybrid mapping. So, the class of hybrid
mappings of $C$ into itself contains the class of firmly nonexpansive mappings of $C$ into itself.
(ii), it is sufficient to show that if every firmly nonexpansive mapping in $C$
To show
into itself has a fixed point in $C$ , then every nonexpansive mapping of $C$ into itself has a fixed
point in $C$ . Let
is a firmly
be a nonexpansive mapping of $C$ into itself. Then,nonexpansive mapping. Further, it is not difficult to show $F(T)=F(S)$ . So, every firmly
nonexpansive mapping in $C$ into itself has a fixed point in $C$ if and only if every nonexpansive口mapping of $C$ into itself has a fixed point in $C$ . This completes the proof.
$(i)\Rightarrow$$T$$S= \frac{1}{2}I+\frac{1}{2}T$

Similarly, we have the following theorem.
Theorem 4.10. Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subset of
H. Then, the following are equivalent:(i) Every nonspreading mapping of $C$ into itself has a fixed point in $C$ ;(ii) $C$ is bounded.

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5

Nonlinear Erdodic Theorems

Baillon [1] proved the first nonlinear ergodic theorem for nonexpansive mappings in a Hilbert
space.
Theorem 5.1. Let $H$ be a Hilbert space, let $C$ be a nonempty closed convex subset of $H$ and
let $T$ be a nonexpansive mapping of $C$ into itself such that $F(T)$ is nonempty. Then, for any$x\in C$ ,$S_{n}x= \frac{1}{n}\sum_{k=0}^{n-1}T^{k}x$converges weakly to an element $z\in F(T)$ .
We can also prove the following nonlinear ergodic theorem [31] for our nonlinear operators
in a Hilbert space. Before proving it, we need, for example, the following result [31].

Lemma 5.2. Let $H$ be a Hilbert space and let $C$ be a nonempty closed convex subset of $H$ .
Let $T:Carrow C$ be a mapping such that
$2\Vert Tx-Ty\Vert^{2}\leq\Vert x-y\Vert^{2}+\Vert Tx-y\Vert^{2}$,$\forall x,$$y\in C$.

Then $T$ is demiclosed, i. e., $x_{n}arrow u$ and $x_{n}-Tx_{n}arrow 0$ imply $u\in F(T)$ .

Proof. Let $\{x_{n}\}\subset C$ be a sequence such that $x_{n}arrow u$ and $x_{n}-Tx_{n}arrow 0$ as $narrow\infty$ . Thenthe sequences
and $\{Tx_{n}\}$ are bounded. Suppose that $u\neq Tu.$ From Opial’s theorem[14], we have$\{x_{n}\}$$\lim_{narrow}\inf_{\infty}\Vert x_{n}-u\Vert^{2}<\lim_{narrow}\inf_{\infty}\Vert x_{n}-Tu\Vert^{2}$$= \lim_{narrow}\inf_{\infty}\Vert x_{n}-Tx_{n}+Tx_{n}-Tu\Vert^{2}$$= \lim_{narrow}\inf_{\infty}\Vert Tx_{n}-Tu\Vert^{2}$$\leq\lim_{narrow}\inf_{\infty}\frac{1}{2}(\Vert x_{n}-u\Vert^{2}+\Vert Tx_{n}-u\Vert^{2})$$= \lim_{narrow}\inf_{\infty}\frac{1}{2}(\Vert x_{n}-u\Vert^{2}+\Vert Tx_{n}-x_{n}+x_{n}-u\Vert^{2})$$= \lim_{narrow}\inf_{\infty}\Vert x_{n}-u\Vert^{2}$.

This is a contradiction. Hence we get the conclusion.

$\square$

Theorem 5.3. Let $H$ be a Hilbert space, let $C$ be a nonempty closed convex subset of $H$ and
let $T$ be a mapping of $C$ into itself such that $F(T)$ is nonempty. Suppose that $T$ satisfies oneof the following conditions:(i) $T$ is nonspreading;(ii) $T$ is hybrid;(iii) $2\Vert Tx-Ty\Vert^{2}\leq\Vert x-y\Vert^{2}+\Vert Tx-y\Vert^{2}$ ,(iv) $3\Vert Tx-Ty\Vert^{2}\leq 2\Vert Tx-y\Vert^{2}+\Vert Ty-x\Vert^{2}$ ,$\forall x,$$y\in C$$\forall x,$;$y\in C$.

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Then, for any $x\in C$ ,$S_{n}x= \frac{1}{n}\sum_{k=0}^{n-1}T^{k}x$converges weakly to an element $z\in F(T)$ .

Proof. Let us prove the case of (iii) by using Lemma 5.2. We first show that $F(T)$ is closed

and convex. It follows from Lemma 5.2 that $F(T)$ is closed. In fact, let $\{x_{n}\}\subset F(T)$ and$x_{n}arrow z$ .
Then, we have $x_{n}arrow z$ and $x_{n}-Tx_{n}=0$ . So, from Lemma 5.2 we have $z=Tz$ .
This implies that $F(T)$ is closed. Let us show that $F(T)$ is convex. Let $x,$ $y\in F(T)$ and let$\alpha\in[0,1]$ . Put $z=\alpha x+(1-\alpha)y$ . Then, we
have from (2.1) that$\Vert z-Tz\Vert^{2}=\Vert\alpha x+(1-\alpha)y-Tz\Vert^{2}$$=\alpha\Vert x-Tz\Vert^{2}+(1-\alpha)\Vert y-Tz\Vert^{2}-\alpha(1-\alpha)\Vert x-y\Vert^{2}$$=\alpha\Vert Tx-Tz\Vert^{2}+(1-\alpha)\Vert Ty-Tz\Vert^{2}-\alpha(1-\alpha)\Vert x-y\Vert^{2}$$\leq\alpha(1-\alpha)^{2}\Vert x-y\Vert^{2}+(1-\alpha)\alpha^{2}\Vert x-y\Vert^{2}-\alpha(1-\alpha)\Vert x-y\Vert^{2}$$=\alpha(1-\alpha)(1-\alpha+\alpha-1)\Vert x-y\Vert^{2}$

$=0$

.

So, we have $Tz=z$ . This implies that $F(T)$ is convex. Let $x\in C$ and let $P$ be the metricprojection of $H$ onto $F(T)$ . Then, we have$\Vert PT^{n}x-T^{n}x\Vert\leq\Vert PT^{n-1}x-T^{n}x\Vert$$=\Vert TPT^{n-1}x-T^{n}x\Vert$$\leq\Vert PT^{n-1}x-T^{n-1}x||$.

This implies that $\{\Vert PT^{n}x-T^{n}x\Vert\}$ is nonincreasing. We also know that for any $v\in C$ and$u\in F(T)$ ,$\langle$v–Pv, $Pv-u\rangle\geq 0$

and hence$\Vert v-Pv\Vert^{2}\leq$$\langle$v–Pv, $v-u\rangle$ .

So, we get$\Vert Pv-u\Vert^{2}=||Pv-v+v-u\Vert^{2}$$=||Pv-v||^{2}-2\langle Pv-v,$ $u-v\rangle+||v-u||^{2}$$\leq\Vert v-u\Vert^{2}-\Vert Pv-v||^{2}$

Let $m,$$n\in \mathbb{N}$.

with $m\geq n$ . Putting $v=T^{m}x$ and $u=PT^{n}x$ , we have$\Vert PT^{m}x-PT^{n}x\Vert^{2}\leq\Vert T^{m}x-PT^{n}x\Vert^{2}-\Vert PT^{m}x-T^{m}x\Vert^{2}$$\leq\Vert T^{n}x-PT^{n}x\Vert^{2}-\Vert PT^{m}x-T^{m}x\Vert^{2}$.

So, $\{PT^{n}x\}$ is a Cauchy sequence. Since $F(T)$ is closed, $\{PT^{n}x\}$ converges strongly to anelement of $F(T)$ . Take $u\in F(T)$ . Then we obtain that for any,$p$$n\in \mathbb{N}$$\Vert S_{n}x-u\Vert\leq\frac{1}{n}\sum_{k=0}^{n-1}\Vert T^{k}x-u\Vert\leq\Vert x-u||$

.

187

So,

is bounded and hence there exists a weakly convergent subsequenceof. If $S_{n_{i}}xarrow v$ , then we have
. In fact, for any $y\in C$ and, we have$\{S_{n}x\}$$\{S_{n}x\}$that$\{S_{n_{i}}x\}$$v\in F^{\urcorner}(T)$$k\in \mathbb{N}\cup\{0\}$$2\Vert T^{k+1}x-Ty\Vert^{2}\leq\Vert T^{k}x-y\Vert^{2}+\Vert T^{k+1}x-y\Vert^{2}$$=\Vert T^{k}x-Ty\Vert^{2}+2\langle T^{k}x-Ty,$ $Ty-y\rangle+\Vert Ty-y\Vert^{2}$$+\Vert T^{k+1}x-Ty\Vert^{2}+2\langle T^{k+1}x-Ty,$ $Ty-y\}+\Vert Ty-y\Vert^{2}$.

So, we obtain that$\Vert T^{k+1}x-Ty\Vert^{2}\leq\Vert T^{k}x-Ty\Vert^{2}+2\langle T^{k}x-Ty,$ $Ty-y\rangle$$+2\langle T^{k+1}x-Ty,$ $Ty-y\rangle+2\Vert Ty-y\Vert^{2}$

Summing these inequalities with respect to $k=0,1,$$\ldots,$.

$n-1$ , we have$\Vert T^{n}x-Ty\Vert^{2}\leq\Vert x-Ty\Vert^{2}+2\langle\sum_{k=0}^{n-1}T^{k}x-nTy,$ $Ty-y\rangle$$+2 \langle\sum_{k=0}^{n-1}T^{k+1}x-nTy,$ $Ty-y\rangle+2n\Vert Ty-y\Vert^{2}$$= \Vert x-Ty\Vert^{2}+4\langle\sum_{k=0}^{n-1}T^{k}x-nTy,$ $Ty-y\rangle$$+2\langle T^{n}x-x,$ $Ty-y\}+2n\Vert

Deviding this inequality by$n$Ty-y||^{2}$ .

, we have$\frac{1}{n}\Vert T^{n}x-Ty\Vert^{2}\leq\frac{1}{n}\Vert x-Ty\Vert^{2}+4\langle S_{n}x-Ty,$ $Ty-y\rangle$$+ \frac{2}{n}\langle T^{n}x-x,$

where
that

$S_{n}x= \frac{1}{n}\sum_{k=0}^{n-1}T^{k}x$. Replacing

$n$

by

$n_{i}$$Ty-y\rangle+2\Vert Ty-y\Vert^{2}$and letting$n_{i}arrow\infty$$0\leq 2\Vert Ty-y\Vert^{2}+4\langle v-Ty,$ $Ty-y\rangle$,, we obtain from $S_{n_{i}}xarrow v$.

Putting $y=v$ , we have $0\leq 2\Vert Tv-v\Vert^{2}+4\langle v-Tv,$ $Tv-v\rangle$ and hence$0\leq\Vert Tv-v\Vert^{2}+2\langle v-Tv,$ $Tv-v\rangle$.

So, we have $0\leq-\Vert Tv-v\Vert^{2}$ and hence $Tv=v$ . To complete the proof of (iii), it is sufficient
to show that if $S_{n_{i}}xarrow v$ and $p= \lim_{narrow\infty}PT^{n}x$ , then $v=p$ . We have that for any $u\in F(T)$ ,$\langle T^{k}x-PT^{k}x,$ $PT^{k}x-u\rangle\geq 0$.

Since $\{\Vert T^{k}x-PT^{k}x\Vert\}$ is nonincreasing, we have$\langle u-p,$ $T^{k}x-PT^{k}x\rangle\leq\langlePT^{k}x-p,$ $T^{k}x-PT^{k}x\rangle$$\leq\Vert PT^{k}x-p\Vert\cdot\Vert T^{k}x-PT^{k}x\Vert$$\leq\Vert PT^{k}x-p\Vert\cdot\Vert x-Px\Vert$

.

188

Adding these inequalities from $k=0$ to $k=n-1$ and dividing$n$, we have$\langle u-p,$ $S_{n}x- \frac{1}{n}\sum_{k=0}^{n-1}PT^{k}x\rangle\leq\frac{\Vert x-Px\Vert}{n}\sum_{k=0}^{n-1}\Vert PT^{k}x-p\Vert$

Since$S_{n_{i}}xarrow v$.

and $PT^{k}xarrow p$ , we have$\langle u-p,$ $v-p\rangle\leq 0$.

$v-p\rangle\leq 0$ and hence $\Vert v-p\Vert^{2}\leq 0$ .
We know $v\in F(T)$ . So, putting $u=v$ , we haveSo, we obtain $v=p$ . This completes the proof of (iii). See [31] for the proofs of (1), (ii) and$\langle v-p,$

(iv).

日

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