c
Applied Mathematics E-Notes, 9(2009), 266-273 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
Strong Convergence Theorems Of Cesàro Mean
Iterations Of Nonexpansive Mappings∗
Yi Shuai Yang, Jing Jing Li and Yun An Cui†
Received 15 October 2008
Abstract
In a reflexive and strictly convex Banach space which has uniformly Gâteaux
differentiable norm, we consider the problem of the convergence of the Cesàro
mean iterations for non-expansive mappings. Under suitable conditions, it was
proved that the iterative sequence converges strongly to a fixed point. The results
presented in this paper also extend and improve some recent results.
1
Introduction
Let X be a real Banach space and T a mapping with domain D(T ) and range R(T ) in
X. T is called non-expansive if for any x, y ∈ D(T ) such that
kT x − T yk ≤ kx − yk.
In 2000, Moudafi [5] introduced viscosity approximation methods and proved that
if X is a real Hilbert space, for given x0 ∈ C, the sequence {xn } generated by the
iteration process
xn+1 = αn f(xn ) + (1 − αn )T xn , n ≥ 0,
where f : C → C is a contraction mapping and {αn } ⊆ (0, 1) satisfies certain conditions, converges strongly to a fixed point of T in C.
In last decades, many mathematical workers studied the iterative algorithms for
various mappings, and obtained a series of good results, see [1, 6, 11].
In 2002, Xu [13] obtained the strong convergence of the iteration sequence {xn }
given as follows:
n
xn+1 = αn x + (1 − αn )
1 X j
S xn
n+1
for n = 0, 1, 2, ... ,
j=0
∗ Mathematics
† Department
Subject Classifications: 47H09, 47H10.
of Mathematics, Harbin University of Science and Technology, Harbin 150080, P. R.
China
266
267
Yang et al.
for a non-expansive mapping in a uniformly convex and uniformly smooth Banach
space. In 2004, Matsushita and Kuroiwa [3] extended the result of Xu to a nonexpansive nonself-mapping in the same space. In 2007, Song and Chen [10] proposed
the following viscosity iterative process {xn } given by
n
xn+1 = αn f(xn ) + (1 − αn )
1 X j
T xn , n ≥ 0,
n+1
j=0
and proved that the explicit process {xn } converges to a fixed point p of T in a uniformly
convex Banach space with weakly sequentially continuous duality mapping and {αn }
satisfies certain conditions. Very recently, Wangkeeree [12] extended Song and Chen’s
result to non-expansive nonself-mapping. The author [12] also extended the result of
Matsushita and Kuroiwa [3] to that of a Banach space. The purpose of this paper is
to study the strong convergence for the iterative sequence {xn } defined by
n
xn+1
1 X j
= αn f(xn ) + βn xn + γn
T xn .
n + 1 j=0
(1)
Our results extend and improve the corresponding ones by [8, 9, 10, 13].
2
Preliminaries
Let X be a real Banach space, and let J denote the normalized duality mapping from
∗
X into 2X given by
J(x) = {f ∈ X ∗ : hx, fi = kxkkfk, kfk = kxk},
∀x ∈ X
(2)
where X ∗ denotes the dual space of X and h·, ·i denotes the generalized duality pairing.
In the sequel, we shall denote the single-valued duality mapping by j, and denote
F (T ) = {x ∈ X : T x = x}. When {xn } is a sequence in X, then xn → x (xn *
x, xn + x) will denote strong (weak, weak star)convergence of the sequence{xn } to x.
Recall that the norm is said to be uniformly Gâteaux differentiable if for each
x ∈ SX := {x ∈ X : kxk = 1} the limit limt→0 kx+tyk−kxk
exists uniformly for x ∈ SX .
t
It is well known that every uniformly smooth Banach space has uniformly Gâteaux
∗
differentiable norm, and this implies that the duality mapping J : X → 2X defined by
(2) is single-valued and uniformly continuous on bounded subset of X from the strong
topology of X to the weak star topology of X ∗ (see e.g., [7]). A Banach space X is
said to be strictly convex if kxk = kyk = 1, x 6= y implies k x+y
k < 1. Now, let us first
2
recall the following lemmas.
LEMMA 1.[4] Let X be a real Banach space. For each x, y ∈ X, the following
conclusion holds:
kx + yk2 ≤ kxk2 + 2hy, j(x + y)i,
∀j(x + y) ∈ J(x + y).
LEMMA 2. [2] Let {an }, {bn }, {cn } be three nonnegative
real sequences satisfying
P∞
a
≤
(1
−
t
)a
+
b
+
c
with
{t
}
⊂
[0,
1],
t
= ∞, bn = o(tn ), and
n+1
n
n
n
n
n
n
n=0
P∞
c
<
∞.
Then
a
→
0.
n
n
n=0
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Convergence Of Cesàro Mean Iterations Of Nonexpansive Mappings
LEMMA 3. [11] Let {xn } and {yn } be bounded sequences in a Banach space X and
let {γn } be a sequence in [0, 1] with 0 < lim inf n→∞ γn ≤ lim supn→∞ γn < 1. Suppose
that xn+1 = γn xn + (1 − γn )yn for all n ∈ N and
lim sup(kyn+1 − yn k − kxn+1 − xn k) ≤ 0.
n→∞
Then limn→∞ kyn − xn k = 0.
3
Main Results
Let X be a Banach space, C a nonempty closed convex subset of X, and T : C → C
a non-expansive mapping with F (T ) 6= ∅ and f : C → C be a contraction with
contraction constant α. For tn ∈ (0, 1), define a mapping Ttfn : C → C by
n
Ttfn (z) = tn f(z) + (1 − tn )
1 X i
T z.
n + 1 i=0
Clearly, for each z ∈ C we have Ttfn is a contractive mapping with contraction constant
t = 1 − tn (1 − α). Hence, it follows from Banach’s contractive principle that Ttfn has a
unique fixed point (say) zn ∈ C, that is,
n
zn = tn f(zn ) + (1 − tn )
1 X i
T zn .
n+1
(3)
i=0
Pn
1
i
Now if we set Tn := n+1
i=0 T , then the mapping T : C → C is said to satisfy
property (A) if C is bounded and for each x ∈ C, limn→∞ kTn x − T (Tn x)k = 0.
There exist non-expansive mappings satisfying property (A).
EXAMPLE 4. Take C = [0, 1] and the norm is the ordinary
distance
Pn Euclidean
1
x
on the line. For each x ∈ C, T x = x2 , then Tn x = n+1
i and T (Tn x) =
i=0
2
Pn
1
x
i=0 2i+1 . Hence we have
n+1
kTn x − T (Tn x)k =
≤
≤
k
n
n
i=0
n
i=0
1 X x
1 X x
−
k
n+1
2i n + 1
2i+1
1 X x
x
k i − i+1 k
n + 1 i=0 2
2
n+1 !
1
1
1−
→0
n+1
2
as n → ∞. That is T satisfies property (A).
LEMMA 5. Let C be a nonempty bounded closed convex subset of Banach space
X and T : CP→ C be a non-expansive mapping. For each x ∈ C and the Cesàro means
n
1
i
Tn x = n+1
i=0 T x,
269
Yang et al.
(i) Tn is non-expansive from C to itself.
(ii) limn→∞ kTn+1 x − Tn xk = 0.
PROOF. (i) It is easy to see that Tn is a mapping from C to itself. We now prove
that Tn is non-expansive. In fact, since T is non-expansive, then for each x, y ∈ C we
have
n
kTn x − Tn yk ≤
n
1 X i
1 X
kT x − T i yk ≤
kx − yk = kx − yk.
n + 1 i=0
n + 1 i=0
(4)
This implies that Tn is non-expansive.
(ii) Since C is bounded, it is easy to prove that the sequence {T i x} is bounded and
there exists a constant M > 0 such that M > max{supx∈C kT i xk : i = 0, 1, ...}. Thus,
kTn+1 x − Tn xk =
≤
n+1
n
1 X i
1 X i
T x−
T xk
n + 2 i=0
n + 1 i=0
Pn+1
Pn
n
X
k i=0 T i x − i=0 T i xk
1
+
k
T i xk
n+2
(n + 2)(n + 1) i=0
k
n
≤
X
1
1
kT n+1 xk +
kT i xk
n+2
(n + 2)(n + 1)
i=0
≤
1
1
2M
M+
M=
→ 0.
n+2
n+2
n+2
Since X has a uniformly Gâteaux differentiable norm, the duality mapping is uniformly continuous on bounded subsets of X from the strong topology of X to the weak
star topology of X ∗ (see e.g.,[7]). By Lemma 5 and Theorem 3.2 of [10], it is easy to
prove the following theorem.
THEOREM 6. Let X be a reflexive and strictly convex Banach space which has
uniformly Gâteaux differentiable norm, C a nonempty closed convex subset of X,
T : C → C a non-expansive mapping with F (T ) 6= ∅ and f : C → C be a contraction
with contraction constant α. For any given z0 ∈ C, let {zn } be the iterative sequence
defined by (3) and limn→∞ tn = 0. Suppose that T satisfies property (A). Then {zn }
converges strongly to some fixed point p of T .
LEMMA 7. Let C be a nonempty closed convex subset of a real Banach space X,
T be a non-expansive self-mapping of C with F (T ) 6= ∅. Let {αn }, {βn }, {γn } be
three
P∞ real sequences in [0, 1] and satisfy (i) αn + βn + γn = 1; (ii) limn→∞ αn = 0 and
βn ≤ lim supn→∞ βn < 1. Let the sequence {xn }
n=0 αn = ∞; (iii) 0 < lim inf n→∞
1 Pn
j
be defined by (1) and Tn = n+1
j=0 T , then we have
(a) {xn } is bounded;
(b) limn→∞ kxn − Tn xn k = 0.
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Convergence Of Cesàro Mean Iterations Of Nonexpansive Mappings
PROOF. (a). From Lemma 5(i) we have Tn is non-expansive. We now show that
the sequence {xn } is bounded. In fact, take u ∈ F (T ),
kxn+1 − uk =
≤
≤
kαn (f(xn ) − u) + βn (xn − u) + γn (Tn xn − u)k
αn kf(xn ) − uk + βn kxn − uk + γn kTn xn − uk
(1 − (1 − α)αn )kxn − uk + αn kf(u) − uk.
It implies by induction that
kxn − uk ≤ max kx0 − uk,
1
kf(u) − uk
1−α
and {xn } is bounded, so are {f(xn )}, {T xn } and {Tn xn }. Next we rewrite the iteration
process (1) as follows
xn+1
=
=
αn f(xn ) + βn xn + γn Tn xn
αn
γn
βn xn + (1 − βn )
f(xn ) +
Tn xn .
1 − βn
1 − βn
Thus, if we set yn = γ̃n f(xn ) + (1 − γ̃n )Tn xn , where γ̃n =
αn
,
1−βn
then we get
xn+1 = βn xn + (1 − βn )yn .
It is easy to check that {yn } is bounded.
Next we show that limn→∞ kxn − T xn k = 0. To see this, we calculate
yn+1 − yn
=
=
γ̃n+1 f(xn+1 ) + (1 − γ̃n+1 )Tn+1 xn+1 − γ̃n f(xn ) − (1 − γ̃n )Tn xn
γ̃n+1 (f(xn+1 ) − f(xn )) + (1 − γ̃n+1 )(Tn+1 xn+1 − Tn+1 xn )
+(1 − γ̃n+1 )(Tn+1 xn − Tn xn ) + (γ̃n+1 − γ̃n )(f(xn ) − Tn xn ).
It follows that
kyn+1 − yn k ≤
αγ̃n+1 kxn+1 − xn k + (1 − γ̃n+1 )kTn+1 xn − Tn xn k
+(1 − γ̃n+1 )kxn+1 − xn k + |γ̃n+1 − γ̃n |kf(xn ) − Tn xn k.
This implies that
kyn+1 − yn k −
≤
kxn+1 − xn k
(α − 1)γ̃n+1 kxn+1 − xn k + |γ̃n+1 − γ̃n |kf(xn ) − Tn xn k
+(1 − γ̃n+1 )kTn+1 xn − Tn xn k.
Thus, we have from (ii) and Lemma 5(ii) that
lim sup(kyn+1 − yn k − kxn+1 − xn k) ≤ 0.
n→∞
Apply Lemma 3 to get kyn − xn k = 0. Again by Eq. (5) we get
lim kxn+1 − xn k = lim (1 − βn )kyn − xn k = 0.
n→∞
n→∞
(5)
271
Yang et al.
This implies from (1) that
kxn − Tn xn k ≤
≤
kxn+1 − xn k + kxn+1 − Tn xn k
kxn+1 − xn k + αn kf(xn ) − Tn xn k + βn kxn − Tn xn k.
It follows that
kxn − Tn xn k ≤
αn
1
kxn+1 − xn k +
kf(xn ) − Tn xn k → 0.
1 − βn
1 − βn
(6)
We give an example concerning {αn }, {βn } and {γn }.
1
EXAMPLE 8. For each n ≥ 1, we set αn = n+4
, γn = 1 − αn − βn and
1
1
if n is odd
3 + n+5
.
βn =
1
1
+
if n is even
4
n+6
Then {αn }, {βn } and {γn } satisfying the assumption of Lemma 7.
THEOREM 9. Let X be a reflexive and strictly convex Banach space which has
uniformly Gâteaux differentiable norm, C a nonempty closed convex subset of X,
T : C → C a non-expansive mapping with F (T ) 6= ∅ and f : C → C be a contraction
with contraction constant α. Let {αn }, {βn }, {γn } be three
sequences in [0, 1]
Preal
∞
and satisfy (i) αn + βn + γn = 1; (ii) limn→∞ αn = 0 and n=0 αn = ∞; (iii) 0 <
lim inf n→∞ βn ≤ lim supn→∞ βn < 1. Let the sequence {xn } be defined by (1). Suppose
T satisfies property (A), then {xn } converges strongly to fixed point of T .
PROOF. By Lemma 7, We have the following assertions:
(I) {xn } is bounded, so are {f(xn )} and {Tn xn };
(II) limn→∞ kTn xn − xn k = 0.
We show that
lim suphp − f(p), j(p − xn )i ≤ 0.
(7)
n→∞
Indeed we can write zm − xn = tm (f(zm ) − xn ) + (1 − tm )(Tm zm − xn ). Putting
Pn (m)
=
(kTm xn − Tn xn k + kTn xn − xn k)(kTm xn − Tn xn k
+kTn xn − xn k + 2kzm − xn k),
then we have from (II) that
lim sup kTm xn − Tn xn k ≤
n→∞
lim sup kTn−m xn − xn k
n→∞
≤
lim sup kTn xn − xn k = 0.
n→∞
It follows that lim supn→∞ Pn (m) = 0 and using Lemma 1, we obtain
kzm − xn k2
≤
≤
≤
(1 − tm )2 kTm zm − xn k2 + 2tm hf(zm ) − xn , J(zm − xn )i
(1 − tm )2 (kTm zm − Tm xn k + kTm xn − Tn xn k + kTn xn − xn k)2
+2tm hf(zm ) − xn , J(zm − xn )i
(1 − tm )2 kzm − xn k2 + Pn (m) + 2tm kzm − xn k2
+2tm hf(zm ) − zm , J(zm − xn )i.
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Convergence Of Cesàro Mean Iterations Of Nonexpansive Mappings
The last inequality implies
hzm − f(zm ), J(zm − xn )i ≤
tm
1
kzm − xn k2 +
Pn (m).
2
2tm
This implies that
lim suphzm − f(zm ), J(zm − xn )i ≤ M
n→∞
tm
2
where M > 0 is a constant such that M ≥ kzm − xn k2 for all m, n ≥ 1. Taking the
lim sup as m → ∞, by Theorem 6 we obtain (7).
Finally we show that xn → p. Apply Lemma 1 and (1) to get
kxn+1 − pk2
=
≤
≤
kαn (f(xn ) − p) + βn (xn − p) + γn (Tn xn − p)k2
kβn (xn − p) + γn (Tn xn − p)k2 + 2αn hf(xn ) − p, J(xn+1 − p)i
(1 − αn )2 kxn − pk2 + 2αn hf(xn ) − f(p), J(xn+1 − p)i
≤
+2αn hf(p) − p, J(xn+1 − p)i
(1 − αn )2 kxn − pk2 + 2ααn kxn − pkkxn+1 − pk
+2αn hf(p) − p, J(xn+1 − p)i
≤
(1 − αn )2 kxn − pk2 + ααn (kxn − pk2 + kxn+1 − pk2 )
+2αn hf(p) − p, J(xn+1 − p)i.
It then follows that
kxn+1 − pk2
≤
≤
≤
1 − αn (2 − α)
α2n
kxn − pk2 +
kxn − pk2
1 − ααn
1 − ααn
2αn
hf(p) − p, J(xn+1 − p)i
+
1 − ααn
α2n
2(1 − α)αn
kxn − pk2 +
kxn − pk2
1−
1 − ααn
1 − ααn
2αn
+
hf(p) − p, J(xn+1 − p)i
1 − ααn
αn
(1 − t̃n )kxn − pk2 +
(αn M + 2γ̃n+1 )
1 − ααn
n
where t̃n = 2(1−α)α
, M > supn kxn − pk2 and γ̃n+1 = max{hf(p) − p, J(xn+1 − p)i, 0}.
1−ααn
It is easily seen that limn→∞ γ̃n+1 = 0. Apply Lemma 2 to conclude that xn → p.
Acknowledgment. The authors would like to thank Prof. Nikhat Bano and the
anonymous referee for his valuable suggestions which helps to improve this manuscript.
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