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CONTINUOUS PSEUDOCONTRACTIONS IN BANACH SPACESANIEFIOK UDOMENE Received 27 June 2005; Revised 21 November 2005; Accepted 28 November 2005 LetE be a reflexive Banach space with a uniforml

CONTINUOUS PSEUDOCONTRACTIONS IN BANACH SPACES

ANIEFIOK UDOMENE

Received 27 June 2005; Revised 21 November 2005; Accepted 28 November 2005

LetE be a reflexive Banach space with a uniformly Gˆateaux differentiable norm, let K

be a nonempty closed convex subset ofE, and let T : K → K be a uniformly continuous

pseudocontraction If f : K → K is any contraction map on K and if every nonempty

closed convex and bounded subset ofK has the fixed point property for nonexpansive

self-mappings, then it is shown, under appropriate conditions on the sequences of real numbers{α n },{μ n }, that the iteration processz1∈ K, z n+1 = μ n(α n Tz n+ (1− α n)z n) + (1− μ n)f (z n), n ∈ N, strongly converges to the fixed point ofT, which is the unique

solution of some variational inequality, provided thatK is bounded.

Copyright © 2006 Aniefiok Udomene This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

LetE be a real Banach space with dual E ∗andK a nonempty closed convex subset of

E Let J : E →2E ∗ denote the normalized duality mapping defined by J(x) := { f ∈ E ∗:

x, f  = x2,  f  = x, x ∈ E} where·,· denotes the generalized duality pairing.

Following Morales [6], a mappingT with domain D(T) and range ᏾(T) in E is called strongly pseudocontractive if for some constant k < 1 and ∀x, y ∈ D(T),

(λ − k)x − y ≤(λI − T)(x) −(λI − T)(y) (1.1)

for allλ > k; while T is called a pseudocontraction if (1.1) holds fork =1 The mappingT is called Lipschitz if there exists L ≥0 such thatTx − T y ≤ Lx − y,∀x, y ∈ D(T) The

mappingT is called nonexpansive if L = 1 and is called a (strict) contraction if L < 1 Every

nonexpansive mapping is a pseudocontraction The converse is not true The example,

T(x) =1− x2/3, 0≤ x ≤1, is a continuous pseudocontraction which is not nonexpansive

It follows from a result of Kato [3] thatT is pseudocontractive if and only if there exists j(x − y) ∈ J(x − y) such that Tx − T y, j(x − y) ≤ x − y2,∀x, y ∈ D(T).

Hindawi Publishing Corporation

Fixed Point Theory and Applications

Volume 2006, Article ID 69758, Pages 1 12

DOI 10.1155/FPTA/2006/69758

In [9], Schu introduced the iterative process (1.2) below and proved the following theorem

Theorem 1.1 [9, Theorem 2.4, page 113] Let K be a nonempty, closed convex, and bounded subset of a Hilbert space H; let T : K → K be a Lipschitz pseudocontractive map with Lipschitz constant L ≥ 0; {λ n } n ∈N ⊂ (0, 1) with lim n →∞ λ n = 1; {α n } n ∈N ⊂ (0, 1) with

limn →∞ α n = 0 such that ( {α n },{μ n } ) has property (A), {(1− μ n)(1− λ n)−1} is bounded, and lim n →∞(1− μ n)/α n = 0, where k n:=(1 +α2

n(1 +L)2)1/2 and μ n:= λ n /k n , for all n ∈ N; fix an arbitrary point w ∈ K, and define that for all n ∈ N,

z n+1:= μ n+1

α n Tz n+

1− α n

z n

+

1− μ n+1

Then {z n } n converges strongly to the unique fixed point of T closest to w.

Here the pair of sequences ({α n } n,{μ n } n)⊂(0,∞)× (0, 1) is said to have property (A)

if and only if the following conditions hold

(i) {α n } nis decreasing;

(ii) {μ n } nis strictly increasing;

(iii) There exists a strictly increasing sequence{β n } n ⊂ Nsuch that

(a) limn(α n − α n+β n)/(1 − μ n)=0;

(b) limn(1− μ n+β n)(1− μ n)−1=1;

(c) limn β n(1− μ n)= ∞

The first iterative process of this nature was introduced by Halpern [2]: for any fixed

w ∈ K and arbitrary z0∈ K,

z n+1 = μ n Tz n+

1− μ n

w, n =0, 1, 2, , (1.3) where{μ n }is a sequence in (0, 1) with limn →∞ μ n =1

In [8], Moudafi proposed a viscosity approximation method of selecting a particu-lar fixed point of a given nonexpansive mapping in Hilbert spaces, where he proved the following theorem

Theorem 1.2 [8, Theorem 2.2, page 48] LetH be a Hilbert space, let T : K → K be a nonexpansive self-mapping of a nonempty closed convex subset K of H, and let f : K → K

be a contraction With an initial z0∈ K, define the sequence {z n } by

z n+1 = 1

1 + n Tz n+  n

1 + n fz n

Supposed that lim n →∞  n = 0,∞

n =1 n = ∞, and lim n →∞ |1/ n+1 −1/ n | = 0 Then {z n } converges strongly to the unique solution of the variational inequality:

find x∈ F(T) such that(I − f ) x,x− x≤0, ∀x ∈ F(T), (1.5)

(i.e., the unique solution of the operator Proj F(T) ◦ f ).

Xu [12] extendedTheorem 1.2to the more general uniformly smooth Banach spaces.

IfΠK denotes the set of all contractions onK, he proved the following theorem.

Theorem 1.3 [12, Theorem 4.2, page 289] LetE be a uniformly smooth Banach space, K

a closed convex subset of E, and T : K → K a nonexpansive mapping with F(T) = ∅, and

f ∈ΠK Assume that {α n } ⊂ (0, 1) satisfies the following conditions:

(i) limn →∞ α n = 0;

(ii)∞

n =0α n = ∞;

(iii) either lim n →∞ α n+1 /α n = 1 or∞

n =0|α n+1 − α n | < ∞.

Then the sequence {z n } generated by z0∈ K,

z n+1:= α n fz n

+

1− α n

Tz n, n =0, 1, 2, , (1.6)

converges strongly to Q( f ), where Q : Π K → F(T) is defined by Q( f ) := σ −limt →0x t , with

x t satisfying

x t = tTx t+ (1− t) fx t

LetK be a nonempty closed convex and bounded subset of a real reflexive Banach

space with a uniformly Gˆateaux differentiable norm Further to Theorems1.2and1.3, the purpose of this paper is to use the following iteration process:z1∈ K,

z n+1 = μ n

α n Tz n+

1− α n

z n

+

1− μ n

fz n

where{μ n } n,{α n } nare sequences in (0, 1) and f : K → K is a contraction map, to

approx-imate the fixed point of a uniformly continuous pseudocontraction, which solves some variational inequality If the mapf is a constant map then we recover the iteration process

(1.2) from (1.8)

2 Preliminaries

LetE be a real normed linear space and let S := {x ∈ E : x =1}.E is said to have a Gˆateaux differentiable norm and E is called smooth if the limit

lim

t →0

x + ty − x

exists for eachx, y ∈ S E is said to have a uniformly Gˆateaux differentiable norm if for

eachy ∈ S the limit is attained uniformly for x ∈ S.

The modulus of smoothness of E is defined by

ρ E(τ) := sup

x + y+x − y

2 −1 :x =1,y = τ , τ > 0. (2.2)

E is equivalently said to be smooth if ρ E(τ) > 0 ∀τ > 0 Every uniformly smooth Banach

space is a reflexive Banach space with a uniformly Gˆateaux differentiable norm An ex-ample given in [7] illustrates that this inclusion is proper

LetE be a linear space and let K be a subset of E Then, for any x ∈ K, the set I K(x) = {x + λ(z − x) : z ∈ K, λ ≥1} is called the inward set of x A mapping T : K → E is said to satisfy the inward condition if Tx ∈ I K(x) for each x ∈ K, and is said to satisfy the weakly inward condition if Tx ∈ cl[I K(x)], the closure of I K(x), for each x ∈ K.

We will let LIM

n be a Banach limit Recall that LIM

n ∈( ∞)∗ such that LIM

n  =1, lim infn →∞ a n ≤LIM

n a n ≤lim supn →∞ a n, and LIM

n a n =LIM

n a n+1for all{a n } n ∈  ∞

The modulus of uniform continuity, δ(), ofT is defined for all  > 0 by

δ()=sup{λ : x − y < λ =⇒ Tx − T y < } (2.3)

andδ(0) =0 By [4, Proposition 3],δ() is nondecreasing, 0≤ δ()≤ ∞, andδ(Tx −

T y)≤ x − y, for allx, y ∈ E Furthermore, [4, Propositions 1 and 2] assert that the function

φ(t) =sup{s : δ(s) ≤ t} (2.4)

called the pseudo-inverse of δ is nondecreasing and right continuous, 0 ≤ φ(t) ≤ ∞for

t ≥0 andTx − T y ≤ φ(x − y)∀x, y ∈ E.

The following lemmas will be needed in the sequel.Lemma 2.1is well known, (see, e.g., [7]) The proof ofLemma 2.2can be deduced from [11, Lemma 2.5]

Lemma 2.1 Let E be an arbitrary real Banach space Then

x + y2≤ x2+ 2

for all x, y ∈ E and for all j(x + y) ∈ J(x + y).

Lemma 2.2 Let {a n } n be a sequence of nonnegative real numbers such that

a n+1 ≤1− α n

where {α n } n ⊂ [0, 1], {β n } n ⊂ [0, 1], and∞

n =0α n = ∞, lim n →∞ β n = 0 Then, lim n →∞ a n = 0.

Lemma 2.3,Proposition 2.4, andLemma 2.5that follow appear in [10] For complete-ness, we present also their proofs

Lemma 2.3 Let E be a Banach space Suppose K is a nonempty closed convex subset of E and T : K → E is a continuous pseudocontraction satisfying the weakly inward condition Then for each contraction map f : K → K, with contraction constant α ∈ [0, 1), there exists

a unique continuous path t → x t ∈ K, t ∈ [0, 1) satisfying

x t = tTx t+ (1− t) fx t

Proof Let f : K → K be a contraction map with constant α ∈[0, 1) Then, for eacht ∈

[0, 1), the mappingT t f :K → E defined by T t f(x) = tTx + (1 − t) f (x) is a continuous

strong pseudocontraction with constantt + (1 − t)α ∈[0, 1), which satisfies the weakly inward condition By [1, Corollary 1],T t f has a unique fixed pointx t ∈ K, that is,

x t = tTx t+ (1− t) fx t

To prove the continuity of the path, we follow the same line of arguments as in [7] Let

t0∈[0, 1) Then for allj(x t − x t0)∈ J(x t − x t0),

x t − x t

0  2

= tTx t − Tx t0,jx t − x t0



+ (1− t)fx t

− fx t0



,jx t − x t0



+

t − t0 

Tx t0− fx t0



,jx t − x t0



≤(t + (1 − t)α)x t − x t

0  2 + t − t0 Tx t

0− f (x t0)x t − x t

0 , (2.9)

so thatx t − x t0 ≤(|t − t0|/(1 − t)(1 − α))Tx t0− f (x t0) Hence the proof 

Proposition 2.4 Let E be a Banach space and let K be a nonempty closed convex subset

of E Let the mapping T : K → E be a pseudocontraction such that for each contraction map,

f : K → K with contraction constant α ∈ [0, 1), the equation

has a solution x t for every t ∈ [0, 1) Then the following hold.

(i) If for some u ∈ K, the path y t = tT y t+ (1− t)u is bounded, then for any contraction map f : K → K, the path {x t } described by ( 2.7 ) is bounded.

(ii) If T has a fixed point in K, then the path {x t } is bounded.

(iii) If x ∗ ∈ F(T), then for all j(x t − x ∗)∈ J(x t − x ∗ ),



x t − fx t

,jx t − x ∗

(iv) If 0 ≤ s ≤ t < 1 then

x t − Tx t  ≤ 1 +α

1− αx s − Tx s. (2.12)

Proof (i) Let the path {y t }given byy t = tT y t+ (1− t)u, for some u ∈ K, be bounded.

Then the set{ f (y t)}is bounded Letj(x t − y t)∈ J(x t − y t) From the estimates

x t − y t 2

= tTx t − T y t,jx t − y t

+ (1− t)fx t

− u, jx t − y t

≤ tx t − y t 2

+ (1− t)f

x t

− ux t − y t, (2.13)

we have thatx t − y t  ≤  f (x t)− u ≤ αx t − y t + f (y t)− u Thus,

x t − y t  ≤ 1

1− αf

y t

Hence,{x t }is bounded

(ii) Letx ∗ ∈ F(T), and let j(x t − x ∗)∈ J(x t − x ∗) Then

x t − x ∗ 2

= tTx t − x ∗,jx t − x ∗

+ (1− t)fx t

− x ∗,jx t − x ∗

≤ tx t − x ∗ 2

+ (1− t)f

x t

− x ∗x t − x ∗ (2.15)

so thatx t − x ∗  ≤  f (x t)− x ∗  ≤ αx t − x ∗ + f (x ∗)− x ∗  Thus,

x t − x ∗  ≤ 1

1− αf

x ∗

Hence,{x t }is bounded

(iii) Letx ∗ ∈ F(T), and let j(x t − x ∗)∈ J(x t − x ∗) Then



x t − fx t

,jx t − x ∗

= tTx t − fx t

,jx t − x ∗

= tTx t − x ∗,jx t − x ∗

+tx ∗ − fx t

,jx t − x ∗

≤ tx t − fx t

,jx t − x ∗

.

(2.17)

Thus,x t − f (x t),j(x t − x ∗) ≤0

(iv) Let 0≤ s ≤ t < 1 Then

x t − Tx t  =1− t

t x t − f

x t

≤1− t t

(1 +α)x t − x s+ s

1− sx s − Tx s

≤1− t t



(1 +α)(t − s)

(1− α)(1 − t)(1 − s)+1− s s



x s − Tx s

≤(1 +α)(1 − t)

(1− α)t



t − s

(1− t)(1 − s)+1− s s



x s − Tx s

=1 +α

1− αx s − Tx s.

(2.18)



Lemma 2.5 Let E be a reflexive Banach space with a uniformly Gˆateaux differentiable norm, let K be a nonempty closed convex subset of E, let T : K → E be a continuous pseudo-contraction satisfying the weakly inward condition, and let f : K → K be a contraction map with constant α ∈ [0, 1) Suppose that every nonempty closed convex and bounded subset of

K has the fixed point property (f.p.p.) for nonexpansive self-mappings If there exists u0∈ K such that the set

B =x ∈ K : Tx = u0+λx − u0



for some λ > 1 (2.19)

is bounded, then the path {x t }, t ∈ [0, 1) described by ( 2.7 ) converges strongly to the fixed point of T, which is the unique solution of the variational inequality

p ∈ F(T) such thatp − f (p), jp − x ∗

≤0, x ∗ ∈ F(T). (2.20)

Proof It follows fromLemma 2.3that for each contraction map f : K → K there exists

a unique continuous patht → x t ∈ K, t ∈[0, 1) satisfying (2.7) Let there existsu0∈ K

such that the setB is bounded Then byProposition 2.4(i), the path{x t }described by (2.7) is bounded It is easy to see that this implies that the set { f (x t) :t ∈[0, 1)} is

bounded The boundedness of the set{Tx t:t ∈[0, 1)}follows fromProposition 2.4(iv) Let supt ∈[0,1)x t  ≤ M Then x t − x s  ≤2M for any t,s ∈[0, 1) Setx n = x t n fort n →

1− Defineψ : K → Rbyψ(x) =LIM

n x n − x2 ∀x ∈ K Since E is reflexive, ψ is

con-vex, continuous andψ(x) → ∞ asx → ∞, we have that the setC := {y ∈ K : ψ(y) =

infx ∈ K ψ(x)}is nonempty, closed and convex We show thatC is bounded Let y ∈ C.

Thenψ(y) ≤LIM

n x n − x02≤4M2, wherex0≡ x t0 Applying the convexity of the func-tional (1/2) · 2:K → R, we deduce that

y2≤2LIM

n x n − y 2

+ 2LIM

n x n 2

≤2ψ(y) + 2M2≤10M2, (2.21)

that is,y ≤ √10M, ∀y ∈ C Thus, C is bounded The mapping J1=(2I − T) −1 is a nonexpansive self-mapping ofK (see [5, Theorem 6]).C is invariant under J1 Indeed, let

y ∈ C Then

ψJ1(y)=LIM

n x n − J1(y) 2

≤LIM

n x n − J1

x n+x n − y 2

≤LIM

n x n − Tx n+x n − y 2

=LIM

n x n − y 2

By hypothesis, J1 has a fixed point p ∈ C Thus, T p = p Let τ ∈(0, 1) Thenψ(p) ≤ ψ((1 − τ)p + τx), x ∈ K, and using Lemma 2.1, we have that 0≤(ψ((1 − τ)p + τx) − ψ(p))/τ ≤ −2LIM

n x − p, j(x n − p − τ(x − p)) Thus

LIM

n



x − p, jx n − p − τ(x − p)≤0. (2.23)

Since, in this setting,J is norm-to-weak ∗uniformly continuous on bounded sets, letting

τ →0, we have that

LIM

n



x − p, jx n − p≤0, x ∈ K. (2.24)

In particular,

LIM

n



Observe that

(1− α)x n − p 2

≤x n − fx n

,jx n − p+

f (p) − p, jx n − p. (2.26) Using Proposition 2.4(iii) and (2.25), we have find that LIM

n x n − p =0 Therefore, there exists a subsequence{x n k}of{x n }such thatx n k → p as k → ∞ Assume that there is another subsequence{x n l }of{x n }such thatx n l → q ∈ F(T) as l → ∞ Withx n k → p and

settingx ∗ = q, it follows fromProposition 2.4(iii) that



Also, withx n l → q and setting x ∗ = p inProposition 2.4(iii), we have that



Inequalities (2.27) and (2.28) yield that

p − q2≤f (p) − f (q), j(p − q)≤ αp − q2, (2.29) which implies thatp = q, since α ∈[0, 1) Thus,x n → p as n → ∞andp ∈ F(T) is unique.

Again, usingProposition 2.4(iii), we observe that



p − f (p), jp − x ∗

Hence,p is the unique solution of the variational inequality (2.20) This concludes the

3 Main results

In the results that follow, if the mapT is uniformly continuous and δ() denotes the modulus of continuity ofT, we will let φ denote the pseudoinverse of δ and will assume

that the set{φ(t)/t : 0 < t < 1}is bounded Observe that ifT is Lipschitz, then it is clear

that the set{φ(t)/t : 0 < t < 1}is bounded

Theorem 3.1 Let K be a nonempty closed convex and bounded subset of a real Banach space E Let T : K → K be a uniformly continuous pseudocontraction and let f : K → K be

a contraction map with contraction constant α ∈ [0, 1) Let {z n } be a sequence generated from an arbitrary z1∈ K by ( 1.8 ), where {μ n }, {α n } are real sequences in (0, 1) satisfying the following conditions:

(i){α n } is decreasing and lim n →∞ α n = 0;

(ii) limn →∞ μ n = 1 and∞

n =0(1− μ n)= ∞;

(iii) (a) limn →∞(1− μ n)/α n = 0,

(b) limn →∞ α2

n /(1 − μ n)= 0,

(c) limn →∞ |μ n − μ n −1|/(1 − μ n)2= 0,

(d) limn →∞(α n −1− α n)/α n −1(1− μ n)= 0.

Then z n − Tz n  → 0 as n → ∞.

Proof We first prove that z n − x n  →0 asn → ∞, where {x n }is a sequence satisfying (2.7)

Sett n = α n /(1 − μ n+α n), ∀n ∈ N Thent n ∈(0, 1) for eachn ∈ N By the given con-dition (iii)(a),t n →1 asn → ∞ It follows fromLemma 2.3 that there exists a unique sequence{x n } ⊂ K satisfying the following conditions:

x n = t n Tx n+

1− t n

fx n

Equation (3.1) can be rewritten as follows:

x n = μ n

α n Tx n+

1− α n

x n

+

1− μ n

fx n

+

1− μ n

α n

Tx n − x n

Using the pseudocontractivity ofT, we make the following estimates:

z n+1 − x n 2

= μ n α n

Tz n − Tx n,jz n+1 − x n

+μ n

1− α n

z n − x n,jz n+1 − x n

+

1− μ n

fz n

− fx n

,jz n+1 − x n

+

1− μ n

α n

x n − Tx n,jz n+1 − x n

= μ n α n

Tz n+1 − Tx n,jz n+1 − x n

+μ n α n Tz n − Tz n+1,jz n+1 − x n

+μ n

1− α n

z n − x n,jz n+1 − x n

+

1− μ n

fz n

− fx n

,jz n+1 − x n

+

1− μ n

α n

x n − Tx n,jz n+1 − x n

≤ μ n α nz n+1 − x n 2

+μ n α nTz n − Tz n+1z n+1 − x n

+μ n

1− α nz n − x nz n+1 − x n+

1− μ nf

z n

− fx nz n+1 − x n

+

1− μ n

α nx n − Tx nz n+1 − x n.

(3.3) Thus, we have that

z n+1 − x n  ≤ μ n α nz n+1 − x n+μ n α nTz n − Tz n+1

+

μ n

1− α n

+

1− μ n

αz n − x n+

1− μ n

α nx n − Tx n

≤ μ n α nz n+1 − x n+μ n α n φz n − z n+1

+

μ n

1− α n

+

1− μ n

αz n − x n+

1− μ n

α nx n − Tx n,

(3.4)

so that

z n+1 − x n  ≤1−(1− α)1− μ n

1− α n μ n



z n − x n −1+x n −1− x n

+ α n

1− α n μ n φz n − z n+1+1− μ n

α n

1− α n μ n

x n − Tx n. (3.5)

Since the mappingJn:=[I + (α n /(1 − μ n))(I − T)] −1is nonexpansive andx n =  J n(f (x n)),

x n − x n −1 =  J n

fx n

− x n −1 =  J n

fx n

−  J n

fx n −1 

+Jn

fx n −1 

− x n −1 

≤f

x n

− fx n −1 +J n

fx n −1



− x n −1 

≤ αx n − x n −1+J n

fx n −1



− x n −1 ,

(3.6)

so that

x n − x n −1 ≤ 1

1− αJ n

fx n −1



− x n −1 

≤ 1

1− α



fx n −1



−

x n −1+ α n

1− μ n



x n −1− Tx n −1 

= 1

1− α

α n −1

1− μ n −1− α n

1− μ n

x n −1− Tx n −1 

= 1

1− α

1− α n

1− μ n

1− μ n −1

α n −1

fx n −1



− x n −1 

= 1

1− α



α n −1− α n

1− μ n

+α n

μ n −1− μ n

α n −1



f

x n −1



− x n −1 

≤ 1

1− α



α n −1− α n

α n −1

+ μ n −1− μ n

1− μ n



f

x n −1



− x n −1 .

(3.7)

We estimatez n − z n+1  Letc :=supn ≥1{(1− μ n)/α n } Since the sequences{z n },{x n }and the set{φ(t)/t : 0 < t < 1}are bounded, letz n − Tz n  ≤ M, x n − Tx n  ≤ M,  f (z n)−

z n  ≤ M,  f (x n)− x n  ≤ M ∀n ∈ Nand sup{φ(t)/t : 0 < t < 1} ≤ M for some constant

M > 0 Then

z n+1 − z n  =  μ n α n

Tz n − z n

+

1− μ n

fz n

− z n

≤ α nTz n − z n+

1− μ nf

z n

− z n

≤α n+

1− μ n

M ≤ α n(1 +c)M,

(3.8)

for alln ∈ N It follows from (3.5) that

z n+1 − x n  ≤1−(1− α)1− μ n

1− α n μ n



z n − x n −1+ 1

1− α



α n −1− α n

α n −1 − μ n − μ n −1

1− μ n



M

+ α n

1− α n μ n φα n(1 +c)M+



1− μ n

α n

1− α n μ n M.

(3.9) There existsN ∈ Nsuch thatα n(1 +c)M < 1 ∀n ≥ N Thus,

z n+1 − x n  ≤1−(1− α)1− μ n

1− α n μ n



z n − x n −1

+



1

1− α



α n −1− α n

α n −1 − μ n − μ n −1

1− μ n



+α2

n(1 +c)M

1− α n μ n +



1− μ n

α n

1− α n μ n



M, ∀n ≥ N.

(3.10)

Since, in this setting,J is norm-to-weak ∗uniformly continuous on bounded sets, letting

τ →0, we have... unique solution of the variational inequality (2.20) This concludes the

3 Main results

In the results that follow, if the mapT is uniformly continuous and δ() denotes... reflexive Banach space with a uniformly Gˆateaux differentiable norm, let K be a nonempty closed convex subset of E, let T : K → E be a continuous pseudo-contraction satisfying the