Uniformly normal structure and uniformly Lipschitzian semigroups

Nonlinear Analysis 73 (2010) 3742–3750

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Uniformly normal structure and uniformly Lipschitzian semigroups Lu-Chuan Ceng a , Hong-Kun Xu b,c,∗ , Jen-Chih Yao b a

Department of Mathematics, Shanghai Normal University, and Scientific Computing Key Laboratory of Shanghai Universities, Shanghai 200234, China

b

Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 80424, Taiwan

c

Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia

article

info

Article history: Received 17 May 2010 Accepted 28 July 2010

abstract Assume that X is a real Banach space with uniformly normal structure and C is a nonempty closed convex subset of X . We show that a κ -uniformly Lipschitzian semigroup of nonlinear self-mappings of C admits a common fixed point if the semigroup has a bounded orbit and if κ is appropriately greater than one. This result applies, in particular, to the framework of uniformly convex Banach spaces. © 2010 Elsevier Ltd. All rights reserved.

MSC: 47H09 47H10 47H20 Keywords: Uniformly normal structure Uniformly Lipschitzian semigroup Fixed point Characteristic of convexity Modulus of convexity

1. Introduction Let C be a nonempty closed convex subset of a real Banach space X . A mapping T : C → C is said to be a Lipschitzian mapping if, for each integer n ≥ 1, there exists a constant κn > 0 such that

kT n x − T n yk ≤ κn kx − yk for all x, y ∈ C .

(1.1)

A Lipschitzian mapping T is said to be a κ -uniformly Lipschitzian mapping if (1.1) holds when κn ≡ κ for all n ≥ 1. The notion of κ -uniformly Lipschitzian mappings was introduced in 1973 by Goebel and Kirk [1]. They proved that if X is a uniformly convex Banach space and C is a closed convex bounded subset of X , then every κ -uniformly Lipschitzian mapping T : C → C has a fixed point provided κ < γ , where γ > 1 is the unique solution of the equation



1 − δX

  1

γ

γ =1

(1.2)

with δX being the modulus of convexity of X . Goebel and Kirk [1] then posed the question whether or not the constant γ > 1 which solves Eq. (1.2) is the largest number for which any κ -uniformly Lipschitzian mapping T : C → C with κ < γ has a fixed point. √ In 1975, Lifschitz [2] obtained a fixed point theorem for κ -uniformly Lipschitzian mappings on C with κ < 2 in some framework of Banach spaces (e.g., Hilbert spaces). Since then, κ -uniformly Lipschitzian mappings have extensively been

∗ Corresponding author at: Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 80424, Taiwan.Tel.: +886 7 5252000 3836; fax: +886 7 5253809. E-mail addresses: [email protected] (L.-C. Ceng), [email protected] (H.-K. Xu), [email protected] (J.-C. Yao). 0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.07.044

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investigated by many authors. Moreover, some of the results for uniformly Lipschitzian mappings have been extended to uniformly Lipschitzian semigroups, and even more general, to Lipschitzian semigroups; see [3–13]. Particularly, in 1993, Tan and Xu [3] answered the question of Goebel and Kirk [1] mentioned above in the negative by proving the following Theorem 1.1 ([3, Theorem 3.5]). Let X be a real uniformly convex Banach space, C a nonempty closed convex subset of X , and T = {Ts : s ∈ G} a κ -uniformly Lipschitzian semigroup on C with κ < α , where α > 1 is the unique solution of the equation

  α 2 −1 1 δ 1− =1 N (X ) X α

(1.3)

where N (X ) > 1 is the normal structure coefficient of X . Suppose there exists an x0 ∈ C such that the orbit {Ts x0 : s ∈ G} is bounded. Then there exists z ∈ C such that Ts z = z for all s ∈ G. It is not hard to prove that γ < α , where γ and α are the solutions of Eqs. (1.2) and (1.3), respectively. Consequently, the constant γ solving Eq. (1.2) is not the biggest number for which every κ -uniformly Lipschitzian mapping T : C → C with κ < γ has a fixed point. Indeed, such a best possible number γ is still unknown, even in the setting of Hilbert spaces. It is therefore an interesting question to find another constant α ∗ which is strictly bigger than α and for which every κ -uniformly Lipschitzian mapping T : C → C with κ < α ∗ has a fixed point. Some years later, Zeng and Yang [10] proved a fixed point result for Lipschitzian semigroups as follows. Theorem 1.2 ([10, Theorem 3.1]). Let C be a nonempty bounded subset of a uniformly convex Banach space X , and let T = {Ts : s ∈ G} be a Lipschitzian semigroup on C with lim inf k|Ts |k < s

p γ0 N (X ),

where

γ0 = inf{γ : γ (1 − δX (1/γ )) ≥ 1/2},

(1.4)

and k|Ts |k is the exact Lipschitzian constant of Ts . Suppose also that there exists a nonempty bounded closed convex subset E of C with the following properties:

(P1 ) x ∈ E implies ωw (x) ⊂ E; where ωw (x) is the weak ω-limit set of T at x, i.e., ωw (x) = {y ∈ X : y = weak- lim Ttα x for some subnet {tα } ⊂ G}. tα

(P2 ) T is asymptotically regular on E; i.e., limt kTt +s x − Tt xk = 0, ∀s ∈ G, x ∈ E. Then there exists z ∈ C such that Ts z = z for all s ∈ G. The purpose of this paper is to continue the investigation on the existence of fixed points of uniformly Lipschitzian semigroups T = {Ts : s ∈ G} in the setting of Banach spaces X under conditions weaker than uniform convexity. More precisely, our contributions are twofold: (1) we shall replace the uniform convexity of X in Theorem 1.1 with the weaker condition of the uniformly normal structure of X ; and (2) we shall remove the asymptotic regularity on E of the semigroup T = {Ts : s ∈ G} in Theorem 1.2. 2. Preliminaries Let X be a real Banach space. Recall that X is strictly convex if its unit sphere does not contain any line segments, that is, X is strictly convex if and only if the following implication holds: x, y ∈ X ,

kxk = kyk = 1 and k(x + y)/2k = 1 ⇒ x = y.

In order to measure the degree of convexity (rotundity) of X , we define its modulus of convexity δX : [0, 2] → [0, 1] by

δX (ε) = inf{1 − k(x + y)/2k : kxk ≤ 1, kyk ≤ 1 and kx − yk ≥ ε}. The characteristic ε0 of convexity of X is defined as

ε0 = ε0 (X ) = sup{ε ∈ [0, 2] : δX (ε) = 0}. The following properties of the modulus δX of convexity of X are quite well known (see [14]): (a) (b) (c) (d) (e)

δX is increasing on [0, 2], and moreover strictly increasing on [ε0 , 2]; δX is continuous on [0, 2) (but not necessarily at ε = 2); δX (2) = 1 if and only if X is strictly convex; δX (0) = 0 and limε→2− δX (ε) = 1 − ε0 /2; [ka − xk ≤ r , ka − yk ≤ r and kx − yk ≥ ε] ⇒ ka − (x + y)/2k ≤ r (1 − δX (ε/r )).

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A Banach space X is said to be uniformly convex if δX (ε) > 0 for all ε ∈ (0, 2], or equivalently ε0 (X ) = 0. It is evident that a uniformly convex Banach space is strictly convex. It is also known that a uniformly convex Banach space is reflexive. By properties (a)–(e) above, we can see that if X is uniformly convex, then δX is strictly increasing and continuous on [0, 2]. Recall that the normal structure coefficient N (X ) of X is the number (see [15])

 inf

diamK rK (K )



,

where the infimum is taken over all bounded closed convex subsets K of X with more than one member, and rK (K ) and diam(K ) are the Chebyshev radii of K relative to itself and the diameter of K , respectively, i.e., rK (K ) = infx∈K supy∈K kx − yk and diam K = supx,y∈K kx − yk. A Banach space X is said to have uniformly normal structure if N (X ) > 1. It is known that a Banach space with uniformly normal structure is reflexive and that all uniformly convex or uniformly smooth Banach √ spaces have uniformly normal structure (see, e.g., [13]). It has also been computed that N (H ) = 2 for a Hilbert space H. The computations of the normal structure coefficient N (X ) for general Banach spaces look however complicated. No exact values of N (X ) are known except for some special cases (e.g., Hilbert spaces and Lp spaces). In general, we have the following (rough) lower bound for N (X ) (see [15–17]): N (X ) ≥

1 1 − δX (1)

.

In particular, X has uniformly normal structure if δX (1) > 0. Other lower bounds for N (X ) in terms of some Banach space parameters or constants can be found in [18,19]. Tan and Xu [3] have also proven that if X is uniformly convex√ and γ > 1 is the unique solution of Eq. (1.2), then N (X ) > γ . √ Note that for a Hilbert space H, we have N (H ) = 2 and γ = 5/2. Suppose that X is a uniformly convex Banach space. Then it is easily seen that the equation

  1 e α 2 δX−1 1 − N (X ) = 1 α

(2.1)

has a unique solution α > 1, where e N (X ) = 1/N (X ). Tan and Xu [3] proved that if γ > 1 and α > 1 are the solutions of (1.2) and (2.1), respectively, then γ < α . Note that for a Hilbert space H one easily calculate that

γ =

√

α = (31/2 − 1)−1/2 > γ .

5/2,

On the other hand, let C be a nonempty closed convex subset of a real Banach space X , and let G be an unbounded subset of [0, ∞) such that t + h ∈ G for all t , h ∈ G and t − h ∈ G for all t , h ∈ G with t > h (e.g., G = [0, ∞) or G = N, the set of nonnegative integers). Then a family T = {Ts : s ∈ G} of self-mappings of C is said to be a Lipschitzian semigroup on C if the following conditions are satisfied: (i) Ts+h x = Ts Th x for all s, h ∈ G and x ∈ C ; (ii) for each x ∈ C , the mapping s 7→ Ts x from G into C is continuous when G has the relative topology of [0, ∞); (iii) for each s ∈ G, there exists a positive number κs > 0 such that

kTs x − Ts yk ≤ κs kx − yk for all x, y ∈ C . In particular, if κs ≡ κ for all s ∈ G in (iii), then T = {Ts : s ∈ G} is said to be a κ -uniformly Lipschitzian semigroup on C . Finally in this section we introduce some notation.

• limt and lim supt always stand for lim t →∞ and lim sup t →∞ , respectively. t ∈G t ∈G • F (T) denotes the set of all common fixed points of T = {Ts : s ∈ G}. • ωw (x) stands for the weak ω-limit set of T at x, i.e., the set {y ∈ X : y = weak- lim Ttα x for some subnet {tα } ⊂ G}. tα

3. Fixed point theorems in Banach spaces with uniformly normal structure We need the notion of asymptotic centers, due to Edelstein [20]. Let C be a nonempty closed convex subset of a Banach space X and let {xt : t ∈ G} be a bounded net of elements of X . Then the asymptotic radius and the asymptotic center of {xt }t ∈G with respect to C are the number rC {xt } = inf lim sup kxt − yk y∈C

t

and respectively, the (possibly empty) set AC ({xt }) = {y ∈ C : lim sup kxt − yk = rC ({xt })}. t

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Lemma 3.1 ([3, Lemma 2.1]). If C is a nonempty closed convex subset of a reflexive Banach space X , then for every bounded net {xt }t ∈G of elements of X , AC ({xt }) is a nonempty bounded closed convex subset of C . In particular, if X is a uniformly convex Banach space, then AC ({xt }) consists of a single point. The following lemma can be proven in exactly the same way as in [21] for sequences and the proof is thus omitted here. Lemma 3.2 ([3, Lemma 2.2]). Suppose that X is a Banach space with uniformly normal structure. Then for every bounded net {xt }t ∈G of elements of X there exists y ∈ co({xt : t ∈ G}) such that lim sup kxt − yk ≤ e N (X )D({xt }), t

where e N (X ) = 1/N (X ), and co(E ) is the closure of the convex hull of a set E ⊂ X and D({xt }) = lim(sup{kxi − xj k : t ≤ i, j ∈ G}) t

is the asymptotic diameter of {xt }. We next present the first result of this paper which weakens the uniform convexity assumption in Theorem 1.1. Theorem 3.3. Suppose that X is a real Banach space with N (X ) > max(1, ε0 ), C is a nonempty closed convex subset of X , and T = {Ts : s ∈ G} is a uniformly κ -Lipschitzian semigroup on C with κ < α∗ . Here ε0 is the characteristic of convexity of X and

     1 1 1 α∗ = sup α : α 2 δX−1 1 − N (X )−1 ≤ 1 and 1 − ∈ 0, 1 − ε0 . α α 2

(3.1)

If {Ts x0 : s ∈ G} is bounded for some x0 ∈ C , then there exists z ∈ C such that Ts z = z for all s ∈ G. Proof. Put e N (X ) = N (X )−1 . Observe that the set



α:α δ

2 −1 X



  1 1 e 1− N (X ) ≤ 1 and 1 − ∈ 0, 1 − ε0 6= ∅. α α 2 1



(3.2)

Indeed, by properties (a), (b), (d) of the modulus δX of convexity of X , we see that the mapping

  1 δX : [ε0 , 2) → δX ([ε0 , 2)) = 0, 1 − ε0 2

is strictly increasing and continuous, and hence a bijection. Thus, we deduce that lim α 2 δX−1

α→1+

 1−

1

α



e N (X ) = ε0e N (X ) < 1, N (X ) = δX−1 (0)e

which implies that there exists α0 > 1 such that α02 δX−1 (1 − α1 )e N (X ) < 1 and 0 1−

1

α0

  1 ∈ δX ((ε0 , 2)) = 0, 1 − ε0 . 2

This verifies our assertion (3.2). Since X has a uniformly normal structure, X is reflexive. Due to the boundedness of {Ts x0 : s ∈ G} and by Lemma 3.1, we get that AC ({Tt x0 }t ∈G ) is a nonempty bounded closed convex subset of C . Then we can choose x1 ∈ AC ({Tt x0 }t ∈G ) such that lim sup kTt x0 − x1 k = inf lim sup kTt x0 − yk. y∈C

t

t

Since T is a κ -uniformly Lipschitzian property, we know that {Tt x1 } remains bounded. Consequently we can choose x2 ∈ AC ({Tt x1 }t ∈G ) such that lim sup kTt x1 − x2 k = inf lim sup kTt x1 − yk. y∈C

t

t

Continuing this process, we can construct a sequence {xn }∞ n=0 in C with the properties: (i) for each n ≥ 0, {Tt xn }t ∈G is bounded; (ii) for each n ≥ 0, xn+1 ∈ AC ({Tt xn }t ∈G ); that is, xn+1 is a point in C such that lim kTt xn − xn+1 k = inf lim kTt xn − yk. t

y∈C

t

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Write rn = rC ({Tt xn }t ∈G ). Then by Lemma 3.2 we have rn = lim sup kTt xn − xn+1 k t

≤e N (X )D({Tt xn }t ∈G ) e = N (X ) lim(sup{kTi xn − Tj xn k : t ≤ i, j ∈ G}) t

≤e N (X ) · κ · d(xn ), that is, rn ≤ κ e N (X )d(xn ),

(3.3)

where d(xn ) = sup{kTt xn − xn k : t ∈ G}. We may assume that d(xn ) > 0 for all n ≥ 0 (since otherwise xn is a common fixed point of the semigroup T and the proof is finished). Let n ≥ 0 be fixed and let ε > 0 be small enough. We can choose j ∈ G such that

kTj xn+1 − xn+1 k > d(xn+1 ) − ε and then choose s0 in G so large that

kTs xn − xn+1 k < rn + ε for all s ≥ s0 . It turns out that, for s ≥ s0 + j,

kTs xn − Tj xn+1 k ≤ κkTs−j xn − xn+1 k ≤ κ(rn + ε). It then follows from property (e) that

  

Ts xn − 1 (xn+1 + Tj xn+1 ) ≤ κ(rn + ε) 1 − δX d(xn+1 ) − ε

2 κ (r + ε) n

for s ≥ s0 + j and hence



rn ≤ lim sup

Ts xn − s

  

d(xn+1 ) − ε (xn+1 + Tj xn+1 ) ≤ κ( r + ε) 1 − δ . n X

2 κ(rn + ε) 1

Taking the limit as ε → 0 we obtain rn ≤ κ rn



1 − δX



d(xn+1 )



κ rn

.

This implies that

δX



d(xn+1 )



κ rn

≤1−

1

κ

.

(3.4)

Now, we claim that d(xn+1 ) ≤ κ rn δX−1

 1−

1



κ

.

(3.5)

Indeed, if d(xn+1 )/(κ rn ) ∈ [0, ε0 ), then noticing that δX : [ε0 , 2) → [0, 1 − ε0 /2) is a bijection and that 1 − κ1 lies in d(x ) [0, 1 − ε0 /2) by the assumption κ < α∗ , we have δX−1 (1 − κ1 ) ≥ ε0 ; hence κnr+1 ≤ δX−1 (1 − κ1 ) and (3.5) follows. If d(xn+1 )/(κ rn ) ∈ [ε0 , 2], then it is clear that and (3.5), we obtain d(xn+1 ) ≤ κ 2e N (X )δX−1

 1−

1

κ



d(xn+1 ) κ rn

n

≤ δX−1 (1 − κ1 ). This also shows that (3.5) is true. Therefore, utilizing (3.3)

d(xn ).

(3.6)

Write A = κ 2e N (X )δX−1 (1 − κ1 ). Then A < 1. Indeed, from the assumption that κ < α∗ it follows that there exists an α˜ > κ such that

  1 1 α˜ 2 δX−1 1 − N (X )−1 ≤ 1 and 1 − ∈ δX ((ε0 , 2)). α˜ α˜

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It then turns out that δX−1 (1 − 1k ) < δX−1 (1 − α1˜ ), and N (X )δX−1 A = κ 2e

 1−

1



κ

  1 N (X )δX−1 1 − < α˜ 2e ≤ 1. α˜

Hence, it follows from (3.6) that d(xn ) ≤ Ad(xn−1 ) ≤ An d(x0 ).

(3.7)

Since

kxn+1 − xn k ≤ lim sup kTt xn − xn+1 k + lim sup kTt xn − xn k t

t

≤ rn + d(xn ) ≤ 2d(xn ), P∞ we get from (3.7) that n=1 kxn+1 − xn k < ∞, and hence {xn } is norm Cauchy. Let z = k · k − limn xn . Finally, we have for each s ∈ G, kz − Ts z k ≤ kz − xn k + kTs xn − xn k + kTs xn − Ts z k ≤ (1 + κ)kz − xn k + d(xn ) → 0 as n → ∞. Hence, Ts z = z for all s ∈ G and the proof is complete.



Remark 3.4. It is known that ε0 (X ) = 0 if and only if X is uniformly convex. Suppose that X is a uniformly convex Banach space. Then the mapping δX : [0, 2] → [0, 1] is strictly increasing and continuous; consequently, the equation α 2 δX−1 (1 − α1 )e N (X ) = 1 has a unique solution (say) αˆ > 1. In this case, we see that





1



1



1



1− N (X ) ≤ 1 and 1 − ∈ 0 , 1 − ε0 α α 2     1 1 = sup α : α 2 δX−1 1 − N (X )−1 ≤ 1 and 1 − ∈ (0, 1) = α. ˆ α α

sup α : α δ

2 −1 X

−1

Therefore, Theorem 1.1 (i.e., [3, Theorem 3.5]) is a special case of Theorem 3.3. Furthermore, Theorem 3.3 also gives a negative answer to the question of Goebel and Kirk [1] (on page 139). Example 3.5. Let β ≥ 1 and consider the James space Xβ = (`2 , | · |β ), where the norm | · |β is given by

|x|β = max{kxk2 , βkxk∞ }. It is known [22] (see also [23,24]) that, for 1 ≤ β ≤

√ 2,

√ • N ( Xβ ) = p 2/β , • ε0 (Xβ ) = 2 β 2 − 1. q √ Therefore, if 1 ≤ β < 1+2 3 , then it is easily verified that N (Xβ ) > max{1, ε0 (Xβ )}; thus Theorem 3.3 applies to Xβ (note that Xβ is uniformly convex if and only if β = 1). Corollary 3.6. Let X be a real Banach space with N (X ) > max(1, ε0 ), C a nonempty bounded closed convex subset of X , and T : C → C a κ -uniformly Lipschitzian mapping with κ < α∗ (here α∗ is as given in Theorem 3.3). Then T has a fixed point. Our next result removes the asymptotic regularity on E of the semigroup T = {Ts : s ∈ G} in Theorem 1.2. Theorem 3.7. Let C be a nonempty bounded subset of a uniformly convex Banach space X , and T = {Ts : s ∈ G} be a κ -uniformly Lipschitzian semigroup on C with

κ<

p

γ0 N (X ),

where γ0 = inf{γ ≥ 1 : γ (1 − δX (1/γ )) ≥ 1/2}.

(3.8)

Suppose also that there exists a nonempty bounded closed convex subset E of C with the following property (P): (P) x ∈ E implies ωw (x) ⊂ E. Then there exists z ∈ E such that Ts z = z for all s ∈ G. Proof. Take an x0 ∈ E and consider, for each t ∈ G, the bounded net {Ts x0 : t ≤ s ∈ G}. According to Lemma 3.2, we have a yt ∈ co{Ts x0 : t ≤ s ∈ G} such that lim sup kTs x0 − yt k ≤ e N (X )D({Ts x0 }t ≤s∈G ), s

(3.9)

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where e N (X ) = N (X )−1 and D({zt }) denotes the asymptotic diameter of the net {zt }, i.e., the number lim(sup{kzi − zj k : t ≤ i, j ∈ G}). t

Since X is reflexive, {yt } admits a subnet {ytβ } converging weakly to some x1 ∈ X . From (3.9) and the weakly lower semicontinuity of the functional lim supt kTt x0 − yk, it follows that lim sup kTt x0 − x1 k ≤ e N (X )D({Tt x0 }t ∈G ).

(3.10)

t

It is also easily seen that x1 belongs to the set

T

t ∈G

co{Ts x0 : t ≤ s ∈ G} and that

kz − x1 k ≤ lim sup kz − Tt x0 k for all z ∈ X .

(3.11)

t

Observing property (P) and the fact that t ∈G co{Ts x0 : t ≤ s ∈ G} = co{ωw (x0 )} which is easy to prove by using the Separation Theorem (see [25]), we know that x1 actually lies in E. So we can repeat the above process and obtain a sequence {xn }∞ n=0 in E with the properties: for all nonnegative integers n ≥ 0,

T

lim sup kTt xn − xn+1 k ≤ e N (X )D({Tt xn }t ∈G ),

(3.12)

kz − xn+1 k ≤ lim sup kz − Tt xn k for all z ∈ X .

(3.13)

t

and t

Write rn = lim supt kTt xn − xn+1 k and d(xn ) = sup{kTt xn − xn k : t ∈ G}. Then in view of (3.12), we have rn = lim sup kTt xn − xn+1 k t

≤e N (X )D({Tt xn }t ∈G ) e = N (X ) lim(sup{kTi xn − Tj xn k : t ≤ i, j ∈ G}) t

≤e N (X ) · κ · d(xn ), that is, rn ≤ κ e N (X )d(xn ),

for n = 0, 1, 2, . . . .

(3.14)

We may assume that d(xn ) > 0 for all n ≥ 0. Let n ≥ 0 be fixed and let ε > 0 be small enough. First choose j ∈ G such that

kTj xn+1 − xn+1 k > d(xn+1 ) − ε and then choose s0 ∈ G so large that

kTs xn − xn+1 k < rn + ε. We then have, for s ≥ s0 + j,

kTs xn − Tj xn+1 k ≤ κkTs−j xn − xn+1 k ≤ κ(rn + ε). It then follows from (e) that, for s ≥ s0 + j,

  

Ts xn − 1 (xn+1 + Tj xn+1 ) ≤ κ(rn + ε) 1 − δX d(xn+1 ) − ε .

2 κ(r + ε) n

Hence from (3.13) (taking z := (xn+1 + Tj xn+1 )/2) we obtain



1

(d(xn+1 ) − ε) < ( T x − x ) j n +1 n +1

2 2



1

≤ lim sup T x − ( x + T x ) t n n + 1 j n + 1

2 t    d(xn+1 ) − ε ≤ κ(rn + ε) 1 − δX . κ(rn + ε) 1

(3.15)

Taking the limit as ε → 0 we have 1 2

d(xn+1 ) ≤ κ rn



1 − δX



d(xn+1 )

κ rn



.

(3.16)

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On the other hand, we easily find by (3.11) that, for each j,

kTj xn+1 − xn+1 k ≤ lim sup kTj xn+1 − Tt xn+1 k ≤ κ rn . t

It turns out that d(xn+1 ) ≤ κ rn .

(3.17)

Combining (3.16) and (3.17) and using the definition of γ0 in (3.8), we infer that (κ rn )/d(xn+1 ) ≥ γ0 . It turns out from (3.14) that d(xn+1 ) ≤

κ2 κ rn ≤ d(xn ). γ0 γ0 N (X )

Consequently, we obtain d(xn ) ≤ Ad(xn−1 ) ≤ An d(x0 ), where A = κ 2 [γ0 N (X )]−1 < 1 by assumption. Noticing that

kxn+1 − xn k ≤ lim sup kTt xn − xn k + lim sup kTt xn − xn+1 k t

t

≤ d(xn ) + rn ≤ (1 + κ e N (X ))d(xn ) e ≤ (1 + κ N (X ))An d(x0 ), (3.18) P∞ we see that the series n=1 kxn+1 − xn k is convergent. This implies that {xn } is strongly convergent. Let z = k·k− limn→∞ xn . Then we have, for each s ∈ G, kz − Ts z k ≤ kz − xn k + kxn − Ts xn k + kTs xn − Ts z k ≤ (1 + κ)kz − xn k + kxn − Ts xn k ≤ (1 + κ)kz − xn k + d(xn ) → 0 as n → ∞. This shows that Ts z = z for each s ∈ G. The proof is therefore complete.



Remark 3.8. Compared with Theorem 1.2 (i.e., [10, Theorem 3.1]), Theorem 3.7 improves and extends Theorem 1.2 in two aspects: (1) the closedness and convexity of C are replaced by the closedness and convexity of the subset E of C , respectively; (2) no assumption of asymptotic regularity of T on C is made in Theorem 3.7. Acknowledgements The first author was supported in part by the National Science Foundation of China (10771141), Doctoral Program Foundation of Ministry of Education of China (20070270004), Science and Technology Commission of Shanghai Municipality grant (075105118), and Shanghai Leading Academic Discipline Project (S30405). The second author was supported in part by NSC 97-2628-M-110-003-MY3 (Taiwan). The third author was supported in part by NSC 98-2115-M-110-001 (Taiwan). The authors are grateful to the referee for his/her comments and suggestions which improved the presentation of this manuscript. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]

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