# Iterative Algorithm for Common Fixed Points of Infinite Family of Nonexpansive Mappings in Banach Spaces

Authors: Songnian He; Jun Guo
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

LetC be a nonempty closed convex subset of a real uniformly smooth Banach space X, {Tk}k=1∞:C→C an infinite family of nonexpansive mappings with the nonempty set of common fixed points ⋂k=1∞Fix(Tk), and f:C→C a contraction. We introduce an explicit iterative algorithm xn+1=αnf(xn)+(1-αn)Lnxn, where Ln=∑k=1n(ωk/sn)Tk,Sn=∑k=1nωk,
and wk>0 with ∑k=1∞ωk=1. Under certain appropriate conditions on {αn}, we prove that {xn} converges strongly to a common fixed point x* of {Tk}k=1∞, which solves the following variational inequality: 〈x*-f(x*),J(x*-p)〉≤0,p∈⋂k=1∞Fix(Tk), where J is the (normalized) duality mapping of X. This algorithm is brief and needs less computational work, since it does not involve W-mapping.


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## Body

## 1. Introduction

LetX be a real Banach space, C a nonempty closed convex subset of X, and X* the dual space of X. The (normalized) duality mapping J:X→2X* is defined by (1.1)J(x)={x*∈X*:〈x,x*〉=‖x‖2,‖x*‖=‖x‖},∀x∈X.
If X is a Hilbert space, then J=I, where I is the identity mapping. It is well known that if X is smooth, then J is single valued.Recall that a mappingf:C→C is a contraction, if there exists a constant α∈[0,1) such that (1.2)‖f(x)-f(y)‖≤α‖x-y‖,∀x,y∈C.
We use ΠC to denote the collection of all contractions on C, that is, (1.3)ΠC={f:fisacontractionon⁡C}.
A mapping T:C→C is said to be nonexpansive, if (1.4)‖Tx-Ty‖≤‖x-y‖,∀x,y∈C.
We use Fix(T) to denote the set of fixed points of T, namely, Fix(T)={x∈C:Tx=x}. One classical way to study nonexpansive mappings is to use contractions to approximate a nonexpansive mapping ([1–11]). Browder [1] first considered the following approximation in a Hilbert space. Fix u∈C and define a contraction Ft from C into itself by(1.5)Ftx=tu+(1-t)Tx,x∈C,
where t∈(0,1). Banach contraction mapping principle guarantees that Ft has a unique fixed point in C. Denote by zt∈C the unique fixed point of Ft, that is,(1.6)zt=tu+(1-t)Tzt.
In the case of T having fixed points, Browder [1] proved the following.Theorem 1.1.
In a Hilbert space, ast→0, zt defined in (1.6) converges strongly to a fixed point of T that is closest to u, that is, the nearest point projection of u onto Fix(T).Halpern [3] introduced an iteration process (discretization of (1.6)) in a Hilbert as follows:(1.7)zn+1=αnu+(1-αn)Tzn,n≥0,
where u,z0∈C are arbitrary (but fixed) and {αn} is a sequence in (0,1). Lions [4] proved the following.Theorem 1.2.
In a Hilbert space, if{αn} satisfies the following conditions:  (K1)
limn→∞αn=0;(K2)
∑n=0∞αn=∞;(K3)
limn→∞|αn-αn-1|/αn+12=0. Then {zn} converges strongly to the nearest point projection of u onto Fix(T).The Banach space versions of Theorems1.1 and 1.2 were obtained by Reich [5]. He proved the following.Theorem 1.3.
In a uniformly smooth Banach spaceX, both zt defined in (1.6) and {zn} defined in (1.7) converge strongly to a same fixed point of T. If one defines Q:C→Fix(T) by
(1.8)Q(u)=lim⁡t→0zt,
then Q is the sunny nonexpansive retraction from C onto Fix(T). Namely, Q satisfies the property:
(1.9)‖Qx-Qy‖2≤〈x-y,J(Qx-Qy)〉,x,y∈C,
where J is the duality mapping of X.Moudafi [6] introduced a viscosity approximation method and proved the strong convergence of both the implicit and explicit methods in Hilbert spaces. Xu [7] extended Moudafi's results in Hilbert spaces. Given a real number t∈(0,1) and a contraction f∈ΠC, define a contraction Ttf:C→C by(1.10)Ttfx=tf(x)+(1-t)Tx,x∈C.
Let xt∈C be the unique fixed point of Ttf. Thus,(1.11)xt=tf(xt)+(1-t)Txt.
Corresponding explicit iterative process is defined by(1.12)xn+1=αnf(xn)+(1-αn)Txn,
where x0∈C is arbitrary (but fixed) and {αn} is a sequence in (0,1). It was proved by Xu [7] that under certain appropriate conditions on {αn}, both xt defined in (1.11) and {xn} defined in (1.12) converged strongly to x*∈C, which is the unique solution of the variational inequality: (1.13)〈(I-f)x*,x-x*〉≥0,x∈Fix⁡(T).
Xu [7] also extended Moudafi's results to the setting of Banach spaces and proved the strong convergence of both the implicit method (1.11) and explicit method (1.12) in uniformly smooth Banach spaces.In order to deal with some problems involving the common fixed points of infinite family of nonexpansive mappings,W-mapping is often used, see [12–20]. Let {Tk}k=1∞:C→C be an infinite family of nonexpansive mappings and let {ξk}k=1∞ be a real number sequence such that 0<ξk<1 for every k∈ℕ. For any n∈ℕ, we define a mapping Wn of C into itself as follows:(1.14)Un,n+1=I,Un,n=ξnTnUn,n+1+(1-ξn)I,Un,n-1=ξn-1Tn-1Un,n+(1-ξn-1)I,⋮Un,k=ξkTkUn,k+1+(1-ξk)I,Un,k-1=ξk-1Tk-1Un,k+(1-ξk-1)I,⋮Un,2=ξ2T2Un,3+(1-ξ2)I,Wn=Un,1=ξ1T1Un,2+(1-ξ1)I.
Such Wn is called the W-mapping generated by {Tk}k=1∞ and {ξk}k=1∞, see [12, 13].Yao et al. [10] introduced the following iterative algorithm for infinite family of nonexpansive mappings. Let X be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm and C a nonempty closed convex subset of X. Sequence {xn} is defined by(1.15)xn+1=αnu+(1-αn)Wnxn,n≥0,
where u,x0∈C are arbitrary (but fixed) and {αn}⊂(0,1). It was proved that under certain appropriate conditions on {αn}, the sequence {xn} generated by (1.15) converges strongly to a common fixed point of {Tk}k=1∞ [13].SinceW-mapping contains many composite operations of {Tk}, it is complicated and needs large computational work. In this paper, we introduce a new iterative algorithm for solving the common fixed point problem of infinite family of nonexpansive mappings. Let X be a real uniformly smooth Banach space, C a nonempty closed convex subset of X, {Tk}k=1∞:C→C an infinite family of nonexpansive mappings with the nonempty set of common fixed points ⋂k=1∞Fix(Tk), and f∈ΠC. Given any x0∈C, define a sequence {xn} by(1.16)xn+1=αnf(xn)+(1-αn)Lnxn,n≥0,
where {αn}⊂(0,1), Ln=∑k=1n(ωk/sn)Tk, Sn=∑k=1nωk and wk>0 with ∑k=1∞ωk=1. Under certain appropriate conditions on {αn}, we prove that {xn} converges strongly to x*∈⋂k=1∞Fix(Tk), which solves the following variational inequality: (1.17)〈x*-f(x*),J(x*-p)〉≤0,p∈⋂k=1∞Fix⁡(Tk),
where J is the duality mapping of X. Because Ln doesn't contain many composite operations of {Tk}, this algorithm is brief and needs less computational wok.We will useM to denote a constant, which may be different in different places.


## 2. Preliminaries

LetB={x∈X:∥x∥=1} denotes the unit sphere of X. A Banach space X is said to be strictly convex, if ∥(x+y)/2∥<1 holds for all x,y∈B, x≠y. A Banach space X is said to be uniformly convex if for each ɛ∈(0,2], there exists a constant δ>0 such that for any x,y∈B, ∥x-y∥≥ɛ implies ∥(x+y)/2∥≤1-δ. It is known that a uniformly convex Banach space is reflexive and strictly convex, see [21].The norm ofX is said to be Gâteaux differentiable if(2.1)lim⁡t→0‖x+ty‖-‖x‖t
exists for each x,y∈B and in this case X is said to be smooth. The norm of X is said to be uniformly Gâteaux differentiable if for each y∈B, the limit (2.1) is attained uniformly for x∈B. The norm of X is said to be Frêchet differentiable, if for each x∈B, the limit (2.1) is attained uniformly for y∈B. The norm of X is said to be uniformly Frêchet differentiable, if the limit (2.1) is attained uniformly for x,y∈B and in this case X is said to be uniformly smooth.LetD be a nonempty subset of C. A mapping Q:C→D is said to be sunny [22] if(2.2)Q(x+t(x-Q(x)))=Q(x),∀x∈C,t≥0,
whenever x+t(x-Q(x))∈C. A mapping Q:C→D is called a retraction if Qx=x for all x∈D. Furthermore, Q is sunny nonexpansive retraction from C onto D if Q is a retraction from C onto D which is also sunny and nonexpansive.A subsetD of C is called a sunny nonexpansive retraction of C if there exits a sunny nonexpansive retraction from C onto D.Lemma 2.1 (see [22]).
LetC be a closed convex subset of a smooth Banach space X. Let D be a nonempty subset of C and Q:C→D be a retraction. Then the following are equivalent.(a)
Q is sunny and nonexpansive.(b)
∥Qx-Qy∥2≤〈x-y,J(Qx-Qy)〉, for all x,y∈C.(c)
〈x-Qx,J(y-Qx)〉≤0, for all x∈C,y∈D.Lemma 2.2 (see [23]).
Let{sn} be a sequence of nonnegative real numbers satisfying
(2.3)sn+1≤(1-γn)sn+γnβn+δn,n≥0,
where {γn}⊂(0,1), {βn} and {δn} satisfy the following conditions:  (A1)
∑n=0∞γn=∞;(A2)
limsupn→∞βn≤0;(A3)
δn≥0 (n≥0), ∑n=1∞δn<∞. Then limn→∞sn=0.Lemma 2.3 (see [24]).
In a Banach spaceX, the following inequality holds:
(2.4)‖x+y‖2≤‖x‖2+2〈y,j(x+y)〉,∀x,y∈X,
where j(x+y)∈J(x+y).Lemma 2.4 (see [25]).
LetC be a closed convex subset of a strictly convex Banach space X. Let {Tn:n∈ℕ} be a sequence of nonexpansive mappings on C. Suppose ⋂n=1∞Fix(Tn) is nonempty. Let {λn} be a sequence of positive numbers with ∑n=1∞λn=1. Then a mapping S on C defined by Sx=∑n=1∞λnTnx for x∈C is well defined, nonexpansive and Fix(S)=⋂n=1∞Fix(Tn) holds.Lemma 2.5 (see [7]).
LetX be a uniformly smooth Banach space, C a closed convex subset of X, T:C→C a nonexpansive mapping with Fix(T)≠∅, and f∈ΠC. Then {xt} defined by
(2.5)xt=tf(xt)+(1-t)Txt
converges strongly to a point in Fix(T). If we define a mapping Q:ΠC→Fix(T) by
(2.6)Q(f):=lim⁡t→0xt,f∈ΠC,
then Q(f) solves the variational inequality:
(2.7)〈(I-f)Q(f),J(Q(f)-p)〉≤0,f∈ΠC,p∈Fix⁡(T).Lemma 2.6.
LetX be a Banach space, {xk} a bounded sequence of X, and {ωk} a sequence of positive numbers with ∑k=1∞ωk=1. Then ∑k=1∞ωkxk is convergent in X.Lemma 2.7.
Let X be Banach space,{Tk:k∈ℕ} a sequence of nonexpansive mappings on X with  ⋂k=1∞Fix(Tk)≠∅, and  {ωk} a sequence of positive numbers with  ∑k=1∞ωk=1. Let  T=∑k=1∞ωkTk,  Lm=∑k=1m(ωk/Sm)Tk, and  Sm=∑k=1mωk. Then  Lm uniformly converges to  T in each bounded subset  S of  X.Proof.
Forallx∈S, we observe that
(2.8)‖Lmx-Tx‖=‖∑k=1mωkSmTkx-∑k=1∞ωkTkx‖=‖∑k=1mωk-ωkSmSmTkx-∑k=m+1∞ωkTkx‖≤‖∑k=1m1-SmSmωkTkx‖+‖∑k=m+1∞ωkTkx‖≤1-SmSm∑k=1mωk‖Tkx‖+∑k=m+1∞ωk‖Tkx‖≤1-SmSmM+M∑k=m+1∞ωk,
where M=supx∈S,k≥1∥Tkx∥<∞. Taking m→∞ in above last inequality, we have that
(2.9)lim⁡m→∞‖Lmx-Tx‖=0
holds uniformly for x∈S and this completes the proof.


## 3. Main Results

Theorem 3.1.
LetC be a nonempty closed convex subset of a real uniformly smooth Banach space X, {Tk}k=1∞:C→C an infinite family of nonexpansive mappings with ⋂k=1∞Fix(Tk)≠∅, and {ωk} a sequence of positive numbers with ∑k=1∞ωk=1. Let Ln=∑k=1n(ωk/Sn)Tk, Sn=∑k=1nωk, and f∈ΠC with coefficient α∈[0,1). Given any x0∈C, let {xn} be a sequence generated by
(3.1)xn+1=αnf(xn)+(1-αn)Lnxn,n≥0,
where {αn}⊂(0,1) satisfies the following conditions:  (A1)
limn→∞αn=0;(A2)
∑n=0∞αn=∞;(A3)
either∑n=0∞|αn+1-αn|<∞ or limn→∞(αn+1/αn)=1. Then {xn} converges strongly to x*∈⋂k=1∞Fix(Tk), which solves the following variational inequality:
(3.2)〈x*-f(x*),J(x*-p)〉≤0,p∈⋂k=1∞Fix⁡(Tk).Proof.
Step 1.
We show that{xn} is bounded.
Noticing nonexpansiveness ofLn, take a p∈⋂k=1∞Fix(Tk) to derive that
(3.3)‖xn+1-p‖=‖αnf(xn)+(1-αn)Lnxn-p‖≤αn‖f(xn)-p‖+(1-αn)‖Lnxn-p‖=αn‖f(xn)-f(p)+f(p)-p‖+(1-αn)‖Lnxn-p‖≤αn‖f(xn)-f(p)‖+αn‖f(p)-p‖+(1-αn)‖Lnxn-p‖≤ααn‖xn-p‖+αn‖f(p)-p‖+(1-αn)‖xn-p‖=(1-(1-α)αn)‖xn-p‖+αn‖(f(p)-p)‖=(1-(1-α)αn)‖xn-p‖+(1-α)αn‖(f(p)-p)‖1-α≤max⁡{‖(f(p)-p)‖1-α,‖xn-p‖}.
By induction, we obtain
(3.4)‖xn+1-p‖≤max⁡{‖(f(p)-p)‖1-α,‖x0-p‖},n≥0,
and {xn} is bounded, so are {Tkxn}, {Lnxn}, and {f(xn)}.Step 2.
We prove thatlimn→∞∥xn+1-xn∥=0.
By (3.1), We have
(3.5)‖xn+1-xn‖=‖αnf(xn)+(1-αn)Lnxn-αn-1f(xn-1)-(1-αn-1)Ln-1xn-1‖=‖αnf(xn)-αnf(xn-1)+αnf(xn-1)-αn-1f(xn-1)+(1-αn)Lnxn-(1-αn)Ln-1xn-1+(1-αn)Ln-1xn-1-(1-αn-1)Ln-1xn-1‖≤ααn‖xn-xn-1‖+|αn-αn-1|‖f(xn-1)‖+(1-αn)‖Lnxn-Ln-1xn-1‖+|αn-αn-1|‖Ln-1xn-1‖=ααn‖xn-xn-1‖+|αn-αn-1|(‖f(xn-1)‖+‖Ln-1xn-1‖)+(1-αn)‖Lnxn-Ln-1xn-1‖≤ααn‖xn-xn-1‖+|αn-αn-1|(‖f(xn-1)‖+‖Ln-1xn-1‖)+(1-αn)‖Lnxn-Lnxn-1‖+(1-αn)‖Lnxn-1-Ln-1xn-1‖≤(1-(1-α)αn)‖xn-xn-1‖+(1-αn)‖Lnxn-1-Ln-1xn-1‖+|αn-αn-1|(‖f(xn-1)‖+‖Ln-1xn-1‖)≤(1-(1-α)αn)‖xn-xn-1‖+(1-αn)‖Lnxn-1-Ln-1xn-1‖+|αn-αn-1|M,
where M=supn≥1(∥f(xn-1)∥+∥Ln-1xn-1∥). At the same time, we observe that
(3.6)∑n=1∞‖Lnxn-1-Ln-1xn-1‖=∑n=1∞‖∑k=1nωkSnTkxn-1-∑k=1n-1ωkSn-1Tkxn-1‖=∑n=1∞‖ωnSnTnxn-1+∑k=1n-1-ωnωkSnSn-1Tkxn-1‖≤∑n=1∞{‖ωnSnTnxn-1‖+‖∑k=1n-1ωnωkSnSn-1Tkxn-1‖}≤∑n=1∞ωnSn‖Tnxn-1‖+∑n=1∞∑k=1n-1ωnωkSnSn-1‖Tkxn-1‖≤∑n=1∞ωnSnM+∑n=1∞ωnSnM=∑n=1∞2MSnωn,
where M=supk≥1,n≥1∥Tkxn-1∥. Applying Lemma 2.6 and compatibility test of series, we have
(3.7)∑n=1∞‖Lnxn-1-Ln-1xn-1‖<∞.
Put
(3.8)γn=(1-α)αn,βn=|αn-αn-1|M(1-α)αn,δn=(1-αn)‖Lnxn-1-Ln-1xn-1‖.
It follows that
(3.9)‖xn+1-xn‖≤(1-γn)‖xn-xn-1‖+γnβn+δn.
It is easily seen from (A2), (A3), and (3.7) that
(3.10)∑n=1∞γn=∞,limsup⁡n→∞βn≤0,∑n=1∞δn<∞.
Applying Lemma 2.2 to (3.9), we obtain
(3.11)lim⁡n→∞‖xn+1-xn‖=0.Step 3.
We show thatlimn→∞∥xn-Txn∥=0.
Indeed we observe that(3.12)‖xn-Txn‖≤‖xn-xn+1‖+‖xn+1-Txn‖=‖xn-xn+1‖+‖αnf(xn)+(1-αn)Lnxn-Txn‖≤‖xn-xn+1‖+αn‖f(xn)-Txn‖+(1-αn)‖Lnxn-Txn‖.
Hence, by (3.11), (A1), and Lemma 2.7, we have
(3.13)lim⁡n→∞‖xn-Txn‖=0.Step 4.
We prove that(3.14)limsup⁡n→∞〈f(x*)-x*,J(xn-x*)〉≤0,
where x*=limt→0xt with xt being the fixed point of the contraction
(3.15)x⟼tf(x)+(1-t)Tx.
From Lemma 2.5, we have x*∈Fix(T) and
(3.16)〈(I-f)x*,J(x*-p)〉≤0,p∈Fix⁡(T).
By Lemma 2.4, we have x*∈⋂k=1∞Fix(Tk) and
(3.17)〈(I-f)x*,J(x*-p)〉≤0,p∈⋂k=1∞Fix⁡(Tk).
By xt=tf(xt)+(1-t)Txt, we have
(3.18)xt-xn=t(f(xt)-xn)+(1-t)(Txt-xn).
It follows from Lemma 2.3 that
(3.19)‖xt-xn‖2=‖t(f(xt)-xn)+(1-t)(Txt-xn)‖2≤(1-t)2‖Txt-xn‖2+2t〈f(xt)-xn,J(xt-xn)〉≤(1-t)2(‖Txt-Txn‖+‖Txn-xn‖)2+2t〈f(xt)-xt,J(xt-xn)〉+2t〈xt-xn,J(xt-xn)〉≤(1-t)2(‖xt-xn‖+‖Txn-xn‖)2+2t〈f(xt)-xt,J(xt-xn)〉+2t〈xt-xn,J(xt-xn)〉≤(1-t)2‖xt-xn‖2+bn(t)+2t〈f(xt)-xt,J(xt-xn)〉+2t‖xt-xn‖2,
where bn(t)=∥Txn-xn∥(2∥xt-xn∥+∥Txn-xn∥)→0 (n→∞). It follows from above last inequality that
(3.20)〈xt-f(xt),J(xt-xn)〉≤t2‖xt-xn‖2+12tbn(t).
Taking n→∞ in (3.20) yields
(3.21)limsup⁡n→∞〈xt-f(xt),J(xt-xn)〉≤t2M,
where M≥∥xt-xn∥2 for all n≥1 and t∈(0,1). Taking t→0 in (3.21), we have
(3.22)limsup⁡t→0limsup⁡n→∞〈xt-f(xt),J(xt-xn)〉≤0.
Noticing the fact that two limits are interchangeable due to the fact the duality mapping J is norm-to-norm uniformly continuous on bounded sets, it follows from (3.22), we have
(3.23)limsup⁡n→∞〈f(x*)-x*,J(xn-x*)〉=limsup⁡n→∞limsup⁡t→0〈xt-f(xt),J(xt-xn)〉=limsup⁡t→0limsup⁡n→∞〈xt-f(xt),J(xt-xn)〉≤0.
Hence (3.14) holds.Step 5.
Finally, we prove thatxn→x* (n→∞).
Indeed we observe that(3.24)‖xn+1-x*‖2=‖αnf(xn)+(1-αn)Lnxn-x*‖2=‖αn(f(xn)-f(x*))+(1-αn)(Lnxn-x*)+αn(f(x*)-x*)‖2≤‖αn(f(xn)-f(x*))+(1-αn)(Lnxn-x*)‖2+2〈αn(f(x*)-x*),J(xn+1-x*)〉≤{αn‖f(xn)-f(x*)‖+(1-αn)‖Lnxn-x*‖}2+2αn〈f(x*)-x*,J(xn+1-x*)〉≤(ααn‖xn-x*‖+(1-αn)‖xn-x*‖)2+2αn〈f(x*)-x*,J(xn+1-x*)〉≤(1-(1-α)αn)2‖xn-x*‖2+2αn〈f(x*)-x*,J(xn+1-x*)〉≤(1-(1-α)αn)‖xn-x*‖2+2αn〈f(x*)-x*,J(xn+1-x*)〉.
By view of (3.14) and condition (A2), it follows from Lemma 2.2 that xn→x*(n→∞). This completes the proof.Corollary 3.2.
LetC be a nonempty closed convex subset of a real uniformly smooth Banach space X, {Tk}k=1∞:C→C an infinite family of nonexpansive mappings with ⋂k=1∞Fix(Tk)≠∅, {ωk} a sequence of positive numbers with ∑k=1∞ωk=1. Let Ln=∑k=1n(ωk/Sn)Tk, Sn=∑k=1nωk, and u∈C. Given any x0∈C, let {xn} be a sequence generated by
(3.25)xn+1=αnu+(1-αn)Lnxn,n≥0,
where {αn}⊂(0,1) satisfies the following conditions:  (A1)
limn→∞αn=0;(A2)
∑n=0∞αn=∞;(A3)
either∑n=0∞|αn+1-αn|<∞ or limn→∞(αn+1/αn=)1. Then {xn} converges strongly to x*∈⋂k=1∞Fix(Tk), which solves the following variational inequality:
(3.26)〈x*-u,J(x*-p)〉≤0,p∈⋂k=1∞Fix⁡(Tk).



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