# On the Strong Convergence of Viscosity Approximation Process for Quasinonexpansive Mappings in Hilbert Spaces

Authors: Kanokwan Wongchan; Satit Saejung
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We improve the viscosity approximation process for approximation of a fixed point of a quasi-nonexpansive mapping in a Hilbert space proposed by Maingé (2010). An example beyond the scope of the previously known result is given.


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## Body

## 1. Introduction

LetH be a real Hilbert space with inner product 〈·,·〉 and the induced norm ∥·∥. In this paper, we denote the strong and weak convergence by → and ⇀, respectively. For a subset C of H, a mapping T:C→C is said to be nonexpansive if ∥Tx-Ty∥≤∥x-y∥ for all x,y∈C; and it is quasinonexpansive if its fixed-point setFix(T):={x∈C:x=Tx} is nonempty and ∥Tx-p∥≤∥x-p∥ for all x∈C and p∈Fix(T). It is clear that every nonexpansive mapping with a nonempty fixed-point set is quasinonexpansive, but the converse is not true. The process for approximation of a fixed point of a nonexpansive or quasinonexpansive mapping is one of interesting problems in mathematics and it has been investigated by many researchers. One of the effective processes for this problem is given by Moudafi [1]. Let C be a closed convex subset of H, and T:C→C is a nonexpansive mapping with a nonempty fixed-point set Fix(T). Moudafi proposed the following scheme which is known as Moudafi's viscosity approximation process: (1.1)x1=x∈Carbitrarilychosen,xn+1=ɛn1+ɛnf(xn)+11+ɛnTxn,
where f:C→C is a contraction; that is, there exists an α∈[0,1) such that ∥f(x)-f(y)∥≤α∥x-y∥ for all x,y∈C and {ɛn} is a sequence in (0,1) satisfying (M1)
limn→∞ɛn=0,(M2)
∑n=1∞ɛn=∞,(M3)
limn→∞(1/ɛn-1/ɛn+1)=0.It was also proved that{xn} converges to an element z∈Fix(T) satisfying the following inequality: (1.2)〈f(z)-z,q-z〉≤0
for all q∈Fix(T).In the literature, Moudafi's scheme has been widely studied and extended (see [2–5] and references therein). For example, Xu [6] improved this result to a Banach space. The interesting improvement of this result given by Maingé [7] is our starting point. His result is given below.Theorem 1.1.
LetC be a closed convex subset of a Hilbert space H, and T:C→C is a quasinonexpansive mapping such that I-T is demiclosed at zero, that is, z∈Fix(T) whenever {zn} is a sequence in C such that zn⇀z and zn-Tzn→0. Suppose that f:C→C is a contraction. Let {xn} be a sequence in C defined by
(1.3)x1=x∈Carbitrarilychosen,xn+1=αnf(xn)+(1-αn)((1-ω)I+ωT)xn,
where ω∈(0,1), I is an identity mapping, and {αn} is a sequence in (0,1) satisfying (C1)
limn→∞αn=0,(C2)
∑n=1∞αn=∞.Then the sequence {xn} converges to an element z∈Fix(T) and the following inequality holds
(1.4)〈f(z)-z,q-z〉≤0
for all q∈Fix(T).It should be noted that Maingé's result is more widely applicable than Moudafi's. However, after a careful reading, we find that there is a small mistake in Maingé's proof. The following fact (see [7, Remark 2.1(i3)]) is used: if T:C→C is quasinonexpansive and Tω:=(1-ω)I+ωT where ω∈(0,1], then (1.5)〈x-Tωx,x-q〉≥ω‖x-Tx‖2
for all x∈C and q∈Fix(T). Note that the inequality above is equivalent to (1.6)〈x-Tx,x-q〉≥‖x-Tx‖2.
But this fails; for example, let us consider the nonexpansive mapping T:ℝ→ℝ defined by Tx=-x for all x∈ℝ. It is clear that Fix(T)={0} and 〈x-Tx,x-q〉=2x2≱4x2=∥x-Tx∥2.Recall the following identities in a Hilbert spaceH: for x,y∈H, ω∈[0,1](i)
∥x+y∥2=∥x∥2+2〈x,y〉+∥y∥2;(ii)
∥(1-ω)x+ωy∥2=(1-ω)∥x∥2+ω∥y∥2-(1-ω)ω∥x-y∥2.The correction of Maingé's result is as follows.Proposition 1.2.
LetC be a subset of a Hilbert space and T:C→C be a mapping with a nonempty fixed-point set Fix(T). Suppose that Tω:=(1-ω)I+ωT where ω∈(0,1]. Then T is quasinonexpansive if and only if
(1.7)〈x-Tωx,x-q〉≥ω2‖x-Tx‖2
for all x∈C and q∈Fix(T).Proof.
Notice thatx-Tωx=ω(x-Tx) and
(1.8)‖Tωx-q‖2=‖(Tωx-x)+(x-q)‖2=‖Tωx-x‖2+2〈Tωx-x,x-q〉+‖x-q‖2=ω2‖x-Tx‖2+2〈Tωx-x,x-q〉+‖x-q‖2.
On the other hand,
(1.9)‖Tωx-q‖2=‖(1-ω)(x-q)+ω(Tx-q)‖2=(1-ω)‖x-q‖2+ω‖Tx-q‖2-(1-ω)ω‖x-Tx‖2.
Hence
(1.10)2〈x-Tωx,x-q〉=ω(‖x-q‖2-‖Tx-q‖2)+ω‖x-Tx‖2.Remark 1.3.
Unfortunately, this effects the main result (see [7, Theorem 3.1]) in Maingé's paper. More precisely, inequality (32) of its proof (page 78, line 22) should read
(1.11)12‖xn+1-z‖2-12‖xn-z‖2+(12-ω)ω(1-αn)‖xn-Txn‖2≤αn(αn‖f(xn)-xn‖2-〈xn-f(xn),xn-z〉)
rather than
(1.12)12‖xn+1-z‖2-12‖xn-z‖2+(1-ω)ω(1-αn)‖xn-Txn‖2≤αn(αn‖f(xn)-xn‖2-〈xn-f(xn),xn-z〉).
Therefore, Theorem 1.1 above is valid for only ω∈(0,1/2) under the same technique.The purpose of this paper is to simultaneously present a correction of the proof of Theorem1.1 which is valid for allω∈(0,1), and extend his scheme to a wider class of mappings including average mappings, that is, mappings of the form (1-ω)I+ωT. Our result is more general than Maingé's theorem. An example of a quasinonexpansive mapping which is not applied by Maingé's theorem but applied by our result is given.


## 2. Result

First, let us recall some lemmas which are needed for proving the main result.Lemma 2.1 (see [8, Lemma 2.3]).
Let{sn} be a sequence of nonnegative real numbers, {αn} a sequence of (0,1) with ∑n=1∞αn=∞, {βn} a sequence of nonnegative real numbers with ∑n=1∞βn<∞, and {γn} a sequence of real numbers with limsupn→∞γn≤0. Suppose that
(2.1)sn+1≤(1-αn)sn+αnγn+βn
for all n∈ℕ. Then limn→∞sn=0.The following nice result was proved by Maingé (see [7, Lemma 2.1]).Lemma 2.2.
Let{sn} be a sequence of nonnegative real numbers. If there exists a subsequence {snj} of {sn} such that snj<snj+1 for all j∈ℕ, then there exists a subsequence {smk} of {sn} such that
(2.2)smk≤smk+1,sk≤smk+1
for all k∈ℕ.For a closed convex subsetC of a Hilbert space H, the metric projectionPC:H→C is defined for each x∈H as the unique element PCx∈C such that (2.3)‖x-PCx‖=inf⁡{‖x-z‖:z∈C}.
It is well known that (see, e.g., [9]) for x∈H and y∈C(2.4)y=PCx⟺〈x-y,y-z〉≥0,∀z∈C.
For x,y∈H, the following inequality is known as the subdifferential inequality: (2.5)‖x+y‖2≤‖x‖2+2〈y,x+y〉.A mappingT:C→C is said to be strongly quasinonexpansive [10] if it is quasinonexpansive and zn-Tzn→0 whenever {zn} is a bounded sequence in C such that limn→∞(∥zn-p∥-∥Tzn-p∥)=0 for some p∈Fix(T). It is known that every metric projection is strongly quasinonexpansive.We are now ready to present our main result.Theorem 2.3.
LetC be a closed convex subset of a Hilbert space H and T:C→C is a strongly quasinonexpansive mapping such that I-T is demiclosed at zero. Suppose that f:C→C is a contraction. Let {xn} be a sequence in C defined by
(2.6)x1=x∈Carbitrarilychosen,xn+1=αnf(xn)+(1-αn)Txn,
where {αn} is a sequence in (0,1) satisfying (C1)
limn→∞αn=0,(C2)
∑n=1∞αn=∞.Then the sequence {xn} converges to an element z∈Fix(T) and the following inequality holds
(2.7)〈f(z)-z,q-z〉≤0
for all q∈Fix(T).Before we give the proof, we note thatFix(T) is closed and convex (see [11] for more general setting). Hence the mapping PFix(T)∘f:C→C is a contraction. Then it follows from the well-known Banach's contraction principle that there exists a unique element z∈C such that z=PFix(T)∘f(z). In particular, z∈Fix(T) and 〈f(z)-z,q-z〉≤0 for all q∈Fix(T).Let us assume that∥f(x)-f(y)∥≤α∥x-y∥ for all x,y∈C where α is a real number in [0,1).Lemma 2.4.
The sequence{xn} is bounded.Proof.
We consider the following inequality:(2.8)‖xn+1-z‖≤αn‖f(xn)-z‖+(1-αn)‖Txn-z‖≤αn‖f(xn)-f(z)‖+αn‖f(z)-z‖+(1-αn)‖Txn-z‖≤(αnα+1-αn)‖xn-z‖+αn‖f(z)-z‖=(1-αn(1-α))‖xn-z‖+αn(1-α)‖f(z)-z‖1-α≤max⁡{‖xn-z‖,‖f(z)-z‖1-α}.
By induction, we conclude that the sequence {∥xn-z∥} is bounded and hence so is the sequence {xn}.Lemma 2.5.
The following inequality holds for alln∈ℕ:
(2.9)‖xn+1-z‖2≤(1-αn)2‖xn-z‖2+2αnα‖xn-z‖‖xn+1-z‖+2αn〈f(z)-z,xn+1-z〉.Proof.
It follows from the subdifferential inequality that(2.10)‖xn+1-z‖2=‖αn(f(xn)-z)+(1-αn)(Txn-z)‖2≤(1-αn)2‖Txn-z‖2+2αn〈f(xn)-z,xn+1-z〉≤(1-αn)2‖xn-z‖2+2αn〈f(xn)-f(z),xn+1-z〉+2αn〈f(z)-z,xn+1-z〉≤(1-αn)2‖xn-z‖2+2αnα‖xn-z‖‖xn+1-z‖+2αn〈f(z)-z,xn+1-z〉.Lemma 2.6.
If there exists a subsequence{xnk} of {xn} such that liminfk→∞(∥xnk+1-z∥-∥xnk-z∥)≥0, then limsupk→∞〈f(z)-z,xnk+1-z〉≤0.Proof.
First, we note thatαnk→0 and let us consider the following inequality:
(2.11)0≤lim⁡inf⁡k→∞(‖xnk+1-z‖-‖xnk-z‖)≤lim⁡inf⁡k→∞(αnk‖f(xnk)-z‖+(1-αnk)‖Txnk-z‖-‖xnk-z‖)=lim⁡inf⁡k→∞(‖Txnk-z‖-‖xnk-z‖)≤limsup⁡k→∞(‖Txnk-z‖-‖xnk-z‖)≤0.
This implies that limk→∞(∥xnk-z∥-∥Txnk-z∥)=0. Since T is a strongly quasinonexpansive mapping, xnk-Txnk→0. In particular, xnk-xnk+1→0. Because {xnk} is bounded, so there exists a subsequence {xnkl} of {xnk} such that xnkl⇀q and
(2.12)lim⁡l→∞〈f(z)-z,xnkl-z〉=limsup⁡k→∞〈f(z)-z,xnk-z〉.
It follows from the demiclosedness of I-T at zero that q∈Fix(T). Then
(2.13)lim⁡l→∞〈f(z)-z,xnkl-z〉=〈f(z)-z,q-z〉≤0.
Hence limsupk→∞〈f(z)-z,xnk+1-z〉=limsupk→∞〈f(z)-z,xnk-z〉≤0, as desired.Proof of Theorem2.3.
Let us consider the following two cases.Case 1.
There exists anN∈ℕ such that ∥xn+1-z∥≤∥xn-z∥ for all n≥N. It follows then that limn→∞∥xn-z∥ exists and hence liminfn→∞(∥xn+1-z∥-∥xn-z∥)=0. This implies that limsupn→∞〈f(z)-z,xn+1-z〉≤0. By Lemma 2.5, for all n≥N,
(2.14)‖xn+1-z‖2≤(1-αn)2‖xn-z‖2+2αnα‖xn-z‖‖xn+1-z‖+2αn〈f(z)-z,xn+1-z〉≤(1-2αn+2αnα)‖xn-z‖2+αn2‖xn-z‖2+2αn〈f(z)-z,xn+1-z〉=(1-2αn(1-α))‖xn-z‖2+2αn(1-α)(αn‖xn-z‖22(1-α)+〈f(z)-z,xn+1-z〉1-α).
Notice that ∑n=N∞2αn(1-α)=∞ and
(2.15)limsup⁡n→∞(αn‖xn-z‖22(1-α)+〈f(z)-z,xn+1-z〉1-α)≤0.
By Lemma 2.1, we have limn→∞∥xn-z∥2=0.Case 2.
There exists a subsequence{∥xnj-z∥} of {∥xn-z∥} such that ∥xnj-z∥<∥xnj+1-z∥ for all j∈ℕ. In this case, it follows from Lemma 2.2 that there exists a subsequence {∥xmk-z∥} of {∥xn-z∥} such that
(2.16)‖xmk-z‖≤‖xmk+1-z‖,‖xk-z‖≤‖xmk+1-z‖
for all k∈ℕ. It follows from liminfk→∞(∥xmk+1-z∥-∥xmk-z∥)≥0 that limsupk→∞〈f(z)-z,xmk+1-z〉≤0. Moreover, by Lemma 2.5, we have
(2.17)‖xmk+1-z‖2≤(1-αmk)2‖xmk-z‖2+2αmkα‖xmk-z‖‖xmk+1-z‖+2αmk〈f(z)-z,xmk+1-z〉≤(1-αmk)2‖xmk+1-z‖2+2αmkα‖xmk+1-z‖2+2αmk〈f(z)-z,xmk+1-z〉.
In particular, it follows that
(2.18)(2-αmk-2α)‖xmk+1-z‖2≤2〈f(z)-z,xmk+1-z〉.
This implies that
(2.19)(2-2α)limsup⁡k→∞‖xmk+1-z‖2≤limsup⁡k→∞2〈f(z)-z,xmk+1-z〉≤0.
Hence
(2.20)limsup⁡k→∞‖xk-z‖2≤limsup⁡k→∞‖xmk+1-z‖2=0.
Then limk→∞∥xk-z∥2=0. This completes the proof.Remark 2.7.
IfC is a convex subset of a Hilbert space and T:C→C is a quasinonexpansive mapping, then the mapping Tω:=(1-ω)I+ωT is strongly quasinonexpansive whenever ω∈(0,1) (see [10]). This means that Maingé's result is included in ours as a special case.Remark 2.8.
There is a strongly quasinonexpansive mappingS such that S is not of the form (1-ω)I+ωT where α∈(0,1/2) and T is a quasinonexpansive mapping. This means that there is an example which is beyond the scope of Maingé's result (see Remark 1.3, Theorem 1.1 with his old proof is valid for onlyα∈(0,1/2)).Example 2.9.
LetA={(x,x):x∈ℝ}. It is clear that A is a closed and convex subset of ℝ2. Notice that S:=PA is a strongly quasinonexpansive mapping and (0,0)∈Fix(S). Suppose that S=(1-ω)I+ωT where ω∈(0,1/2) and T is a quasinonexpansive mapping. Then, by Proposition 1.2, we have
(2.21)〈(2,0)-S(2,0),(2,0)-(0,0)〉≥ω2‖(2,0)-T(2,0)‖2=12ω‖(2,0)-S(2,0)‖2.
It is easy to see that S(2,0)=(1,1). In particular,
(2.22)2=〈(2,0)-(1,1),(2,0)〉≥12ω‖(2,0)-(1,1)‖2=1ω.
That is ω≥1/2, a contradiction.


## 3. Conclusion

We propose a viscosity approximation process for approximation of a fixed point of a quasinonexpansive mapping. This not only corrects Maingé's result but also essentially improves his result to a more general relaxation.



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