Proof.
We firstly show that the sequence {xn} is bounded. For our convenience, we take T:=PC(I-ξA*(I-PQ)A). Then, for any x*∈Γ, we have Tx*=x*. Now, we observe that
(3.2)∥xn+1-x*∥≤∥αnu+βnxn+γnTxn-x*∥≤αn∥u-x*∥+βn∥xn-x*∥+γn∥Txn-x*∥.
Now, we note that the condition 0<ξ<(2/∥A∥2) implies that the operator PC(I-ξA*(I-PQ)A) is averaged. Since I-PQ is firmly nonexpansive mappings and so is (1/2)-average, which is 1-ism. Also observe that A*(I-PQ)A is (1/∥A∥2)-ism so that ξA*(I-PQ)A is (1/ξ∥A∥2)-ism. Further, from the fact that I-ξA*(I-PQ)A is (ξ∥A∥2/2)-averaged and PC is (1/2)-averaged, we may obtain that PC(I-ξA*(I-PQ)A) is χ-averaged, where
(3.3)χ=12+ξ∥A∥22-12·ξ∥A∥22=2+ξ∥A∥24.
This implies that T=χI+(1-χ)S, where χ=(2+ξ∥A∥2/4)∈(0,1) for some nonexpansive mappings S. Note that T is also nonexpansive mappings. Hence, we have
(3.4)∥Txn-x*∥=∥Txn-Tx*∥≤∥xn-x*∥.
From the inequalities (3.2) and (3.4), we have
(3.5)∥xn+1-x*∥≤αn∥u-x*∥+(1-αn)∥xn-x*∥≤max {∥u-x*∥,∥xn-x*∥}.
Continuing inductively, we may obtain that the inequality
(3.6)∥xn+1-x*∥≤max {∥u-x*∥,∥x0-x*∥},
holds for all n≥0. So, {xn} is bounded so does {Txn}.
Next, we will show that lim n→∞∥xn+1-xn∥=0. Observe that
(3.7)∥xn+1-xn∥=∥(αnu+βnxn+γnTxn)-(αn-1u+βn-1xn-1+γn-1Txn-1)∥≤∥(αnu+βnxn+γnTxn)-(αnu+βnxn-1+γnTxn-1)∥+∥(αnu+βnxn-1+γnTxn-1)-(αn-1u+βn-1xn-1+γn-1Txn-1)∥≤(1-αn)∥xn-xn-1∥+|αn-αn-1|∥u∥+|βn-βn-1|∥xn-1∥+|γn-γn-1|∥Txn-1∥.
Since {xn} and {Txn} are bounded, there exists M=sup (∥u∥,∥xn+1∥,∥Txn-1∥)>0 such that
(3.8)∥xn+1-xn∥≤(1-αn)∥xn-xn-1∥+M(|αn-αn-1|+|βn-βn-1|+|γn-γn-1|).
According to Lemma 2.5 and the condition (C3), we have lim n→∞∥xn+1-xn∥=0.
We note that
(3.9)∥xn-Txn∥≤∥xn-xn+1∥+∥xn+1-Txn∥≤∥xn-xn+1∥+∥αnu+βnxn+(1-αn-βn)Txn-Txn∥=∥xn-xn+1∥+∥αn(u-Txn)+βn(xn-Txn)∥≤∥xn-xn+1∥+αn∥(u-Txn)∥+βn∥(xn-Txn)∥=11-βn∥xn+1-xn∥+αn1-βn∥u-Txn∥.
Consequently, by the condition (C1) and (C2), we also have lim n→∞∥xn-Txn∥=0. Next, we will show that
(3.10)lim sup n→∞〈v-z0,xn+1-z0〉≤0, where z0=PΓv.
To show this, we can choose a subsequence {xnk} of {xn} such that
(3.11)lim sup n→∞〈v-z0,Txn-z0〉=lim k→∞〈v-z0,Txnk-z0〉.
As {xn} is bounded, there exists a subsequence {xnk} which converges weakly to z. We may assume without loss of generality that xnk⇀z. Since ∥Txn-xn∥→0, we obtain Txnk⇀z as k→∞. By Lemma 2.3, we obtain that z∈Fix (T)=Γ.
Now from (2.4), observe that
(3.12)lim sup n→∞〈v-z0,xn-z0〉=lim sup n→∞〈v-z0,Txn-z0〉=lim k→∞〈v-z0,Txnk-z0〉=〈v-z0,z-z0〉≤0.
Therefore, we compute
(3.13)∥xn+1-z0∥2=〈αnv+βnxn+γnTxn-z0,xn+1-z0〉=αn〈v-z0,xn+1-z0〉+βn〈xn-z0,xn+1-z0〉+γn〈Txn-z0,xn+1-z0〉≤αn〈v-z0,xn+1-z0〉+12βn(∥xn-z0∥2+∥xn+1-z0∥2)+12γn(∥xn-z0∥2+∥xn+1-z0∥2)≤αn〈v-z0,xn+1-z0〉+12(1-αn)(∥xn-z0∥2+∥xn+1-z0∥2)
which implies that
(3.14)∥xn+1-z0∥2≤(1-αn)(∥xn-z0∥2)+2αn〈v-z0,xn+1-z0〉.
Finally, by (3.12), (3.14), and Lemma 2.5, we conclude that {xn} converges to z0. This completes the proof.