Proof.
Let p=projΘ(0). Then, p∈Fix(W)∩EP(ψ) and Ap∈Fix(S)∩EP(φ). Set zn=UκφAρn, yn=(1-ζn)(ρn+ςA*(SUκφ-I)Aρn) and un=Uιψ[(1-ζn)(ρn+ςA*(SUκφ-I)Aρn)] for all n∈ℕ. Then un=Uιψyn. From Lemma 1, we know that Uιψ and Uκφ are firmly nonexpansive. Thus, we have
(14) ∥zn-Ap∥=∥UκφAρn-Ap∥≤∥Aρn-Ap∥,(15) ∥un-p∥=∥Uιψyn-p∥≤∥yn-p∥,(16) ∥SUκφAρn-Ap∥2=∥SUκφAρn-SUκφAp∥ ≤∥UκφAρn-UκφAp∥2 ≤∥Aρn-Ap∥2-∥UκφAρn-Aρn∥2.
Note that
(17)∥un+1-un∥=∥Uιψyn+1-Uιψyn∥≤∥yn+1-yn∥,(18)∥zn+1-zn∥=∥UκφAρn+1-UκφAρn∥≤∥Aρn+1-Aρn∥.
From (12) and (15), we have
(19)∥ρn+1-p∥=∥Wun-p∥≤∥un-p∥≤∥yn-p∥.
Observe that
(20)∥yn-p∥2=∥(1-ζn) ×(ρn-p+ςA*(Szn-Aρn))-ζnp∥2≤(1-ζn)∥(ρn-p+ςA*(Szn-Aρn))∥2 +ζn∥p∥2=(1-ζn)[∥A*(Szn-Aρn)∥2∥ρn-p∥+2ς ×〈ρn-p,A*(Szn-Aρn)〉 +ς2∥A*(Szn-Aρn)∥2] +ζn∥p∥2.
Since A* is the adjoint of A, we have
(21)〈ρn-p,A*(Szn-Aρn)〉 =〈A(ρn-p),Szn-Aρn〉 =〈Aρn-Ap+Szn-Aρn -(Szn-Aρn),Szn-Aρn〉 =〈Szn-Ap,Szn-Aρn〉-∥Szn-Aρn∥2.
Using parallelogram law, we obtain
(22)〈Szn-Ap,Szn-Aρn〉 =12(∥Szn-Ap∥2+∥Szn-Aρn∥2kkkkkkk-∥Aρn-Ap∥2).
From (16), (21) and (22), we have
(23)〈ρn-p,A*(Szn-Aρn)〉 =12(∥Szn-Ap∥2+∥Szn-Aρn∥2kkkkkkkk-∥Aρn-Ap∥2) -∥Szn-Aρn∥2 ≤12(∥Aρn-Ap∥2-∥zn-Aρn∥2kkkkkkk+∥Szn-Aρn∥2-∥Aρn-Ap∥2) -∥Szn-Aρn∥2 =-12∥zn-Aρn∥2 -12∥Szn-Aρn∥2.
By (20) and (23), we deduce
(24)∥yn-p∥2 ≤(1-ζn)[(-12∥zn-Aρn∥2-12∥Szn-Aρn∥2)∥ρn-p∥2+ς2∥A∥2∥Szn-Aρn∥2 +2ς(-12∥zn-Aρn∥2 -12∥Szn-Aρn∥2)]+ζn∥p∥2 =(1-ζn) ×[∥ρn-p∥2+(ς2∥A∥2-ς)∥Szn-Aρn∥2 -ς∥zn-Aρn∥2]+ζn∥p∥2 ≤(1-ζn)∥ρn-p∥2+ζn∥p∥2.
It follows from (19), we get
(25)∥ρn+1-p∥2≤∥yn-p∥2≤(1-ζn)∥ρn-p∥2+ζn∥p∥2≤max{∥ρn-p∥2,∥p∥2}.
The boundedness of the sequence {ρn} yields.
Set vn=ρn+ςA*(SUκφ-I)Aρn. Then, we have
(26)∥vn+1-vn∥2=∥ρn+1-ρn +ς[A*(Szn+1-Aρn+1)-A*(Szn-Aρn)]∥2=∥ρn+1-ρn∥2 +2ς〈ρn+1-ρn, A*[(Szn+1-Aρn+1)-(Szn-Aρn)]〉 +ς2∥A*(Szn+1-Aρn+1)-A*(Szn-Aρn)A*(Szn+1-Aρn+1)∥2≤∥ρn+1-ρn∥2 +2ς〈Aρn+1-Aρn, Szn+1-Szn-(Aρn+1-Aρn)〉 +ς2∥A∥2∥Szn+1-Szn-(Aρn+1-Aρn)∥2=∥ρn+1-ρn∥2 +ς2∥A∥2∥Szn+1-Szn-(Aρn+1-Aρn)∥2 +2ς〈Szn+1-Szn, Szn+1-Szn-(Aρn+1-Aρn)〉 -2ς∥Szn+1-Szn-(Aρn+1-Aρn)∥2=∥ρn+1-ρn∥2 +ς2∥A∥2∥Szn+1-Szn-(Aρn+1-Aρn)∥2 +ς(∥Szn+1-Szn∥2 +∥Szn+1-Szn-(Aρn+1-Aρn)∥2 -∥Aρn+1-Aρn∥2) -2ς∥Szn+1-Szn-(Aρn+1-Aρn)∥2=∥ρn+1-ρn∥2 +(ς2∥A∥2-ς) ×∥Szn+1-Szn-(Aρn+1-Aρn)∥2 +ς(∥Szn+1-Szn∥2-∥(Aρn+1-Aρn)∥2)≤∥ρn+1-ρn∥2 +(ς2∥A∥2-ς) ×∥Szn+1-Szn-(Aρn+1-Aρn)∥2 +ς(∥zn+1-zn∥2-∥Aρn+1-Aρn∥2).
Since ς∈(0,1/∥A∥2), we derive by virtue of (18) and (26) that
(27) ∥vn+1-vn∥≤∥ρn+1-ρn∥.
According to (17) and (27), we have
(28)∥ρn+1-ρn∥=∥Wun+1-Wun∥≤∥un+1-un∥≤∥yn+1-yn∥=∥(1-ζn+1)vn+1-(1-ζn)vn∥=∥(1-ζn+1)(vn+1-vn)+(ζn-ζn+1)vn∥≤(1-ζn+1)∥vn+1-vn∥+|ζn+1-ζn|∥vn∥≤(1-ζn+1)∥ρn+1-ρn∥+|ζn+1-ζn|∥vn∥.
It follows that
(29)∥ρn+1-ρn∥≤|ζn+1-ζn|ζn+1∥vn∥.
Since {ρn} is bounded, we can deduce {vn} is also bounded. From (29), we have
(30)limn→∞∥ρn+1-ρn∥=0.
Hence,
(31)limn→∞∥ρn-Wun∥=0.
Using the firmly-nonexpansivenessity of Uιψ, we have
(32)∥un-p∥2=∥Uιψyn-p∥2≤∥yn-p∥2-∥Uιψyn-yn∥2=∥yn-p∥2-∥un-yn∥2.
Thus, we get
(33)∥ρn+1-p∥2≤∥un-p∥2≤∥yn-p∥2-∥un-yn∥2≤(1-ζn)∥ρn-p∥2+ζn∥p∥2-∥un-yn∥2.
It follows that
(34)∥un-yn∥2≤∥ρn-p∥2-∥ρn+1-p∥2+ζn∥p∥2≤(∥ρn-p∥+∥ρn+1-p∥2)∥ρn+1-p∥2+ζn∥p∥2.
This together with (30) and (C1) implies that
(35)limn→∞∥un-yn∥=0.
Note that
(36)∥ρn+1-p∥2≤∥yn-p∥2≤(1-ζn)∥ρn-p∥2 +(1-ζn)(ς2∥A∥2-ς)∥Szn-Aρn∥2 -(1-ζn)ς∥zn-Aρn∥2+ζn∥p∥2.
Hence,
(37)(1-ζn)(ς-ς2∥A∥2)∥Szn-Aρn∥2 +(1-ζn)ς∥zn-Aρn∥2 ≤∥ρn-p∥2-∥ρn+1-p∥2+ζn∥p∥2 ≤(∥ρn-p∥+∥ρn+1-p∥)∥ρn+1-ρn∥ +ζn∥p∥2,
which implies that
(38)limn→∞∥Szn-Aρn∥=limn→∞∥zn-Aρn∥=0.
So, we get
(39)limn→∞∥Szn-zn∥=0.
Since
(40)∥yn-pn∥=∥ςA*(SUκφ-I)Aρn+ζnvn∥≤ς∥A∥∥Szn-Aρn∥+ζn∥vn∥,
we get
(41)limn→∞∥ρn-yn∥=0.
From (31), (35), and (41), we get
(42)limn→∞∥ρn-Wρn∥=0.
Now, we show that
(43)limsupn→∞〈p,yn-p〉≥0.
Choose a subsequence {yni} of {yn} such that
(44)limsupn→∞〈p,yn-p〉=limi→∞〈p,yni-p〉.
Notice that {yni} is bounded, we can choose {ynij} of {yni} such that ynij⇀z. Without loss of generality, we assume that yni⇀z. From the above conclusions, we derive that
(45)ρni⇀z, uni⇀z, Aρni⇀Az, zni⇀Az.
By Lemma 2, (39), and (41), we deduce z∈Fix(W) and Az∈Fix(S).
Next, we show that z∈EP(ψ). Since un=Uιψyn, we have
(46)ψ(un,x†)+1ι〈x†-un,un-yn〉≥0, ∀x†∈C.
By the monotonicity of ψ, we have
(47)1ι〈x†-un,un-yn〉≥ψ(x†,un),
and so
(48)〈x†-uni,uni-yniι〉≥ψ(x†,uni).
Since ∥un-yn∥→0, uni⇀z, we obtain (uni-yni)/ι→0. Thus, 0≥ψ(x†,z). For t with 0<t≤1 and x†∈C, let yt=tx†+(1-t)z∈C. We obtain ψ(yt,z)≤0. Hence,
(49)0=ψ(yt,yt)≤tψ(yt,x†)+(1-t)ψ(yt,z)≤tψ(yt,x†).
So, 0≤ψ(yt,x†). And, thus, 0≤ψ(z,x†). This implies that z∈EP(ψ). Similarity, we can prove that Az∈EP(φ). To this end, we deduce z∈Fix(W)∩EP(ψ) and Az∈Fix(S)∩EP(φ). That is to say, z∈Θ. Therefore,
(50)limsupn→∞〈p,yn-p〉=limi→∞〈p,yni-p〉=limi→∞〈p,z-p〉≥0.
Finally, we prove ρn→p. From (12), we have
(51)∥ρn+1-p∥2≤∥yn-p∥2=∥(1-ζn)(vn-p)-ζnp∥2≤(1-ζn)∥vn-p∥2-2ζn〈p,yn-p〉≤(1-ζn)∥ρn-p∥2-2ζn〈p,yn-p〉.
Applying Lemma 3 and (50) to (51), we deduce ρn→p. The proof is completed.