Employing the preliminaries in the previous section, now we will study the strong convergence of the new two-step viscosity iteration (13) for set-valued nonexpansive operators in complete CAT(0) spaces.

Proof. The proof shall be divided into the following four steps.

Step 1. We first prove that sequences {xn}, {f(xn)}, {yn}, and {zn} are bounded. In fact, setting p∈F(T), then from Lemma 3, we know (29)dyn,p≤αndfxn,p+1-αndistzn,Tp≤αndfxn,p+1-αnHTxn,Tp≤αndfxn,p+1-αndxn,p≤αndfxn,fp+αndfp,p+1-αndxn,p≤1-αn1-kdxn,p+αndfp,p,and (30)dxn+1,p≤βndxn,p+1-βndyn,p≤1-αn1-k1-βndxn,p+αn1-k1-βndfp,p1-k≤maxdxn,p,dfp,p1-k.Thus, we obtain (31)dxn,p≤maxdx1,p,dfp,p1-k.Hence, {xn} is bounded, so is {f(xn)}. By (29), it is easy to know that {yn} is bounded. Since d(zn,p)≤H(T(xn,T(p))≤d(xn,p), one can easily know that the sequence {zn} is also bounded.

Step 2. We present that limn→∞d(xn,yn)=0, limn→∞dist(xn,T(xn))=0, limn→∞d(xn,zn)=0, limn→∞d(xn,xn+1)=0, limn→∞d(zn,zn+1)=0, and limn→∞dist(zn,T(zn))=0. Indeed, by applying Lemmas 3 and 4, we have (32)dyn,yn+1≤dαnfxn⊕1-αnzn,αn+1fxn+1⊕1-αn+1zn+1≤dαnfxn⊕1-αnzn,αnfxn⊕1-αnzn+1+dαnfxn⊕1-αnzn+1,αnfxn+1⊕1-αnzn+1+dαnfxn+1⊕1-αnzn+1,αn+1fxn+1⊕1-αn+1zn+1≤1-αndzn,zn+1+αndfxn,fxn+1+αn-αn+1dfxn+1,zn+1≤1-αn1-kdxn,xn+1+αn-αn+1dfxn+1,zn+1,and so (33)dyn,yn+1-dxn,xn+1≤αn-αn+1dfxn+1,zn+1-1-kαndxn,xn+1.From limn→∞αn=0 and the boundedness of {xn}, {f(xn)}, and {zn}, we know (34)lim supn→∞dyn+1,yn-dxn+1,xn≤0.It follows from Lemma 5 that (35)limn→∞dxn,yn=0.Thus, (36)distxn,Txn≤dxn,zn≤dxn,yn+αndfxn,zn⟶0 as n⟶∞.By (36), now we know that (37)limn→∞dzn,xn=0.

Moreover, (38)dxn,xn+1=1-βndxn,yn⟶0and (39)dzn,zn+1≤dxn,xn+1⟶0as n→∞. By (36) and (37), we get (40)distzn,Tzn≤dzn,xn+distxn,Txn+HTxn,Tzn≤2dxn,zn+distxn,Txn⟶0 as n⟶∞.

Step 3. Now, we show that (41)lim supn→∞d2fx~,x~-d2fx~,zn≤0,with x~=PF(T)f(x~) satisfying (42)x~fx~→,xx~→≥0, ∀x∈FT.Above all, since T(x) is compact for any x∈E, then T(x)∈BC(X). It follows from Lemma 7 that F(T) is closed and convex, which implies that PF(T)u is well defined for any u∈X. By Lemma 12 (i), we know that {xα} generated by (10) converges strongly to x~=PF(T)f(x~) as α→0+. Then by Lemma 8, we know that x~ is the unique solution of the following variational inequality: (43)x~fx~→,xx~→≥0, ∀x∈FT.

Next, since {zn} is bounded and limn→∞dist(zn,T(zn))=0, it follows from Lemma 12 (ii) that for all Banach limits μ, (44)d2fx~,x~≤μnd2fx~,zn,and so (45)μnd2fx~,x~-d2fx~,zn≤0.Further, limn→∞d(zn,zn+1)=0 implies that (46)lim supn→∞d2fx~,x~-d2fx~,zn+1-d2fx~,x~-d2fx~,zn=0.By Lemma 11, we have (47)lim supn→∞d2fx~,x~-d2fx~,zn≤0.

Step 4. limn→∞xn=x~ will be verified. In fact, by Lemma 3 and (13), now we know (48)d2xn+1,x~≤βnd2xn,x~+1-βnd2yn,x~-βn1-βnd2xn,yn≤βnd2xn,x~+1-βnd2yn,x~,and (49)d2yn,x~≤αnd2fxn,x~+1-αnd2zn,x~-αn1-αnd2fxn,zn≤1-αnH2Txn,Tx~+αn2d2fxn,zn+αnd2fxn,x~-d2fxn,zn≤1-αnd2xn,x~+αn2d2fxn,zn+αnd2fxn,x~-d2fxn,zn.It follows from (21), Cauchy-Schwarz inequality, and Lemma 9 that (50)αnd2fxn,x~-d2fxn,zn≤2αndfxn,fx~dzn,x~+fx~x~→,znx~→-d2zn,x~≤2αnkdxn,x~dzn,x~+fx~x~→,znx~→-d2zn,x~≤αnkd2xn,x~+d2zn,x~+2αnfx~x~→,znx~→-2αnd2zn,x~≤αnkd2xn,x~+αnd2fx~,x~-d2fx~,zn.From (50) and (49), we know (51)d2yn,x~≤1-αn1-kd2xn,x~+αnd2fx~,x~-d2fx~,zn+αn2d2fxn,zn.Combining (51) and (48), we get (52)d2xn+1,x~≤βnd2xn,x~+1-βn1-αn1-kd2xn,x~+αnd2fx~,x~-d2fx~,zn+αn2d2fxn,zn≤1-1-kαn1-βnd2xn,x~+αn1-βnd2fx~,x~-d2fx~,zn+1-βnαn2d2fxn,zn≤1-1-kαn1-βnd2xn,x~+αn1-βnd2fx~,x~-d2fx~,zn+αn2d2fxn,zn,i.e., (53)un+1≤1-αn′un+αn′βn′, ∀n≥1,where un=d2(xn,x~), αn′=(1-k)αn(1-βn), and (54)βn′=1-βnd2fx~,x~-d2fx~,zn+αnd2fxn,zn1-k1-βn.Thus, from the conditions (L1)-(L3) and the inequality (41), it follows that αn′∈(0,1), and (55)∑n=1∞αn′=∞,lim supn→∞ βn′≤0.Hence, it follows from Lemma 6 that un→0. This implies that the proof is completed.