Our main result is given by the following theorem.

Proof. Let x0 be an arbitrary element of lev S⪯; that is, (19)x0∈X,x0⪯Sx0.Such an element exists from Assumption (A2). From Assumption (A3), we have (20)x1⪰Sx1,where x1=Tx0. Again, from Assumption (A3), we have (21)x2⪯Sx2,where x2=Tx1. Now, let us consider the Picard sequence {xn}⊂X defined by (22)xn+1=Txn, n=0,1,2,….Proceeding as above, by induction we get(23)x2n⪯Sx2n,x2n+1⪰Sx2n+1, n=0,1,2,….Therefore, by Assumption (A4), we have (24)dTx2n,Tx2n+1≤ψdx2n,x2n+1, n=0,1,2,….Again, by Assumption (A4), we have (25)dTx2n+1,Tx2n+2≤ψdx2n+1,x2n+2, n=0,1,2,….As a consequence, we have(26)dxn+1,xn≤ψdxn,xn-1, n=1,2,3,….From (26), since ψ is a nondecreasing function, for every n=1,2,3,…, we have(27)dxn+1,xn≤ψdxn,xn-1≤ψ2dxn-1,xn-2≤⋯≤ψndx1,x0.Suppose that (28)dx1,x0=0. In this case, from (23), we have (29)x0=x1=Tx0,x0⪯Sx0,x0=x1⪰Sx1=Sx0.Since ⪯ is a partial order, this proves that x0∈X is a solution to (1). Now, we may suppose that d(x1,x0)≠0. Let (30)δ=dx1,x0>0.From (27), we have(31)dxn+1,xn≤ψnδ, n=0,1,2,….Using the triangle inequality and (31), for all m=1,2,3,…, we have(32)dxn,xn+m≤dxn,xn+1+dxn+1,xn+2+⋯+dxn+m-1,xn+m≤ψnδ+ψn+1δ+⋯+ψn+m-1δ=∑i=nn+m-1ψiδ≤∑i=n∞ψiδ.On the other hand, since ∑k=0∞ψkδ<∞, we have(33)∑i=n∞ψiδ⟶0 as n→∞,which implies that {xn}={Tnx0} is a Cauchy sequence in (X,d). Then there is some x∗∈X such that(34)limn→∞dxn,x∗=0.On the other hand, from (23), we have (35)x2n∈lev S⪯, n=0,1,2,….Since S:X→X is level closed from the left (from Assumption (A2)), passing to the limit as n→∞ and using (34), we obtain (36)x∗∈lev S⪯,that is,(37)x∗⪯Sx∗.Now, using (23), (37), and Assumption (A4), we obtain (38)dTx2n+1,Tx∗≤ψdx2n+1,x∗, n=0,1,2,…,that is, (39)dx2n+2,Tx∗≤ψdx2n+1,x∗, n=0,1,2,….Passing to the limit as n→∞, using (34), the continuity of ψ at 0, and the fact that ψ(0)=0 (see Lemma 3), we get (40)dx∗,Tx∗=0,that is,(41)x∗=Tx∗.Next, using (37), (41), and Assumption (A3), we obtain (42)x∗=Tx∗⪰STx∗=Sx∗,that is,(43)x∗⪰Sx∗.Since ⪯ is a partial order, inequalities (37) and (43) yield(44)x∗=Sx∗.Further, (41) and (44) yield that x∗∈X is a solution to problem (1). Therefore, (i) is proved.

Suppose now that y∗∈X is another solution to (1) with x∗≠y∗. Using Assumption (A4) and the result (i) in Lemma 3, we obtain (45)dx∗,y∗=dTx∗,Ty∗≤ψdx∗,y∗<dx∗,y∗,which is a contradiction. Therefore, x∗∈X is the unique solution to (1), which proves (ii).

Passing to the limit as m→∞ in (32), we obtain estimate (16). In order to obtain estimate (17), observe that, by (26), we inductively obtain (46)dxn+k,xn+k+1≤ψk+1dxn-1,xn, n≥1, k≥0,and hence, similar to the derivation of (32), we obtain (47)dxn+p,xn≤∑k=1pψkdxn-1,xn, p≥0, n≥1. Now, passing to the limit as p→∞, (17) follows.

The proof is complete.

As a consequence, we have the following result.

We end the paper with the following illustrative example.

Example 8. Let X=[0,∞) and d be the metric on X defined by (53)dx,y=x-y, x,y∈X×X.Then (X,d) is a complete metric space. Let R be the binary relation on X defined by (54)R=x,x:x∈X∪0,2. Consider the partial order on X defined by (55)x,y∈X×X, x⪯y⟺x,y∈R.Let us define the pair of mappings T,S:X→X by (56)Tx=xif x∉0,2,2otherwise,Sx=2if x∈0,2,1if x>2.Observe that, in this case, we have (57)lev S⪯=x∈X:x⪯Sx=0,2,which is nonempty and closed set. Therefore, the operator S:X→X is level closed from the left, and Assumption (A2) is satisfied. Moreover, we have (58)x∈X:Sx⪯x=2.In order to check the validity of Assumption (A3), let x∈X be such that x⪯Sx; that is, x∈{0,2}. If x=0, then Tx=T0=2 and STx=ST0=S2=2. Then STx⪯Tx. If x=2, then Tx=T2=2 and STx=ST2=S2=2. Then STx⪯Tx. Now, let x∈X be such that Sx⪯x; that is, x=2. In this case, we have STx=ST2=S2=2 and Tx=T2=2. Then Tx⪯STx. Therefore, Assumption (A3) is satisfied. Now, let (x,y)∈X×X be such that x⪯Sx and Sy⪯y; that is, x∈{0,2} and y=2. For (x,y)=(0,2), we have (59)dTx,Ty=dT0,T2=d2,2=0≤ψd0,2,for every ψ∈Ψ. For (x,y)=(2,2), we have (60)dTx,Ty=dT2,T2=0≤ψd2,2=ψ0,for every ψ∈Ψ. Therefore, Assumption (A4) is satisfied. Now, applying Theorem 4, we deduce that problem (1) has a unique solution x∗∈X. Clearly, in our case, we have x∗=2.