JFS Journal of Function Spaces 2314-8888 2314-8896 Hindawi Publishing Corporation 10.1155/2016/6759320 6759320 Research Article A New Approach for the Approximations of Solutions to a Common Fixed Point Problem in Metric Fixed Point Theory http://orcid.org/0000-0002-7967-0554 Altun Ishak 1,2 2 http://orcid.org/0000-0001-7802-5070 Al Arifi Nassir 3 Jleli Mohamed 4 Lashin Aref 5,6 6 http://orcid.org/0000-0002-8380-9424 Samet Bessem 4 Cianciaruso Filomena 1 College of Science King Saud University Riyadh Saudi Arabia ksu.edu.sa 2 Department of Mathematics Faculty of Science and Arts Kirikkale University 71450 Yahsihan Kirikkale Turkey kku.edu.tr 3 College of Science Geology and Geophysics Department King Saud University Riyadh 11451 Saudi Arabia ksu.edu.sa 4 Department of Mathematics College of Science King Saud University P.O. Box 2455 Riyadh 11451 Saudi Arabia ksu.edu.sa 5 College of Engineering Petroleum and Natural Gas Engineering Department King Saud University Riyadh 11421 Saudi Arabia ksu.edu.sa 6 Faculty of Science Geology Department Benha University Benha 13518 Egypt bu.edu.eg 2016 27102016 2016 03 07 2016 18 09 2016 2016 Copyright © 2016 Ishak Altun et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We provide sufficient conditions for the existence of a unique common fixed point for a pair of mappings T,S:XX, where X is a nonempty set endowed with a certain metric. Moreover, a numerical algorithm is presented in order to approximate such solution. Our approach is different to the usual used methods in the literature.

King Saud University
1. Introduction and Problem Formulation

Let (X,d) be a complete metric space and T,S:XX be two given operators. In this paper, we are interested on the problem:(1)Find xX such thatx=Tx,x=Sx.We provide sufficient conditions for the existence of one and only one solution to (1). Moreover, we present a numerical algorithm in order to approximate such solution. Our approach is different to the existing methods in the literature.

System (1) arises in the study of different problems from nonlinear analysis. For example, when we deal with the solvability of a system of integral equations, such problem can be formulated as a common fixed point problem for a pair of self-mappings T,S:XX, where T and S are two operators that depend on the considered problem. For some examples in this direction, we refer to  and references therein.

The most used techniques for the solvability of problem (1) are based on a compatibility condition introduced by Jungck . Such techniques are interesting and can be useful for the solvability of certain problems (see  and references therein). However, two major difficulties arise in the use of such approach. At first, the compatibility condition is not always satisfied, and in some cases it is not easy to check such condition. Moreover, the numerical approximation of the common fixed point is constructed via the axiom of choice using certain inclusions, which makes its numerical implementation difficult.

In this paper, problem (1) is investigated under the following assumptions.

Assumption (A1). We suppose that X is equipped with a partial order . Recall that is a partial order on X if it satisfies the following conditions:

xx, for every xX.

xy and yz imply that xz, for every (x,y,z)X×X×X.

xy and yx imply that x=y, for every (x,y)X×X.

Assumption (A2). The operator S:XX is level closed from the left; that is, the set (2)levS=xX:xSxis nonempty and closed.

In order to make the lecture for the reader easy, let us give an example.

Example 1.

Let X=C([0,1];R) be the set of real valued and continuous functions on [0,1]. We consider the metric d on X defined by (3)dx,y=maxxt-yt:t0,1,x,yX×X.We endow X with the partial order given by (4)x,yX×X,xyxtyt,t0,1.Next, define the operator S:XX by (5)Sxt=0txsds,t0,1.Clearly, S:XX is well-defined. Now, consider the set (6)levS=xX:xSx,that is, (7)levS=xX:xt0txsds,t0,1.Let {xn}levS be a sequence that converges to some xX (with respect to d); that is, (8)xnt0txnsds,t0,1,n,dxn,x0 as n.Since the uniform convergence implies the point-wise convergence, for all t[0,1], we have (9)limnxnt=xt.Moreover, for all t0,1,(10)0txnsds-0txsds0tdxn,xdsdxn,x0as n.Therefore,(11)xt0txsds,t0,1,which proves that S:XX is level closed from the left.

Remark 2.

Note that the fact that S:XX is level closed from the left does not imply that (12)levS=xX:xSxis closed. Several counterexamples can be obtained. We invite the reader to check this fact by himself.

Assumption (A3). For every xX, we have (13)xSxTxSTx,xSxTxSTx.

In order to fix our next assumption, we need to introduce the following class of mappings. We denote by Ψ the set of functions ψ:[0,)[0,) satisfying the conditions:

ψ is nondecreasing.

For all t>0, we have (14)μ0tk=0ψkt<.

Here, ψk is the kth iterate of ψ. Any function ψΨ is said to be a (c)-comparison function.

We have the following properties of (c)-comparison functions.

Lemma 3 (see [<xref ref-type="bibr" rid="B2">10</xref>]).

Let ψΨ. Then

ψ(t)<t, for all t>0,

ψ(0)=0,

ψ is continuous at t=0,

μ0 is nondecreasing and continuous at 0.

Our next assumption is the following.

Assumption (A4). There exists a function ψΨ such that, for every (x,y)X×X, we have (15)xSx,ySydTx,Tyψdx,y.

Now, we are ready to state and prove our main result.

2. A Common Fixed Point Result and Approximations

Our main result is given by the following theorem.

Theorem 4.

Suppose that Assumptions (A1)–(A4) are satisfied. Then

for any x0levS, the Picard sequence {Tnx0} converges to some xX, which is a solution to (1),

xX is the unique solution to (1),

the following estimates(16)dTnx0,xμndTx0,x0,n=0,1,2,,(17)dTnx0,xμ1dTn-1x0,Tnx0,n=1,2,3,

hold, where (18)μnt=k=nψkt,t0,n=0,1,2,.

Proof.

Let x0 be an arbitrary element of levS; that is, (19)x0X,x0Sx0.Such an element exists from Assumption (A2). From Assumption (A3), we have (20)x1Sx1,where x1=Tx0. Again, from Assumption (A3), we have (21)x2Sx2,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)x2nSx2n,x2n+1Sx2n+1,n=0,1,2,.Therefore, by Assumption (A4), we have (24)dTx2n,Tx2n+1ψdx2n,x2n+1,n=0,1,2,.Again, by Assumption (A4), we have (25)dTx2n+1,Tx2n+2ψdx2n+1,x2n+2,n=0,1,2,.As a consequence, we have(26)dxn+1,xnψdxn,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,xnψdxn,xn-1ψ2dxn-1,xn-2ψndx1,x0.Suppose that (28)dx1,x0=0. In this case, from (23), we have (29)x0=x1=Tx0,x0Sx0,x0=x1Sx1=Sx0.Since is a partial order, this proves that x0X 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,xnψnδ,n=0,1,2,.Using the triangle inequality and (31), for all m=1,2,3,, we have(32)dxn,xn+mdxn,xn+1+dxn+1,xn+2++dxn+m-1,xn+mψnδ+ψn+1δ++ψn+m-1δ=i=nn+m-1ψiδi=nψiδ.On the other hand, since k=0ψkδ<, we have(33)i=nψiδ0as n,which implies that {xn}={Tnx0} is a Cauchy sequence in (X,d). Then there is some xX such that(34)limndxn,x=0.On the other hand, from (23), we have (35)x2nlevS,n=0,1,2,.Since S:XX is level closed from the left (from Assumption (A2)), passing to the limit as n and using (34), we obtain (36)xlevS,that is,(37)xSx.Now, using (23), (37), and Assumption (A4), we obtain (38)dTx2n+1,Txψdx2n+1,x,n=0,1,2,,that is, (39)dx2n+2,Txψdx2n+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=TxSTx=Sx,that is,(43)xSx.Since is a partial order, inequalities (37) and (43) yield(44)x=Sx.Further, (41) and (44) yield that xX is a solution to problem (1). Therefore, (i) is proved.

Suppose now that yX is another solution to (1) with xy. Using Assumption (A4) and the result (i) in Lemma 3, we obtain (45)dx,y=dTx,Tyψdx,y<dx,y,which is a contradiction. Therefore, xX 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+1ψk+1dxn-1,xn,n1,k0,and hence, similar to the derivation of (32), we obtain (47)dxn+p,xnk=1pψkdxn-1,xn,p0,n1. Now, passing to the limit as p, (17) follows.

The proof is complete.

Observe that Theorem 4 holds true if we replace Assumption (A2) by the following.

Assumption (A2). The operator S:XX is level closed from the right; that is, the set (48)levS=xX:xSxis nonempty and closed.

As a consequence, we have the following result.

Theorem 5.

Suppose that Assumptions (A1) and (A2)′–(A4) are satisfied. Then

for any x0levS, the Picard sequence {Tnx0} converges to some xX, which is a solution to (1),

xX is the unique solution to (1),

the following estimates(49)dTnx0,xμndTx0,x0,n=0,1,2,,dTnx0,xμ1dTn-1x0,Tnx0,n=1,2,3,

hold.

Taking S=IX (the identity operator), we obtain immediately from Theorem 4 (or from Theorem 5) the following fixed point result.

Corollary 6.

Let (X,d) be a complete metric space. Let T:XX be a given mapping. Suppose that there exists some ψΨ such that (50)dTx,Tyψdx,y,x,yX×X. Then

for any x0X, the Picard sequence {Tnx0} converges to some xX, which is a fixed point of T,

xX is the unique fixed point of T,

the following estimates (51)dTnx0,xμndTx0,x0,n=0,1,2,,dTnx0,xμ1dTn-1x0,Tnx0,n=1,2,3,

hold.

Remark 7.

Observe that all the obtained results hold true if we replace the partial order by any binary relation R which is antisymmetric; that is, R satisfies (52)x,yX×X,xRy,yRxx=y.

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,yX×X.Then (X,d) is a complete metric space. Let R be the binary relation on X defined by (54)R=x,x:xX0,2. Consider the partial order on X defined by (55)x,yX×X,xyx,yR.Let us define the pair of mappings T,S:XX by (56)Tx=xif x0,2,2otherwise,Sx=2if x0,2,1if x>2.Observe that, in this case, we have (57)levS=xX:xSx=0,2,which is nonempty and closed set. Therefore, the operator S:XX is level closed from the left, and Assumption (A2) is satisfied. Moreover, we have (58)xX:Sxx=2.In order to check the validity of Assumption (A3), let xX be such that xSx; that is, x{0,2}. If x=0, then Tx=T0=2 and STx=ST0=S2=2. Then STxTx. If x=2, then Tx=T2=2 and STx=ST2=S2=2. Then STxTx. Now, let xX be such that Sxx; that is, x=2. In this case, we have STx=ST2=S2=2 and Tx=T2=2. Then TxSTx. Therefore, Assumption (A3) is satisfied. Now, let (x,y)X×X be such that xSx and Syy; that is, x{0,2} and y=2. For (x,y)=(0,2), we have (59)dTx,Ty=dT0,T2=d2,2=0ψd0,2,for every ψΨ. For (x,y)=(2,2), we have (60)dTx,Ty=dT2,T2=0ψd2,2=ψ0,for every ψΨ. Therefore, Assumption (A4) is satisfied. Now, applying Theorem 4, we deduce that problem (1) has a unique solution xX. Clearly, in our case, we have x=2.

Remark 9.

Note that Theorem 4 (or Theorem 5) provides us just the existence and uniqueness of a common fixed point of the operators T,S:XX. However, the uniqueness of the fixed points of T is not satisfied in general. As we observe in Example 8, the operator T has infinitely many fixed points.

Competing Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors extend their appreciation to Distinguished Scientist Fellowship Program (DSFP) at King Saud University (Saudi Arabia).

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