# A Proposal for Revisiting Ćirić and Caristi Type Theorems in Metric Spaces

Authors: Belkassem Seddoug; Soumia Chaira; Karim Chaira
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

In this paper, we revisit the renowned fixed point theorems of Ćirić and Caristi. We propose some new fixed point theorems in a metric space with partial order. To make our results effective, several examples are presented.


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## Body

## 1. Introduction and Preliminary

This work is motivated by some recent works on the extension of Banach Contraction Principle to metric spaces with a partial order [1]. Caristi’s fixed point theorem is maybe one of the most useful extension of Banach Contraction Principle [2–4]. It has been successfully applied in many topics such as differential equations, convex minimization, operator theory, variational inequalities, and control theory. For known Caristi-type fixed point results in the literature, see [5–13]. Recall that this theorem states that any map T:E⟶E has a fixed point provided that E is complete and there exists a lower semicontinuous map φ:E⟶0,+∞ such that dx,Tx≤φx−φTx, for every x∈E. The proofs given to Caristi’s result vary and use different techniques (see [14, 15]).Using the combined Ćirić–Caristi condition, we introduce new fixed point theorems under hypotheses of the form(1)dTx,Ty,ordTx,xdTx,Ty≤Dominated Function,where the “Dominated Function” can be chosen to be(2)φx−φTxSx,yorφx−φTxNx,yorφx−φTxmax1,Sx,y,for certain functions Sx,y and Nx,y. Other corresponding forms under some advanced settings such as “partial order” are also discussed. To the best of our knowledge, we provide all the possible conditions to make the Caristi-type fixed point theorems appropriate and applicable in most situations.More precisely, the renowned results [14, 16, 17] for a single-valued map are the following.Theorem 1 (see Theorem1 in [16]).
LetE,d be a complete metric space and T:E⟶E be a mapping. Suppose that there exists k∈0,1 such that(3)dTx,Ty≤kNx,y,for all x,y∈E, where(4)Nx,y=maxdx,y,dx,Tx,dy,Ty,dy,Tx,dx,Ty.Then,T has a unique fixed point in E.Definition 1.
LetE,d,< be a partially ordered complete metric space. We say that E verifies the condition (OSC) if for any decreasing sequence xnn∈ℕ in E such that limn⟶+∞xn exists, then there exists infnxn and infnxn=limn⟶+∞xn.Theorem 2 (see Theorem5 in [14]).
LetE,< be a partially ordered set and suppose that there exists a distance d in E such that E,d is a complete metric space satisfying the (OSC) property. Let T:E⟶E be a monotone increasing mapping. Assume there exists a lower semicontinuous function φ:E⟶0,+∞ such that(5)dx,Tx≤φx−φTx,wheneverTx<x.Then,T has a fixed point if and only there exists x0∈E such that Tx0<x0.In this article, we prove new fixed point theorems of Caristi type and Ćirić type. All optional conditions for dominated functions are presented and discussed.


## 2. Main Results

Theorem 3.
LetE,d,< be a partially ordered complete metric space satisfying the (OSC) property and T:E⟶E be a monotonically increasing map such that there exists a function φ:E⟶0,+∞ satisfying(6)∀x,y∈E,x<y⟹dx,TxdTx,Ty≤φx−φTxdx,y.Then, for anyx0∈E such that Tx0<x0, the sequence xnn∈ℕ defined by xn+1=Txn converges to a fixed point of T.Proof.
On the one hand, the case, where there existsn∈ℕ, such that xn=Txn, gives xn as a fixed point of T.
On the other hand, ifxn≠Txn for every n∈ℕ, then, by induction, the sequence xnn∈ℕ is monotonically decreasing, and taking x=xn+1 and y=xn in (6), one obtains, for every n∈ℕ,(7)0<dxn+1,xn+22dxn,xn+1≤φxn+1−φxn+2,∑k=0ndxk+1,xk+22dxk,xk+1≤∑k=0nφxk+1−φxk+2.
Since the sequenceφxnn∈ℕ is necessarily positive and monotonically decreasing, the series ∑n≥0φxn−φxn+1 is convergent and, by the comparison principle, ∑n≥0dxn+1,xn+22/dxn,xn+1 is also convergent. Since, for every k∈ℕ, one has(8)2dxk+1,xk+2≤dxk+1,xk+22dxk,xk+1+dxk,xk+1.
So, for everyn∈ℕ, one obtains the inequality(9)∑k=1n+1dxk,xk+1≤∑k=0ndxk+1,xk+22dxk,xk+1+dx0,x1,concluding the convergence of the series ∑n≥0dxn,xn+1. Hence, xnn∈ℕ is a Cauchy sequence. Set x=limn⟶∞xn. Since E has the (OSC) property, we obtain x<xn for all n∈ℕ. By (6), one has(10)dx,TxdTx,Txn≤φx−φTxdx,xn,and so limn⟶∞dx,TxdTx,Txn=0. Therefore, dx,Tx=0.Example 1.
LetE=0,1 be endowed with the usual distance dx,y=x−y and the order “<” be defined by(11)∀x,y∈E2,x<y⇔x≥y.
LetT and φ be two functions defined on E by(12)Tx=x+12,ifx∈0,1,0,ifx=0,φx=1−x2,ifx∈0,1,where T is monotonically increasing on E.
We need only to consider two cases to check the hypothesis of Theorem3.Case 1: letx,y∈0,1 with x<y. We have(13)1−x2x−y2≤1−x2−1−x4x−y.So,(14)dx,TxdTx,Ty≤φx−φTxdx,y.Case 2:x=0 and y∈0,1. We have(15)d0,T0dT0,Ty=0=φ0−φT0d0,y.Remark 1.
(i)
In Example1, T does not satisfy the Banach Contraction Principle, for which we take x=0 and y=1/3, and we have dT0,T1/3>kd0,1/3for allk∈0,1.(ii)
In Example1, T does not satisfy the inequality of Caristi, for which we take x∈]0,1[, and we have dx,Tx=1−x/2 and φx−φTx=1−x/4. So, dx,Tx=1−x/2>φx−φTx.Corollary 1.
LetE,d,< be a partially ordered complete metric space satisfying the (OSC) property and T:E⟶E be a monotonically increasing map such that there exists a function φ:E⟶0,+∞ satisfying(16)∀x,y∈E,x<y⟹dx,TxdTx,Ty≤φx−φTpxdx,y,for some p∈ℕ∖0. Then, for any x0∈E such that Tx0<x0, the sequence xnn∈ℕ, defined by xn+1=Txn, converges to a fixed point of T.Proof.
If we consider the functionΦ defined on E by Φx=∑k=0p−1φTkxfor allx∈E, we obtain dx,TxdTx,Ty≤Φx−ΦTxdx,y. By applying Theorem 3, we get that T admits a fixed point.Example 2.
LetE=0,+∞ endowed with the usual distance dx,y=x−y and the order “<” defined by(17)∀x,y∈E2,x<y⟺x≥y.
LetT and φ be two functions defined on E by(18)Tx=x+12,ifx∈0,1,x,ifx∈0∩1,+∞,φx=23x,ifx∈0,+∞,0,ifx=0,where E has the condition (OSC) and T is monotonically increasing on E.
We need only to consider five cases to check the hypothesis of Corollary1:Case 1:x,y∈0,1 with x<y. We have T2x=x+3/4 and(19)1−x2x−y2≤231x−4x+3x−y.So,(20)dx,TxdTx,Ty≤φx−φT2xdx,y.Case 2:y∈0,1 and x∈]1,+∞[. We have dx,Tx=0 and φx−φT2x=0, so(21)dx,TxdTx,Ty≤φx−φT2xdx,y.Case 3:y=0 and x∈0,1. We have 1+x3+x≤8 and T2x=x+3/4, so(22)1−x2x+12≤231x−4x+3x.Hence,(23)dx,TxdTx,T0≤φx−φT2xdx,0.Case 4:y=0 and x∈]1,+∞[. We have T2x=x, then dx,Tx=0=φx−φT2x. So,(24)dx,TxdTx,T0≤φx−φT2xdx,0.Case 5:x,y∈]1,+∞[ with x<y. We have T2x=x, then dx,Tx=0=φx−φT2x. So,(25)dx,TxdTx,Ty≤φx−φT2xdx,y.
This shows that inequality (16) is satisfied and all the hypotheses of Corollary 1 are verified. In addition, T has the fixed points 0 and every x∈1,+∞.Remark 2.
(i)
Example2 does not satisfy the inequality for p=1. If we take the third case y=0 and x=2/3, we have d2/3,T2/3dT2/3,T0>φ2/3−φT2/3d2/3,0.(ii)
Example2 does not verify the inequality of Theorem 4 in [11]. If we take y=0 and x=2/3, we have d2/3,T2/3>0 but dT2/3,T0>φ2/3−φT2/3d2/3,0.(iii)
Note thatdT0,T1/2=3/4>d0,1/2=1/2, so the mapping T is not nonexpansive.Definition 2.
LetE,d,< be a partially ordered complete metric space and T:E⟶E be a mapping. We say that T satisfies the condition wc if, for any monotone decreasing sequence xnn∈ℕ in E such that there exists x=limn⟶+∞xn and x<xn for all n∈ℕ, we have limn⟶+∞dx,Txn=0.Theorem 4.
LetE,d,< be a partially ordered complete metric space and a monotonically increasing map T:E⟶E such that there exists a function φ:E⟶0,+∞ satisfying(26)∀x,y∈E,x<y⟹dx,TxdTx,Ty≤φx−φTxSx,y,where(27)Sx,y=dx,y+dx,Tx+dy,Ty.Assume that there exists a pointx0∈E satisfying Tx0<x0. If we further add one of the following hypotheses,(C1)T is continuous(C2) The mapx↦dx,Tx is lower semicontinuous(C3)E has the condition OSC, T has the condition wc, and φ is lower semicontinuous then T has at least one fixed point.Proof.
We define the sequencexnn≥0 by xn+1=Txn for each n∈ℕ. On the one hand, the case where there exists n∈ℕ, such that xn=Txn, gives xn as a fixed point of T. On the other hand, if xn≠Txn for every n∈ℕ, then, by induction, the sequence xnn∈ℕ is monotonically decreasing, and taking x=xn+1 and y=xn in (3), one obtains, for every n∈ℕ,(28)0<dxn+1,xn+22Sxn,xn+1≤φxn+1−φxn+2,where Sxn,xn+1=2dxn,xn+1+dxn+1,xn+2. So, the sequence φxnn∈ℕ is necessarily positive and monotonically decreasing, and therefore the series ∑n≥0φxn−φxn+1 is convergent. Thus, ∑n≥0dxn+1,xn+22/Sxn,xn+1 is also convergent. For every k∈ℕ, one has(29)4dxk+1,xk+2≤4dxk+1,xk+222dxk,xk+1+dxk+1,xk+2+2dxk,xk+1+dxk+1,xk+2.
Therefore, for everyn∈ℕ, one obtains the inequality(30)∑k=1n+1dxk,xk+1≤4∑k=0ndxk+1,xk+222dxk,xk+1+dxk+1,xk+2+2dx0,x1,obtaining the convergence of the series ∑n≥0dxn,xn+1. Hence, xnn∈ℕ is a Cauchy sequence. Set x∗=limn⟶∞xn.
(C1) and (C2) The case whereT is continuous is obvious. If the function x↦dx,Tx is lower semicontinuous, we obtain(31)dx∗,Tx∗≤liminfn⟶+∞dxn,Txn=limn⟶+∞dxn,xn+1=0.
Thus,dx∗,Tx∗=0, i.e., Tx∗=x∗.
(C3) SupposeE has the condition OSC, so there exists infn∈ℕxn and is equal to x∗. We have x∗<xn for all n∈ℕ. Since T is strictly increasing, Tx∗<Txn=xn+1, thus Tx∗<infn∈ℕxn=x∗.
In what follows, we suppose thatTx∗≠x∗ (because otherwise T admits x∗ as a fixed point).
We putEo=x∈E,Tx<xandx<x∗. The set Eo is nonempty because x∗∈Eo. Let x∈Eo, so x<x∗. Since T is monotonically increasing, Tx<Tx∗<x∗, thus TEo⊂Eo.
Takingx=x∗ and y=xn in inequality (26), we have(32)dx∗,Tx∗dTx∗,Txn≤φx∗−φTx∗Sx∗,xn.
Lettingn tend to +∞, we obtain(33)dx∗,Tx∗2≤φx∗−φTx∗dx∗,Tx∗.
By (33), we obtain(34)dx∗,Tx∗≤φx∗−φTx∗.
We define the partial order “≺¯” on Eo by(35)x≺¯y⇔x<yanddx,y≤φy−φx,where “≤” is the usual order on ℝ. It is interesting to see that φ is monotonically increasing from Eo to 0,+∞.
Assume thatFββ∈Γ is a set of totally ordered subset of Eo such that, for any β1,β2∈Γ, Fβ1⊂Fβ2, or Fβ2⊂Fβ1, where Γ is the index set. Let G=∪β∈ΓFβ. For any x,y∈G there exists βx,βy∈Γ such that x∈Fβx and y∈Fβy. We may assume that Fβx⊂Fβy, so x,y∈Fβy, that is, x and y are comparable. Hence, G is totally ordered subset of Eo, and by Zorn’s lemma, Eo has a maximal totally ordered subset.
LetF be a maximal totally ordered subset of Eo. We consider φ0=infx∈Fφx and a sequence ypp∈ℕ⊂F such that φypp∈ℕ is decreasing and convergent to φ0. By monotony of φ on Eo, the sequence ypp∈ℕ is also decreasing in Eo. Therefore, for every integers p≤q, we have dyp,yq≤φyp−φyq, which implies that ypp∈ℕ is a Cauchy’s sequence and there exists a unique y∗∈E such that limp⟶∞yp=y∗. So, infpyp=limp⟶+∞yp=y∗, that is, y∗<yp for all p∈ℕ. Moreover, by the lower semicontinuity of φ, we obtain φy∗≤liminfp⟶+∞φyp=φ0.
Next, we show thaty∗≺¯z for all z∈F. For that we distinguish two cases:Case 1: there existsx∈F satisfying φx=φ0, so we have x≺¯z, for any z∈F. It follows that x<yp and dx,yp≤φyp−φ0 for all p∈ℕ. Letting p tend to infinity, we have dx,y∗=0, which means that y∗=x≺¯z for all z∈F.Case 2: letz∈F, φz≠φ0≤φz, so there exists N>0 such that yp≺¯z whenever p≥N. Hence, y∗<yp<z and dz,yp≤φz−φyp for all p≥N. Letting p tends to infinity, we obtain dz,y∗≤φz−φ0≤φz−φy∗, and so y∗≺¯z.
On the basis of the above discussion, we can claim thaty∗≺¯z for all z∈F. Hence, y∗ is lower bound of F in Eo with respect to the order “≺¯.”
We havey∗<yp for every p∈ℕ. Then, by inequality (26),(36)dy∗,Ty∗dTy∗,Typ≤φy∗−φTy∗dy∗,yp+dy∗,Ty∗+dyp,Typ,and so(37)dy∗,Ty∗dy∗,Ty∗−dy∗,Typ≤φy∗−φTy∗2dy∗,yp+dy∗,Ty∗+dy∗,Typ.
SinceT has the condition wc, and when p tends to infinity, we obtain(38)dy∗,Ty∗2≤φy∗−φTy∗dy∗,Ty∗.
We claim thatTy∗=y∗. Indeed, if Ty∗≠y∗, one has(39)dy∗,Ty∗≤φy∗−φTy∗.
SinceT is monotonically increasing, we have Ty∗<Tx<xfor allx∈F. Particularly, Ty∗<yp for all p∈ℕ, which implies that Ty∗<infpyp=y∗. Thus, y∗∈Eo and, by the fact that TEo⊂Eo, we get Ty∗∈Eo and Ty∗≺¯y∗≺¯z for all z∈F. Since Ty∗≠y∗, we get that Ty∗∈F and Ty∗,y∗∪F is a totally ordered subset of Eo. This contradicts the maximality of F and finishes the proof.Example 3.
LetE=1/n:n∈ℕ∗∪0 endowed with the usual distance dx,y=x−y and the usual order “<.” We put, for all n∈ℕ∖0, xn=1/n and x0=0. Let T and φ be the two functions defined on E by(40)Txn=xn+1,ifn∈ℕ∖0,0,ifn=0,φxn=2xn,ifn∈ℕ∖1,x1,ifn=1,where E,d is a complete metric space and verifies the condition OSC. T is monotone increasing on E and has the condition wc (actually, T is continuous), φ is lower semicontinuous on E, and Tx1≤x1.
We need only to consider two cases to check the hypothesis of Theorem4:Case 1: letn,m∈ℕ∖0 such that m<n and n≠m. We have xn≤xm and 1/n+1+3/m+1≤4/m, i.e., xn+1+3xm+1≤4xm. So,(41)xm+1−xn+1≤2xm−xn+xn−xn+1+xm−xm+1,xn−xn+1xm+1−xn+1≤φxn−φxn+1xm−xn+xn−xn+1+xm−xm+1.Thus,(42)dxn,TxndTxm,Txn≤φxn−φTxnSxm,xn.Case 2:x=x0 and y=xn, n∈ℕ∖0. We have dx0,Tx0=0=φx0−φTx0. So,(43)dx0,Tx0dTx0,Txn=φx0−φTx0Sx0,xn.
This shows that inequality (26) is satisfied and all the hypotheses hold (including condition (C3)). In addition, T has the fixed point 0.
Note thatdx1,Tx1>φx1−φTx1, then T does not satisfy the inequality of Caristi.Corollary 2.
LetE,d,< be a partially ordered complete metric space and T:E⟶E be a monotone increasing map such that there exists a function φ:E⟶0,+∞ satisfying(44)∀x,y∈E,x<y⟹dx,TxdTx,Ty≤φx−φTxNx,y,where(45)Nx,y=maxdx,y,dx,Tx,dx,Ty,dTx,y,dTy,y.Assume that there exists a pointx0∈E satisfying Tx0<x0. If we further add one of the hypotheses C1, C2, or C3 of Theorem 4, then T has at least one fixed point.Corollary 3.
LetE,d,< be a partially ordered complete metric space and let T:E⟶E be a monotone increasing map such that there exists a function φ:E⟶0,+∞ satisfying(46)∀x,y∈E,x<y⟹dx,TxdTx,Ty≤φx−φTpxSx,y,for some p∈ℕ, where(47)Sx,y=dx,y+dx,Tx+dy,Ty.Assume that there exists a pointx0∈E satisfying Tx0<x0. If we further add one of the hypotheses (C1), (C2), or (C3) of Theorem 4, replacing the lower semicontinuity of φ in (C3) with the lower semicontinuity of φ°Ti, for each i∈1,…,p−1, then T has at least one fixed point.Proof.
Letx,y∈E2 such that x<y. By inequality (46), we have(48)dx,TxdTx,Ty≤φx−φTpxSx,y.
If we consider the functionΦ define on E by Φx=∑k=0p−1φTkxfor allx∈E, we obtain dx,TxdTx,Ty≤Φx−ΦTxdx,y, and by hypothesis (C3), the function Φ:E⟶0,+∞ is lower semicontinuous. Applying Theorem 4, we deduce that T admits a fixed point.
Our goal in the rest of this article is to generalize the recent Theorem4 in [11] and to give a short proof of it.Theorem 5.
LetE,d be a complete metric space and T:E⟶E a map such that there exists a function φ:E⟶[0,+∞[ satisfying(49)dx,Tx>0⟹dTx,Ty≤φx−φTxmax1,Sx,y,where(50)Sx,y=dx,y+dx,Tx+dy,Ty.Then,T has a fixed point.Proof.
Letx0∈E and define the sequence xnn∈ℕ by xn+1=Txn. On the one hand, the case where there exists n∈ℕ, such that xn=Txn, gives xn as a fixed point of T. On the other hand, if xn≠Txn for every n∈ℕ, then, taking x=xn and y=xn+1 in (49), one obtains, for every n∈ℕ,(51)0<dxn+1,xn+2max1,Sxn,xn+1≤φxn−φxn+1,where Sxn,xn+1=2dxn,xn+1+dxn+1,xn+2, which shows that the sequence φxnn∈ℕ is necessarily positive and monotonically decreasing. Therefore, the series ∑n≥0φxn−φxn+1 is convergent, and by the comparison principle, ∑n≥0dxn+1,xn+2/max1,Sxn,xn+1 is also convergent and limn⟶+∞dxn+1,xn+2/max1,Sxn,xn+1=0.
Letk∈0,1/3, there exists N∈ℕ such that, for all n≥N,(52)dxn+1,xn+2≤kmax1,Sxn,xn+1≤kmax1,2dxn,xn+1+dxn+1,xn+2.
So,(53)dxn+1,xn+2≤2k1−kmax1,dxn,xn+1≤max1,dxn,xn+1because k∈0,1/3. The sequence dxn,xn+1n is then positive and bounded above by(54)M=max1,dx0,x1,…,dxN,xN+1.
Next, we prove thatxnn≥0 is a Cauchy sequence. For this, we present two methods.Method 1.
Sincedxn,xn+1n≥0 is bounded, we get limsupn⟶+∞dxn,xn+1<∞. We have(55)limsupn⟶+∞dxn+1,xn+2≤limn⟶+∞φxn−φxn+1max1,limsupn⟶+∞Sxn,xn+1=0.
Thus,limn⟶+∞dxn,xn+1=0.
Suppose thatxnn≥0 is not a Cauchy sequence. Then, there exists ε>0 for which we can find subsequences xmkk≥0 and xnkk≥0 with nk>mk>k, and nk is the smallest integer such that(56)dxnk,xmk≥ε,dxmk,xnk−1<ε.
So,(57)ε≤dxmk,xnk≤dxmk,xnk−1+dxnk−1,xnk≤ε+dxnk−1,xnk.
Takingk⟶∞ and using limn⟶+∞dxn,xn+1=0, we obtain(58)limk⟶+∞dxmk,xnk=ε.
Similarly, we have(59)dxnk,xmk≤dxnk,xnk−1+dxnk−1,xmk−1+dxmk−1,xmk,dxnk−1,xmk−1≤dxnk−1,xnk+dxnk,xmk+dxmk,xmk−1.
Lettingk⟶∞ in the above two inequalities and using (56), we obtain(60)limk⟶+∞dxmk−1,xnk−1=ε.
Now, takingy=xmk−1 and x=xnk−1 in (49), one obtains(61)dTxnk−1,Txmk−1≤φxnk−1−φTxnk−1max1,Sxnk−1,xmk−1,dxnk,xmk≤φxnk−1−φxnk×max1,dxnk−1,xmk−1+dxnk−1,xnk+dxmk−1,xmk.
Lettingk⟶∞, we obtain(62)0<ε≤limk⟶+∞φxnk−1−φxnkmax1,ε=0,which is a contradiction. This shows that xnn≥0 is a Cauchy sequence.Method 2.
We havemax1,Sxn,xn+1≤M′=max1,3M for all n∈ℕ. So, for all n∈ℕ,(63)dxn+1,xn+2=dxn+1,xn+2max1,Sxn,xn+1×max1,Sxn,xn+1≤M′dxn+1,xn+2max1,Sxn,xn+1.
Since∑n≥0dxn+1,xn+2/max1,Sxn,xn+1 is convergent, the series ∑n≥0dxn+1,xn+2 is also convergent. Therefore, dxn,xn+1n is a Cauchy sequence. Hence, the sequence xnn≥0 is convergent. Let x∗=limn⟶∞xn. Taking x=xn and y=x∗ in (49), one obtains, for every n∈ℕ,(64)dxn+1,Tx∗≤φxn−φxn+1max1,Sx∗,xn.
Lettingn⟶+∞, we have dx∗,Tx∗=0.Example 4.
LetE=0,1∪1+2/2,+∞ with the usual distance dx,y=x−y. Let T and φ be the functions defined on E by(65)Tx=1,ifx∈0,1,1x+1,ifx∈1+22,+∞,φx=12,ifx∈0,1,3,ifx∈1+22,+∞,0,ifx=1.
We need only to consider five cases to check the hypothesis of Theorem5.Case 1:x,y∈1+2/2,+∞.(66)dTx,Ty=x−y1+x1+y≤52=φx−φTx≤φx−φTxmax1,Sx,y,dTx,Ty≤φy−φTymax1,Sx,y.Case 2:x,y∈0,1.(67)dTx,Ty=0≤12=φx−φTx≤φx−φTxmax1,Sx,y,dTx,Ty≤φy−φTymax1,Sx,y.Case 3:x∈0,1 and y∈1+2/2,+∞. We have y/1+y≤1/2y−y/1+y because y≥1+2/2. So,(68)dTx,Ty≤φx−φTxdy,Ty≤max1,Sx,y,dTx,Ty≤52=φy−φTy≤φy−φTymax1,Sx,y.Case 4:x=1 and y∈1+2/2,+∞. We have d1,T1=0 and(69)dT1,Ty=y1+y≤52=φy−φTy≤φy−φTymax1,Sx,y.Case 5:x=1 and y∈0,1. We have d1,T1=0 and(70)dT1,Ty=0≤12=φy−φTy≤φy−φTymax1,Sx,y.
This shows that inequality (49) is satisfied and all the hypotheses are verified. In addition, T has the fixed point 1.
We notice that, in the third case,x∈0,1 and y∈1+2/2,+∞, and we have dTx,Ty>φx−φTx.Remark 3.
If we take Example1 of [11], T satisfies condition (49) of our theorem and admits a fixed point. However, T does not satisfy the Banach Contraction Principle nor the Ćirić contraction.
Theorem4 in [11] treats the case, where max1,Sx,y in (49), and is replaced by(71)Nx,y=maxdx,y,dx,Tx,dx,Ty,dTx,y,dTy,y.
By triangular inequality, we obtain(72)Nx,y≤max1,Sx,y,which gives rise to the following corollary.Corollary 4.
LetE,d be a complete metric space and T:E⟶E a map such that there exists a function φ:E⟶0,+∞ satisfying(73)dx,Tx>0⟹dTx,Ty≤φx−φTxNx,y,where(74)Nx,y=maxdx,y,dx,Tx,dx,Ty,dTx,y,dTy,y.Then,T has a fixed point.Corollary 5.
LetE,d be a complete metric space and T:E⟶E a map such that there exists a function φ:E⟶0,+∞ satisfying, for all x,y∈E2,(75)dx,Tx>0⟹dTx,Ty≤φx−φTxdx,yα,for some real α∈0,1. Then, T has a fixed point.Proof.
If assumption (75) holds for some real α∈0,1 then (49) holds, since for every x,y∈E2, one has(76)dx,yα≤max1,dx,y≤max1,Sx,y,which allows to conclude that T admits a fixed point.



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*Source: 1853704-2020-07-01.xml*