# Some Remarks on Fixed Point Theorems for Interpolative Kannan Contraction

Authors: Youssef Errai; El Miloudi Marhrani; Mohamed Aamri
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we use interpolation to obtain fixedpoint and common fixed point results for a new type of Kannan contraction mappings in complete metric andb-metric spaces. Our results extend and improve some results on fixed point theory in the literature. We also give some examples to illustrate the given results.


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## Body

## 1. Introduction and Preliminaries

It is well known that fixed point theory played a central role in various scientific fields. The well-known result in this area is undoubtedly the famous Banach contraction principle (see [1]) which motivated researchers to find other forms of contractions. In this line, we cite the well-known Kannan contraction that does not require continuous mapping.Definition 1 (see [2]).
LetE be a metric space. A self-mapping on E is said to be a Kannan contraction if there exists λ∈0,1/2 such that
(1)dTx,Ty≤λdx,Tx+dy,Ty,for all x,y∈E.Kannan obtained the following theorem.Theorem 2 (see [2]).
IfE,d is a complete metric space, then every Kannan contraction on E has a unique fixed point.In 2018, Karapinar published a new type of contraction obtained from the definition of the Kannan contraction by interpolation as follows.Definition 3 (see [3]).
LetE,d be a metric space. A self-mapping T:E→E is said to be an interpolative Kannan-type contraction if there are two constants λ∈0,1 and α∈0,1 such that
(2)dTx,Ty≤λdx,Txαdy,Ty1−α,for all x,y∈E with x≠Tx and y≠Ty.Karapinar obtained the following result.Theorem 4 (see [3]).
LetE,d be a complete metric space and T:E→E be an interpolative Kannan-type contraction mapping. Then, T has a fixed point.This theorem has been generalized by Noorwali [4] who proposed a common fixed point for two maps, and in 2019, Gabba and Karapinar defined the λ,α,β-interpolative Kannan contraction as follows.Definition 5 (see [5]).
A self-mappingT on a metric space E,d is called λ,α,β-interpolative Kannan contraction if λ∈0,1,α,β∈0,1 with α+β≤1 and such that
(3)dTx,Ty≤λdx,Txαdy,Tyβ,for all x,y∈E with x≠Tx and y≠Ty.And they gave the following theorem.Theorem 6 (see [5]).
LetX,d be a complete metric space and let T:X→X be an λ,α,β-interpolative Kannan contraction. Then, T has a fixed point in X.The interpolative method has been used by several researchers to obtain generalizations of other forms of contractions (see [6, 7]).In this paper, we discuss the results of [4], and we give some generalizations for existence of fixed points for λ,α,β-interpolative Kannan contraction on complete metric spaces and complete b-metric spaces.Let us recall some basic results ofb-metric spaces:Definition 7 (see [8, 9]).
LetX be a nonempty set and s≥1 be a given real number. A function d:X×X→ℝ+ is a b-metric if for all x,y,z∈X, the following conditions are satisfied:
(b1) dx,y=0 if and only if x=y
(b2) dx,y=dy,x
(b3) dx,z≤sdx,y+dy,z
TheX,d,s is called a b-metric space.Definition 8 (see [10]).
LetX,d be a b-metric space
(a)
xn is b-convergent in X,d if there exists x∈X such that dxn,x→0 as n→∞. In this case, we write limn→∞xn=x(b)
xn is a b-Cauchy sequence in X,d if dxn,xm→0 as n,m→∞(c)
X,d is b-complete if every b-Cauchy sequence in Xb-convergesDefinition 9 (see [11]).
LetX,d be a b-metric space and g,h:X→X. An element x in X is called a coincidence point of g and hifgx=hxLemma 10 (see [12]).
Letxn be a sequence in a b-metric space X,d,s with s≥1 such that
(4)dxn,xn+1≤λdxn−1,xn,for some λ∈0,1, and each n∈ℕ. Then, xn is a b-Cauchy sequence in X,d,s.Lemma 11 (see Lemma 2.1 in [11]).
LetX,d be a b-metric space with s≥1 and assume that xn and yn are b-convergent to x,y, respectively, then we have
(5)1s2dx,y≤liminfn→∞dxn,yn≤limsupn→∞dxn,yn≤s2dx,y.
In particular, ifx=y, then we have limn→∞dxn,yn=0. Moreover, for each z∈X, we have
(6)1sdx,z≤liminfn→∞dxn,z≤limsupn→∞dxn,z≤sdx,z.


## 2. Discussion and Results

Using theλ,α,β-interpolative Kannan contraction defined as above, we give our first main result.Theorem 12.
LetX,d be a complete metric space and T is self-mapping on X such that
(7)dTx,Ty≤λdx,Txαdy,Tyβ,for all x,y∈X with x≠Tx and y≠Ty, and where λ∈0,1 and α,β∈0,1 such that α+β≥1. If there exists x∈X such that dx,Tx≤1, then T has a fixed point in X.Proof.
For the caseα+β=1, see Karapinar et al. [3]. Assume α+β>1 and define a sequence xn by x0=x and xn+1=Txn, for all integer n and assume that xn≠Txn, for all n.
We have(8)dxn,xn+1≤λdxn,xn−1αdxn,xn+1β,foralln≥1.
Sincedx0,x1≤1, we obtain by (8) dx1,x2≤λ.
Assume that there exists a realγn such that dxn,xn+1≤λγn; we obtain
(9)dxn+1,xn+2≤λdxn,xn+1αdxn+1,xn+2β,which gives
(10)dxn+1,xn+21−β≤λdxn,xn+1α≤λλαγn.
It follows that(11)dxn+1,xn+2≤λγn+1,where γn+1=1/1−β1+αγn, for all n≥1 with γ0=0 and γ1=1.
Sinceα/1−β>1, we have limnγn=+∞. It follows that
(12)∑n=0+∞dxn,xn+1≤∑n=0+∞λγn,which is convergent, and consequently, xn is a Cauchy sequence in X,d. Thus, xn converges to some x¯∈X. Assume that x¯≠Tx¯, we obtain, by (1):
(13)dxn+1,Tx¯≤λdxn,Txnαdx¯,Tx¯β,foralln.
Ifn→+∞, we obtain dx¯,Tx¯=0, which is a contradiction. Then, Tx¯=x¯.Example 13.
LetX=x,y,z,w be set endowed with the metric d defined by
(14)dx,x=dy,y=dz,z=dw,w=0,dx,y=3,dx,z=4,dx,w=52,dy,z=32,dy,w=2,dz,w=32,
and define the self-mappingT on X by
(15)Tx=x,Ty=w,Tz=x,Tw=y.
Forλ=99/100,α=5/8, and β=1/2, we have
(16)dTu,Tv≤λdu,Tuαdv,Tvβ,for all u,v∈X−x. Moreover, T has a fixed point in X.In [4], Noorwali gave the following result.Theorem 14.
LetM,d be a complete metric space and S,T:M→M be self-mappings. Assume that there are two constants λ∈0,1,α∈0,1 such that the condition
(17)dTp,Sq≤λdp,Tpαdq,Sq1−α,is satisfied for all p,q∈M such that Tp≠p and Sq≠q. Then, S and T have a unique common fixed point.The uniqueness is not true in the caseS=T; moreover, in the proof, the author considers the sequence defined by p0∈M,p2n+1=Tp2n, and p2n+2=Sp2n+1. In the case where there is no three consecutive identical terms in the sequence pn, the author uses the inequality
(18)dp2n+1,p2n+2≤λdp2n,p2n+1αdp2n+1,p2n+21−α.We note that the above inequality is not valuable ifp2n=p2n+1 or p2n+1=p2n+2. Moreover, the following example shows that the theorem is not true in this form.Example 15.
IfS is the identity map on M, it is clear that the result is not valid for any mapping T without a fixed point.Example 16.
LetM=p,q,r,s be endowed with the metric defined by the following Table 1 of values.Table 1
Table of metricd.
pqrsp03452q30322r432032s522320And we define two self-mappings onM by the following matrix of values:
(19)S:pqrsqpprandT:pqrspsrsForλ=99/100 and α=3/5, we obtain
(20)dSx,Ty≤λdx,Sxαdy,Ty1−α,for all x,y∈M with Sx≠x and Ty≠y, but S and T have no common fixed point in M.As an alternative of this theorem, we give the following result.Theorem 17.
LetM,d be a complete metric space, S,T:M→M be two self-mappings on M. Assume that there are some λ∈0,1,α∈0,1 such that the followings conditions hold:
(i)
dSp,Tq≤λdp,Spαdq,Tq1−α, for all p,q∈M with Sp≠p and Tq≠q(ii)
dSp,Sq≤λdp,Spαdq,Sq1−α, for all p,q∈M with Sp≠p and Sq≠q(iii)
dTp,Tq≤λdp,Tpαdq,Tq1−α, for all p,q∈M with Tp≠p and Tq≠q
Then,S and T have a common fixed point.Proof.
Letx∈M; we define by induction a sequence xn by x0=x,x2n+1=Sx2n, and x2n+2=Tx2n+1, for all n.
We shall discuss the following cases:(a)
dxn,xn+1>0, for all n. Then, by the same arguments as in the proof of Theorem 2.1 in [4], we can prove that xnn is a Cauchy sequence in M which converges to a common fixed point of S and T. Note that this common fixed point is not necessarily unique(b)
There existsn0 such that xn0=xn0+1=xn0+2. Then, xn0 is a common fixed point of S and T(c)
Assume thatxn is without three consecutive identical terms, but it contains a subsequence xnk such that xnk=xnk+1, for all k∈ℕ. We have the following situations
(c1) Letp∈ℕ such that xp=xp+1; if p=2n for some integer n, we have
(21)Tx2n−1=x2n≠x2n−1andTx2n+1=x2n+2≠x2n+1.
It follows that(22)dx2n+1,x2n+2=dTx2n−1,Tx2n+1≤λdx2n−1,x2nαdx2n+1,x2n+21−α,and then
(23)dx2n+1,x2n+2≤λ1/αdx2n−1,x2n≤λdx2n−1,x2n.
And ifp=2n−1 for some integer n, we have
(24)Sx2n−2=x2n−1≠x2n−2,Sx2n=x2n+1≠x2n.
By (ii), we obtain(25)dx2n,x2n+1=dx2n−1,x2n+1=dSx2n−2,Sx2n≤λdx2n−2,x2n−1αdx2n,x2n+11−α.
It follows that(26)dx2n,x2n+1≤λ1/αdx2n−2,x2n−1.
From (23) and (26) and since λ1/α≤λ, we obtain
(27)xp=xp+1⇒dxp+1,xp+2≤λdxp−1,xp.
(c2) Assume thatxp−1≠xp and xp≠xp+1; if p=2n, for some n∈ℕ, we obtain
(28)Txp−1≠xp−1,Sxp≠xp.i⇒dxp,xp+1=dTxp−1,Sxp≤λdxp,xp+1αdxp−1,xp1−α,which leads to
(29)dxp,xp+1≤λ1/1−αdxp−1,xp≤λdxp−1,xp.
And ifp=2n+1, for some n∈ℕ, we obtain
(30)Txp=xp+1≠xp,Sxp−1≠xp−1.
By (i), we obtain(31)dxp,xp+1=dSxp−1,Txp≤λdxp−1,xpαdxp,xp+11−α,which leads to
(32)dxp,xp+1≤λdxp−1,xp.
The inequalities (29) and (32) imply that
(33)dxp,xp+1≤λdxp−1,xp,where p is such that xp−1≠xp and xp≠xp+1.
In view of (27) and (33), we obtain
(34)dxp,xp+1≤λmaxdxp−2,xp−1,dxp−1,xp,for all p∈ℕ.
Letφ:ℕ→ℕ defined by φp=p if xp≠xp+1 and φp=p+1 if xp=xp+1. It is clear that φ is monotone nondecreasing and the sequences xn and xφn have the same set of values.
Moreover,xφp≠xφp+1, for all p∈ℕ∗, and we have
(35)dxφp,xφp+1≤λdxφp−1,xφp,for all p∈ℕ. It follows that xφn is a Cauchy sequence, and then xn is also a Cauchy sequence. And since M is complete, there exists u∈M such that limndxn,u=0.
(c3) Assume thatSu≠u; we have the following cases:
First case: there existsn0 such that x2n=x2n+1, for all n≥n0. Thus, Tx2n−1≠x2n−1, for all n≥n0. And we obtain
(36)dSu,Tx2n−1≤λdu,Suαdx2n−1,x2n1−α.
It follows thatlimndSu,x2n=dSu,u=0, which is a contradiction.
Second case: there exists a subsequencex2nkk with x2nk≠x2nk+1, for all k; we have Sx2nk≠x2nk, for all k. Thus,
(37)dSu,Sx2nk≤λdu,Suαdx2nk,x2nk+11−α.
It follows thatlimkdSu,x2nk+1=dSu,u=0, which is a contradiction. Then, Su=u.
By identical arguments, we can prove thatTu=u, which ends the proof.Remark 18.
IfS=T, the conditions (i), (ii), and (iii) are identical, and we obtain Theorem 2.2 of [3].Example 19.
LetX=0,+∞; we define on X a metric d and two self-mappings S and T by
(38)dx,y=x+y,ifx≠y,0,ifx=y,Sx=1,ifx∈0,2,e−x,ifx∈2,+∞,Tx=1,ifx∈0,2,1x,ifx∈2,+∞.
Forλ=24/25 and α=1/2, we have
(i)(39)dSx,Ty≤λdx,Sxαdy,Ty1−α.
First case:x,y∈0,2; the inequality (39) is obvious.
Second case:x∈0,2 and y∈2,+∞ with x≠1; we have
(40)dSx,Ty=1+1y≤32,λdx,Sxαdy,Ty1−α=2425x+11/2y+1y1/2≥2425521/2≥32.
Third case:x∈2,+∞ and y∈0,2 with y≠1; we have
(41)dSx,Ty=e−x+1≤e−2+1,λdx,Sxαdy,Ty1−α=2425x+e−x1/2y+11/2≥242521/2≥e−2+1.
Fourth case:x,y∈2,+∞; we have
(42)dSx,Ty=e−x+1y≤e−2+12,λdx,Sxαdy,Ty1−α=2425x+e−x1/2y+1y1/2≥242551/2≥e−2+12.
Then, equation (39) is satisfied.
(ii) To prove(43)dSx,Sy≤λdx,Sxαdy,Sy1−α,we consider the following cases:
First case:x,y∈0,2; the inequality (43) is obvious.
Second case:x∈0,2 and y∈2,+∞ with x≠1; we have
(44)dSx,Sy=1+e−y≤1+e−2,λdx,Sxαdy,Sy1−α=2425x+11/2y+e−y1/2≥242521/2≥1+e−2.
Third case:x,y∈2,+∞; we have dSx,Sy=e−x+e−y≤2e−2 and
(45)λdx,Txαdy,Ty1−α=2425x+e−x1/2y+e−y1/2≥24252≥2e−2.
(iii) By the same arguments, we prove(46)dTx,Ty≤λdx,Txαdy,Ty1−α.
First case:x,y∈0,2; the inequality (46) is obvious.
Second case:x∈0,2 and y∈2,+∞ with x≠1; we have
(47)dTx,Ty=1+1y≤32,λdx,Txαdy,Ty1−α=2425x+11/2y+1y1/2≥2425521/2≥32.
Third case:x,y∈2,+∞; we have dTx,Ty=1/x+1/y≤1 and
(48)λdx,Txαdy,Ty1−α=2425x+1x1/2y+1y1/2≥242552≥1.
The inequalities (39), (43), and (46) show that all conditions of Theorem 17 are satisfied and 1 is a common fixed point of S and T.Theorem 20.
LetX,d,s be a b-complete b-metric space, T a self-mapping on X, and λ:0,+∞→0,1 such that for all z∈0,+∞, limsupx→z+λx<1. Assume that there exists α∈0,1 such that
(49)dTx,Ty≤λdx,ydx,Txαdy,Ty1−α,for all x,y∈X with Tx≠x and Ty≠y. Then, T has a fixed point in X.Proof.
Letx∈X; we define a sequence xn by x0=x and xn+1=Txn, for all n. And assume that xn≠xn+1; otherwise, T has a fixed point in X. We obtain
(50)dxn+1,xn≤λdxn,xn−1dxn,xn+1αdxn−1,xn1−α,for all integer n. The sequence dxn+1,xn is nonincreasing in 0,+∞, then it converges to some positive real l∈0,1. It follows that there exists K∈0,1 such that dxn+1,xn≤K, for large integers. Thus, by Lemma 10, xnn is a b-Cauchy sequence in X. Let u∈X such that limndxn,u=0 and assume that Tu≠u; we have
(51)dxn+1,Tu≤λdxn,udxn,xn+1αdu,Tu1−α≤dxn,xn+1αdu,Tu1−α,for all n. Then, by Lemma 11, we have
(52)1sdu,Tu≤limsupndxn,xn+1αdu,Tu1−α=0,which is a contradiction, and then u is a fixed point of T.Definition 21.
LetX,d,s be a b-metric space and T,g:X→X be self-mappings on X. We say that the mapping T is an g-interpolative Kannan-type contraction if there exist two constants λ∈0,1 and α∈0,1 such that
(53)dTx,Ty≤λdgx,Txαdgy,Ty1−α,for all x,y∈X with Tx≠gx and Ty≠gy.Theorem 22.
LetX,d,s be ab-complete ab-metric space and T a g-interpolative Kannan-type contraction. Assume that TX⊆gX and gX is closed. If there exist λ,α∈0,1 such that (53) holds. Then, T and g have a coincidence point in X.Proof.
Letx∈X; since TX⊆gX, we can define inductively a sequence xn such that x0=x, and gxn+1=Txn, for all integer n.
If there existsn∈0,1,2,⋯ such that gxn=Txn, then xn is a coincidence point of g and T. Assume that gxn≠Txn, for all n. By (53), we obtain
(54)dTxn+1,Txn≤λdgxn+1,Txn+1αdgxn,Txn1−α=λdTxn,Txn+1αdTxn−1,Txn1−α.
Thus,(55)dTxn+1,Txn≤λ1/1−αdTxn,Txn−1≤λdTxn,Txn−1.
Lemma10 implies that Txn is a b-Cauchy sequence, and consequently, gxn is also a b-Cauchy sequence. Let z∈X such that
(56)limndTxn,z=limndgxn+1,z=0.
And sincez∈gX, there exists u∈X such that z=gu. We claim that u is a coincidence point of g and T. For this, if we assume that gu≠Tu, we obtain
(57)dTxn,Tu≤λdgxn,Txnαdgu,Tu1−α.
By Lemma11, we obtain
(58)1sdz,Tu≤λlimsupndgxn,Txnαdgu,Tu1−α=0,which is a contradiction. Then, z=gu=Tu.Example 23.
LetX=0,+∞ and d the b-metric defined on X by
(59)dx,y=x+y2,ifx≠y,0,ifx=y.
X,d is a complete b-metric space.
Define two self-mappingsT and g on X by gx=x2, for all x∈X and
(60)Tx=1,ifx∈0,2,1x,ifx∈2,+∞.
T is a g-interpolative Kannan contraction for λ=3/4 and α=2/5. For this, we discuss the following cases:
(i)
Ifx,y∈0,2, then dTx,Ty=0≤λdgx,Txαdgy,Ty1−α(ii)
Ifx,y∈2,+∞ with x≠y, we have(61)dTx,Ty=1x+1y2≤1,λdgx,Txαdgy,Ty1−α=λx2+1x2αy2+1y21−α≥34922≥dTx,Ty.(iii)
Ifx∈0,2 and y∈2,+∞ with x≠1, we have(62)dTx,Ty=d1,1y=1+1y2≤322,λdgx,Txαdgy,Ty1−α=λx2+12αy2+1y21−α≥34926/5≥dTx,Ty.(iv)
Ifx∈2,+∞ and y∈0,2 with y≠1, we obtain(63)dTx,Ty=d1x,1=1+1x2≤322,λdgx,Txαdgy,Ty1−α=λx2+1x2αy2+121−α≥34924/5≥dTx,Ty.
We conclude thatdTx,Ty≤λdgx,Txαdgy,Ty1−α, for all x,y∈X such that Tx≠x and Ty≠y. And we remark that 1 is a coincidence point of g and T.Remark 24.
In Theorem22, S and g need not have a common fixed point.Example 25.
LetX=0,1,2,3 be endowed with the metric du,v=∣u−v∣. We define S and g by their matrix of values as follows:
(64)S:01231110,g:01231230.
Forλ=1/2 and α=1/2, we have
(65)dSu,Sv≤λdgu,Suαdgv,Sv1−α.
0 and 3 are coincidence points ofg and S, but g and S have no common fixed point.Remark 26.
If we defineX,d,T, and α as in Example 23 and λt=t/1+t, we obtain
(66)dTx,Ty≤λdx,ydx,Txαdy,Ty1−α,for all x,y∈X such that Tx≠x and Ty≠y. And T has a fixed point.



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