Some fixed point theorems for (φ,ψ,p)-contractive maps and (φ,k,p)-contractive maps on a complete metric space are proved. Presented fixed point theorems generalize many results existing in the literature.

1. Introduction and Preliminaries

Branciari [1] established a fixed point result for an integral type inequality, which is a generalization of Banach contraction principle. Kada et al. [2] introduced and studied the concept of w-distance on a metric space. They give examples of w-distances and improved Caristi’s fixed point theorem, Ekeland’s ϵ-variational’s principle, and the nonconvex minimization theorem according to Takahashi (see many useful examples and results on w-distance in [2–5] and in references therein). Kada et al. [2] defined the concept of w-distance in a metric space as follows.

Definition 1.1 (see [<xref ref-type="bibr" rid="B2">2</xref>]).

Let X be a metric space endowed with a metric d. A function p:X×X→[0,∞) is called a w-distance on X if it satisfies the following properties:

p(x,z)≤p(x,y)+p(y,z) for any x,y,z∈X,

p is lower semicontinuous in its second variable, that is, if x∈X and yn→y in X then p(x,y)≤liminfn→∞p(x,yn),

for each ϵ>0, there exists δ>0 such that p(z,x)≤δ and p(z,y)≤δ imply d(x,y)≤ϵ.

We denote by Φ the set of functions φ:[0,+∞)→[0,+∞) satisfying the following hypotheses:

φ is continuous and nondecreasing,

φ(t)=0 if and only if t=0.

We denote by Ψ the set of functions ψ:[0,+∞)→[0,+∞) satisfying the following hypotheses:

ψ is right continuous and nondecreasing,

ψ(t)<t for all t>0.

Let p be a w-distance on metric space (X,d), φ∈Φ and ψ∈Ψ. A map T from X into itself is a (φ,ψ,p)-contractive map on X if for each x,y∈X, φp(Tx,Ty)≤ψφp(x,y).

The following lemmas are used in the next section.

Lemma 1.2 (see [<xref ref-type="bibr" rid="B4">3</xref>]).

If ψ∈Ψ, then limn→∞ψn(t)=0 for each t>0, and if φ∈Φ,{an}⊆[0,∞) and limn→∞φ(an)=0, then limn→∞an=0.

Lemma 1.3 (see [<xref ref-type="bibr" rid="B2">2</xref>]).

Let (X,d) be a metric space and let p be a w-distance on X.

If {xn} is a sequence in X such that limnp(xn,x)=limnp(xn,y)=0, then x=y. In particular, if p(z,x)=p(z,y)=0, then x=y.

If p(xn,yn)≤αnp(xn,y)≤βn for any n∈ℕ, where {αn} and {βn} are sequences in [0,∞) converging to 0, then {yn} converges to y.

Let p be a w-distance on metric space (X,d) and {xn} a sequence in X such that for each ɛ>0 there exist Nɛ∈N such that m>n>Nɛ implies p(xn,xm)<ɛ (or limm,n→∞p(xn,xm)=0), then {xn} is a Cauchy sequence.

Note that if p(a,b)=p(b,a)=0 and p(a,a)≤p(a,b)+p(b,a)=0, then p(a,a)=0 and, by Lemma 1.3, a=b.

In [3], Razani et al. proved a fixed point theorem for (φ,ψ,p)-contractive mappings, which is a new version of the main theorem in [1], by considering the concept of the w-distance.

The main aim of this paper is to present some generalization fixed point Theorems by Kada et al. [2], Hicks and Rhoades [6] and several other results with respect to (φ,ψ,p)-contractive maps on a complete metric space.

In the next theorem we state one of the main results of this paper generalizing Theorem 4 of [2]. In what follows, we use φp to denote the composition of φ with p.

Theorem 2.1.

Let p be a w-distance on complete metric space (X,d),φ∈Φ and ψ∈Ψ. Suppose T:X→X is a map that satisfies
φp(Tx,T2x)≤ψ(φp(x,Tx)),
for each x∈X and that
inf{p(x,y)+p(x,Tx):x∈X}>0
for every y∈X with y≠Ty. Then there exists u∈X such that u=Tu. Moreover, if v=Tv, then p(v,v)=0.

Proof.

Fix x∈X. Set xn+1=Txn with x0=x. Then by (2.1)
φp(xn,xn+1)≤ψφp(xn-1,xn)≤ψ2φp(xn-2,xn-1)≤⋅⋅⋅≤ψn(φp(x0,x1)),
thus limnφp(xn,xn+1)=0 and Lemma 1.2 implies
limn→∞p(xn,xn+1)=0,
and similarly
limn→∞p(xn+1,xn)=0.

Now we proof that {xn} is a Cauchy sequence. By triangle inequality, continuity of φ and (2.4), we have
φp(xn,xn+2)≤ψφ[p(xn,xn+1)+p(xn+1,xn+2)]⟶0,
as n→∞ and so limn→∞φp(xn,xn+2)=0 which concludes
limn→∞p(xn,xn+2)=0.

By induction, for any k>0 we have
limn→∞p(xn,xn+k)=0.

So, by Lemma 1.3, {xn} is a Cauchy sequence, and since X is complete, there exists u∈X such that xn→u inX.

Now we prove that u is a fixed point of T.

From (2.8), for each ɛ>0, there exists Nɛ∈ℕ such that n>Nɛ implies p(xNɛ,xn)<ɛ but xn→u and p(x,·) is lower semicontinuous, thus
p(xNɛ,u)≤limn→∞infp(xNɛ,xn)≤ɛ.
Therefore, p(xNɛ,u)≤ɛ. Set ɛ=1/k,Nɛ=nk and we have
limk→∞p(xnk,u)=0.

Now, assume that u≠Tu. Then by hypothesis, we have0<inf{p(x,u)+p(x,Tx):x∈X}≤inf{p(xn,u)+p(xn,xn+1):n∈N}⟶0as n→∞ by (2.4) and (2.10). This is a contradiction. Hence u=Tu.

If v=Tv, we haveφp(v,v)=φp(Tv,T2v)≤ψφp(v,Tv)=ψφp(v,v)<φp(v,v).

This is a contradiction. So φp(v,v)=0, and by hypothesis p(v,v)=0.

Here we give a simple example illustrating Theorem 2.1. In this example, we will show that Theorem 4 in [2] cannot be applied.

Example 2.2.

Let X={(1/n)∣n∈ℕ}∪{0}, which is a complete metric space with usual metric d of reals. Moreover, by defining p(x,y)=y, p is a w-distance on (X,d). Let T:X→X be a map as T(1/n)=1/(n+1), T0=0. Suppose φ(t)=t1/t is a continuous and strictly nondecreasing map and ψ(t)=(1/3)t, for any t>0. We have
supx∈Xp(Tx,T2x)p(x,Tx)=1,
and so there is not any r∈[0,1) such that p(Tx,T2x)≤rp(x,Tx), and hence Theorem 4 in [2] dose not work. But
φp(Tx,T2x)=p(Tx,T2x)1/p(Tx,T2x)=(1n+2)n+2≤13(1n+1)n+1=13p(x,Tx)1/p(x,Tx)=ψφp(x,Tx),
because for any n∈ℕ we have ((n+1)/(n+2))n+11/(n+2)≤1/3. Also for any n∈ℕ we have 1/n≠T(1/n). So for arbitrary n∈ℕ, inf{p(1/m,1/n)+p(1/m,1/(m+1)):m∈ℕ}=1/n>0, hence T is satisfied in Theorem 2.1. We note that 0 is a fixed point for T.

The next examples show the role of the conditions (2.1) and (2.2).

Example 2.3.

Let X=[-1,1], d(x,y)=|x-y|, and define p:X→X by p(x,y)=|3x-3y|, where x,y∈X. Set ψ(t)=rt and φ(t)=t for all t∈[0,∞). Let us define T:X→X by T0=1 and Tx=x/10 if x≠0. We have
φp(T0,T20)=p(T0,T20)=p(1,110)=3-310≤3=13p(0,T0)=ψφp(0,T0).

If x≠0, then
φp(Tx,T2x)=p(Tx,T2x)=p(x10,x100)=110|3x-3x10|≤13p(x,Tx)=ψφp(x,Tx)
and hence (2.1) holds.

Now, we remark that 0≠T(0), and
infn∈Np(Tn(x),0)+p(Tn(x),TTn(x))=0foreveryx∈X.
Thus, the condition (2.2) is not satisfied, and there is no z∈X with Tz=z. In this case we observe that Theorem 2.1 is invalid without condition (2.2).

Example 2.4.

Let X=[2,∞)∪{0,1}, d(x,y)=|x-y|, x,y∈X, and set p=d. Let ψ,φ be as Example 2.3. Let us define T:X→X by T0=1 and Tx=0 if x≠0. Clearly, T has no fixed point in X. Now, for each x∈X and that
inf{d(x,y)+d(x,Tx):x∈X}>0
for every y∈X with y≠Ty, so condition (2.2) is satisfied. But, for x=0, d(Tx,T2x)>rd(x,Tx) for any r∈[0,1). Hence, condition (2.1) dose not hold. We note that Theorem 2.1 dose not work without condition (2.1).

Suppose θ:ℝ+→ℝ+ is Lebesgue-integrable mapping which is summable and ∫0ɛθ(η)dη>0, for each ɛ>0. Now, in the next corollary, set φ(t)=∫0tθ(η)dη and ψ(t)=ct, where c∈[0,1[. Then, φ∈Φ and ψ∈Ψ. Hence we can conclude the following corollary as a special case.

Corollary 2.5.

Let T be a selfmap of a complete metric space (X,d) satisfying
∫0d(Tx,T2x)θ(t)dt≤c∫0d(x,Tx)θ(t)dt
for all x∈X. Suppose that
inf{d(x,y)+d(x,Tx):x∈X}>0foreveryy∈X
with y≠Ty. Then there exists a u∈X such that Tu=u.

Note that Corollary 2.5 is invalid without condition (2.20). For example, take X={0}∪{1/2n:n≥1}, which is a complete metric space with usual metric d of reals. Define T:X→X by T(0)=1/2 and T(1/2^n)=1/2n-1 for n≥1. Set φ(t)≡1. It is easy to check that ∫0d(Tx,T2x)φ(t)dt≤(1/2)∫0d(x,Tx)φ(t)dt, for any x∈X; however, y≠Ty for any y∈X and inf{d(x,y)+d(x,Tx):x∈X}=0. Clearly, T has got no fixed point in X.

Remark 2.6.

From Theorem 2.1, we can obtain Theorem 4 in [2] as a special case. For this, in the hypotheses of Theorem 2.1, set ψ(t)=rt and φ(t)=t for all t∈[0,∞).

Corollary 2.7.

Let p be a w-distance on complete metric space (X,d), φ∈Φ and ψ∈Ψ. Suppose T is a continuous mapping for X into itself such that (2.1), is satisfied. Then there exists u∈X such that u=Tu. Moreover, if v=Tv, then p(v,v)=0.

Proof.

Assume that there exists y∈X with y≠Ty and inf{p(x,y)+p(x,Tx):x∈X}=0. Then there exists a sequence {xn} such that
p(xn,y)+p(xn,Txn)⟶0
as n→∞. Hence p(xn,y)→0 and p(xn,Txn)→0 as n→∞. Lemma 1.3 implies that Txn→y as n→∞. Now by assumption
φp(Txn,T2xn)≤ψ(φp(xn,Txn))
and so φp(Txn,T2xn)→0 as n→∞. By Lemma 1.2, p(Txn,T2xn)→0 as n→∞. We also have
p(xn,T2xn)≤p(xn,Txn)+p(Txn,T2xn),
hence p(xn,T2xn)→0 as n→∞. By Lemma 1.3, we conclude that {T2xn} converges to y. Since T is continuous, we have
Ty=T(limn→∞Txn)=limn→∞T2xn=y.
This is a contradiction. Therefore, if y≠Ty, then inf{p(x,y)+p(x,Tx):x∈X}>0. So, Theorem 2.1 gives desired result.

In Example 2.3,T is satisfied in condition (2.1), but it is not continuous. So, the hypotheses in Corollary 2.7are not satisfied. We note that T has no fixed point.

It is an obvious fact that, if f:X→X is a map which has a fixed point x∈X, then x is also a fixed point of fn for every natural number n. However, the converse is false. If a map satisfies F(f)=F(fn) for each n∈ℕ, where F(f) denotes a set of all fixed points of f, then it is said to have property P [7, 8]. The following theorem extends and improves Theorem 2 of [7].

Theorem 2.8.

Let (X,d) be a complete metric space with w-distance p on X. Suppose T:X→X satisfies

φp(Tx,T2x)≤ψφp(x,Tx),∀x∈X,
or

with strict inequality, ψ≡1 and for all x∈X, x≠Tx. If F(T)≠∅, then T has property P.

Proof.

We shall always assume that n>1, since the statement for n=1 is trivial. Let u∈F(Tn). Suppose that T satisfies (i). Then,
φp(u,Tu)=φp(Tnu,TTnu)≤ψφp(Tn-1u,TTn-1u)≤⋯≤ψnφp(u,Tu),
and so p(u,Tu)=0. Now from
φp(u,u)=ψφp(u,Tnu)≤∑i=0n-1ψφp(Tiu,Ti+1u)=0,
we have p(u,u)=0. Hence, by Lemma 1.3, we have u=Tu, and u∈F(T). Suppose that T satisfies (ii). If Tu=u, then there is nothing to prove. Suppose, if possible, that Tu≠u. Then a repetition of the argument for case (i) leads to φp(u,Tu)<ψφp(u,Tu), that is a contradiction. Therefore, in all cases, u=Tu and F(Tn)=F(T).

The following theorem extends Theorem 2.1 of [6]. A function G mapping X into the real is T-orbitally lower semicontinuous at z if {xn} is a sequence in O(x,∞) and xn→z implies that G(p)≤liminfn→∞G(xn).

Theorem 2.9.

Let (X,d) be a complete metric space with w-distance p on X. Suppose T:X→X and there exists an x such that
φp(Ty,T2y)≤ψφp(y,Ty),∀y∈O(x,∞).
Then,

limTnx=z exists,

φp(Tnx,z)≤ψn1-ψφp(x,Tx)forn≥1,

p(z,Tz)=0 if and only if G(x)=p(x,Tx) is T-orbitally lower semicontinuous at z.

Proof.

Observe that (i) and (ii) are immediate from the proof of Theorem 2.1. We prove (iii). It is clear that p(z,Tz)=0 impling G(x) is T-orbitally lower semicontinuous at z.

xn=Tnx→z and G is T-orbitally lower semicontinuous at x implies
0≤φp(z,Tz)=φG(z)≤liminfn→∞φG(xn)=liminfn→∞ψφp(xn,Txn)≤liminfn→∞ψnφp(x,Tx)=0.
So, p(z,Tz)=0.

The mapping T is orbitally lower semicontinuous at u∈X if limk→∞Tnkx=u implies that limk→∞Tnk+1x=Tu. In the following, we improve Theorem 2 of [9] that it is correct form Theorem 1 of [7].

Theorem 2.10.

Let p be a w-distance on complete metric space (X,d),φ∈Φ and ψ∈Ψ. Suppose T:X→X is orbitally lower semicontinuous map on X that satisfies
φp(Tx,T2x)≤ψ(φp(x,Tx))
for each x∈X. Then there exists u∈X such that u∈F(T). Moreover, if v=Tv, then p(v,v)=0.

Proof.

Observe that the sequence {xn} is a Cauchy sequence immediate from the proof of Theorem 2.1 and so there exists a point u in X such that xn→u as n→∞. Since T is orbitally lower semicontinuous at u, we have p(u,Tu)≤liminfn→∞p(xn,xn+1)=0. Now, we have
φp(u,Tu)≤φliminfn→∞p(xn,xn+1)=φ(0)=0,
and so p(u,Tu)=0. Similarly, p(Tu,u)=0. Hence, u∈F(T). By Theorem 2.1 we can conclude that if v=Tv, then p(v,v)=0.

The following example shows that Theorem 2 in [9] cannot be applicable. So our generalization is useful.

Example 2.11.

Let =[0,∞) be a metric space with metric d defined by d(x,y)=(40/3)|x-y|,x,y∈X, which is complete. We define p:X→X by p(x,y)=(1/3)|y|. Let φ be as defined before in Corollary 2.5 and ψ(t)=(1/10)t,t>0. Assume that T:X→X by Tx=x/10 for any x∈X. We have, d(Tx,T2x)=(4/3)d(x,Tx),x∈X, and so Theorem 2 in [9] dose not work. But
φp(Tx,T2x)≤ψ(φp(x,Tx))
for each x∈X. Hence by Theorem 2.10 there exists a fixed point for T. We note that 0 is fixed point for T.

In this section we obtain fixed points for (φ,k,p)-contractive maps (i.e., (φ,ψ,p)-contractive maps that ψ(t)=k for all t∈[0,∞), where k∈[0,1)).

In 1969, Kannan [10] proved the following fixed point theorem. Contractions are always continuous and Kannan maps are not necessarily continuous.

Theorem 3.1 (see [<xref ref-type="bibr" rid="B9">10</xref>]).

Let (X,d) be a complete metric space. Let T be a Kannan mapping on X, that is, there exists k∈[0,1/2) such that
d(Tx,Ty)≤k(d(x,Tx)+d(y,Ty))
for all x,y∈X. Then, T has a unique fixed point in X. For each x∈X, the iterative sequence {Tnx}n≥1 converges to the fixed point.

In the next theorem, we generalize this theorem as follows.

Theorem 3.2.

Let (X,d) be a complete metric space. Let T be a (φ,k)-Kannan mapping on X, that is, there exists k∈[0,1/2) such that
φd(Tx,Ty)≤k(φd(x,Tx)+φd(y,Ty))
for all x,y∈X. Then, T has a unique fixed point in X. For each x∈X, the iterative sequence {Tnx}n≥1 converges to the fixed point.

Proof.

Let x∈X and define xn+1=Tnx for any n∈N, and set r=k/(1-k). Then, r∈[0,1),
φd(Tx,T2x)≤k(φd(x,Tx)+φd(Tx,T2x))
and so
φd(Tx,T2x)≤rφd(x,Tx).

Then, from the proof of Theorem 2.1, limTnx=z exists. From (3.4), we have
φd(Tnx,Tz)≤rφd(Tn-1x,z)≤rn1-rφd(x,Tx)forn≥1.
Thus, limTnx=Tz, and so z=Tz. Clearly, z is unique. This completes the proof.

The set of all subadditive functions φ in Φ is denoted by Φ′. In the following theorems, we generalize Theorems 3.4 and 3.5 due to Suzuki and Takahashi [4].

Theorem 3.3.

Let p be a w-distance on complete metric space (X,d),φ∈Φ′ and T be a selfmap. Suppose there exists k∈[0,1/2) such that

φp(Tx,T2x)≤kφp(x,T2x) for each x∈X,

inf{p(x,z)+p(x,Tx):x∈X}>0 for every z∈X with z≠Tz.

Then T has a fixed point in X. Moreover, if v is a fixed point of T, then p(v,v)=0.Proof.

Fix x∈X. Define x0=x and xn=Tnx0 for every n∈ℕ. Put r=k/(1-k). Then, 0≤r<1. By hypothesis, since φ∈Φ′, we have
φp(xn,xn+1)≤kφp(xn-1,xn+1)≤kφp(xn-1,xn)+kφp(xn,xn+1),
for all n∈ℕ. It follows that
φp(xn,xn+1)≤rφp(xn-1,xn)≤⋯≤rnφp(x0,x1),
for all n∈ℕ. Using the similar argument as in the proof of Theorem 2.1, we can prove that the sequence {un} is Cauchy and so there exists u∈X such that xn→u as n→∞. Also, we have u∈F(T). Since
φp(v,v)=φp(Tv,T2v)≤kφp(v,T2v)=kφp(v,v),
we have φp(v,v)=0 and so p(v,v)=0. The proof is completed.

Corollary 3.4.

Let p be a w-distance on complete metric space (X,d),φ∈Φ′ and let T be a continuous map. Suppose there exists k∈[0,1/2) such that
φp(Tx,T2x)≤kφp(x,T2x),
for each x∈X.

Then T has a fixed point in X. Moreover, if v is a fixed point of T, then p(v,v)=0.

Proof.

It suffices to show that inf{p(x,z)+p(x,Tx):x∈X}>0 for every u∈X with u≠Tu. Assume that there exists u∈X with u≠Tu and inf{p(x,u)+p(x,Tx):x∈X}=0. Then there exists a sequence {xn} in X such that limn→∞[p(xn,u)+p(xn,Txn)]=0. It follows that p(xn,u)→0 and p(xn,Txn)→0 as n→∞. Hence, Txn→u. On the other hand, since φ∈Φ′ and (3.9), we have
φp(xn,T2xn)≤φp(xn,Txn)+φp(Txn,T2xn)≤φp(xn,Txn)+kφp(xn,T2xn),
and hence
φp(xn,T2xn)≤11-kφp(xn,Txn),
for all n∈ℕ. Thus, p(xn,T2xn)→0 as n→∞. Therefore, T2xn→u. Since T:X→X is continuous, we have
T(u)=T(limn→∞Txn)=limn→∞T2xn=u,
which is a contradiction. Therefore, using Theorem 3.3, p(v,v)=0. This completes the proof.

Question 1.

Can we generalize Theorems 3.2, 3.3, and Corollary 3.4 for (φ,ψ,p)-contractive maps?

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