Chatterjea type fixed point in Partial $b$-metric spaces

TL;DR

This paper establishes new fixed point theorems of Chatterjea type in partial $b$-metric spaces, extending classical contraction principles and exploring stability and completion properties.

Contribution

It introduces fixed point theorems of Chatterjea type in partial $b$-metric spaces and extends the Banach contraction principle in this context.

Findings

Proved Chatterjea type fixed point theorems in partial $b$-metric spaces.

Extended Banach contraction principle to partial $b$-metric spaces.

Verified $T$-stability of Picard's iteration and discussed the $P$ property.

Abstract

In this paper, we give and prove two Chatterjea type fixed point theorems on partial $b$-metric space. We propose an extension to the Banach contraction principle on partial $b$-metric space which was already presented by Shukla and also study some related results on the completion of a partial metric type space. In particular, we prove a joint Chatterjea-Kannan fixed point theorem. We verify the $T$-stability of Picard's iteration and conjecture the $P$ property for such maps. We also give examples to illustrate our results.

Equations

$$p(x,y)+p(z,z)\le s[p(x,z)+p(z,y)]$$

$$B_p^{\prime }(x,\epsilon ):=\{y\in X:p(x,y)<\epsilon +p(x,x)\},x\in X,\epsilon >0,$$

$$x_n\stackrel{p}{⟶}x.$$

$$\underset{n\to \infty ,m\to \infty }{\lim }p(x_n,x_m)$$

$$\underset{n\to \infty ,m\to \infty }{\lim }p(x_n,x_m)=\underset{n\to \infty }{\lim }p(x,x_n)=p(x,x).$$

$$\underset{n,m\to \infty }{\lim }p(x_n,x_m)=0.$$

$$\underset{n,m\to \infty }{\lim }p(x_n,x_m)=\underset{n\to \infty }{\lim }p(x_n,x)=p(x,x)=0.$$

$$p(T^{n}x,T^{n}y)\le \lambda p(x,y),$$

$$\alpha p_1(x,y)\le p_2(x,y)\le \beta p_1(x,y),\text{for all }x,y,z\in X.$$

$$p(Tx,Ty)\le \lambda p(x,y),$$

$$p(T^{n}x,T^{n}y)\le \lambda p(x,y),$$

$$T^{n+1}x=T(T^{n}x)=Tx=T^n(Tx),$$

$$p(T^{n}x,T^{n}y)\le Kp(x,y),$$

$$K^{\frac{1}{n}}<\frac{1}{\lambda }<1,$$

$$p^{\prime }(x,y)=\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}x,T^{i}y),\text{ whenever }x,y\in X,$$

$$p^{\prime }(Tx,Ty)\le \frac{1}{\lambda }{p}^{\prime }(x,y).$$

$$p^{\prime }(x,y)=p^{\prime }(x,x)=p^{\prime }(y,y).$$

$$\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}x,T^{i}y)=\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}x,T^{i}x)=\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}y,T^{i}y).$$

$$p(T^{i}x,T^{i}x)=p(T^{i}y,T^{i}x)=p(T^{i}y,T^{i}y)\text{ for }i=0,\cdots ,n-1,$$

$$0\le p(T^{i}x,T^{i}x)\le p(T^{i}x,T^{i}y),$$

$$\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}x,T^{i}x)\le \sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}x,T^{i}y)$$

$$p^{\prime }(x,x)\le p^{\prime }(x,y).$$

$$p^{\prime }(x,y)=\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}x,T^{i}y)=\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}y,T^{i}x)=p^{\prime }(y,x),$$

$$p^{\prime }(x,y)=p^{\prime }(y,x)$$

$${\lambda }^i[p(T^{i}x,T^{i}y)+p(T^{i}a,T^{i}a)]\le {\lambda }^{i}s[p(T^{i}x,T^{i}a)+p(T^{i}a,T^{i}y)],$$

$$p^{\prime }(x,y)$$

$$\le \sum _{i=0}^{n-1}{\lambda }^{i}s[p(T^{i}x,T^{i}a)+p(T^{i}a,T^{i}y)]-\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}a,T^{i}a)$$

$$=s\sum _{i=0}^{n-1}{\lambda }^i[p(T^{i}x,T^{i}a)+p(T^{i}a,T^{i}y)]-\sum _{i=0}^{n-1}{\lambda }^{i}p(T^{i}a,T^{i}a)$$

$$=s[p^{\prime }(x,a)+p^{\prime }(a,y)]-p^{\prime }(a,a).$$

$$p^{\prime }(x,y)+p^{\prime }(a,a)\le s[p^{\prime }(x,a)+p^{\prime }(a,y)]$$

$$p^{\prime }(Tx,Ty)=\frac{1}{\lambda }[p^{\prime }(x,y)-p(x,y)]+{\lambda }^{n-1}p(T^{n}x,T^{n}y).$$

$$p^{\prime }(Tx,Ty)$$

Full text

1. Introduction and Preliminaries

In literature, one finds numerous generalizations of metric spaces and Banach contraction principle (BCP). In this line, Czerwik [2] proposed $b$-metric spaces as a generalization of metric spaces and proved the famous BCP in such spaces. In this sequel, Gaba [3] introduced the so-called “metric type space” and proved a common fixed point theorem with the help what he called $\lambda$-sequence in that setting.

After Matthews [10] introduced partial metric spaces as a generalization of the metric space, many authors have studied fixed point theorems on theses spaces (e.g. [1, 11]), in particular, Shukla[12] gave some analog of the Banach contraction principle as well as the Kannan type fixed point theorem in partial $b$-metric spaces.

In this paper, analogs of the Chatterjea fixed point theorem are proved.

First, we recall some definitions from partial $b$-metric spaces.

Definition 1.1**.**

(Compare [10]) A partial metric type on a set $X$ is a function $p:X\times X\to[0,\infty)$ such that:

1. (pm1)

$x=y$ iff $(p(x,x)=p(x,y)=p(y,y)$ whenever $x,y\in X$, 2. (pm2)

$0\leq p(x,x)\leq p(x,y)$ whenever $x,y\in X$, 3. (pm3)

$p(x,y)=p(y,x);$ whenever $x,y\in X$, 4. (pm4)

There exists a real number $s\geq 1$ such that

[TABLE]

for any points $x,y,z\in X$.

The pair $(X,p)$ is called a partial metric type space or a partial $b$-metric space.

It is clear that, if $p(x,y)=0$ , then, from (pm1) and (pm2), $x=y$.

The family $\mathcal{B}^{\prime}$ of sets

[TABLE]

is a basis for a topology $\tau(p)$ on $X$. The topology $\tau(p)$ is $T_{0}$.

Definition 1.2**.**

Let $(X,p)$ be a partial $b$-metric space. Let $(x_{n})_{n\geq 1}$ be any sequence in $X$ and $x\in X$. Then:

1. (1)

The sequence $(x_{n})_{n\geq 1}$ is said to be convergent with respect to $\tau(p)$ (or $\tau(p)$-convergent) and converges to $x$, if $\lim\limits_{n\to\infty}p(x,x_{n})=p(x,x)$. We write

[TABLE] 2. (2)

The sequence $(x_{n})_{n\geq 1}$ is said to be a $p$-Cauchy sequence if

[TABLE]

exists and is finite.

$(X,p)$ is said to be complete if for every $p$-Cauchy sequence $(x_{n})_{n\geq 1}\subseteq X$, there exists $x\in X$ such that:

[TABLE]

We give these additional definitions, useful to characterize some specific complete partial metric type spaces.

Definition 1.3**.**

Let $(X,p)$ be a partial $b$-metric space.

The sequence $(x_{n})_{n\geq 1}\subset X$ is called [math]-Cauchy if

[TABLE]

$(X,p)$ is called [math]-complete if for every [math]-Cauchy sequence $(x_{n})_{n\geq 1}\subseteq X$, there exists $x\in X$ such that:

[TABLE]

2. BCP extension

In this section, we show that if $T$ is a self-map on a partial metric space type space $(X,p)$ and has a power which is a contraction, i.e. there exists $n\in\mathbb{N},n>1$ and $0\leq\lambda<1$ such that

[TABLE]

then there is a transformation $p^{\prime}=\phi(p)$ of $p$ such that $T$ a contraction on $(X,p^{\prime}))$. Moreover, we prove that the partial metric type space $(X,p^{\prime})$ is [math]-complete if $T$ is uniformly continuous.

Ideas for this section are merely copies of the results presented in [4]. We adjust them in the partial metric type setting. We begin with the following definitions.

Definition 2.1**.**

Two partial metrics type $p_{1}$ and $p_{2}$ on a set $X$ are said to be equivalent if there exist $\alpha,\beta\geq 0$ such that

[TABLE]

Definition 2.2**.**

Given two partial metric type spaces $(X,p_{1})$ and $(Y,p_{2})$ , we say that $T:(X,p_{1})\to(Y,p_{2})$ is uniformly continuous if for every real number $\varepsilon>0$ and $\lambda>0$ there exists $\delta=\delta(\lambda)>0$ such that for every $x,y\in X$ with $p_{1}(x,y)<\delta$, we have that $p_{2}(Tx,Ty)<\varepsilon$.

Theorem 2.3**.**

([12, Theorem 1.])

Let $(X,p)$ be a complete partial $b$-metric space with coefficient $s\geq 1$ and let $T:X\to X$ be a mapping such that there exists $\lambda\in[0,1)$ satisfying

[TABLE]

whenever $x,y\in X.$ Then $T$ has a unique fixed point.

We give the following natural corollary:

Corollary 2.4**.**

Let $(X,p)$ be a complete partial metric type space and let $T:X\to X$ be a mapping such that there exists $\lambda\in[0,1)$ satisfying

[TABLE]

for some $n>1$, whenever $x,y\in X.$ Then $T$ has a unique fixed point.

Proof.

By Theorem 2.3, $T^{n}$ has a unique fixed point, say $x\in X$ with $T^{n}x=x$. Since

[TABLE]

it follows that $Tx$ is a fixed point of $T^{n}$, and thus, by the uniqueness of $x$, we have $Tx=x$, that is, $T$ has a fixed point. Since, the fixed point of $T$ is necessarily a fixed point of $T^{n}$, so it is unique. ∎

The main theorem of this section is as follows:

Theorem 2.5**.**

Let $d$ be a partial metric type on a space $X$ and $T:(X,p)\to(X,p)$ a self mapping such that:

[TABLE]

for some $n>1$ and $0<K<1$, whenever $x,y,z\in X.$ If $\lambda$ is a nonnegative real such that

[TABLE]

then the application $p^{\prime}:X^{2}\to[0,\infty)$ defined by :

[TABLE]

satisfies:

• i)

$p^{\prime}$* is a partial metric type on the space $X$;*

• ii)

$T:(X,p^{\prime})\to(X,p^{\prime})$* a self mapping such that:*

[TABLE]

Proof.

We first prove that $p^{\prime}$ is a partial metric type:

1. (pm1)

Indeed for $x,y\in X,$ if $x=y$, then

[TABLE]

Conversely, assume $x,y\in X,$ are such that $p^{\prime}(x,y)=p^{\prime}(x,x)=p^{\prime}(y,y)$, which means

[TABLE]

It is therefore obvious that

[TABLE]

in particular $p(x,x)=p(y,y)=p(x,y)$, i.e. $x=y.$ 2. (pm2)

For all $x,y\in X$ and for all $i=0,\cdots,n-1,$ we have

[TABLE]

and hence

[TABLE]

i.e.

[TABLE] 3. (pm3)

For all $x,y\in X$,

[TABLE]

that is

[TABLE]

for all $x,y\in X.$ 4. (pm4)

For all $x,y,a\in X$, since

[TABLE]

we get

[TABLE]

So

[TABLE]

for any $x,y,a\in X.$

Hence, $p^{\prime}$ is a partial metric type space on $X$.

We now prove that $T:(X,p^{\prime})\to(X,p^{\prime})$ is a contraction with constant $\frac{1}{\lambda}$.

It is readily seen, by a simple computation, that

[TABLE]

Since $T^{n}:(X,p)\to(X,p)$ is a contraction with constant $K$, it follows that

[TABLE]

because of the choice $K^{\frac{1}{n}}<\frac{1}{\lambda}.$ This completes the proof. ∎

As observed in [4, Remark 2.2], under the assumptions of Theorem 2.5, it is readily seen that

[TABLE]

The term $h(x,y):=\sum_{i=0}^{\infty}\lambda^{i}p(T^{i}x,T^{i}y)$ therefore defines a partial metric type, equivalent to $p^{\prime}$, as long as the series happen to converge for some $\lambda>1$.

Next, we establish that whenever the mapping $T:(X,p)\to(X,p)$ is uniformly continuous and the partial metric type $p$ is [math]-complete, then the partial metric type $p^{\prime}$ is also [math]-complete.

Theorem 2.6**.**

We repeat the assumptions of Theorem 2.5. If $T$ is uniformly continuous and the partial metric type $p$ is [math]-complete, then so is the partial metric type $p^{\prime}$.

Proof.

Since $p(x,y)\leq p^{\prime}(x,y)$ for any $x,y\in X$, any [math]-Cauchy sequence in $(X,p^{\prime})$ is also a [math]-Cauchy sequence in $(X,p)$. It is therefore enough to prove that, under uniform continuity of $T$, any convergent sequence $(x_{n})_{n\geq 1}\subset(X,p)$ such that there exists $x^{*}\in X$ with

[TABLE]

is such that there exists $y^{*}\in X$ with

[TABLE]

So let $(x_{n})_{n\geq 1}$ be a sequence in the $G$-metric space $(X,p)$ such that $(x_{n})_{n\geq 1}$ converges to some $\xi\in(X,p).$ and $p(\xi,\xi)=0.$ Set $M=\max\{\lambda^{i},i=1,\cdots,n-1\}$ and observe that

[TABLE]

Since all the powers of $T$ are also uniformly continuous in $(X,d)$, we can write that, for any $\varepsilon>0$, there exists $\eta>0$ such that for all $x,y\in X,$ and $i=1,\cdots,n-1$

[TABLE]

Since $\{x_{n}\}$ converges to some $\xi\in(X,p),$ and $p(\xi,\xi)=0$ there exists $n_{0}\in\mathbb{N}$ such that

[TABLE]

Then

[TABLE]

i.e.

[TABLE]

Thus $(x_{n})_{n\geq 1}$ converges to $\xi$ with respect to the partial metric space $p^{\prime}$ and $p^{\prime}(\xi,\xi)=0.$

This completes the proof.

∎

In concluding this section, we introduce what we call partial ultra-metrics and conjecture that the construction of Frink[5] could be used to obtain a modular metric from an ultra-modular metric. Taking inspiration from the theory of ultra-metric space and that of metric type spaces (see [3]), we can define:

Definition 2.7**.**

A partial ultra-metric on the set $X$ is is a function $p:X\times X\to[0,\infty)$ such that:

1. (pm1)

$x=y$ iff $(p(x,x)=p(x,y)=p(y,y)$ whenever $x,y\in X$, 2. (pm2)

$0\leq p(x,x)\leq p(x,y)$ whenever $x,y\in X$, 3. (pm3)

$p(x,y)=p(y,x);$ whenever $x,y\in X$, 4. (pm4)

here exists a real number $s\geq 1$ such that

[TABLE]

for any points $x,y,z\in X$.

The pair $(X,p)$ is called a partial ultra-metric space .

We are interested in the following question:

Problem 2.8**.**

Given a partial ultra metric $\omega$ on a non empty set $X$, can we construct a partial metric type $\omega^{\prime}$ on $X$ such that $\omega$ and $\omega^{\prime}$ are equivalent? If not, are there conditions which guarantee the existence of such a partial metric type $\omega^{\prime}$ on $X$?

The authors plan to take up this investigation [8] by using “the chain construction” as a tool.

3. Main results

In this section, we present some fixed point results for Chatterjea type mapping in the setting of a partial $b$-metric space. Following theorem is an analog to Chatterjea fixed point theorem in partial $b$-metric space.

Theorem 3.1**.**

Let $(X,p)$ be a complete partial $b$-metric space with coefficient $s\geq 2$ and $T:X\to X$ be a mapping satisfying the following condition:

[TABLE]

for all $x,y\in X$, where $\lambda\in\left[0,\frac{1}{s^{2}}\right)$. Then $T$ has a unique fixed point $u\in X$ and $p(u,u)=0$.

Proof.

Let us first show that if $T$ has a fixed point $u$, then it is unique and $p(u,u)=0$.

From (Ch), we have

[TABLE]

a contradiction, unless $p(u,u)=0.$

Suppose $u,v\in X$ are two distinct fixed points of $T$, that is, $Tu=u$, $Tv=v$ and $u\neq v.$. Then it follows from (Ch) that

[TABLE]

a contradiction, unless $p(u,v)=0$, i.e. $u=v.$ Thus if a fixed point of $T$ exists, then it is unique. For existence of fixed point, let $x_{0}\in X$ be arbitrary; set $x_{n}=T^{n}x_{0}$ and $b_{n}=p(x_{n},x_{n+1})$. Without loss of generality, we may assume that $b_{n}>0$ for all $n\geq 0$ otherwise $x_{n}$ is a fixed point of $T$ for at least one $n\geq 0$.

For any $n\in\mathbb{N}$, it follows from (Ch) that

[TABLE]

therefore $b_{n}\leq\mu b_{n-1}$ where $\mu=\frac{\lambda}{1-\lambda}<1$ (since $\lambda\in\left[0,\frac{1}{s^{2}}\right)\subset\left[0,\frac{1}{2}\right)$). On repeating this, one obtains

[TABLE]

hence $\lim\limits_{n\to\infty}b_{n}=0.$

For $m,n\in\mathbb{N}$ with $m>n$, we obtain

[TABLE]

Using (3.1) in the above inequality,

[TABLE]

As $\lambda\in\left[0,\frac{1}{s^{2}}\right)\subset\left[0,\frac{1}{s}\right)$ and $s>1$, it follows from the above inequality that

[TABLE]

Therefore, $(x_{n})$ is a Cauchy sequence in $X$. By completeness of X there exists $x^{*}\in X$ such that

[TABLE]

We shall show that $x^{*}$ is a fixed point of $T$.

For any $n\in\mathbb{N}$ it follows from (Ch) that

[TABLE]

Taking limit as $n\to\infty$, as $p(x^{*},x^{*})=0,$ we have

[TABLE]

–a contradiction, unless $p(x^{*},Tx^{*})=0,$ that is, $Tx^{*}=x^{*}$. Thus, $x^{*}$ is the unique fixed point of $T$. ∎

Theorem 3.2**.**

Let $(X,p)$ be a complete partial $b$-metric space with coefficient $s>1$ and $T:X\to X$ be a mapping satisfying the following condition:

[TABLE]

for all $x,y\in X$, where $\lambda\in\left[0,\frac{1}{s}\right)$. Then $T$ has a unique fixed point $u\in X$ and $p(u,u)=0$.

Proof.

Let us first show that if $T$ has a fixed point $u$, then it is unique and $p(u,u)=0$.

Suppose $u,v\in X$ are two distinct fixed points of $T$, that is, $Tu=u$, $Tv=v$ and $u\neq v.$. Then it follows from (Ch2) that

[TABLE]

since $p(x,x)\leq p(x,y)$ whenever $x,y\in X.$

At this point, we distinguish between two cases.

1. Case 1.

$\max\{p(x_{n},x_{n-1}),p(x_{n-1},x_{n+1})\}=p(x_{n},x_{n-1})$

[TABLE]

Iterating this process, we get

[TABLE]

for all $n\in\mathbb{N}$.

From the proof of the previous theorem, we can easily establish that for $m,n\in\mathbb{N}$ with $m>n$,

[TABLE]

As $\lambda\in\left[0,\frac{1}{s}\right)$ and $s>1$, it follows from the above inequality that

[TABLE]

Therefore, $(x_{n})$ is a Cauchy sequence in $X$. By completeness of X there exists $x^{*}\in X$ such that

[TABLE]

We shall show that $x^{*}$ is a fixed point of $T$. For any $n\in\mathbb{N}$, we have

[TABLE]

Using (3.3) in the above inequality we obtain $p(x^{*},Tx^{*})=0$, that is, $Tx^{*}=x^{*}$. Thus, $x^{*}$ is the unique fixed point of $T$. 2. Case 2.

If $\max\{p(x_{n},x_{n-1}),p(x_{n-1},x_{n+1})\}=p(x_{n+1},x_{n-1})$, a similar argument as in the Case 1 leads to the existence of a unique fixed point of $T$.

∎

Problem 3.3**.**

Theorem 3.1 advocates for the existence of a fixed point for a Chatterjea contraction in a complete partial $b$-metric space for which the constant $s$ is such that $s\geq 2$. An interesting question/problem could be to investigate if Theorem 3.1 can be formulated for values $1<s<2$ with an appropriate interval for the contraction constant $\lambda.$ Of course Theorem 3.1 remains true for the sharp inequality $s\geq\sqrt{2}$ but the our question remains since we still have to figure out what happens for $1\leq s<\sqrt{2}.$

We conclude this section by presenting a joint Chatterjea-Kannan fixed point leading to the existence of a unique fixed point.

Theorem 3.4**.**

Let $(X,p)$ be a [math]-complete partial $b$-metric space with coefficient $s\geq 1$ and $T:X\to X$ be a self mapping satisfying the following condition:

[TABLE]

for all $x,y\in X$, where $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ are nonnegative real numbers satisfying:

$\lambda_{1}+\lambda_{2}+2s\lambda_{3}+s\lambda_{4}+s\lambda_{5}<1.$* Then $T$ has a unique fixed point $u$ in $X$ and $p(u,u)=0$.*

In proving this theorem, we shall need the following lemma.

Lemma 3.5**.**

Let $(X,p)$ be a partial $b$-metric space with coefficient $s\geq 1$ and $T:X\to X$ be a self mapping. Suppose that $(x_{n})$ is a sequence in $X$ constructed as $x_{n+1}=Tx_{n}$ and such that

[TABLE]

for all $n\in\mathbb{N},$ where $\lambda\in[0,1)$ is a constant. Then $(x_{n})$ is a [math]-Cauchy sequence.

Proof.

Let $x_{0}\in X$ and construct a Picard iterative sequence $(x_{n})$ by $x_{n+1}=Tx_{n},(n\in\mathbb{N})$. We distinguish the following three cases.

1. Case 1.

$\lambda\in[0,\frac{1}{s})\ (s>1)$. By $p(x_{n},x_{n+1})\leq\lambda p(x_{n-1},x_{n})$, we have $p(x_{,}x_{n+1})\leq\lambda^{n}p(x_{0},x_{1}).$ Thus, for any $n>m$ and $n,m\in\mathbb{N}$, we have, by following the proof of Theorem 3.1

[TABLE]

which implies that $(x_{n})$ is a [math]-Cauchy sequence. 2. Case 2.

Let $\lambda\in[\frac{1}{s},1)\ (s>1)$. In this case, we have $\lambda^{n}\to 0$ as $n\to\infty$. So there is $n_{o}\in\mathbb{N}$ such that $\lambda^{n_{o}}<\frac{1}{s}$. Thus, by Case 1, we claim that

[TABLE]

is a [math]-Cauchy sequence. Then $(x_{n})$ is a [math]-Cauchy sequence. 3. Case 3.

Let $s=1$. Similar to the process of Case 1, the claim holds.

∎

Now, we prove the Theorem 3.4.

Proof.

Choose $x_{0}\in X$ and construct a Picard iterative sequence $(x_{n})$ by $x_{n+1}=Tx_{n},(n\in\mathbb{N})$. If there exists $n_{o}\in\mathbb{N}$ such that $x_{n_{o}}=x_{n_{o}+1}$ , then $x_{n_{o}}=x_{n_{o}+1}=Tx_{n_{o}}$ i.e. $x_{n_{o}}$ is a fixed point of $T$. Next, without loss of generality, let $x_{n}\neq x_{n+1}$ for all $n\in\mathbb{N}.$ By (3.6), we have

[TABLE]

In view of axioms (pm2) and (pm4), we have

[TABLE]

i.e.

[TABLE]

We also have

[TABLE]

i.e.

[TABLE]

Hence

[TABLE]

It follows that

[TABLE]

Again, by (3.6), and exploiting the symmetry of $p$, i.e. $p(x_{n},x_{n+1})=p(Tx_{n},Tx_{n-1})$, we are led to

[TABLE]

Adding up (3.4) and (3.5) yields

[TABLE]

Put $\lambda=\frac{2\lambda_{1}+2s\lambda_{3}+s\lambda_{4}+s\lambda_{5}}{2-2\lambda_{2}-2s\lambda_{3}-s\lambda_{4}-s\lambda_{5}}$. In view of $\lambda_{1}+\lambda_{2}+2s\lambda_{3}+s\lambda_{4}+s\lambda_{5}<1,$ then $0\leq\lambda<1.$

Thus, by Lemma 3.5, $(x_{n})$ is a [math]-Cauchy sequence in $X$. Since $(X,p)$ is [math]-complete, then there exists some point $x*\in X$ such that:

[TABLE]

By (3.6), it is easy to see that

[TABLE]

Taking the limit as $n\to\infty$, we get $\lim\limits_{n\to\infty}p(x_{n+1},Tx^{*})=0$

On another side,

[TABLE]

Taking the limit on both sides as $n\to\infty$, we get

[TABLE]

It gives that $Tx^{*}=x^{*}$ . In other words, $x^{*}$ is a fixed point of $T$.

For uniqueness of the fixed point, assume $y^{*}$ is another fixed point of $T$, then by (3.6), it is easy to check that

[TABLE]

Because $\lambda_{1}+\lambda_{2}+2s\lambda_{3}+s\lambda_{4}+s\lambda_{5}<1$ implies $\lambda_{1}+\lambda_{3}<1,$ we conclude that $x^{*}=y^{*}$ since $p(x^{*},y^{*})=0.$

∎

Corollary 3.6**.**

Let $(X,p)$ be a complete partial metric space with coefficient $s\geq 1$ and $T:X\to X$ be a self mapping satisfying the following condition:

[TABLE]

for all $x,y\in X$, where $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ are nonnegative real numbers satisfying:

$\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{4}+\lambda_{5}<1.$* Then $T$ has a unique fixed point in $X$.*

Proof.

Take $s=1$ in Theorem 3.4, thus the claim holds. ∎

Remark 3.7*.*

Take $\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}=0$ in Theorem 3.4 or in Corollary 3.6, then Theorem 3.4 and Corollary 3.6 are reduced to [12, Theorem 2.4] and Banach contraction principle, respectively. From this point of view, our results are genuine generalizations of the previous results.

Recently, Qing and Rhoades [13] established the notion of $T$-stability of Picard’s iteration in metric space. In the following, we modify their definition and introduce the concept of $T$-stability of Picard’s iteration in partial $b$-metric space.

Definition 3.8**.**

Let $(X,p)$ be a partial $b$-metric space, $x_{0}\in X$ and $T:X\to X$ be a mapping with $F(T)\neq\emptyset$, where $F(T)$ denotes the set of all fixed points of $T$. Then Picard’s iteration $x_{n+1}=Tx_{n}$ is said to be $T$-stable with respect to $T$ if $x_{n}\overset{p}{\longrightarrow}q,\ q\in F(T)$ and whenever $(y_{n})$ is a sequence in $X$ with $\lim\limits_{n\to\infty}p(y_{n+1},Ty_{n})=0$, we have $y_{n}\overset{p}{\longrightarrow}q$.

What follows is a useful lemma for the proof of our main result in this section.

Lemma 3.9**.**

[9]** Let $(a_{n})$, $(c_{n})$ be nonnnegative sequences satisfying $a_{n+1}\leq ha_{n}+c_{n}$ for all $n\in\mathbb{N},\ 0\leq h<1,\ \lim\limits_{n\to\infty}c_{n}=0$. Then $\lim\limits_{n\to\infty}a_{n}=0$.

Now we state our main result on $T$-stability.

Theorem 3.10**.**

Under the conditions of Theorem 3.4, if $2s\lambda_{1}+2\lambda_{3}+(s+s^{2})(\lambda_{4}+\lambda_{5})<2$, then Picard’s iteration is $T$-stable.

Proof.

From Theorem 3.4, we know that $T$ has a unique fixed point $x^{*}\in X$ and $p(x^{*},x^{*})=0$. Assume that $(y_{n})$ is a sequence in $X$ with $\lim\limits_{n\to\infty}p(y_{n+1},Ty_{n})=0$. Taking advantage of (3.6), on the one hand, we have

[TABLE]

which means

[TABLE]

On the other hand, owing to the symmetry of $p$, we have

[TABLE]

which yields

[TABLE]

Combining (3.6) and (3.7), we get

[TABLE]

leading to

[TABLE]

If we set $l={s(2\lambda_{1}+s\lambda_{4}+s\lambda_{5})}{2-2\lambda_{3}-s\lambda_{4}-s\lambda_{5}}$, it follows from $2s\lambda_{1}+2\lambda_{3}+(s+s^{2})(\lambda_{4}+\lambda_{5})<2$ that $0\leq l<1$.

In view of Lemma 3.9, set $a_{n}=p(y_{n},x^{*}),\ c_{n}=sp(y_{n+1},Ty_{n})$, and owing to (3.8), we have

[TABLE]

Thus, $\lim\limits_{n\to\infty}p(y_{n},x^{*})=0=p(x^{*},x^{*})$, i.e. have $y_{n}\overset{p}{\longrightarrow}x^{*}$. As a consequence, Picard’s iteration is $T$-stable. ∎

Corollary 3.11**.**

Under the conditions of Corollary 3.6, Picard’s iteration is $T$-stable.

Proof.

Just notice that Corollary 3.6 is a special case of Theorem 3.4 where we take $s=1$.

∎

Corollary 3.12**.**

([12, Theorem 1.]) Let $(X,p)$ be a complete partial $b$-metric space with coefficient $s\geq 1$ and $T:X\to X$ be a mapping satisfying the following condition:

[TABLE]

for all $x,y\in X$, where $\lambda\in\left[0,1\right)$. The Picard’s iteration is $T$-stable.

Proof.

Just notice that Corollary 3.12 is a special case of Corollary 3.6 where we take $\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}=0$. ∎

Problem 3.13**.**

The authors plan, in [8], to study the $T$-stability of both the Kannan and the Chatterjea contractions for the Picard iteration for a self mapping defined on partial $b$-metric space.

The Corollary (2.4) illustrates the idea of the so-called $P$* property*. If a map $T$ satisfies $F(T)=F(T^{n})$ for each $n\in\mathbb{N}$, then it is said to have the $P$ property (see [7]). The following results are generalizations of the corresponding results in partial $b$-metric spaces.

Theorem 3.14**.**

Let $(X,p)$ be a partial $b$-metric space with coefficient $s\geq 1$. Let $T:X\to X$ be a mapping such that $F(T)\neq\emptyset$ and that

[TABLE]

for all $x\in X$, where $0\leq\lambda<1$ is a constant. Then $T$ has the $P$ property.

Proof.

We always assume that $n>1$, since the statement for $n=1$ is trivial. Let $z\in F(T^{n})$. It is clear that

[TABLE]

Hence, $p(z,Tz)=0$, that is., $Tz=z$.

∎

In concluding this section, we make a conjecture with respect to $P$ property with regards to Theorem 3.4 and Corollary 3.6. They are yet to be proved.

Conjecture 3.15**.**

Under the conditions of Theorem 3.4, $T$ has the $P$ property. For the proof, it is enough to check if the mapping $T$ satisfies (3.9).

Also

Conjecture 3.16**.**

Under the conditions of Corollary 3.6, $T$ has the $P$ property.

We conclude this paper by giving examples to illustrate Theorem 3.4.

Example 3.17**.**

Let $X=\{1,2,3,4\}$ and $p:X\times X\to\mathbb{R}$ be defined by

[TABLE]

Then $(X,p)$ is a complete partial $b$-metric space with coefficient $s=4>1$.

Now define the self mapping $TX\to X$ by

[TABLE]

A simple computation gives:

[TABLE]

Then, $T$ satisfies all the conditions of Theorem 3.4, with $\lambda_{1}\in\left[\frac{3}{4},1\right),\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}=0$ and obviously $\lambda_{1}+\lambda_{2}+2s\lambda_{3}+s\lambda_{4}+s\lambda_{5}<1.$ Now, by Theorem 3.4, $T$ has a unique fixed point , which in this case is $1$.

Example 3.18**.**

Let $X=[0,1],\ k>1$ and define a mapping $p:X\times X\to\mathbb{R}^{+}$ by $p(x,y)=|x-y|^{k}$ for all $x,y\in X$. Then $(X,p)$ is a complete partial $b$-metric space with coefficient $s=2^{k}>1$. Define a mapping $T:X\to X$ by $Tx=e^{x-\lambda}$, where $\lambda>1+\ln 2$ is a constant. Then by mean value theorem of differentials, for any $x,y\in X$ and $x\neq y$, there exists some real number $\xi$ belonging to between $x$ and $y$ such that

[TABLE]

Hence

[TABLE]

Then, $T$ satisfies all the conditions of Theorem 3.4, with $\lambda_{1}=(e^{1-\lambda})^{k},\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}=0$ and obviously $\lambda_{1}+\lambda_{2}+2s\lambda_{3}+s\lambda_{4}+s\lambda_{5}<1.$ Now, by Theorem 3.4, $T$ has a unique fixed point in $u\in X$.

In view of $\lambda>1+\ln 2$, then $\lambda_{1}=(e^{1-\lambda})^{k}<2^{1-p}=\frac{1}{s}$, so $2s\lambda_{1}+2\lambda_{3}+(s+s^{2})(\lambda_{4}+\lambda_{5})<2$ and all the conditions of Theorem 3.10 are satisfied. So by Theorem 3.10, the Picard’s iteration is $T$-stable.

To see exactly what this $T$-stability means, consider the sequence $y_{n}=\frac{n}{n+1}u\in X$. It follows that

[TABLE]

Note that $y_{n}=\frac{n}{n+1}u\to u\ (n\to\infty).$

4. Going further

Recently, Zheng et al.[14] introduced the so-called $\theta$-$\phi$ contraction in complete metric spaces and this technique was successfully applied to Kannan type mapping in partial metric spaces (see [6]). The results of the present paper will be applied in future investigations by the authors regarding $\theta$-$\phi$ contraction in complete partial $b$-metric spaces. Hence the continuation of this research is considering $\theta$-$\phi$-Chatterjea type contraction in partial $b$-metric spaces and investigate the existence of fixed points. We have a definition for $\theta$-$\phi$-Chatterjea type contraction and we must verify that it follows the idea of Chatterjea contractions and generalizes them in a way that keeps their properties and their relationship with other contractions. Moreover, a natural question is to check whether this new type of contraction is $T$-stable and has the $P$ property.