1. Introduction

Metric fixed point theory is a vast and expanding branch of functional analysis. It is widely held that the theory has its origin in Banach’s Contraction Principle [1],[2] which was established in the year of 1922. This result has served to establish several important results is different branches of mathematics [2]–[4]. It has several generalizations and extensions which were advanced over the years and still are being actively pursued [3],[5]–[10]. In this paper, we consider two such generalizations, one is the fixed point result of local contraction due to Edelstein [11] and the other is the fixed point result for weak contractions first proved by Alber et al. [12] in Hilbert spaces and subsequently established in the metric spaces by Rhoades [13]. The result in [13] has itself become a source of an area of fixed point theory where weak inequalities are considered [14]–[19]. Our result in this paper combines the above two ideas to lead to a unique fixed point result for uniform local weak contraction mappings that we define here.

2. Preliminaries

In this section we recall some definitions and introduce the notion of local weak contraction.

In what follows, we denote by \(B(x,r)\) the open ball of radius \(r\) centered at the point \(x\), by \(\mathbb{N}\) the set of natural numbers, and by \(\mathbb{R}\) the set of real numbers.

Definition 1 (local contraction).

[11] A map \(T:X\rightarrow X\) is said to be locally (Banach) contractive if, for every \(x\in X\), there exist \(\varepsilon, \lambda\) (\(\in \mathbb R\), with \(\varepsilon >0,\ 0\leq \lambda<1\)), which may depend on \(x\), such that \(\forall \, p,\, q \in B(x,\varepsilon)\),

$$d(T(p),T(q))<\lambda d(p,q).$$

Definition 2 (Uniform Local contraction).

[11] A map \(T:X\rightarrow X\) is said to be uniformly locally contractive if it is locally contractive and both \(\varepsilon\) and \( \lambda\) do not depend on \(x\).

It is usually abbreviated as “\((\varepsilon,\lambda)-\)uniformly locally contractive” when the role of \(\varepsilon\) and \(\lambda\) are to be stressed down.

Definition 3 (\(\varepsilon\)-chainable metric space [11]).

A metric space \(X\) is said to be \(\varepsilon\)-chainable (\(\varepsilon \in \mathbb R, >0\)) if, for any \(x, y \in X\), there is an \(\varepsilon\)-chain from \(x\) to \(y\), that is, there are finite number of points \(a_0,a_1,..., a_n\) in \(X\), with \(x=a_0,\ y=a_n\) such that \(d(a_i, a_{i+1})<\varepsilon,\ \forall i=0,1,...,n-1\).

Note that \(\varepsilon\)-chainable metric spaces have appeared in the works like [11],[20]–[25].

Definition 4 (\(\psi\)-weak contraction).

Let \(\psi:[0,\infty)\rightarrow [0,\infty)\) be a given continuous function which is nondecreasing and satisfies \(\psi(t) >0\) for \(t>0\) and \(\psi(0)=0\). A function \(T:X\rightarrow X\) is said to be a \(\psi\)-weak contraction if \(\forall x,y \in X\), \(d(Tx,Ty)\leq d(x,y)-\psi(d(x,y))\).

Definition 5 (weak contraction [13]).

A map \(T:X\rightarrow X\) is said to be a weak contraction if there is a continuous function \(\psi:[0,\infty)\rightarrow [0,\infty)\) which is nondecreasing and satisfies \(\psi(t) >0\) for \(t>0\) and \(\psi(0)=0\) and \(\lim\limits_{t\rightarrow \infty}\psi(t)=\infty\), such that

$$x,y \in X \implies d(Tx,Ty)\leq d(x,y)-\psi(d(x,y)).$$

Remark 1.

There is a subtle difference between weak contraction and \(\psi\)-weak contraction. While the latter implies the former, it is possible that a weak contraction is not a \(\psi\)-weak contraction for a prior given \(\psi\). If we consider \(X=[0,1]\), with the usual distance, \(T(x)=x-\frac{1}{2}x^2\) and \(\psi(t)=(1-\lambda)t\), where \(0<\lambda <1\), then, although it is a weak contraction [15], but is not a \(\psi\)-weak contraction, for this given \(\psi\).

We now introduce a local version of the weak contraction.

Definition 6 (\((\varepsilon, \psi)\)-uniform local weak contraction).

Let \(\psi:[0,\infty)\rightarrow [0,\infty)\) be a given continuous function which is nondecreasing and satisfies \(\psi(t) >0\) for \(t>0\) and \(\psi(0)=0\). Let \((X,d)\) be a metric space. A function \(T:X\rightarrow X\) is said to be an \((\varepsilon, \psi)\)-uniform local weak contraction if

$$ d(x,y)< \varepsilon \implies d(Tx,Ty)\leq d(x,y)-\psi(d(x,y)).$$
(1)

3. Main Results

Theorem 1.

Let \(\varepsilon >0 \) and \(X\) be a complete, \(\varepsilon\) -chainable metric space. Suppose \(\psi:[0,\infty)\rightarrow [0,\infty)\) be a continuous, nondecreasing function such that \( \psi(t) >0 \) for \(t>0\) and \(\psi(0)=0\) . Suppose also that \(T: X\rightarrow X\) is a self-map which is an \((\varepsilon, \psi) \) -uniform local weak contraction, that is, for each \(x,y \in X\) if \(d(x,y)< \varepsilon\) , then

$$d(Tx,Ty)\leq d(x,y)-\psi(d(x,y)).$$

Then \(T\) has a unique fixed point \(\bar{x}\) in \(X\) .

Proof.

Let \(x\in X\) be an arbitrary element. We construct a sequence \(\{x_{n}\}\) such that

$$ x_0:=x,\ x_i:=T^ix,\ \forall i\in \mathbb{N}.$$
(2)

As \(X\) is \(\varepsilon\)-chainable, let \(x=a_0,a_1,..., a_n=Tx\), be an an \(\varepsilon\)-chain from \(x\) to \(Tx\), where

$$ d(a_i, a_{i+1})<\varepsilon, \ \forall i=0,1,..,n-1.$$
(3)

Since \(d(a_i, a_{i+1})<\varepsilon, \forall i=0,1,..,n-1 \), equation-\((1)\) is also satisfied for every pair of consecutive elements of the chain. Thus, we have

$$d(Ta_i,Ta_{i+1})\leq d(a_i,a_{i+1})-\psi(d(a_i,a_{i+1})) \leq d(a_i,a_{i+ 1})<\varepsilon.$$

Inductively, we obtain \(d(T^ma_i,T^ma_{i+1})<\varepsilon,\) for any \(\ m\in \mathbb{N}\).

Let \(R^i_m :=d(T^ma_i,T^ma_{i+1})\). Then,

$$\begin{aligned} \, R^i_{m+1} =d(T^{m+1}a_i,T^{m+1}a_{i+1}) & \leq d(T^ma_i,T^ma_{i+1})- \psi(d(T^ma_i,T^ma_{i+1})) \\ & \leq d(T^ma_i,T^ma_{i+1}) =R^i_m. \end{aligned}$$

Thus, \(\{R^i_m\}\) is a nonincreasing sequence, and being bounded below (\(0\) is a lower bound), it must be convergent. Suppose \(\lim \limits_{m\rightarrow \infty}R^i_m = R^i\geq 0\), for each \(i=0,1,...,n-1.\)

Again, we have

$$\begin{aligned} \, R^i_{m+1} &\leq R^i_m- \psi(R^i_m) \nonumber\\ &\Rightarrow \lim\limits_{m\rightarrow \infty} R^i_{m+1} \leq \lim\limits_{m\rightarrow \infty} R^i_{m} - \lim\limits_{m\rightarrow \infty}\psi(R^i_m) \nonumber\\ &\Rightarrow R^i \leq R^i- \psi(R^i) \text{\qquad[since $\psi$ is continuous]} \nonumber\\ &\Rightarrow \psi(R^i) =0 \nonumber\\ &\Rightarrow R^i =0. \end{aligned}$$
(4)

Now using triangle inequality, we have

$$ d(x_m,x_{m+1})=d(T^mx,T^m(Tx))\leq \sum_{i=0}^{n-1}d(T^ma_i,T^ma_{i+1})=\sum_{i= 0}^{n-1}R^i_m.$$
(5)

Taking limit \(m \longrightarrow \infty,\) we obtain,

$$ \lim\limits_{m\rightarrow \infty} d(x_m,x_{m+1})\leq \lim\limits_{m\rightarrow \infty}\sum_{i=0}^{n-1}R^i_m=\sum_{i=0}^{n-1}\lim\limits_{m\rightarrow \infty}R^i_m=0.$$
(6)

We will show that \(\{x_{n}\}\) is a Cauchy sequence.

Let \(\varepsilon_1>0\) be chosen arbitrarily and take \(\varepsilon_0= \min\{\varepsilon, \varepsilon_1\}>0.\) As \(\lim\limits_{m\rightarrow \infty}d(x_m,x_{m+1})= 0\), \(\exists k \in \mathbb N\) such that

$$ d(x_k,x_{k+1})<\min\{\frac{\varepsilon_0}{2},\psi(\frac{\varepsilon_0}{2})\}.$$
(7)

We note that if \(d(y,x_k)<\varepsilon_0\leq \varepsilon\), then \((1)\) holds for \(x=x_k\). Thus, if \(y\in B(x_k,\frac{\varepsilon_0}{2})\), we have,

$$\begin{aligned} \, d(Ty,x_k) & \leq d(Ty,Tx_k)+d(Tx_k,x_k) \nonumber\\ & \leq d(y,x_k) -\psi(d(y,x_k))+d(x_k,x_{k+1}) \nonumber \\ & <\frac{\varepsilon_0}{2}-\psi(d(y,x_k))+\frac{\varepsilon_0}{2} \ \ \ \ \ \ \text{[ $\because y\in B(x_k,\frac{\varepsilon_0}{2}) $]} \nonumber\\ &\leq \varepsilon_0-\psi(d(y,x_k))\nonumber \\ & \leq \varepsilon_0. \end{aligned}$$
(8)

\(\therefore Ty\in B(x_k,\varepsilon_0), \forall y \in B(x_k,\frac{\varepsilon_0}{2})\).

Also, if \(\frac{\varepsilon_0}{2}\leq d(y,x_k)\leq \varepsilon_0\), by monotonicity of \(\psi\), we have \(\psi(\frac{\varepsilon_0}{2})\leq \psi(d(y,x_k))\).

$$\begin{aligned} \, d(Ty,x_k) & \leq d(Ty,Tx_k)+d(Tx_k,x_k) \nonumber\\ & \leq d(y,x_k) -\psi(d(y,x_k))+d(x_k,x_{k+1}) \nonumber \\ & \leq d(y,x_k)-\psi(\frac{\varepsilon_0}{2})+\psi(\frac{\varepsilon_0}{2}) \nonumber\\ &\leq d(y,x_k)\nonumber \\ & \leq \varepsilon_0. \end{aligned}$$
(9)

By the above two cases we have \(Ty\in B(x_k,\varepsilon_0), \forall y \in B(x_k,\varepsilon_0)\), which implies \(x_m\in B(x_k,\varepsilon_0) \) for all \(m\geq k\). Hence \(d(x_m,x_k)<\varepsilon_0\leq \varepsilon_1\) for all \(m\geq k\), and thus \(\{x_n\}\) is a Cauchy sequence. \(X\) being complete, \(\{x_n\}\) must converge to some \(\bar{x} \in X\). Again for any \(\varepsilon_2>0\) let \(\delta =\min \{\varepsilon,\varepsilon_2\}\). Now, if \(d(x,y)<\delta\), then \(d(x,y)<\varepsilon\) and thus \(d(Tx,Ty)\leq d(x,y)-\psi(d(x,y)) \leq d(x,y)<\delta\leq \varepsilon_2\), which means that \(T\) is continuous. Therefore, \(\bar{x} =\lim\limits_{n\rightarrow \infty} x_{n+1}=\lim\limits_{n\rightarrow \infty}T(x_n)=T\bar{x} \). Therefore, \( \bar{x} \) is a fixed point of \(T\).

We now prove that \(\bar{x}\) is unique. If not, let \(\exists\ \ \bar{y} (\neq \bar{x} )\in X\) such that \(\bar{y} =T\bar{y}\). Then \(d(\bar{x} ,\bar{y})>0\). Now consider the \(\varepsilon\)-chain from \(\bar x\) to \(\bar y\). Let \(\bar x=b_0,b_1,..., b_p=\bar y\) be an an \(\varepsilon\)-chain from \(\bar x\) to \(\bar y\). Thus, \(d(b_i, b_{i+1})<\varepsilon, \forall i=0,1,..,p-1\). By the same argument as we reached (5) we obtain that

$$ d(\bar x,\bar y)=d(T^m \bar x,T^m \bar y)\leq \sum_{i=0}^{p-1}d(T^mb_i,T^mb_{i+1}).$$
(10)

Taking limit \(m \longrightarrow \infty \) we obtain, \( d(\bar x,\bar y)=0\), which is a contradiction. Hence the fixed point would be unique.

Remark 2.

The above result generalizes the result of Edelstein[11] and the result of Rhoades [13] in the context of \(\varepsilon\)-chainable metric spaces.

The next result shows that instead of assuming the whole space \(X\) to be \(\varepsilon\)-chainable, if \(T(X)\) only is assumed to be so, the conclusion of the previous result still remains valid.

Theorem 2.

Let \(\varepsilon >0\) , and let \(X\) be a complete metric space. Suppose \(\psi:[0,\infty)\rightarrow [0,\infty)\) be a continuous, nondecreasing function such that \( \psi(t) >0 \) for \(t>0\) and \(\psi(0)=0\) . Also let \(T: X\rightarrow X\) be a self-map which is \((\varepsilon, \psi) \) -uniform local weak contraction, that is, for each \(x,y \in X\) if \(d(x,y)< \varepsilon\) , then

$$d(Tx,Ty)\leq d(x,y)-\psi(d(x,y)).$$

If \(T(X)\) is closed and \(\varepsilon\) -chainable, then \(T\) has a unique fixed point.

Proof.

Let \(z\in X\) be arbitrary, then \(Tz \in T(X)\). We write \(x_0 = x=Tz\) and set \(x_n=T^{n}x\), for \(n=1,2,...\). Then repeating the same argument as in Theorem 1, we assure the existence of a fixed point in \(T(X)\) and hence in \(X\). For the uniqueness part, we note that if \(z\) is any fixed point of \(T\), then \(z=T(z)\in T(X)\). Then following the same argument for the uniqueness part as in Theorem 1, we assure the uniqueness of the fixed point.

4. Illustrative Example

In this section we present an example to show that our theorem is applicable to a function which is neither a uniform local (Banach) contraction nor a \(\psi\)-weak contraction on the whole space, for a preassigned \(\psi\).

Example 1.

Let \(X=A\cup B\), where \(A=\big\{(x(t),y(t)) : x(t)={1}/{2}-t, y(t)=0, 0\leq t\leq{1}/{2}\big\}\) and \(B= \big\{(x(s),y(s)): x(s)=0, y(s)=s+{1}/{4}, 0\leq s\leq {1}/{4}\big\}\).

\(X\) is clearly a complete subspace of the metric space \(\mathbb R^2\) with the usual distance \(d\). Further we note that, \(X\) is \(\eta \)-chainable for any \(\eta > {1}/{4}\).

Let \(\psi:[0,\infty)\rightarrow [0,\infty)\) be defined by \(\psi(u)=u^2/2\). Clearly, \(\psi\) is a continuous, nondecreasing function such that \( \psi(t) >0\) for \(t>0\) and \(\psi(0)=0\).

Let \(T: X\rightarrow X\) be defined by

$$T(x(u),y(u))=\begin{cases} \big(\frac{1}{2}-\left(u-\frac{1}{2} u^2\right), 0\big) & if\ (x,y)\in A\\ \big(\frac{1}{2}-\big\{\big(u+\frac{3}{4}\big)-\frac{1}{2} \big(u+\frac{3}{4}\big)^2\big\} , 0\big) & if\ (x,y)\in B. \end{cases}$$

First, we show that \(T\) is not a uniform local (Banach) contraction. If possible, let \(T\) be a uniform local contraction. Then there exist \(\varepsilon, \lambda \) (with \(\varepsilon >0\), \(0\leq \lambda <1\)) such that, for all \(U,V \in X\),

$$ \text{if}\quad d(U,V)<\varepsilon,\qquad \text{then}\quad d(T(U),T(V))<\lambda \,d(U,V).$$
(11)

Let \(t=\min\{\varepsilon,\ {(1-\lambda)}/{2}\}\). Now if we consider the points \(U({1}/{2},0)\) and \(V({1}/{2}-t,0)\) then \(d(U,V)= t\) and

$$\begin{aligned} \, d(TU,TV)=d\left(\left(\frac{1}{2},0\right),\left( \frac{1}{2}-\left\{t-\frac{1}{2}t^2\right\},0\right)\right) =t-\frac{1}{2}t^2. \end{aligned}$$

Since \(d(U,V)= t<\varepsilon\), relation-(11) is satisfied. Thus, we obtain

$$\begin{aligned} \, t-\frac{1}{2}t^2<& \lambda t, \nonumber\\ \text{or}\quad t-\frac{1}{2}t^2<& (1-2t) t \quad [\text{because }t\leq (1-\lambda)/2 \implies \lambda \leq 1-2t ],\nonumber\\ \text{or}\quad -\frac{1}{2}t^2<& - 2t^2. \end{aligned}$$
(12)

Relation (12) is absurd. So we arrive at a contradiction, and thus \(T\) is not a uniform local contraction.

Now, we show that \(T\) is not a \(\psi\)-weak contraction. For this, consider a pair of points \(U({1}/{2},0)\) and \(V(0,{1}/{2})\) of \(X\) corresponding to \(t=0\) and \(s={1}/{4}\), respectively. Then

$$d(U,V)= \frac{\sqrt {2}}{2},\qquad d(U,V) -\psi(d(U,V))= \frac{\sqrt {2}}{2}- \frac{1}{2}\times \frac{1}{2}=\frac{2\sqrt{2}-1}{4}.$$

Again,

$$\begin{aligned} \, T\biggl(\biggl(\frac{1}{2},0\biggr)\biggr)&=\biggl(\frac{1}{2}-\biggl(0-\frac{1}{2}\times 0^2\biggr),0\biggr) =\biggl(\frac{1}{2},0\biggr), \\ T\biggl(\biggl(0,\frac{1}{2}\biggr)\biggr)&=\biggl(0, \frac{1}{2}-\biggl(1-\frac{1}{2}\times 1^2\biggr)\biggr)=(0,0). \end{aligned}$$

Thus,

$$d( T(U),T(V))= \frac{1}{2} \nleq \frac{2\sqrt{2}-1}{4}= d(U,V) -\psi(d(U,V)).$$

Hence relation (1) is not satisfied for this pair of points, which shows that it is not a \(\psi\)-weak contraction.

We now show that \(T\) is a \((\eta,\psi)\)-uniformly local weak contraction map, for some \(\eta >0\).

Let us now consider the following cases.

Case I: Let us consider two points \(P(x(t),y(t))\in A\) and \(Q((x(s),y(s))\in B\), where \(0\leq t\leq{1}/{2}\) and \(0\leq s\leq {1}/{4}\). Then

$$d(P,Q)=\sqrt{\left(\frac{1}{2}-t\right)^2+\left(s+\frac{1}{4}\right)^2}.$$

Thus,

$$d(P,Q)-\psi\left(d(P,Q)\right)=\sqrt{\left(\frac{1}{2}-t\right)^2 +\left(s+\frac{1}{4}\right)^2}- \dfrac{1}{2}\left(\left(\frac{1}{2}-t\right)^2+\left(s+\frac{1}{4}\right)^2\right).$$

Now,

$$\begin{aligned} \, d\left(T(P),T(Q)\right)=& d\left( (\frac{1}{2}-(t-\frac{1}{2} {t}^2),0),\left(\frac{1}{2}-\left\{\left(s+\frac{3}{4}\right)-\frac{1}{2} \left(s+ \frac{3}{4}\right)^2\right\} , 0\right)\right)\\ =& \left\{\left(s+\frac{3}{4}\right)-\frac{1}{2} \left(s+ \frac{3}{4}\right)^2\right\}- \left( t-\frac{1}{2} {t}^2\right). \end{aligned}$$

Consider the function \(F:\mathbb{R\times R}\rightarrow \mathbb R\), defined by

$$\begin{aligned} \, F(t,s)&= \left\{\sqrt{\left(\frac{1}{2}-t\right)^2+\left(s+\frac{1}{4}\right)^2}- \dfrac{1}{2}\left(\left(\frac{1}{2}-t\right)^2+\left(s+ \frac{1}{4}\right)^2\right)\right\} \\*&\qquad{}-\left\{ \left\{\left(s+\frac{3}{4}\right)-\frac{1}{2} \left(s+\frac{3}{4}\right)^2\right\}- \left( t-\frac{1}{2} {t}^2\right)\right\}. \end{aligned}$$

Now \(F({1}/{2},0) >0\) and \(F\) is continuous at \(({1}/{2},0)\). Thus, by the Neighborhood property of Continuous functions, there is a \(\delta-neighborhood\) of the point \(({1}/{2},0)\) where the function assumes only positive values.

Thus, for \(t={1}/{2}\) and \(s=0\), the corresponding pair of points \((0,0) \in A\) and \((0,{1}/{4})\in B\) satisfies the relation (1) and there is some \(\delta\) with \({1}/{4}>\delta >0\), such that, for all pair of points \(R(x(t),y(t))\in A\) and \(S((x(s),y(s))\in B)\), where \(t\in \left({1}/{2} - \delta ,{1}/{2}\right]\) and \(s\in \left[0, \delta\right)\), relation (1) remains satisfied.

Now let us consider \(\alpha \geq 0\) be such that \(\alpha \leq \delta\). The points corresponding to parametric values \(t={1}/{2} - \alpha\) and \(s= \alpha\) are \(P\left(\alpha,0\right)\) and \(Q(0,{1}/{4}+\alpha )\), respectively, and the distance between them is

$$d(P,Q)= d\left(\left( \alpha,0\right) ,\biggl(0,\frac{1}{4}+ \alpha \biggr)\right) =\sqrt{{\alpha}^2+\left(\frac{1}{4}+ \alpha \right)^2}.$$
$$\therefore d(P,Q)-\psi\left(d(P,Q)\right)=\sqrt{{\alpha}^2+\left (\frac{1}{4}+ \alpha \right)^2}-\dfrac{1}{2}\left\{{\alpha}^2+ \left(\frac{1}{4}+ \alpha \right)^2\right\}=g(\alpha) (say).$$

\(d(P,Q)-\frac{1}{2}(d(P,Q))^2=g(\alpha)\) is a strictly increasing function of \(d(P,Q)\) when \(d(P,Q)<1\). Thus, \(g(\alpha)\) is increasing with respect to \(\alpha\), for \(0\leq\alpha <{1}/{4}\). Hence the minimum value of \(g(\alpha)\) in \([0,{1}/{4})\) is \({7}/{32}\), corresponding to \(\alpha =0\). Now,

$$d\left(T(P),T(Q)\right)= \left\{\left(\frac{3}{4}+ \alpha\right)-\frac{1}{2} \left(\frac{3}{4}+ \alpha\right)^2\right\} - \left\{\left(\frac{1}{2}- \alpha\right)-\frac{1}{2} \left(\frac{1}{2}- \alpha\right)^2\right\}=\frac{3}{32}+ \frac{3}{4}\alpha=f(\alpha)$$

(say), which is clearly a continuous and increasing function of \(\alpha\). In particular, for \(\alpha =0.1\), we have \(d\left(T(P),T(Q)\right)={27}/{160} <{7}/{32}\), the minimum value of the R. H. S of (1) in this case. Thus, \(f(\alpha)<f(0.1)={27}/{160}<{7}/{32}=g(0)<g(\alpha)\). Thus, for \(\alpha \leq 0.1\) the inequality (1) is satisfied. Hence we can choose \(\delta=0.1\)

Thus, in this case we see that the pair of points \(P, Q\) (where \(P\in A, Q\in B\)), whose distance is less than

$$\eta=\sqrt{{0.1}^2+\left(\frac{1}{4}+ 0.1 \right)^2} \approx 0.3640,$$

satisfies relation (1).

Case II: \(P(x(s_1),y(s_1)), Q((x(s_2),y(s_2))\in B)\), \(0\leq s_1\leq s_2\leq {1}/{4}\).

$$\begin{aligned} \, d(T(P),T(Q)) =& \left\{\left(s_2 +\frac{3}{4}\right)-\frac{1}{2} \left(s_2+ \frac{3}{4}\right)^2\right\}- \left\{\left(s_1 +\frac{3}{4}\right)-\frac{1}{2} \left(s_1+ \frac{3}{4}\right)^2\right\}\\ =& \left( s_2-s_1 \right) - \frac{1}{2}\left\{\left(s_2+\frac{3}{4}\right)^2 - \left(s_1+ \frac{3}{4}\right)^2 \right\}\\ =& \left( s_2-s_1 \right) - \frac{1}{2}\left\{\left({s_2}^2-{s_1}^2\right)+2\times \frac{3}{4}\left(s_2-s_1\right)\right\} =\frac{1}{4}\left( s_2-s_1 \right) - \frac{1}{2}\left({s_2}^2-{s_1}^2\right). \end{aligned}$$

Now, \(d(P,Q)= s_2-s_1\). Therefore,

$$d(P,Q)-\psi(d(P,Q))=d(P,Q)-\frac{1}{2}(d(P,Q))^2=(s_2-s_1)-\frac{1}{2}(s_2-s_1)^2.$$

Since \(0\leq s_1\leq s_2\), we note that \({s_2}^2-{s_1}^2=(s_2+s_1)(s_2-s_1) \geq (s_2-s_1)(s_2-s_1) = (s_2-s_1)^2\). Thus, we have

$$\frac{1}{4}(s_2-s_1)-\frac{1}{2}\left\{{s_2}^2-{s_1}^2\right\}\leq (s_2-s_1)-\frac{1}{2}(s_2-s_1)^2.$$

Therefore, (1) is satisfied. In particular, if \(d(P,Q) <\eta\), then (1) is also satisfied.

Case III: \(P(x(t_1),y(t_1)), Q((x(t_2),y(t_2))\in A)\), \(0\leq t_1\leq t_2\leq \frac{1}{2}\). Then

$$d(T(P),T(Q))=\left\{t_2-\frac{1}{2}{t_2}^2\right\}-\left\{ t_1-\frac{1}{2}{t_1}^2 \right\}=(t_2-t_1)-\frac{1}{2}\left\{{t_2}^2-{t_1}^2\right\}.$$

Also, \(d(P,Q)=t_2-t_1\). Therefore,

$$d(P,Q)-\psi(d(P,Q))=d(P,Q)-\frac{1}{2}(d(P,Q))^2= (t_2-t_1)-\frac{1}{2}(t_2-t_1)^2.$$

Since \(0\leq t_1\leq t_2\), we note that

$${t_2}^2-{t_1}^2=(t_2+t_1)(t_2-t_1) \geq (t_2-t_1)(t_2-t_1) = (t_2-t_1)^2 .$$

Thus, we have

$$(t_2-t_1)-\frac{1}{2}\left\{{t_2}^2-{t_1}^2\right\}\le (t_2-t_1)-\frac{1}{2}(t_2-t_1)^2.$$

Thus, (1) is satisfied. In particular, if \(d(P,Q) <\eta\), then (1) also holds.

Hence, from the above cases we can conclude that the function \(T\) is an \((\eta, \psi )\)- uniformly locally weak contraction for \(\eta = 0.36\). The space \(X\) is \(0.36\)-chainable. Thus, all the conditions of Theorem 1 are satisfied. The point \(({1}/{2},0)\) is a fixed point of \(T\).

Acknowledgments

The first author acknowledges the Indian Institute of Engineering Science of Technology, Shibpur , for supporting him as JRF.