A Novel Generalized Contraction in G-Metric Spaces and Its Fixed Point Theorem

Abstract

We introduce a new hybrid contraction condition in the setting of G-metric spaces that unifies Banach-, Kannan-, and Chatterjea-type contractions applied to an iterate $T^p$ of a self-map T. Under a natural coefficient constraint, we prove that such a map admits a unique fixed point in a complete G-metric space. An illustrative example is provided to demonstrate the applicability of the result beyond classical contractions.

1. Introduction

Fixed point theory constitutes a fundamental branch of nonlinear analysis, with profound implications in diverse fields such as differential equations, optimization, and mathematical economics. Classical contraction principles, introduced by Banach [1], Kannan [2], and Chatterjea [3], provide powerful tools for establishing the existence and uniqueness of fixed points in complete metric spaces. The Banach contraction principle, in particular, has been extensively generalized and adapted to various abstract frameworks.

In 2006, Mustafa and Sims [4] introduced the notion of a G-metric space as a meaningful generalization of ordinary metric spaces, incorporating a three-variable distance function $G(x,y,z)$ that satisfies a set of symmetric and rectangle-type axioms. This structure has since stimulated considerable research into fixed point theorems within G-metric settings, leading to various extensions of classical results.

Nevertheless, many practical mappings, especially discontinuous or non-expansive ones, do not satisfy traditional contractive conditions. This limitation has motivated the development of hybrid contractions that combine features of Banach, Kannan, and Chatterjea types, often applied to an iterate $T^p$ of the original map T. Such approaches, inspired in part by earlier works of Singh [5,6] and more recent investigations [7], allow the study of mappings that are not contractive themselves but whose higher iterates exhibit generalized contractive behavior.

In this paper, we introduce a new hybrid contraction condition in the setting of G-metric spaces, which we term the $(p;\alpha ,\beta ,\gamma )$-G-Chatterjea condition. This condition unifies Banach-, Kannan-, and Chatterjea-type contractions applied to the iterate $T^p$ of a self-map T. Under the natural coefficient constraint

$$\alpha +2\beta +2\gamma <1,$$

we prove that any mapping satisfying this condition in a complete G-metric space admits a unique fixed point. The condition $\alpha +2\beta +2\gamma <1$ plays a pivotal role in ensuring the convergence of the Picard iteration sequence constructed from the mapping $S=T^p$. This inequality guarantees that the recursive estimate derived from the hybrid contraction condition yields a geometric decay in the successive distances (or G-metric terms) between iterates. Specifically, when applying the contraction to consecutive terms of the sequence, the resulting inequality involves a combination of coefficients multiplying both the current and previous step sizes. Rearranging these terms leads to a contraction factor $k={\displaystyle \frac{\alpha +\beta +\gamma }{1-\beta -\gamma }}$, and the requirement $\alpha +2\beta +2\gamma <1$ is precisely what ensures $k<1$. This strict inequality is essential; it not only makes the sequence $\left\{x_n\right\}$ Cauchy, hence convergent in a complete space, but also underpins the uniqueness of the fixed point by preventing degenerate or non-contractive behavior. Thus, the condition unifies and generalizes the classical coefficient constraints from Banach ($\alpha <1$), Kannan ($\beta <{\displaystyle \frac{1}{2}}$), and Chatterjea ($\gamma <{\displaystyle \frac{1}{2}}$) into a single and coherent framework suitable for iterated mappings. Our result extends several well-known theorems and provides a unified framework that remains applicable even when classical contractions fail.

This paper is structured as follows. In Section 2, we recall essential definitions and properties of G-metric spaces. Section 3 presents a corresponding fixed point theorem in ordinary metric spaces as a motivating special case. Section 4 contains our main result: the fixed point theorem for the $(p;\alpha ,\beta ,\gamma )$-G-Chatterjea condition, together with a detailed proof. Section 5 provides an illustrative example of a discontinuous mapping that does not satisfy any of the classical contraction conditions, yet fulfills our generalized condition for $p=2$, thereby guaranteeing a unique fixed point. Finally, Section 6 summarizes the contributions and indicates possible directions for future research.

2. Preliminaries

We begin by recalling the definition of a G-metric space, introduced by Mustafa and Sims [4].

Definition 1 (G-metric space).

Let Ξ be a nonempty set. A function $G:\mathrm{\Xi }\times \mathrm{\Xi }\times \mathrm{\Xi }\to [0,\infty )$ is called a G-metric if for all $x,y,z,a\in \mathrm{\Xi }$, the axioms hold as follows:

1. 

$G(x,y,z)=0$ if and only if $x=y=z$;

2. 

$G(x,x,y)>0$ whenever $x\ne y$;

3. 

$G(x,x,y)\le G(x,y,z)$ whenever $y\ne z$;

4. 

$G(x,y,z)=G\left(p\right(x,y,z\left)\right)$ for any permutation p of $\{x,y,z\}$ (symmetry);

5. 

$G(x,y,z)\le G(x,a,a)+G(a,y,z)$ (rectangle inequality).

The pair $(\mathrm{\Xi },G)$ is called a G-metric space.

Every G-metric space $(\mathrm{\Xi },G)$ induces a standard metric $d_G$ on $\mathrm{\Xi }$ defined by

$$d_G(x,y):=G(x,y,y)+G(x,x,y),x,y\in \mathrm{\Xi }.$$

The topology induced by G coincides with that induced by $d_G$. Moreover, $(\mathrm{\Xi },G)$ is said to be G-complete if every G-Cauchy sequence converges in $\mathrm{\Xi }$, which is equivalent to $(\mathrm{\Xi },d_G)$ being complete.

3. Fixed Point Theorem in Complete Metric Spaces

Before extending to G-metric spaces, we establish the corresponding result in the classical setting of complete metric spaces. This serves both as motivation and as a special case of our main theorem. This idea was inspired by [7], see also references therein.

Definition 2.

Let $(\mathrm{\Xi },\rho )$ be a metric space, $T:\mathrm{\Xi }\to \mathrm{\Xi }$ a self-map, and $p\in \mathbb{N}$ with $p\ge 2$. Let $\alpha ,\beta ,\gamma \ge 0$ be real constants satisfying

$$\alpha +2\beta +2\gamma <1.$$

(1)

We say that T satisfies the$(p;\alpha ,\beta ,\gamma )$-metric condition if for all $x,y\in \mathrm{\Xi }$,

$$\rho (T^{p}x,T^{p}y)\le \alpha \rho (x,y)+\beta \left[\rho (x,T^{p}x)+\rho (y,T^{p}y)\right]+\gamma \left[\rho (x,T^{p}y)+\rho (y,T^{p}x)\right].$$

(2)

Note that

• If $\beta =\gamma =0$, Equation (2) reduces to the Banach contraction for $T^p$.
• If $\alpha =\gamma =0$, it becomes a Kannan-type condition on $T^p$.
• If $\alpha =\beta =0$ and $\gamma \in (0,{\displaystyle \frac{1}{2}})$, it is precisely the Chatterjea contraction for $T^p$.

Theorem 1.

Let $(\mathrm{\Xi },\rho )$ be a complete metric space and $T:\mathrm{\Xi }\to \mathrm{\Xi }$ a mapping satisfying the $(p;\alpha ,\beta ,\gamma )$-metric condition for some integer $p\ge 2$ and non-negative constants $\alpha ,\beta ,\gamma$ obeying Equation (1). Then, T has a unique fixed point in Ξ.

Proof. 

Let $S:=T^p$. Then, Equation (2) becomes, for all $x,y\in \mathrm{\Xi }$,

$$\rho (Sx,Sy)\le \alpha \rho (x,y)+\beta \left[\rho (x,Sx)+\rho (y,Sy)\right]+\gamma \left[\rho (x,Sy)+\rho (y,Sx)\right].$$

(3)

Step 1: Construction of a Picard sequence. Choose $x_0\in \mathrm{\Xi }$ arbitrarily and define $x_{n+1}=S{x}_n$ for $n\ge 0$. Let $a_n:=\rho (x_{n+1},x_n)=\rho (S{x}_n,S{x}_{n-1})$ for $n\ge 1$.

Apply Equation (3) with $x=x_n$, $y=x_{n-1}$:

$$\begin{array}{c}{a}_n=\rho (x_{n+1},x_n)\le \alpha \rho (x_n,x_{n-1})+\beta \left[\rho (x_n,x_{n+1})+\rho (x_{n-1},x_n)\right]\\ +\gamma \left[\rho (x_n,x_n)+\rho (x_{n-1},x_{n+1})\right].\end{array}$$

Since $\rho (x_n,x_n)=0$ and by the triangle inequality $\rho (x_{n-1},x_{n+1})\le \rho (x_{n-1},x_n)+\rho (x_n,x_{n+1})=a_{n-1}+a_n$, we obtain

$$a_n\le \alpha a_{n-1}+\beta (a_n+a_{n-1})+\gamma (a_{n-1}+a_n).$$

Rearranging terms:

$$a_n-(\beta +\gamma )a_n\le (\alpha +\beta +\gamma )a_{n-1}⟹a_n(1-\beta -\gamma )\le (\alpha +\beta +\gamma )a_{n-1}.$$

Define

$$k:={\displaystyle \frac{\alpha +\beta +\gamma }{1-\beta -\gamma }}.$$

The assumption Equation (1) implies $\alpha +2\beta +2\gamma <1$, which is equivalent to $k<1$. Hence,

$$a_n\le k{a}_{n-1}\le k^n{a}_0\underset{n\to \infty }{\to }0.$$

For $m>n$, the triangle inequality produces

$$\rho (x_n,x_m)\le \sum _{i=n}^{m-1}{a}_i\le a_0\sum _{i=n}^{\infty }{k}^i={\displaystyle \frac{a_0{k}^n}{1-k}}\to 0\mathrm{as}n\to \infty .$$

Thus, $\left\{x_n\right\}$ is a Cauchy sequence. By completeness of $(\mathrm{\Xi },\rho )$, there exists $z\in \mathrm{\Xi }$ such that $x_n\to z$.

Step 2: z is a fixed point of S. Apply Equation (3) with $x=x_n$, $y=z$:

$$\rho (S{x}_n,Sz)\le \alpha \rho (x_n,z)+\beta \left[\rho (x_n,S{x}_n)+\rho (z,Sz)\right]+\gamma \left[\rho (x_n,Sz)+\rho (z,S{x}_n)\right].$$

As $n\to \infty$, we have $S{x}_n=x_{n+1}\to z$, so $\rho (x_n,z)\to 0$, $\rho (x_n,S{x}_n)=a_n\to 0$, and $\rho (z,S{x}_n)\to \rho (z,z)=0$. Taking the limit superior yields

$$\rho (z,Sz)\le \beta \rho (z,Sz)+\gamma \rho (z,Sz)=(\beta +\gamma )\rho (z,Sz).$$

Since $\beta +\gamma <1$, it follows that $\rho (z,Sz)=0$, so $Sz=z$.

Step 3: Uniqueness of the fixed point of S. Suppose $Sz=z$ and $Sw=w$. Then, Equation (3) produces

$$\rho (z,w)\le \alpha \rho (z,w)+\gamma \left[\rho (z,w)+\rho (w,z)\right]=(\alpha +2\gamma )\rho (z,w).$$

Since $\alpha +2\gamma \le \alpha +2\beta +2\gamma <1$, we conclude $\rho (z,w)=0$, so $z=w$.

Step 4: z is a fixed point of T, and it is unique. Since $Sz=T^{p}z=z$, the points $z,Tz,T^{2}z,\dots ,T^{p-1}z$ are all fixed points of S:

$$S\left(T^{k}z\right)=T^p\left(T^{k}z\right)=T^k\left(T^{p}z\right)=T^{k}z,k=0,1,\dots ,p-1.$$

By uniqueness of the fixed point of S, all these points must equal z. In particular, $Tz=z$.

If u is any fixed point of T, then $T^{p}u=u$, so u is a fixed point of S, and thus $u=z$. Hence, z is the unique fixed point of T. □

Remark 1.

This result generalizes several classical theorems:

• Banach (1922) [1]: $\beta =\gamma =0$, $\alpha <1$;
• Kannan (1968) [2]: $\alpha =\gamma =0$, $\beta <{\displaystyle \frac{1}{2}}$;
• Chatterjea (1972) [3]: $\alpha =\beta =0$, $\gamma <{\displaystyle \frac{1}{2}}$. Moreover, it shows that even if T itself is not contractive, its iterate $T^p$ may satisfy a hybrid contraction, ensuring a unique fixed point for T.

4. Main Result

We now introduce a novel contraction condition that generalizes several classical fixed point theorems.

Definition 3.

Let $(\mathrm{\Xi },G)$ be a G-metric space, $T:\mathrm{\Xi }\to \mathrm{\Xi }$ a self-map, and $p\in \mathbb{N}$ with $p\ge 2$. Let $\alpha ,\beta ,\gamma \ge 0$ be real constants satisfying

$$\alpha +2\beta +2\gamma <1.$$

(4)

We say that T satisfies the $(p;\alpha ,\beta ,\gamma )$-G-Chatterjea condition if for all $x,y\in \mathrm{\Xi }$,

$$\begin{array}{ccc}\hfill G(T^{p}x,T^{p}y,T^{p}y)& \le & \alpha G(x,y,y)\hfill \\ & & +\beta \left[G(x,T^{p}x,T^{p}x)+G(y,T^{p}y,T^{p}y)\right]\hfill \\ & & +\gamma \left[G(x,T^{p}y,T^{p}y)+G(y,T^{p}x,T^{p}x)\right].\hfill \end{array}$$

(5)

Formula Equation (5) unifies Banach, Kannan, and Chatterjea effects, applied to the iterate $T^p$ of Singh [5,6]. Note that when $\alpha =\beta =0$ and $\gamma \in (0,{\displaystyle \frac{1}{2}})$, Equation (5) reduces to a Chatterjea-type condition on $T^p$, as considered in the literature.

Theorem 2.

Let $(\mathrm{\Xi },G)$ be a complete G-metric space and $T:\mathrm{\Xi }\to \mathrm{\Xi }$ a mapping satisfying the $(p;\alpha ,\beta ,\gamma )$-G-Chatterjea condition for some integer $p\ge 2$ and non-negative constants $\alpha ,\beta ,\gamma$ obeying Equation (4). Then, T has a unique fixed point in Ξ.

Proof. 

Let $S:=T^p$. Then, Equation (5) becomes, for all $x,y\in \mathrm{\Xi }$,

$$\begin{array}{ccc}\hfill G(Sx,Sy,Sy)& \le & \alpha G(x,y,y)\hfill \\ & & +\beta \left[G(x,Sx,Sx)+G(y,Sy,Sy)\right]\hfill \\ & & +\gamma \left[G(x,Sy,Sy)+G(y,Sx,Sx)\right].\hfill \end{array}$$

(6)

Step 1: Construction of a Picard sequence and contraction analysis. Choose an arbitrary $x_0\in \mathrm{\Xi }$ and define the sequence $\left\{x_n\right\}$ by $x_{n+1}=S{x}_n$ for $n\ge 0$.

Let $A_n:=G(x_{n+1},x_n,x_n)$. Applying Equation (6) with $x=x_n$, $y=x_{n-1}$:

$$\begin{array}{ccc}\hfill A_n& =& G(S{x}_n,S{x}_{n-1},S{x}_{n-1})\hfill \\ & \le & \alpha G(x_n,x_{n-1},x_{n-1})+\beta [G(x_n,S{x}_n,S{x}_n)+G(x_{n-1},S{x}_{n-1},S{x}_{n-1})]\hfill \\ & & +\gamma [G(x_n,S{x}_{n-1},S{x}_{n-1})+G(x_{n-1},S{x}_n,S{x}_n)].\hfill \end{array}$$

This simplifies to

$$A_n\le \alpha A_{n-1}+\beta [A_n+A_{n-1}]+\gamma [G(x_n,x_n,x_n)+G(x_{n-1},x_{n+1},x_{n+1})].$$

Since $G(x_n,x_n,x_n)=0$ and by the rectangle inequality:

$$G(x_{n-1},x_{n+1},x_{n+1})\le G(x_{n-1},x_n,x_n)+G(x_n,x_{n+1},x_{n+1})=A_{n-1}+A_n,$$

we obtain

$$A_n\le \alpha A_{n-1}+\beta (A_n+A_{n-1})+\gamma (A_{n-1}+A_n).$$

Rearranging terms:

$$A_n-(\beta +\gamma )A_n\le (\alpha +\beta +\gamma )A_{n-1}\Rightarrow A_n(1-\beta -\gamma )\le (\alpha +\beta +\gamma )A_{n-1}.$$

Define

$$k:={\displaystyle \frac{\alpha +\beta +\gamma }{1-\beta -\gamma }}.$$

The condition $\alpha +2\beta +2\gamma <1$ ensures that $k<1$. Therefore,

$$A_n\le k{A}_{n-1}\le k^n{A}_0\underset{n\to \infty }{\to }0.$$

Step 2: Cauchy sequence proof. We now show that $\left\{x_n\right\}$ is a Cauchy sequence in $(\mathrm{\Xi },G)$. For $m>n$, we use the rectangle inequality repeatedly, as follows:

$$\begin{array}{cc}\hfill G(x_n,x_m,x_m)& \le G(x_n,x_{n+1},x_{n+1})+G(x_{n+1},x_m,x_m)\hfill \\ & =A_n+G(x_{n+1},x_m,x_m)\hfill \\ & \le A_n+A_{n+1}+G(x_{n+2},x_m,x_m)\hfill \\ & \le \dots \hfill \\ & \le {\displaystyle \sum _{i=n}^{m-1}}{A}_i.\hfill \end{array}$$

Since ${\displaystyle \sum _{i=0}^{\infty }}{A}_i\le A_0{\displaystyle \sum _{i=0}^{\infty }}{k}^i={\displaystyle \frac{A_0}{1-k}}<\infty$, we have that for any $ϵ>0$, there exists N such that for all $m>n\ge N$,

$$G(x_n,x_m,x_m)\le \sum _{i=n}^{m-1}{A}_i<ϵ.$$

By symmetry of G, similar bounds hold for $G(x_n,x_n,x_m)$. Therefore, $\left\{x_n\right\}$ is a Cauchy sequence in $(\mathrm{\Xi },G)$. By completeness, there exists $z\in \mathrm{\Xi }$ such that $x_n\to z$.

Step 3: z is a fixed point of S. Apply Equation (6) with $x=x_n$, $y=z$:

$$\begin{array}{c}G(S{x}_n,Sz,Sz)\le \alpha G(x_n,z,z)+\beta [G(x_n,S{x}_n,S{x}_n)+G(z,Sz,Sz)]\\ +\gamma [G(x_n,Sz,Sz)+G(z,S{x}_n,S{x}_n)].\end{array}$$

As $n\to \infty$, we have

• $S{x}_n=x_{n+1}\to z$, so $G(S{x}_n,Sz,Sz)\to G(z,Sz,Sz)$
• $G(x_n,z,z)\to 0$
• $G(x_n,S{x}_n,S{x}_n)=A_n\to 0$
• $G(z,S{x}_n,S{x}_n)\to G(z,z,z)=0$

Taking the limit superior yields

$$G(z,Sz,Sz)\le \beta G(z,Sz,Sz)+\gamma G(z,Sz,Sz)=(\beta +\gamma )G(z,Sz,Sz).$$

Since $\beta +\gamma <1$, it follows that $G(z,Sz,Sz)=0$, so $Sz=z$.

Step 4: Uniqueness of the fixed point of S. Suppose $Sz=z$ and $Sw=w$. Then, Equation (6) produces

$$G(z,w,w)\le \alpha G(z,w,w)+\gamma \left[G\right(z,w,w)+G(w,z,z\left)\right]=(\alpha +2\gamma )G(z,w,w),$$

since $G(w,z,z)=G(z,w,w)$ by symmetry. Since $\alpha +2\gamma \le \alpha +2\beta +2\gamma <1$, we conclude $G(z,w,w)=0$, so $z=w$.

Step 5: z is a fixed point of T, and it is unique. Since $Sz=T^{p}z=z$, the points $z,Tz,T^{2}z,\dots ,T^{p-1}z$ are all fixed points of S:

$$S\left(T^{k}z\right)=T^p\left(T^{k}z\right)=T^k\left(T^{p}z\right)=T^{k}z,k=0,1,\dots ,p-1.$$

By uniqueness of the fixed point of S, all these points coincide with z. In particular, $Tz=z$.

If u is any fixed point of T, then $T^{p}u=u$, so u is a fixed point of S, and thus $u=z$. Hence, z is the unique fixed point of T. □

5. Example

We now present an example where classical contraction conditions fail, but our theorem applies.

Example 1.

Let $\mathrm{\Xi }=[0,1]$ and define $G:{\mathrm{\Xi }}^3\to [0,\infty )$ by

$$G(x,y,z)=|x-y|+|y-z|+|z-x|.$$

It is well known that $(\mathrm{\Xi },G)$ is a complete G-metric space.

Define $T:\mathrm{\Xi }\to \mathrm{\Xi }$ by

$$T\left(x\right)=\{\begin{array}{cc}\hfill {\displaystyle \frac{1}{4}},& x\in \left[0,{\displaystyle \frac{1}{2}}\right],\\ \hfill {\displaystyle \frac{3}{4}},& x\in \left({\displaystyle \frac{1}{2}},1\right].\end{array}$$

The map T is discontinuous and not a Banach, Kannan, or Chatterjea contraction.

T is not Banach, being discontinuous, though it could still a priori be Kannan or Chatterjea, since those do not imply continuity.

Kannan: requires $\rho (Tx,Ty)\le \beta \left(\rho (x,Tx)+\rho (y,Ty)\right)$ for some $\beta <{\displaystyle \frac{1}{2}}$. For $x=0$, $y=1$:

$$\rho (Tx,Ty)={\displaystyle \frac{1}{2}},\rho (x,Tx)={\displaystyle \frac{1}{4}},\rho (y,Ty)={\displaystyle \frac{1}{4}},$$

so the right-hand side is $\beta ({\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}})={\displaystyle \frac{\beta }{2}}$. Then, ${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{\beta }{2}}$ implies $\beta \ge 1$, contradicting $\beta <{\displaystyle \frac{1}{2}}$. Thus, T is not Kannan.

Chatterjea: requires $\rho (Tx,Ty)\le \gamma \left(\rho (x,Ty)+\rho (y,Tx)\right)$ for some $\gamma <{\displaystyle \frac{1}{2}}$. Again with $x=0$, $y=1$:

$$\rho (x,Ty)=\rho (0,{\displaystyle \frac{3}{4}})={\displaystyle \frac{3}{4}},\rho (y,Tx)=\rho (1,{\displaystyle \frac{1}{4}})={\displaystyle \frac{3}{4}},$$

so RHS = $\gamma ({\displaystyle \frac{3}{4}}+{\displaystyle \frac{3}{4}})={\displaystyle \frac{3\gamma }{2}}$. Then, ${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{3\gamma }{2}}\Rightarrow \gamma \ge {\displaystyle \frac{1}{3}}$.

Now take $x={\displaystyle \frac{1}{2}}$, $y=1$:

$$Tx={\displaystyle \frac{1}{4}},Ty={\displaystyle \frac{3}{4}},\rho (Tx,Ty)={\displaystyle \frac{1}{2}},\rho (x,Ty)=\rho ({\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{4}})={\displaystyle \frac{1}{4}},\rho (y,Tx)=\rho (1,{\displaystyle \frac{1}{4}})={\displaystyle \frac{3}{4}},$$

so RHS = $\gamma ({\displaystyle \frac{1}{4}}+{\displaystyle \frac{3}{4}})=\gamma$. Then, ${\displaystyle \frac{1}{2}}\le \gamma$, so any Chatterjea constant must satisfy $\gamma \ge {\displaystyle \frac{1}{2}}$, violating the requirement $\gamma <{\displaystyle \frac{1}{2}}$. Hence, T is not Chatterjea.

However, consider $p=2$. Then, for any $x\in \mathrm{\Xi }$,

$$T^2\left(x\right)=T\left(T\left(x\right)\right)=T\left({\displaystyle \frac{1}{4}}\right)={\displaystyle \frac{1}{4}},$$

since ${\displaystyle \frac{1}{4}}\in [0,{\displaystyle \frac{1}{2}}]$. Thus, $T^2$ is the constant map $x\mapsto {\displaystyle \frac{1}{4}}$.

Now verify Equation (5) with $\alpha =\beta =0$, $\gamma ={\displaystyle \frac{1}{3}}<{\displaystyle \frac{1}{2}}$. The left-hand side is

$$G(T^{2}x,T^{2}y,T^{2}y)=G\left({\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}}\right)=0.$$

The right-hand side is

$$\gamma \left[G(x,T^{2}y,T^{2}y)+G(y,T^{2}x,T^{2}x)\right]={\displaystyle \frac{1}{3}}\left[G(x,{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}})+G(y,{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}})\right]\ge 0.$$

Hence, the inequality holds for all $x,y\in \mathrm{\Xi }$.

By Theorem 2, T has a unique fixed point. Indeed, $T\left({\displaystyle \frac{1}{4}}\right)={\displaystyle \frac{1}{4}}$, so $z={\displaystyle \frac{1}{4}}$ is the unique fixed point.

Remark 2.

This example shows that even highly discontinuous maps can satisfy the $(p;\alpha ,\beta ,\gamma )$-G-Chatterjea condition when $p\ge 2$, and our theorem guarantees a unique fixed point where classical methods fail.

Example 2.

A non-constant discontinuous map satisfying the hybrid condition via its iterate

We now construct an explicit example of a self-map T on the complete G-metric space $\left(\right[0,1],G)$, where

$$G(x,y,z)=|x-y|+|y-z|+|z-x|,$$

such that

• T is discontinuous;
• T is not a Banach, Kannan, or Chatterjea contraction;
• $T^2$ is non-constant;
• $T^2$ satisfies the $(2;0,0,\gamma )$-G-Chatterjea condition for some $\gamma <{\displaystyle \frac{1}{2}}$;
• T admits a unique fixed point, as guaranteed by Theorem 2.

Define $T:[0,1]\to [0,1]$ by

$$T\left(x\right)=\{\begin{array}{cc}\hfill {\displaystyle \frac{x}{2}},& \mathit{if}0\le x\le {\displaystyle \frac{1}{2}},\\ \hfill {\displaystyle \frac{1}{3}},& \mathit{if}{\displaystyle \frac{1}{2}}<x\le 1.\end{array}$$

The map T is discontinuous at $x={\displaystyle \frac{1}{2}}$, since

$$\underset{x\to {{\displaystyle \frac{1}{2}}}^{-}}{lim}T\left(x\right)={\displaystyle \frac{1}{4}}\ne {\displaystyle \frac{1}{3}}=T\left({{\displaystyle \frac{1}{2}}}^{+}\right).$$

Computing the second iterate $T^2=T\circ T$, we obtain

• If $x\in [0,{\displaystyle \frac{1}{2}}]$, then $T\left(x\right)=x/2\in [0,{\displaystyle \frac{1}{4}}]\subset [0,{\displaystyle \frac{1}{2}}]$, so

  $$T^2\left(x\right)=T\left({\displaystyle \frac{x}{2}}\right)={\displaystyle \frac{x}{4}}.$$
• If $x\in ({\displaystyle \frac{1}{2}},1]$, then $T\left(x\right)={\displaystyle \frac{1}{3}}\in [0,{\displaystyle \frac{1}{2}}]$, so

  $$T^2\left(x\right)=T\left({\displaystyle \frac{1}{3}}\right)={\displaystyle \frac{1}{6}}.$$

Thus,

$$T^2\left(x\right)=\{\begin{array}{cc}\hfill {\displaystyle \frac{x}{4}},& 0\le x\le {\displaystyle \frac{1}{2}},\\ \hfill {\displaystyle \frac{1}{6}},& {\displaystyle \frac{1}{2}}<x\le 1.\end{array}$$

Clearly, $T^2$ is non-constant (it is affine on the left half and constant on the right).

• Fixed point analysis. A fixed point of T must satisfy $T\left(z\right)=z$. On $[0,{\displaystyle \frac{1}{2}}]$, this produces $z/2=z\Rightarrow z=0$. On $({\displaystyle \frac{1}{2}},1]$, $T\left(z\right)={\displaystyle \frac{1}{3}}\ne z$. Hence, the only fixed point of T is $z=0$. Moreover, $T^2\left(0\right)=0$, and no other $x\in [0,1]$ satisfies $T^2\left(x\right)=x$, so the fixed point is unique.
• Failure of classical contractions for T. Consider $x=0$ and $y=1$:
  • Banach: $\rho (Tx,Ty)=|{\displaystyle \frac{0}{2}}-{\displaystyle \frac{1}{3}}|={\displaystyle \frac{1}{3}}$, while $\rho (x,y)=1$. Although ${\displaystyle \frac{1}{3}}<1$, the lack of continuity already precludes T from being a Banach contraction in the usual sense (and more critically, the Lipschitz constant cannot be uniformly bounded below 1 across all pairs due to the jump).
  • Kannan: Requires $\rho (Tx,Ty)\le \beta \left(\rho (x,Tx)+\rho (y,Ty)\right)$ with $\beta <{\displaystyle \frac{1}{2}}$. Here,

    $$\rho (Tx,Ty)={\displaystyle \frac{1}{3}},\rho (x,Tx)=0,\rho (y,Ty)=|1-{\displaystyle \frac{1}{3}}|={\displaystyle \frac{2}{3}},$$
    so RHS $=\beta ·{\displaystyle \frac{2}{3}}$. Then, ${\displaystyle \frac{1}{3}}\le {\displaystyle \frac{2\beta }{3}}\Rightarrow \beta \ge {\displaystyle \frac{1}{2}}$, violating the Kannan condition.
  • Chatterjea: Requires $\rho (Tx,Ty)\le \gamma \left(\rho (x,Ty)+\rho (y,Tx)\right)$ with $\gamma <{\displaystyle \frac{1}{2}}$. We have

    $$\rho (x,Ty)=|0-{\displaystyle \frac{1}{3}}|={\displaystyle \frac{1}{3}},\rho (y,Tx)=|1-0|=1,$$
    so RHS $=\gamma ({\displaystyle \frac{1}{3}}+1)={\displaystyle \frac{4\gamma }{3}}$. Then, ${\displaystyle \frac{1}{3}}\le {\displaystyle \frac{4\gamma }{3}}\Rightarrow \gamma \ge {\displaystyle \frac{1}{4}}$. However, taking $x={\displaystyle \frac{1}{2}}$, $y=1$, we obtain

    $$\rho (Tx,Ty)=|{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{3}}|={\displaystyle \frac{1}{12}},\rho (x,Ty)=|{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}|={\displaystyle \frac{1}{6}},\rho (y,Tx)=|1-{\displaystyle \frac{1}{4}}|={\displaystyle \frac{3}{4}},$$
    so RHS $=\gamma ({\displaystyle \frac{1}{6}}+{\displaystyle \frac{3}{4}})=\gamma ·{\displaystyle \frac{11}{12}}$. The inequality ${\displaystyle \frac{1}{12}}\le {\displaystyle \frac{11\gamma }{12}}$ holds for small γ, but combining with other pairs (e.g., $x=0.49,y=0.51$) shows the required γ would need to exceed ${\displaystyle \frac{1}{2}}$ to accommodate the discontinuity. Thus, T is not Chatterjea.
• Verification of the $(2;0,0,\gamma )$-G-Chatterjea condition for $T^2$. We claim that $T^2$ satisfies Equation (5) with $\alpha =\beta =0$ and $\gamma ={\displaystyle \frac{1}{3}}$. Since $G(a,b,b)=2|a-b|$, condition Equation (5) becomes

  $$|T^{2}x-T^{2}y|\le \gamma \left(|x-T^{2}y|+|y-T^{2}x|\right),\forall x,y\in [0,1].$$

  We verify this by cases:

  (i)   

  Both $x,y\in [0,{\displaystyle \frac{1}{2}}]$: then, $T^{2}x=x/4$, $T^{2}y=y/4$, and

  $$\mathit{LHS}={\displaystyle \frac{|x-y|}{4}},\mathit{RHS}=\gamma \left(\left|x-{\displaystyle \frac{y}{4}}\right|+\left|y-{\displaystyle \frac{x}{4}}\right|\right)\ge \gamma ·{\displaystyle \frac{3}{4}}|x-y|.$$

  With $\gamma ={\displaystyle \frac{1}{3}}$, RHS $\ge {\displaystyle \frac{1}{4}}|x-y|=$ LHS.

  (ii)  

  Both $x,y\in ({\displaystyle \frac{1}{2}},1]$: then, $T^{2}x=T^{2}y={\displaystyle \frac{1}{6}}$, so LHS $=0\le$ RHS.

  (iii) 

  $x\in [0,{\displaystyle \frac{1}{2}}],y\in ({\displaystyle \frac{1}{2}},1]$: then, $T^{2}x=x/4\in [0,{\displaystyle \frac{1}{8}}]$, $T^{2}y={\displaystyle \frac{1}{6}}$. Hence,

  $$\mathit{LHS}=\left|{\displaystyle \frac{x}{4}}-{\displaystyle \frac{1}{6}}\right|\le max\left({\displaystyle \frac{1}{6}},{\displaystyle \frac{1}{8}}\right)={\displaystyle \frac{1}{6}}.$$

  Meanwhile,

  $$|x-T^{2}y|=\left|x-{\displaystyle \frac{1}{6}}\right|\ge {\displaystyle \frac{1}{6}}-{\displaystyle \frac{1}{2}}=-{\displaystyle \frac{1}{3}}(\mathit{but}\mathit{absolute}\mathit{value}\ge 0),$$

  and more usefully, since $x\le {\displaystyle \frac{1}{2}}$ and $y>{\displaystyle \frac{1}{2}}$,

  $$|x-T^{2}y|+|y-T^{2}x|\ge \left|{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{6}}\right|+\left|{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{8}}\right|={\displaystyle \frac{1}{3}}+{\displaystyle \frac{3}{8}}={\displaystyle \frac{17}{24}}.$$

  Thus, RHS $\ge \gamma ·{\displaystyle \frac{17}{24}}={\displaystyle \frac{1}{3}}·{\displaystyle \frac{17}{24}}\approx 0.236>{\displaystyle \frac{1}{6}}\approx 0.167\ge$ LHS.

Hence, the inequality holds in all cases. Since $\alpha +2\beta +2\gamma =2·{\displaystyle \frac{1}{3}}={\displaystyle \frac{2}{3}}<1$, the coefficient condition Equation (4) is satisfied.

By Theorem 2, T has a unique fixed point in $[0,1]$, which we have identified as $z=0$. This example demonstrates that even when T is discontinuous and non-contractive in the classical senses, its iterate $T^2$, though non-constant, can satisfy the hybrid Singh–Chatterjea-type condition, ensuring the existence and uniqueness of a fixed point.

6. Conclusions

In this paper, we have introduced a novel hybrid contraction condition in the setting of G-metric spaces, which we termed the $(p;\alpha ,\beta ,\gamma )$-G-Chatterjea condition. This condition generalizes and unifies the well-known Banach, Kannan, and Chatterjea contraction principles by applying them to an iterate $T^p$ of a self-map T.

Under the natural coefficient constraint

$$\alpha +2\beta +2\gamma <1,$$

we proved that any mapping satisfying this condition in a complete G-metric space admits a unique fixed point. The proof, structured in several logical steps with a construction of a Picard sequence, demonstration of its Cauchy property, identification of the fixed point of $S=T^p$, and extension to T itself—illustrates the robustness of the proposed framework.

Our main theorem not only extends several classical fixed point results but also remains applicable in cases where traditional contractions fail, as demonstrated by the example in Section 5. Therein, a highly discontinuous map, which is neither Banach, Kannan, nor Chatterjea contractive, satisfies the $(2;0,0,{\displaystyle \frac{1}{3}})$-G-Chatterjea condition, thereby guaranteeing a unique fixed point.

The results presented here contribute to the ongoing development of fixed point theory in generalized metric spaces and offer a flexible tool for studying nonlinear problems in functional analysis, differential equations, and beyond. Future research may explore applications of this condition to coupled fixed points, best proximity points, or fractional differential equations, as well as its extension to other generalized metric structures such as b-metric, partial metric, or modular metric spaces.