Common fixed points of R-weakly commuting maps in generalized metric spaces

Abstract

In this paper, using the setting of a generalized metric space, a unique common fixed point of four R-weakly commuting maps satisfying a generalized contractive condition is obtained. We also present example in support of our result.

2000 MSC: 54H25; 47H10; 54E50.

1 Introduction and preliminaries

The study of unique common fixed points of mappings satisfying certain contractive conditions has been at the center of rigorous research activity. Mustafa and Sims [1] generalized the concept of a metric, in which the real number is assigned to every triplet of an arbitrary set. Based on the notion of generalized metric spaces, Mustafa et al. [2–6] obtained some fixed point theorems for mappings satisfying different contractive conditions. Study of common fixed point theorems in generalized metric spaces was initiated by Abbas and Rhoades [7]. Abbas et al. [8] obtained some periodic point results in generalized metric spaces. While, Chugh et al. [9] obtained some fixed point results for maps satisfying property p in G-metric spaces. Saadati et al. [10] studied some fixed point results for contractive mappings in partially ordered G-metric spaces. Recently, Shatanawi [11] obtained fixed points of Φ-maps in G-metric spaces. Abbas et al. [12] gave some new results of coupled common fixed point results in two generalized metric spaces (see also [13]).

The aim of this paper is to initiate the study of unique common fixed point of four R-weakly commuting maps satisfying a generalized contractive condition in G-metric spaces.

Consistent with Mustafa and Sims [2], the following definitions and results will be needed in the sequel.

Definition 1.1. Let X be a nonempty set. Suppose that a mapping G :

X × X × X → R⁺ satisfies:

G₁ : G(x, y, z) = 0 if x = y = z;

G₂ : 0 < G(x, y, z) for all x, y, z ∈ X, with x ≠ y;

G₃ : G(x, x, y) ≤ G(x, y, z) for all x, y, z ∈ X, with y ≠ z;

G₄ : G(x, y, z) = G(x, z, y) = G(y, z, x) = ··· (symmetry in all three variables); and

G₅ : G(x, y, z) ≤ G(x, a, a) + G(a, y, z) for all x, y, z, a ∈ X.

Then G is called a G-metric on X and (X, G) is called a G-metric space.

Definition 1.2. A sequence {x _n } in a G-metric space X is:

• (i) a G-Cauchy sequence if, for any ε > 0, there is an n ₀ ∈ N (the set of natural numbers) such that for all n, m, l ≥ n ₀, G(x _n , x _m , x _l ) < ε,

• (ii) a G-convergent sequence if, for any ε > 0, there is an x ∈ X and an n ₀ ∈ N, such that for all n, m ≥ n ₀, G(x, x _n , x _m ) < ε.

A G-metric space on X is said to be G-complete if every G-Cauchy sequence in X is G-convergent in X. It is known that → 0 as n, m → ∞.

Proposition 1.3. Let X be a G-metric space. Then the following are equivalent:

• (1) {x _n } is G-convergent to x.

• (2) G(x _n , x _m , x) → 0 as n, m → ∞.

• (3) G(x _n , x _n , x) → 0 as n → ∞.

• (4) G(x _n , x, x) → 0 as n → ∞.

Definition 1.4. A G-metric on X is said to be symmetric if G(x, y, y) = G(y, x, x) for all x, y ∈ X.

Proposition 1.5. Every G-metric on X will define a metric d _G on X by

$$d_G\left(x,y\right)=G\left(x,y,y\right)+G\left(y,x,x\right),\forall x,y\in X.$$

(1.1)

For a symmetric G-metric,

$$d_G\left(x,y\right)=2G\left(x,y,y\right),\forall x,y\in X.$$

(1.2)

However, if G is non-symmetric, then the following inequality holds:

$$\frac{3}{2}G\left(x,y,y\right)\le d_G\left(x,y\right)\le 3G\left(x,y,y\right),\forall x,y\in X.$$

(1.3)

It is also obvious that

$$G\left(x,x,y\right)\le 2G\left(x,y,y\right).$$

Now, we give an example of a non-symmetric G-metric.

Example 1.6. Let X = {1, 2} and a mapping G : X × X × X → R⁺ be defined as

$$\begin{array}{cc}\hfill \left(x,y,z\right)\hfill & \hfill G\left(x,y,z\right)\hfill \\ \hfill \left(1,1,1\right),\left(2,2,2\right)\hfill & \hfill 0\hfill \\ \hfill \left(1,1,2\right),\left(1,2,1\right),\left(2,1,1\right)\hfill & \hfill 0.5\hfill \\ \hfill \left(1,2,2\right),\left(2,1,2\right),\left(2,2,1\right)\hfill & \hfill 1.\hfill \end{array}$$

Note that G satisfies all the axioms of a generalized metric but G(x, x, y) ≠ G(x, y, y) for distinct x, y in X. Therefore, G is a non-symmetric G-metric on X.

In 1999, Pant [14] introduced the concept of weakly commuting maps in metric spaces. We shall study R-weakly commuting and compatible mappings in the frame work of G-metric spaces.

Definition 1.7. Let X be a G-metric space and f and g be two self-mappings of X. Then f and g are called R-weakly commuting if there exists a positive real number R such that G(fgx, fgx, gfx) ≤ RG(fx, fx, gx) holds for each x ∈ X.

Two maps f and g are said to be compatible if, whenever {x _n } in X such that {fx _n } and {gx _n } are G-convergent to some t ∈ X, then lim_n→∞G(fgx _n , fgx _n , gfx _n ) = 0.

Example 1.8. Let X = [0, 2] with complete G-metric defined by

$$G\left(x,y,z\right)=max\left\{|x-y|,|x-z|,|y-z|\right\}.$$

Let f, g, S, T : X → X defined by

$$\begin{array}{lll}\hfill fx& =1,x\ge 0,& \hfill \text{(1)}\\ \hfill gx& =\left\{\left.\begin{array}{c}\hfill 1,x\in \left[0,1\right],\hfill \\ \hfill \frac{2-x}{2},x\in \left(1,2\right],\hfill \end{array}\right|\right.,& \hfill \text{(2)}\\ \hfill Sx& =\left\{\left.\begin{array}{c}\hfill 2-x,x\in \left[0,1\right],\hfill \\ \hfill x,x\in \left(1,2\right],\hfill \end{array}\right|\right.,& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

and

$$Tx=\left\{\left.\begin{array}{c}\hfill \frac{3-x}{2},x\in \left[0,1\right],\hfill \\ \hfill \frac{x}{2},x\in \left(1,2\right],\hfill \end{array}\right|\right..$$

Then note that the pairs {f, S} and {g, T} are R-weakly commuting as they commute at their coincidence points. The pair {f, S} is continuous compatible while the pair {g, T} is non-compatible. To see that g and T are non-compatible, consider a decreasing sequence {x _n } in X such that x _n → 1. Then $g{x}_n\to \frac{1}{2}$, $T{x}_n\to \frac{1}{2}$. $gT{x}_n=\frac{4-x_n}{4}\to \frac{3}{4}$ and $Tg{x}_n=\frac{2-x_n}{4}\to \frac{1}{4}$. □

2 Common fixed point theorems

In this section, we obtain some unique common fixed point results for four mappings satisfying certain generalized contractive conditions in the framework of a generalized metric space. We start with the following result.

Theorem 2.1. Let X be a complete G-metric space. Suppose that {f, S} and {g, T} be pointwise R-weakly commuting pairs of self-mappings on X satisfying

$$\begin{array}{c}G(fx,fx,gy)\le h\max \{G(Sx,Sx,Ty),G(fx,fx,Sx),G(gy,gy,Ty),\\ [G(fx,fx,Ty)+G(gy,gy,Sx)]/2\}\end{array}$$

(2.1)

and

$$\begin{array}{c}G(fx,gy,gy)\le h\max \{G(Sx,Ty,Ty),G(fx,Sx,Sx),G(gy,Ty,Ty),\\ [G(fx,Ty,Ty)+G(gy,Sx,Sx)]/2\}\end{array}$$

(2.2)

for all x, y ∈ X, where h ∈ [0, 1). Suppose that fX ⊆ TX, gX ⊆ SX, and one of the pair {f, S} or {g, T} is compatible. If the mappings in the compatible pair are continuous, then f, g, S and T have a unique common fixed point.

Proof. Suppose that f and g satisfy the conditions (2.1) and (2.2). If G is symmetric, then by adding these, we have

$$\begin{array}{c}{d}_G\left(fx,gy\right)\\ \le \frac{h}{2}max\left\{d_G\left(Sx,Ty\right),d_G\left(fx,Sx\right),d_G\left(gy,Ty\right),\left[d_G\left(fx,Ty\right)+d_G\left(gy,Sx\right)\right]∕2\right\}\\ +\frac{h}{2}max\left\{d_G\left(Sx,Ty\right),d_G\left(fx,Sx\right),d_G\left(gy,Ty\right),\left[d_G\left(fx,Ty\right)+d_G\left(gy,Sx\right)\right]∕2\right\}\\ =hmax\left\{d_G\left(Sx,Ty\right),d_G\left(fx,Sx\right),d_G\left(gy,Ty\right),\left[d_G\left(fx,Ty\right)+d_G\left(gy,Sx\right)\right]∕2\right\},\end{array}$$

for all x, y ∈ X with 0 ≤ h < 1, the existence and uniqueness of a common fixed point follows from [14]. However, if X is non-symmetric G-metric space, then by the definition of metric d _G on X and (1.3), we obtain

$$\begin{array}{c}{d}_G\left(fx,gy\right)\\ =G\left(fx,fx,gy\right)+G\left(fx,gy,gy\right)\\ \le \frac{2h}{3}max\left\{d_G\left(Sx,Ty\right),d_G\left(fx,Sx\right),d_G\left(gy,Ty\right),\left[d_G\left(fx,Ty\right)+d_G\left(gy,Sx\right)\right]∕2\right\}\\ +\frac{2h}{3}max\left\{d_G\left(Sx,Ty\right),d_G\left(fx,Sx\right),d_G\left(gy,Ty\right),\left[d_G\left(fx,Ty\right)+d_G\left(gy,Sx\right)\right]∕2\right\}\\ =\frac{4h}{3}max\left\{d_G\left(Sx,Ty\right),d_G\left(fx,Sx\right),d_G\left(gy,Ty\right),\left[d_G\left(fx,Ty\right)+d_G\left(gy,S_X\right)\right]∕2\right\},\end{array}$$

for all x, y ∈ X. Here, the contractivity factor $\frac{4h}{3}$ needs not be less than 1. Therefore, metric d _G gives no information. In this case, let x₀ be an arbitrary point in X. Choose x₁ and x₂ in X such that gx₀ = Sx₁ and fx₁ = Tx₂. This can be done, since the ranges of S and T contain those of g and f, respectively. Again choose x₃ and x₄ in X such that gx₂ = Sx 3 and fx₃ = Tx₄. Continuing this process, having chosen x _n in X such that gx_2n= Sx_2n+1and fx_2n+1= Tx_2n+2, n = 0, 1, 2, .... Let

$$y_{2n}=S{x}_{2n+1}=g{x}_{2n}and{y}_{2n+1}=T{x}_{2n+2}=f{x}_{2n+1}foralln=0,1,2,\dots .$$

For a given n ∈ N, if n is even, so n = 2k for some k ∈ N. Then from (2.1)

$$\begin{array}{l}G(y_{n+1},y_{n+1},y_n)\\ =G(y_{2k+1},y_{2k+1},y_{2k})\\ =G(f{x}_{2k+1},f{x}_{2k+1},g{x}_{2k})\\ \le h\max \{G(S{x}_{2k+1},S{x}_{2k+1},T{x}_{2k}),G(f{x}_{2k+1},f{x}_{2k+1},S{x}_{2k+1}),\\ G(g{x}_{2k},g{x}_{2k},T{x}_{2k}),[G(f{x}_{2k+1},f{x}_{2k+1},T{x}_{2k})+G(g{x}_{2k},g{x}_{2k},S{x}_{2k+1})]/2\}\\ =h\max \{G(y_{2}k,y_{2}k,y_{2}k-1),G(y_{2k+1},y_{2k+1},y_{2k}),\\ G(y_{2}k,y2{}_k,y_{2}k-1),[G(y_{2k+1},y_{2k+1},y_{2k-1})+G(y_{2k},y_{2k},y_{2k})]/2\}\\ \le h\max \{G(y_2{}_k,y_2{}_k,y_2{}_{k-1}),G(y_{2k+1},y_{2k+1},y_{2k}),\\ [G(y_{2k+1},y_{2k+1},y_{2k})+G(y_2{}_k,y_2{}_k,y_2{}_{k-1})]/2\}\\ =h\max \{G(y_n,y_n,y_{n-1}),G(y_{n+1},y_{n+1},y_n)\}.\end{array}$$

This implies that

$$G\left(y_{n+1},y_{n+1},y_n\right)\le hG\left(y_n,y_n,y_{n-1}\right).$$

If n is odd, then n = 2k + 1 for some k ∈ N. In this case (2.1) gives

$$\begin{array}{l}G(y_{n+1},y_{n+1},y_n)\\ =G(y_{2k+2},y_{2k+2},y_{2k+1})\\ =G(f{x}_{2k+2},f{x}_{2k+2}+g{x}_{2k+1})\\ \le h\max \{G(S{x}_{2k+2},S{x}_{2k+2},T{x}_{2k+1}),G(f{x}_{2k+2},f{x}_{2k+2},S{x}_{2k+2}),\\ G(g{x}_{2k+1},g{x}_{2k+1},T{x}_{2k+1}),[G(f{x}_{2k+2},f{x}_{2k+2},T{x}_{2k+1})+G(g{x}_{2k+1},g{x}_{2k+1},S{x}_{2k+2})]/2\}\\ =h\max \{G(y_{2k+1},y_{2k+1},y_{2k}),G(y_{2k+2},y_{2k+2},y_{2k+1}),\\ G(y_{2k+1},y_{2k+1},y_{2k}),[G(y_{2k+2},y_{2k+2},y_{2k})+G(y_{2k+1},y_{2k+1},y_{2k+1})]/2\}\\ \le h\max \{G(y_{2k+1},y_{2k+1},y_{2k}),G(y_{2k+2},y_{2k+2},y_{2k+1}),\\ [G(y_{2k+2},y_{2k+2},y_{2k+1})+G(y_{2k+1},y_{2k+1},y_{2k})]/2\}\\ =h\max \{G(y_{2k+1},y_{2k+1},y_{2k}),G(y_{2k+2},y_{2k+2},y_{2k+1})\}\\ =h\max \{G(y_n,y_n,y_{n-1}),G(y_{n+1},y_{n+1},y_n)\},\end{array}$$

that is,

$$G\left(y_{n+1},y_{n+1},y_n\right)\le hG\left(y_n,y_n,y_{n-1}\right).$$

Continuing the above process, we have

$$G\left(y_{n+1},y_{n+1},y_n\right)\le h^{n}G\left(y_1,y_1,y_0\right).$$

Thus, if y₀ = y₁, we get G(y _n , y_n+1, y_n+1) = 0 for each n ∈ N. Hence, y _n = y_n+1for each n ∈ N. Therefore, {y _n } is G-Cauchy. So we may assume that y₀ ≠ y₁.

Let n, m ∈ N with m > n,

$$\begin{array}{c}G\left(y_n,y_m,y_m\right)\\ \le G\left(y_n,y_{n+1},y_{n+1}\right)+G\left(y_{n+1},y_{n+2},y_{n+2}\right)+\cdots +G\left(y_{m-1},y_m,y_m\right)\\ \le h^{n}G\left(y_0,y_1,y_1\right)+h^{n+1}G\left(y_0,y_1,y_1\right)+\cdots +h^{m-1}G\left(y_0,y_1,y_1\right)\\ =h^{n}G\left(y_0,y_1,y_1\right)\sum _{i=0}^{m-n-1}{h}^i\\ \le \frac{h^n}{1-h}G\left(y_0,y_1,y_1\right),\end{array}$$

and so G(y _n , y _m , y _m ) → 0 as m, n → ∞. Hence {y _n } is a Cauchy sequence in X. Since X is G-complete, there exists a point z ∈ X such that lim_n→∞y _n = z.

Consequently

$$\underset{n\to \infty }{lim}{y}_{2n}=\underset{n\to \infty }{lim}S{x}_{2n+1}=\underset{n\to \infty }{lim}g{x}_{2n}=z$$

and

$$\underset{n\to \infty }{lim}{y}_{2n+1}=\underset{n\to \infty }{lim}T{x}_{2n+2}=\underset{n\to \infty }{lim}f{x}_{2n+1}=z.$$

Let f and S be continuous compatible mappings. Compatibility of f and S implies that lim_n→∞G(fSx_2n+1, fSx_2n+1, Sfx_2n+1) = 0, that is G(fz, fz, Sz) = 0 which implies that fz = Sz. Since fX ⊂ TX, there exists some u ∈ X such that fz = Tu. Now from (2.1), we have

$$\begin{array}{c}G(fz,fz,gu)\le h\max \{G(Sz,Sz,Tu),G(fz,fz,Sz),G(gu,gu,Tu),\\ [G(fz,fz,Tu)+G(gu,gu,Sz)]/2\}\\ =h\max \{G(fz,fz,fz),G(fz,fz,fz),G(gu,gu,fz),\\ [G(fz,fz,fz)+G(gu,gu,fz)]/2\}\\ \text{= }hG\text{(}fz\text{, }gu\text{, }gu\text{)}\text{.}\end{array}$$

(2.3)

Also, from (2.2)

$$\begin{array}{c}G(fz,gu,gu)\le h\max \{G(Sz,Tu,Tu),G(fz,Sz,Sz),G(gu,Tu,Tu),\\ [G(fz,Tu,Tu)+G(gu,Sz,Sz)]/2\}\\ =h\max \{G(fz,fz,fz),G(fz,fz,fz),G(gu,fz,fz),\\ [G(fz,fz,fz)+G(gu,fz,fz)]/2\}\\ \text{= }hG\text{(}fz\text{,}fz\text{, }gu\text{)}\text{.}\end{array}$$

(2.4)

Combining above two inequalities, we get

$$G\left(fz,fz,gu\right)\le h^{2}G\left(fz,fz,gu\right).$$

Since h < 1, so that fz = gu. Hence, fz = Sz = gu = Tu. As the pair {g, T} is R-weakly commuting, there exists R > 0 such that

$$G\left(gTu,gTu,Tgu\right)\le RG\left(gu,gu,Tu\right)=0,$$

that is, gTu = Tgu. Moreover, ggu = gTu = Tgu = TTu. Similarly, the pair {f, S} is R-weakly commuting, there exists some R > 0 such that

$$G\left(fSz,fSz,Sfz\right)\le RG\left(fz,fz,Sz\right)=0,$$

so that fSz = Sfz and ffz = fSz = Sfz = SSz.

Now by (2.1)

$$\begin{array}{c}G(ffz,ffz,fz)=G(ffz,ffz,gu)\\ \le h\max \{G(Sfz,Sfz,Tu),G(ffz,ffz,Sfz),G(gu,gu,Tu),\\ [G(ffz,ffz,Tu)+G(gu,gu,Sfz)]/2\}\\ =h\max \{G(ffz,ffz,gu),G(ffz,ffz,ffz),G(gu,gu,gu),\\ [G(ffz,ffz,gu)+G(gu,gu,ffz)]/2\}\\ =h\max \{G(ffz,ffz,fz),[G(ffz,ffz,fz)+G(fz,fz,ffz)]/2\}\\ =\frac{h}{2}[G(ffz,ffz,fz)+G(fz,fz,ffz)],\end{array}$$

so that

$$G\left(ffz,ffz,fz\right)\le hG\left(fz,fz,ffz\right).$$

(2.5)

Again from (2.2), we have

$$\begin{array}{c}G(ffz,fz,fz)=G(ffz,gu,gu)\\ \le h\max \{G(Sfz,Tu,Tu),G(ffz,Sfz,Sfz),G(gu,Tu,Tu),\\ [G(f{f}_Z,Tu,Tu)+G(gu,Sfz,Sfz)]/2\}\\ =h\max \{G(Sfz,gu,gu),G(ffz,ffz,ffz),G(gu,gu,gu),\\ [G(ffz,gu,gu)+G(gu,ffz,ffz)]/2\}\\ =h\max \{G(ffz,fz,fz),[G(ffz,fz,fz)+G(fz,ffz,ffz)]/2\}\\ =\frac{h}{2}[G(ffz,fz,fz)+G(ffz,ffz,fz)],\end{array}$$

which implies

$$G\left(ffz,fz,fz\right)\le hG\left(ffz,ffz,fz\right).$$

(2.6)

From (2.5) and (2.6), we obtain

$$G\left(ffz,ffz,fz\right)\le h^{2}G\left(ffz,ffz,fz\right),$$

and since h² < 1 so that ffz = fz. Hence, ffz = Sfz = fz, and fz is the common fixed point of f and S. Since gu = fz, following arguments similar to those given above we conclude that fz is a common fixed point of g and T as well. Now we show the uniqueness of fixed point. For this, assume that there exists another point w in X which is the common fixed point of f, g, S and T. From (2.1), we obtain

$$\begin{array}{c}G(fz,fz,w)=G(ffz,ffz,gw)\\ \le h\max \{G(Sfz,Sfz,Tw),G(ffz,ffz,Sfz),G(gw,gw,Tw),\\ [G(ffz,ffz,Tw)+G(gw,gw,Sfz)]/2\}\\ =h\max \{G(fz,fz,w),G(fz,fz,fz),G(w,w,w),\\ [G(fz,fz,w)+G(w,w,fz)]/2\}\\ =\frac{h}{2}[G(fz,fz,w)+G(w,w,fz)],\end{array}$$

which implies that

$$G\left(fz,fz,w\right)\le hG\left(w,w,fz\right).$$

(2.7)

From (2.2), we get

$$\begin{array}{c}G(fz,w,w)=G(ffz,gw,gw)\\ \le h\max \{G(Sfz,Tw,Tw),G(ffz,Sfz,Sfz),G(gw,Tw,Tw),\\ [G(ffz,Tw,Tw)+G(gw,Sfz,Sfz)]/2\}\\ =h\max \{G(fz,w,w),G(fz,fz,fz),G(w,w,w),\\ [G(fz,w,w)+G(w,fz,fz)]/2\}\\ =\frac{h}{2}[G(fz,w,w)+G(w,fz,fz)],\end{array}$$

which implies

$$G\left(fz,w,w\right)\le hG\left(fz,fz,w\right).$$

(2.8)

Now (2.7) and (2.8) give

$$G\left(fz,fz,w\right)\le h^{2}G\left(fz,fz,w\right),$$

and fz = w. This completes the proof.

Example 2.2. Let X = {0, 1, 2} with G-metric defined by

$$\begin{array}{cc}(x,y,z)& G(x,y,z)\\ (0,0,0),(1,1,1),(2,2,2),& 0\\ \text{(0,}\text{0,}\text{1), (0,}\text{1,}\text{0), (1,}\text{0,}\text{0),}\\ \text{(0,}\text{0,}\text{2), (0,}\text{2,}\text{0), (2,}\text{0,}\text{0),}& 1\\ \text{(0,}\text{2,}\text{2), (2,}\text{0,}\text{2), (2,}\text{2,}\text{0),}\\ \text{(0,}\text{1, 1), (1, 0, 1), (1, 1, 0),}\\ \text{(1, 1, 2), (1, 2, 1), (2, 1, 1),}& 2\\ \text{(1, 2, 2), (2, 1, 2), (2, 2, 1),}\\ (0,1,2),(0,2,1),(1,0,2),& 2\\ \text{(1,}\text{2, 0),}\text{(2,}\text{0,}\text{1), (2,}\text{1,}\text{0),}\end{array}$$

is a non-symmetric G-metric on X because G(0, 0, 1) ≠ G(0, 1, 1).

Let f, g, S, T : X → X defined by

$$\begin{array}{ccccc}\hfill x\hfill & \hfill f\left(x\right)\hfill & \hfill g\left(x\right)\hfill & \hfill S\left(x\right)\hfill & \hfill T\left(x\right)\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}$$

Then fX ⊆ TX and gX ⊆ SX, with the pairs {f, S} and {g, T} are R-weakly commuting as they commute at their coincidence points.

Now to get (2.1) and (2.2) satisfied, we have the following nine cases: (I) x, y = 0, (II) x = 0, y = 2, (III) x = 1, y = 0, (IV) x = 1, y = 2, (V) x = 2, y = 0, (VI) x = 2, y = 2. For all these cases, f(x) = g(y) = 0 implies G(fx, fx, gy) = 0 and (2.1) and (2.2) hold.

• (VII) For x = 0, y = 1, then fx = 0, gy = 2, Sx = 0, Ty = 1.

  $$\begin{array}{l}G(fx,fx,gy)\\ =G(0,0,2)=1\\ \le h\max \{1,0,2,1\}\\ =h\max \{G(0,0,1),G(0,0,0),G(2,2,1),[G(0,0,1)+G(2,2,0)]/2\}\\ =h\max \{G(Sx,Sx,Ty),G(fx,fx,Sx),G(gy,gy,Ty),\\ [G(fx,fx,Ty)+G(gy,gy,Sx)]/2\}.\end{array}$$

Thus, (2.1) is satisfied where $h=\frac{4}{5}$.

Also

$$\begin{array}{l}G(fx,gy,gy)\\ =G(0,2,2)=1\\ \le h\max \{2,0,2,1.5\}\\ =h\max \{G(0,1,1),G(0,0,0),G(2,1,1),[G(0,1,1)+G(2,0,0)]/2\}\\ =h\max \{G(Sx,Ty,Ty),G(fx,Sx,Sx),G(gy,Ty,Ty),\\ [G(fx,Ty,Ty)+G(gy,Sx,Sx)]/2\}.\end{array}$$

Thus, (2.2) is satisfied where $h=\frac{4}{5}$.

• (VIII) Now when x = 1, y = 1, then fx = 0, gy = 2, Sx = 2, Ty = 1.

  $$\begin{array}{l}G(fx,fx,gy)\\ =G(0,0,2)=1\\ \le h\max \{2,1,2,0.5\}\\ =h\max \{G(2,2,1),G(0,0,2),G(2,2,1),[G(0,0,1)+G(2,2,2)]/2\}\\ =h\max \{G(Sx,Sx,Ty),G(fx,fx,Sx),G(gy,gy,Ty),\\ [G(fx,fx,Ty)+G(gy,gy,Sx)]/2\}.\end{array}$$

Thus, (2.1) is satisfied where $h=\frac{4}{5}$.

And

$$\begin{array}{l}G(fx,gy,gy)\\ =G(0,2,2)=1\\ \le h\max \{2,1,2,1\}\\ =h\max \{G(2,1,1),G(0,2,2),G(2,1,1),[G(0,1,1)+G(2,2,2)]/2\}\\ =h\max \{G(Sx,Ty,Ty),G(fx,Sx,Sx),G(gy,Ty,Ty),\\ [G(fx,Ty,Ty)+G(gy,Sx,Sx)]/2\}.\end{array}$$

Thus, (2.2) is satisfied where $h=\frac{4}{5}$.

• (IX) If x = 2, y = 1, then fx = 0, gy = 2, Sx = 1, Ty = 1 and

  $$\begin{array}{l}G(fx,fx,gy)\\ =G(0,0,2)=1\\ \le h\max \{0,1,2,1.5\}\\ =h\max \{G(1,1,1),G(0,0,1),G(2,2,1),[G(0,0,1)+G(2,2,1)]/2\}\\ =h\max \{G(Sx,Sx,Ty),G(fx,fx,Sx),G(gy,gy,Ty),\\ [G(fx,fx,Ty)+G(gy,gy,Sx)]/2\}.\end{array}$$

Thus, (2.1) is satisfied where $h=\frac{4}{5}$.

Also

$$\begin{array}{l}G(fx,gy,gy)\\ =G(0,2,2)=1\\ \le h\max \{0,2,2,2\}\\ =h\max \{G(1,1,1),G(0,1,1),G(2,1,1),[G(0,1,1)+G(2,1,1)]/2\}\\ =h\max \{G(Sx,Ty,Ty),G(fx,Sx,Sx),G(gy,Ty,Ty),\\ [G(fx,Ty,Ty)+G(gy,Sx,Sx)]/2\}.\end{array}$$

Thus, (2.2) is satisfied where $h=\frac{4}{5}$.

Hence, for all x, y ∈ X, (2.1) and (2.2) are satisfied for $h=\frac{4}{5}<1$ so that all the conditions of Theorem 2.1 are satisfied. Moreover, 0 is the unique common fixed point for all of the mappings f, g, S and T.

In Theorem 2.1, if we take f = g, then we have the following corollary.

Corollary 2.3. Let X be a complete G-metric space. Suppose that {f, S} and {f, T} be pointwise R-weakly commuting pairs of self-mappings on X satisfying

$$\begin{array}{c}G(fx,fx,fy)\le h\max \{G(Sx,Sx,Ty),G(fx,fx,Sx),G(fy,fy,Ty),\\ [G(fx,fx,Ty)+G(fy,fy,Sx)]/2\}\end{array}$$

(2.9)

and

$$\begin{array}{c}G(fx,fy,fy)\le h\max \{G(Sx,Ty,Ty),G(fx,Sx,Sx),G(fy,Ty,Ty)\}\\ [G(fx,Ty,Ty)+G(fy,Sx,Sx)]/2\}\end{array}$$

(2.10)

for all x, y ∈ X, where h ∈ [0, 1). Suppose that fX ⊆ SX ∪ TX, and one of the pairs {f, S} or {f, T} is compatible. If the mappings in the compatible pair are continuous, then f, S and T have a unique common fixed point.

Also, if we take S = T in Theorem 2.1, then we get the following.

Corollary 2.4. Let X be a complete G-metric space. Suppose that {f, S} and {g, S} are pointwise R-weakly commuting pairs of self-maps on X and

$$\begin{array}{c}G(fx,fx,gy)\le h\max \{G(Sx,Sx,Sy),G(fx,fx,Sx),G(gy,gy,Sy),\\ [G(fx,fx,Sy)+G(gy,gy,Sx)]/2\}\end{array}$$

(2.11)

and

$$\begin{array}{c}G(fx,gy,gy)\le h\max \{G(Sx,Sy,Sy),G(fx,Sx,Sx),G(gy,Sy,Sy),\\ [G(fx,Sy,Sy)+G(gy,Sx,Sx)]/2\}\end{array}$$

(2.12)

hold for all x, y ∈ X, where h ∈ [0, 1). Suppose that fX ∪ gX ⊆ SX and one of the pairs {f, S} or {g, S} is compatible. If the mappings in the compatible pair are continuous, then f, g and S have a unique common fixed point.

Corollary 2.5. Let X be a complete G-metric space. Suppose that f and g are two self-mappings on X satisfying

$$\begin{array}{c}G(fx,fx,gy)\le h\max \{G(x,x,y),G(fx,fx,x),G(gy,gy,y),\\ [G(fx,fx,y)+G(gy,gy,x)]/2\}\end{array}$$

(2.13)

and

$$\begin{array}{c}G(fx,gy,gy)\le h\max \{G(x,y,y),G(fx,x,x),G(gy,y,y),\\ \text{[}G\text{(}fx\text{,}y\text{, }y\text{)+}G\text{(}gy\text{, }x\text{, }x\text{)] /2}}\end{array}$$

(2.14)

for all x, y ∈ X, where h ∈ [0, 1). Suppose that one of f or g is continuous, then f and g have a unique common fixed point.

Proof. Taking S and T as identity maps on X, the result follows from Theorem 2.1.

Corollary 2.6. Let X be a complete G-metric space and f be a self-map on X such that

$$\begin{array}{c}G(fx,fx,fy)\le h\max \{G(x,x,y),G(fx,fx,x),G(fy,fy,y),\\ [G(fx,fx,y)+G(fy,fy,x)]/2\}\end{array}$$

(2.15)

and

$$\begin{array}{c}G(fx,fy,fy)\le h\max \{G(x,y,y),G(fx,x,x),G(fy,y,y),\\ [G(fx,y,y)+G(fy,x,x)]/2\}\end{array}$$

(2.16)

hold for all x, y ∈ X, where h ∈ [0, 1). Then f has a unique fixed point.

Proof. If we take f = g, and S and T as identity maps on X, then from f has a unique fixed point by Theorem 2.1.

3 Application

Let Ω = [0, 1] be bounded open set in ℝ, L²(Ω), the set of functions on Ω whose square is integrable on Ω. Consider an integral equation

$$p\left(t,x\left(t\right)\right)=\underset{\Omega }{\int }q\left(t,s,x\left(s\right)\right)ds$$

(3.1)

where p : Ω × ℝ → ℝ and q : Ω × Ω × ℝ → ℝ be two mappings. Define G : X × X × X → ℝ₊ by

$$G\left(x,y,z\right)=\underset{t\in \Omega }{sup}|x\left(t\right)-y\left(t\right)|+\underset{t\in \Omega }{sup}|y\left(t\right)-z\left(t\right)|+\underset{t\in \Omega }{sup}|z\left(t\right)-x\left(t\right)|.$$

Then X is a G-complete metric space. We assume the following that is there exists a function G : Ω × ℝ → ℝ⁺:

• (i) p(s, v(t)) ≥ ∫_Ω q(t, s, u(s)) ds ≥ G(s, v(t)) for each s, t ∈ Ω..

• (ii) p(s, v(t)) - G(s, v(t)) ≤ h |p(s, v(t)) - v(t)|.

Then integral equation (3.1) has a solution in L²(Ω).

Proof. Define (fx)(t) = p(t, x(t)) and (gx)(t) = ∫_Ωq(t, s, x(s)) ds. Now

$$\begin{array}{lll}\hfill G\left(fx,fx,gy\right)& =2\underset{t\in \Omega }{sup}|\left(fx\right)\left(t\right)-\left(gy\right)\left(t\right)|& \hfill \text{(1)}\\ =2\underset{t\in \Omega }{sup}\left|p\left(t,x\left(t\right)\right)-\underset{\Omega }{\int }q\left(t,s,y\left(t\right)\right)dt\right|& \hfill \text{(2)}\\ \le 2\underset{t\in \Omega }{sup}|p\left(t,x\left(t\right)\right)-G\left(t,x\left(t\right)\right)|& \hfill \text{(3)}\\ \le 2h\underset{t\in \Omega }{sup}|p\left(t,x\left(t\right)\right)-x\left(t\right)|& \hfill \text{(4)}\\ =hG\left(fx,fx,x\right).& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$$

Thus

$$\begin{array}{c}G(fx,fx,gy)\le h\max \{G(x,x,y),G(fx,fx,x),G(gy,gy,y),\\ [G(fx,fx,y)+G(gy,gy,x)]/2\}\end{array}$$

is satisfied. Similarly (2.14) is satisfied. Now we can apply Corollary 2.5 to obtain the solution of integral equation (3.1) in L²(Ω).

Remark 1. Theorems 2.8-2.9 in [3] and Corollaries 2.6-2.8 in [4] are special cases of our results Theorem 2.1 and Corollaries 2.3-2.6.

Remark 2. A G-metric naturally induces a metric d _G given by d _G (x, y) = G(x, y, y) + G(x, x, y). If the G-metric is not symmetric, the inequalities (2.1) and (2.2) do not reduce to any metric inequality with the metric d _G . Hence, our theorems do not reduce to fixed point problems in the corresponding metric space (X, d _G ).