Common fixed points for single-valued and multi-valued maps satisfying a generalized contraction in G-metric spaces

Abstract

In this article, we establish some common fixed point theorems for a hybrid pair {g, T} of single valued and multi-valued maps satisfying a generalized contractive condition defined on G-metric spaces. Our results unify, generalize and complement various known comparable results from the current literature.

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

1. Introduction and preliminaries

Nadler [1] initiated the study of fixed points for multi-valued contraction mappings and generalized the well known Banach fixed point theorem. Then after, many authors studied many fixed point results for multi-valued contraction mappings see [2–13].

Mustafa and Sims [14] introduced the G-metric spaces as a generalization of the notion of metric spaces. Mustafa et al. [15–19] obtained some fixed point theorems for mappings satisfying different contractive conditions. Abbas and Rhoades [20] initiated the study of common fixed point in G-metric spaces. While Saadati et al. [21] studied some fixed point theorems in generalized partially ordered G-metric spaces. Gajić and Crvenković [22, 23] proved some fixed point results for mappings with contractive iterate at a point in G-metric spaces. For other studies in G-metric spaces, we refer the reader to [24–38]. Consistent with Mustafa and Sims [14], the following definitions and results will be needed in the sequel.

Definition 1.1. (See [14]). Let X be a non-empty set, G : X × X × X → ℝ⁺ be a function satisfying the following properties

(G1) G(x, y, z) = 0 if x = y = z,

(G2) 0 < G(x, x, y) for all x, y ∈ 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),

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

Then the function G is called a generalized metric, or, more specially, a G-metric on X, and the pair (X, G) is called a G-metric space.

Definition 1.2. (See [14]). Let (X, G) be a G-metric space, and let (x _n ) be a sequence of points of X, therefore, we say that (x _n ) is G-convergent to x ∈ X if $\underset{n,m\to +\infty }{\text{lim}}G\left(x,x_n,x_m\right)=0$, that is, for any ε > 0, there exists N ∈ ℕ such that G(x, x _n , x _m ) < ε, for all n, m ≥ N. We call x the limit of the sequence and write x _n → x or $\underset{n\to +\infty }{\text{lim}}{x}_n=x.$

Proposition 1.1. (See [14]). Let (X, G) be a G-metric space. The following statements are equivalent:

• (1) (x _n ) is G-convergent to x,

• (2) G(x _n , x _n , x) → 0 as n → +∞,

• (3) G(x _n , x, x) → 0 as n → +∞,

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

Definition 1.3. (See [14]). Let (X, G) be a G-metric space. A sequence (x _n ) is called a G-Cauchy sequence if for any ε > 0, there is N ∈ ℕ such that G(x _n , x _m , x _l ) < ε for all m, n, l ≥ N, that is, G(x _n , x _m , x _l ) → 0 as n, m, l → +∞.

Proposition 1.2. (See [14]). Let (X, G) be a G-metric space. Then the following statements are equivalent:

• (1) the sequence (x _n ) is G-Cauchy,

• (2) for any ε > 0, there exists N ∈ ℕ such that G(x _n , x _m , x _m ) < ε, for all m, n ≥ N.

Definition 1.4. (See [14]). A G-metric space (X, G) is called G-complete if every G-Cauchy sequence is G-convergent in (X,G).

Every G-metric on X defines a metric d _G on X given by

$$d_G\left(x,y\right)=G\left(x,y,y\right)+G\left(y,x,x\right),\text{for all}x,y\in X.$$

(1)

Recently, Kaewcharoen and Kaewkhao [34] introduced the following concepts. Let X be a G-metric space. We shall denote CB(X) the family of all nonempty closed bounded subsets of X. Let H(.,.,.) be the Hausdorff G-distance on CB(X), i.e.,

$$H_G\left(A,B,C\right)=\text{max}\left\{\underset{x\in A}{\text{sup}}G\left(x,B,C\right),\underset{x\in B}{\text{sup}}G\left(x,C,A\right),\underset{x\in C}{\text{sup}}G\left(x,A,B\right)\right\},$$

where

$$\begin{array}{l}G\left(x,B,C\right)=d_G\left(x,B\right)+d_G\left(B,C\right)+d_G\left(x,C\right),\\ d_G\left(x,B\right)=\text{inf}\left\{d_G\left(x,y\right),y\in B\right\},\\ d_G\left(A,B\right)=\text{inf}\left\{d_G\left(a,b\right),a\in A,b\in B\right\}.\end{array}$$

Recall that G(x, y, C) = inf {G(x, y, z), z ∈ C}. A mapping T : X → 2^X is called a multi-valued mapping. A point x ∈ X is called a fixed point of T if x ∈ Tx.

Definition 1.5. Let X be a given non empty set. Assume that g : X → X and T : X → 2^X.

If w = gx ∈ Tx for some x ∈ X, then x is called a coincidence point of g and T and w is a point of coincidence of g and T.

Mappings g and T are called weakly compatible if gx ∈ Tx for some x ∈ X implies gT(x) ⊆ Tg(x).

Proposition 1.3. (see [34]). Let X be a given non empty set. Assume that g : X → X and T : X → 2^X are weakly compatible mappings. If g and T have a unique point of coincidence w = gx∈ Tx, then w is the unique common fixed point of g and T.

In this article, we establish some common fixed point theorems for a hybrid pair {g,T} of single valued and multi-valued maps satisfying a generalized contractive condition defined on G-metric spaces. Also, an example is presented.

2. Main results

We start this section with the following lemma, which is the variant of the one given in Nadler [1] or Assad and Kirk [4]. Its proof is a simple consequence of the definition of the Hausdorff G-distance H _G (A, B, B).

Lemma 2.1. If A, B ∈ CB(X) and a ∈ A, then for each ε > 0, there exists b ∈ B such that G(a,b,b) ≤ H _G (A, B, B) + ε.

The main result of the article is the following.

Theorem 2.1. Let (X, G) be a G-metric space. Set g : X → X and T : X → CB(X). Assume that there exists a function α : [0,+∞) → [0,1) satisfying $\underset{r\to t^{+}}{\text{lim sup}}\alpha \left(r\right)<1$ for every t ≥ 0 such that

$$H_G\left(Tx,Ty,Tz\right)\le \alpha \left(G\left(gx,gy,gz\right)\right)G\left(gx,gy,gz\right),$$

(2)

for all x, y, z ∈ X. If for any x ∈ X, Tx ⊆ g(X) and g(X) is a G-complete subspace of X, then g and T have a point of coincidence in X. Furthermore, if we assume that gp ∈ Tp and gq ∈ Tq implies G(gq, gp, gp) ≤ H _G (Tq, Tp, Tp), then

(i) g and T have a unique point of coincidence.

(ii) If in addition g and T are weakly compatible, then g and T have a unique common fixed point.

Proof. Let x₀ be arbitrary in X. Since Tx₀ ⊆ g(X), choose x₁ ∈ X such that gx₁ ∈ Tx₀. If gx₁= gx₀, we finished. Assume that gx₀ ≠ gx₁, so G(gx₀, gx₁, gx₁) > 0. We can choose a positive integer n₁ such that

$${\alpha }^{n_1}\left(G\left(g{x}_0,g{x}_1,g{x}_1\right)\right)\le \left[1-\alpha \left(G\left(g{x}_0,g{x}_1,g{x}_1\right)\right)\right]G\left(g{x}_0,g{x}_1,g{x}_1\right).$$

By Lemma 2.1 and the fact that Tx₁ ⊆ g(X), there exists gx₂ ∈ Tx₁ such that

$$G\left(g{x}_1,g{x}_2,g{x}_2\right)\le H_G\left(T{x}_0,T{x}_1,T{x}_1\right)+{\alpha }^{n_1}\left(G\left(g{x}_0,g{x}_1,g{x}_1\right)\right).$$

Using the two above inequalities and (2), it follows that

$$\begin{array}{ll}\hfill G\left(g{x}_1,g{x}_2,g{x}_2\right)& \le H_G\left(T{x}_0,T{x}_1,T{x}_1\right)+{\alpha }^{n_1}\left(G\left(g{x}_0,g{x}_1,g{x}_1\right)\right)\\ \le \alpha \left(G\left(g{x}_0,g{x}_1,g{x}_1\right)\right)G\left(g{x}_0,g{x}_1,g{x}_1\right)+\left[1-\alpha \left(G\left(g{x}_0,g{x}_1,g{x}_1\right)\right)\right]G\left(g{x}_0,g{x}_1,g{x}_1\right)\\ =G\left(g{x}_0,g{x}_1,g{x}_1\right).\end{array}$$

If gx₁ = gx₂, we finished. Assume that gx₁ ≠ gx₂. Now we choose a positive integer n₂> n₁ such that

$${\alpha }^{n_2}\left(G\left(g{x}_1,g{x}_2,g{x}_2\right)\right)\le \left[1-\alpha \left(G\left(g{x}_1,g{x}_2,g{x}_2\right)\right)\right]G\left(g{x}_2,g{x}_2,g{x}_2\right).$$

Since Tx₂ ∈ CB(X) and the fact that Tx₂ ⊆ g(X), we may select gx₃ ∈ Tx₂ such that from Lemma 2.1

$$G\left(g{x}_2,g{x}_3,g{x}_3\right)\le H_G\left(T{x}_1,T{x}_2,T{x}_2\right)+{\alpha }^{n_2}\left(G\left(g{x}_1,g{x}_2,g{x}_2\right)\right),$$

and then, similarly to the previous case, we have

$$\begin{array}{ll}\hfill G\left(g{x}_2,g{x}_3,g{x}_3\right)& \le H_G\left(T{x}_1,T{x}_2,T{x}_2\right)+{\alpha }^{n_2}\left(G\left(g{x}_1,g{x}_2,g{x}_2\right)\right)\\ \le \alpha \left(G\left(g{x}_1,g{x}_2,g{x}_2\right)\right)G\left(g{x}_1,g{x}_2,g{x}_2\right)+\left[1-\alpha \left(G\left(g{x}_1,g{x}_2,g{x}_2\right)\right)\right]G\left(g{x}_1,g{x}_2,g{x}_2\right)\\ =G\left(g{x}_1,g{x}_2,g{x}_2\right).\end{array}$$

By repeating this process, for each k ∈ ℕ*, we may choose a positive integer n _k such that

$${\alpha }^{n_k}\left(G\left(g{x}_{k-1},g{x}_k,g{x}_k\right)\right)\le \left[1-\alpha \left(G\left(g{x}_{k-1},g{x}_k,g{x}_k\right)\right)\right]G\left(g{x}_{k-1},g{x}_k,g{x}_k\right).$$

Again, we may select gx _k+1 ∈ Tx _k such that

$$G\left(g{x}_k,g{x}_{k+1},g{x}_{k+1}\right)\le H_G\left(T{x}_{k-1},T{x}_k,T{x}_k\right)+{\alpha }^{n_k}\left(G\left(g{x}_{k-1},g{x}_k,g{x}_k\right)\right).$$

(3)

The last two inequalities together imply that

$$G\left(g{x}_k,g{x}_{k+1},g{x}_{k+1}\right)\le G\left(g{x}_{k-1},g{x}_k,g{x}_k\right),$$

which shows that the sequence of nonnegative numbers {d _k }, given by d _k = G(gx _k-1 , gx _k , gx _k ), k = 1, 2,. . ., is non-increasing. This means that there exists d ≥ 0 such that

$$\underset{k\to +\infty }{\text{lim}}{d}_k=d.$$

Let now prove that the {gx _k } is a G-Cauchy sequence.

Using the fact that, by hypothesis for t = d, $\underset{r\to d^{+}}{\text{lim sup}}\alpha \left(t\right)<1$, it results that there exists a rank k₀ such that for k ≥ k₀, we have α(d _k ) < h, where

$$\underset{t\to d^{+}}{\text{lim sup}}\alpha \left(t\right)<h<1.$$

Now, by (3) we deduce that the sequence {d _k } satisfies the following recurrence inequality

$$d_{k+1}\le H_G\left(T{x}_{k-1},T{x}_k,T{x}_k\right)+{\alpha }^{n_k}\left(d_k\right)\le \alpha \left(d_k\right)d_k+{\alpha }^{n_k}\left(d_k\right),k\ge 1.$$

(4)

By induction, from (4), we get

$$d_{k+1}\le \prod _{i=1}^k\alpha \left(d_i\right)d_1+\sum _{m=1}^{k-1}\prod _{i=m+1}^k\alpha \left(d_i\right){\alpha }^{n_m}\left(d_m\right)+{\alpha }^{n_k}\left(d_k\right),k\ge 1,$$

which, by using the fact that α < 1, can be simplified to

$$d_{k+1}\le \prod _{i=1}^k\alpha \left(d_i\right)d_1+\sum _{m=1}^{k-1}\prod _{i=\text{max}\left\{k_0,m+1\right\}}^k\alpha \left(d_i\right){\alpha }^{n_m}\left(d_m\right)+{\alpha }^{n_k}\left(d_k\right),k\ge 1,$$

Referring to the proof of Theorem 2.1 in [11] or Lemma 3.2 in [12], we may obtain

$$\prod _{i=1}^k\alpha \left(d_i\right)d_1+\sum _{m=1}^{k-1}\prod _{i=\text{max}\left\{k_0,m+1\right\}}^k\alpha \left(d_i\right){\alpha }^{n_m}\left(d_m\right)+{\alpha }^{n_k}\left(d_k\right)\le c{h}^k,$$

where c is a positive constant. We deduce that

$$d_{k+1}=G\left(g{x}_k,g{x}_{k+1},g{x}_{k+1}\right)\le c{h}^k.$$

Now for k ≥ k₀ and m is a positive arbitrary integer, we have using the property (G4)

$$\begin{array}{ll}\hfill G\left(g{x}_k,g{x}_{k+m},g{x}_{k+m}\right)& \le G\left(g{x}_k,g{x}_{k+1},g{x}_{k+1}\right)+G\left(g{x}_{k+1},g{x}_{k+2},g{x}_{k+2}\right)\\ +\cdots +G\left(g{x}_{k+m-2},g{x}_{k+m-1},g{x}_{k+m-1}\right)+G\left(g{x}_{k+m-1},g{x}_{k+m},g{x}_{k+m}\right)\\ \le c\left[h^k+h^{k+1}+\cdots +h^{k+m-1}\right]\\ \le c\frac{h^k}{1-h}\to 0\text{as}k\to +\infty ,\end{array}$$

since 0 < h < 1. This shows that the sequence {gx _n } is G-Cauchy in the complete subspace g(X). Thus, there exists q ∈ g(X) such that, from Proposition 1.1

$$\underset{n\to +\infty }{\text{lim}}G\left(g{x}_n,g{x}_n,q\right)=\underset{n\to +\infty }{\text{lim}}G\left(g{x}_n,q,q\right)=0.$$

(5)

Since q ∈ g(X), then there exists p ∈ X such that q = gp. From (5), we have

$$\underset{n\to +\infty }{\text{lim}}G\left(g{x}_n,g{x}_n,gp\right)=\underset{n\to +\infty }{\text{lim}}G\left(g{x}_n,gp,gp\right)=0.$$

(6)

We claim that gp ∈ Tp. Indeed, from (2), we have

$$\begin{array}{ll}\hfill G\left(g{x}_{n+1},Tp,Tp\right)& \le H_G\left(T{x}_n,Tp,Tp\right)\\ \le \alpha \left(G\left(g{x}_n,gp,gp\right)\right)G\left(g{x}_n,gp,gp\right).\end{array}$$

(7)

Letting n → +∞ in (7) and using (6), we get

$$G\left(gp,Tp,Tp\right)=\underset{n\to +\infty }{\text{lim}}G\left(g{x}_{n+1},Tp,Tp\right)=0,$$

that is, gp ∈ Tp. That is T and g have a point of coincidence. Now, assume that if gp ∈ Tp and gq ∈ Tq, then G(gq, gp, gp) ≤ H _G (Tq, Tp, Tp). We will prove the uniqueness of a point of coincidence of g and T. Suppose that gp ∈ Tp and gq ∈ Tq. By (2) and this assumption, we have

$$\begin{array}{ll}\hfill G\left(gq,gp,gp\right)& \le H_G\left(Tq,Tp,Tp\right)\\ \le \alpha \left(G\left(gq,gp,gp\right)\right)G\left(gq,gp,gp\right),\end{array}$$

(8)

and since α(G(gq, gp, gp)) < G(gq, gp, gp), so necessarily from (8), we have G(gq, gp, gp) = 0, i.e., gp = gq. In view of

$$H_G\left(Tq,Tp,Tp\right)\le \alpha \left(G\left(gq,gp,gp\right)\right)G\left(gq,gp,gp\right)=0,$$

we get Tq = Tp. Thus, T and g have a unique point of coincidence. Suppose that g and T are weakly compatible. By applying Proposition 1.3, we obtain that g and T have a unique common fixed point.

Corollary 2.1. Let (X,G) be a complete G-metric space. Assume that T : X → CB(X) satisfies the following condition

$$H_G\left(Tx,Ty,Tz\right)\le \alpha \left(G\left(x,y,z\right)\right)G\left(x,y,z\right),$$

(9)

for all x, y, z ∈ X, where α : [0,+∞) → [0,1) satisfies $\underset{r\to t^{+}}{\text{lim sup}}\alpha \left(r\right)<1$ for every t ≥ 0. Then T has a fixed point in X. Furthermore, if we assume that p ∈ Tp and q ∈ Tq implies G(q, p, p) ≤ H _G (Tq, Tp, Tp), then T has a unique fixed point.

Proof. It follows by taking g the identity on X in Theorem 2.1.

Corollary 2.2. Let (X, G) be a G-metric space. Assume that g : X → X and T : X → CB(X) satisfy the following condition

$$H_G\left(Tx,Ty,Tz\right)\le kG\left(gx,gy,gz\right),$$

(10)

for all x, y, z ∈ X, where k ∈ [0,1). If for any x ∈ X, Tx ⊆ g(X) and g(X) is a G-complete subspace of X, then g and T have a point of coincidence in X. Furthermore, if we assume that gp ∈ Tp and gq ∈ Tq implies G(gq, gp, gp) ≤ H _G (Tq, Tp, Tp), then

(i) g and T have a unique point of coincidence.

(ii) If in addition g and T are weakly compatible, then g and T have a unique common fixed point.

Proof. It follows by taking α(t) = k, k ∈ [0,1), in Theorem 2.1.

In the case of single-valued mappings, that is, if T : X → X, (i.e., Tx = {Tx} for any x ∈ X), it is obviously that

$$H_G\left(Tx,Ty,Tz\right)=G\left(Tx,Ty,Tz\right),\forall x,y,z\in X.$$

Furthermore, if gp ∈ Tp (i.e., gp = Tp) and gq ∈ Tq (i.e., gq = Tq), then clearly,

$$G\left(gq,gp,gp\right)=G\left(Tq,Tp,Tp\right)=H_G\left(Tq,Tp,Tp\right),$$

that is, the assumption given in Theorem 2.1 is verified.

Also, the single-valued mappings T, g : X → X are said weakly compatible if Tgx = gTx whenever Tx = gx for some x ∈ X.

Now, we may state the following corollaries from Theorem 2.1 and the precedent corollaries:

Corollary 2.3. Let (X, G) be a complete G-metric space. Assume that T : X → X satisfies the following condition

$$G\left(Tx,Ty,Tz\right)\le \alpha \left(G\left(x,y,z\right)\right)G\left(x,y,z\right)$$

(11)

for all x, y, z ∈ X, where α : [0, +∞) → [0, 1) satisfies $\underset{r\to t^{+}}{\text{lim sup}}\alpha \left(r\right)<1$ for every t ≥ 0. Then, T has a unique fixed point.

Corollary 2.4. Let (X, G) be a G-metric space. Assume that g : X → X and T : X → X satisfy the following condition

$$G\left(Tx,Ty,Tz\right)\le \alpha \left(G\left(gx,gy,gz\right)\right)G\left(gx,gy,gz\right)$$

(12)

for all x, y, z ∈ X, where α : [0, +∞) → [0, 1) satisfies $\underset{r\to t^{+}}{\text{lim sup}}\alpha \left(r\right)<1$ for every t ≥ 0. If T(X) ⊆ g(X) and g(X) is a G-complete subspace of X, then

(i) g and T have a unique point of coincidence.

(ii) Furthermore, if g and T are weakly compatible, then g and T have a unique common fixed point.

Now, we introduce an example to support the useability of our results.

Example 2.1. Let X = [0, 1]. Define T : X → CB(X) by $Tx=\left[0,\frac{1}{16}x\right]$ and define g : X → X by $gx=\sqrt{x}$. Define a G-metric on X by G(x, y, z) = max{|x-y|, |x-z|, |y-z|}. Also, define α : [0, +∞) → [0, 1) by $\alpha \left(t\right)=\frac{1}{2}$ Then:

• (1) Tx ⊆ g(X) for all x ∈ X.

• (2) g(X) is a G-complete subspace of X.

• (3) g and T are weakly compatible.

• (4) H _G (Tx, Ty, Tz) ≤ α(G(gx, gy, gz))G(gx, gy, gz) for all x, y, z ∈ X.

Proof. The proofs of (1), (2), and (3) are clear. By (1), we have

$$d_G\left(x,y\right)=G\left(x,y,y\right)+G\left(y,x,x\right)=2\left|x-y\right|\text{ for all }x,y\in X.$$

To prove (4), let x, y, z ∈ X. If x = y = z = 0, then

$$H_G\left(Tx,Ty,Tz\right)=0\le \alpha \left(G\left(gx,gy,gz\right)\right)G\left(gx,gy,gz\right).$$

Thus, we may assume that x, y, and z are not all zero. With out loss of generality, we assume that x ≤ y ≤ z. Then

$$\begin{array}{l}{H}_G\left(Tx,Ty,Tz\right)=H_G\left(\left[0,\frac{1}{16}x\right],\left[0,\frac{1}{16}y\right],\left[0,\frac{1}{16}z\right]\right)\\ =\text{max}\left\{\underset{0\le a\le \frac{1}{16}x}{\text{sup}}G\left(a,\left[0,\frac{1}{16}y\right],\left[0,\frac{1}{16}z\right]\right),\underset{0\le b\le \frac{1}{16}y}{\text{sup}}G\left(b,\left[0,\frac{1}{16}z\right],\left[0,\frac{1}{16}x\right]\right),\right.\\ \left.\underset{0\le c\le \frac{1}{16}z}{\text{sup}}G\left(c,\left[0,\frac{1}{16}x\right],\left[0,\frac{1}{16}y\right]\right)\right\}.\end{array}$$

Since x ≤ y ≤ z, so $\left[0,\frac{1}{16}x\right]\subseteq \left[0,\frac{1}{16}y\right]\subseteq \left[0,\frac{1}{16}z\right]$ This implies that

$$d_G\left(\left[0,\frac{1}{16}x\right],\left[0,\frac{1}{16}y\right]\right)=d_G\left(\left[0,\frac{1}{16}y\right],\left[0,\frac{1}{16}z\right]\right)=d_G\left(\left[0,\frac{1}{16}x\right],\left[0,\frac{1}{16}z\right]\right)=0.$$

For each $0\le a\le \frac{1}{16}x,$ we have

$$G\left(a,\left[0,\frac{1}{16}y\right],\left[0,\frac{1}{16}z\right]\right)=d_G\left(a,\left[0,\frac{1}{16}y\right]\right)+d_G\left(\left[0,\frac{1}{16}y\right],\left[0,\frac{1}{16}z\right]\right)+d_G\left(a,\left[0,\frac{1}{16}z\right]\right)=0.$$

Also, for each $0\le b\le \frac{1}{16}y$, we have

$$\begin{array}{c}G\left(b,\left[0,\frac{1}{16}z\right],\left[0,\frac{1}{16}x\right]\right)=d_G\left(b,\left[0,\frac{1}{16}z\right]\right)+d_G\left(\left[0,\frac{1}{16}z\right],\left[0,\frac{1}{16}x\right]\right)+d_G\left(b,\left[0,\frac{1}{16}x\right]\right)\\ =\{\underset{2b-\frac{x}{8}\text{if}b\ge \frac{x}{16}.}{0\text{if}b\le \frac{x}{16}}\end{array}$$

This yields that

$$\underset{0\le b\le \frac{1}{16}y}{\text{sup}}G\left(b,\left[0,\frac{1}{16}z\right],\left[0,\frac{1}{16}x\right]\right)=\frac{y}{8}-\frac{x}{8}.$$

Moreover, for each $0\le c\le \frac{1}{16}z$, we have

$$\begin{array}{c}G\left(c,\left[0,\frac{1}{16}x\right],\left[0,\frac{1}{16}y\right]\right)=d_G\left(c,\left[0,\frac{1}{16}x\right]\right)+d_G\left(\left[0,\frac{1}{16}x\right],\left[0,\frac{1}{16}y\right]\right)+d_G\left(c,\left[0,\frac{1}{16}y\right]\right)\\ =\left\{\begin{array}{c}0\text{if}c\le \frac{x}{16}\hfill \\ 2c-\frac{x}{8}\text{if}\frac{x}{16}\le c\le \frac{y}{16}\hfill \\ 4c-\frac{x}{8}-\frac{y}{8}\text{if}c\ge \frac{y}{16}.\hfill \end{array}\right.\end{array}$$

This yields that

$$\underset{0\le c\le \frac{1}{16}z}{\text{sup}}G\left(c,\left[\frac{1}{16}c\right],\left[0,\frac{1}{16}y\right]\right)=\frac{z}{4}-\frac{x}{8}-\frac{y}{8}.$$

We deduce that

$$\begin{array}{ll}\hfill H_G\left(Tx,Ty,Tz\right)& =\frac{z}{4}-\frac{x}{8}-\frac{y}{8}\\ \le \frac{1}{4}\left(z-x\right)\\ =\frac{1}{2}\left(\frac{1}{2}\left(z-x\right)\right)\\ \le \frac{1}{2}\left(\frac{z-x}{\sqrt{x}+\sqrt{z}}\right)\\ =\frac{1}{2}\left(\sqrt{z}-\sqrt{x}\right)\end{array}$$

On the other hand, it is obvious that all other hypotheses of Theorem 2.1 are satisfied and so g and T have a unique common fixed point, which is u = 0.

Remark 1 • Theorem 2.1 improves Kaewcharoen and Kaewkhao [[34], Theorem 3.3] (in case b = c = d = 0).

• Corollary 2.3 generalizes Mustafa [[15], Theorem 5.1.7] and Shatanawi [[35], Corollary 3.4].