**3. Common Fixed Point Results for Almost** *Rg***-Geraghty Type Contraction Mappings**

**Lemma 3.** *Let* (*X*, *d*) *be a b*2*-metric space endowed with a binary relation* R *and f* , *g* : *X* → *X such that f*(*X*) ⊆ *g*(*X*), *with* R *is* (*f* , *g*)*-closed and* R|*g*(*X*) *is transitive. Assume that there exists x*<sup>0</sup> ∈ *X such that gx*0R *f x*0. *Define a sequence* {*xn*} *in X by f xn* = *gxn*+<sup>1</sup> *for n* ≥ 0. *Then*

*gxm*R*gxn and f xm*R *f xn for all m*, *n* ∈ **N**<sup>0</sup> *with m* < *n*.

**Proof.** Since there exists *x*<sup>0</sup> ∈ *X* such that *gx*0R *f x*0, *f xn* = *gxn*+1, and R is (*f* , *g*)-closed, we deduce that *gx*0R*gx*1, then *gx*<sup>1</sup> = *f x*0R *f x*<sup>1</sup> = *gx*2. By continuing this process, we get *gxn*R*gxn*+<sup>1</sup> for all *n* ∈ **N**. Suppose that *m* < *n*, so *gxm*R*gxm*+<sup>1</sup> and *gxm*+1R*gxm*+2, by R is *g*-transitive we have *gxm*R*gxm*+2. Again, since *gxm*R*gxm*+<sup>2</sup> and *gxm*+2R*gxm*+3, we get that *gxm*R*gxm*+3. By continuing this process, we obtain *gxm*R*gxn*. for all *m*, *n* ∈ **N** with *m* < *n*. In similar way and since *f*(*X*) ⊆ *g*(*X*), we conclude *f xm*R *f xn* for all *m*, *n* ∈ **N** with *m* < *n*.

In 1973, Geraghty [22] introduced the class of all functions *β* : [0, ∞) → [0, 1) which satisfy that lim*n*→<sup>∞</sup> *<sup>β</sup>*(*tn*) = 1 implies lim*n*→<sup>∞</sup> *tn* <sup>=</sup> 0. In addition, the author proved a fixed point result, generalizing the Banach contraction principle. Afterwards, there are many results about fixed point theorems by using such functions in this class. Ðuki´c et al. [23] obtained fixed point results of this kind in *b*-metric and from [23] we denote Ω to the family of all functions *<sup>β</sup><sup>s</sup>* : [0, <sup>∞</sup>) <sup>→</sup> [0, <sup>1</sup> *<sup>s</sup>* ) for a real number *s* ≥ 1, which satisfy the condition

$$\lim\_{n \to \infty} \beta\_s(t\_n) = \frac{1}{s} \qquad \text{implies} \quad \lim\_{n \to \infty} t\_n = 0.$$

**Definition 15.** *Let* (*X*, *d*) *be a b*2*-metric space and f* , *g* : *X* → *X. Suppose for all x*, *y*, *a* ∈ *X*,

$$M(\mathbf{x}, y, a) = \max \left\{ d(\mathbf{g}\mathbf{x}, \mathbf{g}y, a), d(\mathbf{g}\mathbf{x}, f\mathbf{x}, a), d(\mathbf{g}y, fy, a), \frac{d(\mathbf{g}\mathbf{x}, fy, a) + d(\mathbf{g}y, f\mathbf{x}, a)}{2s} \right\},$$

*and*

$$N(\mathbf{x}, y, a) = \min\left\{ d(\mathbf{g}\mathbf{x}, f\mathbf{x}, a), d(\mathbf{g}y, fy, a), d(\mathbf{g}\mathbf{x}, fy, a), d(\mathbf{g}y, f\mathbf{x}, a) \right\}.$$

*We say that f is almost* R*g-Geraghty type contraction mapping if there exist L* ≥ 0 *and β<sup>s</sup>* ∈ Ω *such that*

$$d(fx, fy, a) \le \beta\_s(M(x, y, a))M(x, y, a) + LN(x, y, a),\tag{1}$$

*for all x*, *<sup>y</sup>*, *<sup>a</sup>* <sup>∈</sup> *<sup>X</sup>*, *with gx*R*gy*, *f x*R *f y*.

**Definition 16.** *Let* (*X*, *d*) *be a b*2*-metric space and f* : *X* → *X. Suppose for all x*, *y*, *a* ∈ *X*,

$$M(\mathbf{x}, y, a) = \max\left\{ d(\mathbf{x}, y, a), d(\mathbf{x}, f \mathbf{x}, a), d(y, fy, a), \frac{d(\mathbf{x}, fy, a) + d(y, f\mathbf{x}, a)}{2s} \right\},$$

*and*

$$N(\mathbf{x}, \mathbf{y}, a) = \min \{ d(\mathbf{x}, f \mathbf{x}, a), d(\mathbf{y}, f \mathbf{y}, a), d(\mathbf{x}, f \mathbf{y}, a), d(\mathbf{y}, f \mathbf{x}, a) \}.$$

*We say that f is almost* R*-Geraghty type contraction mapping if there exist L* ≥ 0 *and β<sup>s</sup>* ∈ Ω *such that*

$$d(f\mathbf{x}, f\mathbf{y}, a) \le \beta\_s(M(\mathbf{x}, \mathbf{y}, a))M(\mathbf{x}, \mathbf{y}, a) + LN(\mathbf{x}, \mathbf{y}, a),\tag{2}$$

*for all x*, *<sup>y</sup>*, *<sup>a</sup>* <sup>∈</sup> *<sup>X</sup>*, *with x*R*y*, *f x*R *f y*.

Now, we present our main result as follows:

**Theorem 1.** *Let* (*X*, *d*) *be a b*2*-metric space endowed with a binary relation* R *and f* , *g* : *X* → *X such that f*(*X*) ⊆ *g*(*X*), *g*(*X*) *is a b*2*-complete subspace of X. Assume that f is almost* R*g-Geraghty type contraction mapping and the following conditions hold:*


**Proof.** Let *x*<sup>0</sup> ∈ *X* such that *gx*0R *f x*0. The proof is finished if *gx*<sup>0</sup> = *f x*<sup>0</sup> and *x*<sup>0</sup> is a coincidence point of *f* and *g*. Let us take *gx*<sup>0</sup> = *f x*0, then since *f*(*X*) ⊆ *g*(*X*) we can choose *x*<sup>1</sup> ∈ *X* such that *f x*<sup>0</sup> = *gx*1. Continuing this process, we can define a sequence {*gxn*} in *X* by *f xn* = *gxn*+1, for all *n* ∈ **N**0.

We divide the proof into three steps as follows.

**Step 1**: We claim that lim*n*→<sup>∞</sup> *<sup>d</sup>*(*gxn*, *gxn*+1, *<sup>a</sup>*) = 0. From Lemma 3, we have {*gxn*} is <sup>R</sup>preserving sequence that is *gxn*R*gxn*+<sup>1</sup> and *f xn*R *f xn*+1, for all *n* ∈ **N**0. If *f xn*<sup>0</sup> = *f xn*0<sup>+</sup>1, for some *n*<sup>0</sup> ∈ **N**0, then *xn*0+<sup>1</sup> is a coincidence point of *f* and *g*. Suppose that *f xn* = *f xn*+1, for all *n* ∈ **N**0. Therefore, from (1), we obtain

$$\begin{aligned} d(\mathcal{g}\mathbf{x}\_{\mathbb{H}+1}, \mathcal{g}\mathbf{x}\_{\mathbb{H}+2}, \mathcal{g}\mathbf{x}\_{\mathbb{H}}) &= & d(f\mathbf{x}\_{\mathbb{H}}f\mathbf{x}\_{\mathbb{H}+1}, \mathcal{g}\mathbf{x}\_{\mathbb{H}}) \\ &\leq & \mathcal{\beta}\_{\mathbb{H}}(M(\mathbf{x}\_{\mathbb{H}}\mathbf{x}\_{\mathbb{H}+1}, \mathcal{g}\mathbf{x}\_{\mathbb{H}}))M(\mathbf{x}\_{\mathbb{H}}\mathbf{x}\_{\mathbb{H}+1}, \mathcal{g}\mathbf{x}\_{\mathbb{H}}) + LN(\mathbf{x}\_{\mathbb{H}}\mathbf{x}\_{\mathbb{H}+1}, \mathcal{g}\mathbf{x}\_{\mathbb{H}}) \rightarrow (\*) \end{aligned}$$

where

$$\begin{split} M(\mathbf{x}\_{n},\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n}) &= \max \left\{ d(\mathbf{g}\mathbf{x}\_{n},\mathbf{g}\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n}), d(\mathbf{g}\mathbf{x}\_{n},f\mathbf{x}\_{n},\mathbf{g}\mathbf{x}\_{n}), d(\mathbf{g}\mathbf{x}\_{n+1},f\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n}) \right\} \\ &\quad \frac{d(\mathbf{g}\mathbf{x}\_{n},f\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n}) + d(\mathbf{g}\mathbf{x}\_{n+1},f\mathbf{x}\_{n},\mathbf{g}\mathbf{x}\_{n})}{2s} \Big\} \\ &= \max \{ d(\mathbf{g}\mathbf{x}\_{n},\mathbf{g}\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n}), d(\mathbf{g}\mathbf{x}\_{n},\mathbf{g}\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n}), d(\mathbf{g}\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n+2},\mathbf{g}\mathbf{x}\_{n}) \} \\ &\quad \frac{d(\mathbf{g}\mathbf{x}\_{n},\mathbf{g}\mathbf{x}\_{n+2},\mathbf{g}\mathbf{x}\_{n}) + d(\mathbf{g}\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n})}{2s} \} \\ &= \quad d(\mathbf{g}\mathbf{x}\_{n+1},\mathbf{g}\mathbf{x}\_{n+2},\mathbf{g}\mathbf{x}\_{n}), \end{split}$$

and

$$\begin{split} N(\mathbf{x}\_{\boldsymbol{\mathsf{n}},\boldsymbol{\mathsf{x}}\_{\boldsymbol{n}+1},\mathbf{g}\mathbf{x}\_{\boldsymbol{n}})} &= \min \{ d(\mathbf{g}\mathbf{x}\_{\boldsymbol{\mathsf{n}},\boldsymbol{f}}\mathbf{x}\_{\boldsymbol{n},\boldsymbol{\mathsf{g}}}\mathbf{g}\mathbf{x}\_{\boldsymbol{n}}), d(\mathbf{g}\mathbf{x}\_{\boldsymbol{n}+1},\mathbf{f}\mathbf{x}\_{\boldsymbol{n}+1},\mathbf{g}\mathbf{x}\_{\boldsymbol{n}}), d(\mathbf{g}\mathbf{x}\_{\boldsymbol{n}},\mathbf{f}\mathbf{x}\_{\boldsymbol{n}+1},\mathbf{g}\mathbf{x}\_{\boldsymbol{n}}), \\ d(\mathbf{g}\mathbf{x}\_{\boldsymbol{n}+1},\mathbf{f}\mathbf{x}\_{\boldsymbol{n}},\mathbf{g}\mathbf{x}\_{\boldsymbol{n}}) &\} = 0. \end{split}$$

If *d*(*gxn*+1, *gxn*+2, *gxn*) = 0 for some *n* ∈ **N**0, then we have (due to (\*))

$$d(\mathcal{g}\mathbf{x}\_{n+1}, \mathcal{g}\mathbf{x}\_{n+2}, \mathbf{g}\mathbf{x}\_n) \quad \leq \quad \beta\_s(d(\mathcal{g}\mathbf{x}\_{n+1}, \mathcal{g}\mathbf{x}\_{n+2}, \mathcal{g}\mathbf{x}\_n))d(\mathcal{g}\mathbf{x}\_{n+1}, \mathcal{g}\mathbf{x}\_{n+2}, \mathcal{g}\mathbf{x}\_n),$$

yielding that

$$d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, \operatorname{gx}\_{n}) - \beta\_{\mathfrak{s}} (d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, \operatorname{gx}\_{n})) d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, \operatorname{gx}\_{n}) \le 0,$$

or

$$d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, \operatorname{gx}\_n)[1 - \beta\_s(d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, \operatorname{gx}\_n))] \le 0 \to (\*\*).$$
 
$$\text{Divide both sides in (\*\*) by } d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, \operatorname{gx}\_n) \ne 0 \text{, we obtain}$$

$$1 - \beta\_s(d(\mathcal{g}\mathbf{x}\_{n+1}, \mathcal{g}\mathbf{x}\_{n+2}, \mathcal{g}\mathbf{x}\_n)) \le 0,$$

or

$$
\beta\_{\mathbb{S}}(d(\operatorname{\mathbf{g}x}\_{n+1}, \operatorname{\mathbf{g}x}\_{n+2}, \operatorname{\mathbf{g}x}\_{n})) \geq 1,
$$

a contradiction [as *<sup>β</sup><sup>s</sup>* : [0, <sup>∞</sup>) <sup>→</sup> [0, <sup>1</sup> *<sup>s</sup>* ) and *<sup>s</sup>* <sup>≥</sup> 1 so *<sup>β</sup>s*(*c*) <sup>&</sup>lt; <sup>1</sup> *<sup>s</sup>* ≤ 1, that is *βs*(*c*) < 1 for all *c* ∈ [0, ∞)]. Therefore, we must have

$$d(\mathcal{g}\mathbf{x}\_{n+1}, \mathcal{g}\mathbf{x}\_{n+2}, \mathbf{g}\mathbf{x}\_n) = 0, \quad \text{for all } n \in \mathbb{N}\_0. \tag{3}$$

Thus, by the rectangle inequality and (3) we get

$$d(\mathcal{g}\mathbf{x}\_{n}\mathcal{g}\mathbf{x}\_{n+2\prime}a) \le \mathbf{s}[d(\mathcal{g}\mathbf{x}\_{n}\mathcal{g}\mathbf{x}\_{n+1\prime}a) + d(\mathcal{g}\mathbf{x}\_{n+1\prime}\mathcal{g}\mathbf{x}\_{n+2\prime}a)],\tag{4}$$

for all *n* ∈ **N**0, *a* ∈ *X*. Using (4), Lemma 3 and (1) we have

$$\begin{aligned} d(\mathcal{g}\mathbf{x}\_{n+1}, \mathcal{g}\mathbf{x}\_{n+2}, a) &= \quad d(f\mathbf{x}\_{n}, f\mathbf{x}\_{n+1}, a) \\ &\leq \quad \beta\_{\mathbb{S}}(M(\mathbf{x}\_{n}, \mathbf{x}\_{n+1}, a))M(\mathbf{x}\_{n}, \mathbf{x}\_{n+1}, a) + LN(\mathbf{x}\_{n}, \mathbf{x}\_{n+1}, a). \end{aligned} \tag{5}$$

Observe that

$$M(\mathbf{x}\_{n\prime}\mathbf{x}\_{n+1\prime}a) = \max\{d(\mathbf{g}\mathbf{x}\_{n\prime}\mathbf{g}\mathbf{x}\_{n+1\prime}a), d(\mathbf{g}\mathbf{x}\_{n+1\prime}\mathbf{g}\mathbf{x}\_{n+2\prime}a)\},$$

and

$$\begin{aligned} N(\mathbf{x}\_{\mathtt{n}}, \mathbf{x}\_{\mathtt{n}+1}, a) &= \min \{ d(\operatorname{gx}\_{\mathtt{n}} f \mathbf{x}\_{\mathtt{n}}, a), d(\operatorname{gx}\_{\mathtt{n}+1} f \mathbf{x}\_{\mathtt{n}+1}, a), d(\operatorname{gx}\_{\mathtt{n}} f \mathbf{x}\_{\mathtt{n}+1}, a), d(\operatorname{gx}\_{\mathtt{n}+1} f \mathbf{x}\_{\mathtt{n}}, a) \} \\ &= \min \{ d(\operatorname{gx}\_{\mathtt{n}} \operatorname{gx}\_{\mathtt{n}+1}, a), d(\operatorname{gx}\_{\mathtt{n}+1} \operatorname{gx}\_{\mathtt{n}+2} a), d(\operatorname{gx}\_{\mathtt{n}} \operatorname{gx}\_{\mathtt{n}+2} a), d(\operatorname{gx}\_{\mathtt{n}+1} \operatorname{gx}\_{\mathtt{n}+1}, a) \} \\ &= 0. \end{aligned}$$

Now, if *M*(*xn*, *xn*+1, *a*) = *d*(*gxn*+1, *gxn*+2, *a*), then from (5) we have

$$d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, a) \le \beta\_5(d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, a))d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, a) < d(\operatorname{gx}\_{n+1}, \operatorname{gx}\_{n+2}, a),$$

a contradiction. Hence, *M*(*xn*, *xn*+1, *a*) = *d*(*gxn*, *gxn*+1, *a*), and

$$d(\mathcal{g}\mathbf{x}\_{n+1}, \mathcal{g}\mathbf{x}\_{n+2}, a) \le \beta\_3 (d(\mathcal{g}\mathbf{x}\_n, \mathcal{g}\mathbf{x}\_{n+1}, a)) d(\mathcal{g}\mathbf{x}\_n, \mathcal{g}\mathbf{x}\_{n+1}, a) \\ < d(\mathcal{g}\mathbf{x}\_n, \mathcal{g}\mathbf{x}\_{n+1}, a), \tag{6}$$

for all *n* ∈ **N**<sup>0</sup> and *a* ∈ *X*, which implies that the sequence {*d*(*gxn*, *gxn*+1, *a*)} is strictly decreasing of positive numbers. Hence, there exists *<sup>δ</sup>* <sup>≥</sup> 0 such that lim*n*→<sup>∞</sup> *<sup>d</sup>*(*gxn*, *gxn*+1, *<sup>a</sup>*) = *<sup>δ</sup>*. Suppose that *δ* > 0. So, taking the limit as *n* → ∞, from (6) we obtain

$$\frac{1}{s}\delta \le \delta \le \lim\_{n \to \infty} \beta\_{\mathfrak{s}}(d(\operatorname{gx}\_{n\prime}\operatorname{gx}\_{n+1\prime}a))\delta \le \frac{1}{s}\delta.$$

Hence,

$$\lim\_{n \to \infty} \beta\_s(d(\operatorname{gx}\_{n'} \operatorname{gx}\_{n+1'} a)) = \frac{1}{s}.$$

From the property of *<sup>β</sup>s*, we conclude that lim*n*→<sup>∞</sup> *<sup>d</sup>*(*gxn*, *gxn*+1, *<sup>a</sup>*) = 0 a contradiction, hence, *δ* = 0 and

$$\lim\_{n \to \infty} d(\operatorname{gx}\_{n\prime} \operatorname{gx}\_{n+1\prime} a) = 0. \tag{7}$$

**Step 2**: We claim that *d*(*gxi*, *gxj*, *gxk*) = 0 for all *i*, *j*, *k* ∈ **N**0. Since {*d*(*gxn*, *gxn*+1, *a*)} is strictly decreasing and *d*(*gx*0, *gx*1, *gx*0) = 0, we conclude that *d*(*gxn*, *gxn*+1, *gx*0) = 0, for all *n* ∈ **N**0.

Since *d*(*gxm*−1, *gxm*, *gxm*) = 0 for all *m* ∈ **N**<sup>0</sup> and {*d*(*gxn*, *gxn*+1, *a*)} is strictly decreasing we obtain that

$$d(\mathcal{g}\mathbf{x}\_{\boldsymbol{n}}, \mathcal{g}\mathbf{x}\_{\boldsymbol{n}+1}, \mathcal{g}\mathbf{x}\_{\boldsymbol{m}}) = 0, \quad \text{for all } \boldsymbol{n} \ge \boldsymbol{m} - 1. \tag{8}$$

For 0 ≤ *n* < *m* − 1, we have *m* − 1 ≥ *n* + 1, so from (8) we have

$$d(\mathcal{g}\mathbf{x}\_{m-1}, \mathcal{g}\mathbf{x}\_{m}, \mathcal{g}\mathbf{x}\_{n+1}) = d(\mathcal{g}\mathbf{x}\_{m-1}, \mathcal{g}\mathbf{x}\_{m}, \mathcal{g}\mathbf{x}\_{n}) = 0. \tag{9}$$

Thus, by the rectangle inequality, *d*(*gxn*, *gxn*+1, *gxn*+1) = 0, and using (9) we obtain

$$\begin{split}d(\operatorname{gx}\_{\boldsymbol{\pi},\boldsymbol{\eta}}\operatorname{gx}\_{\boldsymbol{n}+1},\operatorname{gx}\_{\boldsymbol{m}}) &\leq \operatorname{s}[d(\operatorname{gx}\_{\boldsymbol{n}},\operatorname{gx}\_{\boldsymbol{n}+1},\operatorname{gx}\_{\boldsymbol{m}-1}) + d(\operatorname{gx}\_{\boldsymbol{n}+1},\operatorname{gx}\_{\boldsymbol{m}}\operatorname{gx}\_{\boldsymbol{m}-1}) + d(\operatorname{gx}\_{\boldsymbol{m}},\operatorname{gx}\_{\boldsymbol{n}}\operatorname{gx}\_{\boldsymbol{m}-1})]] \\ &= \operatorname{s}d(\operatorname{gx}\_{\boldsymbol{n}},\operatorname{gx}\_{\boldsymbol{n}+1},\operatorname{gx}\_{\boldsymbol{m}-1}) \\ &\leq \operatorname{s}d(\operatorname{gx}\_{\boldsymbol{n}},\operatorname{gx}\_{\boldsymbol{n}+1},\operatorname{gx}\_{\boldsymbol{n}+1}) = 0. \end{split}$$

Therefore, we get

$$d(\mathcal{g}\mathbf{x}\_{n}, \mathcal{g}\mathbf{x}\_{n+1}, \mathcal{g}\mathbf{x}\_{m}) = 0,\quad \text{for all } 0 \le n < m - 1. \tag{10}$$

Hence, from (8) and (10) we have

$$d(\mathcal{g}\mathbf{x}\_{\mathsf{m}}\,\mathcal{g}\mathbf{x}\_{\mathsf{n}+1},\mathcal{g}\mathbf{x}\_{\mathsf{m}}) = 0,\quad \text{for all } n, m \in \mathbb{N}\_0.$$

Now, for all *<sup>i</sup>*, *<sup>j</sup>*, *<sup>k</sup>* ∈ **<sup>N</sup>**0, *<sup>i</sup>* < *<sup>j</sup>* and *<sup>d</sup>*(*gxi*, *gxj*, *gxj*−1) = *<sup>d</sup>*(*gxk*, *gxj*, *gxj*−1) = 0, applying the rectangle inequality we get

$$\begin{split}d(\operatorname{gx}\_{i},\operatorname{gx}\_{j},\operatorname{gx}\_{k})&\leq\operatorname{s}[d(\operatorname{gx}\_{i},\operatorname{gx}\_{j},\operatorname{gx}\_{j-1})+d(\operatorname{gx}\_{j},\operatorname{gx}\_{k}\operatorname{gx}\_{j-1})+d(\operatorname{gx}\_{k},\operatorname{gx}\_{i},\operatorname{gx}\_{j-1})]\\&=\operatorname{sd}(\operatorname{gx}\_{k},\operatorname{gx}\_{i},\operatorname{gx}\_{j-1})\\&\leq\operatorname{s}^{2}d(\operatorname{gx}\_{k},\operatorname{gx}\_{i},\operatorname{gx}\_{j-2})\leq...\leq\operatorname{s}^{j-i}d(\operatorname{gx}\_{k},\operatorname{gx}\_{i},\operatorname{gx}\_{i})=0.\end{split}$$

Therefore, for all *i*, *j*, *k* ∈ **N**0, we have

$$d(\mathcal{g}\mathbf{x}\_{i\prime}\mathcal{g}\mathbf{x}\_{j\prime}\mathcal{g}\mathbf{x}\_{k}) = 0.\tag{11}$$

**Step 3**: We show that {*gxn*} is a *b*2-Cauchy sequence. Suppose to the contrary that {*gxn*} is not a *b*2-Cauchy sequence. Then there is *ε* > 0 such that for an integer *k* there exist integers *n*(*k*), *m*(*k*) with *n*(*k*) > *m*(*k*) > *k* such that

$$d(g\chi\_{m(k)}, g\chi\_{n(k)}, a) \ge \varepsilon,\tag{12}$$

for every integer *k*, let *n*(*k*) be the least positive integer with *n*(*k*) > *m*(*k*), satisfying (12) and such that

$$d(\mathcal{g}\mathbf{x}\_{m(k)}, \mathcal{g}\mathbf{x}\_{n(k)-1}, a) < \varepsilon. \tag{13}$$

Using the rectangle inequality, (11) and (12) we have

$$\varepsilon \le d(\mathcal{g}\mathbf{x}\_{m(k)}, \mathcal{g}\mathbf{x}\_{n(k)}, a) \le s |d(\mathcal{g}\mathbf{x}\_{m(k)}, \mathcal{g}\mathbf{x}\_{n(k)-1}, a) + d(\mathcal{g}\mathbf{x}\_{n(k)}, \mathcal{g}\mathbf{x}\_{n(k)-1}, a)|.$$

Again, using the rectangle inequality and (11) in the above inequality, it follows that

$$\varepsilon \le \varepsilon^2 [d(\operatorname{gx}\_{m(k)}, \operatorname{gx}\_{m(k)-1}, a) + d(\operatorname{gx}\_{m(k)-1}, \operatorname{gx}\_{n(k)-1}, a)] + \operatorname{sd}(\operatorname{gx}\_{n(k)}, \operatorname{gx}\_{n(k)-1}, a)].$$

In addition,

$$d(\mathcal{g}\mathbf{x}\_{m(k)-1}, \mathbf{g}\mathbf{x}\_{n(k)-1}, a) \le \mathbf{s} [d(\mathcal{g}\mathbf{x}\_{m(k)-1}, \mathbf{g}\mathbf{x}\_{m(k)}, a) + d(\mathcal{g}\mathbf{x}\_{n(k)-1}, \mathbf{g}\mathbf{x}\_{m(k)}, a)].$$

Taking the upper limit as *k* → ∞, in the above three inequalities and from (7) and (13) it follows that

$$\varepsilon \le \limsup\_{k \to \infty} d(\lg x\_{m(k)}, \lg x\_{n(k)}, a) \prec^{\circ} \text{s} \varepsilon \tag{14}$$

$$\frac{\varepsilon}{s^2} \le \limsup\_{k \to \infty} d(\mathcal{g}\mathbf{x}\_{m(k)-1}, \mathcal{g}\mathbf{x}\_{n(k)-1}, a) \prec s\varepsilon,\tag{15}$$

$$\frac{\varepsilon}{s^3} \le \limsup\_{k \to \infty} d(\mathcal{g}\mathbf{x}\_{m(k)}, \mathcal{g}\mathbf{x}\_{n(k)-1}, a) < \varepsilon. \tag{16}$$

Again, using the rectangle inequality, (11) and (12) we get

$$d(\mathcal{g}\mathbf{x}\_{\mathfrak{m}(k)-1}, \mathcal{g}\mathbf{x}\_{\mathfrak{n}(k)}, a) \le \mathbf{s}[d(\mathcal{g}\mathbf{x}\_{\mathfrak{m}(k)-1}, \mathcal{g}\mathbf{x}\_{\mathfrak{n}(k)-1}, a) + d(\mathcal{g}\mathbf{x}\_{\mathfrak{n}(k)}, \mathcal{g}\mathbf{x}\_{\mathfrak{n}(k)-1}, a)],$$

$$\varepsilon \le d(\mathcal{g}\mathbf{x}\_{\mathfrak{m}(k)}, \mathcal{g}\mathbf{x}\_{\mathfrak{n}(k)}, a) \le \mathbf{s}[d(\mathcal{g}\mathbf{x}\_{\mathfrak{m}(k)}, \mathcal{g}\mathbf{x}\_{\mathfrak{m}(k)-1}, a) + d(\mathcal{g}\mathbf{x}\_{\mathfrak{n}(k)}, \mathcal{g}\mathbf{x}\_{\mathfrak{m}(k)-1}, a)].$$

Taking the upper limit as *k* → ∞, in the above two inequalities and from (7) and (15), we get

$$\frac{\varepsilon}{2} \le \limsup\_{k \to \infty} d(\lg x\_{m(k)-1}, \lg x\_{n(k)}, a) < s^2 \varepsilon. \tag{17}$$

Now, from Lemma <sup>3</sup> we have *f xm*(*k*)−1R *f xn*(*k*)−<sup>1</sup> for all *<sup>m</sup>*(*k*), *<sup>n</sup>*(*k*) <sup>∈</sup> **<sup>N</sup>**<sup>0</sup> with *<sup>m</sup>*(*k*) <sup>&</sup>lt; *<sup>n</sup>*(*k*). Hence, from (1) we conclude that

$$\begin{split}d(\mathcal{g}\mathbf{x}\_{\mathbf{m}(k)},\mathcal{g}\mathbf{x}\_{\mathbf{n}(k)},a) &= d(f\mathbf{x}\_{\mathbf{m}(k)-1},f\mathbf{x}\_{\mathbf{n}(k)-1},a) \\ &\leq \ \ \ \ \mathfrak{F}\_{\mathbf{s}}(M(\mathbf{x}\_{\mathbf{m}(k)-1},\mathbf{x}\_{\mathbf{n}(k)-1},a))M(\mathbf{x}\_{\mathbf{m}(k)-1},\mathbf{x}\_{\mathbf{n}(k)-1},a)+LN(\mathbf{x}\_{\mathbf{m}(k)-1},\mathbf{x}\_{\mathbf{n}(k)-1},a), \end{split}\tag{18}$$

where

$$M(\mathbf{x}\_{m(k)-1}, \mathbf{x}\_{n(k)-1}, a) = \max\{d(\operatorname{gx}\_{m(k)-1}, \operatorname{gx}\_{n(k)-1}, a), d(\operatorname{gx}\_{m(k)-1}, f \mathbf{x}\_{m(k)-1}, a), \dots, a - \mathbf{x}\_{n(k)}\}}{d(\operatorname{gx}\_{n(k)-1}, f \mathbf{x}\_{n(k)-1}, a) + \frac{d(\operatorname{gx}\_{n(k)-1}, f \mathbf{x}\_{m(k)-1}, a)}{2s}} \}\_{\prime}$$

$$= \max\left\{d(\operatorname{gx}\_{m(k)-1}, \operatorname{gx}\_{n(k)-1}, a), d(\operatorname{gx}\_{m(k)-1}, \operatorname{gx}\_{m(k)}, a), d(\operatorname{gx}\_{n(k)-1}, \operatorname{gx}\_{n(k)}, a), \dots, \max\{d(\operatorname{gx}\_{n(k)-1}, \operatorname{gx}\_{n(k)}, a)\}\right\}}\_{\prime \prime}$$

$$\frac{d(\operatorname{gx}\_{m(k)-1}, \operatorname{gx}\_{n(k)}, a) + d(\operatorname{gx}\_{n(k)-1}, \operatorname{gx}\_{m(k)}, a)}{2s}\right\},\tag{19}$$

and

$$\begin{split} \mathcal{N}(\mathbf{x}\_{m(k)-1}\mathbf{x}\_{n(k)-1}, a) &= \min\{d(\mathbf{g}\mathbf{x}\_{m(k)-1}, \mathbf{g}\mathbf{x}\_{m(k)}, a), d(\mathbf{g}\mathbf{x}\_{n(k)-1}, \mathbf{g}\mathbf{x}\_{n(k)}, a), d(\mathbf{g}\mathbf{x}\_{m(k)-1}, \mathbf{g}\mathbf{x}\_{n(k)}, a), \\ &\qquad d(\mathbf{g}\mathbf{x}\_{n(k)-1}, \mathbf{g}\mathbf{x}\_{m(k)}, a)\}. \end{split} \tag{20}$$

Taking the upper limit as *k* → ∞, in (19), (20) and using (7), (15)–(17) it follows that

$$\frac{\varepsilon}{s^2} \le \limsup\_{k \to \infty} M(\mathbf{x}\_{m(k)-1}, \mathbf{x}\_{n(k)-1}, a) < \text{ s\varepsilon},\tag{21}$$

and

$$\limsup\_{k \to \infty} \mathcal{N}(\mathbf{x}\_{m(k)-1}, \mathbf{x}\_{n(k)-1}, a) = 0. \tag{22}$$

Now, taking the upper limit as *k* → ∞ in (18) and using (14), (21) and (22), we conclude that

$$\frac{1}{s} = \frac{\varepsilon}{s\varepsilon} \le \frac{\limsup\_{k \to \infty} d(\operatorname{gx}\_{\operatorname{m}(k)}, \operatorname{gx}\_{\operatorname{n}(k)}, a)}{\limsup\_{k \to \infty} \operatorname{M}(\operatorname{x}\_{\operatorname{m}(k)-1}, \operatorname{x}\_{\operatorname{n}(k)-1}, a)} \le \limsup\_{k \to \infty} \beta\_{s}(\operatorname{M}(\operatorname{x}\_{\operatorname{m}(k)-1}, \operatorname{x}\_{\operatorname{n}(k)-1}, a)) \le \frac{1}{s}.$$

Thus, lim sup *k*→∞ *<sup>β</sup>s*(*M*(*xm*(*k*)−1, *xn*(*k*)−1, *<sup>a</sup>*)) = <sup>1</sup> *<sup>s</sup>* . Hence, lim sup *k*→∞ *<sup>M</sup>*(*xm*(*k*)−1, *xn*(*k*)−1, *<sup>a</sup>*) = 0, which is a contradiction. Therefore, {*gxn*} is a *b*2-Cauchy sequence. As *g*(*X*) is *b*2-complete subspace of *X*, then there exist *z* ∈ *X* such that

$$\lim\_{n \to \infty} \lg \mathbf{x}\_{\mathbb{N}} = \lim\_{n \to \infty} f \mathbf{x}\_{\mathbb{N}} = \lg \mathbf{z}.\tag{23}$$

Now, we show that *z* is a point of coincidence of *f* and *g*. From condition (iii), we have R|*g*(*X*) is *<sup>d</sup>*-self closed and (1) holds for all *<sup>x</sup>*, *<sup>y</sup>*, *<sup>a</sup>* <sup>∈</sup> *<sup>X</sup>* with *gx*R*gy* and *f x*R *f y*. As {*gxn*} ⊆ *g*(*X*), {*gxn*} is R|*g*(*X*)-preserving and *gxn* → *gz* so there exists a subsequence {*gxn*(*k*)}⊆{*gxn*} such that *gxn*(*k*)R|*g*(*X*)*gz* for all *k* ∈ **N**<sup>0</sup> and since R is (*f* , *g*)-closed then *f xn*(*k*)R|*g*(*X*) *f z* for all *k* ∈ **N**0.

If *f xn*(*k*) = *f z* for all *k* > *k*0, and *k*0, *k* ∈ **N**0, then lim *<sup>k</sup>*→<sup>∞</sup> *f xn*(*k*) <sup>=</sup> *f z*, and since lim*n*→<sup>∞</sup> *f xn* <sup>=</sup> *gz*, we have *f z* <sup>=</sup> *gz*, that is *<sup>z</sup>* is a coincidence point of *<sup>f</sup>* and *<sup>g</sup>*.

On other hand, if *f xn*(*k*) = *f z* for all *k* > *k*0, and *k*0, *k* ∈ **N**0, then *f xn*(*k*)R|*g*(*X*) *f z* and *f xn*(*k*) <sup>=</sup> *f z* for all *<sup>k</sup>* <sup>&</sup>gt; *<sup>k</sup>*0, and *<sup>k</sup>*0, *<sup>k</sup>* <sup>∈</sup> **<sup>N</sup>**0. Thus, *gxn*(*k*)R|*g*(*X*)*gz* and *f xn*(*k*)R|*g*(*X*) *f z*, and from (1), we have

$$d(\mathcal{g}\mathbf{x}\_{n(k)+1}, fz, a) = d(f\mathbf{x}\_{n(k)}, fz, a) \le \beta\_s(M(\mathbf{x}\_{n(k)}, z, a))M(\mathbf{x}\_{n(k)}, z, a) + LN(\mathbf{x}\_{n(k)}, z, a), \tag{24}$$

where

$$M(\mathbf{x}\_{n(k)}, z, a) \quad = \max \left\{ d(\operatorname{gx}\_{n(k)}, \operatorname{gz}\_{\prime} a), d(\operatorname{gx}\_{n(k)}, \operatorname{gz}\_{n(k)+1} a), d(\operatorname{gz}\_{\prime} f z, a), \quad \forall a \in \mathbb{R}^{n} \right\}, \tag{25}$$

$$\frac{d(\operatorname{gx}\_{n(k)}, f z, a) + d(\operatorname{gz}\_{\prime} \operatorname{gz}\_{n(k)+1}, a)}{2s} \Big\}, \tag{25}$$

and

$$N(\mathbf{x}\_{n(k)}, z, a) \quad = \min \{ d(\mathbf{g}\mathbf{x}\_{n(k)}, \mathbf{g}\mathbf{x}\_{n(k)+1}, a), d(\mathbf{g}z, fz, a), d(\mathbf{g}\mathbf{x}\_{n(k)}, fz, a), d(\mathbf{g}z, \mathbf{g}\mathbf{x}\_{n(k)+1}, a) \}. \tag{26}$$

Letting *k* → ∞ in (25), (26), we get

$$\limsup\_{k \to \infty} M(\mathfrak{x}\_{n(k)}, z, a) = \max \left\{ d(gz, fz, a), \frac{\limsup\_{k \to \infty} d(gz\_{n(k)}, fz, a)}{2s} \right\},$$

and

$$\limsup\_{k \to \infty} \mathcal{N}(\mathfrak{x}\_{n(k)}, z, a) = 0. \tag{27}$$

From Lemma 1, we have

$$\frac{d(gz, fz, a)}{s} \le \limsup\_{k \to \infty} d(gz\_{n(k)}, fz, a) \le s d(gz, fz, a). \tag{28}$$

Thus,

$$\max\{d(gz, fz, a), \frac{d(gz, fz, a)}{2s^2}\} \le \limsup\_{k \to \infty} M(x\_{n(k)}, z, a) \le \max\{d(gz, fz, a), \frac{d(gz, fz, a)}{2}\}.$$

yields,

$$\limsup\_{k \to \infty} M(\mathfrak{x}\_{n(k)}, z, a) = d(\mathfrak{g}z, fz, a), \tag{29}$$

Again, taking the upper limit as *k* → ∞, in (24) and using Lemma 1, (27) and (29), we get

$$\begin{aligned} \frac{d(gz, fz, a)}{s} &\leq \limsup\_{k \to \infty} d(gz\_{n(k)+1}, fz, a) \\ &\leq \limsup\_{k \to \infty} \beta\_s(M(\mathbf{x}\_{n(k)}, z, a)) \limsup\_{k \to \infty} M(\mathbf{x}\_{n(k)}, z, a) \\ &\leq \limsup\_{k \to \infty} \beta\_s(M(\mathbf{x}\_{n(k)}, z, a)) d(gz, fz, a) \\ &\leq \frac{1}{s} d(gz, fz, a) .\end{aligned}$$

Hence, lim sup *k*→∞ *βs*(*M*(*xn*(*k*), *z*, *a*)) = <sup>1</sup> *<sup>s</sup>* , so from the property of *β<sup>s</sup>* we conclude that lim sup *k*→∞ *M*(*xn*(*k*), *z*, *a*) = 0 implies *d*(*gz*, *f z*, *a*) = 0 for all *a* ∈ *X*. That is, *gz* = *f z*. This shows that *f* and *g* have a coincidence point.

The next theorem shows that under some additional hypotheses we can deduce the existence and uniqueness of a common fixed point.

**Theorem 2.** *In addition to the hypotheses of Theorem 1, suppose that f and g are weakly compatible and for all coincidence points u*, *v of f and g*, *there exists w* ∈ *X such that gu*R*gw and gv*R*gw. Then f and g have a unique common fixed point.*

**Proof.** The set of coincidence points of *f* and *g* is not empty due to Theorem 1. Suppose that *u* and *v* are two coincidence points of *f* and *g*, that is, *f u* = *gu* and *f v* = *gv*. We will show that *gu* = *gv*. By our assumption, there exists *w* ∈ *X* such that

$$
\mathfrak{g}u\mathfrak{R}\mathfrak{g}w \quad \text{and} \quad \mathfrak{g}v\mathfrak{R}\mathfrak{g}w.\tag{30}
$$

Now, proceeding similarly to the proof of Theorem 1, we can define a sequence {*wn*} in *<sup>X</sup>* as *f wn* <sup>=</sup> *gwn*+<sup>1</sup> for all *<sup>n</sup>* <sup>∈</sup> **<sup>N</sup>**<sup>0</sup> and *<sup>w</sup>*<sup>0</sup> <sup>=</sup> *<sup>w</sup>*, with lim*n*→<sup>∞</sup> *<sup>d</sup>*(*gwn*, *gwn*+1, *<sup>a</sup>*) = 0. Since *gu*R*gw*<sup>0</sup> (*gv*R*gw*0) and R is (*f* , *g*)-closed, we conclude that *f u*R *f w*0(*f v*R *f w*0). Hence, *gu*R*gw*1(*gv*R*gw*1). By induction, we have

*gu*R*gwn* and *gv*R*gwn*, ∀*n* ∈ **N**0. (31)

From (1) and using (31), we obtain

$$d(\mathcal{g}u\_\prime \mathcal{g}w\_{n+1}, a) \quad = \quad d(fu\_\prime f w\_n, a) \le \beta\_5(M(u, w\_n, a))M(u, w\_n, a) + LN(u, w\_n, a), \tag{32}$$

where

$$\begin{split} M(u, w\_{\mathbb{H}}, a) &= \max \left\{ d(\mathcal{g}u, \mathcal{g}w\_{\mathbb{H}}, a), d(\mathcal{g}u, fu, a), d(\mathcal{g}w\_{\mathbb{H}}, fu\_{\mathbb{H}}, a), \frac{d(\mathcal{g}u, fw\_{\mathbb{H}}, a) + d(\mathcal{g}w\_{\mathbb{H}}, fu, a)}{2s} \right\}, \\ &= \max \left\{ d(\mathcal{g}u, \mathcal{g}w\_{\mathbb{H}}, a), d(\mathcal{g}w\_{\mathbb{H}}, \mathcal{g}w\_{\mathbb{H}+1}, a), \frac{d(\mathcal{g}u, \mathcal{g}w\_{\mathbb{H}+1}, a) + d(\mathcal{g}w\_{\mathbb{H}}, \mathcal{g}u, a)}{2s} \right\}, \end{split}$$

and

$$\begin{aligned} N(u, w\_n, a) &= \min\{d(gu, fu, a), d(gw\_n, fw\_n, a), d(gu, fw\_n, a), d(gw\_n, fu, a)\} \\ &= \min\{d(gu, gu, a), d(gw\_n, gw\_{n+1}, a), d(gu, gw\_{n+1}, a), d(gw\_n, gu, a)\} = 0. \end{aligned}$$

Hence,

$$\begin{aligned} d(\mathcal{g}u, \mathcal{g}w\_{n+1}, a) &\leq \quad \beta\_s(\mathcal{M}(\mu, w\_{n'}a)) \mathcal{M}(\mu, w\_{n'}a) \\ &< \quad \frac{1}{s} \mathcal{M}(\mu, w\_{n'}a) \leq \mathcal{M}(\mu, w\_{n'}a). \end{aligned}$$

Since,

$$\begin{aligned} d(\mathcal{g}u, \mathcal{g}w\_{n+1}, a) &< \mathcal{M}(u, w\_n, a) \\ &= \max\left\{ d(\mathcal{g}u, \mathcal{g}w\_n, a), d(\mathcal{g}w\_n, \mathcal{g}w\_{n+1}, a), \frac{d(\mathcal{g}u, \mathcal{g}w\_{n+1}, a) + d(\mathcal{g}w\_n, \mathcal{g}u, a)}{2s} \right\} \\ &= \max\left\{ d(\mathcal{g}u, \mathcal{g}w\_n, a), d(\mathcal{g}w\_n, \mathcal{g}w\_{n+1}, a) \right\}. \end{aligned}$$

Thus,

$$M(\mathfrak{u}, w\_{\mathfrak{n}}, a) \quad = \max \left\{ d(\mathcal{g}\mathfrak{u}, \mathcal{g}w\_{\mathfrak{n}}, a), d(\mathcal{g}w\_{\mathfrak{n}}, \mathcal{g}w\_{\mathfrak{n}+1}, a) \right\}.$$

(Case1): if *M*(*u*, *wn*, *a*) = *d*(*gu*, *gwn*, *a*), then

$$d(\mathcal{g}u, \mathcal{g}w\_{n+1}, a) \le \beta\_s(d(\mathcal{g}u, \mathcal{g}w\_n, a))d(\mathcal{g}u, \mathcal{g}w\_n, a) < \frac{1}{s}d(\mathcal{g}u, \mathcal{g}w\_n, a) \le d(\mathcal{g}u, \mathcal{g}w\_n, a), \tag{33}$$

it follows that *d*(*gu*, *gwn*+1, *a*) < *d*(*gu*, *gwn*, *a*). Thus, {*d*(*gu*, *gwn*, *a*)} is strictly decreasing. Hence, there exists *<sup>γ</sup>* <sup>≥</sup> 0 such that lim*n*→<sup>∞</sup> *<sup>d</sup>*(*gu*, *gwn*, *<sup>a</sup>*) = *<sup>γ</sup>*. Letting *<sup>n</sup>* <sup>→</sup> <sup>∞</sup> in (33), we obtain

$$\begin{aligned} \frac{\gamma}{s} \le \gamma = \lim\_{n \to \infty} d(gu, gw\_{n+1}, a) &\le \lim\_{n \to \infty} \beta\_s(d(gu, gw\_n, a)) \lim\_{n \to \infty} d(gu, gw\_{n'}, a) \\ &\le \lim\_{n \to \infty} \beta\_s(d(gu, gw\_n, a))\gamma \\ &\le \frac{\gamma}{s} \end{aligned}$$

this implies

$$\frac{1}{s} \le \lim\_{n \to \infty} \beta\_s(d(gu, gw\_{n'}a)) < \frac{1}{s}.$$

Thus,

$$\lim\_{n \to \infty} \beta\_s(d(gu\_\prime gw\_n, a)) = \frac{1}{s}.$$

From the property of *<sup>β</sup>s*, we conclude that lim*n*→<sup>∞</sup> *<sup>d</sup>*(*gu*, *gwn*, *<sup>a</sup>*) = 0. (Case2): If *M*(*u*, *wn*, *a*) = *d*(*gwn*, *gwn*+1, *a*), then

$$d(\mathcal{g}u, \mathcal{g}w\_{n+1}, a) \quad \le \quad \beta\_{\mathcal{S}}(d(\mathcal{g}w\_{n}, \mathcal{g}w\_{n+1}, a))d(\mathcal{g}w\_{n}, \mathcal{g}w\_{n+1}, a).$$

Therefore,

$$\lim\_{n \to \infty} d(\mathcal{g}u, \mathcal{g}w\_{n+1}, a) \le \lim\_{n \to \infty} \beta\_{\mathbb{S}}(d(\mathcal{g}w\_{n}, \mathcal{g}w\_{n+1}, a)) \lim\_{n \to \infty} d(\mathcal{g}w\_{n}, \mathcal{g}w\_{n+1}, a) = 0.$$

This yields lim*n*→<sup>∞</sup> *<sup>d</sup>*(*gu*, *gwn*+1, *<sup>a</sup>*) = 0. Therefore, from all cases we conclude that

$$\lim\_{u \to \infty} d(gu, gw\_{n\prime}a) = 0.\tag{34}$$

Similarly, we can show that

$$\lim\_{n \to \infty} d(\mathcal{g}v, \mathcal{g}w\_n, a) = 0. \tag{35}$$

Hence, from (34) and (35), we obtain *gu* = *gv*. That is, *f* and *g* have a unique point of coincidence. From Lemma 2 *f* and *g* have a unique common fixed point.

Now, we give an example to justify the hypotheses of Theorem 1.

**Example 2.** *Let X* <sup>=</sup> {*p*, *<sup>q</sup>*,*r*, *<sup>t</sup>*} *be a set with b*2*-metric d* : *<sup>X</sup>*<sup>3</sup> → R *defined by*

*d*(*p*, *q*,*r*) = 0, *d*(*p*, *q*, *t*) = 4, *d*(*p*,*r*, *t*) = 1, *d*(*q*,*r*, *t*) = 6,

*with symmetry in all variables and if at least two of the arguments are equal then d*(*x*, *y*, *a*) = 0*. Then* (*X*, *d*) *is a complete b*2*-metric space with s* = <sup>6</sup> <sup>5</sup> . *Define a binary relation* R *on X by*

$$\mathcal{R} = \{ (p,p), (q,q), (r,r), (p,q), (q,r), (p,r), (r,p), (r,q) \}.$$

*Define f* , *g* : *X* → *X and β* : (0, ∞) → [0, 1) *as follows:*

$$f = \begin{pmatrix} p & q & r & t \\ p & p & r & t \end{pmatrix}, \quad \mathbf{g} = \begin{pmatrix} p & q & r & t \\ p & r & q & t \end{pmatrix}, \quad \beta\_s(t) = \frac{1+t}{s(1+2t)}$$

*We show that all the hypotheses of Theorem 1 are satisfied. Clearly,* (*X*, *d*) *is a complete b*2*-metric space and f*(*X*) ⊆ *g*(*X*), *g*(*X*) *is a b*2*-complete subspace of X*. R = R|*g*(*X*) *is transitive. There is r* ∈ *X such that q* = *gr*R *f r* = *r*. *Since* R|*g*(*X*) *is finite, then it is d-self closed. We show that* R *is* (*f* , *g*)*-closed, we study the nontrivial cases:*


*Now, we check the contractive condition 2. The nontrivial cases are when a* = *t*, *gp*R*gr and f p*<sup>R</sup> *f r* , *gr*R*gq and f r*<sup>R</sup> *f q and gq*R*gr and f q*<sup>R</sup> *f r* .

*In all three cases, we get M*(*p*,*r*, *t*) = *M*(*r*, *q*, *t*) = *M*(*q*,*r*, *t*) = 6, *N*(*p*,*r*, *a*) = *N*(*r*, *q*, *t*) = *N*(*q*,*r*, *t*) = 0, *and then*

$$1 = d(fp, fr, t) = d(p, r, t) \le \frac{35}{13} = \beta\_s(6) \\ 6 = \beta\_s(M(p, r, a))M(p, r, a) + LN(p, r, a),$$

$$1 = d(fr, fq, t) = d(r, p, t) \le \frac{35}{13} = \beta\_s(6) \\ 6 = \beta\_s(M(r, q, a))M(r, q, a) + LN(r, q, a),$$

$$1 = d(fq, fr, t) = d(p, r, t) \le \frac{35}{13} = \beta\_s(6) \\ 6 = \beta\_s(M(q, r, a))M(q, r, a) + LN(q, r, a),$$

*Therefore, all the hypotheses of Theorem 1 are satisfied. Then f and g have two coincidence fixed points p and t. Noting that p*, *t are not* R*-comparable so the uniqueness of coincidence point is not fulfilled.*

By taking *g* = *I* in Theorems 1 and 2 we deduce the following result.

**Corollary 1.** *Let* (*X*, *d*) *be a complete b*2*-metric space endowed with a transitive binary relation* R : *X* → *X and f* : *X* → *X*. *Assume that f is almost* R*-Geraghty type contraction mapping and the following conditions hold:*


*Then f has a fixed point. Moreover, if for u*, *v* ∈ *Fix*(*f*)*, there exists w* ∈ *X such that u*R*w and v*R*w*, *then f has a unique fixed point.*
