Note on recent common coupled fixed point results in multiplicative metric spaces

After the appearance of paper [9] in which the notion of multiplicative metric space was introduced, a large number of scientific papers appeared in which several fixed point theorems were proved in such spaces (for more details, see [2, 13, 14, 15 16, 18, 19]). However, recently, it had not yet been a hot topic since some authors appealed to the equivalence of some metric and multiplicative metric fixed point results. They gave some remarks to support the fact if ones had acted some logarithmic transformation to the multiplicative metric (see [3, 8, 11, 12, 17]). Very recently, Jiang and Gu [13] gave some common coupled fixed point theorems for two mappings satisfying ϕ-type contractive condition in multiplicative metric space. Based on [13], throughout this paper, by using several nontrivial methods, we obtain some common coupled fixed point theorems in metric spaces. Furthermore, we claim that all results of [13] can be reduced to the counterpart of metric spaces. Similar coincidences also happen to [15].

For the reader who is unfamiliar with multiplicative metric space, we recall some of its notions and results as follows:

Definition 1.1

[9] Let X be a nonempty set. An operator d_* : X × X → ℝ is called a multiplicative metric on X, if it satisfies:

(m1)

d_*(x,y) ≥ 1 for all x,y ∈ X and d_*(x,y) = 1 if and only if x = y;

(m2)

d_*(x,y) = d_*(y,x) for all x,y ∈ X;

(m3)

d_*(x,z) ≤ d_*(x,y) ⋅ d_*(y,z) for all x,y, z ∈ X.

In this case, the pair (X,d_*) is called a multiplicative metric space.

Definition 1.2

[1] Let X be a nonempty set, F : X × X → X and g : X → X be mappings. An element (x,y) ∈ X × X is called

a coupled coincidence point of F and g if F(x,y) = gx and F(y,x) = gy. In this case, (gx,gy) is called a coupled point of coincidence of F and g;

a common coupled fixed point of F and g if F(x,y) = gx = x and F(y,x) = gy = y.

Remark 1

If g = I_X (identity mapping) in Definition 1.2, then the pair (x,y) is called a coupled fixed point (see [4, 5, 6, 7]).

Definition 1.3

[1] Let X be a nonempty set. We say that the mappings F : X × X → X and g : X → X are called

w-compatible if gF(x,y) = F(gx,gy) whenever F(x,y) = gx and F(y,x) = gy;

w^*-compatible gF(x,x) = F(gx,gx) whenever F(x,x) = gx.

Let Φ₁ denote the set of all functions ϕ₁ : [1,∞)⁵ → [0,∞) satisfying

(a1)

ϕ₁ is nondecreasing and continuous in each coordinate variable;

(a2)

for all t^* > 1,

$\begin{matrix} ψ_{1} (t^{*}) = max {ϕ_{1} (t^{*}, t^{*}, t^{*}, 1, t^{*}), ϕ_{1} (t^{*}, t^{*}, t^{*}, t^{*}, 1), ϕ_{1} (t^{*}, 1, 1, t^{*}, t^{*}), ϕ_{1} (1, t^{*}, 1, t^{*}, 1), ϕ_{1} (1, 1, t^{*}, 1, t^{*})} \\ < t^{*} . \end{matrix}$ $$\begin{array}{} \displaystyle \psi_1(t^*) = \max\{\phi_1(t^*, t^*, t^*, 1, t^*), \phi_1(t^*, t^*, t^*, t^*, 1), \phi_1(t^*, 1, 1, t^*, t^*), \phi_1(1, t^*, 1, t^*, 1), \phi_1(1, 1, t^*, 1, t^*)\}\\\displaystyle \qquad\,\,\,\,\, \lt t^*. \end{array} $$

From now on, unless otherwise stated, we always choose ϕ₁ ∈ Φ₁.

Theorem 1.4

[13] Let (X,d_*) be a multiplicative metric space, F : X × X → Xandg : X → Xbe two mappings. Suppose that there existsλ ∈ (0,1) such that the condition

$\begin{matrix} d_{*} (F (x, y), F (u, v)) \cdot d_{*} (F (y, x), F (v, u)) \leq ϕ_{1} (\begin{matrix} d_{*}^{λ} (g x, g u) \cdot d_{*}^{λ} (g y, g v), \\ d_{*}^{λ} (F (x, y), g x) \cdot d_{*}^{λ} (F (y, x), g y), \\ d_{*}^{λ} (F (u, v), g u) \cdot d_{*}^{λ} (F (v, u), g v), \\ d_{*}^{λ} (F (x, y), g u) \cdot d_{*}^{λ} (F (y, x), g v), \\ d_{*}^{λ} (F (u, v), g x) \cdot d_{*}^{λ} (F (v, u), g y) \end{matrix}) \end{matrix}$ $$\begin{array}{} \displaystyle d_{*}(F(x,y),F(u,v))\cdot d_{*}(F(y,x),F(v,u))\leq \phi _1\left(\begin{array}{c} d_{*}^\lambda (gx,gu)\cdot d_{*}^\lambda (gy,gv), \\ d_{*}^\lambda (F(x,y),gx)\cdot d_{*}^\lambda (F(y,x),gy), \\ d_{*}^\lambda (F(u,v),gu)\cdot d_{*}^\lambda (F(v,u),gv), \\ d_{*}^\lambda (F(x,y),gu)\cdot d_{*}^\lambda (F(y,x),gv), \\ d_{*}^\lambda (F(u,v),gx)\cdot d_{*}^\lambda (F(v,u),gy) \end{array} \right) \end{array} $$

holds for all (x,y), (u,v) ∈ X × X. IfF(X × X) ⊂ g(X), g(X) is a multiplicative complete subspace ofX, andFandgarew^*-compatible, thenFandghave a unique common coupled fixed point of the form (u,u) ∈ X × X.

Remark 2

There are some mistakes in Theorem 1.4. Indeed see [13], page 1884, line 16−: because $\begin{matrix} h = \frac{λ}{1 - λ} \in (0, 1) \Leftrightarrow λ \in (0, \frac{1}{2})) . \end{matrix}$ $\begin{array}{} h=\frac \lambda {1-\lambda }\in \left( 0,1\right)\Leftrightarrow \lambda \in \left( 0,\frac 12\right) ). \end{array} $ Also see [13], page 1886, line 10+: gx = F(x,x) is unreasonable.

In the following, we improve Theorem 1.4 as follows:

Theorem 1.5

Let (X,d_*) be a multiplicative metric space, F : X × X → Xandg : X → Xbe two mappings. Suppose that there exists $\begin{matrix} λ \in (0, \frac{1}{2}) \end{matrix}$ $\begin{array}{} \lambda \in (0,\frac 12) \end{array} $such that the condition

holds for all (x,y), (u,v) ∈ X × X. IfF(X × X) ⊂ g(X), g(X) is a multiplicative complete subspace ofX, andFandgarew-compatible, thenFandghave a unique common coupled fixed point of the form (u,u) ∈ X × X.

Let (X,d) be a metric space, then the set X × X can be endowed with the following three metrics:

$\begin{matrix} d_{+} (Y, V) = d_{+} ((x, y), (u, v)) = d (x, u) + d (y, v), \\ d_{max} (Y, V) = d_{max} ((x, y), (u, v)) = max (d (x, u), d (y, v)), \\ d_{\frac{1}{2} +} (Y, V) = d_{\frac{1}{2} +} ((x, y), (u, v)) = \frac{d (x, u) + d (y, v)}{2} . \end{matrix}$ $$\begin{array}{} \displaystyle \,\,\,\,d_{+}\left( Y,V\right) =d_{+}\left( \left( x,y\right) ,\left( u,v\right)\right) =d\left( x,u\right) +d\left( y,v\right), \\\displaystyle d_{\max }\left( Y,V\right) =d_{\max }\left( \left( x,y\right) ,\left(u,v\right) \right)=\max \left\{ d\left( x,u\right) ,d\left( y,v\right)\right\},\\\displaystyle\, d_{\frac 12+}\left( Y,V\right) =d_{\frac 12+}\left( \left( x,y\right) ,\left(u,v\right) \right) =\frac{d\left( x,u\right) +d\left( y,v\right) }2. \end{array} $$

It is not hard to verify that (X,d) is complete if and only if one of (X × X, d₊), (X × X, d_max) and $\begin{matrix} (X \times X, d_{\frac{1}{2} +}) \end{matrix}$ $\begin{array}{} \left( X \times X,d_{\frac 12+}\right) \end{array} $ is complete.

Remark 3

It is clear that (x,y) is a coupled coincidence point of F : X × X → X and g : X → X if and only if (x,y) is a coincidence point of the mappings T_F : X × X → X × X and T_g : X × X → X × X which are defined by

$\begin{matrix} T_{F} (x, y) = (F (x, y), F (y, x)) and T_{g} (x, y) = (g x, g y) . \end{matrix}$ $$\begin{array}{} \displaystyle T_F\left( x,y\right) =\left( F\left( x,y\right),F\left( y,x\right) \right) \,\text{ and }\,T_g\left( x,y\right) =\left( gx,gy\right). \end{array} $$

According to the above notions we announce a shorter proof than one of [13, Corollary 2.4].

Indeed, by putting d₁ = lnd we get

$\begin{matrix} d_{1} (F (x, y), F (u, v)) + d_{1} (F (y, x), F (v, u)) \\ \leq max (λ (d_{1} (g x, g u) + d_{1} (g y, g v)), \dots, λ (d_{1} (F (u, v), g x) + d_{1} (F (v, u), g y))) \\ = 2 λ max (\frac{d_{1} (g x, g u) + d_{1} (g y, g v)}{2}, \dots, \frac{d_{1} (F (u, v), g x) + d_{1} (F (v, u), g y)}{2}), \end{matrix}$ $$\begin{array}{} \displaystyle \,\,d_1\left( F\left( x,y\right) ,F\left( u,v\right) \right) +d_1\left(F\left( y,x\right) ,F\left( v,u\right) \right) \\\displaystyle \leq \max \left\{ \lambda \left( d_1\left( gx,gu\right) +d_1\left(gy,gv\right) \right) ,\ldots,\lambda \left( d_1\left( F\left( u,v\right),gx\right) +d_1\left( F\left( v,u\right) ,gy\right) \right) \right\} \\\displaystyle =2\lambda \max \left\{ \frac{d_1\left( gx,gu\right) +d_1\left(gy,gv\right) }2,\ldots,\frac{d_1\left( F\left( u,v\right) ,gx\right) +d_1\left(F\left( v,u\right) ,gy\right) }2\right\} , \end{array} $$

$\begin{matrix} \frac{d_{1} (F (x, y), F (u, v)) + d_{1} (F (y, x), F (v, u))}{2} \\ \leq λ max (\frac{d_{1} (g x, g u) + d_{1} (g y, g v)}{2}, \dots, \frac{d_{1} (F (u, v), g x) + d_{1} (F (v, u), g y)}{2}), \end{matrix}$ $$\begin{array}{} \displaystyle \,\,\frac{d_1\left( F\left( x,y\right) ,F\left( u,v\right) \right) +d_1\left(F\left( y,x\right) ,F\left( v,u\right) \right) }2 \\\displaystyle \leq\lambda \max \left\{ \frac{d_1\left( gx,gu\right) +d_1\left(gy,gv\right) }2,\ldots,\frac{d_1\left( F\left( u,v\right) ,gx\right) +d_1\left(F\left( v,u\right) ,gy\right) }2\right\} , \end{array} $$

that is.,

$\begin{matrix} {(d_{1})}_{\frac{1}{2} +} (T_{F} (Y), T_{F} (V)) \leq λ max ({(d_{1})}_{\frac{1}{2} +} (T_{g} (Y), T_{g} (V)), \dots, {(d_{1})}_{\frac{1}{2} +} (T_{F} (V), T_{g} (Y))), \end{matrix}$ $$\begin{array}{} \displaystyle \left( d_1\right) _{\frac 12+}\left( T_F\left( Y\right) ,T_F\left( V\right)\right) \leq \lambda \max \,\left\{ \left( d_1\right) _{\frac 12+}\left(T_g\left( Y\right) ,T_g\left( V\right) \right) ,\ldots,\left( d_1\right)_{\frac 12+}\left( T_F\left( V\right) ,T_g\left( Y\right) \right)\right\} , \end{array} $$

where Y = (x,y), V = (u,v) ∈ X × X.

The last contractive condition is well-known Das-Naik quasi-contractive condition [10]. Namely, it follows that T_F and T_g have a unique point of coincidence. It is clear that the contractive condition from [13, Corollary 2.4] is equivalent to the Das-Naik’s condition [10].

Main results

In this section, by using nontrivial methods, we establish some common coupled fixed point theorems in metric spaces. Moreover, we show that some recent results in multiplicative metric spaces are indeed equivalent to those in usual metric spaces.

To begin with the main results, let Φ denote the set of all functions ϕ: [0,∞)⁵ → [0,∞) satisfying

(b1)

ϕ is nondecreasing and continuous in each coordinate variable;

(b2)

for all t > 0,

$\begin{matrix} ψ (t) = max {ϕ (t, t, t, 0, t), ϕ (t, t, t, t, 0), ϕ (t, 0, 0, t, t), ϕ (0, t, 0, t, 0), ϕ (0, 0, t, 0, t)} < t . \end{matrix}$ $$\begin{array}{} \displaystyle \psi(t) = \max\{\phi(t, t, t, 0, t), \phi(t, t, t, t, 0), \phi(t, 0, 0, t, t), \phi(0, t, 0, t, 0), \phi(0, 0, t, 0, t)\} \lt t. \end{array} $$

It is easy to see that ψ(t) ≤ t for all t ≥ 0. Indeed, we only need to prove ψ(0) = 0. For any t > 0, by (b1) and (b2), we get ψ(0) ≤ ψ(t) < t. Since t > 0 is arbitrary, then ψ(0) = 0.

From now on, unless otherwise stated, we always choose ϕ ∈ Φ.

Theorem 2.1

Let (X,d) be a metric space, F : X × X → Xandg : X → Xbe two mappings. Suppose that there exists $\begin{matrix} λ \in (0, \frac{1}{2}) \end{matrix}$ $\begin{array}{} \lambda \in (0,\frac 12) \end{array} $such that the condition

$\begin{matrix} d (F (x, y), F (u, v)) + d (F (y, x), F (v, u)) \leq ϕ (\begin{matrix} λ (d (g x, g u) + d (g y, g v)), \\ λ (d (F (x, y), g x) + d (F (y, x), g y)), \\ λ (d (F (u, v), g u) + d (F (v, u), g v)), \\ λ (d (F (x, y), g u) + d (F (y, x), g v)), \\ λ (d (F (u, v), g x) + d (F (v, u), g y)) \end{matrix}) \end{matrix}$ $$\begin{array}{} \displaystyle d(F(x,y),F(u,v))+d(F(y,x),F(v,u))\leq \phi \left(\begin{array}{c} \lambda (d(gx,gu)+d(gy,gv)), \\ \lambda (d(F(x,y),gx)+d(F(y,x),gy)), \\ \lambda (d(F(u,v),gu)+d(F(v,u),gv)), \\ \lambda (d(F(x,y),gu)+d(F(y,x),gv)), \\ \lambda (d(F(u,v),gx)+d(F(v,u),gy)) \end{array}\right) \end{array} $$(2.1)

holds for all (x,y), (u,v) ∈ X × X. IfF(X × X) ⊂ g(X), g(X) is a complete subspace ofX, andFandgarew-compatible, thenFandghave a unique common coupled fixed point of the form (u,u) ∈ X × X.

Proof

Let x₀, y₀ ∈ X. By F(X × X) ⊂ g(X), we choose x₁, y₁ ∈ X such that gx₁ = F(x₀, y₀) and gy₁ = F(y₀, x₀). Similarly, we choose (x₂, y₂) ∈ X such that gx₂ = F(x₁, y₁) and gy₂ = F(y₁, x₁). Continuing this process, we construct two sequences {x_n} and {y_n} in X as follows:

$\begin{matrix} g x_{n + 1} = F (x_{n}, y_{n}), g y_{n + 1} = F (y_{n}, x_{n}) . \end{matrix}$ $$\begin{array}{} \displaystyle gx_{n+1} =F(x_n, y_n),\quad gy_{n+1} = F(y_n, x_n). \end{array} $$

Now, by (2.1) we have

$\begin{matrix} d (g x_{n}, g x_{n + 1}) + d (g y_{n}, g y_{n + 1}) \\ = d (F (x_{n - 1}, y_{n - 1}), F (x_{n}, y_{n})) + d (F (y_{n - 1}, x_{n - 1}), F (y_{n}, x_{n})) \\ \leq ϕ (\begin{matrix} λ (d (g x_{n - 1}, g x_{n}) + d (g y_{n - 1}, g y_{n})), \\ λ (d (F (x_{n - 1}, y_{n - 1}), g x_{n - 1}) + d (F (y_{n - 1}, x_{n - 1}), g y_{n - 1})), \\ λ (d (F (x_{n}, y_{n}), g x_{n}) + d (F (y_{n}, x_{n}), g y_{n})), \\ λ (d (F (x_{n - 1}, y_{n - 1}), g x_{n}) + d (F (y_{n - 1}, x_{n - 1}), g y_{n})), \\ λ (d (F (x_{n}, y_{n}), g x_{n - 1}) + d (F (y_{n}, x_{n}), g y_{n - 1})) \end{matrix}) \\ = ϕ (\begin{matrix} λ (d (g x_{n - 1}, g x_{n}) + d (g y_{n - 1}, g y_{n})), \\ λ (d (g x_{n}, g x_{n - 1}) + d (g y_{n}, g y_{n - 1})), \\ λ (d (g x_{n + 1}, g x_{n}) + d (g y_{n + 1}, g y_{n})), \\ λ (d (g x_{n}, g x_{n}) + d (g y_{n}, g y_{n})), \\ λ (d (g x_{n + 1}, g x_{n - 1}) + d (g y_{n + 1}, g y_{n - 1})), \end{matrix}) \\ \leq ϕ (\begin{matrix} λ (d (g x_{n - 1}, g x_{n}) + d (g y_{n - 1}, g y_{n})), \\ λ (d (g x_{n}, g x_{n - 1}) + d (g y_{n}, g y_{n - 1})), \\ λ (d (g x_{n + 1}, g x_{n}) + d (g y_{n + 1}, g y_{n})), \\ 0, \\ λ (d (g x_{n + 1}, g x_{n}) + d (g x_{n}, g x_{n - 1}) + d (g y_{n + 1}, g y_{n}) + d (g y_{n}, g y_{n - 1})) \end{matrix}) \\ \leq ϕ (\begin{matrix} λ (d (g x_{n + 1}, g x_{n}) + d (g x_{n}, g x_{n - 1}) + d (g y_{n + 1}, g y_{n}) + d (g y_{n}, g y_{n - 1})), \\ λ (d (g x_{n + 1}, g x_{n}) + d (g x_{n}, g x_{n - 1}) + d (g y_{n + 1}, g y_{n}) + d (g y_{n}, g y_{n - 1})), \\ λ (d (g x_{n + 1}, g x_{n}) + d (g x_{n}, g x_{n - 1}) + d (g y_{n + 1}, g y_{n}) + d (g y_{n}, g y_{n - 1})), \\ 0, \\ λ (d (g x_{n + 1}, g x_{n}) + d (g x_{n}, g x_{n - 1}) + d (g y_{n + 1}, g y_{n}) + d (g y_{n}, g y_{n - 1})) \end{matrix}) \\ \leq ψ (λ (d (g x_{n + 1}, g x_{n}) + d (g x_{n}, g x_{n - 1}) + d (g y_{n + 1}, g y_{n}) + d (g y_{n}, g y_{n - 1}))) \\ \leq λ (d (g x_{n + 1}, g x_{n}) + d (g x_{n}, g x_{n - 1}) + d (g y_{n + 1}, g y_{n}) + d (g y_{n}, g y_{n - 1})), \end{matrix}$ $$\begin{array}{} \displaystyle \,\,\,\,d(gx_n, gx_{n+1}) + d(gy_n, gy_{n+1}) \\= d (F(x_{n-1}, y_{n-1}), F(x_n, y_n)) + d (F(y_{n-1}, x_{n-1}), F(y_n, x_n))\\\displaystyle \leq \phi\left(\begin{array}{c} \lambda(d(gx_{n-1}, gx_n) + d(gy_{n-1}, gy_n)), \\ \lambda(d(F(x_{n-1}, y_{n-1}), gx_{n-1}) +d(F(y_{n-1}, x_{n-1}), gy_{n-1})), \\ \lambda(d(F(x_n, y_n), gx_n) +d(F(y_n, x_n), gy_n)), \\ \lambda(d(F(x_{n-1}, y_{n-1}), gx_n) + d(F(y_{n-1}, x_{n-1}), gy_n)), \\ \lambda(d(F(x_n, y_n), gx_{n-1}) + d(F(y_n, x_n), gy_{n-1}))\\ \end{array}\right)\\ = \phi\left(\begin{array}{c} \lambda(d(gx_{n-1}, gx_n) +d(gy_{n-1}, gy_n)), \\ \lambda(d(gx_n, gx_{n-1}) +d(gy_n, gy_{n-1})), \\ \lambda( d(gx_{n+1}, gx_n) +d(gy_{n+1}, gy_n)),\\ \lambda(d(gx_n, gx_n) +d(gy_n, gy_n)), \\ \lambda(d(gx_{n+1}, gx_{n-1}) +d(gy_{n+1}, gy_{n-1})),\\ \end{array} \right)\\\displaystyle \leq \phi\left(\begin{array}{c} \lambda(d(gx_{n-1}, gx_n) + d(gy_{n-1}, gy_n)),\\ \lambda(d(gx_n, gx_{n-1}) + d(gy_n, gy_{n-1})),\\ \lambda(d(gx_{n+1}, gx_n)+ d(gy_{n+1}, gy_n)), \\ 0,\\ \lambda(d(gx_{n+1}, gx_n) + d(gx_n, gx_{n-1}) + d(gy_{n+1}, gy_n) + d(gy_n, gy_{n-1}))\\ \end{array} \right)\\\displaystyle \leq \phi\left(\begin{array}{c} \lambda(d(gx_{n+1}, gx_n) + d(gx_n, gx_{n-1}) + d(gy_{n+1}, gy_n) + d(gy_n, gy_{n-1})),\\ \lambda(d(gx_{n+1}, gx_n) + d(gx_n, gx_{n-1}) + d(gy_{n+1}, gy_n) + d(gy_n, gy_{n-1})),\\ \lambda(d(gx_{n+1}, gx_n) + d(gx_n, gx_{n-1}) + d(gy_{n+1}, gy_n) + d(gy_n, gy_{n-1})),\\ 0,\\ \lambda(d(gx_{n+1}, gx_n) + d(gx_n, gx_{n-1}) + d(gy_{n+1}, gy_n) + d(gy_n, gy_{n-1}))\\ \end{array} \right)\\\displaystyle \leq\psi(\lambda(d(gx_{n+1}, gx_n) + d(gx_n, gx_{n-1}) + d(gy_{n+1}, gy_n) + d(gy_n, gy_{n-1})))\\\displaystyle \leq\lambda(d(gx_{n+1}, gx_n) + d(gx_n, gx_{n-1}) + d(gy_{n+1}, gy_n) + d(gy_n, gy_{n-1})), \end{array} $$

which implies that

$\begin{matrix} d (g x_{n}, g x_{n + 1}) + d (g y_{n}, g y_{n + 1}) \leq \frac{λ}{1 - λ} (d (g x_{n}, g x_{n - 1}) + d (g y_{n}, g y_{n - 1})) . \end{matrix}$ $$\begin{array}{} \displaystyle d(gx_n, gx_{n+1})+ d(gy_n, gy_{n+1})\leq \frac{\lambda}{1-\lambda}( d(gx_n, gx_{n-1}) + d(gy_n, gy_{n-1})). \end{array} $$

Let $\begin{matrix} h = \frac{λ}{1 - λ} < 1, \end{matrix}$ $\begin{array}{} h=\frac{\lambda}{1-\lambda} \lt 1, \end{array} $ then 0 < h < 1. Hence,

$\begin{matrix} d (g x_{n}, g x_{n + 1}) + d (g y_{n}, g y_{n + 1}) \leq h (d (g x_{n}, g x_{n - 1}) + d (g y_{n}, g y_{n - 1})) \\ \leq \dots \leq h^{n} (d (g x_{1}, g x_{0}) + d (g y_{1}, g y_{0})) . \end{matrix}$ $$\begin{array}{} \displaystyle d(gx_n, gx_{n+1})+ d(gy_n, gy_{n+1})\leq h( d(gx_n, gx_{n-1}) + d(gy_n, gy_{n-1}))\\\displaystyle \qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\leq \dots \leq h^n ( d(gx_1, gx_0) + d(gy_1, gy_0)). \end{array} $$

Now, for all n ∈ ℕ, n < m, by the triangle inequality we obtain

$\begin{matrix} d (g x_{n}, g x_{m}) + d (g y_{n}, g y_{m}) \leq d (g x_{n}, g x_{n + 1}) + d (g x_{n + 1}, g x_{n + 2}) + \dots + d (g x_{m - 1}, g x_{m}) \\ + d (g y_{n}, g y_{n + 1}) + d (g y_{n + 1}, g y_{n + 2}) + \dots + d (g y_{m - 1}, g y_{m}) \\ \leq (h^{n} + h^{n + 1} + \dots + h^{m - 1}) (d (g x_{1}, g x_{0}) + d (g y_{1}, g y_{0})) \\ \leq \frac{h^{n}}{1 - h} (d (g x_{1}, g x_{0}) + d (g y_{1}, g y_{0})) \to 0 (n \to \infty) . \end{matrix}$ $$\begin{array}{} \displaystyle d(gx_n, gx_m) + d(gy_n, gy_m)\leq d(gx_n, gx_{n+1}) + d(gx_{n+1}, gx_{n+2})+ \dots+ d(gx_{m-1}, gx_m) \\\displaystyle \qquad\qquad\qquad\qquad\qquad\quad\,\,\,\,\,+\,d(gy_n, gy_{n+1})+d(gy_{n+1}, gy_{n+2})+\dots+d(gy_{m-1}, gy_m)\\\displaystyle \qquad\qquad\qquad\qquad\qquad\quad\,\leq (h^n+h^{n+1}+\cdots+h^{m-1})(d(gx_1, gx_0) + d(gy_1, gy_0))\\\displaystyle \qquad\qquad\qquad\qquad\qquad\quad\,\leq \frac{h^n}{1-h}(d(gx_1, gx_0) + d(gy_1, gy_0))\to 0\,(n\to\infty). \end{array} $$

So d(gx_n, gx_m) → 0 and d(gy_n, gy_m) → 0 (n, m → ∞). This means that {gx_n} and {gy_n} are Cauchy sequences in g(X). By the completeness of g(X), there exist gx,gy ∈ g(X) such that {gx_n} and {gy_n} converge to gx and gy, respectively. Next we prove that F(x,y) = gx and F(y,x) = gy.

By using (2.1), we have

$\begin{matrix} d (F (x, y), g x) + d (F (y, x), g y) \\ \leq d (F (x, y), g x_{n + 1}) + d (g x_{n + 1}, g x) + d (F (y, x), g y_{n + 1}) + d (g y_{n + 1}, g y) \\ = d (F (x, y), F (x_{n}, y_{n})) + d (F (y, x), F (y_{n}, x_{n})) + d (g x_{n + 1}, g x) + d (g y_{n + 1}, g y) \\ \leq ϕ (\begin{matrix} λ (d (g x, g x_{n}) + d (g y, g y_{n})), \\ λ ((F (x, y), g x) + d (F (y, x), g y)), \\ λ (d (F (x_{n}, y_{n}), g x_{n}) + d (F (y_{n}, x_{n}), g y_{n})), \\ λ (d (F (x, y), g x_{n}) + d (F (y, x), g y_{n})), \\ λ (d (F (x_{n}, y_{n}), g x) + d (F (y_{n}, x_{n}), g y)) \end{matrix}) + d (g x_{n + 1}, g x) + d (g y_{n + 1}, g y) \\ = ϕ (\begin{matrix} λ (d (g x, g x_{n}) + d (g y, g y_{n})), \\ λ (d (F (x, y), g x) + d (F (y, x), g y)), \\ λ (d (g x_{n + 1}, g x_{n}) + d (g y_{n + 1}, g y_{n})), \\ λ (d (F (x, y), g x_{n}) + d (F (y, x), g y_{n})), \\ λ (d (g x_{n + 1}, g x) + d (g y_{n + 1}, g y)) \end{matrix}) + d (g x_{n + 1}, g x) + d (g y_{n + 1}, g y) . \end{matrix}$ $$\begin{array}{} \displaystyle \,\,d(F(x, y), gx) + d(F(y, x), gy)\\\leq d(F(x, y), gx_{n+1})+ d(gx_{n+1}, gx) + d(F(y, x), gy_{n+1}) + d(gy_{n+1}, gy)\\= d(F(x, y), F(x_n, y_n)) + d(F(y, x), F(y_n, x_n)) + d(gx_{n+1}, gx) + d(gy_{n+1}, gy)\\\displaystyle \leq\phi \left(\begin{array}{c} \lambda(d(gx, gx_n) + d(gy, gy_n)),\\ \lambda((F(x, y), gx) + d(F(y, x), gy)),\\ \lambda( d(F(x_n, y_n), gx_n) + d(F(y_n, x_n), gy_n)),\\ \lambda(d(F(x, y), gx_n) +d(F(y, x), gy_n)),\\ \lambda(d(F(x_n, y_n), gx) + d(F(y_n, x_n), gy))\\ \end{array} \right)+d(gx_{n+1}, gx) + d(gy_{n+1}, gy)\\\\ =\phi\left(\begin{array}{c} \lambda(d(gx, gx_n) +d(gy, gy_n)),\\ \lambda(d(F(x, y), gx) +d(F(y, x), gy)),\\ \lambda(d(gx_{n+1}, gx_n) +d(gy_{n+1}, gy_n)),\\ \lambda(d(F(x, y), gx_n) + d(F(y, x), gy_n)), \\ \lambda(d(gx_{n+1}, gx) +d(gy_{n+1}, gy))\\ \end{array} \right)+d(gx_{n+1}, gx) + d(gy_{n+1}, gy). \end{array} $$

Let n → ∞ in the above inequality, we obtain

$\begin{matrix} d (F (x, y), g x) + d (F (y, x), g y) \\ \leq ϕ (\begin{matrix} 0, \\ λ (d (F (x, y), g x) + d (F (y, x), g y)), \\ 0, \\ λ (d (F (x, y), g x) + d (F (y, x), g y)), \\ 0 \end{matrix}) + 0 + 0 \\ \leq ψ (λ (d (F (x, y), g x) + d (F (y, x), g y))) \\ \leq λ (d (F (x, y), g x) + d (F (y, x), g y)) . \end{matrix}$ $$\begin{array}{} \displaystyle \,\,\,d(F(x, y), gx) + d(F(y, x), gy)\\\displaystyle \leq \phi\left(\begin{array}{c} 0,\\\lambda(d(F(x, y), gx) +d(F(y, x), gy)),\\ 0,\\\lambda(d(F(x, y), gx) + d(F(y, x), gy)), \\ 0\\ \end{array} \right)+0 + 0\\\displaystyle \leq \psi( \lambda(d(F(x, y), gx) + d(F(y, x), gy)))\\\displaystyle \leq\lambda(d(F(x, y), gx)+d(F(y, x), gy)). \end{array} $$

By virtue of $\begin{matrix} λ \in (0, \frac{1}{2}) \end{matrix}$ $\begin{array}{} \lambda \in (0,\frac 12) \end{array} $, so d(F(x,y),gx) + d(F(y,x),gy) = 0, which implies that d(F(x,y),gx) = 0 and d(F(y,x),gy) = 0. Thus, (gx,gy) is a coupled point of coincidence of the mappings F and g.

Now, we claim that the coupled point of coincidence is unique. Suppose that there is another (x^*, y^*) ∈ X × X such that (gx^*, gy^*) is a coupled point of the mappings F and g, then by (2.1) we have

$\begin{matrix} d (g x, g x^{*}) + d (g y, g y^{*}) = d (F (x, y), F (x^{*}, y^{*})) + d (F (y, x), F (y^{*}, x^{*})) \\ \leq ϕ (\begin{matrix} λ (d (g x, g x^{*}) + d (g y, g y^{*})), \\ λ (d (F (x, y), g x) + d (F (y, x), g y)), \\ λ (d (F (x^{*}, y^{*}), g x^{*}) + d (F (y^{*}, x^{*}), g y^{*})), \\ λ (d (F (x, y), g x^{*}) + d (F (y, x), g y^{*})), \\ λ (d (F (x^{*}, y^{*}), g x) + d (F (y^{*}, x^{*}), g y)) \end{matrix}) \\ = ϕ (\begin{matrix} λ (d (g x, g x^{*}) + d (g y, g y^{*})), \\ λ (d (g x, g x) + d (g y, g y)), \\ λ (d (g x^{*}, g x^{*}) + d (g y^{*}, g y^{*})), \\ λ (d (g x, g x^{*}) + d (g y, g y^{*})), \\ λ (d (g x^{*}, g x) + d (g y^{*}, g y)) \end{matrix}) \\ = ϕ (\begin{matrix} λ (d (g x, g x^{*}) + d (g y, g y^{*})), \\ 0, \\ 0, \\ λ (d (g x, g x^{*}) + d (g y, g y^{*})), \\ λ (d (g x^{*}, g x) + d (g y^{*}, g y)) \end{matrix}) \\ \leq ψ (λ (d (g x, g x^{*}) + d (g y, g y^{*}))) \\ \leq λ (d (g x, g x^{*}) + d (g y, g y^{*})) . \end{matrix}$ $$\begin{array}{} \displaystyle d(gx, gx^*) + d(gy, gy^*)= d(F(x, y), F(x^*, y^*)) + d(F(y, x), F(y^*, x^*))\\\displaystyle \qquad\qquad\qquad\qquad\quad\,\,\,\,\leq \phi\left(\begin{array}{c} \lambda(d(gx, gx^*) + d(gy, gy^*)),\\ \lambda(d(F(x, y), gx) + d(F(y, x), gy)), \\ \lambda(d(F(x^*, y^*), gx^*)+ d(F(y^*, x^*), gy^*)),\\ \lambda(d(F(x, y), gx^*) + d(F(y, x), gy^*)), \\ \lambda(d(F(x^*, y^*), gx) +d(F(y^*, x^*), gy))\\ \end{array} \right)\\\displaystyle\qquad\qquad\qquad\qquad\quad\,\,\,\, =\phi\left(\begin{array}{c} \lambda(d(gx, gx^*) + d(gy, gy^*)),\\ \lambda(d(gx, gx) + d(gy, gy)), \\ \lambda( d(gx^*, gx^*) + d(gy^*, gy^*)), \\ \lambda(d(gx, gx^*) + d(gy, gy^*)), \\ \lambda(d(gx^*, gx) + d(gy^*, gy))\\ \end{array} \right)\\\displaystyle \qquad\qquad\qquad\qquad\quad\,\,\,\,=\phi\left(\begin{array}{c} \lambda(d(gx, gx^*) + d(gy, gy^*)),\\ 0,\\ 0,\\ \lambda(d(gx, gx^*) + d(gy, gy^*)), \\ \lambda(d(gx^*, gx) + d(gy^*, gy))\\ \end{array} \right)\\\displaystyle\qquad\qquad\qquad\qquad\quad\,\,\,\, \leq\psi(\lambda(d(gx, gx^*) + d(gy, gy^*)))\\\displaystyle \qquad\qquad\qquad\qquad\quad\,\,\,\,\leq\lambda(d(gx, gx^*) + d(gy, gy^*)). \end{array} $$

In view of $\begin{matrix} λ \in (0, \frac{1}{2}) \end{matrix}$ $\begin{array}{} \lambda \in (0,\frac 12) \end{array} $, then d(gx,gx^*) + d(gy,gy^*) = 0, this means d(gx,gx^*) = 0 and d(gy,gy^*) = 0, so gx = gx^* and gy = gy^*. Hence, (gx, gy) is a unique coupled point of coincidence of the mappings F and g.

In the following we prove that gx = gy. In fact, by (2.1) we have

$\begin{matrix} d (g x, g y) + d (g y, g x) = d (F (x, y), F (y, x)) + d (F (y, x), F (x, y)) \\ \leq ϕ (\begin{matrix} λ (d (g x, g y) + d (g y, g x)), \\ λ (d (F (x, y), g x) + d (F (y, x), g y)), \\ λ (d (F (y, x), g y) + d (F (x, y), g x)), \\ λ (d (F (x, y), g y) + d (F (y, x), g x)), \\ λ (d (F (y, x), g x) + d (F (x, y), g y)) \end{matrix}) \\ = ϕ (\begin{matrix} λ (d (g x, g y) + d (g y, g x)), \\ 0, \\ 0, \\ λ (d (g x, g y) + d (g y, g x)), \\ λ (d (g y, g x) + d (g x, g y)) \end{matrix}) \\ \leq ψ (λ (d (g x, g y) + d (g y, g x))) \\ \leq λ (d (g x, g y) + d (g y, g x)) . \end{matrix}$ $$\begin{array}{} \displaystyle d(gx, gy) + d(gy, gx) = d(F(x, y), F(y, x)) + d(F(y, x), F(x, y))\\\displaystyle\qquad\qquad\qquad\qquad\,\,\,\,\, \leq \phi\left(\begin{array}{c} \lambda(d(gx, gy) + d(gy, gx)),\\ \lambda(d(F(x, y), gx)+ d(F(y, x), gy)),\\ \lambda(d(F(y, x), gy) + d(F(x, y), gx)), \\ \lambda(d(F(x, y), gy) + d(F(y, x), gx)), \\ \lambda(d(F(y, x), gx) + d(F(x, y), gy))\\ \end{array} \right)\\\displaystyle\qquad\qquad\qquad\qquad\,\,\,\,\, =\phi\left(\begin{array}{c} \lambda(d(gx, gy) + d(gy, gx)),\\ 0,\\ 0, \\ \lambda(d(gx, gy) + d(gy, gx)), \\ \lambda(d(gy, gx) + d(gx, gy))\\ \end{array} \right)\\\displaystyle \qquad\qquad\qquad\qquad\,\,\,\,\,\leq \psi(\lambda(d(gx, gy) + d(gy, gx)))\\\displaystyle\qquad\qquad\qquad\qquad\,\,\,\,\, \leq\lambda(d(gx, gy) + d(gy, gx)). \end{array} $$

On account of $\begin{matrix} λ \in (0, \frac{1}{2}) \end{matrix}$ $\begin{array}{} \lambda \in (0,\frac 12) \end{array} $, so d(gx, gy) + d(gy, gx) = 0, this implies d(gx, gy) = 0 and d(gy, gx) = 0, we obtain that gx = gy. Thus, (gx, gx) is a unique coupled point of coincidence of the mappings F and g.

Finally, we show that F and g have a unique common coupled fixed point. For this, let gx = u. By the w-compatibility of F and g, we get

$\begin{matrix} g u = g (g x) = g F (x, y) = F (g x, g y) = F (u, u) . \end{matrix}$ $$\begin{array}{} \displaystyle gu = g(gx) = gF(x, y) = F(gx, gy) = F(u, u). \end{array} $$

Hence, (gu, gu) is a coupled point of coincidence of F and g. By the uniqueness of coupled point of coincidence of F and g, we have gu = gx. Consequently, we obtain u = gx = gu = F(u,u). Therefore, (u,u) is the unique common coupled fixed point of F and g. This completes the proof. □

Example 1

Let X = [0, 1] and (X, d) be a metric space defined by d(x, y) = |x − y| for all x, y ∈ X. Let F : X × X → X and g : X → X be two mappings defined by

$\begin{matrix} F (x, y) = \frac{x + y}{4}, g x = 8 x \end{matrix}$ $$\begin{array}{} \displaystyle F(x,y)=\frac{x+y}4,\quad\quad gx=8x \end{array}$$

for all x, y ∈ X.

Let $\begin{matrix} λ = \frac{1}{4} \end{matrix}$ $\begin{array}{} \lambda = \frac 14 \end{array}$ and ϕ (t₁, t₂, t₃, t₄, t₅) = $\begin{matrix} \frac{1}{3} \end{matrix}$ $\begin{array}{} \frac 13 \end{array}$ (t₁ + t₂ + t₃ + t₄ + t₅). For all (x, y), (u, v) ∈ X × X, we have

$\begin{matrix} ϕ (\begin{matrix} λ (d (g x, g u) + d (g y, g v)), \\ λ (d (F (x, y), g x) + d (F (y, x), g y)), \\ λ (d (F (u, v), g u) + d (F (v, u), g v)), \\ λ (d (F (x, y), g u) + d (F (y, x), g v)), \\ λ (d (F (u, v), g x) + d (F (v, u), g y)) \end{matrix}) \\ = \frac{1}{3} (| 2 x - 2 u | + | 2 y - 2 v | + (\frac{x + y}{16} - 2 x) + (\frac{y + x}{16} - 2 y)) \\ + (\frac{u + v}{16} - 2 u) + (\frac{v + u}{16} - 2 v) + (\frac{x + y}{16} - 2 u) \\ + ((\frac{y + x}{16} - 2 v) + (\frac{u + v}{16} - 2 x) + (\frac{v + u}{16} - 2 y)) \\ \geq \frac{1}{3} (| 2 x - 2 u + 2 y - 2 v + \frac{x + y}{16} - 2 x + \frac{y + x}{16} - 2 y + \frac{u + v}{16} - 2 u \\ + \frac{v + u}{16} - 2 v + \frac{x + y}{16} - 2 u + \frac{y + x}{16} - 2 v + \frac{u + v}{16} - 2 x + \frac{v + u}{16} - 2 y |) \\ = | \frac{23}{12} (u + v) + \frac{7}{12} (x + y) | \\ \geq | \frac{x + y}{2} - \frac{u + v}{2} | \\ = d (F (x, y), F (u, v)) + d (F (y, x), F (v, u)) . \end{matrix}$ $$\begin{array}{} \displaystyle \,\,\,\,\,\,\phi \left( \begin{array}{c} \lambda (d(gx,gu)+d(gy,gv)), \\ \lambda (d(F(x,y),gx)+d(F(y,x),gy)), \\ \lambda (d(F(u,v),gu)+d(F(v,u),gv)), \\ \lambda (d(F(x,y),gu)+d(F(y,x),gv)), \\ \lambda (d(F(u,v),gx)+d(F(v,u),gy)) \end{array} \right)\\\displaystyle=\frac 13\left(|2x-2u|+|2y-2v|+\left|\frac{x+y}{16}-2x\right|+\left|\frac{y+x}{16}-2y\right| \right.\\ \displaystyle\,\,\,\,\,\,+\left|\frac{u+v}{16}-2u\right|+\left|\frac{v+u}{16}-2v\right|+\left|\frac{x+y}{16}-2u\right|\\ \displaystyle\,\,\,\,\,\,+\left.\left|\frac{y+x}{16}-2v\right|+\left|\frac{u+v}{16}-2x\right|+\left|\frac{v+u}{16}-2y\right|\right) \\ \displaystyle\geq \frac 13\Big(\Big|2x-2u+2y-2v+\frac{x+y}{16}-2x+\frac{y+x}{16}-2y+ \frac{u+v}{16}-2u\\ \displaystyle\,\,\,\,\,\,+\frac{v+u}{16}-2v +\frac{x+y}{16}-2u+\frac{y+x}{16}-2v+\frac{u+v}{16}-2x+\frac{v+u}{16}-2y \Big|\Big) \\ \displaystyle=\Big|\frac{23}{12}(u+v)+\frac 7{12}(x+y)\Big| \\ \displaystyle\geq \Big|\frac{x+y}2-\frac{u+v}2\Big|\\\displaystyle=d(F(x,y),F(u,v))+d(F(y,x),F(v,u)).\end{array} $$

Note that F and g are w-compatible, then all the conditions of Theorem 2.1 are satisfied. Therefore, (0, 0) is the unique common coupled fixed point of F and g.

In [11] the following theorem was proved.

Theorem 2.2

[11]Let (X, d_∗) be a multiplicative metric space. Then (X, d) is a metric space whered(x, y) = ln d_∗(x, y) for allx, y ∈ X.Otherwise, if (X, d) is a metric space, then (X, d_∗) is a multiplicative metric space whered_∗(x, y) = e^d(x,y)for allx, y ∈ X.

Eventually, we make a conclusion as follows:

Theorem 2.3

Theorem 1.5 and Theorem 2.1 are equivalent.

Proof

Let condition (1.1) is satisfied. Then by (1.1) and Theorem 2.2, we have

$\begin{matrix} d (F (x, y), F (u, v)) + d (F (y, x), F (v, u)) = \ln d_{*} (F (x, y), F (u, v)) + \ln d_{*} (F (y, x), F (v, u)) \\ = \ln (d_{*} (F (x, y), F (u, v)) \cdot d_{*} (F (y, x), F (v, u))) \\ \leq \ln ϕ_{1} (\begin{matrix} d_{*}^{λ} (g x, g u) \cdot d_{*}^{λ} (g y, g v), \\ d_{*}^{λ} (F (x, y), g x) \cdot d_{*}^{λ} (F (y, x), g y), \\ d_{*}^{λ} (F (u, v), g u) \cdot d_{*}^{λ} (F (v, u), g v), \\ d_{*}^{λ} (F (x, y), g u) \cdot d_{*}^{λ} (F (y, x), g v), \\ d_{*}^{λ} (F (u, v), g x) \cdot d_{*}^{λ} (F (v, u), g y) \end{matrix}) \\ = \ln ϕ_{1} (\begin{matrix} e^{λ \ln d_{*} (g x, g u)} \cdot e^{λ \ln d_{*} (g y, g v)}, \\ e^{λ \ln d_{*} (F (x, y), g x)} \cdot e^{λ \ln d_{*} (F (y, x), g y)}, \\ e^{λ \ln d_{*} (F (u, v), g u)} \cdot e^{λ \ln d_{*} (F (v, u), g v)}, \\ e^{λ \ln d_{*} (F (x, y), g u)} \cdot e^{λ \ln d_{*} (F (y, x), g v)}, \\ e^{λ \ln d_{*} (F (u, v), g x)} \cdot e^{λ \ln d_{*} (F (v, u), g y)} \end{matrix}) \\ = \ln ϕ_{1} (\begin{matrix} e^{λ (d (g x, g u) + d (g y, g v))}, \\ e^{λ (d (F (x, y), g x) + d (F (y, x), g y))}, \\ e^{λ (d (F (u, v), g u) + d (F (v, u), g v))}, \\ e^{λ (d (F (x, y), g u) + d (F (y, x), g v))}, \\ e^{λ (d (F (u, v), g x) + d (F (v, u), g y))} \end{matrix}) . \end{matrix}$ $$\begin{array}{} \displaystyle d(F(x,y),F(u,v))+d(F(y,x),F(v,u)) =\ln d_*(F(x,y),F(u,v))+\ln d_*(F(y,x),F(v,u))\\ \displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\ln(d_*(F(x,y),F(u,v))\cdot d_*(F(y,x),F(v,u)))\\ \displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\leq\ln \phi_1\left( \begin{array}{c} d_{*}^{\lambda}(gx, gu) \cdot d_{*}^{\lambda}(gy, gv), \\ d_{*}^{\lambda}(F(x, y), gx) \cdot d_{*}^{\lambda}(F(y, x), gy), \\ d_{*}^{\lambda}(F(u, v), gu) \cdot d_{*}^{\lambda}(F(v, u), gv),\\ d_{*}^{\lambda}(F(x, y), gu) \cdot d_{*}^{\lambda}(F(y, x), gv), \\ d_{*}^{\lambda}(F(u, v), gx) \cdot d_{*}^{\lambda}(F(v, u), gy) \\ \end{array} \right)\\ \displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\ln \phi_1\left( \begin{array}{c} e^{\lambda\ln d_{*}(gx, gu)} \cdot e^{\lambda\ln d_{*}(gy, gv)}, \\ e^{\lambda \ln d_{*}(F(x, y), gx)} \cdot e^{\lambda \ln d_{*}(F(y, x), gy)}, \\ e^{\lambda \ln d_{*}(F(u, v), gu)} \cdot e^{\lambda \ln d_{*}(F(v, u), gv)},\\ e^{\lambda \ln d_{*}(F(x, y), gu)} \cdot e^{\lambda\ln d_{*}(F(y, x), gv)}, \\ e^{\lambda \ln d_{*}(F(u, v), gx)} \cdot e^{\lambda \ln d_{*}(F(v, u), gy)} \\ \end{array} \right)\\ \displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\ln \phi_1\left( \begin{array}{c} e^{\lambda(d(gx, gu) +d(gy, gv))}, \\ e^{\lambda(d(F(x, y), gx)+d(F(y, x), gy))}, \\ e^{\lambda(d(F(u, v), gu) +d(F(v, u), gv))},\\ e^{\lambda(d(F(x, y), gu) +d(F(y, x), gv))}, \\ e^{\lambda(d(F(u, v), gx) + d(F(v, u), gy))} \\ \end{array} \right). \end{array}$$(2.2)

Denote

$\begin{matrix} \ln ϕ_{1} (e^{t_{1}}, e^{t_{2}}, e^{t_{3}}, e^{t_{4}}, e^{t_{5}}) = ϕ (t_{1}, t_{2}, t_{3}, t_{4}, t_{5}) \end{matrix}$ $$\begin{array}{} \displaystyle \ln \phi_1(e^{t_1}, e^{t_2}, e^{t_3}, e^{t_4}, e^{t_5})=\phi(t_1,t_2,t_3,t_4,t_5) \end{array}$$

then by (2.2), we get (2.1).

Clearly, (a1) implies (b1). We need to prove that (a2) implies (b2). In fact, let (a2) hold and denote t^∗ = e^t, then by (a2), it follows that

$\begin{matrix} ψ (t) & = max {ϕ (t, t, t, 0, t), ϕ (t, t, t, t, 0), ϕ (t, 0, 0, t, t), ϕ (0, t, 0, t, 0), ϕ (0, 0, t, 0, t)} \\ = max {\ln ϕ_{1} (e^{t}, e^{t}, e^{t}, e^{0}, e^{t}), \ln ϕ_{1} (e^{t}, e^{t}, e^{t}, e^{t}, e^{0}), \ln ϕ_{1} (e^{t}, e^{0}, e^{0}, e^{t}, e^{t}), \\ \ln ϕ_{1} (e^{0}, e^{t}, e^{0}, e^{t}, e^{0}), \ln ϕ_{1} (e^{0}, e^{0}, e^{t}, e^{0}, e^{t})} \\ = \ln max {ϕ_{1} (t^{*}, t^{*}, t^{*}, 1, t^{*}), ϕ_{1} (t^{*}, t^{*}, t^{*}, t^{*}, 1), ϕ_{1} (t^{*}, 1, 1, t^{*}, t^{*}), ϕ_{1} (1, t^{*}, 1, t^{*}, 1), ϕ_{1} (1, 1, t^{*}, 1, t^{*})} \\ = \ln ψ_{1} (t^{*}) < \ln t^{*} = t . \end{matrix}$ $$\begin{array}{} \displaystyle \psi(t)\!\!\!\!\!&=\max\{\phi(t,t,t,0,t), \phi(t,t,t,t,0), \phi(t,0,0,t,t), \phi(0,t,0,t,0), \phi(0,0,t,0,t)\}\\ &=\max\{\ln \phi_1(e^t, e^t, e^t, e^0, e^t), \ln \phi_1(e^t, e^t, e^t, e^t, e^0), \ln \phi_1(e^t, e^0, e^0, e^t, e^t), \\&\,\,\,\,\,\,\,\ln\phi_1(e^0, e^t, e^0, e^t, e^0), \ln\phi_1(e^0, e^0, e^t, e^0, e^t)\}\\ &= \ln\max\{\phi_1(t^*, t^*, t^*, 1, t^*), \phi_1(t^*, t^*, t^*, t^*, 1), \phi_1(t^*, 1, 1, t^*, t^*), \phi_1(1, t^*, 1, t^*, 1), \phi_1(1, 1, t^*, 1, t^*)\}\\ &=\ln \psi_1(t^*)<\ln t^* =t. \end{array}$$

So, we get (b2).

On the other hand, let condition (2.1) is satisfied. By using (2.1) and Theorem 2.2, we obtain

$\begin{matrix} \ln (d_{*} (F (x, y), F (u, v)) \cdot d_{*} (F (y, x), F (v, u))) & = \ln d_{*} (F (x, y), F (u, v)) + \ln d_{*} (F (y, x), F (v, u)) \\ = d (F (x, y), F (u, v)) + d (F (y, x), F (v, u)) \\ \leq ϕ (\begin{matrix} λ (d (g x, g u) + d (g y, g v)), \\ λ (d (F (x, y), g x) + d (F (y, x), g y)), \\ λ (d (F (u, v), g u) + d (F (v, u), g v)), \\ λ (d (F (x, y), g u) + d (F (y, x), g v)), \\ λ (d (F (u, v), g x) + d (F (v, u), g y)) \end{matrix}) \\ = ϕ (\begin{matrix} λ (\ln d_{*} (g x, g u) + \ln d_{*} (g y, g v)), \\ λ (\ln d_{*} (F (x, y), g x) + \ln d_{*} (F (y, x), g y)), \\ λ (\ln d_{*} (F (u, v), g u) + \ln d_{*} (F (v, u), g v)), \\ λ (\ln d_{*} (F (x, y), g u) + \ln d_{*} (F (y, x), g v)), \\ λ (\ln d_{*} (F (u, v), g x) + \ln d_{*} (F (v, u), g y)) \end{matrix}) \\ = ϕ (\begin{matrix} \ln (d_{*}^{λ} (g x, g u) \cdot d_{*}^{λ} (g y, g v)), \\ \ln (d_{*}^{λ} (F (x, y), g x) \cdot d_{*}^{λ} (F (y, x), g y)), \\ \ln (d_{*}^{λ} (F (u, v), g u) \cdot d_{*}^{λ} (F (v, u), g v)), \\ \ln (d_{*}^{λ} (F (x, y), g u) \cdot d_{*}^{λ} (F (y, x), g v)), \\ \ln (d_{*}^{λ} (F (u, v), g x) \cdot d_{*}^{λ} (F (v, u), g y)) \end{matrix}) . \end{matrix}$ $$\begin{array}{} \displaystyle \ln (d_*(F(x,y),F(u,v))\cdot d_*(F(y,x),F(v,u))) \!\!\!\!&=\ln d_*(F(x,y),F(u,v))+\ln d_*(F(y,x),F(v,u))\\ &=d(F(x,y),F(u,v))+d(F(y,x),F(v,u))\\&\leq \phi \left( \begin{array}{c} \lambda (d(gx,gu)+d(gy,gv)), \\ \lambda (d(F(x,y),gx)+d(F(y,x),gy)), \\ \lambda (d(F(u,v),gu)+d(F(v,u),gv)), \\ \lambda (d(F(x,y),gu)+d(F(y,x),gv)), \\ \lambda (d(F(u,v),gx)+d(F(v,u),gy)) \end{array}\right)\\ &=\phi\left( \begin{array}{c} \lambda(\ln d_*(gx, gu) + \ln d_*(gy, gv)), \\ \lambda(\ln d_*(F(x, y), gx) + \ln d_*(F(y, x), gy)), \\ \lambda(\ln d_*(F(u, v), gu) +\ln d_*(F(v, u), gv)),\\ \lambda(\ln d_*(F(x, y), gu) +\ln d_*(F(y, x), gv)), \\ \lambda(\ln d_*(F(u, v), gx) +\ln d_*(F(v, u), gy)) \\ \end{array} \right)\\ &=\phi\left(\begin{array}{c} \ln(d_*^{\lambda}(gx, gu)\cdot d_*^{\lambda}(gy, gv)), \\ \ln(d_*^{\lambda}(F(x, y), gx)\cdot d_*^{\lambda}(F(y, x), gy)), \\ \ln(d_*^{\lambda}(F(u, v), gu) \cdot d_*^{\lambda}(F(v, u), gv)),\\ \ln(d_*^{\lambda}(F(x, y), gu)\cdot d_*^{\lambda}(F(y, x), gv)), \\ \ln(d_*^{\lambda}(F(u, v), gx) \cdot d_*^{\lambda}(F(v, u), gy)) \\ \end{array} \right). \end{array}$$

Let

$\begin{matrix} ϕ_{1} (t_{1}, t_{2}, t_{3}, t_{4}, t_{5}) = e^{ϕ (\ln t_{1}, \ln t_{2}, \ln t_{3}, \ln t_{4}, \ln t_{5})}, \end{matrix}$ $$\begin{array}{} \displaystyle \phi_1(t_1,t_2,t_3, t_4,t_5)=e^{\phi(\ln t_1, \ln t_2, \ln t_3, \ln t_4, \ln t_5)}, \end{array}$$

then condition (1.1) is satisfied.

Clearly, (b1) implies (a1). We need to prove that (b2) implies (a2). In fact, let (b2) hold and denote t^∗ = e^t, then by (b2), it follows that

$\begin{matrix} ψ_{1} (t^{*}) & = max {ϕ_{1} (t^{*}, t^{*}, t^{*}, 1, t^{*}), ϕ_{1} (t^{*}, t^{*}, t^{*}, t^{*}, 1), ϕ_{1} (t^{*}, 1, 1, t^{*}, t^{*}), ϕ_{1} (1, t^{*}, 1, t^{*}, 1), ϕ_{1} (1, 1, t^{*}, 1, t^{*})} \\ = max (e^{ϕ (\ln t^{*}, \ln t^{*}, \ln t^{*}, \ln 1, \ln t^{*})}, e^{ϕ (\ln t^{*}, \ln t^{*}, \ln t^{*}, \ln t^{*}, \ln 1)},) \\ (e^{ϕ (\ln t^{*}, \ln 1, \ln 1, \ln t^{*}, \ln t^{*})}, e^{ϕ (\ln 1, \ln t^{*}, \ln 1, \ln t^{*}, \ln 1)}, e^{ϕ (\ln 1, \ln 1, \ln t^{*}, \ln 1, \ln t^{*})}) \\ = e^{ψ (t)} < e^{t} = t^{*} . \end{matrix}$ $$\begin{array}{} \displaystyle \psi_1(t^*)\!\!\!\!&=\max\{\phi_1( t^*, t^*, t^*, 1, t^*), \phi_1( t^*, t^*, t^*, t^*, 1), \phi_1( t^*, 1, 1, t^*, t^*), \phi_1( 1, t^*, 1, t^*, 1), \phi_1( 1, 1, t^*, 1,t^*)\}\\ &= \max\left\{e^{\phi(\ln t^*, \ln t^*, \ln t^*, \ln 1, \ln t^*)}, e^{\phi(\ln t^*, \ln t^*, \ln t^*, \ln t^*, \ln 1)}, \right.\\&\left.\,\,\,\,\,\,\,\,\,e^{\phi(\ln t^*, \ln 1, \ln 1, \ln t^*, \ln t^*)}, e^{\phi(\ln 1, \ln t^*, \ln 1, \ln t^*, \ln 1)}, e^{\phi(\ln 1, \ln 1, \ln t^*, \ln 1, \ln t^*)}\right\}\\ &=e^{\psi(t)} \lt e^t=t^*. \end{array}$$

Hence, we get (a2). The proof is completed. □

Remark 4

By the same approach as Theorem 2.3 one can prove that all coupled fixed point results in the framework of multiplicative metric spaces from [13] are equivalent to the corresponding ones in ordinary metric spaces. Hence, our results show the superiority. Otherwise, the same thing can be dealt with all theoretical results in [19] including the examples and application which support the multiplicative theory. Actually, for instance, the condition (𝓓4) in [19, Definition 2.1] is equivalent to the following:

$\begin{matrix} D_{1} (x, z) + D_{1} (y, y) \leq D_{1} (x, y) + D_{1} (y, z), \end{matrix}$ $$\begin{array}{} \displaystyle \mathcal{D}_1\left( x,z\right) +\mathcal{D}_1\left( y,y\right) \leq \mathcal{ D}_1\left( x,y\right) +\mathcal{D}_1\left( y,z\right), \end{array}$$

where 𝓓₁ = ln 𝓓. The same things also hold for multiplicative metric-like, multiplicative b-metric and multiplicative b-metric like spaces. Therefore, we say that all results in [19] are equivalent to the counterpart of the context of partial metric-like, partial b-metric and partial b-metric like spaces.

eISSN:: 2444-8656
Sprache:: Englisch

Zeitrahmen der Veröffentlichung:: Volume Open
Fachgebiete der Zeitschrift:: Biologie, andere, Mathematik, Angewandte Mathematik, Allgemeines, Physik

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Note on recent common coupled fixed point results in multiplicative metric spaces

Online veröffentlicht: 31. Jan. 2019

Seitenbereich: 659 - 668

Eingereicht: 24. Okt. 2018

Akzeptiert: 24. Jan. 2019

DOI: https://doi.org/10.2478/AMNS.2018.2.00050

Schlüsselwörtermultiplicative metric space, common coupled fixed point, coupled point of coincidence

© 2018 H. Huang et al., published by Sciendo

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

Schlüsselwörter
multiplicative metric space, common coupled fixed point, coupled point of coincidence