G - METRIC SPACES

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Vol. 44, No. 1, 2014, 91-104

A GENERAL FIXED POINT THEOREM FOR PAIRS OF NON WEAKLY COMPATIBLE MAPPINGS IN

G - METRIC SPACES

Valeriu Popa¹ and Alina-Mihaela Patriciu²

Abstract. In this paper a general fixed point theorem for pairs of non weakly compatible pairs of mappings inG- metric spaces is proved. In the case of a single mapping some results from [2], [3], [11], [13], [17], [30]

are obtained.

AMS Mathematics Subject Classification(2010): 54H25

Key words and phrases: G- metric space, common fixed point, implicit relation

1. Introduction

In [4] and [5] Dhage introduced a new class of generalized metric spaces, named D - metric spaces. Mustafa and Sims [14], [15] proved that most of the claims concerning the fundamental structures on D - metric spaces are incorrect and introduced an appropriate notion ofD - metric space, namedG - metric spaces. In fact, Mustafa, Sims and other authors studied many fixed point results for self mappings in G- metric spaces under certain conditions [1], [2], [16], [17], [29] and other papers.

In [19], [20] and other papers, the first author initiated the study of fixed points for mappings satisfying implicit relations. Actually, the method is used in the study of fixed points in metric spaces, symmetric spaces, quasi - metric spaces, Tychonoﬀ spaces, probabilistic metric spaces, convex metric spaces, in two and three metric spaces, for single valued mappings, hybrid pair of mappings and set valued mappings. Quite recently, the method is used in the study of fixed points for mappings satisfying contractive conditions of integral type, in fuzzy metric spaces and intuitionistic metric spaces. There exists a vast literature in this topic which cannot be completely cited here. The method unified diﬀerent types of contractive and extensive conditions, some proofs of fixed points theorems are more simple. Also, the method allows the study of local and global properties of fixed point structures.

Quite recently, the present authors initiated the study of fixed points inG - metric spaces using implicit relations in [21], [23], [24], [25]. In [24] a general fixed point theorem for pairs of weakly compatible mappings in G - metric spaces is proved.

In this paper, a general fixed point theorem for pairs of non weakly compatible mappings inG- metric space is proved. In the case of a single mapping some results from [2], [3], [11], [13], [17] and [30] are obtained.

1“Vasile Alecsandri” University of Bac˘au, Romania, e-mail: [email protected]

2“Vasile Alecsandri” University of Bac˘au, Romania, e-mail: [email protected]

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2. Preliminaries

Definition 2.1 ([15]). Let X be a nonempty set and G : X³ → R+ be a function satisfying the following properties:

(G₁) :G(x, y, z) = 0 ifx=y=z,

(G₂) : 0< G(x, x, y) for all x, y∈X withx̸=y,

(G₃) :G(x, x, y)≤G(x, y, z) for all x, y∈X withz̸=y,

(G₄) : G(x, y, z) = G(y, z, x) = G(z, x, y) = ... (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).

The function G is called a G - metric and the pair (X, G) is called a G- metric space.

Note that if G(x, y, z) = 0 thenx=y=z [15].

Definition 2.2 ([15]). Let (X, G) be a G- metric space. A sequence (xn) in X is said to be:

a)G- convergent if forε >0, there exist anx∈X andk∈Nsuch that for m, n≥k,m, n∈N,G(x, x_n, x_m)< ε.

b)G- Cauchy if for eachε >0, there existk∈Nsuch that for allm, n, p∈N withm, n, p≥k,G(x_n, x_m, x_p)< ε, that isG(x_n, x_m, x_p)→0 asn, m, p→ ∞. AG- metric space is said to beG- complete if everyG- Cauchy sequence isG- convergent.

Lemma 2.3 ([15]). Let (X, G) be a G - metric space. Then the following properties are equivalent:

1) (xn)isG- convergent to x, 2) G(x, xn, xn)→0 asn→ ∞, 3) G(x, x, xn)→0 asn→ ∞, 4) G(xn, xm, x)→0asn, m→ ∞.

Lemma 2.4 ([15]). Let (X, G) be a G - metric space. Then the following properties are equivalent:

1) The sequence (xn)is G- Cauchy;

2) For every ε > 0, there exist k ∈ N such that for all m, n∈ N, m, n ≥ k,G(xn, xm, xm)< ε.

Lemma 2.5 ([15]). Let (X, G) be a G - metric space. Then the function G(x, y, z)is jointly continuous in all three of its variables.

Lemma 2.6 ([15]). Let (X, G) be a G - metric space. Then G(x, x, y) ≤ 2G(x, y, y).

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3. Implicit relations

Definition 3.1. LetFGbe the set of all real continuous functionsF(t1, ..., t6) : R⁶+→Rsuch that

(F1) :F is non - increasing in variablet6,

(F2) : There existsh∈[0,1) such that for allu, v≥0,F(u, v, u, v,0, u+v)≤ 0 impliesu≤hv.

(F3) : There existsg∈[0,1) such that for all t, t^′ >0,F(t, t,0,0, t^′, t)≤0 impliest≤gt^′.

Example 3.2. F(t₁, ..., t₆) =t₁−kmax{t₂, t₃, t₄, t₅, t₆}, wherek∈[ 0,¹₂)

. (F₁) : Obviously.

(F₂) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u−kmaxu, v, u+v= h−k(u+v)≤0. Henceu≤hv, where 0≤h=₁₋^k_k <1.

(F3) : Let t, t^′ >0 and F(t, t,0,0, t^′, t) =t−kmax{t, t^′} ≤ 0. If t > t^′, then t(1−k)≤0, a contradiction. Hence, t≤t^′ which impliest≤gt^′, where 0≤g=k <1.

Example 3.3. F(t1, ..., t6) =t1−at2−bt4−cmax{2t3, t5+t6}, wherea, b, c≥0 and 0< a+b+ 2c <1.

(F1) : Obviously.

(F2) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u−av−bv−cmax{2u, u+v} ≤ 0. If u > v, then u(1−(a+b+ 2c))≤0, a contradiction. Henceu≤v, which impliesu≤hv, where 0≤h=a+b+ 2c <1.

(F₃) : Lett, t^′ >0 andF(t, t,0,0, t^′, t) =t−at−c(t+t^′)≤0 which implies t≤gt^′, where 0≤g= _a+c^c <1.

Example 3.4. F(t₁, ..., t₆) = t₁−kmax{t₃+t₆, t₄+t₅} −at₂, where 0 ≤ a+ 3k <1.

(F1) : Obviously.

(F2) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u−av−k(2u+v)≤0 which impliesu≤hv, where 0≤h= ₁^a+k₋_2k <1.

(F3) : Lett, t^′ >0 andF(t, t,0,0, t^′, t) =t−at−kmax{t, t^′} ≤0. Ift > t^′, then t(1−(a+k))≤ 0, a contradiction. Hence t ≤t^′ which implies t ≤gt^′ where 0≤g=₁₋^k_a <1.

Example 3.5. F(t₁, ..., t₆) = t₁ −kmax {

t₂, t₃, t₄,t3+t5

2 ,t5+t6

2 }

, where k∈[0,1).

(F1) : Obviously.

(F2) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u−kmax{

u, v,^u₂,^u+v₂ }

≤0. Ifu > v, then u(1−k)≤0, a contradiction. Henceu≤v, which implies u≤hv, where 0≤h=k <1.

(F3) : Lett, t^′>0 andF(t, t,0,0, t^′, t) =t−kmax{t,^t₂^′,^t+t₂^′} ≤0. Ift > t^′, then t(1−k)≤0, a contradiction. Hencet ≤t^′, which implies t≤gt^′, where 0≤g=k <1.

Example 3.6. F(t1, ..., t6) =t1−kmax {

t2, t3, t4,t5+ 2t6

3 ,2t3+t4

3 }

, where k∈[

0,³₄) .

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(F1) : Obviously.

(F2) : Letu, v ≥0 andF(u, v, u, v,0, u+v) =u−kmax{u, v,^2(u+v)₃ ,^2u+v₃ }

≤0. Ifu > v, thenu( 1−^4k₃)

≤0, a contradiction. Henceu≤v, which implies u≤hv, where 0≤h= ^4k₃ <1.

(F3) : Lett, t^′>0 andF(t, t,0,0, t^′, t) =t−kmax{t, ,^2t+t₃ ^′} ≤0. Ift > t^′, thent(1−k)≤0, a contradiction. Hencet≤t^′, which implies t≤gt^′, where 0≤g=k <1.

Example 3.7. F(t1, ..., t6) =t1−at2−(b+d)t3−ct4−emax{t3, t5, t6}, where a, b, c, d, e≥0 anda+b+c+d+ 2e <1.

(F1) : Obviously.

(F2) : Letu, v ≥0 andF(u, v, u, v,0, u+v) =u−av−(b+d)u−cv−e(u+v)≤ 0 which impliesu≤hv, where 0≤h= a+c+e

1−b−d−e <1.

(F3) : Lett, t^′ >0 andF(t, t,0,0, t^′, t) =t−at−emax{t, t^′} ≤0. Ift > t^′ then t(1−(a+e)) ≤0, a contradiction. Hence t ≤t^′ which implies t ≤gt^′, where 0≤g= e

1−a <1.

Example 3.8. F(t₁, ..., t₆) = t₁−kmax{t₂, t₃, t₄, t₅, t₆, t₁+t₃, t₃+t₆,2t₅}, wherek∈[

0,¹₃) . (F₁) : Obviously.

(F₂) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u−kmax{u, v, u+v,2u,2u+ v} ≤ 0. If u > v then u(1−3k) >0, a contradiction. Hence u ≤v which impliesu≤hv, where 0≤h= 3k <1.

(F3) : Lett, t^′>0 andF(t, t,0,0, t^′, t) =t−kmax{t, t^′,2t^′} ≤0. Ift > t^′, thent(1−2k)≤0, a contradiction. Hence,t≤t^′ which impliest≤gt^′, where 0≤g= 2k <1.

Example 3.9. F(t₁, ..., t₆) = t₁−at₂−kmax{t₃+t₅+t₆,2t₃+t₄}, where a, k≥0 anda+ 3k <1.

(F₁) : Obviously.

(F2) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u−av−k(2u+v)≤0 which impliesu≤hv, where 0≤h= a+k

1−2k <1.

(F3) : Lett, t^′>0 andF(t, t,0,0, t^′, t) =t−at−k(t+t^′)≤0 which implies t≤gt^′, where 0≤g= ₁₋^k_a₋_k <1.

Example 3.10. F(t₁, ..., t₆) =t₁−at₂−bt₄−kmax{t₁+t₃+t₅+t₆,2t₃+t₆}, wherea, b, k≥0 anda+b+ 4c <1.

(F₁) : Obviously.

(F₂) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u−av−bv−k(3u+v)≤0 which implies u≤hv, where 0≤h=^a+b+k₁₋_3k <1.

(F3) : Lett, t^′>0 andF(t, t,0,0, t^′, t) =t−at−k(2t+t^′)≤0 which implies t≤gt^′, where 0≤g= ₁₋_a^k₋_2k <1.

Example 3.11. F(t1, ..., t6) =t²₁−at²₂−_1+t^bt⁵2^t⁶

3+t²₄, wherea, b >0 anda+b <1.

(F₁) : Obviously.

(F2) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u²−av²≤0, which implies u≤hv, where 0≤h=√

a <1.

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(F3) : Let t, t^′ >0 andF(t, t,0,0, t^′, t) =t²−at²−bt^′t≤0, which implies t≤gt^′, where 0≤g= b

1−a <1.

Example 3.12. F(t₁, ..., t₆) =t₁−at₂−bmax{t₃, t₄}−cmax{t₃, t₅}−dmax{t₂, t3, t4,^t³^+t₂ ⁵} −et6, wherea, b, c, d, e≥0 and 0≤a+b+c+d+ 2e <1.

(F1) : Obviously.

(F2) : Letu, v≥0 andF(u, v, u, v,0, u+v) =u−av−bmax{u, v} −cu− dmax{v, u,^u₂} −e(u+v)≤0. Ifu > v, thenu(1−a−b−c−d−2e)≤0, a contradiction. Henceu≤vwhich impliesu≤hv, where 0≤h= ^a+b+d+e₁₋_c₋_e <1.

(F3) : Lett, t^′ >0 andF(t, t,0,0, t^′, t) =t−at−ct^′−dmax{t,^t₂^′} −et≤0.

If t > t^′ then t[1−(a+c+d+e)]≤0, a contradiction. Hence t≤t^′ which impliest≤gt^′ where 0≤g=a+c+d+e <1.

Example 3.13. F(t1, ..., t6) =t1−kmax{t2, t3+t4, t5+t6}, wherek∈[ 0,¹₂)

. (F1) : Obviously.

(F₂) : Let u, v ≥ 0 and F(u, v, u, v,0, u+v) = u−k(u+v) ≤ 0, which impliesu≤hv, where 0≤h=k <1.

(F₃) : Let t, t^′ > 0 andF(t, t,0,0, t^′, t) = t−k(t+t^′) ≤0 which implies t≤gt^′ where 0≤g= ₁₋^k_k <1.

Example 3.14. F(t1, ..., t6) =t1−at2−kmax{t3, t4, t5, t6}, where a, k ≥0 anda+ 2k <1.

(F1) : Obviously.

(F2) : Let u, v ≥ 0 and F(u, v, u, v,0, u+v) = u−k(u+v) ≤ 0, which impliesu≤hv, where 0≤h= ^a+k₁₋_k <1.

(F₃) : Lett, t^′>0 andF(t, t,0,0, t^′, t) =t−at−kmax{t, t^′} ≤0. Ift > t^′, then t(1−(a+k))≤0, a contradiction. Hence, t≤t^′ which implies t≤gt^′, where 0≤g=₁₋^k_a <1.

Example 3.15. F(t1, ..., t6) =t⁴₁−k^t²³^t_1+t²⁴^+t2²⁵^t²⁶ 2

, wherek∈(0,1).

(F₁) : Obviously.

(F2) : Let u, v ≥0 and F(u, v, u, v,0, u+v) =u⁴−k^u_1+v²^v² ≤0. If u >0, then u²≤k_1+v^v² ≤kv². Hence u≤hv, where 0≤h=√

k <1. Ifu= 0, then u≤hv.

(F3) : Let t, t^′ > 0 and F(t, t,0,0, t^′, t) = t⁴−k^t_1+t^′²^t² ≤ 0, which implies t²≤k_1+t^t^′² ≤kt^′². Hence,t≤gt^′, where 0≤g=√

k <1.

4. Main results

Theorem 4.1. Let(X, G)be aG- complete metric space and let S, T :X → X be two functions satisfying the following inequalities for allx, y∈X:

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







ϕ₁(G(T x, T x, Sy), G(x, x, y), G(x, T x, T x), G(y, Sy, Sy), G(x, Sy, Sy), G(y, T x, T x))≤0

ϕ2(G(Sx, Sx, T y), G(x, x, y), G(x, Sx, Sx), G(y, T y, T y), G(x, T y, T y), G(y, Sx, Sx))≤0 where ϕ1, ϕ2∈FG. ThenS andT have a unique common fixed point.

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Proof. Letx0∈X be an arbitrary point and x2n+1=Sx2n, x2n+2=T x2n+1

forn= 0,1,2, .... By (ϕ1) we have successively:

ϕ1(G(T x2n+1, T x2n+1, Sx2n), G(x2n+1, x2n+1, x2n), G(x2n+1, T x2n+1, T x2n+1), G(x2n, Sx2n, Sx2n), G(x2n+1, Sx2n, Sx2n), G(x2n, T x2n+1, T x2n+1))≤0, ϕ₁(G(x_2n+2, x_2n+2, x_2n+1), G(x_2n+1, x_2n+1, x_2n), G(x_2n+1, x_2n+2, x_2n+2),

G(x_2n, x_2n+1, x_2n+1),0, G(x_2n, x_2n+2, x_2n+2))≤0.

By (G₅) we have

G(x_2n, x_2n+2, x_2n+2)≤G(x_2n, x_2n+1, x_2n+1) +G(x_2n+1, x_2n+2, x_2n+2).

By (F1) and (G4) we obtain

ϕ1(G(x2n+2, x2n+2, x2n+1), G(x2n+1, x2n+1, x2n), G(x2n+2, x2n+2, x2n+1), G(x2n+1, x2n+1, x2n),0, G(x2n+1, x2n+1, x2n) +G(x2n+2, x2n+2, x2n+1))≤0 which implies by (F₂) that

G(x2n+2, x2n+2, x2n+1)≤hG(x2n+1, x2n+1, x2n), whereh= max{h1, h2}. Similarly, by (ϕ2) we have successively

ϕ₂(G(Sx_2n+2, Sx_2n+2, T x_2n+1), G(x_2n+2, x_2n+2, x_2n+1), G(x_2n+2, Sx_2n+2, Sx_2n+2), G(x_2n+1, T x_2n+1, T x_2n+1), G(x_2n+2, T x_2n+1, T x_2n+1), G(x_2n+1, Sx_2n+2, Sx_2n+2))≤0,

ϕ2(G(x2n+3, x2n+3, x2n+2), G(x2n+2, x2n+2, x2n+1), G(x2n+2, x2n+3, x2n+3), G(x2n+1, x2n+2, x2n+2),0, G(x2n+1, x2n+3, x2n+3))≤0.

By (G5) we have

G(x2n+1, x2n+3, x2n+3)≤G(x2n+1, x2n+2, x2n+2) +G(x2n+2, x2n+3, x2n+3).

By (F₁) and (G₄) we obtain

ϕ₂(G(x_2n+3, x_2n+3, x_2n+2), G(x_2n+2, x_2n+2, x_2n+1), G(x_2n+3, x_2n+3, x_2n+2), G(x_2n+2, x_2n+2, x_2n+1), 0, G(x_2n+2, x_2n+2, x_2n+1) +G(x_2n+3, x_2n+3, x_2n+2))≤0 which implies by (F2) that

G(x_2n+3, x_2n+3, x_2n+2)≤hG(x_2n+2, x_2n+2, x_2n+1), whereh= max{h1, h2}.

Hence

G(xn+1, xn+1, xn)≤hG(xn, xn, xn−1) forn= 1,2, ..., which implies

G(xn+1, xn+1, xn)≤hⁿG(x1, x1, x0).

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Then, form, n∈N,m > n, by repeating use of (G5) we have G(xm, xm, xn) ≤ G(xn+1, xn+1, xn) +G(xn+2, xn+2, xn+1) +

+...+G(x_m, x_m, x_m₋₁)

≤ (hⁿ+hⁿ⁺¹+...+h^m⁻¹)G(x1, x1, x0)

≤ hⁿ

1−hG(x1, x1, x0).

This implies that (x_n) is a G - Cauchy sequence. Since (X, G) is G - complete, there exists u∈X such that (xn) isG- convergent tou. We prove that uis a common fixed point forS and T.

By (ϕ1) we have successively

ϕ1(G(T u, T u, Sx2n), G(u, u, x2n), G(u, T u, T u), G(x2n, Sx2n, Sx2n), G(u, Sx2n, Sx2n), G(x2n, T u, T u))≤0,

ϕ₁(G(T u, T u, x_2n+1), G(u, u, x_2n), G(u, T u, T u), G(x_2n, x_2n+1, x_2n+1), G(u, x_2n+1, x_2n+1), G(x_2n, T u, T u))≤0.

Lettingntend to infinity and by (G₄) we obtain

ϕ1(G(T u, T u, u),0, G(T u, T u, u),0,0, G(T u, T u, u))≤0.

By (F2) we obtain G(T u, T u, u) = 0 which impliesu=T u.

Similarly, by (ϕ2) we have successively

ϕ2(G(Su, Su, T x2n+1), G(u, u, x2n+1), G(u, Su, Su), G(x2n+1, T x2n+1, T x2n+1), G(u, T x_2n+1, T x_2n+1), G(x_2n+1, Su, Su))≤0,

ϕ2(G(Su, Su, x2n+2), G(u, u, x2n+1), G(u, Su, Su), G(x2n+1, x2n+2, x2n+2), G(u, x2n+2, x2n+2), G(x2n+1, Su, Su))≤0.

Lettingntend to infinity and (G4) we obtain

ϕ2(G(Su, Su, u),0, G(Su, Su, u),0,0, G(Su, Su, u))≤0.

By (F₂) we obtain G(Su, Su, u) = 0 which impliesu=Su.

Henceu=Su=T uand uis a common fixed point ofS and T.

We prove thatuis the unique common fixed point ofS andT. Letv=Sv=T v be another common fixed point ofS andT.

Then, by (ϕ1) and (G4) we have successively

ϕ1(G(T u, T u, Sv), G(u, u, v), G(u, T u, T u), G(v, Sv, Sv), G(u, Sv, Sv), G(v, T u, T u))≤0 ϕ1(G(u, u, v), G(u, u, v),0,0, G(u, v, v), G(u, u, v))≤0 which implies by (F2) that

G(u, u, v)≤h1G(v, v, u).

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Similarly,

G(v, v, u)≤h₁G(u, u, v).

Hence

G(u, u, v)(1−h²₁)≤0, which implies G(u, u, v) = 0, i.e. u=v.

Remark 4.2. By Theorem 4.1, using examples 3.2 - 3.13 we obtain new partic- ular results.

Theorem 4.3. Let (X, G)be a G- complete metric space and letT :X →X be a function satisfying the following inequality for all x, y∈X:

(4.2) ϕ(G(T x, T x, T y), G(x, x, y), G(x, T x, T x), G(y, T y, T y), G(x, T y, T y), G(y, T x, T x))≤0 whereϕ∈FG. ThenT has a unique fixed point.

Corollary 4.4(Theorem 2.1 [17]). Let(X, G)be aG- complete metric space and let T : X → X be a function satisfying the following inequality for all x, y, z∈X:

(4.3) G(T x, T y, T z)≤kmax{G(x, y, z), G(x, T x, T x), G(y, T y, T y), G(y, T z, T z), G(x, T y, T y), G(y, T z, T z), G(z, T x, T x)) wherek∈[

0,¹₂)

. Then T has a unique fixed point.

Proof. Forz=xwe obtain by (4.3) that

G(T x, T x, T y)≤kmax{G(x, x, y), G(x, T x, T x), G(y, T y, T y), G(x, T y, T y), G(y, T x, T x))≤0.

By Theorem 4.3 and Example 3.2,T has a unique fixed point.

(4.4) G(T x, T y, T z)≤kmax{G(x, T y, T y) +G(y, T x, T x), G(y, T z, T z) +G(z, T y, T y), G(x, T z, T z) +G(z, T x, T x)) wherek∈[

0,¹₂)

G(T x, T x, T y)≤kmax{G(x, T y, T y) +G(y, T x, T x),2G(x, T x, T x)). By Theorem 4.3 and Example 3.3,T has a unique fixed point.

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(4.5) G(T x, T y, T z)≤kmax{G(z, T x, T x) +G(y, T x, T x), G(y, T z, T z) +G(x, T z, T z), G(x, T y, T y) +G(y, T y, T y)) where k∈[

0,¹₃)

. ThenT has a unique fixed point.

G(T x, T x, T y)≤kmax{G(x, T x, T x) +G(y, T x, T x), G(x, T y, T y) +G(y, T y, T y)) . By Theorem 4.3 and Example 3.4,T has a unique fixed point.

Corollary 4.7 (Theorem 2.1 [3]). Let (X, G)be aG- complete metric space and let T : X → X be a function satisfying the following inequality for all x, y, z∈X:

(4.6)

G(T x, T y, T z)≤kmax{G(x, y, z), G(x, T x, T x), G(y, T y, T y), G(z, T z, T z), G(x, T y, T y) +G(z, T x, T x)

2 ,G(x, T y, T y) +G(y, T x, T x)

2 ,

G(y, T z, T z) +G(z, T y, T y)

2 ,G(x, T z, T z) +G(z, T x, T x)

2 }

where k∈[0,1). ThenT has a unique fixed point.

G(T x, T x, T y)≤kmax{G(x, x, y), G(x, T x, T x), G(y, T y, T y), G(x, T y, T y) +G(x, T x, T x)

2 ,G(x, T y, T y) +G(y, T x, T x)

2 } .

Corollary 4.8. Let(X, G)be aG- complete metric space and letT :X →X be a function satisfying the following inequality for all x, y, z∈X:

(4.7)

G(T x, T y, T z)≤kmax{G(x, y, z), G(x, T x, T x), G(y, T y, T y), G(z, T z, T z), G(y, T x, T x) +G(z, T y, T y) +G(y, T z, T z)

3 ,

G(x, T x, T x) +G(y, T y, T y) +G(z, T z, T z)

3 }

where k∈[ 0,³₄)

G(T x, T x, T y)≤kmax{G(x, x, y), G(x, T x, T x), G(y, T y, T y), 2G(y, T x, T x) +G(x, T y, T y)

3 ,2G(x, T x, T x) +G(y, T y, T y)

3 } .

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Remark 4.9. This Corollary is a generalization of Theorem 2.6 [2], whereh∈ [0,¹₂)

.

Corollary 4.10 (Theorem 3.1 [13]). Let (X, G) be a G - complete metric space and letT :X →X be a mapping such that:

(4.8)

G(T x, T y, T z)≤aG(x, y, z) +bG(x, T x, T x) +cG(y, T y, T y) +dG(z, T z, T z)

+emax{G(x, T y, T y), G(y, T x, T x), G(y, T z, T z), G(z, T y, T y), G(z, T x, T x), G(x, T z, T z)}

for all x, y, z ∈X, where a, b, c, d, e≥0 and a+b+c+d+ 2e <1. ThenT has a unique fixed point.

Proof. Forz=xby (4.8) we obtain

G(T x, T x, T y)≤aG(x, x, y) +bG(x, T x, T x) +cG(y, T y, T y) +dG(x, T x, T x)

+emax{G(x, T y, T y), G(y, T x, T x), G(x, T x, T x)}. .

Corollary 4.11 (Theorem 3.7 [13]). Let (X, G) be a G - complete metric space and letT :X →X be a mapping such that:

(4.9)

G(T x, T y, T z)≤kmax{G(x, T x, T x), G(y, T y, T y), G(z, T z, T z), G(x, T y, T y), G(y, T z, T z), G(z, T x, T x), G(x, T z, T z), G(y, T x, T x), G(z, T y, T y), G(x, T y, T y), G(y, T z, T x), G(z, T x, T y), G(x, y, T z), G(y, z, T x), G(z, x, T y), G(x, y, z)}, for allx, y, z∈X, where k∈[

0,¹₃)

G(T x, T x, T y)≤kmax{G(x, T x, T x), G(y, T y, T y), G(x, T y, T y), G(y, T x, T x), G(x, T y, T x), G(x, y, T x), G(x, x, T y), G(x, x, y)}. . By (G5) and Lemma 2.6 we obtain

G(T x, T x, T y)≤kmax{G(x, T x, T x), G(y, T y, T y), G(x, T y, T y), G(y, T x, T x), G(x, T x, T x) +G(T x, T x, T y),

G(x, T x, T x) +G(y, T x, T x),2G(x, T y, T y), G(x, x, y)}. By Theorem 4.3 and Example 3.8,T has a unique fixed point.

Corollary 4.12. Let(X, G)be aG- complete metric space and letT :X →X be a mapping such that for allx, y, z∈X:

(4.10)

G(T x, T y, T z)≤aG(x, y, z) +kmax{G(x, T y, T y) +G(y, T x, T x)+

+G(z, T z, T z), G(y, T z, T z) +G(z, T y, T y) +G(x, T x, T x), G(z, T x, T x) +G(x, T z, T z) +G(y, T y, T y)}, wherea, k≥0 anda+ 3k <1. ThenT has a unique fixed point.

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G(T x, T x, T y)≤aG(x, x, y)

+kmax{G(x, T y, T y) +G(y, T x, T x) +G(x, T x, T x), 2G(x, T x, T x) +G(y, T y, T y)},

Remark 4.13. Ifa= 0, by Corollary 4.12 we obtain Theorem [13].

Corollary 4.14. Let(X, G)be aG- complete metric space and letT :X →X be a mapping such that for all x, y, z∈X:

(4.11)

G(T x, T y, T z)≤aG(x, y, z) +bG(y, T y, T y) +kmax{G(x, T y, T y) +G(y, T x, T x) +G(z, T x, T y),

G(y, T z, T z) +G(z, T y, T y) +G(x, T y, T z), G(z, T x, T x) +G(x, T z, T z) +G(y, T z, T x)}, where a, b, k≥0 anda+b+ 4k <1. ThenT has a unique fixed point.

Proof. Forz=xand (G5) we obtain

G(T x, T x, T y)≤aG(x, x, y) +bG(y, T y, T y) +kmax{G(x, T y, T y) +G(y, T x, T x) +G(x, T x, T y),

2G(x, T x, T x) +G(y, T x, T x)}

≤aG(x, x, y) +bG(y, T y, T y) +kmax{G(x, T y, T y) +G(y, T x, T x)+

G(x, T x, T x) +G(T x, T x, T y), 2G(x, T x, T x) +G(y, T x, T x)}.

.

Remark 4.15. Ifa=b= 0, by Corollary 4.14 we obtain Theorem 3.11 [13].

Corollary 4.16. Let(X, G)be aG- complete metric space and letT :X →X be a mapping such that for all x, y, z∈X:

(4.12)

G(T x, T y, T z)≤aG(x, y, z) +bmax{G(x, T x, T x), G(y, T y, T y)} +cmax{G(x, T z, T z), G(x, T y, T y)}

+dmax{G(x, y, z), G(x, T x, T x), G(y, T y, T y),

G(x,T y,T y)+G(x,T z,T z)

2 }+eG(y, T x, T x),

where a, b, c, d, e ≥0 and a+b+c+d+ 2e < 1. Then T has a unique fixed point.

G(T x, T x, T y)≤aG(x, x, y) +bmax{G(x, T x, T x), G(y, T y, T y)} +cmax{G(x, T x, T x), G(x, T y, T y)}

+dmax{G(x, x, y), G(x, T x, T x), G(y, T y, T y),

G(x,T y,T y)+G(x,T x,T x)

2 }+eG(y, T y, T y).

.

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Remark 4.17. Corollary 4.16 is a generalization of Theorem 3.2 [11] because in this theorema+b+ 2c+ 2d <1 ande= 0.

Corollary 4.18(Corollary 5 [11]). Let(X, G)be aG- complete metric space and letT :X →X be a mapping such that for allx, y, z∈X:

(4.13) G(T x, T y, T z)≤kmax{G(x, y, z), G(x, T x, T x) +G(y, T y, T y), G(x, T y, T y) +G(y, T x, T x)},

wherek∈[ 0,¹₂)

G(T x, T x, T y)≤kmax{G(x, x, y), G(x, T x, T x) +G(y, T y, T y), G(x, T y, T y) +G(y, T x, T x)}. . By Theorem 4.3 and Example 3.13, T has a unique fixed point.

Corollary 4.19. Let(X, G)be aG- complete metric space and letT :X →X be a mapping such that for allx, y, z∈X:

(4.14)

G(T x, T y, T z)≤aG(x, y, z) +kmax{G(x, T x, T x), G(x, T y, T y), G(x, T z, T z), G(y, T y, T y), G(y, T x, T x),

G(y, T z, T z), G(z, T z, T z), G(z, T x, T x), G(z, T y, T y)},

where x, y, z∈X, wherea, k ≥0 and 0< a+ 2k <1. ThenT has an unique fixed point.

G(T x, T x, T y)≤aG(x, x, y) +kmax{G(x, T x, T x), G(y, T y, T y), G(x, T y, T y), G(y, T x, T x)}, . and the proof it follows by Theorem 4.1 and Example 3.14.

Remark 4.20. Ifa= 0 we obtain Theorem 1 [30].

Note. 1) The notions of quasi - metric spaces are introduced in [31].

Some fixed point theorems for mappings in quasi - metric spaces are proved in [6], [7], [8], [12], [26], [27] and other papers.

Let (X, Q) be a quasi - metric space, where q(x, y) is a quasi - metric on X. In the proofs of Theorem 1 [26] and Theorem 1 [27] is used thatd(x, y) = max{q(x, y), q(y, x)}is a metric onX.

Let (X, G) be a G- metric space. It is proved in Theorem 2.1 [9], Lemma 5.1 [22], Lemma 2.4 [25] thatG(x, x, y) is a quasi - metric q(x, y) onX, hence D(x, y) = max{G(x, x, y), G(y, y, x)}is a metric onX.

Using this fact, the study of ([9], Theorem 3.1) and ([28], Theorems 17, 19, 21) is reduced to the study of fixed points in metric spaces.

2) LetF(t₁, ..., t₆)≤0 be a contractive condition inG- metric spaces.

To use a technique as in [9], [28] it is necessary that F(t1, ..., t6) to be non - increasing in variablest2, ..., t6.

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In the present paper,F is non - increasing only in variablet6. In Example 3.11F is non - decreasing in variablest3 andt4and in Example 3.15,F is non - decreasing in variable t₂. Hence, Theorems 4.1 and 4.3 cannot be derived as the results from [9], [28].

3) In [10] there exists a functionF non - increasing in variablest₂, t₃, ...

which is not derived as the results from [9], [28].

Acknowledgement

The authors thank the anonymous referee for his/her suggestion, that im- prove the initial version of the paper.

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Received by the editors February 13, 2013