1. Introduction

The first important result in the theory of fixed point of compatible mappings was obtained by Gerald Jungck in 1986 ^[2] as a generalization of commuting mappings. In 1993 Jungck, Murthy and Cho ^[3] introduced the concept of compatible mappings of type (A) by generalizing the definition of weakly uniformly contraction maps. Pathak and Khan ^[6] introduced the concept of A-compatible and S-compatible by splitting the definition of compatible mappings of type (A). Fixed point results of compatible mappings are found in [1-8]^[1].

Sharma and Sahu ^[8] proved the following theorem.

THEOREM 1.1 Let A, S and T be three continuous mappings of a complete metric space (X, d) into itself satisfying the following conditions:

(i)   A commutes with S and T respectively

(ii)  S (X) ⊆ A(X) and T(X) ⊆ A(X)

(iii) [d(Sx, Tx)]² ≤ a₁d(Ax, Sx)d(Ay, Ty)+a₂d(Ay, Sx)d(Ax, Ty)+a₃d(Ax, Sx)d(Ax, Ty) +a₄d(Ay, Ty)d(Ay, Sx)+a₅d²(Ax, Ay)

For all x, y ∈ X, where a_i ≥ 0, i = 1, 2, 3, 4, 5 and a₁+a₄+ a₅< 1, 2a₁+3a₃+2a₅<2.

Then A, S and T have a unique common fixed point in X.

Murthy ^[6] pointed out that the constraints taken by Sharma and Sahu in condition (iii) of theorem 1.1 is not true and suggested the corrected replacement as max {a₁+2a₃+ a₅, a₁+2a₄+a₅, a₂+ a₅} < 1 and proved a new fixed point theorem.

The aim of this paper is to prove a common fixed point theorem of S-compatible mappings in metric space by considering four self mappings. Further we give another common fixed point theorem of A-compatible mappings.

2. Preliminaries

Following are definitions of types of compatible mappings.

Definition 2.1 ^[2]: Let A and S be mappings from a complete metric space X into itself. The mappings A and S are said to be compatible if d(ASx_n, SAx_n) = 0 whenever {x_n} is a sequence in X such that Ax_n = Sx_n = t for some t ∈ X.

Definition 2.2 ^[3]: Let A and S be mappings from a complete metric space X into itself. The mappings A and S are said to be compatible of type (A) if d(ASx_n, SSx_n) = 0 and d(SAx_n, AAx_n) = 0 whenever {x_n} is a sequence in X such that forAx_n = Sx_n = t for some t ∈ X.

Definition 2.3 ^[5]: Let A and S be mappings from a complete metric space X into itself. The mappings A and S are said to be A-compatible if d(ASx_n, SSx_n) = 0 whenever {x_n} is a sequence in X such that forAx_n = Sx_n = t for some t ∈ X.

Definition 2.4 ^[5]: Let A and S be mappings from a complete metric space X into itself. The mappings A and S are said to be S-compatible if d(SAx_n, AAx_n) = 0 whenever {x_n} is a sequence in X such that forAx_n = Sx_n = t for some t ∈ X.

Proposition 2.5 ^[6]: Let A and S be mappings from a complete metric space (X, d) into itself. If a pair (A, S) is A-compatible on X and St = At for t ∈ X, then ASt = SSt.

Proposition 2.6 ^[6]: Let A and S be mappings from a complete metric space (X, d) into itself. If a pair (A, S) is S-compatible on X and St = At for t ∈ X, then SAt = AAt.

Proposition 2.7 ^[6]: Let A and S be mappings from a complete metric space (X, d) into itself. If a pair (A, S) is A-compatible on X and Ax_n =Sx_n = t for t ∈ X, then SSx_n → At if A is continuous at t.

Proposition 2.8 ^[6]: Let A and S be mappings from a complete metric space (X, d) into itself. If a pair (A, S) is S-compatible on X and Ax_n =Sx_n = t for t ∈ X, then AAx_n → St if S is continuous at t.

Now we prove the following theorem.

LEMMA 2.9 Let A, B, S and T be mapping from a metric space (X, d) into itself satisfying the following conditions:

(1) A(X) ⊆ T(X) and B(X) ⊆ S(X)

(2) [d(Ax, Bx)]² ≤ a₁d(Ax, Sx)d(By, Ty)+a₂d(By, Sx)d(Ax, Ty)+a₃d(Ax, Sx)d(Ax, Ty)+a₄d(By, Ty)d(By, Sx) +a₅d²(Sx, Ty)

where a₁+ a₂ +2a₃ +a₄+ a₅< 1 and a₁, a₂, a₃, a₄, a₅ ≥ 0

(3) Let x₀ ∈ X then by (1) there exists x₁∈ X such that Tx₁ = Ax₀ and for x₁ there exists x₂∈ X such that Sx₂ = Bx₁ and so on. Continuing this process we can define a sequence {y_n} in X such that

then the sequence {y_n} is Cauchy sequence in X.

Proof. By condition (2) and (3), we have

Where

Since a₁+ a₂ +2a₃ +a₄+ a₅< 1 and a₁, a₂, a₃, a₄, a₅ ≥ 0.

In order to satisfy the inequation, one value of λ will be positive and the other will be negative. We also note that the sum and product of the two values of λ is less than 1 and -1 respectively. Neglecting the negative value, we have where 0<p<1.

Hence {y_n} is Cauchy sequence.

3. Main Results

We prove the following theorem.

THEOREM 3.1: Let A, B, S and T be self maps of a complete metric space (X, d) satisfying the following conditions:

(1) A (X) ⊆ T(X) and B(X) ⊆ S(X)

(2) [d(Ax, Bx)]2 ≤ a1d(Ax, Sx)d(By, Ty)+a2d(By, Sx)d(Ax, Ty)+a3d(Ax, Sx)d(Ax, Ty)+a4d(By, Ty)d(By, Sx) +a5d2(Sx, Ty)

where a1+ a2 +2a3 +a4+ a5< 1 and a1, a2, a3, a4, a5 ≥ 0

(3) Let x0 ∈ X then by (1) there exists x₁∈ X such that Tx₁ = Ax₀ and for x₁ there exists x₂∈ X such that Sx₂ = Bx₁ and so on. Continuing this process we can define a sequence {y_n} in X such that

then the sequence {y_n} is Cauchy sequence in X.

(4) One of A, B, S or T is continuous.

(5) [A, S] and [B, T] are S-compatible mappings on X.

Then A, B, S and T have a unique common fixed point in X.

Proof: By lemma 2.9, {y_n} is Cauchy sequence. Since X is complete, there exists a point z∈ X such that lim y_n = z as n → ∞. Consequently subsequences Ax_2n, Sx_2n, Bx_2n-1 and Tx_2n+1 converges to z.

Let S be a continuous mapping. Since A and S are S-compatible mappings on X, then by proposition 2.8., we have AAx_2n → Sz and SAx_2n → Sz as n → ∞.

Now by condition (2) of lemma 2.9, we have

As n→∞, we have

which is a contradiction. Hence Sz = z,

Now

Letting n→∞, we have [d(Az, z)]² ≤ a₃[d(Az, z)]². Hence Az = z.

Now since Az = z, by condition (1), z ∈ T(X). Also T is self map of X so there exists a point u ∈X such that z = Az = Tu. More over by condition (2), we obtain,

i.e., [d(z, Bu)]² ≤ a₄[d(z, Bu)]².

Hence Bu = z i.e., z = Tu = Bu.

By condition (5), we have

Hence d(Tz, Bz) = 0 i.e., Tz = Bz.

Now,

i.e., [d(z, Tz)]² ≤ a₂[d(z, Tz)]² which is a contradiction. Hence z = Tz i.e, z = Tz = Bz.

Therefore z is common fixed point of A, B, S and T. Similarly we can prove that z is a common fixed point of A, B, S and T if any one of A, B or T is continuous.

Finally, in order to prove the uniqueness of z, suppose w be another common fixed point of A, B, S and T Then we have,

which gives [d(z, Tw)]² ≤ a₂ [d(z, Tw)]². Hence z = w.

This completes the proof.

THEOREM 3.2: Let A, B, S and T be self maps of a complete metric space (X, d) satisfying the following conditions:

(1) A (X) ⊆ T(X) and B(X) ⊆ S(X).

(2) [d(Ax, Bx)]² ≤ a₁d(Ax, Sx)d(By, Ty)+a₂d(By, Sx)d(Ax, Ty)+a₃d(Ax, Sx)d(Ax, Ty) +a₄d(By, Ty)d(By, Sx) +a₅d²(Sx, Ty)

where a₁+ a₂ +2a₃ +a₄+ a₅< 1 and a₁, a₂, a₃, a₄, a₅ ≥ 0.

(3) Let x₀ ∈ X then by (1) there exists x₁∈ X such that Tx₁ = Ax₀ and for x₁ there exists x₂∈ X such that Sx₂ = Bx₁ and so on. Continuing this process we can define a sequence {y_n} in X such that

then the sequence {y_n} is Cauchy sequence in X.

(4) One of A, B, S or T is continuous.

(5) [A, S] and [B, T] are A-compatible mappings on X.

Then A, B, S and T have a unique common fixed point in X.

Proof: Similar to theorem 3.1.

Remark:

(i) By taking a₁= a₂ =k₁ and a₃= a₄ =k₂ and a₅=0 and (A, S) and (B, T) as compatible mappings theorem 3.1 reduces to theorem 1 of Bijendra and Chouhan ^[1].

(ii) By taking S = T and (A, S) and (A, T) as commuting mappings or compatible mappings of type (A) theorem 3.1 reduce to results of Murthy ^[6] and Sharma and Sahu ^[8] under certain conditions.

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