Common Fixed Point Results for Mappings with Rational Expressions

Definition 1. A partial metric on a nonempty set X is a function p : X × X → [0, ∞) such that for all x, y, z ∈ X:

• (P₁)

  p(x, x) = p(y, y) = p(x, y) if and only if x = y,

• (P₂)

  p(x, x) ≤ p(x, y),

• (P₃)

  p(x, y) = p(y, x),

• (P₄)

  p(x, z) ≤ p(x, y) + p(y, z) − p(y, y).

Lemma 2 (see [1].)Let (X, p) be a partial metric space; then one has the following.

• (1)

  A sequence {x_n} in a partial metric space (X, p) converges to a point x ∈ X if and only if lim _n→∞p(x, x_n) = p(x, x).

• (2)

  A sequence {x_n} in a partial metric space (X, p) is called a Cauchy sequence if the lim _n,m→∞p(x_n, x_m) exists and is finite.

• (3)

  A partial metric space (X, p) is said to be complete if every Cauchy sequence {x_n} in X converges to a point x ∈ X; that is, p(x, x) = lim _n,m→∞p(x_n, x_m).

• (4)

  A partial metric space (X, p) is complete if and only if the metric space (X, p^s) is complete. Furthermore, lim _n→∞p^s(x_n, z) = 0 if and only if p(z, z) = lim _n→∞p(x_n, z) = lim _n,m→∞p(x_n, x_m).

Remark 3 (see [1].)Let (X, p) be a partial metric space and let A be a nonempty set in (X, p); then $$ if and only if

where $$ denotes the closure of A with respect to the partial metric p. Note A is closed in (X, p) if and only if $$.

Definition 4 (see [24].)Two families of self-mappings $$ and $$ are said to be pairwise commuting if

• (1)

  T_iT_j = T_jT_i,   i, j ∈ {1,2, …, m};

• (2)

  S_kS_l = S_lS_k,   k, l ∈ {1,2, …, n};

• (3)

  T_iS_k = S_kT_i,   i ∈ {1,2, …, m}, k ∈ {1,2, …, n}.

Proposition 5 (see [9].)Let (X, p) be a partial metric space. For any A, B, C ∈ CB^p(X), one has

• (i)

  δ_p(A, A) = sup {p(a, a) : a ∈ A};

• (ii)

  δ_p(A, A) ≤ δ_p(A, B);

• (iii)

  δ_p(A, B) = 0  implies that  A⊆B;

• (iv)

  δ_p(A, B) ≤ δ_p(A, C) + δ_p(C, B) − inf _c∈Cp(c, c).

Proposition 6 (see [9].)Let (X, p) be a partial metric space. For any A, B, C ∈ CB^p(X), one has

• (h₁)

  H_p(A, A) ≤ H_p(A, B);

• (h₂)

  H_p(A, B) = H_p(B, A);

• (h₃)

  H_p(A, B) ≤ H_p(A, C) + H_p(C, B) − inf _c∈Cp(c, c).

Lemma 7 (see [9].)Let A and B be nonempty, closed, and bounded subsets of a partial metric space (X, p) and h > 1. Then, for every a ∈ A, there exists b ∈ B such that p(a, b) ≤ hH_p(A, B).

Lemma 8 (see [10].)Let A and B be nonempty, closed, and bounded subsets of a partial metric space (X, p) and 0 < h ∈ ℝ. Then, for every a ∈ A, there exists b ∈ B such that p(a, b) ≤ H_p(A, B) + h.

Theorem 9. Let S, T : X → X be mappings on a complete PMS (X, p) and x₀, x, y ∈ X and r > 0. Suppose that there exist nonnegative reals α, β, and γ such that α + β + 2γ < 1. If S and T satisfy

for all $$, where λ = (α + γ)/(1 − β − γ). Then there exists a unique point $$ such that u = Su = Tu. Also p(u, u) = 0.

Proof. Let x₀ be an arbitrary point in X and define

We will prove that $$ for all n ∈ ℕ by mathematical induction. Using inequality (6) and the fact that λ = (α + γ)/(1 − β − γ) < 1, we have It implies that $$. Let $$ for some j ∈ N. If j = 2k + 1, where k = 0,1, 2, …(j − 1)/2, so using inequality (5), we obtain as 1 + p(x_2k, x_2k+1) > p(x_2k, x_2k+1), and so which implies that If j = 2k + 2 where k = 0,1, 2, …, (j − 2)/2, one can easily prove that Thus from inequality (11) and (12), we have Now gives $$. Hence $$ for all n ∈ ℕ. One can easily prove that for all n ∈ ℕ. We now show that {x_n} is a Cauchy sequence. Without loss of generality assume that m > n. Then, using (13*) and the triangle inequality for partial metrics (P₄) we have Inductively, we have Thus, By the definition of p^s, we get for any m ∈ ℕ^*, Hence the sequence {x_n} is a Cauchy sequence in $$. By Lemma 2(4), {x_n} is a Cauchy sequence in $$. Therefore there exists a point $$ with lim _n→∞x_n = u. Also lim _n→∞p^s(x_n, u) = 0. Again from Lemma 2(4), we have By the triangle inequality (P₄), we have Letting n → +∞ and using (19), we obtain By (P₁), we concluded that u = Su. It follows similarly that u = Tu. To prove the uniqueness of common fixed point, let $$ be another common fixed point of S and T, that, is u^* = Su^* = Tu^*. Then so that p(u, u^*) ≤ αp(u, u^*) + γp(u, u^*) because 1 + p(u, u^*) > p(u, u^*). Therefore p(u, u^*)≤(α + γ)p(u, u^*) which is a contradiction so that u = u^* (as α + γ < 1). Hence S and T have a unique common fixed point in $$.

Example 10. Let X = [0, +∞) endowed with the usual partial metric p defined by p : X × X → ℝ⁺ with p(x, y) = max {x, y}. Clearly, (X, p) is a partial metric space. Now we define S, T : X → X as

for all x ∈ X. Taking α = 1/5, β = 1/6, γ = 1/8, x₀ = 1/2, and r = 1/2, then $$. Also, we have p(x₀, x₀) = max {1/2,1/2} = 1/2, λ = (α + γ)/(1 − β − γ) = 39/85 with Also if x, y ∈ (1, +∞), then So the contractive condition does not hold on whole of X. Now if $$, then Therefore, all the conditions of Theorem 9 are satisfied. Thus 0 is the common fixed point of S and T and p(0,0) = 0. Moreover, note that for any metric d on X Therefore common fixed points of S and T cannot be obtained from a metric fixed point theorem.

Corollary 11. Let S, T : X → X be mappings on a complete PMS (X, p). Suppose that there exist nonnegative reals α, β, and γ such that α + β + 2γ < 1. If S and T satisfy

for all x, y ∈ X. Then there exists a unique point u ∈ X such that u = Su = Tu. Also p(u, u) = 0. Further S and T have no fixed point other than u.

Corollary 12. Let S, T : X → X be a mappings on complete PMS (X, p). If S and T satisfy

for all x, y ∈ X, α < 1 (α is a nonnegative real). Then S and T have a common fixed point u ∈ X and p(u, u) = 0.

Remark 13. If we impose Banach type contractive condition for a pair S, T : X → X of mappings on a metric space (X, d); that is, d(Sx, Ty) ≤ αd(x, y) for all x, y ∈ X, and then it follows that Sx = Tx, for all x ∈ X (i.e., S and T are equal). Therefore the above condition fails to find common fixed points of S and T. This can be seen as

However the same condition in partial metric space does not assert that S = T. This can be seen as by taking the partial metric same as in Example 10, for any α ≥ 1/3. Hence Corollary 12 cannot be obtained from a metric fixed point theorem.

Remark 14. By equating α, β, γ to 0 in all possible combinations, one can derive a host of corollaries which include Matthews theorem for mappings defined on a complete partial metric space.

Corollary 15. Let T : X → X be a mapping on a complete PMS (X, p) and x₀, x, y ∈ X and r > 0. Suppose that there exist nonnegative reals α, β, and γ such that α + β + 2γ < 1. If T satisfies

for all $$, where λ = (α + γ)/(1 − β − γ). Then there exists a unique point $$ such that u = Tu. Also p(u, u) = 0. Further T has no fixed point other than u.

Corollary 16. Let T : X → X be a mapping on a complete PMS (X, p). Suppose that there exist nonnegative reals α, β, and γ such that α + β + 2γ < 1. If T satisfies

for all x, y ∈ X. Then there exists a unique point u ∈ X such that u = Tu. Also p(u, u) = 0. Further T has no fixed point other than u.

Example 17. Let X = [0,4] endowed with the usual partial metric p defined by p(x, y) = max {x, y}. Clearly, (X, p) is a complete partial metric space. Now we define F : X → X as follows:

for all x ∈ X. Now, let y ≤ x. If x ∈ [0,2) (and so y ∈ [0,2)). Then p(Fx, Fy) = x/3, p(x, y) = x, p(x, Fx) = x, p(y, Fy) = y, p(y, Fx) = x/3, p(x, Fy) = x. Taking α = 1/3, β = 1/15, γ = 2/15, we can prove that all the conditions of Corollary 16 are satisfied. Now if x ∈ [2,4], then p(Fx, Fy) = x/(1 + x), p(x, y) = x, p(x, Fx) = x, p(y, Fy) = y, p(y, Fx) = x/(1 + x), p(x, Fy) = x and taking α = 1/3, β = 1/15, γ = 2/15, one can verify the condition of the above corollary. Thus all the conditions of Corollary 16 are satisfied and u = 0 is a fixed point of the mapping F.

Theorem 18. If $$ and $$ are two pairwise commuting finite families of self-mapping defined on a complete partial metric space (X, p) such that the mappings S and T (with T = T₁T₂ ⋯ T_m and S = S₁S₂ ⋯ S_n) satisfy the contractive condition (5), then the component maps of the two families $$ and $$ have a unique common fixed point.

Proof. From Theorem 9, we can say that the mappings T and S have a unique common fixed point z; that is, Tz = Sz = z. Now our requirement is to show that z is a common fixed point of all the component mappings of both the families. In view of pairwise commutativity of the families $$ and $$, (for every 1 ≤ k ≤ m) we can write T_kz = T_kTz = TT_kz and T_kz = T_kSz = ST_kz which show that T_kz (for every k) is also a common fixed point of T and S. By using the uniqueness of common fixed point, we can write T_kz = z (for every k) which shows that z is a common fixed point of the family $$. Using the same argument one can also show that (for every 1 ≤ k ≤ n) S_kz = z. Thus component maps of the two families $$ and $$ have a unique common fixed point.

Corollary 19. Let F, G : X → X be two commuting self-mappings defined on a complete PMS (X, p) satisfying the condition

for all x, y ∈ X, α + β + 2γ < 1 (α, β, and γ are nonnegative reals). Then F and G have a unique common fixed point.

Corollary 20. Let T : X → X be a mapping defined on a complete PMS (X, p) satisfying the condition

for all x, y ∈ X, α + β + 2γ < 1 (α, β, and γ are nonnegative reals). Then F has a unique fixed point.

Corollary 21. Let F : X → X be a mapping on a complete PMS (X, p). If F satisfies

for all x, y ∈ X, α < 1. Then F has a unique fixed point.

Example 22. Let X = [0,4]. Define the partial metric p : X × X → ℝ by

Then (X, p) is a complete partial metric space. Let F : X → X be defined as follows: Then for x = 0 and y = 1, we get because 0 ≤ α < 1. However, F² satisfies the requirement of Bryant’s theorem and z = 0 is the unique fixed point of F.

Theorem 23. Let (X, p) be a complete partial metric space and let S, T : X → CB^p(X) be mappings such that

for all x, y ∈ X, 0 ≤ α, β, γ with α + β + 2γ < 1. Then S and T have a common fixed point.

Proof. Assume that M = ((α + γ)/(1 − β − γ)). Let x₀ ∈ X be arbitrary but fixed element of X and choose x₁ ∈ S(x₀). By Lemma 8 we can choose x₂ ∈ T(x₁) such that

So we get Since M = ((α + γ)/(1 − β − γ)), so it further implies that By Lemma 8 we can choose x₃ ∈ S(x₂) such that So we get Continuing in this manner, one can obtain a sequence {x_n} in X as x_2n+1 ∈ S(x_2n) and x_2n+2 ∈ T(x_2n+1) such that where M = ((α + β)/(1 − α)) < 1 for all n ≥ 0. Without loss of generality assume that n > m. Then, using (48) and the triangle inequality for partial metrics (P₄), we have

By the definition of p^s, we get,

This yields that {x_n} is a Cauchy sequence in (X, p^s). Since (X, p) is complete, then from Lemma 2(4), (X, p^s) is a complete metric space. Therefore, the sequence {x_n} converges to some x^* ∈ X with respect to the metric p^s; that is, lim _n→+∞p^s(x_n, x^*) = 0. Again, from Lemma 2(4), we get taking limit as n → ∞ and using (51), we get Now x_2n+1∈T(x_2n+2) gives that which implies that On the other hand by (P₄), we have Taking limit as n → +∞ and using (51) and (55), we obtain p(x^*, S(x^*)) = 0. Therefore, from (51) (p(x^*, x^*) = 0), we obtain which from Remark 14 implies that $$. Similarly one can easily prove that x^* ∈ T(x^*). Thus S and T have a common fixed point.

Remark 24. For β = γ = 0 and S = T, Theorem 23 reduces to the following result of Aydi et al. [9].

Corollary 25 (see [9], Theorem 3.2.)Let (X, p) be a partial metric space. If T : X → CB^p(X) is a multivalued mapping such that for all x, y ∈ X, one has

where k ∈ (0,1). Then T has a fixed point.

Theorem 26. Let (X, p) be a complete partial metric space and S, T : X → CB^p(X) be multivalued mappings such that

for all x, y ∈ X, 0 ≤ α, β with 2α + β < 1, and p(x, Ty) + p(y, Sx) ≠ 0. Then S and T have a common fixed point.

Proof. Assume that l = (α + β)/(1 − α). Let x₀ ∈ X be arbitrary but fixed element of X and choose x₁ ∈ S(x₀).

When p(x, Ty) + p(y, Sx) ≠ 0. By Lemma 8 we can choose x₂ ∈ T(x₁) such that

because p(x₀, x₂) < p(x₀, x₂) + p(x₁, x₁) and p(x₁, x₁) < p(x₁, x₁) + p(x₀, x₂). Thus we get It further implies that By Lemma 8 we can choose x₃ ∈ S(x₂) such that because p(x₁, x₃) < p(x₁, x₃) + p(x₂, x₂) and p(x₂, x₂) < p(x₂, x₂) + p(x₁, x₃). Thus we get It further implies that Continuing in this manner, one can obtain a sequence {x_n} in X as x_2n+1 ∈ S(x_2n) and x_2n+2 ∈ T(x_2n+1)    such that where l = ((α + β)/(1 − α)) < 1 for all n ≥ 0. Without loss of generality assume that m > n. Then using (66) and the triangle inequality for partial metrics (P₄), one can easily prove that

By the definition of p^s,

This yields that {x_n} is a Cauchy sequence in (X, p^s). Since (X, p) is complete, then from Lemma 2(4), (X, p^s) is a complete metric space. Therefore, the sequence {x_n} converges to some x^′ ∈ X with respect to the metric p^s; that is, lim _n→+∞p^s(x_n, x^′) = 0. Again, from Lemma 2(4), we get therefore Now x_2n+2∈T(x_2n+1) gives that which implies that On the other hand, we have Taking limit as n → +∞ and using (69) and (73), we obtain p(x^′, S(x^′)) = 0. Therefore, from (69) (p(x^′, x^′) = 0), we obtain which from Remark 14 implies that $$. It follows similarly that x^′ ∈ T(x^′). Thus S and T have a common fixed point.

Example 27. Let X = {1,2, 3} be endowed with usual order and let p be a partial metric on X defined as

Define the mappings S, T : X → CB_p(X) by Note that Sx and Tx are closed and bounded for all x ∈ X with respect to the partial metric p. To show that for all x, y in X, (59) is satisfied with α = 1/11, β = 4/5, we consider the following cases: if x, y ∈ {1,2}, then, and condition (59) is satisfied obviously.