The Existence Theorems of an Optimal Approximate Solution for Generalized Proximal Contraction Mappings

Remark 1. It is easy to see that A₀ and B₀ are nonempty whenever A∩B ≠ ∅. Further, if A and B are closed subsets of a normed linear space such that d(A, B) > 0, then A₀⊆Bdr(A) and B₀⊆Bdr(B), where Bdr(A) is a boundary of A.

Definition 2 (see [21].)A mapping T : A → B is called a proximal contraction of the first kind if there exists α ∈ [0,1) such that, for all a, b, x, y ∈ A,

Remark 3. If T is self-mapping, then T is a proximal contraction of the first kind deduced to T which is a contraction mapping. But a non-self-proximal contraction is not necessarily a contraction.

Definition 4 (see [21].)A mapping T : A → B is said to be a proximal contraction of the second kind if there exists α ∈ [0,1) such that, for all a, b, x, y ∈ A,

The necessary condition for a self-mapping T to be a proximal contraction of the second kind is that

for all x, y in the domain of T. Therefore, every contraction self-mapping is a proximal contraction of the second kind, but the converse is not true (see Example 5).

Example 5. Consider ℝ endowed with the Euclidean metric. Let the self-mapping T : [0,1]→[0,1] be defined as follows:

It is easy to prove that T is a proximal contraction of the second kind. However, T is not a contraction mapping.

Definition 6. Let S : A → B and T : B → A be mappings. The pair (S, T) is said to be

• (1)

  a cyclic contractive pair if d(A, B) < d(x, y)⇒d(Sx, Ty) < d(x, y) for all x ∈ A and y ∈ B;

• (2)

  a cyclic expansive pair if d(A, B) < d(x, y)⇒d(Sx, Ty) > d(x, y) for all x ∈ A and y ∈ B;

• (3)

  a cyclic inequality pair if d(A, B) < d(x, y)⇒d(Sx, Ty) ≠ d(x, y) for all x ∈ A and y ∈ B.

Definition 7. Let S : A → B and T : B → A be mappings. The pair (S, T) is said to satisfy min-max condition if, for all x ∈ A and y ∈ B,

where min (Sx, Ty) and max (Sx, Ty) are defined by

We observe that the cyclic contractive pairs, cyclic expansive pairs, and cyclic inequality pairs satisfy the min-max condition.

Definition 8. Let T : A → B a mapping and g : A → A be an isometry. The mapping T is said to preserve isometric distance with respect to g if

for all x, y ∈ A.

Definition 9. A point x ∈ A is said to be a best proximity point of a mapping T : A → B if it satisfies the condition that

Observe that a best proximity reduces to a fixed point if the underlying mapping is a self-mapping.

Definition 10. A is said to be approximatively compact with respect to B if every sequence {x_n} in A satisfies the condition that d(y, x_n) → d(y, A) for some y ∈ B has a convergent subsequence.

Remark 11. Any nonempty subset of metric space (X, d) is approximatively compact with respect to itself.

Definition 12. Let A, B be nonempty subset of metric space (X, d), T : A → B and 𝒦 : A → [0,1). A mapping T is said to be a generalized proximal contraction of the first kind with respect to 𝒦 if

for all a, b, x, y ∈ A.

Remark 13. If we take 𝒦(x) = α for all x ∈ A, where α ∈ [0,1), then a generalized proximal contraction of the first kind with respect to 𝒦 reduces to a proximal contraction of the first kind (Definition 2). In case of a self-mapping, it is apparent that the class of contraction mapping is contained in the class of generalized proximal contraction of the first kind with respect to 𝒦 mapping.

Example 14. Consider the metric space ℝ² with Euclidean metric. Let A = {(0, y):−1 < y < 1} and B = {(1, y):−1 < y < 1}. Define a mapping T : A → B as follows:

for all (0, y) ∈ A.

It is easy to check that there is no α ∈ [0,1) satisfing

for all a, b, x, y ∈ A. Therefore, T is not a proximal contraction of the first kind.

Consider a function 𝒦 : A → [0,1) defined by

Next, we claim that T is a generalized proximal contraction of the first kind with respect to 𝒦.

If (0, y₁), (0, y₂) ∈ A such that

for all a, b ∈ A, then we have Therefore, it follows that This implies that T is a generalized proximal contraction of the first kind with respect to 𝒦.

Definition 15. Let A, B be nonempty subset of metric space (X, d), T : A → B and 𝒦 : A → [0,1). A mapping T is said to be a generalized proximal contraction of the second kind with respect to 𝒦 if

for all a, b, x, y ∈ A.

Theorem 16. Let (X, d) a complete metric space and A, B be nonempty closed subsets of X such that A₀ and B₀ are nonempty. Suppose that T : A → B, g : A → A, and 𝒦 : A → [0,1) are mappings satisfying the following conditions:

• (a)

  T is a continuous generalized proximal contraction of first kind with respect to 𝒦;

• (b)

  T(A₀)⊆B₀ and A₀⊆g(A₀);

• (c)

  g is an isometry;

• (d)

  𝒦(x) ≤ 𝒦(y), whenever d(gx, Ty) = d(A, B).

Then there exists a unique point x ∈ A such that d(gx, Tx) = d(A, B).

Proof. Let x₀ be a fixed element in A₀. From T(A₀)⊆B₀ and A₀⊆g(A₀), it follows that there exists a point x₁ ∈ A₀ such that

Again, since Tx₁ ∈ T(A₀)⊆B₀ and A₀⊆g(A₀), there exists a point x₂ ∈ A₀ such that Continuing this process, we can construct the sequence {x_n} in A₀ such that for all n ∈ ℕ. Since T is a generalized proximal contraction of the first kind with respect to 𝒦, it follows that for all n ∈ ℕ. Also, since g is an isometry, we have for all n ∈ ℕ. By using (20) and (d), we have for all n ∈ ℕ. By repeating (23), we get for all n ∈ ℕ. Now, we let k : = 𝒦(x₀)∈[0,1). For positive integers m and n with n > m, it follows from (24) that Since k ∈ [0,1), we have (k^m/(1 − k))d(x₁, x₀) → 0 as m → ∞, which implies that {x_n} is a Cauchy sequence in X. Since X is complete, it follows that the sequence {x_n} converges to point x ∈ X. Since T and g are continuous, we get Next, we suppose that x^* is another point in X such that Since T is a generalized proximal contraction of the first kind with respect to 𝒦, by using (26) and (27), we get Since g is an isometry, it follows that which implies that x = x^*. This completes the proof.

Example 17. Consider the complete metric space ℝ² with Euclidean metric. Let A = {(0, y):−1 ≤ y ≤ 1} and B = {(1, y):−1 ≤ y ≤ 1}. Define two mappings T : A → B and g : A → A as follows:

for all (0, y) ∈ A. Then it is easy to see that d(A, B) = 1, A₀ = A, B₀ = B, and the mapping g is an isometry.

Consider a function 𝒦 : A → [0,1) defined by

Next, we claim that T is a generalized proximal contraction of the first kind with respect to 𝒦. If (0, y₁), (0, y₂) ∈ A such that

for all a, b ∈ A, then we have Therefore, it follows that This implies that the non-self-mapping T is a generalized proximal contraction of the first kind with respect to 𝒦. It is easy to see that 𝒦(x) ≤ 𝒦(y) whenever d(gx, Ty) = d(A, B). Moreover, since T is continuous and g is an isometry, all the conditions of Theorem 16 are satisfied, and so T has a unique element (0,0) ∈ A such that

Corollary 18 (see [21], Theorem 3.3.)Let (X, d) be a complete metric space and A, B nonempty closed subsets of X such that A₀ and B₀ are nonempty. Suppose that T : A → B and g : A → A are mappings satisfying the following conditions:

• (a)

  T is a continuous proximal contraction of the first kind;

• (b)

  T(A₀)⊆B₀ and A₀⊆g(A₀);

• (c)

  g is an isometry.

Then there exists a unique element x ∈ A such that d(gx, Tx) = d(A, B).

Proof. Since a proximal contraction of the first kind is a special case of a generalized proximal contraction of the first kind, we can prove this result by applying Theorem 16.

Corollary 19. Let (X, d) a complete metric space and A, B be nonempty closed subsets of X such that A₀ and B₀ are nonempty. Suppose that T : A → B and 𝒦 : A → [0,1) are mappings satisfying the following conditions:

• (a)

  T is a continuous generalized proximal contraction of first kind with respect to 𝒦;

• (b)

  T(A₀)⊆B₀;

• (c)

  𝒦(x) ≤ 𝒦(y), whenever d(x, Ty) = d(A, B).

Then T has a unique best proximity point in A.

Corollary 20 (see [21], Corollary 3.4.)Let (X, d) be a complete metric space and A, B nonempty closed subsets of X such that A₀ and B₀ are nonempty. Let T : A → B be a mapping satisfying the following conditions:

• (a)

  T is a continuous proximal contraction of the first kind;

• (b)

  T(A₀)⊆B₀.

Then T has a unique best proximity point in A.

Proof. Since a proximal contraction of the first kind is a special case of a generalized proximal contraction of the first kind with respect to 𝒦, we can prove this result by applying Corollary 19.

Theorem 21. Let (X, d) a complete metric space and A, B be nonempty closed subsets of X such that A is approximatively compact with respect to B. Suppose that A₀ and B₀ are nonempty and T : A → B, g : A → A, and 𝒦 : A → [0,1) are mappings satisfying the following conditions:

• (a)

  T is a continuous generalized proximal contraction of the second kind with respect to 𝒦;

• (b)

  T(A₀)⊆B₀ and A₀⊆g(A₀);

• (c)

  g is an isometry;

• (d)

  T preserves isometric distance with respect to g;

• (e)

  𝒦(x) ≤ 𝒦(y), whenever d(gx, Ty) = d(A, B).

Then there exists a point x ∈ A such that d(gx, Tx) = d(A, B). Moreover, if x^* is another point in A for which d(gx^*, Tx^*) = d(A, B), then Tx = Tx^*.

Proof. As in the proof of Theorem 16, for fixed x₀ ∈ A₀, we can define a sequence {x_n} in A₀ such that

for all n ∈ ℕ. Since T is a generalized proximal contraction of the second kind with respect to 𝒦, it follows that Since T preserves isometric distance with respect to g, we have for all n ∈ ℕ. By using (36) and (e), we have for all n ∈ ℕ. By repeating (39), we get for all n ∈ ℕ. Now, we let k : = 𝒦(x₀)∈[0,1). For positive integers m and n with n > m, it follows from (40) that Since k ∈ [0,1), we have (k^m/(1 − k))d(Tx₁, Tx₀) → 0 as m → ∞, which implies that {Tx_n} is a Cauchy sequence in B. By completeness of B⊆X, there exists a point y ∈ B such that Tx_n → y as n → ∞. By (36) and the triangle inequality, we have Letting n → ∞ in (42), we get d(y, gx_n) → d(y, A). Since A is approximatively compact with respect to B, it follows that {gx_n} has a convergence subsequence $$; say $$ as k → ∞. Thus we have which implies that z ∈ A₀. Since A₀⊆g(A₀), we have z = gx for some x ∈ A₀. Therefore, $$ as k → ∞. Since g is an isometry, we get $$ as k → ∞. By the continuity of T, we have $$ as k → ∞ and then y = Tx. From (43), we can conclude that

Next, we suppose that x^* is another point in X such that

Since T is a generalized proximal contraction of the second kind with respect to 𝒦, by the virtue of (44) and (45), we get Since T preserves isometric distance with respect to g, it follows that which implies that Tx = Tx^*. This completes the proof.

Corollary 22 (see [21], Theorem 3.1.)Let (X, d) be a complete metric space and A, B nonempty closed subsets of X such that A is approximatively compact with respect to B. Suppose that A₀ and B₀ are nonempty and T : A → B and g : A → A are mappings satisfying the following conditions:

• (a)

  T is a continuous proximal contraction of the second kind;

• (b)

  T(A₀)⊆B₀ and A₀⊆g(A₀);

• (c)

  g is an isometry;

• (d)

  T preserves isometric distance with respect to g.

Then there exists a point x ∈ A such that d(gx, Tx) = d(A, B). Moreover, if x^* is another point in A for which d(gx^*, Tx^*) = d(A, B), then Tx = Tx^*.

Proof. Since a proximal contraction of the second kind is a special case of a generalized proximal contraction of the second kind with respect to 𝒦, we can prove this result by applying Theorem 21.

Corollary 23. Let (X, d) be a complete metric space and A, B nonempty closed subsets of X such that A is approximatively compact with respect to B. Suppose that A₀ and B₀ are nonempty and T : A → B and 𝒦 : A → [0,1) are mappings satisfying the following conditions:

• (a)

  T is a continuous generalized proximal contraction of the second kind with respect to 𝒦;

• (b)

  T(A₀)⊆B₀;

• (c)

  𝒦(x) ≤ 𝒦(y), whenever d(x, Ty) = d(A, B).

Then T has a best proximity point. Moreover, if x^* is another best proximity point of T, then Tx = Tx^*.

Proof. We can prove this result by applying Theorem 21 with g = I_A, where I_A is an identity mapping on A.

Corollary 24 (see [21], Corollary 3.2.)Let (X, d) be a complete metric space and A, B nonempty closed subsets of X such that A is approximatively compact with respect to B. Suppose that A₀ and B₀ are nonempty and T : A → B is mapping satisfying the following conditions:

• (a)

  T is a continuous generalized proximal contraction of the second kind;

• (b)

  T(A₀)⊆B₀.

Then T has a best proximity point. Moreover, if x^* is another best proximity point of T, then Tx = Tx^*.

Proof. Since a proximal contraction of the second kind is a special case of a generalized proximal contraction of the second kind, we can prove this result by applying Corollary 23.

Theorem 25. Let (X, d) be a complete metric space, A and B nonempty closed subsets of X, and 𝒦 : A ∪ B → [0,1). Suppose that S : A → B is a mapping satisfying

for all x, y ∈ A. Then the following holds.

• (A)

  There exists a nonexpansive mapping T : B → A such that (S, T) satisfies the min-max condition whenever S has a best proximity point.

• (B)

  If there exists a nonexpansive mapping T : B → A such that (S, T) satisfies the min-max condition and 𝒦(Sx) ≤ 𝒦(x) and 𝒦(Tx) ≤ 𝒦(x) for all x ∈ A, then S has a best proximity point.

• (C)

  For two any best proximity points z and z^* of   S, we have

  $$ ()

Proof. (A) Let S has a best proximity point a ∈ A. We define a mapping T : B → A by Ty = a for all y ∈ B. Clearly, T is a nonexpansive mapping. It follows from the definition of T that

for all y ∈ B. Thus we can conclude that min (Sx, Ty) = d(A, B) for all x ∈ A and y ∈ B.

Next, we show that (S, T) satisfies the min-max condition. Suppose that x ∈ A and y ∈ B such that d(A, B) < d(x, y). Then we have

which implies that the pair (S, T) satisfies the min-max condition. Therefore, we can find a nonexpansive mapping T : B → A such that (S, T) satisfies the min-max condition.

(B) Fix x₀ ∈ A and define a sequence {x_n} in A ∪ B by

for all n ∈ ℕ. Since T is nonexpansive, it follows from (48) that for all n ∈ ℕ. By repeating the above argument, we have for all n ∈ ℕ, which implies that the sequence {x_2n} is a Cauchy sequence in X. A similar argument asserts that the sequence {x_2n−1} is a Cauchy sequence in X. By the completeness of X, we conclude that {x_2n} converges to a point a ∈ A and {x_2n−1} converges to a point b ∈ B. Since S is continuous, {Sx_2n} converges to Sa, which implies that {x_2n−1} converges to Sa. Thus Sa = b.

Similarly, it is easy to check that Tb = a. Therefore, we have

Now, we can conclude that By the virtue of the min-max condition of (S, T), we get d(a, b) ≤ d(A, B). Since d(A, B) ≤ d(a, b), we have d(a, b) = d(A, B). Therefore, we have which implies that S has a best proximity point in A.

(C) Let z and z^* be best proximity points of S. Then d(z, Sz) = d(A, B) and d(z^*, Sz^*) = d(A, B). Using the triangle inequality and (48), we have

This implies that d(z, z^*)≤(2/(1 − 𝒦(z)))d(A, B). This completes the proof.

Corollary 26 (see [21], Theorem 3.6.)Let (X, d) be a complete metric space and A and B nonempty closed subsets of X. Suppose that S : A → B is a contraction mapping. Then S has a best proximity point if and only if there exists a nonexpansive mapping T : B → A such that (S, T) satisfies the min-max condition.

Moreover, d(z, z^*)≤(2/(1 − α))d(A, B) for some α ∈ [0,1) and any two best proximity points z and z^* of S.

Proof. Since S is a contraction mapping, we have d(Sx, Sy) ≤ αd(x, y) for some α ∈ [0,1) and all x, y ∈ A. Now, we can prove this result by applying Theorem 25 with a function 𝒦 : A ∪ B → [0,1) defined by 𝒦(x) = α for all x ∈ A ∪ B.