Common best proximity points for a pair of mappings with certain dominating property

Lemma 3.1

Let { u n } be a sequence in a metric space ( X , d ) such that

If { u n } is not a Cauchy sequence, there exist subsequences { u m k } and { u n k } of { u n } with m k > n k > k for all k ∈ N such that

for some ε > 0 .

Proof

Assume that { u n } is not a Cauchy sequence. Then, there exist subsequences { u m k } and { u n k } of { u n } with m k > n k > k for all k ∈ N such that

for some ε > 0 . In addition, we choose the smallest n k satisfying (3.1) so that

By using (3.1) and (3.2), we have that

Since lim n → ∞ d ( u n , u n + 1 ) = 0 , taking the limit as k → ∞ in (3.3) implies

It now remains to show that

By the triangular inequality, we obtain

and

As k → ∞ , we obtain

as required.□

Lemma 3.2

Suppose that f : A → B with f ( A 0 ) ⊆ B 0 has ( α g , ℬ ) -dominating property and is α g -proximally commutative. If A 0 ∩ C ( f , g ) ≠ ∅ , then Cℬ ( f , g ) ≠ ∅ .

Proof

Let u ∈ A 0 ∩ C ( f , g ) . Then, we have u ∈ A 0 and f u = g u . Since f ( A 0 ) ⊆ B 0 , there exists x ∗ ∈ A 0 such that

By the commutativity of f and g , we have

Again, since x ∗ ∈ A 0 and f ( A 0 ) ⊆ B 0 , there exists y ∗ ∈ A 0 such that

Hence, (3.6) and (3.7) become

Since α ( f u , g u ) = α ( f u , f u ) ≥ 1 and α ( f x ∗ , g x ∗ ) = α ( f x ∗ , f x ∗ ) ≥ 1 , both u and x ∗ belong to A ( α , f , g ) . Since f is α g -proximal, (3.8) yields

Next, we claim that x ∗ = y ∗ . Suppose that d ( x ∗ , y ∗ ) > 0 . By the dominating property, we have

and hence,

The property of β gives d ( x ∗ , y ∗ ) = 0 , which leads to a contradiction. Thus, x ∗ = y ∗ , and by (3.7), we obtain

Therefore, Cℬ ( f , g ) ≠ ∅ .□

Theorem 3.3

Let ( X , d ) be a complete metric space, let f : A → B with f ( A 0 ) ⊆ B 0 satisfying ( α g , ℬ ) -dominating property and be α g -proximally commutative. Suppose also that the following hold:

1. A 0 is closed and A 0 ∩ A ( α , f , g ) ≠ ∅ ;

2. f ( A 0 ) ⊆ g ( A 0 ) ;

3. either

   1. f and g are continuous; or

   2. for any sequences { x n } and { u n } in A such that

      d ( u n , f x n ) = d ( A , B ) = d ( u n − 1 , g x n ) ,

      if { u n } converges to u ∈ A with α ( u n , u n − 1 ) ≥ 1 for all n , then there exists a subsequence { x n k } of { x n } such that

      d ( u , f x n k ) = d ( A , B ) = d ( u , g x n k ) .

Proof

First, let x 0 ∈ A 0 ∩ A ( α , f , g ) . The assumptions (i) and (ii) inductively give rise to a sequence { x n } in A 0 satisfying

and a sequence { u n } in A 0 satisfying

for all n . Hence, (3.9) and (3.10) yield

Observe, for now, that if u n 0 = u n 0 + 1 for some n 0 , then (3.10) and (3.11) produce

By the commutativity of f and g , we have f ( u n 0 ) = g ( u n 0 + 1 ) = g ( u n 0 ) , which then fulfills all hypotheses in Lemma 3.2. Thus, Cℬ ( f , g ) ≠ ∅ .

Second, we show that

provided that u n ≠ u n + 1 for all n . From (3.11), note that, for all n ≥ 1

Since f is α g -proximal, (3.12) yields

for all n . By the dominating property, there exists β ∈ ℬ such that

for all n . It is clear that lim n → ∞ d ( u n − 1 , u n ) exists. By the property of β , if lim n → ∞ d ( u n − 1 , u n ) were nonzero, then lim n → ∞ β ( d ( u n − 1 , u n ) ) would not be 1, which contradicts (3.14) as n → ∞ .

Third, we claim that { u n } is a Cauchy sequence. Suppose, for a contradiction, that it is not the case. By Lemma 3.1, there exist subsequences { u m k } and { u n k } of { u n } , with m k > n k > k for all k ∈ N such that

for some ε > 0 . Since { u m k } and { u n k } satisfy (3.12), we have

for all k . It is not hard to see that the same procedure as above applies, so that we obtain

Taking k → ∞ in (3.16) yields

and hence,

which is a contradiction.

Next, we prove the existence of a common proximity point of f and g by showing C ( f , g ) ≠ ∅ and applying Lemma 3.2. Since A 0 is a closed subspace of X , let lim n → ∞ u n = u ∈ A 0 . If f and g are continuous, then

which implies that u ∈ C ( f , g ) . If assumption (iii)(b) is satisfied, then there exists a subsequence { x n k } of { x n } such that

and hence, f u = g u by commutativity; that is, C ( f , g ) ≠ ∅ .

Finally, assume Cℬ ( f , g ) ⊆ A ( α , f , g ) . We show the uniqueness of a common best proximity point. Let x ∗ , y ∗ ∈ Cℬ ( f , g ) . Then,

As above, since f is α g -proximal, we have

By the dominating property, we again obtain

Suppose for a contradiction that x ∗ ≠ y ∗ . Then, β ( d ( x ∗ , y ∗ ) ) = 1 , implying d ( x ∗ , y ∗ ) = 0 . This contradicts x ∗ = y ∗ .□

Example 3.4

Let X = R 3 be equipped with the standard Euclidean metric d . Also, let

It is easy to see that A 0 = A , B 0 = B , and d ( A , B ) = 5 . Define the continuous mappings f , g : A → B by

for all ( x , 1 , 2 ) ∈ A , and also define α : R 3 × R 3 → [ 0 , ∞ ) by

Observe that f ( A 0 ) ⊆ g ( A 0 ) .

We show that f is α g -proximally commutative:

1. For any u ∈ A , it is easy to see that α ( f u , f u ) ≥ 1 .

2. Let u = ( x , 1 , 2 ) and v = ( x ′ , 1 , 2 ) be such that

   α ( g u , g v ) = α ( g ( x , 1 , 2 ) , g ( x ′ , 1 , 2 ) ) = α ( ( x , − 2 , 6 ) , ( x ′ , − 2 , 6 ) ) ≥ 1 .

   Then, we have x ≤ x ′ , and hence, ( ln ( 1 + x ) ) ≤ ( ln ( 1 + x ′ ) ) . Thus,

   α ( f u , f v ) = α ( ( ln ( 1 + x ) , − 2 , 6 ) , ( ln ( 1 + x ′ ) , − 2 , 6 ) ) ≥ 1 .

3. Let u = ( x , 1 , 2 ) , v = ( x ′ , 1 , 2 ) , and z = ( x ″ , 1 , 2 ) be such that α ( f z , g z ) ≥ 1 satisfying

   d ( u , f z ) = d ( A , B ) = d ( v , g z ) .

   It follows by school algebra that x = ln ( 1 + x ″ ) and x ′ = x ″ . Thus, α ( u , v ) ≥ 1 .

4. Now it remains to show that f and g proximally commute. Let u = ( x , 1 , 2 ) , v = ( x ′ , 1 , 2 ) , and z = ( x ″ , 1 , 2 ) satisfy

   d ( u , f z ) = d ( A , B ) = d ( v , g z ) .

   Then, x = ln ( 1 + x ″ ) and x ′ = x ″ . Thus,

   f v = ( ln ( 1 + x ″ ) , − 2 , 6 ) = g u .

Note that 1 − arctan t t is close to zero only if t → 0 . Here, ℬ = { β } , and we may write β instead of ℬ for convenience. We show that f has the ( α g , β ) -dominating property. Let

satisfy

Then, u ˆ 1 = ln ( 1 + z ˆ 1 ) , u ˆ 2 = ln ( 1 + z ˆ 2 ) , v ˆ 1 = z ˆ 1 , v ˆ 2 = z ˆ 2 , and z ˆ 1 , z ˆ 2 ∈ [ 0 , 2 ] . Since ln ( 1 + z ˆ 1 ) ≤ z ˆ 1 and ln ( 1 + z ˆ 2 ) ≤ z ˆ 2 , we have

To obtain inequality (2.1), we first assume v 1 ≠ v 2 . Then, d ( v 1 , v 2 ) = ∣ z ˆ 1 − z ˆ 2 ∣ > 0 . Hence,

If v 1 = v 2 , then u 1 = u 2 ; inequality (2.1) clearly holds.

Theorem 3.3 now applies, which guarantees the existence of a common best proximity point of f and g . Let x ∗ ∈ Cℬ ( f , g ) . Then,

which implies that f x ∗ = g x ∗ . Hence, α ( f x ∗ , g x ∗ ) ≥ 1 . That is, Cℬ ( f , g ) ⊆ A ( α , f , g ) . Therefore, x ∗ is the only common best proximity point of f and g . In fact, x ∗ = ( 0 , 1 , 2 ) .

Corollary 3.5

Suppose that all assumptions in Theorem 3.3  hold, and also let ℬ = { β } , where β ≡ k ∈ [ 0 , 1 ) . Then, Cℬ ( f , g ) ≠ ∅ . Moreover, if Cℬ ( f , g ) ⊆ A ( α , f , g ) , then Cℬ ( f , g ) has only one element. In particular, if g is the identity on A = B = X and α ≡ 1 , then we obtain Banach fixed-point theorem.

Proof

The former part of the corollary is obvious. For the latter part, note that if A = B = X , then A 0 = B 0 = X , which implies that all the assumptions of Theorem 3.3 are met. Moreover, if g is the identity mapping, then the dominating property gives rise to a contraction satisfying (2.2).□