Common fixed-point theorems for non-linear non-self contractive mappings in convex metric spaces

Definition 1.1

[24] Let M be a non-empty set and f : M → M be a mapping. A point u ∈ M is said to be a fixed-point of f if f ( u ) = u . The set of fixed-points of f is denoted by F ( f ) .

Definition 1.2

[20] Let M be a non-empty set and f , g : M → M be two self-mappings. A point u ∈ M is called a coincidence point of a pair ( f , g ) if f ( u ) = g ( u ) . The mappings f and g are said to be coincidentally commuting if they commute at their point of coincidence that is if f ( u ) = g ( u ) for some u ∈ M , then f g ( u ) = g f ( u ) .

Definition 1.3

[17] Let ( M , ϱ ) be a metric space and K be a subset of M . Then f , T : K → M is said to be coincidentally commuting if T ( u ) = f ( u ) implies f T ( u ) = T f ( u ) provided that T u , f u ∈ K .

Definition 1.4

[11,16] Let M be a normed space and K be a subset of M . For p ∈ K the set K is called p -starshaped or starshaped with respect to p ∈ K if t u + ( 1 − t ) p ∈ K for each u ∈ K . Note that K is convex if K is starshaped with respect to every u , p ∈ K .

Definition 1.5

[30] Let M be a metric space and K be a non-empty, convex subset of M , and let p ∈ K . A mapping T : K → K is said to be affine with respect to p if:

for all u , p ∈ K and γ ∈ ( 0 , 1 ) .

Definition 1.6

[11] Let M be a normed linear space and K be a non-empty subset of M . A mapping T : K → M is said to be demiclosed provided that if { u n } ⊆ K , u n ⇀ u ∈ K , and T u n → v ∈ M , then T u = v . (The symbol ⇀ is used to denote weak convergence.)

Theorem 1.7

[10] Let M be a metric space of hyperbolic type, K a non-empty closed subset of M , and ∂ K the boundary of K . Let ∂ K be non-empty and let T : K → M and f : K ∩ T ( K ) → M be two non-self mappings satisfying the following condition:

for all u , v ∈ M , where ϕ : [ 0 , ∞ ) → [ 0 , ∞ ) is a real function which has the following properties:

1. ϕ ( t + ) < t for t > 0 and

2. ϕ ( t ) is non-decreasing.

Suppose that f and T have the following additional properties:

1. ∂ K ⊆ T K , f K ∩ K ⊆ T K ;

2. T u ∈ ∂ K ⇒ f u ∈ K ;

3. K ∩ T K is complete.

Then there exists a coincidence point z in M . Moreover, if f and T are coincidentally commuting, then z is the unique common fixed-point of f and T .

Definition 1.8

[13] Let ( M , ϱ ) be a metric space, K a non-empty closed subset of M , and f , T : K → M . If f and T satisfy the condition ϱ ( f u , f v ) ≤ γ ⋅ ω ( u , v ) , where

for all u , v ∈ K , 0 < γ < 1 , σ ≥ 2 − γ , then f is called a generalized contractive mapping of K into M .

Theorem 1.9

[13] Let M be a convex metric metric space, K a non-empty closed subset of M , and ∂ K the boundary of K . Let ∂ K be non-empty and T : K → M and f : K ∩ T K → M be the generalized contractive mapping of K into M , and

1. ∂ K ⊂ T K , f K ∩ K ⊂ T K ;

2. T u ∈ ∂ K ⇒ f u ⊆ K ;

3. f K ∩ K is complete.

Then there exists a coincidence point z in M . Moreover, if f and T are coincidentally commuting, then z is the unique common fixed-point of f and T .

Definition 2.1

Let ( M , ϱ ) be a metric space, K a non-empty closed subset of M , and f , g , S , T : K → M . If f , g , S , and T satisfy the condition ϱ ( f u , g v ) ≤ γ ⋅ ω ( u , v ) , where

for all u , v ∈ K , 0 < γ < 1 , σ ≥ 2 − γ , then ( f , g ) is called a generalized ( T , S ) contractive mapping of K into M .

Theorem 2.2

Let M be a convex metric space, K be a non-empty closed subset of M , and ∂ K the boundary of K . Let ∂ K be non-empty and S , T : K → M and f , g : K → M be the generalized contractive mappings of K into M , and

1. ∂ K ⊂ S K ∩ T K , f K ∩ K ⊂ T K , g K ∩ K ⊂ S K ;

2. S u ∈ ∂ K ⇒ f u ⊆ K , T u ∈ ∂ K ⇒ g u ⊆ K ;

3. S K and T K are complete.

Then there exists a coincidence point z in M. Moreover, if ( f , S ) and ( g , T ) are coincidentally commuting, then z is the unique common fixed-point of ( f , S ) and ( g , T ) .

Proof

Choose any arbitrary point u ∈ ∂ K . Three sequences { u n } and { θ n } in K and a sequence { v n } in f K ∪ g K ⊂ M are constructed as follows:

Choose θ 0 = u . Since θ 0 ∈ ∂ K , by (i) there exist points u 0 ′ , u 0 ″ ∈ K such that S u 0 ′ = θ 0 = T u 0 ″ ∈ ∂ K . u 0 is selected such that S u 0 = θ 0 . Since S u 0 ∈ ∂ K , by (ii), we have f u 0 ⊆ K . Now from (i), f u 0 ⊆ T K . Thus, there exists u 1 ∈ K such that v 1 = T u 1 with v 1 ∈ f K ⊂ K . This implies that f u 0 ∈ f K ∩ K ⊂ T K . Set v 1 = f u 0 , we choose u 1 ∈ K such that T u 1 = f u 0 . Hence, θ 1 = T u 1 = f u 0 = v 1 .

Since v 1 = f u 0 , there exists a point v 2 ∈ g K ∩ K such that v 2 ∈ S K by (i). Let u 1 ∈ K with θ 1 = T u 1 ∈ ∂ K such that θ 2 = S u 2 = g u 1 = v 2 . If g u 1 = v 2 ∉ K , then there exists θ 2 ∈ ∂ K ( θ 2 ≠ v 2 ) such that θ 2 ∈ seg [ v 1 , v 2 ] . Since u 2 ∈ K , then by (i) we have S u 2 = θ 2 . Hence, θ 2 ∈ ∂ K ∩ seg [ v 1 , v 2 ] .

We can choose v 3 ∈ f K ∩ K , and by (i), v 3 ∈ T K . Let u 2 ∈ K such that T u 3 = v 3 = g u 2 . Carrying on the process, three sequences { u n } ⊆ K , { θ n } ⊆ K , and { v n } ⊆ f K ∪ g K ⊂ M are constructed such that

1. v n = f u n − 1 or v n = g u n − 1 ;

2. θ n = T u n or θ n = S u n ;

3. θ n = v n if and only if v n ∈ K ;

4. θ n ≠ v n whenever v n ∉ K and θ n ∈ ∂ K such that

   θ n ∈ ∂ k ∩ seg [ f u n − 2 , g u n − 1 ] .

This proves that f , g , S , and T are non-self mappings.

Note that by (d) if θ n ≠ v n , then θ n ∈ ∂ K and combining conditions (b), (ii), and (a) we obtain θ n + 1 = v n + 1 ∈ K . Likewise θ n − 1 = v n − 1 ∈ K . If θ n − 1 ∈ ∂ K , then it implies θ n = v n ∈ K . This is due to the fact that two consecutive members v n and v n + 1 in { v n } cannot be such that v n ∉ K and v n + 1 ∉ K . Indeed, if v n ∉ K , then θ n is chosen such that θ n ∈ ∂ K and

Since θ n ∈ ∂ K and ∂ K ⊂ S K ∩ T K , then θ n = S u n ∈ ∂ K , or θ n = T u n ∈ ∂ K , and so by (ii), v n + 1 ∈ f u n ∈ K , or v n + 1 ∈ g u n ∈ K , respectively. Thus, it is proved that if θ n ≠ v n , then v n + 1 ∈ K , and so θ n + 1 = v n + 1 . This means that two members in a row in { v n } cannot be in M \ K . Therefore, as v n ∈ M \ K , there must be a scenario where v n − 1 ∈ K , and so θ n − 1 = v n − 1 .

Next we show that θ n ≠ θ n + 1 for all n . From conditions (a–d) we can identify three options as follows:

1. θ n = v n ∈ K and θ n + 1 = v n + 1 ;

2. θ n = v n ∈ K and θ n + 1 ≠ v n + 1 ;

3. θ n ≠ v n ∈ K in which case θ n ∈ ∂ k ∩ seg [ f u n − 2 , g u n − 1 ] .

Following that, we will look at the instances listed below.

Case 1. Let θ n = v n ∈ K and θ n + 1 = v n + 1 . Using equation (2.1) we obtain

where

It is obvious that there are infinitely many n such that at least one of the following cases holds:

1. ϱ ( θ n , θ n + 1 ) ≤ γ ⋅ ϱ ( θ n − 1 , θ n ) 2 ≤ γ ⋅ ϱ ( θ n − 1 , θ n ) ;

2. ϱ ( θ n , θ n + 1 ) ≤ γ ⋅ ϱ ( θ n − 1 , θ n ) ;

3. ϱ ( θ n , θ n + 1 ) ≤ γ ⋅ ϱ ( θ n , θ n + 1 ) ; (a contradiction)

4. ϱ ( θ n , θ n + 1 ) ≤ γ ⋅ ϱ ( θ n − 1 , θ n ) + ϱ ( θ n , θ n + 1 ) σ ≤ γ σ ( ϱ ( θ n − 1 , θ n ) + ϱ ( θ n , θ n + 1 ) ) ,

   1 − γ σ ϱ ( θ n , θ n + 1 ) ≤ γ σ ϱ ( θ n − 1 , θ n ) ,

   ϱ ( θ n , θ n + 1 ) ≤ γ σ − γ ϱ ( θ n − 1 , θ n ) .

Combining the four cases (i), (ii), (iii), and (iv) we obtain

where k = max γ , γ σ − γ .

Case 2. Let θ n = v n ∈ K but θ n + 1 ≠ v n + 1 . Then θ n + 1 ∈ ∂ K ∩ seg [ v n , v n + 1 ] . From equation (1.1) with w = v , we obtain

Therefore, we have

Hence,

Now,

and because

we have

which gives us

but

From Case (1), we obtain

Case 3. Take θ n ≠ v n . Then θ n ∈ ∂ K ∩ seg [ f u n − 2 , g u n − 1 ] , i.e., θ n ∈ ∂ K ∩ seg [ v n − 1 , v n ] . Since two members in a row in { v n } cannot be in K we have θ n + 1 = v n + 1 and θ n − 1 = v n − 1 . This implies that

We shall find v n − 1 and v n + 1 . Since θ n − 1 = v n − 1 , we can conclude that

with respect to Case (2).

Now, we obtain

where

1. ϱ ( v n , v n + 1 ) ≤ γ ϱ ( θ n − 1 , θ n ) 2 ≤ γ ϱ ( θ n − 1 , θ n ) ;

2. ϱ ( v n , v n + 1 ) ≤ γ ϱ ( θ n − 1 , v n ) = γ ϱ ( v n − 1 , v n ) ;

3. ϱ ( v n , v n + 1 ) ≤ γ ϱ ( θ n , θ n + 1 ) ;

4. ϱ ( v n , v n + 1 ) ≤ γ ϱ ( θ n , v n ) = γ { ϱ ( v n − 1 , v n ) − ϱ ( θ n − 1 , θ n ) } ≤ γ ϱ ( θ n − 1 , θ n ) ;

5. ϱ ( v n , v n + 1 ) ≤ γ σ { ϱ ( θ n − 1 , v n + 1 ) + ϱ ( θ n , v n ) } ≤ γ σ { ϱ ( θ n − 1 , θ n ) + ϱ ( θ n , v n + 1 ) + ϱ ( v n − 1 , v n ) − ϱ ( θ n − 1 , θ n ) } ≤ γ σ { ϱ ( v n − 1 , v n ) + ϱ ( θ n , θ n + 1 ) } ,

θ n + 1 = v n + 1 , θ n − 1 = v n − 1 , and ϱ ( v n − 1 , v n ) ≤ γ ϱ ( θ n − 2 , θ n − 1 ) , we obtain

Substituting equation (2.3) and the aforementioned cases (i–v) in equation (2.2) yields the following:

(i) and (iv)

(ii)

(iii)

(v)

From equations (2.4), (2.5), (2.6), (2.7), and (2.8), we obtain

where

Combining Cases (1), (2), and (3) we obtain

where μ n ∈ { ϱ ( u n − 2 , u n − 1 ) , ϱ ( u n − 1 , u n ) } , and

From Assad and Kirk [6] proceedings, it can be easily verified by induction that for n ≥ 1 ,

where μ 2 ∈ { ϱ ( θ 0 , θ 1 ) , ϱ ( θ 1 , θ 2 ) } .

For n > m using equation (2.10) and the triangle inequality we have

Thus, the sequence is Cauchy. Since K is closed, there exists θ ∈ K such that lim n → ∞ θ n = θ . We now show that θ ∈ S K ∩ T K . Suppose that both subsequences { θ n j } and { θ n k } of { θ n } , defined by θ n j = S w n j ∈ g w n j − 1 and by θ n k = T w n k ∈ f w n k − 1 , respectively, are infinite. Then lim j → ∞ S w n j → θ and lim k → ∞ T w n k → θ . Since S K and T K are complete, θ ∈ S K ∩ T K .

Now, assume that one of the subsequences { θ n j } or { θ n k } is finite. Then there exists an infinite subsequence { θ n m } of { θ n } such that θ n m ∈ ∂ K . Since θ n m ∈ ∂ K and θ n m → θ as m → ∞ , it follows that θ ∈ ∂ K . Hence, by condition (i), θ ∈ S K ∩ T K . Thus, we have shown that θ ∈ S K ∩ T K . This implies that there exist some p , q ∈ K such that

Now, we prove that p is a coincidence point for f and S , and that q is a coincidence point for g and T . Since { θ n } is a coincidence point for { θ n j } ∪ { θ n k } , where the subsequences { θ n j } and { θ n k } are defined as above, at least one of them is infinite. Without loss of generality, suppose that { θ n j } is infinite, where θ n j = S u n j = v n j ∈ g u n j − 1 . Since θ n j ∈ g u n j − 1 , from equation (2.1) we have

where

Taking the limit as j → ∞ we obtain ϱ ( f p , θ ) ≤ γ ϱ ( θ , f p ) , γ q ϱ ( θ , f p ) and hence ϱ ( f p , θ ) = 0 . Hence θ ∈ f p , as f p is closed. Thus, by condition (2.11)  S p ∈ f p .

From condition (2.1)

Thus, we have

1. ϱ ( θ , g q ) ≤ γ ϱ ( θ , g q ) + ϱ ( g u n k − 1 , θ ) ≤ γ ϱ ( θ , g q ) .

2. Since γ < 0 it follows that ϱ ( θ , g q ) = 0 . This implies θ = g q .

3. ϱ ( θ , g q ) ≤ γ σ ϱ ( θ , g q ) .

Since γ < q , it follows that ϱ ( θ , g q ) = 0 . Hence, θ = g q .

In all cases we have θ = g q .

So by condition (2.11)  T q ∈ g q .

Thus, we have proved that p is a coincidence point for f and S , and that q is a coincidence point for g and T . Again from condition (2.1), we obtain ϱ ( f p , g q ) = 0 . Hence, f p = g q . Thus, we have proved that S p = T q ∈ f p = g q .□