Coupled measure of noncompactness and functional integral equations

Definition 1.1

[6]. Let μ : ℬ E → ℛ + be a mapping. The family ℬ E is called MNC on Banach space E if the following conditions hold:

1. For each U 1 ∈ ℬ E , μ ( U 1 ) = θ iff U 1 is a precompact set;

2. For each pair ( U 1 , U 2 ) ∈ ℬ E × ℬ E , we have

   U 1 ⊆ U 2 implies μ ( U 1 ) ≤ μ ( U 2 ) ;

3. For each U 1 ∈ ℬ E ,

   μ ( U 1 ) = μ ( U 1 ¯ ) = μ ( conv U 1 ) ,

   where U 1 ¯ represents the closure of U 1 and conv U 1 represents the convex hull of U 1 ;

4. μ ( λ U 1 + ( 1 − λ ) U 2 ) ≤ λ μ ( U 1 ) + ( 1 − λ ) μ ( U 2 ) for λ ∈ [ 0 , 1 ] ;

5. If { υ n } 0 + ∞ ∈ ℬ E is a decreasing sequence of closed sets and lim n → + ∞ μ ( υ n ) = 0 , then U + ∞ 1 = ⋂ n = 0 + ∞ U n 1 ≠ ∅ .

Theorem 1.2

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G be a compact and continuous operator. Then, F has at least one fixed-point.

Theorem 1.3

(Schauder) Let G be a nonempty, closed, and convex subset of a normed space and F be a continuous operator from G into a compact subset of G . Then, F has a fixed-point.

Theorem 1.4

(Darbo) Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G be a continuous operator. Suppose there is λ ∈ [ 0 , 1 ) such that μ ( F U ) ≤ λ μ ( U ) for each U ∈ G . Then, F has a fixed-point.

Theorem 1.5

(Brouwer) Let G be a nonempty, compact, and convex subset of a finite-dimensional normed space and F : G → G be a continuous operator. Then, F has a fixed-point.

Lemma 1.6

[9] Let φ 1 : ℛ + → ℛ + be an upper semicontinuous and nondecreasing function. In this case, the following conditions are equivalent:

1. lim n → + ∞ φ 1 n ( r ) = 0 for every r > 0 ;

2. φ 1 ( r ) < r for every r > 0 .

Definition 2.1

Let E be a Banach space and μ : ℬ E 2 → ℛ + be a mapping. We say that μ is a coupled MNC on E , if it has the following conditions:

1. ker μ = { ( U 1 , U 2 ) ∈ ℬ E 2 : μ ( U 1 , U 2 ) = θ } is nonempty;

2. For every ( U 1 , U 2 ) ∈ ℬ E 2 , μ ( U 1 , U 2 ) = θ ⇔ ( U 1 , U 2 ) is a precompact set;

3. For each ( ( U 1 , U 2 ) , ( U ′ 1 , U ′ 2 ) ) ∈ ℬ E 2 × ℬ E 2 and ( U 1 , U 2 ) ⊆ ( U ′ 1 , U ′ 2 ) , where U 1 ⊆ U ′ 1 and U 2 ⊆ U ′ 2 , we have

   ( U 1 , U 2 ) ⊆ ( U ′ 1 , U ′ 2 ) implies μ ( U 1 , U 2 ) ≤ μ ( U ′ 1 , U ′ 2 ) ;

4. For every ( U 1 , U 2 ) ∈ ℬ E 2 ,

   μ ( U 1 ¯ , U 2 ¯ ) = μ ( U 1 , U 2 ) = μ ( conv ( U 1 , U 2 ) ) ,

   where conv ( U 1 , U 2 ) denotes the convex hull of ( U 1 , U 2 ) ;

5. μ ( λ ( U 1 , U 2 ) + ( 1 − λ ) ( U ′ 1 , U ′ 2 ) ) ≤ λ μ ( U 1 , U 2 ) + ( 1 − λ ) μ ( U ′ 1 , U ′ 2 ) for λ ∈ [ 0 , 1 ] ;

6. If { U n 1 } 0 + ∞ , { U n 2 } 0 + ∞ in ℬ E are decreasing sequences of closed sets and

   lim n → + ∞ μ { ( U n 1 , U n 2 ) } 0 + ∞ = 0 , then ( U + ∞ 1 , U + ∞ 2 ) = ⋂ n = 0 + ∞ ( U n 1 , U n 2 ) ≠ ∅ .

Theorem 2.2

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G be a continuous map such that

for each ∅ ≠ U 1 ⊆ G , ∅ ≠ U 2 ⊆ G , where μ is an arbitrary coupled MNC and φ 1 , φ 2 : ℛ + → ℛ + such that φ 2 is continuous and φ 1 is lower semicontinuous on ℛ + . Furthermore, φ 1 ( 0 ) = 0 and φ 1 ( s ) > 0 for s > 0 . Then, F has at least one fixed-point in G .

Proof

Taking U 0 1 , U 0 2 = G , U n + 1 1 = conv ( F U n 1 ) ¯ , U n + 1 2 = conv ( F U n 2 ) ¯ , for n = 0 , 1 , 2 , … , we obtain U n + 1 1 ⊆ U n 1 , U n + 1 2 ⊆ U n 2 for n = 0 , 1 , … . Therefore, { U n 1 } 0 + ∞ , { U n 2 } 0 + ∞ are decreasing sequences of closed and convex sets. Moreover, from (2.1), we have

for n = 0 , 1 , 2 , … . Since the sequence { μ ( U n 1 , U n 2 ) } is nonnegative and nonincreasing, we deduce that μ ( U n 1 , U n 2 ) → m when n tends to infinity, where m ≥ 0 is a real number. On the other hand, considering equation (2.2), we obtain

This yields φ 2 ( m ) ≤ φ 2 ( m ) − φ 1 ( m ) . Consequently, φ 1 ( m ) = 0 and so m = 0 . Therefore, we infer μ ( U n 1 , U n 2 ) → 0 as n → + ∞ . Now, considering that ( U n + 1 1 , U n + 1 2 ) ⊆ ( U n 1 , U n 2 ) , by Definition 2.1 (6), ( U + ∞ 1 , U + ∞ 2 ) = ⋂ n = 0 + ∞ ( U n 1 , U n 2 ) is nonempty, closed, and convex. Furthermore, the set ( U + ∞ 1 , U + ∞ 2 ) under the operator F is invariant and ( U + ∞ 1 , U + ∞ 2 ) ∈ ker μ . So, by applying Theorem 1.2, the proof is complete.□

Theorem 2.3

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G a continuous map such that

for each ∅ ≠ U 1 ⊆ G , ∅ ≠ U 2 ⊆ G , where μ is an arbitrary coupled MNC and φ 1 : ℛ + → ℛ + is a nondecreasing function with lim n → + ∞ φ 1 n ( s ) = 0 for every s ≥ 0 . Then, F has at least one fixed-point.

Proof

According to the proof of Theorem 2.2, we define the sequences { U n 1 } , { U n 2 } by induction, where U 0 1 , U 0 2 = G , U n + 1 1 = conv ( F U n 1 ) ¯ , U n + 1 2 = conv ( F U n 2 ) ¯ , for n = 0 , 1 , … . Moreover, in the same as the previous method, we can assume μ ( U n 1 , U n 2 ) > 0 for all n = 1 , 2 , … . In addition, by given assumptions, we obtain

This shows that μ ( U n 1 , U n 2 ) → 0 as n → + ∞ . Since the sequence { ( U n 1 , U n 2 ) } is nested, by Definition 2.1 (6), ( U + ∞ 1 , U + ∞ 2 ) = ⋂ n = 0 + ∞ ( U n 1 , U n 2 ) is a nonempty, closed, and convex subset of ( U 1 , U 2 ) . Therefore, we obtain that ( U + ∞ 1 , U + ∞ 2 ) is a member of ker μ . So, ( U + ∞ 1 , U + ∞ 2 ) is compact. Next, note that F maps ( U + ∞ 1 , U + ∞ 2 ) into itself, and considering Theorem 1.2, we deduce that F has fixed-point in ( U + ∞ 1 , U + ∞ 2 ) . So the proof is complete.□

Corollary 2.4

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G be an operator such that

where φ 1 : ℛ + → ℛ + is a nondecreasing function with lim n → + ∞ φ 1 n ( s ) = 0 for any s ≥ 0 . Then, F has a fixed-point in G .

Proof

Let μ : ℬ E 2 → ℛ + and

where diam ( U 1 , U 2 ) = sup { ‖ υ 1 υ 2 − υ ′ 1 υ ′ 2 ‖ : υ 1 , υ ′ 1 ∈ U 1 , υ 2 , υ ′ 2 ∈ U 2 } . It can be easily seen that μ is coupled MNC in E by Definition 2.1. Furthermore, since φ 1 is nondecreasing, then in view of (2.6), we have

which yields that

By using Theorem 2.3, the proof is complete.□

Definition 3.1

Let E be a Banach space and μ : ℬ E 3 → ℛ + be a mapping. We say that μ is a tripled MNC on E , if it has the following conditions:

1. ker μ = { ( U 1 , U 2 , U 3 ) ∈ ℬ E 3 : μ ( U 1 , U 2 , U 3 ) = θ } is nonempty;

2. For every ( U 1 , U 2 , U 3 ) ∈ ℬ E 3 , μ ( U 1 , U 2 , U 3 ) = θ ⇔ ( U 1 , U 2 , U 3 ) is a precompact set;

3. For each ( ( U 1 , U 2 , U 3 ) , ( U ′ 1 , U ′ 2 , U ′ 3 ) ) ∈ ℬ E 3 × ℬ E 3 , ( ( U 1 , U 2 , U 3 ) ⊆ ( U ′ 1 , U ′ 2 , U ′ 3 ) yields U 1 ⊆ U ′ 1 , U 2 ⊆ U ′ 2 and U 3 ⊆ U ′ 3 ), we have

   ( U 1 , U 2 , U 3 ) ⊆ ( U ′ 1 , U ′ 2 , U ′ 3 ) implies μ ( U 1 , U 2 , U 3 ) ≤ μ ( U ′ 1 , U ′ 2 , U ′ 3 ) ;

4. For every ( U 1 , U 2 , U 3 ) ∈ ℬ E 3 , one has

   μ ( U 1 ¯ , U 2 ¯ , U 3 ¯ ) = μ ( U 1 , U 2 , U 3 ) = μ ( conv ( U 1 , U 2 , U 3 ) ) ,

   where conv ( U 1 , U 2 , U 3 ) denotes the convex hull of ( U 1 , U 2 , U 3 ) ;

5. μ ( λ ( U 1 , U 2 , U 3 ) + ( 1 − λ ) ( U ′ 1 , U ′ 2 , U ′ 3 ) ) ≤ λ μ ( U 1 , U 2 , U 3 ) + ( 1 − λ ) μ ( U ′ 1 , U ′ 2 , U ′ 3 ) for λ ∈ [ 0 , 1 ] ;

6. If { U n 1 } 0 + ∞ , { U n 2 } 0 + ∞ and { U n 3 } 0 + ∞ in ℬ E are decreasing sequences of closed sets and lim n → + ∞ μ { ( U n 1 , U n 2 , U n 3 ) } 0 + ∞ = 0 , then ( U + ∞ 1 , U + ∞ 2 , U + ∞ 3 ) = ⋂ n = 0 + ∞ ( U n 1 , U n 2 , U n 3 ) ≠ ∅ .

Theorem 3.2

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G be a continuous map such that

for each ∅ ≠ U 1 ⊆ G , ∅ ≠ U 2 ⊆ G , and ∅ ≠ U 3 ⊆ G , where μ is an arbitrary tripled MNC and φ 1 , φ 2 : ℛ + → ℛ + such that φ 2 is continuous and φ 1 is lower semicontinuous on ℛ + . Furthermore, φ 1 ( 0 ) = 0 and φ 1 ( s ) > 0 for s > 0 . Then, F has at least one fixed-point in G .

Theorem 3.3

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G be a continuous map such that

for each ∅ ≠ U 1 ⊆ G , ∅ ≠ U 2 ⊆ G , and ∅ ≠ U 3 ⊆ G , where μ is an arbitrary tripled MNC and φ 1 : ℛ + → ℛ + is a nondecreasing function with lim n → + ∞ φ 1 n ( s ) = 0 for every s ≥ 0 . Then, F has at least one fixed-point.

Corollary 3.4

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G be an operator such that

for all υ 1 , υ ′ 1 , υ 2 , υ ′ 2 , υ 3 , υ ′ 3 ∈ G , where φ 1 : ℛ + → ℛ + is a nondecreasing function with lim n → + ∞ φ 1 n ( s ) = 0 for any s ≥ 0 . Then, F has a fixed-point in G .

Theorem 4.1

[13] Let ( G , ϱ ) be a complete metric space and F : G → G be a given mapping. Suppose that there exist θ ∈ Θ and ν ∈ ( 0 , 1 ) such that for all ι , ς ∈ G ,

Then, F has a unique fixed-point.

Theorem 4.2

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G , be a continuous map such that

for each ∅ ≠ U 1 ⊆ G , ∅ ≠ U 2 ⊆ G , where θ ∈ Θ , and μ is an arbitrary coupled MNC and functions φ 1 , φ 2 : ℛ + → ℛ + , such that φ 2 is continuous and φ 1 is lower semicontinuous on ℛ + . Furthermore, φ 1 ( 0 ) = 0 and φ 1 ( s ) > 0 for s > 0 . Then, F has at least one fixed-point in G .

Proof

According to the proof of Theorem 2.2, we define the sequences { U n 1 } , { U n 2 } by induction. Moreover, from (4.2), we obtain

for n = 0 , 1 , 2 , … . Since the sequence { μ ( U n 1 , U n 2 ) } is nonnegative and nonincreasing, we deduce that μ ( U n 1 , U n 2 ) → m when n tends to infinity, where m ≥ 0 is a real number. On the other hand, considering equation (4.3), we obtain

which yields that θ ( φ 2 ( m ) ) ≤ θ ( φ 2 ( m ) ) θ ( φ 2 ( φ 1 ( m ) ) ) . Consequently, θ ( φ 2 ( φ 1 ( m ) ) ) = 1 , then φ 2 ( φ 1 ( m ) ) = 0 and φ 1 ( m ) = 0 so m = 0 . Therefore, we infer μ ( U n 1 , U n 2 ) → 0 as n → + ∞ . Now, considering that ( U n + 1 1 , U n + 1 2 ) ⊆ ( U n 1 , U n 2 ) , by Definition 2.1 (6), ( U + ∞ 1 , U + ∞ 2 ) = ⋂ n = 0 + ∞ ( U n 1 , U n 2 ) is nonempty, closed, and convex. Furthermore, the set ( U + ∞ 1 , U + ∞ 2 ) under the operator F is invariant and ( U + ∞ 1 , U + ∞ 2 ) ∈ k e r μ . So, by applying Theorem 1.2, the proof is complete.□

Theorem 4.3

Let G be a nonempty, bounded, closed, and convex subset of a Banach space E and F : G → G a continuous map such that

for each ∅ ≠ U 1 ⊆ G , ∅ ≠ U 2 ⊆ G , where θ ∈ Θ , φ 2 ∈ Ψ , and μ is an arbitrary coupled MNC. Then, F has at least one fixed-point.

Proof

According to the proof of Theorem 2.2, we define the sequences { U n 1 } , { U n 2 } by induction.

If for an integer N ∈ N one has μ ( U N 1 , U N 2 ) = 0 , then ( U N 1 , U N 2 ) is a precompact set. So the Schauder theorem ensures the existence of a fixed-point for F . Therefore, we can assume μ ( U n 1 , U n 2 ) > 0 for all n ∈ N ∪ { 0 } .

Obviously, { ( U n 1 , U n 2 ) } n ∈ N is a sequence of nonempty, bounded, closed, and convex subsets such that

On the other hand,

Thus, { μ ( U n 1 , U n 2 ) } n ∈ N is a convergent sequence. Assume that

By taking the limit from (4.6), lim n → + ∞ θ ( μ ( U n + 1 1 , U n + 1 2 ) ) = 1 . By ( θ 2 ) , we obtain

From Definition 2.1 (6), ( U + ∞ 1 , U + ∞ 2 ) = ⋂ n = 0 + ∞ ( U n 1 , U n 2 ) is a nonempty, closed, and convex subset of ( U 1 , U 2 ) . Therefore, we obtain ( U + ∞ 1 , U + ∞ 2 ) is a member of ker μ . So, ( U + ∞ 1 , U + ∞ 2 ) is compact. Note that F maps ( U + ∞ 1 , U + ∞ 2 ) into itself, and considering Theorem 1.2, we deduce that F has a fixed-point in ( U + ∞ 1 , U + ∞ 2 ) . So the proof is complete.□