On a measure of noncompactness in the space of regulated functions and its applications

Definition 2.1.

Let x : [ a , b ] → ℝ and let t ∈ ( a , b ] (resp. t ∈ [ a , b ) ). We say that the function x satisfies at the point t the left-hand Cauchy condition (resp. the right-hand Cauchy condition) if for all ε > 0 , there exists δ > 0 such that for all u , v ∈ ( t - δ , t ) ∩ [ a , b ] (resp. u , v ∈ ( t , t + δ ) ∩ [ a , b ] ), we have | x ⁢ ( u ) - x ⁢ ( v ) | ≤ ε .

Definition 2.2.

A function x ∈ B ⁢ ( [ a , b ] ) is said to be a regulated function if it has one-sided limits at every point t ∈ ( a , b ) and the limits x ⁢ ( a + ) and x ⁢ ( b - ) exist.

Definition 2.3.

A function x : [ a , b ] → ℝ is called regulated if for each t ∈ ( a , b ) , the limits x ⁢ ( t - ) and x ⁢ ( t + ) exist and are finite, and the limits x ⁢ ( a + ) and x ⁢ ( b - ) exist and are finite too.

Definition 2.4.

Assume that X is a subset of the space R ⁢ ( [ a , b ] ) . We will say that the set X is equiregulated on the interval [ a , b ] if the following two conditions are satisfied:

1. For all ε > 0 , there exists δ > 0 such that for all x ∈ X , t ∈ ( a , b ] and u ∈ ( t - δ , t ) ∩ [ a , b ] , we have | x ⁢ ( u ) - x ⁢ ( t - ) | ≤ ε .

2. For all ε > 0 , there exists δ > 0 such that for all x ∈ X , t ∈ [ a , b ) and u ∈ ( t , t + δ ) ∩ [ a , b ] , we have | x ⁢ ( u ) - x ⁢ ( t + ) | ≤ ε .

Theorem 2.5.

Let X be a bounded subset of the space R ⁢ ( [ a , b ] ) . The set X is relatively compact in R ⁢ ( [ a , b ] ) if and only if X is equiregulated on the interval [ a , b ] .

Theorem 2.6.

Let X be a bounded subset of the space R ⁢ ( [ a , b ] ) . The set X is relatively compact in R ⁢ ( [ a , b ] ) if and only if the following two conditions are satisfied:

1. For all ε > 0 , there exists δ > 0 such that for all x ∈ X , t ∈ ( a , b ] and u , v ∈ ( t - δ , t ) ∩ [ a , b ] , we have | x ⁢ ( u ) - x ⁢ ( v ) | ≤ ε .

2. For all ε > 0 , there exists δ > 0 such that for all x ∈ X , t ∈ [ a , b ) and u , v ∈ ( t , t + δ ) ∩ [ a , b ] , we have | x ⁢ ( u ) - x ⁢ ( v ) | ≤ ε .

Proof.

At first, let us assume that the set X is relatively compact. Then, according to Theorem 2.5, this means that X is equiregulated on [ a , b ] . Further, fix arbitrarily ε > 0 and choose a number δ > 0 to the number ε 2 pursuant to conditions (i) and (ii) of Definition 2.4. Next, take a number t ∈ ( a , b ] and choose arbitrary numbers u , v ∈ ( t - δ , t ) ∩ [ a , b ] . Then, in view of condition (i), we get

Hence, we infer that the set X satisfies condition (a). Similarly, we can show that the set X satisfies also condition (b).

Conversely, suppose that conditions (a) and (b) are satisfied. Fix a number ε > 0 and choose δ > 0 pursuant to conditions (a) and (b). Next, take an arbitrary number t ∈ ( a , b ] or t ∈ [ a , b ) . Assume, for example, that t ∈ ( a , b ] . Then, according to condition (a), for any function x ∈ X and for arbitrary numbers u , v ∈ ( t - δ , t ) ∩ [ a , b ] , we have

Now, let us take an arbitrary sequence ( v n ) such that ( v n ) ⊂ ( t - δ , t ) ∩ [ a , b ] and v n → t - . Then, in view of (2.1), we obtain

for any n = 1 , 2 , … . Hence, applying standard facts from classical analysis, we conclude that

for an arbitrary u ∈ ( t - δ , t ) ∩ [ a , b ] . This shows that for functions belonging to the set X condition (i) of Definition 2.4 is satisfied.

Similarly, we can prove that the set X satisfies also condition (ii).

Combining the above established facts with Theorem 2.5, we complete the proof. ∎

Definition 3.1.

A function μ : 𝔐 E → ℝ + will be called a measure of noncompactness in the Banach space E if it satisfies the following conditions:

1. The family ker ⁡ μ = { X ∈ 𝔐 E : μ ⁢ ( X ) = 0 } is nonempty and ker ⁡ μ ⊂ 𝔑 E .

2. X ⊂ Y ⇒ μ ⁢ ( X ) ≤ μ ⁢ ( Y ) .

3. μ ⁢ ( X ) = μ ⁢ ( X ¯ ) = μ ⁢ ( Conv ⁡ X ) .

4. μ ⁢ ( λ ⁢ X + ( 1 - λ ) ⁢ Y ) ≤ λ ⁢ μ ⁢ ( X ) + ( 1 - λ ) ⁢ μ ⁢ ( Y ) for λ ∈ [ 0 , 1 ] .

5. If ( X n ) is a sequence of closed sets belonging to 𝔐 E , with X n ⊃ X n + 1 ( n = 1 , 2 , … ) and lim n → ∞ ⁡ μ ⁢ ( X n ) = 0 , then the set X ∞ = ⋂ n = 1 ∞ X n is nonempty.

Lemma 3.2.

Let μ : M E → R + be a function satisfying the following conditions:

1. μ ⁢ ( X ) = 0 ⇔ X ∈ 𝔑 E ,

2. X ⊂ Y ⇒ μ ⁢ ( X ) ≤ μ ⁢ ( Y ) ,

3. μ ⁢ ( X ∪ Y ) = max ⁡ { μ ⁢ ( X ) , μ ⁢ ( Y ) } .

Then μ satisfies axiom (5) of Definition 3.1.

Theorem 3.3.

Let Ω be a nonempty, bounded, closed and convex subset of a Banach space E, and let μ be a measure of noncompactness defined on E. Assume that F : Ω → Ω is a continuous operator such that μ ⁢ ( F ⁢ X ) ≤ k ⁢ μ ⁢ ( X ) for any nonempty subset X of Ω, where k ∈ [ 0 , 1 ) is a constant. Then the operator F has at least one fixed point in the set Ω.

Remark 3.4.

It can be shown that the set Fix ⁡ F of all fixed points of the operator F in the set Ω is a member of the family ker ⁡ μ .

Theorem 3.5.

The function μ defined by formula (3.1) is a regular measure of noncompactness in the space R ⁢ ( [ a , b ] ) .

Proof.

First let us observe that, in view of Theorem 2.6, we deduce that the function μ satisfies condition (i) of Lemma 3.2. This implies immediately that the function μ satisfies axiom (1) of Definition 3.1 with ker ⁡ μ = 𝔑 R . Further, let us take into account the fact that the functions ω 0 - ⁢ ( X ) and ω 0 + ⁢ ( X ) , being the components of the function μ defined by (3.1), are defined by using the supremum. Obviously this yields that the function μ satisfies axiom (2) of Definition 3.1 (or, equivalently, condition (ii) of Lemma 3.2). By the same reasoning, we conclude that the function μ has the maximum property, i.e., it satisfies condition (iii) of Lemma 3.2 (equivalently, axiom 8).

In view of the above established facts and Lemma 3.2, we infer that the function μ = μ ⁢ ( X ) satisfies axiom (5) of Definition 3.1.

Next, let us observe that for arbitrary functions x , y ∈ R ⁢ ( [ a , b ] ) and for λ ∈ ℝ , we obtain

Hence, it is not difficult to deduce that for an arbitrary set X ∈ 𝔐 R , we have

Thus, we see that the function μ satisfies axioms (7) and (8), i.e., the function μ is sublinear.

Applying the same reasoning as above we can easily see that

for an arbitrary set X ∈ 𝔐 R , where the symbol conv ⁡ X stands for the convex hull of the set X. Combining the above inequality with the fact that μ satisfies axiom (2), we infer that

for an arbitrary set X ∈ 𝔐 R .

Finally, we show that

for X ∈ 𝔐 R . To this end, observe firstly that the inequality

is a consequence of axiom (2).

To show the converse inequality let us take an arbitrary function x ∈ X ¯ . This means that x is a limit of a sequence ( x n ) of functions belonging to the set X. Thus, we can write

uniformly on the interval [ a , b ] . Further, let us fix arbitrarily ε > 0 . Then, for a fixed t ∈ ( a , b ] and for u , v ∈ ( t - ε , t ) ∩ [ a , b ] , we have

since the sequence ( x n ) is uniformly convergent to the function x on the interval [ a , b ] .

The above established facts allows us to infer that

In the same way we can show also that

Combining the above inequalities with (3.1), we get

The above estimate in conjunction with (3.4) implies (3.3). This means that the function μ satisfies the first part of axiom (3).

The second part of this axiom is a consequence of equality (3.2) and the fact that Conv ⁡ X = conv ⁡ X ¯ for an arbitrary set X ∈ 𝔐 R .

The proof is complete. ∎

Remark 3.6.

Let us observe that in view of the result proved in [4], we have, for an arbitrary set X ∈ 𝔐 R , that the following inequality is satisfied:

where χ denotes the Hausdorff measure of noncompactness in the space R ⁢ ( [ a , b ] ) and B 1 = B ⁢ ( θ , 1 ) . It is not difficult to calculate that ω 0 - ⁢ ( B 1 ) = ω 0 + ⁢ ( B 1 ) = 2 . Thus, from (3.5), we obtain the estimate

for X ∈ 𝔐 R . It is an open question whether there exists a constant q > 0 such that

for X ∈ 𝔐 R (cf. also [10, 20]).

Notice that the answer to the above question has no influence on our further considerations concerning the applicability of the measure of noncompactness μ given by (3.1).

Remark 4.1.

Let us explain that the expression that the functions g ⁢ ( t , x ) and k ⁢ ( t , s ) are regulated with respect to t, (locally) uniformly with respect to the variables x and s, respectively (cf. the above formulated assumptions (ii) and (iii)), is understood in the sense of Definition 2.1.

Theorem 4.2.

Under assumptions (i)–(v), equation (4.1) has at least one solution in the space R ⁢ ( [ a , b ] ) .

Proof.

Let us consider the operator T associated with equation (4.1). This means that T is defined on the space R ⁢ ( [ a , b ] ) by the formula

For further purposes, let us consider the operators G, F and K defined in the following way:

Then the operator T defined by (4.2) can be represented as

where the symbol K ⋅ F is understood as the composition of the operators F and K.

Now, let us fix arbitrarily ε > 0 and t ∈ ( a , b ] . Then, for an arbitrary function x ∈ R ⁢ ( [ a , b ] ) and for arbitrary u , v ∈ ( t - ε , t ) , on the basis of assumption (ii), we obtain

where

In view of assumption (ii), we infer that ω 1 , ∥ x ∥ - ⁢ ( g , t ; ε ) → 0 as ε → 0 . Particularly this implies that the operator G transforms the space R ⁢ ( [ a , b ] ) into itself.

In a similar way, taking into account assumption (iii), we derive

where

Notice that, in view of assumption (iii), we have that ω 1 - ⁢ ( k , t ; ε ) → 0 (similarly, ω 1 + ⁢ ( k , t ; ε ) → 0 ) as ε → 0 for each t ∈ [ a , b ] . This implies that the function t ↦ ∫ a b k ⁢ ( t , s ) ⁢ 𝑑 s (or the function t ↦ ∫ a b | k ( t , s ) | d s ) is regulated on the interval [ a , b ] . Consequently, it follows that the function t ↦ ∫ a b | k ( t , s ) | d s is bounded on the interval [ a , b ] and simultaneously justifies the fact that k ¯ < ∞ , which was stated before.

Now, from (4.4), (4.5), assumption (iv) and taking into account representation (4.3), we conclude that the operator T transforms the space R ⁢ ( [ a , b ] ) into itself. Obviously, the above reasoning can be repeated for any fixed t ∈ [ a , b ) and for u , v ∈ ( t , t + ε ) ∩ [ a , b ] , if we replace the quantity ω - by ω + .

Let us notice that by utilizing our assumptions, for an arbitrarily fixed x ∈ R ⁢ ( [ a , b ] ) and t ∈ [ a , b ] , we have

where we write ∥ x ∥ in place of ∥ x ∥ ∞ and the constants p ¯ , L, g ¯ , a, b, k ¯ were defined previously or imposed in the assumptions.

The above inequality yields the estimate

Hence, keeping in mind assumption (v), we deduce that there exists a positive number r 0 such that for all functions x ∈ B r 0 ⊂ R ⁢ ( [ a , b ] ) , we have that T ⁢ x ∈ B r 0 , i.e., the operator T transforms the ball B r 0 into itself.

Keeping in mind the convenience, we will further accept that

To prove the continuity of the operator T defined by (4.3) on the ball B r 0 let us observe that, in view of assumption (ii), it is sufficient to prove the continuity of the operator K ⋅ F on B r 0 . Thus, fix arbitrarily ε > 0 and take x , y ∈ B r 0 such that ∥ x - y ∥ ≤ ε . Then, for a fixed t ∈ [ a , b ] , we obtain

where

Observe that taking into account the uniform continuity of the function f on the set [ a , b ] × [ - r 0 , r 0 ] (cf. assumption (iv)), we deduce that ω ⁢ ( f , ε ) → 0 as ε → 0 . Combining this fact with (4.8), we infer that the operator K ⋅ F is continuous on the ball B r 0 .

Further on, let us fix an arbitrary nonempty set X ⊂ B r 0 and a number ε > 0 . Then, for x ∈ X and for t ∈ ( a , b ] , let us choose arbitrary numbers u , v ∈ ( t - ε , t ) ∩ [ a , b ] . Then we get the estimate

Hence, in view of estimates (4.4) and (4.6), we obtain

Further, utilizing estimates (4.5) and (4.6), we derive the following inequality:

where ω 1 - ⁢ ( k , t ; ε ) was introduced previously. Now, linking estimates (4.9) and (4.10), we obtain

Next, keeping in mind assumption (i) and the properties of the functions ε ↦ ω - ⁢ ( p , t ; ε ) , ε ↦ ω 1 , r 0 - ⁢ ( g , t ; ε ) and ε ↦ ω 1 - ⁢ ( k , t ; ε ) , and letting ε → 0 , we obtain the estimate

In the same way, we can prove that

Combining (4.11) and (4.12) and taking into account formula (3.1), expressing the measure of noncompactness μ , we get

Observe that, in view of (4.7) and assumption (v), we obtain

Hence, keeping in mind estimate (4.13) and Theorem 3.3, we conclude that the operator T has at least one fixed point x in the ball B r 0 . Obviously, the function x = x ⁢ ( t ) is a desired solution of equation (4.1) belonging to the space R ⁢ ( [ a , b ] ) . The proof is complete. ∎