Solvability of an infinite system of integral equations on the real half-axis

Definition 2.1

A function μ : 𝔐_E → ℝ₊ will be called a measure of noncompactness in the 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).

4. μ(Conv X) = μ(X).

5. μ(λ X + (1 − λ)Y) ⩽ λμ(X) + (1 − λ)μ(Y) for λ ∈ [0, 1].

6. If (X_n) is a sequence of closed sets from 𝔐_E such that X_n+1 ⊂ X_n for n = 1, 2, … and if limn→∞ μ(X_n) = 0 then the set X∞=⋂n=1∞Xn is nonempty.

The family ker μ from axiom (i) will be called the kernel of the measure of noncompactness μ. Let us also observe that the set X_∞ defined in axiom (vi) is an element of the family ker μ. In fact, it follows easily from the inclusion X_∞ ⊂ X_n for n = 1, 2, … and axiom (ii) which implies the inequality μ(X_∞) ⩽ μ(X_n) for any n ∈ ℕ. Hence we have that μ(X_∞) = 0 and consequently, X_∞ ∈ ker μ. This simple conclusion is very crucial in applications.

In what follows assume that μ is a measure of noncompactness in the space E. The measure μ is called sublinear [3] if it satisfies the following additional conditions

1. μ(X + Y) ⩽ μ(X) + μ(Y).

2. μ(λX) = ∣λ∣μ(X) for λ ∈ ℝ.

If the measure of noncompactness μ satisfies the condition

1. μ(X ∪ Y) = max{μ(X), μ(Y)}

then we say that it has the maximum property. Moreover, if kerμ = 𝔑_E we say that μ is full. If μ is a sublinear and full measure of noncompactness with the maximum property, then μ is said to be regular.

Let us recall that from the historical point of view the first measure of noncompactness was defined in 1930 by K. Kuratowski [11], but the most important and useful measure of noncompactness is the so-called Hausdorff measure of noncompactness which was defined in [12, 13] by the formula

where X ∈ 𝔐_E. The importance of the Hausdorff measure χ is caused by the fact that in some Banach spaces like C([a, b]), c₀ and l_p we can give formulas expressing the measure χ in connection with the structure of the mentioned Banach spaces. Let us mention that χ is a regular measure of noncompactness [3, 14, 15].

However, in a lot of Banach spaces we are not in a position to give formulas for the Hausdorff measure of noncompactness χ. Even more, we are not able to provide formulas for full measures of noncompactness [3, 4]. Therefore, in such a situation we restrict ourselves to measures of noncompactness in the sense of Definition 2.1, which are not full (cf. also [3, 4, 14]). In the present paper we will also consider a measure of noncompactness of such a type.

Namely, assume that E is a given Banach space with the norm ∥⋅∥_E and μ is a measure of noncompactness in the space E. Consider the Banach space BC(ℝ₊, E) consisting of all functions x : ℝ₊ → E which are continuous and bounded on the interval ℝ₊. The space BC(ℝ₊, E) will be equipped with the standard supremum norm

Further, take an arbitrary nonempty and bounded subset X of the space BC(ℝ₊, E). Fix x ∈ X and ε > 0. We define the modulus of the uniform continuity of the function x (cf.[2]) by putting

Obviously, limε→0ω∞(x,ε)=0 if and only if the function x is uniformly continuous on the interval ℝ₊.

Further, let us define the following quantities

Next, let us consider the function μ_∞ defined on the family 𝔐_BC(ℝ+,E) in the following way

where μ_T(X) is defined by the formula

for any fixed T > 0.

Finally, for a given T > 0 let us put

and

Now, linking quantities defined by (2.1), (2.2) and (2.3), we can consider the following function μ_a defined on the family 𝔐_BC(ℝ+,E):

It can be shown that the function μ_a is a measure of noncompactness in the space BC(ℝ₊, E) (cf. [2]). The kernel kerμ_a of the measure μ_a consists of all bounded subsets X of the space BC(ℝ₊, E) such that functions from X are uniformly continuous and equicontinuous on ℝ₊ (equivalently we can say that functions from X are equiuniformly continuous on ℝ₊) and tend to zero at infinity with the same rate. Apart from this, all cross-sections X(t) = {x(t) : x ∈ X} of the set X belong to the kernel kerμ of the measure of noncompactness μ in the Banach space E (cf. [2]). The measure μ_a is not full and has the maximum property. If the measure μ is sublinear in E then the measure μ_a defined by (2.4) is also sublinear [2].

Let us notice that in the similar way as above we may define other measures of noncompactness in the space BC(ℝ₊, E) (see [2]). We will not provide the definitions of those measures since we will use only the measure of noncompactness μ_a further on.

In what follows, taking into account our further purposes, we will consider as the Banach space E, the sequence space l_∞ endowed with the standard supremum norm.

Thus, let us consider the Banach space BC(ℝ₊, l_∞) consisting of functions x : ℝ₊ → l_∞ which are continuous and bounded on ℝ₊. Observe, that if x ∈ BC(ℝ₊, l_∞), then we can write this function in the form

for any t ∈ ℝ₊, where the sequence (x_n(t)) is an element of the space l_∞ for any fixed t. The norm of the function x = x(t) = (x_n(t)) is defined by the equality

In our further considerations the space BC(ℝ₊, l_∞) will be denoted by BC_∞.

Now, we provide the formula expressing the measure of noncompactness μ_a defined by (2.4) in the space BC_∞, provided the measure of noncompactness μ in the sequence space l_∞ is defined by the formula [3]

for X ∈ 𝔐_l∞. In such a case the component μ_∞ defined by (2.2) will be denoted by μ¯∞1.

Thus, our measure of noncompactness μ_a defined by (2.4) will be now denoted by μa1 and is defined as a particular case of (2.4) by the formula

where the components on the right hand side of formula (2.5) are defined in the following way:

In what follows we give an other formula expressing the quantity μ_∞ defined by (2.2).

To this end we prove the following lemma.

Lemma 2.2

The following equality is satisfied

where μ_∞ is defined by formula (2.2).

Proof

Obviously, for any fixed T > 0 we have

Hence, we get

To prove the converse inequality, let us denote

Further, fix an arbitrary number ε > 0. Then we can find a number t₀ ∈ ℝ₊ such that

Hence, for T ⩾ t₀ we obtain

Since the function T → sup{μ(X(t)) : t ∈ [0, T]} is nondecreasing, we get

Combining (2.10) and (2.11), we have

Consequently, in view of the arbitrariness of the number ε, we derive the following inequality

Finally, linking (2.9) and (2.12) we obtain the desired equality.□

Now, let us notice that taking into account Lemma 2.2 and formula (2.7) expressing the quantity μ_∞ in the case of the space BC_∞, we obtain the following corollary.

Corollary 2.3

The quantity (2.7) can be expressed by the following formula

At the end of this section we recall a useful fixed point theorem of Darbo type [3, 16].

To this end let us assume that E is a Banach space and μ is a measure of noncompactness (in the sense of Definition 2.1) in the space E.

Theorem 2.4

Assume that Ω is a nonempty, bounded, closed and convex subset of a Banach space E and Q : Ω → Ω is a continuous operator such that there exists a constant k ∈ [0, 1) for which μ(QX) ⩽ kμ(X) for an arbitrary nonempty subset X of Ω. Then there exists at least one fixed point of the operator Q in the set Ω.

Remark 2.5

It can be shown that the set Fix Q of all fixed points of the operator Q belongs to the family ker μ [3].

This simple observation is very essential in characterization of solutions of considered operator equations which are proved with help of Theorem 2.4.

Lemma 3.1

Let the function x(t) = (x_n(t)) be an element of the space BC_∞. Then the sequence (x_n) is equibounded and locally equicontinuous on ℝ₊.

Lemma 3.2

Let the sequence (a_n(t)) be an element of the space BC_∞ such that limt→∞ a_n(t) = 0 uniformly with respect to n ∈ ℕ (cf. assumption (i)). Then the sequence (a_n) is equibounded and equicontinuous on ℝ₊.

Proof

The equiboundedness of the sequence (a_n) on ℝ₊ follows from Lemma 3.1. To prove the equicontinuity of (a_n) on ℝ₊ let us fix ε > 0. Keeping in mind the remaining part of the assumption in our lemma we can find a number T > 0 such that |a_n(t)| ⩽ ε/4 for t ⩾ T and n = 1, 2, …. On the other hand, in view of Lemma 3.1 we deduce that the sequence (a_n) is locally equicontinuous on ℝ₊. Thus, we can find a number δ > 0 such that |a_n(t₂) – a_n(t₁)| ⩽ ε/2 for t₁, t₂ ∈ [0, T] such that |t₂ – t₁| ⩽ δ and for any n = 1, 2, …. Now, let us take arbitrary t₁, t₂ ∈ ℝ₊ such that |t₂ – t₁| ⩽ δ. Without loss of generality we can assume that t₁ < t₂. If t₁, t₂ ∈ [0, T], then in view of the above established fact we have that |a_n(t₂) – a_n(t₁)| ⩽ ε/2 for n = 1, 2, ….

If t₁, t₂ ⩾ T, then we obtain

for any n = 1, 2, ….

Assume now that t₁ < T ⩽ t₂. Then, fixing arbitrarily n ∈ ℕ and taking into account the above derived facts we get

This shows that the sequence (a_n) is equicontinuous on ℝ₊.□

Corollary 3.3

The constant A defined by equality (3.2) is finite.

Theorem 3.4

Under assumptions (i) – (x) the infinite system (3.1) has at least one solution x(t) = (x_n(t)) in the space BC_∞ = BC(ℝ₊, l_∞).

Proof

At the beginning we define three operators F, V, Q on the space BC_∞ in the following way:

At first we show that the operator F transforms the space BC_∞ into itself.

To this end fix the function x = x(t) = (x_n(t)) ∈ BC_∞. Then, in view of assumption (vi) we have the following estimate

for t ∈ ℝ₊ and for n = 1, 2, …, where the functions f_n and l = l(r) were specified in assumption (vi). Hence, on the basis of (3.3) we deduce that

for any x ∈ BC_∞. This shows that the function Fx is bounded on ℝ₊.

In order to prove the continuity of the function Fx on the interval ℝ₊, let us fix ε > 0. Then, from assumption (v) we obtain that there exists δ > 0 such that for t, s ∈ ℝ₊ and |t – s| ⩽ δ the following inequality is satisfied

for any x = (x_i) ∈ l_∞. This implies that

provided t, s ∈ ℝ₊ are such that |t – s| ⩽ δ. But this means that the function Fx is continuous (even uniformly continuous) on ℝ₊. Finally we conclude that the operator F transforms the space BC_∞ into itself.

Now, we are going to show that the operator V acts from the space BC_∞ into itself.

Thus, similarly as above, fix a function x = x(t) = (x_n(t)) ∈ BC_∞. Then, for arbitrarily fixed numbers t ∈ ℝ₊ and n ∈ ℕ, in view of assumptions (iii) and (ix), we get

Particularly, the above estimate yields that the function Vx is bounded on the interval ℝ₊.

Next, fix ε > 0 and choose a number δ > 0 according to assumption (ii). Then, for arbitrarily fixed numbers t₁, t₂ ∈ ℝ₊ such that |t₂ – t₁| ⩽ δ, on the basis of assumptions (ii) and (ix) (assuming additionally that t₁ < t₂), we obtain

where K₂ is a constant appearing in assuption (iv) and ω_k(δ) denotes a common modulus of continuity of the sequence of functions t → k_n(t, s) on the interval ℝ₊ (according to assumption (ii)). Obviously we have that ω_k(δ) → 0 as δ → 0.

Further on, taking into account estimate (3.6) and assumption (ix), we obtain

Hence we deduce that the function Vx is continuous on the interval ℝ₊. Linking boundedness of the function Vx with its continuity on ℝ₊ we conclude that the operator V transforms the space BC_∞ into itself.

Now, keeping in mind the fact that the space BC_∞ = BC(ℝ₊, l_∞) forms a Banach algebra with respect to the coordinatewise multiplication of function sequences and taking into account the definition of the operator Q as well as assumption (i), we infer that for an arbitrarily fixed function x = x(t) ∈ BC_∞ the function (Qx)(t) = ((Q_nx)(t)) = (a_n(t) + (F_nx)(t)(V_nx)(t)) acts from the interval ℝ₊ into the space l_∞. Indeed, in view of the fact that ((F_nx)(t)) ∈ l_∞ for any t ∈ ℝ₊ and in the light of estimate (3.5) we get

Hence, applying (3.3) we deduce that (Qx)(t) = ((Q_nx)(t)) ∈ l_∞ for each t ∈ ℝ₊.

Next, let us notice that the continuity of the function Qx on ℝ₊ is a simple consequence of the fact that both the function Fx and the function Vx are continuous on ℝ₊. Similarly we can derive that the function Qx is bounded on the interval ℝ₊. Indeed, it is only sufficient to make use assumption (i) and Lemma 3.1.

Finally, let us observe that combining all the above established properties of the function Qx we conclude that the operator Q transforms the space BC_∞ into itself.

Further, let us note that based on estimates (3.4) and (3.5), for arbitrarily fixed n ∈ ℕ and t ∈ ℝ₊ we get

From the above estimate and assumption (x) we infer that there exists a number r₀ > 0 such that the operator Q transforms the ball B_r0 into itself.

In what follows we are going to show that the operator Q is continuous on the ball B_r0. Keeping in mind the representation of the operator Q given at the beginning of our proof we see it is sufficient to prove the continuity of the operators F and V, separately.

To this end let us fix ε > 0 and x ∈ B_r0. Next, choose an arbitrary point y ∈ B_r0 such that ∥x – y∥_BC∞ ⩽ ε. Then, for each fixed t ∈ ℝ₊, in virtue of assumption (vii) we get

Particularly this shows that the operator F is continuous at every point of the ball B_r0.

To prove the continuity of the operator V on the ball B_r0 let us define the function δ = δ(ε) by putting

Obviously, in view of assumption (viii) we have that δ(ε) → 0 as ε → 0. Next, taking x, y ∈ B_r0 such that ∥y – x∥_BC∞ ⩽ ε and t ∈ ℝ₊, for arbitrary n ∈ ℕ we obtain

This yields to the estimate

Thus we see that the operator V is continuous on the ball B_r0.

In the sequel let us fix an arbitrary number ε > 0. Next, choose a number δ > 0 according to assumption (v). Further fix a nonempty subset X of the ball B_r0 and take an arbitrary function x ∈ X and n ∈ ℕ. Then, for arbitrary t, s ∈ ℝ₊ such that |t – s| ⩽ δ, in virtue of assumption (v) we get

Now, on the basis of the above estimate we obtain

Further, let us fix a number ε > 0 and choose t₁, t₂ ∈ ℝ₊ such that |t₂ – t₁| ⩽ ε. Without loss of generality we may assume that t₁ < t₂. Then, in view of estimate (3.7) we derive the following inequality

This yields the estimate

Now, taking into account the representation of the operator Q, for an arbitrary function x ∈ X and for arbitrary numbers t, s ∈ ℝ₊, we obtain

where a(t) = (a_n(t)).

Further, fix ε > 0 and assume that |t – s| ⩽ ε. Then, from the above inequality and estimates (3.8), (3.9), (3.4), and (3.5), we get

Now, in view of Lemma 3.2 we infer that ω^∞(a, ε) → 0 as ε → 0. Next, keeping in mind that ω_k(ε) → 0 as ε → 0, from the above obtained estimate we derive the following inequality

In what follows we will consider the second term of the measure of noncompactness μa1 (cf. formula (2.5)) which is denoted by μ¯∞1 and is defined by formula (2.7).

To this end fix a nonempty subset X of the ball B_r0 and choose an arbitrary function x = x(t) ∈ X. Further, take a natural number n and T > 0. Then, for any fixed t ∈ [0, T], in view of the representation of the operator Q and estimates (3.3) and (3.5), we obtain

Now, taking supremum over all x ∈ X, from the above estimate we get

Hence, taking into account assumption (i) and (vi), we derive the following inequality

Finally, taking supremum over t ∈ [0, T] on both sides of the above inequality and next, passing with T → ∞, in virtue of formula (2.7) we get

In order to estimate the last term a_∞ of the measure of noncompactness μa1 (cf. formula (2.5)) expressed by formula (2.8), let us take a nonempty subset X (X ⊂ B_r0) and choose a function x ∈ X. Further, fix arbitrarily T > 0. Then, taking t ⩾ T and keeping in mind the previously obtained inequalities (3.3) and (3.5), we obtain

Now, taking in the above estimate supremum over t ⩾ T and next, over x ∈ X, we get

Further, passing with T → ∞ and taking into account assumptions (i) and (vi), we derive the following estimate

Finally, combining estimates (3.10)-(3.12) and keeping in mind formula (2.5), we obtain the following inequality for an arbitrary nonempty subset X of the ball B_r0:

Hence, in view of the fact that the operator Q is a continuous self-mapping of the ball B_r0, assumption (x) and Theorem 2.4 we conclude that the infinite system of Volterra-Hamerstein integral equations (3.1) has at least one solution x = x(t) in the space BC_∞ = BC(ℝ₊, l_∞) which belongs to the ball B_r0 and is uniformly continuous on the interval ℝ₊.

The proof is complete. □