Measures of noncompactness and superposition operator in the space of regulated functions on an unbounded interval

In this paper, we formulate necessary and sufficient conditions for relative compactness in the space BG(R+,E)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$BG({\mathbb {R}}_+,E)$$\end{document} of regulated and bounded functions defined on R+\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb R}_+$$\end{document} with values in the Banach space E. Moreover, we construct four new measures of noncompactness in the space BG(R+,E)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$BG({\mathbb {R}}_+,E)$$\end{document}. We investigate their properties and we describe relations between these measures. We provide necessary and sufficient conditions so that the superposition operator (Niemytskii) maps BG(R+,E)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$BG({\mathbb {R}}_+,E)$$\end{document} into BG(R+,E)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$BG({\mathbb {R}}_+,E)$$\end{document} and, additionally, be compact.


Introduction
The measures of noncompactness and fixed point theorems are often chosen to investigate solvability of the nonlinear equations. Using a suitable function space together with convenient measures of noncompactness we can obtain elegant existence theorems. In recent years, there have appeared a lot of papers concerning the space G(J , E) of regular functions defined on a bounded interval J and with values in the Banach space E [2,[4][5][6][7][8][9][10][11]14]. On the other hand, there have not been any papers focused on the space of regular functions on an unbounded interval R + . Throughout this paper we are going to fill this gap. We will investigate the space BG(R + , E) of regular and bounded functions defined on R + and with values in the Banach space E.
In Sect. 3 we formulate necessary and sufficient relative compactness conditions in the space BG(R + , E). In the sequel we construct four new and convenient measures of noncompactness in BG(R + , E) and investigate their properties. In Sect. 4 we present necessary and sufficient conditions for superposition operator so that it maps BG(R + , E) into BG(R + , E). Moreover, we formulate necessary and sufficient conditions so that the superposition operator F f : BG(R + , E) → BG(R + , E) is compact.
Finally, Sect. 5 shows the applicability one of the mentioned measures of noncompactness to the existence result for some nonlinear integral equation.

Notation, definitions and auxiliary facts
This section is devoted to recalling some facts which will be used in our further investigations. Assume that E is a real Banach space with the norm · and the zero element θ . Denote by B E (x, r ) the closed ball centered at x and with radius r . The symbol B E (r ) stands for the ball B E (θ, r ). We write X , convX , ConvX to denote the closure, convex hull and the convex closure of a set X , respectively. Further, let M E denote the family of all nonempty and bounded subsets of E and N E its subfamily consisting of all relatively compact sets.
The characteristic function of the set A is denoted by 1 A .
We accept the following definition of the notion of a measure of noncompactness [3].
5 • 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 lim n→∞ μ(X n ) = 0, then the intersection X ∞ := ∞ n=1 X n is nonempty.
In the sequel we will use measures of noncompactness having some additional properties. Namely, a measure μ is said to be sublinear if it satisfies the following two conditions: A sublinear measure of noncompactness μ satisfying the condition (maximum property) and such that kerμ = N E is said to be regular.
For a given nonempty bounded subset X of E we denote by β E (X ) the so-called Hausdorff measure of noncompactness of X . This quantity is defined by the formula Compactness criteria and measures of noncompactness in the space G([0, T ], E) were investigated in several research papers (see [4][5][6]11]). First, we recall the concept of a equiregulated subset of the space G([0, T ], E). Definition 2. 3 We will say that the set X ⊂ G([0, T ], E) is equiregulated on the interval J = [0, T ] if the following two conditions are satisfied: Let us recall the compactness criterion in G([0, T ], E). This result was formulated by Fraňková [11], (see also [4][5][6] . For x ∈ X and ε > 0 let us denote the following quantities: The quantities ω − (x, t, ε) and ω + (x, t, ε) can be interpreted as left-hand and right-hand sided moduli of convergence of the function x at the point t. Further, let us put: Finally, let us define quantity Theorem 2.5 [14] The function μ T given by formula (1) satisfies conditions Further, we define the space mentioned in the title of this paper. Denote by BG(R + , E) the space consisting of all bounded and regulated functions defined on the interval R + with values in a Banach space E, equipped with the norm x ∞ := sup{ x(t) : t ∈ R + }. It is easy to show that BG(R + , E) is a Banach space. In the next section we are going to introduce a simple criterion of compactness in BG(R + , E) and a few measures of noncompactness in this space.

Measures of noncompactness in BG(R + , E)
For fixed x ∈ BG(R + , E) and T > 0 we put Further, we give the following compactness criterion based on ideas from [16].
Proof (⇒) In view of the fact that X is a totally bounded set, we obtain that the sets X (t) for arbitrary t ∈ R + and X | [0,T ] for arbitrary T > 0 have this property as well. Therefore, we obtain (a) and, in virtue of Theorem, 2.4 we have (b). Let us assume that (c) is not fulfilled. Then, there exist ε 0 > 0 and two sequences {x n }, {y n } ⊂ X such that and x n − y n n ≤ 1 n , n ∈ N.
Relative compactness of the set X implies the existence of convergent subsequences {x k n }, {y k n } of the sequences {x n } and {y n }, i.e. x k n → x ∈ BG(R + , E) and y k n → y ∈ BG(R + , E). However, we obtain a contradiction. Namely, taking n → ∞ we obtain from (2) inequality x − y ∞ ≥ ε 0 . On the other hand, in view of (3) we get x − y ∞ = 0.
(⇐) Let us consider an arbitrary sequence {x n } ⊂ X . Conditions (a) and (b) connected with Theorem 2.4 imply that the sequence {x n } has a subsequence {x 1,n } converging on [0, 1] with respect to the pseudonorm · 1 to some regulated function defined on [0, 1]. Similarly, the sequence {x 1,n } has a subsequence {x 2,n } converging on [0, 2] with respect to the pseudonorm · 2 to some regulated function defined on [0, 2]. Therefore, we obtain the sequence of subsequences {x i,n }, i = 1, 2, . . . converging on the interval [0, i] with respect to the pseudonorm · i to some regulated function defined on R + . Next, putting z n := x n,n , n = 1, 2, . . . we obtain the sequence {z n }, uniformly convergent on bounded intervals to some regulated function z : R + → E. Let us fix ε > 0. Further, if T > 0 and δ > 0 are like in (c), then for sufficiently big n we have z − z n T ≤ δ and in virtue of (c) we obtain z − z n ∞ ≤ ε for sufficiently big n. This yields that z ∈ BG(R + , E) and z n → z with respect to the norm · ∞ which implies relative compactness of the set X in BG(R + , E).
Using this theorem we introduce the first measure of noncompactness in the space BG(R + , E). For this purpose we define several symbols. For a given set X ∈ M BG(R + ,E) and δ > 0 let us denote Using the above quantities we define the mapping μ : M BG(R + ,E) → R + given by the

Remark 3.2
In the definition of the quantity ϕ T (X , δ) in the formula (4) the convex hull convX appears. It is caused by the fact that without it the function μ would not fulfil the condition 3 • of Definition 2.1. This condition, in turn, is necessary to use many fixed point theorems.
The main properties of the function μ are contained in the below given theorem. Proof The equality kerμ = N BG(R + ,E) is a simple consequence of Theorem 3.1. We gather several facts concerning the function ϕ ∞ 0 . First, let us notice that the properties are obvious. Now we will show that Let us fix X ∈ M BG(R + ,E) , T > 0, δ > 0 and take x, y ∈ convX such that x − y T < δ. Then there exist {x n }, {y n } ⊂ convX such that x n → x, y n → y. This means that for arbitrary ε > 0 there exists n 0 ∈ N such that for each n ≥ n 0 we have
Since x n − y n → x − y, then we may choose sufficiently big n ∈ N such that x n − y n T < δ. Therefore, in virtue of inequality (10) we obtain and hence Letting δ → 0 and T → ∞ we obtain (in view of arbitrariness of ε) which together with (6) confirms (9). Further, the property (7) in connection with (9) give us Keeping in mind Theorem 2.5 we have that the mapping μ T given by the formula (1) satisfies conditions 1 • -7 • . Hence, to obtain the properties 2 • , 6 • , 3 • for the measure μ, it is sufficient to join (6), (8), (11) to the conditions 2 • , 6 • , 3 • for the measure μ T and to take limit T → ∞.
Further, the property 5 • will be proved. Let us assume that {X n } is a decreasing sequence of closed and bounded subsets of the space BG(R + , E) and lim n→∞ μ(X n ) = 0. Let us take arbitrarily x n ∈ X n . In particular, it follows that lim n→∞ μ T ({x i : i ≥ n}) = 0 for each T > 0. Since the measure μ T satisfies the condition 5 • in the space G([0, T ], E), putting T = 1, 2, . . . and applying the diagonal method we ensure the existence of subsequence {u n } of the sequence {x n }, which is uniformly convergent on bounded intervals to the function In what follows we provide two examples illustrating that the mapping μ does not fulfil the conditions 4 • , 7 • and 8 • .

Example 3.4
Let us fix arbitrarily e ∈ E, e = 1 and put λ := 1 2 , Let us notice that if T > 1, Thus, if x − y T < δ for some x, y ∈ convX , then x − y ∞ < δ and finally ϕ ∞ 0 (X ) = 0. Reasoning similarly, we infer that ϕ ∞ 0 (Y ) = 0. Now, combining it with (12) and (13) we obtain Further, we have Moreover, Hence, we obtain μ( 1 2 X + 1 2 Y ) ≥ 2 which together with (14) proves that the property 4 • is not fulfilled for λ = 1 2 . The above mentioned sets X and Y show that the property 7 • does not hold either. Now, let us consider the set X ∪ Y . We have The above properties yields μ(X ∪ Y ) ≥ 2 which together with (14) gives a contradiction to 8 • .
Open question It is not known if the measure μ satisfies the weak maximum property. In other words, whether the condition Remark 3.5 Although the mapping μ given by the formula (5) does not fulfil some conditions from Definition 2.1, the conditions which are satisfied are sufficient to apply basic fixed point theorems (i.e. Darbo, Sadovskii). Moreover, the fact kerμ = N BG(R + ,E) is undoubtedly an advantage of the function μ. Indeed, this means that μ "catches" every relative compact subset of BG(R + , E).
The below given result presents a relationship between the function μ and the Hausdorff measure of noncompactness in BG(R + , E). As it turns out, in spite of having identical kernels, they are not equivalent.

Theorem 3.6 For each nonempty and bounded set X ⊂ BG(R + , E) we have the inequality
The above estimation is optimal (see Example 3.7). However, there does not exist a constant k > 0 such that Further, we will show that Let us fix t ∈ R + . Since {a i : i = 1, . . . , n} is a compact set, Theorem 2.4 yields that there exists δ > 0 such that Using these considerations we have In the sequel we show that Since . Let us denote the left side of the inequality (18) by g. Hence, there exist sequences {x n }, {y n } ⊂ convX such that and We may further choose an increasing subsequence of natural numbers {k n } such that there exist indexes i and j which give us We will show that Suppose, on the contrary, In connection with (20) we have (for n ∈ N such that k n ≥ t) which under n → ∞ is a contradiction and confirms (22). Combining (21) and (22) we conclude that Letting n → ∞ and applying (19) we have (18). Summing the inequalities (15), (17) and (18), in view of arbitrariness of ε > 0 we obtain the first part of thesis. For further purposes let us assume that there exists k > 0 such that Without loss of generality we may assume that k ≤ 1. Let us fix arbitrarily e ∈ E, e = 1 and put Hence Let us notice that if T > 1, Thus, if x − y T < δ for x, y ∈ convX , then x − y ∞ < 3 k δ and this implies ϕ ∞ 0 (X ) = 0. Combining this fact with (24) we get μ(X ) = k 3 . Hence, in virtue of (23) and (24) we have k 2 ≤ k 3 , which yields a contradiction. Now, we show that upper estimation from Theorem 3.6 is attained.

Example 3.7
Let e ∈ E is such that e = 1 and let us denote by e the function e : R + → E given by the formula e(t) := e, t ∈ R + . Next, let us put .
Further, for arbitrarily fixed T > 0 we define two functions of a real variable t ∈ R + , i.e.
Inspired by some measures in the space BC(R + , E) of bounded and continuous functions (see [3]), we will introduce several new measures of noncompactness in BG(R + , E). For this purpose we will define a few new set functions. Let us take arbitrarily X ∈ M BG(R + ,E) .
So, for a fixed T > 0 let us define Next, notice that there exists the limit Moreover we put and we can also define where diam X (t) is understood as And now let us consider the functions γ a , γ b , γ c defined on the family M BG(R + ,E) as follows The main properties of the above mentioned mappings are contained in the below given theorem.  E) . Moreover, let us notice that each of the functions a ∞ , b ∞ and c satisfies axioms 2 • − 7 • . The proof of this fact is simple and will be omitted. Since, in view of Theorem 2.5 the component μ T = max{ω − T , ω + T }+β T satisfies axioms 2 • −7 • , it is sufficient to add the appropriate conditions for components a ∞ , b ∞ , c and to take limit T → ∞. Now, we prove 5 • . Let us assume that {X n } is a decreasing sequence of closed and bounded subsets of the space BG(R + , E) and lim n→∞ γ (X n ) = 0, where γ is one of the measures γ a , γ b , γ c . Let us take arbitrarily x n ∈ X n . Reasoning similarly as in the proof of Theorem 3.3, we can state that the sequence {x n } contains subsequence which is convergent in the norm · ∞ to some element u ∈ BG(R + , E), and it implies that u ∈ ∞ n=1 X n = ∅.

Remark 3.9
The above mentioned measures γ a , γ b , γ c satisfy several additional conditions (for example 4 • , 7 • ) that are not fulfilled by the measure μ. However, the measure μ "catches" relative compact subsets in a better way than γ a , γ b and γ c , namely ker γ a , ker γ b , ker γ c N BG(R + ,E) . We omit the simple examples showing that there is no equality.
The mutual relations between these five measures β BG(R + ,E) , μ, γ a , γ b and γ c are presented in the following theorem.
As an example we will prove only the second inequality. Namely, let us fix arbitrarily ε > 0 and let T > 0 be such that b T (X ) ≤ b ∞ (X )+ε. Next, let x, y ∈ convX and x − y T ≤ ε. Then we obtain to obtain the inequalities (35). We omit the simple calculations. Now, let us fix ε > 0 and choose T > 0 such that It has been shown (see [14]) that β G( Now, let u a i be a function belonging to the space BG(R + , E) formed by the function u i by extension on the interval (T , ∞) using vector θ . Then the functions {u a i : i = 1, . . . , n} form μ T (X ) + a ∞ (X ) + 2ε-net of the set X in the space BG(R + , E). Indeed, let us fix arbitrarily x ∈ X and let i be such that Then, for t > T we get Letting T → ∞ and ε → 0 we get the first of inequalities (36). Now, we prove the second of the inequalities (36). Let us extend each function u i on the interval (T , ∞) using vector u i (T ). We denote such a function by u b To do this let us take an arbitrary x ∈ X and let i be such that (38) holds. Since We omit similar proof of the last inequality from (36).

Remark 3.11
Below is given an example showing that the estimations from Theorem 3.10 are optimal. Moreover, let us notice that it is not possible to majorize any of the measures γ a , γ b , γ c by β BG(R + ,E) or μ. The lack of such an estimation is a consequence of the sharp inclusions ker γ a , ker γ b , ker γ c N BG(R + ,E) .

Example 3.12
Let us put E := R. We define two sets X , Y ⊂ BG(R + , R) as follows It is not difficult to show that Now let dim E = ∞ and we put We can easily show that Applying (39), (40) and (41) we obtain

Superposition operator in BG(R + , E)
The properties of the superposition operator in the space G([0, T ], E) of regulated functions on bounded interval have been studied by several authors (for instance [2,[5][6][7]13,15]). However, there are not any results concerning this operator acting in the space of regulated functions on unbounded interval. The results obtained in this section fill the gap. Let us consider a function f : R + × E → E. Then, for every function x : R + → E, we can define the function (F f x)(t) := f (t, x(t)), t ∈ R + . Operator F f defined in such a way is called the superposition (or Nemytskii) operator generated by the function f (see [1]).
In connection with the space BG(R + , E), the natural question appears: what properties must the function f satisfy so that operator F f maps the space BG(R + , E) into itself?
A further part of this section is based on the concepts contained in the paper [15]. Inspired by the result obtained in the paper [13], concerning the space G([0, T ], E), we have the following theorem.

Theorem 4.1 The superposition operator F f maps BG(R + , E) into itself if and only if the function f has the following properties:
Proof As it has been shown in [13], the conditions (a) and (b) are equivalent to the fact that for each regulated function x : R + → E (unnecessarily bounded) the function F f x is regulated on R + , but unnecessarily bounded. If the operator F f maps BG(R + , E) into itself, then the condition (c) has to be satisfied, since in the opposite case the function x ∈ BG(R + , E) such that F f x / ∈ BG(R + , E) could be constructed. Conversely, if the conditions (a)-(c) hold, then for x ∈ BG(R + , E) the function F f x is regulated on R + and consequently bounded on bounded intervals. Using this fact together with (c) we infer that F f x is bounded on R + .

Using the notation
the condition (b) from Theorem 4.1 can be described as follows Similarly, we can describe the condition (a) of Theorem 4.1-we omit the details. Let us denote by E E the linear space consisted of all, not necessarily continuous functions h : E → E. In this space we introduce uniform compact convergence, i.e. for a given family {h s } s∈R + ⊂ E E , the uniform compact convergence {h s } to h ∈ E E with s → t ∈ R + means uniform convergence h s to h on all compact subsets in E when s → t.
Furthermore, for a fixed function f : R + × E → E and for each t ∈ R + we shall denote Hence, we have f t ∈ E E for t ∈ R + . Now we can formulate a theorem that gives (in terms of the function f t ) necessary conditions for any Banach space E and the sufficient ones, when dim E < ∞ such that the superposition operator F f maps BG(R + , E) into itself.

Theorem 4.2 Let the superposition operator F f maps BG(R + , E) into itself, where E is a Banach space. Then the family of functions { f t } t∈R + ⊂ E E satisfies the following conditions:
(a) The mapping R + t → f t ∈ E E is a regulated function. space and conditions (a), (b) and (c) are satisfied, then the superposition operator F f maps BG(R + , E) into itself.

Conversely, if additionally E is a finitely dimensional Banach
Proof (⇒) Let us fix t ∈ R + . Using condition (b) of Theorem 4.1 and based on notation (42) we have the following equality In the sequel we will prove (b1). To do this, let us take an arbitrary x ∈ E and ε > 0. Keeping in mind (43) we infer that there are δ > 0 and τ > 0 such that Now, letting s → t + we have g t (x) − g t (v) ≤ ε which proves continuity of g t in x and thereby on E. Reasoning similarly, we can prove (b2). Now, we will prove (a), i.e. that f s tends to g t in the compact convergence topology on E when s → t + . Let us fix a non-empty and compact set K ⊂ E and ε > 0. Then, in view of (43) and already proven continuity of g t , we have that for each x ∈ K there exist δ x > 0, τ x > 0 such that concurrently and From the family {B E (x, δ x )} x∈K covering compact set K we choose a finite subcover Let us denote τ := min{τ x i : i = 1, . . . , n} and fix arbitrary v ∈ K . Then there exists i such that v ∈ B E (x i , δ x i ). Hence, for s ∈ (t, t + τ ) taking into account (44) and (45) we infer that the following estimation holds for any v ∈ K . In other words, we have uniformly convergence on K . Similarly, we can prove the existence of the limit lim s→t − f s in the topology of compact convergence. The condition (c) is a consequence of the Theorem 4.1. Indeed, the condition (c) from Theorem 4.2 is equivalent to (c) from Theorem 4.1.
(⇐) Let us assume that dim E < ∞ and fix t ∈ R + . Condition (a) implies the existence of the limit g t := lim s→t + f s which, based on (b) is continuous on E. Let us fix x ∈ E and ε > 0. Continuity of g t means that for some r > 0 we get Moreover, (a) implies that for a given compact set B E (x, r ) there is a number τ > 0 such that Combining it with (46), for v ∈ B E (x, r ), s ∈ (t, t + τ ) we derive the following estimate i.e. condition (43) is satisfied and thereby (b) in Theorem 4.1 holds. Similarly, we can prove (a) in Theorem 4.1, so actually F f maps BG(R + , E) into BG(R + , E).

Corollary 4.3 If a Banach space E is finitely dimensional then the superposition operator F f maps BG(R + , E) into BG(R + , E) if and only if three conditions (a), (b) and (c) in
Conversely, if the conditions (H1)-(H3) are satisfied and E is a Banach space then the formula (48) gives such a function f (t, x), that operator F f : BG(R + , E) → BG(R + , E) and it is continuous and compact.

Remark 4.5
Obviously the case when all h n functions in the previous theorem are equal to θ , that is when f (t, x) = g(t), or only a finite number of them is not equal to θ is also allowed.
The proof of Theorem 4.4 will be preceded by two lemmas. However, in order to make our considerations more transparent, we will give a useful notation. For fixed x ∈ BG(R + , E) we will put supp x := {t ∈ R + : x(t) = θ }.
In contrast to standard definition of a support we do not require the closure. Obviously u n ∈ BG(R + , E). For n = m we have This means that the sequence {F f u n } is ε 0 -separable, hence F f is not compact, which ends the proof.
We remind that for arbitrary u ∈ E, the symbol u denotes the function u : R + → E given by u(t) ≡ u, t ∈ R + .

Lemma 4.7 If E is a separable Banach space and F f : BG(R + , E) → BG(R + , E) is compact and continuous, then there exists a sequence T = {t
Proof Let A = {a n : n ∈ N} ⊂ E be a countable dense subset of E. Let us put T := ∪ ∞ n=1 supp(F f a n − F f θ). Keeping in mind Lemma 4.6 we conclude that the set T is countable or finite. If there existed x ∈ E such that (50) did not hold, then there would exist s ∈ R + such that s ∈ supp(F f x − F f θ) \ T . Thus (F f x − F f θ)(s) = ε 0 for some ε 0 > 0 and additionally (F f a n − F f θ)(s) = θ for n ∈ N. If we took such a subsequence {a k n } that a k n → x in E, we would have which is in contradiction to the continuity of F f .

Proof of Theorem 4.4 (⇒)
First let us assume that E is a separable Banach space and F f : BG(R + , E) → BG(R + , E) is compact and continuous. Let T = {t n } be the sequence such as in Lemma 4.7. We define the function g : R + → E and the sequence of functions h n : E → E, n = 1, 2, . . . by the following formulas By Lemma 4.7, for each x ∈ E the mapping R + t → f (t, x) − f (t, θ) can be non-zero only on the set T and its formula is given by ∞ n=1 1 t n (t)h n (x). Therefore Since F f is continuous, the functions h n must be also continuous. Moreover, since F f is compact, the functions h n must also be compact. Further, in virtue of Lemma 4.6 we infer that lim n→∞ ( f (t n , x) − f (t n , θ)) = θ for x ∈ E, i.e.
We have only (47) left to be proved. Suppose the contrary. Then there would exist r > 0, a number ε 0 > 0 and such a subsequence of the sequence {h n }, (also denoted by {h n }), that Let k 1 = 1. We choose x 1 ∈ B E (r ) such that h k 1 (x 1 ) ≥ ε 0 2 . In view of (52) we obtain that there exists k 2 ∈ N such that k 2 > k 1 and h i (x 1 ) ≤ ε 0 4 for i ≥ k 2 . Invoking (53) we conclude that there exists x 2 ∈ B E (r ) such that h k 2 (x 2 ) ≥ ε 0 2 . Further, by (52) we know that there exists k 3 ∈ N such that k 3 > k 2 and h i (x 2 ) ≤ ε 0 4 for i ≥ k 3 . Continuing this reasoning we obtain a strictly increasing sequence {k n } ⊂ N and the sequence {x n } ⊂ B E (r ) such that Let n > m. Then we have i.e. {F f x n } is positively separated and F f is not relatively compact which contradicts the assumptions.
(⇐) Now, let us assume that the conditions (H1)-(H3) hold, E is a Banach space and the function f (t, x) is given by the formula (48). First, we are going to show that F f : E). To do this, let us fix x ∈ BG(R + , E), t ∈ R + and the sequence {t j } convergent to t from one side, e.g. t j → t + . Since the sequence {x(t j )} is bounded, then using (47) we get lim j→∞ ∞ n=1 1 t n (t j )h n (x(t j )) = θ and in view of (48) we have (F f x)(t + ) = lim n→∞ (F f x)(t n ) = lim n→∞ g(t n ) = g(t + ). Similarly we prove that (F f x)(t − ) = g(t − ) i.e. F f x is regulated on R + . Moreover, in virtue of (H1) and (H3) F f x is bounded and therefore F f x ∈ BG(R + , E).
In the next step we prove that the operator F f is compact. Let us fix a bounded sequence {x n } ⊂ B BG(R + ,E) (r ), where r > 0 is a number. Since the operator h 1 is compact, we are able to choose such a subsequence {x n,1 } of a sequence {x n } that the sequence {h 1 (x n,1 (t 1 ))} is convergent. Further, the operator h 2 is also compact so we are able to choose a subsequence {x n,2 } such that the sequence {h 2 (x n,2 (t 2 ))} is convergent. Moreover, {h 1 (x n,2 (t 1 ))} is also convergent. Continuing this procedure we have a sequence of sequences {x n,i } ∞ n=1 , i = 1, 2, . . . for which {x n,i+1 } ∞ n=1 is a subsequence of the sequence {x n,i } ∞ n=1 and there exists a limit lim n→∞ h j (x n,i (t j )) for i = 1, 2, . . . , j = 1, 2, . . . , i. Next, using a diagonal method we define the sequence {y n } by y n := x n,n , n = 1, 2, . . .. Obviously {y n } is a subsequence of {x n }. Moreover, the limit lim n→∞ h i (y n (t i )) exists for each i = 1, 2, . . .. This fact together with (F f y n )(t) = g(t) for t / ∈ T implies that the limit lim n→∞ (F f y n )(t) exists for each t ∈ R + . We are able now to introduce the function z : R + → E defined by the formula z(t) := lim n→∞ (F f y n )(t), i.e.
Keeping in mind (47) we get This implies that z is regulated on R + , Moreover, in view of (H1) and (H3) the function z belongs to the space BG(R + , E). Using (48) we obtain Let us fix ε > 0. In virtue of (47) we conclude that there exists m ∈ N such that for i > m we have h i (B E (r )) ≤ ε and therefore sup i>m lim n→∞ h i (y n (t i )) − h i (y k (t i )) ≤ 2ε. Since there exists a limit lim n→∞ h i (y n (t i )), i = 1, 2, . . ., there is k 0 ∈ N such that for k ≥ k 0 and for i = 1, 2, . . . , m we have Combining the above obtained estimation and (54) we get for k ≥ k 0 the inequality z − F f y k ∞ ≤ 2ε which proves that F f y k → z in BG(R + , E) for k → ∞. Continuity of the operator F f is a consequence of condition (H3) and the equalities (47) and (48)-we omit a simple proof of this fact.
for some function g ∈ BG(R + , E).

An application
As an example of application we consider equation x(t) = f (t, x(t)) + ∞ 0 u(t, s, x(s))ds, t ≥ 0.
We will assume that the functions involved in Eq. (55) satisfy the following conditions: (i) the function f : R + × E → E has the following properties: (i1) f is bounded on R + × B E (r ) for r > 0, (i2) there is a constant q ≥ 0 such that for t ∈ R + , x 1 , x 2 ∈ E, (i3) the family of functions { f (·, x) : x ∈ B E (r )} is equiregulated on R + for all r > 0, (ii) the mapping u : R + × R + × E → E has the following properties: (ii1) the function u(t, ·, x(·)) is integrable on R + for any t ∈ R + and for all x ∈ BG(R + , E), (ii2) u is bounded on R + × R + × B E (r ) for all r > 0, (ii3) the mapping u(t, s, x) is continuous with respect to the variable x, (ii4) there exist functions: φ ∈ BG(R + , R + ), g ∈ L 1 (R + , R + ) and nondecreasing function h : R + → R + such that for all t 1 , t 2 , s ∈ R + , x ∈ E, (ii5) the family of functions {u(t, ·, x) : t ∈ R + , x ∈ B E (r )} is equiregulated on R + for all r > 0, (ii6) there is a function k : R + × R + → R + , integrable with respect to s on R + for all t ∈ R + , such that for all t, s ≥ 0, A ⊂ E, A is bounded, (ii7) there are functions: k 1 : R + × R + → R + , integrable with respect to s on R + for all t ∈ R + , and nondecreasing ψ : R + → R + such that for all t, s ∈ R + , x ∈ E, and moreover For our further purposes we will need following fixed point theorem [3].
Theorem 5.1 Let Q be a nonempty bounded closed convex subset of the Banach space E and let S : Q → Q be continuous and such that μ(S X) ≤ qμ(X ) for any nonempty subset X of Q, where q is a constant, q ∈ [0, 1) and μ is a measure of noncompactness. Then S has a fixed point in the set Q.
Now we formulate our theorem as: Keeping in mind the assumptions (ii2), we have U < ∞.
Thus, joining the above inequality with (60) we infer that γ c (S X) ≤ q + sup t∈R + ∞ 0 k(t, s)ds γ c (X ), and taking into account Theorem 5.1 we derive that S has at least one fixed point in the ball B r ⊂ BG(R + , E).
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.