Some inequalities and superposition operator in the space of regulated functions

Abstract: Some inequalities connected to measures of noncompactness in the space of regulated function R(J, E) were proved in the paper. The inequalities are analogous of well known estimations for Hausdor measure and the space of continuous functions.Moreover two su cient and necessary conditions that superposition operator (Nemytskii operator) can act from R(J, E) into R(J, E) are presented. Additionally, su cient and necessary conditions that superposition operator Ff : R(J, E)→ R(J, E) was compact are given.


Introduction
When studying solvability of various non-linear equations, it is signi cant to properly choose the space in which the equation is considered. Knowledge about some properties of the space e.g. easy to calculate formulas for measures of noncompactness or characteristic of superposition operator etc. combined with xed point theorems allow to obtain general conditions for solvability of studied equations.
In the third chapter several theorems dealing with various types of inequalities, integral, but not only integral, that hold in the spaces of regulated functions and are expressed in terms of measures of noncompactness, will be formulated. These inequalities are analogues of known and often used inequalities holding in the class of continuous functions. In the fourth chapter two theorems (Theorem 4.4 and Theorem 4.6) that give su cient and necessary conditions that superposition operator acted from the space of regulated functions into that space, will be presented. Known so far results in this area usually give only su cient conditions and the only known su cient and necessary conditions are actually rather "tautological". Moreover, sucient and necessary conditions that superposition operator is compact in the space of regulated functions will be given (Theorem 4.9).

Notation, de nitions and auxiliary facts
This section is focuses on recalling some facts which will be used in our 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 ball B E (θ, r) will be denoted by B E (r). We write X, ConvX to denote the closure and the convex closure of a set X, respectively. The symbol X will stand for the norm of the set X ⊂ E i.e., we have X := sup{ x : x ∈ X}. Furthermore, 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. We accept the following de nition of a measure of noncompactness [15].  Subsequently, we will use measures of noncompactness having some additional properties. Namely, a measure µ is said to be sublinear if it satis es the following two conditions: A sublinear measure of noncompactness µ satisfying the condition (strong maximum property) o µ(X ∪ Y) = max{µ(X), µ(Y)} and such that kerµ = N E is said to be regular. Except condition o we can also consider the condition (weak maximum property) For a given nonempty bounded subset X of E, we denote by β E (X) the so-called Hausdor measure of noncompactness of X. This quantity is de ned by formula β E (X) := {r > : X has a nite r−net in E}.
The function β E is an example of regular measure of noncompactness in E. Now we recall some facts concerning regulated functions.

De nition 2.2. A function
where E is a topological vector space, is said to be a regulated function if for every t ∈ [a, b) the right-sided limit x(t + ) := lim s→t + x(s) exists and for every t ∈ (a, b] the leftsided limit x(t − ) := lim s→t − x(s) exists.
From now on, real Banach space will be denoted by E. Denote by R(J, E) the space consisting of all regulated functions de ned on the interval J = [a, b] with values in a real Banach space E. Since every regulated function x ∈ R(J, E) is bounded on the interval J, then the space R(J, E) can be normed via the classical supremum norm ||x||∞ := sup{||x(t)|| : t ∈ J}.
It is easy to show that R(J, E) is a real Banach space. Moreover, every regulated function x : J → E is Riemann integrable. Now, we remind a criterion for relative compactness in the space R(J, E). To this end, we introduce the concept of a equiregulated subset of the space R(J, E) (cf. [3,9]).

De nition 2.3.
We will say that the set X ⊂ R(J, E) is equiregulated on the interval J if the following two conditions are satis ed: Theorem 2.4. [3][4][5]9] A nonempty subset X ⊂ R(J, E) is relatively compact in R(J, E) if and only if X is equiregulated on the interval J and the sets X(t) are relatively compact in E for t ∈ J. Now we are going to recall the construction of a measure of noncompactness in the space R(J, E). To this end, let us take a set X ∈ M R(J,E) . For x ∈ X and ε > 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. Furthermore, let us put: Finally, let us de ne the following quantity µm(X) := max{ω − (X), ω + (X)} + sup t∈J β E (X(t)). Remark 2.6. Above construction of the measure (2.1) addresses inaccuracies existent in the construction of measures given in [3,4].

Inequalities including measures of noncompactness
This section we start with the proof of inequality which analogue for equicontinuous family of continuous functions is often used in studying solvability of nonlinear equations. For a xed nonempty subset X ⊂ R(J, E), let us put Theorem 3.1. Let X ⊂ R(J, E) be nonempty, bounded and equiregulated. Then the function J t → β E (X(t)) ∈ R+ is regulated and the following inequality holds Proof. Let us x ε > . The condition of X being equiregulated implies that From this nite subcover we can choose following subcover such that a = s < s < ... < s k = b and additionally every two consecutive intervals from (3.3) have nonempty intersection. Now we choose one point s i belonging to each of those k intersections, i.e. we have the sequence of points For simplicity let us denote them by From (3.1) and (3.2) we yield that which means that the function J t → β E (X(t)) is regulated, thus Riemann integrable. Let us choose arbitrary Hence and by (3.4) we have Let us set functions y j ,...,jn ∈ R(J, E), j i ∈ { , ..., m i }, i = , ..., n by formulas We prove that the set of vectors given by b a y j ,...,jn (t)dt for Because of (3.7) there exists such a sequence j , ..., jn, that Then, using (3.8) and (3.
which for ε → proves the theorem.
Without the assumption about X ⊂ R(J, E) being equiregulated, the function J t → β E (X(t)) does not have to be measurable in Lebesgue sense. However, for countable subsets of the space R(J, E) we have (see [16]). Remark 3.3. Above theorem is also true given weaker assumption that functions xn are strongly measurable [16]. The example from [17] shows that factor from the above theorem cannot be replaced by smaller even for the sequence {xn} of regulated functions.
In some applications of measures of noncompactness the following lemma can be useful.

Lemma 3.4. [18] If E is a Banach then for each non-empty and bounded set X
One can ask what can be an analogue of this lemma for measure µm and space R(J, E) ? The answer is given in two following theorems.
In the general case the assertion of the previous theorem has to be weakened. Theorem 3.6. For each non-empty and bounded set X ⊂ R(J, E) and any ε > , there exists such countable set X ⊂ X, that µm(X) ≤ µm(X ) + ε. In the above estimation factor cannot be replaced by smaller (see Example 3.7).
Let us arbitrarily x ε > . There exists such a number t ∈ J that sup t∈J β E (X(t)) < β E (X(t )) + ε. Using Lemma 3.4 we get that there exists such a sequence {xn} ⊂ X that Let X :=X ∪ {xn : n ∈ N}. Obviously X is countable and Hence By A we denote the indicator function of a subset A. When the subset A = {a} is singleton, we will write a. Additionally, for arbitrary u ∈ E let u denotes the function u : J → E given by means that µm(X ) = and that proves that factor from the above theorem cannot be replaced by smaller.

Superposition operator
Consider a function f : J × E → E. Then, to every function x : J → E, we may assign the function (F f x)(t) := f (t, x(t)), t ∈ J. Operator F f de ned in such way is said to be superposition (or Nemytskii) operator generated by the function f (see [19][20][21]). In connection with the space R(J, E), the natural question appears: what properties must the function f satisfy in order for operator F f to map the space R(J, E) into itself?
In the paper by Aziz [2] and Michalak [12] the following results were obtained.

Theorem 4.1. [12] A superposition operator F f maps R(J, E) into itself if and only if the function f has the following properties: (1) the limit lim [a,t)×E (s,v)→(t,x)
f (s, v) exists for every (t, x) ∈ (a, b] × E, Given the notation g t (x) := lim condition (2) of Theorem 4.1 can be written in an equivalent form using quanti ers Analogically condition (1) of Theorem 4.1 can be written -we omit the details.

Theorem 4.3. [12] A superposition operator F f maps R(J, E) into itself is continuous if and only if a functioñ
Thus we have f t ∈ E E for t ∈ J. Now we can formulate a theorem that gives (in terms of the function f t ) necessary conditions for any space E and su cient ones when dim E < ∞ such that superposition operator F f maps R(J, E) into itself.

Conversely, if additionally E is a nitely dimensional Banach space and conditions (a) and (b) are satis ed then the superposition operator F f maps R(J, E) into itself.
Proof. (⇒) Let us x t ∈ [a, b). Using condition (2) of Theorem 4.1 and based on notation (4.1) we have the following equality g t (x) = lim s→t + f s (x), x ∈ E. First we prove (b1). Let us x x ∈ E and ε > . Because of (4.2) we have the existence of δ > and τ > such that Now going from s → t + we have g t (x) − g t (v) ≤ ε which proves continuity of g t in x and thereby on E.
We will now prove (a) i.e. that f s converges to g t almost uniformly on E when s → t + . Let us x non-empty and compact set K ⊂ E and ε > . Then, because of (4.2) and already proven continuity of g t , we have that for each x ∈ K there exist δx > , τx > such that concurrently and Out of family {B E (x, δx)} x∈K covering compact set K we choose a nite subcover {B E (x i , δx i )} n i= . Let τ := min{τ i : i = , ..., n}. Let us x arbitrary v ∈ K. Then there exists such i that v ∈ B E (x i , δ i ). Thus for s ∈ (t, t + τ) based on (4.3) and (4.4) we have the following estimation for any v ∈ K, i.e. we have uniform convergence on K. Similarly we can prove the existence of the limit lim s→t − f s in the topology of almost uniform convergence. (⇐) Assume that E has a nite dimension and x t ∈ [a, b). Condition (a) assures the existence of the limit g t := lim s→t + f s which, based on (b) is continuous on E. Let us x x ∈ E and ε > . Continuity of g t means that for some r > we have Moreover (a) implies that for a compact set B E (x, r) there exists τ > such that When we combine it with (4.5), for v ∈ B E (x, r), s ∈ (t, t + τ) we have i.e. condition (4.2) is satis ed and thereby (2) in Theorem 4.1 holds. Similarly we can prove (1) in Theorem 4.1, so actually F f acts from R(J, E) into R(J, E).

Corollary 4.5. If E is a nitely dimensional Banach space then superposition operator F f acts from R(J, E) into R(J, E) if and only if both conditions (a) and (b) in Theorem 4.4 are satis ed.
Now we give further su cient and necessary conditions that superposition operator F f acts from R(J, E) into R(J, E).
To this end, let us recall so-called module of continuity of a mapping h : E → E at a point v ∈ E given by For any subset S ⊂ J, in the space R(J, E) we will use a pseudonorm · S given by Now we can give another su cient and necessary criterion that superposition operator F f acts from R(J, E) into R(J, E). ν(f tn , v, δ) = .
(d) The mapping E u → f (·, u) ∈ R(J, E) is continuous in every point v ∈ E in regard to pseudonorm · J\Dv , i.e. for each v ∈ E and each sequence vn → v we have f (·, v) − f (·, vn) J\Dv → when n → ∞.
Before we prove Theorem 4.6 we give technical lemma, necessary in the next part of the paper. Proof. We give a proof by contradiction. Let us assume that F f acts from R(J, E) into R(J, E). Hence we have Let assume that {tn} is convergent to some t ∈ J from one side, for example tn → t + , and moreover {tn} is strictly decreasing (we can have that choosing a proper subsequence). Let us put Now let us de ne next function z ∈ R(J, E) as follows for t ∈ (t n , t n− ], n = , , ..., vn for t ∈ (t n+ , t n ], n = , , ... . However it is in contradiction with (4.10).
Proof of Theorem 4.6 (⇒) Take an arbitrary v ∈ E and put x(t) ≡ v, t ∈ J. Since F f x ∈ R(J, E), then (a) is satis ed. Let us assume that the set Dv is uncountable. Since   such that (4.14) Conversely, if the conditions (H1)-(H3) are satis ed and E is a Banach space then the formula (4.14) gives such a function f (t, x), that operator F f : R(J, E) → R(J, E) and it is continuous and compact.
Remark 4.10. Obviously the case when all hn functions in the previous theorem are equal to θ, that is when f (t, x) = g(t), or only a nite number of them is not equal to θ is also allowed.
The proof of the theorem will be preceded by two lemmas. Before that however we will give a useful notation.
In contrast to standard de nition of a support we do not require the closure.  Proof. Let A = {an : n ∈ N} ⊂ E be a countable dense subset of E. Let us put T := ∪ ∞ n= supp(F f an − F f θ). By the previous Lemma 4.11, the set T is countable or nite. If there existed x ∈ E such that (4.16) did not hold then there would exist s ∈ J such that s ∈ supp(F f x − F f θ) \ T. Thus (F f x − F f θ)(s) = ε for some ε > and additionally (F f an − F f θ)(s) = θ for n ∈ N. If we took such a subsequence {a kn }, that a kn → x in E we would have which is in contradiction with continuity F f . Combining the above and (4.19) we get for n ≥ n the inequality z − F f yn ∞ ≤ ε which proves that F f yn → z in R(J, E). Continuity of the operator F f is a consequence of condition (H3) -we omit a simple proof of this fact. Problem 4.14. Are the conditions (H1)-(H3) of Theorem 4.9 necessary in case when the space E is not separable?