Relaxation Theorem for Stieltjes Differential Inclusions on Infinite Intervals

For a very general first-order differential problem on an infinite-time horizon involving the Stieltjes derivative with respect to a left-continuous non-decreasing function g:[0,∞)→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$g:[0,\infty )\rightarrow \mathbb {R}$$\end{document}1xg′(t)∈F(t,x(t)),t∈[0,∞)x(0)=x0,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \left\{ \begin{array}{l} x'_g(t) \in F(t,x(t)),\; t\in [0,\infty )\\ x(0)=x_0, \end{array} \right. \end{aligned}$$\end{document}we study the possibility to approximate the solutions of the convexified inclusion by the solutions of the non-convexified problem. Via a generalization to this framework of a classical result concerning continuous selection of trajectory, we thus present a relaxation theorem which states that, similarly to the setting of usual differential inclusions, the approximation can be achieved once we allow to the initial value to differ (but remaining close to) from the initial value of the considered solution of the relaxed inclusion.

V. Marraffa and B. Satco MJOM on a compact interval, any solution of the second one can be approximated in the uniform topology with a solution of the first one when F is Lipschitz with respect to the state.In the classical case, such results can be found, e.g., in [2] (see also the references therein).Due to the huge importance in Optimal Control, in Stability Theory for hybrid systems among other fields, they were then generalized in many directions (e.g., [14,19,25,32]).
The study of the same problem with infinite-time horizon has been addressed, as far as the authors know, only for classical differential inclusions in [20] and [10].
On the other hand, differential problems with Stieltjes derivative [30] replacing the usual derivative offer a powerful tool for studying dynamics of systems with mixed continuous and discrete behavior.
This theory got an increasing attention in the last decade (see [18,27,31,35] and the references therein), since it has proved itself useful in modeling the dynamics of various systems where instantaneous changes (occurring at the discontinuity points of g) and stationary intervals of time (related to intervals where g is constant) are involved, such as in [18,23] or [31]; it is related to the theory of measure differential problems [12,13,15,28,30] and it is worthwhile to mention that this setting covers usual differential problems (for an absolutely continuous map g), discrete problems (when g is a sum of step functions), as well as impulsive equations (for g being a combination of these two types of maps).
In the present paper, via an extension to the setting of differential inclusions with Stieltjes derivative of a classical result on continuous selection of solutions of (2) ([9, Theorem 3.1]) and following the steps of proof proposed in [20], we get a relaxation theorem for this very wide framework on an infinitetime interval.The main tool is a Filippov-Ważewski result for g-differential inclusions recently published [25,Theorem 12].
Like in the classical case where g(t) = t, we cannot hope to find a solution x approximating z with the same initial value (see the counterexample in [20, Section 3]), as in the case of finite time intervals.Also, since it was proved [25,Proposition 13] that, on compact intervals, the closure in the uniform norm topology of the solution set of it was expected that, even on infinite-time intervals, at the discontinuity points of g, it is not necessary to consider the closed convex hull of the values of F .
Our motivation is coming from the large number of applications of the relaxation results regarding the stability properties of differential inclusions [1], of hybrid systems [4,5,33] or in control theory [33,39].
Moreover, since dynamic inclusions on time scales [8,15,37] can also be seen as measure differential inclusions and, therefore, as Stieltjes differential problems, it is possible to deduce a new relaxation result on [0, ∞) for dynamic problems on time scales.The same can be said about generalized differential inclusions [22,29,34,36,38].
Let us finally acknowledge that Stieltjes differential inclusions include, in particular, impulsive differential inclusions; thus, a new relaxation result on infinite-time intervals can be inferred for such problems (with possibly set-valued impulses), as in [28] or [26].

Notations and Preliminaries
Let I ⊂ [0, ∞) be an interval containing 0 and g : I → R be a non-decreasing left-continuous function, which on any compact interval has at most a finite number of accumulation points of its discontinuity points.Without any loss of generality, we may suppose that g(0) = 0 and denote by where g(t+) stands for the limit at the right (which is well defined, since g has bounded variation).
The g-topology τ g on I is the topology having as open balls {s ∈ I; |g(t) − g(s)| < r}, r > 0 (see [18]).The dimension of the Euclidean space The symbol L g (I) will denote the σ-algebra of g-measurable sets [18], while the Lebesgue-Stieltjes integrability of an R d -valued function on I means the abstract Lebesgue integrability w.r.t. the Stieltjes measure μ g generated by g.The space of such functions LS-integrable w.r.t.g will be denoted by L 1 g (I, R d ) and its norm by Let us now recall the notion of differentiability related to Stieltjes type integrals introduced in [30] (motivated by [40]). V.
whenever the limit exists.
The set contains the discontinuities of g and notice that if t ∈ D g , then the gderivative f g (t) is well defined if and only if the right limit f (t+) exists and then Definition 2.1 has no meaning whenever t belongs to , see [30].Moreover, we observe that it is possible to define the g-derivative at the points of C g as well, as in [16,Definition 3.7] (in a way which is coherent with the previous definition), thus allowing one to study higher order Stieltjes differential equations.
By elementary properties of the Lebesgue-Stieltjes integrals, if f ∈ L 1 g (I, R d ), then the primitive [0,•) f (τ )dg(τ ) is g-absolutely continuous (see [30]), that is: a function u : I → R d is g-absolutely continuous if, for every ε > 0, there is δ ε > 0, such that for any set {(t j , t j )} m i=1 of non-overlapping subintervals of It was proved in [30,Proposition 5.3] that any g-absolutely continuous function has bounded variation; it is left-continuous on I and continuous at every continuity point of g.
By AC g (I, R d ), we mean the space of g-absolutely continuous maps from I to R d endowed with the (Banach space) norm Remark that g-absolutely continuous functions are obviously uniformly gcontinuous, i.e., for every ε > 0, there is δ ε > 0, such that for t , t ∈ I Moreover, if a function defined on a bounded interval is uniformly g-continuous, it is bounded ([25, Lemma 2]).
The Stieltjes integrals and the Stieltjes derivative are tightly connected by Fundamental Theorems of Calculus.
(b) ([30, Theorem 5.4], see also [18,Theorem 5.1]) Let T > 0. If f : [0, T ] → R d is g-absolutely continuous, then F g exists μ g -a.e., and We recall now some basic notions of set-valued analysis ( [2,3,6], see also [11]).If X and Y be two Banach spaces, by P(X), we denote the space of non-empty closed subsets of (X, • X ) and by P k (X) its subspace consisting in the compact subsets; it is a complete metric space when endowed with the Hausdorff-Pompeiu distance, D, that is where the excess e(A, B) of A ∈ P k (X) over B ∈ P k (X) is defined as For A ∈ P k (X), let coA denote its closed convex hull and |A| = D(A, {0}).
A multifunction Γ : In [25, Lemma 2], a g-uniformly continuous function is showed to be bounded (for an interesting characterization of g-uniform continuity, see [17]).Following the idea of proof of [25, Lemma 2], we get the boundedness of a jointly uniformly continuous multifunction.
in the product topology of τ g with the usual topology of R d (both generated by pseudometrics), then it is bounded.
It is easy to see that there are only a finite number of discontinuity points of g where To simplify the writing, let τ 0 = 0 and Also, the compact Q can be covered by a finite number of balls B(x l , δ 1 ), l = 1, . . ., N 2 .Then, for every t ∈ [0, T ]\{τ i ; i = 1, . . ., N 1 }, one can find i ∈ {0, . . ., N 1 }, j ∈ {1, . . ., N i }, such that t ∈ I i j and for every x ∈ Q, one can find l ∈ {1, . . ., N 2 }, such that x ∈ B(x l , δ 1 ), whence In conclusion, if we denote by

Main Results
The first two results below allow us to foresee some of the difficulties we are facing when the Stieltjes derivative is involved.Proposition 3.1.([24, Proposition 2.5]) Let T > 0 and g : [0, T ] → R be nondecreasing and left-continuous.If t ∈ [0, T ] and f 1 and f 2 are two real-valued functions defined in a neighborhood of t, g-differentiable at t, then the product f 1 • f 2 is g-differentiable as well at this point and

Lemma 3.2. Let T > 0 and g : [0, T ] → R be non-decreasing and leftcontinuous. Then, for every non-negative function
Proof.We prove the assertion by mathematical induction.By Proposition 3.1 so it is valid for n = 1.Suppose now and let us prove that . Again, by Proposition 3.1 Solutions for multivalued Stieltjes differential problems are understood in the following sense: Existence results on a bounded interval I for such problems are already known (e.g., [7] for convex-valued right-hand side or [12,25,27] in the nonnecessarily convex setting).
We recall the following In the case I = [0, ∞), we can easily derive the following.
Proof.Consider an increasing sequence (T k ) k∈N of continuity points of g tending to ∞ with T 0 = 0. We start by applying proposition 3.4 to get a g-absolutely continuous solution x 0 on [T 0 , T 1 ] for the problem x(0) = x 0 .
Then, we apply again [25,Corollary 10] to get a g-absolutely continuous solution x 1 on [T 1 , T 2 ] for the problem MJOM and we repeat this procedure for each k ≥ 0 in order to get a g-absolutely continuous solution x k on [T k , T k+1 ] for Now, we concatenate these solutions to get a solution on [0, ∞).More precisely, the function is a solution of (1), since its g-derivative satisfies the inclusion at μ g -almost every point in [0, ∞)\{T k ; k ≥ 1} and the exceptional set is μ g -null due to the fact that any T k is a continuity point of g.
We also recall the following technical result: ) are continuous, and for every s ∈ R d , the set , and therefore, it admits continuous selections.
It can be easily shown that the previous proposition remains available for the Stieltjes measure μ g .We state it below in this setting.
, and therefore, it admits continuous selections.
We present next a result on continuous selection of trajectory which generalizes the classical [9, Theorem 3.1] to the setting of g-differential inclusions.
Then, for any continuous s → y(•, s) from R d to AC g ([0, T ], R d ), every β y satisfying (4) and ε > 0, one can find a function x Note that, by hypothesis (χ2), (χ4) is equivalent to: (χ4 ) there is β : Proof.The proof is inspired by that of [9,Theorem 3.1].By a change of variable if necessary (see [9, page 325]), we may suppose, for the sake of simplicity, that y(t, s) = 0 and to look for solutions with x(0) = 0.
Let (ε n ) n be a strictly increasing sequence of positive numbers convergent to ε (suppose without restricting the generality that ε 0 g(T ) < ε 1 ) and

s) .
Step I. We construct a sequence (x n ) n∈N : [0, T ] × R d → R d of approximate solutions, such that for every n ≥ 0, x n (0, s) = 0 and: Obviously, x 0 (0, s) = 0 and s e.}, and similarly As a consequence of hypothesis χ4), the multifunction G 0 has non-empty values.
Indeed, one can find a measurable selection and note that Next, consider x 1 (t, s) = [0,t) h 0 (s)(τ )dg(τ ) for which x 1 (0, s) = 0 and (x 1 ) g (t, s) ∈ F (t, x 0 (t, s), s) for μ g -almost every t hold; moreover, due to Theorem 2.2 Suppose we have defined x 0 , x 1 , . . ., x n with the properties (i)-(iii).Then From (iii), it follows using Theorem 2.2 and interchanging the order of integration that: By Lemma 3.2, we get and by Theorem 2.2 Therefore, (5) implies that e.} and By the same reason as before G n is non-empty valued, l.s.c. and decomposable and H n is non-empty valued; therefore, Proposition 3.7 yields that H n has a continuous selection, i.e., h n : Let It satisfies the conditions (i)-(iii); besides, by ( 7) whence, again by interchanging the order of integration and using Lemma 3.2 so the sequence (x n (•, s)) n is Cauchy in AC ; this holds uniformly in s in some neighborhood of some s 0 due to the continuity assumptions on L and β.
Step II.Let us define

It has the property that
Let us now check that t → x(t, s) is a solution of From ( 7), one deduces that ) n is a pointwisely Cauchy sequence, therefore pointwisely convergent to some y(t, s).The sequence satisfies the Lebesgue dominated convergence theorem as, for all n ∈ N, one can write and see that the term on the right-hand side is LS-integrable w.r.t.g, since •) L(ξ,s)dg(ξ) dg(u) are g-absolutely continuous (thus, bounded), while L(•, s) and β(s) are LS-integrable.Therefore
Looking at (8) x g (t, s) ∈ F (t, x(t, s), s), and the assertion is proved.Let us finally prove the inequality (c).By ( 6) The following was given in [20] when g(t) = t.
MJOM Corollary 3.9.Let T > 0 and F : Then, for each and for every ε > 0, there exists a map x : Proof.The assertion follows by applying Theorem 3.8 for F not depending on s and for y(t, s) = y(t) for every s ∈ R d (so, constant w.r.t.s), so that we can take β(s)(t) = 0 identically, because y is a solution of the above-mentioned problem.
We will combine it with a Relaxation Theorem for Stieltjes differential inclusions recently proved in [25].
Theorem 3.10.([25, Theorem 12]) Let T > 0, g : [0, T ] → R be left-continuous, non-decreasing, with the property that its discontinuity points accumulate only finitely many times and let in the product topology of τ g with the usual topology of R d (both generated by pseudometrics) on any set Then, for every and for every ε > 0, there exists a solution y : We proceed with two auxiliary results needed in the main theorem.Lemma 3.11.Let T > 0, ξ 0 ∈ R d and F : [0, T ] × R d → P(R d ) be given and consider on [0, T ] the problems and Suppose y : [0, T ] → R d is a solution of (10) and let ε > 0 and be the ε-tube around the image of y.
If F satisfies the assumption K1) and: (h2) One can find L > 0, such that for μ g -a.e.t ∈ [0, T ] and every ξ, η ∈ B(T , 1) Proof.Since y is g-absolutely continuous, it is bounded; therefore, B(T , 1) is bounded, and so, hypothesis K1) implies (using Lemma 2.3) that there exists α > 0, such that for ξ ∈ B(T , 1) In particular, F has compact values at any (t, ξ) ∈ [0, T ]×B(T , 1).To get the thesis, we combine Corollary 3.9 with the Relaxation Theorem 3.10.To this aim, we need to modify the function F to ensure that the Lipschitz property holds on the whole space R d .

MJOM
We prove now that (H2) holds.To this purpose, let t ∈ [0, T ] satisfy h2) (i.e., μ g -almost everywhere) and let x, y ∈ R d ; we have to consider the following cases: (i) if x, y ∈ B(T , 1) (iii) if x, y / ∈ B(T , 1) then D( F (t, x), F (t, y)) = D({0}, {0}) = 0. Hence, the Lipschitz condition is satisfied by F on the whole Note that F (t, x) and F (t, x) coincide for any (t, x) ∈ [0, T ] × T ; therefore, the solutions of ( 9), (10), and (11) are the same for the inclusions with F (t, x) instead of F (t, x).Now as for any t ∈ [0, T ], y(t) ∈ T , F (t, y(t)) = F (t, y(t)), and so, y is a solution of (10) with F instead of F in this case.
By Theorem 3.10, there exists a solution y of (9) with F instead of F which satisfies for all t ∈ [0, T ] However, as y(t) ∈ T , y is in fact a solution of (9).Now, we apply Corollary 3.9 with y and τ ) .There is a function Denoting by δ = ε 4e [0,T ] L(τ )dg (τ ) , we get for each η ∈ B(ξ 0 , δ) Therefore, for each η ∈ B(ξ 0 , δ) and all t ∈ [0, T ] This implies that for each η ∈ B(ξ 0 , δ), the solution x(•, η) lies in the tube T in which F and F coincide, so they are solutions of the original system.
It follows that the restriction of x to [0, T ] × B(ξ 0 , δ) satisfies the required conditions.
For the main result, we impose the hypotheses below on F : [0, ∞) × R d → P(R d ).
(i) F is uniformly continuous in (t, x) in the product topology of τ g with the usual topology of R d (both being generated by pseudometrics) on any set [0, T ] × Q with T > 0 and Q ⊂ R d compact; (ii) for every T > 0 and R > 0, one can find L T,R > 0, such that for any ξ, η ∈ R d with max(|ξ|, |η|) ≤ R: Let also r : [0, ∞) → R + be continuous and (T k ) k∈N increasingly tend to ∞ with T 0 = 0.
Then, there exists a nonincreasing sequence (δ k ) k of positive numbers and, for every k ∈ N, a sequence (η k j ) j≥1 , such that: ) for all k ∈ N and j ≥ 1; • for any k ≥ 1, if a subsequence of (η k j ) k≥1 , say (η k j l ) l≥1 , tends to some η k , then (η k−1 j l ) l≥1 tends to some η k−1 and there exists a solution x : Proof.We first remark that assumption i) implies (again by Lemma 2.3) that for every T, R > 0, there is α T,R > 0, such that For the sake of completeness, we list here all the steps, similar to those in the proof of [20,Lemma 3.2] For each k ≥ 1, we apply Lemma 3.11 to the problem and, respectively, First, let δ 0 = r 1 .By induction, for every k ≥ 1, the following will be constructed in the same way as in the proof of [20, Lemma 3.2]:

the following properties:
for any η, x k (•, η) is a solution of (13) with initial value x(0) = η; -for any t, x k (t,

Now let us concatenate conveniently these functions to construct (for each k) a map y
and define η 0 0 = z(T 0 ) η 0 1 = y 1 (T 1 ), η Proof.Consider an increasing sequence (T k ) k∈N of continuity points of g tending to ∞ with T 0 = 0. Let (δ k ) k and, for every k ∈ N, (η k j ) j≥1 be given by Lemma 3.12.The key fact in this point of the proof is that (η 0 j ) j≥1 ⊂ B(z(0), r 1 ) which is compact; therefore, it has a subsequence convergent to some η 0 ∈ B(z(0), r 1 ).The corresponding subsequence of (η 1 j ) j≥1 also has a convergent subsequence, since it is contained in the compact B(z(T 1 ), δ 1 ) and so on, and thus, the diagonal sequence converges to some η k ∈ B(z(T k ), δ k ).By Lemma 3.12, there exists a solution Then, the map x : [0, ∞) → R d defined as is a solution of ( 15) and satisfies, due to (12), Let us note that in the convexified problem in the main theorem, the convex hull is not necessary at the points of discontinuity of g since [25, Proposition 13] states that, on finite intervals, any limit of an uniformly convergent sequence of solutions of x g (t) ∈ F (t, x), x(0) = ξ is a solution of x g (t) ∈ coF (t, x(t)), t / ∈ D g , x(0) = ξ.F (t, x(t)), t ∈ D g .This is consistent with the statements of the relaxation results for difference inclusions [21], for set-valued problems for hybrid systems [4], and for dynamic inclusions on time scales [32].Remark 3.14.We cannot hope to get solutions of the non-convex problem which approximate a given solution of the relaxed inclusion with the same initial value (see the counterexample in [20,Section 4] in the particular case g(t) = t).