Properties of set-valued Young integrals and Young differential inclusions generated by sets of Hölder functions

Abstract

The present studies concern properties of set-valued Young integrals generated by families of $\beta$-Hölder functions and differential inclusions governed by such a type of integrals. These integrals differ from classical set-valued integrals of set-valued functions constructed in an Aumann’s sense. Integrals and inclusions considered in the manuscript contain as a particular case set-valued integrals and inclusions driven by a fractional Brownian motion. Our study is focused on topological properties of solutions to Young differential inclusions. In particular, we show that the set of all solutions is compact in the space of continuous functions. We also study its dependence on initial conditions as well as properties of reachable sets of solutions. The results obtained in the paper are finally applied to some optimality problems driven by Young differential inclusions. The properties of optimal solutions and their reachable sets are discussed.

1 Introduction

Differential inclusions employ on an approach that has been developed due to investigations of dynamic systems having velocities not uniquely determined by the state of the system. The theory of differential inclusions has been explored in connection with optimal control problems and it constitutes nowadays an important branch of set-valued analysis (see e.g. monographs [2, 3, 5, 12, 13, 19, 20, 38] and references therein). In particular, the existence of solutions and properties of solutions sets have been studed, see e.g., [8, 9, 37, 39]. Stochastic differential inclusions form a new direction and an extension of the differential inclusions theory. They generalize deterministic differential inclusions as well as single valued stochastic differential equations. Such investigations were initiated independently by Kisielewicz in [22], Hiai in [18] and continued by many authors, among others in [1, 4, 21,22,23, 25, 26, 28, 31, 34]. For a collected presentation of the theory we refer the reader to the monographs [21, 24].

On the other hand, the Young type integrals were introduced more than eight decades ago in [40]. They have been developed and widely used in the theory of differential equations with rough paths by many authors, see e.g., [7, 10, 14,15,16, 27, 29, 35] and references therein. The theory of rough paths establishes a convenient tool to extend stochastic calculus with respect to a large class of integrators out of range of semimartringales. Namely, one can consider a stochastic integration with respect to integrators such as the Mandelbrot fractional Brownian motion which has Hölder continuous sample paths [30]. As already mentioned, control and optimal control problems have inspired an intensive expansion of the differential and stochastic set-valued inclusions theory. Thus, it seems reasonable to investigate also differential inclusions driven by a fractional Brownian motion and Young type integrals as well. Recently, in [6, 11, 32, 33] the authors studied properties of set-valued Young type integrals and the existence of solutions of differential inclusions with such integrals. Once having conditions assuring the existence of solutions, it is natural to ask about their properties. To the best of our knowledge, this topic has not been investigated yet. The paper concerns both the properties of set-valued Young integrals and topological properties of solutions of Young differential inclusions. The latter constitutes the main novelty. In particular, we show that the set of all solutions is compact in the space of continuous functions. We also study its dependence on initial conditions as well as properties of reachable sets of solutions with respect to time. The results obtained in the paper are applied to a similar analysis concerning optimal solutions and their reachable sets. The continuity type dependence of optimal solutions is shown both with respect to initial conditions and functionals determining the optimality criterion.

The paper is organized as follows. In Sect. 2, we present basic notions and auxiliary properties concerning Hölder continuous functions and Young integrals in a single-valued case. Section 3 deals with the properties of set-valued Young type integrals which are based on the bounded and closed sets of integrands. We shall give a number of properties of such set-valued integrals. Some of them are natural set-valued generalizations of those known from the single-valued Young integral theory. In Sect. 4, we present the existence and main topological properties of the set of all solutions of Young differential inclusions and their reachable sets. The last section deals with applications to the analysis of sets of solutions of differential inclusions being optimal with respect to chosen functionals.

2 Basic notions and auxiliary results

Let $(X,\Vert \cdot \Vert )$ be a Banach space. Let $T>0$ and let $\mathcal {C}(X)$ denote the space of continuous functions $f:[0,T]\rightarrow X$ with a sup-norm

$$\begin{aligned} \Vert f\Vert _{\infty }=\sup _{t\in ( 0,T]}\Vert f(t)\Vert _{X}. \end{aligned}$$

Let $\beta \in (0,1]$. By $\mathcal {C}^{\beta }( X)$ we denote the space of $\beta$-Hölder continuous functions $f:[0,T]\rightarrow X$ with a finite norm

$$\begin{aligned} \Vert f\Vert _{\beta }:=\Vert f\Vert _{\infty }+M_{\beta }(f), \end{aligned}$$

where

$$\begin{aligned} M_{\beta }(f)=\sup _{0\le s<t\le T}\frac{\Vert f(t)-f(s)\Vert _{X}}{( t-s) ^{\beta }}. \end{aligned}$$

It can be shown that $( \mathcal {C}^{\beta }( X),\Vert \cdot \Vert _{\beta })$ is a Banach space. For a nonempty set $\mathcal {H}\subset \mathcal {C}( X)$ we shall use symbols $cl_{\mathcal {C}}( \mathcal {H})$ and $cl_{\mathcal {C}^{\beta }}( \mathcal {H})$ to denote the closure of $\mathcal {H}$ with respect to the norm $\Vert \cdot \Vert _{\infty }$ and $\Vert \cdot \Vert _{\beta }$, respectively. Let $0<\beta ^1<\beta \le 1$. Then it holds $M_{\beta ^1}(\cdot )\le T^{\beta -\beta ^1}M_{\beta }(\cdot )$ and thus $\Vert \cdot \Vert _{\beta ^1}\le ( 1+T^{\beta -\beta ^1}) \Vert \cdot \Vert _{\beta }$. Hence $cl_{\mathcal {C}^{\beta }}( \mathcal {H}) \subset cl_{\mathcal {C}^{\beta ^1}}( \mathcal {H}) \subset cl_{\mathcal {C}}( \mathcal {H})$. We recall the following results needed in the sequel (c.f. Lemma 5.29 and Proposition 5.30 in [14]).

Proposition 2.1

Let $( f_{n}) _{n\ge 1}\subset \mathcal {C}({R}^{d})$ and $\Vert f_{n}-f\Vert _{\infty }\rightarrow 0$. Assume

$\sup _{n\ge 1}M_{\beta }( f_{n}) <\infty .$ Then $f_{n}\rightarrow f$ in $\mathcal {C}^{\beta ^1}({R}^{d})$ for any $\beta ^1<\beta$.

Proposition 2.2

Suppose $( f_{n}) _{n\ge 1}$ is a bounded sequence in $\mathcal {C}( {R}^{d})$ and

$\sup _{n\ge 1}M_{\beta }( f_{n}) <\infty$. Then, for any $\beta ^1<\beta$ there exists a subsequence $( f_{n_{k}}) _{k\ge 1}$ of $( f_{n}) _{n\ge 1}$ which is convergent in $\mathcal {C}^{\beta ^1}( {R}^{d})$ to some $f\in \mathcal {C}^{\beta }( {R}^{d})$.

By these properties we have:

Proposition 2.3

Let $\mathcal {H}$ be a nonempty and bounded set in $\mathcal {C}^{\beta }( {R}^{d})$. Then $cl_{\mathcal {C}}(\mathcal {H})$ $=cl_{\mathcal {C}^{\beta ^1}}( \mathcal {H})$ for any $0<\beta ^1<\beta \le 1$. Moreover, $cl_{\mathcal {C}}( \mathcal {H})$ is a bounded subset in $\mathcal {C}^{\beta }( {R}^{d})$.

Proof

Let $f\in cl_{\mathcal {C}}( \mathcal {H})$ be arbitrary. Then there exists a sequence $( f_{n}) _{n\ge 1}\subset \mathcal {H}$ such that $\Vert f_{n}-f\Vert _{\infty }\rightarrow \infty$ as $n\rightarrow \infty$. Then $f_{n}\rightarrow f$ in $\mathcal {C}^{\beta ^1}( {R}^{d})$ for any $\beta ^1<\beta$ by Proposition 2.1. Thus $f\in cl_{\mathcal {C}^{\beta ^1}}( \mathcal {H})$. Hence $cl_{\mathcal {C}}(\mathcal {H}) =cl_{\mathcal {C}^{\beta ^1}}( \mathcal {H})$. Let $P_{1}:=\sup _{h\in \mathcal {H}}\Vert h\Vert _{\infty }$, $P_{2}:=\sup _{h\in \mathcal {H}}M_{\beta }(h)$ and $P=P_{1}+P_{2}$. Since $\mathcal {H}$ is a nonempty and bounded set in $\mathcal {C}^{\beta }( {R}^{d})$ then $P_{1}$ and $P_{2}$ are finite, and $\sup _{n\ge 1}\Vert f_{n}\Vert _{\beta }\le P$. On the other hand we have

$$\begin{aligned} \frac{\Vert f(t)-f(s)\Vert _{{R}^{d}}}{( t-s)^{\beta }}=\lim _{n\rightarrow \infty }\frac{\Vert f_{n}(t)-f_{n}(s)\Vert _{{R}^{d}}}{( t-s) ^{\beta }}\le \sup _{n\ge 1}M_{\beta }(f_{n})\le P_{2} \end{aligned}$$

for any $0\le s<t\le T$. Thus $M_{\beta }(f)\le P_{2}$. Since $\Vert f\Vert _{\infty }=\lim _{k\rightarrow \infty }\Vert f_{n}\Vert _{\infty }$, we obtain $\Vert f\Vert _{\infty }\le P_{1}$. Thus $\Vert f\Vert _{\beta }\le P$ what proves that $cl_{\mathcal {C}}( \mathcal {H})$ is a bounded subset in $\mathcal {C}^{\beta }( {R}^{d})$. $\square$

In order to define the Young integral, let $g\in \mathcal {C} ^{\alpha }( {R}^{1})$ and $f\in \mathcal {C}^{\beta }( {R}^{d})$ when $\alpha ,\beta \in (0,1]$ with $\alpha +\beta >1$ be given functions. For $0\le s<t\le T$, let $f_{s+}(\tau )=( f(\tau )-f(s+)) \textbf{1}_{( s,t) }( \tau )$ and $f_{t-}(\tau )=( f(\tau )-f(t-)) \textbf{1} _{( s,t) }( \tau )$ where $\textbf{1}_{( s,t) }$ denotes a characteristic function of the interval (s, t) . The right-sided and left-sided fractional derivatives of order $0<\rho <1$ for the function $f:[0,T]\rightarrow {R}^{1}$ and $0\le s<t\le T$ are defined by

$$\begin{aligned} D_{s+}^{\rho }f(\tau )=\frac{1}{\Gamma ( 1-\rho ) }\left( \frac{f(\tau )}{( \tau -s) ^{\rho }}+\rho \int _{s}^{\tau }\frac{f(\tau )-f(u)}{( \tau -u) ^{\rho +1}}du \right) \end{aligned}$$

(2.1)

and

$$\begin{aligned} D_{t-}^{\rho }f(\tau )=\frac{( -1) ^{\rho }}{\Gamma ( 1-\rho ) }\left( \frac{f(\tau )}{( t-\tau ) ^{\rho }}+\rho \int _{\tau }^{t}\frac{f(\tau )-f(u)}{( u-\tau ) ^{\rho +1}}du \right) \end{aligned}$$

(2.2)

for $\tau \in [s,t]$. For the partition $\pi :0=t_{0}<t_{1}<...<t_{m}=T$ of the interval [0, T] we consider the Riemann sum of f with respect to g

$$\begin{aligned} S(f,g,\pi ):=\sum \limits _{i=1}^{m}f( t_{i-1}) (g(t_{i})-g(t_{i-1})). \end{aligned}$$

Let $|\pi |:=\max \{t_{i}-t_{i-1}:1\le i\le m-1\}$. Then one can prove the following result (see e.g. [36]).

Proposition 2.4

Suppose that $g\in \mathcal {C}^{\alpha }( {R}^{1})$ and $f\in \mathcal {C}^{\beta }( {R}^{d})$. Then

$$\begin{aligned} \lim _{|\pi |\rightarrow 0}S(f,g,\pi ):=\int _{0}^{T}fdg \end{aligned}$$

(2.3)

exists and

$$\begin{aligned} \int _{s}^{t}fdg=( -1) ^{\rho }\int _{s}^{t}D_{s+}^{\rho }f_{s+}(\tau )D_{t-}^{1-\rho }g_{t-}(\tau )d\tau +f(s)(g(t)-g(s)) \end{aligned}$$

(2.4)

for $\rho \in (1-\alpha ,\beta )$ and $0\le s<t\le T$. Moreover, the following inequality

$$\begin{aligned} \Vert \int _{s}^{t}fdg-f(s)(g(t)-g(s))\Vert _{{R}^{d}}\le C(\alpha ,\beta )M_{\alpha }( g) M_{\beta }(f)(t-s)^{\alpha +\beta } \end{aligned}$$

(2.5)

holds for every $0\le s<t\le T$ where $C(\alpha ,\beta )$ depends only on $\alpha$ and $\beta$.

Due to Proposition 2.4 we get immediately

$$\begin{aligned} \Vert \int _{s}^{t}fdg\Vert _{{R}^{d}}\le M_{\alpha }( g)(t-s)^{\alpha }\big ( \Vert f\Vert _{\infty }+C(\alpha ,\beta )M_{\beta }(f)(t-s)^{\beta }\big ) \end{aligned}$$

(2.6)

for every $0\le s<t\le T$ and hence

$$\begin{aligned} M_{\alpha }( \int _{0}^{\cdot }fdg) \le M_{\alpha }(g) \big ( \Vert f\Vert _{\infty }+C(\alpha ,\beta )M_{\beta }(f)T^{\beta }\big ). \end{aligned}$$

(2.7)

Indeed, let $s,t\in [0,T]$, $s<t$ be fixed. Then by (2.5) we get

$$\begin{aligned}{} & {} \Vert \int _{s}^{t}fdg\Vert _{{R}^{d}}\le \Vert f( s) \Vert _{{R}^{d}}|g( t) -g( s) |+C(\alpha ,\beta )M_{\alpha }( g) M_{\beta }(f)(t-s)^{\alpha +\beta } \\{} & {} \quad \le \Vert f\Vert _{{\infty }}M_{\alpha }( g) (t-s)^{\alpha }+C(\alpha ,\beta )M_{\alpha }( g) M_{\beta }(f)(t-s)^{\alpha +\beta } \end{aligned}$$

which gives (2.6) and (2.7). The following result is a variant of Lemma 1 from [32]. It will be used in next section.

Lemma 2.5

Let $g\in \mathcal {C}^{\alpha }( {R}^{1})$. Then for every $\rho \in (1-\alpha ,\beta )$ there exists a positive constant $C(\rho )$ such that for every $f_{1},f_{2}\in \mathcal {C}^{\beta }( {R}^{d})$, $0\le s<t\le T$ and $\theta \in (0,1]$ it holds

$$\begin{aligned}{} & {} \Vert \int _{s}^{t}f_{1}dg-\int _{s}^{t}f_{2}dg\Vert _{{R}^{d}} \le C(\rho )\big ( \Vert f_{1}-f_{2}\Vert _{\infty }\nonumber \\{} & {} \quad +(M_{\beta }( f_{1}) +M_{\beta }( f_{2}) )\theta ^{\beta }\big ) \theta ^{-\rho }+\Vert f_{1}(s)-f_{2}(s)\Vert _{{R}^{d}}|g(t)-g(s)|, \end{aligned}$$

(2.8)

where $C(\rho )$ is given by formula

$$\begin{aligned} C( \rho ):=\max ( C_{1}( \rho ),C_{2}(\rho ) ), \end{aligned}$$

(2.9)

with

$$\begin{aligned} C_{1}( \rho ):=\frac{2\rho ( 1-\rho ) M_{\alpha }(g)T^{\alpha }}{\Gamma (1-\rho )\Gamma (\rho )( \rho +\alpha -1)\rho ( 1-\rho ) },\\ C_{2}( \rho ):=\frac{\rho ( 1-\rho ) M_{\alpha }(g)T^{\alpha +\beta }}{\Gamma (1-\rho )\Gamma (\rho )( \rho +\alpha -1) (\beta -\rho )(\beta -\rho +1)}. \end{aligned}$$

Proof

The proof can be given as a slight modification of the proof of Lemma 1 of [32], and therefore, we omit it. $\square$

At the end of this section we recall the notion of a space of $\beta$-Hölder continuous set-valued functions. Denote by $Cl_{b}( X)$ and Comp(X) the families of all nonempty, closed and bounded and nonempty and compact subsets of a Banach space X, respectively. The Hausdorff metric $H_{X}$ in $Cl_{b}( {X})$ is defined by

$$\begin{aligned} H_{X}( B,C) =\max \left\{ \overline{H}_{X}( B,C),\overline{H}_{X}( C,B) \right\} , \end{aligned}$$

where $\overline{H}_{X}( B,C) =\sup _{b\in B}\textrm{dist}_{X}( b,C)$ and $\textrm{dist}_{X}(b,C) =\inf _{c\in C}\Vert b-c \Vert _{X}$ (see e.g., [19] for details). We shall use the notation

$$\begin{aligned} \Vert A\Vert _{X}:=H_{X}( A,\{0\} ) =\sup _{a\in A}\Vert a\Vert _{X}\hbox { for }A\in Cl_{b}( X). \end{aligned}$$

For $B,C,D,E\in Comp( X)$ it holds

$$\begin{aligned} H_{X}( B+C,D+E) \le H_{X}( B,D) +H_{X}( C,E), \end{aligned}$$

(2.10)

where $B+C:=\left\{ b+c:b\in B,c\in C\right\}$ denotes the Minkowski sum of B and C. Let $F:[0,T]\rightarrow Comp({X)}$ be a set-valued function. It is called $\beta$-Hölder continuous if

$$\begin{aligned} M_{\beta }(F):=\sup _{0\le s<t\le T}\frac{H_{X}(F(t),F(s)) }{(t-s)^{\beta }}<\infty . \end{aligned}$$

Then

$$\begin{aligned} \Vert F\Vert _{{\infty }}:=\sup _{t\in (0,T]}\Vert F(t)\Vert _{X}\le \Vert F(0)\Vert _{X}+T^{\beta }M_{\beta }(F)<\infty . \end{aligned}$$

By $\mathcal {C}^{\beta }( Comp({X)})$ we denote the space of such set-valued functions with a norm given by formula $\Vert F\Vert _{{\beta }}:=\Vert F\Vert _{{\infty }}+M_{\beta }(F).$

3 Set-valued Young integral

Let $\mathcal {H}$ be a nonempty subset of $\mathcal {C}^{\beta }( R^{d})$. Similarly as in the previous part we assume that $g\in \mathcal {C}^{\alpha }( R^{1})$ where $\alpha ,\beta \in (0,1]$ with $\alpha +\beta >1$. Define a set-valued Young integral $\int _{0}^{t}\mathcal {H}dg$ of $\mathcal {H}$ with respect to g over the interval $[s,t]\subset [0,T]$, i.e.,

$$\begin{aligned} \int _{s}^{t}\mathcal {H}dg:= \left\{ \int _{s}^{t}fdg:f\in \mathcal {H} \right\} . \end{aligned}$$

(3.1)

It is easy to see that $\int _{s}^{t}\mathcal {H}dg$ is a convex subset of $R^{d}$ if $\mathcal {H}$ is convex. Moreover, we have the following properties of such set-valued Young integral.

Proposition 3.1

Let $0\le s<t\le T$. Let $\mathcal {H}$ be a nonempty subset of $\mathcal {C}^{\beta }( R^{d})$. Then it holds

$$\begin{aligned}{} & {} \Vert \int _{s}^{t}\mathcal {H}dg\Vert _{R^{d}}\le M_{\alpha }( g) (t-s)^{\alpha }\big ( \sup _{f\in \mathcal {H} }\Vert f\Vert _{\infty }+C(\alpha ,\beta )\sup _{f\in \mathcal {H}}M_{\beta }( f) (t-s)^{\beta }\big )\nonumber \\{} & {} \quad \le M_{\alpha }( g) (t-s)^{\alpha }\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }( 1+C(\alpha ,\beta )(t-s)^{\beta }). \end{aligned}$$

(3.2)

Proof

Indeed, let $0\le s<t\le T$ and let

$f\in \mathcal {H}$ be chosen arbitrarily. Then by (2.6) we get

$$\begin{aligned} \Vert \int _{s}^{t}fdg\Vert _{R^{d}}\le M_{\alpha }( g) (t-s)^{\alpha }\big \Vert f\Vert _{\infty }+C(\alpha ,\beta )M_{\beta }(f)(t-s)^{\beta }\big ). \end{aligned}$$

Hence, taking the supremum over

$f\in \mathcal {H}$, we get inequalities (3.2). $\square$

By (3.2) it follows that $\int _{s}^{t}\mathcal {H}dg$ is a bounded subset of $R^{d}$ for every nonempty and bounded set $\mathcal {H}\subset \mathcal {C}^{\beta }( R^{d})$. If $\mathcal {H}\subset \mathcal {C}^{\beta }( R^{d})$ and $\beta ^1<\beta$, then $\mathcal {H}\subset \mathcal {C}^{\beta ^1}( R^{d})$ and one can consider a set-valued Young integral $\int _{s}^{t}cl_{\mathcal {C}^{\beta ^1}}(\mathcal {H)}dg$ as long as $\alpha +\beta ^1>1$. If the set $\mathcal {H}$ is bounded in $\mathcal {C}^{\beta }( R^{d})$, then by Proposition 2.3 it follows that $cl_{\mathcal {C}}(\mathcal {H})$ is a bounded set in $\mathcal {C}^{\beta }( R^{d})$ and $cl_{\mathcal {C}^{\beta ^1}}(\mathcal {H)}=cl_{\mathcal {C}}(\mathcal {H)\subset C}^{\beta }( R^{d})$. Thus $\int _{s}^{t}cl_{\mathcal {C}^{\beta ^1}}(\mathcal {H)}dg=\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H)}dg$ for any $0<\beta ^1<\beta$. Therefore, we can take a closure of $\mathcal {H}\subset \mathcal {C}^{\beta }( R^{d})$ in $\mathcal {C}( R^{d})$. In this case we have the following result.

Corollary 3.2

Let $\mathcal {H}$ be a nonempty and bounded set in $\mathcal {C}^{\beta }( R^{d})$. Then

$$\begin{aligned} \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg=\left\{ \int _{s}^{t}fdg:f\in cl_{\mathcal {C}}(\mathcal {H)} \right\} \end{aligned}$$

is a bounded and closed subset of $R^{d}$ for any $0\le s<t\le T$. Thus the set $\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg$ is compact in $R^{d}$.

Proof

By Proposition 2.3, $cl_{\mathcal {C}}(\mathcal {H)}$ is bounded in $\mathcal {C}^{\beta }( R^{d})$. Thus $\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg$ is a bounded subset in $R^{d}$, by (3.2). It remains to prove the closedness of $\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg$ in $R^{d}$. Let $z_{n}\in \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg$ and suppose $z_{n}\rightarrow z$ in $R^{d}$. Then there exists a sequence $( f_{n})\subset cl_{\mathcal {C}}(\mathcal {H})$ (depending on s and t) such that $z_{n}=\int _{s}^{t}f_{n}dg$ for $n\ge 1$. Since a sequence $(f_{n})$ is bounded in $\mathcal {C}^{\beta }( R^{d})$, then due to Proposition 2.2 there exists a subsequence $( f_{n_{k}})$ of $( f_{n})$ and $f\in \mathcal {C}^{\beta }( R^{d})$ such that $f_{n_{k}}\rightarrow f$ in $\mathcal {C}^{\beta ^1}( {R}^{d})$ for any $\beta ^1<\beta$. Thus $\Vert f_{n_{k}}-f\Vert _{\infty }\rightarrow 0$ and $f\in cl_{\mathcal {C}}(\mathcal {H})$. Since $\alpha +\beta >1$, let us choose $\beta ^1<\beta$ and such that $\alpha +\beta ^1>1$. Since simultaneously $\ f\in \mathcal {C}^{\beta ^1}( R^{d})$, we have by (2.6)

$$\begin{aligned}{} & {} \Vert z_{n_{k}}-\int _{s}^{t}fdg\Vert _{R^{d}}=\Vert \int _{s}^{t}(f_{n_{k}}-f)dg\Vert _{R^{d}}\\{} & {} \quad \le M_{\alpha }( g) (t-s)^{\alpha }\big ( \Vert f_{n_{k}}-f\Vert _{\infty }+C(\alpha ,\delta )M_{\beta ^1}(f_{n_{k}}-f)(t-s)^{\beta ^1}\big ). \end{aligned}$$

Since $f_{n_{k}}\rightarrow f$ in $\mathcal {C}^{\beta ^1}( R^{d})$, then the right side of the above inequality tends to zero as $k\rightarrow \infty$. Thus $z_{n_{k}}\rightarrow \int _{s}^{t}fdg$ in $R^{d}$ and $z=$ $\int _{s}^{t}fdg\in \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H)}dg$ what proves the closedness of $\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H)}dg$ in $R^{d}$. $\square$

It is worth to note the following natural connection between integrals $\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H)}dg$ and $\int _{s}^{t}\mathcal {H}dg$.

Corollary 3.3

Under the same assumptions as in Corollary 3.2 it holds

$$\begin{aligned}{} & {} \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H)}dg=cl_{R^{d}}\left( \int _{s}^{t}\mathcal {H}dg \right) ,\\{} & {} \quad \hbox { i.e., }\left\{ \int _{s}^{t}fdg:f\in cl_{\mathcal {C}}(\mathcal {H)}\right\} =cl_{R^{d}}\left\{ \int _{s}^{t}fdg:f\in \mathcal {H}\right\} . \end{aligned}$$

Proof

By Corollary 3.2 we have

$$\begin{aligned} \left\{ \int _{s}^{t}fdg:f\in cl_{\mathcal {C}}(\mathcal {H)}\right\} \supset cl_{R^{d}}\left\{ \int _{s}^{t}fdg:f\in \mathcal {H}\right\} . \end{aligned}$$

Let $z\in \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H)}dg$. Hence $z=\int _{s}^{t}fdg$ for some $f\in cl_{\mathcal {C}}(\mathcal {H)}$. Then, there exists a sequence $( f_{n}) _{n\ge 1}\subset \mathcal {H}$ and such that $\Vert f_{n}-f\Vert _{\infty }\rightarrow \infty$. Similarly as in the proof of Corollary 3.2, let us choose $\beta ^1<\beta$ and such that $\beta ^1+\alpha >1$. Thanks to Proposition 2.1 we deduce $f_{n}\rightarrow f$ in $\mathcal {C}^{\beta ^1}( R^{d})$ for any $\beta ^1<\beta$ and by (2.6) we obtain

$$\begin{aligned}{} & {} \Vert \int _{s}^{t}f_{n}dg-z\Vert _{R^{d}}=\Vert \int _{s}^{t}(f_{n}-f)dg\Vert _{R^{d}}\\{} & {} \quad \le M_{\alpha }( g) (t-s)^{\alpha }\big ( \Vert f_{n}-f\Vert _{\infty }+C(\alpha ,\delta )M_{\beta ^1}(f_{n}-f)(t-s)^{\beta ^1}\big ). \end{aligned}$$

Thus $\Vert \int _{s}^{t}f_{n}dg-z\Vert _{R^{d}}\rightarrow 0$ as $n\rightarrow \infty$, and we get

$$\begin{aligned} \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H)}dg\subset cl_{R^{d}}\left( \int _{s}^{t}\mathcal {H}dg \right) . \end{aligned}$$

$\square$

Let us note that a set-valued Young integral can be treated as a set-valued mapping itself. It depends on t, $\mathcal {H}$ and g. Below, we present the nature of these dependences.

Proposition 3.4

Suppose that $g\in \mathcal {C}^{\alpha }( R^{1})$ and $\mathcal {H\subset C}^{\beta }( R^{d})$ is a nonempty and bounded set. Then for every $s\in [ 0,T)$ the set-valued function

$$\begin{aligned}{}[ s,T] \ni t\rightarrow \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg\in Comp( R^{d}) \end{aligned}$$

is $\alpha$-Hölder continuous and

$$\begin{aligned} M_{\alpha }\left( \int _{s}^{\cdot }cl_{\mathcal {C}}(\mathcal {H})dg \right) \le M_{\alpha }( g) \max \{1,C(\alpha ,\beta )T^{\beta }\}\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }. \end{aligned}$$

Proof

It follows by Corollary 3.3 that for any $s\le \tau <t\le T$

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg,\int _{s}^{\tau }cl_{\mathcal {C}}(\mathcal {H})dg\right) =H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}dg,\int _{s}^{\tau }\mathcal {H}dg\right) . \end{aligned}$$

Thus

$$\begin{aligned} M_{\alpha }\left( \int _{s}^{\cdot }cl_{\mathcal {C}}(\mathcal {H})dg \right) =M_{\alpha }\left( \int _{s}^{\cdot }\mathcal {H}dg\right) . \end{aligned}$$

Therefore, it is sufficient to show that

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}dg,\int _{s}^{\tau }\mathcal {H}dg\right) \le M_{\alpha }( g) \max \{1,C(\alpha ,\beta )T^{\beta }\}\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }( t-\tau )^{\alpha } \end{aligned}$$

for all $s\le \tau <t\le T$. Let $z_{1}\in \int _{s}^{t}\mathcal {H}dg$. Then there exists $f_{1}\in \mathcal {H}$ such that $z_{1}=\int _{s}^{t}f_{1}dg$. Let $f_{2}\in \mathcal {H}$ be chosen arbitrarily. By (2.6) we have

$$\begin{aligned}{} & {} \Vert z_{1}-\int _{s}^{\tau }f_{2}dg\Vert _{R^{d}}\le \Vert \int _{\tau }^{t}f_{1}dg\Vert _{R^{d}}+\Vert \int _{s}^{\tau }(f_{1}-f_{2})dg\Vert _{R^{d}}\\{} & {} \quad \le M_{\alpha }( g) ( t-\tau ) ^{\alpha }\big (\Vert f_{1}\Vert _{\infty }+C(\alpha ,\beta )M_{\beta }(f_{1})T^{\beta }\big )+\Vert \int _{s}^{\tau }(f_{1}-f_{2})dg\Vert _{R^{d}} \\{} & {} \quad \le M_{\alpha }( g) \max \{1,C(\alpha ,\beta )T^{\beta }\}\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }( t-\tau ) ^{\alpha }+\Vert \int _{s}^{\tau }(f_{1}-f_{2})dg\Vert _{R^{d}}. \end{aligned}$$

Let us notice that $\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }<\infty$ because $\mathcal {H}\subset \mathcal {C}^{\beta }( R^{d})$ is a nonempty and bounded set. Thus

$$\begin{aligned}{} & {} \hbox {dist}_{R^{d}}\big ( z_{1},\int _{s}^{\tau }\mathcal {H}dg\big ) =\inf _{f\in \mathcal {H}}\Vert z_{1}-\int _{s}^{\tau }fdg\Vert _{R^{d}}\\{} & {} \quad \le M_{\alpha }( g) \max \{1,C(\alpha ,\beta )T^{\beta }\}\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }( t-\tau ) ^{\alpha }+\inf _{f\in \mathcal {H}}\Vert \int _{s}^{\tau }(f_{1}-f)dg\Vert _{R^{d}}. \end{aligned}$$

Since last term of the above inequality vanishes, then we get

$$\begin{aligned} \bar{H}_{R^{d}}\left( \int _{s}^{t}\mathcal {H}dg,\int _{s}^{\tau }\mathcal {H}dg\right) \le M_{\alpha }( g) \max \{1,C(\alpha ,\beta )T^{\beta }\}\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }( t-\tau ) ^{\alpha }. \end{aligned}$$

The same inequality holds for $\bar{H}_{R^{d}}\big ( \int _{s}^{\tau }\mathcal {H}dg,\int _{s}^{t}\mathcal {H}dg\big )$ by symmetry. Thus

$$\begin{aligned}{} & {} M_{\alpha }\left( \int _{s}^{\cdot }\mathcal {H}dg\right) =\sup _{s\le \tau <t\le T}\frac{H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}dg,\int _{s}^{\tau }\mathcal {H}dg\right) }{( t-\tau ) ^{\alpha }}\\{} & {} \quad \le M_{\alpha }( g) \max \{1,C(\alpha ,\beta )T^{\beta }\}\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta } \end{aligned}$$

for every $s\in [0,T)$. $\square$

By $H_{\infty }$ we denote the Hausdorff distance in the space $Cl_{b}(\mathcal {C}( R^{d}) )$.

Theorem 3.5

For every $\rho \in (1-\alpha ,\beta )$ there exists a positive constant $C(\rho )$ such that for every $\theta \in (0,1]$, $0\le s<t\le T$ and every nonempty and bounded $\mathcal {H}_{1},\mathcal {H}_{2}\subset \mathcal {C}^{\beta }( R^{d})$, it holds

$$\begin{aligned}{} & {} H_{R^{d}}\left( \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}_{1})dg,\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}_{2})dg\right) \\{} & {} \quad \le C(\rho )\big ( H_{\infty }( \mathcal {H}_{1},\mathcal {H}_{2})+(\sup _{f_{1}\in \mathcal {H}_{1}}M_{\beta }( f_{1})+\sup _{f_{2}\in \mathcal {H}_{2}}M_{\beta }( f_{2}) )\theta ^{\beta }\big ) \theta ^{-\rho }\\{} & {} \quad \quad +M_{\alpha }( g) T^{\alpha }H_{\infty }( \mathcal {H}_{1},\mathcal {H}_{2}). \end{aligned}$$

Proof

Let $f_{1}\in \mathcal {H}_{1}$, $f_{2}\in \mathcal {H}_{2}$, $\rho \in (1-\alpha ,\beta )$, $t\in [0,T]$ and $\theta \in (0,1]$ be given. By Lemma 2.5 we have

$$\begin{aligned}{} & {} \Vert \int _{s}^{t}f_{1}dg-\int _{s}^{t}f_{2}dg\Vert _{R^{d}}\\{} & {} \quad \le C(\rho )\big ( \Vert |f_{1}-f_{2}\Vert |_{\infty }+(M_{\beta }( f_{1}) +M_{\beta }( f_{2}) )\theta ^{\beta }\big ) \theta ^{-\rho }+\Vert f_{1}-f_{2}\Vert _{\infty }M_{\alpha }(g)T^{\alpha }, \end{aligned}$$

where $C(\rho )$ is given by formula (2.9).

Thus

$$\begin{aligned}{} & {} \hbox {dist}_{R^{d}}\left( \int _{s}^{t}f_{1}dg,\int _{s}^{t}\mathcal {H}_{2}dg \right) \\{} & {} \quad \le C(\rho )\big ( \hbox {dist}_{\infty }( f_{1},\mathcal {H}_{2}) +(\sup _{f_{1}\in \mathcal {H}_{1}}M_{\beta }( f_{1})+\sup _{f_{2}\in \mathcal {H}_{2}}M_{\beta }( f_{2}) )\theta ^{\beta }\big ) \theta ^{-\rho }\\{} & {} \quad \quad +\textrm{dist}_{\infty }( f_{1},\mathcal {H}_{2})M_{\alpha }(g)T^{\alpha }. \end{aligned}$$

Hence

$$\begin{aligned}{} & {} \bar{H}_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{1}dg,\int _{s}^{t}\mathcal {H}_{2}dg\right) \\{} & {} \quad \le C(\rho )\left( H_{\infty }( \mathcal {H}_{1},\mathcal {H}_{2})+(\sup _{f_{1}\in \mathcal {H}_{1}}M_{\beta }( f_{1})+\sup _{f_{2}\in \mathcal {H}_{2}}M_{\beta }( f_{2}) )\theta ^{\beta }\right) \theta ^{-\rho }\\{} & {} \quad \quad +M_{\alpha }( g) T^{\alpha }H_{\infty }( \mathcal {H}_{1},\mathcal {H}_{2}). \end{aligned}$$

In a similar way one can prove the same estimation for second Hausdorff submetric $\bar{H}_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{2}dg,\int _{s}^{t}\mathcal {H}_{1}dg\right)$. This completes the proof because by Corollary 3.3 we get

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}_{1})dg,\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}_{2})dg \right) =H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{2}dg,\int _{s}^{t}\mathcal {H} _{1}dg \right) . \end{aligned}$$

$\square$

It also true that a set-valued Young integral defined by formula (3.1) is Lipschitz continuous with respect to integrators from $\mathcal {C}^{\alpha }( R^{1})$.

Proposition 3.6

Let $\mathcal {H}$ be a nonempty and bounded set in $\mathcal {C}^{\beta }( R^{d})$. Then for each $g_{1},g_{2}\in \mathcal {C}^{\alpha }( R^{1})$ and $0\le s<t\le T$, it holds

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg_{1},\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg_{2}\right) \le \hat{C}(\alpha ,\beta ,T,\mathcal {H})\Vert g_{1}-g_{2}\Vert _{\alpha } \end{aligned}$$

(3.3)

where $\hat{C}(\alpha ,\beta ,T,\mathcal {H})$ is some constant depending on $\alpha ,\beta ,T$ and the set $\mathcal {H}$.

Proof

Let $z_{1}\in \int _{s}^{t}\mathcal {H}dg_{1}$. Then $z_{1}=\int _{s}^{t}f_{1}dg_{1}$ for some $f_{1}\in \mathcal {H}$. Let $f_{2}\in \mathcal {H}$ be chosen arbitrarily. Then by (2.6)

$$\begin{aligned}{} & {} \Vert z_{1}-\int _{s}^{t}f_{2}dg_{2}\Vert _{R^{d}}\le \Vert \int _{s}^{t}f_{1}d(g_{1}-g_{2})\Vert _{R^{d}}+\Vert \int _{s}^{t}(f_{1}-f_{2})dg_{2}\Vert _{R^{d}}\\{} & {} \quad \le M_{\alpha }( g_{1}-g_{2}) T^{\alpha }\big (\Vert f_{1}\Vert _{\infty }+C(\alpha ,\beta )M_{\beta }(f_{1})T^{\beta }\big )+\Vert \int _{s}^{t}(f_{1}-f_{2})dg_{2}\Vert _{R^{d}}\\{} & {} \quad \le M_{\alpha }( g_{1}-g_{2}) T^{\alpha }\max \{1,T^{\beta }C(\alpha ,\beta )\}(\Vert f_{1}\Vert _{\infty }+M_{\beta }(f_{1}))+\Vert \int _{s}^{t}(f_{1}-f_{2})dg_{2}\Vert _{R^{d}}\\{} & {} \quad \le \Vert g_{1}-g_{2}\Vert _{\alpha }T^{\alpha }\max \{1,T^{\beta }C(\alpha ,\beta )\}\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }+\Vert \int _{s}^{t}(f_{1}-f_{2})dg_{2}\Vert _{R^{d}}. \end{aligned}$$

Let

$$\begin{aligned} \hat{C}(\alpha ,\beta ,T,\mathcal {H}):=T^{\alpha }\max \{1,T^{\beta }C(\alpha ,\beta )\}\sup _{f\in \mathcal {H}}\Vert f\Vert _{\beta }. \end{aligned}$$

(3.4)

Since $\mathcal {H}$ is bounded in $\mathcal {C}^{\beta }( R^{d})$, $\hat{C}(\alpha ,\beta ,T,\mathcal {H})$ is finite. Thus

$$\begin{aligned}{} & {} \hbox {dist}_{R^{d}}\big ( z_{1},\int _{s}^{t}\mathcal {H}dg_{2}\big )\nonumber \\{} & {} \quad \le \hat{C}(\alpha ,\beta ,T,\mathcal {H})\Vert g_{1}-g_{2}\Vert _{\alpha }+\inf _{f\in \mathcal {H}}\Vert \int _{s}^{t}(f_{1}-f)dg_{2}\Vert _{R^{d}}. \end{aligned}$$

(3.5)

Since second term on the right side of inequality above equals to zero, it follows by (3.5) that

$$\begin{aligned} \bar{H}_{R^{d}}\left( \int _{s}^{t}\mathcal {H}dg_{1},\int _{s}^{t}\mathcal {H}dg_{2}\right) \le \hat{C}(\alpha ,\beta ,T,\mathcal {H})\Vert g_{1}-g_{2}\Vert _{\alpha }. \end{aligned}$$

The same holds for $\bar{H}_{R^{d}}( \int _{s}^{t}\mathcal {H}dg_{2},\int _{s}^{t}\mathcal {H}dg_{1})$ by symmetry. Thus

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}dg_{1},\int _{s}^{t}\mathcal {H}dg_{2}\right) \le \hat{C}(\alpha ,\beta ,T,\mathcal {H})\Vert g_{1}-g_{2}\Vert _{\alpha } \end{aligned}$$

and we obtain (3.3) by Corollary 3.3. $\square$

Using Theorem 3.5 and Proposition 3.6 we get the following continuity property of set-valued Young integral.

Corollary 3.7

Let $(\mathcal {H}_{n}) _{n\ge 1}$ be a bounded sequence of nonempty sets in $\mathcal {C}^{\beta }(\!R^{d}).$ Let $(g_{n}) _{n\ge 1}\subset \mathcal {C}^{\alpha }( R^{1})$. Suppose that for some bounded $\mathcal {H\subset C}^{\beta }( R^{d})$ and $g\in \mathcal {C}^{\alpha }(R^{1})$ it holds

$$\begin{aligned} H_{\infty }( \mathcal {H}_{n},\mathcal {H}) \rightarrow 0\hbox { and }\Vert g_{n}-g\Vert _{\alpha }\rightarrow 0\hbox { as }n\rightarrow \infty . \end{aligned}$$

Then for each $0\le s<t\le T$

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{n}dg_{n},\int _{s}^{t}\mathcal {H}dg\right) \rightarrow 0\hbox { as }n\rightarrow \infty \end{aligned}$$

(3.6)

and thus

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}_{n})dg_{n},\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H)}dg\right) \rightarrow 0\hbox { as }n\rightarrow \infty . \end{aligned}$$

(3.7)

Proof

Let us notice that for every $n\ge 1$ we have

$$\begin{aligned}{} & {} H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{n}dg_{n},\int _{s}^{t}\mathcal {H}dg\right) \\{} & {} \quad \le H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{n}dg_{n},\int _{s}^{t}\mathcal {H}_{n}dg) +H_{R^{d}}(\int _{s}^{t}\mathcal {H}_{n}dg,\int _{s}^{t}\mathcal {H}dg\right) . \end{aligned}$$

By Proposition 3.6 we get

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{n}dg_{n},\int _{s}^{t}\mathcal {H}_{n}dg\right) \le \hat{C}(\alpha ,\beta ,T,\mathcal {H}_{n})\Vert g_{n}-g\Vert _{\alpha }, \end{aligned}$$

(3.8)

where $\hat{C}(\alpha ,\beta ,T,\mathcal {H}_{n})=T^{\alpha }\max \{1,T^{\beta }C(\alpha ,\beta )\}\sup _{f\in \mathcal {H}_{n}}\Vert f\Vert _{\beta }$ by (3.4). Since $( \mathcal {H}_{n}) _{n\ge 1}$ is uniformly bounded in $\mathcal {C}^{\beta }( R^{d})$, then

$$\begin{aligned} \tilde{C}(\alpha ,\beta ,T):=\sup _{n}\hat{C}(\alpha ,\beta ,T,\mathcal {H}_{n})<\infty . \end{aligned}$$

From Theorem 3.5 it follows that for every $\rho \in (1-\alpha ,\beta )$ there exists a positive constant $C(\rho )$ such that for every $\theta \in (0,1]$ the inequality

$$\begin{aligned}{} & {} H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{n}dg,\int _{s}^{t}\mathcal {H}dg\right) \nonumber \\{} & {} \quad \le C(\rho )\big ( H_{\infty }( \mathcal {H}_{n},\mathcal {H}) +(\sup _{f_{1}\in \mathcal {H}_{n}}M_{\beta }( f_{1})+\sup _{f_{2}\in \mathcal {H}}M_{\beta }( f_{2}) )\theta ^{\beta }\big ) \theta ^{-\rho }\nonumber \\{} & {} \quad \quad +M_{\alpha }( g) T^{\alpha }H_{\infty }( \mathcal {H}_{n},\mathcal {H}) \end{aligned}$$

(3.9)

holds. Then, due to assumptions imposed on $( \mathcal {H}_{n}) _{n\ge 1}$, $( g_{n}) _{n\ge 1}$ and by virtue of (3.8), (3.9), we obtain

$$\begin{aligned}{} & {} \lim \sup _{n}H_{R^{d}}\left( \int _{s}^{t}\mathcal {H}_{n}dg_{n},\int _{s}^{t}\mathcal {H}dg\right) \\{} & {} \quad \le C(\rho )\big ( \sup _{n\ge 1}\sup _{f_{1}\in \mathcal {H}_{n}}M_{\beta }( f_{1}) +\sup _{f_{2}\in \mathcal {H}}M_{\beta }(f_{2}) \big ) \theta ^{\beta -\rho }. \end{aligned}$$

Since $\beta >\rho$ and $\theta \in (0,1]$ is arbitrary, we get (3.6). Hence, (3.7) is also satisfied by Corollary 3.3. $\square$

From now we consider a special choice of the set $\mathcal {H}$ in formula (3.1). Let $( f_{n}) _{n\ge 1}\subset \mathcal {C}^{\beta }( R^{d})$ be an arbitrary bounded sequence. Let $\mathcal {H}:=\{f_{n}:n\ge 1\}$. Then, by Proposition 2.3 and Corollary 3.3, a set-valued Young integral (3.1) satisfies the following property

Proposition 3.8

Let $( f_{n}) _{n\ge 1}\subset \mathcal {C}^{\beta }( R^{d})$ be such that $\sup _{n\ge 1}\Vert f_{n}\Vert _{\beta }<\infty$ and let $\mathcal {H}=\{f_{n}:n\ge 1\}$. Then $cl_{\mathcal {C}}\mathcal {H}$ is bounded in $\mathcal {C}^{\beta }( R^{d})$ and

$$\begin{aligned} \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg:=cl_{R^{d}}\left\{ \int _{s}^{t}f_{n}dg:n\ge 1\right\} \end{aligned}$$

(3.10)

for every $0\le s<t\le T$.

By Proposition 3.1 and Proposition 3.8 we get also

Corollary 3.9

(i) Let $( f_{n}) _{n\ge 1}\subset \mathcal {C}^{\beta }( R^{d})$ be such that $\sup _{n\ge 1}\Vert f_{n}\Vert _{\beta }<\infty$ and let $\mathcal {H}=\{f_{n}:n\ge 1\}$. Then

$$\begin{aligned} \Vert \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H})dg\Vert _{R^{d}}\le M_{\alpha }( g) \sup _{n\ge 1}\Vert f_{n}\Vert _{\beta }( (t-s)^{\alpha }+C(\alpha ,\beta )(t-s)^{\alpha +\beta }) \end{aligned}$$

for $0\le s<t\le T$.

(ii) Let $\mathcal {H}_{i}=\{f_{n}^{( i) }:n\ge 1\}$ with $( f_{n}^{(i)}) _{n\ge 1}\subset \mathcal {C}^{\beta }( R^{d})$ and assume that $\sup _{n\ge 1}\Vert f_{n}^{(i)}\Vert _{\beta }<\infty$ for $i=1,2$. Then, for every $\rho \in (1-\alpha ,\beta )$ there exists a positive constant $C(\rho )$ such that for every $\theta \in (0,1]$ and for $0\le s<t\le T$, it holds

$$\begin{aligned}{} & {} H_{R^{d}}\left( \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}_{1})dg,\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}_{2})dg\right) \\{} & {} \quad \le C(\rho )\left( \sup _{n\ge 1}\Vert f_{n}^{(1) }-f_{n}^{( 2) }\Vert _{\infty } +(\sup _{n\ge 1}M_{\beta }( f_{n}^{( 1) }) +\sup _{n\ge 1}M_{\beta }( f_{n}^{( 2) }) )\theta ^{\beta }\right) \theta ^{-\rho }\\{} & {} \quad \quad +M_{\alpha }( g) T^{\alpha }\sup _{n\ge 1}\Vert f_{n}^{( 1) }-f_{n}^{( 2) }\Vert _{\infty }. \end{aligned}$$

Remark 3.10

The set-valued Young integral $\int _{s}^{t}\mathcal {H}dg$ considered in the manuscript is not of the type studied by Coutin-Marie-De Fitte in the paper [11]. However, considering a set-valued integral $\int _{0}^{t}\mathcal {H}dg$ of a bounded and countable set of functions $\mathcal {H}\subset \mathcal {C}^{\beta }( R^{d})$ we are able to introduce a set-valued function F by the formula

$$\begin{aligned} F(t) = cl_{R^d}\{f(t), f \in \mathcal {H}\} \end{aligned}$$

with the Castaing representation $\mathcal {H}$ and define the integral $\int _{s}^{t}F(s)dg(s)$ in the spirit of [11] by the formula

$$\begin{aligned} \mathcal {I}_{CMDF}(F,g,r)=\left\{ \int _{0}^{T}f(t)dg(t):f\in S(F,\beta ,r)\right\} \end{aligned}$$

with

$$\begin{aligned} S(F,\beta ,r)=\left\{ f:f\hbox { is }\beta \hbox {-H}\ddot{\textrm{o}}\hbox {lder selection of }F, \, M_{\beta }(f)\le r\right\} . \end{aligned}$$

If a given $\beta$-Hölder multifunction $F:[0,T]\rightarrow Comp (R^d)$ admits two Castaing representations $\mathcal {H}_1$ and $\mathcal {H}_2$ in $\mathcal {C}^{\beta }( R^{d})$ then it is clear that the integrals $\int _{s}^{t}\mathcal {H}_1dg$ and $\int _{s}^{t}\mathcal {H}_2dg$ are different despite they define the same set-valued function F as the below Example shows.

Example 3.11

Let $f_1,f_2, h_1, h_2:[0,1]\rightarrow R^1$ be defined by formulas:

$$\begin{aligned}{} & {} f_1(t)=t, \;f_2(t)=1/2 \hbox { for every }t\in [0,1],\\{} & {} \quad h_1(t)=t \text{ for } t\in [0,1/2] \text{ and } h_1(t)=1/2 \text{ for } t\in [1/2,1],\\{} & {} \quad h_2(t)=1/2 \text{ for } t\in [0,1/2] \text{ and } h_2(t)=t \text{ for } t\in [1/2,1]. \end{aligned}$$

Let $F(t)=\{f_1(t),f_2(t)\}=\{h_1(t),h_2(t)\}$ while $\mathcal {H}_1=\{f_1,f_2\}$ and $\mathcal {H}_2=\{h_1,h_2\}$. Then $\mathcal {H}_1$ and $\mathcal {H}_2$ are two different Castaing representations of the same Lipschitz set-valued function F. Let $g(t)=t$. An easy calculation shows that $\int _0^1\mathcal {H}_1dt=\{\int _0^1tdt,\int _0^11/2dt\}=\{1/2\}$ and $\int _0^1\mathcal {H}_2dt=\{\int _0^1h_1dt,\int _0^1h_2dt\}=\{3/8,5/8\}$. Both integrals differ from the Coutin-Marie-De Fitte integral

$$\begin{aligned} \mathcal {I}_{CMDF}(F, t, 1) = \left\{ \int _0^1 f_1dt,\int _0^1 f_2dt,\int _0^1 h_1dt,\int _0^1 h_2dt \right\} =\{3/8,1/2,5/8\}, \end{aligned}$$

because the set $\{ f_1,f_2,h_1,h_2\}$ forms the set of all 1-Hölder selections of F.

Remark 3.12

It is also worth noting in this point, that in the case of the integral $\mathcal {I}_{CMDF}(F,g,r)$, only a part of information on a set-valued function F is in use. It follows from the fact that in the definition of $\mathcal {I}_{CMDF}(F,g,r)\;\beta$-Hölder selections f of F with $M_{\beta }(f)\le r$ are considered. It means that for different $\beta$-Hölder set-valued functions $F_{1}$ and $F_{2}$, the set-valued Young integrals $\mathcal {I}_{CMDF}(F_{1},g,r)$ and $\mathcal {I}_{CMDF}(F_{2},g,r)$ can be equal. Indeed, let $\gamma _{1}(t)=2t$ for $t\in [0,\frac{1}{2}]$ and $\gamma _{1}(t)=1$ for $t\in (\frac{1}{2},1]$. Let $\gamma _{2}(t)=4t$ for $t\in [0,\frac{1}{4}]$ and $\gamma _{2}(t)=1$ for $t\in (\frac{1}{4},1]$. Now, let

$$\begin{aligned} F_{1}(t)=[0,\gamma _{1}(t)]\hbox { and }F_{2}(t)=[0,\gamma _{2}(t)] \end{aligned}$$

Then $F_{1}$ and $F_{2}$ are Lipschitz, $F_{1}$ is strictly contained in $F_{2}$ and for $r=2$ it holds

$$\begin{aligned} S(F_{1},1,2)=S(F_{2},1,2) \end{aligned}$$

because only Lipschitz selections ($\beta =1$) with the slope less or equal to 2 can be chosen. Hence $\mathcal {I}_{CMDF}(F_{1},g,2)=\mathcal {I}_{CMDF}(F_{2},g,2).$ The above indicates that we are dealing with a completely different situation than in the case of Aumann integration, where the collection of $L^p$-integrable selectors $S_{L^p}(F)$ uniquely (up to sets of measure zero) defines a multifunction F.

Hovewer, the following result holds true.

Proposition 3.13

Let $F:[0,T]\rightarrow Comp\left( R^{d}\right)$ be an $\beta$-Hölder set-valued function such that $S(F,\beta ,r)\ne \emptyset$. Then there exists a minimal with respect to inclusion “$\subset$” $\beta$-Hölder multifunction $F_{\min }:[0,T]\rightarrow Comp\left( R^{d}\right)$ such that

$$\begin{aligned} S(F_{\min },\beta ,r)=S(F,\beta ,r)\hbox { and }F_{\min }\subset F. \end{aligned}$$

Proof

Let $\mathcal {H}:=S(F,\beta ,r)$ and let us define

$$\begin{aligned} F_{\min }(t):=cl_{R^{d}}\{f(t):f\in \mathcal {H}\}\hbox { for }t\in (0,T]. \end{aligned}$$

We will show that

$$\begin{aligned} S(F_{\min },\beta ,r)=\mathcal {H},\;F_{\min }\subset F \end{aligned}$$

and for any other $\beta$-Hölder multifunction $G:[0,T]\rightarrow Comp\left( R^{d}\right)$ with $S(G,\beta ,r)=\mathcal {H}\hbox { and }G\subset F$ it holds $F_{\min }\subset G$. First, let us note that $F_{\min }$ is $\beta$-Hölder with nonempty and compact values. Moreover, it holds

$$\begin{aligned} \left\| F_{\min }\right\| _{\infty }\le \sup \{\left\| f\right\| _{\infty }:f\in \mathcal {H\}\le }\left\| F\right\| _{\infty }<\infty \end{aligned}$$

and

$$\begin{aligned} M_{\beta }(F_{\min })\le \sup \{M_{\beta }(f):f\in \mathcal {H\}\le }r. \end{aligned}$$

Thus $F_{\min }$ is $\beta$-Hölder with values in $Comp\left( R^{d}\right)$ and

$$\begin{aligned} \left\| F_{\min }\right\| _{\beta }\le \left\| F\right\| _{\infty }+r. \end{aligned}$$

Since for every $f\in \mathcal {H}$ we have $f(t)\in F_{\min }(t)$ for $t\in [0,T]$, then $\mathcal {H}\subset S(F_{\min },\beta ,r)$. Thus $S(F_{\min },\beta ,r)\ne \emptyset$. On the other hand $F_{\min }(t)\subset F(t)$ for $t\in [0,T]$, thus $S(F_{\min },\beta ,r)\subset \mathcal {H}$ and therefore, $S(F_{\min },\beta ,r)=\mathcal {H}$. In order to show that $F_{\min }$ is minimal for F, let

$$\begin{aligned} \mathcal {F}_{F}\left( \mathcal {H}\right) :=\{G:[0,T]\rightarrow Comp\left( R^{d}\right) :S(G,\beta ,r)=\mathcal {H}\hbox { and }G\subset F\}. \end{aligned}$$

Then $F\in \mathcal {F}_{F}\left( \mathcal {H}\right)$ as well as $F_{\min }\in \mathcal {F}_{F}\left( \mathcal {H}\right)$. Moreover, for any $G_{1}$, $G_{2}\in \mathcal {F}_{F}\left( \mathcal {H}\right)$ and such that $G_{1}\subset G_{2}$ we have $\mathcal {F}_{G_{1}}\left( \mathcal {H}\right) \subset \mathcal {F}_{G_{2}}\left( \mathcal {H}\right) \subset \mathcal {F}_{F}\left( \mathcal {H}\right)$. Let $G\in \mathcal {F}_{F}\left( \mathcal {H}\right)$. Since $F_{\min }(t)=cl_{R^{d}}\{g(t):g\in S(G,\beta ,r)\}\subset G(t)$ for $t\in [0,T]$, then $F_{\min }\in \mathcal {F}_{G}\left( \mathcal {H}\right)$ for each $G\in \mathcal {F}_{F}\left( \mathcal {H}\right)$. Hence $F_{\min }\in \mathcal {F}_{F_{\min }}\left( \mathcal {H}\right) \subset \bigcap \limits _{G\in \mathcal {F}_{F}\left( \mathcal {H}\right) }\mathcal {F}_{G}\left( \mathcal {H}\right) \subset \mathcal {F}_{F_{\min }}\left( \mathcal {H}\right)$ and therefore, $F_{\min }\in \mathcal {F}_{F_{\min }}\left( \mathcal {H}\right) =\bigcap \limits _{G\in \mathcal {F}_{F}\left( \mathcal {H}\right) }\mathcal {F}_{G}\left( \mathcal {H}\right)$. Finally, let us note that $\mathcal {F}_{F_{\min }}\left( \mathcal {H}\right) =\{F_{\min }\}$. Indeed, suppose $\tilde{G}\in$ $\mathcal {F}_{F_{\min }}\left( \mathcal {H}\right)$. Then $\tilde{G}(t)\subset F_{\min }(t)$ for $t\in [0,T]$ and $S(\tilde{G},\beta ,r)=S(F_{\min },\beta ,r)=\mathcal {H}$. Moreover,

$$\begin{aligned} F_{\min }(t)=cl_{R^{d}}\{g(t):g\in S(\tilde{G},\beta ,r)\}\subset \tilde{G}(t)\subset F_{\min }(t) \end{aligned}$$

for every $t\in [0,T]$. Thus $\mathcal {F}_{F_{\min }}\left( \mathcal {H}\right) =\{F_{\min }\}$ what means that $F_{\min }$ is $\beta$-Hölder minimal for F. $\square$

4 Young differential inclusions and their solutions

Let $\alpha ,\beta \in (0,1]$ with $\alpha +\beta >1$ and $\alpha >\frac{1}{2}$. Let $\xi \in R^d$ and $g\in \mathcal {C}^{\alpha }( R^{1})$. Assume $F:[0,T]\times R^{d}\rightarrow Conv( R^{d})$ is a given set-valued function and $\mathcal {H}=\{h_{n}:n\ge 1\}$ is a family of functions $h_{n}:[0,T]\times R^{d}\rightarrow R^{d}$. Let us consider a Young differential inclusion

$$\begin{aligned} dx\in F(x)dt+(\mathcal {H}\circ x)dg, x(0)=\xi \end{aligned}$$

(4.1)

which is interpreted in the integral form as follows

$$\begin{aligned} \left\{ \begin{array}{l} x(t)-x(s)\in \int _{s}^{t}F(\tau ,x(\tau ))d\tau +\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}\circ x)dg \\ x( 0) =\xi \end{array} \right. \end{aligned}$$

(4.2)

for $0\le s\le t\le T$, where for every $x\in \mathcal {C}^{\alpha }(R^{d})$, $\mathcal {H}\circ x:=\left\{ h_{n}\circ x:n\ge 1\right\}$ and $( h_{n}\circ x) ( t) =h_{n}(t,x(t))$, $t\in [0,T]$ and provided both integrals in (4.2) exist.

Moreover, assume that F and $\mathcal {H}$ satisfy the following conditions:

(H1) (i) $F:[0,T]\times R^{d}\rightarrow Conv( R^{d})$ is measurable and $\Vert F(\cdot ,0)\Vert _{R^{d}}\in L^{\frac{1}{1-\alpha }}(R^{1})$,

(ii) there exist a constant $L>0$ and $\eta \in (0,1]$ such that

$$\begin{aligned} H_{R^{d}}(F(t,u),F(t,v))\le L\Vert u-v\Vert _{R^{d}}^{\eta } \end{aligned}$$

for every $t\in [ 0,T]$, $u,v\in R^{d}$.

(H2) (i) The set $\mathcal {H}=\{h_{n}:n\ge 1\}$ consists of functions $h_{n}:[0,T]\times R^{d}\rightarrow R^{d}$ such that there exist $\gamma \in ((1-\alpha )/\alpha ,1]$ and a sequence of positive constants $(L_{n})_{n\ge 1}$ with $\tilde{L}:=\sup _{n}L_{n}<\infty$ and such that

$$\begin{aligned} \Vert h_{n}(t,u)-h_{n}(s,v)\Vert _{R^{d}}\le L_{n}( |t-s|^{\beta }+\Vert u-v\Vert _{R^{d}}^{\gamma }) \end{aligned}$$

for every $s,t\in [0,T]$, $u,v\in R^{d}$, $n\ge 1$,

(ii) $C:=\sup _{n}\Vert h_{n}(0,0)\Vert _{R^{d}}<\infty$.

Definition 4.1

A function $x\in \mathcal {C}^{\alpha }(R^{d})$ is a solution to differential inclusion (4.1) if it satisfies (4.2) for every $0\le s\le t\le T$. By $S(\xi ,g, F,\mathcal {H})$ we will denote the set of all solutions of this inclusion.

Let us notice that for every continuous function $x:[0,T]\rightarrow R^{d}$ and for set-valued function F satisfying (H1) it holds

$$\begin{aligned} \Vert F(t,x(t))\Vert _{R^{d}}\le L\Vert x(t)\Vert _{R^{d}}^{\eta }+\Vert F(t,0)\Vert _{R^{d}}\le L\Vert x(t)\Vert _{R^{d}}+b(t) \end{aligned}$$

(4.3)

for every $t\in [0,T]$, where $b(\cdot ):=L+\Vert F(\cdot ,0)\Vert _{R^{d}}\in L^{1/(1-\alpha )}(R^d)$. Last inequality holds because $a^{\eta }\le 1+a$ for every $a>0$ and $\eta \in (0,1]$. Therefore, the set-valued function $F(\cdot ,x(\cdot ))$ is measurable and integrably bounded. Thus the Aumann integral $\int _{s}^{t}F(\tau ,x(\tau ))d\tau$ in (4.2) is well defined (see e.g., [5]). Since $b\in L^{1/(1-\alpha )}(R^{1})$, it follows by the Hölder inequality that $\Vert \int _{s}^{t}b(\tau )d\tau \Vert \le \Vert b\Vert _{L^{1/(1-\alpha )}(R^{1})}(t-s)^{\alpha }$ for every $0\le s\le t\le T$. Thus the function $[0,T]\ni t\rightarrow \int _{0}^{t}b(\tau )d\tau$ is $\alpha$-Hölder continuous, which by (4.3) yields

$$\begin{aligned} \Vert \int _{s}^{t}F(\tau ,x(\tau ))d\tau \Vert _{R^{d}}\le L\int _{s}^{t}\Vert x(\tau )\Vert _{R^{d}}d\tau +\Vert b\Vert _{L^{1/(1-\alpha )}(R^d)}(t-s)^{\alpha } \end{aligned}$$

(4.4)

for every $0\le s\le t\le T$ and

$$\begin{aligned} M_{\alpha }\left( \int _{0}^{\cdot }F(\tau ,x(\tau ))d\tau \right) \le LT^{1-\alpha }\Vert x\Vert _{\infty }+\Vert b\Vert _{L^{1/(1-\alpha )}(R^d)}. \end{aligned}$$

(4.5)

Hence $\int _{0}^{\cdot }F(\tau ,x(\tau ))d\tau \in \mathcal {C}^{\alpha }( Conv( R^{d}) )$. As a consequence of (H2), we have

$$\begin{aligned} K(u):=\sup _{n\ge 1}\Vert h_{n}(0,u)\Vert _{R^{d}}\le \tilde{L}\Vert u\Vert _{R^{d}}^{\gamma }+C \end{aligned}$$

(4.6)

for $u\in R^{d}$ and

$$\begin{aligned} \Vert (h_{n}\circ x)(t)-(h_{n}\circ x)(s)\Vert _{R^{d}}\le L_{n}\max \{1,(M_{\alpha }( x))^{\gamma } \}(|t-s|^{\beta }+|t-s|^{\alpha \gamma }) \nonumber \\ \end{aligned}$$

(4.7)

for every $s,t\in [0,T]$, $n\ge 1$ and $x\in \mathcal {C}^{\alpha }( R^{d})$. Thus, for each $n\ge 1$, $h_{n}\circ x\in \mathcal {C}^{\delta }( R^{d})$, where $\delta =\min \{\beta ,\alpha \gamma \}$. Since $\delta \in (0,1]$ and $\delta +\alpha >1$, then the Young integral $\int _s^t(h_{n}\circ x)dg$ is well defined for each $n\ge 1$. Moreover, for each $x\in \mathcal {C}^{\alpha }( R^{d})$ with $x(0)=\xi$, it follows by (H2) and (4.7)

$$\begin{aligned} \sup _{n\ge 1}\Vert h_{n}\circ x\Vert _{\infty }\le K( \xi ) +\tilde{L}\max \{1,(M_{\alpha }( x))^{\gamma }\}( T^{\beta }+T^{\alpha \gamma }) \end{aligned}$$

(4.8)

and

$$\begin{aligned} \sup _{n\ge 1}M_{\delta }( h_{n}\circ x) \le \tilde{L}\max \{1,(M_{\alpha }( x))^{\gamma } \}( T^{\beta -\delta }+T^{\alpha \gamma -\delta }). \end{aligned}$$

(4.9)

Thus, for any $x\in \mathcal {C}^{\alpha }( R^{d})$ with $x(0)=\xi$, we have $\sup _{n\ge 1}\Vert h_{n}\circ x\Vert _{\delta }<\infty$. Therefore, due to Proposition 3.8, the set $cl_{\mathcal {C}}\{h_{n}\circ x:n\ge 1\}$ is bounded in the space $\mathcal {C}^{\delta }( R^{d})$. Moreover, a set-valued Young integral $\int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}\circ x)dg$ from (4.2) is well defined and

$$\begin{aligned} \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}\circ x)dg=cl_{R^{d}}\left\{ \int _{s}^{t}(h_{n}\circ x)dg:n\ge 1\right\} \end{aligned}$$

for every $0\le s\le t\le T$. Let $A( x):=K( \xi )+\tilde{L}\max \{1,(M_{\alpha }( x))^{\gamma } \}( T^{\beta }+T^{\alpha \gamma })$ and $B( x):=\tilde{L}\max \{1,(M_{\alpha }( x))^{\gamma } \}( T^{\beta -\delta }+T^{\alpha \gamma -\delta })$, i.e., A(x) and B(x) are right-hand sides of (4.8) and (4.9), respectively. Then, by Corollary 3.9 with $f_{n}:=h_{n}\circ x$, we get

Corollary 4.2

Let $\mathcal {H}=\{h_{n}:n\ge 1\}$ satisfy condition (H2) and let $x\in \mathcal {C}^{\alpha }( R^{d})$. Then

$$\begin{aligned} \Vert \int _{s}^{t}cl_{\mathcal {C}}(\mathcal {H}\circ x)dg\Vert _{R^{d}}\le M_{\alpha }( g) \left( A( x)+B( x) \right) \big ( (t-s)^{\alpha }+C(\alpha ,\delta )(t-s)^{\alpha +\delta }\big ). \nonumber \\ \end{aligned}$$

(4.10)

Let $J_{\mathcal {H},T}:\mathcal {C}^{\alpha }( R^{d})\rightarrow R_{+}$ be given by the formula

$$\begin{aligned} J_{\mathcal {H},T}(x):=\sup _{0\le s<t\le T}\frac{\Vert \int _{s}^{t}cl_{ \mathcal {C}}(\mathcal {H}\circ x)dg\Vert _{R^{d}}}{( t-s) ^{\alpha }}. \end{aligned}$$

Then $J_{\mathcal {H},T}(D)$ is a bounded set in $R_{+}$ for every bounded set $D\subset \mathcal {C}^{\alpha }( R^{d})$.

Theorem 4.3

Let $F:[0,T]\times R^{d}\rightarrow Conv( R^{d})$ and $\mathcal {H}=\{h_{n}:n\ge 1\}$ satisfy conditions (H1) and (H2), respectively. Then the set $S(\xi ,g, F,\mathcal {H})$ of all solutions of inclusion (4.1) is a nonempty and closed subset of $\mathcal {C}^{\alpha }( R^{d})$.

Proof

For $A\in Conv( R^{d})$ and $y\in R^{d}$, let $\sigma :Conv( R^{d}) \times R^{d}\rightarrow R^{1}$ be a support function, i.e., $\sigma (A,y):=\sup _{a\in A}\left\langle a,y\right\rangle$, where $\langle \cdot ,\cdot \rangle$ denotes an inner product in $R^{d}$. Let $\Sigma ^{d-1}$ be the unit sphere in $R^{d}$ and let $\omega$ denote a measure on $\Sigma ^{d-1}$ proportional to the Lebesgue measure and such that $\omega ( \Sigma ^{d-1}) =1$. Let $St:Conv( R^{d}) \rightarrow R^{d}$ given by

$$\begin{aligned}{} & {} St(A)=\frac{\sigma (A,1)-\sigma (A,-1)}{2}, \hbox { for }d=1\\{} & {} \quad St(A)=d\int _{\Sigma ^{d-1}}x\sigma (A,x)\omega (dx), \hbox { for }d\ge 2 \end{aligned}$$

be a Steiner point of a set A (see [5]). It is a linear function on the space $Conv( R^{d})$ by Theorem 9.4.1 of [5] and by Proposition 2.2 of [11] it is Lipschitz continuous, i.e., $St(A)\in A$, $St(\alpha _{1}A_{1}+\alpha _{2}A_{2})=\alpha _{1}St(A_{1})+\alpha _{2}St(A_{2})$ and

$$\begin{aligned} \Vert St(A_{1})-St(A_{2})\Vert _{R^{d}}\le \big (2(d+1)\pi ^{-1} \big )^{1/2}H_{R^{d}}(A_{1},A_{2}) \end{aligned}$$

for every A, $A_{1}$, $A_{2}\in Conv( R^{d})$ and $\alpha _{1}$, $\alpha _{2}\in R$. Let $f(t,u)=St(F(t,u))$ for $(t,u)\in [0,T]\times R^{d}$. Then by (H1) it holds

$$\begin{aligned} \Vert f(t,u)\Vert _{R^{d}}\le L\Vert u\Vert _{R^{d}}+b(t) \end{aligned}$$

for every $t\in [0,T]$, $u\in R^{d}$. Let us choose an arbitrary function $h\in \mathcal {H}$ and consider the equation

$$\begin{aligned} x(t)=\xi +\int _{0}^{t}f(\tau ,x(\tau ))d\tau +\int _{0}^{t}h(\tau ,x(\tau ))dg(\tau ). \end{aligned}$$

(4.11)

Thanks to Theorem 4.1 of [35], there exists at least one $\alpha$-Hölder continuous solution to equation (4.11) which is obviously a solution to inclusion (4.2).

In order to show that $S( \xi ,g, F,\mathcal {H})$ is a closed subset of $\mathcal {C}^{\alpha }( R^{d})$ assume that $( x_{k}) \subset S(\xi ,g, F,\mathcal {H})$ and $x_{k}\rightarrow x$ in $\mathcal {C}^{\alpha }( R^{d})$ where $x\in \mathcal {C}^{\alpha }( R^{d})$. Thus $\Vert x_{k}-x\Vert _{\infty }\rightarrow 0$ as $n\rightarrow \infty$, $x(0)=\xi$ and $\sup _{k}M_{\alpha }(x_{k}) <\infty$. We will show that $x\in S(\xi ,g, F,\mathcal {H})$. By properties of the Aumann integral (see e.g., [5, 17]) and using (H2), we get

$$\begin{aligned}{} & {} H_{_{R^{d}}}\left( \int _{s}^{t}F(\tau ,x_{k}(\tau ))d\tau ,\int _{s}^{t}F(\tau ,x(\tau ))d\tau \right) \\{} & {} \quad \le \int _{s}^{t}H_{_{R^{d}}}( F(\tau ,x_{k}(\tau )),F(\tau ,x(\tau ))) d\tau \le L(t-s)\Vert x_{k}-x\Vert _{\infty }^{\eta }. \end{aligned}$$

Thus

$$\begin{aligned} H_{_{R^{d}}}\left( \int _{s}^{t}F(\tau ,x_{k}(\tau ))d\tau ,\int _{s}^{t}F(\tau ,x(\tau ))d\tau \right) \rightarrow 0\hbox { as } k\rightarrow \infty . \end{aligned}$$

(4.12)

On the other hand we get by (H2)

$$\begin{aligned} \sup _{n\ge 1}\Vert (h_{n}\circ x_{k})-(h_{n}\circ x)\Vert _{\infty }\le \Vert x_{k}-x\Vert _{\infty }^{\gamma }\sup _{n\ge 1}L_{n} \end{aligned}$$

for every $k\ge 1$ and thus

$$\begin{aligned} \sup _{n\ge 1}\Vert (h_{n}\circ x_{k})-(h_{n}\circ x)\Vert _{\infty }\rightarrow 0\hbox { as } k\rightarrow \infty . \end{aligned}$$

(4.13)

In order to show that $x\in S(\xi ,g, F,\mathcal {H})$ it suffices to claim that

$$\begin{aligned} \hbox {dist}_{R^{d}}\big ( x(t)-x(s),\int _{s}^{t}F(\tau ,x(\tau ))d\tau +\int _{s}^{t}(\mathcal {H}\circ x)dg\big ) =0 \end{aligned}$$

for every $0\le s\le t\le T$. Indeed, let us note that the following inequality holds

$$\begin{aligned}{} & {} \hbox {dist}_{R^{d}}\big ( x(t)-x(s),\int _{s}^{t}F(\tau ,x(\tau ))d\tau +\int _{s}^{t}(\mathcal {H}\circ x)dg\big )\\{} & {} \quad \le 2\Vert x-x_{k}\Vert _{\infty }+\hbox {dist}_{R^{d}}\big ( x_{k}(t)-x_{k}(s),\int _{s}^{t}F(\tau ,x_{k}(\tau ))d\tau +\int _{s}^{t}(\mathcal {H}\circ x_{k})dg\big )\\{} & {} \quad \quad +H_{R^{d}}\left( \int _{s}^{t}F(\tau ,x_{k}(\tau ))d\tau ,\int _{s}^{t}F(\tau ,x(\tau ))d\tau \right) \\{} & {} \quad \quad +H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{k})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) . \end{aligned}$$

Since $( x_{k}) \subset S(\xi ,g, F,\mathcal {H})$, the second term on the right side above equals to zero while first one tends to zero as $k\rightarrow \infty$. The same is true for the third one because of (4.12). Thus it remains to show that

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{k})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) \rightarrow 0\hbox { as } k\rightarrow \infty . \end{aligned}$$

(4.14)

Let $k\ge 1$ be fixed. By Corollary 3.9(ii) with $f_{n}^{1}:=h_{n}\circ x_{k}$, $f_{n}^{2}:=h_{n}\circ x$ and $\beta :=\delta$ (recall $\delta +\alpha >1)$ we get

for every $\rho \in (1-\alpha ,\delta )$, there exists a constant $C(\rho )$ such that for every $\theta \in (0,1]$ it holds

$$\begin{aligned}{} & {} H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{k})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) \\{} & {} \quad \le C(\rho )\left( \sup _{n\ge 1}\Vert h_{n}\circ x_{k}-h_{n}\circ x\Vert _{\infty }\right. \\{} & {} \left. \quad \quad +(\sup _{n\ge 1}M_{\delta }( h_{n}\circ x_{k})+\sup _{n\ge 1}M_{\delta }( h_{n}\circ x) )\theta ^{\delta } \right) \theta ^{-\rho }\\{} & {} \quad \quad +M_{\alpha }( g) T^{\alpha }\sup _{n\ge 1}\Vert h_{n}\circ x_{k}-h_{n}\circ x\Vert _{\infty }. \end{aligned}$$

But $\sup _{n}M_{\delta }( h_{n}\circ x) <\infty$ by (4.9) and $\sup _{k}\sup _{n}M_{\delta }( h_{n}\circ x_{k}) <\infty$ because of $\sup _{k}M_{\alpha }( x_{k}) <\infty$. Thus we obtain by (4.13)

$$\begin{aligned}{} & {} \lim \sup _{k}H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{k})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) \\{} & {} \quad \le C(\rho )\left( \sup _{k\ge 1}\sup _{n\ge 1}M_{\delta }( h_{n}\circ x_{k}) +\sup _{n\ge 1}M_{\delta }( h_{n}\circ x) \right) \theta ^{\delta }{}^{-\rho }. \end{aligned}$$

Since $\delta >\rho$ and $\theta \in (0,1]$ is arbitrary, we get (4.14) what finishes the proof of closedness of $S(\xi ,g, F,\mathcal {H})$. $\square$

Now we consider the problem of boundedness of solutions set to inclusion (4.1). For this aim we shall use more convenient and natural counterparts of previous notations. For $0\le s<t\le T$ and $x\in \mathcal {C}^{\alpha }( R^{d})$ let

$$\begin{aligned} \Vert x\Vert _{\alpha ,[s,t]}=\Vert x\Vert _{\infty ,[s,t]}+M_{\alpha ,[s,t]}(x), \end{aligned}$$

where

$$\begin{aligned} \Vert x\Vert _{\infty ,[s,t]}=\sup _{u\in [s,t]}\Vert x( u) \Vert _{R^{d}}\hbox { and } M_{\alpha ,[s,t]}(x)=\sup _{s\le u<v\le t}\frac{\Vert x( v)-x(u)\Vert _{R^{d}}}{( u-v) ^{\alpha }}. \end{aligned}$$

If $0=T_{0}<T_{1}<...<T_{k-1}<T_{k}=T$ is a given partition of [0, T], then by the inequality (5) of [27] we get

$$\begin{aligned} M_{\alpha ,[0,T]}(x)\le k^{1/\alpha }\max \{ M_{\alpha ,[T_{i},T_{i+1}]}(x):i=0,1,...,k-1\}. \end{aligned}$$

(4.15)

Let $\Delta :=\{( s,t):0\le s\le t\le T)\}$ and let $\varphi _{1},\varphi _{2}:\Delta \rightarrow R_{+}$ be functions defined by formulas

$$\begin{aligned}{} & {} \varphi _{1}(s,t):=M_{\alpha ,[0,T]}(g)\tilde{L}\big ( ( t-s) ^{\beta }+( t-s) ^{\alpha \gamma }+( t-s) ^{\beta -\delta }+( t-s) ^{\alpha \gamma -\delta }\big )\nonumber \\{} & {} \quad \cdot (1+C(\alpha ,\delta )( t-s) ^{\delta }) \end{aligned}$$

(4.16)

and

$$\begin{aligned} \varphi _{2}(s,t):=\varphi _{1}(s,t)+L( t-s), \end{aligned}$$

(4.17)

where $C(\alpha ,\delta )$ is a constant from Corollary 4.2 and $\tilde{L}=\sup _{n}L_{n}$ (see condition (H2)). By (4.16) and (4.17) it is easy to see that functions $\varphi _{1}(s,\cdot )$, $\varphi _{2}(s,\cdot ):[s,T]\rightarrow R_{+}$ are strictly increasing and continuous for any fixed $s\in [0,T)$. Moreover, $\varphi _{2}(s,t)=\varphi _{2}(0,t-s)$ for $(s,t)\in \Delta$ and $\varphi _{2}(0,0)=0$. Then there exists a finite partition: $0=T_{0}<T_{1}<...<T_{k-1}<T_{k}=T$ such that $\varphi _{2}(T_{i},T_{i+1})<1$ for $i=0,1,...,k-1$.

Recall that $K(u):=\sup _{n\ge 1}\Vert h_{n}(0,u)\Vert _{R^{d}}\le \tilde{L}\Vert u\Vert _{R^{d}}^{\gamma }+C$ for $u\in R^{d}$ by (4.6).

Lemma 4.4

Let $(s,t)\in \Delta$ be such that $\varphi _{2}(s,t)<1$. Then for every $x\in S( \xi ,g, F,\mathcal {H})$ it holds

$$\begin{aligned}{} & {} M_{\alpha ,[s,t]}( x) \le (1-\varphi _{2}(s,t))^{-1}\nonumber \\{} & {} \quad \cdot \big (\Vert b\Vert _{L^{1/(1-\alpha )}(R^d)}+\Vert x(s)\Vert _{R^{d}}L( t-s) ^{1-\alpha }\nonumber \\{} & {} \quad +M_{\alpha ,[0,T]}(g) K( x(s))( 1+C(\alpha ,\delta ))+\varphi _{1}(s,t)\big ). \end{aligned}$$

(4.18)

Proof

Let $x\in S( \xi ,g,F,\mathcal {H})$. Similarly as in inequality (4.5) one obtains

$$\begin{aligned}{} & {} M_{\alpha ,[s,t]}\left( \int _{s}^{\cdot }F(\tau ,x(\tau ))d\tau \right) \le L( t-s) ^{1-\alpha }\Vert x\Vert _{\infty ,[s,t]}+\Vert b\Vert _{L^{1/(1-\alpha ) }(R^d)}\\{} & {} \quad \le \Vert b\Vert _{L^{1/(1-\alpha ) }(R^d)}+\Vert x(s)\Vert _{R^{d}}L( t-s)^{1-\alpha }+L( t-s) M_{\alpha ,[s,t]}( x). \end{aligned}$$

Next, by (4.8), (4.9), Corollary 4.2 together with inequality $a^{\gamma }\le 1+a$ for $a\ge 0$ and $\gamma \in (0,1]$, recall that in our case $\gamma \in (\frac{1-\alpha }{\alpha },1]$, we get

$$\begin{aligned}{} & {} \sup _{s\le u<v\le t}\frac{\Vert \int _{u}^{v}(\mathcal {H}\circ x)dg\Vert _{R^{d}}}{( v-u) ^{\alpha }}\le M_{\alpha ,[0,T]}( g) K( x(s)) ( 1+C(\alpha ,\delta ))\\{} & {} \quad +\varphi _{1}(s,t)+\varphi _{1}(s,t)M_{\alpha ,[s,t]}( x). \end{aligned}$$

Since $x\in S( \xi ,g, F,\mathcal {H})$, then it holds

$$\begin{aligned}{} & {} M_{\alpha ,[s,t]}( x) \le M_{\alpha ,[s,t]}\left( \int _{s}^{\cdot }F(\tau ,x(\tau ))d\tau \right) +\sup _{s\le u<v\le t}\frac{\Vert \int _{u}^{v}(\mathcal {H}\circ x)dg\Vert _{R^{d}}}{( v-u) ^{\alpha }}\\{} & {} \quad \le \Vert b\Vert _{L^{1/(1-\alpha ) }(R^d)}+\Vert x(s)\Vert _{R^{d}}L( t-s)^{1-\alpha }\\{} & {} \quad \quad +M_{\alpha ,[0,T]}( g) K( x(s)) ( 1+C(\alpha ,\delta )) +\varphi _{1}(s,t)+\varphi _{2}(s,t)M_{\alpha ,[s,t]}(x). \end{aligned}$$

because of $\varphi _{2}(s,t)=\varphi _{1}(s,t)+L(t-s)$. Hence we obtain formula (4.18). $\square$

Theorem 4.5

The set $S( \xi , g, F,\mathcal {H})$ of all solutions of the inclusion (4.1) is bounded in the space $\mathcal {C}^{\alpha }( R^{d})$.

Proof

Let $x\in S( \xi ,g, F,\mathcal {H})$. There exists a finite partition $0=T_{0}<T_{1}<...<T_{k-1}<T_{k}=T$ such that $\varphi _{2}(T_{i},T_{i+1})<1$ for $i=0,1,...,k-1$ for some $k\ge 1$. First, let us note that for every $t\in (0,T]$ it holds

$$\begin{aligned} \Vert x\Vert _{\infty ,[0,t]}\le \Vert \xi \Vert _{R^{d}}+t^{\alpha }M_{\alpha ,[0,t]}( x). \end{aligned}$$

(4.19)

By Lemma 4.4 we have

$$\begin{aligned}{} & {} M_{\alpha , [0,T_{1}]}( x) \le (1-\varphi _{2}(0,T_{1}))^{-1} \cdot \big (\Vert b\Vert _{L^{1/(1-\alpha )}(R^d)}+\Vert \xi \Vert _{R^{d}}L( T_1) ^{1-\alpha }\\{} & {} \quad +M_{\alpha ,[0,T]}(g) K( \xi )( 1+C(\alpha ,\delta ))+\varphi _{1}(0,T_{1})\big ):=U_1. \end{aligned}$$

Let us notice that $U_{1}$ does not depend on x but only on $\xi$, F, $\left\{ h_{n}:n\ge 1\right\}$, g, T and parameters $\alpha$, $\beta$, $\gamma$. By the above inequality together with (4.19) we obtain

$$\begin{aligned} \Vert x\Vert _{\alpha ,[0,T_{1}]}\le \Vert \xi \Vert _{R^{d}}+( 1+T_{1}^{\alpha }) U_{1}:=P_{1}. \end{aligned}$$

(4.20)

The constant $P_{1}$ depends on the same quantities as $U_{1}$. Similarly, by (4.19) and (4.15) we get

$$\begin{aligned} \Vert x\Vert _{\infty ,[0,T_{2}]}\le \Vert \xi \Vert _{R^{d}}+T_{2}^{\alpha }2^{1/\alpha }( M_{\alpha ,[0,T_{1}]}( x) +M_{\alpha ,[T_{1},T_{2}]}( x) ). \end{aligned}$$

(4.21)

Using again Lemma 4.4 we obtain

$$\begin{aligned}{} & {} M_{\alpha , [T_{1},T_{2}]}( x) \le (1-\varphi _{2}(T_{1},T_{2}))^{-1} \cdot \big (\Vert b\Vert _{L^{1/(1-\alpha )}(R^d)}+\Vert x(T_1)\Vert _{R^{d}}L( T_{2}-T_1) ^{1-\alpha }\nonumber \\{} & {} \quad +M_{\alpha ,[0,T]}(g) K( x(T_1))( 1+C(\alpha ,\delta ))+\varphi _{1}(T_{1},T_{2})\big ). \end{aligned}$$

(4.22)

The norm $\Vert x(T_{1})\Vert _{R^{d}}\le P_{1}$ and thus $K( x(T_{1})) \le \tilde{L}P_{1}^{\gamma }+C$ by (4.20). Therefore, there exists a constant $U_{2}$, depending on $P_{1}$ but independent of x, which estimates the right side of inequality (4.22). Thus we get by (4.21)

$$\begin{aligned} \Vert x\Vert _{\alpha ,[0,T_{2}]}\le \Vert \xi \Vert _{R^{d}}+( 1+T_{2}^{\alpha }2^{1/\alpha }) (U_{1}+U_{2}):=P_{2}. \end{aligned}$$

Repeating this procedure k-times along the partition $0=T_{0}<T_{1}<...<T_{k-1}<T_{k}=T$ one can find a positive constant $P_k$ which also does not depend on x but only on $\xi$, F, $\{ h_{n}:n\ge 1\}$, g, T and parameters $\alpha$, $\beta$, $\gamma$ and such that $\Vert x\Vert _{\alpha ,[0,T]}\le P_k$. This finishes the proof. $\square$

Remark 4.6

Let $\mathcal {B}_{\alpha }(0,r)$ denote a closed ball in $\mathcal {C}^{\alpha }( R^{d})$ with a radius $r>0$ and centered in the origin. The procedure used in the proof of Theorem 4.5 shows that $S( \xi ,g,F,\mathcal {H}) \subset \mathcal {B}_{\alpha }(0,r)$ with $r=P_k$, where $P_k$ dependents continuously on $\Vert \xi \Vert _{R^{d}}$ and $M_{\alpha ,[0,T]}(g)$.

One may prove the following result.

Theorem 4.7

Let $( \xi _{l}) _{l\ge 1}\subset R^{d}$ and $(g_l)\subset \mathcal {C}^{\alpha }( R^1)$be bounded sequences. Then $\bigcup \limits _{l\ge 1}S( \xi _{l},g_l, F,\mathcal {H})$ is a bounded subset in $\mathcal {C}^{\alpha }(R^{d})$.

Proof

The proof is almost the same as the proof of Theorem 4.5. It is enough to take $\sup _{l\ge 1}M_{\alpha ,[0,T]}(g_l)$ and $\sup _{l\ge 1}\Vert \xi _l\Vert _{R^{d}}$ instead of $M_{\alpha ,[0,T]}(g)$ and $\Vert \xi \Vert _{R^{d}}$, respectively. $\square$

Until now, it is assumed that $\alpha ,\beta \in (0,1]$ with $\alpha +\beta >1$ and $\alpha >1/2$. Let us additionally assume that $\beta <\alpha$. Hence we obtain $(1-\alpha )/\alpha<\beta /\alpha <1$. Let us consider $\gamma \in ( \beta /\alpha ,1)$ in assumption (H2). Then we have the following result.

Theorem 4.8

Let $\alpha ,\beta \in (0,1]$ with $\alpha +\beta >1$, $\beta <\alpha$ and $\alpha >1/2$. Suppose $g\in \mathcal {C}^{\alpha }(R^1)$, $F:[0,T]\times R^{d}\rightarrow Conv( R^{d})$ satisfies condition (H1) and $\mathcal {H}=\{h_{n}:n\ge 1\}$ satisfies condition (H2) for $\gamma \in ( \beta /\alpha ,1)$. Then $S(\xi ,g, F,\mathcal {H})$ is closed in the topology of uniform convergence in the space $\mathcal {C}( R^{d})$.

Proof

By Theorems 4.3 and 4.5, $S( \xi ,g, F,\mathcal {H})$ is a nonempty and bounded set in $\mathcal {C}^{\alpha }( R^{d})$. Suppose $(x_{k}) _{k\ge 1}\subset S( \xi ,g,F,\mathcal {H})$ and $\Vert x_{k}-x\Vert _{\infty }\rightarrow 0$. Hence $\sup _{k\ge 1}M_{\alpha }( x_{k}) <\infty$. But since $\beta <\alpha$ and $\gamma \in ( \beta /\alpha ,1)$, it follows that $\delta :=\min \{\beta ,\alpha \gamma \}=\beta$. Then $\sup _{k}\sup _{n}M_{\beta }( h_{n}\circ x_{k}) <\infty$ by (4.9). Since $\sup _{k}M_{\alpha }( x_{k}) <\infty$, $\Vert x_{k}-x\Vert _{\infty }\rightarrow 0$, and $\beta <\alpha$ then it follows by Proposition 2.1 that $x_{k}\rightarrow x$ in $\mathcal {C}^{\beta }( R^{d})$. Therefore, $M_{\beta }( x) <\infty$. Moreover, $x\in \mathcal {C}^{\alpha }( R^{d})$ by Proposition 2.2. Next, we get

$$\begin{aligned} \Vert (h_{n}\circ x)(t)-(h_{n}\circ x)(s)\Vert _{R^{d}}\le 2L_{n}\max \{1,M_{\beta }( x) \}|t-s|^{\beta }. \end{aligned}$$

by (H2). Hence $\sup _{n}M_{\beta }( h_{n}\circ x) \le 2\tilde{L}\max \{1,M_{\beta }( x) \}$. Now proceeding similarly as in the proof of Theorem 4.3 we obtain

$$\begin{aligned} H_{_{R^{d}}}\left( \int _{s}^{t}F(\tau ,x_{k}(\tau ))d\tau ,\int _{s}^{t}F(\tau ,x(\tau ))d\tau \right) \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{k})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) \rightarrow 0\hbox { as }k\rightarrow \infty . \end{aligned}$$

We deduce from these inequalities

$$\begin{aligned} \mathrm{{dist}}_{R^{d}}\big ( x(t)-x(s),\int _{s}^{t}F(\tau ,x(\tau ))d\tau +\int _{s}^{t}(\mathcal {H}\circ x)dg\big ) =0 \end{aligned}$$

what means that $x\in S( \xi ,g,F,\mathcal {H})$. $\square$

By virtue of Theorem 4.8, Proposition 2.1 and Proposition 2.2 the following statement holds true.

Corollary 4.9

Under conditions of Theorem 4.8, the set $S(\xi ,g, F,\mathcal {H})$ is a compact subset of the space $\mathcal {C}( R^{d})$.

Proof

Indeed, let $( x_{k}) _{k\ge 1}\subset S( \xi ,g,F,\mathcal {H})$. Then $\sup _{k}M_{\alpha }( x_{k}) <\infty$. Since $\beta <\alpha$ then there exists a subsequence $(x_{k_{l}}) _{l\ge 1}$ of $( x_{k}) _{k\ge 1}$ and $x\in \mathcal {C}^{\alpha }( R^{d})$ such that $x_{k_{l}}\rightarrow x$ in $\mathcal {C}^{\beta }( R^{d})$ by Proposition 2.2. Thus $x\in S( \xi ,g,F,\mathcal {H})$, because $S( \xi ,g,F,\mathcal {H})$ is a closed subset of $\mathcal {C}( R^{d})$ by Theorem 4.8. $\square$

We know by Theorem 4.3 and Theorem 4.5 that the set $S( \xi ,g,F,\mathcal {H})$ is a nonempty, bounded and closed set in the space $\mathcal {C}^{\alpha }( R^{d})$. By Corollary 4.9 it is also compact in $\mathcal {C}( R^{d})$. In fact $S(\xi ,g,F,\mathcal {H})$ is a compact set not only in the space $\mathcal {C}( R^{d})$ but also in $\mathcal {C}^{\alpha ^1}( R^{d})$ for any $\alpha ^1<\alpha$.

Corollary 4.10

Under conditions of Theorem 4.8, $S(\xi ,g, F,\mathcal {H})$ is a compact subset of the space $\mathcal {C}^{\alpha ^1}( R^{d})$ for every $0<\alpha ^1<\alpha$.

Proof

Let $( x_{k}) _{k\ge 1}\subset S( \xi ,g,F,\mathcal {H})$ be arbitrary. By Theorem 4.5, $S( \xi ,g,F,\mathcal {H})$ is a bounded set in $\mathcal {C}^{\alpha }( R^{d})$. Hence $\sup _{k}M_{\alpha }( x_{k}) <\infty$. Let $\alpha ^1<\alpha$. By Proposition 2.2 there exist a subsequence $( x_{k_{l}}) _{l\ge 1}$ of $( x_{k}) _{k\ge 1}$ and $x\in \mathcal {C}^{\alpha }( R^{d})$ such that $x_{k_{l}}\rightarrow x$ in $\mathcal {C}^{\alpha ^1}( R^{d})$. We will show that $x\in S( \xi ,g,F,\mathcal {H})$. Since $\gamma \in ( \beta /\alpha ,1)$, it follows that $\delta :=\min \{\beta ,\alpha \gamma \}=\beta$. We have $\sup _{l}M_{\alpha }( x_{k_{l}})<\infty$ and therefore, $\sup _{l}\sup _{n}M_{\beta }( h_{n}\circ x_{k_{l}}) <\infty$ by (4.9). We have assumed that $\beta <\alpha$ and we know that $x\in \mathcal {C}^{\alpha }( R^{d})$. Thus $M_{\beta }( x) <\infty$ and we get by (H2)

$$\begin{aligned} \Vert (h_{n}\circ x)(t)-(h_{n}\circ x)(s)\Vert _{R^{d}}\le 2L_{n}\max \{1,M_{\beta }( x) \}|t-s|^{\beta }. \end{aligned}$$

Hence $\sup _{n}M_{\beta }( h_{n}\circ x) \le 2\tilde{L}\max \{1,M_{\beta }( x) \}$. Now, proceeding similarly as in the proof of Theorem 4.3, we obtain

$$\begin{aligned} H_{_{R^{d}}}\left( \int _{s}^{t}F(\tau ,x_{k_{l}}(\tau ))d\tau ,\int _{s}^{t}F(\tau ,x(\tau ))d\tau \right) \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{k_{l}})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) \rightarrow 0\hbox { as }l\rightarrow \infty \end{aligned}$$

what leeds to equality

$$\begin{aligned} \textrm{dist}_{R^{d}}\big ( x(t)-x(s),\int _{s}^{t}F(\tau ,x(\tau ))d\tau +\int _{s}^{t}(\mathcal {H}\circ x)dg\big ) =0 \end{aligned}$$

It means that $x\in S( \xi ,g,F,\mathcal {H})$. Hence $( x_{k}) _{k\ge 1}\subset S( \xi ,g,F,\mathcal {H})$ admits a subsequence $( x_{k_{l}}) _{l\ge 1}$ convergent to $x\in S( \xi ,g,F,\mathcal {H})$ in $\mathcal {C}^{\alpha ^1}( R^{d})$ for every $\alpha ^1<\alpha$. $\square$

Although the compactness of the set $S( \xi ,g,F,\mathcal {H})$ in the space $\mathcal {C}^{\alpha ^1}( R^{d})$ for $\alpha ^1<\alpha$ is stronger than its compactness in $\mathcal {C}( R^{d})$, however $\mathcal {C}^{\alpha ^1}( R^{d})$ is not a separable Banach space. Because the separability is necessary in last part of the paper, we will use more often Corollary 4.9 instead of Corollary 4.10 in our further considerations.

Due to Corollary 4.9 one can consider solution sets as a set-valued function $S( \cdot ,\cdot , F,\mathcal {H} ):R^{d}\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(\mathcal {C}( R^{d}) ),$ $(\xi ,g)\mapsto S(\xi ,g, F,\mathcal {H})$ and ask about its continuity. This set-valued function will be called a solution map of inclusion (4.1). We shall show that it is upper semicontinuous in the sense of set-valued functions. For this aim let us recall needed details ( [20]).

Definition 4.11

Let X and Y be topological Hausdorff spaces and let $G:X\rightarrow Cl_b(Y)$ be a given set-valued function. G is said to be upper semicontinuous (u.s.c.) at $x_{0}\in X$ if for every open set $U\subset Y$ and $G(x_{0})\subset U$ there exists an open neighborhood V of $x_{0}$ such that $G(x)\subset U$ for every $x\in V$. A set-valued function $G:X\rightarrow Cl_b(Y)$ is called upper semicontinuous (u.s.c.) on X if it is u.s.c. at every $x\in X$.

In the case of metric spaces X and Y and for compact-valued set-valued functions the following characterization holds (see Theorem II.2.2. of [20]).

Proposition 4.12

Let X and Y be metric spaces. A set-valued function $G:X\rightarrow Comp(Y)$ is upper semicontinuous if and only if for every $x\in X$, every sequence $( x_{l})\subset X,$ $x_l\rightarrow x$ and every sequence $( y_{l})$ of $\;Y$ with $y_{l}\in G(x_{l})$, there is a convergent subsequence of $( y_{l})$ whose limit belongs to G(x) .

It can be also proved that a set-valued function $G:X\rightarrow Comp(Y)$ is u.s.c. on X if and only if G is H-u.s.c. on X, i.e., for every $x\in X$ and $\epsilon >0$, there exists $\delta >0$ such that $G(\mathcal {B}_{X}(x,\delta ))\subset V(G(x),\epsilon )$ where $V(K,\epsilon ):=\{y\in Y:\textrm{dist}_{Y}(y,K)<\epsilon \}$ for $K\in Comp(Y)$. For such G its upper semicontinuity is equivalent to the following condition: for every $( x_{l}) \subset X$,

$$\begin{aligned} x_l\rightarrow x\Rightarrow \bar{H}_{Y}( G(x_{l}),G(x)) \rightarrow 0\hbox { as } l \rightarrow \infty , \end{aligned}$$

(4.23)

where $\bar{H}_{Y}$ is a Hausdorff semi-metric in Comp(Y), see e.g., [20] pp. 28-29.

Theorem 4.13

Under assumptions of Theorem 4.8 the solution map

$$\begin{aligned} S( \cdot ,\cdot , F,\mathcal {H} ):R^{d}\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(\mathcal {C}( R^{d}) ) \end{aligned}$$

is upper semicontinuous.

Proof

Let $\xi \in R^{d}$, $g\in \mathcal {C}^{\alpha }( R^1)$ and let $( \xi _{l}) _{l\ge 1}\subset R^{d}$ and $(g_l) \subset \mathcal {C}^{\alpha }(R^1)$ be sequences such that $\xi _{l}\rightarrow \xi$ in $R^d$ and $g_l\rightarrow g$ in $\mathcal {C}^{\alpha }( R^1)$ as $l\rightarrow \infty$. Let $x_{l}\in S( \xi _{l},g_l,F,\mathcal {H})$ for $l\ge 1$. Then $\{x_{l}:l\ge 1\}\subset \mathcal {S}:=\bigcup \limits _{l\ge 1}S(\xi _{l},g_l,F,\mathcal {H})$. Thus due to Theorem 4.7 we have $\sup _{l\ge 1}\Vert x_{l}\Vert _{\alpha }<\infty$. Since $\beta <\alpha$ then there exists a subsequence $( x_{l_{k}})_{k\ge 1}$ of $( x_{l}) _{l\ge 1}$ and $x\in \mathcal {C}^{\alpha }( R^{d})$ such that $x_{k_{l}}\rightarrow x$ in $\mathcal {C}^{\beta }( R^{d})$ by Proposition 2.2. Then $\Vert x_{l_{k}}-x\Vert _{\infty }$ $\rightarrow 0$ as $k\rightarrow \infty$. But $x_{l_{k}}(0)=\xi _{l_{k}}\rightarrow \xi$ and therefore, $x(0)=\xi$. We will show that $x\in S( \xi ,g,F,\mathcal {H})$. Similarly as in the proof of Theorem 4.3 it holds

$$\begin{aligned}{} & {} \textrm{dist}_{R^{d}}( x(t)-x(s),\int _{s}^{t}F(\tau ,x(\tau ))d\tau +\int _{s}^{t}(\mathcal {H}\circ x)dg)\nonumber \\{} & {} \quad \le 2\Vert x-x_{l_{k}}\Vert _{\infty }+\textrm{dist}_{R^{d}}\left( x_{l_{k}}(t)-x_{l_{k}}(s),\int _{s}^{t}F(\tau ,x_{l_{k}}(\tau ))d\tau +\int _{s}^{t}(\mathcal {H}\circ x_{l_{k}})dg\right) \nonumber \\{} & {} \quad \quad +H_{R^{d}}\left( \int _{s}^{t}F(\tau ,x_{l_{k}}(\tau ))d\tau ,\int _{s}^{t}F(\tau ,x(\tau ))d\tau \right) \nonumber \\{} & {} \quad \quad +H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{l_{k}})dg_{l_k},\int _{s}^{t}(\mathcal {H}\circ x_{l_k})dg\right) \nonumber \\{} & {} \quad \quad + H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{l_{k}})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg \right) \end{aligned}$$

(4.24)

for every $0\le s\le t\le T$. Since $x_{l_{k}}\in S( \xi _{l_{k}},g_{l_k},F,\mathcal {H})$, then the term “$\textrm{dist}_{R^{d}}(....)$” vanishes. Moreover,

$$\begin{aligned} H_{_{R^{d}}}\left( \int _{s}^{t}F(\tau ,x_{l_{k}}(\tau )\right) d\tau ,\int _{s}^{t}F(\tau ,x(\tau ))d\tau ) \rightarrow 0\hbox { as }k\rightarrow \infty \end{aligned}$$

(4.25)

because of the inequality

$$\begin{aligned} H_{_{R^{d}}}\left( \int _{s}^{t}F(\tau ,x_{l_{k}}(\tau )\right) d\tau ,\int _{s}^{t}F(\tau ,x(\tau ))d\tau ) \le L(t-s)\Vert x_{l_{k}}-x\Vert _{\infty }^{\eta }. \end{aligned}$$

We deduce by Corollary 3.9, that for every $\rho \in (1-\alpha ,\beta )$, there exists a constant $C(\rho )$ such that for every $\theta \in (0,1]$

$$\begin{aligned}{} & {} H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{l_{k}})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) \nonumber \\{} & {} \quad \le C(\rho )\left( \sup _{n\ge 1}\Vert h_{n}\circ x_{l_{k}}-h_{n}\circ x\Vert _{\infty } \right. \nonumber \\{} & {} \qquad \left. +(\sup _{k\ge 1}\sup _{n\ge 1}M_{\beta }( h_{n}\circ x_{l_{k}}) +\sup _{n\ge 1}M_{\beta }( h_{n}\circ x) )\theta ^{\beta }\right) \theta ^{-\rho }\nonumber \\{} & {} \qquad +M_{\alpha }( g) T^{\alpha }\sup _{n\ge 1}\Vert h_{n}\circ x_{l_{k}}-h_{n}\circ x\Vert _{\infty }. \end{aligned}$$

(4.26)

Next we have

$$\begin{aligned} \sup _{n\ge 1}\Vert (h_{n}\circ x_{l_{k}})-(h_{n}\circ x)\Vert _{\infty }\le \tilde{L}\Vert x_{l_{k}}-x\Vert _{\infty }^{\gamma } \end{aligned}$$

for every $k\ge 1$ by (H2). Thus

$$\begin{aligned} \sup _{n\ge 1}\Vert (h_{n}\circ x_{l_{k}})-(h_{n}\circ x)\Vert _{\infty }\rightarrow 0\hbox { as }k\rightarrow \infty . \end{aligned}$$

(4.27)

Finally, we get by (4.9)

$$\begin{aligned} \sup _{n\ge 1}M_{\beta }( h_{n}\circ x_{l_{k}}) \le \tilde{L}\max \{1,M_{\alpha }^{\gamma }( x_{l_{k}}) \}( 1+T^{\alpha \gamma -\beta }) \end{aligned}$$

and from $\sup _{k\ge 1}M_{\alpha }( x_{l_{k}}) <\infty$, together with $x\in \mathcal {C}^{\alpha }( R^{d})$ we obtain

$$\begin{aligned} K_{1}:=\sup _{k\ge 1}\sup _{n\ge 1}M_{\beta }( h_{n}\circ x_{l_{k}})<\infty \hbox { and }K_{2}:=\sup _{n}M_{\beta }(h_{n}\circ x) <\infty . \end{aligned}$$

(4.28)

Combining (4.26) with (4.27) and (4.28) we have

$$\begin{aligned} \lim \sup _{k}H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{l_{k}})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) \le C(\rho )(K_{1}+K_{2}) \theta ^{\beta -\rho }. \end{aligned}$$

Since $\theta \in (0,1]$ is arbitrary and $\rho <\beta$ then

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{l_{k}})dg,\int _{s}^{t}(\mathcal {H}\circ x)dg\right) \rightarrow 0\hbox { as }k\rightarrow \infty . \end{aligned}$$

(4.29)

By Proposition 3.6 we obtain

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{l_k})dg_{l_k},\int _{s}^{t}(\mathcal {H}\circ x_{l_k})dg\right) \le \hat{C}(\alpha ,\beta ,T,(\mathcal {H}\circ x_{l_k}))\Vert g_{l_k}-g\Vert _{\alpha },\nonumber \\ \end{aligned}$$

(4.30)

where $\hat{C}(\alpha ,\beta ,T,(\mathcal {H}\circ x_{l_k}))=T^{\alpha }\max \{1,T^{\beta }C(\alpha ,\beta )\}\sup _{n\ge 1}\Vert h_n\circ x_{l_k}\Vert _{\beta }.$ Using formulas (4.8) and (4.9) once again we get

$$\begin{aligned}{} & {} \sup _{n\ge 1}\Vert h_n\circ x_{l_k}\Vert _{\beta }\le T^{\alpha }\max \{1,T^{\beta }C(\alpha ,\beta )\}\cdot [K(\xi )+\tilde{L}\max \{1,(M_{\alpha }( x_{l_{k}}))^{\gamma } \}\\{} & {} \quad \cdot (T^{\beta }+T^{\alpha \gamma })+\tilde{L}\max \{1,(M_{\alpha }( x_{l_{k}})^{\gamma } \}( 1+T^{\alpha \gamma -\beta })] \end{aligned}$$

because $\delta =\beta$ here.

Since the set $\{x_{l_k}\}$ is bounded in $\mathcal {C}^{\alpha }(R^d)$ then

$$\begin{aligned} \hat{C}(\alpha ,\beta ,T):=\sup _{k\ge 1}\hat{C}(\alpha ,\beta ,T,(\mathcal {H}\circ x_{l_k}))< \infty . \end{aligned}$$

Thus by (4.30) we get

$$\begin{aligned} H_{R^{d}}\left( \int _{s}^{t}(\mathcal {H}\circ x_{l_{k}})dg_{l_k},\int _{s}^{t}(\mathcal {H}\circ x_{l_k})dg\right) \rightarrow 0\hbox { as }k\rightarrow \infty . \end{aligned}$$

(4.31)

Combining (4.31), (4.29) and (4.25) in view of (4.24) we have

$$\begin{aligned} \textrm{dist}_{R^{d}}\left( x(t)-x(s),\int _{s}^{t}F(\tau ,x(\tau ))d\tau +\int _{s}^{t}(\mathcal {H}\circ x)dg\right) =0 \end{aligned}$$

for every $0\le s\le t\le T$, what means that $x\in S( \xi ,g,F,\mathcal {H})$. By Proposition 4.12 we deduce the desired u.s.c. property of a set-valued function $S( \cdot ,\cdot , F,\mathcal {H} )$. $\square$

Due to the previous theorem we have

Corollary 4.14

Let assumptions of Theorem 4.8 be satisfied.

(i) If $\xi _{n}\rightarrow \xi$ in $R^{d}$ and $g _{n}\rightarrow g$ in $\mathcal {C}^{\alpha }(R^1)$, then

$$\begin{aligned} \bar{H}_{\infty }( S( \xi _{n},g_n,F,\mathcal {H}),S( \xi ,g,F,\mathcal {H}) ) \rightarrow 0\hbox { as }n\rightarrow \infty . \end{aligned}$$

(ii) Let $\Gamma _1 \subset R^{d}$ and $\Gamma _2 \subset \mathcal {C}^{\alpha }(R^1)$ be compact sets. Then the set

$$\begin{aligned} S( \Gamma _1,\Gamma _2,F,\mathcal {H}):=\bigcup \limits _{(\xi ,g)\in \Gamma _1\times \Gamma _2}S( \xi ,g,F,\mathcal {H}) \end{aligned}$$

is bounded in $\mathcal {C}^{\alpha }( R^{d})$ and compact in $\mathcal {C}( R^{d})$.

Proof

First property follows by the comments below Proposition 4.12. The boundedness of $S( \Gamma _1,\Gamma _2,F,\mathcal {H})$ in $\mathcal {C}^{\alpha }( R^{d})$ can be proved similarly as it was done in the proof of Theorem 4.7. Since the set-valued function $S( \cdot ,\cdot , F,\mathcal {H} )$ is u.s.c. by Theorem 4.13, the compactness of the set $S( \Gamma _1,\Gamma _2,F,\mathcal {H})$ in $\mathcal {C}( R^{d})$ follows by Proposition II.2.3. in [20]. $\square$

5 Applications

Let us take an arbitrary continuous functional $\mathcal {J}:\mathcal {C} ( R^{d}) \rightarrow R_{+}$. For instance $\mathcal {J}$ can be taken as $\mathcal {J}(x):=\Vert x\Vert _{\infty }$ or $\mathcal {J}(x):=\min \left\{ \Vert x(t_{i})\Vert _{R^{d}}:i=1,2,...,k\right\}$ where $\left\{ t_{i}:i=1,2,...,k\right\} \subset [0,T]$ is a fixed collection of points, or

$$\begin{aligned} \mathcal {J}(x):=\int _{0}^{T}\Vert p(t,x(t))\Vert _{R^{d}}dt\hbox { for }x\in \mathcal {C}( R^{d}), \end{aligned}$$

where $p:[0,T]\times R^{d}\rightarrow R^{d}$ is a given bounded and continuous function. Then we can consider the following optimality problem for Young differential inclusion (4.1)

$$\begin{aligned}{} & {} \mathcal {J}(x)\rightarrow \max \\{} & {} \quad \hbox { for }x\in S(\xi ,g, F,\mathcal {H}). \end{aligned}$$

Definition 5.1

By an optimal solution to Young differential inclusion (4.1) we mean a function $x\in S(\xi ,g, F,\mathcal {H})$ such that

$$\begin{aligned} \mathcal {J}(x)=\sup \{ \mathcal {J}(y):y\in S(\xi ,g, F,\mathcal {H})\}. \end{aligned}$$

(5.1)

The set

$$\begin{aligned}{} & {} S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})\nonumber \\{} & {} \quad :=\{ x\in S(\xi ,g, F,\mathcal {H})\!:\!\mathcal {J}( x) \!=\!\sup \{\mathcal {J}(y):y\in S(\xi ,g, F,\mathcal {H})\} \} \end{aligned}$$

(5.2)

is called a set of $\mathcal {J}$-optimal solutions to differential inclusion (4.1), while a set-valued function

$$\begin{aligned} S_{\mathcal {J}}( \cdot ,\cdot ,F,\mathcal {H}):R^{d}\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(\mathcal {C}( R^{d}) ) \hbox { given by }(\xi ,g)\mapsto S_{\mathcal {J}}(\xi ,g, F,\mathcal {H}), \end{aligned}$$

is called a $\mathcal {J}$-optimal solution map of inclusion (4.1).

Under conditions of Corollary 4.9 it is clear that optimal solutions exist. So the set of $\mathcal {J}$-optimal solutions to differential inclusion (4.1) is nonempty for every $(\xi ,g)\in R^{d}\times \mathcal {C}^{\alpha }( R^1)$, because the functional $\mathcal {J}$ is assumed to be continuous. Moreover, $S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})\in Comp(\mathcal {C}( R^{d}) )$ for every $\xi \in R^{d}$ and $g\in \mathcal {C}^{\alpha }( R^1).$

Each $x\in S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})$ depends somehow on initial value $\xi \in R^{d}$ and $g\in \mathcal {C}^{\alpha }( R^1)$. In fact it will be shown that it depends on $\xi$ and g in a measurable way. We shall use the following auxiliary lemma. See [5] for details on measurability of set-valued functions.

Lemma 5.2

Let X be a separable metric space and let $U:X\rightarrow R$ be a continuous function. Then the mappings

$$\begin{aligned}{} & {} V:Comp(X)\rightarrow R \hbox { given by } V(K)=\sup _{x\in K}U(x),\\{} & {} \quad W:Comp(X)\rightarrow Comp(X) \hbox { given by } W(K)=\{ x\in K:U(x)=V(K)\} \end{aligned}$$

are measurable.

Theorem 5.3

Under assumptions of Theorem 4.8, there exists a sequence of measurable functions (a Castaing representation) $s_{\mathcal {J}}^{n}:R^{d}\times \mathcal {C}^{\alpha }( R^1)\rightarrow \mathcal {C}( R^{d}), n\ge 1$, such that $s_{\mathcal {J}}^{n}( \xi ,g)$ is an J-optimal solution to inclusion (4.1) for every $n\ge 1$, and such that

$$\begin{aligned} S_{\mathcal {J}}(\xi ,g,F, \mathcal {H})=cl_{\mathcal {C}}\{ s_{\mathcal {J}}^{n}(\xi ,g ):n\ge 1\} \end{aligned}$$

for every $\xi \in R^{d}$ and $g\in \mathcal {C}^{\alpha }( R^1)$.

Proof

By Theorem 4.13, $S( \cdot ,\cdot , F,\mathcal {H} ):R^{d}\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(\mathcal {C}( R^{d}) )$ is u.s.c. and hence by Proposition 8.2.1 of [5] it is measurable. In fact, in this case “measurable” means that $S( \cdot ,\cdot , F,\mathcal {H} )$ is measurable as the function from $R^{d}\times \mathcal {C}^{\alpha }( R^1)$ to a metric space $(Comp(\mathcal {C}(R^{d}),H_{\infty })$ (c.f., Theorem II.3.2. in [20]).

In the notation of Lemma 5.2, letting $X:=\mathcal {C}(R^{d})$ and $U:=\mathcal {J}$ we get $S_{\mathcal {J}}(\cdot ,\cdot ,F,\mathcal {H})=W\circ S(\cdot ,\cdot ,F,\mathcal {H} )$ which implies the measurability of the set $S_{\mathcal {J}}(\cdot ,\cdot ,F,\mathcal {H} )$. Finally, applying Theorem 1 of [17], we get the desired sequence of measurable selections $(\xi ,g)\mapsto s^n_{\mathcal {J}}(\xi ,g)$ of $S_{\mathcal {J}}(\cdot ,\cdot ,F,\mathcal {H} )$ and such that $S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})=cl_{\mathcal {C}}\{ s_{\mathcal {J}}^{n}(\xi ,g):n\ge 1\}$ for every $\xi \in R^{d}$ and $g\in \mathcal {C}^{\alpha }( R^1)$. $\square$

Once having properties of solutions sets $S(\xi ,g, F,\mathcal {H})$ and $S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})$ in (5.2) we can use them for the analysis of reachable sets of solutions in time $t\in [0,T]$. For $t\in [0,T]$, $\xi \in R^{d}$ and $g\in \mathcal {C}^{\alpha }( R^1)$, let

$$\begin{aligned} \mathcal {A}(t,\xi ,g):=\{ x(t):x\in S( \xi ,g,F,\mathcal {H}) \} \end{aligned}$$

and

$$\begin{aligned} \mathcal {A}_{\mathcal {J}}( t,\xi ,g):=\{ x(t):x\in S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})\} \end{aligned}$$

denote reachable and $\mathcal {J}$-optimal reachable sets of inclusion (4.1), respectively. Then $\mathcal {A}( t,\xi ,g ),\mathcal {A}_{\mathcal {J}}( t,\xi ,g) \in Comp(R^{d})$ for any $t\in [0,T]$, $\xi \in R^{d}$ and $g\in \mathcal {C}^{\alpha }( R^1)$ by Corollary 4.9.

Moreover, we have

Corollary 5.4

Let conditions of Theorem 4.8 be satisfied. Then the reachable sets of solutions to differential inclusion (4.1) admit the following properties:

(i) For every fixed $\xi \in R^{d}$ and $g\in \mathcal {C}^{\alpha }( R^1)$ set-valued functions

$$\begin{aligned} \mathcal {A}( \cdot ,\xi ,g),\mathcal {A}_{\mathcal {J}}( \cdot ,\xi ,g):[0,T]\rightarrow Comp(R^d) \end{aligned}$$

are $\alpha$-Hölder continuous on [0, T],

(ii) For every fixed $t\in [0,T]$ a set-valued function

$$\begin{aligned} \mathcal {A}( t,\cdot ,\cdot ):R^{d}\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(R^d) \end{aligned}$$

is upper semicontinuous, while

$$\begin{aligned} \mathcal {A}_{\mathcal {J}}( t,\cdot ,\cdot ):R^{d}\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(R^{d}) \end{aligned}$$

is measurable.

Proof

(i). Let $\xi \in R^{d}$, $g\in \mathcal {C}^{\alpha }( R^1)$ and $0\le s<t\le T$ be given. Suppose $z\in \mathcal {A}( t,\xi ,g)$. Then $z=x(t)$ for some $x\in S( \xi ,g,F,\mathcal {H})$. Hence

$$\begin{aligned}{} & {} \textrm{dist}_{R^{d}}( z,\mathcal {A}( s,\xi ,g)) \le \Vert x(t)-x(s)\Vert _{R^{d}}\\{} & {} \quad \le \Vert \int _{s}^{t}F(\tau ,x(\tau ))d\tau \Vert _{R^{d}}+\Vert \int _{s}^{t}(\mathcal {H}\circ x)dg\Vert _{R^{d}}. \end{aligned}$$

Thanks to (4.4), (4.5) applied to first term, and Corollary 4.2 applied to second term above, we get

$$\begin{aligned}{} & {} \textrm{dist}_{R^{d}}( z,\mathcal {A}( s,\xi ,g)) \le ( LT^{1-\alpha }\Vert x\Vert _{\infty }+\Vert b\Vert _{L^{\frac{1}{1-\alpha }}(R^d)}) ( t-s) ^{\alpha }\\{} & {} \quad +M_{\alpha }( g) ( A( x) +B( x)) ( 1+C(\alpha ,\beta )T^{\beta }) ( t-s)^{\alpha }, \end{aligned}$$

where

$$\begin{aligned} A( x):=K( \xi ) +\tilde{L}\max \{1,(M_{\alpha }( x))^{\gamma } \}( T^{\beta }+T^{\alpha \gamma }) \end{aligned}$$

and

$$\begin{aligned} B( x):=\tilde{L}\max \{1,(M_{\alpha }( x))^{\gamma }\}( 1+T^{\alpha \gamma -\beta }). \end{aligned}$$

Recall that the parameter $\delta$ from the formula (4.10) is equal to $\beta$ now, because of assumptions of Theorem 4.8. Moreover, by Theorem 4.5, the set $S(\xi ,g, F,\mathcal {H})$ is bounded in $\mathcal {C}^{\alpha }( R^{d})$. Thus $\sup \{ \Vert x\Vert _{\infty }: x\in S( \xi ,g,F,\mathcal {H})\}<\infty$ and $\sup \left\{ M_{\alpha }( x)^{\gamma }:x\in S( \xi ,g,F,\mathcal {H}) \right\} <\infty .$

Therefore, there exists a positive constant $C_{\mathcal {A}}(\xi ,g )$ such that

$$\begin{aligned} \textrm{dist}_{R^{d}}( z,\mathcal {A}( s,\xi ,g) ) \le C_{\mathcal {A}}(\xi ,g ) ( t-s) ^{\alpha }. \end{aligned}$$

Hence we get $\bar{H}_{_{R^{d}}}( \mathcal {A}( t,\xi ,g ),\mathcal {A}( s,\xi ,g) ) \le C_{\mathcal {A}}(\xi ,g ) (t-s) ^{\alpha }$. The same inequality holds for a semimetric $\bar{H}_{_{R^{d}}}( \mathcal {A}( s,\xi ,g),\mathcal {A}( t,\xi ,g ) )$. Thus

$$\begin{aligned} H_{R^{d}}( \mathcal {A}( t,\xi ,g ),\mathcal {A}( s,\xi ,g) ) \le C_{\mathcal {A}}(\xi ,g ) (t-s) ^{\alpha }. \end{aligned}$$

Moreover, since $\Vert \mathcal {A}( 0,\xi ,g) \Vert _{R^{d}}=\Vert \xi \Vert _{R^{d}}$ we get

$$\begin{aligned} \sup _{t\in [0,T]}\Vert \mathcal {A}( t,\xi ,g )\Vert _{R^{d}}\le \Vert \xi \Vert _{R^{d}}+C_{\mathcal {A}}(\xi ,g ) T^{\alpha }<\infty . \end{aligned}$$

In a similar way one can prove analogous inequalities for a set-valued function $\mathcal {A}_{\mathcal {J}}( \cdot ,\xi ,g)$. That ends the proof of part (i).

(ii). Let $\xi _{n}\rightarrow \xi$, $g _{n}\rightarrow g$ and let $z_{n}\in \mathcal {A}( t,\xi _{n},g_n)$ for $n\ge 1$. Then there exists a sequence $x_{n}\in S(\xi _{n},g_n,F,\mathcal {H})$ and such that $z_{n}=x_{n}(t)$ for $n\ge 1$. Since the set-valued function $S(\cdot ,\cdot ,F,\mathcal {H} )$ is u.s.c., it follows by Proposition 4.12, that there exists a subsequence $( x_{n_{k}}) _{k\ge 1}$ of a sequence $( x_{n}) _{n\ge 1}$ convergent in $\Vert \mathcal {\cdot }\Vert _{\infty }$-norm to some $x\in S(\xi ,g, F,\mathcal {H})$. Hence $z_{n_{k}}\rightarrow x(t)\in \mathcal {A}( t,\xi ,g )$ what finishes the proof of upper semicontinuity of $\mathcal {A}( t,\cdot ,\cdot )$. The measurability of $\mathcal {A}_{\mathcal {J}}(t,\cdot ,\cdot ):R^{d}\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(R^{d})$ follows by measurability of the $\mathcal {J}$-optimal solution map $S_{\mathcal {J}}(\cdot ,\cdot ,F,\mathcal {H} )$ and continuity of the projection map $\Pi _t: \mathcal {C}^{\alpha }( R^{d})\rightarrow R^{d}$, $\Pi _t(x)=x(t)$ for $t\in [0,T]$, because of $\mathcal {A}_{\mathcal {J}}(t,\cdot ,\cdot )=\Pi _t \circ S_{\mathcal {J}}(\cdot ,\cdot ,F,\mathcal {H})$. $\square$

Applying similar methods one can show the following property.

Corollary 5.5

Under conditions of Theorem 4.8 a reachable map

$$\begin{aligned} \mathcal {A}( \cdot ,\cdot ,\cdot ):[0,T]\times R^d\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(R^{d}) \end{aligned}$$

is upper semicontinuous on $[0,T]\times R^{d}\times \mathcal {C}^{\alpha }( R^1)$, while

$$\begin{aligned} \mathcal {A}_{\mathcal {J}}( \cdot ,\cdot ,\cdot ):[0,T]\times R^d\times \mathcal {C}^{\alpha }( R^1)\rightarrow Comp(R^{d}) \end{aligned}$$

is of a type of Carathéodory, i.e., $\mathcal {A}_{\mathcal {J}}$ is measurable in $(\xi ,g)$ for every fixed $t\in [0,T]$ and continuous in t for every fixed $(\xi ,g)\in R^d\times \mathcal {C}^{\alpha }(R^1)$.

At the end of this section we focus once again on the set of optimal solutions to differential inclusion (4.1), i.e., on the set $S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})$ defined in (5.2). Since $\mathcal {J}:\mathcal {C}( R^{d}) \rightarrow R_{+}$ is continuous by assumption, then $\mathcal {J}\in \mathcal {C}\left( \mathcal {C}\left( R^{d}\right) ,R_{+}\right)$, where $\mathcal {C}\left( \mathcal {C}( R^{d}),R_{+}\right)$ denotes the space of all continuous functions form $\mathcal {C}(R^{d})$ to $R_{+}$. We consider this space with a topology of uniform convergence on compacts (uc topology), i.e., for $\mathcal {J}_{n},\mathcal {J}\in \mathcal {C}\left( \mathcal {C}( R^{d}),R_{+}\right)$, $n\ge 1$

$$\begin{aligned} \mathcal {J}_{n}\rightarrow ^{uc}\mathcal {J}\hbox { if and only if }\sup _{x\in K}|\mathcal {J}_{n}(x)-\mathcal {J}(x)|\rightarrow 0\hbox { as }n\rightarrow \infty \end{aligned}$$

for every $K\in Comp \left( \mathcal {C}(R^{d}\right)$. Let us consider set-valued functions

$$\begin{aligned} \mathcal {C}\left( \mathcal {C}(R^{d}),R_{+}\right) \ni \mathcal {J}\longrightarrow S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})\in Comp\left( \mathcal {C}(R^{d}) \right) \end{aligned}$$

and

$$\begin{aligned} \mathcal {C}\left( \mathcal {C}(R^{d}),R_{+}\right) \ni \mathcal {J}\longrightarrow \mathcal {A}_{\mathcal {J}}(t,\xi ,g )\in Comp(R^{d}). \end{aligned}$$

We show that these set-valued functions are upper semicontinuous. In view of comments stated after Proposition 4.12 it is equivalent to the following result.

Theorem 5.6

Let assumptions of Theorem 4.8 be satisfied. Suppose that functionals $\mathcal {J}_{n},\mathcal {J}\in \mathcal {C}\left( C(R^{d}),R_{+}\right)$, $n\ge 1$ satisfy $\mathcal {J}_{n}\rightarrow ^{uc}\mathcal {J}$. Then

$$\begin{aligned} \bar{H}_{\infty }\left( S_{\mathcal {J}_{n}}(\xi ,g, F,\mathcal {H}),S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})\right) \rightarrow 0\hbox { as }n\rightarrow \infty . \end{aligned}$$

(5.3)

$$\begin{aligned} \bar{H}_{R^d }\left( \mathcal {A}_{\mathcal {J}_{n}}(t,\xi ,g ),\mathcal {A}_{\mathcal {J}}(t, \xi ,g )\right) \rightarrow 0\hbox { as }n\rightarrow \infty . \end{aligned}$$

(5.4)

Proof

We will prove (5.3). To simplify the notation let $K_{0}:=S(\xi ,g, F,\mathcal {H})$ denote the set of all solutions to differential inclusion (4.1). By Corollary 4.9$K_{0}$ is compact in $\mathcal {C}( R^{d})$ and for each $n\ge 1$ the sets

$$\begin{aligned} S_{\mathcal {J}_{n}}(\xi ,g, F,\mathcal {H})=\{ x\in K_{0}:\mathcal {J}_{n}( x) =\sup \{ \mathcal {J}_{n}(y):y\in K_{0}\} \} \end{aligned}$$

(5.5)

and

$$\begin{aligned} S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})=\{ x\in K_{0}:\mathcal {J}( x) =\sup \{ \mathcal {J}(y):y\in K_{0}\} \} \end{aligned}$$

(5.6)

are compact as well. Let $2^{K_0}$ denote a family of all subsets of a set $K_0$. We define set-valued functions $G_{n}:R_{+}\rightarrow 2^{K_{0}}$ and $G:R_{+}\rightarrow 2^{K_{0}}$ by formulas

$$\begin{aligned} G_{n}(u):=\{ y\in K_{0}:\mathcal {J}_{n}(y)=u\} \text{ and } G(u):=\{ y\in K_{0}:\mathcal {J}(y)=u\}. \end{aligned}$$

(5.7)

In this notation if $u_{\mathcal {J}_{n}}:=\sup \{ \mathcal {J}_{n}(y):y\in K_{0}\}$, then we have $G_{n}(u_{\mathcal {J}_{n}})=S_{\mathcal {J}_{n}}(\xi ,g, F,\mathcal {H})$ and similarly if $u_{\mathcal {J}}:=\sup \left\{ \mathcal {J}(y):y\in K_{0}\right\}$, then $G(u_{\mathcal {J}})=S_{\mathcal {J}}(\xi ,g, F,\mathcal {H}).$ Thus in order to show (5.3) it is sufficient to show that

$$\begin{aligned} \bar{H}_{\infty }\left( G_{n}(u_{\mathcal {J}_{n}}),G(u_{\mathcal {J}})\right) \rightarrow 0\hbox { as }n\rightarrow \infty \end{aligned}$$

(5.8)

provided $\mathcal {J}_{n}\rightarrow ^{uc}\mathcal {J}$. Therefore, assume $\mathcal {J}_{n}\rightarrow ^{uc}\mathcal {J}$. Then for each $K\in Comp\left( \mathcal {C}(R^{d})\right)$ we get

$$\begin{aligned} \sup _{x\in K}\mathcal {J}_{n}(x)\rightarrow \sup _{x\in K}\mathcal {J}(x) \text{ and } \text{ in } \text{ particular } u_{\mathcal {J}_{n}}\rightarrow u_{\mathcal {J}}\hbox { as }n\rightarrow \infty . \end{aligned}$$

(5.9)

Suppose that (5.8) does not hold. Then there exists $\epsilon >0$ such that for every $k\ge 1$ one can select $n_{k}\ge k$ satisfying

$$\begin{aligned} \sup _{y\in G_{n_{k}}(u_{\mathcal {J}_{n_{k}}})}\textrm{dist}_{\mathcal {C}}(y,G(u_{\mathcal {J}}))\ge \epsilon , \end{aligned}$$

where $\textrm{dist}_{\mathcal {C}}(y,A)=\inf _{x\in A}\Vert y-x\Vert _{\infty }$. Thus one can choose a sequence $\left( y_{n_{k}}\right)$ such that $y_{n_{k}}\in G_{n_{k}}(u_{\mathcal {J}_{n_{k}}})$ and

$$\begin{aligned} \textrm{dist}_{\mathcal {C}}(y_{n_{k}},G(u_{\mathcal {J}}))\ge \epsilon /2\hbox { for every }k\ge 1. \end{aligned}$$

(5.10)

On the other hand, since $( y_{n_{k}}) _{k\ge 1}$ is contained in a compact set $K_{0}$ then we may assume (passing if needed to a subsequence) that $( y_{n_{k}}) _{k\ge 1}$ uniformly converges to some $y_{0}\in K_{0}$. But $y_{n_{k}}\in G_{n_{k}}(u_{\mathcal {J}_{n_{k}}})$ and therefore, we obtain $\mathcal {J}_{n_{k}}(y_{n_{k}})=u_{\mathcal {J}_{n_{k}}}$ for every $k\ge 1$ by (5.7). Hence by (5.9) we get $\mathcal {J}_{n_{k}}(y_{n_{k}})\rightarrow u_{\mathcal {J}}$ as $k\rightarrow \infty$. Since $\mathcal {J}_{n}\rightarrow ^{uc}\mathcal {J}$, then we claim $\mathcal {J}_{n_{k}}(y_{n_{k}})\rightarrow \mathcal {J}(y_{0})$. Therefore, $\mathcal {J}(y_{0})=u_{\mathcal {J}}$ what means that $y_{0}\in G(u_{\mathcal {J}})$. Recall that for a metric space X and a fixed nonempty and closed set $A\subset X$ the function

$$\begin{aligned} X\ni x\rightarrow \textrm{dist}(x,A)\in R_{+} \end{aligned}$$

is Lipschitz continuous with a Lipschitz constant one. Thus

$$\begin{aligned} \textrm{dist}_{\mathcal {C}}(y_{n_{k}},G(u_{\mathcal {J}}))\rightarrow \textrm{dist}_{\mathcal {C}}(y_{0},G(u_{\mathcal {J}}))=0\hbox { as }k\rightarrow 0 \end{aligned}$$

what contradicts with (5.10). This proves the formula (5.3).

From (5.3) together with (4.23), it follows that the set-valued function $J\rightarrow S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})$ is u.s.c. Therefore, $J\rightarrow \mathcal {A}_{\mathcal {J}}(t,\xi ,g )$ is u.s.c. also, because $\mathcal {A}_{\mathcal {J}}(t, \xi ,g )=\Pi _t \circ S_{\mathcal {J}}(\xi ,g, F,\mathcal {H})$. Then formula (5.4) follows by (4.23). This finishes the proof. $\square$