Nemytskii Operator in the Space of Set-Valued Functions of Bounded φ-Variation ()
1. Introduction
In [1], it was proved that every globally Lipschitz Nemytskii operator
mapping the space 
into itself admits the following representation:
where 
is a linear continuous set-valued function and 
is a set-valued function belonging to the space
. The first such theorem for singlevalued functions was proved in [2] on the space of Lipschitz functions. A similar characterization of the Nemytskii operator has also been obtained in [3] on the space of set-valued functions of bounded variation in the classical Jordan sense. For single-valued functions it was proved in [4]. In [5,6], an analogous theorem in the space of set-valued functions of bounded 
-variation in the sense of Riesz was obtained. Also, they proved a similar result in the case in which that the Nemytskii operator N maps the space of functions of bounded 
-variation in the sense of Riesz into the space of set-valued functions of bounded 
-variation in the sense of Riesz, where
, and 
is globally Lipschitz. In [7], they showed a similar result in the case where the Nemytskii operator 
maps the space 
of setvalued functions of bounded 
-variation in the sense of Riesz into the space 
of set-valued functions of bounded 
-variation in the sense of Riesz and 
is globally Lipschitz.
While in [8], we generalize article [6] by introducing a weight function. Now, we intend to generalize [7] in a similar form we did in [8], i.e., the propose of this paper is proving an analogous result in which the Nemytskii operator 
maps the space 
of setvalued functions of bounded 
-variation in the sense of Riesz with a weight 
into the space 
of set-valued functions of bounded 
-variation in the sense of Riesz with a weight 
and 
is globally Lipschitz.
2. Preliminary Results
In this section, we introduce some definitions and recall known results concerning the Riesz 
-variation.
Definition 2.1 By a 
-function we mean any nondecreasing continuous function 
such that 
if and only if
, and 
as
.
Let 
be the set of all convex continuous functions that satisfy Definition 2.1.
Definition 2.2 Let 
be a normed space and 
be a 
-function. Given 
be an arbitrary (i.e., closed, half-closed, open, bounded or unbounded) fixed interval and 
a fixed continuous strictly increasing function called a it is weight. If
, we define the (total) generalized 
-variation 
of the function 
with respect to the weight function 
in two steps as follows (cf. [9]). If 
is a closed interval and 
is a partition 
of the interval I (i.e.,
), we set
Denote by 
the set of all partitions of
, we set
If 
is any interval in
, we put
The set of all functions of bounded generalized 
- variation with weight 
will be denoted by
.
If
, and
, 
, 
, the 
-variation
, also written as
, is the classical 
-variation of 
in the sense of Riesz [10], showing that 
if and only if 
(i.e., 
is absolutely continuous) and its almost everywhere derivative 
is Lebesgue 
-summable on
. Recall that, as it is well known, the space 
with I, 
and 
as above and endowed with the norm 
is a Banach algebra for all
.
Riesz’s criterion was extended by Medvedev [11]: if
, then 
if and only if 
and
. Functions of bounded generalized 
-variation with 
and 
(also called functions of bounded Riesz-Orlicz 
-variation) were studied by Cybertowicz and Matuszewska [12]. They showed that if
, then
and that the space
is a semi-normed linear space with the LuxemburgNakano (cf. [13,14]) seminorm given by
.
Later, Maligranda and Orlicz [15] proved that the space 
equipped with the norm
is a Banach algebra.
3. Generalization of Medvedev Lemma
We need the following definition:
Definition 3.1 Let 
be a 
-function. We say 
satisfies condition 
if
(1)
For φ convex, (1) is just
. Clearlyfor 
the space 
coincides with the classical space 
of functions of bounded variation. In the particular case when 
and 
, we have the space 
of functions of bounded Riesz 
-variation. Let 
be a measure space with the Lebesgue-Stieltjes measure defined in 
-algebra 
and
Moreover, let 
be a function strictly increasing and continuous in
. We say that 
has 
- measure 0, if given 
there is a countable cover 
by open intervals of
, such that
.
Since 
is strictly increasing, the concept of “
measure
” coincides with the concept of “measure 0” of Lebesgue. [cf. [16], 
25].
Definition 3.2 (Jef) A function 
is said to be absolutely continuous with respect to
, if for every
, there exists 
such that
for every finite number of nonoverlapping intervals
, 
with 
and
.
The space of all absolutely continuous functions
, with respect to a function 
strictly increasing, is denoted by
. Also the following characterization of [17,18] is well-known:
Lemma 3.3 Let
. Then 
exists and is finite in
, except on a set of 
-measure
.
Lemma 3.4 Let
. Then 
is integrable in the sense Lebesgue-Stieltjes and
Lemma 3.5 Let 
such that satisfies the 
condition. If
, then 
is 
-absolutely continuous in
, i.e.,
Also the following is a generalization of Medvedev Lemma [11]:
Theorem 3.6 (Generalization a Medvedev Lemma) Let 
such that satisfies the 
condition,
. Then 1) If 
is 
-absolutely continuous on 
and
then
and
.
2) If 
(i.e.,
), then 
is 
-absolutely continuous on 
and
.
Proof.
) Since 
is 
absolutely continuous, there exists 
a.e. in 
by Lemma 3.3. Let
, 
by Lemma 3.4 and 
is strictly increasing
using the generalized Jenssen’s inequality
Let 
be any partition of interval
; then
and we have
.
Thus
.
) Let
. Then 
is 
-absolutely continuous on 
by Lemma 3.5 and 
exist a.e. on
.
For every
, we consider
a partition of the interval 
define by
,
.
Let 
be a sequence of step functions, defined by 
converge to 
a.e. on
. It is sufficient to prove 
in those points where 
is 
- differentiable and different from
, 
for
, i.e., in
For
, and each
, there exists 
such that
, so
Therefore, 
is a convex combination of points
Now if
, then 
and 
and since 
is 
-differentiable for
, the expressions
tend 
to which is 
-differentiable from 
in
. So results
Since 
is continuous, we have
Using the Fatou’s Lemma and definition of 
sequence, results that
By definition from 
which is what we wished to demonstrate.
Corollary 3.7 Let 
such that satisfies the 
condition, then 
if and only if 
is 
-absolutely continuous on 
and
.
Also
Corollary 3.8 Let 
such that satisfies the 
condition. If
, then 
is 
-absolutely continuous on 
and
4. Set-Valued Function
Let 
be the family of all non-empty convex compact subsets of 
and 
be the Hausdorff metric in
, i.e.,
where
, or equivalently,
where
(2)
Definition 4.1 Let
, 
a fixed continuous strictly increasing function and
. We say that 
has bounded 
-variation in the sense of Riesz if
(3)
where the supremum is taken over all partitions 
of
.
Definition 4.2 Denote by
(4)
and
(5)
both equipped with the metric
(6)
where
Now, let
, 
be two normed spaces and 
be a convex cone in
. Given a set-valued function 
we consider the Nemytskii operator 
generated by
, that is the composition operator defined by:
We denote by 
the space of all setvalued function
, i.e., additive and positively homogeneous, we say that 
is linear if
.
In the proof of the main results of this paper, we will use some facts which we list here as lemmas.
Lemma 4.3 ([19]) Let 
be a normed space and let 
be subsets of
. If 
are convex compact and 
is non-empty and bounded, then
(7)
Lemma 4.4 ([20]) Let
, 
be normed spaces and 
be a convex cone in
. A set-valued function 
satisfies the Jensen equation
(8)
if and only if there exists an additive set-valued function 
and a set 
such that
,
.
We will extend the results of Aziz, Guerrero, Merentes and Sánchez given in [8] and [21] to set-valued functions of 
-bounded variation with respect to the weight function
.
5. Main Results
Lemma 5.1 If 
such that satisfies the 
condition and
then 
is continuous.
Proof. Since
, exists 
such that
(9)
for all partitions of
, in particular given
, we have
(10)
Since 
is convex 
-function, from the last inequality, we get
(11)
By (1),
(12)
This proves the continuity of 
at
. Thus 
is continuous on
.
Now, we are ready to formulate the main result of this work.
Main Theorem 5.2 Let
, 
be normed spaces, 
be a convex cone in 
and 
be two convex 
-functions in
, strictly increasing, that satisfy 
condition and such that there exists constants 
and 
with 
for all
. If the Nemitskii operator 
generated by a set-valued function 
maps the space
into the space
and if it is globally Lipschitz, then the set-valued function 
satisfies the following conditions:
1) For every 
there exists
, such that
(13)
2) There are functions 
and 
such that
(14)
Proof. 1) Since 
is globally Lipschitz, there exists a constant 
such that
(15)
Using the definitions of the operator 
and metric 
we have
where
. In particular,
for all 
and
, 
, where
.
Since 
and 
satisfy
(16)
we obtain
Therefore
(17)
Define the auxiliary function 
by:
(18)
Then 
and
Let us fix 
and define the functions 
by:
(19)
Then the functions 
and
(20)
From the definition of 
and
, we have
(21)
From (16), we get
(22)
Hence,
(23)
Hence, substituting in inequality (5) the particular functions 
defined by (19) and taking 
in (23), we obtain
(24)
for all
.
By Lemma 4.3 and the inequality (24), we have
for all
.
Now, we have to consider the case
. Define the function 
by
(25)
Then the function 
and
Let us fix 
and define the functions 
by
(26)
Then the functions 
(i = 1,2) and
Substituting 
and
, and consider
, we obtain
(27)
for all
, where
By Lemma 4.3 and the above inequality, we get
for all
. Define the function 
by
Hence
and, consequently, for every 
the function 
is continuous.
This completes the proof of part 1).
Now we shall prove that 
satisfies equality 2).
Let us fix 
such that
. Since the Nemytskii operator 
is globally Lipschitzian, there exists a constant
, such that
(28)
where
. Define the function 
by
The function
.
Let us fix 
and define the functions 
by
(29)
The functions 
and
.
Hence, substituting in the inequality (28) the particular functions 
defined by (29), we obtain
(30)
Since 
maps
into
then 
is continuous for all
. Hence letting 
in the inequality (30), we get
(31)
for all 
and
.
Thus for all
, we have
(32)
Since 
is convex, we have
(33)
for all
. Thus for all
, the set-valued function 
satisfies the Jensen Equation (33). Now by Lemma 4.4, there exists an additive set-valued function 
and a set
, such that
(34)
Substituting 
into inequality (13), we deduce that for all 
there exists
, such that
consequently, for every 
the set-valued function 
is continuous, and
.
Since 
is additive and
, then 
for all
, thus
.
The Nemytskii operator 
maps the space 
into the space
, then
.
Consequently the set-valued function 
has to be of the form
where 
and
.
Theorem 5.3 Let
, 
be normed spaces, 
a convex cone in 
and 
be two convex 
-functions in
, strictly increassing, satisfying 
condition and
. If the Nemytskii operator 
generated by a set-valued function
maps the space 
into the space 
and if it is globally Lipschizian, then the set-valued function 
satisfies the following condition
i.e., the Nemytskii operator is constant.
Proof. Since the Nemytskii operator 
is globally Lipschizian between 
and the space
, then there exists a constant
, such that
(35)
Let us fix 
such that
. Using the definitions of the operator 
and of the metric
, we have
(36)
Define the auxiliary function 
by
The function 
and
Let us fix 
and define the functions 
by
(37)
The functions 
and
Hence, substituting in the inequality (36) the auxiliary functions 
defined by (37), we obtain
By Lemma 4.3 and the above inequality, we get
Since
, letting 
in the above inequality, we have
Thus for all 
and for all
, we get
Theorem 5.4 Let
, 
be normed spaces, 
a convex cone in 
and 
be a convex 
- function in 
satisfying the 
condition. If the Nemytskii operator 
generated by a set-valued function 
maps the space 
into the space 
and if it is globally Lipschizian, then the left regularization 
of the function 
defined by
satisfies the following conditions:
• for all 
there exists
, such that
• 
• 
, where 
is a linear continuous set-valued function, and
.
Proof. We take
, and define the auxiliary function 
by:
The function 
and
Let us fix 
and define the functions 
by
(38)
The functions 
and
(39)
From the definition of 
and
, we obtain
(40)
Since the Nemytskii operator 
is globally Lipschitzian between
and
then there exists a constant
, such that
for
. By Lemma 4.3, substituting the particular functions 
defined by (38) in the above inequality, we obtain
(41)
for all
. By Lemma 4.3, we get
(42)
for all 
and
.
In the case where
, by a similar reasoning as above, we obtain that there exists a constant
, such that
(43)
Define the function 
by
(44)
Hence,
By passing to the limit in the inequality (41) by the inequality (43) and the definition of 
we have for all 
that there exists
, such that
Now we shall prove that 
satisfies the following equality
where 
is a linear continuous set-valued functions, and
.
Let us fix 
such that
. Define the partition 
of the interval 
by
The Nemytskii operator 
is globally Lipschitzian between 
and
, then there exists a constant
, such that
(45)
where
and
.
We define the function 
in the following way:
The function 
and
.
Let us fix 
and define the functions 
by:
(46)
The functions 
and
.
Substituting in the inequality (45) the particular functions 
defined in (46), we obtain
(47)
Since the Nemytskii operator 
maps the spaces 
into
, then for all
, the function
. Letting 
in the inequality (47), we get
for all 
and
. By passing to the limit when
, we get
Since 
is a convex function, then
Thus for every
, the set-valued function 
satisfies the Jensen equation. By Lemma 4.4 and by the property (a) previously established, we get that for all 
there exist an additive set-valued function 
and a set
, such that
By the same reasoning as in the proof of Theorem 5.2, we obtain that
and
.
6. Acknowledgements
This research was partly supported by CDCHTA of Universidad de Los Andes under the project NURR-C- 547-12-05-B.