Articles

Recepción: 01 Junio 2020
Revisado: 24 Febrero 2021
Publicación: 01 Septiembre 2021
DOI: https://doi.org/10.15388/namc.2021.26.24367
Abstract: We present a novel generalization of the Hyers–Ulam–Rassias stability definition to study a generalized cubic set-valued mapping in normed spaces. In order to achieve our goals, we have applied a brand new fixed point alternative. Meanwhile, we have obtained a practicable example demonstrating the stability of a cubic mapping that is not defined as stable according to the previously applied methods and procedures.
Keywords: stability, orthogonal set, cubic mapping, fixed point, incomplete metric space.
1 Introduction and literature reviews
The study for the set-valued dynamics in Banach spaces has been developed in the last decades. The pioneering published papers by Aumann [2] and Debreu [9] were inspired by some problems arising in the control theory and mathematical economics. We refer to the articles by Arrow and Debreu [1], McKenzie [29], and the survey by Hess [18].
The stability of functional equations was first introduced by Ulam [38] in 1940. He proposed the following problem: Given a group G1, a metric group
, and a positive number
, does there exist a
such that if a mapping
satisfies the inequality
for all
, then there is a homomorphism
such that
for all
. If the answer is positive, we say that the homomorphisms from
to
are stable. In 1941, Hyers [19] gave a partial solution of Ulam’s problem for the case of approximate additive mappings under the assumption that
and
are Banach spaces. In 1978, a generalized version of the theorem of Hyers by considering the stability problem with unbounded Cauchy differences was given by Rassias [36]. This phenomenon of stability that was introduced by Rassias [36] is called the Hyers–Ulam–Rassias stability of functional equations.
Theorem 1. Let
be a mapping from a normed vector space E into a Banach space E′ subject to the inequality

where ϵ and p are constants with
, and
. Then there exists a unique additive mapping
such that in the case of
,

while in the case of
,

The solution to this problem was obtained by Gajda [13] for
, and the problem for
was solved by Rassias [36]. Rassias and Semrl [37] proved that the stability does not occur for
. The result of the Rassias theorem was generalized by Forti [11] and Gávruta [14], who permitted the Cauchy difference to become arbitrary unbounded.
The stability problems of several functional equations have been extensively investigated by many mathematicians. The results of these kinds of problems have been extensively studied. We refer, for instance, to [6, 12, 15–17, 19, 27] and also [11, 13, 14, 22, 25, 36, 37] and references therein.
A stability problem of Ulam for the cubic functional equation

was established by Jun and Kim [22] for mapping
, where
is a normed space, and
a Banach space. Also, they solved the stability problem of Ulam for the generalized Euler–Lagrange-type cubic functional equation

for fixed integer a with
, and

for fixed integers a, b with
, and
, and the equations being equivalent to (1). Afterwards, referring to [7], Chu et al. extended the cubic functional equation to the following generalized form:

where
is an integer, and they also investigated the Hyers–Ulam stability. Moreo ver, in [25], Jung and Chang investigated a generalized Hyers–Ulam–Rassias stability for a cubic functional equation by using the fixed point alternative. The first systematic study of the interative methods in the stability of mappings is due to Isac and Rassias [20]
The stability of the set-valued functional equations has been widely examined by a number of authors (see [8, 21, 30–32]), and the Hyers–Ulam stability of the set-valued functional equations was proved in [21, 26, 28]. Also, there are many interesting stability results concerning this problem (see [8, 23, 24]).
Quite recently, Eshaghi et al. [10] and M. Ramezani et al. [35] introduced the notion of orthogonal sets and gave a real generalization of the Banach fixed point theorem in incomplete metric spaces. The main result of [10] is the following theorem.
Theorem 2.(See [10].) Let
be an O-complete orthogonal metric space (not necessarily complete metric space) and
. Let
be O-continuous,
-contraction with Lipschitz constant
, and
-preserving. Then
has a unique fixed point
. Also,
is a Picard operator, that is, l
for all
.
For more details about the orthogonal space, we refer the reader to [3–5, 10, 34, 35].
The aim of this paper is to offer a new generalized Hyers–Ulam–Rassias stability result for the functional equation (2) for the set-valued mappings in normed spaces, which are not necessarily Banach spaces, by using the fixed point alternative [10] as in [3]. Examplewise, we present a special case of our results, which is a real extension of the previous results as of this literature.
At first, we recall some basic definitions and our main tools.
Definition 1. (See [10].) Let
and
be a binary relation. If
satisfies the following condition be a binary relation. If satisfies the following condition

then
is called an orthogonal relation, and the pair
– an orthogonal set (briefly, O- set)
Note that in the above definition, we say that
is an orthogonal element. Also, we say that elements
are
-comparable either
or
.
Definition 2. (See [10, 35].) Let
be O-set. A sequence
is called
(i) an orthogonal sequence (briefly, O-sequence) if

(ii) an strongly orthogonal sequence (briefly, SO-sequence) if

Every SO-sequence is an O-sequence. But the converse is not true in general.
Definition 3. (See [10, 35].) Let
be an orthogonal metric space (
is an O-set, and
a metric space).
is
orthogonal complete (briefly, O-complete) if every Cauchy O-sequence is convergent;
strongly orthogonal complete (briefly, SO-complete) if every Cauchy SO-sequence is convergent.
It is easy to see that every complete metric space is O-complete and every O-complete metric space is SO-complete. In [3, 35], the authors proved that the converse is not true in general.
Definition 4. (See [10, 35].) Let
be an orthogonal metric space. Then
is
orthogonal continuous (briefly, O-continuous) at
if for each O-sequence
in
implies
.
strongly orthogonal continuous (briefly, SO-continuous) at
if for each SO-sequence
in
implies
.
Also,
is O-continuous (SO-continuous) on
if
is O-continuous (SO-continuous) in each
.
It is obvious that every continuous mapping is O-continuous and every O-continuous mapping is SO-continuous, but the converse is not hold in general (see [3, 35]).
Definition 5. (See [3].) Let
be an O-set. A mapping
is said to be
-preserving if
whenever
and
.
Theorem 3.Let
be an SO-complete orthogonal metric space (not necessarily complete metric space) and
. Let
be SO-continuous,
-preserving, and
-contraction with Lipschitz constant
. Then f has a unique fixed point
. Also,
is a Picard operator, that is,
for all
.
Proof. The proof of this result uses the same ideas in Theorem 3.11 of [10], and it suffices to replace the O-sequence by SO-sequence.
By the aforementioned results we can conclude that Theorem 3 is a real generalization of Theorem 2. So, in the next steps, we are going to prove the stability of functional equation (2) in the SO-complete normed spaces.
2 An incomplete distance on subsets of a set
Before introducing the main results, we recall some notations and definitions.
Let
be a normed space (not necessarily a Banach space), and let
be anorthogonal relation on
such that
is an orthogonal metric space, wheredisthe induced metric by
.
We say that
is
-preserving whenever
implies
for each
. See the next example.
Example 1. Let
, and let two relations
and
on
be defined as

It is obvious that an orthogonal element of
and
is zero. However,
is not
-preserving. To see this, if
, and
, then
, while
. Notice that it is easy to see that
is
-preserving.
Let
be the set of all nonempty, closed, convex, and bounded subsets of
. Consider the addition and the scalar multiplication as follows:

where
) and
. One can show that

for all
and
. We consider
on pairs of elements in
by

where
and
. Pathak and Shahzad in [33] proved that
is a metric on
. We define the relation
on
as

The following proposition can be proved from some properties of the distance
.
Proposition 1 .(See[33].) For any
and
, the following properties hold:



;
, where
for all positive real number r;

for each positive real number r.
Given
. We define the relation
between A and B as follows:

If
is an orthogonal element of
, then the singleton
is an orthogonal element for
.
Theorem 4. If
is an SO-complete .not necessarily complete. metric space, then
with orthogonal relation
is SO-complete.
Proof. Let
be a Cauchy SO-sequence in
. We need to show that
converges to some element in
.
Let A be the set of limit points of sequences
with
for all
. Our aim is to prove that
and
converges to A. To see end, let us to divide the proof in the following steps.
Step 1:Ais closed. Let
. Definition of A ensures that we can choose the sequence
in A converging to a. This leads to for all
, there exists
in
such that for any
and
as
. Let
be a strictly increasing sequence of positive integers such that for any,
. We observe that

As
, the right-hand of above inequality converges to zero, which implies
.
Step 2:A is convex. Let
and
. Take two sequences
and
such that for each
,
and
and
as
. Since for any
,
is a convex set, then
. The closeness of A implies that
.
Step 3: A is nonempty. We observe from
is a Cauchy sequence that there exists a strictly increasing sequence
such that for all
and
,
. Definition of
ensures that for each
, there exists
for which kank
. This results show that the sequence
is Cauchy.
On the other hand, since
is an SO-sequence, it follows that

Therefore,
is a Cauchy SO-sequence in
. Since
is SO-complete and
for each
, the conclusion follows easily.
Step 4:
. Fix
. There exists a positive integer
such that for all
. By definition of
and condition (v) of
Proposition 1 we see that for all
and
, where
. Let
and
be a sequence such that
for all i and
converges to a. We observe that for all
, and the continuity of D implies that
for all
. This results show that
for all
.
On the other hand, we can choose a positive integer
such that for all
and a strictly increasing sequence
of positive integers such that
and
for all
.
Assume
and
. It follows from
that there is
for whichky
. Similarly, for each i, since 
, then there is
for which
. We easily see that
is a Cauchy sequence. Arguing in the Step 3, we obtain that
is an SO-sequence in
and so converges to an element
. Moreover, for all
,

For large enough numbers of i,
, which implies that
, and hence,
for all
.
Now, take
, then condition (vi) of Proposition 1 ensures that
for each
. This completes the proof of Step 4.
3 New generalized Hyers–Ulam–Rassias stability
Throughout this section, we assume
and
are two normed spaces. Also,
and
are the same orthogonal relations on Y and
as defined in the previous section, respectively. We consider the relation
as
-persevering and d as the metric induced by
.
Definition 6. Let
be a set-valued mapping.
(i) The n-dimensional cubic set-valued functional equation is defined by

for every
, where
is an integer.
(ii) Every solution of the n-dimensional cubic set-valued functional equation is called an n-dimensional cubic set-valued mapping.
Theorem 5.Let
be an integer,
and
be an SO-complete metric space (not necessarily a complete metric space). Assume that
is a set-valued mapping such that
and
are
-comparable for each
and
, and also, there exist two functions
and
satisfying the following conditions:

for all
and also
(A1) 
(A2) For all
,

(A3)For all
,

Then there exist an n-dimensional cubic set-valued mapping
and a subset
in
with card
such that for some positive real number
, we have

for all
. In particular, if
, then the mapping
is unique.
Proof. We denote by
the set

and the generalized metric D on
as follows:

Consider the set
. Putting
in (A2) yields that
, and by using (3) we observe that
. Hence S is a nonempty set.
Now, let
be a function as given by
for all
. We must show that T is a self-adjoint mapping, that is,
. To see this, put 
and
in inequality (3). Since the range of f is convex and applying (A2), we have

and so,

for all
. Dividing by 8m in (5), we get

for all
. Replacing x by
in (6) and applying (A2), we have

for all
. This ensures that
. On the other hand, if
, definition of D conclude that
, and the triangle inequality implies that
, that is,
. Consider

for all
. Define the relation
on S as the following:

It follows from Theorem 4 that
is an SO-complete metric space. Since the relation
is
-preserving, definition of
and
imply that T is
-preserving. By using the hypothesis we obtain

for all
and
. From
-preserving of
and definition of T we get

for all
and
. This means that

for all
. It follows from
-preserving of T that

That is,
and consequently
for all
are SO-sequences in S and
, respectively. In order to show that the SO-sequence
is Cauchy, replacing
and multiplying by 8r in (7) and using (A2) and (A3), we get

all
and
. Considering

we obtain that

for all
and
with
. Since
, letting
in the above inequality, we deduce that the sequence
is a Cauchy sequence for each
. By SO-completeness of
we obtain that for every
, there exists an element
, which is a limit point of
. That is,
is well defined and given by

for all . On the other hand, since, then there exist and such that for all . Put

It follows from
that
. Also, if
is an arbitrary nonzero point of
, then by using (A2) we can easily see that

So, there exists a natural number
for which

and this means that
belongs to
. This implies that card
. Now, were place
by
in definition of
. For
, we have the following implications

Hence, we see that
for all
. It follows from
that T is a contraction. Consequently, T is an SO-continuous mapping and is a contraction on
-comparable elements with Lipschitz constant L. Since
is SO-complete and T is also S-preserving, then from Theorem 3 we conclude that T has a unique fixed point and T is a Picard operator. This means that the sequence
is convergent to the fixed point of T . It follows from (8) that F is a unique fixed point of T . Moreover,

Therefore,
. Relation (7) ensures that inequality (4) holds. Finally, we need to show that F is an n-dimensional cubic set-valued mapping. To this end, let
be fixed elements of
. Since
is a nonnegative and decreasing sequence, then there is
for which
as
. Taking into account (A1), we have
, so there exist
and
such that for all
. Consider the positive integer N such

that for
. By virtue of (3) we obtain
Therefore, F is an n-dimensional cubic set-valued mapping as desired.
Corollary 1. Let
be an integer and
. Let Y be a Banach space and
be a mapping such that there exists a function
satisfying

for all
. If there exists a positive real number
such that

for all
, then there exists a unique n-dimensional cubic mapping
, which satisfies the inequality

for all
. The mapping F is given by

Proof. For every
, define
if and only if
. It is clear that
is an O-set. Moreover, we can consider
as a closed subset of
,which d is the metric induced by
. Since Y is a Banach space, so
is an SO-complete metric space. From definition of
follows that

It is enough to pick
for all
. The result is an immediate consequence of Theorem 5.
Theorem 6. Let
be an integer,
, and
be an SO-complete metric space (not necessarily complete metric space). Suppose that
is a set-valued mapping such that
and
are
-comparable for each
and
, and there exists a function
satisfying equation (3) of Theorem 5 and the following property:
(B1)
if and only if
for all
, and 
is an increasing sequence for all
that are not all zero. Also,

is an unbounded sequence for some 
.
If
is a mapping, which satisfies relation (A1) of Theorem 5 and the following conditions:
(B2) For all
that are not all zero,

(B3) For every nonzero element x of
,

Then there exist an n-dimensional cubic set-valued mapping
and a sub- set
in
with card
such that for some positive real number
, we have

or all
. Moreover, if
, then F is unique.
Proof. By the same reasoning as in the proof of Theorem 5, there exist
and
such that
for each
. Set

As a result of (B1), we can easily see that for some 
, the sequence

Is a decreasing sequence which converges to zero. This concludes that card
. By the same argument of Theorem 5 one can show that the mapping
defined by
for all
is a
-preserving mapping and is a contraction with Lipschitz constant L on
. Define
by
for all
. Replacing
by
in definition of S0 and applying Theorem 3,weobtain that
is a unique fixed point of T. It follows from (6) that
and so

and consequently,

That is, inequality (10) holds. To show that the function F is an n-dimensional set-valued mapping on
, let
be fixed elements of
, which are not all zero. Since

is a nonnegative and decreasing sequence, so the rest of the proof is similar to the proof of Theorem 5.
Corollary 2. Let
be an integer and Y be a Banach space. Suppose that
is a mapping such that there exists a function
satisfying conditions (B1) of Theorem 6 and, in addition

for all
. If there exists a positive real number
such that

For all
, then for every
,there exists a unique n-dimensional cubic mapping
, which satisfies the inequality

for all
. The mapping F is given by

Proof. Take the same metric d and the orthogonal relation of Corollary 1. By the same argument of Corollary 1 one can show that
is an SO-complete metric space and
and
are
-comparable for each
and
. Putting
for all
and applying Theorem 6, we can easily obtain the results.
Corollary 3. Suppose that Y is a Banach space and
and
are fixed. Assume that
is a function satisfies the functional inequality

for all
. Then there exists a unique n-dimensional cubic mapping
such that the inequality

holds for all
, where
, or the inequality

holds for all
, where
.
Proof. Take the same metric d and the orthogonal relation of Corollary 1. By the same argument of Corollary 1 one can show that
is an SO-complete metric space. Moreover, definition of
ensures that
and
. are
-comparable for each
and
. Similarly,
and
are
-comparable for each
and
.
We define 
It follows that

for all
, where
. Set
for all
. This ensures that
and relations (A1) and (A3) of Theorem 5 hold. Applying Theorem 5, we see that inequality (4) holds with
,which yields inequality (12). On the other hand, the function φ satisfies properties (B1), (B2) and also

for all
, where
. Putting
for every
, it is easily seen that
and conditions (A1) and (B3) are hold. Employing Theorem 6, we see that inequality (10) holds with
. This implies inequality (13).
The next example shows that Theorem 6 is a real extension of Corollary 1.
Example 2. Let
be an integer and
, and Y be a Banach space. Let
be a sequence defined by
for all natural number p with
. It is easy to see that
is a strictly increasing sequence of real numbers. Suppose that
is a mapping satisfying

for all
. Define a mapping
by

and the function
) as

Then the following hold:
(i) For every
.
(ii) For every
,

(iii) For every
,

(iv) For every positive real number s, there exist a constant
and an n-dimensional cubic mapping
such that

for all . with
.
Proof. Take the same metric d and the orthogonal relation of Corollary 1. By the same argument of Corollary 1 one can show that
is an SO-complete metric space and
and
) are
-comparable for each
and
. Let us take
and
and let p be the smallest natural number such that
. Then

We observe that

This follows that there exists
, which

Assume that k is the smallest natural number satisfying the above condition. Clearly,
and

Now, we suppose that q is the smallest natural number that
then
. Since.
then
, and we conclude
. This implies that

That is, condition (i) holds. From definition
it is easily seen that
is a nondecreasing mapping.
Finally, it follows from
that for every
, there exists
such that

for all x with
. By the same proof of Theorem 5 we prove (iv).
Notice that there is no
such that inequality (9) holds, and hence, the stability of
does not imply by Corollary 1.
Now, we observe in the following example that our results go further than the stability on Banach spaces.
Example 3. Let
and
be given. Consider
(the set all of continuous functions on [0,1]) with norm
, where 1
. Suppose that
is a mapping satisfying inequality (11) and the following condition:

Then there exists a unique n-dimensional cubic mapping
such that inequality (12) holds for all
, where
, or inequality (13) holds for all
, where
.
Proof. Let q be the conjugate of
. For all
, define

And
. .We claim that
is an SO-complete metric space. Indeed, let
be a Cauchy SO-sequence in Y, and for all
. The relation
ensures that for all
,

We distinguish two cases.
Case 1. There exists a subsequence
of
such that
a.e. for all k. This implies that
.
Case 2. For all sufficiently large
. Take
such that for all
. It follows from (15) that for all
, there exists
for which
. It leads to

for each
. As
, the right-hand side of the above inequality tends to 0.Therefore,
is a Cauchy sequence in
. Assume that
as
. Put
It follows that
and for all
,

This implies that
as
. Note that the case
for all
is in a similar way.
By virtue of (14) and definition of
we obtain that
and
are
- comparable elements for each x
and
. Moreover, putting
in (14), we can also see that
and
are
-comparable elements in Y for all
and
. The rest of the proof is similar to the proof of Corollary 3.
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