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Superstability of Generalized Multiplicative Functionals
Journal of Inequalities and Applications volume 2009, Article number: 486375 (2009)
Abstract
Let be a set with a binary operation
such that, for each
, either
, or
. We show the superstability of the functional equation
. More explicitly, if
and
satisfies
for each
, then
for all
, or
for all
. In the latter case, the constant
is the best possible.
1. Introduction
It seems that the stability problem of functional equations had been first raised by S. M. Ulam (cf. [1, Chapter VI]). "For what metric groups is it true that an
-automorphism of
is necessarily near to a strict automorphism? (An
-automorphism of
means a transformation
of
into itself such that
for all
.)" D. H. Hyers [2] gave an affirmative answer to the problem: if
and
is a mapping between two real Banach spaces
and
satisfying
for all
, then there exists a unique additive mapping
such that
for all
. If, in addition, the mapping
is continuous for each fixed
, then
is linear. This result is called Hyers-Ulam stability of the additive Cauchy equation
. J. A. Baker [3, Theorem
] considered stability of the multiplicative Cauchy equation
: if
and
is a complex valued function on a semigroup
such that
for all
, then
is multiplicative, or
for all
. This result is called superstability of the functional equation
. Recently, A. Najdecki [4, Theorem
] proved the superstability of the functional equation
: if
,
is a (real or complex valued) functional from a commutative semigroup
and
is a mapping from
into itself such that
for all
, then
holds for all
, or
is bounded.
In this paper, we show that superstability of the functional equation holds for a set
with a binary operation
under an additional assumption.
2. Main Result
Theorem 2.1.
Let and
a set with a binary operation
such that, for each
, either

If satisfies

then for all
, or
for all
. In the latter case, the constant
is the best possible.
Proof.
Let be a functional satisfying (2.2). Suppose that
is bounded. There exists a constant
such that
for all
. Set
. By (2.2), we have, for each
,
, and therefore

Thus, . Now it is easy to see that
. Consequently, if
is bounded, then
for all
. The constant
is the best possible since
for
satisfies
for each
. It should be mentioned that the above proof is essentially due to P. Å emrl [5, Proof of Theorem
and Proposition
] (cf. [6, Proposition
]).
Suppose that is an unbounded functional satisfying the inequality (2.2). Since
is unbounded, there exists a sequence
such that
. Take
arbitrarily. Set

By (2.1), . Thus either
or
is an infinite subset of
. First we consider the case when
is infinite. Take
arbitrarily. Choose
with
. Since
is assumed to be infinite, for each
there exists
such that
. Then
is a subsequence of
with
for every
. By the choice of
, we have

Thus we may and do assume that for every
. By (2.2) we have, for each
and
,
. According to (2.5), we have

Consequently, we have, for each

Since , we have
for every
. Applying (2.7), we have

By (2.2) and (2.5), we have

Consequently, we have by (2.8) and (2.7)

Next we consider the case when is infinite. By a quite similar argument as in the case when
is infinite, we see that there exists a subsequence
such that
for every
. Then

In the same way as in the proof of (2.7), we have

for every . According to (2.2) and (2.11), we have

Since for every
, (2.11) and (2.12) show that

Consequently, if is unbounded, then
for all
.
Remark 2.2.
Let be a mapping from a commutative semigroup
into itself. We define the binary operation
by
for each
. Then
satisfies (2.1) since

for all . Therefore, Theorem 2.1 is a generalization of Najdecki [4, Theorem
] and Baker [3, Theorem
].
Remark 2.3.
Let be a set, and
. Suppose that
has a binary operation
such that, for each
, either

If satisfies (2.2) for some
, then by quite similar arguments to the proof of Theorem 2.1, we can prove that
for all
, or
for all
. Thus, Theorem 2.1 is still true under the weaker condition (2.16) instead of (2.2). This was pointed out by the referee of this paper. The condition (2.16) is related to that introduced by Kannappan [7].
Example 2.4.
Let and
be mappings from a semigroup
into itself with the following properties.
(a) for every
.
(b).
(c) for every
.
If we define for each
, then we have
for every
. In fact, if
, then we have

as claimed.
Let be a ring homomorphism from
into itself, that is,
and
for each
. It is well known that there exist infinitely many such homomorphisms on
(cf. [8, 9]). If
is not identically
, then we see that
for every
, the field of all rational real numbers. Thus, if we consider the case when
,
a nonzero ring homomorphism, and
, then
satisfies the conditions (a), (b), and (c).
If we define for each
, then
holds for every
. In fact,

Example 2.5.
Let and, let
. We define the binary operation
by

for each . Then
satisfies the condition (2.1). In fact, let
.
(a)If , then we have

(b)If , then

(c)If and
, then

(d)If and
, then

(e)If , then we have

Therefore, satisfies the condition (2.1). On the other hand, if
, then

Thus, in general. In the same way, we see that if
, then
need not to be true.
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Acknowledgments
The authors would like to thank the referees for valuable suggestions and comments to improve the manuscript. The first and fourth authors were partly supported by the Grant-in-Aid for Scientific Research.
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Miura, T., Takagi, H., Tsukada, M. et al. Superstability of Generalized Multiplicative Functionals. J Inequal Appl 2009, 486375 (2009). https://doi.org/10.1155/2009/486375
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DOI: https://doi.org/10.1155/2009/486375
Keywords
- Similar Argument
- Binary Operation
- Weak Condition
- Commutative Semigroup
- Infinite Subset