Stability of Pexider Equations on Semigroup with No Neutral Element

Abstract

Let be a commutative semigroup with no neutral element, a Banach space, and the set of complex numbers. In this paper we prove the Hyers-Ulam stability for Pexider equation for all , where . Using Jung’s theorem we obtain a better bound than that usually obtained. Also, generalizing the result of Baker (1980) we prove the superstability for Pexider-exponential equation for all , where . As a direct consequence of the result we also obtain the general solutions of the Pexider-exponential equation for all , a closed form of which is not yet known.

1. Introduction

Throughout this paper, we denote by a commutative semigroup with no neutral element, a Banach space, and the set of complex numbers and . A function is called an additive function provided that for all , and is called an exponential function provided that for all . The Hyers-Ulam stability problems of functional equation have been originated by Ulam 1960 [1]. One of the first assertions to be obtained is the following result, essentially due to Hyers [2] that gives an answer for the question of Ulam.

Theorem 1. Let satisfy the functional inequality for all . Then there exists a unique additive function such that for all .

If has a neutral element, then as an easy consequence of Theorem 1 we have the following stability theorem for Pexider equation [3].

Theorem 2. Suppose that has a neutral element and satisfy for all . Then there exist a unique additive function and such that for all .

As a Hyers-Ulam stability theorem for the exponential functional equation, Baker proved that if satisfies the exponential functional inequality for all , then either is a bounded function satisfying for all or is an exponential function (see [4, 5]).

If has a neutral element, we obtain the following stability theorem for Pexider-exponential equation [3].

Theorem 3. Suppose that has a neutral element and satisfy the functional inequality for all . Then either there exist positive constants such that or else there exist an unbounded exponential function and nonzero constants such that

There are numerous results on the stability theorem for Pexider equations [6–12]. It is also very likely that there are some results on the stability of the inequalities (3) and (6) when has no neutral element. However, in the scope of the author, no results are found. In this paper, we prove the stability theorem for the functional inequalities (3) and (6) when has no neutral element.

Also, for inequality (3), using Jung’s theorem which shows the ratio of the diameter of a bounded set in the Euclidean space and the radius of smallest circle enclosing the set, we show how to improve the bound when the Euclidean space is the target space of functions in given functional inequalities.

2. Stability of Pexider Equation

Throughout this section we assume that is a commutative semigroup with no neutral element. We prove the Hyers-Ulam stability of (3). We denote .

Lemma 4. Assume that satisfies the functional inequality for all . Then there exist a unique additive function and such that for all .

Proof. Replacing by in (9) we have From (9) and (11) using the triangle inequality we have Fix and let with . Then we have for all . By Theorem 1, there exists a unique additive function such that for all . This completes the proof.

Theorem 5. Assume that satisfy for all . Then there exist a unique additive function and such that for all , .

Proof. Using (15) and the triangle inequality we have for all . By Lemma 4, there exist a unique additive function and such that for all . Changing the role of and in (15) we have for some additive function and . Replacing by in (15) and using the triangle inequality with (15) and the result we have Fix in (20). Then from (18), (19), and (20) using the triangle inequality we have for all , which implies . From (15), (18), and (19) using the triangle inequality we have for all . This completes the proof.

As a particular case of inequality (15), we have the following.

Theorem 6. Assume that satisfy for all . Then there exist a unique additive function and such that for all .

Proof. Using (23) and the triangle inequality we have for all . By Lemma 4, there exist a unique additive function and such that for all . From (23) and (26) using the triangle inequality we have for all . This completes the proof.

As a direct consequence of the above result we have the following stability theorem for Jensen functional equation when is  2-divisible.

Corollary 7. Assume that satisfy for all . Then there exist a unique additive function and such that for all .

Theorem 8. Assume that satisfy Then there exist a unique additive function and such that for all .

Proof. Replacing by in (30), using the triangle inequality with (30) and the result, putting , and letting we have for all . From (30) and (32) using the triangle inequality we have for all . By Theorem 1, there exists a unique additive function such that for all . On the other hand, using (30) and the triangle inequality we have for all . By Theorem 1, there exists a unique additive function such that for all . From (32), (34), and (36) using the triangle inequality we have for all , which implies . This completes the proof.

Using Jung’s theorem (see [13] for more details) we slightly improve the bound in Theorem 8 when is -dimensional Euclidean space.

Lemma 9. Let be a bounded subset of  , the -dimensional Euclidean space, and let be the diameter of . Then there exists a closed ball with radius such that .

Theorem 10. Let satisfy for all . Then there exist a unique additive function and such that for all .

Proof. Replacing by in (40) and using the triangle inequality with (40) and the result we have for all . Let . Then inequality (42) says that . Thus, by Lemma 9, there exists a closed ball of radius containing . Let be the center of the ball. Then we have for all . From (40) and (43) using the triangle inequality we have for all . By Theorem 1, there exists a unique additive function such that for all . This completes the proof.

As a direct consequence of the above result we have the following.

Corollary 11. Let satisfy for all . Then there exist a unique additive function and such that for all .

Remark 12. The author wants to know if the bounds in the above theorems can be replaced by smaller ones. Also, the author guesses that Jung’s theorem can be applied to obtain smaller bounds when dealing with some other functional inequalities.

3. Stability of Pexider-Exponential Equation

Throughout this section we assume that is a semigroup with no neutral element. Let . We consider the stability of the functional inequality for all . Hereafter, we exclude the trivial cases or .

We need the following lemma.

Lemma 13. Let . Then satisfies the functional equation if and only if either there exist nonzero constant and an exponential function such that or else In particular, if is -divisible or has a neutral element, then every solution of (49) is given by (50).

Proof. Multiplying in (49) we have If (51) fails, then there exist such that . Putting in (52) we have Thus, we have Putting and in (52) and dividing the result by we have where , which gives (50). Obviously, both (50) and (51) are solutions of (49). If in particular, is -divisible or has a neutral element, then we have . Thus, case (51) does not occur since . This completes the proof.

Theorem 14. Assume that satisfy inequality (48). Then satisfies one of the following.(i)There exist positive constants such that (ii)There exist an unbounded exponential function and nonzero constants such that (iii)There exists with such that

Proof. Replacing by and by in (48), respectively, and using triangle inequality we have for all . Since we exclude the trivial cases when or , it follows from inequality (59) that there exist constants such that for all . It follows from (60) that is bounded if and only if is bounded. Assume that is bounded. Then by (48) and (60) we get (56). Assume that is unbounded. Dividing both sides of (59) by we have for all , where . Since is unbounded, we have for all . Putting and in (61) and using the triangle inequality with the resulting inequalities we have for all and . Since is unbounded, from (62) we have . Thus, is independent of . Thus, we have for all . Furthermore, since both and are unbounded, it follows from (59) that if and only if . Thus, (63) holds for all . Now, using the triangle inequality we have for all and . Since is unbounded, it follows from (64) that for all . By Lemma 13, there exists such that for all , or for all . If (66) holds, putting (63) and (66) in (48) we have for all , which gives (57) with . If (67) holds, it remains to determine the values of for . Let . Then using (48) and (63) we have Using the triangle inequality we have This completes the proof.

As a direct consequence of the above result we have the following.

Corollary 15. Assume that is  2-divisible or has a neutral element. Suppose that satisfy the inequality (48). Then either there exist positive constants such that or else there exist an unbounded exponential function and nonzero constants such that

Remark 16. In particular, if is a group with identity 0, it is shown in [14] that every bounded function satisfying (48) satisfies for all , where and .

As a consequence of Theorem 14 we obtain the general solutions of the functional equation as follows:

Corollary 17. Let be unbounded functions. Then satisfies the functional equation (74) if and only if either there exist a nonzero exponential function and nonzero constants such that or else satisfies and and satisfy, respectively, for some , and In particular, is  2-divisible or has a neutral element; then the nonzero solutions of (74) are given by (75).

Remark 18. In particular, if is a group, it follows from Remark 16 that every bounded solution of (74) has the form for all . Furthermore, is  2-divisible and ; it can be shown that for all .

We call the solutions of form (75) regular solutions and call the solutions of forms (76), (77), and (78) irregular solutions.

Example 19. Let be a positive integer and let be the set of all integers . Let satisfy (74). We first exhibit the solutions of form (75). Let be a nonzero exponential function. If for some , then for each we have Thus, we have for all . Since for all we have Thus, we have Putting in (83) and multiplying the results in both sides we have Now, since we have From (84) and (85) we have

Therefore, the regular solutions of (74) are given by for some .

Now, we exhibit the irregular solutions of (74). By (76) we have for all , and if for some , then we have Thus, if for some , then , which implies for all . Now, if , then we have Therefore the irregular solutions of (74) are given by where , , are arbitrary complex numbers.

Finally, we consider the stability of functional equation (49).

Theorem 20. Let satisfy for all . Then either is bounded or there exist a nonzero constant and a nonzero exponential function such that or else In particular, is  2-divisible or has a neutral element; then every solution of (91) is bounded or given by (92).

Proof. In view of (91) we can write for all such that . Since is unbounded, from (94) we have for all . Using Lemma 13, we get the result. This completes the proof.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work was supported by Basic Science Research Program through the National Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (MEST) (no. 2012R1A1A008507). The author is very thankful to the referee for valuable suggestions that improved the presentation of the paper.