## Recent Developments in Integral Transforms, Special Functions, and Their Extensions to Distributions Theory

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# On the Stability of Trigonometric Functional Equations in Distributions and Hyperfunctions

**Academic Editor:**Adem Kılıçman

#### Abstract

We consider the Hyers-Ulam stability for a class of trigonometric functional equations in the spaces of generalized functions such as Schwartz distributions and Gelfand hyperfunctions.

#### 1. Introduction

Hyers-Ulam stability problems of functional equations go back to 1940 when Ulam proposed the following question [1].

Let be a mapping from a group to a metric group with metric such that

Then does there exist a group homomorphism and such that for all ?

This problem was solved affirmatively by Hyers [2] under the assumption that is a Banach space. After the result of Hyers, Aoki [3] and Bourgin [4, 5] treated with this problem; however, there were no other results on this problem until 1978 when Rassias [6] treated again with the inequality of Aoki [3]. Generalizing Hyers' result, he proved that if a mapping between two Banach spaces satisfies with , then there exists a unique additive function such that for all . In 1951 Bourgin [4, 5] stated that if is symmetric in and with for each , then there exists a unique additive function such that for all . Unfortunately, there was no use of these results until 1978 when Rassias [7] treated with the inequality of Aoki [3]. Following Rassias’ result, a great number of papers on the subject have been published concerning numerous functional equations in various directions [6–10, 10–25]. In 1990 Székelyhidi [24] has developed his idea of using invariant subspaces of functions defined on a group or semigroup in connection with stability questions for the sine and cosine functional equations. We refer the reader to [9, 10, 18, 19, 25] for Hyers-Ulam stability of functional equations of trigonometric type. In this paper, following the method of Székelyhidi [24] we consider a distributional analogue of the Hyers-Ulam stability problem of the trigonometric functional inequalities where and is a continuous function. As a distributional version of the inequalities (4), we consider the inequalities for the generalized functions , where and denote the pullback and the tensor product of generalized functions, respectively, and denotes a continuous infraexponential function of order (resp., a function of polynomial growth). For the proof we employ the tensor product of -dimensional heat kernel

For the first step, convolving in both sides of (5) we convert (5) to the Hyers-Ulam stability problems of *trigonometric-hyperbolic type* functional inequalities, respectively,
for all , , where are the Gauss transforms of , respectively, given by
which are solutions of the heat equation, and

For the second step, using similar idea of Székelyhidi [24] we prove the Hyers-Ulam stabilities of inequalities (7). For the final step, taking initial values as for the results we arrive at our results.

#### 2. Generalized Functions

We first introduce the spaces of Schwartz tempered distributions and of Gelfand hyperfunctions (see [26–29] for more details of these spaces). We use the notations: , , , , and , for , , where is the set of nonnegative integers and .

*Definition 1 (see [29]). *One denotes by or the Schwartz space of all infinitely differentiable functions in such that
for all , , equipped with the topology defined by the seminorms . The elements of are called rapidly decreasing functions, and the elements of the dual space are called tempered distributions.

*Definition 2 (see [26]). *One denotes by or the Gelfand space of all infinitely differentiable functions in such that
for some . One says that as if as for some , and one denotes by the dual space of and calls its elements Gelfand hyperfunctions.

It is well known that the following topological inclusions hold:

It is known that the space consists of all infinitely differentiable functions on which can be extended to an entire function on satisfying for some , , and (see [26]).

By virtue of Theorem 6.12 of [27, p. 134] we have the following.

*Definition 3. *Let for , with , and let be a smooth function such that, for each , the Jacobian matrix of at has rank . Then there exists a unique continuous linear map such that when is a continuous function. One calls the pullback of by which is often denoted by .

In particular, let be defined by , . Then in view of the proof of Theorem 6.12 of [27, p. 134] we have

*Definition 4. *Let , . Then the tensor product of and , defined by
for , belongs to .

For more details of pullback and tensor product of distributions we refer the reader to Chapter V-VI of [27].

#### 3. Main Theorems

Let be a Lebesgue measurable function on . Then is said to be an *infraexponential function of order 2 (resp., a function of polynomial growth)* if for every there exists (resp., there exist positive constants , and ) such that
for all . It is easy to see that every infraexponential function of order (resp., every function of polynomial growth) defines an element of via the correspondence
for .

Let . We prove the stability of the following functional inequalities: where and denote the pullback and the tensor product of generalized functions, respectively, denotes a continuous infraexponential functional of order (resp. a continuous function of polynomial growth) with , and means that for all .

In view of (14) it is easy to see that the -dimensional heat kernel belongs to the Gelfand space for each . Thus the convolution is well defined for all . It is well known that is a smooth solution of the heat equation in and as in the sense of generalized functions that is, for every ,

We call the Gauss transform of .

A function from a semigroup to the field of complex numbers is said to be *an additive function* provided that , and is said to be *an exponential function* provided that .

For the proof of stabilities of (19) and (20) we need the following.

Lemma 5 (see [15]). *Let be a semigroup and the field of complex numbers. Assume that satisfy the inequality; for each there exists a positive constant such that
**
for all . Then either is a bounded function or is an exponential function.*

*Proof. *Suppose that is not exponential. Then there are such that . Now we have
and hence

In view of (23) the right hand side of (25) is bounded as a function of . Consequently, is bounded.

Lemma 6 (see [30, p. 122]). *Let be a solution of the heat equation. Then satisfies
**
for some , if and only if
**
for some bounded measurable function defined in . In particular, in as .*

We discuss the solutions of the corresponding trigonometric functional equations in the space of Gelfand hyperfunctions. As a consequence of the results [8, 31, 32] we have the following.

Lemma 7. *The solutions of (28) and (29) are equal, respectively, to the continuous solutions of corresponding classical functional equations
**
The continuous solutions of the functional equation (30) are given by one of the following:*(i)* and is arbitrary,*(ii)*, for some , ,*(iii)* and for some , , .**
Also, the continuous solutions of the functional equation (31) are given by one of the following:*(i)* and for some ,*(ii)* and for some .*

For the proof of the stability of (19) we need the followings.

Lemma 8. *Let be an Abelian group and let satisfy the inequality; there exists a nonnegative function such that
**
for all , , , . Then either there exist , , not both are zero, and such that
**
or else
**
for all , , , .*

*Proof. *Suppose that inequality (33) holds only when . Let
and choose and satisfying . Now it can be easily calculated that
where and . By (35) we have

Also by (35) and (36) we have

From (37) and (38) we have

Since is bounded by , if we fix , , , and , the right hand side of (39) is bounded by a constant , where

So by our assumption, the left hand side of (39) vanishes, so is the right hand side. Thus we have

Now by the definition of we have
Hence the left hand side of (41) is bounded by . So if we fix , , , and in (41), the left hand side of (41) is a bounded function of and . Thus by our assumption. This completes the proof.

In the following lemma we assume that is a continuous function such that exists and satisfies the conditions and or

Lemma 9. *Let be continuous functions satisfying
**
for all , , and there exist , not both are zero, and such that
**Then satisfies one of the followings:*(i)*, is arbitrary,*(ii)* and are bounded functions,*(iii)* for some , and , and is a continuous function; in particular, there exists with as such that
* *for all , , and satisfies the condition; there exists satisfying
* *for all ,*(iv)* for some , , and is a continuous function; in particular, there exists with as such that
* *for all , , and satisfies one of the following conditions; there exists such that
* *for all , or there exists such that
* *for all .*

*Proof. *If , is arbitrary which is case (i). If is a nontrivial bounded function, in view of (46) is also bounded which gives case (ii). If is unbounded, it follows from (47) that and
for some and a bounded function . Putting (53) in (46) we have
for all , . Replacing by and using the triangle inequality, we have, for some ,
for all , . By Lemma 5, is an exponential function. If , putting in (54) we have
Thus we have since is unbounded. Given the continuity of and we have
for some , with . Putting and in (54), dividing , and using the triangle inequality we have
for all , .

From (58) and the continuity of it is easy to see that
exists. Putting and replacing and by in (54) we have
for all .

Fixing , putting letting so that in (54), and using the triangle inequality and (60) we have
for all , . Letting in (61) we have
for all . From (61) the continuity of can be checked by a usual calculus. Letting in (60) we see that . Letting in (54) we have
for all . Putting in (63) and replacing by we have
for all .

Replacing by and using (64) and the triangle inequality we have
for all . Now we divide (65) into two cases: and . First we consider the case . Replacing by and by in (65) we have
for all . From (65) and (66), using the triangle inequality and dividing we have
for all such that . Choosing so that and putting in (67) we have
where , which gives (iii). Now we consider the case . It follows from (65) that
for all . By the well-known results in [3], there exists a unique additive function given by
such that
if , and there exists a unique additive function given by
such that
if . Now by the continuity of and inequality (61), it is easy to see that is continuous. In view of (70) and (72), , , are Lebesgue measurable functions. Thus there exist such that and for all , which gives (iv). This completes the proof.

In the following we assume that satisfies (44) or (45).

Theorem 10. *Let satisfy (19). Then satisfies one of the followings:*(i)*, and is arbitrary,*(ii)* and are bounded measurable functions,*(iii)* for some , , where is a continuous function satisfying the condition; there exists * *for all ,*(iv)* for some , where is a continuous function satisfying one of the following conditions; there exists such that
* *for all , or there exists such that
* *for all ,*(v)*, , for some , .*

*Proof. *Convolving in (19) the tensor product of -dimensional heat kernels in both sides of inequality (19) we have

Similarly we have
where are the Gauss transforms of , respectively. Thus we have the following inequality:
for all , , , where

By Lemma 8 there exist , not both are zero, and such that
or else satisfy
for all , . Assume that (81) holds. Applying Lemma 9, case (i) follows from (i) of Lemma 9. Using (ii) of Lemma 9, it follows from Lemma 7 the initial values of as are bounded measurable functions, respectively, which gives (ii). For case (iii), it follows from (50) that, for all ,
Thus we have in . Letting in (iii) of Lemma 9 we get case (iii). Finally we assume that (82) holds. Letting in (82) we have
By Lemma 6 the solutions of (84) satisfy (i), (iv), or (v). This completes the proof.

Let . Then satisfies the conditions assumed in Theorem 10. In view of (44) and (45) we have if , and if . Thus as a direct consequence of Theorem 10 we have the following.

Corollary 11. *Let or . Suppose that satisfy
**
Then satisfies one of the followings:*(i)*, and is arbitrary,*(ii)* and are bounded measurable functions,*(iii)* for some , , where is a continuous function satisfying the condition; there exists * *for all ,*(iv)* for some , where is a continuous function satisfying the conditions; there exists such that
* *for all ,*(v)*, , for some , .*

Now we prove the stability of (20). For the proof we need the following.

Lemma 12. *Let satisfy the inequality; there exists a such that
**
for all , . Then either there exist , not both are zero, and such that
**
or else
**
for all , .*

*Proof. *As in Lemma 9, suppose that is bounded only when , and let

Since we may assume that is nonconstant, we can choose and satisfying . Now it can be easily got that
where and . From the definition of and the use of (94), we have the following two equations:

By equating (95) and (96), we have

In (97), when , , , and are fixed, the right hand side is bounded; so by our assumption we have

Here, we have

Considering (98) as a function of and for all fixed , and again, we have . This completes the proof.

In the following lemma we assume that is a continuous function such that exists and satisfies the condition .

Lemma 13. *Let be continuous functions satisfying
**
for all , and there exist , not both zero, and such that
**
Then satisfies one of the followings*(i) * and are bounded functions in ,*(ii) * for some , , and is a continuous function; in particular, there exists with as such that
* *for all , and satisfies
* *for all .*

*Proof. *If is bounded, then in view of inequality (100), for each , is also bounded. It follows from Lemma 5 that is (101). If is bounded, case (i) follows. If is a nonzero exponential function, then by the continuity of we have
for some , . Putting (105) in (101) and using the triangle inequality we have for some
for all . In view of (106) it is easy to see that , . Thus is bounded on . If is unbounded; then in view of (101) is also unbounded, hence and
for some and a bounded function . Putting (107) in (101), replacing by , and using the triangle inequality we have

From Lemma 5 we have
for some exponential function . From (107) and (109), is continuous, and we have
for some , . If , we have