Stability of the Exponential Functional Equation in Riesz Algebras

and Applied Analysis 3 (iv) If f is bounded, then f ≡ 0 or f ≡ 1. (v) If f ̸ ≡ 0 then f(−x) = 1/f(x) for x ∈ G. Our main result reads us the following. Theorem 10. Let (G, +) be an Abelian 2-divisible group and let L be an Archimedean Riesz space with a strong unit e ∈ L + . We assume thatL is e-uniformly complete. If a functionF : G → L satisfies 󵄨󵄨󵄨󵄨F (x + y) − F (x) F (y) 󵄨󵄨󵄨󵄨 ≤ u 2 − u for x, y ∈ G, (6) with given u ∈ L + , then there exists an exponential function E : G → L such that |F (x) − E (x)| ≤ u + e for x ∈ G. (7) Proof. The idea of the proof is based on the use of the Yosida Spectral Representation Theorem which enables us to apply Theorem 1 of Baker. The proof runs in four steps. Step 1. Consider F : G → L satisfying (6). According to the Yosida Spectral Representation Theorem, for every x, y ∈ G, we have F(x), F(y), F(x+y), u2−u ∈ C(X).Therefore, by (6), one has 󵄨󵄨󵄨󵄨F (x + y) (s) − F (x) (s) F (y) (s) 󵄨󵄨󵄨󵄨 ≤ u 2 (s) − u (s) for x, y ∈ G, s ∈ X. (8) It means that, for any s ∈ X, F(⋅)(s) satisfies all the assumptions of Theorem 1. By Theorem 1 either F(⋅)(s) is bounded on G with |F(x)(s)| ≤ u(s) for x ∈ G or F(⋅)(s) is exponential on the whole G. Let B := {s ∈ X : |F (x) (s)| ≤ u (s) for x ∈ G} , E :={s∈X : F(⋅)(s) is unbounded and exponential on G}. (9) Of course we haveX =B ∪E andB ∩E = 0. We will prove that E is an open subset of X. For the indirect proof consider s ∈ E and suppose that each neighbourhood U of s has a nonempty intersection with B. Let m k,x ∈ C(X) be given by m k,x (t) := F (kx) (t) − u (t) for k ∈ Z, x ∈ G, t ∈ X. (10) Since s ∈ E, there exist x ∈ G \ {0} and k ∈ Z such that m k,x (s) > 0. On the other hand, according to the supposition, in each neighbourhoodU of s there exists t withm k,x (t) ≤ 0, which brings a contradiction with the continuity ofm k,x . Step 2. For given x ∈ G we define E(x) ∈ C(X) by E (x) (s) = { F (x) (s) , if s ∈ E 1, if s ∈B for s ∈ X. (11) We shall prove the continuity of E(x). First consider the case s ∈ E. Take an arbitrary neighbourhoodV of E(x)(s). Since E is open, there exists a neighbourhood U 1 of s with U 1 ⊂ E. By the choice of s we have E(x)(s) = F(x)(s) and by the continuity of F(x) at s there exists a neighbourhood U 2 of s such that F(x)(U 2 ) ⊂ V. Then U := U 1 ∩ U 2 forms a neighbourhood of s with E(x)(U) ⊂V. Thus, it remains to consider s ∈ B. For arbitrary ε > 0 let V := (1 − ε, 1 + ε) be a neighbourhood of 1 = E(x)(s). We will prove that there exists a neighbourhoodU of s such that F(x)(U ∩ E) ⊂ V. Contrary, suppose that in each neighbourhoodU of s there exists t ∈ U ∩ E with F(x)(t) > 1+ε orF(x)(t) < 1−ε. Consider the caseF(x)(t) < 1−ε.Then, taking into account the positivity of F(x)(t), which follows from the fact that F(⋅)(t) is exponential and unbounded on G, we have


Introduction
In 1979 Baker et al. (cf.[1]) proved that the exponential functional equation  ( + ) =  ()  () for ,  ∈  (1) in the class of functions mapping a vector space  to the real numbers R is superstable; that is, any function  satisfying, with given  > 0, the inequality      ( + ) −  ()  ()     ≤  for ,  ∈ is either bounded or exponential (satisfies (1)).Then Baker generalized this famous result in [2].We quote this theorem here since it will be used in the sequel.
After that the stability of the exponential functional equation has been widely investigated (cf., e.g., [3][4][5][6]).This paper will primarily be concerned with the question if similar result holds true in the class of functions taking values in Riesz algebra  with the common notion of the absolute value || = sup{, −} of an element  ∈  stemming from the order structure of .
The main aim of the present paper is to show that the superstability phenomenon does not hold in such an order setting.However, we prove that the exponential functional equation ( 1) is stable in the Ulam-Hyers sense; that is, for any given  :  →  satisfying inequality (3) there exists an exponential function  :  →  which approximates  uniformly on  in the sense that the set {|()−()| :  ∈ } is bounded in .
As a method of investigation we apply spectral representation theory for Riesz spaces; to be more precise, we use the Yosida Spectral Representation Theorem for Riesz spaces with a strong order unit.

Preliminaries
Throughout the paper N, Z, R, and R + are used to denote the sets of all positive integers, integers, real numbers and nonnegative real numbers, respectively.
For the readers convenience we quote basic definitions and properties concerning Riesz spaces (cf.[13]).
There are several types of convergence that may be defined according to the order structure.One of them is the relatively uniform convergence defined as follows.
Definition 4 (cf.[13, Definition 39.3]).A Riesz space  is called -uniformly complete (with a given  ∈  + ) whenever every -uniform Cauchy sequence has a -uniform limit in .Furthermore  is called uniformly complete if it is -uniformly complete with any  ∈  + .
There is a large class of spaces satisfying the above conditions.In particular every Dedekind -complete space (see Definition 5 below) is an Archimedean and uniformly complete Riesz space.
Definition 5 (cf.[13, Definition 1.1]).We say that a Riesz space  is Dedekind -complete if any non-empty at most countable subset which is bounded above has a supremum.Definition 6 (cf.[13,Definition 21.4]).The element  ∈  + is called a strong unit if for every  ∈  there exists  ∈ R such that || ≤ .
For more detailed information and, in particular, examples of Riesz spaces posessing the above properties we refer the interested reader to [13].
In further considerations the Yosida Spectral Representation Theorem, which is quoted below, will be used.Theorem 7. Yosida Spectral Representation Theorem (cf.[13,Theorem 45.3]).Let  be an Archimedean Riesz space with a strong unit  ∈  + .Then there exists a topological space  and a Riesz subspace L of the space () of all real continuous functions on  (with the pointwise order and pointwise operations of addition and scalar multiplication) and a Riesz isomorphism of  onto L.
We will not distinguish  ∈  and its Yosida representant, if no confusion can occur.
Directly from the construction of the Yosida repesentatives one can deduce that the Yosida representative of a strong unit  ∈  + is a constant function  ≡ 1.We omit the formal details here as they exceed the scope of the paper.
In general, the space L of Yosida representatives is a Riesz subspace of ().The following theorem gives us conditions under which L is the whole ().
Theorem 8 (cf.[13,Theorem 45.4]).Let  be an Archimedean Riesz space with a strong unit  ∈  + and let L and () be as in the previous Theorem.The following conditions are now mutually equivalent.
(i)  is uniformly complete.
(iii) Every -uniform Cauchy sequence in  has an uniform limit in .
Hence, if  is a Dedekind -complete space with a strong unit, then L = ().
In the case where L = (), the Yosida representation L of  is not only a Riesz space but also a Riesz algebra with respect to the pointwise multiplication of functions in ().But then, since  and L are isomorphic as Riesz spaces, we may introduce ring multiplication for the elements ,  ∈  induced by the multiplication of representatives, that is, where  :  → () is the Riesz isomorphism.Notice that ℎ ∈  given by ( 5) is uniquely determined.Such a multiplication makes  into a commutative Riesz algebra with a unit element (a strong unit  ∈  + is an algebra unit element, that is,  =  =  for  ∈ ).
From now on a multiplication in a Riesz space  will be construed in the above sense.

The Main Result
We start with some, easy to prove, properties of exponential real functions on a 2-divisible group.
Our main result reads us the following.
Proof.The idea of the proof is based on the use of the Yosida Spectral Representation Theorem which enables us to apply Theorem 1 of Baker.
The proof runs in four steps.
Of course we have  = B ∪ E and B ∩ E = 0. We will prove that E is an open subset of .For the indirect proof consider  ∈ E and suppose that each neighbourhood U of  has a nonempty intersection with B. Let  , ∈ () be given by  , () :=  () () −  () for  ∈ Z,  ∈ ,  ∈ . ( Since  ∈ E, there exist  ∈  \ {0} and  ∈ Z such that  , () > 0. On the other hand, according to the supposition, in each neighbourhood U of  there exists  with  , () ≤ 0, which brings a contradiction with the continuity of  , .
Step 2. For given  ∈  we define () ∈ () by We shall prove the continuity of ().First consider the case  ∈ E. Take an arbitrary neighbourhood V of ()().Since E is open, there exists a neighbourhood U 1 of  with U 1 ⊂ E. By the choice of  we have ()() = ()() and by the continuity of () at  there exists a neighbourhood U 2 of  such that ()(U 2 ) ⊂ V. Then U := U 1 ∩ U 2 forms a neighbourhood of  with ()(U) ⊂ V. Thus, it remains to consider  ∈ B. For arbitrary  > 0 let V := (1 − , 1 + ) be a neighbourhood of 1 = ()().We will prove that there exists a neighbourhood U of  such that ()(U ∩ E) ⊂ V. Contrary, suppose that in each neighbourhood U of  there exists  ∈ U ∩ E with ()() > 1+ or ()() < 1−.Consider the case ()() < 1−.Then, taking into account the positivity of ()(), which follows from the fact that (⋅)() is exponential and unbounded on , we have Let  > 0 be fixed.Then there exists  = () ∈ N such that  , () >  or  ,− () >  depending on the case where ()() > 1 +  or ()() < 1 − , respectively.On the other hand, by the continuity of  , and the fact that  , () ≤ 0 <  there exists a neighbourhood U 1 of  such that  , () <  for  ∈ U 1 .By the same reasons, there exists a neighbourhood This completes the proof that () ∈ ().
Therefore, by Theorem 8 one may treat () as an element of .Since  ∈  has been chosen arbitrarily, in fact formula (11) defines a function  :  → .
Step 3. We will prove that function  given by ( 11) is exponential.

Final Remarks
Let us recall the following theorem, which provides us with the condition under which a Riesz homomorphism (as a homomorphism between Riesz spaces) is multiplicative.approximate  uniformly on R, that is, satisfy (7).

Theorem 10 .
Let (, +) be an Abelian 2-divisible group and let  be an Archimedean Riesz space with a strong unit  ∈  + .We assume that  is -uniformly complete.If a function  :  →  satisfies [14,rem 11 (cf.[14,Proposition353P]).Let  be an Archimedean -algebra with multiplicative identity .If  is another Archimedean -algebra with multiplicative identity   , and  :  →  is a positive linear operator such that () =   , then  is a Riesz homomorphism if and only if ( ⋅ V) = () ⋅ (V) for all , V ∈ .By the above theorem and the Yosida Spectral Representation Theorem one can obtain the following corollary.Let (, +) be an Abelian 2-divisible group and let  be an Archimedean -algebra with a multiplicative identity  ∈  + which is a strong order unit.We assume that  is -uniformly complete.If a function  :  →  satisfies      ( + ) −  ()  ()     ≤  2 −   ,  ∈ , (15) Corollary 12) states that the exponential functional equation (1) in Riesz algebras is stable in the Ulam-Hyers sense.Taking into account Theorem 1 it is natural to ask if (1) is superstable in the sense of Baker.It appears that the superstability phenomenon in Riesz algebras fails to hold.In the next example we show that there exists a group , an -algebra  satisfying all the assertions of Theorem 10 and a function  :  →  which fulfills (6) but is neither exponential nor bounded.