A variant of d'Alembert functional equation on monoids
aa r X i v : . [ m a t h . G M ] O c t A VARIANT OF D’ALEMBERT FUNCTIONAL EQUATION ONMONOIDS
CHAHBI ABDELLATIF AND ELQORACHI ELHOUCIEN
Abstract.
In this paper, we determine the complex-valued solutions of the functionalequation f ( xσ ( y )) + f ( τ ( y ) x ) = 2 f ( x ) f ( y )for all x, y ∈ M , where M is a monoid, σ : M −→ M is an involutive automorphism and τ : M −→ M is an involutive anti-automorphism. The solutions are expressed in termsof multiplicative functions, and characters of 2-dimensional irreducible representationsof M . Introduction
It is well-known that d’Alembert’s functional equation f ( x + y ) + f ( x − y ) = 2 f ( x ) f ( y ) (1.1)for all x, y ∈ R has its continuous solutions f : R −→ R the functions f ( x ) = 0 and f ( x ) = cosh( cx ), where c ∈ R . Acz´el and Dhombres in [1] wrote several chapters coveringand connecting the Cauchy, d’Alembert and the trigonometric functional equations.The domain of definition of the solutions of (1.1) have progressively been extented from R via abelian groups and monoids (See Acz´el and Dhombres [1], Davison [5,6] and Stetkær [24]for details and references), and the functional equation (1.1) take the general forme f ( xy ) + µ ( y ) f ( xσ ( y )) = 2 f ( x ) f ( y ) (1.2)for all x, y ∈ S , where S is a semigroup, σ : S −→ S is an involutive anti-automorphism of S ,i.e., σ ( xy ) = σ ( y ) σ ( x ) and σ ( σ ( x )) = x for all x, y ∈ S , and µ : S −→ C is a multiplicativefunction on S : µ ( xy ) = µ ( x ) µ ( y ) for all x, y ∈ S , which satisfies µ ( xσ ( x )) = 1 for all x ∈ S. The latest result about the functional equation (1.2) is due to Stetkær [20].By combining an algebraic small dimension lemma with methods by Davison [5, 6] onmonoids, Stetkær [20] described the solution of (1.2) on semigroups in terms of multiplica-tive functions and 2-dimensional irreducible representations, like they have been in previousstudies of solutions on groups and monoids [5, 6].Recently, a number of Mathematicians have studied other versions of d’Alembert’s func-tional equations f ( xσ ( y )) + f ( τ ( y ) x ) = 2 f ( x ) g ( y ) (1.3) Key words and phrases.
Topological group, monoid, d’Alembert’s equation, automorphism, involution.2020 Mathematics Subject Classification. 39B32. for all x, y ∈ S on smigroups and monoids, and where σ, τ : S −→ S are respectivelyinvolutives anti-automorphisms, resp. σ, τ : S −→ S are involutives automorphisms. (See[7], [10], [11], [14], [15], [17] and [19]). Since in two situations σ ◦ τ is an automorphism, theysuccefully used some steps and ideas of Stetkær [23] to eliminate σ and τ from the equation(1.3) and brought the equation to a trigonometric equation, and consequently non-abelianphenomena like representations crop up.In this paper, we solve the functional equation f ( xσ ( y )) + f ( τ ( y ) x ) = 2 f ( x ) f ( y ) (1.4)for all x, y ∈ M , where M is a monoid, σ : M −→ M is an involutive automorphism,and τ : M −→ M is involutive anti-automorphism. The crucial idea used in the paper ofStetkær [23] can not carry over to the our situation, in which σ ◦ τ is an anti-autmorphismof M. The functional equation (1.4) is a natural generalization of the variant of Wilson’sfunctional equation equation f ( xy ) + f ( y − x ) = 2 f ( x ) g ( y ) , x, y ∈ G (1.5)studied by Stetkær [26] and solved by Ebanks and Stetkær [8] on groups.Davison in [5], [6, Theorem 4.12] and Stetkær [24, Theorem 8.26] obtained continuoussolutions of the pre-d’Alembert functional equation g ( xyz ) + g ( xzy ) = 2 g ( x ) g ( yz ) + 2 g ( y ) g ( xz ) + 2 g ( xy ) g ( z ) − g ( x ) g ( y ) g ( z )for all x, y, z ∈ M on a topological groups and monoids.In Proposition 3.2 and Proposition 3.3 we prove some properties of the solutions of our func-tional equation (1.4). In particular, the solutions are central and satisfies the pre-d’Alembertfunctional equation (Proposition 3.3). By using this result we give a full description of func-tions f : M → C satisfying the equation (1.4) in terms of multiplicative functions, andcharacters of 2-dimensional irreducible representations of M. Thus functional equation (1.4)makes sense to solve other functional equations by expressing their solutions in terms ofsolutions of (1.4). This is what we shall do for generalized of variant of Van Vleck’s f ( y − xz ) − f ( xσ ( y ) z ) = 2 f ( x ) f ( y ) , x, y ∈ G. (1.6)and for generalized of variant Kannappan’s functional equation f ( xσ ( y ) z ) + f ( τ ( y ) xz ) = 2 f ( x ) f ( y ) , x, y ∈ M, (1.7)where z is a fixed element in M. Some information, applications and numerous referencesconcerning Van Vleck’s and Kannappan’s functional equation and their further generaliza-tions can be found e.g. in [3], [12], [13], [18] and [22]- [30].2.
Set Up and Notation
Throughout this paper S designed a semigroup, M a monoid, and G a group with neutralelement e. The map σ : M → M denotes an involutive homomorphism and the map τ aninvolutive anti-automorphism. That σ, τ : M → M are involutive means that τ ( τ ( x )) = σ ( σ ( x )) = x for all x ∈ M. We need the following basic definitions.
Definition 2.1. (1) A function f : S → C is central provided that f ( xy ) = f ( yx ) forall x, y ∈ S. VARIANT OF D’ALEMBERT FUNCTIONAL EQUATION 3 (2) A function f : S → C is abelian means f ( x x . . . x n ) = f ( x ǫ (1) x ǫ (2) . . . x ǫ ( n ) )for every n ≥ ǫ of n elements of S .(3) A function f : S → C is σ -even if f ( σ ( x )) = f ( x ) for all x ∈ S, and σ -odd if f ( σ ( x )) = − f ( x ) for all x ∈ S. (4) A function χ : S −→ C is said to be multiplicative if χ ( xy ) = χ ( x ) χ ( y ) for all x, y ∈ S .For any vector space V , we let L ( V ) denote the algebra of linear operators of V into V .A representation of M on non-zero vector space V is a map π : M → L ( V ) such that π ( xy ) = π ( x ) π ( y ) for all x, y ∈ M. The space V is called the representation space of π. Wefurthermore assume that π ( e ) = I : The identity map.Let dim V < + ∞ . The dimension of the representation π is d π = dim V and the matrixcoefficient χ π : M → C defined by χ π ( x ) := tr ( π ( x )) , x ∈ M is called the character of π .If M is a topological space, then we let C ( M ) denote the algebra of continuous functionsfrom M into C . Let V be a 2-dimensional vector space. To any linear map A ∈ L ( V ) we define its adjugate adj ( A ) ∈ L ( V ) from linear algebra. The properties of adjugation can easily be derived fromits matrix form, and in particular we have adj : L ( V ) → L ( V ) is linear, A + adj ( A ) = ( trA ) I. (2.1)Furthermore, Aadj ( A ) = adj ( A ) A = ( detA ) I, adj ( AB ) = adj ( B ) adj ( A ) , and adj ( adj ( A )) = A for all A, B ∈ L ( V ) . Some properties of equation (1.4)
In this section, we prove some properties of the functional equation (1.4) on semigroupsand monoids. In particular, we prove that solutions of of equation (1.4) are solutions ofpre-d’Alembert functional equation.
Lemma 3.1.
Let f : M → C be a non-zero solution of the functional equation (1.4). Then ( f ( e ) = 1 .f ( σ ( x )) + f ( τ ( x )) = 2 f ( x ) (3.1) Proof.
Take y = e in (1.4) we get f ( x ) = f ( e ) f ( x ) . Since f = 0 then f ( e ) = 1 . Putting x = e in (1.4) we get the following identity f ( σ ( y )) + f ( τ ( y )) = 2 f ( e ) f ( y ) . Using that f ( e ) = 1 we get f ( σ ( y )) + f ( τ ( y )) = 2 f ( y )for all y ∈ M. In the following proposition we eliminate the involution anti-automorphism τ from the functional equation (1.4). (cid:3) A. CHAHBI AND E. ELQORACHI
Proposition 3.2.
Let f : S → C be a non-zero solution of the functional equation (1.4).Then f ( σ ( a ) σ ( x ) σ ( y ))+ f ( σ ( a ) σ ( y ) σ ( x )) = 2 f ( σ ( a )) f ( yx )+2 f ( x ) f ( σ ( a ) σ ( y ))+2 f ( σ ( a ) σ ( x )) f ( y ) − f ( σ ( a )) f ( x ) f ( y ) , (3.2) for all a, x, y ∈ S. Proof.
For the pair ( σ ( a ) σ ( x ) , y ) the functional equation (1.4 ) can be written as follows f ( σ ( a ) σ ( x ) σ ( y )) + f ( τ ( y ) σ ( a ) σ ( x )) = 2 f ( σ ( a ) σ ( x )) f ( y ) . (3.3)By applying (1.4 ) to the pair ( τ ( y ) σ ( a ) , x ) we obtain f ( τ ( y ) σ ( a ) σ ( x ))+ f ( τ ( x ) τ ( y ) σ ( a )) = 2 f ( τ ( y ) σ ( a )) f ( x ) = 2 f ( x )[2 f ( σ ( a )) f ( y ) − f ( σ ( a ) σ ( y ))] . (3.4)By using (1.4), we reformulate the second term on the left hand side of (3.4) as follows f ( τ ( x ) τ ( y ) σ ( a )) = f ( τ ( yx ) σ ( a )) = 2 f ( σ ( a )) f ( yx ) − f ( σ ( a ) σ ( yx )) , which turns the identity (3.4) into we obtain f ( τ ( y ) σ ( a ) σ ( x ))+2 f ( σ ( a )) f ( yx ) − f ( σ ( a ) σ ( y ) σ ( x )) = 4 f ( σ ( a )) f ( x ) f ( y ) − f ( x ) f ( σ ( a ) σ ( y )) . Subtracting this from (3.3) we get after some simplifications that f ( σ ( a ) σ ( x ) σ ( y ))+ f ( σ ( a ) σ ( y ) σ ( x )) = 2 f ( σ ( a )) f ( yx )+2 f ( x ) f ( σ ( a ) σ ( y ))+2 f ( σ ( a ) σ ( x )) f ( y ) − f ( σ ( a )) f ( x ) f ( y )(3.5)for all a, x, y ∈ S. This ends the proof. (cid:3)
Proposition 3.3.
Let f : S → C be a non-zero solution of the functional equation (1.4).Then(1) f is central.(2) f ( σ ( x )) + f ( τ ( x )) = λf ( x ) for all x ∈ S , and for same λ ∈ C ∗ . (3) f is a solution of the pre-d’Alembert functional equation i.e., f ( axy ) + f ( ayx ) = 2 f ( a ) f ( xy ) + 2 f ( x ) f ( ay ) + 2 f ( ax ) f ( y ) − f ( a ) f ( x ) f ( y ) for all a, x, y ∈ S. Proof. (1) According to Proposition 3.2 f satisfies (3.2). Interchange of x and y in(3.2) and comparing the result obtained with (3.2) we find that f ( σ ( a )) f ( xy ) = f ( σ ( a )) f ( yx ) for all a, x, y ∈ S. Since f = 0 , then f is central.(2) Replace x by σ ( x ) in (1.4), we obtain f ( σ ( x ) σ ( y )) + f ( τ ( y ) σ ( x )) = 2 f ( σ ( x )) f ( y ) , x, y ∈ S. (3.6)Substitute x by τ ( x ) in in (1.4), we obtain f ( τ ( x ) σ ( y )) + f ( τ ( y ) τ ( x )) = 2 f ( τ ( x )) f ( y ) , x, y ∈ S. (3.7)Adding (3.6) and (3.7) and use that f is central we obtain[ f ( σ ( y ) σ ( x )) + f ( τ ( x ) σ ( y ))] + [ f ( τ ( y ) σ ( x )) + f ( τ ( x ) τ ( y ))]= f ( x )( f ( σ ( y )) + f ( τ ( y ))) = f ( y )( f ( σ ( x )) + f ( τ ( x ))) VARIANT OF D’ALEMBERT FUNCTIONAL EQUATION 5 for all x, y ∈ S . Since f = 0 then there exists y ∈ S such that f ( y ) = 0 , and that f ( σ ( x )) + f ( τ ( x )) = λf ( x ) for all x ∈ S , and where λ = f ( σ ( y ))+ f ( τ ( y )) f ( y ) . Now, if λ = 0 , we get f ( σ ( x )) = − f ( τ ( x )) for all x ∈ S. By replacing x by σ ( x ), y by τ ( y ) respectively x by τ ( x ) and y by σ ( y ) in (1.4) we get ( f ( σ ( x ) σ ( τ ( y ))) + f ( yσ ( x )) = 2 f ( σ ( x )) f ( τ ( y )) f ( τ ( x ) y ) + f ( τ ( σ ( y )) τ ( x )) = 2 f ( τ ( x )) f ( σ ( y ))Since f ( σ ( x )) = − f ( τ ( x )) , x ∈ S and f is central, so by adding the two functionalequations of the above system we get − f ( τ ( x ) y ) − f ( yσ ( x )) + f ( yσ ( x )) + f ( τ ( x ) y ) = − f ( σ ( x )) f ( σ ( y )) = 0for all x, y ∈ S . Since σ = I, so f = 0 , which is a contradiction with f = 0 . (3) Let a, x, y ∈ S be arbitrary. For the pair ( τ ( x ) τ ( a ) , y ) equation (1.4 implies that f ( τ ( x ) τ ( a ) σ ( y )) + f ( τ ( y ) τ ( x ) τ ( a )) = 2 f ( τ ( x ) τ ( a )) f ( y ) . (3.8)By applying (1.4 ) to the pair ( τ ( a ) σ ( y ) , x ) we obtain f ( τ ( a ) σ ( y ) σ ( x ))+ f ( τ ( x ) τ ( a ) σ ( y )) = 2 f ( τ ( a ) σ ( y )) f ( x ) = 2 f ( x )[2 f ( τ ( a )) f ( y ) − f ( τ ( y ) τ ( a ))] . (3.9)By using (1.4 ) we reformulate the first term on the left hand side of (3.9) as follows f ( τ ( a ) σ ( yx )) = 2 f ( τ ( a )) f ( yx ) − f ( τ ( x ) τ ( y ) τ ( a )) , which turns the identity (3.9) into2 f ( τ ( a )) f ( yx ) − f ( τ ( x ) τ ( y ) τ ( a ))+ f ( τ ( x ) τ ( a ) σ ( y )) = 4 f ( τ ( a )) f ( x ) f ( y ) − f ( x ) f ( τ ( y ) τ ( a )) . Subtracting this from (3.8) we get after some simplifications that f ( τ ( axy ))+ f ( τ ( ayx )) = 2 f ( τ ( a )) f ( yx )+2 f ( x ) f ( τ ( ay ))+2 f ( τ ( ax )) f ( y ) − f ( τ ( a )) f ( x ) f ( y )(3.10)By adding (3.2) and (3.10), and using Proposition 3.3 (2), we conclude that f is asolution of pre-d’Alembert functional equation on semigroup S. This completes theproof. (cid:3)
Now, we are ready to prove the main result.
Theorem 3.4.
Abelian solutions of equation (1.4) .The abelian solutions f : M → C of (1.4) are the functions of the form f = χ + χ ◦ σ ◦ τ , where χ : M → C is a multiplicative function such that: (i) χ ◦ σ ◦ τ = χ ◦ τ ◦ σ, and (ii) χ is σ -even and/or τ -even.Proof. It is elementary to check that the functions stated in the Theorem 3.4 define abeliansolutions, so it is left to show that any abelian solution f : M → C of (1.4) can be writtenas in the form in Theorem 3.4. We have from proposition 3.3 that f is solution of pre-d’Alembert functional equation on a Monoid M, and which satisfies f ( e ) = 1, (see Lemma A. CHAHBI AND E. ELQORACHI f is an abelian solution by [24, Theorem 8.13.] then there exist multiplicativefunctions χ , χ : M → C such that f = χ + χ . We will discuss two cases.Case 1. If χ = χ , then letting χ := χ = χ we have f = χ. Substituting f = χ into (1.4)we get that χ ◦ σ + χ ◦ τ = 2 χ. So χ = χ ◦ σ = χ ◦ τ. Then f has the desired form.Case 2. If χ = χ , substituting f = ( χ + χ ) / χ ( x )[ χ ( σ ( y )) + χ ( τ ( y )) − χ ( y ) − χ ( y )]+ χ ( x )[ χ ( σ ( y )) + χ ( τ ( y )) − χ ( y ) − χ ( y )]] = 0for all x, y ∈ M. Since χ = χ , we get from the theory of multiplicative functions (see forinstance ( [24, Theorem 3.18]) that both terms are 0 , so ( χ ( x )[ χ ( σ ( y )) + χ ( τ ( y )) − χ ( y ) − χ ( y )] = 0 χ ( x )[ χ ( σ ( y )) + χ ( τ ( y )) − χ ( y ) − χ ( y )] = 0 (3.11)for all x, y ∈ M. Since χ = χ at least one of χ and χ is not zero.Now if χ = 0 and χ = 0 . It is allowed that χ ( σ ( y )) + χ ( τ ( y )) = χ ( y ) this implies that χ ( σ ( y )) = 0 or χ ( τ ( y )) = 0 , in either case χ = 0 because σ = τ = id. We have now χ = 0 and χ = 0 . From (3.11), we get χ + χ = χ σ + χ ◦ τ = χ ◦ σ + χ ◦ τ. Using that χ = χ , we obtain that χ = χ ◦ σ ◦ τ = χ ◦ τ ◦ σ now as we see that χ + χ = χ σ + χ ◦ τ this implies that χ is is σ -even or τ -even. Finally, we deduce that f has the form stated in Theorem 3.4 with χ = χ . (cid:3) Theorem 3.5.
Non-abelian solutions of equation (1.4) .The non-abelian solutions f M −→ C of (1.4) are of the form f ( x ) = trπ ( x ) for all x ∈ M ,where π is an irreducible, -dimensional representation of M for which adj ( π ◦ σ ) = π ◦ τ, tr ( π ( x )) = tr ( π ( σ ( x ))) = tr ( π ( τ ( x ))) = tr ( π ( σ ◦ τ ( x ))) = tr ( π ( τ ◦ σ ( x ))) , for all x ∈ M. If π ′ is any irreducible representation of M on a finite dimensional vector space, such that f ( x ) = trπ ′ ( x ) for all x ∈ M , then π ′ and π are equivalent.Proof. By Proposition 3.3 (3), f is a non-abelian solution of the pre-d’Alembert functionalequation on M. Then from [24, Theorem 8.26] f has the form f = tr ( π ), where π is anirreducible, 2-dimensional representation of M . Substituting f into the functional equation(1.4) and using that f is central we get tr (cid:18) π ( x )( π ( σ ( y ) + π ( τ ( y )) − tr ( π ( y )) I )) (cid:19) = 0 , x, y ∈ M. Now, Burnside Theorem’s [16] shows that tr (cid:18) A ( π ( σ ( y ) + π ( τ ( y )) − tr ( π ( y )) I ) (cid:19) = 0 , for all A ∈ L ( V ) and for all y ∈ M, and consequently, we get π ( σ ( y ) + π ( τ ( y )) = tr ( π ( y )) I (3.12) VARIANT OF D’ALEMBERT FUNCTIONAL EQUATION 7 for all y ∈ M. Replacing y by σ ( y ) respectively by τ ( y ) in (3.12) we obtain ( π ( y ) + π ( τ ( σ ( y ))) = tr ( π ( σ ( y ))) I π ( y ) + π ( σ ( τ ( y ))) = tr ( π ( τ ( y ))) I (3.13)By adding the two functional equation of the above system and using that f ( σ ( y )) + f ( τ ( y )) = 2 f ( y ) for all y ∈ M, we obtain π ( y ) + π ( τ ( σ ( y ))) + π ( σ ( τ ( y )))2 = tr ( π ( y )) I , for all y ∈ M. From the properties of adjugation (2.1) we get adj ( π ( y )) = π ( τ ( σ ( y )))+ π ( σ ( τ ( y )))2 , for all y ∈ M. Since adj ( π ( xy )) = adj ( π ( x ) π ( y )) = adj ( π ( y )) adj ( π ( x )) then π ( τ ( σ ( y )))( π ( τ ( σ ( x ))) − π ( σ ( τ ( x )))) = π ( σ ( τ ( y )))( π ( τ ( σ ( x ))) − π ( σ ( τ ( x )))) , x, y ∈ M. (3.14)Now by using (3.13) we get( π ( τ ( σ ( y ))) − π ( σ ( τ ( y )))) = ( tr ( π ( σ ( y ))) − tr ( π ( τ ( y )))) I , y ∈ M, since (3.14) then adj ( π ( y )) = π ( τ ( σ ( y ))) = π ( σ ( τ ( y ))) , this implies that adj ( π ( σ ( y ))) = π ( τ ( y )) . By the condition f ( σ ( y )) + f ( τ ( y )) = 2 f ( y ) we see that tr ( π ( σ ( y ))) + tr ( π ( τ ( y ))) =2 tr ( π ( y ) , so we get tr ( π ( σ ( y ))) = tr ( π ( y ) . Finally we obtain the desired non-abelian solution. (cid:3)
Remark . By using similar proof (see Proposition 3.3), we show the centrality of thesolutions of the functional equation (1.4), when σ is an involutive anti-automorphism and τ is in involutive automorphism. So, the order of σ and τ in the functional equation (1.4) isnot important.The following corollaries are immediate consequences of Theorem 3.4 and Theorem 3.5. Corollary 3.7.
Let G be a topological group and σ : G −→ G be a continuous involutiveautomorphism of G . There are two non-zero solutions f : G → C of the functional equation f ( xσ ( y )) + f ( y − x ) = 2 f ( x ) f ( y ) , x, y ∈ G, abelian and non-abelian: (1) Every abelian continuous solution f is of the form f ( x ) = χ ( x ) + χ ( σ ( x − )2 , where χ : G → C is a continuous multiplicative function such that χ is σ -evenand/or χ ( x ) = χ ( x − ) . (2) Every non-abelian continuous solution f is of the form f ( x ) = 12 trπ ( x ) , x ∈ G, where π is a continuous irreducible unitary representation of dimension for which adj ( π ( σ ( x ))) = π ( x − ) , and tr ( π ( σ ( x ))) = tr ( π ( x )) .Proof. It suffices to take τ ( x ) = x − for all x ∈ G in Theorem 3.4 and Theorem 3.5. (cid:3) A. CHAHBI AND E. ELQORACHI
Corollary 3.8.
Let M be a topological monoid and τ be a continuous involutive anti-automorphism of M . There are two non-zero continuous solutions f : M → C of thefunctional equation f ( xy ) + f ( τ ( y ) x ) = 2 f ( x ) f ( y ) , x, y ∈ G, abelian and non-abelian: (1) Every abelian continuous solution f is of the form f = χ + χ ◦ τ , where χ : M → C is a continuous multiplicative function. (2) Every non-abelian continuous solution f is of the form f ( x ) = 12 trπ ( x ) , x ∈ M, where π is a continuous irreducible unitary representation of dimension for which adj ( π ) = π ◦ τ. Proof.
It suffices to take σ ( x ) = x for all x ∈ M in Theorem 3.4 and Theorem 3.5. (cid:3) Corollary 3.9.
Let G be a topological group. There are two non-zero continuous solutions f : G → C of the functional equation f ( xy ) + f ( y − x ) = 2 f ( x ) f ( y ) , x, y ∈ G, abelian and non-abelian: (1) Every abelian continuous solution f is of the form f ( x ) = χ ( x ) + χ ( x − )2 , x ∈ G where χ : G → C ∗ is a continuous multiplicative function. (2) Every non-abelian continuous solution f is of the form f ( x ) = 12 trπ ( x ) , x ∈ G, where π is a continuous irreducible unitary representation of dimension for which adj ( π ( x )) = π ( x − ) . Proof.
It suffices to take σ ( x ) = x and τ ( x ) = x − for all x ∈ G in Theorem 3.4 and Theorem3.5. (cid:3) Some applications and examples
In this section, we determine solutions of the functional equations (1.6) and (1.7). Weshow that the solutions are closely related to the solutions of (1.4).
Proposition 4.1.
Let z in Z ( G ) the center of G be given. Let f : G → C be a non-zerosolution of the functional equation (1.6). Then(1) f ( e ) = 0 . (2) f ( y − z ) = f ( σ ( y ) z ) , y ∈ G. (3) f ( y − ) = − f ( σ ( y )) , y ∈ G. VARIANT OF D’ALEMBERT FUNCTIONAL EQUATION 9 (4) f ( yσ ( z ) z ) = − f ( y ) , y ∈ G. (5) f ( σ ( z ) z ) = 0 . (6) f ( z ) = 1 . (7) f ( z ) = 0 . Proof. (1) Substituting y = e into (1.6) and using that f = 0 , we get that f ( e ) = 0 . (2) Take x = e in (1.6) and using that f ( e ) = 0 we obtain f ( y − z ) = f ( σ ( y ) z ) for all y ∈ G. (3) Replacing x by σ ( x − ) in (1.6) we get f ( y − σ ( x − ) z ) − f ( σ ( x − ) σ ( y ) z ) = 2 f ( σ ( x − )) f ( y ) , x, y ∈ G. This implies that f (( σ ( x ) y ) − z ) − f (( σ ( y − ) σ ( x )) − z ) = 2 f ( σ ( x − )) f ( y ) , x, y ∈ G. Using (2) we get f ( xσ ( y ) z ) − f ( y − xz ) = 2 f ( σ ( x − )) f ( y ) , x, y ∈ G. (4.1)From (1.6) and that f = 0 we obtain f ( x − ) = − f ( σ ( x )) for all x ∈ G. (4) Taking y = σ ( x ) z into (2) and using that z ∈ Z ( G ) we get f ( σ ( x − )) = f ( xσ ( z ) z )since f ( x − ) = − f ( σ ( x )) , then f ( xσ ( z ) z ) = − f ( x ) . (5) Putting y = z in (2) we get 0 = f ( e ) = f ( σ ( z ) z ) . (6) Replacing y by z in (1.6) and using z ∈ Z ( G ) we have f ( x ) − f ( xσ ( z ) z ) =2 f ( x ) f ( z ) , since f ( xσ ( z ) z ) = − f ( x ) , so 2 f ( x ) = 2 f ( x ) f ( z ) , but f = 0 then f ( z ) = 1 . (7) Replacing y by z − in (2) and using (3) we obtain f ( z ) = f ( σ ( z − ) z ) = f ( σ ( z − σ ( z )) = − f ( σ ( z ) − z ) then f ( z ) = 0 (cid:3) Lemma 4.2.
Let z in Z ( G ) be given. Let f : G → C be a non-zero solution of thefunctional equation (1.6). Then(1) f ( z − ) = f ( σ ( z )) = 1 . (2) f ( z ) = f ( z − ) = − f ( σ ( z )) . Proof. (1) If we take x = y = z − in (1.6) and using z ∈ Z ( G ) we get f ( z ) − f ( σ (( z ) − )) = 2 f (( z ) − ) Using that f ( z ) = 1 and f ( y − ) = − f ( σ ( y )) , y ∈ G, we get f ( z − ) = f ( σ ( z )) =1 . (2) Replacing y by (( z ) ) − in Proposition 4.3 (2) and using and Proposition 4.3 (4)we get f ( z ) = f ( σ ( z ) z ) = f ( σ ( z ) σ ( z ) z ) = − f ( σ ( z )) . This ends the proof. (cid:3)
Theorem 4.3.
Let G be a topological group and let z ∈ Z ( G ) be given. Let f : G → C bea continuous solution of the functional equation (1.6).(1) If σ ( z ) z = e , then (1.6) has no non-zero solutions.(2) σ ( z ) z = e, then the solutions have the following forms (a) f ( x ) = χ ( x ) − χ ( x − )2 χ ( z ) , x ∈ G , where χ : G → C ∗ is a continuous multiplicativefunction such that χ ( σ ( x )) = χ ( x ) and χ ( z ) = − . (b) f ( x ) = χ ( x ) − χ ( σ ( x ))2 χ ( z ) , where χ : G → C ∗ is a continuous multiplicative functionsuch that χ ( x ) = χ ( x − ) and χ ( σ ( z ) z ) = − . Proof. (1) If σ ( z ) z = e, by Proposition 4.1 (4) f ( xσ ( z ) z ) = − f ( x ) , x ∈ G, so wehave f ( x ) = 0 for all x ∈ G. (2) From now, we assume that σ ( z ) z = e. Replacing x by xz and y by yz in (1.6)and using that z ∈ Z ( G ) we get f ( y − xz ) − f ( xσ ( y ) z σ ( z ) z ) = 2 f ( xz ) f ( yz )since f ( xσ ( z ) z ) = − f ( x ) , y ∈ G, then f ( y − xz ) + f ( xσ ( y ) z ) = 2 f ( xz ) f ( yz ) , x, y ∈ G. Consider g ( x ) = f ( xz ) , x ∈ G, then the function g satisfies (1.4) with τ ( x ) = x − and g ( e ) = 1 , so g = 0. As g is a solution of equation (1.4) then by Proposition3.3 (3) g is a solution of pre-d’Alembert functional equation. From Proposition4.1 (7) we get g ( z ) = 0 and by Lemma 4.2 we have d ( z ) = 2 g ( z ) − g ( z ) =0 − f ( z ) = 0 − ( − f ( σ ( z )) = f ( σ ( z )) = 0 . So, we get g ( z ) = d ( z ) . Accordingto [24, Proposition 8.14(a)] we have g is abelian. Furthermore, from Corollary 3.6 (1) g ( x ) = χ ( x )+ χ ( σ ( x − ))2 for all x ∈ G, where χ is a non-zero continuous multiplicativefunction such that χ is σ -even and/or χ ( x ) = χ ( x − ). Since g ( z ) = 0 , so χ ( z ) = − χ ( σ ( z − )), which implies that χ ( z σ ( z )) = − . We have f ( x ) = g ( xz − ) and χ ( z σ ( z )) = −
1, so we get that f ( x ) = χ ( x ) − χ ( σ ( x − ))2 χ ( z ) . This completes the proof. (cid:3)
Theorem 4.4.
Let M be a topological monoid. Let z be an arbitrarily fixed element in M. There are two non-zero continuous solutions f : G → C of (1.7), abelian and non-abelian. (1) Every abelian continuous solution f is of the form f = f ( e ) χ + χ ◦ σ ◦ τ , where χ : G → C is a continuous multiplicative function such that: (i) χ ◦ σ ◦ τ = χ ◦ τ ◦ σ, (ii) χ is σ -even and/or τ -even, and (iii) χ ( σ ( τ ( z ))) = χ ( z ) = f ( e ) . (2) Every non-abelian continous solution f is of the form f ( x ) = f ( e ) χ π ( x )2 , x ∈ G where π is a continuous irreducible representation of dimension for which adj ( π ◦ σ ) = π ◦ τ, tr ( π ( x )) = tr ( π ( σ ( x ))) , x ∈ M, and π ( z ) = f ( e ) I Proof.
Taking y = e in (1.7), we get that f ( xz ) = f ( x ) f ( e ) , x ∈ M. (4.2) VARIANT OF D’ALEMBERT FUNCTIONAL EQUATION 11
We have f = 0 , so f ( e ) = 0 . Using (1.7) and (4.2) twe get that the function g ( x ) = f ( x ) f ( z ) is a solution of (1.4). Therefore, by Theorem 3.4, Theorem 3.5 and the identity (4.2) it iseasy to get the form of solutions of (1.7).Conversely, It is easy to check that the all of the possibilities listed in Theorem 4.4 definesolutions of (1.7). (cid:3) Example 4.5.
For a non-abelian example of Monoid, we consider G = M (2 , C ) and take σ as an automorphism and τ as an anti-automorphism such that σ (cid:18) a bc d (cid:19) = J (cid:18) a bc d (cid:19) J = (cid:18) d cb a (cid:19) where J = (cid:18) (cid:19) and τ (cid:18) a bc d (cid:19) = (cid:18) a − b − c d (cid:19) We indicate here the corresponding continuous solutions of (1.4). Simple computations showthat σ ◦ τ = τ ◦ σ Note that T = (cid:18) a cb d (cid:19) . Abelian solutions .The continuous non-zero multiplicative functions on M (2 , C ) are given (see [9, Example5.6]): µ = 1 , or else ( χ ( T ) = | det( T ) | λ − n ( det ( T )) n when det ( T ) = 00 when det ( T ) = 0 , where λ ∈ C with Re λ > n ∈ Z . It is clear that χ ( σ ( τ ( T ))) = χ ( τ ( σ ( T ))) ,χ ( σ ( T )) = χ ( τ ( T )) = χ ( T ) . It follows that the abelian continuous solutions are f ( T ) = | ad − bc | λ − n ( ad − bc ) n . Non-abelian solutions .It is clear that tr ( T ) = tr ( σ ( T )), Adj( σ ( T )) = τ ( T ) and σ ◦ τ ( T ) = τ ◦ σ ( T ) . By usingTheorem 3.4, the non-abelian non-zero continuous solutions f : M (2 , C ) → C of (3.3) are f = tr. Example 4.6.
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Ibn Zohr University, Faculty of Sciences, Department of Mathematics, Agadir, Morocco
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