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## versión impresa ISSN 0716-0917

### Proyecciones (Antofagasta) vol.36 no.4 Antofagasta dic. 2017

#### http://dx.doi.org/10.4067/S0716-09172017000400641

Articles

On the solution of functional equations of Wilson's type on monoids

1Ibn Tofail University, Faculty of Sciences, Department of Mathematics, Kenitra, Morocco, e-mail : izidd-math@hotmail.fr

2Ibn Tofail University, Faculty of Sciences, Department of Mathematics, Kenitra, Morocco, e-mail : abdellatifchahbi@gmail.com

3Ibn Tofail University, Faculty of Sciences, Department of Mathematics, Kenitra, Morocco, e-mail : samkabbaj@yahoo.fr

Abstract:

Let S be a monoid, C be the set of complex numbers, and let σ,τAntihom(S,S) satisfy τ ○ τ =σ ○ σ= id. The aim of this paper is to describe the solution ⨍,g: SC of the functional equation

in terms of multiplicative and additive functions.

Keywords : Wilson's functional equation; monoids; multiplicative function.

1.Notation and terminology

Throughout the paper we work in the following framework: A monoid is a semi-group S with an identity element that we denote e, that is an element such that ex = xe = x for all xS (a semi-group is an algebraic structure consisting of a set together with an associative binary operation) and σ, τ: S →to S are two anti-homomorphisms (briefly σ, τ ∈ Antihom(S,S)) satisfying τ ○ τ = σ ○ σ = id.

For any function ⨍: S → to C we say that ⨍ is σ-even (resp. τ-even) if ⨍ ○ σ= ⨍ (resp. ⨍ ○ τ = ⨍), also we use the notation (x)=( x-1 ) in the case S is a group.

We say that a function χ :S → to C is multiplicative, if χ(xy)=χ(x)χ(y) for all x,yS.

If χ: S → C is multiplicative and χ ≠ 0, then

Iχ:= {x ∈ S|χ(x)=0} is either empty or a proper subset of S.

If S is a topological space, then we let C(S) denote the algebra of continuous functions from S into C.

2. Introduction

Wilson's functional equation on a group G is of the form

where ⨍, g: G → to C are two unknown functions.

Special cases of Wilson's functional equation are d'Alembert's functional equation

and Jensen's functional equation

In (3) Ebanks and Stetkær studied the solutions ⨍, g: G → to C of Wilson's functional equation (2.1) and the following variant of Wilson's functional (see (8))

They solve (2.3) and they obtained some new results about (2.1). We refer also to Wilson's first generalization of d'Alembert's functional equation

For more about the functional equation (2.4) see Aczél (1). The solutions formulas of equation (2.4) for abelian groups are known.

In the same year Stetkær in (10) obtained the complex valued solution of the following variant of d'Alembert's functional equation

where is a semi-group and σ is an involutive homomorphism of . The difference between d'Alembert's standard functional equation

and the variant (2.5) is that τ is an anti-homomorphism (on a group typically the group inversion). Some information, applications and numerous references concerning (2.5) and their further generalizations can be found e.g. in (6),(9),(10).

Some general properties of the solutions of equation

on a topological monoid M equipped with a continuous involution σ : MM can be found in (9).

Stetkær (11) proved a natural interesting relation between Wilson's functional equation (2.6) and d'Alembert's functional equation (2.2) and for σ(x) = x -1. That is if ⨍ ≢ 0 is a solution of equation (2.6), then g is a solution of equation (2.2). In (2) Chahbi et al. give a generalization of the symmetrized multiplicative Cauchy equation. Recently, EL-Fassi et al. (5) obtained the solution of following functional equation

where is a semi-group and σ, τ are two anti-homomorphisms of such that σ ○ σ = τ ○ τ = id.

The main purpose of this paper is to solve the functional equation

where S is a monoid and σ, τ ∈ Antihom(S,S) such that σ ○ σ = τ ○ τ = id. This equation is a natural generalization of (2.7) and of the following new functional equations

where (S, ∙) is a minoid and σ, τ ∈ Antihom(S,S) such that σ ○ σ = τ ○ τ = id. Clearly, if S is a group and = ⨍ with σ (x) = x -1, then functional equation (2.10) becomes the symmetrized multiplicative Cauchy equation (see for instance (7) or ((9),Theorem 3.21). By elementary methods we find all solutions of (2.8) on monoid in terms of multiplicative functions. Finally, we note that the sine addition law on minoid given in (4), (9) is a key ingredient of the proof of our main result (Theorem 3.1).

3. Solution of the functional equation (2.8)

In this section we obtain the solution of the functional equation (2,8) on monoid. The following lemma will be used in the proof of Theorem 3.1.

Lemma 3.1. Let S be a monoid and σ ∈ Antihom(S,S). Let ⨍, g: SC be a solution of the functional equation

Then g is a multiplicative function.

Proof. For all x, y, zS, we have

then g(yz) = g(y) g(z) for all x, y, zS. This implies that g is a multiplicative function.

Theorem 3.1. Let S be a monoid with identity element e, and σ, τ ∈ Antihom(S,S) such that σ ○ σ = τ ○ τ = id (where id denotes the identity map). The solutions ⨍, g: S → C of (2.8) are the following pairs of functions, where χ: SC denotes a multiplicative function such that χ(e) = 1:

Proof. It is elementary to check that the cases stated in the Theorem define solutions, so it is left to show that any solution ⨍, g: S→ C of (2.8) falls into one of these case. We use in the proof similar Stetkaer's computations (10). Let x, y, zS be arbitrary. If we replace x by x σ(y) and y by z in (2.8), we get

On the other hand if we replace x by τ (z) x in (2.8), we infer that

Replacing y by z y in (2.8), we obtain

It follows from (3.4) that (3.3) become

Subtracting this from (3.2) we get after some simplifications that

With the notation

equation (3.6) can be written as follows

This shows that the pair (⨍a, g) satisfies the sine addition law for any a ∈ S. From the Known solution of the sine addition formula (see for example (4), Lemma 3.4), we have the following possibilities.

If ⨍ ≡ 0 we deal with case (1) in the Theorem. So during the rest of the proof we will assume that ⨍ ≢ 0. If we replace (x, y) by (e, σ(x)) in (3.7), we get

⦁Suppose that ⨍x = 0 for all x ∈ S, then ⨍e = 0, i.e., ⨍(x) = ⨍(e) g ○ σ (x) for all x ∈ S, and hence ⨍(e)≠ 0. Indeed, ⨍(e) = 0 would entail ⨍ ≡ 0, contradicting our assumption. From Lemma 3.1, we see that g is a multiplication function. Substituting ⨍ = ⨍(e) g ○ σ into (2.8), we infer that g = g ○ σ ○ τ . We may thus write g = (g + g ○ σ ○ τ)/2 which is the form claimed in the (2).

⦁Now suppose that ⨍x ≠ 0 for some xS.

If ⨍e ≠ 0 then, from ((4), Lemma 3.4), we see that there exist two multiplicative functions χ12: SC such that

Case (3): If χ1 ≠ χ2, then ⨍ e = c(χ1 - χ2) for some constant c ∈ C\{0}. From equality (3.9), we find after a reduction that

where α = (2c + ⨍(e))/2 and β = ⨍(e)-α. Substituting ⨍ and g in (2.8), we get after some simplification that

for all x, yS. Since χ1 ≠ χ2 we get from the theory of multiplicative functions (see for instance ((9) Theorem 3.18) that both terms are 0, so

for all x, yS. Since ⨍ ≢ 0 at least one of α and β is not zero.

Subcase (3.i): If α = 0 and β ≠ 0, by (3.10) and ⨍ ≢ 0, for this to be the case we must have χ1 ≠ 0, χ2 ≠ 0 and χ2 ○ σ ○ τ (y) = χ1(y) for all yS, then ⨍ and g have the desired form (3.i) with χ2:=χ.

If α ≠ 0 and β = 0, by (3.10) and ⨍ ≢ 0, for this to be the case we must have χ1 ≠ 0, χ2 ≠ 0 and χ1 ○ σ ○ τ (y) = χ2(y) for all yS, then ⨍ and g have the desired form (3.i) with χ1:=χ.

Subcase (3.ii): If α ≠ 0 and β ≠ 0, by (3.10) and ⨍ ≢ 0, for this to be the case we must have χ1 ≠ 0, χ2 ≠ 0 , χ1 ○ σ ○ τ (y) = χ2(y) and χ2 ○ σ ○ τ (y) = χ1(y) for all yS, with χ1 =χ, after some simplification, we obtain χ ○ σ ○ τ = χ ○ τ ○ σ and the desired form (3.ii) of ⨍and g.

Case 4: If χ12 then letting χ := χ1 we have g = χ. If S is generated by its squares, then there exists an additive function A: S\Iχ C for which

Subcase (4.i): If x ∈ S\ , then by (3.9) and (3.11), we get

for all x ∈ S\ ,. Substituting ⨍ and g in (2.8), we get after some simplification that

for all x, y ∈ S\ . Suppose that χ ○ σ ○ τ ≠ χ. From (3.12) we infer that A ≡ 0, this contradicts with ⨍e ≠ 0 on S. So χ ○ σ ○ τ = χ and A ○ σ ○ τ = -A.

Subcase (4.ii): If x , then g(x) = ⨍(x)= 0.

If ⨍e = 0, then ⨍(x)= ⨍(e) g ○ σ(x) for all xS, and hence ⨍(e)≠ 0. Replacing (x,y) by (e,x) in (2.8), we get

From (3.13) and ⨍(x)= ⨍(e) g ○ σ (x) for all xS, we obtain ⨍(x)= ⨍(e) g ○ τ (x) for all xS. So, g is a solution of the functional equation

Similar to the proofs of ((10), Theorem 2.1), we find that g = (χ + χ ○ σ ○ τ)/2, where χ: SC is multiplicative and χ ○ σ ○ τ = χ ○ τ ○ σ. Hence we are in case (2) or (3).

The continuity statement follows from ((9) Theorem 3.18 (d)). This completes the proof of Theorem.

4. Some consequences

As immediate consequences of Theorem 3.1, we have the following corollaries.

Corollary 4.1. Let S be a monoid with identity element e, and σ, τ ∈ Antihom(S,S) such that σ ○ σ = τ ○ τ = id. The solutions ⨍, g: SC of the functional equation

are the following pairs of functions, where χ : S → C denotes a multiplicative function such that χ(e) = 1:

(1) ⨍ ≡ 0 and g arbitrary.

(2) g =χ and ⨍ = ⨍(e) χ ○ σ, where χ≠ 0 and ⨍(e) ∈ C\{0}.

Furthermore, if S is a topological monoid and ⨍, gC(S), then χ, χ ○ σ ∈ C(S).

Proof. It suffices to take τ(x) =σ(x) for all xS in Theorem 3.1.

Corollary 4.2. (5)Let S be a monoid with identity element e, and σ, τ ∈ Antihom(S,S) such that σ ○ σ = τ ○ τ = id. The solutions ⨍: SC of the functional equation

are the functions of the form ⨍ = (χ + χ ○ σ ○ τ)/2, where χ: S→ C is a multiplicative such that:

(i) χ ○ σ ○ τ = χ ○ τ ○ σ, and

(ii) χ is σ-even or/and τ-even.

Furthermore, if S is a topological monoid and ⨍ ∈ C(S), then χ, χ ○ σ ○ τ ∈ C(S).

Proof. It suffices to take g(x) = ⨍(x) for all xS in Theorem 3.1..

Corollary 4.3. Let S be a monoid with identity element e, and σ, τ∈ Antihom(S,S) such that σ ○ σ = τ ○ τ = id. The solutions ⨍: SC of the functional equation

are the functions of the form:

(1) ⨍ ≡ 0.

(2) ⨍ = ⨍(e), where ⨍(e)∈ C\{0}.\$

(3) If S is generated by its squares, then

(i) ⨍(x)= A ○ σ(x) + ⨍(e) for xS\ ,

(ii) ⨍(x)= 0 for x ,

where A: S\IχC is an additive function such that A ○ σ ○ τ = -A ≠ 0.

Furthermore, if S is a topological monoid, and ⨍ ∈ C(S), then A ○ σ ∈ C( S\Iχ ).

Proof. It suffices to take g(x) =1 for all xS in Theorem 3.1.

Corollary 4.4. Let S be a monoid with identity element e, and σ ∈ Antihom(S,S) such that σ ○ σ = id. The solutions ⨍: SC of the functional equation

are the functions of the form ⨍ =χ, where χ: S → C is a multiplicative such that χ is σ-even.

Proof. It suffices to take g(x) =⨍(x) and τ(x) =σ(x) for all xS in Theorem 3.1.

Acknowledgments

The author would like to express his most sincere gratitude to the two reviewers for a number of valuable comments which have led to essential improvement of the paper.

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Received: December 2016; Accepted: September 2017 This is an open-access article distributed under the terms of the Creative Commons Attribution License