The Integral Sine Addition Law 205 Characters on G Is a Linearly Independent Subset of the Vector Space of All Complex-Valued Functions on G (See [8, Corollary 3.20])

Home , Functional equation

Proyecciones Journal of Mathematics Vol. 38, No 2, pp. 203-219, June 2019. Universidad Cat´olica del Norte Antofagasta - Chile

Theintegralsineadditionlaw

D. Zeglami Moulay Ismail University, Morocco M. Tial Ibn Tofail University, Morocco and S. Kabbaj Ibn Tofail University, Morocco Received : February 2017. Accepted : January 2019

Abstract In the present paper we determine, in terms of characters and additive functions, the solutions of the integral functional equation for the sine addition law

f(xyt)dµ(t)=f(x)g(y)+g(x)f(y),x,y G, ∈ ZG where G is a locally compact Hausdorﬀ group and µ is a regular, compactly supported, complex-valued Borel measure on G. Some con- sequences of this result and an example are presented.

Subjclass: 39B32, 39B52.

Keywords: Functional equation, Sine and cosine addition laws, Char- acter, Borel measure. 204 D. Zeglami, M. Tial and S. Kabbaj

1. Introduction

Let G be a topological group with neutral element e. The trigonometric addition and subtraction formulas have been studied in the context of functional equations by a number of mathematicians. The monographs by Acz´el [1], by Kannappan [6], by Stetkær [8] and by Sz´ekelyhidi [11] have references and detailed discussions of the classic results. Chung, Kannappan and Ng [2] solved on any group G, the functional equation

f(xy)=f(x)g(y)+f(y)g(x)+h(x)h(y),x,y G. ∈ Poulsen and Stetkær [7] found the complete set of continuous solutions of each of the functional equations

(1.1) g(xy)=g(x)g(y) f(x)f(y),x,y G, − ∈ (1.2) f(xy)=f(x)g(y)+f(y)g(x),x,y G. ∈ In the paper [12], Van Vleck studied the continuous solutions f : R R of the functional equation →

(1.3) f(x y + z0) f(x + y + z0)=2f(x)f(y),x,yR, − − ∈ where z0 > 0isﬁxed. Extensions of (1.3) and related functional equations from R to locally compact groups can be found e.g. in ([4, 5, 9, 13, 14, 15]). Results of [10] where Stetkær ﬁnd the solutions of the cosine addition law with an additional term

(1.4) g(xy)=g(x)g(y) f(x)f(y)+αf(xy), − (on semigroups) for a ﬁxed complex constant α, and of [12] have been an inspiration for this work in their treatments of (1.4) and (1.3), respectively. To formulate our results we introduce the following notations and as- sumptions that will be used throughout the paper: C(G)denotestheal- gebra of continuous, complex valued functions on G. The set of homomor- phisms a : G (C, +) will be called the additive maps and denoted by (G). → A A character χ of G is a homomorphism χ : G C∗,whereC∗ denotes the multiplicative group of non-zero complex numbers.→ So characters need not be unitary in the present paper. It is well known that the set of The integral sine addition law 205 characters on G is a linearly independent subset of the vector space of all complex-valued functions on G (see [8, Corollary 3.20]). Let G be a locally compact Hausdorﬀ group, and let MC(G)denotethe space of all regular, compactly supported, complex-valued Borel measures on G.Forµ MC(G), we use the notation ∈ µ(f)= f(t)dµ(t), ZG for all f C(G). ∈ Let µ MC (G). The purpose of the present paper is to give an explicit description∈ of the continuous solutions f,g : G C of the following integral version of the sine addition law →

(1.5) f(xyt)dµ(t)=f(x)g(y)+g(x)f(y),x,y G. G ∈ Z TosolveEq.(1.5)wereduceittoEqs.(1.1),(1.2)andthefollowing functional equation

f(xy)=f(x)g(y)+g(x)f(y) g(xy),x,yG, − ∈ the continuous solutions of which, in terms of characters and additive functions, are given in Theorem 2.1. As an important consequence, we solve the following functional equation

(1.6) f(xyz0)=f(x)g(y)+g(x)f(y),x,yG, ∈ where z0 G is an arbitrarily ﬁxed element. Eq. (1.6) results from (1.5) ∈ by replacing µ by δz0 : the Dirac measure concentrated at z0.

Let n N, α1 , ,αn C,andz1, ,zn G be arbitrarily ﬁxed elements.∈ As a more··· general∈ consequence··· of our result,∈ we obtain all continuous solutions of the following version of the sine addition formula

n αif(xyzi)=f(x)g(y)+g(x)f(y),x,yG. i=1 ∈ X In Proposition 2.6 we characterize the solutions of the integral functional equation for the sine subtraction law

f(xyt)dµ(t)=f(x)g(y) g(x)f(y),x,y G. G − ∈ Z In the last section, we provide a concrete example to show that non- trivial continuous solutions of (1.6) occur in real life. 206 D. Zeglami, M. Tial and S. Kabbaj

Note that the following integral versions of the addition and subtraction formulas for cosine and sine:

g(xyt)dµ(t)=g(x)g(y) f(x)f(y),x,y G, G − ∈ Z f(xσ(y)t)dµ(t)=f(x)g(y) g(x)f(y),x,y G, G − ∈ Z where σ denotes an involution of G,i.e.,σ(xy)=σ(x)σ(y)andσ(σ(x)) = x for all x, y G, were solved in [16]. ∈

2. The solutions of the integral sine addition law

The purpose of this section is ﬁrst to give an explicit description of the solutions of the functional equation

(2.1) f(xy)=f(x)g(y)+g(x)f(y) g(xy),x,y G, − ∈ where f,g : G C are unknown functions. And secondly to determine the solutions f,g :→G C of the functional equation (1.5), namely → (2.2) f(xyt)dµ(t)=f(x)g(y)+g(x)f(y),x,y G, G ∈ Z

where µ MC(G), in terms of characters and additive maps of G. ∈ In the following theorem we exhibit the solutions of the functional equation (2.1) which were solved by Chung, Kannappan and Ng [2], but their results were formulated diﬀerently from what we need. So we present, using [10, Theorem 6.3], a modiﬁedproofofthesolutionformulas.

Theorem 2.1. Let G be a topological group. The pair f,g C(G) is a solution of the functional equation (2.1) if and only if one of the∈ possibilities listed below takes place:

(i) There exist a constant q C and a continuous character χ of G such that ∈ χ χ f =(1 q + δ) and g =(1+q + δ) , − 2 2

where δ = 1+q2. ± p The integral sine addition law 207

(ii) There exist a constant q C and two diﬀerent continuous characters ∈ ∗ χ1 and χ2 of G such that

χ1 + χ2 χ1 χ2 χ1 + χ2 χ1 χ2 f = +(δ q) − and g = +(δ + q) − , 2 − 2 2 2

where δ = 1+q2. ± (iii) There exist ap continuous character χ of G and a non-zero continuous and additive function a : G C such that →

f = χ and g = χ(1 + a) or f =(1+a)χ and g = χ.

1 Proof. We deﬁne the function H := 2 (f + g). Then the equation (2.1) becomes

H(xy)= (g(x) H(x))(g(y) H(y)) + H(x)H(y),x,y G. − − − ∈ Then we obtain the functional equation

H(xy)=H(x)H(y) K(x)K(y),x,y G, − ∈ where K := g H, the solutions of which were given in [10, Theorem 6.3]. Applying [10, Theorem− 6.3] we infer that there are only the following cases: (i) There exist a constant q C and a continuous character χ of G such that ∈ χ χ K = q and H =(1 1+q2) , 2 ± 2 q which shows that χ χ f =(1 q + δ) and g =(1+q + δ) , − 2 2 where δ = 1+q2. (ii) There± exist a constant q C and two diﬀerent continuous charac- p ∈ ∗ ters χ1 and χ2 of G such that

χ1 χ2 χ1 + χ2 χ1 χ2 K = q − and H = 1+q2 − , 2 2 ± 2 q which gives 208 D. Zeglami, M. Tial and S. Kabbaj

χ1 + χ2 χ1 χ2 χ1 + χ2 χ1 χ2 f = +( q + δ) − and g = +(q + δ) − , 2 − 2 2 2 where δ = 1+q2. ± (iii) Therep exist a continuous character χ of G and a non-zero continuous function A (G) such that ∈ A K = χA and H = χ(1 A), ± and so we obtain

f = χ(1 A + φ)andg = χ(1 + A + φ), − where φ = A.Thenf = χ and g = χ(1 + a)orf =(1 a)χ and g = χ where a :=± 2A. − Conversely, simple computations prove that the formulas above for f and g deﬁne solutions of (2.1). 2

Remark 2.2. i) The continuity of characters and the additive functions in [10, Theorem 6.3 ] and consequently in Theorem 2.1 results from [10, Proposition 5.2]. ii) Eq. (2.1) is symmetric i.e. if the pair of functions (ϕ, ψ) is a solution of (2.1) then (ψ, ϕ) is also a solution of (2.1). This is clearly seen in the form of solutions in Theorem 2.1.

Now we are in the position to describe all solutions of the functional equation (2.2).

Theorem 2.3. Let G be a locally compact Hausdorﬀ group and µ ∈ MC (G). Assume that the pair f,g C(G) is a solution of Eq. (2.2). Then we have the following possibilities:∈

1. g =0and f is any function in C(G) such that G f( t)dµ(t)=0or f =0and g is arbitrary in C(G). · R 2. There exist a constant γ C and a continuous character χ of G, with µ(χ) =0, such that ∈ 6 µ(χ) f = γχ and g = χ. 2 The integral sine addition law 209

3. There exist a constant γ C and two continuous characters χ1 and ∈ χ2 of G, with µ(χ2)= µ(χ1) =0, such that − 6

γ µ(χ1) f = (χ1 + χ2) and g = (χ1 χ2). 2 2 − 4. There exist a constant c C and two diﬀerent continuous characters ∈ χ1 and χ2 of G, with µ(χ2)=µ(χ1) =0, such that 6

µ(χ1) f = c(χ1 χ2) and g = (χ1 + χ2). − 2 5. There exist constants γ,β,q C and two continuous characters χ1 ∈2+2δ 2 2δ and χ2 of G, with µ(χ1)=β 1 q+δ and µ(χ2)=β 1+−q δ , such that − −

χ1 + χ2 χ1 χ2 f = γ +( q + δ) − 2 2 ½ − ¾ χ1 + χ2 χ1 χ2 and g = β +(q + δ) − , 2 2 ½ ¾

where δ := 1+q2. ± 6. There exist ap non-zero continuous function a (G) and a continuous character χ of G, with µ(aχ)=0and µ(χ) =0∈ A, such that 6

f = χa and g = µ(χ)χ.

7. There exist a constant γ C, a continuous function a (G) and a continuous character χ ∈of G,withµ(χ)=µ(aχ) =0, such∈ A that 6

f = γ(1 + a)χ and g = µ(χ)χ.

Conversely, the formulas above for f and g deﬁne solutions of (2.2).

Proof. Let the pair f,g be a solution of the equation (2.2). Letting y = e in (2.2) we get that

(2.3) f(xt)dµ(t)=βf(x)+γg(x),x G, G ∈ Z 210 D. Zeglami, M. Tial and S. Kabbaj where β = g(e)andγ = f(e). So, using (2.3), we can reformulate the form of Eq. (2.2) as

(2.4) βf(xy)=f(x)g(y)+g(x)f(y) γg(xy),x,y G. − ∈ Case 1: Suppose that β = 0 then (2.4) gives

γg(xy)=f(x)g(y)+g(x)f(y) x, y G. ∈ If γ = 0 then equation (2.4) becomes

f(x)g(y)= g(x)f(y) x, y G, − ∈ thus by a small computation we show that we are in the case (1) of our statement. If γ = 0 then from (2.4) we obtain the functional equation 6 g(xy)=F (x)g(y)+g(x)F (y),x,y G, ∈ 1 where F = γ f, the solutions of which were given in [7, Proposition III.1]. We work our way through the possibilities (a)-(d) presented by [7, Propo- sition III.1] to see what the properties (2.3), that g(e)=0andf(e) =0 entail. We use the notation of [7, Proposition III.1] with m replaced here6 by χ. (a) g =0andf is any continuous function. Using (2.3) we get that

G f( t)dµ(t) = 0. So we are in (1). (b)· g = cχ and f = γ χ .Herewehavef(e)=γ = γ , which implies R 2 2 that γ = 0. So this case is excluded. γ (c) g = c(χ1 χ2)andf = 2 (χ1 + χ2). If c =0orχ1 = χ2 then g =0 so we are in (1).− Otherwise, using (2.3) and the linear independence of diﬀerent characters we infer that µ(χ1)=2c and µ(χ2)= 2c.Soweare in (3). − (d) g = χa and f = γχ. Using(2.3)wegeta = µ(χ)=0theng =0. So we are in (1). Case 2: Suppose that β = 0. We can reformulate the form of Eq. (2.4) as 6 f(xy)=f(x)G(y)+f(y)G(x)+γG(xy),x,y G, ∈ 1 where G := β g, the solutions of which, in the subcase γ =0,aregivenin [7, Proposition III.1]. When we analyse them we ﬁnd, using the notation in [7, Proposition III.1] with m replaced here by χ,that: The integral sine addition law 211

(a) f =0andg is any function. So we are in (1). χ β (b) f = cχ and g = β 2 . Here we have g(e)=β = 2 , which implies that β = 0. So this case is excluded. β (c) f = c(χ1 χ2)andg = 2 (χ1 + χ2). If c =0orχ1 = χ2 then f =0 so we are in (1) and− otherwise a computation based on (2.3) and the linear independence of diﬀerent characters shows that

cµ(χ1)=cβ and cµ(χ2)=cβ , which gives µ(χ1)=µ(χ2)=β = 0 thus we are in (4). (d) f = χa and g = βχ. Using6 (2.3) we get that µ(χ)=β =0and µ(aχ) = 0. This requires a =0.Sowearein(6). 6 For the subcase γ = 0, the6 equation (2.4) becomes 6 F (xy)=F (x)G(y)+G(x)F (y) G(xy) x, y G, − ∈ 1 1 where G = β g and F = γ f . Applying Theorem 2.1 we infer that there are only the following 3 cases: (i) There exist a constant q C and a continuous character χ of G such that ∈ χ χ f = γ(1 q + δ) and g = β(1 + q + δ) , − 2 2 where δ = 1+q2.Sincef(e)=γ and g(e)=β then 1 q + δ = ± − 1+q + δ =2(i.ep q =0andδ =1)andsof = γχ and g = βχ.Using (2.3) we get that µ(χ)=2β = 0. So we are in the case (2) of our statement. 6 (ii) There exist a constant q C and two diﬀerent continuous charac- ∈ ∗ ters χ1 and χ2 of G such that

χ1 + χ2 χ1 χ2 f = γ +( q + δ) − 2 − 2 µ χ χ¶ = γ(1 q + δ) 1 + γ(1 + q δ) 2 − 2 − 2 and

χ1 + χ2 χ1 χ2 g = β +(q + δ) − 2 2 µ χ ¶χ = β(1 + q + δ) 1 + β(1 q δ) 2 , 2 − − 2 212 D. Zeglami, M. Tial and S. Kabbaj where δ = 1+q2. A computation based on (2.3) and the linear inde- ± pendence of dipﬀerent characters shows that

γ(1 q + δ)µ(χ1)=γβ(1 q + δ)+γβ(1 + q + δ) − − = γβ(2 + 2δ), and

γ(1 + q δ)µ(χ2)=γβ(1 + q δ)+γβ(1 q δ) − − − − = γβ(2 2δ), − Since q =0,thenq 1 = δ = q +1andsoweﬁnd, as we have γ =0, that 6 − 6 6 6

2+2δ 2 2δ µ(χ )=β and µ(χ )=β − . 1 1 q + δ 2 1+q δ − − So we are in the case (5) of our statement. (iii) There exist a continuous character χ of G and a non-zero continuous function a (G) such that ∈ A

f = γχ and g = βχ(1 + a)or f = γ(1 + a)χ and g = βχ.

µ(χ) By (2.3), the first case gives a =0andβ = 2 . Soweareinthepoint (2). For the case f = γ(1 + a)χ and g = βχ we find, using (2.3), that µ(aχ)=µ(χ)=β = 0. So we are in the case (7) of our statement. 6 Conversely, some computations prove that the formulas above for f and g define solutions of (2.2). 2 In the following corollary we solve the functional equation

(2.5) f(xyz0)=f(x)g(y)+g(x)f(y),x,y G, ∈ for a ﬁxed constant z0 G. ∈ Corollary 2.4. Let G be a topological group. Assume that the pair f,g C(G) is a solution of Eq. (2.5). Then we have the following possibilities:∈

1) f =0and g is arbitrary in C(G). The integral sine addition law 213

2) There exist a constant γ C and a continuous character χ of G such that ∈

χ(z ) f = γχ and g = 0 χ. 2

3) There exist a constant γ C and two continuous characters χ1 and ∈ χ2 of G, with χ2(z0)= χ1(z0), such that −

γ χ1(z0) f = (χ1 + χ2) and g = (χ1 χ2). 2 2 − 4) There exist a constant c C and two diﬀerent continuous characters ∈ χ1 and χ2 of G, with χ2(z0)=χ1(z0), such that

χ1(z0) f = c(χ1 χ2) and g = (χ1 + χ2). − 2 5) There exist constants γ,β,q C and two continuous characters χ1 ∈ 2+2δ 2 2δ and χ2 of G, with χ1(z0)=β 1 q+δ and χ2(z0)=β 1+−q δ , such that − −

χ1 + χ2 χ1 χ2 f = γ +( q + δ) − 2 2 ½ − ¾ χ1 + χ2 χ1 χ2 and g = β +(q + δ) − , 2 2 ½ ¾

where δ := 1+q2. ± 6) There exist ap continuous function a (G) and a continuous char- ∈ A acter χ of G, with a(z0)=0, such that

f = χa and g = χ(z0)χ.

7) There exist a constant γ C, a continuous function a (G) and ∈ ∈ A a continuous character χ of G, with a(z0)=1, such that

f = γ(1 + a)χandg= χ(z0)χ.

Conversely, the formulas above for f and g deﬁne solutions of (2.5). 214 D. Zeglami, M. Tial and S. Kabbaj

Proof. As the proof of Theorem 2.3 with µ = δz0 and the fact that χ(z0) =0. 2 6 Let n N, α1 , ,αn C,andz1, ,zn G be arbitrarily ﬁxed elements. As∈ another··· consequence∈ of our r···esult, in∈ the following corollary, we obtain all continuous solutions of the following version of the sine addition formula

n (2.6) αif(xyzi)=f(x)g(y)+g(x)f(y),x,yG. i=1 ∈ X Corollary 2.5. Let G be a topological group. Assume that the pair f,g C(G) is a solution of Eq. (2.6). Then we have the following possibilities:∈

1) f =0and g is arbitrary in C(G) or g =0and f is any function in n C(G) such that αif (xzi)=0for all x G. i=1 ∈ 2) There exist a constantP γ C and a continuous character χ of G n ∈ with αix (zi) =0such that i=1 6 P n 1 αiχ(zi) f = γχ and g = i=1 χ. 2 P 2

3) There exist a constant γ C and two diﬀerent continuous characters n∈ n χ1 and χ2 of G, with αiχ2(zi)= αiχ1(zi) =0, such that i=1 − i=1 6 P n P γ i=1 αiχ1(zi) f = (χ1 + χ2) and g = (χ1 χ2). 2 P 2 − 4) There exist a constant c C and two diﬀerent continuous characters n∈ n χ1 and χ2 of G, with αiχ2(zi)= αiχ1(zi) =0, such that i=1 i=1 6 P n P i=1 αiχ1(zi) f = c(χ1 χ2) and g = (χ1 + χ2). − P 2 5) There exist constants γ,β,q C and two continuous characters χ1 n ∈ 2+2δ n and χ2 of G, with i=1 αiχ1(zi)=β 1 q+δ and i=1 αiχ2(zi)= 2 2δ − β 1+−q δ , such that P P −

χ1 + χ2 χ1 χ2 f = γ +( q + δ) − and 2 2 ½ − ¾ The integral sine addition law 215

χ1 + χ2 χ1 χ2 g = β +(q + δ) − , 2 2 ½ ¾

where δ := 1+q2. ± p 6) There exist a non-zero continuous function a (G) and a continuous n ∈ A n character χ of G, with i=1 αia(zi)χ(zi)=0and i=1 αiχ1(zi) =0, such that 6 P P

n f = χa and g = αiχ(zi)χ. i=1 X 7) There exist a constant γ C, a continuous function a (G) and a ∈ n n ∈ A continuous character χ of G, with i=1 αiχ(zi)= i=1 αia(zi)χ(zi) = 0, such that 6 P P

n f = γ(1 + a)χ and g = αiχ(zi)χ. i=1 X Conversely, the formulas above for f and g deﬁne solutions of (2.6).

n Proof. As the proof of Theorem 2.3 with µ = i=1 αiδzi . 2 In the following proposition we show that the integral functional equa- P tion for the sine subtraction law

(2.7) f(xyt)dµ(t)=f(x)g(y) g(x)f(y),x,y G, G − ∈ Z has only ”trivial” solutions.

Proposition 2.6. Let G be a locally compact Hausdorﬀ group. The pair f,g C(G) is a solution of Eq. (2.7) if and only if it has one of the following forms:∈

(i) f =0and g is arbitrary in C(G).

(ii) There exists a constant α C such that g = αf and f is any function in C(G) such that f( t)∈dµ(t)=0. G · R 216 D. Zeglami, M. Tial and S. Kabbaj

Proof. Let the pair f,g be a solution of the equation (2.7). It is easy to see that

(2.8) f(yxt)dµ(t)= f(xyt)dµ(t) for all x, y G. G − G ∈ Z Z Letting y = e in (2.8) we get that G f(xt)dµ(t) = 0 for all x G. Thus we get that ∈ R f(x)g(y)=g(x)f(y), for all x, y G. ∈ If f = 0 then we are in (i) and if f =0theng = αf where α = g(a) 6 f(a) for some a G such that f(a) = 0 then we are in (ii). Conversely, the formulas above∈ for f and g deﬁne6 solutions of (2.7). 2

3. An example

Example 3.1. Let G =(R, +),z0 R be a ﬁxed element, and let µ = ∈ δz0 . We indicate here the corresponding continuous solutions of Eq. (2.2) (i.e. (2.5)) by the help of Corollary 2.4. The continuous characters on R are known to be χ(x)=eλx,x R, where λ ranges over C (see for instance [8, Example 3.7(a)]). ∈ The continuous additive functions on R are the functions of the form a(x)=αx, x R, where the constant α ranges over C (see for instance [8, ∈ Corollary 2.4]). In the point (6) of Corollary 2.4 we have a(z0)=0which reduces to α =0i.e. a =0. So this case does not occur here. In the point λ z λ z (4) of Corollary 2.4 we have χ2(z0)=χ1(z0) which reduces to e 1 0 = e 2 0 for some λ1,λ2 R then λ1 = λ2 i.e. χ1 = χ2. So this case does not occur here either. ∈ In conclusion, by help of Corollary 2.4 we ﬁnd that the continuous solutions f,g : R C of the functional equation (2.2), which is here →

f(x + y + z0)=f(x)g(x)+f(y)g(x),x,y R, ∈ are 1. f =0and g is any continuous function. 2. 1 f(x)=γeλx and g(x)= eλ(x+z0),x R, 2 ∈ for some constants γ C,λ C. ∈ ∈ The integral sine addition law 217

3. γ 1 f(x)= (eλ1x + eλ2x) and g(x)= eλ1(x+z0) + eλ2(x+z0) ,x R, 2 2 ∈ ³ ´ (λ λ )z for some constants γ C and λ1,λ2 C such that e 2 1 0 = 1. ∈ ∈ − − 4. eλ1x + eλ2x eλ1x eλ2x f(x)=γ +(δ q) − and Ã 2 − 2 !

eλ1x + eλ2x eλ1x eλ2x g(x)=β +(δ + q) − , Ã 2 2 !

2 for some constants γ,β,q C , λ1,λ2 C and δ := 1+q such λ1z0 2+2δ ∈λ2z0 2 2δ∈ ± that e = β 1 q δ and e = β 1+−q δ . p − − − 5. x f(x)=γ(1 + )eλx and g(x)=eλ(x+z0),x R, z0 ∈ for some constants γ C,λ C. ∈ ∈ Acknowledgement. The authors would like to express their most sincere gratitude to the referee for a number of constructive comments which have led to essential improvement of the paper.

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D. Zeglami, Department of Mathematics, E. N. S. A. M., Moulay Ismail University , B. P. : 15290 Al Mansour, Meknes, Morocco e-mail : [email protected]

M. Tial, Department of Mathematics, Faculty of Sciences, Ibn Tofail University, BP: 14000. Kenitra, Morocco e-mail : [email protected] and

S. Kabbaj Department of Mathematics, Faculty of Sciences, Ibn Tofail University, BP: 14000. Kenitra, Morocco e-mail : [email protected]