A generalization of d'Alembert's functional equation on semigroups

TL;DR

This paper characterizes solutions to a generalized d'Alembert functional equation on semigroups with involutive automorphisms and multiplicative functions, extending classical results to a broader algebraic setting.

Contribution

It provides a comprehensive solution framework for a generalized functional equation on semigroups with specific automorphism and multiplicative conditions, broadening the understanding of such equations.

Findings

Explicit solutions are derived for the functional equation.

The results extend classical d'Alembert equations to semigroups with involutive automorphisms.

Conditions on the multiplicative function  are identified for solution existence.

Abstract

Given a semigroup $S$ generated by its squares equipped with an involutive automorphism $\sigma$ and a multiplicative function $\mu:S\to\mathbb{C}$ such that $\mu(x\sigma(x))=1$ for all $x\in S$, we determine the complex-valued solutions of the following functional equation.

Equations

$$f(xy)-\mu (y)f(\sigma (y)x)=g(x)h(y),x,y\in S,$$

$$\cos (x+y)+\cos (x-y)=2\cos x\cos y,x,y\in \mathbb{R},$$

$$g(x+y)+g(x-y)=2g(x)g(y),x,y\in \mathbb{R},$$

$$g(x+y)+g(x-y)=2g(x)g(y),x,y\in G,$$

$$g(xy)+g(x\tau (y))=2g(x)g(y),x,y\in G,$$

$$f(xy)+f(\sigma (y)x)=2f(x)f(y),x,y\in S,$$

$$f(xy)-f(\sigma (y)x)=g(x)h(y)x,y\in M,$$

$$f(xy)-\mu (y)f(\sigma (y)x)=g(x)h(y),x,y\in M,$$

$${\mathcal{N}}_{\mu }(\sigma ,S):=\{\theta :S\to \mathbb{C}\mid \theta (xy)-\mu (y)\theta (\sigma (y)x)=0,x,y\in S\}.$$

$$F^{\ast }(x)=\mu (x)F(\sigma (x)),x\in S.$$

$$\mu (y)k(x\sigma (y))=k(x)l(y)-k(y)l(x),x,y\in S.$$

$$\left\{\begin{array}[]{r c l}k&=&\chi\,A,\quad$$

l=\chi(1+cA)$$\text{on}$$S\setminus I_{\chi},

$$\\ k&=&0,\quad$$

l=0$$\text{on}$$I_{\chi},

$$\\ \end{array}\right.$$

$$l^o=ck$$

$$k(xy)=k(x)l^e(y)+k(y)l^e(x),x,y\in S,$$

$$\left\{\begin{array}[]{r c l}k&=&\chi\,A,\quad$$

l^{e}=\chi$$\text{on}$$S\setminus I_{\chi},

$$\\ k&=&0,\quad$$

l^{e}=0$$\text{on}$$I_{\chi},

$$\\ \end{array}\right.$$

$$\left\{\begin{array}[]{r c l}l&=&\chi+c\chi\,A=\chi(1+cA)\quad\text{on}\quad$$

S\setminus I_{\chi},

$$\\ l&=&0,\quad\text{on}\quad$$

I_{\chi}.

$$\\ \end{array}\right.$$

$$\theta (xy)=\nu (y)\theta (\tau (y)x)=\nu (y)\nu (x)\theta (\tau (x)\tau (y))=\nu (xy)\theta (\tau (xy))={\theta }^{\ast }(xy),$$

$$h(xy)=g^{\ast }(x)l(y)-g(y)l^{\ast }(x)$$

$$f^e(xy)-\mu (y)f^e(\sigma (y)x)=g^o(x)h(y)$$

$$f^o(xy)-\mu (y)f^o(\sigma (y)x)=g^e(x)h(y)$$

$$2f^o(xy)=g^e(x)h(y)+g^e(y)h(x)$$

$$f(xyz)-\mu (yz)f(\sigma (yz)x)=g(x)h(yz),$$

$$f(\sigma (yz)x)-\mu (\sigma (z)x)f(z\sigma (xy))=g(\sigma (y))h(\sigma (z)x),$$

$$f(z\sigma (xy))-\mu (\sigma (xy))f(xyz)=g(z)h(\sigma (xy)).$$

$$\mu (yz)f(\sigma (yz)x)-\mu (xy)f(z\sigma (xy))=\mu (yz)g(\sigma (y))h(\sigma (z)x),$$

$$\mu (xy)f(z\sigma (xy))-f(xyz)=\mu (xy)g(z)h(\sigma (xy)).$$

$$g(x)h(yz)+\mu (yz)g(\sigma (y))h(\sigma (z)x)+\mu (xy)g(z)h(\sigma (xy))=0.$$

$$\begin{array}{rl}h(yz)& =-g(x_0{)}^{-1}\mu (yz)g(\sigma (y))h(\sigma (z)x_0)-g(x_0{)}^{-1}\mu (x_{0}y)g(z)h(\sigma (x_{0}y))\\ & =g^{\ast }(y)(-g(x_0{)}^{-1}\mu (z)h(\sigma (z)x_0))+g(z)(-g(x_0{)}^{-1}\mu (x_{0}y)h(\sigma (x_{0}y))).\end{array}$$

$$h(yz)=g^{\ast }(y)l(z)+g(z)l_1(y)$$

$$\begin{array}{rl}0& =g(x)[g^{\ast }(y)l(z)+g(z)l_1(y)]+\mu (yz)g(\sigma (y))[\mu (\sigma (z))g(z)l(x)\\ & +g(x)l_1(\sigma (z))]+\mu (xy)g(z)[\mu (\sigma (x))g(x)l(\sigma (y))+g(\sigma (y))l_1(\sigma (x))]\\ & =g(x)g(\sigma (y))[\mu (y)l(z)+\mu (yz)l_1(\sigma (z))]+g(x)g(z)[l_1(y)+l^{\ast }(y)]\\ & +g(\sigma (y))g(z)[\mu (y)l(x)+\mu (xy)l_1(\sigma (x))],\end{array}$$

Full text

[TABLE]

2010 Mathematics Subject Classification. Primary 39B52; Secondary 39B32

1. Introduction

The identity

[TABLE]

also called the cosine functional equation, is a starting point of d’Alembert’s functional equation

[TABLE]

where $g:\mathbb{R}\to\mathbb{C}$ is the unknown function. The functional equation (1.1) go back to d’Alembert’s investigations [4], [5] and [6]. It is well known that the continuous solutions $g\neq 0$ of (1.1) are the functions $g(x)=\dfrac{e^{\alpha x}+e^{-\alpha x}}{2},\,x\in\mathbb{R}$, where $\alpha\in\mathbb{C}$ is a constant.

The obvious extension of (1.1) from $\mathbb{R}$ to an abelian group $(G,+)$ is the functional equation

[TABLE]

where $g:\mathbb{G}\to\mathbb{C}$ is the unknown function. The nonzero solutions of (1.2) are the functions of the form $g(x)=\dfrac{\chi(x)+\chi(-x)}{2},\,x\in G$, where $\chi$ is a character of $G$. The functional equation

[TABLE]

where $G$ is a group that need not be abelian, $\tau:G\to G$ is an involution (i.e. $\tau(xy)=\tau(y)\tau(x)$ and $\tau(\tau(x))=x$ for all $x,y\in G$) is a generalization of (1.2).

The first result for non-abelian groups was obtained by Kannappan [9]. With $\tau(x)=x^{-1}$ for all $x\in G$, the abelian solutions of equation (1.3) (i.e. those satisfy the Kannappan condition: $g(xyz)=g(yxz)$ for all $x,y,z\in G$) are of the form $g(x)=\dfrac{m(x)+m(-x)}{2},\,x\in G$ where $m:G\to\mathbb{C}$ is a multiplicative function. The solutions of (1.3) were obtained by Davison [3] for general groups, even monoids.

The complex valued-solutions of the following variant of d’Alembert’s functional equation

[TABLE]

where $S$ is a semigroup and $\sigma:S\to S$ is an involutive automorphism, was determined by Stetkær [13].

At the same time Ebanks and Stetkær [8] determined the complex-valued solutions of the functional equation

[TABLE]

where $M$ is a monoid generated by its squares and $\sigma:M\to M$ is an involutive automorphism.

Bouikhalene and Elqorachi [2] solved the functional equation

[TABLE]

where $M$ is a monoid generated by its squares or a group, $\sigma:M\to M$ is an involutive automorphism and $\mu:M\to\mathbb{C}$ is a multiplicative function such that $\mu(x\sigma(x))=1$ for all $x\in M$.

For more details about (1.3) we refer to [10], [3, Proposition 2.11], [14, Lemme IV.4] and [15, Proposition 4.2].

The identity element of the monoid or the group is used in the proofs of [2, Theorem 2.1, Proposition 2.2, Theorem 3.2] and [8, Proposition 4.1, Theorem 4.3].

Similar functional equations were treated in Chapter 13 of the book [1] by Aczél and Dhombres.

The present paper shows that the identity element is not crucial by given proofs in the setting of general semigroups. Emphasize that $\sigma$ in the present paper is a homomorphism, not an anti-homomorphism like the group inversion found in other papers.

Although we use similar computations to the ones in [2] and [8] the general setting of (1.4) is for $S$ to be a semigroup generated by its squares, because to formulate the functional equation equation (1.4) and obtain some key properties of solutions like centrality and parity we need only an associative composition in $S$ and the assumption that the semigroup $S$ is generated by its squares, not an identity element and inverses. The main new feature of the present paper is that we do not assume the underlying semigroup has an identity. That makes the exposition more involved and explains why our proofs are longer that those of previous papers about the same functional equations. Thus we study in the present paper (1.4) extending works in which $S$ is a group or a monoid.

The organization of the paper is as follows. In the next section we give notations and terminology. In the third section we give preliminary results that we need in the paper. In section 4 we prove our main results.

The explicit formulas for the solutions are expressed in terms of multiplicative and additive functions.

2. Notations and Terminology

Definition 2.1**.**

Let $f:S\to\mathbb{C}$ be a function on a semigroup $S$. We say that

$f$ is additive if $f(xy)=f(x)+f(y)$ for all $x,y\in S$.

$f$ is multiplicative if $f(xy)=f(x)f(y)$ for all $x,y\in S$.

$f$ is central if $f(xy)=f(yx)$ for all $x,y\in S$.

If $S$ is a semigroup, $\sigma:S\to S$ an involutive automorphism and $\mu:S\to\mathbb{C}$ a multiplicative function such that $\mu(x\sigma(x))=1$ for all $x\in S$, we define the nullspace

[TABLE]

If $\chi:S\to\mathbb{C}$ is a multiplicative function and $\chi\neq 0$, then $I_{\chi}:=\{x\in S\mid\chi(x)=0\}$ is either empty or a proper subset of $S$. $I_{\chi}$ is a two sided ideal in $S$ if not empty and $S\setminus I_{\chi}$ is a subsemigroup of $S$. Notice that $I_{\chi}$ is also a subsemigroup of $S$.

For any function $F:S\to\mathbb{C}$ we define the function

[TABLE]

Let $f:S\to\mathbb{C}$. We call $f^{e}:=\dfrac{f+f^{*}}{2}$ the even part of $f$ and $f^{o}:=\dfrac{f-f^{*}}{2}$ its odd part. The function $f$ is said to be even if $f=f^{*}$, and $f$ is said to be odd if $f=-f^{*}$.

If $g,h:S\to\mathbb{C}$ are two functions we define the function $(g\otimes h)(x,y):=g(x)h(y),\,x,y\in S$.

Blanket assumption: Throughout this paper $S$ denotes a semigroup (a set with an associative composition) generated by its squares. The map $\sigma:S\to S$ denotes an involutive automorphism. That $\sigma$ is involutive means that $\sigma(\sigma(x))=x$ for all $x\in S$. We denote by $\mu:S\to\mathbb{C}$ a multiplicative function such that $\mu(x\sigma(x))=1$ for all $x\in S$.

3. $\mu$-sine subtraction law on a semigroup generated by its squares

In this section we extend the results obtained in [2, Theorem 2.1, Proposition 2.2], [12, Theorem 4.12] and [8, Theorem 3.2 and Lemma 3.4] on groups and monoids to semigroups generated by their squares by solving the $\mu$-sine subtraction law

[TABLE]

Lemma 3.1**.**

Let $G$ be a semigroup such that $G=\{xy\,|\,x,y\in G\}$. Let $f$ and $F$ be functions on $G$ such that $f(xy)=F(yx)$ for all $x,y\in G$. Then $f=F$. In particular $f$ is central.

Proof.

For any $x,y,z\in G$ we have $f(xyz)=f((xy)z)))=F(zxy)=F((zx)y)=f(y(zx))=f((yz)x)=F(x(yz))=F(xyz)$. Applying the assumption on $G$ twice we see that any element of $G$ can be written as $xyz$, where $x,y,z\in G$. So $f=F$ and then $f$ is central. This finishes the proof. ∎

Remark 3.2**.**

The proof of Lemma 3.1 works also for if $G$ is regular semigroup, which by definition means that for each $a\in G$ there exists an element $x\in G$ such that $a=axa$.

Proposition 3.3**.**

*The solutions of the functional equation (3.1) with $k\neq 0$ are the followings pairs

(1) $k=c_{1}\frac{\chi-\chi^{*}}{2}$, $l=\frac{\chi+\chi^{*}}{2}+c_{2}\frac{\chi-\chi^{*}}{2}$ where $\chi:S\to\mathbb{C}$ is a multiplicative function and $c_{1}\in\mathbb{C}\setminus\{0\}$, $c_{2}\in\mathbb{C}$ are constants such that $\chi^{*}\neq\chi$,

(2)*

[TABLE]

where $c\in\mathbb{C}$ is a constant, $\chi:S\to\mathbb{C}$ is a nonzero multiplicative function and $A:S\setminus I_{\chi}\to\mathbb{C}$ is a nonzero additive function such that $\chi^{*}=\chi$ and $A\circ\sigma=-A$.

Proof.

Let $x,y\in S$ be arbitrary. By interchanging $x$ and $y$ in (3.1) we get the identity $\mu(y)k(x\sigma(y))=-\mu(x)k(y\sigma(x))$, which, applied to the pair $(x,\sigma(y))$, read $\mu(\sigma(y))k(xy)=-\mu(x)k(yx)$. Multiplying this by $\mu(y)$ and using that $\mu:S\to\mathbb{C}$ is a multiplicative function and that $\mu(y\sigma(y))=1$, we get that $k(xy)=-k^{*}(yx)$. So, $x$ and $y$ being arbitrary, we deduce, according to Lemma 3.1, that $k=-k^{*}$ and $k$ is central.

On the other hand, by using the same computation used by Bouikhalene and Elqorachi in the proof of [2, Theorem 2.1] we get that there exists a constant $c\in\mathbb{C}$ such that

[TABLE]

and that the pair $(k,l^{e})$ satisfies the sine addition law

[TABLE]

hence, according to [2, Proposition 1.1], that pair falls into two categories:

(i) $k=c_{1}\frac{\chi_{1}-\chi_{2}}{2}$ and $l^{e}=\frac{\chi_{1}+\chi_{2}}{2}$, where $\chi_{1},\chi_{2}:S\to\mathbb{C}$ are different multiplicative functions and $c_{1}\in\mathbb{C}\setminus\{0\}$ is a constant. Since $k^{*}=-k$, $\mu l^{e}=l^{e}$, and $c_{1}\neq 0$ a small computation shows that $\chi_{2}=\chi_{1}^{*}$. Defining $\chi:=\chi_{1}$ and $c_{2}:=c_{1}c\in\mathbb{C}$, and using that $l=l^{e}+l^{o}$ we get that $l=\frac{\chi+\chi^{*}}{2}+c_{2}\frac{\chi-\chi^{*}}{2}$ , where $\chi:S\to\mathbb{C}$ is a multiplicative function such that $\chi\neq\chi^{*}$ and $c_{1}\in\mathbb{C}\setminus\{0\}$, $c_{2}\in\mathbb{C}$ are constants. The result occurs in (1) of Proposition 3.3.

(ii)

[TABLE]

where $\chi:S\to\mathbb{C}$ is a nonzero multiplicative function and $A:S\setminus I_{\chi}\to\mathbb{C}$ is a nonzero additive function. So, taking (3.2) into account, we get that $l^{o}=c\chi\,A$ on $S\setminus I_{\chi}$ and $l^{o}=0$ on $I_{\chi}$. Hence

[TABLE]

Notice that $l^{e}(x)=\chi(x)$ for all $x\in S$. As $\mu l^{e}\circ\sigma=l^{e}$ we have $\chi^{*}=\chi$ and then $\sigma(S\setminus I_{\chi})=S\setminus I_{\chi}$. So, using that $k^{*}=-k$ and the fact that $\chi(x)\neq 0$ for all $x\in S\setminus I_{\chi}$, we get that $A\circ\sigma=-A$. The result occurs in part (2) of Proposition 3.3.

Conversely, if $k$ and $l$ are of the forms (1)-(2) in Proposition (3.3) we check by elementary computations that the pair $(k,l)$ is a solution of Eq. (3.1) with $k\neq 0$. This completes the proof of Proposition 3.3. ∎

Remark 3.4**.**

[2, Proposition 1.1]** corresponds to [8, Lemma 3.4]. So, like in [8, Remark 3.5], the Proposition 3.3 is also valid if $S=\{xy\,|\,x,y\in S\}$, or if $S$ is a regular semigroup.

4. Solutions of Eq. (1.4) on a semigroup generated by its squares

The following Lemma will be used later.

Lemma 4.1**.**

*Let $G$ be a semigroup, $\tau:G\to G$ be an involutive automorphism and $\nu:G\to\mathbb{C}$ be a multiplicative function such that $\nu(x\tau(x))=1$ for all $x\in G$, and $\theta\in\mathcal{N}_{\nu}(\tau,G)$.

(1) $\theta=\theta^{*}$ on $\{xy\,|\,x,y\in G\}$, i.e., $\theta$ is even on products.

(2) If $\theta$ is odd, i.e., $\theta=-\theta^{*}$, then $\theta=0$ on $\{xy\,|\,x,y\in G\}$. In particular $\theta=0$ if $G$ is a regular semigroup, or if $G$ is generated by its squares.*

Proof.

(1) For all $x,y\in G$ we have

[TABLE]

which proves (1).

(2) If $\theta$ is odd then $\theta=-\theta^{*}$. So, according to Lemma 4.1(1), $\theta=0$ on $\{xy\,|\,x,y\in G\}$. If $G$ is a regular semigroup or if $G$ is generated by its squares, then any element of $G$ can be written as $xy$ where $x,y\in G$. Hence $\theta=0$. This finishes the proof. ∎

In Lemma 4.2 below we give some key properties of solutions of the functional equation (1.4).

Lemma 4.2**.**

*Let $f,g,h:S\to\mathbb{C}$ be a solution of the functional equation (1.4). Suppose that $g\neq 0$ and $h\neq 0$. Then

(1) There exists a function $l:S\to\mathbb{C}$ such that*

[TABLE]

*for all $x,y\in S$.

(2) $h^{*}=-h$ and $h$ is central.

(3)*

[TABLE]

*for all $x,y\in S$.

(4)*

[TABLE]

*for all $x,y\in S$.

(5)*

[TABLE]

*for all $x,y\in S$. In particular $f^{o}$ is central.

(6) There exists a constant $b\in\mathbb{C}$ such that $g^{o}=bh$.*

Proof.

(1) We use similar computations to those of the proof of Proposition 3 in [8].

Let $x,y,z\in S$ be arbitrary. By applying Eq (1.4) to the pairs $(x,yz)$, $(\sigma(y),\sigma(z)x)$ and $(z,\sigma(xy))$ we obtain

[TABLE]

[TABLE]

[TABLE]

By multiplying (4.6) by $\mu(yz)$, and (4.7) by $\mu(xy)$ we get that

[TABLE]

[TABLE]

By adding (4.5), (4.8) and (4.9) we obtain

[TABLE]

As $g\neq 0$ we chose $x_{0}\in S$ such that $g(x_{0})\neq 0$. So, by putting $x=x_{0}$ we deduce from (4.10) that

[TABLE]

So, $y$ and $z$ being arbitrary, we get from the last identity, by defining the functions $l(z):=-g(x_{0})^{-1}\mu(z)h(\sigma(z)x_{0})$ and $l_{1}(z):=-g(x_{0})^{-1}\mu(x_{0}z)h(\sigma(x_{0}z))$ for all $z\in S$, that

[TABLE]

for all $y,z\in S$.

Using (4.11) and (4.10) we obtain

[TABLE]

for all $x,y,z\in S$. By putting $x=x_{0}$ and $y=\sigma(x_{0})$ in the identity above we get that

[TABLE]

As $g(x_{0})\mu(x_{0})\neq 0$ we deduce from the identity above that there exists a constant $c\in\mathbb{C}$ such that

[TABLE]

for all $z\in S$. From (4.12) and (4.13) we get, by a small computation, that

[TABLE]

for all $x,y,z\in S$, which implies that $c=0$ because $g\neq 0$. Then, using that $\mu(z\sigma(z))=1$ for all $z\in S$, we infer from (4.13), that $l_{1}=-l^{*}$. Hence, (4.11) becomes

[TABLE]

for all $y,z\in S$, which occurs in (1) Lemma 4.2.

(2) Let $x,y\in S$ be arbitrary. By applying (4.1) to the pair $(\sigma(y),\sigma(x))$ and multiplying the identity obtained by $\mu(xy)$ we get that

[TABLE]

Proceeding exactly as in the proof of the Proposition 3.3 we deduce that $h^{*}=-h$ and $h$ is central. This is the result (2) of Lemma 4.2.

(3) Let $x,y\in S$ be arbitrary. We have

[TABLE]

which implies, since $h^{*}=-h$, that $f^{e}(xy)-\mu(y)f^{e}(\sigma(y)x)=\dfrac{[g(x)-g^{*}(x)]h(y)}{2}$. So that $f^{e}(xy)-\mu(y)f^{e}(\sigma(y)x)=g^{o}(x)h(y)$ for all $x,y\in S$. This is part (3).

(4) By subtracting (4.2) from (1.4) a small computation shows that

[TABLE]

for all $x,y\in S$. This is the result (4).

(5) By applying (4.3) to the pair $(\sigma(y),x)$ and multiplying the identity obtained by $\mu(y)$ we get that

[TABLE]

Since $\mu f^{o}\circ\sigma=-f^{o}$ and $\mu g^{e}\circ\sigma=g^{e}$ the identity above implies that

[TABLE]

When we add this to (4.3) we obtain the functional equation (4.4).

(6) By applying (4.2) to the pair $(x,\sigma(y))$ and multiplying the identity obtained by $\mu(y)$, and using the fact that $\mu(y\sigma(y))=1$ and $h^{*}=-h$, we get that

[TABLE]

The identity (4.2) applied to the pair $(y,x)$ gives

[TABLE]

By adding the two last identities, and seeing that $\mu f^{e}\circ\sigma=f^{e}$ and $\mu(y\sigma(y))=1$ for all $y\in S$, we obtain

[TABLE]

for all $x,y\in S$. So $g^{o}$ and $h$ are linearly dependent. As $h\neq 0$ by assumption, we get that there exists a constant $b\in\mathbb{C}$ such that $g^{o}=bh$, which occurs in part (6). This completes the proof of Lemma 4.2. ∎

In theorem 4.3 we solve the functional equation (1.4) on semigroups generated by their squares.

Theorem 4.3**.**

*The solutions $f,g,h:S\to\mathbb{C}$ of the functional equation (1.4) can be listed as follows:

(1) $f=\theta$, $g=0$ and $h$ is arbitrary, where $\theta\in\mathcal{N}_{\mu}(\sigma,S)$.

(2)$f=\theta$, $g$ is arbitrary and $h=0$, where $\theta\in\mathcal{N}_{\mu}(\sigma,S)$.

(3) $f=\theta+\alpha\frac{\chi+\chi^{*}}{2}+\beta\frac{\chi-\chi^{*}}{2}$ and $g\otimes h=2[\beta\frac{\chi+\chi^{*}}{2}+\alpha\frac{\chi-\chi^{*}}{2}]\otimes\frac{\chi-\chi^{*}}{2}$ where $\alpha,\beta\in\mathbb{C}$ are constants, $\theta\in\mathcal{N}_{\mu}(\sigma,S)$ and $\chi:S\to\mathbb{C}$ is a multiplicative function such that $(\alpha,\beta)\neq(0,0)$ and $\chi^{*}\neq\chi$.

(4)*

[TABLE]

*where $\alpha,\beta\in\mathbb{C}$ are constants, $\theta\in\mathcal{N}_{\mu}(\sigma,S)$, $\chi:S\to\mathbb{C}$ is a nonzero multiplicative function and $A:S\setminus I_{\chi}\to\mathbb{C}$ is a nonzero additive function such that $(\alpha,\beta)\neq(0,0)$, $\chi^{*}=\chi$ and $A\circ\sigma=-A$.

Proof.

We check by elementary computations that if $f,g$ and $h$ are of the forms (1)-(4) then $(f,g,h)$ is a solution of (1.4) , so left is that any solution $(f,g,h)$ of (1.4) fits into (1)-(4).

Let $f,g,h:S\to\mathbb{C}$ satisfy the functional equation (1.4). The parts (1)and (2) are trivial. So, in what remains of the proof we assume that $g\neq 0$ and $h\neq 0$. According to Lemma 4.2(6) there exists a constant $b\in\mathbb{C}$ such that

[TABLE]

We split the discussion into the cases $b\neq 0$ and $b=0$.

Case A: $b\neq 0$. Then $g^{0}\neq 0$. From (4.2) we get that the triple $(f^{e},g^{o},h)$ satisfies the functional equation (1.4). Hence, according to Lemma 4.2(1), there exists a function $L:S\to\mathbb{C}$ such that

[TABLE]

for all $x,y\in S$. Since $\mu g^{o}\circ\sigma=-g^{o}$ we get, taking (4.14) into account, that

[TABLE]

Applying this to the pair $(x,\sigma(y))$ and multiplying the identity obtain by $\mu(y)$, and taking into account that $\mu h\circ\sigma=-h$, we obtain

[TABLE]

for all $x,y\in S$. Defining $m:=-bL^{*}$ we deduce from the identity above that the pair $(h,m)$ satisfies the $\mu$-subtraction law

[TABLE]

for all $x,y\in S$.

On the other hand, the identity (4.3) implies that the triple $(f^{o},g^{e},h)$ satisfies the functional equation (1.4). Let $x,y,z\in S$ be arbitrary. As in the proof of Lemma 4.2 we deduce that the pair $(g^{e},h)$ satisfies the identity (4.10), i.e.,

[TABLE]

for all $x,y,z\in S$. Sine $\mu g^{e}\circ\sigma=g^{e}$ and $h^{*}=-h$ and $h$ is central, we get from the identity above that

[TABLE]

which implies, by replacing $y$ by $\sigma(y)$, multiplying both sides by $\mu(y)$ and using that $\mu g^{e}\circ\sigma=g^{e}$, that

[TABLE]

Using that the pair $(h,m)$ satisfies the $\mu$-subtraction law we get from the last identity that $g^{e}(x)[h(z)m(y)-h(y)m(z)]=g^{e}(z)[h(x)m(y)-h(y)m(x)]-g^{e}(y)[h(x)m(z)-h(z)m(x)]$. So that

[TABLE]

Now, let $y,z\in S$ be arbitrary in (4.16). Since the function $g^{e}$ is even and $h$ is odd, we get from (4.16) that

[TABLE]

for all $x\in S$. Since $h\neq 0$ we get, seeing the formulas of the solutions of the $\mu$-subtraction law given by the Proposition 3.3, that the functions $h$ and $m$ are linearly independent. So that there exist $y_{0},z_{0}\in S$ such that $h(z_{0})m(y_{0})-h(y_{0})m(z_{0})\neq 0$.

By putting $(y,z)=(y_{0},z_{0})$ in (4.17) we deduce that there exists a constant $c_{3}\in\mathbb{C}$ such that

[TABLE]

Moreover, taking into account (4.15), we get from Proposition 3.3 that the pair $(h,m)$ falls into two categories (i) and (ii) that we deal with separately:

(i)

[TABLE]

and

[TABLE]

where $c_{1}\in\mathbb{C}\setminus\{0\}$, $\alpha\in\mathbb{C}$ are constants and $\chi:S\to\mathbb{C}$ is a multiplicative function such that $\chi^{*}\neq\chi$. Then $m^{e}=\frac{\chi+\chi^{*}}{2}$. So, in view of (4.14) and (4.18), we get that

[TABLE]

and

[TABLE]

where $c_{2}:=bc_{1}\in\mathbb{C}\setminus\{0\}$ is a constant. Hence

[TABLE]

By substituting (4.19) and (4.21) in (4.2) we get that

[TABLE]

for all $x,y\in S$. So that

[TABLE]

for all $x,y\in S$. Hence, the function $\theta:=f^{e}-\frac{c_{1}c_{2}}{4}(\chi+\chi^{*})$ belongs to $\mathcal{N}_{\mu}(\sigma,S)$ and we obtain

[TABLE]

On the other hand, by substituting (4.19) and (4.22) in (4.3) we get that

[TABLE]

for all $x,y\in S$. So that

[TABLE]

for all $x,y\in S$. Hence the function $f^{o}-\frac{c_{1}c_{3}}{4}(\chi-\chi^{*})$ belongs to $\mathcal{N}_{\mu}(\sigma,S)$. As $f^{o}-\frac{c_{1}c_{3}}{4}(\chi-\chi^{*})$ is an odd function we derive, according to Lemma 4.1(2), that $f^{o}-\frac{c_{1}c_{3}}{4}(\chi-\chi^{*})=0$ and then

[TABLE]

Combining this with (4.23) and using the fact that $f=f^{e}+f^{o}$ we obtain

[TABLE]

which gives with (4.19) and (4.23), by putting $\alpha:=\frac{c_{1}c_{2}}{2}$ and $\beta:=\frac{c_{1}c_{3}}{2}$, the result in part (3) of Theorem 4.3.

(ii)

[TABLE]

and

[TABLE]

where $c\in\mathbb{C}$ is a constant, $\chi:S\to\mathbb{C}$ is a nonzero multiplicative function and $A:S\setminus I_{\chi}\to\mathbb{C}$ is a nonzero additive function such that $\chi^{*}=\chi$ and $A\circ\sigma=-A$. As in (ii) of the proof of Proposition 3.3 we have $\sigma(S\setminus I_{\chi})=S\setminus I_{\chi}$ and $\sigma(I_{\chi})=I_{\chi}$. Hence, by a small computation using that $\chi^{*}=\chi$ and $A\circ\sigma=-A$, we get from (4.26) that $m^{e}=\chi$. So, in view of (4.14) and (4.18), we get that

[TABLE]

and

[TABLE]

where $c_{2}:=b\in\mathbb{C}\setminus\{0\}$ is a constant. Hence

[TABLE]

Recall that $\sigma(S\setminus I_{\chi})=S\setminus I_{\chi}$ and $\sigma(I_{\chi})=I_{\chi}$. Then

on the subsemigroup $S\setminus I_{\chi}$, we get, by using (4.27), (4.25) and (4.2), that

[TABLE]

for all $x,y\in S\setminus I_{\chi}$, which implies that

[TABLE]

for all $x,y\in S\setminus I_{\chi}$.

Defining $\theta_{1}:=f^{e}-\frac{c_{2}}{4}\chi\,A^{2}$ we see that $\theta_{1}\in\mathcal{N}_{\mu}(\sigma,S\setminus I_{\chi})$ and

[TABLE]

Similarly, by using (4.28), (4.25) and (4.3), we get that

[TABLE]

for all $x,y\in S\setminus I_{\chi}$, which implies that

[TABLE]

for all $x,y\in S\setminus I_{\chi}$. Hence $f^{o}-\frac{c_{3}}{2}\chi\,A\in\mathcal{N}_{\mu}(\sigma,S\setminus I_{\chi})$. As $\chi^{*}=\chi$ and $A\circ\sigma=-A$ we get that the function $f^{o}-\frac{c_{3}}{2}\chi\,A$ is odd on the subsemigroup $S\setminus I_{\chi}$, which is also generated by its squares. Thus, according to Lemma 4.1(2), $f^{o}-\frac{c_{3}}{2}\chi\,A=0$ on $S\setminus I_{\chi}$ and then

[TABLE]

Hence, from (4.30) and (4.31) we deduce that

[TABLE]

On the subsemigroup $I_{\chi}$ we have $h=0$ by (4.25). Then we get for $x\in S$ and $y\in I_{\chi}$ that

[TABLE]

which implies that

[TABLE]

Defining $\theta:=\theta_{1}\cup\theta_{2}$ we see that $\theta\in\mathcal{N}_{\mu}(\sigma,S)$. Indeed, for $x,y\in S$,

If $xy\in S\setminus I_{\chi}$, then $x,y\in S\setminus I_{\chi}$ and $\sigma(y)x\in S\setminus I_{\chi}$ because $\chi^{*}=\chi$. So that

[TABLE]

If $xy\in I_{\chi}$, then $x\in I_{\chi}$ or $y\in I_{\chi}$, and $\sigma(y)x\in I_{\chi}$ because $\chi^{*}=\chi$. So, taking (4.25), (4.29) and (4.33), we get that

[TABLE]

Now, by using (4.32) and (4.33), we obtain

[TABLE]

Combining (4.25), (4.29) and (4.34), and putting $\alpha:=c_{3}$ and $\beta:=c_{2}$, we obtain part (4) of Theorem 4.3.

Case B: $b=0$. In this case we have, taking (4.14) into account, $g^{o}=0$. So, the functional equation (4.2) becomes

[TABLE]

for all $x,y\in S$. Hence, by defining $\theta:=f^{e}$ we have

[TABLE]

with $\theta\in\mathcal{N}_{\mu}(\sigma,S)$.

Moreover $g=g^{e}+g^{o}=g^{e}$. Then $g$ is an even function, i.e., $\mu g\circ\sigma=g$. Hence the identity (4.1) reduces to

[TABLE]

for all $x,y\in S$. According to Lemma 4.2(2) $h$ is central, so (4.36) implies that

[TABLE]

for all $x,y\in S$. So,

[TABLE]

for all $x,y\in S$, which implies that the functions $g$ and $l^{e}$ are linearly dependent. Since $g\neq 0$ we deduce that there exists a constant $c\in\mathbb{C}$ such that

[TABLE]

Now, let $x,y\in S$ be arbitrary. By applying (4.36) to the pair $(\sigma(y),x)$, multiplying the identity obtained by $\mu(y)$ and taking into account that $\mu g\circ\sigma=g$ and $\mu(y\sigma(y))=1$, we get that

[TABLE]

By subtracting (4.38) from (4.36) we obtain

[TABLE]

which implies, using (4.37), that

[TABLE]

So, $x$ and $y$ being arbitrary, the triple $(h,2g,l^{o})$ satisfies the functional equation (1.4), i.e.,

[TABLE]

for all $x,y\in S$.

By using the fact that $\mu(x\sigma(x))=1$ for all $x\in S$, $\mu f^{o}\circ\sigma=-f^{o}$ and $g=g^{e}$ the functional equation (4.3) becomes

[TABLE]

for all $x,y\in S$. Since $h\neq 0$ there exists $y_{0}\in S$ such that $h(y_{0})\neq 0$. Let $x,y\in S$ be arbitrary. When we apply (4.40) to the pairs $(xy,y_{0})$ and $(x\sigma(y),y_{0})$ we deduce that

[TABLE]

and

[TABLE]

which implies, taking into account that $\mu$ is multiplicative and $\mu(y\sigma(y))=1$, that

[TABLE]

The identity (4.4) shows that $f^{o}$ is central. Hence, by applying (4.40), we get that

[TABLE]

Similarly,

[TABLE]

So, by using (4.40), we get that

[TABLE]

From (4.41), (4.42) and (4.43) we obtain

[TABLE]

Moreover, by applying (4.36) to the pairs $(y_{0},x)$ and $(y_{0},\sigma(x))$ we get that

[TABLE]

and

[TABLE]

By adding the first identity above to the second one multiplied by $\mu(x)$, and seeing that $g^{*}(x)=g(x)$, we obtain

[TABLE]

which implies, taking (4.37) into account, that

[TABLE]

Thus

[TABLE]

Combining this with (4.44) we get that

[TABLE]

for all $x,y\in S$, where $\lambda:=\frac{2l^{o}(y_{0})}{h(y_{0})}$.

On the other hand, since $g\neq 0$ there exists $x_{0}\in S$ such that $g(x_{0})\neq 0$. Let $x,y\in S$ be arbitrary. By applying (4.40) to the pair $(x_{0},xy)$ and using that $f^{o}$ is central we get that

[TABLE]

By applying the identity (4.4) to the pairs $(x,yx_{0})$ and $(x,y\sigma(x_{0}))$, and using that $g^{e}=g$ and Lemma 4.2(2), the identity above gives

[TABLE]

which implies, by using (4.39) and (4.45), that

[TABLE]

So, $x$ and $y$ being arbitrary, the pair $(h,\frac{\lambda}{2}g)$ satisfies the sine addition law

[TABLE]

for all $x,y\in S$. Notice that $\lambda\neq 0$ because $h\neq 0$ and $S$ is generated by its squares. According to [2, Proposition 1.1], and proceeding as (i) and (ii) of the proof of Proposition 3.3, we get that the pair $(h,g)$ falls into two categories:

(i)

[TABLE]

and

[TABLE]

where $c_{1},c_{2}\in\mathbb{C}\setminus\{0\}$ are constants and $\chi:S\to\mathbb{C}$ is a multiplicative function such that $\chi^{*}\neq\chi$.

By substituting (4.47) and (4.48) in the functional equation (4.3), and taking into account that $g^{e}=g$, and proceeding exactly as in part (i) of Case A to compute $f^{o}$, we obtain

[TABLE]

which gives with (4.35)

[TABLE]

By putting $\beta:=\frac{c_{1}c_{2}}{2}\in\mathbb{C}\setminus\{0\}$, we get from (4.47), (4.48) and (4.49) the part (3) of Theorem 4.3 corresponding to $\alpha=0$.

(ii)

[TABLE]

and

[TABLE]

where $c_{1}\in\mathbb{C}\setminus\{0\}$ is a constant, $\chi:S\to\mathbb{C}$ is a nonzero multiplicative function and $A:S\setminus I_{\chi}\to\mathbb{C}$ is a nonzero additive function such that $\chi^{*}=\chi$ and $A\circ\sigma=-A$.

By using (4.50), (4.51) and similar computations to the ones made to compute $f^{o}$ on $S\setminus I_{\chi}$ and $I_{\chi}$ in part (ii) of Case A, we obtain

[TABLE]

which gives with (4.35)

[TABLE]

By putting $\alpha:=\frac{c_{1}}{2}\in\mathbb{C}\setminus\{0\}$, we get, from (4.50), (4.51) and (4.52), part (4) of Theorem 4.3 corresponding to $\beta=0$. This completes the proof of Theorem 4.3. ∎

Remark 4.4**.**

Let $\chi:S\to\mathbb{C}$ be a nonzero multiplicative function. If $S=\{xy\,|\,x,y\in S\}$ then $S\setminus I_{\chi}=\{xy\,|\,x,y\in S\setminus I_{\chi}\}$. If $S$ is a regular semigroup so is the subsemigroup $S\setminus I_{\chi}$. Then, in view of Remark 3.2 and Remark 3.4, the proof of Theorem 4.3 works for each type of semigroups. It follows that our results are valid if $S=\{xy\,|\,x,y\in S\}$, and if $S$ is a regular semigroup.