A note on Levi-Civita functional equation

TL;DR

This paper investigates solutions to a generalized Levi-Civita functional equation on monoids, expanding understanding of its structure with multiple multiplicative function components.

Contribution

It characterizes solutions to a complex functional equation involving sums of products of functions on monoids, generalizing classical results.

Findings

Derived explicit solutions for the functional equation.

Extended the class of known solutions to include multiple multiplicative functions.

Provided conditions under which solutions exist.

Abstract

In this paper we find the solutions of the functional equation $$f(xy) = g(x)h(y) + \sum_{j=1}^n g_j(x)h_j(y), \;x,y \in M,$$ where $M$ is a monoid, $n\geq 2$, and $g_j$ (for $j=1,...,n$) are linear combinations of at least $2$ distinct nonzero multiplicative functions.

Equations

$$f(xy)=g(x)h(y)+\sum _{j=1}^n{g}_j(x)h_j(y),x,y\in M,$$

$$f(xy)=\sum _{j=1}^N{g}_j(x)h_j(y),x,y\in M.$$

$$f(xy)=f(x)g(y)+f(y)g(x)+h(x)h(y),x,y\in G,$$

$$f(xy)=k(x)L(y)+g(x)h(y),x,y\in S,$$

$$f(xy)=g_1(x)h_1(y)+\mu (x)h_2(y),x,y\in G,$$

$$f(xy)=g_1(x)h_1(y)+g(x)h_2(y),x,y\in M,$$

$$f(xy)=g(x)h(y)+\sum _{j=1}^n{g}_j(x)h_j(y),x,y\in M,$$

$$g_j=\sum _{i=n_j}^{m_j}{b}_i{\mu }_i,$$

$${\mu }_{n_1}={\mu }_1,...,{\mu }_{m_1};{\mu }_{n_2},...,{\mu }_{m_2};{\mu }_{n_j},...,{\mu }_{m_j};...;{\mu }_{n_n},...,{\mu }_{m_n}={\mu }_m$$

$$\{1,2,...,\hat{q},...,n\}:=\{1,...,n\}\backslash \{q\}.$$

$$\sum _{j=1}^n{g}_j(x)h_j(y)=\sum _{j=1}^n\sum _{i=n_j}^{m_j}{b}_i{\mu }_i(x)h_j(y)=\sum _{i=1}^m{b}_i{\mu }_i(x)h_{J(i)}(y).$$

$$f(xy)=g(x)h(y)+\sum _{i=1}^m{b}_i{\mu }_i(x)h_{J(i)}(y),x,y\in G.$$

$$f=ab\chi ,g=b\chi -\sum _{k=1}^n{c}_k{g}_k,h=a\chi ,h_j=a{c}_j\chi ,$$

$$f=h(e)g+\sum _{j=1}^n{h}_j(e)g_j.$$

$$h(e)g(xy)+\sum _{j=1}^n{h}_j(e)g_j(xy)=g(x)h(y)+\sum _{j=1}^n{g}_j(x)h_j(y),$$

$$g(xy)=\frac{1}{h(e)}[g(x)h(y)+\sum _{j=1}^n{g}_j(x)h_j(y)-\sum _{j=1}^n{h}_j(e)g_j(xy)].$$

$$f((xy)z)=g(xy)h(z)+\sum _{j=1}^n{g}_j(xy)h_j(z),$$

$$f(x(yz))=g(x)h(yz)+\sum _{j=1}^n{g}_j(x)h_j(yz).$$

$$g(xy)h(z)+\sum _{j=1}^n{g}_j(xy)h_j(z)=g(x)h(yz)+\sum _{j=1}^n{g}_j(x)h_j(yz).$$

$$g(x)\left[h(y)\frac{h(z)}{h(e)}-h(yz)\right]=\sum _{j=1}^n{g}_j(x)\left[h_j(yz)-h_j(y)\frac{h(z)}{h(e)}\right]+\sum _{j=1}^n{g}_j(xy)\left[h_j(e)\frac{h(z)}{h(e)}-h_j(z)\right]$$

$$=\sum _{j=1}^n\left(\sum _{i=n_j}^{m_j}{b}_i{\mu }_i(x)\left[h_j(yz)-h_j(y)\frac{h(z)}{h(e)}\right]+\sum _{i=n_j}^{m_j}{b}_i{\mu }_i(xy)\left[h_j(e)\frac{h(z)}{h(e)}-h_j(z)\right]\right)$$

$$=\sum _{j=1}^n\sum _{i=n_j}^{m_j}{b}_i{\mu }_i(x)\left[h_j(yz)-h_j(y)\frac{h(z)}{h(e)}+{\mu }_i(y)\left(h_j(e)\frac{h(z)}{h(e)}-h_j(z)\right)\right].$$

$$=\sum _{i=1}^m{b}_i{\mu }_i(x)\left[h_{J(i)}(yz)-h_{J(i)}(y)\frac{h(z)}{h(e)}+{\mu }_i(y)\left(h_{J(i)}(e)\frac{h(z)}{h(e)}-h_{J(i)}(z)\right)\right].$$

$$h(yz)-h(y)\frac{h(z)}{h(e)}=0,$$

$$h_{J(i)}(yz)-h_{J(i)}(y)\frac{h(z)}{h(e)}+{\mu }_i(y)(h_{J(i)}(e)\frac{h(z)}{h(e)}-h_{J(i)}(z))=0,$$

$$h_{J(i)}(z)=h_{J(i)}(e)\frac{h(z)}{h(e)}=a{c}_{J(i)}\chi (z),\text{ where }{c}_{J(i)}:=\frac{h_{J(i)}(e)}{a}.$$

$$h_j(z)=h_j(e)\frac{h(z)}{h(e)}=a{c}_j\chi (z),\text{ where }{c}_j:=\frac{h_j(e)}{a}.$$

$$g=b\chi -\sum _{k=1}^n{c}_k{g}_k,\text{ where }b:=g(e)+\sum _{k=1}^n{c}_k{g}_k(e)\in K.$$

$$g=-\sum _{k=1}^n{c}_k{g}_k=-\sum _{i=1}^m{c}_{J(i)}{b}_i{\mu }_i.$$

$$f(xy)=\sum _{j=1}^n{g}_j(x)h_j(y),x,y\in M.$$

Full text

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where $M$ is a monoid, $n\geq 2$, and $g_{j}$ (for $j=1,...,n$) are linear combinations of at least $2$ distinct nonzero multiplicative functions.

Key words and phrases. Sine addition formula, Levi-Civita functional equation, Monoid, functional equation, multiplicative function.

2000 Mathematics Subject Classification. 39B52, 39B32

1. introduction

A solution of Levi-Civita functional equation on a monoid $M$ is an ordered set of functions f,g_{1},...,g_{N},h_{1},...,h_{N}$$:M\longrightarrow\mathbb{C} satisfying Levi-Civita functional equation

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Shulman [6, Lemma 4] described the measurable solutions of (1.1) on locally compact groups in terms of group representations under the additional assumption that both $g_{1},...,g_{N}$ and $h_{1},...,h_{N}$ are linearly independent. A description of the (nondegeneated) solutions on abelian groups can be found in Sz kelyhidi [8] for all $N$. They turn out to be exponential polynomials, that is, sums of products of polynomials of additive functions and of solutions of $\varphi(x+y)=\varphi(x)\varphi(y)$. That is in general not so for non-abelian groups as Stetkær [3, Exemple 1] reveals. We refer also to [4, Theorem 5.2].

Chung, Kannappan and Ng [5] solved the Levi-Civita functional equation

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where $G$ is a group.

The functional equation (1.1) was for each $N\leq 4$ thoroughly worked out on abelian groups with implicit formulas for the solutions by Sz kelyhidi [[7], Theorem 10.4]. Ebanks [1] studied the functional equation

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for four unknown central functions $f,g,h,k$ on certain non abelian semigroups $S$, where $L$ is a fixed multiplicative function on S. Stetkær [3] removed the restriction that the solutions of (1.3) should be central and solved the functional equation

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where $G$ is a group (that need not to be abelian), and $\mu:G\longrightarrow\mathbb{C}$ is a character of $G$.

The motivation of the present work is the recent paper, by Ebank and Stetkær [2] in which they derive explicit formulas for the solutions $f,g_{1},h_{1},h_{2}:M\longrightarrow K$ of the functional equation

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where $M$ is a monoid (that need not to be abelian), $K$ is a field, $g$ is a linear combination of $n\geq 2$ distinct nonzero multiplicative functions with nonzero coefficients.

We wish to see what makes the paper [2] work by analyzing a more general Levi-Civita functional equation

[TABLE]

where $M$ is a monoid, $K$ is a field, and $f,g,h,h_{j}:M\longrightarrow K$ are the unknown functions, and where for all $j\in\{1,..,n\}$, $g_{j}$ is defined by

[TABLE]

where $n_{1}=1$, $m_{j}=n_{j+1}-1$ for $j\in\{1,2,...,n-1\}$, $m:=m_{n}$, $m_{j}-n_{j}\geq 1$, for $j\in\{1,2,...,n\}$, $b_{i}\neq 0$ for all $i\in\{1,...,m\}$, and $\{\mu_{1},\mu_{2},...,\mu_{m}\}$ are distinct nonzero multiplicative functions on $M$. Thus we have grouped the elements of the sequence $\mu_{1},\mu_{2},...,\mu_{m}$ into $n$ consecutive, disjoint intervals by

[TABLE]

in accordance with the sequence $g_{1};g_{2};...;g_{j};...;g_{n}$. We note here that $g_{j}$ has at least $2$ terms on the right hand side like in the motivating paper [2].

Equation (1.6) is an extension of the functional equation (1.5) (Take $n=2$ and $h_{2}=0$ in (1.6)).

In our present paper:

1. (1)

We get explicit solution formulas involving multiplicative functions. 2. (2)

The solutions are abelian, except when they blatantly need not be abelian, so that non-abelian phenomena essentially do not occur, except for some arbitrary functions. This contrasts the formally simpler functional equation studied in the paper [3] (It is due to $m_{j}-n_{j}\geq 1$). 3. (3)

The proofs are elementary algebraic manipulations (no homological algebra and the link). No analysis or geometry come into play.

The key elements of our set up are

1. (1)

The functions are defined on a monoid $M$ with an identity element $e$, and map into a field $K$. 2. (2)

Whenever we refer to the functional equation (1.6, then the functions $g_{j}$, $j=1,2,...n$ have the forms (1.7). 3. (3)

The multiplicative functions $\mu_{i}$, appearing in (1.6) are nonzero and distinct and their coefficients $b_{i}$ are nonzero.

Our results are organized as follows. We discuss two cases according to whether $f\neq 0$ and the set $\{g_{1},\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly independent (Proposition 2.4) or not (Proposition 2.5). The main results are given in Theorem 2.6.

Throughout this paper $M$ denotes a monoid: A semigroup $S$ with a neutral element $e$, and $K$ is a field. Let $K^{\ast}=K\setminus\{0\}$ denote the subset of nonzero element. A multiplicative function $\mu:M\longrightarrow K$ is a function such that $\mu(xy)=\mu(x)\mu(y)$ for all $x,y\in M$. Let $M(n,\mathbb{C})$ denote the algebra of all complex $n\times n$ matrices, and let $A^{T}$ denote the transpose matrix of a matrix $A\in M(n,\mathbb{C})$, and $A^{-1}$ its inverse. Let $n\in\mathbb{N}$ such that $n\geq 2$, and let $q\in\{1,...,n\}$, we introduce the following notation

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The fact that distinct multiplicative functions are independent is frequently used in the present paper. The following result is taken from Proposition 1(a) in [2].

Proposition 1.1**.**

*Let $n\in\mathbb{N}$, let $\mu_{1},\mu_{2},...,\mu_{n}:S\longrightarrow K$ be $n$ distinct multiplicative functions, and let $a_{1},a_{2},...,a_{n}\in K$.

If $a_{1}\mu_{1}+a_{2}\mu_{2}+...+a_{n}\mu_{n}=0$, then $a_{1}\mu_{1}=a_{2}\mu_{2}=...=a_{n}\mu_{n}=0$. Thus any set of distinct nonzero multiplicative functions is linearly independent.*

Remark 1.2*.*

Any $i\in\{1,2,...,m\}$ in the form (1.7), belongs to exactly one of the $n$ disjoint subintervals $[n_{j},m_{j}]$ of $[1,m]$, and so the number $j\in[1,n]$ of that subinterval is uniquely determined by $i$. Let $J(i):=j\in[1,n]$, when $j$ is the number of the subinterval containing $i$. So $i\longmapsto J(i)$, $i=1,2,...,m$, is a function of $i$. Note that $J:[1,m]\longrightarrow[1,n]$ is surjective, that $i\in[n_{J(i)},m_{J(i)}]$ for $i=1,2,...,m$, and that $J^{-1}(\{k\})=[n_{k},m_{k}]$ for all $k\in\{1,2,...,n\}.$

Hence, we have

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Thus, the functional equation (1.6) has another form

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2. General solution of the functional equation (1.6)

The following proposition gives a partial solution of the functional equation (1.6).

Proposition 2.1**.**

Let $(f,g,h,h_{1},h_{2},...,h_{n})$ be a solution of the functional equation (1.6) such that $f\neq 0$, $h(e)\neq 0$, and such that the set $\{g,\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly independent. Then there exist a nonzero multiplicative function $\chi:M\longrightarrow K$ such that $\chi\neq\mu_{i}$ for all $i\in\{1,2,...,m\}$, and $a,b\in K^{\ast},$ and $c_{j}\in K$ with $j\in\{1,2,...,n\}$ such that

[TABLE]

where $j\in\{1,2,...,n\}$.

Proof.

Taking $y=e$ in (1.6) we get

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Substituting (2.1) in equation (1.6) we get

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from which we deduce that

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Let $x,y,z$ be in $M$, using (1.6) we have

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[TABLE]

Using the associativity of the monoid operation, we deduce that

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Replacing $g(xy)$ in this equation by its form in (2.2) we obtain after computations

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[TABLE]

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Since by hypothesis, $\{g,\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly independent and all $b_{i}$ are nonzero we deduce that

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and

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for all $i=1,2,...,m.$

Equation (2.3) implies that $h=a\chi$, where $\chi:=h/h(e):M\longrightarrow K$ is a nonzero multiplicative function and $a=h(e)\neq 0.$ Using the facts that $J^{-1}(\{j\})=[n_{j},m_{j}]$, $\mu_{n_{j}},...,\mu_{m_{j}}$ are distinct, and $m_{j}-n_{j}\geq 1$ for all $j\in\{1,2,...,n\},$ we get from (2.4) that

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Since the function $J$ is surjective we deduce that for all $j\in\{1,2,...,n\}$

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Taking $x=e$ in (2.2) we find that

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If $b=0$ we get that

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This means that $\{g,\mu_{1},...,\mu_{m}\}$ is a linear dependent set, which is not true by hypothesis. Thus, $b\neq 0$. Using the same argument we prove that $\chi$ is distinct from $\mu_{1},\mu_{2},...,\mu_{m}$.

To find $f$ we replace the functions $g,h,h_{1},...,h_{n}$ by their forms in (2.1) and we obtain that $f=ab\chi$. ∎

The following lemmas will be useful in the proofs of our main results.

Lemma 2.2**.**

Let $(f,h_{1},h_{2},...,h_{n})$ be a solution of the functional equation

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Then $f=h_{j}=0$ for all $j\in\{1,...,n\}.$

Proof.

Assume for contradiction that $f\neq 0$. Taking $y=e$ in (2.6) we get that

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If $h_{j}(e)=0$ for all $j\in\{1,...,n\}$ then we get from (2.7) that $f=0$ which contradicts the hypothesis that $f\neq 0$. Thus, there exists $k\in\{1,...,n\}$ such that $h_{k}(e)\neq 0$. We may take $k=n$, and the equation (2.6) becomes

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The set $\{g_{n},\mu_{1},...,\mu_{m_{n-1}}\}$ is linearly independent. In fact, if it is not, we will have

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with $(c_{1},c_{2},...,c_{n})\neq(0,0,...,0)$. Using the form of $g_{n}$, equation (2.8) can be written as follows

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Since $n_{n}=m_{n-1}+1$ the last equation is equivalent to $\sum_{i=1}^{m}a_{i}\mu_{i}(x)=0$, where $a_{i}=c_{i}$ for $i=1,2,...,m_{n-1}$, and $a_{i}=c_{n}b_{i}$ for $i=n_{n},...,m$. The multiplicative functions $\mu_{1},\mu_{2},...,\mu_{m}$ are distinct and nonzero, so according to Proposition 1.1 we deduce that $a_{i}=0$ for all $i\in\{1,2,...,m\}$, and hence $c_{i}=0$ for $i=1,2,...,m_{n-1}$, and $c_{n}=0$. This means that $(c_{1},c_{2},...,c_{n})=(0,0,...,0)$ and this contradicts the hypothesis that the set $\{g_{n},\mu_{1},...,\mu_{m_{n-1}}\}$ is linearly dependent. Thus $\{g_{n},\mu_{1},...,\mu_{m_{n-1}}\}$ is a linearly independent set. Now, using the fact that $f\neq 0$, $h_{n}(e)\neq 0$, and that $\{g_{n},\mu_{1},...,\mu_{m_{n-1}}\}$ is linearly independent, we get from Proposition 2.1 that there exist a nonzero multiplicative function $\chi:M\longrightarrow K$ such that $\chi\neq\mu_{i}$ for all $i\in\{1,2,...,m_{n-1}\}$, and $a,b\in K^{\ast}$, and $c_{j}\in K,$ for $j=1,2,...,n$ such that

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for all $j=1,2,...,n-1$. We get from (2.9) that

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This means that the set $\{\chi,\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly dependent which is not possible according to Proposition 1.1 because $\chi,\mu_{1},\mu_{2},...,\mu_{m}$ are distinct multiplicative functions. We deduce that $f=0$. Hence, we get from the functional equation (2.6) that

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Since $\{\mu_{1},...,\mu_{m}\}$ is linearly independent we deduce that $b_{i}h_{J(i)}=0$. Using the fact that the coefficients $b_{i}$ are nonzero we deduce that $h_{J(i)}=0$. The subjectivity of the function $J$ implies that $h_{j}=h_{J(i)}=0$ for all $j\in\{1,...,n\}.$ This completes the proof. ∎

Lemma 2.3**.**

Let $(f,g,h,h_{1},...,h_{n})$ be a solution of the functional equation (1.6) such that $f\neq 0$. If the set $\{g,\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly independent, then the set $\{f,\mu_{1},\mu_{2},...,\mu_{m}\}$ is also linearly independent.

Proof.

If $h=0$ then we have

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Applying Lemma 2.2 we deduce that $f=0$, which contradicts the hypothesis that $f\neq 0$. Hence $h\neq 0$. Now assume that $\{f,\mu_{1},...,\mu_{m}\}$ is linearly dependent. This means that $f$ can be written as follows

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where $\alpha_{i}\in K$ for all $i\in\{1,...,m\}.$ Substituting the last expression of $f$ in equation (1.9) we get

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Since $h\neq 0$ there exists $y_{0}$ such that $h(y_{0})\neq 0$. Thus, we obtain

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This means that $\{g,\mu_{1},\mu_{2},...,\mu_{m}\}$ is a linearly dependent set. This completes the proof. ∎

Now, using Lemma 2.2 and Lemma 2.3 we can omit the condition $h(e)\neq 0$ from Proposition 2.1.

Proposition 2.4**.**

Let $(f,g,h,h_{1},h_{2},...,h_{n})$ be a solution of the functional equation (1.6) such that $f\neq 0$. If the set $\{g,\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly independent, then there exist a nonzero multiplicative function $\chi:M\longrightarrow K$ such that $\chi\neq\mu_{i}$ for all $i\in\{1,2,...,m\}$, and $a,b\in K^{\ast}$, and $c_{j}\in K$ for $j=1,2,...,n$, such that

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Conversely, the formulas (2.11) define solutions of equation (1.6) .

Proof.

The case of $h(e)\neq 0$ is Proposition 2.1. Suppose that $h(e)=0$. The functional equation (2.1) becomes

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This means that $\{f,\mu_{1},...,\mu_{m}\}$ is linearly dependent. According to Lemma 2.3 we deduce that $\{g,\mu_{1},...,\mu_{m}\}$ is linearly dependent. This contradicts the hypothesis that $\{g,\mu_{1},...,\mu_{m}\}$ is linearly independent. Thus $h(e)\neq 0$.

The converse can be proved easily by substituting the formulas (2.11) in (1.6). ∎

Proposition 2.5 is the main result concerning the solutions of (1.6), when $f\neq 0$ and $\{g,\mu_{1},...,\mu_{m}\}$ is linearly dependent.

Proposition 2.5**.**

Let $(f,g,h,h_{1},...,h_{n})$ be a solution of the functional equation (1.6) such that $f\neq 0$. If the set $\{g,\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly dependent, then there exist a unique $q\in\{1,..,n\}$ and constants $a,b,c,d,c_{j}\in K$ for $j\in\{1,..,\widehat{q},...,n\}$, with $c\neq d$ and $(a,b)\neq(0,0)$ such that

• •

If $m_{q}-n_{q}=1$, then

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[TABLE]

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[TABLE]

• •

If $m_{q}-n_{q}\geq 2$, then

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[TABLE]

where $k\in\{n_{q},...,m_{q}\}$ and $a\neq 0$.

Proof.

That the set $\{g,\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly dependent means that

[TABLE]

where $\beta_{0},\beta_{1},...,\beta_{m}$ are constants such that $(\beta_{0},\beta_{1},...,\beta_{m})\neq(0,0,...,0).$ Since the set $\{\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly independent, then $\beta_{0}\neq 0$ and hence, we can write

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If we take $y=e$ in (1.9) we get

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Replacing $g$ by its form (2.18) in (1.9), and using (2.19) we get

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This can be written as follows

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Since the set $\{\mu_{1},\mu_{2},...,\mu_{m}\}$ is linearly independent we deduce that

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Here we discuss two cases

Case 1 : Suppose that there exists $q\in\{1,...,n\}$ such that $a_{k}b_{s}\neq a_{s}b_{k}$ for some pair $k,s\in\{n_{q},...,m_{q}\}.$

Replacing $i$ in (2.20) by $k$ and $s$ and using that $J(s)=J(k)=q$ we get respectively

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[TABLE]

Since $n\geq 2$, the set $\{1,...,\widehat{q},..,n\}$ is not empty. Let $j\in\{1,...,\widehat{q},..,n\}$ and let $p\in\{n_{j},,...,m_{j}\}$. If we replace $i$ in (2.20) by $p$ and use that $J(p)=j$ we get that

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This three equations can be written in a matrix form:

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where

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and

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Since $a_{k}b_{s}\neq a_{s}b_{k}$ and $b_{p}\neq 0$ we have

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Hence, the matrix $A$ has an inverse:

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where $\alpha=a_{k}b_{s}-a_{s}b_{k}$.

We get after some computations that there exist $a,b\in K$ such that

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[TABLE]

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where $\alpha_{j}\in K$ for all $j\in\{1,...,\widehat{q},...,n\}$. If we put $d:=a_{s}/b_{s}$, $c:=a_{k}/b_{k}$, $c_{j}:=a_{p}/b_{p}$ we get that

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for all $j\in\{1,...,\widehat{q},...,n\}$. Since $a_{s}b_{k}\neq a_{k}b_{s}$, we have $d\neq c$.

If $a=b=\alpha_{j}=0$ for all $j\in\{1,...,\widehat{q},...,n\}$ then $h=h_{q}=h_{j}=0$ which gives that $f=0$. Since $f$ is a nonzero function we deduce that $(a,b,\alpha_{1},...,\widehat{\alpha_{q}},..,\alpha_{n})\neq(0,0,0,...,0)$.

Since we have $m_{j}-n_{j}\geq 1$ for all $j\in\{1,...,\widehat{q},...,n\}$, then the set $\{n_{j},...,\widehat{p},...,m_{j}\}$ is not empty. Hence we can choose $l\in\{n_{j},...,\widehat{p},...,m_{j}\}$, where $j\in\{1,...,\widehat{q},...,n\}$. Replacing $i$ in equation (2.20) by $l$ we get

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Deriving the expressions of $h(e)$ and $h_{j}(e)$ from equations (2.32) and (2.33) and replacing the functions $h$, and $h_{j}$ by their forms, the equation (2.34) can be written as follows

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Hence

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The set $\{\mu_{s},\mu_{k},\mu_{p},\mu_{l}\}$ is linearly independent because $\mu_{s},\mu_{k},\mu_{p},\mu_{l}$ are distinct.

Thus we get that

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Since the coefficients $b_{i}$ are nonzero for all $i=1,2,...,m$ we get from the first relation of (2.35) that $\alpha_{j}=0$ for all $j\in\{1,..,\widehat{q},...,n\}$. Hence $(a,b)\neq(0,0)$ because $(a,b,\alpha_{1},..,\alpha_{n})\neq(0,0,0,..,0)$. Thus we get from the second and the third equation of (2.35) that $a_{l}=c_{j}b_{l}$ for all $l\in\{n_{j},...,\widehat{p},...,m_{j}\}$. Since $c_{j}=a_{p}/b_{p}$ we get that $a_{i}=c_{j}b_{i}$ for all $i\in\{n_{j},...,m_{j}\}$ where $j\in\{1,2,...,\widehat{q},...,n\}$. This means that $a_{p}/b_{p}=a_{l}/b_{l}=c_{j}$, and hence $a_{p}b_{l}=a_{l}b_{p}$ for all $p,l\in\{n_{j},...,m_{j}\}$, and for all $j\in\{1,2,...,\widehat{q},...,n\}$. This proves the uniqueness of $q\in\{1,2,...,n\}$.

Consequently, using the form of $g$ we obtain

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Hence

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Here, we treat two cases:

Case 1.1: Suppose that $m_{q}-n_{q}=1$. We may take $s=n_{q}$ and $k=n_{q}+1=m_{q}$. In this case we have

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and we get from (2.32) that

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Using those forms we obtain that

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Since $a_{n_{q}}=db_{n_{q}}$ and $a_{m_{q}}=cb_{m_{q}}$ we obtain after simplification that

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Case 1.2 : Suppose that $m_{q}-n_{q}\geq 2$. In this case we can choose an element $r\in\{n_{q},...,m_{q}\}$ such that $r\neq s$ and $r\neq k$. Substituting the forms of $h$, $h_{q}$, $h(e)$, and $h_{q}(e)$ in (2.20) and replacing $i$ by $r$ we find

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which can be written as follows

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Since $\mu_{s},\mu_{k},\mu_{r}$ are distinct the set $\{\mu_{r},\mu_{k},\mu_{r}\}$ is linearly independent. This implies that

[TABLE]

Taking in account that $(a,b)\neq(0,0)$ and $c\neq d$ we deduce that either $a\neq 0$ and $b=0$, or $b\neq 0$ and $a=0$.

Assume that $a\neq 0$ and $b=0$. In this case we find that $a_{r}=db_{r}$. Noting that $a_{s}=db_{s}$ we deduce that $a_{i}=db_{i}$ for all $i\in\{n_{q},...,\widehat{k},...,m_{q}\}$. Thus, using equation (2.36) we get that

[TABLE]

Using that $b=0$ we deduce from (2.32) that $h=a\mu_{k},\;h_{q}=-da\mu_{k},\;h_{j}=-c_{j}a\mu_{k},$ for all $j\in\{1,...,\widehat{q},...n\}$,

[TABLE]

So we obtain solution (a).

The case of $a=0$ and $b\neq 0$ can be treated in a similar way.

Case 2 : Suppose that $a_{s}b_{k}=a_{k}b_{s}$ for all $k,s\in\{n_{j},...,m_{j}\}$ and all $j=1,2,...,n$. This means that there exists $v_{j}\in K$ such that $a_{i}=v_{j}b_{i}$ for all $i\in\{n_{j},...,m_{j}\}$. Hence we have

[TABLE]

[TABLE]

Substituting the last form of $g$ in (2.1) we get that

[TABLE]

This equation has the form of the functional equation (2.6). According to Lemma 2.2 we deduce that $f=0$, which is not true by hypothesis. Thus this case cannot occur. This completes the proof. ∎

Theorem 2.6 gives all solutions of the functional equation (1.6).

Theorem 2.6**.**

Let $(f,g,h,h_{1},h_{2},...,h_{n})$ be a solution of the functional equation (1.6). Then the solutions have one of the following forms:

1. (1)

[TABLE]

where $\chi:M\longrightarrow K$ is any nonzero multiplicative function such that $\chi\neq\mu_{i}$ for all $i\in\{1,2,...,m\}$, and $a,b\in K^{\ast}$, and $c_{j}\in K$ for $j=1,2,...,n$. 2. (2)

there exists a unique $q\in\{1,..,n\}$ and there exist constants $a,b,c,d,c_{j}\in K$ for $j\in\{1,..,\widehat{q},...,n\}$, with $c\neq d$ and $(a,b)\neq(0,0)$ such that

• •

If $m_{q}-n_{q}=1$, then

[TABLE]

[TABLE]

[TABLE]

[TABLE]

• •

If $m_{q}-n_{q}\geq 2$, then

[TABLE]

[TABLE]

where $k\in\{n_{q},...,m_{q}\}$ and $a\neq 0$. 3. (3)

$f=0,\;g=\sum_{k=1}^{n}c_{k}g_{k}$, $h:M\longrightarrow K$ is any nonzero function, $h_{j}=-c_{j}h$, for all $j\in\{1,..,n\}$ where $c,c_{j}\in K.$ 4. (4)

$f=0$, $g:M\longrightarrow K$ is any function, $h=h_{j}=0$ for all $j\in\{1,..,n\}$.

Proof.

The statements (1),(2) and (3) are proved in Proposition 2.5 and Proposition 2.1. It is left to treat the case of $f=0$.

Assume that $f=0$. The functional equation (1.6) reduces to

[TABLE]

Here we discuss two cases:

If $h=0$ then we get that

[TABLE]

Using the form (1.7) of $g$ this can be written as follows

[TABLE]

Since $\mu_{1},...,\mu_{m}$ are distinct multiplicative functions, the set $\{\mu_{1},...,\mu_{m}\}$ is linearly independent, hence $h_{J(i)}=0$ because $b_{i}\neq 0$ for all $i=1,...,m$. Using the fact that $J$ is surjective we deduce that $h_{j}=0$ for all $j\in\{1,2,...,n\}$. This is the solution (5) of Theorem 2.6.

If $h\neq 0,$ then there exists $y_{0}\in K$ such that $h(y_{0})\neq 0$. Taking $y=y_{0}$ in (2.44) we deduce that

[TABLE]

where $c_{j}=-h_{j}(y_{0})/h(y_{0})$. Replacing $g$ in (2.44) by its new form (2.45) we get

[TABLE]

Proceeding as in the first case we deduce that $h_{j}=-c_{j}h$ for all $j\in\{1,2,...,n\}$. This is the solution (4), and this completes the proof. ∎

The following theorem gives the solutions of the functional equation (1.6) when $n=2$.

Theorem 2.7**.**

Let $(f,g,h,h_{1},h_{2})$ be a solution of the functional equation

[TABLE]

*such that $g_{1}=\sum_{i=1}^{N}b_{i}\mu_{i}$, $g_{2}=\sum_{i=N+1}^{m}b_{i}\mu_{i}$, where $N\geq 2$, $m-N\geq 2$, $\{\mu_{1},...,\mu_{m}\}$ are distinct nonzero multiplicative functions and $b_{1},...,b_{m}$ are all nonzero.

Then we have one of the following forms:*

1. (1)

[TABLE]

where $\chi:M\longrightarrow K$ is any nonzero multiplicative function such that $\chi\neq\mu_{i}$ for all $i\in\{1,2,...,m\}$, and $a,b\in K^{\ast}$, and $c_{1},c_{2}\in K.$ 2. (2)

[TABLE]

[TABLE]

where $a,b,c,d,c_{2}\in K$ such that $c\neq d$ and $(a,b)\neq(0,0)$. 3. (3)

[TABLE]

[TABLE]

where $k\in\{1,...,N\}$, $N\geq 3$ and $a,b,c,d,c_{2}\in K$ such that $c\neq d$ and $a\neq 0$. 4. (4)

[TABLE]

[TABLE]

where $m-N=2$, $a,b,c,d,c_{1}\in K$ such that $c\neq d$ and $(a,b)\neq(0,0)$. 5. (5)

[TABLE]

[TABLE]

where $m-N\geq 3$, $k\in\{N+1,...,m\}$ and $a,b,c,d,c_{1}\in K$ such that $c\neq d$ and $a\neq 0$. 6. (6)

$f=0,\;g=c_{1}g_{1}+c_{2}g_{2}$, $h:M\longrightarrow K$ is any nonzero function, $h_{1}=-c_{1}h,\;h_{2}=-c_{2}h$, where $c_{1},c_{2}\in K.$ 7. (7)

$f=0$, $g:M\longrightarrow K$ is any function, $h=h_{1}=h_{2}=0$.