2 Special cases of the original equation

On a class of linear functional equations without range condition

Eszter Gselmann, Gergely Kiss and Csaba Vincze March 20, 2019

Abstract

The main purpose of this work is to provide the general solutions of a class of linear functional equa- tions. Letn≥ 2 be an arbitrarily fixed integer, let furtherXandY be linear spaces over the fieldKand let αi, βi∈K,i=1, . . . ,nbe arbitrarily fixed constants. We will describe all those functions f,f_i,j:X×Y →K, i, j=1, . . . ,nthat fulfill functional equation

f









n

X

i=1

αixi,

n

X

i=1

βiyi







=

n

X

i,j=1

f_i,j(xi, yj) (xi ∈X, yi ∈Y,i=1, . . . ,n).

Additionally, necessary and sufficient conditions will also be given that guarantee the solutions to be non- trivial.

Dedicated to Professor J´anos Acz´el on the occasion of his95^thbirthday.

1 Introduction

As J´anos Acz´el wrote in his famous and pioneering monograph [1]: ‘Functional equations have a long history and occur almost everywhere. Their influence and applications can be felt in every field, and all fields benefit from their contact, use, and technique.’ Almost the same can be said about the class of linear functional equations. This area is one of the most investigated topic in this field, several authors studied this class, see e.g. [2, 3, 4, 5, 6, 7, 8, 9, 10, 14, 18, 19, 20].

The main purpose of this paper is to describe the general solutions of a class of linear functional equations.

More precisely, we are interested in the following problem. Letn≥ 2 be an arbitrarily fixed integer, let further X andY be linear spaces over the fieldKand letαi, βi ∈K,i=1, . . . ,nbe arbitrarily fixed constants. Assume further that for the functions f, fi,j: X×Y →K,i, j=1, . . . ,n, functional equation

f









n

X

i=1

αixi,

n

X

i=1

βiyi







 =

n

X

i,j=1

fi,j(xi, yj) (xi ∈X, yi ∈Y,i=1, . . . ,n) (1) is fulfilled.

This equation belongs to the class of linear functional equations, that was thoroughly investigated by L. Sz´ekelyhidi in [15, 16, 17]. For the sake of completeness, here we briefly recall the main results from Sz´ekelyhidi [15].

Definition 1. IfG,S are groups andnis a positive integer, then a functionA:Gⁿ→ S is said to ben-additive if it is a homomorphism in each variable. Let F:Gⁿ →S be a function, then the functionϕ: G → S defined by

ϕ(x)= F(x, . . . ,x) (x∈G) is said to be thediagonal of F and it is denoted by diag(F). Further, let

Ak(x, y)= A(x, . . . ,| {z }x

ktimes

, y, . . . , y| {z }

n−ktimes

) (x, y∈G).

arXiv:1903.07974v1 [math.CA] 19 Mar 2019

Remark. LetG,S be groups,nbe a positive integer andA: Gⁿ → S be ann-additive function. Then for all k∈Zand for arbitraryi∈ {1, . . . ,n}we have

A(x₁, . . . ,x_i−1,kxi,xi+1, . . . ,xn)=kA(x₁, . . . ,x_i−1,xi,xi+1, . . . ,xn) (x₁, . . . ,xn ∈G). For a function f, rng(f) denotes the range of f.

Definition 2. LetG,S be Abelian groups, letnbe a non-negative integer. The function f: G → S is said to beof degree n, if there exist functions fi:G→ S and homomorphismsϕi, ψi: G→Gsuch that

rng(ϕi)⊂rng(ψi) (i=1,2, . . . ,n+1) (R1) and functional equation

f(x)+

n+1

X

i=1

fi(ϕi(x)+ψi(y))= 0 (x, y∈G) (2)

holds.

Definition 3. LetG,S be Abelian groups, letnbe a non-negative integer. The function f:G → S is called a(generalized) polynomial degree n, if for allk = 0,1, . . . ,nthere exists a k-additive mapping Ak: G^k → S such that

f =

n

X

k=0

diag(Ak), where 0-additive functions are to be understood constant functions.

Theorem 1 (Theorem 3.6 of [15]). Let G,S be Abelian groups and suppose that G is divisible. Let n be a non-negative integer. The function f: G→S is of degree n if and only if it is a polynomial of degree n.

Theorem 2(Theorem 3.9 of [15]). Let G,S be Abelian groups and suppose that G is divisible and S is torsion free. Let n∈Nbe a non-negative integer and letϕi, ψibe homomorphisms of G onto itself such that

rng

ψj◦ψ⁻¹_i −ϕj◦ϕ⁻¹_i

=G (i, j, i, j= 1, . . . ,n+1). (R2) The functions fi: G→ S (i= 0,1, . . . ,n+1)satisfy functional equation

f0(x)+

n+1

X

i=1

fi(ϕi(x)+ψi(y))= 0 (x, y∈G)

if and only if for all k= 0,1, . . . ,n and i=0,1, . . . ,n+1there exist symmetric k-additive functions A⁽ⁱ⁾_k : G^k → S such that

fi =

n

X

k=0

diag A⁽ⁱ⁾_k

(i=0,1, . . . ,n+1) and the equations

A⁽⁰⁾_k,j(x,0)+

n+1

X

i=1

A⁽ⁱ⁾_k,j(ϕi(x), ψi(y))= 0 (x, y∈G) hold for all j=0,1, . . . ,n and k= j, j+1, . . . ,n.

Observe that equation (1) can be reduced to the form (2). Indeed, suppose thatn=2 (or substitute zero in place of the variables except a distinguished pair) and consider the following family of homomorphisms

ϕα,β(x, y)= α 0 0 β

!

· x y

!

(x∈X, y∈Y, α, β∈K).

With these notations (1) can be re-written as f

ϕα1,β1(u)+ϕα2,β2(v)

=

f_1,1(ϕ_1,1(u)+ϕ_0,0(v))+ f_1,2(ϕ_1,0(u)+ϕ_0,1(v))+ f_2,1(ϕ_0,1(u)+ϕ_1,0(v))+ f_2,2(ϕ_0,0(u)+ϕ_1,1(v)) (u,v∈X×Y). At the same time (as it can be seen in the following subsection), wecannotstate that the functions involved are polynomials. This is because the fact that the homomorphismsϕ_α,βdefined above in general do not fulfill range condition (R1), neither fulfill range condition (R2). What is more, they are injective if and only if α, β , 0 and in such a situationϕ⁻¹_α,β = ϕ_α⁻¹_,β⁻¹. Notice that equation (1) involves the projections ϕ1,0, ϕ0,1 and ϕ_0,0. None of these are injective. This shows that Theorems 1 and 2 cannot be applied in our situation.

2 Special cases of the original equation

2.1 The one-variable sub-case

In this sub-case letn ∈N,n ≥ 2 be arbitrarily fixed, Xbe a linear space over the fieldKand suppose that for the functions f, f1, . . . , fn: X →Kfunctional equation

f









n

X

i=1

αixi







 =

n

X

i=1

fi(xi) (x₁, . . . ,xn ∈X) (3)

holds with certain constantsα₁, . . . , αn∈K. Observe that without loss of generality

f(0)= f1(0)= . . .= fn(0)=0 (∗)

can be assumed. Otherwise we consider the functions ef(x) = f(x)− f(0) ef₁(x) = f₁(x)− f₁(0)

...

efn(x) = fn(x)− fn(0)

(x∈X).

They clearly vanish at zero and they also fulfill the above functional equation. Therefore from now on we always suppose that (∗) holds.

As we will see, the solutions of equation (3) heavily depend on whether or not there are zeros among the parametersα₁, . . . , αn. We may (and also do) assume that these parameters are arranged in the following way:

there exists a non-negative integerk≤ nsuch thatαi , 0 fori=1, . . . ,k, butαi = 0 for alli=k+1, . . . ,n.

Proposition 1. Let n ∈ N,n ≥ 2be arbitrarily fixed, X be a linear space over the field Kand suppose that for the functions f, f₁, . . . , fn: X →Cfunctional equation(3)holds with certain constantsα₁, . . . , αn ∈Kand assume that(∗)is also satisfied. Suppose further thatαi , 0for i=1, . . . ,k, butαi = 0for all i= k+1, . . . ,n.

Then

(i) in case k = 0, all the functions f₁, . . . , fn are identically zero and f: X → C is any function fulfilling f(0)=0,

(ii) in case k = 1, all the functions f₂, . . . , fn are identically zero and f, f1: X → C are any functions vanishing at zero and fulfilling

f (α1x)= f1(x) (x∈X), (iii) otherwise, there exists an additive functionχ: X →Csuch that

f(x)= χ(x) and fi(x)=χ(αix) for i= 1, . . . ,k and the functions fk+1, . . . , fnare identically zero.

Conversely, the mappings f, f₁, . . . , fn: X→ Cvanish at zero and they also fulfill(3).

Proof. In casek= 0 equation (3) reduces to

n

X

i=1

fi(xi)=0 (x₁, . . . ,xn∈X).

Since we have independent variables, this immediately yields that the involved functions have to be constant functions. In view of (∗) this means that they have to be identically zero and the only information we get for the function f is that f(0)= 0.

In casek≥ 1, our equation can be written as f









k

X

i=1

αixi







=

n

X

j=1

fj(xj) (x₁, . . . ,xn ∈X). With the substitution

x1= . . .= xk =0 we obtain that

0=

n

X

j=k+1

fj(xj) (xk+1, . . . ,xn ∈X),

which (similarly as above) yields that the functions fk+1, . . . , fn are identically zero. Using this, the functions f, f₁, . . . , fk fulfill

f









k

X

i=1

αixi







=

k

X

i=1

fi(xi) (x₁, . . . ,xk ∈X). (4)

Ifk =1, this is nothing but

f(α₁x)= f₁(x) (x∈X), showing that in this case there is nothing to prove.

Assume that k ≥ 2 and let i, j ∈ {1, . . . ,k} be different integers. Then equation (4) with xl = 0 for l∈ {1, . . . ,k} \ {i, j}is

f(αixi+αjxj)= fi(xi)+ fj(xj)

xi,xj ∈X , which, after introducing the functions

efl(x)= fl

x αl

!

(x∈X,l= 1, . . . ,k) can be reduced to the system of Pexider equations

f(xi+xj)=efi(xi)+efj(xj)

xi,xj ∈X,i, j∈ {1, . . . ,k},i, j . This means that there exists an additive functionχ: X →Csuch that

f(x)= χ(x) and fi(x)=χ(αix) for i= 1, . . . ,k.

3 The two-variable case with n = 2

In this section we will focus on functional equation f(α₁x₁+α₂x₂, β₁y₁+β₂y₂)

= f_1,1(x₁, y₁)+ f_1,2(x₁, y₂)+ f_2,1(x₂, y₁)+ f_2,2(x₂, y₂) (x₁,x₂∈X, y₁, y₂ ∈Y), (5)

where f, f1,1, f1,2, f2,1, f2,2: X×Y →Kdenote the unknown functions andα₁, α₂, β₁, β₂ ∈Kare given constants.

Observe that without loss of generality

f(0,0)= f_i,j(0,0)= 0 can be supposed. Otherwise we consider the functions

ef(x, y) = f(x, y)− f(0,0)

ff_i,j(x, y) = f_i,j(x, y)− f_i,j(0,0) (x∈X, y∈Y).

They clearly vanish at the point (0,0) and they also fulfill the same functional equation. Similarly as previously, from now on we always suppose that all the involved functions vanish at the point (0,0).

This section will be divided into two parts. At the first one, we will consider the so-called degenerate cases, where at least one of the parametersα₁, α₂, β₁, β₂ is zero. After that the non-degenerate case will follow, that is, when none of the above parameters are zero.

3.1 Degenerate cases

3.1.1 The homogeneous caseα₁ =α₂ =β₁ =β₂ =0 In caseα1 =α2 =β1 =β2= 0 equation (5) reduces to

f_1,1(x₁, y₁)+ f_1,2(x₁, y₂)+ f_2,1(x₂, y₁)+ f_2,2(x₂, y₂)=0 (x₁,x₂∈X, y₁, y₂ ∈Y).

Proposition 2. Let X and Y be linear spaces over the field Kand f1,1, f1,2, f2,1, f2,2: X×Y → Kbe functions such that

f1,1(x₁, y₁)+ f1,2(x₁, y₂)+ f2,1(x₂, y₁)+ f2,2(x₂, y₂)=0 (x₁,x₂∈X, y₁, y₂ ∈Y). (6) Then and only then for all i, j = 1,2 there exist functionsχi,j: X → Kand ζi,j: Y → K vanishing at0such that

fi,j(x,z)= χi,j(x)+ζi,j(z) (x∈X,z∈Y,i, j=1,2) as well as

χ_1,2(x)+χ_1,1(x) = 0 χ2,2(x)+χ2,1(x) = 0 ζ2,1(z)+ζ1,1(z) = 0 ζ_2,2(z)+ζ_1,2(z) = 0

(x∈X,z∈Y).

Proof. Fori, j∈ {1,2}let us define the functionsχi,j: X →Kandζi,j: Y →Kthrough χ_i,j(x)= f_i,j(x,0) and ζ_i,j(z)= f_i,j(0,z) (x∈X,z∈Y). With the notation

E(x₁,x₂, y₁, y₂)= f_1,1(x₁, y₁)+ f_1,2(x₁, y₂)+ f_2,1(x₂, y₁)+ f_2,2(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂∈Y), identities

E(x,0,0,0) = 0 E(0,x,0,0) = 0 E(0,0,z,0) = 0 E(0,0,0,z) = 0

(x∈X,z∈Y)

give that

χ_1,2(x)+χ_1,1(x) = 0 χ2,2(x)+χ2,1(x) = 0 ζ2,1(z)+ζ1,1(z) = 0 ζ_2,2(z)+ζ_1,2(z) = 0

(x∈X,z∈Y).

Moreover, equations

E(x₁,0, y₁,0) = 0 E(0,x₂, y₁,0) = 0 E(x1,0,0, y2) = 0 E(0,x2,0, y₂) = 0

(x₁,x₂ ∈X, y₁, y₂ ∈Y)

yield that

f_i,j(x,z)= χ_i,j(x)+ζ_i,j(z) (x∈X,z∈Y,i, j= 1,2),

where we used the previously proved identities, too.

3.1.2 The caseα₁ =α₂ =β₁ =0andβ₂ ,0 In such a situation (5) reduces to

f(0, β₂y₂)= f1,1(x₁, y₁)+ f1,2(x₁, y₂)+ f2,1(x₂, y₁)+ f2,2(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂ ∈Y).

Obviously, β₂ = 1 can be assumed, otherwise we consider the functions ff_1,2, ff_2,2: X ×Y → K defined through

ff1,2(x,z) = f1,2

x,_β^z

2

ff_2,2(x,z) = f_2,2

x,_β^z

2

(x∈X,z∈Y).

Proposition 3. Let X and Y be linear spaces over the fieldKand f, f1,1, f1,2, f2,1, f2,2: X×Y →Kbe functions such that

f(0, y₂)= f_1,1(x₁, y₁)+ f_1,2(x₁, y₂)+ f_2,1(x₂, y₁)+ f_2,2(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂ ∈Y).

Then and only then for all i, j = 1,2 there exist functionsχi,j: X → Kand ζi,j: Y → K vanishing at0such that

f_i,j(x,z)= χ_i,j(x)+ζ_i,j(z) (x∈X,z∈Y,i, j=1,2) as well as

χ1,2(x)+χ1,1(x) = 0 χ_2,2(x)+χ_2,1(x) = 0 ζ_2,1(z)+ζ_1,1(z) = 0 ζ2,2(z)+ζ1,2(z) = f(0,z)

(x∈X,z∈Y).

Proof. Fori, j∈ {1,2}let us define the functionsχ_i,j: X →Kandζ_i,j: Y →Kthrough χi,j(x)= fi,j(x,0) and ζi,j(z)= fi,j(0,z) (x∈X,z∈Y). Furthermore, let

E(x₁,x₂, y₁, y₂)

= f(0, y₂)− f1,1(x₁, y₁)− f1,2(x₁, y₂)− f2,1(x₂, y₁)− f2,2(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂ ∈Y) to obtain the following system of equations

E(x₁,0, y₁,0) = 0 E(x₁,0,0, y₂) = 0 E(0,x2, y1,0) = 0 E(0,x₂,0, y₂) = 0

(x₁,x₂ ∈X, y₁, y₂∈Y),

or equivalently

f_1,1(x₁, y₁)+ f_1,2(x₁,0)+ f_2,1(0, y₁) = 0 f1,2(x₁, y₂)+ f1,1(x₁,0)+ f2,2(0, y₂) = f(0, y₂) f2,1(x2, y1)+ f2,2(x2,0)+ f1,1(0, y1) = 0 f_2,2(x₂, y₂)+ f_2,1(x₂,0)+ f_1,2(0, y₂) = f(0, y₂)

(x₁,x₂ ∈X, y₁, y₂ ∈Y).

Finally, the system of equations

E(x₁,0,0,0) = 0 E(0,x₂,0,0) = 0 E(0,0, y1,0) = 0 E(0,0,0, y₂) = 0

(x₁,x₂ ∈X, y₁, y₂ ∈Y)

yields that

f1,2(x₁,0)+ f1,1(x₁,0) = 0 f2,2(x2,0)+ f2,1(x2,0) = 0 f2,1(0, y₁)+ f1,1(0, y₁) = 0 f_2,2(0, y₂)+ f_1,2(0, y₂) = f (0, y₂)

(x₁,x₂∈X, y₁, y₂ ∈Y),

which in view of the above definitions completes the proof.

3.1.3 The caseα₁, α₂ ,0andβ₁, β₂ =0 In such a situation (5) implies that

f(α₁x₁+α₂x₂,0)= f_1,1(x₁,0)+ f_1,2(x₁,0)+ f_2,1(x₂,0)+ f_2,2(x₂,0) (x₁,x₂ ∈X) because the left hand side does not depend ony₁andy₂.

Obviously,α1, α2 = 1 can be assumed, otherwise we consider the functions ff1,2, ff2,2: X×Y → Kdefined through

ff_1,1(x,z) = f_{1,1 x}

α1,z ff_1,2(x,z) = f_{1,2 x}

α1,z ff2,1(x,z) = f2,1

_x

α2,z ff2,2(x,z) = f2,2

_x

α₂,z

(x∈X,z∈Y).

Proposition 4. Let X and Y be linear spaces over the fieldKand f, f1,1, f1,2, f2,1, f2,2: X×Y →Kbe functions such that

f(x₁+x₂,0)= f1,1(x₁,0)+ f1,2(x₁,0)+ f2,1(x₂,0)+ f2,2(x₂,0) (x₁,x₂ ∈X). Then and only then there exists an additive functionχ: X →Ksuch that

f(x,0) = χ(x) f_1,1(x,0)+ f_1,2(x,0) = χ(x) f2,1(x,0)+ f2,2(x,0) = χ(x)

(x∈X).

Proof. Consider the functionsχ, ϕ, ψ: X→ Kdefined through χ(x) = f(x,0)

ϕ(x) = f_1,1(x,0)+ f_1,2(x,0) ψ(x) = f2,1(x,0)+ f2,2(x,0)

(x∈X)

to get the following Pexider equation

χ(x₁+x₂)=ϕ(x₁)+ψ(x₂) (x₁,x₂∈X).

Since all the functionsχ, ϕ, ψvanish at zero, we get thatϕ≡ ψ≡χand the functionχhas to be additive.

To finish the discussion of equation (5) in this special case, apply Proposition 2 to the functions ef_i,j(x, y)= f_i,j(x, y)− f_i,j(x,0), (x∈X, y∈Y)

where fi,j(x,0) are given in Proposition 4.

3.1.4 The caseα₁, β₁ , 0andα₂, β₂ =0 In such a situation (5) implies that

f(α1x1, β1y1)= f1,1(x1, y1)+ f1,2(x1,0)+ f2,1(0, y1) (x₁∈X, y1 ∈Y) because the left hand side does not depend onx2andy₂.

Obviously, due to similar reasons as previously, α₁, β₁ = 1 can be assumed. The proof of the following proposition is a straightforward calculation, so we omit it.

Proposition 5. Let X and Y be linear spaces over the field Kand f, f_1,1, f_1,2, f_2,1: X×Y → Kbe functions.

Functional equation

f(x1, y1)= f1,1(x1, y1)+ f1,2(x1,0)+ f2,1(0, y1) (x₁ ∈X, y1 ∈Y). is fulfilled if and only if there exist functionsχ: X→ Kandζ: Y → Ksuch that

f_1,2(x,0) = χ(x) f2,1(0,z) = ζ(z) f(x,z)− f1,1(x,z) = χ(x)+ζ(z)

(x∈X,z∈Y).

To finish the discussion of equation (5) in this special case, apply Proposition 2 to the functions ef1,1(x, y) = 0,

ef1,2(x, y) = f1,2(x, y)− f1,2(x,0), ef_2,1(x, y) = f_2,1(x, y)− f_2,1(0, y), ef2,2(x, y) = f2,2(x, y),

(x∈X, y∈Y),

where f_1,2(x,0) and f_2,1(0, y) were determined in Proposition 5.

3.1.5 The caseα₁, α₂, β₁ ,0andβ₂ =0 In such a situation (5) implies that

f(α1x1+α2x2, β1y1)= f1,1(x1, y1)+ f1,2(x1,0)+ f2,1(x2, y1)+ f2,2(x2,0) (x₁∈ X, y1 ∈Y), because the left hand side does not depend ony2.

Obviously, due to similar reasons as previously,α₁, α₂, β₁ =1 can be assumed.

Proposition 6. Let X and Y be linear spaces over the fieldKand f, f1,1, f1,2, f2,1, f2,2: X×Y →Kbe functions.

Functional equation

f(x₁+x₂, y₁)= f1,1(x₁, y₁)+ f1,2(x₁,0)+ f2,1(x₂, y₁)+ f2,2(x₂,0) (x₁,x₂∈ X, y₁ ∈Y)

is fulfilled if and only if there exist a mapping A: X×Y →Kadditive in its first variable and there are functions χ, χ1,1, χ2,1: X→ Kandζ, ζ1,1, ζ2,1:Y →Kvanishing at zero so thatχis additive and

f(x,z) = A(x,z)+χ(x)+ζ(z) f1,1(x,z) = A(x,z)+χ1,1(x)+ζ1,1(z) f_2,1(x,z) = A(x,z)+χ_2,1(x)+ζ_2,1(z)

(x∈X,z∈Y) and also

χ(x) = f(x,0) χ_1,1(x) = f_1,1(x,0) χ2,1(x) = f2,1(x,0)

ζ(z) = ζ1,1(z)+ζ2,1(z) f_1,2(x,0) = χ(x)−χ_1,1(x) f2,2(x,0) = χ(x)−χ2,1(x)

(x∈X,z∈Y)

hold.

Proof. With the substitutiony₁ =0 our equation yields that

f(x₁+x₂,0)= f1,1(x₁,0)+ f1,2(x₁,0)+ f2,1(x₂,0)+ f2,2(x₂,0) (x₁,x₂∈X,). From this we immediately get that

ef(x₁+x2, y1)= ef1,1(x₁, y1)+ ef2,1(x₂, y1) (x₁,x2 ∈X, y1 ∈Y), where the functions ef, ef1,1, ef2,1: X×Y →Kare defined by

ef(x, y) = f(x, y)− f(x,0) ef1,1(x, y) = f1,1(x, y)− f1,1(x,0) ef2,1(x, y) = f2,1(x, y)− f2,1(x,0)

(x∈X, y∈Y).

This means that the functions ef, ef1,1, ef2,1fulfill a Pexider equation onX for any fixedy∈Y. Thus there exists a mappingA: X×Y →Kadditive in its first variable and functionsζ, ζ_1,1, ζ_2,1: Y → Kso that

ζ(z)= ζ_1,1(z)+ζ_2,1(z) (z∈Y) and

ef(x,z) = A(x,z)+ζ(z) ef_1,1(x,z) = A(x,z)+ζ_1,1(z) ef2,1(x,z) = A(x,z)+ζ2,2(z)

(x∈X,z∈Y). In terms of the functions f, f_1,1, f_2,1this means that

f(x,z) = A(x,z)+χ(x)+ζ(z) f_1,1(x,z) = A(x,z)+χ_1,1(x)+ζ_1,1(z) f2,1(x,z) = A(x,z)+χ2,1(x)+ζ2,2(z)

(x∈X,z∈Y),

where

χ(x) = f(x,0) χ_1,1(x) = f1,1(x,0) χ2,1(x) = f2,1(x,0)

(x∈X).

Observe thatχis additive. Indeed, using the above the forms of f, f1,1 and f2,2, our equation withy₁ = 0 and the fact thatAis additive in its first variable, we obtain that

χ(x₁+x₂)=χ_1,1(x₁)+ f_1,2(x₁,0)+χ_2,1(x₂)+ f_2,2(x₂,0) (x₁,x₂ ∈X),

that is,χfulfills a Pexider equation. Sinceχ(0) = 0, this means thatχ has to be additive. Thus, using again the form of the functions f, f_1,1, f_2,1and our equation with x₂ = 0, we get that

f1,2(x,0)= χ(x)−χ_1,1(x) (x∈X). Similarly, our equation withx1 = 0 implies that

f2,2(x,0)= χ(x)−χ2,1(x) (x∈X).

To finish the discussion of equation (5) in this special case, apply Proposition 2 to the functions

ef1,1(x, y) = 0,

ef1,2(x, y) = f1,2(x, y)− f1,2(x,0), ef_2,1(x, y) = 0,

ef2,2(x, y) = f2,2(x, y)− f2,2(x,0),

(x∈X, y∈Y)

where f1,2(x,0) and f2,2(x,0) are given in Proposition 6.

3.2 The non-degenerate case

After making clear the degenerate cases, now we can focus on the case α1, α2, β1, β2 , 0 and provide the general solution of the functional equation

f(α1x1+α2x2, β1y1+β2y2)

= f1,1(x1, y1)+ f1,2(x1, y2)+ f2,1(x2, y1)+ f2,2(x2, y2) (x₁,x2∈X, y1, y2 ∈Y), where f, f_1,1, f_1,2, f_2,1, f_2,2: X×Y →Kdenote the unknown functions andα₁, α₂, β₁, β₂ ∈Kare given constants.

Obviously, it is enough to consider the case α₁ = α₂ = β₁ = β₂ = 1, that is, to consider the following functional equation

f (x₁+x₂, y₁+y₂)= f_1,1(x₁, y₁)+ f_1,2(x₁, y₂)+ f_2,1(x₂, y₁)+ f_2,2(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂ ∈Y). In this subsection we always assume that the characteristic of the fieldKisdifferentfrom 2.

Proposition 7. Let X and Y be linear spaces over the fieldK. Then the functions f, f1,1, f1,2, f2,1, f2,2: X×Y → Ksatisfy functional equation

f (x₁+x2, y₁+y₂)= f1,1(x₁, y₁)+ f1,2(x₁, y₂)+ f2,1(x₂, y₁)+ f2,2(x₂, y₂) (x₁,x2 ∈X, y₁, y₂∈Y). (7) if and only if

f(x,z) = A(x,z)+χ(x)+ζ(z)

f_i,j(x,z) = A(x,z)+χ_i,j(x)+ζ_i,j(z) (x∈X,z∈Z),

where the mapping A: X ×Y → K is a bi-additive function and for i, j ∈ {1,2} χ, χ_i,j: X → Kas well as ζ, ζi,j: Y →Kare functions such thatχandζare additive functions andχi,j andζi,j vanish at the point(0,0) and χ(x) = χ1,1(x)+χ1,2(x)=χ2,1(x)+χ2,2(x)

ζ(z) = ζ_1,1(z)+ζ_2,1(z)= ζ_1,2(z)+ζ_2,2(z) (x∈X,z∈Y) are also fulfilled.

Proof. Assume that the functions f, f_1,1, f_1,2, f_2,1, f_2,2: X×Y →Kfulfill functional equation (7) for anyx₁,x₂∈ X andy₁, y₂ ∈Y. With the substitutiony₂ =0 we obtain that

f (x₂+x1, y1)= f2,1(x2, y1)+ f2,2(x2,0)+ f1,1(x1, y1)+ f1,2(x1,0) (x₁,x2∈X, y1 ∈Y), which immediately implies that

f (x₂+ x₁, y₁)= ff2,1(x₂, y₁)+ ff1,1(x₁, y₁) (x₁,x₂ ∈X, y₁ ∈Y), where the functions ff_1,1, ff_2,1 are defined by

ff1,1(x,z) = f1,1(x,z)+ f1,2(x,0)

ff_2,1(x,z) = f_2,1(x,z)+ f_2,2(x,0) (x∈X,z∈Y).

This means that there exists a functionA⁽¹⁾: X×Y → Kwhich is additive in its first variable and a function ζ: Y →Kvanishing at zero such that

f(x,z)= A⁽¹⁾(x,z)+ζ(z) (x∈X,z∈Y).

Substituting this form into equation (7), with x₂= 0 and with a similar argument we receive that A⁽¹⁾(x,z)= A(x,z)+χ(x) (x∈X,z∈Y),

whereA: X×Y →Kis a bi-additive mapping andχ: X→ Kis a function that vanishes at zero.

All in all this means that

f(x,z)= A(x,z)+χ(x)+ζ(z) (x∈X,z∈Y).

Additionally, equation (7), first withy₁ =y₂ =0 yields thatχhas to be additive and secondly, withx₁ = x₂ =0 we receive that the functionζalso has to be additive, too.

Define the functions F_1,1,F_1,2,F_2,1,F_2,2onX×Y through F_1,1(x,z) = f_1,1(x,z)−A(x,z)− χ(x)

2 − ζ(z) 2 F1,2(x,z) = f1,2(x,z)−A(x,z)− χ(x)

2 − ζ(z) 2 F2,1(x,z) = f2,1(x,z)−A(x,z)− χ(x)

2 − ζ(z) 2 F2,2(x,z) = f2,2(x,z)−A(x,z)− χ(x)

2 − ζ(z) 2

(x∈X,z∈Y)

to deduce that they fulfill functional equation (6). Due to Proposition 2, for alli, j =1,2 there exist functions eχ_i,j: X→ Kandeζ_i,j: Y → Kvanishing at zero such that

Fi,j(x,z)=eχi,j(x)+eζi,j(z) (x∈X,z∈Y,i, j= 1,2), that is, for the functions f_i,j we have

fi,j(x,z)= A(x,z)+χi,j(x)+ζi,j(z) (i, j∈ {1,2},x∈X,z∈Y). Finally, using the equations in Proposition 2 for the functionseχi,jandeζi,j, identities

f(x,0) = χ(x)= f_1,1(x,0)+ f_1,2(x,0)=χ_1,1(x)+χ_1,2(x) f(x,0) = χ(x)= f2,1(x,0)+ f2,2(x,0)=χ2,1(x)+χ2,2(x)

f(0,z) = ζ(z)= f1,1(0,z)+ f2,1(0,z)= ζ1,1(z)+ζ2,1(z) f(0,z) = ζ(z)= f_1,2(0,z)+ f_2,2(0,z)= ζ_1,2(z)+ζ_2,2(z)

(x∈ X,z∈Y)

complete the proof.

3.3 Related equations

3.3.1 The functional equation of bi-additivity

As a trivial consequence of the results of the previous section we get the following.

Corollary 1. Let X and Y be linear spaces over the field Kwithchar(K) , 2. The mapping f: X ×Y → K fulfills the functional equation of bi-additivity, that is,

f(x₁+x₂, y₁+y₂) = f(x₁, y₁)+ f(x₁, y₂)+ f(x₂, y₁)+ f(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂ ∈Y) if and only if f is bi-additive.

3.3.2 The rectangle equation

LetXandY be linear spaces over the fieldKand let f: X×Y → Kbe a function.

Then functional equation

f(x+u, y+v)+ f(x+u, y−v)+ f(x−u, y+v)+ f(x−u, y−v)=4f(x, y) (x, y∈X,u, v∈Y), or equivalently (provided that char(K),2)

4f

x1+ x2

2 ,y1+y2

2

= f(x₁, y₁)+ f(x₁, y₂)+ f(x₂, y₁)+ f(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂∈Y). is called therectangle equation.

Indeed, both the above equations express the following: the value of f at the center of any rectangle, with parallel sides to the coordinate axes, equals the mean of the values of f at the vertices.

0000 00 1111 11

00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0

11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 1

00000000000000000000000000000000 11111111111111111111111111111111

y₂

y1

x₁ x₂

_x

1+x₂ 2 ,^y1+₂^y2

x y

This equation as well as its generalization were investigated (among others) in [2, 5, 14].

With the aid of the results of the previous section, we obtain the following straightaway.

Proposition 8. Let X and Y be linear spaces over the field K with char(K) , 2 and f: X × Y → K be a function. The function f fulfills the rectangle equation, i.e.,

4f

x1+ x2

2 ,y1+y2

2

= f(x1, y1)+ f(x1, y2)+ f(x2, y1)+ f(x2, y2) (x₁,x2 ∈X, y1, y2∈Y),

if and only if there exists a bi-additive mapping A: X×Y →Kand additive functionsχ: X→ Kandζ: Y →K such that

f(x,z)= A(x,z)+χ(x)+ζ(z) (x∈X,z∈Y). Remark. In case char(K)= 2, the rectangle equation reduces to equation

f(x+u, y+v)+ f(x+u, y−v)+ f(x−u, y+v)+ f(x−u, y−v)= 0 (x, y∈X,u, v∈Y), or equivalently

f(x₁, y₁)+ f(x₁, y₂)+ f(x₂, y₁)+ f(x₂, y₂)= 0 (x₁,x₂∈X, y₁, y₂ ∈Y). Thus, Proposition 2 yields that there exist functionsχ: X →Kandζ: Y →Ksuch that

f(x, y)= χ(x)+ζ(y) (x∈X, y∈Y). The Cauchy equation onX×Y

Proposition 9. Let X and Y be linear spaces over the field K and f, g,h: X × Y → K be functions. Then functional equation

f(x₁+x₂, y₁+y₂)=g(x₁, y₁)+h(x₂, y₂) (x₁,x₂∈X, y₁, y₂ ∈Y) (8) holds if and only if there exist additive functionsχ: X→ K,ζ: Y → Ksuch that

f(x, y) = χ(x)+ζ(z) g(x, y) = χ(x)+ζ(z)

h(x, y) = χ(x)+ζ(z) (x∈X,z∈Y).

4 On the reduction of equations with n > 2 to the two-variable case

In this section we intend to investigate the following problem. Let X and Y be linear spaces over the field K, let further αi, βi ∈ K, i = 1, . . . ,n be arbitrarily fixed constants. Assume further that for the functions

f, fi,j: X×Y → K,i, j=1, . . . ,n, functional equation

f









n

X

i=1

αixi,

n

X

i=1

βiyi







 =

n

X

i,j=1

f_i,j(xi, yj) (xi ∈X, yi ∈Y,i=1, . . . ,n) (9) is fulfilled.

We will show that in case n> 2, the results of the previous section can be applied. Indeed, letλ, κ, µ, ν∈ {1, . . . ,n}such thatλ,κandµ,ν, but otherwise arbitrary. In this case equation (9) with the substitutions

xi = 0 ifi,λ, κ and yj =0 if j,µ, ν yields that

f

αλxλ+ακxκ, βµyµ+βνyν

= fλ,µ(xλ, yµ)+ fλ,ν(xλ, yν)+ fκ,µ(xκ, yµ)+ fκ,ν(xκ, yν) + X

j,µ,ν

fλ,j(xλ,0)+ X

j,µ,ν

fκ,j(xκ,0)+ X

i,λ,κ

fi,µ(0, yµ)+X

i,λ,κ

fi,ν(0, yν)

for anyxλ,xκ ∈Xandyµ, yν ∈Y. Consider the functionsgfλ,µ,ffκ,ν: X×Y →Kdefined by gfλ,µ(x,z)= fλ,µ(x,z)+ X

j,µ,ν

fλ,j(x,0)+ X

i,λ,κ

fi,µ(0,z) (x∈X,z∈Y) and

ff_κ,ν(x,z)= f_κ,ν(x,z)+ X

j,µ,ν

f_κ,j(x,0)+X

i,λ,κ

f_i,ν(0,z) (x∈X,z∈Y) to receive that

f

α_λxλ+α_κxκ, β_µy_µ+β_νy_ν

=gfλ,µ(x_λ, y_µ)+ fλ,ν(x_λ, y_ν)+ fκ,µ(x_κ, y_µ)+ ffκ,ν(x_κ, y_ν) (10) is satisfied for any xλ,xκ ∈ Xandyµ, yν ∈Y. This equation can however be handled with the aid of the results of Section 3.

5 The case of a single unknown function in the equation — existence of non-trivial solutions

LetXandY be linear spaces over the same fieldKand consider the following functional equation f(α₁x₁+α₂x₂, β₁y₁+β₂y₂)

= γ_1,1f(x₁, y₁)+γ_1,2f(x₁, y₂)+γ_2,1f(x₂, y₁)+γ_2,2f(x₂, y₂) (x₁,x₂∈X, y₁, y₂ ∈Y), (11) where f: X×Y → Kdenotes the unknown function andα1, α2, β1, β2 ∈Kandγ1,1, γ1,2, γ2,1, γ2,2 ∈Kare given constants.

Recall that due to the linearity of the above equation we may (and we also do) suppose that f(0,0)=0

holds. Otherwise the function

ef(x, y)= f(x, y)− f(0,0) (x∈X, y∈Y)

can be considered. This function clearly vanishes at the point (0,0) and it fulfills the same functional equation, too.

Furthermore, the linearity of the investigated equation implies that the identically zero function is always a solution. In this section we would like to study under what conditions admits equation (11) anon-identically zerosolution. Clearly, in every case the results of the previous sections can be applied with the choice

fi,j(x, y)= γi,jf(x, y) (x∈X, y∈Y).

This means that the assumption that the function f is not identically zero will imply algebraic conditions for the involved parametersα₁, α₂, β₁, β₂∈Kandγ_1,1, γ_1,2, γ_2,1, γ_2,2 ∈K.

Similarly as before, first we consider the so-called degenerate cases.

5.1 Degenerate cases

5.1.1 The caseα₁ =α₂ =β₁ =β₂ =0

In caseα₁ =α₂ =β₁ =β₂= 0 equation (11) reduces to

γ1,1f(x1, y1)+γ1,2f(x1, y2)+γ2,1f(x2, y1)+γ2,2f(x2, y2)= 0 (x₁,x2 ∈X, y1, y2 ∈Y), whereγ_i,j ∈Kfor anyi, j∈ {1,2}.

Proposition 10. Let X and Y be linear spaces over the fieldK,γ_i,j ∈Kbe given constants such that not all of them are zero and f: X×Y →Kbe a function such that

γ_1,1f(x₁, y₁)+γ_1,2f(x₁, y₂)+γ_2,1f(x₂, y₁)+γ_2,2f(x₂, y₂)= 0 (x₁,x2 ∈X, y₁, y₂ ∈Y). (12) Then and only then there exist functionsχ: X→ Kandζ: Y →Kvanishing at zero such that

f(x,z)=χ(x)+ζ(z) (x∈X,z∈Y). Furthermore

(i) either the following system of linear equations

γ_1,1+γ_1,2 = 0 γ2,1+γ2,2 = 0 is fulfilled or the functionχis identically zero.

(ii) either the following system of linear equations

γ_1,1+γ_2,1 = 0 γ1,2+γ2,2 = 0 is fulfilled or the functionζ is identically zero.

Proof. In view of Proposition 2 we get that there exist functionsχ: X → Kandζ: Y → Kvanishing at zero such that

f(x,z)=χ(x)+ζ(z) (x∈X,z∈Y). Using this representation of the function f, equation (12) yields that

γ_1,1(χ(x₁)+ζ(y₁))+γ_1,2(χ(x₁)+ζ(y₂))+γ_2,1(χ((x₂)+ζ(y₁))+γ_2,2(χ(x₂)+ζ(y₂))=0

(x₁,x2∈X, y1, y2 ∈Y), or equivalently

χ(x₁) γ_1,1+γ_1,2+χ(x₂) γ_2,1+γ_2,2+ζ(y₁) γ_1,1+γ_2,1+ζ(y₂) γ_1,2+γ_2,2= 0

(x₁,x₂∈X, y₁, y₂ ∈Y).

Since we have independent variables we get that

γ1,1+γ1,2 = 0 γ2,1+γ2,2 = 0 is fulfilled or the functionχis identically zero. Similarly,

γ_1,1+γ_2,1 = 0 γ1,2+γ2,2 = 0

holds or the functionζ is identically zero.

5.1.2 The caseα₁ =α₂ =β₁ =0andβ₂ ,0 In such a situation (11) reduces to

f(0, β2y2)= γ1,1f(x1, y1)+γ1,2f(x1, y2)+γ2,1f(x2, y1)+γ2,2f(x2, y2) (x1,x2 ∈X, y1, y2 ∈Y). In view of Proposition 3, the proof of the following proposition is straightforward and similar to that of Proposition 10. The basic step is to consider f as the sum of single variable functions (Proposition 3) and substitute such a special form of f into the functional equation.

Proposition 11. Let X and Y be linear spaces over the fieldK,β2, γi,j ∈Kbe given constants such that not all of them are zero, and f: X×Y →Kbe a function such that

f(0, β2y2)= γ1,1f(x₁, y1)+γ1,2f(x₁, y2)+γ2,1f(x₂, y1)+γ2,2f(x₂, y2) (x₁,x2 ∈X, y1, y2 ∈Y). (13) Then and only then there exist functionsχ: X→ Kandζ: Y →Kvanishing at zero such that

f(x,z)=χ(x)+ζ(z) (x∈X,z∈Y). Furthermore

(i) either the following system of linear equations

γ1,1+γ1,2 = 0 γ2,1+γ2,2 = 0 is fulfilled or the functionχis identically zero.

(ii) either the following system of equations

γ_1,1+γ_2,1 = 0

ζ(β₂z) = γ1,2+γ2,2ζ(z) (z∈Y) is fulfilled or the functionζ is identically zero.

5.1.3 The caseα₁, α₂ ,0andβ₁, β₂ =0 In such a situation (11) reduces to

f(α₁x₁+α₂x₂,0)= γ1,1f(x₁, y₁)+γ1,2f(x₁, y₂)+γ2,1f(x₂, y₁)+γ2,2f(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂ ∈Y). As before, taking f as the sum of single variable functions (Proposition 4), substitute into the functional equation.

Proposition 12. Let X and Y be linear spaces over the field K, α₁, α₂, γi,j ∈ Kbe given constants such that not all of them are zero and f: X×Y → Kbe a function such that

f(α₁x₁+α₂x₂,0)=γ1,1f(x₁, y₁)+γ1,2f(x₁, y₂)+γ2,1f(x₂, y₁)+γ2,2f(x₂, y₂) (x₁,x₂∈X, y₁, y₂ ∈Y). (14) Then and only then there exists an additive function a: X → Kand a function ζ: Y → Kvanishing at zero such that

f(x, y)=a(x)+ζ(y) (x∈X, y∈Y). Furthermore the above additive function a has to fulfill

a(α1x1+α2x2)= (γ1,1+γ1,2)a(x1)+(γ2,1+γ2,2)a(x2)

for arbitrary x1,x2 ∈X and for the mappingζalternative (ii) of Proposition 10 is fulfilled.

5.1.4 The caseα₁, β₁ , 0andα₂, β₂ =0 In such a situation (11) reduces to

f(α₁x₁, β₁y₁)= γ1,1f(x₁, y₁)+γ1,2f(x₁, y₂)+γ2,1f(x₂, y₁)+γ2,2f(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂ ∈Y). To prove the following result, consider f as the sum of single variable functions (Proposition 5) and substitute into the functional equation.

Proposition 13. Let X and Y be linear spaces over the fieldK,α₁, α₂, γi,j ∈Kbe given constants and f: X× Y → Kbe a function such that

f(α₁x₁, β₁y₁)= γ1,1f(x₁, y₁)+γ1,2f(x₁, y₂)+γ2,1f(x₂, y₁)+γ2,2f(x₂, y₂) (x₁,x₂ ∈X, y₁, y₂ ∈Y). (15) Then and only then

(A) γ_1,2, γ_2,1, γ_2,2 =0and f: X×Y →Kis an arbitrary function fulfilling f(α1x, β1y)= γ1,1f(x, y) (x∈X, y∈Y), (B) or there exist functionsχ: X →Kandζ: Y →Kvanishing at zero such that

f(x, y)=χ(x)+ζ(y) (x∈X,z∈Y). Furthermore the mappingsχandζalso fulfill

χ(α1x) = (γ1,1+γ1,2)χ(x)

ζ(β₁z) = (γ_1,1+γ_2,1)ζ(z) (x∈X,z∈Y) and

(i) either

γ2,1+γ2,2 =0 orχis identically zero;

(ii) either

γ_1,2+γ_2,2 =0 orζis identically zero.

5.1.5 The caseα1, α2, β1 ,0andβ2 =0 In such a situation (11) reduces to

f(α1x1+α2x2, β1y1)=γ1,1f(x1, y1)+γ1,2f(x1, y2)+γ2,1f(x2, y1)+γ2,2f(x2, y2) (x₁,x2 ∈X, y1, y2 ∈Y). Proposition 14. Let X and Y be linear spaces over the fieldK,α₁, α₂, β₁, γi,j ∈K, i, j=1,2be given constants and f: X×Y →Kbe a function such that

f(α₁x₁+α₂x₂, β₁y₁)

= γ1,1f(x1, y1)+γ1,2f(x1, y2)+γ2,1f(x2, y1)+γ2,2f(x2, y2) (x₁,x2∈X, y1, y2 ∈Y). (16) Then and only then there exists a mapping A: X × Y → K additive in its first variable, further there are functionsχ: X→ Kandζ: Y → Kvanishing at zero such thatχis additive and

f(x, y)= A(x, y)+χ(x)+ζ(y) (x∈X, y∈Y). Furthermore, we have that

ζ(β₁y)=(γ1,1+γ2,1)ζ(y) (y∈Y) and also

γ1,2+γ2,2ζ(y)= 0 (y∈Y),

yielding thatγ1,2+γ2,2 =0orζis identically zero. Additionally, the alternatives below also hold