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ON THE SOLUTIONS OF SOME CLASSES OF FUNCTIONAL EQUATIONS WITH TRANSFORMED ARGUMENT NGUYEN VAN MAU Abstract We deal with some functional equations with transformed arguments in real plane

ON THE SOLUTIONS OF SOME CLASSES OF FUNCTIONAL EQUATIONS WITH TRANSFORMED ARGUMENT

NGUYEN VAN MAU Abstract

We deal with some functional equations with transformed arguments

in real plane By an algbraic approach we solve some kinds of functional equations with the reflection arguments

f (x, y)±f (2p−x, y)±f (x, 2q−y)+f (2p−x, 2q−y) = h(x, y), (x, y) ∈ Ω,

(0.1) where (p, q) is the symmetric center of the set Ω ⊂ R × R, h(x, y) is given

In applications, we formulate the necessary and sufficient condition for solvability of the following functional equations

f (xy)±f ((1−x)y±f (x(1−y))+f ((1−x)(1−y)) = h(xy), ∀x, y ∈ (0, 1) and

f (x + y) ± f (−x + y) ± f (x − y) + f (−x − y)) = h(x + y), ∀x, y ∈ (−1, 1) and describe the formulae of the general solution f (xy) and f (x + y), respectively

1

September 25, 2013

functions with reflection argument

In this section we will describe some classes of two-variable functions with transfromed argument Namely, we deal with two-variable functions being skew symmetric about a given point (p, q)

Definition 1.1 Let be given a set Ω := P × Q ⊂ R × R and the point (p, q)

is the center (centeral point) of Ω Function f (x, y) defined in Ω is said to be even-even (or even in both variables or skew symmetric) about the point (p, q) iff

f (2p − x, y) = f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω

Definition 1.2 Let be given a set Ω := P × Q ⊂ R × R and the point (p, q)

is the center (centeral point) of Ω Function f (x, y) defined in Ω is said to be even-odd about the point (p, q) iff

f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω

Remark 1 Similarly, we have the definitions of odd-even and odd-odd fun-tions

In a special case, we have

Definition 1.3 Function f (x, y) defined in R × R is said to be even-even iff

f (−x, y) = f (x, y) and f (x, −y) = f (x, y), ∀x, y ∈ R

The following natural questions arise:

∗

2000 Mathematics Subject Classification Primary 39B99, 39B62, 39B22, 39B32, 39B52.

Problem 1.1 How to describe the two-variable function f (x, y) in the cases

f (x, y) is even-even in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω (1.1) Solution Note that

f (2p − x, 2q − y) = f (x, 2q − y) = f (x, y), ∀x, y ∈ Ω,

so we can write

f (x, y) = 1

4[f (x, y) + f (x, 2q − y) + f (2p − x, y) + f (2p − x, 2q − y)] (1.2) Now we prove that the function f (x, y) is even in both variables (x, y) if and only if there exists a function g(x, y) defined in R × R such that

f (x, y) = 1

4[g(x, y) + g(x, 2q − y) + g(2p − x, y) + g(2p − x, 2q − y)]. (1.3) Indeed, if f (x, y) is of the form (1.3) then it is easy to check the conditions (1.1) are saistified and if f (x, y) is even then it has the form (1.2) and then the form (1.3) with g = f

Corollary 1.1 The two-variable function f (x, y) is even in both variables (x, y), i.e

f (−x, y) = f (x, y) and f (x, −y) = f (x, y), ∀(x, y) ∈ R (1.4) iff it is of the form

f (x, y) = 1

4[g(x, y) + g(x, −y) + g(−x, y) + g(−x, −y)]. (1.5) where g(x, y) is an arbitrary function defined in R × R

Similarly, we have

Problem 1.2 How to describe the two-variable function f (x, y) in the cases

f (x, y) is even-odd in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω (1.6) Solution Note that

f (2p − x, 2q − y) = f (x, 2q − y) = −f (x, y), ∀x, y ∈ Ω,

so we can write

f (x, y) = 1

4[f (x, y) + f (2p − x, y) − f (x, 2q − y) − f (2p − x, 2q − y)] (1.7) Now we prove that the function f (x, y) is even-odd in both variables (x, y) if and only if there exists a function g(x, y) defined in R × R such that

f (x, y) = 1

4[g(x, y) + g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y)]. (1.8) Indeed, if f (x, y) is of the form (1.8) then it is easy to check the conditions (1.6) are saistified and if f (x, y) is even-odd then it has the form (1.7) and then the form (1.8) with g = f

Similarly, we can formulate the following representations

Theorem 1.1 The two-variable function f (x, y) in the cases f (x, y) is even-even in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω (1.9)

is of the form

f (x, y) = 1

4[g(x, y) + g(2p − x, y) + g(x, 2q − y) + g(2p − x, 2q − y)] (1.10) Theorem 1.2 The two-variable function f (x, y) in the cases f (x, y) is even-odd in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω (1.11)

is of the form

f (x, y) = 1

4[g(x, y) + g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y)] (1.12) Theorem 1.3 The two-variable function f (x, y) in the cases f (x, y) is odd-even in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = −f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω (1.13)

is of the form

f (x, y) = 1

4[g(x, y) − g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y)] (1.14)

Theorem 1.4 The two-variable function f (x, y) in the cases f (x, y) is odd-ood in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = −f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω (1.15)

is of the form

f (x, y) = 1

4[g(x, y) − g(2p − x, y) − g(x, 2q − y) + g(2p − x, 2q − y)] (1.16) Now we consider the special case of two-variable function f (x, y) defined in the set Ω0 = (0, 1) × (0, 1) in the cases f (x, y) is even - even in both variables (x, y) about the point0,1

2

 So theorem 2.1 can be formulate in the following form

Corollary 1.2 The two-variable function f (x, y) in the case when f (x, y) is even-even in both variables (x, y) about the point 1

2,

1 2

 , i.e

f (1 − x, y) = f (x, y) and f (x, 1 − y) = f (x, y), ∀x, y ∈ (0, 1) (1.17)

is of the form

f (x, y) = 1

4[g(x, y) + g(1 − x, y) + g(x, 1 − y) + g(1 − x, 1 − y)]. (1.18) Corollary 1.3 The two-variable function f (x, y) in the case when f (x, y) is odd-odd in both variables (x, y) about the point 1

2,

1 2

 , i.e

f (1 − x, y) = −f (x, y) and f (x, 1 − y) = −f (x, y), ∀x, y ∈ (0, 1) (1.19)

is of the form

f (x, y) = 1

4[g(x, y) − g(1 − x, y) − g(x, 1 − y) + g(1 − x, 1 − y)]. (1.20)

fun-tions induced by involufun-tions

In this section we will solve the following functional equations

f (x, y) + f (2p − x, y) + f (x, 2q − y) + f (2p − x, 2q − y) = h(x, y), ∀(x, y) ∈ Ω

(2.1)

f (x, y) − f (2p − x, y) − f (x, 2q − y) + f (2p − x, 2q − y) = h(x, y), ∀(x, y) ∈ Ω,

(2.2) where (p, q) is the symmetric center of the set Ω ⊂ R × R, h(x, y) is given Denote by X the set of all functions defined on X and X = L0(X, X), where L0(X, X) denotes the linear space of all linear operators A : X → X with dom A = X It is easy to check that X is an algebra (linear ring) over field R

Consider the following linear elements (operators) V and W in X as follows (V f )(x, y) = f (2p − x, y), (W f )(x, y) = f (x, 2q − y), f ∈ X (2.3)

It is easy to see that V and W are involution elements, i.e V2 = I and

W2 = I, where I is an identity element of X Moreover, they are commutative, i.e V W = W V

Now we rewrite (2.1) in the form

Kf := (I + V + W + V W )f = h (2.4) Lemma 2.1 Operator K defined by (2.4) is an algebraic element with chara-teristic polynomial

Proof Note that the following identities hold

V K = K, W K = K, V W K = K

So K2 = 4K and K is not a scalar operator, which toghether imply PK(t) =

t2− 4t, which was to be proved

Theorem 2.1 The general solution of the homogeneous equation Kf = 0 is

of the form

f (x, y) = 1

4[3g(x, y)−g(2p−x, y)−g(x, 2q−y)−g(2p−x, 2q−y)], g ∈ X (2.6) Proof By Lemma 2.1, from equality (K − K)f = 0, ∀f ∈ X, we find 14K2− 4K



f = 0 ⇔ (K2 − 4K)f = 0 ⇔ K(K − 4I)f = 0 Hence (K − 4I)X ⊂ ker K On the other hand, if ϕ ∈ ker K then Kϕ = 0 and K(K − 4I)ϕ = (K − 4I)Kϕ = 0 It follows ϕ ∈ =(4I − K), which was to be proved

Theorem 2.2 The non-homogeneous equation (2.1) (Kf = h) is solvable if and only if the following condition

If it is the case, then the general solution of (2.1) is of the form

f (x, y) = 3g(x, y) − g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y) (2.8) +1

4[3h(x, y) − h(2p − x, y) − h(x, 2q − y) − h(2p − x, 2q − y)], g ∈ X. Proof Suppose that the equation (2.1) is solvable and f0 is a solution Then

By Lemma 2.1 from equality Kf0 = h it follows K2f0 = Kh ⇔ 4Kf0 = Kh ⇔ 4h = Kh

Suppose that the condition (2.7) is satisfied Write the non-homogeneous equation (2.1) in the form Kf = 14Kh or in the equivalent form

Kf −1

4h



Theorem 2.1 gives the general solution of (2.9) in the form

f = 1

4h + (4I − K)ψ ⇔ f =

1

4h + (4I − K)ψ, ψ ∈ X, i.e it has the form (2.8)

Now we consider (2.2) Write it in the form

Lf := (I − V − W + V W )f = h (2.10) Lemma 2.2 Operator L defined by (2.4) is an algebraic element with chara-teristic polynomial

Proof The proof follows from the following identities

−V L = L, −W L = L, V W L = L

So L2 = 4L and L is not a scalar operator, which toghether imply PL(t) =

t2− 4t, which was to be proved

Theorem 2.3 The general solution of the homogeneous equation Kf = 0 is

of the form

f (x, y) = 1

4[3g(x, y) + g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y)], g ∈ X.

(2.12)

Proof By the same method as for theorem2.1.

Theorem 2.4 The non-homogeneous equation (2.1) (Lf = h) is solvable if and only if the following condition

If it is the case, then the general solution of (2.1) is of the form

f (x, y) = 3g(x, y) + g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y) (2.14) +1

4[3h(x, y) + h(2p − x, y) + h(x, 2q − y) − h(2p − x, 2q − y)], g ∈ X. Proof Suppose that the equation (2.2) is solvable and f0 is a solution Then

By Lemma 2.2 from equality Lf0 = h it follows L2f0 = Lh ⇔ 4Lf0 = Lh ⇔ 4h = Lh

Suppose that the condition (2.13) is satisfied Write the non-homogeneous equation (2.2) in the form Lf = 14Lh or in the equivalent form

Lf −1

4h



Theorem 2.1 gives the general solution of (2.15) in the form

f = 1

4h + (4I − L)ψ ⇔ f =

1

4h + (4I − L)ψ, ψ ∈ X, i.e it has the form (2.14)

Now we consider some special cases of equation when q = p = 1

2 In that case, the center point of Ω is 1

2,

1 2

 and (2.1) is of the form

f (x, y)+f (1−x, y)+f (x, 1−y)+f (1−x, 1−y) = h(x, y), ∀x, y ∈ (0, 1) (3.1) and

f (x, y)−f (1−x, y)−f (x, 1−y)+f (1−x, 1−y) = h(x, y), ∀x, y ∈ (0, 1) (3.2)

In this case, the role of x and y in the left side of (3.1) are the same

Now return to the function f (t), we can formulate the following

Theorem 3.1 The function f (t) satisfying the conditions

f ((1 − x)y) = f (xy), ∀x, y ∈ (0, 1) (3.3)

if and only if there exists a function g(t) such that

f (xy) = 1

4[g(xy) + g((1 − x)y) + g(x(1 − y)) + g((1 − x)(1 − y))]. (3.4) Proof From (3.7) we find f (x(1 − y)) = f (xy) and

f ((1 − x)(1 − y)) = f (x(1 − y)) = f (xy), ∀x, y ∈ (0, 1)

So we can write f (t) in the form

f (xy) = 1

4[f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y))]. Last equality gives us the proof of the theorem

Theorem 3.2 The functional equation

f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y))] = h(xy) (3.5)

is solvable if and only if there exists h(t) satisfying the following condition

h(x(1 − y)) = h(xy), ∀x, y ∈ (0, 1) (3.6)

So now we shall examine the equation (3.6)

Putting x = t, y = 1

2t, where t ∈

1

2, 1

 into (3.6), we get h



t − 1 2



= h

1 2

 , ∀t ∈

1

2, 1

 or

h(x) = h1

2

 , ∀x ∈0,1

2



Similarly, putting xy = t, then x(1 − y) = t1 − y

y and

h(t) = h1 − y

y t



(3.7) and (3.8) together inply h(t) ≡ const

So, the necessary condition for solvability of (3.1) if

h(x, y) = h(y, x), ∀x, y ∈ (0, 1) (3.9) Now we formulate the similar results as in the previous section

Theorem 3.3 The general solution of the homogeneous equation

f (x, y) + f (1 − x, y) + f (x, 1 − y) + f (1 − x, 1 − y) = 0, ∀x, y ∈ (0, 1) (3.10)

is of the form

f (x, y) = 1

4[3g(x, y) − g(1 − x, y) − g(x, 1 − y) − g(1 − x, 1 − y)], g ∈ X (3.11) Theorem 3.4 The non-homogeneous equation

f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y)) = h(xy), ∀x, y ∈ (0, 1)

(3.12)

is solvable iff h(t) ≡ const in (0, 1) If it is the case, then the general solution

of (3.14) is of the form

f (xy) = 1

4[c+3g(xy)−g((1−x)y)−g(x(1−y))−g((1−x)(1−y))], c ∈ R, g ∈ X

(3.13) Proof

Theorem 3.5 The function f (t) is a general solution of the homogeneous equation

f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y)) = 0, ∀x, y ∈ (0, 1) (3.14)

if and only if there exists a function g(t) such that

f (xy) = 1

4[3g(xy)−g((1−x)y)−g(x(1−y))−g((1−x)(1−y))], g ∈ X (3.15) Remark 2 The homogeneous equation (3.14) was posed by Sahoo and Sander

in 1990 (see [3]-[4]) The continuous solutions were found by Z Daroczy and

A Jarai in [5]

Now we deal with the equation induced by addition of arguments We consider the special cases of equation when q = p = 0 In that case, the center point of Ω is (−1, 1) and (2.1) is of the form

f (x, y) + f (−x, y) + f (x, −y) + f (−x, −y) = h(x, y), ∀x, y ∈ (−1, 1) (3.16) Theorem 3.6 The function f (t) is a general solution of the homogeneous equation

f (x + y) + f (−x + y) + f (x − y) + f (−x − y)) = 0, ∀x, y ∈ (−1, 1) (3.17)

if and only if f (x) is an odd function in (−2, 2)

Proof If f is odd in (−2, 2), then f (x − y) = −f (−x + y) and f (−x − y) =

−f (x + y) Hence,

f (x + y) + f (−x + y) + f (x − y) + f (−x − y)) = 0, ∀x, y ∈ (−1, 1) Conversely, suppose f is a solution of (3.17) Putting x = 0, y = 0 into (3.17)

we find f (0) = 0 Similarly, putting y = x, into (3.15) and ussing the equality

f (0) = 0, we find f (−2x) = −f (2x), i.e f is odd in (−2, 2)

Theorem 3.7 The non-homogeneous equation

f (x+y)+f (−x+y)+f (x−y)+f (−x−y)) = h(x+y), ∀x, y ∈ (−1, 1) (3.18)

is solvable iff h(t) ≡ const in (−2, 2) If it is the case, then the general solution

of (3.14) is of the form

f (t) = 1

where g is an arbitrary odd function in (−2, 2)

Proof Suppose equation (3.18) is solvable and f is its solution Putting y = x and y = −x into (3.18), we find h(2x) = h(0), x ∈ (−1, 1), i.e h(t) ≡ h(0) in (−2, 2)

If h(t) ≡ h(0) in (−2, 2) then we can reduce equation (3.18) to the equation ϕ(x + y) + ϕ(−x + y) + ϕ(x − y) + ϕ(−x − y)) = 0, ∀x, y ∈ (−1, 1), (3.20)

where ϕ(t) = f (x)−1

4 Hence, the solution (3.19) follows from theorem 3.6. Now we return to the equation

f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y)) = 0 (3.21)

By the same way as previous equations, we have

Theorem 3.8 The function f (t) satisfying the conditions

f ((1 − x)y) = −f (xy), ∀x, y ∈ (0, 1) (3.22)

if and only if there exists a function g(t) such that

f (xy) = 1

4[g(xy) − g((1 − x)y) − g(x(1 − y)) + g((1 − x)(1 − y))]. (3.23)

Proof From (3.23) we find f (x(1 − y)) = −f (xy) and

f ((1 − x)(1 − y)) = −f (x(1 − y)) = f (xy), ∀x, y ∈ (0, 1)

So we can write f (t) in the form

f (xy) = 1

4[f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y))]. Last equality gives us the proof of the theorem

Theorem 3.9 The functional equation

f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y))] = h(xy) (3.24)

is solvable if and only if there exists h(t) ≡ 0

Theorem 3.10 The non-homogeneous equation

f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y)) = h(xy), ∀x, y ∈ (0, 1)

(3.25)

is solvable iff h(t) ≡ 0 in (0, 1) If it is the case, then the general solution of (3.14) is of the form

f (xy) = 1

4[3g(xy)+g((1−x)y)+g(x(1−y))−g((1−x)(1−y))], g ∈ X (3.26) Theorem 3.11 The function f (t) is a general solution of the homogeneous equation

f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y)) = 0, ∀x, y ∈ (0, 1) (3.27)

if and only if there exists a function g(t) such that

f (xy) = 1

4[3g(xy)+g((1−x)y)+g(x(1−y))−g((1−x)(1−y))], g ∈ X (3.28) Remark 3 The equation (3.27) was posed firstly by K Lajko in [8] for X = R and then by Sahoo and Sander in 1990 (see [3]-[4]) The differentiable solutions were found by C.J Eliezer in [6]

Theorem 3.12 The non-homogeneous equation

f (x+y)−f (−x+y)−f (x−y)+f (−x−y)) = h(x+y), ∀x, y ∈ (−1, 1) (3.29)

is solvable iff h(t) ≡ const in (−2, 2) If it is the case, then the general solution

of (3.27) is of the form

f (t) = c + 1

2(g(t) − g(−t)), t ∈ (−2, 2), (3.30) where g is an arbitrary function in (−2, 2), c = f (0)