Conditional Equations Related to Drygas Functional Equations

Sadegh Izadi; Sedigheh Jahedi; Mehdi Dehghanian

doi:10.2478/amsil-2025-0015

1. Preliminaries

Recall that a function A : ℝ → ℝ is additive if the equation A(x + y) = A(x) + A(y) holds for all x, y ∈ ℝ.

Kuczma [13] proved that any additive function A : ℝ → ℝ is ℚ-homogeneous, that is,

A (sx) = sA (x),

for all x ∈ ℝ and s ∈ ℚ. A function h : ℝ → ℝ is called quadratic if the equation

h (x + y) + h (x - y) = 2 h (x) + 2 h (y)

holds for all x, y ∈ ℝ.

For instance, consider the additive functions A₁, A₂ : ℝ → ℝ. It is easy to see that A₁(x)A₂(x) and A₁(x²), x ∈ ℝ, are quadratic.

A function B : ℝ × ℝ → ℝ is named symmetric biadditive if B is additive in each variable and satisfies B(x, y) = B(y, x) for all x, y ∈ ℝ.

In 1965, Aczél [1] showed that a quadratic function h : ℝ → ℝ can be associated with a symmetric biadditive function B : ℝ × ℝ → ℝ given by the following formula

(1.1)

\begin{matrix} B (x, y) = \frac{1}{2} [h (x + y) - h (x) - h (y)], & x, y \in ℝ . \end{matrix}

Aczél and Dhombres [2] proved that the function h : ℝ → ℝ is quadratic if and only if, there is a symmetric biadditive function B : ℝ × ℝ → ℝ such that h(x) = B(x, x) for all x ∈ ℝ. This B is unique (see [2]). Moreover, the ℚ-homogeneity of biadditive functions yields

\begin{matrix} B (rx, sy) = rsB (x, y), & h (rx) = B (rx, rx) = r^{2} h (x), \end{matrix}

for all x, y ∈ ℝ and r, s ∈ ℚ. By using (1.1) and induction on n, one can show that

h (\sum_{i = 0}^{n} ω_{i}) = \sum_{i = 0}^{n} h (ω_{i}) + 2 \sum_{0 \leq j < k \leq n} B (ω_{j}, ω_{k}),

for all n ∈ ℕ and ω₀, . . . , ω_n ∈ ℝ.

Some mathematicians have investigated additive functions A that satisfy the conditional equation yA(x) = xA(y) for the pairs (x, y) ∈ ℝ² under the condition P (x, y) = 0 for some fixed polynomial P of two variables. For some special polynomials P this assumption implies that A is continuous (see for example [4 12, 14, 15]).

Recently, Z. Boros and E. Garda-Mátyás [5] and [6], E. Garda-Mátyás [11] studied quadratic functions h : ℝ → ℝ that satisfy the additional condition

y^{2} h (x) = x^{2} h (y),

where (x, y) are arbitrary points on a specified curve.

J. Brzdęk and A. Mureńko [7] established the Gołąb-Schinzel equation under certain additional conditions.

The functional equation

(1.2)

f (x + y) + f (x - y) = 2 f (x) + f (y) + f (- y),

which was considered by Drygas [9] is known as the Drygas equation and its solutions as Drygas functions. It is a generalization of the quadratic functional equation. In [10], Ebanks et al. obtained the general solution of the Drygas functional equation as

(1.3)

\begin{matrix} f (x) = A (x) + B (x, x), & x \in ℝ, \end{matrix}

where A : ℝ → ℝ is an additive function and B : ℝ × ℝ → ℝ is a symmetric biadditive function. The continuous solutions of functional equation (1.2) on ℝ are of the form f (x) = αx + βx², where α, β ∈ ℝ are constants (see [16]).

Consider the sets

\begin{array}{l} \begin{array}{l} Δ_{0} = {(x, y) \in ℝ^{2} : x > 0 and y = x^{n}}, & n \in ℤ, | n | \geq 2, \end{array} \\ Δ_{1} = {(x, y) \in ℝ^{2} : y = x^{2}}, \\ Δ_{2} = {(x, y) \in ℝ^{2} : x > 0 and y = log (x)}, \\ Δ_{3} = {(x, y) \in ℝ^{2} : y = exp (x)} . \end{array}

Motivated by the results of [5], this paper is devoted to finding Drygas functions f₁, f₂ : ℝ → ℝ satisfying the equation

(1.4)

(y^{2} + y) f_{1} (x) = (x^{2} + x) f_{2} (y),

for the pairs (x, y) ∈ Δ_j, where j = 0, 1, 2, 3.

M. Dehghanian et al. [8] investigated Drygas functions f : ℝ → ℝ that satisfy the conditional equation (1.4) on the graph of a power function.

Lemma 1.1.

[5] Let m ∈ ℕ and 𝔽 be a field. Suppose that Ω is a set, Γ ⊂ 𝔽 contains at least m + 1 elements, and the functions Λ_j : Ω → 𝔽 (j = 0, 1, . . . , m) satisfy

\sum_{j = 0}^{m} Λ_{j} (x) s^{j} = 0,

for all x ∈ Ω and s ∈ Γ. Then Λ_j(x) = 0 for all x ∈ Ω and 0 ≤ j ≤ m.

This paper contains results for the Drygas functions that satisfy the equation (1.4) for (x, y) ∈ Δ_j, where j = 0, 1, 2, 3.

2. Main results

In the following theorem, we apply Lemma 1.1 with Ω = ℝ₊, 𝔽 = ℝ and Γ = ℚ₊, where ℝ₊ and ℚ₊ are the sets of positive real and positive rational numbers, respectively.

Theorem 2.1

Suppose that f : ℝ → ℝ is a Drygas function. Then f fulfills the conditional equation

(2.1)

(x^{2} + x) f (y) = (y^{2} + y) f (x),

for (x, y) ∈ Δ₀ if and only if

\begin{matrix} f (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}

where α is a real constant.

Proof

First, assume that f fulfills (2.1), x ∈ ℝ₊ and n ≥ 2. In this case, the equation (2.1) becomes

(2.2)

\begin{matrix} (x + 1) f (x^{n}) = (x^{2 n - 1} + x^{n - 1}) f (x), & x \in ℝ_{+} . \end{matrix}

Substituting x + s, s ∈ ℚ₊, for x in (2.2), we get

\begin{matrix} (x + s + 1) f ({(x + s)}^{n}) = ({(x + s)}^{2 n - 1} + {(x + s)}^{n - 1}) f (x + s), & x \in ℝ_{+} . \end{matrix}

By expanding the binomial terms, we obtain

(2.3)

(x + s + 1) f (\sum_{m = 0}^{n} (\begin{matrix} n \\ m \end{matrix}) x^{m} s^{n - m}) = \sum_{l = 0}^{2 n - 1} (\begin{matrix} 2 n - 1 \\ l \end{matrix}) x^{l} s^{2 n - l - 1} + \sum_{k = 0}^{n - 1} (\begin{matrix} n - 1 \\ k \end{matrix}) x^{k} s^{n - k - 1}] f (x + s) .

By (1.3), there exist an additive function A : ℝ → ℝ and a symmetric biadditive function B : ℝ × ℝ → ℝ such that f (x) = A(x) + B(x, x) for all x ∈ ℝ. Thus, the equation (2.3) takes the form

(2.4)

\begin{array}{l} (x + s + 1) \sum_{m = 0}^{n} ((\begin{matrix} n \\ m \end{matrix}) s^{n - m} A (x^{m}) + {(\begin{matrix} n \\ m \end{matrix})}^{2} s^{2 n - 2 m} B (x^{m}, x^{m})) + 2 \sum_{0 \leq i < j \underline{<} n} (\begin{matrix} n \\ i \end{matrix}) (\begin{matrix} n \\ j \end{matrix}) s^{2 n - i - j} B (x^{i}, x^{j})] \\ - \sum_{k = 0}^{n - 1} (\begin{matrix} n - 1 \\ k \end{matrix}) x^{k} s^{n - k - 1} (f (x) + sA (1) + s^{2} B (1, 1) + 2 sB (x, 1)) \\ - \sum_{l = 0}^{2 n - 1} (\begin{matrix} 2 n - 1 \\ l \end{matrix}) x^{l} s^{2 n - l - 1} (f (x) + sA (1) + s^{2} B (1, 1) + 2 sB (x, 1)) = 0. \end{array}

Hence, for any fixed x ∈ ℝ₊, we obtain the polynomial (2.4) of degree at most 2n + 1 in ℝ₊ which is equal to zero for all s ∈ ℚ₊. By Lemma 1.1, every coefficient of (2.4) has to be equal to zero. Since the coefficient of s²ⁿ⁺¹ is B(1, 1) − B(1, 1) = 0, then the degree of the polynomial (2.4) is less than 2n + 1. Furthermore, from the coefficient of s²ⁿ, we deduce that

A (1) + 2 B (x, 1) + (2 n - 1) xB (1, 1) - xB (1, 1) - B (1, 1) - 2 nB (x, 1) = 0,

for all x ∈ ℝ₊. Put x = 1 in the above equality, we have

A (1) = B (1, 1) = \frac{f (1)}{2}

. Therefore, as |n| ≥ 2, we get

(2.5)

\begin{matrix} B (x, 1) = xB (1, 1), & x \in ℝ_{+} . \end{matrix}

From the coefficient of s²ⁿ⁻¹, we arrive at

\begin{matrix} 0 = f (x) + (2 n - 1) x [A (1) + 2 B (x, 1)] + (\begin{matrix} 2 n - 1 \\ 2 \end{matrix}) x^{2} B (1, 1) \\ - {(\begin{matrix} n \\ 1 \end{matrix})}^{2} B (x, x) - 2 n (x + 1) B (x, 1) - 2 (\begin{matrix} n \\ 2 \end{matrix}) B (x^{2}, 1) \\ = f (x) + (2 n - 1) (1 + 2 x) xB (1, 1) + (2 n - 1) (n - 1) x^{2} B (1, 1) \\ - n^{2} B (x, x) - 2 n x^{2} (x + 1) B (1, 1) - n (n - 1) x^{2} B (1, 1) . \end{matrix}

Now with A(1) = B(1, 1), and (2.5), we obtain

(2.6)

f (x) = (x^{2} + x) B (1, 1) + n^{2} B (x, x) - n^{2} x^{2} B (1, 1),

for all x ∈ ℝ₊. Thus,

\begin{matrix} B (x, x) = \frac{f (x) + f (- x)}{2} = x^{2} B (1, 1) + n^{2} B (x, x) - x^{2} B (1, 1), & x \in ℝ_{+} . \end{matrix}

As n ≥ 2,

(2.7)

\begin{matrix} B (x, x) = x^{2} B (1, 1) = \frac{f (1)}{2} x^{2}, & x \in ℝ_{+} . \end{matrix}

By equations (2.6) and (2.7), we conclude that

\begin{matrix} f (x) = \frac{f (1)}{2} (x + x^{2}), & x \in ℝ_{+} . \end{matrix}

Hence,

A (x) = f (x) - B (x, x) = \frac{f (1)}{2} x

for all x ∈ ℝ₊. Also, for x = 0 above equation holds, because f (0) = 0.

Now, for x = −u < 0,

\begin{array}{l} f (x) & = A (- u) + B (- u, - u) = - A (u) + {(- 1)}^{2} B (u, u) \\ = \frac{f (1)}{2} (- u + {(- u)}^{2}) = \frac{f (1)}{2} (x + x^{2}) . \end{array}

Therefore,

\begin{matrix} f (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}

where

α = \frac{f (1)}{2}

Finally, for the case n ≤ −2, take p = −n ≥ 2 in (2.2) to obtain

(2.8)

(x + 1) f (\frac{1}{x^{p}}) = (\frac{1}{x^{2 p + 1}} + \frac{1}{x^{p + 1}}) f (x) = (\frac{1 + x^{p}}{x^{2 p + 1}}) f (x),

for x ∈ ℝ₊. Substitute x^−p for x in (2.8) to gain

(\frac{1}{x^{p}} + 1) f (x^{p^{2}}) = (x^{2 p^{2} + p} + x^{p^{2} + p}) f (\frac{1}{x^{p}}) .

By (2.8), we obtain

\frac{1 + x^{p}}{x^{p}} f (x^{p^{2}}) = \frac{x^{2 p^{2} + p} + x^{p^{2} + p}}{x + 1} (\frac{1 + x^{p}}{x^{2 p + 1}}) f (x),

(2.9)

(x + 1) f (x^{p^{2}}) = (x^{2 p^{2} - 1} + x^{p^{2} - 1}) f (x) .

In (2.9), set p² = k ∈ ℕ, and use a similar proof as in the previous case.

Obviously, the converse holds.

The additive function θ : ℝ → ℝ is named a derivation if θ(xy) = xθ(y) + yθ(x) for all x, y ∈ ℝ. Thus, every derivation θ satisfies θ(x²) = 2xθ(x) for all x ∈ ℝ. Moreover, there exist nontrivial derivations on ℝ (see [13, Theorem 14.2.2]). Also, θ(x²) and (θ(x))² are quadratic functions (see [3]).

A functional ℋ: ℝ² → ℝ is named a bi-derivation if the mappings

\begin{matrix} s \mapsto ℋ (s, x) & and & s \mapsto ℋ (x, s), & s \in ℝ, \end{matrix}

are derivations for every x ∈ ℝ.

The set of derivations of order 2, denoted by 𝒟₂(ℝ), is the set of the additive functions θ : ℝ → ℝ that can be written as

θ (xy) - x θ (y) - θ (x) y = H (x, y),

for some bi-derivation ℋ on ℝ².

In the case n = 1, condition (2.1) has the form

(x + x^{2}) f (x) = (x + x^{2}) f (x),

whence f can be discontinuous as well.

Equation (2.1) for pairs of (x, y) ∈ ℝ² that fulfill condition xy = 1 is as follows

(2.10)

\begin{matrix} f (x) = x^{3} f (\frac{1}{x}), & x \in ℝ \ {0} . \end{matrix}

Now, by giving a counterexample, we show that there exists a discontinuous Drygas function that satisfies condition (2.10).

Assume that θ : ℝ → ℝ is a nontrivial derivation. Then

\begin{matrix} θ (\frac{1}{x}) = - \frac{1}{x^{2}} θ (x), & x \in ℝ \ {0} . \end{matrix}

Therefore,

f (x) = - θ (x) + \frac{1}{2} θ (x^{2})

is a discontinuous Drygas function that fulfills (2.10) for every x ∈ ℝ.

Lemma 2.2 ([5])

Assume that δ : ℝ → ℝ is an additive function. Then δ ∈ 𝒟₂(ℝ) if and only if

δ (x^{4}) = 6 x^{2} δ (x^{2}) - 8 x^{3} δ (x),

for every x ∈ ℝ.

Theorem 2.3

Drygas functions f₁, f₂ : ℝ → ℝ fulfill the condition (1.4) for (x, y) ∈ Δ₁ if and only if there exists an additive function δ : ℝ → ℝ such that

\begin{array}{l} δ (x^{4}) = 6 x^{2} δ (x^{2}) - 8 x^{3} δ (x) + 3 x^{4} δ (1), \\ f_{1} (x) = (x + 1) 2 δ (x) - x δ (1)], \\ \begin{array}{l} f_{2} (x) = \frac{1}{4} (x + 1) 6 δ (x) - \frac{1}{x} δ (x^{2}) - x δ (1)], & x \in ℝ \ {0} \\ f_{2} (0) = 0. \end{array} \end{array}

In particular, f₁(1) = 0 if and only if δ ∈ 𝒟₂(ℝ).

Proof

Since f₁, f₂ : ℝ → ℝ are Drygas functions, by (1.3), there exist additive functions A₁, A₂ : ℝ → ℝ and symmetric biadditive functions B₁, B₂ : ℝ × ℝ → ℝ such that

\begin{matrix} f_{1} (x) = A_{1} (x) + B_{1} (x, x) & and & f_{2} (x) = A_{2} (x) + B_{2} (x, x) \end{matrix},

for all x ∈ ℝ. Put y = x² in (1.4), to obtain

\begin{matrix} (x^{2} + x^{4}) f_{1} (x) = (x + x^{2}) f_{2} (x^{2}), & x \in ℝ . \end{matrix}

By dividing both sides by x ≠ 0 (since f₁(0) = f₂(0) = 0), we have

(2.11)

\begin{matrix} (x + 1) f_{2} (x^{2}) = (x^{3} + x) f_{1} (x), & x \in ℝ . \end{matrix}

Set x = −1 in (2.11), then f₁(−1) = −A₁(1) + B₁(1, 1) = 0. Thus,

A_{1} (1) = B_{1} (1, 1) = \frac{f_{1} (1)}{2} .

Let s ∈ ℚ. Substituting x + s for x in (2.11), we get

(2.12)

\begin{matrix} (x + s + 1) f_{2} ({(x + s)}^{2}) = ({(x + s)}^{3} + x + s) f_{1} (x + s), & x \in ℝ . \end{matrix}

By expanding the powers of sums on both side of this equation and by the ℚ-homogeneity of A₁, A₂, B₁ and B₂, equation (2.12) becomes

\begin{array}{l} x A_{2} (x^{2}) + s A_{2} (x^{2}) + A_{2} (x^{2}) + 2 s x A_{2} (x) + 2 s^{2} A_{2} (x) + 2 s A_{2} (x) \\ + s^{2} x A_{2} (1) + s^{3} A_{2} (1) + s^{2} A_{2} (1) + x B_{2} (x^{2}, x^{2}) + s B_{2} (x^{2}, x^{2}) \\ + B_{2} (x^{2}, x^{2}) + 4 s^{2} x B_{2} (x, x) + 4 s^{3} B_{2} (x, x) + 4 s^{2} B_{2} (x, x) + s^{4} x B_{2} (1, 1) \\ + s^{5} B_{2} (1, 1) + s^{4} B_{2} (1, 1) + 4 s x B_{2} (x^{2}, x) + 4 s^{2} B_{2} (x^{2}, x) + 4 s B_{2} (x^{2}, x) \\ + 2 s^{2} x B_{2} (v) + 2 s^{3} B_{2} (x^{2}, 1) + 2 s^{2} B_{2} (x^{2}, 1) + 4 s^{3} x B_{2} (x, 1) \\ + 4 s^{4} B_{2} (x, 1) + 4 s^{3} B_{2} (x, 1) \\ = x^{3} A_{1} (x) + 3 s x^{2} A_{1} (x) + 3 s^{2} x A_{1} (x) + s^{3} A_{1} (x) + x A_{1} (x) + s A (x) \\ + s x^{3} A_{1} (1) + 3 s^{2} x^{2} A_{1} (1) + 3 s^{3} x A_{1} (1) + s^{4} A_{1} (1) + s x A_{1} (1) + s^{2} A (1) \\ + x^{3} B_{1} (x, x) + 3 s x^{2} B_{1} (x, x) + 3 s^{2} x B_{1} (x, x) + s^{3} B_{1} (x, x) + x B_{1} (x, x) \\ + s B_{1} (x, x) + s^{2} x^{3} B_{1} (1, 1) + 3 s^{3} x^{2} B_{1} (1, 1) + 3 s^{4} x B_{1} (1, 1) + s^{5} B_{1} (1, 1) \\ + s^{2} x B_{1} (1, 1) + s^{3} B_{1} (1, 1) + 2 s x^{3} B_{1} (x, 1) + 6 s^{2} x^{2} B_{1} (x, 1) + 6 s^{3} x B_{1} (x, 1) \\ + 2 s^{4} B_{1} (x, 1) + 2 s x B_{1} (x, 1) + 2 s^{2} B_{1} (x, 1), \end{array}

for all x ∈ ℝ. Hence,

\begin{array}{l} 0 = & B_{1} (1, 1) - B_{2} (1, 1)] s^{5} \\ + A_{1} (1) + 3 x B_{1} (1, 1) + 2 B_{1} (x, 1) - x B_{2} (1, 1) + - 4 B_{2} (x, 1) - B_{2} (1, 1)] s^{4} \\ + [f_{1} (x) + 3 x A_{1} (1) + 6 x B_{1} (x, 1) + B_{1} (1, 1) + 3 x^{2} B_{1} (1, 1) \\ - 4 x B_{2} (x, 1) - A_{2} (1) - 4 B_{2} (x, x) - 2 B_{2} (x^{2}, 1) - 4 B_{2} (x, 1)] s^{3} \\ + [3 x f_{1} (x) + 3 x^{2} A_{1} (1) + A_{1} (1) + x^{3} B_{1} (1, 1) + x B_{1} (1, 1) \\ + 6 x^{2} B_{1} (x, 1) + 2 B_{1} (x, 1) - A_{2} (1) - 2 A_{2} (x) - x A (1) \\ - 4 x B_{2} (x, x) - 4 B_{2} (x, x) - 4 B_{2} (x^{2}, x) - 2 x B_{2} (x^{2}, 1) - 2 B_{2} (x^{2}, 1)] s^{2} \\ + [3 x^{2} f_{1} (x) + f_{1} (x) + x^{3} A_{1} (1) + x A_{1} (1) + 2 x^{3} B_{1} (x, 1) + 2 x B_{1} (x, 1) \\ - f_{2} (x^{2}) - 2 x A_{2} (x) - 2 A_{2} (x) - 4 x B_{2} (x^{2}, x) - 4 B_{2} (x^{2}, x)] s \\ + x^{3} f_{1} (x) + x f_{1} (x) - x f_{2} (x^{2}) - f_{2} (x^{2})] . \end{array}

By Lemma 1.1, the coefficients of sⁿ for n = 0, 1, 2, 3, 4, 5 are equal to zero. The coefficient of s⁵ implies B₁(1, 1) = B₂(1, 1). So, by taking x = 1 in (2.11), we obtain

A_{1} (1) = B_{1} (1, 1) = A_{2} (1) = B_{2} (1, 1) = \frac{f_{1} (1)}{2} .

According to the coefficient of s⁴ we see that

(2.13)

\begin{matrix} 2 B_{2} (x, 1) = x B_{1} (1, 1) + B_{1} (x, 1), & x \in ℝ . \end{matrix}

From the coefficient of s³ and (2.13), we conclude that

(2.14)

f_{1} (x) = 2 B_{1} (x, 1) - x B_{1} (1, 1) - 4 x B_{1} (x, 1) + B_{1} (x^{2}, 1) + 4 B_{2} (x, x),

for all x ∈ ℝ. Hence, by (2.14),

(2.15)

A_{1} (x) = \frac{f_{1} (x) - f_{1} (- x)}{2} = 2 B_{1} (x, 1) - x B_{1} (1, 1),

and

(2.16)

B_{1} (x, x) = \frac{f_{1} (x) + f_{1} (- x)}{2} = 4 B_{2} (x, x) + B_{1} (x^{2}, 1) - 4 x B_{1} (x, 1),

for all x ∈ ℝ.

Replacing x with −x in (2.11) yields

(2.17)

\begin{matrix} (- x + 1) f_{2} (x^{2}) = - (x^{3} + x) f_{1} (- x), & x \in ℝ . \end{matrix}

Adding both sides of (2.11) and (2.17) gives us

\begin{matrix} f_{2} (x^{2}) = (x^{3} + x) A_{1} (x), & x \in ℝ, \end{matrix}

and hence,

(2.18)

\begin{matrix} f_{1} (x) = (x + 1) A_{1} (x) = A_{1} (x) + B_{1} (x, x), & B_{1} (x, x) = x A_{1} (x) . \end{matrix}

for all x ∈ ℝ. Thus,

(2.19)

\begin{matrix} f_{2} (x^{2}) = (x^{2} + 1) B_{1} (x, x), & x \in ℝ . \end{matrix}

From (2.15) and (2.18), we have

(2.20)

\begin{matrix} B_{1} (x, x) = 2 x B_{1} (x, 1) - x^{2} B_{1} (1, 1), & x \in ℝ . \end{matrix}

Combining (2.16) and (2.20) yields

(2.21)

B_{2} (x, x) = \frac{3}{2} x B_{1} (x, 1) - \frac{1}{4} B_{1} (x^{2}, 1) - \frac{1}{4} x^{2} B_{1} (1, 1),

for all x ∈ ℝ. For x ∈ ℝ and s ∈ ℚ, if we write sx instead of x in equation (2.19), then

\begin{matrix} f_{2} (s^{2} x^{2}) = (s^{2} x^{2} + 1) B_{1} (sx, sx), & x \in ℝ . \end{matrix}

Thus,

A_{2} (x^{2}) - B_{1} (x, x)] s^{2} + B_{2} (x^{2}, x^{2}) - x^{2} B_{1} (x, x)] s^{4} = 0.

From Lemma 1.1 we have

A_{2} (x^{2}) = B_{1} (x, x), B_{2} (x^{2}, x^{2}) = x^{2} B_{1} (x, x) .

So, B₂ x², x² = x²A₂ x² for all x ∈ ℝ. Setting x² = t, we have t > 0 and B₂(t, t) = tA₂(t). It follows that B₂(x, x) = xA₂(x) for all x > 0.

Now, for x = −t < 0,

B_{2} (x, x) = B_{2} (- t, - t) = B_{2} (t, t) = t A_{2} (t) = - t A_{2} (- t) = x A_{2} (x) .

Therefore,

A_{2} (x) = \frac{1}{x} B_{2} (x, x)

for all x ∈ ℝ\{0}.

From the above equality and (2.21), we obtain

A_{2} (x) = \frac{3}{2} B_{1} (x, 1) - \frac{1}{4 x} B_{1} (x^{2}, 1) - \frac{1}{4} x B_{1} (1, 1) .

Define the additive function δ : ℝ → ℝ by

\begin{matrix} δ (x) = B_{1} (x, 1), & x \in ℝ . \end{matrix}

Therefore,

f_{1} (x) = (x + 1) 2 δ (x) - x δ (1)],

and

f_{2} (x) = \frac{1}{4} (x + 1) 6 δ (x) - \frac{1}{x} δ (x^{2}) - x δ (1)],

for all x ∈ ℝ\{0}.

Next, f₁(1) = 0 if and only if δ(1) = 0, or equivalently, if and only if

δ (x^{4}) = 6 x^{2} δ (x^{2}) - 8 x^{3} δ (x),

for all x ∈ ℝ. By Lemma 2.2, this is equivalent to δ ∈ 𝒟₂(ℝ).

The only if part is trivial.

In Theorem 2.3, if we suppose that δ is a derivation, then f₁(x) = 2f₂(x) for all x ∈ ℝ.

Example 2.4

Let 0 ≠ a ∈ ℝ. Define f₁, f₂ : ℝ → ℝ by

\begin{matrix} f_{1} (x) = 2 a (x + 1) θ (x), & f_{2} (x) = a (x + 1) θ (x), \end{matrix}

for all x ∈ ℝ, where θ : ℝ → ℝ is a nontrivial derivation. Then f₁, f₂ are discontinuous Drygas functions and satisfy the conditions of Theorem 2.3 with δ(x) = aθ(x) for all x ∈ ℝ.

Theorem 2.5

Drygas functions f₁, f₂ : ℝ → ℝ satisfy the conditional equation (1.4) on ℝ₊ for (x, y) ∈ Δ₂ and $f_{1} (x) = x^{3} f_{1} (\frac{1}{x})$ for all x ∈ ℝ₊ if and only if

(2.22)

\begin{matrix} f_{1} (x) = f_{2} (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}

where α is a real constant.

Proof

The conditional equation (1.4) for y = log(x) is

(2.23)

\begin{matrix} {(log (x))}^{2} + log (x)] f_{1} (x) = (x^{2} + x) f_{2} (log (x)), & x \in ℝ_{+} \cdot \end{matrix}

Replacing x with $\frac{1}{x}$ in (2.23), we arrive, by using the fact that $f_{1} (x) = x^{3} f_{1} (\frac{1}{x})$ , at

(2.24)

\begin{matrix} {(log (x))}^{2} - log (x)] f_{1} (x) = (x^{2} + x) f_{2} (- log (x)), & x \in ℝ_{+} \cdot \end{matrix}

Substituting x² for x in (2.23) and applying properties of logarithmic and Drygas functions, we see that

(2.25)

\begin{array}{l} 4 {(log (x))}^{2} + 2 log (x)] f_{1} (x^{2}) = (x^{4} + x^{2}) f (2 log (x)) \\ = (x^{4} + x^{2}) 3 f_{2} (log (x)) + f_{2} (- log (x))], \end{array}

for all x ∈ ℝ₊.

From (2.23), (2.24) and (2.25) we deduce that

4 {(log (x))}^{2} + 2 log (x)] f_{1} (x^{2}) = \frac{x^{4} + x^{2}}{x^{2} + x} 4 {(log (x))}^{2} + 2 log (x)] f_{1} (x),

which implies

(2.26)

(x + 1) f_{1} (x^{2}) = (x^{3} + x) f_{1} (x),

for all

x \in ℝ_{+} \ 1, exp (- \frac{1}{2})\}

Obviously, (2.26) holds for x = 1.

Putting x = exp(1) in (2.24), we have f₂(−1) = 0. So, A₂(1) = B₂(1, 1), where A₂ : ℝ → ℝ is an additive function and B₂ : ℝ × ℝ → ℝ is a symmetric biadditive function and f₂(x) = A₂(x) + B₂(x, x).

Taking $x = exp (- \frac{1}{2})$ in (2.24), we get

\begin{array}{l} \frac{3}{4} f_{1} (exp (- \frac{1}{2})) & = (exp (- 1) + exp (- \frac{1}{2})) f_{2} (\frac{1}{2}) \\ = (exp (- 1) + exp (- \frac{1}{2})) A_{2} (\frac{1}{2}) + B_{2} (\frac{1}{2}, \frac{1}{2})] \\ = \frac{3}{4} (exp (- 1) + exp (- \frac{1}{2})) A (1) \\ = \frac{3}{4} (exp (- 1) + exp (- \frac{1}{2})) \frac{f_{2} (1)}{2} . \end{array}

Hence,

(2.27)

f_{2} (1) = \frac{2}{(exp (- 1) + exp (- \frac{1}{2}))} f_{1} (exp (- \frac{1}{2})) .

Setting x = exp(−1) in (2.24), we obtain

(2.28)

2 f_{1} (exp (- 1)) = (exp (- 2) + exp (- 1)) f_{2} (1) .

It follows from (2.27) and (2.28) that

\begin{array}{l} (exp (- \frac{1}{2}) + 1) f (exp (1)) \\ = (exp (- \frac{3}{2}) + exp (- \frac{1}{2})) f_{1} (exp (- \frac{1}{2})) . \end{array}

Therefore,

(x + 1) f_{1} (x^{2}) = (x^{3} + x) f_{1} (x),

for all x ∈ ℝ₊. By Theorem 2.1,

\begin{matrix} f_{1} (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}

where

α = \frac{f_{1} (1)}{2}

. By replacing f₁(x) in (2.23), we have

\begin{matrix} f_{2} (log (x)) = α [log (x))^{2} + log (x)], & x \in ℝ_{+}, \end{matrix}

where

α = \frac{f_{1} (1)}{2}

. Consequently

\begin{matrix} f_{2} (x) = α (x + x^{2}) = f_{1} (x), & x \in ℝ, \end{matrix}

where

α = \frac{f_{1} (1)}{2}

One can easily verify the sufficiency of (2.22).

As a consequence, Theorem 2.5 can be generalized to the case of exponential functions, that is (x, y) ∈ Δ₃, because the logarithmic and exponential functions of the same basis are inverses of each other.

Corollary 2.6

Drygas functions f₁, f₂ : ℝ → ℝ satisfy the conditional equation (1.4) for (x, y) ∈ Δ₃ and $f_{2} (x) = x^{3} f_{2} (\frac{1}{2})$ for all x ∈ ℝ₊ if and only if

\begin{matrix} f_{1} (x) = f_{2} (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}

where α is a real constant.

Remark 1

Theorem 2.5 and Corollary 2.6 also hold if y = log_a(x) or y = a^x for a ∈ ℝ₊\{1}.

Conditional Equations Related to Drygas Functional Equations

Full Article

1. Preliminaries

(1.1)

(1.2)

(1.3)

(1.4)

Lemma 1.1.

2. Main results

Theorem 2.1

(2.1)

Proof

(2.2)

(2.3)

(2.4)

(2.5)

(2.6)

(2.7)

(2.8)

(2.9)

(2.10)

Lemma 2.2 ([5])

Theorem 2.3

Proof

(2.11)

(2.12)

(2.13)

(2.14)

(2.15)

(2.16)

(2.17)

(2.18)

(2.19)

(2.20)

(2.21)

Example 2.4

Theorem 2.5

(2.22)

Proof

(2.23)

(2.24)

(2.25)

(2.26)

(2.27)

(2.28)

Corollary 2.6

Remark 1