Conditional Equations Related to Drygas Functional Equations

Sadegh Izadi; Sedigheh Jahedi; Mehdi Dehghanian

doi:10.2478/amsil-2025-0015

Full Article

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, $A (sx) = sA (x),$ A(sx) = sA(x), for all x ∈ ℝ and s ∈ ℚ. A function h : ℝ → ℝ is called quadratic if the equation $h (x + y) + h (x - y) = 2 h (x) + 2 h (y)$ h(x + y) + h(x - y) = 2h(x) + 2h(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) $\begin{matrix} B (x, y) = \frac{1}{2} [h (x + y) - h (x) - h (y)], & x, y \in ℝ . \end{matrix}$ \matrix{{B(x,y) = {1 \over 2}[h(x + y) - h(x) - h(y)],} & {x,y \in {\mathbb R}.} \cr}

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 $\begin{matrix} B (rx, sy) = rsB (x, y), & h (rx) = B (rx, rx) = r^{2} h (x), \end{matrix}$ \matrix{{B(rx,sy) = rsB(x,y),} & {h(rx) = B(rx,rx) = {r^2}h(x),} \cr} for all x, y ∈ ℝ and r, s ∈ ℚ. By using (1.1) and induction on n, one can show that $h (\sum_{i = 0}^{n} ω_{i}) = \sum_{i = 0}^{n} h (ω_{i}) + 2 \sum_{0 \leq j < k \leq n} B (ω_{j}, ω_{k}),$ h\left({\sum\limits_{i = 0}^n {{\omega_i}}} \right) = \sum\limits_{i = 0}^n {h({\omega_i})} + 2\sum\limits_{0 \le j < k \le n} {B({\omega_j},{\omega_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 $y^{2} h (x) = x^{2} h (y),$ {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) $f (x + y) + f (x - y) = 2 f (x) + f (y) + f (- y),$ f(x + y) + f(x - y) = 2f(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) $\begin{matrix} f (x) = A (x) + B (x, x), & x \in ℝ, \end{matrix}$ \matrix{{f(x) = A(x) + B(x,x),} & {x \in {\mathbb R},} \cr} 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 $\begin{array}{l} \begin{array}{l} Δ_{0} = {(x, y) \in ℝ^{2} : x > 0 and y = x^{n}}, & n \in ℤ, | n | \geq 2, \end{array} \\ Δ_{1} = {(x, y) \in ℝ^{2} : y = x^{2}}, \\ Δ_{2} = {(x, y) \in ℝ^{2} : x > 0 and y = log (x)}, \\ Δ_{3} = {(x, y) \in ℝ^{2} : y = exp (x)} . \end{array}$ \matrix{{\matrix{{{\Delta_0} = \{(x,y) \in {{\mathbb R}^2}:x > 0\,\,{\rm{and}}\,\,y = {x^n}\} ,} \hfill & {n \in {\mathbb Z},\,\,|n| \ge 2,} \hfill \cr}} \hfill \cr {{\Delta_1} = \{(x,y) \in {{\mathbb R}^2}:y = {x^2}\} ,} \hfill \cr {{\Delta_2} = \{(x,y) \in {{\mathbb R}^2}:x > 0\,\,{\rm{and}}\,\,y = \log (x)\} ,} \hfill \cr {{\Delta_3} = \{(x,y) \in {{\mathbb R}^2}:y = \exp (x)\} .} \hfill \cr}

Motivated by the results of [5], this paper is devoted to finding Drygas functions f₁, f₂ : ℝ → ℝ satisfying the equation (1.4) $(y^{2} + y) f_{1} (x) = (x^{2} + x) f_{2} (y),$ ({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 $\sum_{j = 0}^{m} Λ_{j} (x) s^{j} = 0,$ \sum\limits_{j = 0}^m {{\Lambda_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) $(x^{2} + x) f (y) = (y^{2} + y) f (x),$ ({x^2} + x)f(y) = ({y^2} + y)f(x), for (x, y) ∈ Δ₀ if and only if $\begin{matrix} f (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}$ \matrix{{f(x) = \alpha (x + {x^2}),} & {x \in {\mathbb R}}, \cr} 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) $\begin{matrix} (x + 1) f (x^{n}) = (x^{2 n - 1} + x^{n - 1}) f (x), & x \in ℝ_{+} . \end{matrix}$ \matrix{{(x + 1)f({x^n}) = ({x^{2n - 1}} + {x^{n - 1}})f(x),} & {x \in {{\mathbb R}_ +}.} \cr} Substituting x + s, s ∈ ℚ₊, for x in (2.2), we get $\begin{matrix} (x + s + 1) f ({(x + s)}^{n}) = ({(x + s)}^{2 n - 1} + {(x + s)}^{n - 1}) f (x + s), & x \in ℝ_{+} . \end{matrix}$ \matrix{{(x + s + 1)f({{(x + s)}^n}) = ({{(x + s)}^{2n - 1}} + {{(x + s)}^{n - 1}})f(x + s),} & {x \in {{\mathbb R}_ +}.} \cr} By expanding the binomial terms, we obtain (2.3) $(x + s + 1) f (\sum_{m = 0}^{n} (\begin{matrix} n \\ m \end{matrix}) x^{m} s^{n - m}) = \sum_{l = 0}^{2 n - 1} (\begin{matrix} 2 n - 1 \\ l \end{matrix}) x^{l} s^{2 n - l - 1} + \sum_{k = 0}^{n - 1} (\begin{matrix} n - 1 \\ k \end{matrix}) x^{k} s^{n - k - 1}] f (x + s) .$ (x + s + 1)f\left({\sum\limits_{m = 0}^n {\left({\matrix{n \cr m \cr}} \right){x^m}{s^{n - m}}}} \right) = \left[ {\sum\limits_{l = 0}^{2n - 1} {\left({\matrix{{2n - 1} \cr l \cr}} \right){x^l}{s^{2n - l - 1}}} + \sum\limits_{k = 0}^{n - 1} {\left({\matrix{{n - 1} \cr k \cr}} \right)} {x^k}{s^{n - k - 1}}} \right]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) $\begin{array}{l} (x + s + 1) \sum_{m = 0}^{n} ((\begin{matrix} n \\ m \end{matrix}) s^{n - m} A (x^{m}) + {(\begin{matrix} n \\ m \end{matrix})}^{2} s^{2 n - 2 m} B (x^{m}, x^{m})) + 2 \sum_{0 \leq i < j \underline{<} n} (\begin{matrix} n \\ i \end{matrix}) (\begin{matrix} n \\ j \end{matrix}) s^{2 n - i - j} B (x^{i}, x^{j})] \\ - \sum_{k = 0}^{n - 1} (\begin{matrix} n - 1 \\ k \end{matrix}) x^{k} s^{n - k - 1} (f (x) + sA (1) + s^{2} B (1, 1) + 2 sB (x, 1)) \\ - \sum_{l = 0}^{2 n - 1} (\begin{matrix} 2 n - 1 \\ l \end{matrix}) x^{l} s^{2 n - l - 1} (f (x) + sA (1) + s^{2} B (1, 1) + 2 sB (x, 1)) = 0. \end{array}$ \matrix{{(x + s + 1)\left[ {\sum\limits_{m = 0}^n {\left({\left({\matrix{n \cr m \cr}} \right){s^{n - m}}A({x^m}) + {{\left({\matrix{n \cr m \cr}} \right)}^2}{s^{2n - 2m}}B({x^m},{x^m})} \right)} + 2\sum\limits_{0 \le i < j\underline < n} {\left({\matrix{n \cr i \cr}} \right)\left({\matrix{n \cr j \cr}} \right){s^{2n - i - j}}B({x^i},{x^j})}} \right]} \hfill \cr {- \sum\limits_{k = 0}^{n - 1} {\left({\matrix{{n - 1} \cr k \cr}} \right){x^k}{s^{n - k - 1}}(f(x) + sA(1) + {s^2}B(1,1) + 2sB(x,1))}} \hfill \cr {- \sum\limits_{l = 0}^{2n - 1} {\left({\matrix{{2n - 1} \cr l \cr}} \right){x^l}{s^{2n - l - 1}}(f(x) + sA(1) + {s^2}B(1,1) + 2sB(x,1))} = 0.} \hfill \cr} 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 $A (1) + 2 B (x, 1) + (2 n - 1) xB (1, 1) - xB (1, 1) - B (1, 1) - 2 nB (x, 1) = 0,$ A(1) + 2B(x,1) + (2n - 1)xB(1,1) - xB(1,1) - B(1,1) - 2nB(x,1) = 0, for all x ∈ ℝ₊. Put x = 1 in the above equality, we have $A (1) = B (1, 1) = \frac{f (1)}{2}$ A(1) = B(1,1) = {{f(1)} \over 2} . Therefore, as |n| ≥ 2, we get (2.5) $\begin{matrix} B (x, 1) = xB (1, 1), & x \in ℝ_{+} . \end{matrix}$ \matrix{{B(x,1) = xB(1,1),} & {x \in {{\mathbb R}_ +}.} \cr} From the coefficient of s²ⁿ⁻¹, we arrive at $\begin{matrix} 0 = f (x) + (2 n - 1) x [A (1) + 2 B (x, 1)] + (\begin{matrix} 2 n - 1 \\ 2 \end{matrix}) x^{2} B (1, 1) \\ - {(\begin{matrix} n \\ 1 \end{matrix})}^{2} B (x, x) - 2 n (x + 1) B (x, 1) - 2 (\begin{matrix} n \\ 2 \end{matrix}) B (x^{2}, 1) \\ = f (x) + (2 n - 1) (1 + 2 x) xB (1, 1) + (2 n - 1) (n - 1) x^{2} B (1, 1) \\ - n^{2} B (x, x) - 2 n x^{2} (x + 1) B (1, 1) - n (n - 1) x^{2} B (1, 1) . \end{matrix}$ \matrix{{0 = f(x) + (2n - 1)x[A(1) + 2B(x,1)] + \left({\matrix{{2n - 1} \cr 2 \cr}} \right){x^2}B(1,1)} \cr {- {{\left({\matrix{n \cr 1 \cr}} \right)}^2}B(x,x) - 2n(x + 1)B(x,1) - 2\left({\matrix{n \cr 2 \cr}} \right)B({x^2},1)} \cr {= f(x) + (2n - 1)(1 + 2x)xB(1,1) + (2n - 1)(n - 1){x^2}B(1,1)} \cr {- {n^2}B(x,x) - 2n{x^2}(x + 1)B(1,1) - n(n - 1){x^2}B(1,1).} \cr} Now with A(1) = B(1, 1), and (2.5), we obtain (2.6) $f (x) = (x^{2} + x) B (1, 1) + n^{2} B (x, x) - n^{2} x^{2} B (1, 1),$ 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, $\begin{matrix} B (x, x) = \frac{f (x) + f (- x)}{2} = x^{2} B (1, 1) + n^{2} B (x, x) - x^{2} B (1, 1), & x \in ℝ_{+} . \end{matrix}$ \matrix{{B(x,x) = {{f(x) + f(- x)} \over 2} = {x^2}B(1,1) + {n^2}B(x,x) - {x^2}B(1,1),} & {x \in {{\mathbb R}_ +}.} \cr} As n ≥ 2, (2.7) $\begin{matrix} B (x, x) = x^{2} B (1, 1) = \frac{f (1)}{2} x^{2}, & x \in ℝ_{+} . \end{matrix}$ \matrix{{B(x,x) = {x^2}B(1,1) = {{f(1)} \over 2}{x^2},} & {x \in {{\mathbb R}_ +}.} \cr} By equations (2.6) and (2.7), we conclude that $\begin{matrix} f (x) = \frac{f (1)}{2} (x + x^{2}), & x \in ℝ_{+} . \end{matrix}$ \matrix{{f(x) = {{f(1)} \over 2}(x + {x^2}),} & {x \in {{\mathbb R}_ +}.} \cr} Hence, $A (x) = f (x) - B (x, x) = \frac{f (1)}{2} x$ A(x) = f(x) - B(x,x) = {{f(1)} \over 2}x for all x ∈ ℝ₊. Also, for x = 0 above equation holds, because f (0) = 0.

Now, for x = −u < 0, $\begin{array}{l} f (x) & = A (- u) + B (- u, - u) = - A (u) + {(- 1)}^{2} B (u, u) \\ = \frac{f (1)}{2} (- u + {(- u)}^{2}) = \frac{f (1)}{2} (x + x^{2}) . \end{array}$ \matrix{{f(x)} \hfill & {= A(- u) + B(- u, - u) = - A(u) + {{(- 1)}^2}B(u,u)} \hfill \cr {} \hfill & {= {{f(1)} \over 2}(- u + {{(- u)}^2}) = {{f(1)} \over 2}(x + {x^2}).} \hfill \cr} Therefore, $\begin{matrix} f (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}$ \matrix{{f(x) = \alpha (x + {x^2}),} & {x \in {\mathbb R},} \cr} where $α = \frac{f (1)}{2}$ \alpha = {{f(1)} \over 2} .

Finally, for the case n ≤ −2, take p = −n ≥ 2 in (2.2) to obtain (2.8) $(x + 1) f (\frac{1}{x^{p}}) = (\frac{1}{x^{2 p + 1}} + \frac{1}{x^{p + 1}}) f (x) = (\frac{1 + x^{p}}{x^{2 p + 1}}) f (x),$ (x + 1)f\left({{1 \over {{x^p}}}} \right) = \left({{1 \over {{x^{2p + 1}}}} + {1 \over {{x^{p + 1}}}}} \right)f(x) = \left({{{1 + {x^p}} \over {{x^{2p + 1}}}}} \right)f(x), for x ∈ ℝ₊. Substitute x^−p for x in (2.8) to gain $(\frac{1}{x^{p}} + 1) f (x^{p^{2}}) = (x^{2 p^{2} + p} + x^{p^{2} + p}) f (\frac{1}{x^{p}}) .$ \left({{1 \over {{x^p}}} + 1} \right)f\left({{x^{{p^2}}}} \right) = \left({{x^{2{p^2} + p}} + {x^{{p^2} + p}}} \right)f\left({{1 \over {{x^p}}}} \right). By (2.8), we obtain $\frac{1 + x^{p}}{x^{p}} f (x^{p^{2}}) = \frac{x^{2 p^{2} + p} + x^{p^{2} + p}}{x + 1} (\frac{1 + x^{p}}{x^{2 p + 1}}) f (x),$ {{1 + {x^p}} \over {{x^p}}}f({x^{{p^2}}}) = {{{x^{2{p^2} + p}} + {x^{{p^2} + p}}} \over {x + 1}}\left({{{1 + {x^p}} \over {{x^{2p + 1}}}}} \right)f(x), or (2.9) $(x + 1) f (x^{p^{2}}) = (x^{2 p^{2} - 1} + x^{p^{2} - 1}) f (x) .$ (x + 1)f\left({{x^{{p^2}}}} \right) = \left({{x^{2{p^2} - 1}} + {x^{{p^2} - 1}}} \right)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 $\begin{matrix} s \mapsto ℋ (s, x) & and & s \mapsto ℋ (x, s), & s \in ℝ, \end{matrix}$ \matrix{{s \mapsto {{\cal H}(s,x)}} & {{\rm{and}}} & {s \mapsto {{\cal H}(x,s),}} & {s \in {\mathbb R},} \cr} 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 $θ (xy) - x θ (y) - θ (x) y = H (x, y),$ \theta (xy) - x\theta (y) - \theta (x)y = {{\cal H}(x,y),} for some bi-derivation ℋ on ℝ².

In the case n = 1, condition (2.1) has the form $(x + x^{2}) f (x) = (x + x^{2}) f (x),$ (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) $\begin{matrix} f (x) = x^{3} f (\frac{1}{x}), & x \in ℝ \ {0} . \end{matrix}$ \matrix{{f(x) = {x^3}f\left({{1 \over x}} \right)\;,} & {x \in {\mathbb R}\backslash \{0\} .} \cr} 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 $\begin{matrix} θ (\frac{1}{x}) = - \frac{1}{x^{2}} θ (x), & x \in ℝ \ {0} . \end{matrix}$ \matrix{{\theta \left({{1 \over x}} \right) = - {1 \over {{x^2}}}\theta (x),} & {x \in {\mathbb R}\backslash \{0\} .} \cr} Therefore, $f (x) = - θ (x) + \frac{1}{2} θ (x^{2})$ f(x) = - \theta (x) + {1 \over 2}\theta ({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 $δ (x^{4}) = 6 x^{2} δ (x^{2}) - 8 x^{3} δ (x),$ \delta ({x^4}) = 6{x^2}\delta ({x^2}) - 8{x^3}\delta (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 $\begin{array}{l} δ (x^{4}) = 6 x^{2} δ (x^{2}) - 8 x^{3} δ (x) + 3 x^{4} δ (1), \\ f_{1} (x) = (x + 1) 2 δ (x) - x δ (1)], \\ \begin{array}{l} f_{2} (x) = \frac{1}{4} (x + 1) 6 δ (x) - \frac{1}{x} δ (x^{2}) - x δ (1)], & x \in ℝ \ {0} \\ f_{2} (0) = 0. \end{array} \end{array}$ \matrix{{\delta ({x^4}) = 6{x^2}\delta ({x^2}) - 8{x^3}\delta (x) + 3{x^4}\delta (1),} \hfill \cr {{f_1}(x) = (x + 1)\left[ {2\delta (x) - x\delta (1)} \right],} \hfill \cr {\left\{{\matrix{{{f_2}(x) = {1 \over 4}(x + 1)\left[ {6\delta (x) - {1 \over x}\delta ({x^2}) - x\delta (1)} \right],} \hfill & {x \in {\mathbb R}\backslash \{0\}} \hfill \cr {{f_2}(0) = 0.} \hfill & {} \hfill \cr}} \right.} \hfill \cr} 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 $\begin{matrix} f_{1} (x) = A_{1} (x) + B_{1} (x, x) & and & f_{2} (x) = A_{2} (x) + B_{2} (x, x) \end{matrix},$ \matrix{{{f_1}(x) = {A_1}(x) + {B_1}(x,x)} & {{\rm{and}}} & {{f_2}(x) = {A_2}(x) + {B_2}(x,x)} \cr} , for all x ∈ ℝ. Put y = x² in (1.4), to obtain $\begin{matrix} (x^{2} + x^{4}) f_{1} (x) = (x + x^{2}) f_{2} (x^{2}), & x \in ℝ . \end{matrix}$ \matrix{{({x^2} + {x^4}){f_1}(x) = (x + {x^2}){f_2}({x^2}),} & {x \in {\mathbb R}.} \cr} By dividing both sides by x ≠ 0 (since f₁(0) = f₂(0) = 0), we have (2.11) $\begin{matrix} (x + 1) f_{2} (x^{2}) = (x^{3} + x) f_{1} (x), & x \in ℝ . \end{matrix}$ \matrix{{(x + 1){f_2}({x^2}) = ({x^3} + x){f_1}(x),\;} & {x \in {\mathbb R}.} \cr} Set x = −1 in (2.11), then f₁(−1) = −A₁(1) + B₁(1, 1) = 0. Thus, $A_{1} (1) = B_{1} (1, 1) = \frac{f_{1} (1)}{2} .$ {A_1}(1) = {B_1}(1,1) = {{{f_1}(1)} \over 2}. Let s ∈ ℚ. Substituting x + s for x in (2.11), we get (2.12) $\begin{matrix} (x + s + 1) f_{2} ({(x + s)}^{2}) = ({(x + s)}^{3} + x + s) f_{1} (x + s), & x \in ℝ . \end{matrix}$ \matrix{{(x + s + 1){f_2}\left({{{(x + s)}^2}} \right) = \left({{{(x + s)}^3} + x + s} \right){f_1}(x + s),} & {x \in {\mathbb R}.} \cr} 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 $\begin{array}{l} x A_{2} (x^{2}) + s A_{2} (x^{2}) + A_{2} (x^{2}) + 2 s x A_{2} (x) + 2 s^{2} A_{2} (x) + 2 s A_{2} (x) \\ + s^{2} x A_{2} (1) + s^{3} A_{2} (1) + s^{2} A_{2} (1) + x B_{2} (x^{2}, x^{2}) + s B_{2} (x^{2}, x^{2}) \\ + B_{2} (x^{2}, x^{2}) + 4 s^{2} x B_{2} (x, x) + 4 s^{3} B_{2} (x, x) + 4 s^{2} B_{2} (x, x) + s^{4} x B_{2} (1, 1) \\ + s^{5} B_{2} (1, 1) + s^{4} B_{2} (1, 1) + 4 s x B_{2} (x^{2}, x) + 4 s^{2} B_{2} (x^{2}, x) + 4 s B_{2} (x^{2}, x) \\ + 2 s^{2} x B_{2} (v) + 2 s^{3} B_{2} (x^{2}, 1) + 2 s^{2} B_{2} (x^{2}, 1) + 4 s^{3} x B_{2} (x, 1) \\ + 4 s^{4} B_{2} (x, 1) + 4 s^{3} B_{2} (x, 1) \\ = x^{3} A_{1} (x) + 3 s x^{2} A_{1} (x) + 3 s^{2} x A_{1} (x) + s^{3} A_{1} (x) + x A_{1} (x) + s A (x) \\ + s x^{3} A_{1} (1) + 3 s^{2} x^{2} A_{1} (1) + 3 s^{3} x A_{1} (1) + s^{4} A_{1} (1) + s x A_{1} (1) + s^{2} A (1) \\ + x^{3} B_{1} (x, x) + 3 s x^{2} B_{1} (x, x) + 3 s^{2} x B_{1} (x, x) + s^{3} B_{1} (x, x) + x B_{1} (x, x) \\ + s B_{1} (x, x) + s^{2} x^{3} B_{1} (1, 1) + 3 s^{3} x^{2} B_{1} (1, 1) + 3 s^{4} x B_{1} (1, 1) + s^{5} B_{1} (1, 1) \\ + s^{2} x B_{1} (1, 1) + s^{3} B_{1} (1, 1) + 2 s x^{3} B_{1} (x, 1) + 6 s^{2} x^{2} B_{1} (x, 1) + 6 s^{3} x B_{1} (x, 1) \\ + 2 s^{4} B_{1} (x, 1) + 2 s x B_{1} (x, 1) + 2 s^{2} B_{1} (x, 1), \end{array}$ \matrix{{\,\,\,\,\,x{A_2}({x^2}) + s{A_2}({x^2}) + {A_2}({x^2}) + 2sx{A_2}(x) + 2{s^2}{A_2}(x) + 2s{A_2}(x)} \hfill \cr {\,\,\,\,\, + {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})} \hfill \cr {\,\,\,\,\, + {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)} \hfill \cr {\,\,\,\,\, + {s^5}{B_2}(1,1) + {s^4}{B_2}(1,1) + 4sx{B_2}({x^2},x) + 4{s^2}{B_2}({x^2},x) + 4s{B_2}({x^2},x)} \hfill \cr {\,\,\,\,\, + 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)} \hfill \cr {\,\,\,\,\, + 4{s^4}{B_2}(x,1) + 4{s^3}{B_2}(x,1)} \hfill \cr {= {x^3}{A_1}(x) + 3s{x^2}{A_1}(x) + 3{s^2}x{A_1}(x) + {s^3}{A_1}(x) + x{A_1}(x) + sA(x)} \hfill \cr {\,\,\,\,\, + 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) + sx{A_1}(1) + {s^2}A(1)} \hfill \cr {\,\,\,\,\, + {x^3}{B_1}(x,x) + 3s{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)} \hfill \cr {\,\,\,\,\, + 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)} \hfill \cr {\,\,\,\,\, + {s^2}x{B_1}(1,1) + {s^3}{B_1}(1,1) + 2s{x^3}{B_1}(x,1) + 6{s^2}{x^2}{B_1}(x,1) + 6{s^3}x{B_1}(x,1)} \hfill \cr {\,\,\,\,\, + 2{s^4}{B_1}(x,1) + 2sx{B_1}(x,1) + 2{s^2}{B_1}(x,1),} \hfill \cr} for all x ∈ ℝ. Hence, $\begin{array}{l} 0 = & B_{1} (1, 1) - B_{2} (1, 1)] s^{5} \\ + A_{1} (1) + 3 x B_{1} (1, 1) + 2 B_{1} (x, 1) - x B_{2} (1, 1) + - 4 B_{2} (x, 1) - B_{2} (1, 1)] s^{4} \\ + [f_{1} (x) + 3 x A_{1} (1) + 6 x B_{1} (x, 1) + B_{1} (1, 1) + 3 x^{2} B_{1} (1, 1) \\ - 4 x B_{2} (x, 1) - A_{2} (1) - 4 B_{2} (x, x) - 2 B_{2} (x^{2}, 1) - 4 B_{2} (x, 1)] s^{3} \\ + [3 x f_{1} (x) + 3 x^{2} A_{1} (1) + A_{1} (1) + x^{3} B_{1} (1, 1) + x B_{1} (1, 1) \\ + 6 x^{2} B_{1} (x, 1) + 2 B_{1} (x, 1) - A_{2} (1) - 2 A_{2} (x) - x A (1) \\ - 4 x B_{2} (x, x) - 4 B_{2} (x, x) - 4 B_{2} (x^{2}, x) - 2 x B_{2} (x^{2}, 1) - 2 B_{2} (x^{2}, 1)] s^{2} \\ + [3 x^{2} f_{1} (x) + f_{1} (x) + x^{3} A_{1} (1) + x A_{1} (1) + 2 x^{3} B_{1} (x, 1) + 2 x B_{1} (x, 1) \\ - f_{2} (x^{2}) - 2 x A_{2} (x) - 2 A_{2} (x) - 4 x B_{2} (x^{2}, x) - 4 B_{2} (x^{2}, x)] s \\ + x^{3} f_{1} (x) + x f_{1} (x) - x f_{2} (x^{2}) - f_{2} (x^{2})] . \end{array}$ \matrix{{0 =} \hfill & {\left[ {{B_1}(1,1) - {B_2}(1,1)} \right]{s^5}} \hfill \cr {} \hfill & {+ \left[ {{A_1}(1) + 3x{B_1}(1,1) + 2{B_1}(x,1) - x{B_2}(1,1) + - 4{B_2}(x,1) - {B_2}(1,1)} \right]{s^4}} \hfill \cr {} \hfill & {+ [{f_1}(x) + 3x{A_1}(1) + 6x{B_1}(x,1) + {B_1}(1,1) + 3{x^2}{B_1}(1,1)} \hfill \cr {} \hfill & {\,\,\, - 4x{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}} \hfill \cr {} \hfill & {+ [3x{f_1}(x) + 3{x^2}{A_1}(1) + {A_1}(1) + {x^3}{B_1}(1,1) + x{B_1}(1,1)} \hfill \cr {} \hfill & {\,\,\, + 6{x^2}{B_1}(x,1) + 2{B_1}(x,1) - {A_2}\left(1 \right) - 2{A_2}(x) - xA(1)} \hfill \cr {} \hfill & {\,\,\, - 4x{B_2}(x,x) - 4{B_2}(x,x) - 4{B_2}({x^2},x) - 2x{B_2}({x^2},1) - 2{B_2}({x^2},1)]{s^2}} \hfill \cr {} \hfill & {+ [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) + 2x{B_1}(x,1)} \hfill \cr {} \hfill & {\,\,\, - {f_2}({x^2}) - 2x{A_2}(x) - 2{A_2}(x) - 4x{B_2}({x^2},x) - 4{B_2}({x^2},x)]s} \hfill \cr {} \hfill & {+ \left[ {{x^3}{f_1}(x) + x{f_1}(x) - x{f_2}({x^2}) - {f_2}({x^2})} \right].} \hfill \cr} 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 $A_{1} (1) = B_{1} (1, 1) = A_{2} (1) = B_{2} (1, 1) = \frac{f_{1} (1)}{2} .$ {A_1}(1) = {B_1}(1,1) = {A_2}(1) = {B_2}(1,1) = {{{f_1}(1)} \over 2}. According to the coefficient of s⁴ we see that (2.13) $\begin{matrix} 2 B_{2} (x, 1) = x B_{1} (1, 1) + B_{1} (x, 1), & x \in ℝ . \end{matrix}$ \matrix{{2{B_2}(x,1) = x{B_1}(1,1) + {B_1}(x,1),} & {x \in {\mathbb R}.} \cr} From the coefficient of s³ and (2.13), we conclude that (2.14) $f_{1} (x) = 2 B_{1} (x, 1) - x B_{1} (1, 1) - 4 x B_{1} (x, 1) + B_{1} (x^{2}, 1) + 4 B_{2} (x, x),$ {f_1}(x) = 2{B_1}(x,1) - x{B_1}(1,1) - 4x{B_1}(x,1) + {B_1}({x^2},1) + 4{B_2}(x,x), for all x ∈ ℝ. Hence, by (2.14), (2.15) $A_{1} (x) = \frac{f_{1} (x) - f_{1} (- x)}{2} = 2 B_{1} (x, 1) - x B_{1} (1, 1),$ {A_1}(x) = {{{f_1}(x) - {f_1}(- x)} \over 2} = 2{B_1}(x,1) - x{B_1}(1,1), and (2.16) $B_{1} (x, x) = \frac{f_{1} (x) + f_{1} (- x)}{2} = 4 B_{2} (x, x) + B_{1} (x^{2}, 1) - 4 x B_{1} (x, 1),$ {B_1}(x,x) = {{{f_1}(x) + {f_1}(- x)} \over 2} = 4{B_2}(x,x) + {B_1}({x^2},1) - 4x{B_1}(x,\;1), for all x ∈ ℝ.

Replacing x with −x in (2.11) yields (2.17) $\begin{matrix} (- x + 1) f_{2} (x^{2}) = - (x^{3} + x) f_{1} (- x), & x \in ℝ . \end{matrix}$ \matrix{{(- x + 1){f_2}({x^2}) = - ({x^3} + x){f_1}(- x),} & {x \in {\mathbb R}.} \cr} Adding both sides of (2.11) and (2.17) gives us $\begin{matrix} f_{2} (x^{2}) = (x^{3} + x) A_{1} (x), & x \in ℝ, \end{matrix}$ \matrix{{{f_2}({x^2}) = ({x^3} + x){A_1}(x),} & {x \in {\mathbb R},} \cr} and hence, (2.18) $\begin{matrix} f_{1} (x) = (x + 1) A_{1} (x) = A_{1} (x) + B_{1} (x, x), & B_{1} (x, x) = x A_{1} (x) . \end{matrix}$ \matrix{{{f_1}(x) = (x + 1){A_1}(x) = {A_1}(x) + {B_1}(x,x),} & {\;{B_1}(x,x) = x{A_1}(x).} \cr} for all x ∈ ℝ. Thus, (2.19) $\begin{matrix} f_{2} (x^{2}) = (x^{2} + 1) B_{1} (x, x), & x \in ℝ . \end{matrix}$ \matrix{{{f_2}({x^2}) = ({x^2} + 1){B_1}(x,x),} & {x \in {\mathbb R}.} \cr} From (2.15) and (2.18), we have (2.20) $\begin{matrix} B_{1} (x, x) = 2 x B_{1} (x, 1) - x^{2} B_{1} (1, 1), & x \in ℝ . \end{matrix}$ \matrix{{{B_1}(x,x) = 2x{B_1}(x,1) - {x^2}{B_1}(1,1),} & {x \in {\mathbb R}.} \cr} Combining (2.16) and (2.20) yields (2.21) $B_{2} (x, x) = \frac{3}{2} x B_{1} (x, 1) - \frac{1}{4} B_{1} (x^{2}, 1) - \frac{1}{4} x^{2} B_{1} (1, 1),$ {B_2}(x,x) = {3 \over 2}x{B_1}(x,1) - {1 \over 4}{B_1}({x^2},1) - {1 \over 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 $\begin{matrix} f_{2} (s^{2} x^{2}) = (s^{2} x^{2} + 1) B_{1} (sx, sx), & x \in ℝ . \end{matrix}$ \matrix{{{f_2}({s^2}{x^2}) = ({s^2}{x^2} + 1){B_1}(sx,sx),} & {x \in {\mathbb R}.} \cr} Thus, $A_{2} (x^{2}) - B_{1} (x, x)] s^{2} + B_{2} (x^{2}, x^{2}) - x^{2} B_{1} (x, x)] s^{4} = 0.$ \left[ {{A_2}({x^2}) - {B_1}(x,x)} \right]{s^2} + \left[ {{B_2}({x^2},{x^2}) - {x^2}{B_1}(x,x)} \right]{s^4} = 0. From Lemma 1.1 we have $A_{2} (x^{2}) = B_{1} (x, x), B_{2} (x^{2}, x^{2}) = x^{2} B_{1} (x, x) .$ {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, $B_{2} (x, x) = B_{2} (- t, - t) = B_{2} (t, t) = t A_{2} (t) = - t A_{2} (- t) = x A_{2} (x) .$ {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, $A_{2} (x) = \frac{1}{x} B_{2} (x, x)$ {A_2}(x) = {1 \over x}{B_2}(x,x) for all x ∈ ℝ\{0}.

From the above equality and (2.21), we obtain $A_{2} (x) = \frac{3}{2} B_{1} (x, 1) - \frac{1}{4 x} B_{1} (x^{2}, 1) - \frac{1}{4} x B_{1} (1, 1) .$ {A_2}(x) = {3 \over 2}{B_1}(x,1) - {1 \over {4x}}{B_1}({x^2},1) - {1 \over 4}x{B_1}(1,1). Define the additive function δ : ℝ → ℝ by $\begin{matrix} δ (x) = B_{1} (x, 1), & x \in ℝ . \end{matrix}$ \matrix{{\delta (x) = {B_1}(x,\;1),} & {x \in {\mathbb R}.} \cr} Therefore, $f_{1} (x) = (x + 1) 2 δ (x) - x δ (1)],$ {f_1}(x) = (x + 1)\left[ {2\delta (x) - x\delta (1)} \right], and $f_{2} (x) = \frac{1}{4} (x + 1) 6 δ (x) - \frac{1}{x} δ (x^{2}) - x δ (1)],$ {f_2}\left(x \right) = {1 \over 4}\left({x + 1} \right)\left[ {6\delta \left(x \right) - {1 \over x}\delta \left({{x^2}} \right) - x\delta \left(1 \right)} \right], for all x ∈ ℝ\{0}.

Next, f₁(1) = 0 if and only if δ(1) = 0, or equivalently, if and only if $δ (x^{4}) = 6 x^{2} δ (x^{2}) - 8 x^{3} δ (x),$ \delta \left({{x^4}} \right) = 6{x^2}\delta \left({{x^2}} \right) - 8{x^3}\delta \left(x \right), 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 $\begin{matrix} f_{1} (x) = 2 a (x + 1) θ (x), & f_{2} (x) = a (x + 1) θ (x), \end{matrix}$ \matrix{{{f_1}\left(x \right) = 2a\left({x + 1} \right)\theta \left(x \right),} & {{f_2}\left(x \right) = a\left({x + 1} \right)\theta \left(x \right),} \cr} 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 $f_{1} (x) = x^{3} f_{1} (\frac{1}{x})$ {f_1}\left(x \right) = {x^3}{f_1}\left({{1 \over x}} \right) for all x ∈ ℝ₊ if and only if (2.22) $\begin{matrix} f_{1} (x) = f_{2} (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}$ \matrix{{{f_1}\left(x \right) = {f_2}\left(x \right) = \alpha \left({x + {x^2}} \right),} & {x \in {\mathbb R},} \cr} where α is a real constant.

Proof

The conditional equation (1.4) for y = log(x) is (2.23) $\begin{matrix} {(log (x))}^{2} + log (x)] f_{1} (x) = (x^{2} + x) f_{2} (log (x)), & x \in ℝ_{+} \cdot \end{matrix}$ \matrix{{\left[ {{{(\log \;(x))}^2} + \log \;(x)} \right]{f_1}\left(x \right) = \left({{x^2} + x} \right){f_2}\left({\;\log \;\left(x \right)} \right),} & {x \in {{\mathbb R}_ +}.} \cr}

Replacing x with $\frac{1}{x}$ {1 \over x} in (2.23), we arrive, by using the fact that $f_{1} (x) = x^{3} f_{1} (\frac{1}{x})$ {f_1}\left(x \right) = {x^3}{f_1}\left({{1 \over x}} \right) , at (2.24) $\begin{matrix} {(log (x))}^{2} - log (x)] f_{1} (x) = (x^{2} + x) f_{2} (- log (x)), & x \in ℝ_{+} \cdot \end{matrix}$ \matrix{{\left[ {{{(\log \;(x))}^2} - \log \;(x)} \right]{f_1}\left(x \right) = ({x^2} + x){f_2}\left({- \;\log \;(x)} \right),} & {x \in {{\mathbb R}_ +}.} \cr} Substituting x² for x in (2.23) and applying properties of logarithmic and Drygas functions, we see that (2.25) $\begin{array}{l} 4 {(log (x))}^{2} + 2 log (x)] f_{1} (x^{2}) = (x^{4} + x^{2}) f (2 log (x)) \\ = (x^{4} + x^{2}) 3 f_{2} (log (x)) + f_{2} (- log (x))], \end{array}$ \matrix{{\left[ {4{{(\log (x))}^2} + 2\log (x)} \right]{f_1}({x^2}) = ({x^4} + {x^2})f\left({2\log (x)} \right)} \hfill \cr {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = ({x^4} + {x^2})\left[ {3{f_2}(\log (x)) + {f_2}\left({- \;\log (x)} \right)} \right],} \hfill \cr} for all x ∈ ℝ₊.

From (2.23), (2.24) and (2.25) we deduce that $4 {(log (x))}^{2} + 2 log (x)] f_{1} (x^{2}) = \frac{x^{4} + x^{2}}{x^{2} + x} 4 {(log (x))}^{2} + 2 log (x)] f_{1} (x),$ \left[ {4{{(\log (x))}^2} + 2\log (x)} \right]{f_1}({x^2}) = {{{x^4} + {x^2}} \over {{x^2} + x}}\left[ {4{{(\log (x))}^2} + 2\log (x)} \right]{f_1}(x), which implies (2.26) $(x + 1) f_{1} (x^{2}) = (x^{3} + x) f_{1} (x),$ (x + 1){f_1}({x^2}) = ({x^3} + x){f_1}(x), for all $x \in ℝ_{+} \ 1, exp (- \frac{1}{2})\}$ x \in {{\mathbb R}_ +}\backslash \left\{{1,\exp \;\left({- \;{1 \over 2}} \right)} \right\} .

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 $x = exp (- \frac{1}{2})$ x = \;\exp \;\left({- {1 \over 2}} \right) in (2.24), we get $\begin{array}{l} \frac{3}{4} f_{1} (exp (- \frac{1}{2})) & = (exp (- 1) + exp (- \frac{1}{2})) f_{2} (\frac{1}{2}) \\ = (exp (- 1) + exp (- \frac{1}{2})) A_{2} (\frac{1}{2}) + B_{2} (\frac{1}{2}, \frac{1}{2})] \\ = \frac{3}{4} (exp (- 1) + exp (- \frac{1}{2})) A (1) \\ = \frac{3}{4} (exp (- 1) + exp (- \frac{1}{2})) \frac{f_{2} (1)}{2} . \end{array}$ \matrix{{{3 \over 4}{f_1}\left({\;\exp \;\left({- {1 \over 2}} \right)} \right)} \hfill & {= \left({\;\exp \;\left({- 1} \right) + \;\exp \;\left({- {1 \over 2}} \right)} \right){f_2}\left({{1 \over 2}} \right)} \hfill \cr {} \hfill & {= \left({\;\exp \;\left({- 1} \right) + \;\exp \;\left({- {1 \over 2}} \right)} \right)\left[ {{A_2}\left({{1 \over 2}} \right) + {B_2}\left({{1 \over 2},\;{1 \over 2}} \right)} \right]} \hfill \cr {} \hfill & {= {3 \over 4}\left({\;\exp \;\left({- 1} \right) + \;\exp \;\left({- {1 \over 2}} \right)} \right)A\left(1 \right)} \hfill \cr {} \hfill & {= {3 \over 4}\left({\;\exp \;\left({- 1} \right) + \;\exp \;\left({- {1 \over 2}} \right)} \right){{{f_2}\left(1 \right)} \over 2}.} \hfill \cr} Hence, (2.27) $f_{2} (1) = \frac{2}{(exp (- 1) + exp (- \frac{1}{2}))} f_{1} (exp (- \frac{1}{2})) .$ {f_2}\left(1 \right) = {2 \over {\left({\;\exp \;\left({- 1} \right) + \;\exp \;\left({- {1 \over 2}} \right)} \right)}}{f_1}\left({\;\exp \;\left({- {1 \over 2}} \right)} \right). Setting x = exp(−1) in (2.24), we obtain (2.28) $2 f_{1} (exp (- 1)) = (exp (- 2) + exp (- 1)) f_{2} (1) .$ 2\;{f_1}\left({\;\exp \;\left({- 1} \right)} \right) = \left({\;\exp \;\left({- 2} \right) + \;\exp \;\left({- 1} \right)} \right){f_2}\left(1 \right). It follows from (2.27) and (2.28) that $\begin{array}{l} (exp (- \frac{1}{2}) + 1) f (exp (1)) \\ = (exp (- \frac{3}{2}) + exp (- \frac{1}{2})) f_{1} (exp (- \frac{1}{2})) . \end{array}$ \matrix{{\left({\;\exp \;\left({- {1 \over 2}} \right) + 1} \right)f\left({\;\exp \;\left(1 \right)} \right)} \hfill \cr {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left({\;\exp \;\left({- {3 \over 2}} \right) + \;\exp \;\left({- {1 \over 2}} \right)} \right){f_1}\left({\;\exp \;\left({- {1 \over 2}} \right)} \right).} \hfill \cr} Therefore, $(x + 1) f_{1} (x^{2}) = (x^{3} + x) f_{1} (x),$ \left({x + 1} \right){f_1}\left({{x^2}} \right) = \left({{x^3} + x} \right){f_1}\left(x \right), for all x ∈ ℝ₊. By Theorem 2.1, $\begin{matrix} f_{1} (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}$ \matrix{{{f_1}\left( x \right) = \alpha \left( {x + {x^2}} \right),} & {x \in {\mathbb R},} } where $α = \frac{f_{1} (1)}{2}$ \alpha = {{{f_1}\left(1 \right)} \over 2} . By replacing f₁(x) in (2.23), we have $\begin{matrix} f_{2} (log (x)) = α [log (x))^{2} + log (x)], & x \in ℝ_{+}, \end{matrix}$ \matrix{{{f_2}\left( {\;\log \;\left( x \right)} \right) = \alpha [\left( {\;\log \;{\left( x \right))}^2 + \;\log \;\left( x \right)} \right],} & {x \in {{\mathbb R}_+ },}} where $α = \frac{f_{1} (1)}{2}$ \alpha = {{{f_1}\left(1 \right)} \over 2} . Consequently $\begin{matrix} f_{2} (x) = α (x + x^{2}) = f_{1} (x), & x \in ℝ, \end{matrix}$ \matrix{{{f_2}\left( x \right) = \alpha \left( {x + {x^2}} \right) = {f_1}\left( x \right),} & {x \in {\mathbb R},}} where $α = \frac{f_{1} (1)}{2}$ \alpha = {{{f_1}\left(1 \right)} \over 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 $f_{2} (x) = x^{3} f_{2} (\frac{1}{2})$ {f_2}\left(x \right) = {x^3}{f_2}\left({{1 \over 2}} \right) for all x ∈ ℝ₊ if and only if $\begin{matrix} f_{1} (x) = f_{2} (x) = α (x + x^{2}), & x \in ℝ, \end{matrix}$ \matrix{{{f_1}\left(x \right) = {f_2}\left(x \right) = \alpha \left({x + {x^2}} \right),} & {x \in {\mathbb R},} \cr} 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

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