On alienation of two functional equations of quadratic type

We deal with an alienation problem for an Euler–Lagrange type functional equation f(αx+βy)+f(αx-βy)=2α2f(x)+2β2f(y)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} f(\alpha x + \beta y) + f(\alpha x - \beta y) = 2\alpha ^2f(x) + 2\beta ^2f(y) \end{aligned}$$\end{document}assumed for fixed nonzero real numbers α,β,1≠α2≠β2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha ,\beta ,\, 1 \ne \alpha ^2 \ne \beta ^2$$\end{document}, and the classic quadratic functional equation g(x+y)+g(x-y)=2g(x)+2g(y).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} g(x+y) + g(x-y) = 2g(x) + 2g(y). \end{aligned}$$\end{document}We were inspired by papers of Kim et al. (Abstract and applied analysis, vol. 2013, Hindawi Publishing Corporation, 2013) and Gordji and Khodaei (Abstract and applied analysis, vol. 2009, Hindawi Publishing Corporation, 2009), where the special case g=γf\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$g = \gamma f$$\end{document} was examined.

No word has been said what happens in the case where the coefficient α fails to be rational.The following example shows that, in general, in the case of irrational α the assertion mentioned is no longer valid even in the case where X = Y = R-the real line.
Example.Let f : R −→ R be a nonzero additive function such that f (πx) = π 2 f (x), x ∈ R. (The existence of such functions may be derived from Kuczma's monograph [5, Theorem 4.12.1 ], for instance; clearly each such function is discontinuous.)Since f is odd (as an additive function) it fails to be quadratic because, otherwise, being even it would vanish identically on X, a contradiction.On the other hand, for every x, y ∈ R and each arbitrarily fixed β ∈ R one has which proves that f yields a solution of Eq. ( 3) with α = π and γ = −β 2 .
It seems worthwhile to remark also that the norm structures in real linear spaces X and Y in Kim-Han-Shim's result are superfluous.
Observe that Eq. (3) results from ( 1) and ( 2) with g = γf by summing them side by side.In what follows we will examine a more general equation (4) with a quadratic difference of an arbitrary function g instead of that of γf .
Our chief concern will be to answer the question when Eq. ( 4) forces the functions f and g to satisfy Eqs. ( 1) and ( 2), respectively; in other words, whether Eqs. ( 1) and ( 2) are alien.The Example above shows that in the case where α is irrational, in general, this is not the case.
Let us start with establishing the following two facts.Fact 1.Since Eq. ( 4) is linear the even parts e f , e g of f and g, as well as the odd parts o f , o g of f and g, are solutions of (4); here where h : X −→ Y is a function.Fact 2. Since Eq. ( 4) gives the equality the functions f − f (0) and g − g(0) satisfy Eq. ( 4) whenever so do the functions f and g.Therefore, without loss of generality, we may assume that we deal with Eq. ( 4) jointly with f (0) = 0 and g(0) = 0.

Lemma. If an odd function
for all x, y ∈ X where γ is a real number with 0 = γ 2 = 1, then ϕ is a polynomial function of at most third degree.More precisely, there exist a symmetric map A 3 : X 3 −→ Y enjoying the property that is additive in each variable, and an additive map a : X −→ Y such that where A 3 (x):=A 3 (x, x, x), x ∈ X.
Proof.Since γ 2 = 1 Eq.(γ) may be rewritten in the form On account of (a modified) Theorem 9.5 from Székelyhidi's monograph [7] we infer that ϕ is a polynomial function of at most 3-rd degree, i.e.
Due to the oddness of ϕ we have ϕ(0) = 0 whence c = 0 and the summand A 2 disappears.Consequently, ( Now, applying representation (5) jointly with the well known addition formula and (γ), on account of some simple calculations, we deduce that (A 3 ) holds true for all x, y ∈ X, as claimed.Note that the equality (A 3 ) is always satisfied provided that γ is rational.
Proof.On setting x = 0 in (4), we get or, equivalently, for all x, y ∈ X.In particular, with y = 0 Eq. ( * * ) implies that Since α = 0 Eq. ( * * ) may be written in the following form: whence with the aid of * * * , by setting γ := β/α, we infer that In view of the inequality β = 0 this states nothing else but (γ) with ϕ = f .Therefore, by means of the Lemma, there exist a symmetric map A 3 : X 3 −→ Y enjoying the property (A 3 ), that is additive in each variable, and an additive map a : X −→ Y such that where A 3 (x) := A 3 (x, x, x), x ∈ X.In view of the definition of γ condition (A 3 ) may equivalently be rewritten as which gives the second part of (6).
Finally, from (5f) and * * * we have ϕ(x) valid for all x ∈ X.Clearly, for all x ∈ X, we have also which states that ϕ vanishes identically on X, i.e.
and a(αx) = α 2 a(x), x ∈ X. (7) To get the first part of ( 6) it remains to apply representation (5f) in ( * * ) and to use the addition formula for A 3 jointly with (7).
To finish the proof it remains to perform a mechanical calculation showing that each pair (f, g) of functions described above yields a solution to Eq. ( 4).
Remark 2. In the case where both nonzero coefficients α and β are rational and α = 1, the first condition in (6) forces the function to vanish.In particular, A 3 = 0 and, consequently, the only odd solution of Eq. ( 4) is just the zero function because the additive summand a in representation (5f) has to satisfy (7) and being rationally homogeneous it has to vanish as well.Proposition 2. If functions f and g satisfying Eq. (4) jointly with f (0) = g(0) = 0 are even then both f and g are quadratic provided that 1 = α 2 = β 2 = 0 and α is a rational number.
Proof.On setting y = 0 (resp.x = 0) in (4), we get with h := g + β 2 f.On account of (a modified) Theorem 9.5 from Székelyhidi's monograph [7] we infer that F is a polynomial function of at most 4-th degree, i.e.
here A k (x) = A k (x, . . ., x), x ∈ X, where A k : X k −→ Y is a symmetric map that is additive in each variable, k ∈ {1, 2, 3, 4}.The evenness of f and g imply the evenness of F whence A 3 = A 1 = 0. Consequently, in view of (9), we get With the aid of the well known addition formulas for A 4 and A 2 : +6A 4 (x, x, y, y) + 4A 4 (x, y, y, y) + A 4 (y), x,y ∈ X, and a simple calculation shows that the equality is valid for all x, y ∈ X.Now, Eq. ( 4) assumes the form Replacing here x by αx and y by αy and taking the first equality in (8) into account along with the fact that α is a nonzero rational number we obtain The latter two equations imply that the equality (α 2 − 1)A 4 (x, x, y, y) = 0 is satisfied for all x, y ∈ X. Since, by assumption, α 2 = 1 we infer that A 4 (x, x, y, y) = 0 for all x, y ∈ X.Consequently, one has Putting here βx and αy in place of x and y, respectively, we obtain the equality which, on account of (8) and the fact that α = 0, gives now or, equivalently, Interchanging here the roles of x and y and applying (8) jointly with the evenness of f we infer that Therefore, in view of (10), we conclude that the equality holds true for all x, y ∈ X.By assumption α 2 = β 2 which forces f to be quadratic.Thus the proof has been completed because the quadraticity of g results now easily from ( 4), (10) and the quadraticity of f .Theorem 1.Let X, Y be two real linear spaces and let α, β be two nonzero real numbers such that If α is rational then equations and are alien in the class of all mappings f, g : X −→ Y vanishing at zero, i.e. equation is satisfied if and only if both Eqs.(1) and (2) are fulfilled provided that f (0) = g(0) = 0.Moreover, if that is the case, then f is quadratic.
Proof.Let a pair (f, g) of mappings from X into Y be a solution of Eq. ( 4).Since the pairs (o f , o g ), (e f , e g ) yield also solutions of (4) (cf.Fact 1), Proposition 1 guarantees the existence of a symmetric map A 3 : X 3 −→ Y enjoying the properties (6), that is additive in each variable, and an additive map a : X −→ Y satisfying the condition a(αx) = α 2 a(x) for all x ∈ X such that where A 3 (x) = A 3 (x, x, x), x ∈ X.Moreover, conditions (7) are satisfied which jointly with the rationality of the coefficient α / ∈ {0, 1} imply that A 3 = a = 0. Consequently, o f = o g = 0, i.e. both f and g are even.
On the other hand, due to the rationality of α, Proposition 2 states then that both f and g are quadratic.In particular, g satisfies Eq. ( 2) which jointly with (4) implies the validity of (1).Therefore we are faced with the alienation of Eqs. ( 1) and ( 2), as claimed.
Since the converse implication is trivial the proof has been completed.As an easy consequence we get now the following Theorem 2. Let X, Y be two real linear spaces and let α, β be two nonzero real numbers such that If α is rational then Eqs.(2 ) Moreover, if that is the case and g(0) = 0, then α 2 + β 2 = 1 and f , with quadratic mappings F and G.
It is a straightforward matter to verify the remaining statements.
Corollary.Fix a real constant γ and put g = γf in Eq. (4) to get an Euler-Lagrange type equation which was examined in paper [4] by Chang Il Kim, Giljun Han and Seong-A.Shim.The basic results on solutions of (HKS) established in [4] become special cases of Theorem 1 from the present paper.
Remark 3. Equation (HKS) admits nontrivial solutions that are not quadratic (see Proposition 1 and the Example above).They were omitted in [4] because the authors were dealing mainly with rational coefficients α in (HKS).
Remark 4. I presented the results of the present paper during the 20th Debrecen-Katowice Winter Seminar on Functional Equations and Inequalities that was held in Hajdúszoboszló (Hungary) from January 29 till February 1, 2020.At that time my proofs were entirely different.In particular, Theorem 4 from Kim-Han-Shim paper [4] and, indirectly, Theorem 2.1 from Jun-Kim-Chang's paper [3] were used as proof tools.
During the discussion after my talk, a Hungarian mathematician Mihály Bessenyei (see [1]) asked me the question whether such approach was unavoidable since in that case my generalization of results from [4] is just formal only.During the meeting I was unable to answer Bessenyei's question but it mobilized me to work on that problem.Finally, I succeeded to get a positive answer using a celebrated theorem of Székelyhidi from [7].