On the generalized Fréchet functional equation with constant coefficients and its stability

We study a generalization of the Fréchet functional equation, stemming from a characterization of inner product spaces. We show, in particular, that under some weak additional assumptions each solution of such an equation is additive. We also obtain a theorem on the Ulam type stability of the equation. In its proof we use a fixed point result to show the existence of an exact solution of the equation that is close to a given approximate solution.


Introduction
In this paper we study the following functional equation (with constant coefficients) where A 1 , . . . , A 7 ∈ K and K ∈ {R, C} (R and C denote the fields of real and complex numbers, respectively), in the class of functions F : X → Y , where (X, +) is a commutative monoid (i.e., a semigroup with a neutral element denoted by 0) and Y is a Banach space over the field K.

This equation is a generalization of the Fréchet functional equation
F (x + y + z) + F (x) + F (y) + F (z) = F (x + y) + F (x + z) + F (y + z). (2) Namely, in case A i = A j = 0 for i, j ∈ {1, . . . , 7}, Eq. (1) can be easily reduced to Eq. (2). Therefore, we will be interested mainly in the case where A i = A j for some i, j.
Fréchet [17] proved that a normed space (X, · ) is an inner product space if and only if for all x, y, z ∈ X J. Brzdęk et al. AEM x + y + z 2 + x 2 + y 2 + z 2 = x + y 2 + x + z 2 + y + z 2 . (3) This means that a normed space X is an inner product space if and only if the function F , given by F (x) ≡ x 2 , satisfies Eq. (2). For further similar results we refer to [1,4,5,7,13,15,20,[27][28][29]. The first part of this paper contains some results on solutions of Eq. (1) in the case where (X, +) is a monoid and Y is a Banach space. It is known that, if X is a group and Y is an abelian group divisible by 2, then the general solution of Eq. (2) is the sum of a quadratic and an additive function (see [23, pp. 249-250]). An analogous result is true for Eq. (1), but then each solution is a sum of a quadratic, an additive and a constant function (see [4,Proposition 7]). We will show that under the assumption that at least two coefficients A i are not equal, every solution F of Eq. (1), with F (0) = 0, is an additive function.
In the second part of this paper we will consider the problem of Ulam stability of Eq. (1). A stability result concerning Eq. (2) can be found in [5]. Analogous outcomes for Eq. (1) were proved in [26] (see also [4,Corollary 6]), under the assumptions that (X, +) is a commutative group, A 1 = 0 and A 2 + A 3 + A 4 = A 5 + A 6 + A 7 , which has motivated us to study a bit further the dependence of the stability of Eq. (1) on the values of the coefficients A 1 , . . . , A 7 . Moreover, we weaken the assumptions on X by assuming 'only' that it is a commutative monoid. In this way we also complement the recent outcomes in [4].
(H1) S is a nonempty set, E is a Banach space, and functions f 1 , . . . , f k : S → S and l 1 , . . . , l k : S → R + are given. (H2) T : E S → E S is an operator satisfying the inequality Assume that functions ε : S → R + and ϕ : S → E fulfil the following two conditions Then there exists a unique fixed point ψ of T with Moreover, Using this theorem we prove that an operator defined in the space Y X determines an exact solution of Eq. (1) as the limit of a sequence of its iterates on an approximate solution of this equation. Such a method has been used in, e.g., [4][5][6]10,26,30,33,34]. Moreover, the results that we provide correspond to the outcomes in [3,9,13,16,18,21,24,25,31,32] (for more details see, e.g., [8,19,22]) and complement [4, Corollary 6].

Additive solutions of the generalized Fréchet functional equation
In this section we present some auxiliary observations on solutions of Eq. (1). The main result of this part says that, under a natural additional assumption concerning the coefficients A i , each solution F of Eq. (1), with F (0) = 0, is an additive function.
Throughout this section X is a monoid, Y is a vector space over the field K ∈ {R, C}, and A 1 , . . . , A 7 ∈ K. Let us recall that every solution of Eq. (2) satisfies the equality F (0) = 0. But this is not necessarily the case for Eq. (1). The following result gives a sufficient condition for the equality F (0) = 0.

Proposition 2. If the condition
is fulfilled, then each solution F : X → Y of Eq. (1) satisfies the condition F (0) = 0.
From now on we are interested in solutions of Eq. (1) which satisfy the relation F (0) = 0.
Proof. Putting y = 0, z = 0 in (1), we get Thus since F is a nonzero function. In a similar way we obtain the other two relations in (9). More precisely, putting x = 0, z = 0 and x = 0, y = 0 we get Proof. Let a : X → Y be an additive function. Then By (9) we obtain Hence using the additivity of the function a once again we get Thus by (10) and (11) we get i.e. a satisfies Eq. (1). Proof. Let a : X → Y be a nonzero additive function. Assume that a satisfies Eq. (1). Then by Proposition 3 relations (9) hold, since a(0) = 0. The converse implication follows directly from Proposition 4.
Proof. Assume that Eq. (1) has a nonzero solution F : X → Y such that F (0) = 0. Then by Proposition 3 relations (9) hold. Hence by Corollary 5 every nonzero additive function satisfies Eq. (1). Moreover, it is easily seen that the function a : X → Y , given by a(x) = 0 for all x ∈ X, is a solution of Eq. (1).
Thus we have shown that the set of all nonzero solutions F : X → Y of Eq. (1) such that F (0) = 0, if non-empty, contains the set of all additive functions a : X → Y . Under the assumption that relations (9) hold we obtain by Proposition 4 that the considered set of solutions of Eq. (1) is non-empty. The next result states that, under a suitable assumption on the coefficients A i , the two sets coincide.
First let us assume that A 1 = A 5 , i.e., B = 0. Putting z = 0 into (1) we get By (12) Eq. (13) can be written in the form Now we proceed to the case where A 1 = A 5 . Then by (12), A 6 = A 2 and A 7 = A 3 and Eq. (1) has the form Hence with x = 0 we obtain that Next, from (9) we get that Hence since by (12) we have A 7 = A 3 and A 6 = A 2 . Consequently, Eq. (14) can be written in the form Now we consider the following two subcases: (16) we obtain that F is additive. Now let us assume that A 1 = A 3 . Then from (12) and (15) we obtain that Hence putting y = 0 we get Directly from Theorem 7 we obtain the following result.

Corollary 8. If there exists a nonadditive nonzero solution
Moreover, from Theorem 7 and Proposition 2 we get the following description of the set of solutions of Eq. (1). Now we proceed to the general case, without the assumption that F (0) = 0.
where a : X → Y is an additive function and c = F (0). Proof. Let F satisfy Eq. (1). If F (0) = 0, then from Theorem 7 we obtain that (17) holds with c = 0. Now, let us assume that F (0) = 0. Define the function F 0 : X → Y by the formula Then F 0 is a solution of Eq. (1), since F satisfies Eq. (1) and consequently condition (7) holds. On account of Theorem 7 the function F 0 is additive. Thus At the end of this section let us recall that in the only remaining case A 1 = · · · = A 7 there exists a nonadditive solution F of Eq. (1) such that F (0) = 0. Namely, without loss of generality we can assume that A 1 = · · · = A 7 = 1 (we exclude the trivial case A 1 = · · · = A 7 = 0). Next, let (X, · ) be an inner product space and Y = R. Then the function F : X → R, given by F (x) := x 2 , is a solution of Eq. (1), where the norm is derived from the inner product (cf. [17]). Moreover, in [26] it was proved that if F : X → R, F (x) = x 2 is a solution of (1), then A 1 = · · · = A 7 (which corresponds to Corollary 8).
We prove a stability result for the generalized Fréchet functional equation. As a corollary we obtain that near an approximate solution of Eq. (1) there exists an additive function. The proof of this fact is based on the results of the previous section.
Let us recall that the following theorem, concerning the stability of Eq. (1), was proved in [26].
Theorem 12. Let A 1 = 0 and Assume that f : X → Y , c : Z\{0} → [0, ∞) and L : X → [0, ∞) satisfy the following three conditions: The main theorem of this section corresponds to Theorem 12 and reads as follows.

Theorem 13. Let
Vol. 92 (2018) On the generalized Fréchet functional equation 363 If f : X → Y fulfils the condition then there exists a unique function F : X → Y satisfying (1) such that F (0) = 0 and where with Proof. In the first part of the proof we define an operator T and show that T satisfies the assumptions of Theorem 1. Taking y = z = x in (20) we obtain Hence, for each x ∈ X, where In particular, for x = 0 we have Let us note that the operator T is linear. From (23) and (24) we get directly that which means that condition (4) holds. In particular, on account of (23) and (25) with x = 0, we have

Now we will show that condition (H2) of Theorem 1 is satisfied with
Let us fix ξ, μ ∈ Y X . Then for every x ∈ X we have Hence by the triangle inequality we obtain that Actually, we use this relation for x ∈ X\{0}. In case x = 0 we have a bit more: which means that condition (H2) holds. Define an operator Λ : R + X → R + X as in (H3) by Vol. 92 (2018) On the generalized Fréchet functional equation 365 for every η ∈ R + X . Then for each η ∈ R + X we have and Λη(0) := β 0 η(0).
Let us note that the operator Λ is nondecreasing, i.e., Λη ≤ Λζ for all η, ζ ∈ R + X with η ≤ ζ. Moreover, by (26) and (27) T Now we will show that condition (5) is satisfied, i.e., the function series Fix an x ∈ X\{0}. In view of (28) and (18), we have By induction one can show that the monotonicity and linearity of Λ implies Consequently, for each x ∈ X\{0} we receive the following estimate: .
In case x = 0 we have Hence by induction we obtain Therefore Thus we have shown that By Theorem 1 (with S = X and E = Y ), there exists a function F : X → Y satisfying the equation such that Moreover, Let us note that condition (34) means that Eq. (1) is satisfied for x = y = z.
Next we will show that the function F satisfies Eq. (1) for all x, y, z ∈ X. To obtain this fact we will prove by induction that for all (x, y, z) ∈ X 3 , n ∈ N 0 := N ∪ {0} we have where λ := max{β, β 0 }. For n = 0 condition (35) follows directly from (20). Assume that (35) holds for some n ∈ N 0 and all (x, y, z) ∈ X 3 . Then by (24) we have Vol. 92 (2018) On the generalized Fréchet functional equation 367 for every (x, y, z) ∈ X. Hence by (18) for (x, y, z) ∈ X. Finally, by (25), which ends the proof of (35). Letting n → ∞ in (35), we obtain Thus, we have proved that, there exists a function F : X → Y satisfying Eq. (1) for which Now we will show that F (0) = 0. From (24) we get by induction that since β 0 < 1. Consequently, we have F (0) = lim n→∞ T n f (0) = 0. It remains to prove the statement concerning the uniqueness of F . To get this fact we first show by induction that for all ξ, μ ∈ Y X , n ∈ N By (29) condition (39) holds for n = 1. Fix ξ, μ ∈ Y X and assume that (39) holds for n ∈ N. Then by (29) Hence by the inductive hypothesis and the monotonicity of Λ we obtain Corollary 16. Let A i = A j for some i, j ∈ {1, . . . , 7}, A 2 + A 3 + A 4 = 0 and relations (9) hold. Let a function L : 3 are given by (19).
If f : X → Y fulfils condition (20), then there exists a unique additive solution a : X → Y of Eq. (1) and a constant c ∈ Y such that where ρ L (x) is given by (42).
Let us note that under the assumptions of Theorem 13 and Corollary 14 we get that − 3A 1 + 2(A 5 + A 6 + A 7 ) = 0, in the case where the solution of Eq. (1) occurring in each of these results is a nonzero function. Indeed, from Proposition 3 and Corollary 5, respectively, we obtain that relations (9) are satisfied. Hence we get that A 2 + A 3 + A 4 = −3A 1 + 2(A 5 + A 6 + A 7 ).