Solution and Stability of the Multiquadratic Functional Equation

Proposition 1 (see [11].)A function f : V → W is quadratic if and only if there exists a unique symmetric biadditive function B : V × V → W such that f(x) = B(x, x) for any x ∈ V. The biadditive function B is given by

Theorem 2. A mapping f : Vⁿ → W is multiquadratic if and only if there exists a multiadditive mapping M : V²ⁿ → W such that

for all x₁, …, x_n ∈ V, and M satisfies the following symmetric condition for all x_ij ∈ V, where i ∈ {1,2, …, n} and j ∈ {1,2}. Moreover, the mapping M is given by where x_ij ∈ V, i ∈ {1,2, …, n}, j ∈ {1,2}.

Proof. We prove this theorem by using induction on n. Clearly, Theorem 2 is true for n = 1 thanks to Proposition 1. Now, we assume that the present theorem is true for some n ≥ 2, and we consider the case for n + 1.

We first assume that there exists a multiadditive mapping M : V²⁽ⁿ⁺¹⁾ → W such that

for all x₁, …, x_n, x_n+1 ∈ V, and M satisfies the following symmetric condition: for all x_ij ∈ V, where i ∈ {1,2, …, n, n + 1} and j ∈ {1,2}. Then, for each i ∈ {1,2, …, n, n + 1}, we have that

for all $$. Thus, f is multiquadratic.

Conversely, we assume that f : Vⁿ⁺¹ → W is a multiquadratic function. We need to find the desired multiadditive function M : V²⁽ⁿ⁺¹⁾ → W. For this, we give the following notations.

For each fixed z ∈ V, define the mapping g_z : Vⁿ → W by

Then g_z is a multiquadratic mapping (as f : Vⁿ⁺¹ → W is multiquadratic). By induction, we let M_z : V²ⁿ → W denote the corresponding multiadditive mapping for g_z; that is, M_z satisfies the symmetric condition (5) and

for all x₁, …, x_n ∈ V. Moreover, the mapping M_z : V²ⁿ → W is given by for all x_ij ∈ V, where i ∈ {1,2, …, n} and j ∈ {1,2}.

On the other hand, for any fixed elements x₁₁, x₁₂, …, x_n1, x_n2 ∈ V, define $$ by

for all x ∈ V. It can be verified that $$ is a quadratic mapping. Thus, it follows from Proposition 1 that there exists a symmetric biadditive mapping $$ such that

for all x ∈ V. The mapping $$ is given by

for all x, y ∈ V.

Now, we define the mapping M : V²⁽ⁿ⁺¹⁾ → W by

for all x_ij ∈ V, i ∈ {1,2, …, n + 1}, j ∈ {1,2}. In the following, we will show that M is the desired function for f : Vⁿ⁺¹ → W. First, we show that M is multiadditive. Indeed, by the definition of M (see (16)) and noting that for any z ∈ V the function M_z is multiadditive, one can obtain that for each i ∈ {1,2, …, n}, for all $$. Moreover, by the definition of M in (16) and the notations we gave in (13) and (15), we have that for all $$. Similarly, we can see that M is additive in the other variables. Thus, we have shown that M is multiadditive.

Furthermore, since f is multiquadratic, we obtain that

for all x₁, …, x_n, x_n+1 ∈ V. Thus, by the definition of M in (16) and the notations we gave in (10) and (11), one has for all x₁, …, x_n, x_n+1 ∈ V.

Now, we verify the expression of the mapping M. By the definition of M again and the notations we gave in (10) and (12), also noting that f is multiquadratic, one can obtain that

for all x_ij ∈ V, i ∈ {1,2, …, n + 1}, j ∈ {1,2}.

Finally, we check the symmetric property of M. Fix any x_ij ∈ V, where i ∈ {1,2, …, n + 1} and j ∈ {1,2}. Since f is multiquadratic, it follows that f is an even mapping in each variable. Then by (21), it is easy to verify that

Moreover, due to the symmetric property of $$ and $$ and from the definition of M (see (16)) we can get for each i ∈ {1,2, …, n}. So the desired symmetric property of M is proved. Thus, we have shown that M : V²⁽ⁿ⁺¹⁾ → W is the desired multiadditive mapping for the multiquadratic mapping f : Vⁿ → W. The proof is complete.

Theorem 3. Let V be a commutative semigroup with the identity element 0, and let W be a linear space. A mapping f : Vⁿ → W is multiquadratic if and only if

for all (x₁₁, …, x_n1), (x₁₂, …, x_n2) ∈ Vⁿ.

Proof. Assume that f : Vⁿ → W satisfies (24). Putting

in (24) we get 2ⁿf(0, …, 0) = 2²ⁿf(0, …, 0), and consequently, we have f(0, …, 0) = 0. Next, fix j ∈ {1, …, n}, x_j1 ∈ V, and put $$, where i_k ∈ {1,2}, for k ∈ {1, …, n}∖{j}. Then, by (24), and thus f(0, …, 0, x_j1, 0, …, 0) = 0. Continuing in this fashion, we obtain that f(x) = 0 for any x ∈ Vⁿ with at least one component which is equal to 0.

Now, fix j ∈ {1, …, n}, x₁₁, …, x_n1, x_j2 ∈ V and put x_k2 = 0 for k ∈ {1, …, n}∖{j} in (24). Then

and thus which proves that f is multiquadratic.

Conversely, we assume that f is multiquadratic, and we prove (24) by mathematical induction. If f : V → W is quadratic, then f(x₁₁ + x₁₂) + f(x₁₁ − x₁₂) = 2f(x₁₁) + 2f(x₁₂) for all x₁₁, x₁₂ ∈ V. So, (24) holds for n = 1. It is easy to verify that (24) holds for n = 2. Indeed,

for all x₁₁, x₁₂, x₂₁, x₂₂ ∈ V. Assume that (24) holds for some positive integer n > 2. Then, Thus, (24) holds for n + 1, and this completes the proof.

Theorem 4. Assume that for every i ∈ {1, …, n}, φ_i : Vⁿ⁺¹ → [0, ∞) is a mapping such that for any (x₁, …, x_n+1) ∈ Vⁿ⁺¹

If f : Vⁿ → W is a function satisfying for all $$, i ∈ {1, …, n}, then for every i ∈ {1, …, n} there exists a multiquadratic mapping F_i : Vⁿ → W such that for any (x₁, …, x_n) ∈ Vⁿ one has For every i ∈ {1, …, n} the function F_i is given by for all (x₁, …, x_n) ∈ Vⁿ.

Proof. Fix x₁, …, x_n ∈ V, j ∈ N ∪ {0} (where N denotes the set of the positive integers) and i ∈ {1, …, n}. Putting $$ in (32), we get

Hence Dividing both sides of the above inequality by 4^j and replacing x_i by 2^jx_i, we obtain and consequently for any nonnegative integers l and m with l < m, we obtain Therefore, it follows from (31) that $$ is a Cauchy sequence. Since the space W is complete, this sequence is convergent, and we define F_i : Vⁿ → W by (34). Putting l = 0, letting m → ∞ in (38), and using (31), we see that (33) holds.

Finally, fix also $$, j ∈ N, and notice that according to (32) we have

Next, fix k ∈ {1, …, n}∖{i}, $$, and assume that k < i (the same arguments apply to the case where k > i). Then, it follows from (32) that Letting j → ∞ in the above two inequalities and using (31), we see that the mapping F_i is multiquadratic.

Theorem 5. Assume that φ : V²ⁿ → [0, ∞) is a mapping such that

for all (x₁₁, x₁₂, …, x_n1, x_n2) ∈ V²ⁿ. If f : Vⁿ → W is a function satisfying for all (x₁₁, x₁₂, …, x_n1, x_n2) ∈ V²ⁿ and letting f(x) = 0 for any x ∈ Vⁿ with one component which is equal to 0, then there exists a unique multiquadratic mapping F : Vⁿ → W such that for all (x₁₁, …, x_n1) ∈ Vⁿ. The function F is given by for all (x₁₁, …, x_n1) ∈ Vⁿ.

Proof. Fix (x₁₁, …, x_n1) ∈ Vⁿ and j ∈ N ∪ {0}. Putting x_i2 : = x_i1 for i ∈ {1, …, n} in (42), we get

Dividing both sides of the above inequality by 4^n(j+1) and replacing x_i1 by 2^jx_i1 for i ∈ {1, …, n}, we see that and consequently for any nonnegative integers l and m with l < m we obtain Therefore, it follows from (41) that {1/4^njf(2^jx₁₁, …, 2^jx_n1)} _j∈N is a Cauchy sequence. Since the space W is complete, this sequence is convergent, and we define F : Vⁿ → W by (44). Putting l = 0, taking m → ∞ in (47), and using (41), we can see that the inequality (43) holds.

Next, fix also (x₁₂, …, x_n2) ∈ Vⁿ, and note that according to (42) we have

Letting j → ∞ in the above inequality and using (41), we see that F satisfies (24). By Theorem 3, we obtain that F is multiquadratic.

Finally, assume that F^′ : Vⁿ → W is another multiquadratic mapping satisfying (43). Fix k ∈ N ∪ {0}. Since F and F^′ are multiquadratic mappings, it is easy to verify that

Then, using (41) and (43), we have hence letting k → ∞ we obtain F = F^′.