On a functional equation that has the quadratic-multiplicative property

Theorem 1.1

[10] Let G be an abelian group and E be a real Banach space. If a mapping f : G → E satisfies the inequality

for some δ > 0 and all x , y ∈ G , then there exists a unique quadratic mapping q : G → E such that

Theorem 1.2

[12] Let E 1 and E 2 be a real normed space and a real Banach space, respectively, and let p ≠ 2 be a positive constant. If a mapping f : E 1 → E 2 satisfies the inequality

for some ε > 0 and all x , y ∈ E 1 , then there exists a unique quadratic mapping q : E 1 → E 2 such that

Theorem 2.1

[13] A function f : ℝ → ℝ satisfies the system of the following functional equations:

if and only if there exists an additive-multiplicative function w : ℂ → ℂ such that f ( x ) = | w ( x ) | 2 for all x ∈ ℝ .

Theorem 2.2

[14] A function f : ℝ → ℝ is additive-multiplicative, i.e., f satisfies the system of the following functional equations:

if and only if f ≡ 0 or f ( x ) = x for all x ∈ ℝ .

Theorem 2.3

If a function ϕ : ℝ → ℝ satisfies ( 1.3 ) for all s , t , u , v ∈ ℝ , then ϕ ≡ 0 ,  2 ϕ ≡ 1 or ϕ ( t ) = t 2 for all t ∈ ℝ .

Proof

Let ϕ be a constant function. Then, ϕ ≡ 0 or 2 ϕ ≡ 1 . We assume that ϕ is not constant. Letting t = u = v = 0 in (1.3), we obtain

This implies that ϕ ( 0 ) = 0 , since ϕ is not constant. Letting u = v = 0 in (1.3), we get

Putting t = 1 in (2.1), we get ϕ ( 1 ) = 1 , since ϕ is not constant. Letting t = v = 1 in (1.3) and using ϕ ( 1 ) = 1 , we obtain

By Theorem 2.1, there exists an additive-multiplicative function w : ℂ → ℂ such that ϕ ( t ) = | w ( t ) | 2 for all t ∈ ℝ . Since ϕ ( 1 ) = 1 , letting u = v = 1 in (1.3), we get

Then, w ( s ) ∈ ℝ or w ( t ) ∈ ℝ . This implies w ( ℝ ) ⊆ ℝ . Therefore, w : ℝ → ℝ is an additive-multiplicative function, and by Theorem 2.2 we conclude w ( t ) = t for all t ∈ ℝ . Hence, ϕ ( t ) = t 2 for all t ∈ ℝ .□

Corollary 2.4

If a nonzero mapping ϕ : ℝ → ℝ with ϕ ( 0 ) = 0 satisfies ( 1.3 ) for all s , t , u , v ∈ ℝ , then ϕ ( t ) = t 2 for all t ∈ ℝ .

Proposition 2.5

Let ϕ : ℝ → ℝ be a derivation and satisfy ( 1.3 ) for all s , t , u , v ∈ ℝ . Then, ϕ ≡ 0 .

Proof

Since ϕ is a derivation, we get ϕ ( 0 ) = ϕ ( 1 ) = 0 . Letting u = v = 0 in (1.3), we get ϕ ( s t ) = ϕ ( s ) ϕ ( t ) for all s , t ∈ ℝ . Hence, ϕ ( s ) = ϕ ( s ) ϕ ( 1 ) = 0 for all s ∈ ℝ .□

Theorem 3.1

Let ε ∈ { − 1 , 1 } and ϑ : ℝ 4 → [ 0 , ∞ ) be a function such that

Proof

We assume that ε = 1 . Letting u = v = 0 in (3.1), we get

Putting t = 1 in (3.3), we get

Letting t = v = 1 in (3.1), we get

Then, it follows from (3.4) and (3.5) that

where ψ ( s , u ) : = ϑ ( s , 1 , u , 1 ) + 2 ϑ ( s , 1 , 0 , 0 ) + 2 ϑ ( u , 1 , 0 , 0 ) . By [15], there exists a unique quadratic function Q : ℝ → ℝ such that

Then from (3.1) and (3.3), we obtain

Since Q ( 0 ) = 0 , we infer Q ( 1 ) ∈ { 0 , 1 } . Hence by Theorem 2.3, we get Q ( s ) = s 2 for all s ∈ ℝ if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) = 0 . Then, we obtain (3.2).

Similarly, we can obtain the result for ε = − 1 . Hence, the proof of the theorem is now complete.□

Corollary 3.2

Suppose that ϕ : ℝ → ℝ is a function satisfying

for some ε ⩾ 0 . Then, we have the following assertions:

1. If ϕ ( 0 ) ≠ 0 , then ϕ is bounded.

2. If ϕ ( 0 ) = 0 , then there exists a unique quadratic function Q : ℝ → ℝ , which satisfies ( 1.3 ) and

where Q ( s ) = s 2 if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) ≠ 1 .

Proof

Let ϕ ( 0 ) ≠ 0 . Letting t = u = v = 0 in (3.6), we obtain

Then, we get

For the case ϕ ( 0 ) = 0 , the result follows from Theorem 3.1.□

Corollary 3.3

Let ε ⩾ 0 and ϑ : ℝ 4 → [ 0 , ∞ ) be a function. Suppose that ϕ : ℝ → ℝ is a function satisfying

Let { e 1 , e 2 , e 3 , e 4 } be the standard basis in ℝ 4 and ϑ i : ℝ → ℝ be a function defined by ϑ i ( s ) = ϑ ( s e i ) for a given i ∈ { 1 , 2 , 3 , 4 } . If ϕ ( 0 ) ≠ 0 and ϑ i is bounded for some i, then ϕ is bounded.

Corollary 3.4

Fix ε ⩾ 0 and p ∈ ( 0 , 2 ) . Suppose that ϕ : ℝ → ℝ is a function satisfying

Then, we have the following assertions:

1. If ϕ ( 0 ) ≠ 0 , then

   ϕ ( s ) − 1 2 ⩽ ε | s | p , s ∈ ℝ .

2. If ϕ ( 0 ) = 0 , then there exists a unique quadratic function Q : ℝ → ℝ which satisfies ( 1.3 ) and

where Q ( s ) = s 2 if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) ≠ 1 .

Theorem 3.5

Fix ε ⩾ 0 and p > 4 . Suppose that ϕ : ℝ → ℝ is a function satisfying

Then, we have the following assertions:

1. If ϕ ( 0 ) ≠ 0 , then

   ϕ ( s ) − 1 2 ⩽ ε | s | p , s ∈ ℝ .

2. If ϕ ( 0 ) = 0 , then there exists a unique quadratic function Q : ℝ → ℝ , which satisfies (1.3) and

where Q ( s ) = s 2 if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) ≠ 1 .

Proof

It suffices to prove (ii). Letting u = v = 0 in (3.7), we get

Putting v = t and u = s in (3.7), we obtain

Using (3.9) and (3.10), we obtain

Replacing s and t by 2 2 ( n + 1 ) / 2 and 1 2 ( n + 1 ) / 2 , respectively, in the last inequality and multiplying the resulting inequality by 4 n , we obtain

Then,

This implies that the sequence {4 n ϕ ( s 2 n ) } is a Cauchy sequence and so the sequence is convergent for all s ∈ ℝ . We define a function Q : ℝ → ℝ by

Since ϕ ( 0 ) = 0 , we get Q ( 0 ) = 0 . By Assumption (3.7) and Inequality (3.9), we obtain that Q satisfies (1.3) and Q ( s t ) = Q ( s ) Q ( t ) for all s , t ∈ ℝ . Hence by Theorem 2.3, we get Q ( s ) = s 2   ( s ∈ ℝ ) if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) = 0 . Letting n = 0 and m → ∞ in (3.12), we obtain (3.8).□

Theorem 4.1

[17] Let ( X , d ) be a generalized complete metric space and Λ : X → X be a strictly contractive function with the Lipschitz constant L < 1 . Suppose that for a given element a ∈ X there exists a nonnegative integer k such that d ( Λ k + 1 a , Λ k a ) < ∞ . Then,

1. the sequence { Λ n a } n = 1 ∞ converges to a fixed point b ∈ X of Λ ;

2. b is the unique fixed point of Λ in the set Y = { y ∈ X :   d ( Λ k a , y ) < ∞ } ;

3. d ( y , b ) ⩽ 1 1 − L d ( y , Λ y ) for all y ∈ Y .

Theorem 4.2

Let 0 < L < 1 and ϑ : ℝ 4 → [ 0 , ∞ ) be a function such that

where ψ ( s , t ) ≔ ϑ ( s , 1 , t , 1 ) + 2 ϑ ( s , 1 , 0 , 0 ) + 2 ϑ ( t , 1 , 0 , 0 ) . Suppose that ϕ : ℝ → ℝ is a function satisfying ϕ ( 0 ) = 0 and

Then, there exists a unique quadratic function Q : ℝ → ℝ , which satisfies ( 1.3 ) and

where Q ( s ) = s 2 if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) ≠ 1 .

Proof

It is clear that lim n → ∞ ψ ( 2 n s , 2 n t ) 4 n = 0 for all s , t ∈ ℝ . Using a similar argument of the proof of Theorem 3.1, we obtain

where ψ ( s , u ) ≔ ϑ ( s , 1 , u , 1 ) + 2 ϑ ( s , 1 , 0 , 0 ) + 2 ϑ ( u , 1 , 0 , 0 ) . Hence by [18, Theorem 4], there exists a unique quadratic function Q : ℝ → ℝ such that

From the definition of Q and using (4.1) and (4.2), we obtain that Q satisfies (1.3). Since Q ( 0 ) = 0 , it follows from Theorem 2.3 that Q ( s ) = s 2   ( s ∈ ℝ ) if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) = 0 . Hence, we get (4.3), which ends the proof.□

Theorem 4.3

Let 0 < L < 1 and ϑ : ℝ 4 → [ 0 , ∞ ) be a function such that

where ψ ( s , t ) ≔ ϑ ( s , 1 , t , 1 ) + 2 ϑ ( s , 1 , 0 , 0 ) + 2 ϑ ( t , 1 , 0 , 0 ) . Suppose that ϕ : ℝ → ℝ is a function satisfying ϕ ( 0 ) = 0 and

Then, there exists a unique quadratic function Q : ℝ → ℝ , which satisfies ( 1.3 ) and

where Q ( s ) = s 2 if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) ≠ 1 .

Theorem 4.4

Let d > 0 and ε ⩾ 0 . Assume that a function ϕ : ℝ → ℝ with ϕ ( 0 ) = 0 satisfies the inequality

for all s , t , u , v ∈ ℝ with max { | s | , | t | , | u | , | v | } ⩾ d . Then, there exists a unique quadratic function Q : ℝ → ℝ , which satisfies ( 1.3 ) and

where Q ( s ) = s 2 if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) ≠ 1 .

Proof

Letting u = v = 0 in (4.5), we get

Setting v = t in (4.5), we obtain

It follows from (4.6) and (4.7) that

Letting t = d and replacing s and u by s d and u d in (4.8), respectively, we get

Hence by [12, Theorem 1], there is a unique quadratic function Q : ℝ → ℝ such that

Applying (4.5) and (4.9), we obtain

Since Q ( 0 ) = 0 , by Theorem 2.3, we get Q ( s ) = s 2 if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) = 0 .□

Corollary 4.5

Assume that a nonzero function ϕ : ℝ → ℝ with ϕ ( 0 ) = 0 satisfies

Then, ϕ ( s ) = s 2 for all s ∈ ℝ .

Proof

Let { ε n } be a sequence of positive real numbers which decreases monotonically to zero. Then, by (4.10) there exists an increasing sequence { d n } of positive real numbers such that lim n → + ∞ d n = + ∞ and

According to Theorem 4.4, there exists a unique quadratic function Q n : ℝ → ℝ satisfying (1.3) and | ϕ ( s ) − Q n ( s ) | ⩽ ε n for all s ∈ ℝ . Then, the quadratic function Q n : ℝ → ℝ satisfies | ϕ ( s ) − Q n ( s ) | ⩽ ε 1 for all s ∈ ℝ and all n ⩾ 1 . Hence, the uniqueness implies Q n = Q 1 for all n. Therefore, | ϕ ( s ) − Q 1 ( s ) | ⩽ ε n for all s ∈ ℝ and all n ⩾ 1 . Letting n → + ∞ , we get ϕ ( s ) = Q 1 ( s ) for all s ∈ ℝ . So ϕ satisfies (1.3) and ϕ ( 0 ) = 0 . According to Corollary 2.4, we get ϕ ( s ) = s 2 for all s ∈ ℝ .□

Theorem 4.6

Let d > 0 and ε ⩾ 0 . Assume that a function ϕ : ℝ → ℝ with ϕ ( 0 ) = 0 satisfies one of the following inequalities:

1. | D ϕ ( s , t , u , v ) | ⩽ ε , s , t , u , v ∈ ℝ with | s | ⩾ d ;

2. | D ϕ ( s , t , u , v ) | ⩽ ε , s , t , u , v ∈ ℝ with | t | ⩾ d ;

3. | D ϕ ( s , t , u , v ) | ⩽ ε , s , t , u , v ∈ ℝ with | u | ⩾ d ;

4. | D ϕ ( s , t , u , v ) | ⩽ ε , s , t , u , v ∈ ℝ with | v | ⩾ d ;

5. | D ϕ ( s , t , u , v ) | ⩽ ε , s , t , u , v ∈ ℝ with | s | + | t | + | u | + | v | ⩾ d .

Then, there exists a unique quadratic function Q : ℝ → ℝ , which satisfies (1.3) and

where Q ( s ) = s 2 if Q ( 1 ) = 1 and Q ≡ 0 if Q ( 1 ) ≠ 1 .

Corollary 4.7

Assume that a nonzero function ϕ : ℝ → ℝ with ϕ ( 0 ) = 0 satisfies one of the following conditions:

1. lim sup | s | → + ∞ | D ϕ ( s , t , u , v ) | = 0 , s , t , u , v ∈ ℝ ;

2. lim sup | t | → + ∞ | D ϕ ( s , t , u , v ) | = 0 , s , t , u , v ∈ ℝ ;

3. lim sup | u | → + ∞ | D ϕ ( s , t , u , v ) | = 0 , s , t , u , v ∈ ℝ ;

4. lim sup | v | → + ∞ | D ϕ ( s , t , u , v ) | = 0 , s , t , u , v ∈ ℝ ;

5. lim sup | s | + | t | + | u | + | v | → + ∞ | D ϕ ( s , t , u , v ) | = 0 , s , t , u , v ∈ ℝ .

Then, ϕ ( s ) = s 2 for all s ∈ ℝ .