Inequalities
Volume 13, Number 1 (2019), 95–104 doi:10.7153/jmi-2019-13-07
THE STABILITY OF AN ADDITIVE
(ρ1,ρ2)–FUNCTIONAL INEQUALITY IN BANACH SPACES
CHOONKILPARK
(Communicated by J. K. Kim)
Abstract. In this paper, we introduce and solve the following additive (ρ1,ρ2)-functional in- equality
f(x+y)−f(x)−f(y)ρ1(f(x+y) +f(x−y)−2f(x)) (1) +
ρ2
2f
x+y 2
−f(x)−f(y) , whereρ1andρ2arefixed nonzero complex numbers with√
2|ρ1|+|ρ2|<1 .
Using thefixed point method and the direct method, we prove the Hyers-Ulam stability of the additive (ρ1,ρ2)-functional inequality (1) in complex Banach spaces.
1. Introduction and preliminaries
The stability problem of functional equations originated from a question of Ulam [26] concerning the stability of group homomorphisms. Hyers [10] gave afirst affir- mative partial answer to the question of Ulam for Banach spaces. Hyers’ Theorem was generalized by Aoki [2] for additive mappings and by Rassias [24] for linear mappings by considering an unbounded Cauchy difference. A generalization of the Rassias the- orem was obtained by G˘avruta [9] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’ approach. The stability of quadratic functional equation was proved by Skof [25] for mappings f :E1→E2, where E1is a normed space andE2is a Banach space. Cholewa [6] noticed that the theorem of Skof is still true if the relevant domainE1is replaced by an Abelian group.
Park [21, 22] defined additive ρ-functional inequalities and proved the Hyers- Ulam stability of the additive ρ-functional inequalities in Banach spaces and non- Archimedean Banach spaces. The stability problems of various functional equations have been extensively investigated by a number of authors (see [1,8,12,13,14,15,16, 20]).
We recall a fundamental result infixed point theory.
Mathematics subject classification(2010): Primary 39B62, 47H10, 39B52.
Keywords and phrases: Hyers-Ulam stability, additive (ρ1,ρ2)-functional inequality, fixed point method, direct method, Banach space.
c , Zagreb
Paper JMI-13-07 95
THEOREM1.1 [3,7]Let (X,d) be a complete generalized metric space and let J: X→X be a strictly contractive mapping with Lipschitz constantα<1. Then for each given element x∈X , either
d(Jnx,Jn+1x) =∞
for all nonnegative integers n or there exists a positive integer n0such that (1) d(Jnx,Jn+1x)<∞, ∀nn0;
(2)the sequence{Jnx} converges to afixed point y∗of J ;
(3) y∗ is the uniquefixed point of J in the set Y ={y∈X|d(Jn0x,y)<∞}; (4) d(y,y∗)1−α1 d(y,Jy)for all y∈Y .
In 1996, Isac and Rassias [11] were thefirst to provide applications of stability the- ory of functional equations for the proof of newfixed point theorems with applications.
By usingfixed point methods, the stability problems of several functional equations have been extensively investigated by a number of authors (see [4,5,18,19,23]).
In Section 2, we solve the additive(ρ1,ρ2)-functional inequality (1) and prove the Hyers-Ulam stability of the additive(ρ1,ρ2)-functional inequality (1) in Banach spaces by using thefixed point method. In Section 3, we prove the Hyers-Ulam stability of the additive(ρ1,ρ2)-functional inequality (1) in Banach spaces by using the direct method.
Throughout this paper, let X be a real or complex normed space with norm · andY a complex Banach space with norm · . Assume that ρ1 and ρ2 are fixed nonzero complex numbers with√
2|ρ1|+|ρ2|<1.
2. Additive(ρ1,ρ2)-functional inequality(1): afixed point method In this section, we solve and investigate the additive(ρ1,ρ2)-functional inequality (1) in complex Banach spaces.
LEMMA2.1 If a mapping f:X→Y satisfies f(0) =0 and
f(x+y)−f(x)−f(y)ρ1(f(x+y) +f(x−y)−2f(x)) (2) +
ρ2
2f
x+y 2
−f(x)−f(y) for all x,y∈X , then f is additive.
Proof. Assume that f satisfies the inequality (2).
Letting y=x in (2), we get f(2x)−2f(x)|ρ1| · f(2x)−2f(x) and so f(2x) =2f(x)for allx∈X, since |ρ1|<1. Thus
fx 2
=1
2f(x) (3)
for allx∈X.
It follows from (2) and (3) that
f(x+y)−f(x)−f(y)ρ1(f(x+y) +f(x−y)−2f(x)) +
ρ2
2f
x+y 2
−f(x)−f(y)
=ρ1(f(x+y) +f(x−y)−2f(x)) +ρ2(f(x+y)−f(x)−f(y)) and so
(1− |ρ2|)f(x+y)−f(x)−f(y)|ρ1| · f(x+y) +f(x−y)−2f(x) (4) for allx,y∈X.
Letting z=x+yandw=x−yin (4), we get (1− |ρ2|)
f(z)−f z+w
2
−f z−w
2
|ρ1| ·
f(z) +f(w)−2f z+w
2
and so 1
2(1− |ρ2|)f(z+w) +f(z−w)−2f(z)|ρ1| · f(z+w)−f(z)−f(w) for allz,w∈X.
It follows from (4) and (5) that 1
2(1− |ρ2|)2f(x+y)−f(x)−f(y)|ρ1|2f(x+y)−f(x)−f(y) for allx,y∈X. Since√
2|ρ1|+|ρ2|<1, f(x+y) =f(x)+f(y)for allx,y∈X. Thus f is additive.
Using thefixed point method, we prove the Hyers-Ulam stability of the additive (ρ1,ρ2)-functional inequality (2) in complex Banach spaces.
THEOREM2.2 Let ϕ:X2→[0,∞)be a function such that there exists an L<1 with ϕ2x,y
2
L
2ϕ(x,y) (5)
for all x,y∈X . Let f:X→Y be a mapping satisfying f(0) =0and
f(x+y)−f(x)−f(y)ρ1(f(x+y) +f(x−y)−2f(x)) (6) +
ρ2
2f
x+y
2
−f(x)−f(y)
+ϕ(x,y) for all x,y∈X . Then there exists a unique additive mapping A:X→Y such that
f(x)−A(x) L
2(1−L)(1− |ρ1|)ϕ(x,x) for all x∈X .
Proof. Lettingy=xin (6), we get
(1− |ρ1|)f(2x)−2f(x)ϕ(x,x) (7) for allx∈X.
Consider the set
S:={h:X→Y, h(0) =0} and introduce the generalized metric onS:
d(g,h) =inf{μ∈R+:g(x)−h(x)μϕ(x,x), ∀x∈X}, where, as usual, infφ= +∞. It is easy to show that(S,d)is complete (see [17]).
Now we consider the linear mapping J:S→S such that Jg(x):=2gx
2 for allx∈X.
Let g,h∈S be given such thatd(g,h) =ε. Then g(x)−h(x)εϕ(x,x) for allx∈X. Hence
Jg(x)−Jh(x)=2gx 2
−2hx
22εϕx2,x 2
2εL2ϕ(x,x) =Lεϕ(x,x)
for allx∈X. So d(g,h) =ε implies that d(Jg,Jh)Lε. This means that d(Jg,Jh)Ld(g,h)
for allg,h∈S.
It follows from (7) that f(x)−2fx
2 1
1− |ρ1|ϕx2,x 2
L
2(1− |ρ1|)ϕ(x,x) for allx∈X. So d(f,J f)2(1−|ρL 1|).
By Theorem1.1, there exists a mappingA:X→Y satisfying the following:
(1)Ais afixed point ofJ, i.e.,
A(x) =2Ax 2
(8)
for allx∈X. The mappingAis a uniquefixed point ofJ in the set M={g∈S:d(f,g)<∞}.
This implies thatAis a unique mapping satisfying (8) such that there exists aμ∈(0,∞) satisfying
f(x)−A(x)μϕ(x,x) for allx∈X;
(2)d(Jlf,A)→0 asl→∞. This implies the equality
l→∞lim2nfx 2n
=A(x)
for allx∈X;
(3)d(f,A)1−L1 d(f,J f), which implies f(x)−A(x) L
2(1−L)(1− |ρ1|)ϕ(x,x) for allx∈X.
It follows from (5) and (6) that A(x+y)−A(x)−A(y)
=lim
n→∞2n f
x+y 2n
−fx 2n
−fy
2n
lim
n→∞2n|ρ1| f
x+y 2n
+f
x−y 2n
−2fx 2n
+lim
n→∞2n|ρ2| 2f
x+y
2n+1
−f x 2n
−fy
2n
+lim
n→∞2nϕ2xn, y 2n
=ρ1(A(x+y) +A(x−y)−2A(x))+ ρ2
2A
x+y 2
−A(x)−A(y) for allx,y∈X. So
A(x+y)−A(x)−A(y)ρ1(A(x+y) +A(x−y)−2A(x)) +
ρ2
2A
x+y
2
−A(x)−A(y) for allx,y∈X. By Lemma2.1, the mappingA:X→Y is additive.
COROLLARY2.3 Let r>1andθ be nonnegative real numbers, and let f:X→Y be a mapping satisfying f(0) =0 and
f(x+y)−f(x)−f(y)ρ1(f(x+y) +f(x−y)−2f(x)) (9) +
ρ2
2f
x+y
2
−f(x)−f(y)
+θ(xr+yr)
for all x,y∈X . Then there exists a unique additive mapping A:X→Y such that f(x)−A(x) 2θ
(2r−2)(1− |ρ1|)xr for all x∈X .
Proof. The proof follows from Theorem2.2by taking ϕ(x,y) =θ(xr+yr) for allx,y∈X. ChoosingL=21−r, we obtain the desired result.
THEOREM2.4 Let ϕ:X2→[0,∞)be a function such that there exists an L<1 with ϕ(x,y)2Lϕ2x,y
2
for all x,y∈X . Let f :X→Y be a mapping satisfying f(0) =0 and(6). Then there exists a unique additive mapping A:X→Y such that
f(x)−A(x) 1
2(1−L)(1− |ρ1|)ϕ(x,x) for all x∈X .
Proof. Let(S,d)be the generalized metric space defined in the proof of Theorem 2.2.
Now we consider the linear mapping J:S→S such that Jg(x):=1
2g(2x) for allx∈X.
It follows from (7) that f(x)−1
2f(2x)
1
2(1− |ρ1|)ϕ(x,x) for allx∈X.
The rest of the proof is similar to the proof of Theorem2.2.
COROLLARY2.5 Let r<1 and θ be positive real numbers, and let f :X→Y be a mapping satisfying f(0) =0 and(9). Then there exists a unique additive mapping A:X→Y such that
f(x)−A(x) 2θ
(2−2r)(1− |ρ1|)xr for all x∈X .
Proof. The proof follows from Theorem2.4by taking ϕ(x,y) =θ(xr+yr) for allx,y∈X. ChoosingL=2r−1, we obtain the desired result.
REMARK2.6. Ifρ is a real number such that √
2|ρ1|+|ρ2|<1 andY is a real Banach space, then all the assertions in this section remain valid.
3. Additive (ρ1,ρ2)-functional inequality(1): a direct method
In this section, we prove the Hyers-Ulam stability of the additive(ρ1,ρ2)-functional inequality (2) in complex Banach spaces by using the direct method.
THEOREM3.1 Let ϕ:X2→[0,∞)be a function such that Ψ(x,y):=
∑
∞j=1
2jϕ2xj, y 2j
<∞ (10)
for all x,y∈X . Let f :X→Y be a mapping satisfying f(0) =0 and(6). Then there exists a unique additive mapping A:X→Y such that
f(x)−A(x) 1
2(1− |ρ1|)Ψ(x,x) (11) for all x∈X .
Proof. Lettingy=xin (6), we get
(1− |ρ1|)f(2x)−2f(x)ϕ(x,x) (12) for allx∈X. So
f(x)−2fx
2 1
1− |ρ1|ϕx2,x 2
for allx∈X. Hence 2lfx 2l
−2mf x
2mm−1
∑
j=l
2jfx 2j
−2j+1f x 2j+1 m−1
∑
j=l
2j
2(1− |ρ1|)ϕ2j+1x , x 2j+1
(13)
for all nonnegative integersmandl withm>l and allx∈X. It follows from (13) that the sequence{2kf(2xk)} is Cauchy for all x∈X. Since Y is complete, the sequence {2kf(2xk)} converges. So one can define the mappingA:X→Y by
A(x):=lim
k→∞2kfx 2k
for allx∈X. Moreover, lettingl=0 and passing the limitm→∞in (13), we get (11).
It follows from (6) and (10) that A(x+y)−A(x)−A(y)
=lim
n→∞2n f
x+y 2n
−fx 2n
−fy
2n
lim
n→∞2n|ρ1| f
x+y
2n
+f
x−y
2n
−2fx 2n
+lim
n→∞2n|ρ2| 2f
x+y
2n+1
−f x 2n
−fy
2n
+lim
n→∞2nϕ2xn, y 2n
=ρ1(A(x+y) +A(x−y)−2A(x))+ ρ2
2A
x+y 2
−A(x)−A(y) for allx,y∈X. So
A(x+y)−A(x)−A(y)ρ1(A(x+y) +A(x−y)−2A(x)) +
ρ2
2A
x+y 2
−A(x)−A(y) for allx,y∈X. By Lemma2.1, the mappingA:X→Y is additive.
Now, let T:X→Y be another additive mapping satisfying (11). Then we have A(x)−T(x) =2qAx
2q
−2qTx 2q 2qAx
2q
−2qfx
2q+2qT x 2q
−2qfx 2q
2q
1− |ρ1|Ψ x 2q, x
2q ,
which tends to zero asq→∞for allx∈X. So we can conclude thatA(x) =T(x)for allx∈X. This proves the uniqueness of A.
COROLLARY3.2 Let r>1andθ be nonnegative real numbers, and let f:X→Y be a mapping satisfying f(0) =0 and(9). Then there exists a unique additive mapping A:X→Y such that
f(x)−A(x) 2θ
(2r−2)(1− |ρ1|)xr for all x∈X .
THEOREM3.3 Let ϕ :X2→[0,∞) be a function and let f :X →Y be a mapping satisfying f(0) =0,(6)and
Ψ(x,y):=
∑
∞j=0
1
2jϕ(2jx,2jy)<∞
for all x,y∈X . Then there exists a unique additive mapping A:X→Y such that f(x)−A(x) 1
2(1− |ρ1|)Ψ(x,x) (14) for all x∈X .
Proof. It follows from (12) that f(x)−1
2f(2x)
1
2(1− |ρ1|)ϕ(x,x) for allx∈X. Hence
1
2l f(2lx)− 1
2mf(2mx) m−1
∑
j=l
1 2jf
2jx
− 1 2j+1f
2j+1x m−1
∑
j=l
1
2j+1(1− |ρ1|)ϕ(2jx,2jx) (15) for all nonnegative integersmandl withm>l and allx∈X. It follows from (15) that the sequence{21nf(2nx)} is Cauchy for all x∈X. SinceY is complete, the sequence {21nf(2nx)} converges. So one can define the mapping A:X→Y by
A(x):=lim
n→∞
1 2nf(2nx)
for allx∈X. Moreover, lettingl=0 and passing the limitm→∞in (15), we get (14).
The rest of the proof is similar to the proof of Theorem3.1.
COROLLARY3.4 Let r<1 and θ be positive real numbers, and let f :X→Y be a mapping satisfying f(0) =0 and(9). Then there exists a unique additive mapping A:X→Y such that
f(x)−A(x) 2θ
(2−2r)(1− |ρ1|)xr for all x∈X .
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(Received August 31, 2016) Choonkil Park
Research Institute for Natural Sciences Hanyang University Seoul 04763, Republic of Korea e-mail:baak@@hanyang.ac.kr
Journal of Mathematical Inequalities www.ele-math.com