We prove the generalized Hyers-Ulam stability of the following additive-quadratic-cubic-quartic functional equation f(x+2y)+f(x−2y)=4f(x+y)+4f(x−y)−6f(x)+f(2y)+f(−2y)−4f(y)−4f(−y) in various complete random normed spaces.
1. Introduction
The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers' theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The paper of Rassias [4] has provided a lot of influence in the development of what we call generalized Hyers-Ulam stability or as Hyers-Ulam-Rassias stability of functional equations. A generalization of the Rassias theorem was obtained by Găvruţa [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias approach.
The functional equation f(x+y)+f(x-y)=2f(x)+2f(y)
is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. A generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Cholewa [6] for mappings f:X→Y, where X is a normed space and Y is a Banach space. Czerwik [7] proved the generalized Hyers-Ulam stability of the quadratic functional equation. The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem (see [8–12]).
In [13], Jun and Kim consider the following cubic functional equation:f(2x+y)+f(2x-y)=2f(x+y)+2f(x-y)+12f(x).
It is easy to show that the function f(x)=x3 satisfies the functional equation (1.2), which is called a cubic functional equation, and every solution of the cubic functional equation is said to be a cubic mapping.
Considered the following quartic functional equationf(2x+y)+f(2x-y)=4f(x+y)+4f(x-y)+24f(x)-6f(y).
It is easy to show that the function f(x)=x4 satisfies the functional equation, which is called a quartic functional equation, and every solution of the quartic functional equation is said to be a quartic mapping. One can easily show that an odd mapping f:X→Y satisfies the additive-quadratic-cubic-quadratic functional equation
f(x+2y)+f(x-2y)=4f(x+y)+4f(x-y)-6f(x)+f(2y)+f(-2y)-4f(y)-4f(-y)
if and only if it is an additive-cubic mapping, that is, f(x+2y)+f(x-2y)=4f(x+y)+4f(x-y)-6f(x).
It was shown in Lemma 2.2 of [14] that g(x):=f(2x)-2f(x) and h(x):=f(2x)-8f(x) are cubic and additive, respectively, and that f(x)=(1/6)g(x)-(1/6)h(x).
One can easily show that an even mapping f:X→Y satisfies (1.4) if and only if it is a quadratic-quartic mapping, that is, f(x+2y)+f(x-2y)=4f(x+y)+4f(x-y)-6f(x)+2f(2y)-8f(y).
Also g(x):=f(2x)-4f(x) and h(x):=f(2x)-16f(x) are quartic and quadratic, respectively, and f(x)=(1/12)g(x)-(1/12)h(x).
For a given mapping f:X→Y, we define Df(x,y)∶=f(x+2y)+f(x-2y)-4f(x+y)-4f(x-y)+6f(x)-f(2y)-f(-2y)+4f(y)+4f(-y)
for all x,y∈X.
Let X be a set. A function d:X×X→[0,∞] is called a generalized metric on X if d satisfies
d(x,y)=0 if and only if x=y,
d(x,y)=d(y,x) for all x,y∈X,
d(x,z)≤d(x,y)+d(y,z) for all x,y,z∈X.
We recall the fixed-point alternative of Diaz and Margolis.
Theorem 1.1 (see [15, 16]).
Let (X,d) be a complete generalized metric space and let J:X→X be a strictly contractive mapping with Lipschitz constant L<1, then for each given element x∈X, either
d(Jnx,Jn+1x)=∞
for all nonnegative integers n or there exists a positive integer n0 such that
d(Jnx,Jn+1x)<∞ for all n≥n0,
the sequence {Jnx} converges to a fixed point y* of J,
y* is the unique fixed point of J in the set Y={y∈X∣d(Jn0x,y)<∞},
d(y,y*)≤(1/(1-L))d(y,Jy) for all y∈Y.
In 1996, Isac and Rassias [17] were the first to provide applications of stability theory of functional equations for the proof of new fixed-point theorems with applications. By using fixed point methods, the stability problems of several functional equations have been extensively investigated by a number of authors (see [18–21]).
2. Preliminaries
In the sequel, we adopt the usual terminology, notations, and conventions of the theory of random normed spaces, as in [22–26]. Throughout this paper, Δ+ is the space of all probability distribution functions, that is, the space of all mappings F:ℝ∪{-∞,+∞}→[0,1], such taht F is left continuous, nondecreasing on ℝ, F(0)=0 and {F(+∞)=1}. D+ is a subset of Δ+ consisting of all functions F∈Δ+ for which l-F(+∞)=1, where l-f(x) denotes the left limit of the function f at the point x, that is, l-f(x)=limt→x-f(t). The space Δ+ is partially ordered by the usual pointwise ordering of functions, that is, F≤G if and only if F(t)≤G(t) for all t in ℝ. The maximal element for Δ+ in this order is the distribution function ɛ0 given byɛ0(t)={0,ift≤0,1,ift>0.
A triangular norm (shortly t-norm) is a binary operation on the unit interval [0,1], that is, a function T:[0,1]×[0,1]→[0,1], such that for all a,b,c∈[0,1] the following four axioms satisfied:
T(a,b)=T(b,a) (commutativity),
T(a,(T(b,c)))=T(T(a,b),c) (associativity),
T(a,1)=a (boundary condition),
T(a,b)≤T(a,c) whenever b≤c (monotonicity).
Basic examples are the Łukasiewicz t-norm TL,TL(a,b)=max(a+b-1,0) for all a,b∈[0,1] and the t-norms TP,TM,TD, where TP(a,b):=ab, TM(a,b):=min{a,b}, TD(a,b):={min(a,b),ifmax(a,b)=1,0,otherwise.
If T is a t-norm, then xT(n) is defined for every x∈[0,1] and n∈N∪{0} by 1, if n=0 and T(xT(n-1),x) if n≥1. A t-normT is said to be of Hadžić type (we denote by T∈ℋ) if the family (xT(n))n∈N is equicontinuous at x=1 (cf. [27]).
Other important triangular norms are the following (see [28]):
The Sugeno-Weber family {TλSW}λ∈[-1,∞] is defined by T-1SW=TD, T∞SW=TP and
TλSW(x,y)=max(0,x+y-1+λxy1+λ)
if λ∈(-1,∞).
The Domby family {TλD}λ∈[0,∞] is defined by TD if λ=0, TM if λ=∞, and
TλD(x,y)=11+(((1-x)/x)λ+((1-y)/y)λ)1/λ
if λ∈(0,∞).
The Aczel-Alsina family {TλAA}λ∈[0,∞] is defined by TD if λ=0, TM if λ=∞ and
TλAA(x,y)=e-(|logx|λ+|logy|λ)1/λ
if λ∈(0,∞).
A t-norm T can be extended (by associativity) in a unique way to an n-array operation taking for (x1,…,xn)∈[0,1]n the value T(x1,…,xn) defined by Ti=10xi=1,Ti=1nxi=T(Ti=1n-1xi,xn)=T(x1,…,xn).
T can also be extended to a countable operation taking for any sequence (xn)n∈N in [0,1] the valueTi=1∞xi=limn→∞Ti=1nxi.
The limit on the right side of (6.4) exists since the sequence (Ti=1nxi)n∈ℕ is nonincreasing and bounded from below.
Proposition 2.1 (see [28]).
We have the following.
For T≥TL, the following implication holds:
limn→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)<∞.
If T is of Hadžić type, then
limn→∞Ti=1∞xn+i=1
for every sequence (xn)n∈N in [0,1] such that limn→∞xn=1.
If T∈{TλAA}λ∈(0,∞)∪{TλD}λ∈(0,∞), then
limn→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)α<∞.
If T∈{TλSW}λ∈[-1,∞), then
limn→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)<∞.
Definition 2.2 (see [26]).
A Random normed space (briefly, RN-space) is a triple (X,μ,T), where X is a vector space, T is a continuous t-norm, and μ is a mapping from X into D+ such that, the following conditions hold:
(RN1) μx(t)=ɛ0(t) for all t>0 if and only if x=0,
(RN2) μαx(t)=μx(t/|α|) for all x∈X, and α≠0,
(RN3) μx+y(t+s)≥T(μx(t),μy(s)) for all x,y∈X and t,s≥0.
Definition 2.3.
Let (X,μ,T) be an RN-space.
A sequence {xn} in X is said to be convergent to x in X if, for every ϵ>0 and λ>0, there exists positive integer N such that μxn-x(ϵ)>1-λ whenever n≥N.
A sequence {xn} in X is called a Cauchy sequence if, for every ϵ>0 and λ>0, there exists positive integer N such that μxn-xm(ϵ)>1-λ whenever n≥m≥N.
An RN-space (X,μ,T) is said to be complete if and only if every Cauchy sequence in X is convergent to a point in X. A complete RN-space is said to be random Banach space.
Theorem 2.4 (see [25]).
If (X,μ,T) is an RN-space and {xn} is a sequence such that xn→x, then limn→∞μxn(t)=μx(t) almost everywhere.
The theory of random normed spaces (RN-spaces) is important as a generalization of deterministic result of linear normed spaces and also in the study of random operator equations. The RN-spaces may also provide us with the appropriate tools to study the geometry of nuclear physics and have important application in quantum particle physics. The generalized Hyers-Ulam stability of different functional equations in random normed spaces, RN-spaces, and fuzzy normed spaces has been recently studied [20, 24, 29–39].
3. Non-Archimedean Random Normed Space
By a non-Archimedean field, we mean a field 𝒦 equipped with a function (valuation) |·| from K into [0,∞) such that |r|=0 if and only if r=0, |rs|=|r||s|, and |r+s|≤max{|r|,|s|} for all r,s∈𝒦. Clearly, |1|=|-1|=1 and |n|≤1 for all n∈ℕ. By the trivial valuation, we mean the mapping |·| taking everything but 0 into 1 and |0|=0. Let X be a vector space over a field 𝒦 with a non-Archimedean nontrivial valuation |·|. A function ∥·∥:X→[0,∞) is called a non-Archimedean norm if it satisfies the following conditions:
(NAN1) ∥x∥=0 if and only if x=0,
(NAN2) for any r∈𝒦 and x∈X, ∥rx∥=|r|∥x∥,
(NAN3) the strong triangle inequality (ultrametric), namely,
‖x+y‖≤max{‖x‖,‖y‖}(x,y∈X),
then (X,∥·∥) is called a non-Archimedean normed space. Due to the fact that ‖xn-xm‖≤max{‖xj+1-xj‖:m≤j≤n-1}(n>m),
a sequence {xn} is a Cauchy sequence if and only if {xn+1-xn} converges to zero in a non-Archimedean normed space. By a complete non-Archimedean normed space, we mean one in which every Cauchy sequence is convergent.
In 1897, Hensel [40] discovered the p-adic numbers of as a number theoretical analogues of power series in complex analysis. Fix a prime number p. For any nonzero rational number x, there exists a unique integer nx∈ℤ such that x=(a/b)pnx, where a and b are integers not divisible by p. Then |x|p:=p-nx defines a non-Archimedean norm on ℚ. The completion of ℚ with respect to the metric d(x,y)=|x-y|p is denoted by ℚp, which is called the p-adic number field.
Throughout the paper, we assume that X is a vector space and Y is a complete non-Archimedean normed space.
Definition 3.1.
A non-Archimedean random normed space (briefly, non-Archimedean RN-space) is a triple (X,μ,T), where X is a linear space over a non-Archimedean field 𝒦, T is a continuous t-norm, and μ is a mapping from X into D+ such that the following conditions hold:
(NA-RN1) μx(t)=ɛ0(t) for all t>0 if and only if x=0,
(NA-RN2) μαx(t)=μx(t/|α|) for all x∈X, t>0, and α≠0,
(NA-RN3) μx+y(max{t,s})≥T(μx(t),μy(s)) for all x,y,z∈X and t,s≥0.
It is easy to see that if (NA-RN3) holds, then so is
(RN3)μx+y(t+s)≥T(μx(t),μy(s)).
As a classical example, if (X,∥.∥) is a non-Archimedean normed linear space, then the triple (X,μ,TM), where μx(t)={0,t≤‖x‖,1,t>‖x‖,
is a non-Archimedean RN-space.
Example 3.2.
Let (X,∥·∥) be a non-Archimedean normed linear space. Define
μx(t)=tt+‖x‖(x∈X,t>0),
then (X,μ,TM) is a non-Archimedean RN-space.
Definition 3.3.
Let (X,μ,T) be a non-Archimedean RN-space. Let {xn} be a sequence in X, then {xn} is said to be convergent if there exists x∈X such that
limn→∞μxn-x(t)=1
for all t>0. In that case, x is called the limit of the sequence {xn}.
A sequence {xn} in X is called a Cauchy sequence if for each ɛ>0 and each t>0 there exists n0 such that for all n≥n0 and all p>0, we have μxn+p-xn(t)>1-ɛ.
If each Cauchy sequence is convergent, then the random norm is said to be complete and the non-Archimedean RN-space is called a non-Archimedean random Banach space.
Remark 3.4 (see [41]).
Let (X,μ,TM) be a non-Archimedean RN-space, then
μxn+p-xn(t)≥min{μxn+j+1-xn+j(t):j=0,1,2,…,p-1}.
So, the sequence {xn} is a Cauchy sequence if for each ɛ>0 and t>0 there exists n0 such that for all n≥n0,
μxn+1-xn(t)>1-ɛ.
4. Generalized Ulam-Hyers Stability for a Quartic Functional Equation in Non-Archimedean RN-Spaces of Functional Equation (1.4): An Odd Case
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦.
Next, we define a random approximately AQCQ mapping. Let Ψ be a distribution function on X×X×[0,∞) such that Ψ(x,y,·) is nondecreasing and Ψ(cx,cx,t)≥Ψ(x,x,t|c|)(x∈X,c≠0).
Definition 4.1.
A mapping f:X→Y is said to be Ψ-approximately AQCQ if
μDf(x,y)(t)≥Ψ(x,y,t)(x,y∈X,t>0).
In this section, we assume that 2≠0 in 𝒦 (i.e., characteristic of 𝒦 is not 2). Our main result, in this section, is the following.
We prove the generalized Hyers-Ulam stability of the functional equation Df(x,y)=0 in non-Archimedean random spaces, an odd case.
Theorem 4.2.
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦 and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦. Let f:X→Y be an odd mapping and Ψ-approximately AQCQ mapping. If for some α∈ℝ, α>0, and some integer k, k>3 with |2k|<α,
Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),limn→∞Tj=n∞M(2x,αjt|8|kj)=1(x∈X,t>0),
then there exists a unique cubic mapping C:X→Y such that
μf(x)-2f(x/2)-C(x/2)(t)≥Ti=1∞M(x,αi+1t|8|ki)
for all x∈X and t>0, where
M(x,t)∶=Tk-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2k-1x2,2k-1x2,t|4|),Ψ(2k-1x,2k-1x2,t)](x∈X,t>0).
Proof.
Letting x=y in (4.2), we get
μf(3y)-4f(2y)+5f(y)(t)≥Ψ(y,y,t)
for all y∈X and t>0. Replacing x by 2y in (4.2), we get
μf(4y)-4f(3y)+6f(2y)-4f(y)(t)≥Ψ(2y,y,t)
for all y∈X and t>0. By (4.7) and (4.8), we have
μf(4y)-10f(2y)+16f(y)(t)≥T(μ4(f(3y)-4f(2y)+5f(y))(t),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))=T(μf(3y)-4f(2y)+5f(y)(t|4|),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))≥T(Ψ(y,y,t|4|),Ψ(2y,y,t))
for all y∈X and t>0. Letting y:=x/2 and g(x):=f(2x)-2f(x) for all x∈X in (4.9), we get
μg(x)-8g(x/2)(t)≥T(Ψ(x2,x2,t|4|),Ψ(x,x2,t))
for all x∈X and t>0. Now, we show by induction on j that for all x∈X, t>0 and j≥1,
μg(2j-1x)-8jg(x/2)(t)≥Mj(x,t)∶=T2j-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2j-1x2,2j-1x2,t|4|),Ψ(2j-1x,2j-1x2,t)].
Putting j=1 in (4.11), we obtain (4.10). Assume that (4.11) holds for some j≥1. Replacing x by 2jx in (4.10), we get
μg(2jx)-8g(2j-1x)(t)≥T(Ψ(2j-1x,2j-1x,t|4|),Ψ(2jx,2j-1x,t)).
Since |8|≤1,
μg(2jx)-8j+1g(x/2)(t)≥T(μg(2jx)-8g(2j-1x)(t),μ8g(2j-1x)-8j+1g(x/2)(t))=T(μg(2jx)-8g(2j-1x)(t),μg(2j-1x)-8jg(x/2)(t|8|))≥T2(Ψ(2j-1x,2j-1x,t|4|),Ψ(2jx,2j-1x,t),Mj(x,t))=Mj+1(x,t)for all x∈X and t>0. Thus, (4.11) holds for all j≥2. In particular,
μg(2k-1x)-8kg(x/2)(t)≥M(x,t)(x∈X,t>0).
Replacing x by 2-(kn+k-1)x in (4.14) and using inequality (4.3), we obtain
μg(x/2kn)-8kg(x/2k(n+1))(t)≥M(2x2k(n+1),t)(x∈X,t>0,n=0,1,2,…).
Then
μ8kng(x/2kn)-8k(n+1)g(x/2k(n+1))(t)≥M(2x,αn+1|8k(n+1)|t)(x∈X,t>0,n=0,1,2,…).
Hence
μ8kng(x/2kn)-8k(n+p)g(x/2k(n+p))(t)≥Tj=nn+p(μ8kjg(x/2kj)-8k(j+p)g(x/2k(j+p))(t))≥Tj=nn+pM(2x,αj+1|(8k)j+1|t)≥Tj=nn+pM(2x,αj+1|(8k)j+1|t)(x∈X,t>0,n=0,1,2,…).
Since
limn→∞Tj=n∞M(2x,αj+1|(8k)j+1|t)=1(x∈X,t>0),
then
{8kng(x2kn)}n∈Nis a Cauchy sequence in the non-Archimedean random Banach space (Y,μ,T). Hence we can define a mapping C:X→Y such that
limn→∞μ(88k)ng(x/2kn)-C(x)(t)=1(x∈X,t>0).
Next for each n≥1, x∈X and t>0,
μg(x)-(88k)ng(x/2kn)(t)=μ∑i=0n-1(88k)ig(x/2ki)-(88k)i+1g(x/2k(i+1))(t)≥Ti=0n-1(μ(88k)ig(x/2ki)-(88k)i+1g(x/2k(i+1))(t))≥Ti=0n-1M(2x,αi+1t|8k|i+1).Therefore,
μg(x)-C(x)(t)≥T(μg(x)-(88k)ng(x/2kn)(t),μ(88k)ng(x/2kn)-C(x)(t))≥T(Ti=0n-1M(2x,αi+1t|8k|i+1),μ(88k)ng(x/2kn)-C(x)(t)).
By letting n→∞, we obtain
μg(x)-C(x)(t)≥Ti=1∞M(2x,αi+1t|8k|i+1).
So,
μf(x)-2f(x/2)-C(x/2)(t)≥Ti=1∞M(x,αi+1t|8k|i+1).
This proves (4.5). From Dg(x,y)=Df(2x,2y)-2Df(x,y), by (4.2), we deduce that
μDf(2x,2y)(t)≥Ψ(2x,2y,t),μ-2Df(x,y)(t)=μDf(x,y)(t|2|)≥μDf(x,y)(t)≥Ψ(x,y,t),
and so, by (NA-RN3) and (4.2), we obtain
μDg(x,y)(t)≥T(μDf(2x,2y)(t),μ-2Df(x,y)(t))≥T(Ψ(2x,2y,t),Ψ(x,y,t))∶=N(x,y,t).
It follows that
μ8knDg(x/2kn,y/2kn)(t)=μDg(x/2kn,y/2kn)(t|8|kn)≥N(x2kn,y2kn,t|8|kn)≥⋯≥N(x,y,αn-1t|8|k(n-1))for all x,y∈X, t>0, and n∈ℕ. Since
limn→∞N(x,y,αn-1t|8|k(n-1))=1for all x,y∈X and t>0, by Theorem 2.4, we deduce that
μDC(x,y)(t)=1for all x,y∈X and t>0. Thus, the mapping C:X→Y satisfies (1.4).
Now, we have
C(2x)-8C(x)=limn→∞[8ng(x2n-1)-8n+1g(x2n)]=8limn→∞[8n-1g(x2n-1)-8ng(x2n)]=0for all x∈X. Since the mapping x→C(2x)-2C(x) is cubic (see Lemma 2.2 of [14]), from the equality C(2x)=8C(x), we deduce that the mapping C:X→Y is cubic.
Corollary 4.3.
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦 under a t-norm T∈ℋ. Let f:X→Y be an odd and Ψ-approximately AQCQ mapping. If, for some α∈ℝ, α>0, and some integer k, k>3, with |2k|<α,
Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),
then there exists a unique cubic mapping C:X→Y such that
μf(x)-2f(x/2)-C(x/2)(t)≥Ti=1∞M(x,αi+1t|8|ki)
for all x∈X and t>0.
Proof.
Since
limn→∞M(x,αjt|8|kj)=1(x∈X,t>0)
and T is of Hadžić type, from Proposition 2.1, it follows that
limn→∞Tj=n∞M(x,αjt|8|kj)=1(x∈X,t>0).Now, we can apply Theorem 4.2 to obtain the result.
Example 4.4.
Let (X,μ,TM) be non-Archimedean random normed space in which
μx(t)=tt+‖x‖(x∈X,t>0).
And let (Y,μ,TM) be a complete non-Archimedean random normed space (see Example 3.2). Define
Ψ(x,y,t)=t1+t.
It is easy to see that (4.3) holds for α=1. Also, since
M(x,t)=t1+t,
we have
limn→∞TM,j=n∞M(x,αjt|8|kj)=limn→∞(limm→∞TM,j=nmM(x,t|8|kj))=limn→∞limm→∞(tt+|8k|n)=1(x∈X,t>0).
Let f:X→Y be an odd and Ψ-approximately AQCQ mapping. Thus, all the conditions of Theorem 4.2 hold, and so there exists a unique cubic mapping C:X→Y such that
μf(x)-2f(x/2)-C(x/2)(t)≥tt+|8k|.
Theorem 4.5.
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦. Let f:X→Y be an odd mapping and Ψ-approximately AQCQ mapping. If for some α∈ℝ, α>0, and some integer k, k>1 with |2k|<α,
Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),limn→∞Tj=n∞M(2x,αjt|2|kj)=1(x∈X,t>0),
then there exists a unique additive mapping A:X→Y such that
μf(x)-8f(x/2)-A(x/2)(t)≥Ti=1∞M(x,αi+1t|2|ki)
for all x∈X and t>0, where
M(x,t)∶=Tk-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2k-1x2,2k-1x2,t|4|),Ψ(2k-1x,2k-1x2,t)](x∈X,t>0)
Proof.
Letting y:=x/2 and g(x)∶=f(2x)-8f(x) for all x∈X in (4.9), we get
μg(x)-2g(x/2)(t)≥T(Ψ(x2,x2,t|4|),Ψ(x,x2,t))
for all x∈X and t>0.
The rest of the proof is similar to the proof of Theorem 4.2.
Corollary 4.6.
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦 under a t-norm T∈ℋ. Let f:X→Y be an odd and Ψ-approximately AQCQ mapping. If, for some α∈ℝ,α>0, and some integer k,k>1, with |2k|<α,
Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),
then there exists a unique additive mapping A:X→Y such that
μf(x)-8f(x/2)-A(x/2)(t)≥Ti=1∞M(x,αi+1t|2|ki)
for all x∈X and t>0.
Proof.
Since
limn→∞M(x,αjt|2|kj)=1(x∈X,t>0)
and T is of Hadžić type, from Proposition 2.1, it follows that
limn→∞Tj=n∞M(x,αjt|2|kj)=1(x∈X,t>0).Now, we can apply Theorem 4.5 to obtain the result.
Example 4.7.
Let (X,μ,TM) non-Archimedean random normed space in which
μx(t)=tt+‖x‖(x∈X,t>0),and let (Y,μ,TM) be a complete non-Archimedean random normed space (see Example 3.2). Define
Ψ(x,y,t)=t1+t.
It is easy to see that (4.3) holds for α=1. Also, since
M(x,t)=t1+t,
we have
limn→∞TM,j=n∞M(x,αjt|2|kj)=limn→∞(limm→∞TM,j=nmM(x,t|2|kj))=limn→∞limm→∞(tt+|2k|n)=1(x∈X,t>0).
Let f:X→Y be an odd and Ψ-approximately AQCQ mapping. Thus, all the conditions of Theorem 4.2 hold, and so there exists a unique additive mapping A:X→Y such that
μf(x)-8f(x/2)-A(x/2)(t)≥tt+|2k|.
5. Generalized Hyers-Ulam Stability of the Functional Equation (1.4) in Non-Archimedean Random Normed Spaces: An Even Case
Now, we prove the generalized Hyers-Ulam stability of the functional equation Df(x,y)=0 in non-Archimedean Banach spaces, an even case.
Theorem 5.1.
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦. Let f:X→Y be an even mapping, f(0)=0, and Ψ-approximately AQCQ mapping. If for some α∈ℝ, α>0, and some integer k, k>4 with |2k|<α,
Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),limn→∞Tj=n∞M(2x,αjt|16|kj)=1(x∈X,t>0),
then there exists a unique quartic mapping Q:X→Y such that
μf(x)-4f(x/2)-Q(x/2)(t)≥Ti=1∞M(x,αi+1t|16|ki)
for all x∈X and t>0, where
M(x,t)∶=Tk-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2k-1x2,2k-1x2,t|4|),Ψ(2k-1x,2k-1x2,t)](x∈X,t>0).
Proof.
Letting x=y in (4.2), we get
μf(3y)-6f(2y)+15f(y)(t)≥Ψ(y,y,t)
for all y∈X and t>0. Replacing x by 2y in (4.2), we get
μf(4y)-4f(3y)+4f(2y)+4f(y)(t)≥Ψ(2y,y,t)
for all y∈X and t>0. By (5.4) and (5.5), we have
μf(4y)-20f(2y)+64f(y)(t)≥T(μ4(f(3y)-4f(2y)+5f(y))(t),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))=T(μf(3y)-4f(2y)+5f(y)(t|4|),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))≥T(Ψ(y,y,t|4|),Ψ(2y,y,t))
for all y∈X and t>0. Letting y∶=x/2 and g(x)∶=f(2x)-4f(x) for all x∈X in (5.6), we get
μg(x)-16g(x/2)(t)≥T(Ψ(x2,x2,t|4|),Ψ(x,x2,t))
for all x∈X and t>0.
The rest of the proof is similar to the proof of Theorem 4.2.
Corollary 5.2.
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦 under a t-norm T∈ℋ. Let f:X→Y be an even, f(0)=0, and Ψ-approximately AQCQ mapping. If, for some α∈ℝ, α>0, and some integer k, k>4, with |2k|<α,
Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),
then there exists a unique quartic mapping Q:X→Y such that
μf(x)-4f(x/2)-Q(x/2)(t)≥Ti=1∞M(x,αi+1t|16|ki)
for all x∈X and t>0.
Proof.
Since
limn→∞M(x,αjt|16|kj)=1(x∈X,t>0)
and T is of Hadžić type, from Proposition 2.1, it follows that
limn→∞Tj=n∞M(x,αjt|16|kj)=1(x∈X,t>0).
Now, we can apply Theorem 5.1 to obtain the result.
Example 5.3.
Let (X,μ,TM) be non-Archimedean random normed space in which
μx(t)=tt+‖x‖(x∈X,t>0).
And let (Y,μ,TM) be a complete non-Archimedean random normed space (see Example 3.2). Define
Ψ(x,y,t)=t1+t.
It is easy to see that (4.3) holds for α=1. Also, since
M(x,t)=t1+t,
we have
limn→∞TM,j=n∞M(x,αjt|16|kj)=limn→∞(limm→∞TM,j=nmM(x,t|16|kj))=limn→∞limm→∞(tt+|16k|n)=1(x∈X,t>0).
Let f:X→Y be an even, f(0)=0, and Ψ-approximately AQCQ mapping. Thus all the conditions of Theorem 5.1 hold, and so there exists a unique quartic mapping Q:X→Y such that
μf(x)-4f(x/2)-Q(x/2)(t)≥tt+|16k|.
Theorem 5.4.
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦 and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦. Let f:X→Y be an even mapping, f(0)=0 and Ψ-approximately AQCQ mapping. If for some α∈ℝ, α>0, and some integer k, k>2 with |2k|<α,
Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),limn→∞Tj=n∞M(2x,αjt|4|kj)=1(x∈X,t>0),
then there exists a unique quadratic mapping Q:X→Y such that
μf(x)-16f(x/2)-Q(x/2)(t)≥Ti=1∞M(x,αi+1t|4|ki)
for all x∈X and t>0, where
M(x,t)∶=Tk-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2k-1x2,2k-1x2,t|4|),Ψ(2k-1x,2k-1x2,t)](x∈X,t>0).
Proof.
Letting y:=x/2 and g(x)∶=f(2x)-16f(x) for all x∈X in (5.6), we get
μg(x)-4g(x/2)(t)≥T(Ψ(x2,x2,t|4|),Ψ(x,x2,t))
for all x∈X and t>0.
The rest of the proof is similar to the proof of Theorem 5.1.
Corollary 5.5.
Let 𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦 under a t-norm T∈ℋ. Let f:X→Y be an even, f(0)=0, and Ψ-approximately AQCQ mapping. If, for some α∈ℝ, α>0, and some integer k, k>2, with |2k|<α,
Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),
then there exists a unique quadratic mapping Q:X→Y such that
μf(x)-16f(x/2)-Q(x/2)(t)≥Ti=1∞M(x,αi+1t|4|ki)
for all x∈X and t>0.
Proof.
Since
limn→∞M(x,αjt|4|kj)=1(x∈X,t>0)
and T is of Hadžić type, from Proposition 2.1, it follows that
limn→∞Tj=n∞M(x,αjt|4|kj)=1(x∈X,t>0).Now, we can apply Theorem 5.4 to obtain the result.
Example 5.6.
Let (X,μ,TM) be a non-Archimedean random normed space in which
μx(t)=tt+‖x‖(x∈X,t>0).
And let (Y,μ,TM) be a complete non-Archimedean random normed space (see Example 3.2). Define
Ψ(x,y,t)=t1+t.
It is easy to see that (4.3) holds for α=1. Also, since
M(x,t)=t1+t,
we have
limn→∞TM,j=n∞M(x,αjt|4|kj)=limn→∞(limm→∞TM,j=nmM(x,t|4|kj))=limn→∞limm→∞(tt+|4k|n)=1(x∈X,t>0).
Let f:X→Y be an even, f(0)=0, and Ψ-approximately AQCQ mapping. Thus, all the conditions of Theorem 5.4 hold, and so there exists a unique quadratic mapping Q:X→Y such that
μf(x)-16f(x/2)-Q(x/2)(t)≥tt+|4k|.
6. Latticetic Random Normed Space
Let ℒ=(L,≥L) be a complete lattice, that is, a partially ordered set in which every nonempty subset admits supremum and infimum, and 0ℒ=infL, 1ℒ=supL. The space of latticetic random distribution functions, denoted by ΔL+, is defined as the set of all mappings F:ℝ∪{-∞,+∞}→L such that F is left continuous and nondecreasing on ℝ, F(0)=0ℒ, F(+∞)=1ℒ.
DL+⊆ΔL+ is defined as DL+={F∈ΔL+:l-F(+∞)=1ℒ}, where l-f(x) denotes the left limit of the function f at the point x. The space ΔL+ is partially ordered by the usual pointwise ordering of functions, that is, F≥G if and only if F(t)≥LG(t) for all t in ℝ. The maximal element for ΔL+ in this order is the distribution function given byɛ0(t)={0L,ift≤0,1L,ift>0.
In Section 2, we defined t-norms on [0,1], and now we extend t-norms on a complete lattice.
Definition 6.1 (see [42]).
A triangular norm (t-norm) on L is a mapping 𝒯:(L)2→L satisfying the following conditions:
Let {xn} be a sequence in L converges to x∈L (equipped order topology). The t-norm 𝒯 is said to be a continuous t-norm if
limn→∞T(xn,y)=T(x,y)
for all y∈L.
A t-norm 𝒯 can be extended (by associativity) in a unique way to an n-array operation taking for (x1,…,xn)∈Ln the value 𝒯(x1,…,xn) defined by Ti=10xi=1,Ti=1nxi=T(Ti=1n-1xi,xn)=T(x1,…,xn).
𝒯 can also be extended to a countable operation taking for any sequence (xn)n∈N in L the valueTi=1∞xi=limn→∞Ti=1nxi.
The limit on the right side of (6.4) exists since the sequence (𝒯i=1nxi)n∈ℕ is nonincreasing and bounded from below.
Note that we put 𝒯=T whenever L=[0,1]. If T is a t-norm, then xT(n) is defined for every x∈[0,1] and n∈N∪{0} by 1 if n=0 and T(xT(n-1),x) if n≥1. A t-norm T is said to be of Hadžić type, (we denote by T∈ℋ) if the family (xT(n))n∈N is equicontinuous at x=1 (cf. [27]).
Definition 6.2 (see [42]).
A continuous t-norm 𝒯 on L=[0,1]2 is said to be continuous t–representable if there exist a continuous t-norm * and a continuous t-conorm ⋄ on [0,1] such that, for all x=(x1,x2), y=(y1,y2)∈L,
T(x,y)=(x1*y1,x2⋄y2).
For example,
T(a,b)=(a1b1,min{a2+b2,1}),M(a,b)=(min{a1,b1},max{a2,b2})
for all a=(a1,a2), b=(b1,b2)∈[0,1]2 are continuous t-representable. Define the mapping 𝒯∧ from L2 to L by
T∧(x,y)={x,ify≥Lx,y,ifx≥Ly.
Recall (see [27, 28]) that if {xn} is a given sequence in L, (𝒯∧)i=1nxi is defined recurrently by (𝒯∧)i=11xi=x1 and (𝒯∧)i=1nxi=𝒯∧((𝒯∧)i=1n-1xi,xn) for all n≥2.
A negation on ℒ is any decreasing mapping 𝒩:L→L satisfying 𝒩(0ℒ)=1ℒ and 𝒩(1ℒ)=0ℒ. If 𝒩(𝒩(x))=x, for all x∈L, then 𝒩 is called an involutive negation. In the following, ℒ is endowed with a (fixed) negation 𝒩.
Definition 6.3.
A latticetic random normed space (in short LRN-space) is a triple (X,μ,𝒯∧), where X is a vector space and μ is a mapping from X into DL+ such that the following conditions hold:
(LRN1) μx(t)=ɛ0(t) for all t>0 if and only if x=0,
(LRN2) μαx(t)=μx(t/|α|) for all x in X, α≠0 and t≥0,
(LRN3) μx+y(t+s)≥L𝒯∧(μx(t),μy(s)) for all x,y∈X and t,s≥0.
We note that from (LPN2) it follows that μ-x(t)=μx(t) for all x∈X and t≥0.
Example 6.4.
Let L=[0,1]×[0,1] and operation ≤L be defined by
L={(a1,a2):(a1,a2)∈[0,1]×[0,1],a1+a2≤1},(a1,a2)≤L(b1,b2)⟺a1≤b1,a2≥b2,∀a=(a1,a2),b=(b1,b2)∈L.
then (L,≤L) is a complete lattice (see [42]). In this complete lattice, we denote its units by 0L=(0,1) and 1L=(1,0). Let (X,∥·∥) be a normed space. Let 𝒯(a,b)=(min{a1,b1},max{a2,b2}) for all a=(a1,a2), b=(b1,b2)∈[0,1]×[0,1] and μ be a mapping defined by
μx(t)=(tt+‖x‖,‖x‖t+‖x‖)(t∈R+),
then (X,μ,𝒯) is a latticetic random normed spaces.
If (X,μ,𝒯∧) is a latticetic random normed space, then
V={V(ɛ,λ):ɛ>L0L,λ∈L∖{0L,1L}},V(ɛ,λ)={x∈X:Fx(ɛ)>LN(λ)},
is a complete system of neighborhoods of null vector for a linear topology on X generated by the norm F.
Definition 6.5.
Let (X,μ,𝒯∧) be a latticetic random normed spaces.
A sequence {xn} in X is said to be convergent to x in X if, for every t>0 and ε∈L∖{0ℒ}, there exists a positive integer N such that μxn-x(t)>L𝒩(ε) whenever n≥N.
A sequence {xn} in X is called a Cauchy sequence if, for every t>0 and ε∈L∖{0ℒ}, there exists a positive integer N such that μxn-xm(t)>L𝒩(ε) whenever n≥m≥N.
A latticetic random normed spaces (X,μ,𝒯∧) is said to be complete if and only if every Cauchy sequence in X is convergent to a point in X.
Theorem 6.6.
If (X,μ,𝒯∧) is a latticetic random normed space and {xn} is a sequence such that xn→x, then limn→∞μxn(t)=μx(t).
Proof.
The proof is the same as classical random normed spaces, see [25].
7. Generalized Hyers-Ulam Stability of the Functional Equation (1.4): An Odd Case via Fixed-Point Method
Using the fixed point method, we prove the generalized Hyers-Ulam stability of the functional equation Df(x,y)=0 in random Banach spaces: an odd case.
Theorem 7.1.
Let X be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and Φ let be a mapping from X2 to DL+ (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<1/8,
Φ2x,2y(t)≤LΦx,y(αt)(x,y∈X,t>0).
Let f:X→Y be an odd mapping satisfying
μDf(x,y)(t)≥LΦx,y(t)
for all x,y∈X and t>0. Then
C(x)∶=limn→∞8n(f(x2n-1)-2f(x2n))
exists for each x∈X and defines a cubic mapping C:X→Y such that
μf(2x)-2f(x)-C(x)(t)≥LT∧(Φx,x(1-8α5αt),Φ2x,x(1-8α5αt))
for all x∈X and t>0.
Proof.
Letting x=y in (7.2), we get
μf(3y)-4f(2y)+5f(y)(t)≥LΦy,y(t)
for all y∈X and t>0. Replacing x by 2y in (7.2), we get
μf(4y)-4f(3y)+6f(2y)-4f(y)(t)≥LΦ2y,y(t)
for all y∈X and t>0. By (7.5) and (7.6),
μf(4y)-10f(2y)+16f(y)(5t)≥LT∧(μ4(f(3y)-4f(2y)+5f(y))(4t),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))=T∧(μf(3y)-4f(2y)+5f(y)(t),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))≥LT∧(Φy,y(t),Φ2y,y(t))
for all y∈X and t>0. Letting y:=x/2 and g(x)∶=f(2x)-2f(x) for all x∈X, we get
μg(x)-8g(x/2)(5t)≥LT∧(Φx/2,x/2(t),Φx,x/2(t))
for all x∈X and t>0.
Consider the set
S∶={h:X⟶Y,h(0)=0}
and introduce the generalized metric on S:
d(h,k)=inf{u∈R+:μh(x)-k(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t)),∀x∈X,∀t>0}
where, as usual, inf∅=+∞. It is easy to show that (S,d) is complete (see the proof of Lemma 2.1 of [24]).
Now, we consider the linear mapping J:S→S such that
Jh(x)∶=8h(x2)
for all x∈X, and we prove that J is a strictly contractive mapping with the Lipschitz constant 8α.
Let h,k∈S be given such that d(h,k)<ɛ. Then
μh(x)-k(x)(ɛt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. Hence
μJh(x)-Jk(x)(8αɛt)=μ8h(x/2)-8k(x/2)(8αɛt)=μh(x/2)-k(x/2)(αɛt)≥T∧(Φx/2,x/2(αt),Φx,x/2(αt))≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(h,k)<ɛ implies that
d(Jh,Jk)≤α8ɛ.
This means that
d(Jh,Jk)≤α8d(h,k)
for all h,k∈S. It follows from (7.8) that
μg(x)-8g(x/2)(5αt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5α≤5/8.
By Theorem 1.1, there exists a mapping C:X→Y satisfying the following:
C is a fixed point of J, that is,
C(x2)=18C(x)
for all x∈X. Since g:X→Y is odd, C:X→Y is an odd mapping. The mapping C is a unique fixed point of J in the set
M={h∈S:d(h,g)<∞}.
This implies that C is a unique mapping satisfying (7.17) such that there exists a u∈(0,∞) satisfying
μg(x)-C(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.
d(Jng,C)→0 as n→∞. This implies the equality
limn→∞8ng(x2n)=C(x)
for all x∈X.
d(h,C)≤(1/(1-8α))d(h,Jh) with h∈M, which implies the inequality
d(g,C)≤5α1-8α,
from which it follows that
μg(x)-C(x)(5α1-8αt)≥LT∧(Φx,x(t),Φ2x,x(t)).
This implies that the inequality (7.4) holds. From Dg(x,y)=Df(2x,2y)-2Df(x,y), by (7.2), we deduce that
μDf(2x,2y)(t)≥LΦ2x,2y(t),μ-2Df(x,y)(t)=μDf(x,y)(t2)≥LΦx,y(t2)
and so, by (LRN3) and (7.1), we obtain
μDg(x,y)(3t)≥LT∧(μDf2x,2y(t),μ-2Df(x,y)(2t))≥LT∧(Φ2x,2y(t),Φx,y(t))≥LΦ2x,2y(t).
It follows that
μ8nDg(x/2n,y/2n)(3t)=μDg(x/2n,y/2n)(3t8n)≥Φx/2n-1,y/2n-1(t8n)≥L⋯≥LΦx,y(18t(8α)n-1)
for all x,y∈X, t>0 and n∈ℕ.
Since limn→∞Φx,y((3/8)(t/(8α)n-1))=1 for all x,y∈X and t>0, by Theorem 2.4, we deduce that
μDC(x,y)(3t)=1L
for all x,y∈X and t>0. Thus the mapping C:X→Y satisfies (1.4).
Now, we have
C(2x)-8C(x)=limn→∞[8ng(x2n-1)-8n+1g(x2n)]=8limn→∞[8n-1g(x2n-1)-8ng(x2n)]=0
for all x∈X. Since the mapping x→C(2x)-2C(x) is cubic (see Lemma 2.2 of [14]), from the equality C(2x)=8C(x), we deduce that the mapping C:X→Y is cubic.
Corollary 7.2.
Let θ≥0 and let p be a real number with p>3. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an odd mapping satisfying
μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Note that (X,μ,TM) is a complete LRN-space, in which L=[0,1], then
C(x)∶=limn→∞8n(f(x2n-1)-2f(x2n))
exists for each x∈X and defines a cubic mapping C:X→Y such that
μf(2x)-2f(x)-C(x)(t)≥(2p-8)t(2p-8)t+5(1+2p)θ‖x‖p
for all x∈X and t>0.
Proof.
The proof follows from Theorem 7.1 by taking
Φx,y(t)∶=tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2-p, and we get
μf(2x)-2f(x)-C(x)(t)≥min((1-23-p)t(1-23-p)t+5⋅2-pθ(2‖x‖p),(1-23-p)t(1-23-p)t+5⋅2-pθ(‖2x‖p+‖x‖p))≥(1-23-p)t(1-23-p)t+5⋅2-pθ(‖2x‖p+‖x‖p)=(2p-8)t(2p-8)t+5⋅(2p+1)θ‖x‖p,
which is the desired result.
Theorem 7.3.
Let X be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+ (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<8,
Φx/2,y/2(t)≤LΦx,y(αt)(x,y∈X,t>0).
Let f:X→Y be an odd mapping satisfying (7.2), then
C(x)∶=limn→∞18n(f(2n+1x)-2f(2nx))
exists for each x∈X and defines a cubic mapping C:X→Y such that
μf(2x)-2f(x)-C(x)(t)≥LT∧(Φx,x(8-α5t),Φ2x,x(8-α5t))
for all x∈X and t>0.
Proof.
Let (S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Consider the linear mapping J:S→S such that
Jh(x)∶=18h(2x)
for all x∈X, and we prove that J is a strictly contractive mapping with the Lipschitz constant α/8.
Let h,k∈S be given such that d(h,k)<ɛ, then
μh(x)-k(x)(ɛt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. Hence
μJh(x)-Jk(x)(α8ɛt)=μ(1/8)h(2x)-(1/8)k(2x)(α8ɛt)=μh(2x)-k(2x)(αɛt)≥LT∧(Φ2x,2x(αt),Φ4x,2x(αt))≥T∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(h,k)<ɛ implies that
d(Jh,Jk)≤α8ɛ.
This means that
d(Jh,Jk)≤α8d(h,k)
for all g,h∈S. Letting g(x)∶=f(2x)-2f(x) for all x∈X, from (7.8), we get that
μg(x)-(1/8)g(2x)(58t)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5/8.
By Theorem 1.1, there exists a mapping C:X→Y satisfying the following:
C is a fixed point of J, that is,
C(2x)=8C(x)
for all x∈X. Since g:X→Y is odd, C:X→Y is an odd mapping. The mapping C is a unique fixed point of J in the set
M={h∈S:d(h,g)<∞}.
This implies that C is a unique mapping satisfying (7.42) such that there exists a u∈(0,∞) satisfying
μg(x)-C(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.
d(Jng,C)→0 as n→∞. This implies the equalit
limn→∞18ng(2nx)=C(x)
for all x∈X.
d(h,C)≤(1/(1-α/8))d(h,Jh) for every h∈M, which implies the inequality
d(g,C)≤58-α,
from which it follows that
μg(x)-C(x)(58-αt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. This implies that the inequality (7.35) holds.
From
μDg(x,y)(3t)≥LT∧(Φ2x,2y(t),Φx,y(t))≥LT∧(Φ2x,2y(t),Φx,y(t8)),
by (7.33), we deduce that
μ8-nDg(2nx,2ny)(3t)=μDg(2nx,2ny)(3⋅8nt)≥LΦ2nx,2ny(8n-1t)≥L⋯≥Φx,y((8α)n-1tα)
for all x,y∈X, t>0, and n∈ℕ. As n→∞, we deduce that
μDC(x,y)(3t)=1L
for all x,y∈X and t>0. Thus the mapping C:X→Y satisfies (1.4).
Now, we have
C(2x)-8C(x)=limn→∞[18ng(2n+1x)-18n-1g(2nx)]=8limn→∞[18n+1g(2n+1x)-18ng(2nx)]=0
for all x∈X. Since the mapping x→C(2x)-2C(x) is cubic (see Lemma 2.2 of [14]), from the equality C(2x)=8C(x), we deduce that the mapping C:X→Y is cubic.
Corollary 7.4.
Let θ≥0 and let p be a real number with 0<p<3. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an odd mapping satisfying (7.28), then
C(x)∶=limn→∞18n(f(2n+1x)-2f(2nx))
exists for each x∈X and defines a cubic mapping C:X→Y such that
μf(2x)-2f(x)-C(x)(t)≥(8-2p)t(8-2p)t+5(1+2p)θ‖x‖p
for all x∈X and t>0. Note that (X,μ,TM) is a complete LRN-space, in which L=[0,1].
Proof.
The proof follows from Theorem 7.3 by taking
μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2p, and we get the desired result.
Theorem 7.5.
Let X be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+ (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<1/2,
Φ2x,2y(t)≤LΦx,y(αt)(x,y∈X,t>0).
Let f:X→Y be an odd mapping satisfying (7.2), then
A(x)∶=limn→∞2n(f(x2n-1)-8f(x2n))
exists for each x∈X and defines an additive mapping A:X→Y such that
μf(2x)-8f(x)-A(x)(t)≥LT∧(Φx,x(1-2α5αt),Φ2x,x(1-2α5αt))
for all x∈X and t>0.
Proof.
Let (S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Letting y∶=x/2 and g(x)∶=f(2x)-8f(x) for all x∈X in (7.7), we get
μg(x)-2g(x/2)(5t)≥LT∧(Φx/2,x/2(t),Φx,x/2(t))
for all x∈X and t>0.
Now, we consider the linear mapping J:S→S such that
Jh(x)∶=2h(x2)
for all x∈X. It is easy to see that J is a strictly contractive self-mapping on S with the Lipschitz constant 2α.
It follows from (7.58) and (7.55) that
μg(x)-2g(x/2)(5αt)≥TM(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5α<∞.
By Theorem 1.1, there exists a mapping A:X→Y satisfying the following:
A is a fixed point of J, that is,
A(x2)=12A(x)
for allx∈X. Sinceg:X→Y is odd,A:X→Y is an odd mapping. The mappingA is a unique fixed point ofJ in the set
M={h∈S:d(h,g)<∞}.
This implies that A is a unique mapping satisfying (7.61) such that there exists a u∈(0,∞) satisfying
μg(x)-A(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.
d(Jng,A)→0 as n→∞. This implies the equality
limn→∞2ng(x2n)=A(x)
for all x∈X.
d(h,A)≤(1/(1-2α))d(h,Jh) for each h∈M, which implies the inequality
d(g,A)≤5α1-2α.
This implies that the inequality (7.57) holds. Since μDg(x,y)(3t)≥LΦ2x,2y(t), it follows that
μ2nDg(x/2n,y/2n)(3t)=μDg(x/2n,y/2n)(3t2n)≥Φx/2n-1,y/2n-1(t2n)≥L⋯≥LΦx,y(12t(2α)n-1)
for all x,y∈X, t>0, and n∈ℕ. As n→∞, we deduce that
μDA(x,y)(3t)=1L
for all x,y∈X and t>0. Thus, the mapping A:X→Y satisfies (1.4).
Now, we have
A(2x)-2A(x)=limn→∞[2ng(x2n-1)-2n+1g(x2n)]=2limn→∞[2n-1g(x2n-1)-2ng(x2n)]=0
for all x∈X. Since the mapping x→A(2x)-8A(x) is additive (see Lemma 2.2 of [14]), from the equality A(2x)=2A(x), we deduce that the mapping A:X→Y is additive.
Corollary 7.6.
Let θ≥0 and let p be a real number with p>1. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an odd mapping satisfying (7.28), then
A(x)∶=limn→∞2n(f(x2n-1)-8f(x2n))
exists for each x∈X and defines an additive mapping A:X→Y such that
μf(2x)-8f(x)-A(x)(t)≥(2p-2)t(2p-2)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].
Proof.
The proof follows from Theorem 7.5 by taking
μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2-p, and we get the desired result.
Theorem 7.7.
Let X be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+ (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<2,
Φx,y(αt)≥LΦx/2,y/2(t)(x,y∈X,t>0).
Let f:X→Y be an odd mapping satisfying (7.2), then
A(x)∶=limn→∞12n(f(2n+1x)-8f(2nx))
exists for each x∈X and defines an additive mapping A:X→Y such that
μf(2x)-8f(x)-A(x)(t)≥LT∧(Φx,x(2-α5αt),Φ2x,x(2-α5αt))
for all x∈X and t>0.
Proof.
Let (S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Consider the linear mapping J:S→S such that
Jh(x)∶=12h(2x)
for all x∈X. It is easy to see that J is a strictly contractive self-mapping on S with the Lipschitz constant α/2. Let g(x)=f(2x)-8f(x), from (7.58), it follows that
μg(x)-1/2g(2x)(52t)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5/2. By Theorem 1.1, there exists a mapping A:X→Y satisfying the following:
A is a fixed point of J, that is,
A(2x)=2A(x)
for all x∈X. Since h:X→Y is odd, A:X→Y is an odd mapping. The mapping A is a unique fixed point of J in the set
M={h∈S:d(h,g)<∞}.
This implies that A is a unique mapping satisfying (7.77) such that there exists a u∈(0,∞) satisfying
μg(x)-A(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.
d(Jng,A)→0 as n→∞. This implies the equality
limn→∞12ng(2nx)=A(x)
for all x∈X.
d(h,A)≤(1/(1-α/2))d(h,Jh), which implies the inequality
d(g,A)≤52-α.
This implies that the inequality (7.74) holds.
Proceeding as in the proof of Theorem 7.5, we obtain that the mapping A:X→Y satisfies (1.4). Now, we have
A(2x)-2A(x)=limn→∞[12ng(2n+1x)-12n-1g(2nx)]=2limn→∞[12n+1g(2n+1x)-12ng(2nx)]=0
for all x∈X. Since the mapping x→A(2x)-8A(x) is additive (see Lemma 2.2 of [14]), from the equality A(2x)=2A(x), we deduce that the mapping A:X→Y is additive.
Corollary 7.8.
Let θ≥0 and let p be a real number with 0<p<1. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an odd mapping satisfying (7.28), then
A(x)∶=limn→∞12n(f(2n+1x)-8f(2nx))
exists for each x∈X and defines an additive mapping A:X→Y such that
μf(2x)-8f(x)-A(x)(t)≥(2-2p)t(2-2p)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].
Proof.
The proof follows from Theorem 7.7 by taking
μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2p, and we get the desired result.
8. Generalized Hyers-Ulam Stability of the Functional Equation (1.4): An Even Case via Fixed-Point Method
Using the fixed point method, we prove the generalized Hyers-Ulam stability of the functional equation Df(x,y)=0 in random Banach spaces, an even case.
Theorem 8.1.
Let X be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+ (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<1/16,
Φx,y(αt)≥LΦ2x,2y(t)(x,y∈X,t>0).
Let f:X→Y be an even mapping satisfying f(0)=0 and (7.2), then
Q(x)∶=limn→∞16n(f(x2n-1)-4f(x2n))
exists for each x∈X and defines a quartic mapping Q:X→Y such that
μf(2x)-4f(x)-Q(x)(t)≥LT∧(Φx,x(1-16α5αt),Φ2x,x(1-16α5αt))
for all x∈X and t>0.
Proof.
Letting x=y in (7.2), we get
μf(3y)-6f(2y)+15f(y)(t)≥LΦy,y(t)
for all y∈X and t>0. Replacing x by 2y in (7.2), we get
μf(4y)-4f(3y)+4f(2y)+4f(y)(t)≥LΦ2y,y(t)
for all y∈X and t>0. By (8.4) and (8.5),
μf(4x)-20f(2x)+64f(x)(5t)≥LT∧(μ4(f(3x)-6f(2x)+15f(x))(4t),μf(4x)-4f(3x)+4f(2x)+4f(x)(t))≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. Letting g(x)∶=f(2x)-4f(x) for all x∈X, we get
μg(x)-16g(x/2)(5t)≥LT∧(Φx/2,x/2(t),Φx,x/2(t))
for all x∈X and t>0. Let (S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Now we consider the linear mapping J:S→S such that Jh(x)∶=16h(x/2) for all x∈X. It is easy to see that J is a strictly contractive self-mapping on S with the Lipschitz constant 16α. It follows from (8.7) that
μg(x)-16g(x/2)(5αt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So,
d(g,Jg)≤5α≤516<∞.
By Theorem 1.1, there exists a mapping Q:X→Y satisfying the following:
Q is a fixed point of J, that is,
Q(x2)=116Q(x)
for allx∈X. Sinceg:X→Y is even withg(0)=0,Q:X→Y is an even mapping withQ(0)=0. The mappingQ is a unique fixed point ofJ in the set
M={h∈S:d(h,g)<∞}.
This implies that Q is a unique mapping satisfying (8.10) such that there exists a u∈(0,∞) satisfying
μg(x)-Q(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.
d(Jng,Q)→0 as n→∞. This implies the equality
limn→∞16ng(x2n)=Q(x)
for all x∈X.
d(h,Q)≤(1/(1-16α))d(h,Jh) for every h∈M, which implies the inequality
d(g,Q)≤5α1-16α.
This implies that the inequality (8.3) holds.
Proceeding as in the proof of Theorem 7.1, we obtain that the mapping Q:X→Y satisfies (1.4). Now, we have
Q(2x)-16Q(x)=limn→∞[16ng(x2n-1)-16n+1g(x2n)]=16limn→∞[16n-1g(x2n-1)-16ng(x2n)]=0
for all x∈X. Since the mapping x→Q(2x)-4Q(x) is quartic, we get that the mapping Q:X→Y is quartic.
Corollary 8.2.
Let θ≥0 and let p be a real number with p>4. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an even mapping satisfying f(0)=0 and (7.28), then
Q(x)∶=limn→∞16n(f(x2n-1)-4f(x2n))
exists for each x∈X and defines a quartic mapping Q:X→Y such that
μf(2x)-4f(x)-Q(x)(t)≥(2p-16)t(2p-16)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].
Proof.
The proof follows from Theorem 8.1 by taking
μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2-p, and we get the desired result.
Theorem 8.3.
Let X be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+ (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<16,
Φx,y(αt)≥Φx/2,y/2(t)(x,y∈X,t>0).
Let f:X→Y be an even mapping satisfying f(0)=0 and (7.2), then
Q(x)∶=limn→∞116n(f(2n+1x)-4f(2nx))
exists for each x∈X and defines a quartic mapping Q:X→Y such that
μf(2x)-4f(x)-Q(x)(t)≥LT∧(Φx,x(16-α5t),Φ2x,x(16-α5t))
for all x∈X and t>0.
Proof.
In the generalized metric space (S,d) defined in the proof of Theorem 7.1, we consider the linear mapping J:S→S such that
Jh(x)∶=116h(2x)
for all x∈X. It is easy to see that J is a strictly contractive self-mapping on S with the Lipschitz constant α/16.
Letting g(x)∶=f(2x)-4f(x) for all x∈X, by (8.7), we get
μg(x)-(1/16)g(2x)(516t)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5/16.
By Theorem 1.1, there exists a mapping Q:X→Y satisfying the following:
Q is a fixed point of J, that is,
Q(2x)=16Q(x)
for all x∈X. Since g:X→Y is even with g(0)=0, Q:X→Y is an even mapping with Q(0)=0. The mapping Q is a unique fixed point of J in the set
M={h∈S:d(h,g)<∞}.
This implies that Q is a unique mapping satisfying (8.24) such that there exists a u∈(0,∞) satisfying
μg(x)-Q(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.
d(Jng,Q)→0 as n→∞. This implies the equality
limn→∞116ng(2nx)=Q(x)
for all x∈X.
d(g,Q)≤(16/(16-α))d(g,Jg) for each h∈M, which implies the inequality
d(g,Q)≤5/(16-α).
This implies that the inequality (8.21) holds.
Proceeding as in the proof of Theorem 7.3, we obtain that the mapping Q:X→Y satisfies (1.4). Now, we have
Q(2x)-16Q(x)=limn→∞[116ng(2n+1x)-116n-1g(2nx)]=16limn→∞[116n+1g(2n+1x)-116ng(2nx)]=0
for all x∈X. Since the mapping x→Q(2x)-4Q(x) is quartic, we get that the mapping Q:X→Y is quartic.
Corollary 8.4.
Let θ≥0 and let p be a real number with 0<p<4. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an even mapping satisfying f(0)=0 and (7.28), then
Q(x)∶=limn→∞116n(f(2n+1x)-4f(2nx))exists for each x∈X and defines a quartic mapping Q:X→Y such that
μf(2x)-4f(x)-Q(x)(t)≥(16-2p)t(16-2p)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].
Proof.
The proof follows from Theorem 8.3 by taking
μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2p, and we get the desired result.
Theorem 8.5.
Let Xbe a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+ (Φ(x,y) is by denoted Φx,y) such that, for some 0<α<1/4,
Φx,y(αt)≥LΦ2x,2y(t)(x,y∈X,t>0).
Let f:X→Y be an even mapping satisfying f(0)=0 and (7.2), then
T(x)∶=limn→∞4n(f(x2n-1)-16f(x2n))
exists for each x∈X and defines a quadratic mapping T:X→Y such that
μf(2x)-16f(x)-T(x)(t)≥LT∧(Φx,x(1-4α5αt),Φ2x,x(1-4α5αt))
for all x∈X and t>0.
Proof.
Let (S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Letting g(x)∶=f(2x)-16f(x) for all x∈X in (8.6), we get
μg(x)-4g(x/2)(5t)≥LT∧(Φx/2,x/2(t),Φx,x/2(t))
for all x∈X and t>0. It is easy to see that the linear mapping J:S→S such that
Jh(x)∶=4h(x2)
for all x∈X, is a strictly contractive self-mapping with the Lipschitz constant 4α.
It follows from (8.36) that
μg(x)-4g(x/2)(5αt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5α<∞.
By Theorem 1.1, there exists a mapping T:X→Y satisfying the following:
T is a fixed point of J, that is,
T(x2)=14T(x)
for all x∈X. Since g:X→Y is even with g(0)=0, T:X→Y is an even mapping with T(0)=0. The mapping T is a unique fixed point of J in the set M={h∈S:d(h,g)<∞}. This implies that T is a unique mapping satisfying (8.39) such that there exists a u∈(0,∞) satisfying
μg(x)-T(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.
d(Jng,T)→0 as n→∞. This implies the equality
limn→∞4ng(x2n)=T(x)
for all x∈X.
d(h,T)≤(1/(1-4α))d(h,Jh) for each h∈M, which implies the inequality
d(g,T)≤5α1-4α.
This implies that the inequality (8.35) holds.
Proceeding as in the proof of Theorem 7.1, we obtain that the mapping T:X→Y satisfies (1.4). Now, we have
T(2x)-4T(x)=limn→∞[4ng(x2n-1)-4n+1g(x2n)]=4limn→∞[4n-1g(x2n-1)-4ng(x2n)]=0
for all x∈X. Since the mapping x→T(2x)-16T(x) is quadratic, we get that the mapping T:X→Y is quadratic.
Corollary 8.6.
Let θ≥0 and let p be a real number with p>2. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an even mapping satisfying f(0)=0 and (7.28), then
T(x)∶=limn→∞4n(f(x2n-1)-16f(x2n))
exists for each x∈X and defines a quadratic mapping T:X→Y such that
μf(2x)-16f(x)-T(x)(t)≥(2p-4)t(2p-4)t+5(1+2p)θ‖x‖p
for all x∈X and t>0.
Proof.
The proof follows from Theorem 8.5 by taking
Φx,y(t)∶=tt+θ(‖x‖p+‖y‖p)
for all x,y∈X. Then we can choose α=2-p, and we get the desired result.
Theorem 8.7.
Let X be a linear space, let (Y,μ,TM) be a complete RN-space, and let Φ be a mapping from X2 to D+ (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<4,
Φx,y(αt)≥Φx/2,y/2(t)(x,y∈X,t>0).
Let f:X→Y be an even mapping satisfying f(0)=0 and (7.2), then
T(x)∶=limn→∞14n(f(2n+1x)-16f(2nx))
exists for each x∈X and defines a quadratic mapping T:X→Y such that
μf(2x)-16f(x)-T(x)(t)≥TM(Φx,x(4-α5t),Φ2x,x(4-α5t))
for all x∈X and t>0.
Proof.
Let (S,d) be the generalized metric space defined in the proof of Theorem 7.1.
It is easy to see that the linear mapping J:S→S such that
Jh(x)∶=14h(2x)
for all x∈X is a strictly contractive self-mapping with the Lipschitz constant α/4.
Letting g(x)∶=f(2x)-16f(x) for all x∈X, from (8.36), we get
μg(x)-1/4g(2x)(54t)≥TM(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5/4.
By Theorem 1.1, there exists a mapping T:X→Y satisfying the following:
(1) T is a fixed point of J, that is,
T(2x)=4T(x)
for all x∈X. Since g:X→Y is even with g(0)=0, T:X→Y is an even mapping with T(0)=0. The mapping T is a unique fixed point of J in the set
M={h∈S:d(h,g)<∞}.
This implies that T is a unique mapping satisfying (8.52) such that there exists a u∈(0,∞) satisfying
μg(x)-T(x)(ut)≥TM(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.
(2) d(Jng,T)→0 as n→∞. This implies the equality
limn→∞14ng(2nx)=T(x)
for all x∈X.
(3) d(h,T)≤(1/(1-α/4))d(h,Jh) for each h∈M, which implies the inequality
d(g,T)≤5/(4-α).
This implies that the inequality (8.49) holds.
Proceeding as in the proof of Theorem 2.3, we obtain that the mapping Q:X→Y satisfies (1.4). Now, we have
T(2x)-4T(x)=limn→∞[14ng(2n+1x)-14n-1g(2nx)]=4limn→∞[14n+1g(2n+1x)-14ng(2nx)]=0
for all x∈X. Since the mapping x→T(2x)-16T(x) is quadratic, we get that the mapping T:X→Y is quadratic.
Corollary 8.8.
Let θ≥0 and let p be a real number with 0<p<2. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an even mapping satisfying f(0)=0 and (7.28). Then
T(x)∶=limn→∞14n(f(2n+1x)-16f(2nx))
exists for each x∈X and defines a quadratic mapping T:X→Y such that
μf(2x)-16f(x)-T(x)(t)≥(4-2p)t(4-2p)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].
Proof.
The proof follows from Theorem 8.5 by taking
Φx,y(t)∶=tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2p, and we get the desired result.
Acknowledgments
The authors are grateful to the area Editor Professor Yeong-Cheng Liou and the reviewer for their valuable comments and suggestions. Y. J. Cho was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant no. 2011-0021821).
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