# Stability of Pexiderized Quadratic Functional Equation in Random 2-Normed Spaces

Authors: Mohammed A. Alghamdi
Journal: Journal of Function Spaces
 (2015)

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## Abstract

The aim of this paper is to investigate the stability of Hyers-Ulam-Rassias type theorems by considering the pexiderized quadratic functional equation in the setting of random 2-normed spaces (RTNS), while the concept of random 2-normed space has been recently studied by Goleţ (2005).


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## Body

## 1. Introduction and Preliminaries

In 1940, Ulam [1] proposed the famous “Ulam stability problem,” which was solved by Hyers [2], in 1941, for additive mappings. In 1950, Aoki [3] solved this Ulam problem for weaker additive mappings; for some historical comments regarding the work of Aoki we refer to [4]. In 1978, Rassias [5] generalized the theorem of Hyers for linear mappings in which the Cauchy difference is allowed to be unbounded by replacing ϵ with a function depending on x and y in the Hyers theorem. The generalization of Hyers theorem was also presented by Rassias [6–9] in 1982–1989. Some important Ulam stability problems on Cauchy equation on semigroups, approximately additive mappings, and Jensen equation have been investigated by Gajda [10], Găvruta [11], and Jung [12], respectively. Until now, the stability problems for different types of functional equations in various spaces have been extensively studied, for instance, by Mirmostafaee and Moslehian [13, 14], Rassias [15], Chang et al. [16, 17], Xu et al. [18], Jun and Kim [19], Mursaleen et al. [20–22], and many others. Also very interesting results on additive, quadratic, and cubic functional equations have been achieved by Mohiuddine et al. [23–29]. This paper is inspired from the work of Alotaibi and Mohiuddine [30] in which they solved stability problem for cubic functional equation in random 2-normed spaces.The pexiderized quadratic functional equation is of the formf(x+y)+f(x-y)=2g(x)+2h(y). For f=g=h, it is called the quadratic functional equation.The terminology and notations used below are standard as in [31–33].A functionf:R→R0+ is called a distribution function if it is nondecreasing and is left continuous with inf⁡t∈Rf(t)=0 and sup⁡t∈Rf(t)=1. By D+, we denote the set of all distribution functions such that f(0)=0.Ifa∈R0+, then Ha∈D+, where (1)Hat=0ift≤a;1ift>a.It is obvious that H0≥f for all f∈D+.At-norm is a continuous mapping ∗:[0,1]×[0,1]→[0,1] such that ([0,1],∗) is abelian monoid with unit one and c∗d≥a∗b if c≥a and d≥b for all a,b,c,d∈[0,1]. A triangle function τ is a binary operation on D+ which is commutative and associative and τ(f,H0)=f for every f∈D+.Gähler [34] presented the following notion of 2-normed space.LetX be a linear space of a dimension d (2≤d<∞). A function ·,·:X×X→R is called 2-normed on X if it satisfied the following conditions: for every x,y∈X, (i) x,y=0 if and only if x and y are linearly dependent; (ii) x,y=y,x; (iii) αx,y=|α|x,y for every α∈R; and (iv) x+y,z≤x,z+y,z for every x,y,z∈X. In this case, (X,·,·) is called a 2-norm space.Goleţ [35] defined and studied the notion of random 2-normed space with the help of 2-norm of Gähler [34]. Recently, the notion of statistical convergence and lacunary statistical convergence have been studied by Mursaleen [36] and Mohiuddine and Aiyub [37], respectively, in random 2-normed spaces.LetX be a linear space of a dimension greater than one and let τ be a triangle function. A function F:X×X→D+ is called aprobabilistic 2-norm on X if it satisfies the following conditions:(i)
Fx,y(t)=H0(t) (∀x,y∈X) if x and y are linearly dependent,(ii)
Fx,y(t)≠H0(t) if x and y are linearly independent,(iii)
Fx,y(t)=Fy,x(t),(iv)
Fαx,y(t)=Fx,y(t/|α|) for all t>0, α≠0 and x,y∈X,(v)
Fx+y,z≥τ(Fx,z,Fy,z) whenever x,y,z∈X,where Fx,y(t) denotes the value of Fx,y at t∈R and the triple (X,F,τ) is called aprobabilistic2-normed space. If we replaced (v) by (v′)
Fx+y,z(t1+t2)≥Fx,z(t1)∗Fy,z(t2), for all x,y,z∈X and t1,t2∈R0+,then triple (X,F,∗) is called arandom 2-normed space (RTNS).Example A.
Let(X,·,·) be a 2-normed space with x,z=x1z2-x2z1, x=(x1,x2), z=(z1,z2), and a∗b=ab for a,b∈[0,1]. For all x∈X, t>0, and nonzero z∈X, consider (2)Fx,zt=tt+x,zift>00ift≤0.Then (X,F,∗) is a RTNS.We remark that every 2-normed space(X,·,·) can be made RTNS by considering Fx,y(t)=H0(t-x,y), for every x,y∈X, t>0, and a∗b=min⁡{a,b}, where a,b∈[0,1].The notions of convergence and Cauchy sequences have been recently studied by Alotaibi and Mohiuddine [30] in the setting of RTNS.Let(X,F,∗) be a RTNS. Then, a sequence x=(xk) is said to be(i)
convergent in (X,F,∗) (F-convergent) to L if for every ϵ>0 and θ∈(0,1) there exists k0∈N such that Fxk-L,z(ϵ)>1-θ whenever k≥k0 and nonzero z∈X. In this case we write F-limk→∞xk=L;(ii)
Cauchy sequence in (X,F,∗) (F-Cauchy) if for every ϵ>0, θ>0, and nonzero z∈X there exists a number N=N(ϵ,z) such that limFxk-xl,z(ϵ)>1-θ for all k,l≥N. We say that RTNS is complete if every F-Cauchy sequence is F-convergent. A complete RTNS is called random 2-Banach space.


## 2. Main Results

Throughout the paper, byY, (Z,F′,∗), and (Y,F,∗), we denote linear space, random 2-normed space, and random 2-Banach space, respectively. Firstly, we prove the stability of the pexiderized quadratic functional equation in RTNS for an odd case.Letφ be a function from X×X to Z. A mapping f:X→Y is said to be φ-approximately pexiderized quadratic function if there exist mappings g,h:X→Y such that (3)Ff(x+y)+f(x-y)-2g(x)-2h(y),zt≥Fφ(x,y),z′t,for all x,y∈X, t>0, and nonzero z∈X.Theorem 1.
Suppose thatf,g and h are odd functions from X to Y satisfying (3). If for some real number α with 0<|α|<2(4)φ2x,2y=αφx,y,for all x,y∈X, then there exists a unique additive mapping T:X→X such that (5)Ffx-Tx,zt≥Fx,z′′2-α4t,Fgx+hx-Tx,zt≥Fx,z′′6-3α14-αt,where (6)Fx,z′′t=Fφx,x,z′t3∗Fφx,0,z′t3∗Fφ0,x,z′t3,for all x∈X, t>0, and nonzero z∈X.Proof.
Replacingx by y and y by x in (3), we obtain (7)Ff(x+y)-f(x-y)-2g(y)-2h(x),zt≥Fφ(y,x),z′tfor all x,y∈X, t>0, and nonzero z∈X. It follows from (3) and (7) that (8)Ff(x+y)-g(x)-h(y)-g(y)-h(x),zt≥Fφ(x,y),z′t∗Fφ(y,x),z′t.Substituting y=0 in (8), we get (9)Ff(x)-g(x)-h(x),zt≥Fφ(x,0),z′t∗Fφ(0,x),z′t.From (8) and (9), we conclude that (10)Ffx+y-fx-fy,z3t≥Fφ(x,y),z′t∗Fφ(y,x),z′t∗Fφ(x,0),z′t∗Fφ(0,x),z′t∗Fφ(y,0),z′t∗Fφ(0,y),z′t,for every x,y∈Xt>0 and nonzero z∈X. Then, by our assumption, (11)F2nx,z′′t=Fx,z′′tαn.Taking x=y in (10), for all x∈X, t>0, and nonzero z∈X, we get (12)Ff(2x)-2f(x),zt≥Fx,z′′t.Putting x=2nx in (12), we have(13)Ff(2n+1x)/2n+1-f2nx/2n,zt=Ff(2n+1x)-f(2nx),z2nt≥F2nx,z′′2nt≥Fx,z′′2αnt.Thus, (14)Ff2n+1x/2n+1-f2nx/2n,zα2nt≥Fx,z′′t.Therefore, for each n>m≥0, (15)Ff2nx/2n-f2mx/2m,z∑k=m+1n‍α2k-1t=F∑k=m+1n‍(f2kx/2k-f2k-1x,z/2k-1)∑k=m+1n‍α2k-1t≥∏k=m+1n‍Ff2kx/2k-f2k-1x/(2k-1),zα2k-1t≥Fx,z′′t,where ∏j=1n‍aj=a1∗a2∗⋯∗an. Let ϵ>0 and t0>0 be given. With the help of the definition of RTNS, we have Fx,z′′(t)=1 and, therefore, we can find some t1>t0 such that Fx,z′′(t1)>1-ϵ. The convergence of the series ∑n=1∞‍(α/2)nt1 gives some n0∈N such that for each n>m≥n0, ∑k=m+1n‍(α/2)k-1t1<t0. Therefore, (16)Ff2nx/2n-f2mx/2m,zt0≥Ff(2nx)/2n-f(2mx)/2m,z∑k=m+1n‍α2k-1t1≥Fx,z′′(t1)>1-ϵ.It follows that (f(2nx)/2n) is a Cauchy sequence in (Y,F,∗). Since (Y,F,∗) is complete RTNS, this sequence converges to some point in Y; that is, T(x)∈Y. Therefore, a mapping T from X to Y is defined by T(x)=F-limn→∞(f2nx/2n). Fix x,y∈X and t>0. From (10), we get that(17)Ff(2n(x+y))/2n-f(2nx)/2n-f(2ny)/2n,zt4=Ff2nx+y-f2nx-f2ny,z2nt4≥Fφx,y,z′2nt12αn∗Fφy,x,z′2nt12αn∗Fφx,0,z′2nt12αn∗Fφ0,x,z′2nt12αn∗Fφy,0,z′2nt12αn∗Fφ0,y,z′2nt12αnfor all n. Moreover, (18)FTx+y-Tx-Ty,zt≥FTx+y-f2nx+y/2n,zt4∗FTx-f2nx/2n,zt4∗FTy-f2ny/2n,zt4∗Ff(2n(x+y))/2n-f(2nx)/2n-f(2ny)/2n,zt4for all n. From (17) and (18), we obtain (19)FT(x+y)-T(x)-T(y),zt=1.Thus, T(x+y)=T(x)+T(y). Now by taking (15) with m=0, we get(20)FT(x)-f(x),zt≥FT(x)-f(2nx)/2n,zt2∗Ff(2nx)/2n-f(x),zt2≥FT(x)-f(2nx)/2n,zt2∗Fx,z′′t2∑k=1n‍α/2k-1≥Fx,z′′t2∑k=1∞‍α/2k-1=Fx,z′′2-α4t.It follows from (9) and (20) that (21)Fg(x)+h(x)-T(x),z14-α12t≥Ff(x)-T(x),zt∗Fg(x)+h(x)-f(x),z2-α12t≥Fx,z′′2-α4t∗Fφ(x,0),z′2-α12t∗Fφ(0,x),z′2-α12t≥Fx,z′′2-α4t.Thus, we obtained (5). Now we will prove the uniqueness of T. For this, we assume that T′ is another additive mapping from X into Y, which satisfies the required inequality. Since, for each n∈N, T(2nx)=2nT(x) and T′(2nx)=2nT′(x), then (22)FT(x)-T′(x),zt=FT(2nx)-T′(2nx),z2nt≥FT′(2nx)-f(2nx),z2nt2∗Ff(2nx)-T(2nx),z2nt2≥F2nx,z′′2-α2nt8=Fx,z′′2/αn2-αt8.We obtain with the help of the definition of RTNS that (23)Fx,z′′2/αn2-αt8=1.Therefore, FT(x)-T′(x),z(t)=1, for all x∈X, t>0, and nonzero z∈X. Hence, T(x)=T′(x) for all x∈X.Now, we are going to prove the stability of the pexiderized quadratic functional equation in RTNS for an even case.Theorem 2.
If (4) holds for 0<|α|<4, let f, g, and h be three even functions from X to Y such that f(0)=g(0)=h(0)=0 and satisfies (3). Then there is a unique quadratic mapping C:X→Y such that, for every x∈X, t>0, and nonzero z∈X, (24)FCx-fx,zt≥Fx,z′′4-α16t,FCx-gx,zt≥Fx,z′′12-3α52-αt,FCx-hx,zt≥Fx,z′′12-3α52-αt,where Fx,z′′(t) is defined by (6).Proof.
Substitutex by y and y by x in (3). Then, for all x,y∈X, t>0, and nonzero z∈X, we obtain (25)Ff(x+y)+f(x-y)-2g(y)-2h(x),zt≥Fφ(y,x),z′t.Again substituting y=x in (3), we get (26)Ff(2x)-2g(x)-2h(x),zt≥Fφ(x,x),z′t.Putting x=0 in (3), we get (27)F2f(y)-2h(y),zt≥Fφ0,y,z′t.For y=0, (3) becomes (28)F2f(x)-2g(x),zt≥Fφ(x,0),z′t.It follows from (25), (27), and (28) that (29)Ffx+y-fx-y-2fx-2fy,zt≥Fφ(x,y),z′t3∗Fφ(x,0),z′t3∗Fφ(0,y),z′t3.By substituting y=x in (29), we get (30)F2f(x)-4f(x),zt≥Fx,z′′t.From (4), we obtain (31)F2nx,z′′t=Fx,z′′tαn,for every x∈X, nonzero z∈X and for each n≥0. It follows from (30) and (31) that (32)Ff(2n+1x)-4f(2nx),zt≥Fx,z′′tαn.From (32), we obtain (33)Ff(2n+1x)/4n+1-f(2nx)/4n,zt=Ff(2n+1x)-4f(2nx),z4n+1t≥Fx,z′′4n+1tαnor, equivalently, (34)Ff(2n+1x)/4n+1-f(2nx)/4n,zαnt4n+1≥Fx,z′′t.Therefore, for all x∈X, t>0, and nonzero z∈X and for each n>m≥0(35)Ff(2nx)/4n-f(2mx)/4m,z∑k=m+1n‍αk-1t4k=F∑k=m+1n‍(f(2kx)/4k-f(2k-1x)/4k-1),z∑k=m+1n‍αk-1t4k≥∏k=m+1n‍Ff(2kx)/4k-f(2k-1x)/4k-1,zαk-1t4k≥Fx,z′′t,where ∏‍ is the same as Theorem 1. Given ϵ>0 and t0>0, since Fx,z′′(t)=1, there is some t1>t0 such that Fx,z′′(t1)>1-ϵ. By the convergence of ∑k=1∞‍(αk-1/4k)t1, we can find some n0 such that ∑k=m+1n‍(αk-1/4k)t1<t0 for each n>m≥n0. This gives that(36)Ff(2nx)/4n-f(2mx)/4m,zt0≥Ff(2nx)/4n-f(2mx)/4m,z∑k=m+1n‍αk-14kt1≥Fx,z′′t0>1-ϵ.We see that (f(2nx)/4n) is a Cauchy sequence in (Y,F,∗) and so it is convergent to some point C(x)∈Y. Therefore, a mapping C from X to Y is defined by C(x)=F-limn→∞(f(2nx)/4n). Fix x,y∈X and t>0. Thus, (29) gives that(37)Ff(2n(x+y))/4n+f(2n(x-y))/4n-2(f(2nx)/4n)-2(f(2ny)/4n),zt5=Ff(2n(x+y))+f(2n(x-y))-2f(2nx)-2f(2ny),z4nt5≥Fφ(x,y),z′4nt15αn∗Fφ(x,0),z′4nt15αn∗Fφ(0,y),z′4nt15αn,for all n. Furthermore,(38)FC(x+y)+C(x-y)-2C(x)-2C(y),zt≥FC(x+y)-f(2n(x+y))/4n,zt5∗FC(x-y)-f(2n(x-y))/4n,zt5∗F2C(x)-2(f(2nx)/4n),zt5∗F2C(y)-2(f(2ny)/4n),zt5∗Ff(2n(x+y))/4n+f(2n(x-y))/4n-2(f(2nx)/4n)-2(f(2ny)/4n),zt5.Equations (37) and (38) give that (39)FC(x+y)+C(x-y)-2C(x)-2C(y),zt=1,for all x,y∈X, t>0, and nonzero z∈X. Thus, C(x+y)+C(x-y)=2C(x)+2C(y). Using (35) with m=0, we get (40)FC(x)-f(x),zt≥FC(x)-f(2nx)/4n,zt2∗Ff(2nx)/4n-f(x),zt2≥FC(x)-f(2nx)/4n,zt2∗Fx,z′′4t2∑k=1n‍α/4k-1≥Fx,z′′4t2∑k=0∞‍α/4k=Fx,z′′4-α16tfor sufficiently large n. From (28) and (40), we conclude that (41)FC(x)-g(x),z52-α48t≥FC(x)-f(x),zt∗Ff(x)-g(x),z4-α48t≥Fx,z′′4-α16t∗Fφ(x,0),z′4-α48t≥Fx,z′′4-α16t.Thus, (42)FC(x)-g(x),zt≥Fx,z′′12-3α52-αt.Similarly, one can show that the above inequality also holds for h. We obtain the uniqueness assertion of this theorem by proceeding the same lines as in Theorem 1.Theorem 3.
Suppose that (4) holds with 0<|α|<4. If a map f:X→Y satisfies (43)Ff(x+y)+f(x-y)-2f(x)-2f(y),zt≥Fφ(x,y),z′t,for all x,y∈X, t>0, and nonzero z∈X with f(0)=0. Then, there are unique mappings T,C:X→Y such that T is additive, C is quadratic, and (44)Ff(x)-T(x)-C(x),zt≥Mx,z2-α8∗4-α32t,for all x∈X, t>0, and nonzero z∈X, where (45)Mx,zt=Fφ(x,x),z′t3∗Fφ(-x,-x),z′t3∗Fφ(x,0),z′t3∗Fφ(0,x),z′t3∗Fφ(-x,0),z′t3∗Fφ(0,-x),z′t3.Proof.
Passing to the odd partf∘ and even part fe of f, we deduce from (43) that (46)Ff∘(x+y)+f∘(x-y)-2f∘(x)-2f∘(y),zt≥Fφ(x,y),z′t∗Fφ(-x,-y),z′t.On the other hand,(47)Ffe(x+y)+fe(x-y)-2fe(x)-2fe(y),zt≥Fφ(x,y),z′t∗Fφ(-x,-y),z′t.With the help of the proofs of Theorems 1 and 2, we obtain unique additive and quadratic mappings T and C, respectively, satisfying (48)Ff∘x-Tx,zt≥Mx,z2-α4t,Ffex-Cx,zt≥Mx,z4-α16t.Therefore, (49)Ff(x)-T(x)-C(x),zt≥Ff∘-T(x),zt2∗Ffe-C(x),zt2≥Mx,z2-α8t∗Mx,z4-α32t=Mx,z2-α8∗4-α32t,for all x∈X, t>0, and nonzero z∈X.Remark 4.
Let(X,〈·,·∣·〉) be a 2-inner product space. We can define a 2-norm on X×X by x1,x2=〈x1,x1∣x2〉 for all x1,x2∈X. In this case, parallelogram law is given by (50)x1+x2,x32+x1-x2,x32=2x1,x32+2x2,x32,for all x1,x2,x3∈X (for more details of 2-inner product space we refer to [38]).Now we give the following illustrative example.Example 5.
Let(X,〈·,·∣·〉) be a 2-inner product space. Let Y be a 2-normed space such that x,z=x1z2-x2z1, where x=(x1,x2) and z=(z1,z2). Suppose that a∗b=ab for all a,b∈[0,1]. Suppose that F and F′ are two random 2-norms on Y and R, respectively, which are given by Example A. Suppose that the random 2-norm F makes Y into an random 2-Banach space. Fixing x∘,y∘,z∘∈Y and a∈X, we define (51)fx=x,a∣s1x∘+x,s12y∘+x,s1z∘,gx=x,a∣s1x∘+x,s12y∘,hx=x,s12y∘+x,s1z∘,φx,s2=x+s2,s1+x-s2,s1hhi-2s2,s1z∘,for each x,s1,s2∈X. Using parallelogram law, one can easily verify that(52)fx+s2+fx-s2-2gx-2hs2=x+s2,s1+x-s2,s1h-2s2,s1z∘,for all x,s1,s2∈X. Therefore,(53)Ff(x+s2)+f(x-s2)-2g(x)-2h(s2),zt=Fφ(x,s2),z′t,for each x,s2∈X, t∈R, and nonzero z∈X. Moreover, φ(2x,2s2)=2φ(x,s2) for each x,s2∈X. We can see that the conditions of Theorems 1 and 2 for f,g,h and |α|=2<2 are satisfied. It follows that odd and even parts of f can be approximated by linear and quadratic functions, respectively. In fact f∘, the odd part of f and f∘(x)=〈x,a∣s1〉x∘, is linear. The even part of f is fe, and fe(x)=x,s12y∘+x,s1z∘ contains a quadratic C(x)=x,s12y∘. Also (54)Ffex-Cx,zt=Fx,s1|z∘|,z′t≥Fx,z′′4-216t.



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