亚洲免费av电影一区二区三区,日韩爱爱视频,51精品视频一区二区三区,91视频爱爱,日韩欧美在线播放视频,中文字幕少妇AV,亚洲电影中文字幕,久久久久亚洲av成人网址,久久综合视频网站,国产在线不卡免费播放

        ?

        APPROXIMATE SOLUTION OF P-RADICAL FUNCTIONAL EQUATION IN 2-BANACH SPACES?

        2019-05-31 03:38:54MuaadhALMAHALEBI

        Muaadh ALMAHALEBI

        Department of Mathematics,Faculty of Sciences,University of Ibn Tofail,Kenitra,Morocco E-mail:muaadh1979@hotmail.fr Abdellatif CHAHBI

        Department of Mathematics,Faculty of Sciences,University of Ibn Zohr,Agadir,Morocco E-mail:ab 1980@live.fr

        Abstract The aim of this paper is to introduce and solve the p-radical functional equation

        Key words stability;hyperstability;2-Banach spaces;p-radical functional equations

        1 Introduction

        Throughout this paper,we will denote the set of natural numbers by N,the set of real numbers by R and R+=[0,∞)the set of nonnegative real numbers.By Nm,m∈N,we will denote the set of all natural numbers greater than or equal to m.

        The notion of linear 2-normed spaces was introduced by G?hler[23,24]in the middle of 1960s.We need to recall some basic facts concerning 2-normed spaces and some preliminary results.

        De finition 1.1Let X be a real linear space with dimX>1 and k·,·k:X ×X ?→ [0,∞)be a function satisfying the following properties:

        (1)kx,yk=0 if and only if x and y are linearly dependent,(2)kx,yk=ky,xk,

        (3)kλx,yk=|λ|kx,yk,

        (4)kx,y+zk≤kx,yk+kx,zk

        for all x,y,z ∈ X and λ ∈ R.Then the function k·,·k is called a 2-norm on X and the pair(X,k·,·k)is called a linear 2-normed space.Sometimes condition(4)called the triangle inequality.

        Example 1.2For x=(x1,x2),y=(y1,y2)∈X=R2,the Euclidean 2-norm kx,ykR2is de fined by

        Lemma 1.3Let(X,k·,·k)be a 2-normed space.If x ∈ X and kx,yk=0 for all y ∈ X,then x=0.

        De finition 1.4A sequence{xk}in a 2-normed space X is called a convergent sequence if there is an x∈X such that

        for all y∈X.If{xk}converges to x,write xk?→x with k?→∞ and call x the limit of{xk}.In this case,we also write

        De finition 1.5A sequence{xk}in a 2-normed space X is said to be a Cauchy sequence with respect to the 2-norm if

        for all y∈X.If every Cauchy sequence in X converges to some x∈X,then X is said to be complete with respect to the 2-norm.Any complete 2-normed space is said to be a 2-Banach space.

        Now,we state the following results as lemma(see[28]for the details).

        Lemma 1.6Let X be a 2-normed space.Then,

        (2)if kx,zk=0 for all z∈X,then x=0,

        (3)for a convergent sequence xnin X,

        for all z∈X.

        The concept of stability for a functional equation arises when de fining,in some way,the class of approximate solutions of the given functional equation,one can ask whether each mapping from this class can be somehow approximated by an exact solution of the considered equation.Namely,when one replaces a functional equation by an inequality which acts as a perturbation of the considered equation.The first stability problem of functional equation was raised by Ulam[32]in 1940.This included the following question concerning the stability of group homomorphisms.

        Let(G1,?1)be a group and let(G2,?2)be a metric group with a metric d(·,·).Given ε >0,does there exist a δ>0 such that if a mapping h:G1→ G2satis fies the inequality

        for all x,y∈G1,then there exists a homomorphism H:G1→G2with

        for all x∈G1?

        If the answer is affirmative,we say that the equation of homomorphism is stable.Since then,this question attracted the attention of many researchers.In 1941,Hyers[25]gave a first partial answer to Ulam’s question and introduced the stability result as follows:

        Theorem 1.7(see[25]) Let E1and E2be two Banach spaces and f:E1→E2be a function such that

        for some δ>0 and for all x,y ∈ E1.Then the limit

        exists for each x∈E1,and A:E1→E2is the unique additive function such that

        for all x∈E1.Moreover,if f(tx)is continuous in t for each fixed x∈E1,then the function A is linear.

        Later,Aoki[9]and Bourgin[10]considered the problem of stability with unbounded Cauchy di ff erences.Rassias[29]attempted to weaken the condition for the bound of the norm of Cauchy di ff erence

        and proved a generalization of Theorem 1.7 using a direct method(cf.Theorem 1.8).

        Theorem 1.8(see[29]) Let E1and E2be two Banach spaces.If f:E1→E2satis fies the inequality

        for some θ≥0 for some p∈R with 0≤p<1,and for all x,y∈E1,then there exists a unique additive function A:E1→E2such that

        for each x∈E1.If,in addition,f(tx)is continuous in t for each fixed x∈E1,then the function A is linear.

        After then,Rassias[30,31]motivated Theorem 1.8 as follows.

        Theorem 1.9(see[30,31]) Let E1be a normed space,E2be a Banach space,and f:E1→E2be a function.If f satis fies the inequality

        for some θ≥ 0,for some p∈ R withand for all x,y∈ E1?{0E1},then there exists a unique additive function A:E1→E2such that

        for each x∈E1?{0E1}.

        Note that Theorem 1.9 reduces to Theorem 1.7 when p=0.For p=1,the analogous result is not valid.Also,Brzdk[11]showed that estimation(1.2)is optimal for p ≥ 0 in the general case.

        Theorem 1.10(see[13]) Let X be a nonempty set,(Y,d)be a complete metric space,andbe a non-decreasing operator satisfying the hypothesis

        for every sequence{δn}n∈Nin YXwith

        Suppose that T:YX→YXis an operator satisfying the inequality

        and

        for all x∈X,then the limit

        exists for each x ∈X.Moreover,the function ψ ∈ YXde fined by

        is a fixed point of T with

        for all x∈X.

        Theorem 1.11(see[14]) Let X be a nonempty set,(Y,d)be a complete metric space,f1,···,fr:X → X and L1,···,Lr:X → R+be given mappings.Suppose that T:YX→ YXandare two operators satisfying the conditions

        for all ξ,μ ∈ YX,x ∈ X and

        If there exist functions ε:X → R+and ? :X → Y such that

        and

        for all x∈X,then the limit(1.7)exists for each x∈X.Moreover,the function(1.8)is a fixed point of T with(1.9)for all x∈X.

        Then by using this theorem,Brzdk[14]improved,extended and complemented several earlier classical stability results concerning the additive Cauchy equation(in particular Theorem 1.9).Over the last few years,many mathematicians have investigated various generalizations,extensions and applications of the Hyers-Ulam stability of a number of functional equations(see,for instance,[1–5,16,17]and references therein);in particular,the stability problem of the radical functional equations in various spaces was proved in[6–8,21,22,26,27].

        Now,we state an analogue of Theorem 1.11 in 2-Banachspaces.Another version of Theorem 1.11 in 2-Banach space can be found in[20].

        Theorem 1.12Let X be a nonempty set,?Y,k·,·k?be a 2-Banach space,g:X → Y be a surjective mapping and let f1,···,fr:X → X and L1,···,Lr:X → R+be given mappings.Suppose thatandare two operators satisfying the conditions

        for all ξ,μ ∈ YX,x,z ∈ X and

        If there exist functions ε:X ×X → R+and ? :X → Y such that

        and

        for all x,z∈X,then the limit

        exists for each x∈X.Moreover,the function ψ:X→Y de fined by

        is a fixed point of T with

        for all x,z∈X.

        ProofFirst,we use the mathematical induction to show that,for every n∈N0,

        Clearly,by(1.16),the case n=0 is trivial.Now,for n∈N1,suppose that(1.21).Then,using(1.14)and inductive assumption,for every x,z∈X,we obtain

        completing the proof of(1.21).Therefore,for n,k∈N0,k>0,

        then

        Putting n=0 in(1.23),we obtain that(1.20)holds for all x,z∈X.In view of(1.14),we conclude

        for all n∈N0and all x,z∈X.Letting n→∞ in in the last inequality,we get T(ψ)=ψ.For the uniqueness of ψ,suppose that ψ1,ψ2∈ YXare two fixed points of T such that

        For every m∈N0,we show that

        For the case m=0,we have

        Next,we assume that(1.24)is valid for some m∈N1.Then,by using(1.14),for every x,z∈X,we get the following inequality

        Therefore,we prove that(1.24)holds for any m∈N0.Now,letting m→∞,on account of(1.17),we get ψ1= ψ2.

        In 2016,Aiemsomboon and Sintunavarat[7]used Theorem 1.11 to investigate a new type of stability for the radical quadratic functional equation of the form

        where f is a self-mapping on the set of real numbers and they generalized some earlier classical results concerning the Hyers-Ulam stability for that functional equations.

        In this paper,we achieve the general solutions of the following p-radical functional equation:

        and discuss the generalized Hyers-Ulam-Rassias stability problem in 2-Banach spaces by using Theorem 1.12.

        2 Solution of Equation(1.25)

        In this section,we give the general solution of functional equation(1.25).The next theorem can be derived from([19],Corollary 2.3 and Proposition 2.4(a)).However,for the convenience of readers we present it with a direct proof.

        Theorem 2.1Let Y be a linear space.A function f:R→Y satis fies the functional equation(1.25)if and only if

        with some additive function F:R→Y.

        ProofIndeed,It is not hard to check without any problem that if f:R→Y satis fies(2.1),then it is a solution to(1.25).On the other hand,if f:R→Y is a solution of(1.25),then we consider the following two cases.

        3 Approximation of the p-Radical Functional Equation(1.25)

        In the following two theorems,we use Theorem 1.12 to investigate the generalized Hyers-Ulam stability of the p-radical functional equation(1.25)in 2-Banach spaces.

        Hereafter,we assume that?Y,k·,·k?is a 2-Banach space.

        Theorem 3.1Let h1,h2:R2→R+be two functions such that

        is an in finite set,where

        for all n∈N,where i=1,2.Assume that f:R→Y satis fies the inequality

        for all x,y,z∈R.Then there exists a unique p-radical function F:R→Y such that

        for all x,z∈R,where

        ProofReplacing y with mx,where x∈R and m∈N,in inequality(3.3),we get

        for all x,z∈R.For each m∈N,we de fine the operator Tm:YR→YRby

        Further put

        and observe that

        Then inequality(3.5)takes the form

        Furthermore,for every x,z ∈ R,ξ,μ ∈ YR,we obtain

        This brings us to de fine the operatorby

        For each m∈N,the above operator has the form described in(1.15)withf2(x)=mx and L1(x)=L2(x)=1 for all x∈R.By induction,we will show that for each x,z∈R,n∈N0,and m∈U we have

        where

        From(3.7)and(3.8),we obtain that inequality(3.11)holds for n=0.Next,we will assume that(3.11)holds for n=k,where k∈N.Then we have

        for all x,z∈R,m∈U.This shows that(3.11)holds for n=k+1.Now we can conclude that inequality(3.11)holds for all n∈N0.Hence,we obtain

        for all x,z∈R,m ∈U.Therefore,according to Theorem 1.12 with ?=f and X=R and using the surjectivity of g,we get that the limit

        exists for each x∈R and m∈U,and

        To prove that Fmsatis fies the functional equation(1.25),just prove the following inequality

        for every x,y,z∈R,n∈N0,and m∈U.Since the case n=0 is just(3.3),take k∈N and assume that(3.13)holds for n=k and every x,y,z∈R,m∈U.Then,for each x,y,z∈R and m∈U,we get

        Thus,by induction,we show that(3.13)holds for every x,y,z∈R,n∈N0,and m∈U.Letting n→∞in(3.13),we obtain the equality

        This implies that Fm:R→Y,de fined in this way,is a solution of the equation

        Next,we will prove that each p-radical function F:R→Y satisfying the inequality

        with some L>0,is equal to Fmfor each m∈U.To this end,we fix m0∈U and F:R→Y satisfying(3.16).From(3.12),for each x,z∈R,we get

        The case j=0 is exactly(3.17).We fix k∈N and assume that(3.18)holds for j=k.Then,in view of(3.17),for each x,z∈R,we get

        This shows that(3.18)holds for j=k+1.Now we can conclude that inequality(3.18)holds for all j∈N0.Now,letting j→∞in(3.18),we get

        Thus,we have also proved that Fm=Fm0for each m∈U,which(in view of(3.12))yields

        This implies(3.4)with F=Fm0and(3.19)con firms the uniqueness of F.

        By a similar way we can prove the following theorem.

        Theorem 3.2Let h:R2→R+be a function such that

        is an in finite set,where

        for all n∈N.Assume that f:R→Y satis fies the inequality

        for all x,y,z∈R.Then there exists a unique p-radical function F:R→Y such that

        for all x,z∈R,where

        ProofReplacing in(3.23)y by mx,where x∈R and m∈N,we get

        for all x,z∈R.For each m∈N,we de fine

        As in Theorem 3.1,we observe that(3.25)takes form

        Λmhas the form described in(1.15)and(1.14)is valid for every ξ,μ ∈ YR,x,z ∈ R.It is not hard to show that

        for all x,z∈R and n∈N0.Therefore

        for all x,z∈R and m∈U.Also the remaining reasonings are analogous as in the proof of Theorem 3.1. ?

        The following theorem concerns the η-hyperstability of(1.25)in 2-Banach spaces.Namely,We consider functions f:R→Y ful filling(1.25)approximately,i.e.,satisfying the inequality

        with η:R3→R+is a given mapping.Then we find a unique p-radical function F:R→Y which is close to f.Then,under some additional assumptions on η,we prove that the conditional functional equation(1.25)is η-hyperstable in the class of functions f:R → Y,i.e.,each f:R → Y satisfying inequality(3.31),with such η,must ful fil equation(1.25).

        Theorem 3.3Let h1,h2and U be as in Theorem 3.1.Assume that

        Then every f:R→Y satisfying(3.3)is a solution of(1.25).

        ProofSuppose that f:R→Y satis fies(3.3).Then,by Theorem 3.1,there exists a mapping F:R→Y satis fies(1.25)and

        for all x,z∈R,where

        Since,in view of(3.32),λ0=0.This means that f(x)=F(x)for all x∈R,whence

        which implies that f satis fies the functional equation(1.25)on R.

        4 Some Particular Cases

        According to above theorems,we derive some particular cases from our main results.

        Corollary 4.1Let h1,h2:R2→ (0,∞)be as in Theorem 3.1 such that

        Assume that f:R→Y satis fies(1.25).Then there exist a unique p-radical function F:R→Y and a unique constant κ∈R+with

        ProofBy the de finition of λi(n)in Theorem 3.1,we observe that

        and

        Combining inequalities(4.3)and(4.5),we get

        Write

        From(4.1),there is a subsequence{γnk}of a sequence{γn}such thatthat is,

        From(4.5)and(4.6),we find that

        This implies that

        and hence

        which means that λ0de fined in Theorem 3.1 is equal to κ.

        By a similar method,we can prove the following corollary where κ=1.

        Corollary 4.2Let h:R→(0,∞)be as in Theorem 3.2 such that

        Assume that f:R→Y satis fies(1.25).Then there exist a unique p-radical function F:R→Y such that

        Corollary 4.3Let θ≥ 0,r≥ 0,s,t∈R such that s+t<0.Suppose that f:R → Y such that f(0)=0 satisfy the inequality

        where p is odd,or f is even function satis fies(4.10)where p is even integer.Then f satis fies(1.25)on R.

        ProofThe proof follows from Theorem 3.1 by de finingand h1(0,z)=h2(0,z)=0 with θ1,θ2∈ R+,r1,r2∈ R+and s,t∈ R such that θ1θ2= θ,r1+r2=r and s+t<0.

        For each n∈N,we have

        Also,we have λ2(n)=ntfor all n ∈ N.Clearly,we can find n0∈ N such that

        According to Theorem 3.1,there exists a unique p-radical function F:R→Y such that

        for all x,z∈R,where

        On the other hand,Since s+t<0,one of s,t must be negative.Assume that t<0.Then

        Thus by Theorem 3.3,we get the desired results.

        The next corollary prove the hyperstability results for the inhomogeneous p-radical functional equation.

        Corollary 4.4Let θ,s,t,r ∈ R such that θ≥ 0 and s+t<0.Assume that G:R2→ Y and f:R→Y such that f(0)=0 and satisfy the inequality

        where p is odd,or f is even function satis fies(4.14)where p is even integer.If the functional equation has a solution f0:R→Y,then f is a solution to(4.15).

        ProofFrom(4.14)we get that the function K:R→Y de fined by K:=f?f0satis fies(4.10).Consequently,Corollary 4.3 implies that K is a solution to p-radical functional equation(1.25).Therefore,

        which means f is a solution to(4.15).

        美女高潮流白浆视频在线观看| 黑人大荫道bbwbbb高潮潮喷| 亚洲男人天堂2019| 亚洲高清精品50路| 看国产亚洲美女黄色一级片 | 欧美日韩国产高清| 日韩精品免费视频久久| 又大又长粗又爽又黄少妇视频| 国外亚洲成av人片在线观看| 99热国产在线| 亚洲国产精品久久性色av| 无码av天天av天天爽| 国产97在线 | 免费| 亚洲欧美日韩中文v在线| 免费蜜桃视频在线观看| 国产爆乳美女娇喘呻吟| 97久久天天综合色天天综合色hd | 亚洲xxxx做受欧美| 九九99久久精品在免费线97| 在线视频一区二区国产| 麻豆av一区二区三区| 久久久国产精品ⅤA麻豆| 久久久亚洲女精品aa| 极品尤物精品在线观看| 国产精品国产三级国av在线观看| 日韩www视频| 在线日本高清日本免费| 国产一区亚洲二区三区| 88久久精品无码一区二区毛片| 一区二区免费电影| 精品人妻在线一区二区三区在线| 日本丰满熟妇videossex一| 久久99精品久久久久久野外| 天堂av在线一区二区| 亚洲天堂二区三区三州| 午夜精品一区二区三区的区别| 国产成人免费高清激情明星| 午夜免费观看一区二区三区| 亚洲男同gay在线观看| 亚洲综合性色一区| 99久久免费精品高清特色大片|