Remarks on the stability of Drygas’ equation and the

Aequationes Math.68(2004)108–116 0001-9054/04/010108-9DOI10.1007/s00010-003-2722-6c Birkh¨a user Verlag,Basel,2004 Aequationes MathematicaeRemarks on the stability of Drygas’equation and the Pexider-quadratic equationDilian YangSummary.In this note,we improve the results proved by S.-M.Jung[8],and S.-M.Jung and P. K.Sahoo[9]on the stability of the Pexider-quadratic functional equation and Drygas’equation, respectively.We give sharper bounds on more general settings by making use of different and easier ways.An example is also given to illustrate that the quadratic equation(and so the Pexider-quadratic equation)is not stable on the free group generated by two elements.Mathematics Subject Classification(2000).Primary39B62,39B82;Secondary20F xx. Keywords.Pexider-quadratic equation,Drygas’equation,quadratic equation,stability.1.IntroductionThe stability theory of functional equations started with the talk of S.M.Ulam held at the Wisconsin University in1940.Since then,the stability problems of functional equations have been studied extensively by many mathematicians (cf.[2]–[12],[14]and[15],and the references therein).Among them,references [8]and[9]studied the stabilities of the Drygas’equation and Pexider-quadratic equation,respectively.The main aim of this note is to give sharper estimations on more general settings by making use of different and easier ways.Moreover, motivated by the idea of Theorem8in[3],an example is also given to show that the quadratic functional equation(and so Pexider-quadratic functional equation) is not stable on the free group generated by two elements.In this note,we assume that G is an abelian group and B is a Banach space. Let f:G→B be any function.The symbols f e(resp.f o)denote the even(resp. odd)part of f.Notice that f o(0)=0and f e(0)=f(0).Vol.68(2004)Remarks on the stability 1092.Preliminary lemmasConsider the functional equationf 1(x +y )+f 2(x −y )=f 3(x )+f 4(y )for all x,y ∈G,(1)where f i (i =1,2,3,4)are functions mapping G to B .As usual,a function A :G →B is said to be additive (resp.quadratic )if it satisfies A (x +y )=A (x )+A (y )(resp.Q (x +y )+Q (x −y )=2Q (x )+2Q (y ))for all x,y ∈G .Theorem 4and Corollary 6in [1](or Theorem 6in [13])imply the following directly.Lemma 1.Let G be a 2-divisible abelian group.The general solution f i :G →B (i =1,2,3,4)of equation (1)is given by ⎧⎪⎪⎨⎪⎪⎩f 1(x )=Q (x )−(A 1−A 2)(x )+b 1,f 2(x )=Q (x )−(A 1+A 2)(x )+b 2,f 3(x )=2Q (x )−2A 1(x )+b 3,f 4(x )=2Q (x )+2A 2(x )+b 4,with b 1+b 2=b 3+b 4,where Q :G →B is a quadratic function and A 1,A 2:G →B are additive functions.In particular,if f satisfies Drygas’functional equation,i.e.,f (x +y )+f (x −y )=2f (x )+f (y )+f (−y )for all x,y ∈G,(2)then f (x )=˜Q (x )+˜A (x )for some quadratic function ˜Q :G →B and an additive function ˜A:G →B ,where G need not be 2-divisible.Let X,Y be two real normed spaces and δ≥0be a constant.Theorem 1in [10]implies that if f :X →Y satisfies 2f x +y 2 −f (x )−f (y ) ≤δ,then there exists a unique additive function F :X →Y such that f (x )−F (x ) ≤δ+ f (0) .We notice that the norm of X plays no role in the proof of this special ing a completely similar procedure as to Theorem 1in [10],we can get the stability of Jensen’s functional equation f (x +y )+f (x −y )=2f (x ),where f maps G to B .Notice that there is a slight difference between the form of Jensen’s functional equation here and that in [10].Lemma 2.If f :G →B satisfiesf (x +y )+f (x −y )−2f (x ) ≤ (3)for some ≥0and for all x,y ∈G ,then there is a unique additive function A :G →B such thatf (x )−A (x )−f (0) ≤(4)for all x ∈G .The following result,proved by I.Feny˝o ,states the stability of the quadratic functional equation (cf.[2]or Theorem 1in [9]).110 D.Yang AEMLemma3.If f:G→B satisfiesf(x+y)+f(x−y)−2f(x)−2f(y) ≤ (5) for some ≥0and for all x,y∈G,then there is a unique quadratic function Q:G→B such thatf(x)−Q(x) ≤1( + f(0) )(6)3for all x∈G.3.Main resultsWe are now ready to give our main Theorem.Theorem1.Let G be a2-divisible abelian group and B a Banach space.Suppose that functions f i:G→B(i=1,2,3,4)satisfyf1(x+y)+f2(x−y)−f3(x)−f4(y) ≤ (7) for some ≥0and for all x,y∈G.Then there are a unique quadratic function Q:G→B and exactly two additive functions A1,A2:G→B such that,(8) f1(x)−Q(x)−(A1−A2)(x)−f1(0) ≤476,(9) f2(x)−Q(x)−(A1+A2)(x)−f2(0) ≤476,(10)f3(x)−2Q(x)−2A1(x)−f3(0) ≤203(11)f4(x)−2Q(x)+2A2(x)−f4(0) ≤263for all x∈G.Proof.Replace x by−x and y by−y in(7)to getf1(−x−y)+f2(−x+y)−f3(−x)−f4(−y) ≤ .(12) Adding(resp.subtracting)the argument of the norm of(12)to(resp.from)that of(7),then taking the norm and dividing the expression obtained by2,we havef e1(x+y)+f e2(x−y)−f e3(x)−f e4(y) ≤ ,(13)(resp. f o1(x+y)+f o2(x−y)−f o3(x)−f o4(y) ≤ ).(14) Letting y=0,x=0,and x=y=0consecutively in(7),we getf1(x)+f2(x)−f3(x)−f4(0) ≤ ,(15)f1(y)+f2(−y)−f3(0)−f4(y) ≤ ,(16)f1(0)+f2(0)−f3(0)−f4(0) ≤ ,(17)Vol.68(2004)Remarks on the stability111and sof e1(x)+f e2(x)−f e3(x)−f4(0) ≤ ,(18)f o1(x)+f o2(x)−f o3(x) ≤ ,(19)f e1(y)+f e2(−y)−f3(0)−f e4(y) ≤ (20) for all x,y∈G.Inequalities(18)and(20)result inf e3(x)−f e4(x)−f3(0)+f4(0) ≤2 (21) since f e2(y)=f e2(−y)for all y∈G.Consecutively making use of(18)twice,(13) twice and then(21),we getf e3(x+y)+f e3(x−y)−2f e3(x)−2f e3(y)+2f3(0)≤ f e3(x+y)−f e1(x+y)−f e2(x+y)+f4(0)+ f e3(x−y)−f e1(x−y)−f e2(x−y)+f4(0)+ f e1(x+y)+f e2(x−y)−f e3(x)−f e4(y)+ f e1(x−y)+f e2(x+y)−f e3(x)−f e4(−y)+ −2f e3(y)+f e4(y)+f e4(−y)−2f4(0)+2f3(0)≤ + + + +2×2 =8 .Applying Lemma3to f e3−f3(0),we get a unique quadratic function Q:G→B such thatf e3(x)−Q(x)−f3(0) ≤13×8 =83.(22)Similarly,using(19)twice and(13)twice yieldsf o3(x+y)+f o3(x−y)−2f o3(x)≤ f o3(x+y)−f o1(x+y)−f o2(x+y)+ f o3(x−y)−f o1(x−y)−f o2(x−y)+ f o1(x+y)+f o2(x−y)−f o3(x)−f o4(y)+ f o1(x−y)+f o2(x+y)−f o3(x)−f o4(−y)≤ + + + =4 .Applying Lemma2to f o3we get a unique additive function A1:G→B satisfyingf o3(x)−A1(x) ≤4 .(23)It follows from(22)and(23)thatf3(x)−Q(x)−A1(x)−f3(0) ≤83+4=203.(24)Combining(21)and(22)givesf e4(x)−Q(x)−f4(0) ≤ f e4(x)−f e3(x)+f3(0)−f4(0) + f e3(x)−Q(x)−f3(0)≤2 +83 =143.(25)112 D.Yang AEMNow we use the notationh(x):=f1(x)−f2(x),(26)k(x):=f4(x)−f4(−x)(27) for all x∈G.Replace y by−y in(7)to getf1(x−y)+f2(x+y)−f3(x)−f4(−y) ≤ .(28) Subtracting the argument of the norm(28)from(7)and then taking the norm, with(26)and(27),we obtainh(x+y)−h(x−y)−k(y) ≤2 .(29) Now we use repeatedly(29)for the estimatek(x+y)+k(x−y)−2k(x)= −h(z+x+y)+h(z−x−y)+k(x+y)−h(z+x−y)+h(z−x+y)+k(x−y)+h(z+x+y)−h(z−x+y)−k(x)+h(z+x−y)−h(z−x−y)−k(x)≤ −h(z+x+y)+h(z−x−y)+k(x+y)+ −h(z+x−y)+h(z−x+y)+k(x−y)+ h(z+x+y)−h(z−x+y)−k(x)+ h(z+x−y)−h(z−x−y)−k(x)≤4×2 =8 .Note that k(0)=0and apply Lemma2to the function k in order to get a unique additive function A2:G→B such thatk(x)−A2(x) ≤8 .(30) In view of(27)and(30)we havef o4(x)−12A2(x)≤4 .(31)Inequalities(31)and(25)lead tof4(x)−Q(x)−12A2(x)−f4(0)≤263.(32)By(15),(24),and(17)we obtainf1(x)+f2(x)−Q(x)−A1(x)−f1(0)−f2(0)≤ f1(x)+f2(x)−f3(x)−f4(0)+ f3(x)−Q(x)−A1(x)−f3(0)+ f4(0)+f3(0)−f1(0)−f2(0)≤ +203 + =263.(33)Vol.68(2004)Remarks on the stability113 Letting x=0and x=y consecutively in(29)yieldsh(y)−h(−y)−k(y) ≤2 and h(2x)−h(0)−k(x) ≤2 .Noting that h(y)−h(−y)=2h o(y),k is odd,and G is2-divisible,we get with the aid of the above two inequalitiesh o(x)−12k(x)≤ and h e(x)−h(0) ≤2 .Thush(x)−12k(x)−h(0)≤3 .(34)It follows from(30)and(34)thath(x)−12A2(x)−h(0)≤7 .(35)Substituting h by f1−f2(cf.(26))in(35)yieldsf1(x)−f2(x)−12A2(x)−f1(0)+f2(0)≤7 .(36)From(33)and(36)we havef1(x)−12Q(x)−12A1(x)−14A2(x)−f1(0)≤476,(37)f2(x)−12Q(x)−12A1(x)+14A2(x)−f2(0)≤476.(38)Therefore we have proved inequalities(24),(32),(37),and(38).To get(8)-(11) it suffices to replace Q,A1,and A2by2Q,2A1,and−4A2,respectively.The proof of the rest of the theorem is the same as[8],and so we omit it here.Immediately we can get the stability of Drygas’equation(2)by Theorem1, where f1(x)=f2(x)=:f(x),f3(x)=2f(x),and f4(x)=f(x)+f(−x)for each x∈G.It is easy to see that our estimate is sharper than that in[9].Indeed,from (7),we have 2f(0) ≤ .By(10)we getf3(x)−2Q(x)−2A1(x) ≤233,implyingf(x)−Q(x)−A1(x) ≤236for all x∈G.It is interesting that we can get much sharper estimate by use of Lemmas2and3only.Moreover,unlike Theorem1,G need not be2-divisible.114 D.Yang AEMProposition1.If f:G→B satisfiesf(x+y)+f(x−y)−2f(x)−f(y)−f(−y) ≤ (39) for some ≥0and for all x,y∈G,then there is a unique quadratic function Q:G→B and a unique additive function A:G→B such thatf(x)−Q(x)−A(x) ≤32(40) for all x∈G.In fact,replace x by−x and y by−y in(39)to getf(−x−y)+f(−x+y)−2f(−x)−f(−y)−f(y) ≤ .(41) As before,from(39)and(41)we getf e(x+y)+f e(x−y)−2f e(x)−2f e(y) ≤ ,(resp. f o(x+y)+f o(x−y)−2f o(x) ≤ ).By Lemma3(resp.Lemma2),there exists a unique quadratic(resp.additive) function Q(resp.A):G→B such thatf e(x)−Q(x) ≤13( + f(0) )(42)(resp. f o(x)−A(x) ≤ ).(43)From(39),wefind f(0) ≤bining(42)and(43)clearly leads to(40),which ends our proof. Remark1.In Theorem1(resp.Proposition1),if we only require that G is a 2-divisible group(resp.group),not necessarily abelian,such that G satisfies Kan-nappan’s condition,i.e.,f(xyz)=f(xzy)for all x,y,z∈G,then the correspond-ing conclusions still remain true.But from next Section we can see that in general this is not the case.4.A counter-exampleMotivated by the idea of Theorem8in[3],we can construct an example to show that the quadratic functional equation is not stable on the free group generated by2elements.Vol.68(2004)Remarks on the stability115Proposition2.Let G:= a,b be the free group generated by two elements a and b.Then the quadratic functional equationq(xy)+q(xy−1)=2q(x)+2q(y)for all x,y∈G(44) is not stable,where q:G→R is a function.More precisely,there is an f:G→R such that|f(xy)+f(xy−1)−2f(x)−2f(y)|≤4for all x,y∈G;(45) however,there is no q:G→R satisfying(44)so that f−q is bounded.Here,“f−q is bounded”means|f(x)−q(x)|≤K for all x∈G and for some non-negative constant K(independent of x).Proof.Let x∈G and assume that x is a reduced word.Define f:G→R byf(x)=r(x)+s(x)for all x∈G,(46) where r(x)(resp.s(x))denotes the number of pairs of the form ab(resp.b−1a−1)contained in x.(We should notice that f in(46)is different from that of Theorem 8in[3].)So f(x−1)=f(x)for all x∈G(i.e.,f is even)since f(x−1)=r(x−1)+ s(x−1)=s(x)+r(x)=f(x)by(46).It is easy to see that r(xy)−r(x)−r(y)∈{0,1}and s(xy)−s(x)−s(y)∈{0,1}.Thusf(xy)−f(x)−f(y)∈{0,1,2},for all x,y∈G.Therefore we getf(xy)+f(xy−1)−2f(x)−2f(y)=(f(xy)−f(x)−f(y))+(f(xy−1)−f(x)−f(y))=(f(xy)−f(x)−f(y))+(f(xy−1)−f(x)−f(y−1))∈{0,1,2,3,4},implying that(45)is true.To the contrary,suppose that there is a quadratic function q:G→R(i.e.,q satisfies equation(44))such that f−q is bounded. It follows from(46)that f(a n)=0and f(b n)=0for each integer n.Thus we have q(a n)and q(b n)are also bounded.It is well-known that(cf.also[16]) q(a n)=n2q(a)and q(b n)=n2q(b).This implies that q(a)=q(b)=0.Now consider f((ab)n)−q((ab)n).Clearly from(46)f((ab)n)=n for each non-negative integer n.The identity f((ab)n)−q((ab)n)=n−n2q(ab)implies a contradiction, because if q(ab)=0(resp.=0)then f((ab)n)−q((ab)n)=n(resp.n(1−nq(ab))). Therefore,no such a quadratic function q exists,which concludes the proof. Remark2.In Proposition2,obviously the free group G generated by the two elements a and b is not2-divisible.Also it is easy to see that f defined via(46)does not satisfy Kannappan’s condition.For instance,let x=a,y=b,and z=a−1. 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Dilian YangDepartment of Pure MathematicsUniversity of WaterlooWaterloo,OntarioCanada N2L3G1e-mail:dyang@math.uwaterloo.caManuscript received:July14,2003and,infinal form,October23,2003.。