Stability of a Monomial Functional Equation on a Restricted Domain

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1 mathematics Article Stability of a Moomial Fuctioal Equatio o a Restricted Domai Yag-Hi Lee Departmet of Mathematics Educatio, Gogju Natioal Uiversity of Educatio, Gogju 32553, Korea; yaghi2@hamail.et Academic Editor: Hari Moha Srivastava Received: 24 August 2017; Accepted: 8 October 2017; Published: 18 October 2017 Abstract: I this paper, we prove the stability of the followig fuctioal equatio C i ( 1 i f (ix + y! f (x = 0 o a restricted domai by employig the direct method i the sese of Hyers. Keywords: stability; moomial fuctioal equatio; direct method MSC: 39B82; 39B52 1. Itroductio Let V ad W be real vector spaces, X a real ormed space, Y a real Baach space, N (the set of atural umbers, ad f : V W a give mappig. Cosider the fuctioal equatio C i ( 1 i f (ix + y! f (x = 0 (1 for all x, y V, where C i :=!. The fuctioal Equatio (1 is called a -moomial fuctioal i!( i! equatio ad every solutio of the fuctioal Equatio (1 is said to be a moomial mappig of degree. The fuctio f : R R give by f (x := ax is a particular solutio of the fuctioal Equatio (1. I particular, the fuctioal Equatio (1 is called a additive (quadratic, cubic, quartic, ad quitic, respectively fuctioal equatio for the case = 1 ( = 2, = 3, = 4, ad = 5, respectively ad every solutio of the fuctioal Equatio (1 is said to be a additive (quadratic, cubic, quartic, ad quitic, respectively mappig for the case = 1 ( = 2, = 3, = 4, ad = 5, respectively. A mappig A : V W is said to be additive if A(x + y = A(x + A(y for all x, y V. It is easy to see that A(rx = ra(x for all x V ad all r Q (the set of ratioal umbers. A mappig A : V W is called -additive if it is additive i each of its variables. A mappig A is called symmetric if A (x 1, x 2,..., x = A (x π(1, x π(2,..., x π( for every permutatio π : {1, 2,..., } {1, 2,..., }. If A (x 1, x 2,..., x is a -additive symmetric mappig, the A (x will deote the diagoal A (x, x,..., x for x V ad ote that A (rx = r A (x wheever x V ad r Q. Such a mappig A (x will be called a moomial mappig of degree (assumig A 0. Furthermore, the resultig mappig after substitutio x 1 = x 2 =... = x l = x ad x l+1 = x l+2 =... = x = y i A (x 1, x 2,..., x will be deoted by A l, l (x, y. A mappig p : V W is called a geeralized polyomial (GP mappig of degree N provided that there exist A 0 (x = A 0 W ad i-additive symmetric mappigs A i : V i W (for 1 i such that p(x = Ai (x, for all x V ad A 0. For f : V W, let h be the differece operator defied as follows: h f (x = f (x + h f (x Mathematics 2017, 5, 53; doi: /math

2 Mathematics 2017, 5, 53 2 of 11 for h V. Furthermore, let 0 h f (x = f (x, 1 h f (x = h f (x ad h h f (x = +1 h f (x for all N ad all h V. For ay give N, the fuctioal equatio +1 h f (x = 0 for all x, h V is well studied. I explicit form we ca have h f (x = C i ( 1 i f (x + ih. The followig theorem was proved by Mazur ad Orlicz [1,2] ad i greater geerality by Djoković (see [3]. Theorem 1. Let V ad W be real vector spaces, N ad f : V W, the the followig are equivalet: (1 +1 h f (x = 0 for all x, h V. (2 x1 x2... x+1 f (x 0 = 0 for all x 0, x 1, x 2,..., x +1 V. (3 f (x = A (x + A 1 (x + + A 2 (x + A 1 (x + A 0 (x for all x V, where A 0 (x = A 0 is a arbitrary elemet of W ad A i (x(i = 1, 2,..., is the diagoal of a i-additive symmetric mappig A i : V i W. I 2007, L. Cădariu ad V. Radu [4] proved a stability of the moomial fuctioal Equatio (1 (see also [5 7], i particular, the followig result is give by the author i [6]. Theorem 2. Let p be a o-egative real umber with p =, let θ > 0, ad let f : X Y be a mappig such that C i ( 1 i f (ix + y! f (x θ( x p + y p (2 for all x, y X. The there exist a positive real umber K ad a uique moomial fuctio of degree F : X Y such that holds for all x X. The mappig F : X Y is give by for all x X. f (x F(x K x p (3 F(x := lim s f (2 s x 2 s The cocept of stability for the fuctioal Equatio (1 arises whe we replace the fuctioal Equatio (1 by a iequality (2, which is regarded as a perturbatio of the equatio. Thus, the stability questio of fuctioal Equatio (1 is whether there is a exact solutio of (1 ear each solutio of iequality (2. If the aswer is affirmative with iequality (3, we would say that the Equatio (1 is stable. The direct method of Hyers meas that, i Theorem 2, F(x satisfyig iequality (3 is costructed by the limit of the sequece { f (2s x 2 s } s N as s. Historically, i 1940, Ulam [8] proposed the problem cocerig the stability of group homomorphisms. I 1941, Hyers [9] gave a affirmative aswer to this problem for additive mappigs betwee Baach spaces, usig the direct method. Subsequetly, may mathematicias came to deal with this problem (cf. [10 17]. I 1998, A. Giláyi dealt with the stability of moomial fuctioal equatio for the case p = 0 (see [18,19] ad he proved for the case whe p is a real costat (see [20]. Thereafter, C.-K. Choi proved stability theorems for may kids of restricted domais, but his theorems are maily coected

3 Mathematics 2017, 5, 53 3 of 11 with the case of p = 0. If p < 0 i (2, the the iequality (2 caot hold for all x X, so we have to restrict the domai by excludig 0 from X. The mai purpose of this paper is to geeralize our previous result (Theorem 2 by replacig the real ormed space X with a restricted domai S of a real vector space V ad by replacig the cotrol fuctio θ( x p + y p with a more geeral fuctio ϕ : S 2 [0,. 2. Stability of the Fuctioal Equatio (1 o a Restricted Domai I this sectio, for a give mappig f : V W, we use the followig abbreviatio for all x, y V. D f (x, y := C i ( 1 i f (ix + y! f (x Lemma 1. The equalities ad C i = C j C k ( 1 i+k (i = 0,, (4 j+k=2i 0j,k C i = j+k=2i 1 0j,k C j C k ( 1 k (i = 1,, (5 hold. Proof. From the equalities C i ( 1 i x 2i = (x 2 1, (x 2 1 = (x + 1 (x 1, (x + 1 (x 1 = we get the equality ( ( C j x j C k ( 1 k x k = C i ( 1 i x 2i = C j C k ( 1 k x j+k, C j C k ( 1 k x j+k (6 for all x R. Sice the coefficiet of the term x 2i of the left-had side i (6 is C i ( 1 i ad the coefficiet of the term x 2i of the right-had side i (6 is C j C k ( 1 k, we get the Equality (4. j+k=2i 0j,k We easily kow that the coefficiet of the term x 2i 1 of the left-had side i (6 is 0 ad the coefficiet of the term x 2i 1 of the right-had side i (6 is C j C k ( 1 k. So we get the Equality (5. j+k=2i 1 0j,k We rewrite a refiemet of the result give i [6]. Lemma 2. (Lemma 1 i [6] The equality C j D f (x, jx + y D f (2x, y =!( f (2x 2 f (x (7

4 Mathematics 2017, 5, 53 4 of 11 holds for all x, y V. I particular, if D f (x, y = 0 for all x, y V, the Proof. By (4, (5, ad the equality C j = 2, we get the equalities C j D f (x, jx + y D f (2x, y f (2x = 2 f (x (8 = C j ( 1 k C k f ((j + kx + y = j+k=2i 0j,k + = C i ( 1 i f (2ix + y +! f (2x i=1 j+k=2i 1 0j,k C j ( 1 k C k f (2ix + y C j! f (x C j ( 1 k C k f ((2i 1x + y C j! f (x +! f (2x C j! f (x C i ( 1 i f (2ix + y +! f (2x for all x, y V. Lemma 3. If f satisfies the fuctioal equatio D f (x, y = 0 for all x, y V\{0} with f (0 = 0, the f satisfies the fuctioal equatio D f (x, y = 0 for all x, y V. Proof. Sice D f (0, 0 = 0, D f (0, y = 0 for all y V\{0}, ad D f (x, 0 = ( 1 D f ( x, x ( 1 D f ( x, ( + 1x + D f (x, x for all x V\{0}, we coclude that f satisfies the fuctioal equatio D f (x, y = 0 for all x, y V. We rewrite a refiemet of the result give i [7]. Theorem 3. (Corollary 4 i [7] A mappig f : V W is a solutio of the fuctioal Equatio (1 if ad oly if f is of the form f (x = A (x for all x V, where A is the diagoal of the -additive symmetric mappig A : V W. Proof. Assume that f satisfies the fuctioal Equatio (1. We get the equatio +1 x f (y = D f (x, x + y D f (x, y = 0 for all x, y V. By Theorem 1, f is a geeralized polyomial mappig of degree at most, that is, f is of the form f (x = A (x + A 1 (x + + A 2 (x + A 1 (x + A 0 (x for all x V, where A 0 (x = A 0 is a arbitrary elemet of W ad A i (x(i = 1, 2,..., is the diagoal of a i-additive symmetric mappig A i : V i W. O the other had, f (2x = 2 f (x holds for all x V by Lemma 2, ad so f (x = A (x. Coversely, assume that f (x = A (x for all x V, where A (x is the diagoal of the -additive symmetric mappig A : V W. From A (x + y = A (x + 1 i=1 C i A i,i (x, y,

5 Mathematics 2017, 5, 53 5 of 11 A (rx = r A (x, A i,i (x, ry = r i A i,i (x, y (x, y V, r Q, we see that f satisfies (1, which completes the proof of this theorem. Theorem 4. Let S be a subset of a real vector space V ad Y a real Baach space. Suppose that for each x V\{0} there exists a real umber r x > 0 such that rx S for all r r x. Let ϕ : S 2 [0, be a fuctio such that 2 k ϕ(2 k x, 2 k y < (9 for all x, y S. If the mappig f : S Y satisfies the iequality D f (x, y ϕ(x, y (10 for all x, y S, the there exists a uique moomial mappig of degree F : V Y such that for all x S, where f (x F(x Φ(x := Φ(2 k x! 2 (k+1 (11 C j ϕ(x, jx + x + ϕ(2x, x. I particular, F is represeted by for all x V. f (2 F(x = m x lim m 2 m Proof. Let x V\{0} ad m be a iteger such that 2 m r x. It follows from (7 i Lemma 2 ad (10 that! f (2 m+1 x 2 f (2 m x = C j D f (2 m x, 2 m (jx + x D f (2 m+1 x, 2 m x C j D f (2 m x, 2 m (j + 1x + D f (2 m+1 x, 2 m x C j ϕ(2 m x, 2 m (j + 1x + ϕ(2 m+1 x, 2 m x = Φ(2 m x. From the above iequality, we get the followig iequalities f (2 m x 2 m f ( 2 m+1 x 2 (m+1 Φ(2m x! 2 (m+1 ad ( f (2 m x 2 m f 2 m+m x 2 (m+m m+m 1 k=m Φ(2 k x! 2 (k+1 (12

6 Mathematics 2017, 5, 53 6 of 11 for all m N. So the sequece { f (2m x 2 m } m N is a Cauchy sequece for all x V\{0}. Sice lim m f (2 m 0 2 m = 0 ad Y is a real Baach space, we ca defie a mappig F : V Y by f (2 F(x = m x lim m 2 m for all x V. By puttig m = 0 ad lettig m i the iequality (12, we obtai the iequality (11 if x S. From the iequality (10, we get D f (2 m x, 2 m y 2 m ϕ (2m x, 2 m y 2 m. for all x, y V\{0}, where 2 m r x, r y. Sice the right-had side i the above equality teds to zero as m, we obtai that F satisfies the iequality (1 for all x, y V\{0}. By Lemma 3 ad F(0 = 0, F satisfies the Equality (1 for all x, y V. To prove the uiqueess of F, assume that F is aother moomial mappig of degree satisfyig the iequality (11 for all x S. The equality F (x = F (2 m x 2 m follows from the Equality (8 i Lemma 2 for all x V\{0} ad m N. Thus we ca obtai the iequalities F (x f (2m x 2 m = F (2 m x 2 m f (2m x 2 m Φ(2 k+m x! 2 (k+m+1 = k=m Φ(2 k x! 2 (k+1, for all x V\{0}, where 2 m r x. Sice Φ(2 k x k=m 0 as m ad F (0 = 0, F (x =! 2 (k+1 lim m 2 m f (2 m x for all x V, i.e., F(x = F (x for all x V. This completes the proof of the theorem. We ca give a geeralizatio of Theorems 2 ad 5 i [6] as the followig corollary. Corollary 1. Let p ad r be real umbers with p < ad r > 0, let X be a ormed space, ε > 0, ad f : X Y be a mappig such that D f (x, y ε( x p + y p (13 for all x, y X with x, y > r. The there exists a uique moomial mappig of degree F : X Y satisfyig f (x F(x ε(2 p C j (j+1 p!(2 2 p x p (for x > r, ((k+1 p +k p ε x p + (k p + i=2 C i (ik + i k p +!(k + 1 p ε(2p C j (j+1 p!(2 2 p x p (for kx > r. I particular, if p < 0, the f is a moomial mappig of degree itself. Proof. If we set ϕ(x, y := ε( x p + y p ad S = {x X x > r}, the there exists a uique moomial mappig of degree F : X Y satisfyig f (x F(x ε(2p C j (j + 1 p!(2 2 p x p (14

7 Mathematics 2017, 5, 53 7 of 11 for all x X with x > r by Theorem 4. Notice that if F is a moomial mappig of degree, the D F((k + 1x, kx = 0 for all x X ad k R. Hece the equality ( 1 ( f (x F(x =D f ((k + 1x, kx + ( 1 (F f ( kx + i=2 C i ( 1 i (F f (i(k + 1x kx!(f f ((k + 1x holds for all x X ad k R. So if F : X Y is the moomial mappig of degree satisfyig (14, the F : X Y satisfies the iequality with a real umber k f (x F(x ((k + 1p + k p ε x p + (k p + i=2 C i (ik + i k p +!(k + 1 p ε(2p C j (j + 1 p!(2 2 p x p (15 for all kx > r. Moreover, if p < 0, the lim k (k p + i=2 C i (ik + i kp +!(k + 1 p = 0 ad lim k (k p + (k + 1 p = 0. Hece we get f (x = F(x for all x X\{0} by (15. Sice lim k k p = 0 ad the iequality f (0 F(0 1 D ( f F(kx, kx + ( 1 (F f ( kx + i=2 C i ( 1 i (F f ((i 1kx!(F f (kx [ ] p + 2 1!(2 2 p (1 + C i+1 i p +! k p ε x p holds for ay fixed x X\{0} with x > r ad all atural umbers k, we get f (0 = F(0. Theorem 5. Let S be a subset of a real vector space V ad Y a real Baach space. Suppose that for each x V there exists a real umber r x > 0 such that rx S for all r r x. Let ϕ : V 2 [0, be a fuctio such that i=1 2 k ϕ(2 k x, 2 k y < (16 for all x, y S. Suppose that a mappig f : V Y satisfies the iequality (10 for all x, y S, where ix + y S for all i = 0, 1,...,. The there exists a uique moomial mappig of degree F : V Y such that 2 f (x F(x k! Φ 2 k+1 (17 for all x with x S, where Φ(x is defied as i Theorem 4. I particular, F is represeted by F(x = lim m 2m f (2 m x

8 Mathematics 2017, 5, 53 8 of 11 for all x V. Proof. Let x V ad m be a iteger such that 2 m ( + 1 r x. It follows from (7 i Lemma 2 ad (10 that! f 2 m 2 f = 2 m+1 =Φ C j D f ( x C j D f C j ϕ 2 m+1 (j + 1x, 2m+1 2 m+1 (j + 1x, 2m+1 2 m+1 (j + 1x, 2m+1 2 m+1 for all x V. From the above iequality, we get the iequality D f 2 m, x 2 m+1 ( x + D f 2 m, x 2 m+1 + ϕ 2 m, x 2 m+1 2m f ( 2 m 2 (m+m x m+m 1 2 f k 2 m+m! Φ k=m 2 k+1 (18 for all x V ad m N. So the sequece {2 m f 2 m }m N is a Cauchy sequece by the iequality (16. From the completeess of Y, we ca defie a mappig F : V Y by F(x = lim m 2m f 2 m for all x V. Moreover, by puttig m = 0 ad lettig m i (18, we get the iequality (17 for all x S with ( + 1x S. From the iequality (10, if m is a positive iteger such that ix+y 2 m S for all i = 0, 1,...,, the we get ( x 2 m D f 2 m, y ( x 2 m 2 m ϕ 2 m, y 2 m, for all x, y V. Sice the right-had side i this iequality teds to zero as m, we obtai that F is a moomial mappig of degree. To prove the uiqueess of F assume that F is aother moomial mappig of degree satisfyig the iequality (17 for all x S with ( + 1x S. So the equality F (x = 2 m F 2 m holds for all x V by (8 i Lemma 2. Thus, we ca ifer that F (x 2 m ( f 2 m = 2 m F x 2 m 2 m f k=m ( 2 (m+k Φ 2 k Φ 2 k+1 x 2 m+k+1 2 m ( for all positive itegers m, where (+1x 2 m r x. Sice k=m 2k Φ x 0 as m, we kow that 2 k+1 F (x = lim m 2 m f ( x 2 m for all x V. This completes the proof of the theorem. We ca give a geeralizatio of Theorem 3 i [6] as the followig corollary.

9 Mathematics 2017, 5, 53 9 of 11 Corollary 2. Let p ad r be real umbers with p > ad r > 0, ad X a ormed space. Let f : X Y be a mappig satisfyig the iequality (13 for all x, y with x, y < r. The there exists a uique moomial mappig of degree F : X Y satisfyig for all x X with x < f (x F(x ε(2p C j (j + 1 p!(2 2 p x p r +1. The followig example shows that the assumptio p = caot be omitted i Corollarys 1 ad 2. This example is a extesio of the example of Gajda [21] for the moomial fuctioal iequality (13 (see also [22]. Example 1. Let ψ : R R be defied by ψ(x = { x, for x < 1, 1, for x 1. (19 Cosider that the fuctio f : R R is defied by f (x = for all x R. The f satisfies the fuctioal iequality ( + 1 m ψ(( + 1 m x (20 m=0 C i ( 1 i f (ix + y! f (x 4 ( + 1!( + 12 ( x + y. (21 for all x, y R, but there do ot exist a moomial mappig of degree F : R R ad a costat d > 0 such that f (x F(x d x for all x R. Proof. It is clear that f is bouded by 2 o R. If x + y x + y 1 (+1, the = 0, the f satisfies (21. Ad if D f (x, y 2 2 ( + 1! 4 ( + 1!( + 1 ( x + y, which meas that f satisfies (21. Now suppose that 0 < x + y < 1 (+1. The there exists a oegative iteger k such that 1 ( + 1 (k+2 x + y < 1. (22 ( + 1 (k+1 Hece ( + 1 k x < 1 (+1, ( + 1 k y < 1 (+1, ( + 1 m (x + iy < 1, ad ( + 1 m y < 1 for all m = 0, 1,..., k 1. Hece, for m = 0, 1,..., k 1, C i ( 1 i ψ(( + 1 m (ix + y!ψ(( + 1 m x = 0. (23

10 Mathematics 2017, 5, of 11 From the defiitio of f, the iequality (22, ad the iequality (23, we obtai that D f (x, y = ( + 1 m( C i ( 1 i ψ(( + 1 m (ix + y!ψ(( + 1 m x m=0 + 1 m=0( m C i ( 1 i ψ(( + 1 m (ix + y!ψ(( + 1 m x + 1 m=k( m C i ( 1 i ψ(( + 1 m (ix + y!ψ(( + 1 m x m=k ( + 1 m 2 ( + 1! 4 ( + 1 k ( + 1! 4 ( ( + 1!( x + y. (24 Therefore, f satisfies (21 for all x, y R. Now, we claim that the fuctioal Equatio (1 is ot stable for p = i Corollarys 1 ad 2. Suppose o the cotrary that there exists a moomial mappig of degree F : R R ad costat d > 0 such that f (x F(x d x for all x R. Notice that F(x = x F(1 for all ratioal umbers x. So we obtai that f (x (d + F(1 x (25 for all x Q. Let k N with k + 1 > d + F(1. If x is a ratioal umber i (0, ( + 1 k, the ( + 1 m x (0, 1 for all m = 0, 1,..., k, ad for this x we get f (x = m=0 ( + 1 m ψ(( + 1 m x =(k + 1x > (d + F(1 x k ( + 1 m (( + 1 m x m=0 which cotradicts ( Coclusios The advatage of this paper is that we do ot eed to prove the stability of additive quadratic, cubic, ad quartic fuctioal equatios separately. Istead we ca apply our mai theorem to prove the stability of those fuctioal equatios simultaeously. Ackowledgmets: The author would like to thak referees for their valuable suggestios ad commets. Coflicts of Iterest: The author declares o coflict of iterest. Refereces 1. Mazur, S.; Orlicz, W. Grudlegede eigeschafte der polyomische operatioe. Erste mitteilug. Stud. Math. 1934, 5, Mazur, S.; Orlicz, W. Grudlegede eigeschafte der polyomische operatioe. Zweite mitteilug. Stud. Math. 1934, 5, Djoković, D.Z. A represetatio theorem for (X 1 1(X 2 1 (X 1 ad its applicatios. A. Polo. Math , Cădariu, L.; Radu, V. The fixed poits method for the stability of some fuctioal equatios. Caepathia J. Math. 2007, 23, Choi, C.-K. Stability of a Moomial Fuctioal Equatio o Restricted Domais of Lebesgue Measure Zero. Results Math. 2017, 1 11, doi: /s Lee, Y.-H. O the stability of the moomial fuctioal equatio. Bull. Korea Math. Soc. 2008, 45,

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