In this paper, the authors characterize the solutions of the functional equation F(y)−F(x)=(y−x)[f(αx+βy)+f(βx+αy)] for real functions F,f  and real coefficients α,β . Some of the results obtained by the authors are also valid on integral domains

Koclęga-Kulpa, Barbara

Szostok, Tomasz

Annales Mathematicae Silesianae

English

Repozytorium Uniwersytetu Śląskiego RE-BUŚ

             Title: On a functional equation connected to Gauss quadrature rule   Author: Barbara Koclęga-Kulpa, Tomasz Szostok  Citation style: Koclęga-Kulpa Barbara, Szostok Tomasz. (2008). On a functional equation connected to Gauss quadrature rule. "Annales Mathematicae Silesianae" ([Nr] 22 (2008), s. 27-40). Annales Mathematicae Silesianae 22 (2008), 27–40Prace Naukowe Uniwersytetu Śląskiego nr 2703, KatowiceON A FUNCTIONAL EQUATION CONNECTED TO GAUSSQUADRATURE RULEBarbara Koclęga-Kulpa, Tomasz SzostokAbstract. We consider the functional equationF (y)− F (x) = (y − x)[f(αx+ βy) + f(βx+ αy)]stemming from Gauss quadrature rule. In previous results equations of thistype with rational only coefficients α and β were considered. In this paper weallow these numbers to be irrational. We find all solutions of this equationfor functions acting on R. However, some results are valid also on integraldomains.1. IntroductionFunctional equations connected to well known quadrature rules were stud-ied by many authors. In the monograph [7] the equation stemming fromSimpson quadrature ruleF (y)− F (x) = (y − x)[16f(x) +23f(x+ y2)+16f(y)]was considered for functions acting on R. This equation, in a bit more gen-eral form, for functions transforming an integral domain into itself has beensolved in [3].Received: 13.10.2008. Revised: 12.01.2009.(2000) Mathematics Subject Classification: 39B52.Key words and phrases: functional equations on integral domains, quadrature rules.28 Barbara Koclęga-Kulpa, Tomasz SzostokOn the other hand, M. Sablik [9] during the 7th Katowice-Debrecen Win-ter Seminar on Functional Equations and Inequalities presented the generalsolution of the equation(1) g(x)− f(y) = (x− y)[h(x) + k(sx+ ty) + k(tx+ sy) + h(y)]in the case s, t ∈ Q without any regularity assumptions concerning the func-tions considered. Some other equations of this type were solved in [4]. How-ever, these results are valid only for equations with rational coefficients on theright-hand side.In the current paper we are going to consider an equation connected tothe simplest quadrature rule with irrational coefficient - the Gauss quadraturerule with two nodes. Thus we shall find the solutions of(2) F (y)− F (x) = (y − x)[f(αx+ βy) + f(βx+ αy)].In our method we shall need a lemma, which was established by M. Sablik([8, Lemma 2.3]) and improved by I. Pawlikowska (cf. [6]).By a polynomial function of order n in this paper we mean a solution of thefunctional equation ∆n+1h f(x) = 0, where ∆nh stands for the n–th iterate ofthe difference operator ∆hf(x) = f(x+h)− f(x). Observe that a continuouspolynomial function of order n is a polynomial of degree at most n (see [5,Theorem 4, p. 398]).Let us now quote Sablik’s result. First we need some notations. LetG,H be Abelian groups and SA0(G,H) := H, SA1(G,H) := Hom(G,H)(i.e. the group of all homomorphisms from G into H), and for i ∈ N, i ≥ 2,let SAi(G,H) be the group of all i–additive and symmetric mappings fromGi into H. Furthermore, let P := {(α, β) ∈ Hom(G,G)2 : α(G) ⊂ β(G)}.Finally, for x ∈ G let xi = (x, . . . , x)︸ ︷︷ ︸i, i ∈ N.Lemma 1. Fix N ∈ N ∪ {0} and let I0, . . . , IN be finite subsets of P.Suppose that H is uniquely divisible by N ! and that the functions ϕi : G →SAi(G,H) and ψi,(α,β) : G→ SAi(G,H), (α, β) ∈ Ii, i = 0, . . . , N , satisfy(3) ϕN (x)(yN ) +N−1∑i=0ϕi(x)(yi) =N∑i=0∑(α,β)∈Iiψi,(α,β)(α(x) + β(y))(yi)On a functional equation connected to Gauss quadrature rule 29for every x, y ∈ G. Then ϕN is a polynomial function of order at most k− 1,wherek =N∑i=0card( N⋃s=iIs).2. ResultsFirst we need some auxiliary lemma.Lemma 2. Let P be an integral domain with a unit element 1 such that2 = 1 + 1 6= 0 and 3 = 1 + 1 + 1 6= 0. Let the functions f, F : P → P satisfythe equation(4) F (x)− F (y) = (x− y)[b1f(α1x+ β1y) + · · ·+ bnf(αnx+ βny)]for all x, y ∈ P and some b1, . . . , bn, α1, . . . , αn, β1, . . . , βn ∈ P. Further, let fbe a function of the form(5) f(x) = a3(x) + a2(x) + a1(x) + a0, x ∈ P,where ai, i ∈ {1, 2, 3} is a diagonalization of some i−additivie and symmetricfunction and a0 ∈ P is a constant.Then all functions ai, i ∈ {1, 2, 3} also satisfy (4) with some Fi.Proof. We may assume that F (0) = 0, so substituting y = 0 in (4) weobtainF (x) = x[b1f(α1x) + · · ·+ bnf(αnx)], x ∈ P.Using this formula in (4) we get(6) x[b1f(α1x) + · · ·+ bnf(αnx)]− y[b1f(α1y) + · · ·+ bnf(αny)]= (x− y)[b1f(α1x+ β1y) + · · ·+ bnf(αnx+ βny)], x, y ∈ P.It is easy to see that if f, g are solutions to this equation, then also f + gsatisfies this equation. Moreover, if f satisfies (6) and α ∈ P is given, thenalso αf is a solution. On the other hand, if αf satisfies (6) and α 6= 0, then f30 Barbara Koclęga-Kulpa, Tomasz Szostokis a solution of this equation. Now take a 6= 0 and substitute ax, ay in placeof x and y, We obtain (after canceling by a) thatx[b1f(α1ax) + · · ·+ bnf(αnax)]− y[b1f(α1ay) + · · ·+ bnf(αnay)]= (x− y)[b1f(α1ax+ β1ay) + · · ·+ bnf(αnax+ βnay)], x, y ∈ P,which means that the function x 7→ f(ax) also satisfies our equation (if fdoes).Now consider f of the form (5), and putg1(x) := 8f(x)− f(2x), x ∈ P.Then g1 is a solution of (6) and on the other hand(7) g1(x) = 4a2(x) + 6a1(x) + 7a0.Define g2 by the formulag2(x) := 4g1(x)− g1(2x) = 12a1(x) + 21a0.Similarly as before g2 satisfies (6), constant function 21a0 is clearly a solutionof (6), which means that also 12a1 satisfies this equation thus also a1 satisfies(6) since 2, 3 6= 0. Further, in view of (7) we easily obtain that also a2 is asolution of (6). Finally, from (5) it follows that a3 satisfies our equation. Remark 1. Let P be an integral domain such that 2, 3 6= 0. If functionsAi, Bi : P → P, i = 1, 2, 3, satisfy(8) A3(x) +A2(x) +A1(x) = B3(x) +B2(x) +B1(x), x ∈ P,and Ai, Bi are i-homogeneous, thenAi = Bi, i = 1, 2, 3.Indeed, if we take −x in place of x, in (8), then we obtain(9) −A3(x) +A2(x)−A1(x) = −B3(x) +B2(x)−B1(x), x ∈ P.This, together with (8) means that A2 = B2 and, consequently(10) A3(x) +A1(x) = B3(x) +B1(x), x ∈ P.On a functional equation connected to Gauss quadrature rule 31Moreover, taking here 2x instead of x, we get(11) 8A3(x) + 2A1(x) = 8B3(x) + 2B1(x), x ∈ P.Multiplying (10) by 2 and substracting it from (11) we arrive at6A3(x) = 6B3(x), x ∈ P,which means that A3 = B3 and obviously also A1 = B1.Now we shall prove some lemmas concerning solutions of a simplified ver-sion of equation (2). However, at this moment we are concerned only withsolutions, which are diagonalizations of symmetric and i−additive functions.Lemma 3. Let P be a integral domain with a unit element 1 such that2, 3 6= 0. If functions G, f : P → P satisfy the equation(12) G(y)−G(x) = (y − x)[f(x+ γy) + f(γx+ y)], x, y ∈ P,and f(x) := C(x, x, x) for some 3−additive and symmetric function C : P →P, thenf((γ + 1)x) = 2cx3for some c ∈ P and all x ∈ P.Proof. Let f be given by the formula f(x) = C(x, x, x), x ∈ P , where Cis 3-additive and symmetric.We may assume that G(0) = 0, so substituting x = 0 in (12) we get(13) G(y) = y[f(γy) + f(y)], y ∈ P.Using this formula in (12), we obtain for all x, y ∈ P(14) y[f(γy) +f(y)]− x[f(γx) +f(x)] = (y − x)[f(x+ γy) + f(γx+ y)].On the other hand, using here the form of f after some calculation wearrive at(15) x[f(γy) + f(y)]− y[f(γx) + f(x)] = 3(y − x)[C(x, x, γy)+ C(x, γy, γy) + C(y, y, γx) + C(y, γx, γx)]for all x, y ∈ P.32 Barbara Koclęga-Kulpa, Tomasz SzostokNow we define a new function g(x) := f(x) + f(γx), x ∈ P. Then in viewof (15) we havexg(y)− yg(x) = 3(y − x)[C(x, x, γy) + C(x, γy, γy)(16)+ C(y, y, γx) + C(y, γx, γx)].Further, taking here x = 1 we obtain the following formula(17) g(y)−yg(1) = 3(y−1)[C(1, 1, γy)+C(1, γy, γy)+C(y, y, γ)+C(y, γ, γ)].Using Remark 1, we may compare here the terms of order one and weobtain that(18) yg(1) = 3C(y, γ, γ) + 3C(1, 1, γy), y ∈ P,further comparing expressions of order two in (17) we get(19) 3y[C(y, γ, γ) + C(1, 1, γy)] = 3[C(1, γy, γy) + C(y, y, γ)], y ∈ P,thus, using here (18), we may write(20) y2g(1) = 3[C(1, γy, γy) + C(y, y, γ)], y ∈ P.Finally, comparing terms of order three in (17), we obtaing(y) = 3y[C(1, γy, γy) + C(y, y, γ)], y ∈ P,which, in view of (20), gives us(21) g(y) = y3g(1), y ∈ P,so we may write thatf(x) + f(γx) = cx3, x ∈ P,(where c = g(1)) and, more precisely,(22) C(x, x, x) + C(γx, γx, γx) = cx3, x ∈ P.Now, putting 2x in place of y in (16), we obtain6cx4 = 3x[2C(x, x, γx) + 4C(x, γx, γx) + 4C(x, x, γx) + 2C(x, γx, γx)],On a functional equation connected to Gauss quadrature rule 33thus(23) 3C(x, x, γx) + 3C(x, γx, γx) = cx3, x ∈ P.Adding equations (22) and (23), we may writeC(x, x, x) + 3C(x, x, γx) + 3C(x, γx, γx) + C(γx, γx, γx) = 2cx3, x ∈ P,i.e.f((1 + γ)x) = 2cx3, x ∈ P. Lemma 4. Let P be an integral domain with a unit element 1 such that2 6= 0. If functions G, f : P → P satisfy the equation (12) and f(x) := B(x, x)for some 2−additive and symmetric function B : P 2 → P, then2f((γ + 1)x) = 3bx2for some b ∈ P and all x ∈ P.Proof. Let f(x) = B(x, x), x ∈ P, where B is biadditive and symmetric.We may assume, without loss of generality, that G(0) = 0 and putting x = 0in (12) we obtain(24) G(y) = y [f(y) + f(γy)] , y ∈ P.Consequently, equation (12) takes formy [f(y) + f(γy)]−x [f(x) + f(γx)] = (y−x)[f(x+γy)+f(γx+y)], x, y ∈ P.Further, using here the form of f, we may write for all x, y ∈ Py [f(y) + f(γy)]− x [f(x) + f(γx)] = (y − x)[f(x) + 2B(x, γy) + f(γy)+ f(γx) + 2B(γx, y) + f(y)].Now we define a new function g(x) := f(x) + f(γx), x ∈ P, then we obtainyg(y)− xg(x) = (y − x)[g(x) + g(y) + 2B(x, γy) + 2B(γx, y)], x, y ∈ P,and, after some simple calculations,(25) xg(y)− yg(x) = 2(y − x)[B(x, γy) +B(γx, y)], x, y ∈ P.34 Barbara Koclęga-Kulpa, Tomasz SzostokPutting here x = 1 we getg(y)− yg(1) = 2(y − 1)[B(1, γy) +B(γ, y)], x, y ∈ P.Similarly as in the proof of the previous lemma we compare now the termsof the same order. Thus we getg(y) = 2y[B(1, γy) +B(γ, y)], y ∈ P,andyg(1) = 2[B(1, γy) +B(γ, y)], y ∈ P,which means thatg(y) = by2, y ∈ P,where b := g(1). Further, from the obtained form of g we havef(x) + f(γx) = bx2, x ∈ P,i.e.(26) 2B(x, x) + 2B(γx, γx) = 2bx2, x ∈ P.On the other hand, taking in (25) 2x in place of y, and using form of g,we get4B(x, γx) = bx2, x ∈ P.Adding this equality to (26), we may write2B(x, x) + 2B(γx, γx) + 4B(γx, x) = 3bx2,thus2f((1 + γ)x) = 3bx2, x ∈ P. Lemma 5. Let P be an integral domain and let f,G : P → P be solutionsof (12) such that f is an additive function. Thenf((γ + 1)x) = axfor some a ∈ P and all x ∈ P.On a functional equation connected to Gauss quadrature rule 35Proof. Similarly as before we easily obtain that f satisfies (14). Since fis additive, we gety[f(γy) + f(y)]− x[f(γx) + f(x)] = (y − x)[f(x) + f(γy) + f(γx) + f(y)]for all x, y ∈ P and, consequently,x[f(γy) + f(y)] = y[f(γx) + f(x)], x, y ∈ P.Taking here y = 1 we arrive atx[f(γ) + f(1)] = f(γx) + f(x), x ∈ P,thus putting a := f(γ) + f(1) we havef(γx) + f(x) = ax.Using here the additivity of f we obtain f((γ + 1)x) = ax for all x ∈ P. Now we are in position to obtain the result concerning the form of solutionsof equation (2) for functions acting on an integral domain.Theorem 1. Let P be an integral domain divisible by 6. If functions f, F :P → P satisfy (12) with some γ ∈ P, such that P is divisible by γ, γ + 1 andγ − 1, thenf(x) = cx3 + bx2 + ax+ d, x ∈ P,for some a, b, c, d ∈ P.Proof. Substituting in (12) x−yγ+1 in place of x andx+γyγ+1 in place of y, weobtainG(x+ γyγ + 1)−G(x− yγ + 1)= y[f((x+ (γ − 1)y) + f(x)], x, y ∈ P,and, consequently,yf(x) = G(x+ γyγ + 1)−G(x− yγ + 1)− yf((x+ (γ − 1)y), x, y ∈ P.36 Barbara Koclęga-Kulpa, Tomasz SzostokUsing here Lemma 1 with N = 1, I0 :={(1γ+1 id,γγ+1 id),(1γ+1 id,−1γ+1 id)}and I1 = {(id, (γ− 1)id)} we obtain that f is a polynomial function of degreeat most 3. Now, since P is divisible by 6, we havef(x) = C(x, x, x) +B(x, x) +A(x) + d, x ∈ P,where C : P 3 → P,B : P 2 → P , and A : P → P are symmetric and 3, 2, 1–additive, respectively (see for example [5]).Now using Lemma 2 we obtain that functions f3(x) := C(x, x, x), f2(x) :=B(x, x), and f1(x) := A(x) also satisfy (12), which in view of Lemmas 3, 4and 5 means that f3((γ+1)x) = c0x3, f2((γ+1)x) = b0x2, and f1((γ+1)x) =a0x for some a0, b0, c0 ∈ P. However, P is divisible by γ+1 and, consequently,taking c := c0(γ+1)3 , b :=b0(γ+1)2 and a :=a0γ+1 , we arrive atf(x) = cx3 + bx2 + ax+ d, x ∈ P. Now we shall deal with functions defined and taking values in R. In thiscase we shall obtain a general solution of equation (2).Theorem 2. Let functions f, F : R → R satisfy (2) with some constantsα, β ∈ R.(i) If α = β 6= 0, then f(x) = ax+ d, x ∈ R.(ii) If α = β = 0, then f may be any function.(iii) If α = 0, β 6= 0 or α 6= 0, β = 0, then f(x) = ax+ d, x ∈ R.(iv) If α = −β 6= 0, then f(x) = h(x)+ A(x)x for all x 6= 0 and some functionsh,A : R→ R which are odd and additive, respectively.(v) If α 6= −β and α2 + β2 6= 4αβ, then f(x) = ax+ d, x ∈ R.(vi) If α 6= −β and α2 + β2 = 4αβ, then f(x) = cx3 + bx2 + ax+ d, x ∈ R.In each case, function F is expressed by formulaF (x) = x[f(αx) + f(βx)] + F (0), x ∈ R.Conversely, in each case functions given by the above formulas satisfyequation (2).Proof. Let us first consider the case α = β 6= 0. Then equation (2) takesthe formF (y)− F (x) = 2(y − x)f(α(x+ y)).On a functional equation connected to Gauss quadrature rule 37Put f˜(t) := 2f(αt), t ∈ R. Then we get the equation solved by J. Aczél in [1]and consequently we obtain f˜(x) = a˜x+ d˜, which means that f(x) = ax+ dwhere a := a˜2α and d :=d˜2 .In the case (iii) we get (after a simple substitution)F (y)− F (x) = (y − x)[h(x) + h(y)]and we obtain our result from the paper of Sh. Haruki [2].Now we consider the case α = −β 6= 0, this means that equation (2) takesthe formF (y)− F (x) = (y − x)[f(α(x− y)) + f(α(y − x))].Define a new function f1 : R→ R byf1(x) := f(αx).We obtain the equation(27) F (y)− F (x) = (y − x)[f1(x− y) + f1(y − x)], x, y ∈ R.Take in (27) x = 0 to obtain(28) F (y) = y[f1(−y) + f1(y)] + F (0), y ∈ R,which used in (27) gives usF (y)− F (x) = F (y − x)− F (0), x, y ∈ R.Substituting here x+ y in place of y we getF (x+ y)− F (x) = F (y)− F (0), x, y ∈ R.This means that function A1 := F−F (0) is additive, thus from (28) we obtainx[f1(−x) + f1(x)] = A1(x)and furtherf1(−x) + f1(x) = A1(x)x, x 6= 0.38 Barbara Koclęga-Kulpa, Tomasz SzostokNow we put H(x) := f1(x)−f1(−x)2 , which gives usf1(x) =f1(x)− f1(−x)2+f1(x) + f1(−x)2= H(x) +A1(x)2xfor all x ∈ R \ {0}. And from the definition of f we getf(x) = f1(xα)= H(xα)+αA1(xα)2x.To finish this part of the proof it suffices to take h(x) := H(xα)and A(x) :=α2A1(xα).Finally we consider the case α 6= −β. Note that both numbers α, β aredifferent from zero. Now we put in (2) 1αx and1αy instead of x and y, respec-tively. ThusF(1αy)− F(1αx)=1α(y − x)[f(x+βαy)+ f(βαx+ y)].Multiplying this equation by α and taking G(x) := αF(1αx), γ := βα we obtainthat f satisfiesG(y)−G(x) = (y − x)[f(x+ γy) + f(γx+ y)].Thus we have got the equation (12) which in view of Theorem 1 means thatf(x) = cx3 + bx2 + ax+ d, x ∈ R.To finish this part of the proof we assume that c 6= 0 and we are going toshow thatα2 + β2 = 4αβ.Indeed, if function f(x) = cx3 + bx2 + ax+ d satisfies (2) then from Lemma2 we know that also f3(x) := cx3 satisfies (2). Thus we haveG3(y)−G3(x) = (y − x)[f3(αx+ βy) + f3(βx+ αy)], x, y ∈ Rfor some function G3. Using the form of f and equalityG3(y) = y[f3(βy) + f3(αy)], y ∈ R,On a functional equation connected to Gauss quadrature rule 39we get after some simple calculations(β3 + α3)(y4 − x4) = (y − x)[(α3 + β3)(x3 + y3) + 3αβ(α+ β)(xy2 + x2y)],i.e.(β3+α3)(y3+x3+xy2+x2y) = (α3+β3)(x3+ y3)+3αβ(α+β)(xy2+x2y),and further(β3 + α3) = 3αβ(α+ β),thus4αβ = α2 + β2.Similarly, if we assume that b 6= 0 we obtain the same equality for α and β.On the other hand, it is easy to show that in every of the above casesobtained functions satisfy equation (2). Remark 2. Let us consider the following quadrature rule with two nodes∫ yxf(t)dt ≈ 12(y − x)[f(αx+ (1− α)y) + f((1− α)x+ αy)],which leads to a functional equationF (y)− F (x) = (y − x)[f(αx+ (1− α)y) + f((1− α)x+ αy)].If we addititionally assume that this equation is satisfied by f(x) = x3thenα =3−√36.This means that from our result we obtain that the only two-nodes quadraturerule satisfied exactly by polynomials of degree 3 (or 2) is the well known Gaussquadrature rule. One should emphasize that we assume here no regularity offunctions f, F .40 Barbara Koclęga-Kulpa, Tomasz SzostokReferences[1] Aczél J., A mean value property of the derivative of quadratic polynomials—withoutmean values and derivatives, Math. Mag. 58 (1985), no. 1, 42–45.[2] Haruki Sh., A property of quadratic polynomials, Amer. Math. Monthly 86 (1979), no.7, 577–579.[3] Koclęga-Kulpa B., Szostok T., On some equations connected to Hadamard inequalities,Aequationes Math. 75 (2008), 119–129.[4] Koclęga-Kulpa B., Szostok T., Wąsowicz Sz., Some functional equations characterizingpolynomials, Tatra Mt. Math. Publ. (to appear).[5] M. Kuczma, An Introduction to the Theory of Functional Equations and Inequalities.Cauchy’s Equation and Jensen’s Inequality, Państwowe Wydawnictwo Naukowe (Pol-ish Scientific Publishers) and Uniwersytet Śląski, Warszawa–Kraków–Katowice, 1985.[6] Pawlikowska I., Solutions of two functional equations using a result of M. Sablik,Aequationes Math. 72 (2006), 177–190.[7] Riedel T., Sahoo P.K.,Mean value theorems and functional equations, World Scientific,Singapore–New Jersey–London–Hong Kong, 1998.[8] Sablik M., Taylor’s theorem and functional equations, Aequationes Math. 60 (2000),258–267.[9] Sablik M., On a problem of P.K. Sahoo – joint work with Arkadiusz Lisak, talk at the7th KDWS, Będlewo, Poland, January 31 – February 3, 2007.[10] Report of Meeting. The Fifth Katowice–Debrecen Winter Seminar on Functional Equa-tions and Inequalities, Ann. Math. Sil. 19 (2005), 65–78.Institute of MathematicsSilesian UniversityBankowa 1440-007 KatowicePolande-mail: koclega@math.us.edu.ple-mail: szostok@math.us.edu.pl

On a functional equation connected to Gauss quadrature rule

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On a functional equation connected to Gauss quadrature rule

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