General Mathematics Vol. 17, No. 1 (2009), 23–32 Order of approximation of functions of two variables by new type gamma operators1 Aydın İzgi Abstract The theorems on weighetd approximation and order of approximation of continuous functions of two variables by new type Gamma operators on all positive square region are established. 2000 Mathematics Subject Classification: 41A25,41A36 Key words and phrases: Order of approximation,weighted approximation,new type Gamma operators. 1 Introduction Lupaş and Müller[11] introduced the sequence of linear positive operators {Gn }, Gn : C(0, ∞) → C(0, ∞) defined by Gn (f ; x) = Z∞ n gn (x, u)f ( )du u 0 Gn is called Gamma operator, where gn (x, u) = 1 xn+1 −xu n e u , x > 0. n! Received 15 February, 2008 Accepted for publication (in revised form) 21 March, 2008 23 24 Aydın İzgi Mazhar [12] used same gn (x, u) of Gamma operator and introduced the following sequence of linear positive operators: Z∞ Fn (f ; x) = gn (x, u)du 0 Z∞ gn−1 (u, t)f (t)dt 0 (2n)!xn+1 = n!(n − 1)! Z∞ tn−1 f (t)dt, n > 1, x > 0 (x + t)2n+1 0 for any f which the last integral is convergent. Now we will modify the operators Fn (f ; x) as the following operators An (f ; x) (see [9]) which confirm An (t2 ; x) = x2 . Many linear operators Ln (f ; x) confirm, Ln (1; x) = 1, Ln (t; x) = x but don’t confirm Ln (t2 ; x) = x2 (see [2],[10]). An (f ; x) = Z∞ gn+2 (x, u)du gn (u, t)f (t)dt 0 0 = Z∞ (2n + 3)!xn+3 n!(n + 2)! Z∞ tn f (t)dt, x > 0 (x + t)2n+4 0 If we choose Kn (x, t) = (2n + 3)! xn+3 tn , x, t ∈ (0, ∞), n!(n + 2)! (x + t)2n+4 we can show An (f ; x) as the following form: (1) An (f ; x) = Z∞ Kn (x, t)f (t)dt. 0 In this study we will investigate approximation and order of approximation of the following operators which defined for two variables functions. (2) An,m (f ; x, y) = Z∞ 0 Kn,m (x, t; y, u)f (t, u)dtdu Order of approximation... 25 where Kn,m (x, t; y, u) = Kn (x, t) × Km (y, u). It can be easily to see that: An,m (f ; x, y) = An (f1 ; x) + Am (f2 ; y) ; if f (t, u) = f1 (t) + f2 (u) An,m (f ; x, y) = An (f1 ; x) × Am (f2 ; y) ; if f (t, u) = f1 (t) × f2 (u) Thus for any p, q ∈ N, p ≤ n and q ≤ m, we can see the following equalities (see [9]): (3) (4) (5) (6) An,m (1; x, y) = 1 An,m (t + u; x, y) = x + y − y x − n+2 m+2 An,m (t2 + u2 ; x, y) = x2 + y 2 An,m (t3 + u3 ; x) = x3 + y 3 + (7) An,m (t4 +u4 ; x) = x4 +y 4 + 3 3 3 x + y3 n m 4(2n + 3) 4 4(2m + 3) 4 x + y , n > 1, m > 1. n(n − 1) m(m − 1) (8) An,m ({t + u} − {x + y}); x, y) = − x y − n+2 m+2 (9) An,m ({(t − x)2 + (u − y)2 }; x, y) = 2y 2 2x2 + n+2 m+2 (10) An,m ({(t − x)4 + (u − y)4 }; x, y) = 12(n + 4) x4 (n + 2)n(n − 1) 12(m + 4)y 4 + , n > 1, m > 1 (m + 2)m(m − 1) 26 Aydın İzgi Let C(R2 ) be the set of all real-valued functions of two variables continuous on R2 := {(x, y) : x ≥ 0, y ≥ 0}, σ(x, y) = 1 + x2 + y 2 , −∞ < x, y < ∞ and Bσ be sets of all functions f defined on R2 satisfying the condition (11) | f (x, y)| ≤ Mf σ(x, y) where Mf is a constant depending only on f and the norm is defined by kf kσ = sup (x,y)∈R2 | f (x, y)| . σ(x, y) Cσ denotes the subspaces of all continuous functions which belonging to Bσ and Cσk denotes the subspaces of all functions belonging to Cσ with f (x, y) = k < ∞, x,y→∞ σ(x, y) lim where k is a constant depending only on f. The approximation theorems for two variables are proved by Volkov[13]. He proved the theorem: Theorem A ([13]). If {Tn } is a sequence of linear positive operators satisfying the conditions lim kTn (1; x1 , x2 ) − 1kC(X) = 0 . n→∞ lim kTn (ti ; x1 , x2 ) − xi kC(X) = 0, i = 1, 2 . lim Tn (t21 + t22 ; x1 , x2 ) − (x21 + x22 )C(X) = 0 . n→∞ n→∞ then for any function f ∈ C(X), which is bounded in R2 lim kTn (f ; x1 , x2 ) − f (x1 , x2 )kC(X) = 0 . n→∞ where X is a compact set. Gadzhiev proved the following theorem for one variable functions. Theorem B ([3, 4]). {Tn } be the sequence of linear positive operators which mapping from Cρ into Bρ satisfying the conditions lim kTn (tv ; x) − xv kρ = 0, v = 0, 1, 2. n→∞ Order of approximation... 27 Then , for any f ∈ Cρk , lim kTn f − f kρ = 0. n→∞ and there exist a function f ∈ Cρ \ Cρk such that lim kTn f − f kρ ≥ 1. n→∞ Analogously as in Theorem B, the theorems on weighted approximation for functions of several variables are proved by Gadzhiev [5]. Applying Theorem B to the operators Tn (f ; x) = ( Vn (f ; x), if x ∈ [0, an ] f (x), if x > an one then also has the following theorem. Theorem C ([6]). Let (an ) be a sequence with lim an = ∞ and {Vn } be a n→∞ sequence of linear positive operators taking Cρ [0, an ] into Bρ [0, an ]. If for v = 0, 1, 2 lim kVn (tv ; x) − xv kρ,[0,an ] = 0, n→∞ then for any f ∈ Cρk [0, an ] lim kVn f − f kρ,[0,an ] = 0, n→∞ where Bρ [0, an ], Cρ [0, an ] and Cρk [0, an ] denote the same as Bρ , Cρ and Cρk respectively, but the functions taken on [0, an ] instead of R and the norm is taken as |f (x)| . kf kρ,[0,an ] = sup x∈[0,an ] ρ(x) 28 Aydın İzgi 2 Approximation of An,m Let (bn ) is be a sequence has positive terms, increasing and has the following conditions, (12) lim bn = ∞ n→∞ and b2n =0 n→∞ n lim We will denote the rectangular region (0, bn ] × (0, bm ] by Dn,m and let Bσ (Dn,m ) be sets of all functions f defined on Dn,m satisfying the condition (11) By the using (3) and (5), we have An,m (σ(t, u); x, u) = σ(x, y). Therefore, kAn,m (f ; x)kσ(Dn,m ) is uniformly bounded on Dn,m . Hence {An,m } is a sequence of linear positive operators taking Cσ (Dn,m ) into Bσ (Dn,m ) . Theorem 1. Let f ∈ Cσk , then lim kAn,m (f ; x, y) − f (x, y)kσ(Dn,m ) = 0. n,m→∞ Proof. lim kAn,m (1; x, y) − 1kσ(Dn,m ) = 0. n,m→∞ x n + 2 = 0. lim kAn,m (t; x, y) − xkσ = lim sup n,m→∞ n,m→∞ (x,y)∈(D σ(x, y) n,m ) y m + 2 = 0. lim kAn,m (u; x, y) − ykσ = lim sup n,m→∞ n,m→∞ (x,y)∈(D σ(x, y) n,m ) lim An,m (t2 + u2 ; x, y) − (x2 + y 2 )σ(Dn,m ) = 0 n,m→∞ Similar to Theorem B we obtain the desired results. Order of approximation... 3 29 The Order of Approximation of An,m We now want to find the degree of approximation of functions f ∈ Cσk by the operators An,m on Dn,m . It is well- known that the usual first modulus of continuity o n p ̟(f ; δ) = sup |f (t, u) − f (x, y)| : (t − x)2 + (u − y)2 ≤ δ; t, u, x, y ∈ [a, b] don’t tend to zero, as δ → 0, on any infinite interval and any infinite area, respectively. In [8] was defined the weighted modulus of continuity for f ∈ Cσk as the following (see also [1]): Λ(f ; δ, η) = sup |f (x + t, y + u) − f (x, y)| : x, y ∈ R2 , |t| ≤ δ, |u| ≤ η σ(x, y)σ(t, u) Λ(f ; δ, η) is having the following properties: lim Λ(f ; δ, η) = 0. δ,η→0 Λ(f ; λ1 δ, λ2 η) ≤ 4(1 + λ1 )(1 + λ2 )Λ(f ; δ, η), for λ1 > 0, λ2 > 0 and (13) |f (t, u) − f (x, y)| ≤ 8(1 + x2 + y 2 )Λ(f ; δn , δm )(1 + |t − x| |u − y| )(1 + ) δn δm ×(1 + (t − x)2 )(1 + (u − y)2 ) Theorem 2. For every f ∈ Cσk the inequality sup (x,y)∈Dn,m |An,m (f ; x, y) − f (x, y)| ≤ 4232Λ(f ; (1 + x2 + y 2 ) is true for all n, m sufficiently large. s 2b2n n+2 , s 2b2m ) m+2 30 Aydın İzgi Proof. If we use (13) and (3) we have |An,m (f ; x, y) − f (x, y)| ≤ 8(1 + x2 + y 2 )Λ(f ; δn , δm ) |t − x| |u − y| ×An,m ((1 + )(1 + )(1 + (t − x)2 )(1 + (u − y)2 ); x, y) δn δm ≤ 8(1 + x2 + y 2 )Λ(f ; δn , δm ) |u − y| |t − x| )(1 + (t − x)2 ); x) × Am ((1 + )(1 + (u − y)2 ); y) ×An ((1 + δn δm a2 + b 2 We know that a.b ≤ is hold for all positive real numbers a and 2 b. Thus, apply equalities (3), (9), (10) and Cauchy-Schwarz inequalities, we will get |An,m (f ; x, y) − f (x, y)| ≤ (1 + x2 + y 2 ) # " r 1 2x2 1 1 2x2 12(n + 4) Λ(f ; δn , δm ) × 1 + (1 + + + x4 ) 2δn n + 2 δn n + 2 2δn (n + 2)n(n − 1) " # r 1 2y 2 12(m + 4) 1 2y 2 1 × 1 + (1 + ) + + y4 2δm m + 2 δm m + 2 2δm (m + 2)m(m − 1) r r 2b2n 2b2m and δm = and consider δn ≤ 1, Choosing δn = n+2 m+2 b2 δm ≤ 1 for all n, m sufficiently large since lim n = 0 , we obtain the n→∞ n desired result. References [1] N. I. 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Harran Universtiy Science and Arts Faculty Department of Mathematics Şanlıurfa-Turkey E-mail: aydinizgi@yahoo.com /a izgi@harran.edu.tr