PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 99(113) (2016), 265–279 DOI: 10.2298/PIM1613265G DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS Vijay Gupta and Neha Malik Abstract. Several modifications of the discrete operators are available in the literature. In the recent years, certain modifications of the well-known Baskakov and Szász–Mirakyan operators have been discussed based on certain parameters. We propose mixed summation-integral type operators and estimate the quantitative asymptotic formula and a global direct result for the special case. For general case, we establish moments and some direct convergence results in ordinary approximation, which includes pointwise approximation, asymptotic formula and a direct result in terms of modulus of continuity. 1. Introduction In order to generalize the Baskakov operators, Mihesan [15] proposed the following operators based on a non-negative constant a, independent of n as Bna (f ; x) = (1.1) ∞ X ban,k (x)f k=0 k n , where ban,k (x) =e ax − 1+x Pk i=0 k i (n)i ak−i xk , k! (1 + x)n+k and the rising factorial is given by (n)i = n(n + 1) · · · (n + i − 1), (n)0 = 1. Also, in order to generalize the Szász–Mirakyan operators, Jain [13] introduced the following operators Snβ (f, x) = ∞ X (β) Ln,k (x)f k=0 k , n x ∈ [0, ∞) 2010 Mathematics Subject Classification: 41A25; 41A30. Key words and phrases: Baskakov operators, Szász-Mirakyan operators, direct results, modulus of continuity. Communicated by Gradimir Milovanović. 265 266 GUPTA AND MALIK where 0 6 β < 1 and the basis function is defined as nx(nx + kβ)k−1 −(nx+kβ) e . k! P∞ P∞ (β) It was seen in [15] and [13] that k=0 ban,k (x) = k=0 Ln,k (x) = 1. The integral modification of the classical Bernstein polynomial was introduced by Durrmeyer [6]. Gupta et al. [10] proposed the hybrid Durrmeyer type operators by taking the gen(β) eral basis function Ln,k (x) under summation sign, the actual Durrmeyer operators for this basis function were recently considered by Gupta and Greubel [16]. In order to approximate Lebesgue integrable functions on the interval [0, ∞), for a non-negative parameter a and 0 6 β < 1, we propose the mixed hybrid Durrmeyer type operators as follows: (β) Ln,k (x) = Dna,β (f, x) = (1.2) (β) ∞ X hLn,k (t), f (t)i k=0 R∞ ban,k (x) (β) hLn,k (t), 1i where hf, gi = 0 f (t) g(t) dt. Some approximation properties for the particular case β = 0 were recently discussed by Agrawal et al. [1]. For the special case of a = β = 0, these operators reduce to the Baskakov–Szász operators introduced about twenty years ago by Gupta and Srivastava [12], which are defined as Dn (f, x) := Dn0,0 (f, x) = (0) ∞ X hLn,k (t), f (t)i k=0 ∞ X =n (0) hLn,k (t), 1i b0n,k (x) k=0 where b0n,k (x) n+k−1 xk = , k (1 + x)n+k Z ∞ 0 (0) b0n,k (x) (0) Ln,k (t)f (t)dt, Ln,k (t) = e−nt (nt)k . k! (β) We may point out here that hLn,k (t), 1i is dependent of k, while for β = 0 this term is just 1/n. Dragomir [3] established interesting approximation results of continuous linear functionals in real normed spaces. Very recently, Gupta and Agarwal [9] presented convergence estimates on several linear positive operators. We also mention for the readers some of the related work as [2, 7, 8, 11, 17, 18] etc. Here, we first estimate the quantitative asymptotic formula for the special case and a global direct result. For general case, we establish moments and some direct results in ordinary approximation, which includes pointwise approximation, asymptotic formula, error-estimation in terms of first and second order modulus of continuity. 2. Approximation for the case a = β = 0 Very recently Luo–Milovanovic–Agarwal [14] established some results on the extended beta and extended hypergeometric functions. We observe that in the special case, if we take a = β = 0, then the moments can be obtained in terms of DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 267 hypergeometric series as follows: Using the identities: k! = (1)k and Γ(k + m + 1) = Γ(m + 1)(m + 1)k , we have Z ∞ ∞ X Dn (tm , x) = n b0n,k (x) L0n,k (x)tm dt =n = k=0 ∞ X 0 (n)k xk k!(1 + x)n+k k=0 ∞ X 1 nm k=0 Z 0 ∞ e−nt (nt)k m t dt k! k (n)k x Γ(k + m + 1) k!(1 + x)n+k (1)k ∞ (1 + x)−n X (n)k (m + 1)k Γ(m + 1)xk = nm k!(1 + x)k (1)k k=0 m!(1 + x)−n x = F n, m + 1; 1; . 2 1 nm 1+x Applying the well known transformation 2 F1 (a, b; c; x) we obtain at once = (1 − x)−a 2 F1 a, c − b; c; Dn (tm , x) = x , x−1 m! 2 F1 (n, −m; 1; −x). nm Furthermore, we obtain Dn (t, x) = x + 1 , n Dn (t2 , x) = 2 + 4nx + n(n + 1)x2 n2 and 2 + 2nx + nx2 1 , Dn ((t − x)2 , x) = . n n2 In general, using the similar approach, one can show that: (2.2) Dn ((t − x)r , x) = O n−[(r+1)/2] , (2.1) Dn (t − x, x) = where [α] denotes the integral part of α. Let Bx2 [0, ∞) be the set of all functions f defined on R+ satisfying the condition |f (x)| 6 Mf (1+x2 ) with some constant Mf , depending only on f , but independent of x. Bx2 [0, ∞) is called weighted space and it is a Banach space endowed with the norm f (x) kf kx2 = sup . + 1 + x2 x∈R Let Cx2 [0, ∞) = C[0, ∞) ∩ Bx2 [0, ∞) and by Cxk2 [0, ∞), we denote subspace of f (x) all continuous functions f ∈ Bx2 [0, ∞) for which limx→∞ 1+x 2 is finite. We know that usual first modulus of continuity ω(δ) does not tend to zero, as δ → 0 on infinite interval. Thus, we use weighted modulus of continuity Ω(f, δ) 268 GUPTA AND MALIK defined on infinite interval R+ (see [17]). Let Ω(f, δ) = sup |h|<δ, x∈R+ |f (x + h) − f (x)| (1 + h2 )(1 + x2 ) for each f ∈ Cx2 [0, ∞). Now, some elementary properties of Ω(f, δ) are collected in the following lemma. Lemma 2.1. Let f ∈ Cxk2 [0, ∞). Then, i) Ω(f, δ) is a monotonically increasing function of δ, δ > 0. ii) For every f ∈ Cxk2 [0, ∞), limδ→0 Ω(f, δ) = 0. iii) For each λ > 0, (2.3) Ω(f, λδ) 6 2(1 + λ)(1 + δ 2 )Ω(f, δ). From the inequality (2.3) and definition of Ω(f, δ), we get |t − x| (1 + δ 2 )Ω(f, δ) (2.4) |f (t) − f (x)| 6 2(1 + x2 ) 1 + (t − x)2 1 + δ for every f ∈ Cx2 [0, ∞) and x, t ∈ R+ . The following estimate is quantitative Voronovskaya type asymptotic formula: Theorem 2.1. Let f ′′ ∈ Cxk2 [0, ∞), a = β = 0 and x > 0. Then we have 1 x(x + 2) ′′ f ′′ (x) 1 f (x) 6 +8(1+x2 )O(n−1 )Ω f ′′ , √ . Dn (f, x)−f (x)− f ′ (x)− 2 n 2n n n Proof. By Taylor’s formula, there exist η lying between x and y such that f (y) = f (x) + f ′ (x)(y − x) + where f ′′ (x) (y − x)2 + h(y, x)(y − x)2 , 2 f ′′ (η) − f ′′ (x) 2 and h is a continuous function which vanishes at 0. Applying the operator Dn to above equality, we get f ′ (x) f ′′ (x) h 2 + 2nx + nx2 i Dn (f, x) − f (x) = + + Dn h(y, x)(y − x)2 , x . 2 n 2 n Also, we can write that f ′ (x) f ′′ (x) 2x + x2 f ′′ (x) − + Dn |h(y, x)|(y − x)2 , x Dn (f, x) − f (x) − 6 2 n 2 n n To estimate last inequality using (2.4) and the inequality |η − x| 6 |y − x|, we can write |y − x| |h(y, x)| 6 (1 + (y − x)2 )(1 + x2 ) 1 + (1 + δ 2 )Ω(f ′′ , δ). δ Also, ( 2(1 + x2 )(1 + δ 2 )2 Ω(f ′′ , δ), |y − x| < δ |h(y, x)| 6 |y−x| 2 2 2 ′′ (1 + (y − x) )(1 + x ) 1 + δ (1 + δ )Ω(f , δ), |y − x| > δ h(y, x) := DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 269 Now choosing δ < 1, we have (y − x)4 (1 + δ 2 )2 Ω(f ′′ , δ) |h(y, x)| 6 2(1 + x2 ) 1 + δ4 (y − x)4 6 8(1 + x2 ) 1 + Ω(f ′′ , δ). δ4 Using (2.2), we deduce that Dn (|h(y, x)|(y − x)2 , x) = 8(1 + x2 )Ω(f ′′ , δ) Dn ((t − x)2 , x) + δ −4 Dn ((t − x)6 , x) = 8(1 + x2 )Ω(f ′′ , δ) O(n−1 ) + δ −4 O(n−3 ) . √ Choosing δ = 1/ n we have √ Dn (|h(y, x)|(y − x)2 , x) 6 8(1 + x2 )O(n−1 )Ω f ′′ , 1/ n . By CB [0, ∞), we mean the class of all real valued continuous and bounded functions f on [0, ∞). The second order Ditzian–Totik modulus of smoothness is defined by ωϕ2 (f, δ) = sup sup 06h6δ x±hϕ(x)∈[0,∞) ϕ(x) = |f (x + hϕ(x)) − 2f (x) + f (x − hϕ(x))|, p (x + 1)(x + 2), x > 0. The corresponding K-functional is K2,ϕ (f, δ 2 ) = inf {kf − hk + δ 2 kϕ2 h′′ k}, 2 (ϕ) h∈W∞ 2 where W∞ (ϕ) = {h ∈ CB [0, ∞) : h′ ∈ ACloc [0, ∞) : ϕ2 h′′ ∈ CB [0, ∞)}. By [5, Thm. 2.1.1], it follows that C −1 ωϕ2 (f, δ) 6 K2,ϕ (f, δ 2 ) 6 Cωϕ2 (f, δ) for some absolute constant C > 0. Also, the Ditzian–Totik modulus of the first order is given by − → ω ϕ (f, δ) = sup sup |f (x + hϕ(x)) − f (x)|, 06h6δ x±hϕ(x)∈[0,∞) where ϕ is an admissible step-weight function on [0, ∞). Theorem 2.2. If f ∈ CB [0, ∞) and n ∈ N, then we have the inequality √ kDn (f, x) − f (x)k 6 4ωϕ2 f, 1/ n + ω~ϕ f, 1/n . p Proof. We set ϕ(x) = (x + 1)(x + 2), Wϕ2 [0, ∞) = {g ∈ ACloc [0, ∞) : ϕ2 g ′′ ∈ CB [0, ∞)} D̃n (f, x) = Dn (f, x) − f (x + 1/n) + f (x). |t−x| ˜ Then ϕ|t−u| 2 (u) 6 ϕ2 (x) for u between x and t, Dn (t − x, x) 2 g ∈ Wϕ [0, ∞), by Taylor’s formula, we have g(t) = g(x) + g ′ (x)(t − x) + Z = 0 by using (2.1), and for t x g ′′ (u)(t − u) du. Applying the operator Dn to above equality and then taking modulus, we get 270 GUPTA AND MALIK |D̃n (g, x) − g(x)| Z Z t ′′ 6 Dn (t − u) g (u) du, x + 1 x+ n 1 ′′ x + − u g (u)du n x x 1 Z x+ n |x + n1 − u| Dn ((t − x)2 , x) + kϕ2 g ′′ k du 6 kϕ2 g ′′ k (x + 1)(x + 2) (x + 1)(x + 2) x n 1 1 2 o 2 6 kϕ2 g ′′ k + 6 kϕ2 g ′′ k. n n n Now for f ∈ CB [0, ∞), we have |Dn (f, x) − f (x)| 1 = |D̃n (f − g, x) − (f − g)(x)| + |D̃n (g, x) − g(x)| + f x + − f (x) n p 1 2 6 4kf − gk + kϕ2 g ′′ k + f x + (x + 1)(x + 2) p − f (x) n n (x + 1)(x + 2) n o 1 1 6 4 kf − gk + kϕ2 g ′′ k + sup f t + ϕ(t) p − f (t) n n (x + 1)(x + 2) t>0 n o 1 1 6 4 kf − gk + kϕ2 g ′′ k + ~ωϕ f, p n n (x + 1)(x + 2) n o 1 1 6 4 kf − gk + kϕ2 g ′′ k + ~ωϕ f, . n n Hence, by definition of K2,ϕ (f, δ 2 ), we have the inequality 1 1 1 1 − → kDn (f, x) − f (x) 6 4K2,ϕ f, +− ω ϕ f, 6 4ωϕ2 f, √ +→ ω ϕ f, . n n n n 3. Moments for the general case For the general case, we obtain moments as follows: Lemma 3.1. For the operators defined by (1.1), if we denote ∞ k m X µan,m (x) = ban,k (x) , n k=0 then, we have the following recurrence relation: x(1 + x) a ax a µan,m+1 (x) = [µn,m (x)]′ + x + µ (x). n n(1 + x) n,m Proof. Using the identity x(1 + x)2 [ban,k (x)]′ = [(k − nx)(1 + x) − ax]ban,k (x), we have x(1 + x)2 [µan,m (x)]′ = ∞ X k=0 k m [(k − nx)(1 + x) − ax]ban,k (x) n DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 271 = n(1 + x) ∞ X k=0 ∞ k m k m+1 X − [nx(1 + x) + ax] ban,k (x) ban,k (x) n n k=0 = n(1 + x)µan,m+1 (x) − [nx(1 + x) + ax]µan,m (x). Remark 3.1. Using Lemma 3.1, the first few moments are as follows: ax µan,0 (x) = 1, µan,1 (x) = x + , n(1 + x) µan,2 (x) = x2 + µan,3 (x) = x3 + + + µan,4 (x) = x4 + + + + 1h 2ax2 i 1 h a2 x2 ax i x + x2 + + 2 + , n (1 + x) n (1 + x)2 (1 + x) 1h 3 3ax3 i 3x + 3x2 + n (1 + x) h 1 3ax3 6ax2 3a2 x3 i 3 2 2x + 3x + x + + + n2 (1 + x) (1 + x) (1 + x)2 h 2 2 1 3a x ax a3 x3 i + + n3 (1 + x)2 (1 + x) (1 + x)3 h 1 4ax4 i 6x4 + 6x3 + n (1 + x) 1h 12ax4 18ax3 6a2 x4 i 4 3 2 11x + 18x + 7x + + + n2 (1 + x) (1 + x) (1 + x)2 h 1 8ax4 18ax3 14ax2 4 3 2 6x + 12x + 7x + x + + + n3 (1 + x) (1 + x) (1 + x) 3a2 x3 (x2 + 4x + 3) 9a2 x3 4a3 x4 3a2 x4 i + + + + (1 + x)3 (1 + x)2 (1 + x)3 (1 + x)2 h 2 2 3 3 4 4 i 1 7a x ax 6a x a x + + + . 4 2 3 n (1 + x) (1 + x) (1 + x) (1 + x)4 Lemma 3.2. [16, Lemma 2] For 0 6 β < 1, we have (β) hLn,k (t), tr i (β) hLn,k (t), 1i = Pr (k; β), R∞ where hf, gi = 0 f (t) g(t) dt and Pr (k; β) is a polynomial of order r in the variable k. In particular P0 (k; β) = 1, 1h 1 i (1 − β)k + , P1 (k; β) = n 1−β h 1 2! i P2 (k; β) = 2 (1 − β)2 k 2 + 3k + , n 1−β 1h (11 − 8β)k 3! i P3 (k; β) = 3 (1 − β)3 k 3 + 6(1 − β)k 2 + + , n 1−β 1−β 272 GUPTA AND MALIK 1h 10(5 − 3β)k 4! i (1 − β)4 k 4 + 10(1 − β)2 k 3 + 5(7 − 4β)k 2 + + , 4 n 1−β 1−β 1h P5 (k; β) = 5 (1 − β)5 k 5 + 15(1 − β)3 k 4 + 5(1 − β)(17 − 8β)k 3 n 15(15 − 20β + 6β 2 )k 2 (274 − 144β)k 5! i + + + . 1−β 1−β 1−β P4 (k; β) = Lemma 3.3. If the r-th order moment with monomials er (t) = tr , r = 0, 1, . . . of the operators (1.2) be defined by a,β Tn,r (x) := Dna,β (er , x) = (β) ∞ X hLn,k (t), tr i (β) k=0 The first few are a,β Tn,0 (x) = 1, hLn,k (t), 1i ban,k (x) = a,β Tn,1 (x) = x(1 − β) + ∞ X Pr (k; β)ban,k (x). k=0 ax(1 − β) 1 + , n(1 + x) n(1 − β) i 1h 2 2ax2 (1 − β)2 x (1 − β)2 + x(1 − β)2 + + 3x n 1+x 1 h a2 x2 (1 − β)2 ax(1 − β)2 3ax 2 i + 2 + + + , n (1 + x)2 1+x 1+x 1−β a,β Tn,2 (x) = x2 (1 − β)2 + a,β Tn,3 (x) = x3 (1 − β)3 i 1h 3 3ax3 (1 − β)3 + 3x (1 − β)3 + 3x2 (1 − β)3 + + 6x2 (1 − β) n (1 + x) 1h 3 3ax3 (1 − β)3 + 2 2x (1 − β)3 + 3x2 (1 − β)3 + x(1 − β)3 + n (1 + x) 6ax2 (1 − β)3 3a2 x3 (1 − β)3 + + + 6x2 (1 − β) (1 + x) (1 + x)2 12ax2 (1 − β) (11 − 8β)x i + 6x(1 − β) + + (1 + x) (1 − β) h 2 2 3 3 3 3 3 1 3a x (1 − β) ax(1 − β) a x (1 − β) + 3 + + n (1 + x)2 (1 + x) (1 + x)3 6a2 x2 (1 − β) 6ax(1 − β) (11 − 8β)ax 6 i + + + + (1 + x)2 (1 + x) (1 − β)(1 + x) (1 − β) a,β Tn,4 (x) = x4 (1 − β)4 i 1h 4 4ax4 (1 − β)4 + 6x (1 − β)4 + 6x3 (1 − β)4 + + 10x3 (1 − β)2 n (1 + x) h 1 12ax4 (1 − β)4 + 2 11x4 (1 − β)4 + 18x3 (1 − β)4 + 7x2 (1 − β)4 + n (1 + x) 3 4 2 4 4 18ax (1 − β) 6a x (1 − β) + + + 30x3 (1 − β)2 (1 + x) (1 + x)2 DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 273 + 30x2 (1 − β)2 + i 30ax3 (1 − β)2 + (35 − 20β)x2 (1 + x) 1h 4 6x (1 − β)4 + 12x3 (1 − β)4 + 7x2 (1 − β)4 + x(1 − β)4 n3 8ax4 (1 − β)4 18ax3 (1 − β)4 14ax2 (1 − β)4 + + + (1 + x) (1 + x) (1 + x) 2 3 2 4 2 3 3a x (x + 4x + 3)(1 − β) 9a x (1 − β)4 4a3 x4 (1 − β)4 + + + (1 + x)3 (1 + x)2 (1 + x)3 2 4 4 3a x (1 − β) + + 20x3 (1 − β)2 + 30x2 (1 − β)2 + 10x(1 − β)2 (1 + x)2 30ax3 (1 − β)2 60ax2 (1 − β)2 30a2 x3 (1 − β)2 + + + (1 + x) (1 + x) (1 + x)2 (70 − 40β)ax2 (50 − 30β)x i + (35 − 20β)x2 + (35 − 20β)x + + (1 + x) (1 − β) h 2 2 4 4 3 3 4 4 4 1 7a x (1 − β) ax(1 − β) 6a x (1 − β) a x (1 − β)4 + + + + 4 n (1 + x)2 (1 + x) (1 + x)3 (1 + x)4 2 3 3 2 2 2 2 30a x (1 − β) 10ax(1 − β) 10a x (1 − β) + + + (1 + x)2 (1 + x) (1 + x)3 2 2 (35 − 20β)a x (35 − 20β)ax (50 − 30β)ax 24 i + + + + . (1 + x)2 (1 + x) (1 − β)(1 + x) (1 − β) + a,β Proof. Obviously by (1.2), we have Tn,0 (x) = 1. Next, by definition of a,β Tn,r (x), we have a,β Tn,r (x) = (β) ∞ X hLn,k (t), tr i (β) k=0 hLn,k (t), 1i ban,k (x) = ∞ X Pr (k; β) ban,k (x). k=0 Thus, using Lemma 3.1 and Lemma 3.2, we have ∞ ∞ X X 1h 1 i a,β Tn,1 (x) = ban,k (x)P1 (k; β) = ban,k (x) (1 − β)k + n 1−β k=0 k=0 1 ax(1 − β) 1 µa (x) = x(1 − β) + + n(1 − β) n,0 n(1 + x) n(1 − β) ∞ ∞ X X 1 h 2 i a,β Tn,2 (x) = ban,k (x)P2 (k; β) = ban,k (x) 2 (1 − β)2 k 2 + 3k + n 1−β = (1 − β)µan,1 (x) + k=0 k=0 3 2 µa (x) = (1 − + µan,1 (x) + 2 n n (1 − β) n,0 h i x2 x a2 x2 2ax2 ax = (1 − β)2 x2 + + + 2 + + n n n (1 + x)2 n(1 + x) n2 (1 + x) h i 3 ax 2 + x+ + 2 n n(1 + x) n (1 − β) β)2 µan,2 (x) 274 GUPTA AND MALIK i 2ax2 (1 − β)2 1h 2 x (1 − β)2 + x(1 − β)2 + + 3x n 1+x 1 h a2 x2 (1 − β)2 ax(1 − β)2 3ax 2 i + 2 + + + . n (1 + x)2 1+x 1+x 1−β = x2 (1 − β)2 + a,β A continuation of this process will provide Tn,r (x) for cases of r > 3. a,β Remark 3.2. If we denote the central moment as Un,r (x) = Dna,β ((t − x)r , x), then ax(1 − β) 1 a,β Un,1 (x) = −xβ + + , n(1 + x) n(1 − β) 1h 2ax2 β(1 − β) 2x i 3x + x(1 + x)(1 − β)2 − − n 1+x 1−β 1 h x2 a2 (1 − β)2 ax(1 − β)2 3ax 2 i + 2 + + + , n (1 + x)2 1+x 1+x 1−β a,β Un,2 (x) = x2 β 2 + a,β Un,3 (x) = x3 (1 − β)3 − x3 − 3x3 (1 − β)2 + 3x3 (1 − β) 1h 3 3ax3 (1 − β)3 + 3x (1 − β)3 + 3x2 (1 − β)3 + n (1 + x) 6ax3 (1 − β)2 + 6x2 (1 − β) − 3x3 (1 − β)2 − (1 + x) 3 3ax (1 − β) 3x2 i − 3x2 (1 − β)2 − 9x2 + + (1 + x) (1 − β) 1 h 3 3ax3 (1 − β)3 + 2 2x (1 − β)3 + 3x2 (1 − β)3 + x(1 − β)3 + n (1 + x) 6ax2 (1 − β)3 3a2 x3 (1 − β)3 + + + 6x2 (1 − β) (1 + x) (1 + x)2 12ax2 (1 − β) (11 − 8β)x + 6x(1 − β) + + (1 + x) (1 − β) 3a2 x3 (1 − β)2 3ax2 (1 − β)2 9ax2 6x i − − − − (1 + x)2 (1 + x) (1 + x) (1 − β) h 2 2 1 3a x (1 − β)3 ax(1 − β)3 a3 x3 (1 − β)3 + 3 + + n (1 + x)2 (1 + x) (1 + x)3 2 2 6a x (1 − β) 6ax(1 − β) (11 − 8β)ax 6 i + + + + (1 + x)2 (1 + x) (1 − β)(1 + x) (1 − β) a,β Un,4 (x) = x4 (1 − β)4 + x4 + 6x4 (1 − β)2 − 4x4 (1 − β) − 4x4 (1 − β)3 1h 4 4ax4 (1 − β)4 + 6x (1 − β)4 + 6x3 (1 − β)4 + n 1+x 12ax4 (1 − β)2 + 10x3 (1 − β)2 + 6x4 (1 − β)2 + 1+x DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 275 4x3 4ax4 (1 − β) − 1+x 1−β i 4 12ax (1 − β)3 − 12x4 (1 − β)3 − 12x3 (1 − β)3 − − 24x3 (1 − β) 1+x 1 h 12ax4 (1 − β)4 + 2 11x4 (1 − β)4 + 18x3 (1 − β)4 + 7x2 (1 − β)4 + n 1+x 18ax3 (1−β)4 6a2 x4 (1−β)4 + + + 30x3 (1−β)2 + 30x2 (1−β)2 1+x (1 + x)2 30ax3 (1−β)2 6a2 x4 (1−β)2 6ax3 (1−β)2 + + (35 − 20β)x2 + + 1+x (1 + x)2 1+x 2 3 12x 18ax + − 8x4 (1 − β)3 − 12x3 (1 − β)3 − 4x2 (1 − β)3 + 1+x 1−β 12ax4 (1 − β)3 24ax3 (1 − β)3 12a2 x4 (1 − β)3 − − − 1+x 1+x (1 + x)2 3 48ax (1−β) 4(11 − 8β)x2 i − 24x3 (1−β) − 24x2 (1−β) − − 1+x 1−β h 8ax4 (1−β)4 1 + 3 6x4 (1−β)4 + 12x3 (1−β)4 + 7x2 (1−β)4 + x(1−β)4 + n 1+x 2 4 2 3 2 3 4 18ax (1 − β) 14ax (1 − β) 3a x (x + 4x + 3)(1 − β)4 + + + 1+x 1+x (1 + x)3 9a2 x3 (1 − β)4 4a3 x4 (1 − β)4 3a2 x4 (1 − β)4 + + + + 20x3 (1 − β)2 2 3 (1 + x) (1 + x) (1 + x)2 30ax3 (1 − β)2 60ax2 (1 − β)2 + 30x2 (1 − β)2 + 10x(1 − β)2 + + 1+x 1+x 30a2 x3 (1 − β)2 (70 − 40β)ax2 2 + + (35 − 20β)x + (35 − 20β)x + (1 + x)2 1+x (50 − 30β)x 12a2 x3 (1 − β)3 4ax2 (1 − β)3 4a3 x4 (1 − β)3 + − − − 1−β (1 + x)2 1+x (1 + x)3 2 3 2 2 24a x (1 − β) 24ax (1 − β) 4(11 − 8β)ax 24x i − − − − (1 + x)2 1+x (1 − β)(1 + x) 1 − β h 2 2 1 7a x (1 − β)4 ax(1 − β)4 6a3 x3 (1 − β)4 a4 x4 (1 − β)4 + 4 + + + n (1 + x)2 1+x (1 + x)3 (1 + x)4 2 2 2 2 3 3 2 30a x (1 − β) 10ax(1 − β) 10a x (1 − β) + + + 2 (1 + x) 1+x (1 + x)3 (35 − 20β)a2 x2 (35 − 20β)ax (50 − 30β)ax 24 i + + + . + (1 + x)2 1+x (1 − β)(1 + x) 1 − β + 6x3 (1 − β)2 + 18x3 − 4. Direct estimates for general case In this section, we establish the following direct results: 276 GUPTA AND MALIK Proposition 4.1. Let f be a continuous function on [0, ∞) and β = βn → 0 as n → ∞, the sequence {Dna,β (f, x)} converges uniformly to f (x) in [a, b] ⊂ [0, ∞). Proof. Using Lemma 3.3, we have Dna,β (e0 , x) = 1, Dna,β (e1 , x) = x(1 − β) + and ax(1 − β) 1 + n(1 + x) n(1 − β) i 1h 2 2ax2 (1 − β)2 x (1 − β)2 + x(1 − β)2 + + 3x n 1+x 1 h a2 x2 (1 − β)2 ax(1 − β)2 3ax 2 i + 2 + + + . n (1 + x)2 1+x 1+x 1−β Dna,β (e2 , x) = x2 (1 − β)2 + Obviously, Dna,β (e0 , x), Dna,β (e1 , x) and Dna,β (e2 , x) converges uniformly to 1, x and x2 respectively on every compact subset of [0, ∞). Thus, the required result follows from the well known Bohman–Korovkin theorem. Theorem 4.1. Let f be a bounded integrable function on [0, ∞) and has the second derivative at a point x ∈ [0, ∞), then with the condition β = βn → 0 as n → ∞, it follows that h x2 + 2x i h ax i ′ f (x) + f ′′ (x). lim n[Dna,βn (f, x) − f (x)] = 1 + n→∞ 1+x 2 Proof. By Taylor’s expansion of f , we have 1 (4.1) f (t) = f (x) + f ′ (x)(t − x) + f ′′ (x)(t − x)2 + r(t, x)(t − x)2 , 2 where r(t, x) is the remainder term and limt→x r(t, x) = 0. Operating Dna,β to the equation (4.1), we obtain Dna,β (f, x) − f (x) = Dna,β (t − x, x)f ′ (x) + Dna,β ((t − x)2 , x) f ′′ (x) 2 + Dna,β (r(t, x)(t − x)2 , x) Using the Cauchy–Schwarz inequality, we have q q (4.2) Dna,β (r(t, x)(t − x)2 , x) 6 Dna,β (r2 (t, x), x) Dna,β ((t − x)4 , x). As r2 (x, x) = 0 and r2 (t, x) ∈ C2∗ [0, ∞), we have (4.3) lim Dna,βn (r2 (t, x), x) = r2 (x, x) = 0 n→∞ uniformly with respect to x ∈ [0, A], for some A > 0. Now from (4.2), (4.3) and from Remark 3.2, we get limn→∞ nDna,βn (r(t, x)(t − x)2 , x) = 0. Thus lim n(Dna,βn (f, x) − f (x)) h i 1 = lim n Dna,βn (t−x, x)f ′ (x)+ f ′′ (x)Dna,βn ((t−x)2 , x)+Dna,βn (r(t, x)(t−x)2 , x) n→∞ 2 h h x2 + 2x i ax i ′ = 1+ f (x) + f ′′ (x). 1+x 2 n→∞ DIRECT ESTIMATIONS OF NEW GENERALIZED BASKAKOV–SZÁSZ OPERATORS 277 We denote the norm kf k = supx∈[0,∞) |f (x)|. For f ∈ CB [0, ∞) and δ > 0, the m-th order modulus of continuity is defined as ωm (f, δ) = sup sup |∆m h f (x)|, 06h6δ x∈[0,∞) where ∆ is the forward difference. In case m = 1, we mean the usual modulus of continuity denoted by ω(f, δ). Peetre’s K-functional is defined as K2 (f, δ) = inf 2 [0,∞) g∈CB 2 {kf − gk + δkg ′′ k : g ∈ CB [0, ∞)}, where 2 CB [0, ∞) = {g ∈ CB [0, ∞) : g ′ , g ′′ ∈ CB [0, ∞)}. Theorem 4.2. Let f ∈ CB [0, ∞) and 0 < β < 1, then ax(1 − β) p 1 |Dna,β (f, x) − f (x)| 6 Cω2 f, δn + ω f, + − xβ n(1 + x) n(1 − β) where C is a positive constant and δn is given by 1h 4ax2 β(1 − β) 2x 2xβ i δn = 2x2 β 2 + 3x + x(x + 1)(1 − β)2 − − − n 1+x 1−β 1−β h i 2 2 2 2 5ax ax(1 − β) 2 1 1 2a x (1 − β) + 2 + + + + . 2 2 n (1 + x) 1+x 1+x 1−β (1 − β) with β = βn → 0 as n → ∞. Proof. We introduce the auxiliary operators D̄na,β : CB [0, ∞) → CB [0, ∞) by ax(1 − β) 1 (4.4) D̄na,β (f, x) = Dna,β (f, x) − f x(1 − β) + + + f (x). n(1 + x) n(1 − β) These operators are linear and preserve the linear functions in view of Lemma 3.3. 2 Let g ∈ CB [0, ∞) and x, t ∈ [0, ∞). By Taylor’s expansion, we have Z t g(t) = g(x) + (t − x) g ′ (x) + (t − u) g ′′ (u) du, x Hence, |D̄na,β (g, x) − g(x)| Z t Z t a,β ′′ a,β ′′ 6 D̄n (t − u)g (u)du, x 6 Dn (t − u)g (u)du, x x x Z + x(1−β)+ x ax(1−β) 1 + n(1−β) n(1+x) ax(1 − β) 1 ′′ x(1 − β) + + − u g (u)du n(1 + x) n(1 − β) 6 Dna,β ((t − x)2 , x)kg ′′ k Z x(1−β)+ ax(1−β) + 1 n(1+x) n(1−β) ax(1 − β) 1 + + − xβ dukg ′′ k n(1 + x) n(1 − β) x 278 GUPTA AND MALIK Next, using Remark 3.2, we have h |D̄na,β (g, x) − g(x)| 6 Dna,β ((t − x)2 , x) ax(1 − β) 2 i (4.5) 1 + + − xβ kg ′′ k = δn kg ′′ k. n(1 + x) n(1 − β) Since |Dna,β (f, x)| 6 (β) ∞ X hLn,k (t), |f (t)|i k=0 (β) hLn,k (t), 1i ban,k (x) 6 kf k (β) ∞ X hLn,k (t), 1i ban,k (x) (β) k=0 hLn,k (t), 1i 6 kf k. Now by (4.4), we have (4.6) |kD̄na,β (f, x)|| 6 k|Dna,β (f, x)k + 2kf k 6 3kf k, f ∈ CB [0, ∞). 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