Journal of Inequalities in Pure and Applied Mathematics

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1 Journal of Inequalities in Pure and Applied Mathematics ON SIMULTANEOUS APPROXIMATION FOR CERTAIN BASKAKOV DURRMEYER TYPE OPERATORS VIJAY GUPTA, MUHAMMAD ASLAM NOOR AND MAN SINGH BENIWAL School of Applied Sciences Netaji Subhas Institute of Technology Sector 3 Dwarka New Delhi 1175, India vijay@nsit.ac.in Mathematics Department COMSATS Institute of Information Technology Islamabad, Pakistan noormaslam@hotmail.com Department of Applied Science Maharaja Surajmal Institute of Technology C-4, Janakpuri, New Delhi , India man_s_5@yahoo.co.in Department of Mathematics Ch Charan Singh University Meerut 54, India mkgupta@hotmail.com c Victoria University ISSN (electronic): volume 7, issue 4, article 15, 6. Received 6 January, 6; accepted 16 August, 6. Communicated by: N.E. Cho Abstract Home Page

2 Abstract In the present paper, we study a certain integral modification of the well known Baskakov operators with the weight function of Beta basis function. We establish pointwise convergence, an asymptotic formula an error estimation and an inverse result in simultaneous approximation for these new operators. Mathematics Subject Classification: 41A3, 41A36. Key words: Baskakov operators, Simultaneous approximation, Asymptotic formula, Pointwise convergence, Error estimation, Inverse theorem. The work carried out when the second author visited Department of Mathematics and Statistics, Auburn University, USA in fall 5. The authors are thankful to the referee for making many valuable suggestions, leading to the better presentation of the paper. 1 Introduction Basic Results Direct Theorems Inverse Theorem References Page of 33

3 1. Introduction For f C γ [, ) {f C[, ) : f(t) Mt γ for some M >, γ > } we consider a certain type of Baskakov-Durrmeyer operator as (1.1) where and B n (f(t), x) = = p n,k (x) W n (x, t)f(t)dt b n,k (t)f(t)dt + (1 + x) n f() ( ) n + k 1 x k p n,k (x) = k (1 + x), n+k 1 b n,k (t) = B(n + 1, k) t k 1 (1 + t) n+k+1 W n (x, t) = p n,k (x)b n,k (t) + (1 + x) n δ(t), δ(t) being the Dirac delta function. The norm- γ on the class C γ [, ) is defined as f γ = sup f(t) t γ. t< The operators defined by (1.1) are the integral modification of the well known Baskakov operators with weight functions of some Beta basis functions. Very Page 3 of 33

4 recently Finta [] also studied some other approximation properties of these operators. The behavior of these operators is very similar to the operators recently introduced in [6], [9] and also studied in [8]. These operators reproduce not only the constant functions but also the linear functions, which is the interesting property of such operators. The other usual Durrmeyer type integral modification of the Baskakov operators [5] reproduce only the constant functions, so one can not apply the iterative combinations easily to improve the order of approximation for the usual Baskakov Durrmeyer operators. For recent work in this area we refer to [7]. In the present paper we study some direct results which include pointwise convergence, asymptotic formula, error estimation and inverse theorem in the simultaneous approximation for the unbounded functions of growth of order t γ. Page 4 of 33

5 . Basic Results In this section we mention certain lemmas which will be used in the sequel. Lemma.1 ([3]). For m N {}, if the m th order moment be defined as U n,m (x) = k= ( ) m k p n,k (x) n x, then U n, (x) = 1, U n,1 (x) = and nu n,m+1 (x) = x(1 + x)(u (1) n,m(x) + mu n,m 1 (x)). Consequently we have U n,m (x) = O ( n [(m+1)/]). Lemma.. Let the function T n,m (x), m N {}, be defined as ( T n,m (x) = B n (t x) m x ) = p n,k (x) b n,k (t)(t x) m dt + (1 + x) n ( x) m. Then T n, (x) = 1, T n,1 =, T n, (x) = x(1+x) and also there holds the recurrence n 1 relation (n m)t n,m+1 (x) = x(1 + x) T (1) n,m(x) + mt n,m 1 (x) + m(1 + x)t n,m (x). Page 5 of 33

6 Proof. By definition, we have T (1) n,m(x) = p (1) Using the identities and we have n,k (x) b n,k (t)(t x) m dt m p n,k (x) b n,k (t)(t x) m 1 dt n(1 + x) n 1 ( x) m m(1 + x) n ( x) m 1. x(1 + x)p (1) n,k (x) = (k nx)p n,k(x) t(1 + t)b (1) n,k (t) = [(k 1) (n + )t]b n,k(t), x(1 + x) [ T n,m(x) (1) + mt n,m 1 (x) ] = p n,k (x) (k nx)b n,k (t)(t x) m dt + n(1 + x) n ( x) m+1 = p n,k (x) [ (k 1) (n + )t + (n + )(t x) + (1 + x) ] b n,k (t)(t x) m dt + n(1 + x) n ( x) m+1 Page 6 of 33

7 = p n,k (x) t(1 + t)b (1) n,k (t)(t x)m dt + (n + )[T n,m+1 (x) (1 + x) n ( x) m+1 ] + (1 + x)[t n,m (x) (1 + x) n ( x) m ] + n(1 + x) n ( x) m+1 = (m + 1)(1 + x)[t n,m (x) (1 + x) n ( x) m ] Thus, we get (m + )[T n,m+1 (1 + x) n ( x) m+1 ] mx(1 + x)[t n,m 1 (x) (1 + x) n ( x) m 1 ] + (n + )[T n,m+1 (1 + x) n ( x) m+1 ] + (1 + x)[t n,m (x) (1 + x) n ( x) m ] + n(1 + x) n ( x) m+1. (n m)t n,m+1 (x) = x(1 + x)[t (1) n,m(x) + mt n,m 1 (x)] + m(1 + x)t n,m (x). This completes the proof of recurrence relation. From the above recurrence relation, it is easily verified for all x [, ) that T n,m (x) = O ( n [(m+1)/]). Remark 1. It is easily verified from Lemma.1 that for each x (, ) B n (t i, x) = (n + i 1)!(n i)! x i (n + i )!(n i)! +i(i 1) x i 1 +O(n ). Page 7 of 33

8 Corollary.3. Let δ be a positive number. Then for every γ >, x (, ), there exists a constant M(s, x) independent of n and depending on s and x such that t x >δ W n (x, t)t γ dt C[a,b] M(s, x)n s, s = 1,, 3,... Lemma.4. There exist the polynomials Q i,j,r (x) independent of n and k such that {x(1 + x)} r D r[ p n,k (x) ] = n i (k nx) j Q i,j,r (x)p n,k (x), where D d dx. i+j r i,j By C, we denote the class of continuous functions on the interval (, ) having a compact support and C r is the class of r times continuously differentiable functions with C r C. The function f is said to belong to the generalized Zygmund class Liz(α, 1, a, b), if there exists a constant M such that ω (f, δ) Mδ α, δ >, where ω (f, δ) denotes the modulus of continuity of nd order on the interval [a, b]. The class Liz(α, 1, a, b) is more commonly denoted by Lip (α, a, b). Suppose G (r) = {g : g C r+, supp g [a, b ] where [a, b ] (a, b)}. For r times continuously differentiable functions f with supp f [a, b ] the Peetre s K-functionals are defined as K r (ξ, f) = inf g G (r) [ f (r) g (r) C[a,b ] + ξ { g (r) C[a,b ] + g (r+) C[a,b ] < ξ 1. }], Page 8 of 33

9 For < α <, C r (α, 1, a, b) denotes the set of functions for which sup ξ α/ K r (ξ, f, a, b) < C. <ξ 1 Lemma.5. Let < a < a < b < b < b < and f (r) C with supp f [a, b ] and if f C r (α, 1, a, b ), we have f (r) Liz(α, 1, a, b ) i.e. f (r) Lip (α, a, b ) where Lip (α, a, b ) denotes the Zygmund class satisfying K r (δ, f) Cδ α/. Proof. Let g G (r), then for f C r (α, 1, a, b ), we have δ f (r) (x) δ (f (r) g (r) )(x) + δ g (r) (x) δ (f (r) g (r) ) C[a,b ] + δ g (r+) C[a,b ] 4M 1 K r (δ, f) M δ α. Lemma.6. If f is r times differentiable on [, ), such that f (r 1) = O(t α ), α > as t, then for r = 1,, 3,... and n > α + r we have B n (r) (f, x) = Proof. First B (1) n (f, x) = (n + r 1)!(n r)! p (1) p n+r,k (x) k= b n r,k+r (t)f (r) (t)dt. n,k (x) b n,k (t)f(t)dt n(1 + x) n 1 f(). Page 9 of 33

10 Now using the identities (.1) (.) p (1) n,k (x) = n[p n+1,k 1(x) p n+1,k (x)], b (1) n,k (t) = (n + 1)[b n+1,k 1(t) b n+1,k (t)]. for k 1, we have B (1) n (f, x) = n[p n+1,k 1 (x) p n+1,k (x)] = np n+1, (x) + n b n,1 (t)f(t)dt n(1 + x) n 1 f() p n+1,k (x) b n,k (t)f(t)dt n(1 + x) n 1 f() [b n,k+1 (t) b n,k (t)]f(t)dt = n(1 + x) n 1 (n + 1)(1 + t) n f(t)dt n(1 + x) n 1 f() + n p n+1,k (x) Integrating by parts, we get B (1) n 1 n b(1) n 1,k+1 (t)f(t)dt. (f, x) = n(1 + x) n 1 f() + n(1 + x) n 1 (1 + t) n 1 f (1) (t)dt Page 1 of 33

11 = n(1 + x) n 1 f() + p n+1,k (x) k= p n+1,k (x) b n 1,k+1 (t)f (1) (t)dt. b n 1,k+1 (t)f (1) (t)dt Thus the result is true for r = 1. We prove the result by induction method. Suppose that the result is true for r = i, then B (i) n (f, x) = (n + i 1)!(n i)! p n+i,k (x) k= Thus using the identities (.1) and (.), we have B (i+1) n (f, x) (n + i 1)!(n i)! = (n + i)[p n+i+1,k 1 (x) p n+i+1,k (x)] b n i,k+i (t)f (i) (t)dt. b n i,k+i (t)f (i) (t)dt (n + i 1)!(n i)! + ( (n + i)(1 + x) n i 1 ) b n i,i (t)f (i) (t)dt (n + i)!(n i)! = p n+i+1, (x) b n i,i+1 (t)f (i) (t)dt (n + i)!(n i)! p n+i+1, (x) b n i,i (t)f (i) (t)dt Page 11 of 33

12 = + (n + i)!(n i)! p n+i+1,k (x) (n + i)!(n i)! p n+i+1, (x) 1 (n + i)!(n i)! + p n+i+1,k (x) [b n i,k+i+1 (t) b n i,k+i (t)]f (i) (t)dt (n i) b(1) n i 1,i+1 (t)f (i) (t)dt 1 (n i) b(1) n i 1,k+i+1 (t)f (i) (t)dt. Integrating by parts, we obtain B (i+1) n (f, x) = (n + i)!(n i 1)! p n+i+1,k (x) k= This completes the proof of the lemma. b n i 1,k+i+1 (t)f (i+1) (t)dt. Page 1 of 33

13 3. Direct Theorems In this section we present the following results. Theorem 3.1. Let f C γ [, ) and f (r) exists at a point x (, ). Then we have B n (r) (f, x) = f (r) (x) as n. Proof. By Taylor expansion of f, we have f(t) = r i= where ε(t, x) as t x. Hence B (r) n (f, x) = = f (i) (x) (t x) i + ε(t, x)(t x) r, i! W n (r) (t, x)f(t)dt r f (i) (x) i! i= + =: R 1 + R. W n (r) (t, x)(t x) i dt W n (r) (t, x)ε(t, x)(t x) r dt First to estimate R 1, using the binomial expansion of (t x) i and Remark 1, we Page 13 of 33

14 have r f (i) (x) i ( ) i i v r R 1 = ( x) W i! v x r n (t, x)t v dt i= v= = f [ ] (r) (x) d r (n + r 1)!(n r)! x r + terms containing lower powers of x r! dx r [ ] (n + r 1)!(n r)! = f (r) (x) f (r) (x) as n. Next applying Lemma.4, we obtain R = R i+j r i,j n i b n,k (t) ε(t, x) t x r dt + W n (r) (t, x)ε(t, x)(t x) r dt, Q i,j,r (x) {x(1 + x)} r k nx j p n,k (x) (n + r + 1)! (1 + x) n r ε(, x) x r. (n 1)! The second term in the above expression tends to zero as n. Since ε(t, x) as t x for a given ε > there exists a δ such that ε(t, x) < ε whenever < t x < δ. If α max{γ, r}, where α is any integer, then we can find a constant M 3 >, ε(t, x)(t x) r M 3 t x α, for t x δ. Page 14 of 33

15 Therefore R M 3 i+j r i,j =: R 3 + R 4. n i k= { ε p n,k (x) k nx j t x <δ b n,k (t) t x r dt + t x δ } b n,k (t) t x α dt Applying the Cauchy-Schwarz inequality for integration and summation respectively, we obtain R 3 εm 3 i+j r i,j n i { p n,k (x)(k nx) j } 1 { p n,k (x) Using Lemma.1 and Lemma., we get R 3 = ε O(n r/ )O(n r/ ) = ε o(1). b n,k (t)(t x) r dt Again using the Cauchy-Schwarz inequality, Lemma.1 and Corollary.3, we } 1. Page 15 of 33

16 get R 4 M 4 M 4 M 4 = i+j r i,j i+j r i,j i+j r i,j i+j r i,j n i n i { t x δ n i { { p n,k (x) p n,k (x) k nx j p n,k (x) k nx j { t x δ t x δ } 1 b n,k (t)(t x) α dt p n,k (x)(k nx) j } 1 b n,k (t)(t x) α dt b n,k (t) t x α dt } 1 b n,k (t)dt } 1 n i O(n j/ )O(n α/ ) = O(n (r α)/ ) = o(1). Collecting the estimates of R 1 R 4, we obtain the required result. Theorem 3.. Let f C γ [, ). If f (r+) exists at a point x (, ). Then lim n [ B n (r) (f, x) f (r) (x) ] n = r(r 1)f (r) (x) + r(1 + x)f (r+1) (x) + x(1 + x)f (r+) (x). Page 16 of 33

17 Proof. Using Taylor s expansion of f, we have f(t) = r+ i= f (i) (x) (t x) i + ε(t, x)(t x) r+, i! where ε(t, x) as t x and ε(t, x) = O((t x) γ ), t for γ >. Applying Lemma., we have n [ B n (r) (f(t), x) f (r) (x) ] ] First, we have r+ E 1 = n i= = n [ r+ f (i) (x) i! i= + n =: E 1 + E. f (i) (x) i! i j= W n (r) (t, x)(t x) i dt f (r) (x) W n (r) (t, x)ε(t, x)(t x) r+ dt ( ) i ( x) i j j = f (r) (x) n [ B n (r) (t r, x) (r!) ] r! + f (r+1) (x) (r + 1)! n [ (r + 1)( x)b n (r) [ + f (r+) (x) (r + )! n W n (r) (t, x)t j dt nf (r) (x) (t r, x) + B n (r) (t r+1, x) ] (r + )(r + 1) x B n (r) (t r, x) Page 17 of 33

18 + (r + )( x)b n (r) (t r+1, x) + B n (r) (t r+, x) ]. Therefore, by Remark 1, we have [ ] (n + r 1)!(n r)! E 1 = nf (r) (x) 1 + nf [ { } (r+1) (x) (n + r 1)!(n r)! (x 1)( x) (r + 1)! { } (n + r)!(n r 1)! (n + r 1)!(n r 1)! + (r + 1)!x + (r + 1)r r! + nf [ ] (r+) (x) (r + )(r + 1) (n + r 1)!(n r)! x r! (r + )! { (n + r)!(n r 1)! + (r + )( x) x(r + 1)! } (n r 1)!(n r 1)! +(r + 1)r r! { (n + r + 1)!(n r )! (r + )! + x }] (n + r)!(n r )! + (r + )(r + 1) (r + 1)!x + O(n ). In order to complete the proof of the theorem it is sufficient to show that E as n which easily follows proceeding along the lines of the proof of Theorem 3.1 and by using Lemma.1, Lemma. and Lemma.4. Page 18 of 33

19 Lemma 3.3. Let < α <, < a < a < a < b < b < b <. If f C with supp f [a, b ] and (f, ) f (r) C[a,b] = O(n α/ ), then B n (r) K r (ξ, f) = M 5 { n α/ + nξk r (n 1, f) }. Consequently K r (ξ, f) M 6 ξ α/, M 6 >. Proof. It is sufficient to prove K r (ξ, f) = M 7 {n α/ + nξk r (n 1, f)}, for sufficiently large n. Because supp f [a, b ] therefore by Theorem 3. there exists a function h (i) G (r), i = r, r +, such that B (r) n (f, ) h (i) M C[a,b ] 8n 1. Therefore, K r (ξ, f) 3M 9 n 1 + B (r) n (f, ) f (r) C[a,b ] { B (r) + ξ n (f, ) + C[a,b ] B (r+) n (f, ) } C[a.,b ] Next, it is sufficient to show that there exists a constant M 1 such that for each g G (r) (3.1) B (r+) n (f, ) C[a,b ] M 1n{ f (r) g (r) C[a,b ] +n 1 g (r+) C[a,b ]. Page 19 of 33

20 Also using the linearity property, we have (3.) B n (r+) (f, ) C[a,b ] Applying Lemma.4, we get r+ x W n(x, t) r+ dt B (r+) n (f g, ) + C[a,b ] B (r+) n (g, ). C[a,b ] i+j r+ i,j p n,k (x) n i k nx j Q i,j,r+ (x) {x(1 + x)} r+ b n,k (t)dt + dr+ dx r+ [(1 + x) n ]. Therefore by the Cauchy-Schwarz inequality and Lemma.1, we obtain (3.3) B n (r) (f g, ) M C[a,b ] 11n f (r) g (r), C[a,b ] where the constant M 11 is independent of f and g. Next by Taylor s expansion, we have r+1 g (i) (x) g(t) = (t x) i + g(r+) (ξ) i! (r + )! (t x)r+, i= where ξ lies between t and x. Using the above expansion and the fact that m W x m n (x, t)(t x) i dt = for m > i, we get (3.4) B (r+) n (g, ) C[a,b ] M 1 g (r+) r+ C[a,b ] x W n(x, t)(t x) r+ dt. r+ C[a,b ] Page of 33

21 Also by Lemma.4 and the Cauchy-Schwarz inequality, we have E r+ x W n(x, t) r+ (t x)r+ dt n i p n,k (x) k nx j Q i,j,r+ (x) b {x(1 + x)} r+ n,k (t)(t x) r+ dt i+j r+ i,j Hence + dr+ dx r+ [( x)r+ (1 + x) n ] ( Q i,j,r+ (x) {x(1 + x)} r+ i+j r+ i,j ( p n,k (x) + dr+ p n,k (x)(k nx) j ) 1 b n,k (t)(t x) r+4 dt dx r+ [( x)r+ (1 + x) n ] = n i Q i,j,r+ (x) {x(1 + x)} r+ O(nj/ )O i+j r+ i,j ) 1 ( ( n (1+ r ) ). (3.5) B (r+) n (g, ) C[a,b ] M 13 g (r+) C[a,b ]. ) 1 b n,k (t)dt Combining the estimates of (3.)-(3.5), we get (3.1). The other consequence follows form [1]. This completes the proof of the lemma. Page 1 of 33

22 Theorem 3.4. Let f C γ [, ) and suppose < a < a 1 < b 1 < b <. Then for all n sufficiently large, we have B n (r) (f, ) f (r) max { ( ) } C[a1,b 1 M ] 14 ω f (r), n 1, a, b + M 15 n 1 f γ, where M 14 = M 14 (r), M 15 = M 15 (r, f). Proof. For sufficiently small δ >, we define a function f,δ (t) corresponding to f C γ [, ) by f,δ (t) = δ δ δ δ δ ( f(t) η f(t) ) dt 1 dt, where η = t 1+t, t [a, b] and ηf(t) is the second forward difference of f with step length η. Following [4] it is easily checked that: (i) f,δ has continuous derivatives up to order k on [a, b], (ii) f (r),δ C[a 1,b 1 ] M 1 δ r ω (f, δ, a, b), (iii) f f,δ C[a1,b 1 ] M ω (f, δ, a, b), (iv) f,δ C[a1,b 1 ] M 3 f γ, where M i, i = 1,, 3 are certain constants that depend on [a, b] but are independent of f and n [4]. Page of 33

23 We can write B (r) n (f, ) f (r) C[a1,b 1 ] B (r) n (f f,δ, ) C[a1,b 1 ] + B (r) n =: H 1 + H + H 3. (f,δ, ) f (r) + f (r) f (r) C[a1,b 1 ],δ,δ C[a1,b 1 ] Since f (r),δ = ( f (r)),δ (t), by property (iii) of the function f,δ, we get H 3 M 4 ω (f (r), δ, a, b). Next on an application of Theorem 3., it follows that H M r+ 5 n 1 f (j),δ j=r. C[a,b] Using the interpolation property due to Goldberg and Meir [4], for each j = r, r + 1, r +, it follows that f (j) M } 6 { f,δ C[a,b] + f (r+). C[a,b],δ C[a1,b 1 ] Therefore by applying properties (iii) and (iv) of the of the function f,δ, we obtain H M 7 4 n 1{ f γ + δ ω (f (r), δ) }.,δ Page 3 of 33

24 Finally we shall estimate H 1, choosing a, b satisfying the conditions < a < a < a 1 < b 1 < b < b <. Suppose ψ(t) denotes the characteristic function of the interval [a, b ], then H 1 B (r) n (ψ(t)(f(t) f,δ (t)), ) C[a1,b 1 ] =: H 4 + H 5. + B (r) n ((1 ψ(t))(f(t) f,δ (t)), ) C[a1,b 1 ] Using Lemma.6, it is clear that ( B n (r) ψ(t)(f(t) f,δ (t)), x ) (n + r 1)!(n r)! = p n+r,k (x) k= b n r,k+r (t)ψ(t)(f (r) (t) f (r),δ (t))dt. Hence B (r) n (ψ(t)(f(t) f,δ (t)), ) M f (r) C[a1,b 1 ] 8 f (r),δ C[a,b ] Next for x [a 1, b 1 ] and t [, ) \ [a, b ], we choose a δ 1 > satisfying t x δ 1. Therefore by Lemma.4 and the Cauchy-Schwarz inequality, we have I B (r) n ((1 ψ(t))(f(t) f,δ (t), x). Page 4 of 33

25 and I i+j r i,j n i Q i,j,r (x) {x(1 + x)} r + p n,k (x) k nx j b n,k (t)(1 ψ(t)) f(t) f,δ (t) dt (n + r 1)! (1 + x) n r (1 ψ()) f() f,δ (). (n 1)! For sufficiently large n, the second term tends to zero. Thus I M 9 f γ i+j r i,j M 9 f γ δ m 1 M 9 f γ δ m 1 n i i+j r i,j n i p n,k (x) k nx j t x δ 1 b n,k (t)dt ( b n,k (t)(t x) 4m dt i+j r i,j { p n,k (x) n i { p n,k (x) k nx j ( ) 1 p n,k (x)(k nx) j } 1 b n,k (t)(t x) 4m dt } 1. ) 1 b n,k (t)dt Page 5 of 33

26 Hence by using Lemma.1 and Lemma., we have I M ) 1 f γ δ1 m O (n (i+ j m) M 11 n q f γ, where q = m r. Now choosing m > satisfying q 1, we obtain I M 11 n 1 f γ. Therefore by property (iii) of the function f,δ (t), we get H 1 M f (r) 8 f (r) + M 11 n 1 f γ C[a,b ],δ M 1 ω (f (r), δ, a, b) + M 11 n 1 f γ. Choosing δ = n 1, the theorem follows. Page 6 of 33

27 4. Inverse Theorem This section is devoted to the following inverse theorem in simultaneous approximation: Theorem 4.1. Let < α <, < a 1 < a < b < b 1 < and suppose f C γ [, ). Then in the following statements (i) (ii) (i) B (r) n (f, ) C[a1,b 1 ] = O(n α/ ), (ii) f (r) Lip (α, a, b ), where Lip (α, a, b ) denotes the Zygmund class satisfying ω (f, δ, a, b ) Mδ α. Proof. Let us choose a, a, b, b in such a way that a 1 < a < a < a < b < b < b < b 1. Also suppose g C with suppg [a, b ] and g(x) = 1 on the interval [a, b ]. For x [a, b ] with D d, we have dx B n (r) (fg, x) (fg) (r) (x) = D r (B n ((fg)(t) (fg)(x)), x) = D r (B n (f(t)(g(t) g(x)), x)) + D r (B n (g(x)(f(t) f(x)), x)) =: J 1 + J. Using the Leibniz formula, we have J 1 = r x r W n (x, t)f(t)[g(t) g(x)]dt Page 7 of 33

28 = r ( r ) i i= r 1 = i= =: J 3 + J 4. W (i) n (x, t) r i [f(t)(g(t) g(x))]dt xr i ( r i ) g (r i) (x)b (i) n (f, x) + Applying Theorem 3.4, we have i= W n (r) (x, t)f(t)(g(t) g(x))dt r 1 ( r ) J 3 = g (r i) (x)f (i) (x) + O ( ) n α, i uniformly in x [a, b ]. Applying Theorem 3., the Cauchy-Schwarz inequality, Taylor s expansions of f and g and Lemma., we are led to r g (i) (x)f (r i) (x) ( ) J 4 = r! + o n 1 i!(r i)! i= r ( r ) = g (i) (x)f (r i) (x) + o ( ) n α, i i= uniformly in x [a, b ]. Again using the Leibniz formula, we have r ( r ) J = W n (i) (x, t) r i [g(t)(f(t) f(x))]dt i i= xr i r ( r ) = g (r i) (x)b n (i) (f, x) (fg) (r) (x) i i= Page 8 of 33

29 = r i= = O ( n α ), ( r i ) g (r i) (x)f (i) (x) (fg) (r) (x) + o(n α/ ) uniformly in x [a, b ]. Combining the above estimates, we get B (r) n (fg, ) (fg) (r) = O ( ) C[a n α,b ]. Thus by Lemma.4 and Lemma.5, we have (fg) (r) Lip (α, a, b ) also g(x) = 1 on the interval [a, b ], it proves that f (r) Lip (α, a, b ). This completes the validity of the implication (i) (ii) for the case < α 1. To prove the result for 1 < α < for any interval [a, b ] (a 1, b 1 ), let a, b be such that (a, b ) (a, b ) and (a, b ) (a 1, b 1). Letting δ > we shall prove the assertion α <. From the previous case it implies that f (r) exists and belongs to Lip(1 δ, a 1, b 1). Let g C be such that g(x) = 1 on [a, b ] and supp g (a, b ). Then with χ (t) denoting the characteristic function of the interval [a 1, b 1], we have B (r) n (fg, ) (fg) (r) C[a,b ] D r [B n (g( )(f(t) f( )), )] C[a,b ] =: P 1 + P. + D r [B n (f(t)(g(t) g( )), )] C[a,b ] Page 9 of 33

30 To estimate P 1, by Theorem 3.4, we have P 1 = D r [B n (g( )(f(t), )] (fg) (r) C[a,b ] r ( r ) = g (r i) ( )B n (i) (f, ) (fg) (r) i i= C[a,b ] r ( r ) = g (r i) ( )f (i) (fg) (r) + O(n i i= C[a α/ ),b ] = O ( ) n α. Also by the Leibniz formula and Theorem 3., have r ( r ) P g (r i) ( )B n (f, ) + B n (r) (f(t)(g(t) g( ))χ (t), ) i i= + O(n 1 ) =: P 3 + P 4 C[a,b ] + O(n 1 ). Then by Theorem 3.4, we have r 1 ( r ) P 3 = g (r i) (x)f (i) (x) + O ( ) n α, i i= C[a,b ] Page 3 of 33

31 uniformly in x [a, b ]. Applying Taylor s expansion of f, we have P 4 = = i= W n (r) (x, t)[f(t)(g(t) g(x))χ (t)dt r f (i) (x) i! i= + W (r) n W n (r) (x, t)(t x) i (g(t) g(x))dt (x, t) (f (r) (ξ) f (r) (x)) (t x) r (g(t) g(x))χ (t)dt, r! where ξ lying between t and x. Next by Theorem 3.4, the first term in the above expression is given by r ( r g m) (m) f (r m) (x) + O ( ) n α, m= uniformly in x [a, b ]. Also by mean value theorem and using Lemma.4, we can obtain the second term as follows: W n (r) (x, t) (f (r) (ξ) f (r) (x)) (t x) r (g(t) g(x))χ (t)dt r! C[a,b ] r n m+s Q m,s,r (x) W n (x, t) t x δ+r+1 x(1 + x) m+s r m,s ( ) = O n δ, f (r) (ξ) f (r) (x) g (η) χ (t)dt r! C[a,b ] Page 31 of 33

32 choosing δ such that δ α. Combining the above estimates we get B (r) n (fg, ) (fg) (r) = O ( ) C[a n α,b ]. Since suppfg (a, b ), it follows from Lemma.4 and Lemma.5 that (fg) (r) Liz(α, 1, a, b ). Since g(x) = 1 on [a, b ], we have f (r) Liz(α, 1, a, b ). This completes the proof of the theorem. Remark. As noted in the first section, these operators also reproduce the linear functions so we can easily apply the iterative combinations to the operators (1.1) to improve the order of approximation. Page 3 of 33

33 References [1] H. BERENS AND G.G. LORENTZ, Inverse theorem for Bernstein polynomials, Indiana Univ. Math. J., 1 (197), [] Z. FINTA, On converse approximation theorems, J. Math. Anal. Appl., 31(1) (5), [3] G. FREUD AND V. POPOV, On approximation by Spline functions, Proceeding Conference on Constructive Theory Functions, Budapest (1969), [4] S. GOLDBERG AND V. MEIR, Minimum moduli of ordinary differential operators, Proc. London Math. Soc., 3 (1971), [5] V. GUPTA, A note on modified Baskakov type operators, Approximation Theory Appl., 1(3) (1994), [6] V. GUPTA, M.K. GUPTA AND V. VASISHTHA, Simultaneous approximation by summation-integral type operators, Nonlinear Funct. Anal. Appl., 8(3), (3), [7] V. GUPTA AND M.A. NOOR, Convergence of derivatves for certain mixed Szasz Beta operators, J. Math. Anal. Appl., 31(1) (6), 1 9. [8] N. ISPIR AND I. YUKSEL, On the Bezier variant of Srivastava-Gupta operators, Applied Math. E Notes, 5 (5), [9] H.M. SRIVASTAVA AND V. GUPTA, A certain family of summation integral type operators, Math. Comput. Modelling, 37 (3), Page 33 of 33