Báo cáo hóa học: "Research Article On the Generalized Favard-Kantorovich and Favard-Durrmeyer Operators in Exponential Function Spaces" ppt

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 75142, 12 pages doi:10.1155/2007/75142 Research Article On the Generalized Favard-Kantorovich and Favard-Durrmeyer Operators in Exponential Function Spaces Grzegorz Nowak and Aneta Sikorska-Nowak Received 18 January 2007; Revised 12 June 2007; Accepted 14 November 2007 Recommended by Ulrich Abel We consider the Kantorovich- and the Durrmeyer-type modifications of the generalized Favard operators and we prove an inverse approximation theorem for functions f such that w σ f ∈ L p (R), where 1 ≤ p ≤∞and w σ (x) =ex p(−σx 2 ), σ>0. Copyright © 2007 G. Nowak and A. Sikorska-Nowak. This is an open access article dis- tributed under the Creative Commons Attribution License, which per mits unrestricted use, distribution, and reproduction in any medium, provided the original work is prop- erly cited. 1. Preliminaries Let L p,σ (R) =  f :   w σ f   p < ∞  for 1 ≤ p ≤∞ (1.1) be the weighted function space, where w σ (x) =ex p(−σx 2 ), σ>0, g p =   ∞ −∞   g(x)   p dx  1/p if 1 ≤ p<∞, g ∞ = essup x∈R   g(x)   . (1.2) We define the generalized Favard operators F n for functions f : R→R by F n f (x) = ∞  k=−∞ f (k/n)p n,k (x; γ)(x ∈R, n ∈ N), (1.3) 2 Journal of Inequalities and Applications where N ={1,2, }, p n,k (x; γ) = 1 nγ n √ 2π exp  − 1 2γ 2 n  k n −x  2  (1.4) and γ = (γ n ) ∞ n=1 is a positive sequence convergent to zero (see [1]). In the case where γ 2 n = ϑ/(2n) with a positive constant ϑ, F n become the known Favard operators intro- duced by Favard [2]. Some approximation properties of the classical Favard operators for continuous functions f on R are presented in [3, 4]. Some approximation properties of their generalization can be found, for example, in [1, 5]. Denote by F ∗ n the Kantorovich- type modification of operators F n ,definedby F ∗ n f (x) = n ∞  k=−∞ p n,k (x; γ)  (k+1)/n k/n f (t)dt (x ∈R, n ∈ N), (1.5) and by  F n the Durrmeyer-type modification of operators F n  F n f (x) = n ∞  k=−∞ p n,k (x; γ)  ∞ −∞ p n,k (t;γ) f (t)dt (x ∈R, n ∈ N), (1.6) where f ∈ L p,σ (R). Some estimates concerning the rates of pointwise convergence of the operators F ∗ n f and  F n f can be found in [6, 7]. Recently, several autors investigated the conditions under which global smoothness of a function f , as measured by its modulus of continuity ω( f ; ◦), is retained by the elements of approximating sequences (L n f ) (see, e.g., [8, 9]). For example, Kratz and Stadtm ¨ uller considered in [10]awideclassofdiscreteoperatorsL n and derived estimates of the form ω  L n f ;t  ≤ Kω( f ;t)(t>0), (1.7) with a positive constant K independent of f ,n,andt. For bounded functions f ∈ C(R) and operators F n satisfying γ 2 1 ≥ 1 2 π −2 log2, n 2 γ 2 n ≥ 1 2 π −2 logn if n ≥2, (1.8) they obtained the inequality ω(F n f ;t) ≤ 140ω( f ;t)+16π·tf  (t>0), (1.9) where f =sup{|f (x)| : x ∈ R}. Forbounded functions f ∈ C m (R) ={f : w m f  ∞ < ∞}, w m (x) = (1 + x 2m ) −1 , m ∈ N and for operators F n satisfying nγ 2 n ≥ c>0foralln ∈ N, Pych-Taberska [5]obtainedthe inequality ω 2  F n f ;t  m ≤ K  (1 +t 2 0 )ω 2 ( f ;t) m + t 2 f  m  (0 <t≤ t 0 ) (1.10) for all n ∈ N, n ≥n c where n c ∈ N and K is a constant. G. Nowak and A. Sikorska-Nowak 3 In this paper, we obtained an analogous inequality for the rth weighted modulus of smoothness of the function f ∈ L p,σ (R),σ>0,1 ≤ p ≤∞, ω r ( f ;t) σ,p = sup 0<h≤t   w σ Δ r h f  p (r ∈ N), (1.11) where Δ r h f (x) = r  i=0  r i  (−1) i f  x + h  r/2 −i  . (1.12) Namely, suppose that (γ n ) is a positive null sequence satisfying nγ r/2+1 n ≥ c max n∈N {γ r/2−1 n } > 0foralln ∈ N and σ 1 >σ>0. Then there exist positive constants, K,K 1 , such that for all n ≥K 1 and for arbitrary positive number t 0 ω r  L n f ,t  σ 1 ,p ≤ K  (1 +t 2 0 )ω r ( f ,t) σ,p + t r   w σ f   p   0 <t≤ t 0  , (1.13) where L n denotes the Favard-Kantorovich operator or the Favard-Durrmeyer operator. Throughout the paper, the sy mbols K(σ,σ 1 , ), K j (σ,σ 1 , )(j = 1,2, )willmean some positive constants, not necessarily the same at each occurrence, depending only on the parameters indicated in parentheses. 2. Preliminary results Let γ = (γ n ) ∞ n=1 be a positive sequence and let nγ 2 n ≥ c for all n ∈N, with a positive abso- lute constant c.Asisknown[5], for v ∈ N 0 ={0}∪N, n ∈ N, x ∈ R, ∞  k=−∞     k n −x     v p n,k (x; γ) ≤ 15A c  2 e  v/2  (2v)!γ v n , (2.1) where A c = max {1,(2cπ 2 ) −1 }. A simple calculation and the known Schwarz inequality lead to  ∞ −∞     k n −t     v p n,k (t;γ)dt ≤  (2v)!! γ v n n  k ∈ Z =  0,±1,±2,  . (2.2) Let us choose n ∈ N, j ∈ N 0 and let us write G ∗ n, j f (x) = n ∞  k=−∞ p n,k (x; γ)  k n −x  j  (k+1)/n k/n f (t)dt, (2.3)  G n, j f (x) = n ∞  k=−∞ p n,k (x; γ)  k n −x  j  ∞ −∞ p n,k (t;γ) f (t)dt, (2.4) where f ∈ L p,σ (R), 1 ≤ p ≤∞, σ>0. Obviously, G ∗ n,0 f (x) = F ∗ n f (x)and  G n,0 f (x) =  F n f (x). 4 Journal of Inequalities and Applications Lemma 2.1. Let γ = (γ n ) ∞ n=1 beapositivesequenceconvergentto0andletnγ 2 n ≥ c for all n ∈ N, with a positive absolute constant c.Thenforj ∈N 0 , f ∈ L p,σ (R), σ>0, 1 ≤ p ≤∞, and σ 1 >σ,   w σ 1 G ∗ n, j f   p ≤ 15A c exp  σ 1  σ + σ 1  σ 1 −σ    2 j  !2 j/2 γ j n   w σ f   p (2.5) for all n ∈ N such that γ 2 n ≤ (σ 1 −σ)/(4σ(σ + σ 1 )),   w σ 1  G n, j f   p ≤ 30A c  (2 j)!2 j/2 γ j n   w σ f   p (2.6) for all n ∈ N such that γ n ≤ max{(σ 1 −σ)/(2 √ σ(σ + σ 1 ));( √ σ 1 −σ)/( √ 2(σ + σ 1 ))}. Proof. In view of definition (2.3), exp  − σ 1 x 2  | G ∗ n, j f (x)|≤n ∞  k=−∞ exp  − σ 1 x 2  p n,k (x; γ)     k n −x     j × exp  σ(|k|+1  2 /n 2   (k+1)/n k/n exp  − σt 2  | f (t)|dt. (2.7) Using the inequality (u +v) 2 ≤ σ + σ 1 2σ u 2 + σ + σ 1 σ 1 −σ v 2 (u ∈ R, v ∈ R), (2.8) we can easily observe, that p n,k (x; γ)exp  − σ 1 x 2  exp  σ  k +1 n  2  ≤ √ 2exp  σ 1 (σ + σ 1 ) σ 1 −σ  p n,k  x; √ 2γ  , p n,k (x; γ)exp  − σ 1 x 2  exp  σ  k n  2  ≤ √ 2p n,k  x; √ 2γ  , (2.9) for n ∈ N such that γ 2 n ≤ (σ 1 −σ)/(4σ(σ + σ 1 )) (see [9]), where the symbol √ 2γ means the sequence ( √ 2γ n ) ∞ n=1 . Therefore, exp( −σ 1 x 2 )   G ∗ n, j f (x)   ≤ exp  σ 1  σ + σ 1  σ 1 −σ  √ 2n ∞  k=−∞ p n,k  x; √ 2γ  ×     k n −x     j  (k+1)/n k/n exp  − σt 2  | f (t)|dt. (2.10) From (2.2), we have   w σ 1 G ∗ n, j f   1 ≤ exp  σ 1  σ + σ 1  σ 1 −σ    2 j  !!( √ 2) j+1 γ j n   w σ f   1 . (2.11) G. Nowak and A. Sikorska-Nowak 5 Instead, for p =∞,from(2.1) it follows that   w σ 1 G ∗ n, j f   ∞ ≤ √ 2exp  σ 1  σ + σ 1  σ 1 −σ    w σ f   ∞ essup x∈R  ∞  k=−∞ p n,k  x; √ 2γ      k n −x     j  ≤ 15 √ 2A c exp  σ 1  σ + σ 1  σ 1 −σ   2 j  2 j  !γ j n w σ f  ∞ . (2.12) Finally, by Riesz-Thorin theorem, we have (2.5). In view of definition (2.4) and the inequality p n,k (x; γ)p n,k (t;γ)exp(−σ 1 x 2 )exp(σt 2 ) ≤ 2p n,k (x; √ 2γ)p n,k (t; √ 2γ), (2.13) for n ∈ N such that γ n ≤ max{(σ 1 −σ)/(2 √ σ(σ + σ 1 )); √ σ 1 −σ/( √ 2(σ + σ 1 ))} (see [6]), we have exp  − σ 1 x 2     G n, j f (x)   ≤ 2n ∞  k=−∞ p n,k  x; √ 2γ      k n −x     j  ∞ −∞ exp  − σt 2  p n,k  √ 2t;γ    f (t)   dt. (2.14) Applying (2.1)and(2.2), we get w σ 1  G n, j f  1 ≤ 30A c  (2 j)!!γ j n 2 j/2 w σ f  1 , w σ 1  G n, j f  ∞ ≤ 30A c  (2 j)!γ j n 2 j/2 w σ f  ∞ . (2.15) Finally, by Riesz-Thorin theorem, we have (2.6). Further , for δ>0, x ∈ R,andr ∈N we define Stieklov function of f f (δ,2r) (x) = 1 δ 2r 2  2r r   δ/2 −δ/2 ···  δ/2 −δ/2 r  i=1  2r r −i  (−1) i−1 f  x + i  t 1 + ···+ t 2r  dt 1 ···dt 2r . (2.16)  Lemma 2.2. For all r = 1,2, , 0 <δ≤1, σ 1 >σ>0, 1 ≤ p ≤∞,andx ∈ R,   w σ 1 f (r) (δ,2r)   p ≤ K  r,σ,σ 1  1 δ r ω r ( f ;δ) σ,p , (2.17)   w σ 1  f (δ,2r) − f    p ≤ K  r,σ,σ 1  ω r ( f ;δ) σ,p . (2.18) Proof. It is easy to see by induction that f (r) (δ,2r) (x) = 2  2r r  r  i=1 (−1) i−1  2r r −i  1 (iδ) 2r ×  iδ/2 −iδ/2 ···  iδ/2 −iδ/2 Δ r iδ f  x + u 1 + ···+ u r  du 1 ···du r . (2.19) 6 Journal of Inequalities and Applications Let σ 2 = (2σ 1 + σ)/3. In view of the inequality exp  − σ 1 x 2 + σ 2 (x + u) 2  ≤ exp  σ 2 σ 1 σ 1 −σ 2 u 2  , (2.20) where 0 <δ ≤ 1andu = u 1 + ···+ u r ,(u ≤r 2 /2), we have   w σ 1 f (r) (δ,2r)   p ≤ 2  2r r  exp  σ 1 σ 2 σ 1 −σ 2 r 4 4  r  i=1  2r r −i  1 (iδ) r w σ 2 Δ r iδ f  p . (2.21) Applying the Minkowski inequality and the fact that for 0 ≤ l i ≤ i −1(0≤i ≤ r), 0 <h≤ 1, exp  − σ 2 x 2 + σ  x + h  l 1 + ···+ l r − r(i −1) 2  2  ≤ exp  σσ 2 σ 2 −σ  r(i −1) 2  2  , (2.22) we obtain   w σ 2 Δ r iδ f   p =sup |h|≤δ   ∞ −∞     exp  − σ 2 x 2  i−1  l 1 =0 ··· i−1  l r =0 Δ r h f  x+h  l 1 + ···+ l r − r(i −1) 2      p dx  1/p ≤ exp  σσ 2 σ 2 −σ r 2 (i −1) 2 4  i r ω r ( f ;δ) σ,p . (2.23) So (2.17) is evident. It is easy to see that f (δ,2r) (x) − f (x) = (−1) r−1 δ 2r 1  2r r   δ/2 −δ/2 ···  δ/2 −δ/2 Δ 2r t 1 +···+t 2r f (x)dt 1 ···dt 2r . (2.24) By Minkowski inequality, for 1 ≤ p ≤∞,wehave(2.18).  Lemma 2.3. Suppose that γ = (γ n ) ∞ n=1 is a positive sequence convergent to 0 and that nγ r/2+1 n ≥ cK(r),wherer ∈ N, r ≥ 2, K(r) = max n∈N {γ r/2−1 n }, c is a positive absolute con- stant and let a r = 1 for even r and a r = 2 for odd r.Thenfor f ∈L p,σ (R), σ>0, 1 ≤ p ≤∞ and σ 1 >σ,wehave   w σ 1  F ∗ n f  (r) −(n/a r ) r F ∗ n Δ r a r /n f    p ≤ K(σ,σ 1 ,c,r)   w σ f   p (2.25) for all n ∈ N such that γ 2 n ≤ (σ 1 −σ)/(4σ(σ + σ 1 )) and nγ n > 4a 2 r r 2 ,and   w σ 1   F n f  (r) −(n/a r ) r  F n Δ r a r /n f    p ≤ K(σ,σ 1 ,c,r)   w σ f   p (2.26) for all n ∈ N such that γ n ≤ max{(σ 1 − σ)/(2 √ σ(σ + σ 1 )); √ σ 1 −σ/( √ 2(σ + σ 1 ))} and nγ n >r 2 /4. G. Nowak and A. Sikorska-Nowak 7 Proof. We consider an even r.Letr = 2r 1 , r 1 ∈ N, x ∈ R.Then n r F ∗ n  Δ r 1/n f (x)  = n 2r 1 +1 ∞  k=−∞ p n,k (x; γ) 2r 1  i=0  2r 1 i  (−1) i  (k+r 1 −i+1)/n (k+r 1 −i)/n f (t)dt = n 2r 1 +1 r 1 −1  i=0  2r 1 i  (−1) i × ∞  k=−∞  p n,k−(r 1 −i) (x; γ)+p n,k+(r 1 −i) (x; γ)   (k+1)/n k/n f (t)dt + n 2r 1 +1 ∞  k=−∞ ⎛ ⎝ 2r 1 r 1 ⎞ ⎠ (−1) r 1 p n,k (x; γ)  (k+1)/n k/n f (t)dt. (2.27) It is easy to see that p n,k−(r 1 −i) (x; γ)+p n,k+(r 1 −i) (x; γ) =p n,k (x; γ)  exp  r 1 −i nγ 2 n  k n −x  − (r 1 −i) 2 2n 2 γ 2 n  +exp  − r 1 −i nγ 2 n  k n −x  − (r 1 −i) 2 2n 2 γ 2 n  = p n,k (x; γ) ∞  l=1 (−1) l l! [l/2]  j=0  l 2 j  2 2j+1−l  k n −x  2j n 2j−2l γ −2l n (r 1 −i) 2l−2 j +2p n,k (x; γ). (2.28) Consequently, using definition (2.3), we get n r F ∗ n  Δ r 1/n f (x)  = 2r 1  l=1 [l/2]  j=0 n 2(r 1 +j−l) γ −2l n (−1) l 2 2j+1−l  2 j  !  l −2 j  ! × r 1 −1  i=0  2r 1 i  (−1) i  r 1 −i  2l−2 j G ∗ n,2j f (x) + ∞  l=2r 1 +1 [l/2]  j=0 n 2(r 1 +j−l) γ −2l n (−1) l 2 2j+1−l  2 j  !  l −2 j  ! × r 1 −1  i=0  2r 1 i  (−1) i  r 1 −i  2l−2 j G ∗ n,2j f (x) = S n,1 f (x)+S n,2 f (x). (2.29) 8 Journal of Inequalities and Applications In view of (2.5) and using Stirling formula, we obtain   w σ 1 S n,2 f   p ≤ K 1  σ,σ 1 ,c    w σ f   p 4 r 1 n 2r 1 ∞  l=2r 1 +1 r 2l 1 n 2l γ 2l n 2 l [l/2]  j=0  (4 j  !2 j  2 j  !  l −2 j  ! n 2j γ 2j n 4 j r −2j 1 ≤ K 2  σ,σ 1 ,c,r    w σ f   p n 2r 1 ∞  l=2r 1 +1  r 2 1 /2) l (n 2 γ 2 n ) l [l/2]  j=0  n 2 γ 2 n  j 64 j ≤ K 3  σ,σ 1 ,c,r    w σ f   p   16r 2 1  2r 1 +1 n 2 γ 2r 1 +2 n + n 2r 1 ∞  l=2r 1 +2  16r 2 1 nγ n  l  . (2.30) Assuming (16r 2 1 )/(nγ n ) < 1 and using the condition nγ r 1 +1 n ≥ cK(r), we get w σ 1 S n,2 f  p ≤ K 4 (σ,σ 1 ,c,r)w σ f  p . (2.31) Now observe that r 1 −1  i=0  2r 1 i  (−1) i  r 1 −i  2s = ⎧ ⎨ ⎩ 0if0<s<r 1 , (2r 1 )!/2ifs = r 1 . (2.32) The equality follows simply from properties of finite differences since the left-hand side of the equation is a half of the finite difference of the polynomial (r 1 −x) 2s . Therefore, S n,1 f (x) = 2r 1  l=r 1 (−1) l 2 2j+1−l l!n 2l−2 j−2r 1 γ 2l n  l 2 j  r 1 −1  i=0  2r 1 i  (−1) i  r 1 −i  2l−2 j G ∗ n,2j f (x) = r 1  l=0 l −1  j=0 (−1) r 1 +l 2 2j+1−l−r 1 (r 1 + l)!n 2l−2 j γ 2l+2r 1 n  r 1 + l 2 j  r 1 −1  i=0  2r 1 i  (−1) i  r 1 −i  2r 1 +2l−2 j G ∗ n,2j f (x) + r 1  l=0 (−1) 2r 1 −l γ 4r 1 −2l n  2r 1  ! 2 l l!  2r 1 −2l  ! G ∗ n,2j f (x). (2.33) It is easy to see, by the method of induction, that p (v) n,k (x; γ) = p n,k (x; γ) [v/2]  i=0 v!(−1) i (v −2i)!(2i)!! 1 γ 2v−2i n  k n −x  v−2i , v ∈ N. (2.34) Therefore, S n,1 f (x) = r 1  l=0 l −1  j=0 (−1) r 1 +l 2 2j+1−l−r 1  r 1 + l  !n 2l−2 j γ 2l+2r 1 n  r 1 + l 2 j  r 1 −1  i=0  2r 1 i  (−1) i  r 1 −i  2r 1 +2l−2 j G ∗ n,2j f (x) +  F ∗ n f (x)  (2r 1 ) . (2.35) G. Nowak and A. Sikorska-Nowak 9 Consequently, from (2.29)   (F ∗ n f ) (2r 1 ) (x) −n 2r 1 F ∗ n Δ 2r 1 1/n f (x)   ≤ K 5 (r) r 1 −1  j=0 r 1  l=j+1 n 2j (nγ n ) 2l γ 2r 1 n   G ∗ n,2j f (x)   +   S n,2 f (x)   . (2.36) The condition nγ r 1 +1 n ≥ cK(r) and the boundedness of the sequence (γ n )leadto    F ∗ n f  (2r 1 ) (x) −n 2r 1 F ∗ n Δ 2r 1 1/n f (x)   ≤ K 6 (r,c) r 1 −1  j=0 γ −2j n   G ∗ n,2j f (x)|+   S n,2 f (x)   . (2.37) Collecting the results we get estimate (2.25)forevenr, immediately. Now, we will prove inequality (2.25)foroddr.Namely,letr = 2r 2 +1,r 2 ∈ N, x ∈R. Then n r F ∗ n  Δ r 2/n f (x)  = n 2r 2 +2 r 2  i=0 ∞  k=−∞  2r 2 +1 i  (−1) i ×  p n,k−(2r 2 +1−2i) (x; γ) − p n,k+(2r 2 +1−2i) (x; γ)   (k+1)/n k/n f (t)dt. (2.38) It is easy to see that p n,k−(2r 2 +1−2i) (x; γ) − p n,k+(2r 2 +1−2i) (x; γ) = p n,k (x; γ) ∞  l=1 (−1) l+1 l! [(l−1)/2]  j=0  l 2 j +1  2 2j+2−l  k n −x  2j+1 n 2j+1−2l γ 2l n  2r 2 +1−2i  2j−2l+1 . (2.39) Consequently, n r F ∗ n (Δ r 2/n f (x)) = 2r 2 +1  l=1 n 2r 2 +2 [(l −1)/2]  j=0 n 2j−2l γ −2l n (−1) l+1 2 2j+2−l (2 j +1)!(l −2 j −1)! × r 2  i=0  2r 2 +1 i  (−1) i (2r 2 +1−2i) 2l−2 j−1 G ∗ n,2j+1 f (x) + ∞  l=2r 2 +2 n 2r 2 +2 [(l −1)/2]  j=0 n 2j−2l γ −2l n (−1) l+1 2 2j+2−l (2 j +1)!(l −2 j −1)! × r 2  i=0  2r 2 +1 i  (−1) i (2r 2 +1−2i) 2l−2 j−1 G ∗ n,2j+1 f (x) = S ∗ n,1 f (x)+S ∗ n,2 f (x). (2.40) Some simple calculation, Stirling formula and (2.5)give   w σ 1 S ∗ n,2 f   p ≤ K 7 (σ,σ 1 ,c,r)   w σ f   p (2.41) 10 Journal of Inequalities and Applications for n ∈ N such that (16r 2 )/(nγ n ) < 1.Next,inviewof(2.25) and the equality r 2  i=0  2r 2 +1 i  (−1) i  r 2 −i+1/2  2s−1 = ⎧ ⎨ ⎩ 0if0<s<r 2 +1,  2r 2 +1  !/2ifs = r 2 +1 (2.42) we obtain S ∗ n,1 f (x) = r 2  l=0 l −1  j=0 (−1) r 2 +l 2 2j+1−l−r 2 (2 j +1)!  l + r 2 −2j  ! n 2j−2l γ −2l−2r 2 −2 n × r 2  i=0  2r 2 +1 i  (−1) i  2r 2 +1−2i  2r 2 +2l−2 j+1 ×G ∗ n,2j+1 f (x)+2 2r 2 +1 (F ∗ n f ) (2r 2 +1) (x) . (2.43) Using (2.40) and the condition nγ r 2 +3/2 n ≥ cK(r), we have   (F ∗ n f ) (2r 2 +1) (x) −(n/2) 2r 2 +1 F ∗ n Δ 2r 2 +1 2/n f (x)   ≤ K 8 (r,c) r 2 −1  j=0 1 γ 2j+1 n   G ∗ n,2j+1 f (x)   +   S ∗ n,2 f (x)   . (2.44) Applying (2.5), we get (2.25)foroddr. Therefore, inequality (2.25)isproved. Now we will prove (2.26). Let r = 2r 1 , r 1 ∈ N. A simple calculation and the equality p n,k (t −(r 1 −i)/n;γ) = p n,k+r 1 −i (t;γ)give n r  F n  Δ r 1/n f (x)  = n 2r 1 +1 r 1 −1  i=0 ∞  k=−∞  2r 1 i  (−1) i  p n,k−(r 1 −i) (x; γ)+p n,k+(r 1 −i) (x; γ)  ×  ∞ −∞ p n,k (t;γ) f (t)dt + n 2r 1 +1 ∞  k=−∞  2r 1 r 1  (−1) i p n,k (x; γ) ×  ∞ −∞ p n,k (t;γ) f (t)dt. (2.45) The estimate (2.26) follows now the same way as ( 2.25).  3. Main result Theorem 3.1. Suppose that r ∈ N, (γ n ) is a positive null sequence satisfying nγ r/2+1 n ≥ cK(r) for all n ∈ N with some c>0 where K(r) = max n∈N {γ r/2−1 n }. Then there exists a constant K>0, such that for all f ∈ L p,σ (R), σ 1 >σ>0, 1 ≤ p ≤∞,andforanarbitrary positive number t 0 , ω r  F ∗ n f ,t  σ 1 ,p ≤ K  σ,σ 1 ,r,c  1+t 2 0  ω r ( f ,t) σ,p + t r   w σ f   p  0 <t≤ t 0  (3.1) for all n ∈ N such that γ 2 n ≤ (σ 1 −σ)/(4σ(σ + σ 1 )) and nγ n > 16r 2 ,and ω r   F n f ,t  σ 1 ,p ≤ K  σ,σ 1 ,r,c  1+t 2 0  ω r ( f ,t) σ,p + t r   w σ f   p  0 <t≤ t 0  (3.2) [...]... Commentarii Mathematici, vol 29, pp 103–112, 2001 [8] G A Anastassiou, C Cottin, and H H Gonska, “Global smoothness of approximating functions,” Analysis, vol 11, no 1, pp 43–57, 1991 [9] G Nowak, “Direct theorems for generalized Favard-Kantorovich and Favard-Durrmeyer operators in exponential function spaces,” to appear in Ukrainian Mathematical Journal [10] W Kratz and U Stadtm¨ ller, On the uniform... Pych-Taberska, On the generalized Favard operators, ” Functiones et Approximatio Commentarii Mathematici, vol 26, pp 265–273, 1998 [6] P Pych-Taberska and G Nowak, “Approximation properties of the generalized FavardKantorovich operators, ” Commentationes Mathematicae, vol 39, pp 139–152, 1999 [7] G Nowak and P Pych-Taberska, “Some properties of the generalized Favard-Durrmeyer operators, ” Functiones et Approximatio... Consequently by (2.17), (2.18) and assuming now 0 < t ≤ t0 , we have ∗ ωr Fn f ,t σ 1 ,p ≤ K σ,σ 1 ,r,c r 1 + t0 ωr ( f ;t)σ,p + t r wσ f p (3.10) On the same way we can prove (3.2) for Fn f , using (2.6) and (2.26) References [1] W Gawronski and U Stadtm¨ ller, “Approximation of continuous functions by generalized u Favard operators, ” Journal of Approximation Theory, vol 34, no 4, pp 384–396, 1982... “Sur les multiplicateurs d’interpolation,” Journal de Math´matiques Pures et Appliqu´es, e e vol 23, pp 219–247, 1944 [3] M Becker, P L Butzer, and R J Nessel, “Saturation for Favard operators in weighted function spaces,” Studia Mathematica, vol 59, no 2, pp 139–153, 1976 [4] M Becker, “Inverse theorems for Favard operators in polynomial weight spaces,” Commentationes Mathematicae, vol 22, no 2, pp... modulus of continuity of certain discrete approxu imation operators, ” Journal of Approximation Theory, vol 54, no 3, pp 326–337, 1988 Grzegorz Nowak: Higher School of Marketing and Management, Ostroroga 9a, 64-100 Leszno, Poland Email address: grzegnow@amu.edu.pl Aneta Sikorska-Nowak: Faculty of Mathematics and Computer Science, ´ Adam Mickiewicz University, Umultowska 87, 61-614 Poznan, Poland Email...G Nowak and A Sikorska-Nowak 11 √ √ √ for all n ∈ N such that γn ≤ max {(σ 1 − σ)/(2 σ(σ + σ 1 )); σ 1 − σ/( 2(σ + σ 1 ))} and nγn > r 2 /4 Proof Let σ 2 = (3σ 1 + σ)/4 In view of the inequality exp − σ 1 x2 + σ 2 (x + u)2 ≤ exp σ 2σ 1 2 u σ1 − σ2 (u ∈ R) (3.3) and the generalized Minkowski inequality it is easy to see that for 0 < h ≤ 1 wσ 1 Δr f h... (2.25), and (3.4) Let (r) p ∗ Fn f(δ,2r) ≤ K σ 2 ,σ 3 ,r,c (r) ∗ − n/ar )Fn Δr r /n f(δ,2r) a wσ 3 f(δ,2r) p (r) + wσ 3 f(δ,2r) p p ∗ + nr wσ 2 Fn Δr r /n f(δ,2r) a p (3.8) Using (2.5) for j = 0 and (3.7) we have ∗ ωr Fn f ,t σ 1 ,p ≤ K σ,σ 1 ,r,c wσ 3 ( f − f(δ,2r) ) p (r) + t r wσ 3 f(δ,2r) p + t r wσ 3 f(δ,2r) p (3.9) 12 Journal of Inequalities and Applications Consequently by (2.17), (2.18) and assuming... 2 f (3.5) p Applying these inequalities, we get wσ 1 Δr f h p ≤ wσ 1 Δr f − f(δ,2r) h r2 σ 2σ 1 ≤ 2r exp 4 σ1 − σ2 + wσ 1 Δr f(δ,2r) h p wσ 2 ( f − f(δ,2r) ) p p (r) + hr wσ 2 f(δ,2r) p (3.6) , where f(δ,2r) (x) (δ > 0, x ∈ R, r ∈ N) is defined by (2.16) ∗ Hence, applying this inequality for Fn f we have ∗ ωr Fn f ,t σ 1 ,p ≤ 2r exp ∗ Hence, wσ 2 (Fn f(δ,2r) )(r) σ 3 = (2σ 1 + σ)/3, then ∗ wσ 2 Fn f(δ,2r) . oper- ators in exponential function spaces,” to appear in Ukrainian Mathematical Journal. [10] W. Kratz and U. Stadtm ¨ uller, On the uniform modulus of continuity of certain discrete approx- imation operators, ”. by Ulrich Abel We consider the Kantorovich- and the Durrmeyer-type modifications of the generalized Favard operators and we prove an inverse approximation theorem for functions f such that w σ f. approximation properties of the classical Favard operators for continuous functions f on R are presented in [3, 4]. Some approximation properties of their generalization can be found, for example, in