Applied Mathematics and Computation 244 (2014) 683–694
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Pointwise approximation by Bernstein type operators in mobile interval Hee Sun Jung a, Naokant Deo b,⇑, Minakshi Dhamija b a b
Department of Mathematics Education, Sungkyunkwan University, Seoul 110-745, Republic of Korea Department of Applied Mathematics, Delhi Technological University (Formerly Delhi College of Engineering), Bawana Road, Delhi 110042, India
a r t i c l e
i n f o
a b s t r a c t In the present paper, we study pointwise approximation by Bernstein–Durrmeyer type h i 1 , with use of Peetre’s K-functional and operators in the mobile interval x 2 0; 1 nþ1
Keywords: Bernstein operators Pointwise approximation Rate of convergence
x2uk ðf ; tÞ ð0 6 k 6 1Þ, we give its properties and obtain the direct and inverse theorems for these operators. Ó 2014 Published by Elsevier Inc.
1. Introduction and axillary results ~ n , were introduced and studied by Deo et al. [6] and defined as: In the year 2008, the operators B
~ n ðf ; xÞ ¼ B
n X k ; pn;k ðxÞf n k¼0
ð1:1Þ
where
n nk n k 1 1 x 1 pn;k ðxÞ ¼ 1 þ x n nþ1 k
ð1:2Þ
h i 1 . If n is sufficient large then operators (1.1) convert in the classical Bernstein operators: and x 2 0; 1 nþ1
Bn ðf ; xÞ ¼
n X n k k x ð1 xÞnk f : n k k¼0
ð1:3Þ
Now we consider Durrmeyer type operators
V n ðf ; xÞ ¼
n ðn þ 1Þ2 X pn;k ðxÞ n k¼0
Z
n nþ1
pn;k ðtÞf ðtÞdt:
ð1:4Þ
0
In 2003, a very interesting general sequence of linear positive operators was introduced by Srivastava and Gupta [18] and investigated as well as estimated the rate of convergence. Then the faster rate of convergence was studied by Deo [2] for ⇑ Corresponding author. E-mail addresses:
[email protected] (H.S. Jung),
[email protected] (N. Deo),
[email protected] (M. Dhamija). http://dx.doi.org/10.1016/j.amc.2014.07.034 0096-3003/Ó 2014 Published by Elsevier Inc.
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these operators and similar type modification have given and studied simultaneous approximation for these operators (1.1) in [4]. Ditzian [8] used x2uk ðf ; tÞ and gave an interesting direct estimate for the Bernstein polynomials. Felten [10] studied local and global approximation theorems for positive linear operators, later on similar type results studied for Durrmeyer operators by Guo et al. [15] and gave the direct and inverse theorems for pointwise approximation by Bernstein–Durrmeyer operators via Ditzian–Totik moduli x2uk ðf ; tÞ. In the year 1998, Guo et al. [14] studied pointwise estimate for Szász–Durrmeyer operators with the help of Ditzian–Totik modulus of smoothness xruk ðf ; tÞ in the interval ½0; 1Þ. Deo [3] studied pointwise estimate for modified Baskakov type operator. Recently Abel et al. [1] used properties of the Jacobi polynomials in order to give a new proof of geometric series of Bernstein operators. Very recently, Gupta and Agarwal [12] have given last two decades, literature on positive linear operators in their book and some interesting results were given by researchers [5,11,13,17] on approximation operators. In a similar manner, in this research work, we give the direct and the inverse theorem for pointwise approximation by h i 1 Bernstein type operators by Ditzian–Totik modulus of smoothness x2uk ðf ; tÞ in the mobile interval 0; 1 nþ1 . h i h i 1 1 be the set of continuous and bounded functions on 0; 1 nþ1 and First we give some notations. Let C 0; 1 nþ1
2 sup Dhuk f ðxÞ; 0 0 such that c1 y 6 x 6 cy). Now we give some basic properties of the operators (1.1) as follow: h i 1 ~ n verify the following: and n 2 N. The operators B Lemma 1.1. Let ei ðtÞ ¼ t i ; i ¼ 0; 1; 2, then for x 2 0; 1 nþ1
~ n ðe0 ; xÞ ¼ 1; B
ð1:9Þ
~ n ðe1 ; xÞ ¼ x þ 1 x; B n
ð1:10Þ
2 ~ n ðe2 ; xÞ ¼ ðn þ 1Þ ðn 1Þ x2 þ n þ 1 x: B n3 n2
ð1:11Þ
We give next Lemma along the line of proposition 1:2 p. 326 of Derriennic [7]. h i 1 and n 2 N we have Lemma 1.2 [7]. Let es ðtÞ ¼ t s ; s ¼ 0; 1; 2; . . . with the properties s 6 n, then for x 2 0; 1 nþ1
ðV n es ÞðxÞ ¼
s X s s! ðn!Þ2 nsr xr : ðn þ s þ 1Þ! r¼0 r r! ðn rÞ!ðn þ 1Þsr1
ð1:12Þ
Proof. In the account of (1.4), we obtain
Z 0
n nþ1
n Z n nk n nþsþ1 nþ1 n n 1 1 1 n pn;k ðtÞts dt ¼ 1 þ t kþs 1 t dt ¼ 1 þ bðk þ s þ 1; n k þ 1Þ n nþ1 n nþ1 k k 0 sþ1 n n! ðk þ sÞ! ¼ : nþ1 ðn þ s þ 1Þ! k!
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685
Therefore from (1.4), we obtain
ns
ðV n es ÞðxÞ ¼
s1
ðn þ 1Þ
n X n! ðk þ sÞ! p ðxÞ : ðn þ s þ 1Þ! k¼0 n;k k!
ð1:13Þ
For any x; y 2 R and any s; n 2 N satisfying the inequality s 6 n we have n
X n k nk ðk þ sÞ! @s s xy x ðx þ yÞn ¼ : s k! @x k k¼0
ð1:14Þ
By Leibnitz formula, the left side of (1.14), we have s sr s r X X s @ s s! @s s n! n n s @ x ðx þ yÞ x ðx þ yÞ ¼ xr ðx þ yÞnr : ¼ @xs r @xsr @xr r r! ðn rÞ! r¼0 r¼0
ð1:15Þ
1 From (1.14) and (1.15), with y ¼ 1 nþ1 x s X s s! r¼0
r
nr n X n n! n n ðk þ sÞ! xr ¼ p ðxÞ : r! ðn rÞ! nþ1 n þ 1 k¼0 n;k k!
ð1:16Þ
Now from (1.13), with the help of (1.16), we obtain
ðV n es ÞðxÞ ¼
s X s s! ðn!Þ2 nsr xr : ðn þ s þ 1Þ! r¼0 r r! ðn rÞ!ðn þ 1Þsr1
h i 1 and n 2 N. The operators V n verify the following: Lemma 1.3. Let ei ðtÞ ¼ ti ; i ¼ 0; 1; 2, then for x 2 0; 1 nþ1
V n ðe0 ; xÞ ¼ 1; V n ðe1 ; xÞ ¼
V n ðe2 ; xÞ ¼
ð1:17Þ
n n þ x; ðn þ 1Þðn þ 2Þ ðn þ 2Þ 2n2 2
ðn þ 1Þ ðn þ 2Þðn þ 3Þ
þ
ð1:18Þ
4n2 nðn 1Þ xþ x2 : ðn þ 2Þðn þ 3Þ ðn þ 1Þðn þ 2Þðn þ 3Þ
Proof. From (1.16), we obtain these assertions for s ¼ 0; 1; 2.
ð1:19Þ
h
h i 1 , we have Lemma 1.4. For x 2 0; 1 nþ1
V n ððe1 xÞ; xÞ ¼
n 2ðn þ 1Þx ; ðn þ 1Þðn þ 2Þ
ð1:20Þ
h i 2 n2 þ ðn þ 1Þ2 ðn 3Þu2 ðxÞ 2nðn 3Þ 2ðn 3Þ V n ððe1 xÞ2 ; xÞ ¼ x x2 ¼ : þ ðn þ 2Þðn þ 3Þ ðn þ 1Þ2 ðn þ 2Þðn þ 3Þ ðn þ 1Þ2 ðn þ 2Þðn þ 3Þ ðn þ 1Þðn þ 2Þðn þ 3Þ 2n2
ð1:21Þ Proof. From Lemma 1.3, we obtain these assertions.
h
2. Direct estimates In this section we give rate of convergence. h i 1 Let f 2 C 0; 1 nþ1 , the modulus of continuity of f, defined as:
xðf ; dÞ ¼
sup
jf ðtÞ f ðxÞj:
ð2:1Þ
1 ;jtxj6d x;t2½0;1nþ1
In the year 1990, Lenze [16] introduced the Lipschitz type maximal functions of order a as follows:
^ a ðf ; dÞ ¼ x
lim 1 t–x;t2½0;1nþ1
jf ðtÞ f ðxÞj ; j t xj a
x 2 0; 1
1 ; 0 < a 6 1: nþ1
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^ a ðf ; dÞ is equivalent to f 2 LipM ðaÞ. The remarkable is that the roundedness of x h i 1 ; V n ðf ; xÞ be given by (1.4), then we have Theorem 2.1. Let f 2 C 0; 1 nþ1
pffiffiffi pffiffiffiffiffi kV n ðf ; :Þ f k 6 1 þ b x f ; kn ;
where b ¼
ð2:2Þ
pffiffiffiffiffiffi ffi n . nþ1
Proof. By Popoviciu’s technique we can write
jf ðtÞ f ðxÞj 6 xðf ; dÞ
j t xj þ1 ; d
for any d > 0:
ð2:3Þ
h i 1 , we have Using positivity and linearity properties of the operators V n ðf ; xÞ and for n 2 N and x 2 0; 1 nþ1
jV n ðf ; xÞ f ðxÞj 6 V n ðjf ðtÞ f ðxÞj; xÞ;
ð2:4Þ
From (2.3) and (2.4), we get
jV n ðf ; xÞ f ðxÞj 6 xðf ; dÞ
V n ðjt xj; xÞ þ1 ; d
ð2:5Þ
Now using Cauchy–Schwartz inequality and from Lemma 1.4, we get
jV n ðf ; xÞ f ðxÞj 6 xðf ; dÞ 6 xðf ; dÞ If we choose d ¼
1=2 ðV n ððtxÞ2 ;xÞÞ d
pffiffiffiffiffi kn x d
þ1 ; ð2:6Þ
þ1 :
h i pffiffiffiffiffi 1 , we get the required result (2.2). h kn in (2.6) and take maximum over x 2 0; 1 nþ1
Motivated from Guo et al. [15], now we give direct theorem. h i 1 ; 0 6 k 6 1, then we have Theorem 2.2. For f 2 C 0; 1 nþ1
2 jV n ðf ; xÞ f ðxÞj 6 C x2uk f ; n1=2 d1k n ðxÞ þ x f ; 1=n :
ð2:7Þ
Proof. From Lemma 1.4, we have
2
V n ððt xÞ ; xÞ ¼
h i 2 n2 þ ðn þ 1Þ2 ðn 3Þu2 ðxÞ ðn þ 1Þ2 ðn þ 2Þðn þ 3Þ
6 Cn1 d2n ðxÞ:
Let
n 2ðn þ 1Þx Mn ðf ; xÞ ¼ f ðxÞ f x þ ðn þ 1Þðn þ 2Þ and
Ln ðf ; xÞ ¼ V n ðf ; xÞ þ M n ðf ; xÞ then we obtain
jn þ 2ðn þ 1Þxj 6 x ðf ; 1=nÞ: jM n ðf ; xÞj 6 x f ; ðn þ 1Þðn þ 2Þ Ln ð1; xÞ ¼ 1; Ln ððt xÞ; xÞ ¼ 0; Ln ðt xÞ2 ; x 6 Cn1 d2n ðxÞ and kLn k 6 3: 2 , for a fixed x and k, we get From (1.7) and (1.8), we choose g ¼ g n;x;k 2 D k
kf g k 6 C x2uk f ; n1=2 dn1k ðxÞ ;
2 2k 00
u g 6 C x2 k f ; n1=2 d1k ðxÞ ; n1=2 d1k n ðxÞ n u
1=2 1k 4=ð2kÞ 00
dn ðxÞ n kg k 6 C x2uk f ; n1=2 d1k n ðxÞ :
ð2:8Þ
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687
From ([9], p. 141), for t < u < x, we have
jt uj
u2k ðuÞ
6
j t xj
u2k ðxÞ
jt uj
and
d2k n ðuÞ
6
jt xj d2k n ðxÞ
ð2:9Þ
:
So that
jLn ðf ; xÞ f ðxÞj 6 jLn ððf gÞ; xÞj þ jf ðxÞ gðxÞj þ jLn ðg; xÞ gðxÞj 6 4kf g k þ jLn ðg; xÞ gðxÞj Using Taylor expansion with integral reminder
f ðtÞ ¼ f ðxÞ þ f 0 ðxÞðt xÞ þ
Z
t
ðt uÞf 00 ðuÞdu
ð2:10Þ
x
(see [9], p. 134) and by (2.8) and (2.9), we have
Z t Z t ðt uÞg 00 ðuÞdu; x 6 V n ðt uÞgðuÞdu; x jLn ðg; xÞ gðxÞj ¼ Ln x
x
Z nþ2ðnþ1Þx ! xþðnþ1Þðnþ2Þ n þ 2ðn þ 1Þx ðt xÞ2 00 þ u gðuÞdu 6 C d2k xþ g ; x V n n x ðn þ 1Þðn þ 2Þ d2k n ðxÞ
2k 00 n þ 2ðn þ 1Þx 2 2 2k 00 d g þ n2 d2k ðxÞd2k g 00 6 C n1=2 d1k þ d2k n ðxÞ dn g n ðxÞ n n n ðn þ 1Þðn þ 2Þ
2 00 : 6 C n1=2 dn1k ðxÞ d2k n g
h i 1 k ; d2k For x 2 0; 1 nþ1 n ¼ 1=n and from (1.8) we get
ð2:11Þ
2 2
n1=2 dn1k ðxÞ u2k g 00 6 CK uk f ; n1=2 d1k x2uk f ; n1=2 d1k n ðxÞ n ðxÞ :
From (2.11), we have
jLn ðf ; xÞ f ðxÞj 6 4kf g k þ Cn1 dn2ð1kÞ ðxÞu2k g 00 6 C kf g k þ n1 dn2ð1kÞ ðxÞu2k g 00 6 C x2uk f ; n1=2 d1k n ðxÞ : h i 1 , we obtain Hence, for f 2 C 0; 1 nþ1
2 jV n ðf ; xÞ f ðxÞj 6 jLn ðf ; xÞ f ðxÞj þ jLn ðf ; xÞj 6 C x2uk f ; n1=2 d1k n ðxÞ þ x f ; 1=n :
ð2:12Þ
This completes the proof. h
3. Inverse theorem In this section we give inverse theorem. h i 1 2 ; 0 < a < 2k Theorem 3.1. For f 2 C 0; 1 nþ1 ; 0 6 k 6 1, then the following statements are equivalent:
a ; jV n ðf ; xÞ f ðxÞj ¼ O n1=2 d1k n ðxÞ
ð3:1Þ
x2uk ðf ; tÞ ¼ Oðta Þ and x ðf ; tÞ ¼ O tað1k=2Þ ; pffiffiffi pffiffiffi where dn ðxÞ ¼ uðxÞ þ 1= n max uðxÞ; 1= n . To prove the inverse theorem we need the following notations. Let us denote
k f k0 ¼
C a;k ¼
sup 1 x2½0;1nþ1
n o daðk1Þ ðxÞf ðxÞ ; n
f 2 C 0; 1
k f k1 ¼
1 ; k f k0 < 1 ; nþ1 o n ð 2 aÞð1kÞ 2k sup ðxÞf 0 ðxÞ ; dn
1 x2½0;1nþ1
ð3:2Þ
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C 1a;k ¼ f 2 C a;k ; kf k1 < 1 ; n o d2þaðk1Þ ðxÞ f 00 ðxÞ ; k f k2 ¼ sup n 1 x2½0;1nþ1
C 2a;k
¼ f 2 C a;k ; f 0 2 A:C:loc ; kf k2 < 1 ;
K 1a;k ðf ; tÞ ¼ inf kf g k0 þ t kg k1 ; g2C 1a;k
K 2a;k ðf ; tÞ ¼ inf kf g k0 þ t kg k2 : g2C 2a;k
Lemma 3.2. If 0 6 k 6 1; 0 < a < 2, then
kV n f k1 6 Cn1=2 k f k0 ðf 2 C a;k Þ;
ð3:3Þ
kV n f k1 6 C k f k1 f 2 C 1a;k ;
ð3:4Þ
kV n f k2 6 Cnk f k0 ðf 2 C a;k Þ;
ð3:5Þ
kV n f k2 6 C k f k2 f 2 C 2a;k :
ð3:6Þ
Proof. The proof of Lemma 3.2 will be given in Section 4. Lemma 3.3 [15]. For 0 < t < 18 ;
Z
t uk ðxÞ=2
t uk ðxÞ=2
t 2
h
h i 1 1 ; b < 2, we have 6 x 6 1 nþ1 2t ; x 2 0; 1 nþ1
b db n ðx þ uÞdu 6 CðbÞtdn ðxÞ:
h i 1 1 Lemma 3.4 [15]. For 0 < t < 14 ; t 6 x 6 1 nþ1 t; x 2 0; 1 nþ1 ; 0 6 b 6 2, we have
Z
t uk ðxÞ=2
t uk ðxÞ=2
Z
t uk ðxÞ=2
tuk ðxÞ=2
2 b db n ðx þ u þ v Þdudv 6 Ct dn ðxÞ:
Proof. First, we let a new K-functional as
K auk ðf ; t r Þ ¼ infr kðf ðxÞ gðxÞÞk0 þ tr kg ðrÞ ðxÞkr : g2C k
By this definition we may choose g 2 C rk such that
kðf ðxÞ gðxÞÞk0 þ nr=2 kg ðrÞ ðxÞkr 6 CK auk f ; nr=2 :
Suppose that for f 2 C½0; 1,
kM n ðf ; xÞ f ðxÞk0 6 Cna=2 : Then we will prove
xruk ðf ; tÞ ¼ Oðta Þ: ðrÞ a=2 K auk ðf ; t r Þ 6 kðf ðxÞ M n ðf ; xÞÞk0 þ t r MðrÞ þ t r M ðrÞ n ðf ; xÞ 6 Cn n ðf g; xÞ þ M n ðg; xÞ r r r
a=2
r=2 tr a
r a=2 r=2 þ t n kðf ðxÞ gðxÞÞk0 þ kgðxÞkr 6 C n þ r=2 K uk f ; n ; 6C n n which implies
K auk ðf ; t r Þ 6 Ct a :
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On the other hand, notice that for i ¼ 1; 2; . . . ; r; x rt uk ðxÞ=2 2 ð0; 1Þ. Let 0 6 r 6 2. For g 2 C 0k ,
! !að1kÞ=2r r r X X r að1kÞ r 2r r k r k r d d x þ j t u ðxÞ x þ j tu ðxÞ 6 CkgðxÞk0 Dtuk gðxÞ 6 kgðxÞk0 2 2 j n j n j¼0 j¼0 6 CkgðxÞk0 dnað1kÞ ðxÞ: aðk1Þ For g with g ðr1Þ 2 AC loc ; kg ðrÞ ðxÞdrþ ðxÞk < 1 and for 0 < tuk ðxÞ < 1=8r and rtuk ðxÞ=2 6 x 6 1 rt uk ðxÞ=2 n
Z tuk ðxÞ Z tuk2ðxÞ 2 r ðrÞ . . . g ðx þ u þ þ u Þdu du Dhuk gðxÞ 6 1 r 1 r t uk ðxÞ tuk ðxÞ 2 2 k k Z tu ðxÞ Z tu ðxÞ 2 2 rþað1kÞ 6 Ct r dðrþaÞð1kÞ ðxÞkgðxÞk : 6 kgkr . . . d ðx þ u þ þ u Þdu du 1 r 1 r r n n tuk ðxÞ tuk ðxÞ 2 2
From the above, for 0 < tuk ðxÞ < 1=8r and rtuk ðxÞ=2 6 x 6 1 rt uk ðxÞ=2 and choosing appropriate g we obtain
r r r aÞð1kÞ ðxÞkgkr Dhuk f ðxÞ 6 Dhuk ðf ðxÞ gðxÞÞ þ Dhuk gðxÞ 6 C kf gk0 dnað1kÞ ðxÞ þ tr dðrþ n ! tr 6 Ct a : 6 Cdnað1kÞ ðxÞ kf gk0 þ t r dnrð1kÞ ðxÞkgkr 6 Cdnað1kÞ ðxÞK auk f ; rð1kÞ dn ðxÞ
4. Proof of Lemma 3.2 n To prove Lemma 3.2, we let a ¼ nþ1 and we will use the following notations, in this section: Let
~n;k ðxÞ :¼ ð nk Þxk ð1 xÞnk : q Then we know for pn;k ðxÞ defined in (1.2) that
pn;k ðxÞ ¼ an
n k ~n;k ðyÞ; x ða xÞnk ¼ q k
x y¼ ; a
0 6 y 6 1:
Let us denote
n o k f kr;½0;1 ¼ sup dnrþaðk1Þ ðxÞf ðrÞ ðxÞ ; x2ð0;1Þ
n
o C a;k;½0;1 ¼ f 2 C r ½0; 1; k f kr;½0;1 < 1 ; r
and
pffiffiffi ~dn ðxÞ ¼ u ~ ðxÞ þ 1= n;
~ ðxÞ ¼ u
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xð1 xÞ;
Then we know
pffiffiffi ~dn ðxÞ max u ~ ðxÞ; 1= n ; and
Z n X ~ n ðf ; xÞ ¼ ðn þ 1Þ q ~n;k ðxÞ Q k¼0
1
~n;k ðtÞf ðtÞdt: q
0
Then we have the following results: Lemma 4.1. Let r 2 N. If 0 6 k 6 1; 0 < a < r, then we have for f 2 C 0a;k;½0;1
~rþaðk1Þ ~ ðrÞ ðxÞQ n ðf ; xÞ 6 Cnr=2 kf k0;½0;1 ; dn
that is,
~ Q n ðf ; xÞ
r;½0;1
6 Cnr=2 kf k0;½0;1 ;
ð4:1Þ
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Lemma 4.2. Let r 2 N. If 0 6 k 6 1; 0 < a < r, then we have for f 2 C ra;k;½0;1
~rþaðk1Þ ~ ðrÞ ðxÞQ n ðf ; xÞ 6 Ckf kr ; dn that is,
~ Q n ðf ; xÞ 6 Ckf kr : r
Proof of Lemma 3.2. Let ~f ðxÞ ¼ f ðaxÞ. Then we see the followings;
Z a n n x Z 1 X n þ 1X ~ n ~f ; x ¼ ðn þ 1Þ q ~n;k ~n;k ðtÞ~f ðtÞdt ¼ q pn;k ðxÞ pn;k ðsÞf ðsÞdt ¼ V n ðf ; xÞ; Q a a 0 a k¼0 0 k¼0 d d ~ 0 ~ 1 V n ðf ; xÞ ¼ Q f;y ; dx dy n a and
r r r d d ~ 0 ~ 1 f ; y V ðf ; xÞ ¼ ; Q r n r a dx dy n
x y¼ : a
Moreover, since
uðxÞ ¼
x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ~ ; xða xÞ ¼ au a
we have
x x pffiffiffi pffiffiffi ~ þ 1= n ~dn ¼ ~dn ðyÞ: dn ðxÞ ¼ uðxÞ þ 1= n ¼ au a a Now, we prove Lemma 3.2. For 0 6 x 6 a and y ¼ x=a,
r r d ~ ~ 1 rþaðk1Þ ~rþaðk1Þ ðrÞ ðxÞV n ðf ; xÞ dn ðyÞ r Q n f ; y : dn a dy From Lemma 4.1, we have
r r r rþaðk1Þ aðk1Þ aðk1Þ d ~ ~ d ~ d r=2 ~ r=2 r=2 r ~ ~d : Cn Q d f ; y f ðyÞk f ðyÞ ð y Þ 6 Cn k ¼ Cn sup ðyÞ a sup d ðxÞ f ðxÞ n r r r 0;½0;1 n n n dy dy dx y2ð0;1Þ y2ð0;1Þ Therefore, we have for 0 6 x 6 a
rþaðk1Þ r=2 ðxÞV ðrÞ kf k0 ; dn n ðf ; xÞ 6 Cn that is,
kV n ðf ; xÞkr 6 Cnr=2 kf k0 : Similarly, we can prove using Lemma 4.2
rþaðk1Þ ðxÞV ðrÞ dn n ðf ; xÞ 6 Ckf kr ; that is,
ðrÞ V n ðf ; xÞ 6 Ckf kr : r
In the following we give the proof of Lemma 4.1. Proof of Lemma 4.1. From (4.1) and differentiation r-times on x, we know
~ ðrÞ ðf ; xÞ ¼ ð1Þr Q n
Z 1 nr ðn þ 1Þ!n! X ~nr;k ðxÞ ~ðrÞ q q nþr;kþr ðtÞf ðtÞdt: ðn rÞ!ðn þ rÞ! k¼0 0
ð4:2Þ
H.S. Jung et al. / Applied Mathematics and Computation 244 (2014) 683–694
691
Then we obtain
~ ðrÞ Q n ðf ; xÞ 6
Z 1 nr ðn þ 1Þ!n! X ðrÞ ~nr;k ðxÞ ~nþr;kþr ðtÞ~dnað1kÞ ðtÞdtkf k0;½0;1 : q q ðn rÞ!ðn þ rÞ! k¼0 0
Since we easily see ðrÞ ~n;k q ðtÞ ¼
r r n! X ~nr;krþi ðtÞ; q ð1Þi ðn rÞ! i¼0 i
we have the following relation: ðrÞ ~nþr;kþr q ðtÞ ¼
r r ðn þ rÞ! X ~n;kþi ðtÞ q ð1Þi n! i i¼0
ð4:3Þ
Using (4.3), we know
Z 1 r X X r nr ~ ðrÞ r ~nr;k ðxÞ ~n;kþi ðtÞ~dnað1kÞ ðtÞdtkf k0;½0;1 : q ðn þ 1Þq Q n ðf ; xÞ 6 Cn i k¼0 0 i¼0 Let
p¼
2r
and
að1 kÞ
1 2r að1 kÞ ¼ : q 2r
ð4:4Þ
Then since 1=p þ 1=q ¼ 1, using Hölder inequality, we have
!1=q !1=p Z 1 Z 1 r nr nr X X X r ~ ðrÞ r 2r ~ ~ ~ ~ ~ qn;kþi ðtÞdt qn;kþi ðtÞdn ðtÞdt ðn þ 1Þ qnr;k ðxÞ ðn þ 1Þ qnr;k ðxÞ : Q n ðf ; xÞ 6 Cn kf k0;½0;1 i 0 0 i¼0 k¼0 k¼0 Then we first know
Z nr X ~nr;k ðxÞ ðn þ 1Þ q
!1=q
1
~n;kþi ðtÞdt q
6
nr X ~nr;k ðxÞ ðn þ 1Þ q
0
k¼0
k¼0
1 nþ1
!1=q ¼ 1:
Since we see
n n k k nk 2r 2r ~n;k ðtÞ ¼ u ~ ðtÞq ~ ðtÞ ð n þ 2rk þ r Þt kþr ð1 tÞnþrk t ð1 tÞ u ¼ n þ 2r k kþr ðk þ 1Þ ðk þ rÞðn k þ 1Þ ðn k þ rÞ ~nþ2r;kþr ðtÞ; q ¼ ðn þ 1Þðn þ 2Þ ðn þ 2rÞ then we know
Z
1
~n;k ðtÞdt ¼ ~ 2r ðtÞq u
0
ðk þ 1Þ ðk þ rÞðn k þ 1Þ ðn k þ rÞ :¼ aðn; k; rÞ: ðn þ 1Þðn þ 2Þ ðn þ 2rÞðn þ 2r þ 1Þ
Then secondly we have
Z nr X ~nr;k ðxÞ ðn þ 1Þ q 0
k¼0
1
!1=p ~n;kþi ðtÞ~d2r q n ðtÞdt
nr X ~nr;k ðxÞ aðn; k þ i; rÞ þ 6 C ðn þ 1Þ q k¼0
!1=p 1 : nr ðn þ 1Þ
For 0 6 k 6 2r or n 3r 6 k 6 n r, 2r X k¼0
þ
nr X
! ~nr;k ðxÞaðn; k þ i; rÞ 6 2 ðn þ 1Þq
k¼n3r
ð4rÞr nr
and for 2r þ 1 6 k 6 n 3r 1
nr n3r1 n3r1 X X k ~nr;k ðxÞaðn; k þ i; rÞ 6 xr ð1 xÞr ~n3r;kr ðxÞðn þ 1Þaðn; k þ i; rÞ 6 CðrÞxr ð1 xÞr ; q ðn þ 1Þq n 3r k¼2rþ1 k¼2rþ1 kr
692
H.S. Jung et al. / Applied Mathematics and Computation 244 (2014) 683–694
because
nr
!
ðn 3r þ 1Þ ðn rÞ k ! ðn þ 1Þaðn; k þ i; rÞ ¼ n 3r ðk r þ 1Þ kðn k 2r þ 1Þ ðn k rÞ kr ðk þ i þ 1Þ ðk þ i þ rÞðn k i þ 1Þ ðn k i þ rÞ ðn þ 2Þ ðn þ 2rÞðn þ 2r þ 1Þ rþi rþi rþi 1þ 1 þ 6 1þ krþ1 krþ2 k 2r i 2r i 2r i 1þ 1þ 1 þ n k 2r þ 1 n k 2r þ 2 nkr r r rþi 2r i 1þ 6 CðrÞ: 6 1þ k nkr
Therefore, we have for x 2 ð0; 1Þ,
Z nr X ~nr;k ðxÞ ðn þ 1Þ q
1 0
k¼0
!1=p ~n;kþi ðtÞ~d2r q n ðtÞdt
6 C nr þ xr ð1 xÞr ~d2r n :
ð4:5Þ
Thus, for x 2 ð0; 1=nÞ [ ð1 1=n; 1Þ
~ ðrÞ að1kÞ : Q n ðf ; xÞ 6 Ckf k0;½0;1 nr ~dnað1kÞ kf k0;½0;1 nr=2 ~drþ n This implies that
~rþaðk1Þ ~ ðrÞ ðxÞQ n ðf ; xÞ 6 Ckf k0;½0;1 nr=2 : dn pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ~n;k ðxÞ, we have the following expressions: ~ 2 ðxÞ. Differentiating q xð1 xÞ ¼ u
dn ðxÞ Let x 2 ð1=n; 1 1=nÞ. Then ~ r ~ðrÞ q n;k ðxÞ ¼ ðxð1 xÞÞ
i r X k ~n;k ðxÞ; Q i ðr; n; xÞni x q n i¼0
ð4:6Þ
where Q i ðx; nÞ is a polynomial in nxð1 xÞ of degree ½ðr iÞ=2 with polynomial coefficients such that
ðxð1 xÞÞr Q i ðx; nÞni 6 C
n xð1 xÞ
rþi 2 ð4:7Þ
:
Using the expression (4.6), we know
~ ðrÞ ðf ; xÞ ¼ ðn þ 1Þ Q n
Z n X ~ðrÞ q n;k ðxÞ
1
~n;k ðtÞf ðtÞdt ¼ ðxð1 xÞÞr q
0
k¼0
i Z 1 r n X X k ~n;k ðxÞ ~n;k ðtÞf ðtÞdt: q Q i ðr; n; xÞni ðn þ 1Þ x q n 0 i¼0 k¼0
Then we have by (4.7)
i rþi Z 1 X r n X 2 n k ~ ðrÞ ~n;k ðxÞ ~n;k ðtÞ~dnað1kÞ ðtÞdtkf k0;½0;1 : q ðn þ 1Þ x q Q n ðf ; xÞ 6 xð1 xÞ n 0 i¼0 k¼0 Using Hölder inequality with p and q defined in (4.4), we obtain
i Z n X k ~n;k ðxÞ ðn þ 1Þ x q n k¼0
0
1
!1=q qi Z 1 n X k ~n;k ðxÞ ~n;k ðtÞdt q ðn þ 1Þ x q n 0 k¼0 !1=p Z 1 n X 2r ~ ~n;k ðxÞ ~n;k ðtÞdn ðtÞdt q ðn þ 1Þq :
~n;k ðtÞ~dnað1kÞ ðtÞdt 6 C q
k¼0
0
}lder inequality for 2m=iq and 2m=ð2m iqÞ, we have For the first term in the above equation, using Ho
!1=q !1=q !i=2m qi qi 2m Z 1 n n i=2m X X k k k x q ~n;k ðxÞ ~n;k ðtÞdt ~ ~ q q ðn þ 1Þ x q ¼ ðxÞ 6 x ðxÞ 6 nm ~d2m ¼ ni=2 ~din : n;k n;k n n n n 0 k¼0 k¼0 k¼0
n X
693
H.S. Jung et al. / Applied Mathematics and Computation 244 (2014) 683–694
For the second term, we know by the same procedure as the proof of (4.5)
Z n X ~n;k ðxÞ ðn þ 1Þq
6 C ~d2r=p ~dnað1kÞ : n
~n;k ðtÞ~d2r q n ðtÞdt
0
k¼0
!1=p
1
Then since we know
n xð1 xÞ
rþi 2
!rþi 2
n ~d2
;
n
we have
r X n ~ ðrÞ Q n ðf ; xÞ 6 Ckf k0;½0;1 ~d2 n i¼0
!rþi 2 að1kÞ ni=2 ~din ~dnað1kÞ 6 Ckf k0;½0;1 nr=2 ~drþ ; n
that is,
~rþaðk1Þ ~ ðrÞ ðxÞQ n ðf ; xÞ 6 Cnr=2 kf k0;½0;1 : dn
Proof of Lemma 4.2. From (4.2), we know
Z nr X ~ ðrÞ ~nr;k ðxÞ Q ðf ; xÞ 6 C ðn þ 1Þq n
0
k¼0
1
~nþr;kþr ðtÞ~dnrþað1kÞ dtkf kr;½0;1 : q
Then we see
Z nr X ~nr;k ðxÞ ðn þ 1Þq
1
0
k¼0
að1kÞ ~nþr;kþr ðtÞ~drþ q dt 6 n
Z n X ~nr;k ðxÞ ðn þ 1Þq k¼0
6
nr X ~nr;k ðxÞ ðn þ 1Þq k¼0
~nþr;kþr ðtÞ~dn2r dt q
0
Z
!1=p
1
!1=p
1 r
r
r
~nþr;kþr ðtÞ minfn ; t ð1 tÞ gdt q
;
0
where p ¼ 2r=ðr að1 kÞÞ. Here, we first know
Z nr X ~nr;k ðxÞ ðn þ 1Þq
1
~nþr;kþr ðtÞnr dt ¼ nr q
0
k¼0
nþ1 nþrþ1
and we secondly have
Z nr X ~nr;k ðxÞ ðn þ 1Þq
1
~nþr;kþr ðtÞtr ð1 tÞr dt ¼ q
0
k¼0
Z nr X ~nþr;kþr ðxÞxr ð1 xÞr ðn þ 1Þq k¼0
because
~nþr;kþr ðtÞtr ð1 tÞr ¼ q
nþr
ðn þ rÞ! k!ðn k rÞ! kþr ~nr;k ðtÞ ¼ ~nr;k ðtÞ: q q nr ðk þ rÞ!ðn kÞ! ðn rÞ! k
Thus, we have
Z nr X ~nr;k ðxÞ ðn þ 1Þq 0
k¼0
1
að1kÞ ~nþr;kþr ðtÞ~drþ q dt 6 C ~dnrþað1kÞ : n
This implies
~ ðrÞ Q n ðf ; xÞ 6 C ~dnrþað1kÞ kf kr;½0;1 ; that is,
~ Q n ðf ; xÞ
r;½0;1
6 Ckf kr;½0;1 :
0
1
~nr;k ðtÞdt ¼ q
n þ 1 r x ð1 xÞr ; nrþ1
694
H.S. Jung et al. / Applied Mathematics and Computation 244 (2014) 683–694
Acknowledgments The authors are extremely thankful to the referees for valuable suggestions that greatly improved this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]
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