doi:10.4236/am.2011.24047 Published Online April 2011 (http://www.SciRP.org/journal/am)
Pointwise Approximation Theorems for Combinations of
Bernstein Polynomials with Inner Singularities
Wenming Lu1, Lin Zhang2
1Department of Mathematics, Hangzhou Dianzi University, Hangzhou, China 2Department of Mathematics, Zhejiang University, Hangzhou, China E-mail: lu_wenming@163.com, [email protected], godyalin@163.com Received October 8, 2010; revised January 14, 2011; accepted January 17, 2011
Abstract
It is well-known that Bernstein polynomials are very important in studying the characters of smoothness in theory of approximation. A new type of combinations of Bernstein operators are given in [1]. In this paper,
we give the Bernstein-Markov inequalities with step-weight functions w x
for combinations of Bernsteinpolynomials with inner singularities as well as direct and inverse theorems.
Keywords:Bernstein Polynomials, Inner Singularities, Pointwise Approximation, Bernstein-Markov
Inequalities, Direct and Inverse Theorems
1. Introduction
The set of all continuous functions, defined on the inter-val I, is denoted by C I
. For any f C
0,1
, the corresponding Bernstein operators are defined as fol-lows:
,
0
, : n n k
k
B f x f k p x
n
,where
, : C 1
n k k k
n k n
p x x x , k0,1, , n, x
0,1 .Approximation properties of Bernstein operators have been studied very well (see [2-7], for example). In order to approximate the functions with singularities, Della Vecchia et al. [8] introduced some kinds of modified Bernstein operators. Throughout the paper, C denotes
a positive constant independent of n and x, which may be different in different cases. Ditzian and Totik extended the method of combinations and defined the following combinations of Bernstein operators:
1
,
0
, : r i ,
n r i n
i
B f x C n B f x
,with the conditions:
a) n n 0n1nr1Cn,
b) ri 10C ni( ) C
,c) ri 01C ni( ) 1
,d) r10
k 0i i
i C n n
, for k1, , r1.For any positive integer r, we consider the determi-nant
1 1 1 1
2 1 2 2 2 3 4 1
2 2 1 2 1 2 1 2 2 2 3 4 4 1 :
2 2 1 3 2 2 4 2 3 2 2 4 1
r
r r r r
r r r r r r r r
A
r r r r r
We obtain 2
2 ! 0
r
r j
1 2 2 1
1 2 2 1
1 2 2 1
1 2 2 1
1,
2 1 2 2 4 1 0,
2 1 2 2 2 2 1 4 4 1 0,
2 1 ! 3 2 1 2 2 4 1 0.
r
r
r
r
a a a
r a r a r a
r r a r r a r r a
r a r a r r a
(1.1)
Let
2 1 2 2 4 1
1 2 2 1 1, 0 1,
: 0, 0,
1, 1.
r r r
r
a x a x a x x
x x
x
with the coefficients a a1, , ,2 a2r1 satisfying (1.1).
From (1.1), we see that (2 ) ( , )
( )x C r
, 0
x 1 for 0 x 1. Moreover, it holds that
1 1, i
0 0 , i0,1, , 2 r and i
1 0, i1, , 2 r. Let
1
1
, : r i i i
H f x f x l x
,and
11,
1 1,
:
r
j j i j
i r
j j i i j
x x l x
x x
,
1 2
in r i
x
n
, 1, 2, ,i r1.
Further, let
1
2 ,
n n
x
n
x2 n n ,
n
3 ,
n n
x
n
4
2
n n
x
n
and
11
2 1
x x x
x x
,
3 2
4 3
x x x
x x
. Set
1 2
1 2
, : 1
1 .
n
F f x F x f x x x
x x H x
We have
5 2 3 2
1 1 5 2 3 2
3 2 1 2
2 2 1 2 3 2
, 0, ,1 ,
1 , , ,
,
, , ,
1 , , .
r r
r r
n
r r
r r
f x x x x
f x x x H x x x x
F f x
H x x x x
H x x x f x x x x
Obviously, F f xn
,
is linear, reproduces polynomi- als of degree r, and
,
2r
0,1
n
F f x C , provided
that fC 2r
0,1
. Now, we can define our new com- binations of Bernstein operators as follows:
1
, ,
0
, , r i ,
n r n r n i n n
i
B f x B F x C n B F x
(1.2)where C ni
satisfy the conditions (a)-(d).2. The Main Results
Let : 0,1
R, 0 be an admissible step-weight function of the Ditzian-Totik modulus of smoothness, that is, satisfies the following conditions:1) For every proper subinterval
a b, 0,1 there exists a constant C C a b1
, 0 such that
1
1 1
C x C
for x
a b, .2) There are two numbers
0 0 and
1 0 for which
0
1
, as 0 , 1 , as 1 .
x x
x
x x
~
(X~Y means C-1Y ≤ X≤ CY for some C).
Combining conditions (I) and (II) on , we can de-duce that
1
2 2
C x x C x , x
0,1 ,where 2
x x 0
1 x
1
W. M. LU ET AL. 391
Let w x
x ,0 1, 0 and
0,1 \ : lim 0
wx
C f C wf x
.
The norm in Cw is defined as
0 1
: sup
w
w C
x
f wf wf x
. Define
1
: : r 0,1 , r
r r
w
W f C f AC w f ,
1
, : : 0,1 ,
r r
r r
w
W f C f AC w f .
For f Cw, we define the weighted modulus of smoo-
thness by
0 0 1
, : sup sup
r r
h x w
h t x
W f t w x f x
,
where
0
1 +
2
r k
r k
h r
k
r
f x C f x k h x
,
0
1 C +
r k
r k
h r
k
f x f x r k h
.Recently Felten showed the following two theorems in [4]:
Theorem A. Let
x = x
1x
and let
: 0,1 R
, 0 be an admissible step-weight func-tion of the Ditzian-Totik modulus of smoothness ([3]) such that 2 and 2 2 are concave. Then, for
0,1fC and 0 2
,
,
2 , 1 2
n
x
B f x f x C f n
x
.
Theorem B. Let
x = x
1x
and let
: 0,1 R
, 0 be an admissible step-weight func-tion of the Ditzian-Totik modulus of smoothness such that 2 and 2 2are concave. Then, for fC
0,1and 0 2,
,
1 2
n
x
B f x f x C n
x
implies 2
f t, O t
.
Our main results are the following:
Theorem 2.1. For any 0,
1min 0 , 1 , 2 f Cw
, we have
2, 1 ,
r r
r n r
w x x B f x Cn wf . (2.1)
Theorem 2.2. For any 0, f Wr, we have
, 1 ,
r r
r r
n r
w x x B f x
C
w f . (2.2)Theorem 2.3. For f Cw, 0,
1 min 0 , 1
2
,
0 0,r
, n
x
x 1n
, we have
0
0 , 1
1/ 2
,
,
n r
n r
w
w x f x B f x
x
O f t O t
n x
(2.3)
3. Lemmas
Lemma 3.1. For any non-negative real u and v, we have
1 , 1
1
u v
n
v u
n k k
n n
p x Cx x
k n k
. (3.1)Lemma 3.2. If R, then
2
, 0
n n k k
p x k nx Cn x
. (3.2)Lemma 3.3. For any fWr, 0, we have r r
r r
n
w F C w f . (3.3)
Proof. We first prove x xr5 2,xr3 2 (The same
as the others), we have
: 1 2r r
r r
n
r r
n
w x x F x w x x f x
w x x f x F x I I
Obviously
1
r r
C w f
I
For I2, we have
0 2
2
r r
n i
r r i
r
n i
w x x f x F x
w x x n f x F x
I
By [3], we have
5 2 3 2
5 2 3 2 5 2 3 2
,
2 2
, , .
r r
r r r r
r i n
x x
r i i
r
x x x x
f x F x
C n f H n f
5 2 3 2
5 2, 3 2 1
2 2 ,
2
:
r r
r r
r r
x x r
r
x x
Cn w x x f H
Cw x x f T
I
T
By Taylor expansion, we have
1
0
1
! 1
d . 1 !
i u r
u i
i u
x r r
i x
x x
f x f x
u
x s f s s
r
(3.4)
It follows from (3.4) and the identity
1
r
v v
i i i
x l x Cx
, v0,1, , r. we have
1 0
1 1
1 0 1
1 1
,
! 1
d 1 !
1
d , 1 !
i
i
u r r
u i
i i u
r x
r r i x i
i
r u r
u v
u v v
u i i
u v i
r x
r r i x i
i
x x
H f x f x l x
u
l x x s f s s
r
f x C f x C x x l x
l x x s f s s
r
which implies that
1
1
,
1 d ,
1 !
i r
r x
r r
r
i x i
i
w x x f x H f x
w x x l x x s f s s
r
Since
i
l x C, for x xr5 2,xr3 2, i1, , r.
It follows from
1 1
r r
i i
x s x x
w s w x
,
s between xi and x, then
1 1
/ 2
,
d
.
i
r
r x
r r r
i x i r r
r
w x x f x H f x
Cw x x x s f s s
C
n w f
So
2
r r
C w f
I .
Then, the lemma is proved.
According to methods of Lemma 3.3, we can easily get:
Lemma 3.4. If f Wr, 0, then
,
r r
n x r
w x g x H g x C w g
n x
(3.5)
Lemma 3.5. For any 0, f Cw, we have
, 1n r
wB f C wf . (3.6)
Proof. By (1.2), we have
, 1 , 1 1
1
, 0 1
1
,0 0
1
, 0
1 2 3
, ,
.
0
1
:
i
i
i
i i
n r n r n
n r
i n n k
i k i
r
i n n
i r
i n n n
i
w x B f x w x B F x
k
w x C n F p x
n
w x C n F p x
w x C n F p
I I
x
I
Now, the theorem can be proved easily.
Lemma 3.6. Let min
0 , 1
1 2 , then for rN,
1 0
8
t r
and 1
2 2
rt x rt, we have
2 2
1 1
2 2
d d
t t r
r r r
k r
t t
k
x u u u Ct x
(3.7)Lemma 3.7. Let n
:
n k,
k n n
A x w x p x
, then
/ 2n
A x Cn for 0 1 and 0.
Proof. If x 3
n
, then the statement is trivial.
Hence assume 0 x 3 n
( the case 3 x 1
n
can be treated similarly). Then for a fixed x the maxi-mum of pn k,
x is attained for kkn:n n. By using Stirling’s formula, we get
,
1 n
n
n n n
n
n k k
n k k n k
n n
n n
n
nx x
e
p x C
k k n k n k
e e
1 1 .
n n
k n k
n n
n n
k nx k nx
C
k n k
n
W. M. LU ET AL. 393
Now from the inequalities
1 1 , 2 nk nx n n nx n x
n n x
and 2 1 2
1 u e u u , 1 u eu, u0.
We have that the second inequality is valid. To prove the
first one we consider the function
21
2 1
u u
u e u
.
Here
0 0,
2 1 2
1 u u 1
u u e
,
0 0,
u u u
2
e u 12u2 0 , whence
u 0 for 0u .
Hence
2 , 1 exp 2 n n nn k n n
n n
k nx k nx
C
p x k k nx
k k n
2 2exp 2 Cn x n n k nx C e k n .
Thus
Cn x2 nA x C x e . An easy calcula- tion shows that here the maximum is attained when
C x
n
and the lemma follows.
Lemma 3.8. For 0 1, , 0, we have
2
,
n k k n n
w x p x k nx Cn x
(3.8)Proof. By (3.2) and the lemma 3.7, we have
2 1 2 1 2 , 1 2 2 2 , ( ) . n n n n k k n nn n n k
k n n
w x w x p x
p x k nx Cn x
Lemma 3.9. For any 0, f Wr,we have
, 1
r r
n r
wB f Cn wf . (3.9)
Proof. We first prove x 0,1 n
(The same as
1 1 ,1 x n
, now
, 1 2 1 , 0 0 2 ,0 0 0
2 ,0 0 0 2 0 0 , ! ! i i i i i i r n r n r r r i
i n n r k
i i k n i
n r
r r
r j
i r n n r k
i k j i
r r
r j
i r n n r
i j i
r r
r j i
i r n
i j i
w x B f x
n k
w x C n F p x
n r n
k r j
Cw x n C F p x
n
r j
Cw x n C F p x
n
n j
Cw x n C F
n
, 1 2 ,0 1 0
1 2 3
: .
i i
i
i n r n r
n r
r r
r j
i r n n r k
i k j i
p x
k r j
Cw x n C F p x
n
H H H
We have
2 1 ,0 0 0 2 0 0 1 . i i r r ri n n r
i j i
r r
n r i
r
i j i
r
r j
H Cw x n F p x
n
n x
Cn wf x
r j n Cn wf
Similarly, we can get H2Cn wfr , and
3
r
H Cn wf .
When x 1,1 1
n n
, according to [3], we have
2 3, 1 , 1
2 2 0 0 , 2 2 0 0
, 1 2
, , , , : i i i i r r
n r n r n
r r
r j
j i i i
i j j
n n k
k n A i i
r r
r j
j i i i
i j j
n k
x k n x i i
w x B f x w x B F x
w x x Q x n C n n
k k
x F p x
n n
w x x Q x n C n n
k k
x H p x
n n
where A: 0,
x2
x3,1 , H is a linear function. Ifi
k A
n , when
1 i 2 ii
w x
C n k n x
w k n
, we have
2
i i
n
k n , also
,
1
2r j
j i i
Q x n n x x , and
2
,
2
2r j
r j
j i i i
x Q x n n C n x
2 2
1 2 ,
0 0 0
2 2
2
, 1 2 2
0 0 0
1 : .
i
i
i
i
r j
j n
r r i
n n k
i j k i i
r j
j n
r r i
i i n k
i j k i
n k k
Cw x x F p x
n n
x
n k
C wf n k n x x p x I I
n x
By a simple calculation, we have I1Cn wfr . By (3.2), then
2 2
2
2 2 ,
0 0 0
i
i
r j n
r r j j
r i
i i n k
i j k
n
I C wf n k n x p x Cn wf
x
.We note that
1 4
max , :
i
k
H H x H x H a
n
.
if x
x x1 , 4
,we have w x
w a
.So, if x
x x1 , 4
, then 2Cn w a H ar
Cn wfr .If x
x x1 , 4
, then w a
ni 2
, by (3.8), we have
2 3
2 2
2
2 2
0 0
, .
i i
r j
r r
i i i j j
r n k
x k n x i
n
Cw a H a w x n
x
k
x p x Cn wf
n
It follows from combining the above inequalities that the lemma is proved.
4. Proof of Theorems
4.1. Proof of Theorem 2.1When f Cw, min
0 , 1
1 2, we discuss it asfollows:
Case 1. If 0
x 1 n
, by (3.9), we have
( ) , 1
2 , 1
,
,
r r
n r
r r
r r
n r r
w x x B f x
x
C x w x B f x Cn wf
x
(4.1)
Case 2. If
x 1 n , we have
, 1 , 1
2 2
0 0
, ,
, .
r r
n r n r n
r r
r j
j i i i
i j
B f x B F x
x Q x n C n n
, 0i
i j
n
n n k
k i i
k k
x F p x
n n
,where
,
1
2r j
j i i
Q x n n x x
, and
2
,
2
2r j
r j
j i i i
x Q x n n C n x
So
, 1
2 2
, 2
0 0 0
2 2
2
0 0
2 2
, 2
0 0
,
i
i
i
i
r
r r
n r r j
j n
r r
i
n n k
i j k i i
r j
j
r r
r i
n
i j k n A i i
r j
r r
r i
n k
i j
w x x B f x Cw x x
n k k
x F p x
n n
x
n k k
Cw x x x F
n n
x
n
p x Cw x x
x
k x
n
2 3
, : 1 2.
i i
j
n k
x k n x i i
k
H p x
n
(4.2) where A: 0,
x2
x3,1 , we can easily get2 1
r
Cn wf
, and 2
2
r
Cn wf
. By bringing these facts together, the theorem is proved.
4.2. Proof of Theorem 2.2 When f Wr
, by [3], we have
2
, 1 1 ,
0 0
, i i
i n r r
r r r
n r n i i n n r k
i k n i
k
B F x C n n F p x
n
W. M. LU ET AL. 395
If 0 k ni r, we have
1
1 0i d
i
r r
r r n
n i n
i i
n
k k
F Cn F u u
n n
. (4.4)
If k0, we have
1
1 0 0i d
i
r
r
r n r
n n
n
F C
u F u u . (4.5)
Similarly
1
1 2
1 1 1 d .
i i
r r
r i r
r
n i n
n i
n
n r
F Cn u F u u
n
(4.6)
By (4.3), we have
, 1 2
1 ,
0 0 1 2
1 ,
0 1 2
1 ,0 0
2
1 ,
0
,
0
1 .
i
i i
i
i i
i i
i i i
r r
n r
n r r
r r r
i n n r k
i k n i
n r r
r r r
i n n r k
i k n i
r
r r r
i n n r
i n
r
r r r
i n n r n r
i n
w x x B f x
k
Cw x x n F p x
n
k
Cw x x n F p x
n
Cw x x n F p x
Cw x x n F p x
(4.7)
which combining with (4.4)-(4.6) give
, 1 ,
r r
r r
n r
w x x B f x C w f .
Combining with the theorem 2.1 and theorem 2.2, we can obtain
Corollary For any 0, 0 1, we have
, 1
1 2 1
2
,
,
max , , ,
, .
{
r r
n r
r r
r
w r
r r
w
w x x B f x
Cn n x wf f C
C w f f W
(4.8)
4.3. Proof of Theorem 2.3 4.3.1. The Direct Theorem
We know
1
1
d
1 !
n n n
t r r
n x
F t F x F t x
t u F u u
r
(4.9)
, 1 , 0
k n r
B x x , k1, 2, , r1 (4.10)
According to the definition of Wr
, for any g W r, we have , 1
,
, 1
,
r r
n r n r n
B g x B G g x , and
n n r, 1
n,
n r, 1
r
n, , ,
w x G x B G x w x B R G t x x ,
there of
, ,
t
r1 r
dr n x n
R G t x
t u G u u, we have
, 1
1 , 1
1 1 2 , 1 2
1 1 2 , 1 2
,
,
d ,
d , .
n n r n
r t r
r
n n r x r
r t r
r
n n r x r
r t n r x
w x G x B G x
t u
C w G w x B du x
w u u
t u
C w G w x B u x
u
t u
B u x
w u
(4.11)
also
1 1
2 d 2 , 2 d 2
r r r r
t t
r r
x x
t u t x t u t x
u C u C
u x w u w x
.(4.12) By (3.2), (3.3) and (4.12), we have
, 1
, 1
,
,
n n r n
r r
r r
n n r
r
r
n r
n
w x G x B G x
C w G x B t x x
x
C w G
n x
. (4.13)
By (3.3), (3.5) and (4.13), when g Wr
, then
, 1
, 1
, ,
, , ,
δ δ
.
n r n
n n r
r r
r r
n r n r
n
w x g x B g x w x g x G g x
w x G g x B g x w x g x H g x
x x
C w G C w g
n x n x
(4.14) For f Cw, we choose proper g W r, by (3.6) and (4.14), then
, 1
, 1 , 1
,
, ,
δ
, .
n r
n r n r
n r
w
w x f x B f x w x f x g x
w x B f g x w x g x B g x
x
C f
n x
4.3.2. The Inverse Theorem
, , inf :
r
r r r r
r w g
K f t w f g t w g g W .
By [3], we have
1
,
, , ,
r r r
r
w w w
C f t K f t C f t . (4.15)
Proof. Let 0, by (4.15), we choose proper g so
that
r
,
w
w f g C f , w gr r C rr
f,
w
.
(4.16)
For r N , 0 1 8
t r
and 1
2 2
rt x rt, we have
0
, 1 , 1 , 1
1
2 2 2
, 1 1
0 2 2 1
, , ,
2
, d d
2
r r r r
h h n r h n r h n r
h x h x n
r r
r j
r h x h x n r k r
j k
h
w x f x w x f x B f x w x B f g x w x B g x
r
x j h x
C n w x B f g x u u u
r
x j h x
w x
2 2
, 1 1 1 2 3
1 2 2
, d d :
h x h x r
r
n r k r
h x x
k
B g x u u u J J J
(4.17)
Obviously
0
1 1/ 2
n x
J C
n x
. (4.18)
By (3.9) and (4.16), we have
2 2
2 1
2 2
d d ,δ
h x h x
r r r r r
r
h x h x w
J Cn w f g u u Cn h x f
(4.19)By the first inequality of (4.8) and (4.16), we let 1, then
2 2
2 2
2 1
1 2 2
d d ,δ .
h x h x
r r r
r r r r r
k r
h x h x w
k
J Cn w f g x u u u Cn h x x f
(4.20)By (3.7) and (4.16), we have
1
2 2
3 1
1 1
2 2
d d δ ,δ
h x h x r r
r
r r r r r
k k r
h x h x w
k k
J C w g w x w x u x u u u Ch f
(4.21)Now, by (4.17)-(4.21), there exists a constant M 0 so that
0
0
2 1/ 2
1 2 1/ 2
min , , ,
, , .
r r n
r r r r r r r r
h r w w
r
n r r n r r r r
w w
x x
w x f x C n n x h f h f
n x x
x x
C h M n f h f
x
n x
When n2, we have
1 1 1
2 2 2
1
1 2
n n n
n x n x n x ,
Choosing proper x, δ, n N , so that
1 1
1
2 n x 1 2 n x
n n
x x
,
Therefore
r
0 r r r
,
h w
w x f x C h f
which implies
,
0
,
r r r r
w w
f t C h f
W. M. LU ET AL. 397
So, by Berens-Lorentz lemma in [3], we get
, 0r w
f t Ct
.
5. References
[1] D. S. Yu, “Weighted Approximation of Functions with Singularities by Combinations of Bernstein Operators,”
Journal of Applied Mathematics and Computation, Vol. 206, No. 2, 2008, pp. 906-918.
[2] Z. Ditzian, “A Global Inverse Theorem for Combinations of Bernstein Polynomials,” Journal of Approximation Theory, Vol. 26, No. 3, 1979, pp. 277-292.
doi:10.1016/0021-9045(79)90065-0
[3] Z. Ditzian and V. Totik, “Moduli of Smoothness,” Springer- Verlag, Berlin, 1987.
[4] M. Felten, “Direct and Inverse Estimates for Bernstein
Polynomials,” Constructive Approximation, Vol. 14, No. 3, 1989, pp. 459-468. doi:10.1007/s003659900084
[5] S. S. Guo, C. X. Li and X. W. Liu, “Pointwise Approxi-mation for Linear Combinations of Bernstein Operators,”
Journal of Approximation Theory, Vol. 107, No. 1, 2000, pp. 109-120. doi:10.1006/jath.2000.3504
[6] G. G. Lorentz, “Bernstein Polynomial,” University of Toronto Press, Toronto, 1953.
[7] J. J. Zhang and Z. B. Xu, “Direct and Inverse Approxi-mation Theorems with Jacobi Weight for Combinations and Higher Derivatives of Baskakov Operators,” Journal of Systems Science and Mathematical Sciences, In Chi-nese, Vol. 28, No. 1, 2008, pp. 30-39.