A POINTWISE APPROXIMATION THEOREM FOR
LINEAR COMBINATIONS OF BERNSTEIN POLYNOMIALS
SHUNSHENG GUO, SHUJIE YUE, CUIXIANG LI, GE YANG AND YIGUO SUN
Abstract. Recently, Z. Ditzian gave an interesting direct estimate for
Bernstein polynomials. In this paper we give direct and inverse results of
this type for linear combinations of Bernstein polynomials.
1. Introduction
For the Bernstein polynomial
(1.1)
Bn (f, x) =
n
k
pnk (x),
n
f
k=0
pnk (x) =
Berens and Lorentz showed in [1] that
(1.2)
Bn (f, x) − f (x) = O
α 1
√ δn (x)
n
n k
x (1 − x)n−k ,
k
⇐⇒ ω 1 (f, t) = O(tα ),
where 0 < α < 1, δn (x) = ϕ(x) + √1n , ϕ(x) = x(1 − x).
Recently, Ditzian [3] gave the following interesting result.
1
|Bn (f, x) − f (x)| ≤ Cωϕ2 λ (f, n− 2 ϕ(x)1−λ ), 0 ≤ λ ≤ 1.
However, Ditzian did not consider the inverse result in [3]. We did give such
an inverse result in [6], where we obtained the following equivalence.
1
Bn (f, x) − f (x) = O((n− 2 ϕ(x)1−λ )α ) ⇐⇒ ωϕ2 λ (f, t) = O(tα ), 0 < α < 2.
In this paper we consider linear combinations of Bernstein polynomials, that
is
(1.3)
Bn,r (f, x) =
r−1
i=0
Ci (n)Bni (f, x),
1991 Mathematics Subject Classification. 41A25, 41A36, 41A17.
Key words and phrases. Pointwise approximation, Bernstein polynomial.
Received: July 2, 1996.
c
1996
Mancorp Publishing, Inc.
397
398 SHUNSHENG GUO, SHUJIE YUE, CUIXIANG LI, GE YANG AND YIGUO SUN
where ni and Ci satisfy [5]
(a) n = n0 < n1 < · · · < nr−1 ≤ Kn;
r−1
|Ci (n)| < C
i=0
(1.4)
(c)
r−1
Ci (n) = 1;
r−1
(d)
i=0
Ci (n)n−ρ
i = 0,
i=0
We recall that
(1.5)
(b)
ωϕr λ (f, t) = sup
sup
0<h≤t x± r hϕλ (x)∈[0,1]
ρ = 1, 2, · · · , r − 1.
r
∆hϕλ (x) f (x)
2
is equivalent to the K-functional
(1.6)
Kϕλ (f, tr ) =
inf
g (r−1) ∈A.Cloc
(f − gc[0,1] + tr ϕrλ g (r) C[0,1] ).
We write ωϕr λ (f, t) ∼ Kϕλ (f, tr ), i.e., there exists a constant C such that
C −1 Kϕλ (f, tr ) ≤ ωϕr λ (f, t) ≤ CKϕλ (f, tr ).
(1.7)
Now we state our main result.
Theorem. For f ∈ C[0, 1], 0 < α < r, 0 ≤ λ ≤ 1, we have
(1.8)
1
Bn,r (f, x) − f (x) = O((n− 2 δn1−λ (x))α ) ⇐⇒ ωϕr λ (f, t) = O(tα ).
Remark. In the case r = 1 and λ = 0 our result is (1.2) of Berens and
Lorentz [1].
2. Direct theorem
In this section we prove the direct part of (1.8). We need the K-functional
(see [5], p. 24):
(2.1)
K ϕλ (f, tr ) =
λ
{f − g + tr ϕrλ g (r) + tr/(1− 2 ) g (r) }.
inf
g (r−1) ∈A.C
which is also equivalent to
loc
ωϕr λ (f, t).
Theorem 1. For f ∈ C[0, 1], 0 ≤ λ ≤ 1,
(2.2)
1
|Bn,r (f, x) − f (x)| ≤ Aωϕλ (f, n− 2 δn (x)1−λ ).
Proof. By (2.1), we may choose gn ≡ gn.x.λ for a fixed x and λ such that
1
(2.3)
f − gn ≤ A1 ωϕr λ (f, n− 2 δn1−λ (x)),
(2.4)
(n− 2 δn1−λ (x))r ϕrλ gn(r) ≤ A2 ωϕr λ (f, n− 2 δn1−λ (x)),
(2.5)
(n− 2 δn1−λ (x))2r/(2−λ) gn(r) ≤ A3 ωϕr λ (f, n− 2 δn1−λ (x)).
1
1
1
1
We recall that [5, p. 134]
(2.6)
Bn,r ((· − x)k , x) = 0,
k = 1, 2, · · · , r − 1
POINTWISE APPROXIMATION
399
and that [5, p. 138]
Bn ((· − x)2j , x) =
j−1
ϕ(x)2(j−m) −2m
n
qm (x),
j−m
n
m=0
where qm (x) are fixed bounded polynomials. Therefore
Bn ((· − x)2j , x) ≤ M n−j δn2j (x).
(2.7)
From the definition of the Bn,r we have
|Bn,r (f, x) − f (x)| ≤ |Bn,r (f − gn , x)| + |f (x) − gn (x)| + |Bn,r (gn , x) − gn (x)|
(2.8)
≤ (C + 1)f − gn + |Bn,r (gn , x) − gn (x)|.
As in [4, Lemma 5.3], we obtain
(2.9)
t
r−1 t
rλ
(r)
(t − u)r−1 g (r) (u)du ≤ (t − x)
δ
(u)g
(u)du
,
n
n
n
δ rλ (x)
x
x
n
(2.10)
t
(t − x)r−1 t rλ
r−1
(r)
(r)
(t − u)
≤
g
(u)du
ϕ
(u)g
(u)du
.
n
n
ϕrλ (x)
x
x
Using (2.6), (2.7) and (2.9), we get
|Bn,r (gn , x) − gn (x)|
t
1
(t − u)r−1 gn(r) (u)du, x)
= Bn,r (
(r − 1)! x
(2.11)
≤
≤
r−1
i=0
r−1
i=0
|Ci (n)|δnrλ gn(r) ∞
1
Bn
(r − 1)! i
|t − x|r
,x
δnrλ (x)
r
|Ci (n)|δnrλ gn(r) ∞ δn−rλ (x)M1 n− 2 δnr (x)
r
≤ CM1 n− 2 δnr(1−λ) (x)δnrλ gn(r) ∞ .
Similarly, by (2.10) we have
(2.12)
r
|Bn,r (gn , x) − gn (x)| ≤ CM1 n− 2 δnr (x)ϕ−rλ (x)ϕrλ gn(r) .
If x ∈ En = [ n1 , 1 − n1 ], then δn (x) ∼ ϕ(x), and, by (2.4) and (2.12),
1
(2.13)
|Bn,r (gn , x) − gn (x)| ≤ M2 (n− 2 δn1−λ (x))r ϕrλ gn(r) 1
≤ M2 A2 ωϕr λ (f, n− 2 δn1−λ (x)).
400 SHUNSHENG GUO, SHUJIE YUE, CUIXIANG LI, GE YANG AND YIGUO SUN
If x ∈ Enc = [0, n1 ) (1 − n1 , 1], then δn (x) ∼
(2.11) we obtain
√1 ,
n
and by (2.4), (2.5) and
1
|Bn,r (gn , x) − gn (x)| ≤ M3 (n− 2 δn1−λ (x))r (ϕrλ gn(r) + n−
rλ
2
gn(r) )
1
≤ M4 ((n− 2 δn1−λ (x))r ϕrλ gn(r) (2.14)
1
+ (n− 2 δn1−λ (x))2r/(2−λ) gn(r) )
1
≤ M5 ωϕr λ (f, n− 2 δn1−λ (x)).
From (2.3), (2.8), (2.13) and (2.14) we get (2.2).
3. Inverse Theorem
In this section we prove the inverse part of (1.8).
Theorem 2. For f ∈ C[0, 1], 0 < α < r, 0 ≤ λ ≤ 1, if
1
|Bn,r (f, x) − f (x)| ≤ B(n− 2 δn1−λ (x))α
then
ωϕr λ (f, t) = O(tα ).
(3.1)
To prove Theorem 2 we need some new notation and some lemmas. We use
the following notation.
C0 = {f ∈ C[0, 1], f (0) = f (1) = 0},
f 0 = sup |δnα(λ−1) (x)f (x)|,
x∈(0,1)
Cλ0
= {f ∈ C0 , f 0 < ∞},
f r = sup |δnr+α(λ−1) (x)f (r) (x)|,
x∈(0,1)
Cλr = {f ∈ C0 , f r < ∞, f (r−1) ∈ A.Cloc }.
For f ∈ C0 we define the K -functional as follows:
Kλα (f, tr ) = infr {f − g0 + tr gr }.
(3.2)
g∈Cλ
We also need the following lemmas which will be proved in the next section.
Lemma 3.1. If n ∈ N, 0 < α < r, then
r
Bn f r ≤ B1 n 2 f 0 (f ∈ Cλ0 ).
(3.3)
and
(3.4)
Bn f r ≤ B2 f r
Lemma 3.2. For 0 < t <
(3.5)
t
2
− 2t
···
t
2
− 2t
Now we prove (3.1).
1
8r ,
δn−β (x +
rt
2 ≤
r
k=1
(f ∈ Cλr ).
x≤1−
rt
2
and 0 ≤ β ≤ r, we have
uk )du1 · · · dur ≤ C(β)tr δn−β (x).
POINTWISE APPROXIMATION
401
Proof of (3.1). Since Bn (f, x) preserves linear functions, we consider only
f ∈ C0 for r > 1. If r = 1, f (x) = ax + b then ωϕ1 λ (ax + b, t) = aϕλ (x)t ≤
atα (0 < α < 1).
So, we may assume f ∈ C0 . From (3.2) we have
Kλα (f, tr ) ≤ Bn,r (f ) − f 0 + tr Bn,r f r
(3.6)
and we may choose g ∈ Cλr such that
(3.7)
1
f − g0 ≤ 2Kλα (f, n− 2 )
and
r
1
n− 2 gr ≤ 2Kλα (f, n− 2 ).
By the assumption of Theorem 2, one has
α
Bn,r (f, x) − f (x)0 ≤ Bn− 2 .
(3.8)
Using Lemma 3.1 and (3.7) we have
Bn,r (f )r ≤ Bn,r (f − g)r + Bn,r gr
r
≤ M (n 2 f − g0 + gr )
(3.9)
r
r
≤ 2M n 2 Kλα (f, n− 2 ).
From (3.6), (3.8) and (3.9) we obtain
α
r
r
Kλα (f, tr ) ≤ M1 (n− 2 + tr n 2 Kλα (f, n− 2 ))
and this implies, via the Berens-Lorentz lemma [1], that if α < r then
Kλα (f, tr ) ≤ M2 tα .
(3.10)
α(1−λ)
On the other hand, notice that δn
(3.11)
(x) is concave. So, we have, for g ∈ Cλr ,
r
r
r α(1−λ)
r
δn
x + (j − )tϕλ (x)
∆tϕλ (x) g(x) ≤ g0
2
j
j=0
≤ g0 2r δnα(1−λ) (x).
Using Lemma 3.2 for g ∈ Cλr , 0 < tϕλ (x) <
we have
(3.12)
1
8r
and
rtϕλ (x)
2
r
∆tϕλ (x) g(x)
t λ
t ϕλ (x)
2 ϕ (x)
2
≤
···
g (r) (x + u1 + · · · + ur )du1 · · · dur t λ
− t ϕλ (x)
− 2 ϕ (x)
2
t ϕλ (x)
≤ gr
···
2
− 2t ϕλ (x)
t λ
ϕ (x)
2
− 2t ϕλ (x)
λ
≤ x ≤ 1 − rtϕ2 (x) ,
δn−r+α(1−λ) (x + u1 + · · · + ur )du1 · · · dur
≤ M3 tr ϕrλ (x)δn−r+α(1−λ) (x)gr ≤ M3 tr δn(α−r)(1−λ) (x)gr .
402 SHUNSHENG GUO, SHUJIE YUE, CUIXIANG LI, GE YANG AND YIGUO SUN
1
From (3.9)-(3.12), 0 < tϕλ (x) < 8r
,
choosing an appropriate g , we have
rtϕλ (x)
2
≤ x ≤ 1−
rtϕλ (x)
2
r
∆tϕλ (x) f (x) ≤ ∆rtϕλ (x) (f − g)(x) + ∆rtϕλ (x) g(x)
≤ M4 δnα(1−λ) (x) f − g0 + tr δnr(λ−1) (x)gr
2M4 δnα(1−λ) Kλα
≤
≤ 2M4 δnα(1−λ) (x)
α
f,
r(1−λ)
α(1−λ)
δn
tr
δn
tα
and by
(x)
(x)
= M5 t .
The proof of (3.1) is complete.
4. Proofs of the lemmas
Proof of Lemma 3.1. We first prove (3.3). Suppose that En = [ n1 , 1 − n1 ].
For x ∈ Enc = (0, n1 ) (1 − n1 , 1), we have [5] by Hölder’s inequality
n−r
n!
k
(r)
r
$ 1f
pn−r,k (x)
∆
Bn (f, x) = (n − r)!
n
n
k=0
r
n−r
r
k+r−j
≤ Lnr f 0
k=0 j=0

≤ L1 nr f 0 
δnα(1−λ)
j
r
n−r
k=0 j=0
δn2r
n
k+r−j
n
pn−r,k (x)
α(1−λ)/2r
pn−r,k (x)
.
For n > 4r we have
n−r
ϕ
2r
k=0
k+1
pn−r,k (x)
n

=
2r
+
k=0
n−3r
n−r
+
k=2r+1

 ϕ2r
k=n−3r+1
k+1
pn−r,k (x)
n
:= I1 + I2 + I3 .
Obviously I1 + I3 ≤ 2((3r + 1)!)r /nr and, by simple computation, we have
n−3r
I2 =
k=2r+1
k+1
n
= xr (1 − x)r
·
r n−k−1
n
r
(n − r)!
xk (1 − x)n−r−k
(n − r − k)!k!
n−3r
(k + 1)r
(n − r) · · · (n − 3r + 1)
·
n2r
k · · · (k − r + 1)
k=2r+1
(n − k − 1)r
pn−3r,k−r (x)
(n − r − k) · · · (n − 2r − k + 1)
≤ 2 × 2r × 3r ϕ2r (x).
POINTWISE APPROXIMATION
403
1 r
2r k+1
By this and δn2r ( k+r−j
n ) ∼ ϕ ( n ) + ( n ) , one has
(r)
Bn (f, x) ≤ L2 nr f 0 δnα(1−λ) (x).
Recalling that x ∈ Enc implies δn (x) ∼
√1 ,
n
we see that
r
δnr+α(λ−1) (x) Bn(r) (f, x) ≤ L3 n 2 f 0 .
(4.1)
For x ∈ En we use the expression (cf. [5])
Bn(r) (f, x) = (x(1 − x))−r
r
Qi (x, n)ni
i=0
n k
n
k=0
i
−x
k
pn,k (x)
n
f
with Qi (n, x) a polynomials in nx(1−x) of degree [(r−i)/2] with nonconstant
bounded coefficients. Thus,
−r
i
(x(1
−
x))
Q
(x,
n)n
≤
L
i
4
n
x(1 − x)
(r+i)/2
for x ∈ En .
If x ∈ En then δn (x) ∼ ϕ(x) and, recalling that δn2 ( nk ) ∼ ϕ2 ( nk ) + n1 , we have
by using Hölder inequality twice
i
(r+i)/2 r n n
(r)
k − x δ α(1−λ) k pn,k (x)f 0
Bn (f, x) ≤ L5
n
n
ϕ2 (x)
n
i=0
k=0
1
(r+i)/2 2i
n r 2
n
k
≤ L5 f 0
n ϕ2 (x)
i=0
ϕ
2
k=0
k
n
1
+
n
r
Proceeding as in (4.1), we obtain
n k=0
ϕ2
k
n
+
1
n
−x
n
k=0
pn,k (x)
α(1−λ)
2r
pn,k (x)
r
.
pn,k (x) ≤ Cδn2r (x).
It is known that for m ∈ N [5]
2m
n k
−x
pn,k (x) ≤ L6 n−m ϕ2m (x)
n
for x ∈ En .
k=0
Consequently, with x ∈ En , δn (x) ∼ ϕ(x) and
(4.2)
(r+i)/2
r n
(r)
Bn (f, x) ≤ L6 f 0
ϕ2 (x)
i=0
r/2
≤ L6 (r + 1)
n
ϕ2 (x)
ϕ2i (x)
ni
1
2
ϕα(1−λ) (x)
ϕα(1−λ) (x)f 0 .
Hence, using δn (x) ∼ ϕ(x) we obtain (3.1) for x ∈ En . We complete our
proof by using (4.1) and (4.2).
404 SHUNSHENG GUO, SHUJIE YUE, CUIXIANG LI, GE YANG AND YIGUO SUN
Proof of (3.4). We recall that [5]
n−r
k (r)
r
r
$
Bn (f, x) ≤ n
pn−r,k (x).
∆ 1 f
(4.3)
For 0 < k < n − r, by [5, p. 155],
k
k
1−
n
n
therefore δn
k
n
k
+y
≤R
n
≤ R1 δ n
k
n
≤ R2 n
k
1− −y ,
n
0<y<
r
,
n
+ y . Hence we have, as in [5],
r
$ 1 f ( k ) ≤ R2 n−r+1
∆
n n (4.4)
n
n
k=0
−r+1
(r) k
du
f
+
u
n
0
r
r
n
n
f r
0
δn−r+α(1−λ)
k
+ u du
n
r
k
.
≤ R2 n−r+1 f r δn−r+α(1−λ)
n
n
For k = 0, note that u ∈ (0, nr ) implies δn (u) ∼
r
$ 1 f (0) ≤ R3
∆
n
r
n
≤ R3 f r
≤ R4 n
Thus, we have
ur−1 |f (r) (u)|du
0
(4.5)
√1 .
n
r
n
ur−1 δn−r+α(1−λ) (u)du
0
−(r+α(1−λ))/2
f r .
Similarly for k = n − r we have
r
$ 1 f ( n − r ) ≤ R5 n−(r+α(1−λ))/2 f r .
∆
n
n (4.6)
k+1
From (4.3)-(4.6) we get, with δn ( nk ) ∼ δn ( n−r+2
),
r+α(λ−1)
(x)Bn(r) (f, x)
δ n
(4.7)
≤ R6 nr f r δnr+α(λ−1) (x) n
+
n−r−1
k=1
n−r δn−r+α(λ−1)
−r+α(λ−1)
2
(pn−r,0 (x) + pn−r,n−r (x))
k
pn−r,k (x) .
n
By a simple computation, it is easy to get (cf. [2])
n−r−1
k=1
and
n−r−1
k=1
n
k
n
n−k
r
r
pn−r,k (x) ≤ C
pn−r,k (x) ≤ C
1
,
xr
1
.
(1 − x)r
POINTWISE APPROXIMATION
Hence
n−r−1
ϕ
−2r
k=1
(4.8)
≤ 2r
k
pn−r,k (x)
n
n−r−1
k=1
Note that
δn ( nk )
405
n
k
r
+
n
n−k
r pn−r,k (x) ≤ C1 ϕ−2r (x).
∼ max{ϕ( nk ), √1n }, and that by (4.7) and (4.8) we have
r+α(λ−1)
(x)Bn(r) (f, x)
δn
≤ R6 f r δnr+α(λ−1) (x) nr (pn−r,0 (x) + pn−r,n−r (x))
+
n−r−1
k=1
δn−2r
r−α(1−λ)
2r
k
pn−r,k (x)
n
≤ R7 f r δnr+α(λ−1) (x) min nr ,
≤
n−r−1
k=1
r+α(λ−1)
−r+α(1−λ)
R8 f r δn
(x)δn
(x) =
ϕ−2r
r−α(1−λ)
2r
k
pn−r,k (x)
n
R8 f r .
Now we have proved the inequality (3.4). This finishes the proof of Lemma
3.1.
Proof of Lemma 3.2. It is known that, for 0 < t <
(cf. [7]),
t
2
− 2t
···
t
2
− 2t
1 rt
8r , 2
≤ x ≤ 1−
rt
2
ϕ−r (x + u1 + · · · + ur )du1 · · · dur ≤ Ctr ϕ−r (x).
Using this and Hölder’s inequality we obtain
t
2
− 2t
···
≤ C1
···
t
2
− 2t
δn−r+α(1−λ) (x + u1 + · · · + ur )du1 · · · dur
t
2
− 2t
t
2
− 2t
min ϕ
−r
(x + u1 + · · · + ur ), n
r
≤ C2 tr min ϕ−r (x), n 2
r−α(1−λ)
r
r
2
du1 · · · dur
r−α(1−λ)
r
tα(1−λ)
tα(1−λ)
≤ C3 tr δn−r+α(1−λ) (x),
which is the stated result.
References
[1] H. Berens and G. G. Lorentz, Inverse theorems for Bernstein polynomials, Indiana
Univ. Math. J. 21 (1972), 693-708.
406 SHUNSHENG GUO, SHUJIE YUE, CUIXIANG LI, GE YANG AND YIGUO SUN
[2] Z. Ditzian, A Global inverse theorem for combinations of Bernstein polynomials, J.
Approx. Th. 26 (1979), 277-292.
[3] Z. Ditzian, Direct estimate for Bernstein polynomials, J. Approx. Th. 79 (1994), 165166.
[4] Z. Ditzian and K. Ivanov, Bernstein-type operators and their derivatives, J. Approx.
Th. 56 (1989), 72-90.
[5] Z. Ditzian and V. Totik, Moduli of Smoothness, Springer-Verlag, Berlin-New York,
1987.
[6] S. Guo and C. Li, Inverse estimate for Bernstein polynomials, to appear.
[7] L. Xie, On the Linear Combination of Bernstein Polynomials, J. Math. Res. Expos.
15 (1995), 577-582 (in Chinese).
Department of Mathematics
Hebei Teacher’s University
Shijiazhuang 050016
People’s Republic of China