VOL. 80 1999 NO. 1
APPROXIMATION BY LINEAR COMBINATION OF SZ ´ ASZ–MIRAKIAN OPERATORS
BY
H. S. K A S A N A (PATIALA)
ANDP. N. A G R A W A L (ROORKEE)
Introduction. To approximate continuous functions on the interval [0, ∞), O. Sz´ asz and G. Mirakian generalized the Bernstein polynomials as follows:
S n (f ; x) =
∞
X
ν=0
φ n,ν (x)f (ν/n), where
φ n,ν (x) = e −nx (nx) ν /ν!, f ∈ C[0, ∞).
Singh [8] has obtained an estimate for bounded continuous functions in simultaneous approximation involving higher derivatives by these operators.
Sun [9] has tried to extend this estimate to functions of bounded variation with O(t αt ) growth of the derivatives and has remarked that unfortunately, for continuous derivatives his estimate does not include the case f 0 ∈ Lip 1 on every finite subinterval of [0, ∞). In this case he only obtains
S (r) n (f ; x) − f (r) (x) = O(log n/n), r = 0, 1, 2, . . .
This degree is worse than the usual degree 1/n. He put up the question of whether a unified approach can be developed which may improve this estimate for the class f 0 ∈ Lip 1 on every finite subinterval of [0, ∞).
In this paper we present a unified approach which improves the estimate of Sun [9] for continuous functions and moreover, it makes the results of Singh [8] applicable to unbounded functions.
In the sequel ha, bi denotes an open interval in [0, ∞) containing the closed interval [a, b] and k · k [a,b] means the sup norm on the space C[a, b].
The mth moment of the Sz´ asz–Mirakian operator is defined as V n,m (x) =
∞
X
ν=0
φ n,ν (x) ν n − x
m
, m = 0, 1, 2, . . .
1991 Mathematics Subject Classification: Primary 41A25; Secondary 41A28.
[123]
Let d 0 , d 1 , . . . , d k be arbitrary but fixed distinct positive integers. Then, following Kasana and Agrawal [5], the linear combinations S n (f, k, x) of S d
jn (f ; x), j = 0, 1, . . . , k, are introduced as
S n (f, k, x) = 1
∆
S d
0n (f ; x) d −1 0 d −2 0 . . . d −k 0 S d
1n (f ; x) d −1 1 d −2 1 . . . d −k 1 . . . . S d
kn (f ; x) d −1 k d −2 k . . . d −k k ,
where ∆ is the Vandermonde determinant obtained by replacing the oper- ator column of the determinant by the entries 1. On simplification this is reduced to
S n (f, k, x) =
k
X
j=0
C(j, k)S d
jn (f ; x), where
C(j, k) =
k
Y
i=0 i6=j
d j
d j − d i , k 6= 0, C(0, 0) = 1;
and this is the form of linear combinations considered by May [7].
1. To prove the main theorem we need the following auxilliary results.
Lemma 1.1. For V n,m (x), we have the recurrence relation nV n,m+1 (x) = xV n,m 0 (x) + mxV n,m−1 (x), m ≥ 1.
Gr¨ of [2] has proved that:
(a) V n,0 = 1, V n,1 = 0;
(b) V n,m (x) is a polynomial in x of degree [m/2] and in n −1 of degree m − 1, m > 1;
(c) for all finite x, V n,m (x) = O(n −[(m+1)/2] ).
Lemma 1.2. Let f (t) = O(t αt ) as t → ∞ with α > 0, and δ be a positive number. Then
X
|ν/n−x|>δ
φ n,ν (x)f (ν/n) = O(e −γn ), where γ is a constant depending on f , x and δ.
This lemma is due to Hermann [3]. A better estimate can also be found in [1].
Corollary 1.3. For δ > 0 and s = 0, 1, . . . , we have
k X
|ν/n−x|>δ
φ n,ν (x)(ν/n) αν/n k [a,b] ≤ K s n −s ,
where K s is a constant depending on s.
Lemma 1.4. If C(j, k), j = 0, 1, . . . , k, are defined as in the previous section then
k
X
j=0
C(j, k)d −m j = 1, m = 0, 0, m = 1, . . . , k.
May [7] has proved this lemma using Lagrange polynomials. A simpler exposition can be seen as Lemma 2 of Kasana [4].
Lemma 1.5. There exist polynomials T p,q,r (x) independent of n and ν such that
x r d r
dx r φ n,ν (x) = X
2p+q≤r p,q≥0
n p (ν − nx) q T p,q,r (x)φ n,ν (x).
This can be proved by induction; for a detailed proof we refer the reader to Kasana et al. [6].
2. We state and prove our main result as follows.
Theorem. Let f be bounded on every finite subinterval of [0, ∞) and f (t) = O(t αt ) as t → ∞, for some α > 0. If f (r+1) ∈ Cha, bi, then, for n sufficiently large,
kS n (r) (f, k, ·) − f (r) k [a,b] ≤ C 1 n −1/2 ω(f (r+1) ; n −1/2 ) + C 2 n −(k+1) , where C 1 = C 1 (k, r), C 2 = C 2 (k, r, f ) and ω(f (r+1) ; δ) is the modulus of continuity of f (r+1) on ha, bi defined as
ω(f (r+1) ; δ) = sup
x∈ha,bi
sup
|h|≤δ
|∆ h f (r+1) (x)|.
P r o o f. Write f (t) =
r+1
X
i=0
f (i) (x)
i! (t − x) i + f (r+1) (ξ) − f (r+1) (x)
(r + 1)! (t − x) r+1 χ(t) + ε(t, x)(1 − χ(t)),
where ξ lies between t and x and χ(t) is the characteristic function of ha, bi. As
S n (r) (f, k, x) =
k
X
j=0
C(j, k)S d (r)
jn (f ; x)
we have
S d (r)
jn (f ; x) =
∞
X
ν=0
φ (r) d
jn,ν (x)f
ν d j n
=
∞
X
ν=0
φ (r) d
j
n,ν (x)
r+1 X
i=0
f (i) (x) i!
ν d j n − x
i
+ f (r+1) (ξ) − f (r+1) (x) (r + 1)!
ν d j n − x
r+1
χ
ν d j n
+ ε
ν d j n , x
1 − χ
ν d j n
.
Thus,
S n (r) (f, k, x) =
k
X
j=0
∞
X
ν=0
C(j, k)φ (r) d
jn,ν (x)
r+1
X
i=0
f (i) (x) i!
ν d j n − x
i
+ f (r+1) (ξ) − f (r+1) (x) (r + 1)!
ν d j n − x
r+1
χ
ν d j n
+ ε
ν d j n , x
1 − χ
ν d j n
,
= I n,1 + I n,2 + I n,3 (say).
Now, I n,1 =
r+1
X
i=0
f (i) (x) i!
k
X
j=0
C(j, k)
∞
X
ν=0
φ (r) d
jn,ν (x)
ν d j n − x
i
=
k
X
j=0
C(j, k)
r+1
X
i=0
f (i) (x) i!
i
X
l=0
i l
(−x) i−l
∞
X
ν=0
φ (r) d
j
n,ν (x)
ν d j n
l
=
k
X
j=0
C(j, k)
r+1
X
i=0
f (i) (x) i!
i
X
l=0
i l
(−x) i−l
∞
X
ν=0
S d (r)
jn (t l ; x).
But S d
jn (t l ; x) is a polynomial in x of degree exactly l and the coefficient of x l is 1. So, for 0 ≤ l < r, S d
jn (t l ; x) = 0 and, for l = r, we have S d
jn (t l ; x) = r!. Further,
I n,1 =
k
X
j=0
C(j, k)
f (r) (x) + f (r+1) (x) (r + 1)!
r + 1 r
(−x)S d (r)
j
n (t r ; x) + r + 1
r + 1
(−x) 0 S d (r)
jn (t r+1 ; x)
= f (r) (x) +
k
X
j=0
C(j, k)
× f (r+1) (x)
r + 1)! {(−x)(r + 1)! + S d (r)
j
n (t (r+1) ; x)}
= f (r) (x) + f (r+1) (x)
k
X
j=0
C(j, k)
−x + 1
(r + 1)! S d (r)
jn (t (r+1) ; x)
= f (r) (x) + f (r+1) (x)
k
X
j=0
C(j, k)
×
−x + 1
(r + 1)!
(r + 1)!x + r(r + 1) 2d j n r!
= f (r) (x) + f (r+1) (x)
k
X
j=0
C(j, k)
−x +
x + r 2d j n
= f (r) (x) + f (r+1) (x) r 2n
k
X
j=0
C(j, k)
d j n = f (r) (x),
since P C(j, k)/(d j n) = 0, by Lemma 1.4. Thus, if S n (r) (f, k, x) = I n,1 + I n,2 + I n,3 , then S n (r) (f, k, x) − f (r) (x) = I n,2 + I n,3 .
To estimate I n,2 it is sufficient to consider it without the linear combi- nation. Let
I n,2 ≡
∞
X
ν=0
φ (r) n,ν (x) f (r+1) (ξ) − f (r+1) (x) (r + 1)!
ν n − x
r+1
χ ν n
. Then, using Lemmas 1.5 and 1.2, we get for t ∈ ha, bi and δ > 0,
I n,2 ≤
∞
X
ν=0
X
2p+q≤r p,q≥0
n p |ν − nx| q |T p,q,r (x)|
x r φ n,ν (x)
× |f (r+1) (ξ) − f (r+1) (x)|
(r + 1)!
ν n − x
r+1
χ ν n
≤ X
2p+q≤r p,q≥0
n p
∞
X
ν=0
φ n,ν (x)|ν − nx| q |T p,q,r (x)|
(r + 1)!x r
×
1 + |ν/n − x|
δ
ω(f (r+1) ; δ) ν n − x
r+1
≤ M 1 (r)ω(f (r+1) ; δ) X
2p+q≤r p,q≥0
n p+q
∞
X
ν=0
φ n,ν (x)
×
ν n − x
q+r+1
+ |ν/n − x| q+r+2 δ
, where
M 1 (r) = sup
a≤x≤b
sup
2p+q≤r
p;q≥0