Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2012, Article ID 546784, 28 pages
doi:10.1155/2012/546784
Research Article
Approximation by Lupas-Type Operators and
Szász-Mirakyan-Type Operators
Hee Sun Jung1 and Ryozi Sakai2
1
2
Department of Mathematics Education, Sungkyunkwan University, Seoul 110-745, Republic of Korea
Department of Mathematics, Meijo University, Nagoya 468-8502, Japan
Correspondence should be addressed to Hee Sun Jung, hsun90@skku.edu
Received 28 July 2011; Accepted 5 January 2012
Academic Editor: Yuantong Gu
Copyright q 2012 H. S. Jung and R. Sakai. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Lupas-type operators and Szász-Mirakyan-type operators are the modifications of Bernstein polynomials to infinite intervals. In this paper, we investigate the convergence of Lupas-type operators
and Szász-Mirakyan-type operators on 0, ∞.
1. Introduction and Main Results
For f ∈ C0, 1, Bernstein operator Bn fx is defined as follows: Let
pn,k x n
k
xk 1 − xn−k ,
0 k n,
1.1
and then we define
n
k
Bn f x .
pn,k xf
n
k0
1.2
Derriennic 1 gave a modified operator of Bn f such as
Mn∗ f
1
n
x n 1 pn,k x pn,k tftdt,
k0
0
1.3
2
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and obtained the result that for f ∈ C2 0, 1,
lim
Mn∗ f x − fx 1 − 2xf x x1 − xf x.
n→∞
1.4
Lupas investigated a family of linear positive operators which mapped the class of all bounded and continuous functions on 0, ∞ into C0, ∞ such that
Ln f x ∞
nk−1
k0
xk
1 x
k
f
nk
k
,
n
x ∈ 0, ∞.
1.5
Moreover, Sahai and Prasad 2 modified Lupas operators as follows: Let f be integrable on
0, ∞ and let n be a positive integer. Then we define
∞
Mn f x n − 1 Pn,k x
∞
Pn,k y f y dy,
x ∈ 0, ∞,
0
k0
1.6
where
nk−1
Pn,k x k
xk
1 xnk
.
1.7
In this paper, we assume that n is a positive integer. Then they obtained the following;
Theorem 1.1 see 2, Theorem 1. If f is integrable on 0, ∞ and admits its r 1th and r 2th
derivatives, which are bounded at a point x ∈ 0, ∞, and f r x Oxα (α is a positive integer
2) as x → ∞, then
r
lim n Mn f
x − f r x r 11 − 2xf r1 x x1 − xf r2 x.
n→∞
1.8
Theorem 1.1 holds only for bounded x K, so it does not mean the norm convergence
on 0, ∞. In this paper, we improve Theorem 1.1 with respect to the norm convergence on
0, ∞.
Let 0 < p ∞ and let w be a positive weight, that is, wx 0 for x ∈ R. For a function
g on 0, ∞, we define the norm by
g :
Lp 0,∞
⎧
⎪
⎪
⎪
⎨
1/p
0,∞
gtp dt
,
⎪
⎪
⎪
⎩ sup gt,
0,∞
0<p<∞
1.9
p ∞.
For convenience, for nonnegative integers n 2, r, and n − r − 2 0, we let
An,r :
n − 1!n − 2!
.
n − r − 2!n r − 1!
1.10
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3
Then we have the following results:
Theorem 1.2. Let 0 < p ∞. Let α and r be nonnegative integers and n − r − 2 0. Let f ∈
Cr1 0, ∞ satisfy
r f x O1x 1α ,
r1 x O1x 1α2 .
f
1.11
Then we have uniformly for f and n,
r
1
x − f r x O 1/3 x 1α2 .
An,r Mn f
n
1.12
In particular, if x 1α2 wxLp 0,∞ < ∞, then we have uniformly for n,
r
x − f r x wx
An,r Mn f
Lp 0,∞
O
1
n1/3
1.13
.
Remark 1.3. a We see that for nonnegative integers n 2, r, and n − r − 2 0,
n − 1!n − 2!
−→ 1 as n −→ ∞.
n − r − 2!n r − 1!
1.14
b The following weight is useful.
⎧
1
⎪
1 ⎨λ > α 2,
p
wλ x 1 xλ ⎪
⎩λ α 2,
0 < p < ∞,
1.15
p ∞.
Let
ψx :
1
.
1x
1.16
Theorem 1.4. Let r and β be nonnegative integers and n − r − 2 0. Let f ∈ Cr2 0, ∞ satisfy
r1
xψ 2β1 x
f
r2
xψ 2β x
f
< ∞.
1.17
1
r1
r2
2β1
2β
O
f
.
xψ
x
xψ x
f
L∞ 0,∞
L∞ 0,∞
n
1.18
L∞ 0,∞
< ∞,
L∞ 0,∞
Then we have uniformly for f and n,
r
x − f r xψ 2β2 x
An,r Mn f
4
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Let us define the weighted modulus of smoothness by
ωk f; η; t : sup Δkh f·η·
L∞ 0,∞
0ht
t 0, k 1, 2,
,
1.19
where
Δ1h fx fx h − fx,
Δ2h fx fx − 2fx h fx 2h.
1.20
Theorem 1.5. Let β and r be nonnegative integers and n − r − 2 0. Let f ∈ Cr 0, ∞. Then we
have uniformly for f and n,
r
x − f r x ψ 2β2 x
An,r Mn f
L∞ 0,∞
1
1
1
ω2 f r ; ψ 2β ; √
.
C √ ω1 f r ; ψ 2β1 ; √
n
n
n
1.21
The Szász-Mirakyan operators are also generalizations of Bernstein polynomials on infinite intervals. They are defined by:
∞
k
,
Sn f x Sn,k xf
n
k0
1.22
where
Sn,k x e−nx nxk
.
k!
1.23
In 3, the class of Szász-Mirakyan operators Sn;r,q f; x was defined as follows:
Sn;r,q f; x :
∞
rk
1 nxrk
,
f
Ar nx k0 rk!
nq
x ∈ 0, ∞,
1.24
where q > 0 and
Ar t ∞
trk
,
rk!
k0
t ∈ 0, ∞.
1.25
Theorem 1.6 see 3. Let q > 0 and r ∈ N be fixed numbers. Then there exists Mq,r const. > 0
depending only on q and r such that, for every uniformly continuous and bounded function fxe−qx
on 0, ∞, the following inequalities hold;
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5
a
Sn;q,r f; x − fx ϕxe−qx L∞ 0,∞
1 f xe−qx f xe−qx ;
Mq,r
L∞ 0,∞
L∞ 0,∞
nq
1.26
b
Sn;q,r f; x − fx ϕxe−qx L∞ 0,∞
Mq,r
1 δn,q ω1 f; e−qx ; δn,q ω2 f; e−qx ; δn,q ,
nq
1.27
where δn,q : n q−1/2 .
c for every fixed x ∈ 0, ∞, we have for every continuous f with f j xe−qx , j 0, 1, 2,
bounded on 0, ∞,
x
lim n Sn;q,r f; x − fx −qxf x f x.
2
n→∞
1.28
Now, we modify the Szász-Mirakyan operators as follows: let f be integrable on 0, ∞,
then we define
∞
Qn,β f x n β
Sn,k x
k0
∞
Snβ,k y f y dy,
x ∈ 0, ∞,
0
1.29
where β is a nonnegative integer. Then we have the following results:
Theorem 1.7. Let α, β and r be nonnegative integers. Let f ∈ Cr1 0, ∞ satisfies
r f x O1eβx x 1α ,
r1 x O1eβx x 1α2 .
f
1.30
Then one has uniformly for f and n,
n β r
r
1
r
Qn,β f
x − f x O 1/3 eβx x 1α2 .
n
n
1.31
In particular, let 0 < p ∞. If one supposes eβx x 1α2 wxLp 0,∞ < ∞, then one has uniformly for f and n,
r
n β r
1
r
O
.
−
f
wx
Q
f
x
x
n,β
n
n1/3
Lp 0,∞
1.32
6
Journal of Applied Mathematics
Remark 1.8. a We note that for nonnegative integers β and r,
nβ
n
r
−→ 1
1.33
as n −→ ∞.
b The following weight is useful.
wλ,β x e−βx wλ x,
1.34
where wλ x is defined in Remark 1.3.
Theorem 1.9. Let β, γ, and r be nonnegative integers. Let f ∈ Cr2 0, ∞ satisfies
r1
xe−βx ψ 2γ1 x
f
L∞ 0,∞
< ∞,
r2
xe−βx ψ 2γ x
f
L∞ 0,∞
< ∞.
1.35
Then one has uniformly for f and n,
n β r
r
r
−βx 2γ2
Qn,β f
x − f x e ψ
x
n
1
r1
r2
−βx 2γ1
−βx 2γ
O
xe ψ
x
f
xe ψ x
.
f
L∞ 0,∞
L∞ 0,∞
n
1.36
Theorem 1.10. Let β, γ, and r be nonnegative integers. Then one has for f ∈ Cr 0, ∞,
n β r
r
r
−βx 2γ2
Qn,β f
x − f x e ψ
x
n
L∞ 0,∞
1
1
1
ω2 f; e−βx ψ 2γ x; √
.
C √ ω1 f; e−βx ψ 2γ1 x; √
n
n
n
1.37
2. Proofs of Results
First, we will prove results for Lupas-type operators such as Theorems 1.2, 1.4, and 1.5. To
prove theorems, we need some lemmas.
Lemma 2.1. Let m and r be nonnegative integers and n > m r 1. Let
Tn,m,r x : n − r − 1
∞
∞
m
Pnr,k x
Pn−r,kr y y − x dy.
k0
Then
i Tn,0,r x 1,
0
2.1
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7
ii
r 11 2x
,
n − r 2
Tn,1,r x 2n − 1x1 x
r 1r 21 2x2
;
Tn,2,r x n − r − 2n − r − 3
n − r − 2n − r − 3
2.2
iii for m 1,
n − m − r − 2Tn,m1,r x
x1 x Tn,m,r
x 2mTn,m−1,r x m r 11 2xTn,m,r ,
2.3
where n > m r 2;
iv for m 0,
Tn,m,r x O
qn,m,r x,
nm1/2
1
2.4
where qn,m,r x is a polynomial of degree m such that the coefficients are bounded independently of n and they are positive for n > m r 1.
Proof. i, ii, and iii have been proved in 2, Lemma 1. So we may show only the part of
2.4. For m 1, 2, 2.4 holds. Let us assume 2.4 for m 2. We note
Tn,m,r
x
O
1
nm1/2
qm,r
x,
qm,r
x ∈ Pm−1 .
2.5
So, we have by the assumption of induction,
n − m − r − 2Tn,m1,r x
x1 x T n,m,r x 2mTn,m−1,r x m r 11 2xTn,m,r
1
1
x1 x C m1/2 qm,r x 2m m/2 qm−1,r x
n
n
m r 11 2x
2.6
1
qm,r x.
nm1/2
Here, if m is even, then
m
m2
m2
m1
1
1
,
2
2
2
2
m
m2
m2
1
1
,
2
2
2
2
m
2.7
8
Journal of Applied Mathematics
and if m is odd, then
m1
m2
m1
1
1
,
2
2
2
m−1
m1
m2
1
1
.
2
2
2
2
m
2.8
Hence, we have
Tn,m1,r x O
qn,m1,r x,
nm2/2
1
2.9
and here we see that qn,m1,r x is a polynomial of degree m 1 such that the coefficients of
qn,m1,r x are bounded independently of n. Moreover, we see from 2.6 that the coefficients
of qn,m1,r x are positive for n > m r 2.
Lemma 2.2 see 2, Lemma 2. Let r be a nonnegative integer and n − r − 2 0. Then one has for
f ∈ Cr 0, ∞:
∞
r
n − r − 1!n r − 1! Pnr,k x
Mn f
x n − 1!n − 2!
k0
∞
Pn−r,kr y f r y dy.
0
2.10
Let
∞
∞
r
Pn−r,kr y f r y dy.
Mn f
x n − r − 1 Pnr,k x
k0
0
2.11
Then we have
r
r
n f
M
x An,r Mn f
x,
2.12
where An,r is defined by 1.10.
Proof of Theorem 1.2. Let |y − x| 1. By the second inequality in 1.11,
r y − f r x y − xf r1 ξ
f
O1y − xξα2 O1y − xx 1α2 .
2.13
Journal of Applied Mathematics
9
Let ε : n−γ , 0 < γ < 1,
r
r
M
n f
−
f
x
x
∞
n − r − 1 Pnr,k x
k0
×
|y−x|<ε
Pn−r,kr y f r y − f r xdy r r
Pn−r,kr y f
y − f xdy |y−x|ε
: A B.
2.14
First, we see by 2.13 and Lemma 2.1,
∞
α2
A O1n − r − 1 Pnr,k x
Pn−r,kr y y − x x 1 dy
|y−x|<ε
k0
O1ε|Tn,0,r x|x 1
α2
O1εx 1
α2
2.15
.
Next, we estimate B. By the first inequality in 1.11,
∞
r r B Cn − r − 1 Pnr,k x
Pn−r,kr y f
y f x dy
|y−x|>ε
k0
∞
α
α
Cn − r − 1 Pnr,k x
Pn−r,kr y y 1 x 1 dy.
|y−x|>ε
k0
2.16
Here, using
α
α α α i
y1 y−x x1 y − x x 1α−i
i
i0
2.17
and the notation:
i
⎧
⎨1,
i : odd
⎩0,
i : even,
2.18
we have
∞
B C|n−r − 1| Pnr,k x
k0
|y−x|>ε
Pn−r,kr y ×
α
α i
α−i
α
y − x x 1 2x 1 dy
i
i1
10
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∞
Pnr,k x
C|n − r − 1|
k0
|y−x|>ε
∞
C|n − r − 1| Pnr,k x
k0
Pn−r,kr y ×
α
α i1
i y − x i
α−i
dy
y−x x 1
ε i
y−x 2
Pn−r,kr y
x 1α dy
ε
|y−x|>ε
: B1 B2 .
2.19
Then, we obtain
α
α 1 i
α−i
B1 C
Tn,ii
,r
x 1
ε
i
i1
α
α
nγi
qn,ii
,r xx 1α−i
O
C
ii
1/2
n
i
i1
1
1
αi
O ii
1/2−γi
x 1
O 1−γ x 1α1 .
n
n
2.20
Here, we used the following that for i 1,
i i
1
− γi
1 − γ,
2
2.21
because
⎧
i1
⎪
⎨
− γ,
i i
1
2
− γi
⎪
2
⎩i,
2
i : odd,
2.22
i : even.
And we know that
2
1 2 α
1
1
B2 C|Tn,2,r x|
xα O 3/2 qn,2,r x
x
ε
ε
n
n2γ
1
qn,2,r xxα O
O
x 1α2 .
n3/2
n1−2γ
2.23
Thus, we obtain
BO
1
n1−2γ
x 1α2 .
2.24
Journal of Applied Mathematics
11
Therefore, we have uniformly on n,
r
1
1
α2
r
M
n f
x − f x O γ x 1 O 1−2γ x 1α2 .
n
n
2.25
Here, if we let γ 1/3, then we have
r
1
r
M
n f
x − f x O 1/3 x 1α2 ,
n
2.26
that is, 1.12 is proved. So, we also have a norm convergence 1.13.
Proof of Theorem 1.4. We know that for f ∈ Cr2 0, ∞,
f
r
t f
r
x f
r1
xt − x t
t − uf r2 udu,
2.27
x
t
1 x2β 1 t2β t − x2 ,
t − uf r2 udu Cf r2 xψ 2β x
x
L∞ 0,∞
2.28
where ψt 1/1 x. Then we obtain from 2.10 and 2.27,
r
n f
M
x
f
r
x f
r1
xTn,1,r x n − r − 1
∞
Pnr,k x
k0
∞
Pn−r,kr y
0
y
y − u f r2 udu dy
x
2.29
and from 2.28,
∞
∞
y
r2
Pn−r,kr y
y−u f
udu dy
n − r − 1 Pnr,k x
0
x
k0
f r2 xψ 2β x
L∞ 0,∞
∞
∞
2β 2 Pn−r,kr y 1 y
y − x dy .
1 x |Tn,2,r x| n − r − 1 Pnr,k x
0
k0
×
2β
2.30
Using 1 y2β Cy − x2β 1 x2β , we have
∞
∞
2β 2 Pn−r,kr y 1y
y−x dy
n−r −1 Pnr,k x
0
k0
C Tn,2β2,r x 1x2β |Tn,2,r x| .
2.31
12
Journal of Applied Mathematics
Therefore, we have
r
r
M
n f
x − f xψ 2β2 x
f r1 xψ 2β1 x|Tn,1,r x| ψx
Cf r2 xψ 2β x
1 x2β |Tn,2,r x|ψ 2β2 x
L∞ 0,∞
Cf r2 xψ 2β x
L∞ 0,∞
Tn,2β2,r xψ 2β2 x
1 r1
O
xψ 2β1 xgn,1,r xψx
f
n
1 r2
O
xψ 2β x
1 x2β gn,2,r xψ 2β2 x
f
L∞ 0,∞
n
1
gn,2β2,r xψ 2β2 x.
O β1 f r2 xψ 2β x
L∞ 0,∞
n
2.32
Since we know that for x ∈ 0, ∞,
gn,1,r xψx C,
1 x2β gn,2,r xψ 2β2 x C,
gn,2β2,r xψ 2β2 x C,
2.33
we have
r
r
M
n f
x − f xψ 2β2 x
1
r1
r2
2β1
2β
f
.
O
xψ
x
xψ x
f
L∞ 0,∞
L∞ 0,∞
n
2.34
Lemma 2.3. Let r and β be nonnegative integers and n − r − 2 0. Let f ∈ Cr 0, ∞ satisfies
r 2β f ψ L∞ 0,∞
< ∞.
2.35
Then one has uniformly for n, f and x ∈ 0, ∞,
r 2β
M
n f
.
xψ x Cf r ψ 2β L∞ 0,∞
2.36
Journal of Applied Mathematics
13
Proof. Using 1 y2β Cy − x2β 1 x2β , we have
∞
∞
2β Pn−r,kr y 1 y dy C ψ −2β x Tn,2β,r x .
n − r − 1 Pnr,k x
0
k0
2.37
The assumption 2.35 means
2β
r y C 1y .
f
2.38
Then we can obtain by 2.10,
r M
n f
x
∞
∞
2β Cn − r − 1 Pnr,k x
Pn−r,kr y 1 y dyf r ψ 2β L∞ 0,∞
0
k0
C ψ
−2β
x Tn,2β,r x f r ψ 2β C ψ −2β x O
1
nβ
2.39
L∞ 0,∞
r 2β qn,2β,r x f ψ L∞ 0,∞
.
Consequently, since |qn,2β,r x|ψ 2β x is uniformly bounded on 0, ∞, we have the result.
The Steklov function fh x for f ∈ C0, ∞ is defined as follows:
4
f h x : 2
h
h/2
2fx s t − fx 2s t ds dt,
x 0, h > 0.
2.40
0
Then for the Steklov function fh x with respect to f ∈ C0, ∞, we have the following properties.
Lemma 2.4 cf.4. Let fx ∈ C0, ∞ and ηx be a positive and nonincreasing function on
0, ∞. Then (i) fh x ∈ C2 0, ∞;
(ii)
f x − fx ηx h
L
∞ 0,∞
h
ω2 f; η;
;
2
2.41
(iii)
f h xηx
L∞ 0,∞
ηx
ηx
1 4
h
ω1 f; η; h
;
ω1 f; η;
h
2 ηx h/2 h
ηx h
2.42
14
Journal of Applied Mathematics
(iv)
f h xηx
L∞ 0,∞
1 4
h
ω
2ω
f;
η;
f;
η;
h
.
2
2
2
4
h2
2.43
Proof. i For f ∈ C0, ∞, we have the Steklov functions fh x and fh x as follows. We
note
4
f h x 2
h
h/2 xh/2
0
2fu tdu −
x
xh
x
1
fu 2tdu dt,
2
x 0, h > 0.
2.44
Then, we can see from 2.44,
h/2 1
h
2 f x t − fx t − fx h 2t − fx 2t dt
2
2
0
1
4 h/2
2Δ1h/2 fx t − Δ1h fx 2t dt.
2
2
h 0
4
f h x 2
h
2.45
Similarly to 2.44, we know
xh/2 1 xh h
4
− fu du −
f u
f h x 2 2
fu h − fu du .
2
4 x
h
x
2.46
Therefore, we have from 2.46,
h
4
1
fx − fx 2h − 2fx h fx
f h x 2 2 fx h − 2f x 2
4
h
2.47
1 2
4
2
2 2Δh/2 fx − Δh fx .
4
h
Therefore, i is proved.
ii We easily see from 2.44 that
h/2
2
fx − f x ηx 4
Δ
fxηxds
dt
st
h
h2 0
ω2
h
.
f; η;
2
2.48
Journal of Applied Mathematics
15
iii From 2.46, we have
4 h/2 ηx
1
dt
2 Δh/2 fx tηx t
f h xηx 2
h 0
ηx t 4 h/2 1 ηx
1
2
Δh fx 2tηx 2t
dt
h 0 2
ηx 2t ηx
ηx
4
1 h
ω1 f; η;
ω1 f; η; h
.
h
2 ηx h/2 h
ηx h
2.49
iv From 2.47, we have
4
1 h
ω2 f; η; h .
f h xηx 2 2ω2 f; η;
2
4
h
2.50
Proof of Theorem 1.5. We know that for fx ∈ Cr 0, ∞,
r
f h x f r x,
h
r1
f h x f r x,
r2
f h x f r x.
h
h
2.51
Then, we have
r
r
2β2
M
n f
x − f x ψ
x
L∞ 0,∞
r
2β2
n f − f
M
xψ
x
h
L∞ 0,∞
2.52
r
r
2β2
n fh
M
x
−
f
ψ
x
h x
L∞ 0,∞
f r h x − f r x ψ 2β2 x
L∞ 0,∞
.
From 2.51 and 2.41 of Lemma 2.4,
r
2β2
M
n f − f
xψ
x
h
L∞ 0,∞
r f r x − f h x ψ 2β2 x
L∞ 0,∞
f r x − f r x ψ 2β2 x
h
ω2 f r ; ψ 2β2 ; h .
L∞ 0,∞
2.53
16
Journal of Applied Mathematics
Here, we suppose 0 < h 1 and then we know that
ψx
2,
ψx h
ψx
2.
ψx h/2
2.54
From Theorem 1.4, 2.51, 2.42, and 2.43 of Lemma 2.4, we have
r
r
2β2
M
n f
x − fh x ψ
x
h
L∞ 0,∞
1
r1
r2
O
f h xψ 2β x
f h xψ 2β1 x
L∞ 0,∞
L∞ 0,∞
n
1
1
1
ω1 f r ; ψ 2β1 ; h 2 ω2 f r ; ψ 2β ; h .
O
n
h
h
2.55
Therefore, we have
Mn fr x − f r x ψ 2β2 x
L∞ 0,∞
1 r 2β1 1
1
r
2β
ω1 f ; ψ
O
; h 2 ω2 f ; ψ ; h
ω2 f r ; ψ 2β2 ; h .
n
h
h
2.56
√
If we let h 1/ n, then
r
r
2β2
M
n f
x − f x ψ
x
L∞ 0,∞
1
1
1
ω2 f r ; ψ 2β ; √
,
C √ ω1 f r ; ψ 2β1 ; √
n
n
n
2.57
√
√
because ω2 f r ; ψ 2β2 ; 1/ n ω2 f r ; ψ 2β ; 1/ n.
From now on, we will prove Theorems 1.7, 1.9, and 1.10, which are the results for the
Szász-Mirakyan operators, analogously to the case of Lupas-type operators.
Lemma 2.5. Let r be a nonnegative integer. Then one has for f ∈ Cr 0, ∞,
r
Qn,β f
x n β
n
nβ
r ∞
k0
Sn,k x
∞
0
Snβ,kr y f r y dy,
x ∈ 0, ∞. 2.58
Journal of Applied Mathematics
17
Proof. We know that
r
Sn,k x
−ny r−i k i
r
ny
r e i0
k!
i
−ny i
r
r e r
Sn,k x i
i0
ny
k r−i
k!
i
i0
r
r
−1r −1i nr Sn,k−i x,
2.59
r
r
i0
i
−1i nr Sn,k−ri x.
Therefore, we have
r
Qn,β f
x
nβ
∞
r
Sn,k x
k0
nβ
∞
Snβ,k y f y dy
0
r ∞
r
i0 ki
nβ
nβ
n
nβ
n
nβ
i
−1r −1i nr Sn,k−i x
∞
Snβ,k y f y dy
0
r ∞
Sn,k x
k0
∞ r
r
0
i0
i
r
−1r −1i n β Snβ,ki y f y dy
2.60
∞
r ∞
r
Sn,k x
−1r Snβ,kr y f y dy
k0
0
∞
r ∞
n
nβ
Sn,k x
Snβ,kr y f r y dy.
n β k0
0
Lemma 2.6. Let a, b, and m be nonnegative integers.
Rn,m,r a, b; x : n b
∞
∞
m
Sna,k x
Snb,kr y y − x dy.
k0
0
2.61
Then one has
i Rn,0,r a, b; x 1 and Rn,1,r a, b; x a − bx r 1/n b;
ii For m 1
n bRn,m1,r a, b; x
1
xRn,m,r a, b; x a − bx m r 1Rn,m,r a, b; x 2xmRn,m−1,r a, b; x;
2.62
18
Journal of Applied Mathematics
iii
Rn,m,r a, b; x O
1
nm1/2
gn,m,r a, b; x,
2.63
where gn,m,r a, b; x is a polynomial of degree m such that the coefficients of gn,m,r a,
b; x are bounded independently of n.
Proof. Let Rn,m,r x : Rn,m,r a, b; x. Then i
∞
∞
Rn,0,r x n b Sna,k x
Snb,kr y dy 1,
0
k0
∞
Rn,1,r x n b Sna,k x
∞
Snb,kr y y − x dy
0
k0
∞
kr1
−x
Sna,k x
nb
k0
∞
k
r1
−x
Sna,k x
n
b
n
b
k0
n ax r 1
a − bx r 1
−x .
nb
nb
nb
2.64
1
ii Using xSn,k x k − nxSn,k x, we obtain
1
x Rn,m,r x mRn,m−1,r x
n b
∞
∞
m
1
xSna,k x
Snb,kr y y − x dy
0
k0
n b
2.65
∞
∞
m
Sn,k x
k − n axSnb,kr y y − x dy.
k0
0
Here, we see
k − n axSnb,kr y
k r − n by − r a − bx n b y − x Snb,kr y
1
ySnb,kr y − r a − bxSnb,kr y n bSnb,kr y y − x .
2.66
Journal of Applied Mathematics
19
Then substituting 2.66 for 2.65, we consider the following;
1
x Rn,m,r x mRn,m−1,r x
∞
∞
1
n b Sn,k x
ySnb,kr y − r a − bxSnb,kr y
0
k0
m
n bSnb,kr y y − x y − x dy
: .
1
2
2.67
3
Then, we have
1
∞
∞
m
1
n b Sn,k x
ySnb,kr y y − x dy
0
k0
∞
∞
m1
1
n b Sna,k x
Snb,kr y y − x
dy
0
k0
∞
∞
m
1
xn b Sna,k x
Snb,kr y y − x dy
k0
2.68
0
−m 1Rn,m,r x − xmRn,m−1,r x.
Here the last equation follows by parts of integration. Furthermore, we have
1
−r a − bxRn,m,r x n bRn,m1,r x.
2.69
2
Therefore, we have
1
n bRn,m1,r x xRn,m,r x a − bx m r 1Rn,m,r x 2xmRn,m−1,r x.
2.70
iii It is proved by the same method as the proof of Lemma 2.1 iv.
Proof of Theorem 1.7. Let |y − x| 1. By the second inequality in 1.30,
r y − f r x y − xf r1 ξ
f
Cy − xeβξ ξα2 Cy − xeβx x 1α2 .
2.71
20
Journal of Applied Mathematics
Let ε n−γ , 0 < γ < 1,
n β r
r
r
Qn,β f
x − f x
n
∞
nβ
Sn,k x
Snβ,kr y f r y − f r xdy
|y−x|<ε
k0
r y − f r xdy Snβ,kr y f
|y−x|ε
2.72
: A B.
First, we see that by 2.71 and Lemma 2.6i,
∞
AC nβ
Sn,k x
k0
|y−x|<ε
Snβ,kr y y − xdy eβx x 1α2
Cεeβx x 1α2
1
O γ eβx x 1α2 .
n
2.73
Next, to estimate B, we split it into two parts:
∞
r r
B nβ
Sn,k x
Snβ,kr y f
y − f xdy
|y−x|ε
k0
nβ
∞
Sn,k x
k0
∞
nβ
k0
|y−x|ε
Sn,k x
|y−x|ε
Snβ,kr y f r y f r x dy
2.74
α
Snβ,kr y eβy y 1 eβx x 1α dy
: B1 B2 .
First, we estimate
∞
Sn,k x
B1 n β
k0
|y−x|>ε
α
Snβ,kr y eβy y 1 dy.
2.75
Then, using the following facts:
βy
n β kr
,
Snβ,kr y e Sn,kr y
n
2.76
Journal of Applied Mathematics
21
nβ r
Sn,k x Snβ,k xeβx ,
n
α
α α α i
y1 y−x x1 y − x x 1α−i ,
i
i0
nβ
n
kr
2.77
2.78
we have
∞
n β r
α
B1 Ce n β
Snβ,k x
Sn,kr y y 1 dy
n
|y−x|>ε
k0
βx
∞
n β r1 Ceβx n
Snβ,k x
n
k0
α
α i
Sn,kr y
×
y − x x 1α−i dy
i
|y−x|>ε
i1
∞
n β r
Ceβx n
Snβ,k x
Sn,kr y x 1α dy
n
|y−x|>ε
k0
2.79
: Ceβx B11 B12 .
Then, using 2.18 and Lemma 2.6, we have
B11
α
α i
Sn,kr y
y − x x 1α−i dy
i
|y−x|>ε
i1
k0
∞
α
α i y − x i
x 1α−i dy
Sn,kr y
n Snβ,k x
y−x ε
i
|y−x|>ε
i1
k0
∞
n Snβ,k x
∞
n Snβ,k x
k0
α
α ii
1 i
Sn,kr y
y−x
x 1α−i dy
ε
i
|y−x|>ε
i1
2.80
α
α
1 i
Rn,ii
r β, 0; x
x 1α−i
ε
i
α
α
i
O ii
1/2 qn,ii
,r β, 0; x x 1α−i .
n
i
i1
i1
Then by 2.21 we have
B11
O 1−γ x 1α1 .
n
1
2.81
22
Journal of Applied Mathematics
For B12 , we have
B12
∞
n Snβ,k x
k0
|y−x|>ε
Sn,kr y x 1α dy
y−x 2
Sn,kr y
dyx 1α
ε
|y−x|>ε
k0
1 2
n2γ
α
qn,2,r β, 0; x x 1α
Rn,2,r β, 0; x
x 1 O
3/2
ε
n
1
O 1−2γ x 1α2 .
n
∞
n Snβ,k x
2.82
From 2.81, 2.82 and 2.79, we have
B1 O
1
n1−γ
eβx x 1α2 .
2.83
We estimate
∞
B2 n β
Sn,k x
k0
|y−x|>ε
Snβ,kr y eβx x 1α dy.
2.84
Then we can estimate B2 by the same method as B12 ,
∞
B2 n β
Sn,k x
k0
|y−x|>ε
Snβ,kr y eβx x 1α dy
y−x 2
nβ
Sn,k x
Snβ,kr y
dy eβx x 1α
ε
|y−x|>ε
k0
∞
2.85
1 2 βx
Rn,2,r 0, β; x
e x 1α ,
ε
so we have
B2 O
n2γ
n3/2
qn,2,r 0, β; x eβx x 1α O
1
n1−2γ
eβx x 1α2 .
2.86
Consequently, we obtain from 2.83 and 2.86,
1
1
1
α2
βx
B O 1−γ O 1−2γ
O 1−2γ eβx x 1α2 .
e x 1
n
n
n
2.87
Journal of Applied Mathematics
23
Therefore, from 2.73 and 2.87, we have uniformly on n,
n β r
r
1
1
r
βx α2
Qn,β f
x − f x O γ e x O 1−2γ eβx xα2 .
n
n
n
2.88
Here, if we let γ 1/3, then we have
n β r
r
1
r
O
eβx xα2 ,
−
f
Q
f
x
x
n,β
n
n1/3
2.89
that is, 1.31 is proved. So, we also have a norm convergence 1.32.
Lemma 2.7. Let m and b be nonnegative integers. Let
∞
∞
m
Un,m,r b; x : n b Sn,k x
Snb,kr y y − x eby dy.
0
k0
2.90
Then one has
Un,m,r b; x nb
n
r1
ebx Rn,m,r b, 0; x.
2.91
Proof. From 2.76 and 2.77 we have
n b kr
,
n
n b kr
nb r
Sn,k x Snb,k xebx .
n
n
Snb,kr xebx Sn,kr x
2.92
2.93
We have from 2.92, 2.93, and noting 2.61,
∞
∞
m
Un,m,r b; x n b Sn,k x
Snb,kr y y − x eby dy
k0
0
∞
∞ m
n b kr
n b
Sn,k x
Sn,kr y y − x dy
n
0
k0
∞
∞
m
n b r bx n b
e
Snb,k x
Sn,kr y y − x dy
n
0
k0
nb
n
r1
ebx Rn,m,r b, 0; x.
2.94
24
Journal of Applied Mathematics
Proof of Theorem 1.9. We prove this theorem, similarly to the proof of Theorem 1.4. Using
2.93 and 2.27, we have for f ∈ Cr2 0, ∞,
nβ
n
r
r
Qn,β f
x
∞
nβ
Sn,k x
f
Snβ,kr y f r y dy
0
k0
r
∞
x f
r1
x n β
∞
nβ
Sn,k x
∞
∞
Sn,k x
∞
0
k0
y
Snβ,kr y
0
k0
Snβ,kr y y − x dy
r2
y−u f
udu dy
2.95
x
f r x f r1 xRn,1,r 0, β; x
∞
∞
y
nβ
Sn,k x
Snβ,kr y
y − u f r2 udu dy.
0
k0
x
We estimate the last term. We note the given condition:
r2
xe−βx ψ 2γ x
f
L∞ 0,∞
< ∞.
2.96
Using the inequality 1 t2γ C1 x2γ t − x2γ and Lemma 2.7, we have
∞
: n β
Sn,k x
∞
2
Snβ,kr y eβy ψ −2γ y eβx ψ −2γ x y − x dy
0
k0
∞
∞
C nβ
Sn,k x
Snβ,kr y
0
k0
× eβy ψ −2γ x eβy
y−x
2γ
eβx ψ −2γ x
2
y − x dy
C ψ −2γ xUn,2,r β; x Un,2γ2,r β; x eβx ψ −2γ xRn,2,r 0, β; x
C e ψ
βx
−2γ
n β r1
xRn,2,r β, 0; x
n
e Rn,2γ2,r
βx
n β r1
βx −2γ
β, 0; x
e ψ xRn,2,r 0, β; x .
n
Then, we can estimate as follows:
∞
∞
y
r2
Sn,k x
Snβ,kr y
y−u f
udu dy
nβ
0
x
k0
2.97
Journal of Applied Mathematics
f r2 xe−βx ψ 2γ x
25
L∞ 0,∞
Cf r2 xe−βx ψ 2γ x
L∞ 0,∞
×
eβx
n β r1
n β r1
ψ xRn,2,r β, 0; x
Rn,2γ2,r β, 0; x
n
n
ψ −2γ xRn,2,r 0, β; x .
−2γ
2.98
Then, we have by iv of Lemma 2.6,
n β r
r
r
Qn,β f
x − f xe−βx ψ 2γ2 x
n
f r2 xe−βx ψ 2γ x
L∞ 0,∞
1
1
ψ 2 xgn,2,r β, 0; x O γ gn,2γ2,r β, 0; x ψ 2γ2 x
× O
n
n
1
ψ 2 xgn,2,r 0, β; x
O
n
1 r1
O
xe−βx ψ 2γ1 xgn,1,r 0, β; x ψx.
f
n
2.99
Consequently, we have
n β r
r
r
−βx 2γ2
Q
f
ψ
−
f
e
x
x
x
n,β
n
2.100
1
r1
r2
−βx 2γ1
−βx 2γ
O
,
f
xe ψ
x
xe ψ x
f
L∞ 0,∞
L∞ 0,∞
n
since we know that |gn,2,r β, 0; xψ 2 x|, |gn,2,r 0, β; xψ 2 x|, |gn,2γ2,r β, 0; xψ 2γ2 x|, and
|gn,1,r 0, β; xψx| are uniformly bounded on 0, ∞.
Theorem 2.8. Let β and γ be nonnegative integers and r > 0 be a positive integer. Then one has uniformly for n, f and x ∈ 0, ∞,
r −βx 2γ
n β
r
.
x e ψ x C
Qn,β f
f xe−βx ψ 2γ x
L∞ 0,∞
n
2.101
26
Journal of Applied Mathematics
Proof. Using 1 y2γ C1 x2γ y − x2γ , we have
∞
∞
βy 2γ Sn,k x
Snβ,kr y e 1 y dy
nβ
0
k0
∞
∞
βy 2γ 2γ
nβ
dy
Sn,k x
Snβ,kr y e 1 x y − x
0
k0
2.102
C Un,0,r β; x ψ −2γ x Un,2γ,r β; x .
By Lemma 2.7 and i of Lemma 2.6, we know
Un,0,r β; x nβ
n
r1
eβx Rn,0,r β, 0; x nβ
n
r1
eβx ,
n β r1 βx
e Rn,2γ,r β, 2γ; x
n
n β r1 βx
1
O γ
e gn,2γ,r β, 2γ; x .
n
n
Un,2γ,r β; x 2.103
Therefore, we have
∞
∞
βy 2γ Sn,k x
Snβ,kr y e 1 y dy
nβ
0
k0
C
nβ
n
r1
1
eβx ψ −2γ x O γ gn,2γ,r β, 2γ; x .
n
2.104
Since |gn,2γ,r β, 2γ; xψ 2γ x| is uniformly bounded on 0, ∞, we have
n β r
r
−βx 2γ
ψ
Q
f
xe
x
n,β
n
∞
∞
βy 2γ −βx 2γ
nβ
Sn,k x
Snβ,kr y e 1 y dye ψ x
0
k0
× f r xe−βx ψ 2γ x
C
nβ
n
L∞ 0,∞
r1 1
r
.
1O γ
f xe−βx ψ 2γ x
L∞ 0,∞
n
Therefore, we have the result.
2.105
Journal of Applied Mathematics
27
Proof of Theorem 1.10. We will prove this theorem by the same method as the proof of
Theorem 1.5. First, we split it as follows:
n β r
r
r
−βx 2γ2
Qn,β f
x − f x e ψ
x
n
L∞ 0,∞
r
nβ r
−βx 2γ2
ψ
Q
f
−
f
xe
x
n,β
h
n
L∞ 0,∞
r
nβ r
r
−βx 2γ2
Qn,β f h
x − f h x e ψ
x
n
L∞ 0,∞
r
f h x − f r x e−βx ψ 2γ2 x
.
2.106
L∞ 0,∞
Then for the first term, we have, using Theorem 2.8 and 2.41,
Qn,β f − fh r xe−βx ψ 2γ2 x
L∞ 0,∞
C
n β
r
Cω2 f r ; e−βx ψ 2γ2 x; h .
f x − f r h x e−βx ψ 2γ2 x
L∞ 0,∞
n
2.107
For the second term, we have from Theorem 1.9,
n β r
r
r
−βx 2γ2
ψ
−
f
e
Q
f
x
x
x
n,β h
h
n
L∞ 0,∞
2.108
1
r2
r1
−βx 2γ1
−βx 2γ
O
fh xe ψ x
.
x
fh xe ψ
L∞ 0,∞
L∞ 0,∞
n
Here, we suppose 0 < h 1 and then we know that e−βx ψx/e−βxh ψx h and
e−βx /e−βxh are uniformly bounded on 0, ∞. Therefore, we have from 2.42 and 2.43
of Lemma 2.4,
r1
fh xe−βx ψ 2γ1 x
L∞ 0,∞
1 C ω1 f; e−βx ψ 2γ1 x; h ,
h
L∞ 0,∞
1
C 2 ω2 f; e−βx ψ 2γ x; h .
h
r2
fh xe−βx ψ 2γ x
2.109
Therefore, we have
n β r
r
r
−βx 2
ψ
−
f
e
Q
f
x
x
x
n,β h
h
n
L∞ 0,∞
1
1
1
−βx 21
−βx 2γ
ω1 f; e ψ x; h 2 ω2 f; e ψ x; h .
O
n
h
h
2.110
28
Journal of Applied Mathematics
Consequently, we have
n β r
r
r
−βx 2γ2
Q
f
ψ
−
f
e
x
x
x
n,β
n
1
1
1 O
ω1 f; e−βx ψ 2γ1 x; h 2 ω2 f; e−βx ψ 2γ x; h
n
h
h
Cω2 f r ; e−βx ψ 2γ2 xx; h .
2.111
√
If we let h 1/ n, then
n β r
r
r
−βx 2γ2
ψ
−
f
e
Q
f
x
x
x
n,β
n
1
1
1
C √ ω1 f; e−βx ψ 2γ1 x; √
ω2 f; e−βx ψ 2γ x; √
,
n
n
n
2.112
√
√
since ω2 f r ; e−βx ψ 2γ2 x; 1/ n ω2 f; e−βx ψ 2γ x; 1/ n.
3. Conclusion
In this paper, Lupas-type operators and Szász-Mirakyan-type operators are treated and the
various weighted norm convergence on 0, ∞ of these operators are investigated. Moreover,
this paper proves theorems on degree of approximation of f ∈ Cr 0, ∞ by these operators
using the modulus of smoothness of f.
Acknowledgments
The authors thank the referees for many valuable suggestions and comments. H. S. Jung was
supported by SEOK CHUN Research Fund, Sungkyunkwan University, 2010.
References
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