RATES OF CONVERGENCE OF
CHLODOVSKY-KANTOROVICH POLYNOMIALS IN CLASSES OF LOCALLY INTEGRABLE FUNCTIONS
Paulina Pych-Taberska
Faculty of Mathematics and Computer Science Adam Mickiewicz University
Umultowska 87, 61–614 Pozna´ n, Poland e-mail: ppych@amu.edu.pl
Dedicated to Professor Micha l Kisielewicz on his 70
thbirthday
Abstract
In this paper we establish an estimation for the rate of pointwise convergence of the Chlodovsky-Kantorovich polynomials for functions f locally integrable on the interval [0, ∞). In particular, corresponding estimation for functions f measurable and locally bounded on [0, ∞) is presented, too.
Keywords and phrases: Chlodovsky polynomial, Kantorovich poly- nomial, rate of convergence.
2000 Mathematics Subject Classification: 41A25.
1. Introduction
Let f be a function defined on the interval [0, ∞) and let N = {1, 2 . . .}.
The Bernstein-Chlodovsky polynomials C
nf of the function f are defined as C
nf (x) :=
n
X
k=0
f kb
nn
P
n,kx b
nfor x ∈ [0, b
n], n ∈ N, (1)
where P
n,k(t) :=
nkt
k(1−t)
n−kfor t ∈ [0, 1] and (b
n) is a positive increasing sequence satisfying the properties
n→∞
lim b
n= ∞, lim
n→∞
b
nn = 0.
(2)
These polynomials were first introduced by I. Chlodovsky in 1937 as a gen- eralization of the classical Bernstein polynomials
B
nf (x) :=
n
X
k=0
f k n
P
n,k(x), 0 ≤ x ≤ 1,
of functions f defined on the interval [0, 1] (see [5] or [8], Chap. II). The well-known Chlodovsky theorem states that if
n→∞
lim sup
0≤t≤bn
|f(t)| exp
−α n b
n= 0 for every α > 0, (3)
then lim
n→∞C
nf (x) = f (x) at every point x of continuity of f . In 1960 J. Albrycht and J. Radecki [1] proved the Voronovskaya-type theorem for operators (1). Some other approximation properties of the Chlodovsky poly- nomials can be found e.g. in [3, 7].
For functions f Lebesgue-integrable on the interval [0, 1] the classical Kantorovich polynomial of order n is defined as
B
?nf (x) := d
dx B
n+1F (x) ≡ (n + 1)
n
X
k=0
P
n,k(x) Z
n+1k+1k n+1
f (t)dt, 0 ≤ x ≤ 1,
where F is an indefinite integral of f . It is well known that lim
n→∞B
n?f (x) = f (x) at any point x of (0, 1) where f is the derivative of its indefinite integral (see e.g. [8], Chap. II).
In this paper we consider the Kantorovich-type modification of the Chlodovsky operators (1). Namely, assuming that f ∈ L
loc[0, ∞), that is f is locally integrable on [0, ∞), and denoting
F (x) = Z
x0
f (t)dt for x > 0,
we define the Chlodovsky-Kantorovich polynomial of degree n − 1 as K
n−1f (x) := d
dx C
nF (x), n ∈ N.
It is easy to verify that
K
n−1f (x) = n b
nn−1
X
k=0
P
n−1,kx b
nZ
(k+1)bnn
kbn n
f (t)dt, 0 ≤ x ≤ b
n,
(4)
(see [3], Section 4).
In order to formulate our first result let us consider those points x ∈ (0, ∞) at which
h→0
lim 1 h
Z
h0
(f (x + t) − f(x)) dt = 0 (5)
and let us introduce the pointwise characteristic w
x(δ; f ) := sup
0<|h|≤δ
1 h
Z
h0
(f (x + t) − f(x)) dt
, δ > 0.
(6)
Clearly, w
x(δ; f ) is a non-decreasing function of δ > 0 and lim
δ→0+w
x(δ; f )
= 0 almost everywhere on [0, ∞), that is at every point x ∈ (0, ∞) at which (5) is satisfied.
Theorem 1. Let f ∈ L
loc[0, ∞) and let at a fixed point x ∈ (0, ∞) condition (5) be fulfilled. Then, for all integers n such that b
n> 2x, pn/b
n≥ 3, we have
|K
n−1f (x) − f(x)|
≤ c(q) 1 + ϕ
q/2n(x) x
q! b
nn
q−12 [n/bn]X
k=1
k
q−32w
xx
√ k ; f
+ c(r)
x
rϕ
r/2n(x) b
nn
r/2|f(x)|
+ c(r) x
rs b
nx(b
n− x) ϕ
r/2n(x) b
nn
r−12Z
bn0
|f(t)|dt exp
− nx 8b
n, where q, r are arbitrary positive integers, c(q) and c(r) are positive numbers depending only on the indicated parameter q and r, respectively, ϕ
n(x) = x(1 −
bxn) +
bnnand [n/b
n] denotes the greatest integer not greater than n/b
n. Taking into acount fundamental assumptions (2) and choosing in Theorem 1, q = 3, r = 2 we easily get
Corollary 1. If f ∈ L
loc[0, ∞) and if
n→∞
lim Z
bn0
|f(t)|dt exp
−α n b
n= 0 for every α > 0,
then
n→∞
lim K
n−1f (x) = f (x) almost everywhere on [0, ∞).
Now, let us consider the subclass M
loc[0, ∞) consisting of all measurable functions f locally bounded on [0, ∞). In this case
w
x(δ; f ) ≤ osc(f; I
x(δ)) ≡ sup
u,v∈Ix(δ)
|f(u) − f(v)|, where 0 ≤ δ ≤ x, I
x(δ) := [x − δ, x + δ].
Theorem 2. Let f ∈ M
loc[0, ∞) and let at a fixed point x ∈ (0, ∞) the one- sided limits f (x+), f (x−) exist. Then, for all integers n such that b
n> 2x, pn/b
n≥ 3, we have
|K
n−1f (x) − 1
2 (f (x+) + f (x−)) |
≤ c(q) 1 + ϕ
q/2n(x) x
q! b
nn
q−12 [n/bn]X
k=1
k
q−32osc
g
x; I
xx
√ k
+ c
s b
nx(b
n− x) ϕ
1/2n(x)M (b
n; f ) exp
− nx 8b
n+ 2 r b
nn s
b
nx(b
n− x) |f(x+) − f(x−)|,
where M (b
n; f ) = sup
0≤t≤bn|f(t)|, ϕ
n(x) = x
1 −
bxn+
bnn,
g
x(t) :=
f (t) − f(x+) if t > x,
0 if t = x,
f (t) − f(x−) if 0 ≤ t < x, (7)
q is an arbitrary positive integer, c(q) is a positive constant depending only
on q and c is a positive absolute constant.
The function g
xis continuous at x. Hence lim
δ→0+osc(g
x; I
x(δ)) = 0. Con- sequently, Theorem 2 yields the following
Corollary 2. If f ∈ M
loc[0, ∞) and if at a fixed point x ∈ (0, ∞) the limits f (x+), f (x−) exist, then under the Chlodovsky assumption (3), we have
n→∞
lim K
n−1f (x) = 1
2 (f (x+) + f (x−)) . (8)
Remark. In particular, let us consider the class BV
Φ[0, ∞) of functions of bounded variation in the Young sense on the interval [0, ∞) (for the definition see e.g. [4, 10]). If f ∈ BV
Φ[0, ∞), then M(b
n; f ) ≤ M (M = const.). The estimation given in Theorem 2 and the relation (8) hold true at every point x ∈ (0, ∞).
2. Auxiliary results
We now present certain results which will be used in the proof of our main theorems. For this, let us introduce the notation: given any fixed x ∈ [0, b
n] and any non-negative integer q, we will write
µ
n,q(x) :=
n
X
k=0
kb
nn − x
qP
n,kx b
n,
|µ
n,q|(x) :=
n
X
k=0
kb
nn − x
q
P
n,kx b
n.
Moreover, we will use the notation c
j(p), j = 1, 2, . . . , for positive constants, not necessarily the same at each occurrence, depending only on the indicated parameter p.
Lemma 1. Let n ∈ N, x ∈ [0, b
n].
(i) µ
n,0(x) = 1, µ
n,1(x) = 0, µ
n,2(x) = b
nn x
1 − x
b
n. (ii) If s ∈ N, n ≥ 2, then
µ
n,2s(x) ≤ c
1(s) b
nn
sx
1 − x
b
nx(1 − x b
n) + b
nn
s−1.
(iii) If q ∈ N, q ≥ 2, n ≥ 2, then
|µ
n,q|(x) ≤ c
2(q) b
nn
q2x
1 − x
b
nx
1 − x
b
n+ b
nn
q2−1. P roof. Formulas (i) follow by easy calculation. Suppose s > 1 and put y := x/b
n. Then y ∈ [0, 1] and
µ
n,2s(x) = b
nn
2s nX
k=0
(k − ny)
2sP
n,k(y).
Applying the known represetation formula for the above sum (see [6], Lemma 3.6 with c = −1) we obtain
µ
n,2s(x) = b
nn
2s sX
j=1
β
j,s(ny(1 − y))
j,
where β
j,sare real numbers independent of y and bounded uniformly in n.
Now, let us observe that for y ∈ [0,
1n] or y ∈ [1 −
1n, 1] one has ny(1 − y) ≤
n−1n
< 1 and
s
X
j=1
β
j,s(ny(1 − y))
j≤ ny(1 − y)
s
X
j=1
|β
j,s|.
If y ∈
1n
, 1 −
n1then (ny(1 − y))
−1≤
n−1n≤ 2 and
s
X
j=1
β
j,s(ny(1 − y))
j≤ (ny(1 − y))
ss
X
j=1
|β
j,s| (ny(1 − y))
j−s≤ (ny(1 − y))
ss
X
j=1
2
s−j|β
j,s|.
Consequently, for all y ∈ [0, 1] (that is for all x ∈ [0, b
n]) we have µ
n,2s(x) ≤ c
1(s) b
nn
2sny(1 − y) (1 + ny(1 − y))
s−1with
c
1(s) ≥
s
X
j=1
2
s−j|β
j,s|.
Inequality (ii) follows by taking y = x/b
n. The same estimation holds true for |µ
n,q|(x) with even q (q = 2s). If q is odd (q = 2s + 1), then
|µ
n,q|(x) ≤ (µ
n,4s(x))
12(µ
n,2(x))
12by Cauchy-Schwarz inequality, and the proof is complete.
Lemma 2. If n ∈ N, 0 < x < b
n, then x
b
n1 − x
b
nK
n−1f (x)
= n b
2nn
X
k=0
kb
nn − x
P
n,kx b
nZ
kbnn −x 0
f (x + t)dt.
(9)
P roof. By (4) and by partial summation, we find that
K
n−1f (x) = n b
nn−1
X
k=0
P
n−1,kx b
nZ
(k+1)n bn
kbn n
f (t)dt =
= n
b
nP
n−1,n−1x b
nZ
bn0
f (t)dt
+ n b
nn−1
X
k=1
P
n−1,k−1x b
n− P
n−1,kx b
nZ
kbnn
0
f (t)dt.
Putting y = x/b
nand observing that
y(1 − y) (P
n−1,k−1(y) − P
n−1,k(y)) = k n − y
P
n,k(y)
for k = 1, 2, . . . , n − 2 and
y(1 − y)n P
n−1,n−1(y) = yP
n,n−1(y) = n(1 − y)P
n,n(y), we easily get
x b
n1 − x
b
nK
n−1f (x) = n b
nn
X
k=0
k n − x
b
nP
n,kx b
nZ
kbnn
0
f (t)dt.
Now, it is enough to recall that
n
X
k=0
kb
nn − x
P
n,kx b
n= µ
n,1(x) = 0
(Lemma 1 (i)). Consequently, x
b
n(1 − x
b
n)K
n−1f (x) = n b
2nn
X
k=0
kb
nn − x
P
n,kx b
nZ
kbnnx
f (t)dt
and the proof is complete.
Note that a corresponding representation like in the formula (9) for the classical Kantorovich polynomials is given in [2].
Lemma 3. If 0 < δ ≤ x < b
nthen X
|kbnn −x|≥δ
P
n,kx b
n≤ 2 exp
− nδ
24xb
nfor all n ∈ N such that b
n≥
3x−δ3x2.
The proof of Lemma 3 runs as in [1] and is based on the known Chlodovsky inequality ([8], Theorem 1.5.3)):
X
|k−nt|≥2z
√
nt(1−t)
P
n,k(t) ≤ 2 exp −z
2,
provided that 0 ≤ t ≤ 1, 0 ≤ z ≤
32pnt(1 − t).
Lemma 4. Let 0 < x < b
nand let n ≥ 2.
(i) If 0 ≤ k ≤ n − 1, then
P
n−1,k( x b
n) ≤ 1
√ e r b
nn
s b
nx(b
n− x) .
(ii)
X
nx bn<k≤n
P
n−1,kx b
n− 1 2
≤ 0.82 √ 2 r b
nn s
b
nx(b
n− x) .
P roof. Estimation (i) follows from the result by X.M. Zeng [11] (Theo- rem 1): if 0 ≤ k ≤ n and y ∈ (0, 1), then
P
n,k(y) ≤ 1
√ 2e 1 pny(1 − y) .
Inequality (ii) is an immediate consequence of the Berry-Ess´een Theorem:
X
k n>y
P
n,k(y) − 1 2
< 0.82
pny(1 − y) , 0 < y < 1
(see e.g., [12], Lemma 2).
3. Proofs of theorems
Proof of Theorem 1. In view of Lemma 1 (i) one can write x
b
n1 − x
b
nf (x) = n
b
2nµ
n,2(x)f (x)
= n b
2nn
X
k=0
kb
nn − x
P
n,kx b
nZ
kbnn −x0
dtf (x).
The above identity and the representation (9) lead to x
b
n1 − x
b
n(K
n−1f (x) − f(x))
= n b
2nn
X
k=0
kb
nn − x
P
n,kx b
nZ
kbnn −x
0
(f (x + t) − f(x)) dt
≡ X
k∈Λ
+ X
k∈Ω
,
(10)
where Λ and Ω are the sets of indices k ∈ {0, 1, . . . , n} such that |
kbnn−x| ≤ x and
kbnn− x > x, respectively.
For the sake of brevity let us introduce the notation: d
n= pb
n/n, m = [pn/b
n], w
x(δ; f ) = w
x(δ). Consider the sum P
k∈Λ
and divide the set Λ in the following manner: Λ = S
mj=0
Λ
j, where Λ
jare the sets of indices k such that
0 ≤
kb
nn − x
≤ xd
nif j = 0, jxd
n<
kb
nn − x
≤ (j + 1)xd
nif j = 1, 2, . . . , m − 1, mxd
n<
kb
nn − x
≤ x if j = m.
In view of definition (6),
X
k∈Λ
≤
m−1
X
j=0
T
n,j(x)w
x((j + 1)xd
n) + T
n,m(x)w
x(x),
where
T
n,j(x) := n b
2nX
k∈Λj
kb
nn − x
2P
n,kx b
n.
From Lemma 1 (i) one has T
n,0(x) ≤ n
b
2nµ
n,2(x) = x b
n1 − x
b
n.
Next, given any positive integer q, we have
T
n,j(x) ≤ n b
2n1 (jxd
n)
qn
X
k=0
kb
nn − x
q+2
P
n,kx b
nfor j = 1, 2, . . . , m. Hence Lemma 1 (iii) yields T
n,j(x) ≤ c
1(q)
j
qx
qx b
n1 − x
b
nϕ
q/2n(x)
where ϕ
n(x) = x
1 −
bxn+
bnn. Consequently,
X
k∈Λ
≤ x b
n1 − x
b
n1 + c
1(q)
x
qϕ
q/2n(x)
m−1
X
j=1
w
x((j + 1)xd
n)
j
q+ w
x(x) m
q
. Clearly,
m−2
X
j=1
w
x((j + 1)xd
n)
j
q≤ 3
qd
q−1nZ
mdn2dn
w
x(xt) t
qdt
≤ 3
qd
q−1nZ
m21
( √
s)
q−3w
xx
√ s
ds
≤ c
2(q)d
q−1nm2−1
X
k=1
( √
k + 1)
q−3w
xx
√ k
and w
x(x)
(m − 1)
q+ w
x(x)
m
q≤ 2
(m − 1)
qw
x(x) ≤ 3
qd
q−1nw
x(x).
Hence
X
k∈Λ
≤ c
3(q) x b
n1 − x
b
n1 + ϕ
q/2n(x) x
q! d
q−1nm2−1
X
k=1
√ k
q−3w
xx
√ k
. (11)
Now, let us consider the sum P
k∈Ω
in formula (10). Given any positive integer r, we have
X
k∈Ω
≤ n
b
2nx
rX
k∈Ω
kb
nn − x
r+2
P
n,kx b
n|f(x)|
+ n
b
2nx
rX
k∈Ω
kb
nn − x
r+1
P
n,kx b
nZ
bn0
|f(t)|dt
≤ n
b
2nx
r|µ
n,r+2|(x) |f(x)|
+ n
b
2nx
rZ
bn0
|f(t)|dt (µ
n,2r+2(x))
1/2X
k∈Ω
P
n,kx b
n!
1/2.
Applying Lemmas 1 and 3 we then get
X
k∈Ω
≤ c
4(r) x
rx b
n1 − x
b
nb
nn
r/2ϕ
r/2n(x)|f(x)|
+ c
4(r) x
rx b
n1 − x
b
nb
nn
r2−12s b
nx(b
n− x) ϕ
r/2n(x) Z
bn0
|f(t)|dt exp
− nx 8b
n.
This gives the desired conclusion when combined with (10) and (11).
Proof of Theorem 2 . Let f ∈ M
loc[0, ∞) and let the limits f(x+), f(x−) exist at a fixed point x > 0. Consider the function g
xdefined by (7). It is easily seen that
f (t) − f (x+) + f (x−)
2 = g
x(t) + f (x+) − f(x−)
2 sgn
x(t) +
f (x) − f (x+) + f (x−) 2
δ
x(t),
where sgn
x(t) = sgn(t − x), δ
x(t) = 1 if t = x, δ
x(t) = 0 otherwise (see e.g.
[9]). Hence
K
n−1f (x) − f (x+) + f (x−) 2
= K
n−1g
x(x) + f (x+) − f(x−)
2 K
n−1sgn
x(x).
(12)
The function g
xis continuous at x and g
x(x) = 0. So, K
n−1g
x(x) = K
n−1g
x(x) − g
x(x) can be estimated as in the proof of Theorem 1. Namely, using formula (10) in which f is replaced by g
xand observing that
w
x(δ; g
x) ≤ osc (g
x; I
x(δ)) for 0 < δ ≤ x we get the estimation for
P
k∈Λ
as in (11) with w
x √x kreplaced by osc
g
x; I
x √x k. Indeed, we estimate the sum P
k∈Ω
as follows:
X
k∈Ω
≤ 2n
b
2nM (b
n; f ) X
k∈Ω
kb
nn − x
2P
n,kx b
n,
where M (b
n; f ) = sup
0≤t≤bn|f(t)|. Next, the Cauchy-Schwarz inequality and Lemmas 1, 3 lead to
X
k∈Ω
≤ 2M(b
n; f ) n
b
2n(µ
n,4(x))
1/22 exp
− nx 4b
n 1/2≤ c x b
n1 − x
b
ns b
nx(b
n− x) ϕ
1/2n(x)M (b
n; f ) exp
− nx 8b
n, where c is an absolute positive constant. Consequently,
|K
n−1g
x(x)| ≤
≤ c(q) 1 + ϕ
q/2n(x) x
q! b
nn
q−12 [n/bn]X
k=1
( √
k)
q−3osc
g
x; I
xx
√ k
+ c
s b
nx(b
n− x) ϕ
1/2n(x)M (b
n; f ) exp
− nx 8b
n,
where q is arbitrary positive integer, c(q) is a positive constant depending only on q and c is an absolute constant.
Now it is enough to estimate the term K
n−1sgn
x(x). Choose the integer l such that x ∈ [
nlb
n,
l+1nb
n). It is clear that
K
n−1sgn
x(x) = X
k>l
P
n−1,kx b
n− X
k<l
P
n−1,kx b
n+ n
b
nP
n−1,lx b
n2 l
n b
n+ b
nn − 2x
= 2 X
k>l
P
n−1,kx b
n− 1 + 2P
n−1,lx b
nn b
nl + 1 n b
n− x
.
Therefore,
|K
n−1sgn
x(x)| ≤ 2
X
k>l