On the rates of convergence of

Paulina Pych-Taberska, Harun Karsli

On the rates of convergence of

Chlodovsky-Kantorovich operators and their B´ezier variant

Abstract. In the present paper we consider the B´ezier variant of Chlodovsky- Kantorovich operators Kn−1,αf for functions f measurable and locally bounded on the interval [0,∞). By using the Chanturiya modulus of variation we estimate the rate of pointwise convergence of Kn−1,αf(x) at those x > 0 at which the one-sided limits f(x+) , f(x−) exist.

2000 Mathematics Subject Classification: 41A25, 41A35.

Key words and phrases: rate of convergence, Chlodovsky-Kantorovich operator, B´ezier basis, Chanturiya’s modulus of variation, pth power variation.

1. Introduction. Let f be a function defined on the interval [0, ∞) and let ℕ := {1, 2, ...} . The Bernstein-Chlodovsky polynomials Cⁿf applied to f are defined as

(1) Cnf (x) :=

Xn k=0

f

kbn

n

 Pn,k

x bn



for x∈ [0, bⁿ], n ∈ ℕ,

where Pn,k(t) =

 n k



t^k(1 − t)^n−k for t ∈ [0, 1] and (bⁿ) is a positive increasing sequence satisfying the properties

(2) lim

n→∞bn= ∞ and lim_n→∞bn

n = 0

(see [5,11,15]). They were introduced by I. Chlodovsky [5] in 1937 as a generalization of the classical Bernstein polynomials Bnf of a function f on the interval [0, 1] ( defined by (1) in which bn = 1 for all n ∈ ℕ).

Recently, Butzer and Karsli [3] considered the Kantorovich variant of the Chlodovsky- Bernstein polynomials as

(3) Knf (x) := n + 1 bn+1

Xn k=0

Pn,k

 x bn+1

 ^(k+1)bn+1Zⁿ⁺¹

kbn+1 n+1

f (u)du if 0 ¬ x ¬ bⁿ⁺¹,

where f ∈ L^loc[0, ∞) i.e. f is locally integrable on the interval [0, ∞) and (bⁿ) satisfies conditions (2). Note that the classical Kantorovich polynomials K_n^∗f of a function f integrable on the interval [0, 1] are defined as in (3) in which bn= 1 for all n ∈ ℕ. Several authors have studied approximation properties of those classical Kantorovich polynomials (see e.g. [1, 6, 9, 11, 12, 17, 18] etc.). It is easy to see that the operators Kn and K_n^∗ are described in the following way. If F denotes an indefinite integral of f, then K_n^∗f (x) = _dx^d Bn+1F (x) and Knf (x) = _dx^dCn+1F (x). As is well- known, if f is Lebesgue integrable on [0, 1] then

nlim→∞K_n^∗f (x) = f (x) almost everywhere on [0, 1].

Analogous relation holds true for operators (3). Namely (see [14]) if f ∈ L^loc[0, ∞) and if

nlim→∞

bn

Z

0

|f(u)| du exp(−σn bn

) = 0 for each σ > 0,

then

nlim→∞Knf (x) = f (x) almost everywhere on [0,∞),

i.e. at every x > 0 at which F⁰(x) = f(x). Some modified Chlodovsky-Kantorovich operators are considered also in [8].

In this paper we introduce the B´ezier variant of the Chlodovsky-Kantorovich operators (3) of order (n − 1) for f ∈ L^loc[0, ∞) as

(4) Kn−1,αf (x) := n bn

nX−1 k=0

Q^(α)_n−1,k

x bn

 ^(k+1)bnZⁿ

kbnn

f (u)du if 0¬ x ¬ bⁿ,

where α > 0, Q^(α)_n−1,k(t) = J_n^α_−1,k(t) − Jn^α−1,k+1(t) and Jn−1,k(t) are the B´ezier basis functions defined for t ∈ [0, 1] as

Jn−1,k(t) =

n−1

X

j=k

Pn−1,j(t) if k = 0, 1, . . . , n − 1,

Jn−1,n(t) = 0 (see [2]). Clearly, if α = 1 then Kn−1,αf reduce to operators (3) with n replaced by n− 1. Our paper is concerned with the rate of pointwise convergence of operators (4) when f ∈ M^loc[0, ∞),i.e. f is measurable and locally bounded

on [0, ∞). By using the Chanturiya modulus of variation we present estimates for the rate of convergence of K_n−1,αf (x) at the points x of continuity of f and at the discontinuity points of the first kind of f. We will formulate our results for K_n−1,αf with α > 0. The corresponding estimates for the Chlodovsky-Kantorovich polynomials K_n−1f follow immediately as a special case α = 1.

At first, we give the following definition.

Definition 1.1 . Let g be a bounded function on a compact interval I = [a, b].

The modulus of variation νk(g; [a, b]) of the function g is defined for nonnegative integers k as follows:

ν0(g; [a, b]) := 0 and for k ­ 1

νk(g; [a, b]) := sup

Πk

k−1

X

j=0

|g(x^2j+1) − g(x^2j)| ,

where Πk is an arbitrary system of k disjoint intervals (x2j, x2j+1), j = 0, 1, ..., k −1 i.e., a ¬ x⁰< x1¬ x²< x3...¬ x2k−2< x_2k−1¬ b ( see [4,10] ).

Some properties of the modulus of variation and its applications can be found e.g. in [4] ( see [10,13] also). In particular, νk(g; I) is a non-decreasing function of k∈ ℕ,

νk(g; I) ¬ 2k sup

t∈I|g(t)| , νk(g; Z) ¬ ν^k(g; I) for any compact interval Z contained in I.

Let p ­ 1. The pth power variation of a function g on a compact interval I = [a, b]

is denoted by Vp(g, I) and is defined as the upper bound of the set of numbers



X

j

|g(s^j) − g(t^j)|^p





1/p

over all finite systems of non-overlapping intervals (sj, tj) ⊂ I. If f is of bounded pth power variation on I, then for every k∈ ℕ,

(5) νk(g; I) ¬ k^1−1/pVp(g, I),

(see [4]). The class of all functions of bounded pth power variation on every compact interval contained in [0, ∞) will be denoted by BVloc^p [0, ∞).

Throughout this paper we will assume that the sequence (bn) satisfies the funda- mental conditions (2). The symbols cγ and cγ,ρwill denote some positive constants, not necessarily the same at each occurrence, depending only on the indicated para- meter γ and parameters γ, ρ, respectively. The symbol [a] will denote the greatest integer not greater than a. Moreover, if x ∈ (0, ∞) the intervals H^x(u) = [x−u, x+u]

for 0 < u ¬ x will be used.

Theorem 1.2 Let f ∈ M^loc[0, ∞) and let the one-sided limits f(x+), f(x−) exist at a fixed point x ∈ (0, ∞). Then, for all integers n ­ 2 such that bⁿ> 2x, n/bn­ N (α, x) one has

K^n−1,αf (x)− 1

2^αf (x+)− (1 − 1

2^α)f(x−) 

 ¬ 2v¹(gx; Hx(xp bn/n))

+cαϕn(x) x²





m−1X

j=1

vj(gx; Hx(jxp bn/n))

j³ +vm(gx; Hx(x)) m²





+cα,qM (bn; f) x^2q

bn

n

q

ϕ^q_n(x) + 4κα

rbn

n

s bn

x (bn− x)|f(x+) − f(x−)| , where m := [p

n/bn], M(b; f) := sup

0¬t¬b|f(t)| , ϕⁿ(x) = x 1 − b^xn

+^b_nⁿ,

(6) gx(t) :=





f (t)− f(x+) if t > x,

0 if t = x,

f (t)− f(x−) if 0 ¬ t < x,

κα = max {1, α} , q is an arbitrary positive integer, N(α, x) = 4 if α ­ 1 and N (α, x) = max{4, 50/x} if 0 < α < 1. If α ­ 1 then c^α= α and cα,q= αcq. Retaining the notation used in Theorem 1.2 and using inequality (5), we get

Theorem 1.3 Let f ∈ BVloc^p [0, ∞), p ­ 1, and let x ∈ (0, ∞). Then, for all integers n­ 2 such that bⁿ> 2x, n/bn­ N(α, x) one has

K^n−1,αf (x)− 1

2^αf (x+)− (1 − 1

2^α)f(x−)   ¬



2 + cαϕn(x) x²



Vp(g_x; H_x(xp bn/n))

+cαϕn(x) x²

rbn

n

!_{1+1/p [n/bn]} X

k=1

Vp(gx; Hx(x/√

√ k)

k1−1/p +cα,qM (bn; f) x^2q

bn

n

q

ϕ^q_n(x)

+4κα

rbn

n

s bn

x (bn− x)|f(x+) − f(x−)| .

Here we note that in some subclasses of Mloc[0, ∞) the right-hand sides of the inequalities mentioned in the hypotheses of our theorems tend to zero as n → ∞.

In view of (3) we have m = hp n/bn

i → ∞ as n → ∞. So, in Theorem 1.2 it is enough to consider only the term

Λm(x) =

mX−1 j=1

vj(gx; Hx(jxdn))

j³ where dn =p

bn/n.

Clearly,

Λm(x) ¬

mX−1 j=1

v1(gx; Hx(jxdn)) j² ¬ 4dⁿ

mdZ n

dn

v1(gx; Hx(xt)) t² dt

¬ 4dⁿ

m+1Z

1

v1(gx; Hx(x

s))ds ¬ 4 m

Xm k=1

v1(gx; Hx(x k)).

Since the function gxis continuous at x and v₁(gx; Hx(x/k)) denotes the oscillation of gxon the interval Hx(x/k), we have

klim→∞v1(gx; Hx(x k)) = 0 and consequently

mlim→∞Λm(x) = 0.

As regards Theorem 1.3, it is easy to verify that in view of continuity of gx at x,

mlim→∞

1 m^1+1/p

(m+1)²

X

k=1

√ 1

k1−1/pVp(gx; Hx( x

√k)) = 0.

So, we get the following approximation theorem.

Corollary 1.4 Suppose that f ∈ M^loc[0, ∞) ( in particular, f ∈ BVloc^p [0, ∞) , p­ 1) and that there exists a positive integer q such that

nlim→∞

bn

n

q

M (bn; f) = 0.

Then, at every point x ∈ (0, ∞) at which the limits f(x+), f(x−) exist, we have

n→∞limKn−1,αf (x) = 1

2^αf (x+) + (1− 1

2^α)f(x−).

Obviously, the above relations hold true for every measurable function f bounded on [0, ∞), in particular for every function f of bounded pth power variation (p ­ 1) on the whole interval [0, ∞).

2. Auxiliary results. We now mention certain results which are necessary to prove our main theorem. For this, let us introduce the following notation. Given any x ∈ [0, bⁿ] and any non-negative integer q we will write

ψ^q_x(t) := (t − x)^q f or t∈ [0, ∞),

µn−1,q(x) := Kn−1ψ^q_x(x) ≡ n bn

n−1

X

k=0

Pn−1,k

x bn

 ^(k+1)bnZⁿ

kbnn

(t − x)^qdt.

Lemma 2.1 If n ∈ ℕ (n ­ 2) , x ∈ [0, bⁿ], then

µn−1,0(x) = 1, µn−1,1(x) =bn− 2x 2n , µn−1,2(x) = bn

n

n− 2 n x

 1 − x

bn

 + bn

3n



¬bn

nϕn(x) and, for q > 1,

(7) µn−1,2q(x) ¬ c^q

bn

n

q

ϕ^q_n(x), where

ϕn(x) = x

 1 − x

bn

 +bn

n.

Proof Formulas for µn−1,0(x), µn−1,1(x), µn−1,2(x) follow by simple calculation.

Now, we will prove (7). Suppose q > 1 and put y := x/bn. Then y∈ [0, 1] and

µn−1,2q(x) = n bn

nX−1 k=0

Pn−1,k(y)

(k+1)bn

Zn

kbnn

(t − ybⁿ)^2qdt

= nb^2q_n

nX−1 k=0

Pn−1,k(y)

k+1

Zn k n

(s − y)^2qds = b^2q_nK_n^∗₋₁ψ^2q_y (y) , (8)

where K_n−1^∗ is the classical Kantorovich operator of order n − 1. Clearly, K_n^∗₋₁ψ^2q_y (y) = d

dyBnG(y), where G is an indefinite integral of ψy, i.e.

G(y) = Zy

0

(t − y)^2qdt for y ∈ [0, 1]

and BnG is the Bernstein polynomial of order n of the function G.

Hence (see [6, formula 9.5.14]), K_n−1^∗ ψ_y^2q(y) = 1

2q + 1 d

dyBnψ_y^2q+1(y) + Bnψ_y^2q(y)

= 1

n^2q

 1

n (2q + 1) d

dyTn,2q+1(y) + Tn,2q(y)

 (9) ,

where

Tn,r(y) = Xn k=0

Pn,k(y) (k − ny)^r, y∈ [0, 1] .

As is known (see [6, Lemma 9.4.3] or [7, Lemma 3.6]), for Tn,r(y) there holds the recursion relation

Tn,0(y) = 1, Tn,1(y) = 0,

Tn,r+1(y) = nry(1 − y)Tn,r−1(y) + y(1 − y) d

dyTn,r(y) for r ­ 2.

From this relation there follow by induction the following representation formulas

Tn,2q(y) = Xq j=1

βj,2q(n) (ny(1 − y))^j,

Tn,2q+1(y) = (1 − 2y) Xq j=1

βj,2q+1(n) (ny(1 − y))^j,

where βj,2q(n), βj,2q+1(n) are real numbers independent of y and bounded uniformly in n ( see [7, Corollary 3.7]).

Observing that

d

dyy(1− y) = 1 − 2y and that

(1 − 2y)²= 1 − 4y(1 − y) we easily get from (9) that

K_n−1^∗ ψ_y^2q(y) = 1 n^2q

Xq j=0

ηj,q(n) (ny(1 − y))^j,

where |η^j,q(n)| ¬ γ^j,q and γj,q are non-negative numbers depending only on j and q. Now, it is enough to use the following inequalities:

a ) if y ∈0,_n¹

or y ∈

1 − ¹n, 1

, then ny(1− y) ¬ (n − 1)/n < 1 and 

Xq j=0

ηj,q(n) (ny(1 − y))^j   ¬

Xq j=0

γj,q,

b ) if y ∈₁

n, 1−n¹

, then (ny(1− y))⁻¹¬ n/(n − 1) ¬ 2 and 

Xq j=0

ηj,q(n) (ny(1 − y))^j  

¬ (ny(1 − y))^q Xq j=0

2^q−jγj,q.

Consequently,

K_n−1^∗ ψ_y^2q(y) ¬ cq

n^2q (1 + (ny(1 − y))^q)

with

cq = Xq j=0

γj,q2^q−j, and by (8)

µn−1,2q(x) ¬ c^q

bn

n

2q 1 +nx

bn

 1 − x

bn

q

.

This inequality is equivalent to the one to be proved. _■ Note that inequality (7) can be obtained also by using the representation formulas for T_n,2q(y) and T_n,2q+1(y) given in [6, Lemma 9.5.5].

Lemma 2.2 Let 0 < x < bn, α > 0and let

Φn−1,α(x, t) := n bn

nX−1 k=0

Q^(α)_n−1,k(x bn

)χn,k(t),

where χn,k(t)are the characteristic functions of the intervals In,k:= [^kb_nⁿ,^(k+1)b_n ⁿ), k = 0, 1, ..., n − 1.

(i) If 0 < s < x then

(10)

Zs

0

Φ_n−1,α(x, t) dt ¬ καcq

(x − s)^2q

bn

n

^q ϕ^q_n(x),

(ii) If x < s < bn then

(11)

bn

Z

s

Φn−1,α(x, t) dt ¬ cα,q

(s − x)^2q

bn

n

q

ϕ^q_n(x).

In the above inequalities

ϕn(x) = x

 1 − x

bn

 +bn

n,

qis an arbitrary positive integer, κα= max {1, α} . If α ­ 1 then c^α,q = αcq, c1= 1.

Proof (i) Let 0 < s < x < bn. If α­ 1 then

Q^(α)_n−1,k

x bn



= J_n^α_−1,k

x bn



− Jn^α−1,k+1

x bn



¬ αPn−1,k

x bn

 . Hence

Zs

0

Φn−1,α(x, t) dt ¬ αn bn

nX−1 k=0

Pn−1,k(x bn

) Zs

0

χn,k(t)dt ¬ α

(x − s)^2qµn−1,2q(x).

Consider now the case 0 < α < 1. Let s ∈ [k^∗bn/n, (k^∗+ 1) bn/n). Then we can write s = (k^∗+ ε) bn/n where 0¬ ε < 1 and

Zs

0

Φn−1,α(x, t) dt = n bn

nX−1 k=0

Q^(α)_n−1,k(x bn

) Zs

0

χn,k(t)dt

=

kX^∗−1 k=0

Q^(α)_n−1,k(x bn

) + n bn

Q^(α)_n−1,k(x bn

)

(k^∗+ε)bZ n/n

k^∗bn/n

dt

=

kX^∗−1 k=0

Q^(α)_n−1,k(x bn

) + εQ^(α)_n−1,k∗(x bn

)

= 1 −

n−1X

k=k^∗

Q^(α)_n−1,k(x bn

) + εQ^(α)_n−1,k∗(x bn

)

= 1 − (1 − ε) Jn^α−1,k^∗(x

bn) − εJn^α−1,k^∗+1(x bn

)

¬ 1 − (1 − ε) Jⁿ−1,k^∗(x

bn) − εJⁿ−1,k^∗+1(x bn

)

=

kX^∗−1 k=0

Q⁽¹⁾_n−1,k(x bn

) + εQ⁽¹⁾_n−1,k∗(x bn

)

= Zs

0

Φn−1,1(x, t) dt

¬ 1

(x − s)^2qµn−1,2q(x)

(see [13,18]). Thus, estimate (10) follows by Lemma 2.1.

(ii) Let 0 < x < s < bn. If α­ 1, by using the same method in the proof of (i), one has the same bound, namely

bn

Z

s

Φn−1,α(x, t) dt ¬ α

(x − s)^2qµn−1,2q(x) ¬ αcq

(x − s)^2q

bn

n

q

ϕ^q_n(x).

If 0 < α < 1 then retaining the symbol k^∗ used in the proof of (i) we obtain

bn

Z

s

Φn−1,α(x, t) dt = n bn

n−1

X

k=0

Q^(α)_n−1,k(x bn

)

bn

Z

(k^∗+ε)bn/n

χn,k(t)dt

= n

bnQ^(α)_n−1,k∗(x

bn)_(k∗+ 1) bn

n − s

+

n−1

X

k=k^∗+1

Q^(α)_n−1,k(x bn)

¬ n

bn

P_n−1,k^α ∗(x bn

)_(k∗+ 1) bn

n − s

+

n−1

X

k=k^∗+1

Pn−1,k(x bn

)

!^α

¬ 2^1−α n bn

Pn−1,k^∗(x bn

)_(k∗+ 1) bn

n − s

+

n−1

X

k=k^∗+1

Pn−1,k(x bn

)

!α

¬ 2^1−α

(s − x)^2q∗_n

bn

α





^Pn−1,k∗(

x bn)

(k∗+1)bn

Zn s

|t − x|^2q/αdt

+

n−1

X

k=k^∗+1

Pn−1,k(x bn

)

(k+1)bn

Zn

kbnn

|t − x|^2q/αdt





α

¬ 2^1−α

(s − x)^2q m_n−1,2q/α(x)
α

,

where

mn−1,γ(x) = n bn

n−1

X

k=0

Pn−1,k(x bn

)

(k+1)bn

Zn

kbnn

|t − x|^γdt.

Write l = 2/α and denote by [l] the integral part of l. As in [18, Lemma 6] choose the numbers

p = 2[l]

2[l] + 2 − l, p⁰ = 2[l]

l− 2, r = 2

p, r⁰= 2v

p⁰, v = [l] + 1.

Clearly, l > 2, p > 1, p⁰ > 1, 1/p + 1/p⁰ = 1 and l = r + r⁰. Applying twice the

H¨older inequality (first for integrals, next for sums) we obtain

m_n−1,2q/α(x) ¬ n bn

n−1X

k=0

P_n^1/p_−1,k(x bn

)P_n^1/p_−1,k⁰ (x bn

)

(k+1)bn

Zn

kbn n

|t − x|^qr|t − x|^qr0dt

¬



_bⁿ_n

n−1

X

k=0

Pn−1,k(x bn

)

(k+1)bn

Zn

kbnn

|t − x|^qrpdt





1/p



_bⁿ_n

n−1

X

k=0

Pn−1,k(x bn)

(k+1)bn

Zn kbn

n

|t − x|^qr0p0dt





1/p⁰

= (µn−1,2q(x))^1/p(µn−1,2vq(x))^1/p0.

Applying Lemma 2.1 and observing that 1/p + v/p⁰= 1/α we get

mn−1,2q/α(x) ¬

 cq

bn

n

q

ϕ^q_n(x)

^1/p cvq

bn

n

vq

ϕ^vq_n (x)

^1/p0

= cα,q

bn

nϕn(x)

q/α

.

Thus inequality (11) follows. _■

In order to formulate the next lemma we introduce the following intervals. If x > 0 we will write

Ix(u) := [x + u, x] ∩ [0, ∞) if u < 0 Ix(u) := [x, x + u] if u > 0.

Lemma 2.3 Let f ∈ M^loc[0, ∞) and let at a fixed point x ∈ (0, ∞) the one-sided limits f(x+), f(x−) exist. Consider the function g^x defined by (6) and write dn :=

pbn/n. If h = −x or h = x, then for all integers n such that n/bⁿ­ 4 we have 

Z

Ix(h)

gx(t)Φ_n−1,α(x, t) dt  

 ^{¬ v1(gx; Ix(hdn))}

+cαϕn(x) x²

m−1X

j=1

vj(gx; Ix(jhdn))

j³ +vm(gx; Ix(h)) m²

! ,

where m := [p

n/bn], ϕn(x) = x (1 − x/bⁿ) + bn/n.If α ­ 1 then c^α= 8α.

Proof Restricting the proof to h = −x we define the points t^j = x + jhdn for j = 0, 1, 2, ..., m and we denote tm+1= 0. Put Tj= [tj, x] for j = 1, 2, ..., m + 1 and write

Z

Ix(h)

gx(t)Φn−1,α(x, t) dt = Zx

t1

gx(t)Φn−1,α(x, t) dt + Xm j=1

gx(tj)

tj

Z

tj+1

Φn−1,α(x, t) dt

+ Xm j=1

tj

Z

tj+1

(gx(t) − g^x(tj)) Φn−1,α(x, t) dt

= : I1(n, x) + I2(n, x) + I3(n, x), say.

Clearly,

|I¹(n, x)| ¬ Zx

t₁

|g^x(t) − g^x(x)| Φn−1,α(x, t) dt

¬ v¹(gx; T1)

bn

Z

0

Φ_n−1,α(x, t) dt = v1(gx; T1).

By the Abel lemma on summation by parts and by (10) (with q = 1) we have

|I²(n, x)| ¬ |g^x(t1)|

t1

Z

0

Φ_n−1,α(x, t) dt

+

mX−1 j=1

|g^x(tj+1) − g^x(tj)|

tj+1

Z

0

Φn−1,α(x, t) dt

¬ καϕn(x) x²



|g^x(t₁) − g^x(x)| +

mX−1 j=1

|g^x(t_j+1) − g^x(tj)| 1 (j + 1)²





= καϕn(x) x²



|g^x(t1) − g^x(x)| +

mX−2 j=1

Xj k=1

|g^x(tk+1) − g^x(tk)|

 1

(j + 1)² − 1 (j + 2)²

 + 1

m² Xm k=1

|g^x(t_k+1) − g^x(tk)|

!

¬ καϕn(x) x²



v₁(gx; T1) + 2

m−2X

j=1

vj+1(gx; T_j+1)

(j + 1)³ +vm(gx; Tm) m²





¬ καϕn(x) x²



2

mX−1 j=1

vj(gx; Tj)

j³ +vm(gx; Tm+1) m²



 .

Next, in view of (10) (with q = 1) and the Abel transformation,

|I3(n, x)| ¬ Xm j=1

v1(gx; [t_j+1,tj])

tj

Z

tj+1

Φn−1,α(x, t) dt

¬ καϕn(x) x²

Xm j=1

v1(gx; [tj+1,tj]) j²

= καϕn(x) x²

Xm k=1

v1(gx; [tk+1,tk]) m²

+

mX−1 j=1

Xj k=1

v1(gx; [tk+1,tk])

1

j²− 1 (j + 1)²





¬ καϕn(x) x²



vm(gx; T_m+1)

m² + 6

m−1X

j=1

vj(gx; T_j+1) (j + 1)³





¬ καϕn(x) x²



vm(gx; Tm+1)

m² + 6

Xm j=2

vj(gx; Tj) j³



 .

Collecting the results and observing that Tj = Ix(jhdn) we get the desired estimate for h = −x. In case h = x the proof runs analogously; we use inequality (11) instead

of (10). _■

Lemma 2.4 Let α > 0 and let

sgn^(α)_x (t) :=





2^α− 1 if t > x,

0 if t = x,

−1 if 0 ¬ t < x.

Then 

Kn−1,αsgn^(α)_x (x) ¬ 2^α   

n−1X

k=k⁰+1

P_n−1,k(x bn

)

!^α

− 1 2^α

+ 2^αQ^(α)_n−1,k∗(x bn

),

where k⁰= [(n − 1)x/bⁿ], k^∗= [nx/bn].

Proof Since x ∈ [k^∗bn/n, (k^∗+ 1) bn/n) we have

Kn−1,αsgn^(α)_x (x) = (2^α− 1)

n−1

X

k=k^∗+1

Q^(α)_n−1,k(x bn) −

kX^∗−1 k=0

Q^(α)_n−1,k(x bn

)

+n bn

Q^(α)_n−1,k∗(x bn

)



(2^α− 1)

(k^∗+ 1) bn

n − x



−



x−k^∗bn

n



,

i.e.

Kn−1,αsgn^(α)_x (x) = 2^α

nX−1 k=k^∗+1

Q^(α)_n−1,k(x

bn) − 1 + 2^αQ^(α)_n−1,k∗(x bn

)n bn ∗

∗

(k^∗+ 1) bn

n − x

 . The above identity can be rewritten in the form

Kn−1,αsgn^(α)_x (x) = 2^α

n−1

X

k=k^∗

Q^(α)_n−1,k(x

bn) − 1 − 2^αQ^(α)_n−1,k∗(x bn

)



x−k^∗bn

n

 .

Observing that k^∗= k⁰ or k^∗= k⁰+ 1 and that

nX−1 k=v

Q^(α)_n−1,k(x bn

) =

nX−1 k=v

Pn−1,k(x bn

)

!^α

we easily get the estimate forKⁿ−1,αsgn^(α)x (x) as desired. ■

Lemma 2.5 Let x > 0, α > 0 and let k⁰ = [(n − 1)x/bⁿ]. Then 

n−1X

k=k⁰+1

P_n−1,k

x bn

!^α

− 1 2^α

  ¬√

2κα

rbn

n

s bn

x (bn− x),

where κα = max {1, α} and n ­ 2 is so large that bⁿ > 2x, n/bn > n0(α, x) with n0(α, x) = 0if α ­ 1, n⁰(α, x) = 21/x if 0 < α < 1.

Proof Let y = x/bn. Then k⁰ = [(n−1)y]. In view of the well-known Berry-Ess´een theorem 

X

ny<k¬n

Pn,k(y) − 1 2  

¬ 0.8 pny(1− y) (see e.g. [17, Lemma 2]). Consequently, assuming α ­ 1 we have 

n−1X

k=k⁰+1

Pn−1,k

x bn

!^α

− 1 2^α

 ¬ α

X

k>(n−1)y

Pn−1,k(y) − 1 2   

¬ 0.8α

p(n − 1) y(1 − y)¬√ 2α

rbn

n

s bn

x (bn− x). If 0 < α < 1 then



 X

ny<k¬n

Pn,k(y)





α

− 1 2^α

¬ 1 pny(1− y)

for all n > _25y(1−y)²⁵⁶ (see [18, Lemma 1]). Hence 

nX−1 k=k⁰+1

Pn−1,k(y)

!^α

− 1 2^α

   =



 X

k>(n−1)y

Pn−1,k(y)





α

− 1 2^α

¬ 1

p(n − 1) y(1 − y)¬√ 2

rbn

n

s bn

x (bn− x) for all n such that n − 1 > _25y(1−y)²⁵⁶ , i.e. for all n such that bn > 2x and

n/bn > 21/x. _■

Lemma 2.6 Let x > 0, α > 0 and let k^∗= [nx/bn].Then

Q^(α)_n−1,k∗

x bn



¬ κ^α rbn

n

s bn

x (bn− x),

where κα = max {1, α} and n ­ 2 is so large that bⁿ > 2x, n/bn > n1(α, x) with n1(α, x) = 0if α ­ 1, n¹(α, x) = 50/x if 0 < α < 1.

Proof If α ­ 1 then Q^(α)n−1,k^∗(x/bn) ¬ αPn−1,k^∗(x/bn) . The desired estimate follows with κα= α by the known inequality

(12) Pn,k(y) ¬ 1

√2ep

ny(1− y) f or y∈ (0, 1) , k ∈ ℕ (see [16, Theorem 1]). Indeed, putting y = x/bn we have

Q^(α)_n−1,k∗

x bn



¬ αPⁿ−1,k^∗(y) ¬ α

√2ep

(n − 1) y(1 − y) < α rbn

n

s bn

x (bn− x). Suppose now 0 < α < 1. Using the mean-value theorem we get

Q^(α)_n−1,k∗(y) = αγ_n^α_−1,k⁻¹ ∗(y) (Jn−1,k^∗(y) − Jⁿ−1,k^∗+1(y))

= αγ_n^α_−1,k⁻¹ ∗(y)P_n−1,k∗(y)

where Jn−1,k^∗+1(y) < γn−1,k^∗(y) < Jn−1,k^∗(y) . Denoting k⁰ = [(n − 1) y] we have k^∗= k⁰ or k^∗ = k⁰+ 1.

If k^∗= k⁰ then

γn−1,k^∗(y) > Jn−1,k⁰+1(y) = X

k>(n−1)y

Pn−1,k(y) > 1 4

for all n such that n − 1 > _25y(1−y)²⁵⁶ (see [18, inequality (12)]).