EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 2, 2018, 457-467
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On approximation properties of generalised
(p, q)-Bernstein operators

Döne Karahan1,∗, Aydn Izgi 1

1 Mathematics Department, Science and Letter Faculty, Harran University, Sanlıurfa, Turkey

Abstract. In this study, a (p, q)–analogue of Bernstein operators is introduced and approximation
properties of (p, q)–Bernstein operators are investigated. Some basic theorems are proved. The
rate of approximation by modulus of continuity is estimated.

2010 Mathematics Subject Classifications: 41A25, 41A36

Key Words and Phrases: Korovkin theorem, Bernstein operator, (p, q)-integers, modulus of
continuity

1. Introduction

In 1912, for a function f(x) defined on the closed interval [0, 1], the expression

Bn(f ;x) =
n∑

k=0

f

(
k

n

)(
n

k

)
xk(1− x)n−k (1)

was called the Bernstein polynomial of order n of the function f(x) in [26]. Later, the var-
ious generalizations of Bernstein polynomials (1) were investigated in [2], [3], [10]-[13]. In
recent years, the development of q-calculus has allowed to be made of new generalizations
of approximation theory. Firstly, Lupaş [4] introduced the q-analogue of the Bernstein
operators and investigated its approximation properties in 1987. After then, the various
applications of q-Bernstein operators were handled by Phillips [7], [8]. The approximation
properties of q-generalization of other operators were studied in [1], [5], [9], [23], [24], [27],
[28] .

Recently, Mursaleen et al. applied (p, q)-in calculus approximation theory and intro-
duced the (p, q)-analogue of Bernstein operators and other operators [12], [14]–[22].

In [3], Izgi introduced a class of new type Bernstein polynomials and investigated its
approximation properties:

∗Corresponding author.

Email addresses: dkarahan@harran.edu.tr (D. Karahan), aydinizgi@yahoo.com (A. Izgi)

http://www.ejpam.com 457 c© 2018 EJPAM All rights reserved.



D. Karahan, A. Izgi / Eur. J. Pure Appl. Math, 11 (2) (2018), 457-467 458

Fn,a,b(f ;x) =
n∑

k=0

f

(
k(n+ a)

n(n+ b)

)
qn,k,a,b(x), 0 ≤ x ≤ n+ a

n+ b
, (2)

where a, b ∈ N, 0 ≤ a ≤ b,

qn,k,a,b(x) =

(
n+ b

n+ a

)n(n
k

)
xk
(
n+ a

n+ b
− x
)n−k

. (3)

The aim of this paper is to introduce (p, q)-analogue of generalized Bernstein operator
(2) and is to study approximation properties for (p, q)-Bernstein operator.

Now we remember certain notations of (p, q)-calculus.
For any p > 0 and q > 0, the (p, q) integers [n]p,q are defined by

[n]p,q = pn−1+pn−2q+pn−3q2+...+pqn−2+qn−1 =



pn−qn
p−q , when p 6= q 6= 1

n pn−1, when p = q 6= 1

[n]q, when p = 1
n, when p = q = 1

where [n]q denotes the q-integers and n = 0, 1, 2, · · · .
Let p, q > 0 be given. We define a (p, q)-factorial, [n]p,q! of k ∈ N, as

[n]p,q! =

{
[1]p,q[2]p,q...[n]p,q, n ∈ N
1, n = 0.

(4)

The (p, q)-binomial coefficient

[
n
r

]
p,q

by

[
n
r

]
p,q

=
[n]p,q!

[n− r]p,q![r]p,q!
. (5)

We recall that (p, q)-derivative operator Dp,q is given by

Dp,qf(x)=
f(px)− f(qx)

(p− q)x
, x 6= 0; Dp,qf(0) = lim

x→0
Dp,qf(x). (6)

For any polynomial f(x) of degree Nand any number c, we have the following (p, q)-
Taylor expansion:

f(x) =

N∑
j=0

(
Dj

p,qf
)

(c)
(n− c)jp,q

[j]p,q!
(7)

where (x− c)np,q is (p, q)-analogue of (x− c)n and

(x− c)np,q =

{
1, n = 0,
(x− c)(px− qc)...(pn−1x− qn−1c), n ≥ 1

(8)



D. Karahan, A. Izgi / Eur. J. Pure Appl. Math, 11 (2) (2018), 457-467 459

Now, we give construction of our operators and some properties of them. For f ∈
C
[
0,

[n+a]p,q
[n+b]p,q

]
,

F p,q
n,a,b(f ;x) =

1

p
n(n−1)

2

n∑
k=0

f

(
[k]p,q[n+ a]p,q

pk−n[n]p,q[n+ b]p,q

)
qp,qn,k,a,b(x), 0 ≤ x ≤ [n+ a]p,q

[n+ b]p,q
, (9)

where a, b ∈ N, 0 ≤ a ≤ b,

qp,qn,k,a,b(x) =

(
[n+ b]p,q
[n+ a]p,q

)n [
n
k

]
p,q

p
k(k−1)

2 xk
n−k−1∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)
. (10)

2. Main Results

Lemma 1. For ∀x ∈
[
0,

[n+a]p,q
[n+b]p,q

]
and ∀n ∈ N, (p, q)-Bernstein operators (9) are satisfied

the following equalities:

i. F p,q
n,a,b(1;x) = 1,

ii. F p,q
n,a,b(t;x) = x,

iii. F p,q
n,a,b(t

2;x) =
pn−1[n+a]p,q
[n]p,q [n+b]p,q

x+
q[n−1]p,q
[n]p,q

x2.

Proof.

F p,q
n,a,b(1;x) =

1

p
n(n−1)

2

n∑
k=0

qp,qn,k,a,b(x)

=

(
[n+ b]p,q
[n+ a]p,q

)n n∑
k=0

[
n
k

]
p,q

p
k(k−1)

2 xk
n−k−1∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)

=1

F p,q
n,a,b(t;x) =

1

p
n(n−1)

2

n∑
k=0

[k]p,q[n+ a]p,q
pk−n[n]p,q[n+ b]p,q

qp,qn,k,a,b(x)

=
1

p
n(n−3)

2

(
[n+ b]p,q
[n+ a]p,q

)n−1 n∑
k=1

[k]p,q
[n]p,q

[
n
k

]
p,q

p
k(k−3)

2 xk
n−k−1∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)

=
1

p
n(n−3)

2

(
[n+ b]p,q
[n+ a]p,q

)n−1 n∑
k=1

[
n− 1
k − 1

]
p,q

p
k(k−3)

2 xk
n−k−1∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)

=
x

p
(n−1)(n−2)

2

(
[n+ b]p,q
[n+ a]p,q

)n−1 n−1∑
k=0

[
n− 1
k

]
p,q

p
(k+1)(k−2)

2 xk
n−k−2∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)



D. Karahan, A. Izgi / Eur. J. Pure Appl. Math, 11 (2) (2018), 457-467 460

= x

F p,q
n,a,b(t

2;x) =
1

p
n(n−1)

2

n∑
k=0

(
[k]p,q[n+ a]p,q

pk−n[n]p,q[n+ b]p,q

)2

qp,qn,k,a,b(x)

=
1

p
n(n−5)

2

(
[n+ b]p,q
[n+ a]p,q

)n−2 n−1∑
k=0

[k + 1]p,q
[n]p,q

[
n− 1
k

]
p,q

p
(k+1)(k−4)

2 xk+1

×
n−k−2∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)

=
1

p
n(n−5)

2 [n]p,q

(
[n+ b]p,q
[n+ a]p,q

)n−2 n−1∑
k=0

[
n− 1
k

]
p,q

p
(k+1)(k−4)

2 xk+1(pk + q[k]p,q)

×
n−k−2∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)

=
1

p
n(n−5)

2 [n]p,q

(
[n+ b]p,q
[n+ a]p,q

)n−2
{

n−1∑
k=0

[
n− 1
k

]
p,q

p
k2−k−4

2 xk+1

×
n−k−2∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)

+ q[n− 1]p,q

n−2∑
k=0

[
n− 2
k

]
p,q

p
(k−2)(k−3)

2 xk+2

×
n−k−3∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)}

=
pn−1x

[n]p,q

1

p
(n−1)(n−2)

2

(
[n+ b]p,q
[n+ a]p,q

)n−2 n−1∑
k=0

[
n− 1
k

]
p,q

p
k(k−1)

2 xk

×
n−k−2∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)

+
q[n− 1]p,qx

2

[n]p,q

1

p
(n−2)(n−3)

2

(
[n+ b]p,q
[n+ a]p,q

)n−2

×
n−2∑
k=0

[
n− 2
k

]
p,q

p
k(k−1)

2 xk
n−k−3∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qsx
)

=
pn−1[n+ a]p,q
[n]p,q[n+ b]p,q

x+
q[n− 1]p,q

[n]p,q
x2

Theorem 1. Let 0 < qn < pn ≤ 1 and

lim
n→∞

pn = 1, lim
n→∞

qn = 1.

If ∀f ∈ C
[
0,

[n+a]p,q
[n+b]p,q

]
, then

lim
n→∞

F pn,qn
n,a,b (f ;x) = f(x)



D. Karahan, A. Izgi / Eur. J. Pure Appl. Math, 11 (2) (2018), 457-467 461

is uniformly on
[
0,

[n+a]p,q
[n+b]p,q

]
.

Proof. The proof is based on Korovkin theorem, so it is enough to prove the conditions

lim
n→∞

∥∥F pn,qn
n,a,b (tm;x)− xm

∥∥ = 0, m = 0, 1, 2

uniformly on
[
0,

[n+a]p,q
[n+b]p,q

]
. From Lemma 1, we get

lim
n→∞

∥∥F pn,qn
n,a,b (1;x)− 1

∥∥ = 0,

lim
n→∞

∥∥F pn,qn
n,a,b (t;x)− x

∥∥ = 0.

Now, we show that
lim
n→∞

∥∥F pn,qn
n,a,b (t2;x)− x2

∥∥ = 0.

From Lemma 1, we obtain

max
x∈

[
0,

[n+a]pn,qn
[n+b]pn,qn

] |F pn,qn
n,a,b (t2;x)− x2|

= max
x∈

[
0,

[n+a]pn,qn
[n+b]pn,qn

]
∣∣∣∣pn−1n [n+ a]p,q
[n]p,q[n+ b]p,q

x+
qnx

2[n− 1]pn,qn
[n]pn,qn

− x2
∣∣∣∣

≤

∣∣∣∣∣
(

[n+ a]pn,qn
[n+ b]pn,qn

)2 pn−1n

[n]pn,qn

∣∣∣∣∣+

∣∣∣∣∣
(

[n+ a]pn,qn
[n+ b]pn,qn

)2(qn[n− 1]pn,qn
[n]pn,qn

− 1

)∣∣∣∣∣
=

(
[n+ a]pn,qn
[n+ b]pn,qn

)2 2pn−1n

[n]pn,qn
.

Then, we get ∥∥F pn,qn
n,a,b (t2;x)− x2

∥∥ ≤ ( [n+ a]pn,qn
[n+ b]pn,qn

)2 2pn−1n

[n]pn,qn
.

Thus, we have

lim
n→∞

max
x∈

[
0,

[n+a]
[n+b]

]|F pn,qn
n,a,b (tm;x)− xm| = 0, m = 0, 1, 2.

In accordance with the Bohman- Korovkin theorem [25], we obtained the desired result.

Lemma 2. Let k − th degree moment for the polynomials (9) defined by

T p,q
n,k(x) = F p,q

n,a,b

(
(t− x)k;x

)
, k = 0, 1, 2. (11)



D. Karahan, A. Izgi / Eur. J. Pure Appl. Math, 11 (2) (2018), 457-467 462

Then we have T p,q
n,0(x) = 1, T p,q

n,1(x) = 0 and

T p,q
n,2(x) =

pn−1[n+ a]p,q
[n]p,q[n+ b]p,q

x+

(
q[n− 1]p,q

[n]p,q
− 1

)
x2. (12)

Moreover, let the sequence {pn}, {qn} satisfying 0 < qn < pn ≤ 1 such that pn → 1, qn →
1 and pnn → α, qnn → β as n→∞, where 0 ≤ α, β < 1. Then

lim
n→∞

[n]pn,qnF
pn,qn
n,a,b

(
(t− x)2;x

)
= λx− αx2 (13)

is uniformly on
[
0,

[n+a]p,q
[n+b]p,q

]
, where 0 < λ ≤ 1.

Proof. It is clear that T p,q
n,0(x) = 1 and T p,q

n,1(x) = 0 hold. From (11), we obtain

T p,q
n,2(x) = F p,q

n,a,b

(
(t− x)2;x

)
=

n∑
k=0

(
[k]p,q[n+ a]p,q

pk−n[n]p,q[n+ b]p,q
− x
)2

qp,qn,k,a,b(x)

=
n∑

k=0

(
[k]p,q[n+ a]p,q
p2k−2n[n][n+ b]

)2

qp,qn,k,a,b(x)

− 2x
n∑

k=0

[k]p,q[n+ a]p,q
pk−n[n]p,q[n+ b]p,q

qp,qn,k,a,b(x) + x2
n∑

k=0

qp,qn,k,a,b(x)

=
pn−1[n+ a]p,q
[n]p,q[n+ b]p,q

x+
q[n− 1]p,q

[n]p,q
x2 − 2x2 + x2

=
pn−1[n+ a]p,q
[n]p,q[n+ b]p,q

x+

(
q[n− 1]p,q

[n]p,q
− 1

)
x2.

Using the equality (12) we have

lim
n→∞

[n]pn,qnF
pn,qn
n,a,b

(
(t− x)2;x

)
= lim

n→∞

(
pn−1n [n+ a]p,q

[n+ b]p,q
x+ (qn[n− 1]pn,qn − [n]pn,qn)x2

)
= lim

n→∞

(
pn−1n [n+ a]p,q

[n+ b]p,q
x

)
+ lim

n→∞

(
qn
pn−1n − qn−1n

pn − qn
− pnn − qnn
pn − qn

)
x2

= λx+ lim
n→∞

pn−1n (qn − pn)

pn − qn
x2

= λx− αx2.



D. Karahan, A. Izgi / Eur. J. Pure Appl. Math, 11 (2) (2018), 457-467 463

Lemma 3.

Fn,a,b(f ; 0) = f(0) and F p,q
n,a,b

(
f ;

[n+ a]p,q
[n+ b]p,q

)
= f

(
[n+ a]p,q
[n+ b]p,q

)
. (14)

Proof. Taking x = 0 into equation (9), we get

F p,q
n,a,b(f ; 0) =

1

p
(n−1)n

2

(
[n+ b]p,q
[n+ a]p,q

)n

f

(
[0]p,q[n+ a]p,q

p0−n[n]p,q[n+ b]p,q

)[
n
0

]
p,q

× p
0(0−1)

2 x0
n−1∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qs0
)

+ 0 + 0 + ...

=
1

p
(n−1)n

2

(
[n+ b]p,q
[n+ a]p,q

)n

f(0)
n−1∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps
)

=
1

p
(n−1)n

2

f(0)
n−1∏
s=0

ps

= f(0).

Similarly, taking x =
[n+a]p,q
[n+b]p,q

into equation (9), we get

F p,q
n,a,b

(
f ;

[n+ a]p,q
[n+ b]p,q

)
=

1

p
(n−1)n

2

(
[n+ b]p,q
[n+ a]p,q

)n

f

(
[0]p,q[n+ a]p,q

p0−n[n]p,q[n+ b]p,q

)[
n
0

]
p,q

p
0(0−1)

2

(
[n+ a]p,q
[n+ b]p,q

)0

×
n−1∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qs [n+ a]p,q
[n+ b]p,q

)
+ ...

+
1

p
(n−1)n

2

(
[n+ b]p,q
[n+ a]p,q

)n

f

(
[n+ a]p,q
[n+ b]p,q

)[
n
n

]
p,q

p
n(n−1)

2

(
[n+ a]p,q
[n+ b]p,q

)n

×
−1∏
s=0

(
[n+ a]p,q
[n+ b]p,q

ps − qs [n+ a]p,q
[n+ b]p,q

)

in above expansion, the terms corresponding to k = 0, 1, ..., n− 1, becomes zero, because

for k = 0, we find
(
[n+a]p,q
[n+b]p,q

− [n+a]p,q
[n+b]p,q

)
as the first factor of each product. It is accepted∏−1

s=0

(
[n+a]p,q
[n+b]p,q

ps − qs [n+a]p,q
[n+b]p,q

)
= 1 so we get

F p,q
n,a,b

(
f ;

(
[n+ a]p,q
[n+ b]p,q

))
= f

(
[n+ a]p,q
[n+ b]p,q

)
.



D. Karahan, A. Izgi / Eur. J. Pure Appl. Math, 11 (2) (2018), 457-467 464

Theorem 2. If f ∈ C
[
0,

[n+a]p,q
[n+b]p,q

]
, then the following inequality holds.

|F p,q
n,a,b(f ;x)− f(x)| ≤

(
1 +

[n+ a]p,q
[n+ b]p,q

)
ω

(
f ;

√
2pn−1

[n]p,q

)
(15)

Proof. From the well-known properties of modulus of continuity we have

|f(t)− f(x)| ≤
(

1 +
|t− x|
δn

)
ω (f ; δn) ,

where δn is any sequences of positive numbers. Since the polynomials F p,q
n,a,b(f ;x) also

linear positive operators, we have

|F p,q
n,a,b(f ;x)− f(x)| ≤

(
1 +

1

δn

√
F p,q
n,a,b ((t− x)2;x)

)
ω (f ; δn) .

Use Cauchy-Schwartz inequality and Lemma 1 , then we obtain

|Fn,a,b(f ;x)− f(x)| ≤

1 +
1

δn

√(
[n+ a]p,q
[n+ b]p,q

)2 2pn−1

[n]p,q

ω (f ; δn)

=

(
1 +

1

δn

[n+ a]p,q
[n+ b]p,q

√
2pn−1

[n]p,q

)
ω (f ; δn) .

Put δn =
√

2pn−1

[n]p,q
, then we get inequality (15).

Theorem 3. (Voronovskaya Type Theorem) Let the sequence {pn}, {qn} satisfying 0 <
qn < pn ≤ 1 such that pn → 1, qn → 1 and pnn → α, qnn → β as n → ∞, where

0 ≤ α, β < 1. For ∀f ∈ C2
[
0,

[n+a]p,q
[n+b]p,q

]
, we have

lim
n→∞

[n]pn,qn

(
F pn,qn
n,a,b (f ;x)− f(x)

)
=
x(λ− αx)

[2]p,q
D2

p,q (f(x)) , 0 < λ ≤ 1. (16)

Proof. Let f ∈ C2
[
0,

[n+a]p,q
[n+b]p,q

]
, that is f,Dp,q(f),D2

p,q(f) ∈ C
[
0,

[n+a]p,q
[n+b]p,q

]
. Define

ψ(t, x) =


f(t)−f(x)−(t−x)Dp,q(f)− 1

[2]p,q
(t−x)2p,qD2

p,q(f)

(t−x)2p,q
, t 6= x

0, t = x.

Then, it is clear that ψ(x, x) = 0 and ψ(., x) ∈ C
[
0,

[n+a]p,q
[n+b]p,q

]
. Hence, from Taylor’s

theorem we have

f(t) = f(x) + (t− x)Dp,q(f) +
1

[2]p,q
(t− x)2p,qD2

p,q(f) + (y − x)2qψ(y, x).



REFERENCES 465

Form Lemma 2,

[n]pn,qn

(
F pn,qn
n,a,b (f ;x)− f(x)

)
= [n]pn,qnF

pn,qn
n,a,b ((t− x);x)Dpn,qn(f)

+
[n]pn,qn
[2]pn,qn

F pn,qn
n,a,b

(
(t− x)2;x

)
D2

pn,qn(f)

+[n]pn,qnF
pn,qn
n,a,b

(
(t− x)2ψ(t, x);x

)
. (17)

If we apply the Cauchy-Schwartz inequality for the last term on the right hand side of
(17), we conclude that

[n]pn,qnF
pn,qn
n,a,b

(
(t− x)2ψ(t, x);x

)
≤
(

[n]2pn,qnF
pn,qn
n,a,b

(
(t− x)4;x

)) 1
2

×
(
F pn,qn
n,a,b

(
ψ2(t, x);x

)) 1
2
. (18)

Let η(t, x) := ψ2(t, x). So, we get η(x, x) = 0 and η(., x) ∈ C2
[
0,

[n+a]p,q
[n+b]p,q

]
. From Theorem

1, we have

lim
n→∞

F pn,qn
n,a,b

(
ψ2(t, x);x

)
= lim

n→∞
F pn,qn
n,a,b (η(t, x);x) = η(x, x) = 0 . (19)

Then taking limit as n→∞ in (17) and using (18), (19) and Lemma 2

lim
n→∞

[n]pn,qn

(
F pn,qn
n,a,b (f ;x)− f(x)

)
=
λx− αx2

[2]p,q
D2

p,q(f)

uniformly with respect to x ∈
[
0,

[n+a]p,q
[n+b]p,q

]
.

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