Order of Approximation of Functions of Two Variables by New Type Gamma Operators1 Aydın Izgi˙

General Mathematics Vol. 17, No. 1 (2009), 23–32

Order of approximation of functions of two variables by new type gamma operators1 Aydın Izgi˙

Abstract

The theorems on weighetd approximation and order of approximation of continuous functions of two variables by new type Gamma operators on all positive square region are established.

2000 Mathematics Subject Classiﬁcation: 41A25,41A36 Key words and phrases: Order of approximation,weighted approximation,new type Gamma operators.

1 Introduction

Lupa¸sand M¨uller[11] introduced the sequence of linear positive operators {Gn},Gn : C(0, ∞) → C(0, ∞) deﬁned by

∞ n G (f; x)= g (x, u)f( )du n n u Z0

xn+1 G is called Gamma operator, where g (x, u)= e−xuun,x> 0. n n n! 1Received 15 February, 2008 Accepted for publication (in revised form) 21 March, 2008

23 24 Aydın Izgi˙

Mazhar [12] used same gn(x, u) of Gamma operator and introduced the following sequence of linear positive operators:

∞ ∞

Fn(f; x) = gn(x, u)du gn−1(u, t)f(t)dt Z0 Z0 ∞ (2n)!xn+1 tn−1 = f(t)dt, n > 1,x> 0 n!(n − 1)! (x + t)2n+1 Z0 for any f which the last integral is convergent. Now we will modify the operators Fn(f; x) as the following operators An(f; x) (see [9]) which confirm 2 2 An(t ; x) = x . Many linear operators Ln (f; x) confirm, Ln (1; x) = 1, 2 2 Ln (t; x)= x but don’t confirm Ln (t ; x)= x (see [2],[10]).

∞ ∞

An(f; x) = gn+2(x, u)du gn(u, t)f(t)dt Z0 Z0 ∞ (2n + 3)!xn+3 tn = f(t)dt, x > 0 n!(n + 2)! (x + t)2n+4 Z0 If we choose (2n + 3)! xn+3tn K (x, t)= , x,t ∈ (0, ∞), n n!(n + 2)! (x + t)2n+4 we can show An(f; x) as the following form:

∞

(1) An(f; x)= Kn(x, t)f(t)dt.

Z0 In this study we will investigate approximation and order of approximation of the following operators which deﬁned for two variables functions.

∞

(2) An,m(f; x, y)= Kn,m(x, t; y, u)f(t, u)dtdu

Z0 Order of approximation... 25

where Kn,m(x, t; y, u)= Kn(x, t) × Km(y, u). It can be easily to see that:

An,m(f; x, y) = An(f1; x)+ Am(f2; y) ; if f(t, u)= f1(t)+ f2(u)

An,m(f; x, y) = An(f1; x) × Am(f2; y) ; if f(t, u)= f1(t) × f2(u)

Thus for any p,q ∈ N, p ≤ n and q ≤ m, we can see the following equalities (see [9]):

(3) An,m(1; x, y) = 1

x y (4) A (t + u; x, y)= x + y − − n,m n + 2 m + 2

2 2 2 2 (5) An,m(t + u ; x, y)= x + y

3 3 (6) A (t3 + u3; x)= x3 + y3 + x3 + y3 n,m n m

4(2n + 3) 4(2m + 3) (7) A (t4 +u4; x)= x4 +y4 + x4 + y4, n> 1,m> 1. n,m n(n − 1) m(m − 1)

x y (8) A ({t + u} − {x + y}); x, y)= − − n,m n + 2 m + 2

2x2 2y2 (9) A ({(t − x)2 +(u − y)2}; x, y)= + n,m n + 2 m + 2

12(n + 4) (10) A ({(t − x)4 +(u − y)4}; x, y)= x4 n,m (n + 2)n(n − 1)

12(m + 4)y4 + , n> 1,m> 1 (m + 2)m(m − 1) 26 Aydın Izgi˙

Let C(R2) be the set of all real-valued functions of two variables continuous on R2 := {(x, y) : x ≥ 0, y ≥ 0}, σ(x, y)=1+ x2 + y2, −∞

(11) | f(x, y)| ≤ Mf σ(x, y) where Mf is a constant depending only on f and the norm is deﬁned by | f(x, y)| kfkσ = sup . (x,y)∈R2 σ(x, y)

Cσdenotes the subspaces of all continuous functions which belonging to k Bσ and Cσ denotes the subspaces of all functions belonging to Cσ with f(x, y) lim = k < ∞, x,y→∞ σ(x, y) where k is a constant depending only on f. The approximation theorems for two variables are proved by Volkov[13]. He proved the theorem:

Theorem A ([13]). If {Tn} is a sequence of linear positive operators satisfying the conditions

lim kTn(1; x1, x2) − 1k = 0 . n→∞ C(X)

lim kTn(ti; x1, x2) − xik = 0, i = 1, 2 . n→∞ C(X) 2 2 2 2 lim Tn(t + t ; x1, x2) − (x + x ) = 0 . n→∞ 1 2 1 2 C(X) then for any function f ∈ C(X), which is bounded in R2

lim kTn(f; x1, x2) − f(x1, x2)k = 0 . n→∞ C(X) where X is a compact set. Gadzhiev proved the following theorem for one variable functions.

Theorem B ([3, 4]). {Tn} be the sequence of linear positive operators which mapping from Cρ into Bρ satisfying the conditions

v v lim kTn(t ; x) − x k = 0, v = 0, 1, 2. n→∞ ρ Order of approximation... 27

k Then , for any f ∈ Cρ ,

lim kTnf − fk = 0. n→∞ ρ

k and there exist a function f ∈ Cρ \ Cρ such that

lim kTnf − fk ≥ 1. n→∞ ρ

Analogously as in Theorem B, the theorems on weighted approximation for functions of several variables are proved by Gadzhiev [5]. Applying Theorem B to the operators

Vn(f; x), if x ∈ [0,an] Tn(f; x)= ( f(x), if x>an one then also has the following theorem.

Theorem C ([6]). Let (an) be a sequence with lim an = ∞ and {Vn} be a n→∞ sequence of linear positive operators taking Cρ[0,an] into Bρ[0,an]. If for v = 0, 1, 2

v v lim kVn(t ; x) − x k = 0, n→∞ ρ,[0,an]

k then for any f ∈ Cρ [0,an]

lim kVnf − fk = 0, n→∞ ρ,[0,an]

k k where Bρ[0,an],Cρ[0,an] and Cρ [0,an] denote the same as Bρ,Cρ and Cρ respectively, but the functions taken on [0,an] instead of R and the norm is taken as |f(x)| f . k kρ,[0,an] = sup x∈[0,an] ρ(x) 28 Aydın Izgi˙

2 Approximation of An,m

Let (bn) is be a sequence has positive terms, increasing and has the following conditions,

2 bn (12) lim bn = ∞ and lim = 0 n→∞ n→∞ n

We will denote the rectangular region (0, bn] × (0, bm] by Dn,m and let

Bσ (Dn,m) be sets of all functions f deﬁned on Dn,m satisfying the condition (11) By the using (3) and (5), we have

An,m(σ(t, u); x, u)= σ(x, y).

Therefore, kA (f; x)k is uniformly bounded on D . Hence n,m σ(Dn,m) n,m {An,m} is a sequence of linear positive operators taking Cσ(Dn,m) into

Bσ (Dn,m) .

k Theorem 1. Let f ∈ Cσ , then

lim kAn,m(f; x, y) − f(x, y)k = 0. n,m→∞ σ(Dn,m)

Proof.

lim kAn,m(1; x, y) − 1k = 0. n,m→∞ σ(Dn,m) x n + 2 lim kAn,m(t; x, y) − xk = lim sup = 0. n,m→∞ σ n,m→∞ (x,y)∈(Dn,m) σ(x, y) y m + 2 lim kAn,m(u; x, y) − yk = lim sup = 0. n,m→∞ σ n,m→∞ (x,y)∈(Dn,m) σ(x, y) 2 2 2 2 lim An,m(t + u ; x, y) − (x + y ) = 0 n,m→∞ σ(Dn,m)

Similar to Theorem B we obtain the desired results. Order of approximation... 29

3 The Order of Approximation of An,m

k We now want to ﬁnd the degree of approximation of functions f ∈ Cσ by the operators An,m on Dn,m. It is well- known that the usual ﬁrst modulus of continuity

̟(f; δ) = sup |f(t, u) − f(x, y)| : (t − x)2 +(u − y)2 ≤ δ; t, u, x, y ∈ [a, b] n p o don’t tend to zero, as δ → 0, on any infinite interval and any infinite area, respectively. k In [8] was defined the weighted modulus of continuity for f ∈ Cσ as the following (see also [1]):

|f(x + t, y + u) − f(x, y)| Λ(f; δ, η) = sup : x, y ∈ R2, |t| ≤ δ, |u| ≤ η σ(x, y)σ(t, u) Λ(f; δ, η) is having the following properties:

lim Λ(f; δ, η) = 0. δ,η→0

Λ(f; λ1δ, λ2η) ≤ 4(1 + λ1)(1 + λ2)Λ(f; δ, η), for λ1 > 0,λ2 > 0

and (13) 2 2 |t − x| |u − y| |f(t, u) − f(x, y)| ≤ 8(1 + x + y )Λ(f; δn, δm)(1 + )(1 + ) δn δm

×(1+(t − x)2)(1+(u − y)2)

k Theorem 2. For every f ∈ Cσ the inequality

|A (f; x, y) − f(x, y)| 2b2 2b2 n,m f n , m sup 2 2 ≤ 4232Λ( ; ) (x,y)∈Dn,m (1 + x + y ) sn + 2 sm + 2 is true for all n, m suﬃciently large. 30 Aydın Izgi˙

Proof. If we use (13) and (3) we have

2 2 |An,m(f; x, y) − f(x, y)| ≤ 8(1 + x + y )Λ(f; δn, δm)

|t − x| |u − y| 2 2 ×An,m((1 + )(1 + )(1+(t − x) )(1+(u − y) ); x, y) δn δm 2 2 ≤ 8(1 + x + y )Λ(f; δn, δm)

|t − x| 2 |u − y| 2 ×An((1 + )(1+(t − x) ); x) × Am((1 + )(1+(u − y) ); y) δn δm a2 + b2 We know that a.b ≤ is hold for all positive real numbers a and 2 b. Thus, apply equalities (3), (9), (10) and Cauchy-Schwarz inequalities, we will get |A (f; x, y) − f(x, y)| n,m ≤ (1 + x2 + y2)

2 2 1 2x 1 2x 1 12(n + 4) 4 Λ(f; δn, δm) × 1+(1+ ) + + x " 2δn n + 2 δn rn + 2 2δn (n + 2)n(n − 1) # 1 2y2 1 2y2 1 12(m + 4) × 1+(1+ ) + + y4 " 2δm m + 2 δm rm + 2 2δm (m + 2)m(m − 1) # 2b2 2b2 Choosing δ = n and δ = m and consider δ ≤ 1, n n + 2 m m + 2 n r r 2 bn δm ≤ 1 for all n, m suﬃciently large since lim = 0 , we obtain the n→∞ n desired result.

References

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[12] S.M.Mazhar, Approximation By Positive Operators On inﬁnitenﬁnite Intervals, Vol.5, 1991, Fas. 2, 99-104 32 Aydın Izgi˙

[13] V. I. Volkov, On the convergence of sequences of linear positive operators in the space of continuous functions of two variable, Math. Sb. N. S. 43(85) (1957) 504 (in Russian).

Harran Universtiy Science and Arts Faculty Department of Mathematics S¸anlıurfa-Turkey E-mail: [email protected] /a [email protected]