Key words and phrases: positive linear operator, rate of convergence, Voronovskaya theorem, modulus of continuity, modulus of smoothness, mixed modulus of smooth- ness, limit problem.

Eugeniusz Wachnicki

Approximation by bivariate Mazhar-Totik operators

Abstract. The aim of this paper is to study a bivariate version of the operator inve- stigated in [2], [4]. We shall present Voronovskaya type theorem and theorems giving a rate of convergence of this operator. Some applications for the limit problem are indicated.

2000 Mathematics Subject Classification: 41A36, 41A25, 35K15.

Key words and phrases: positive linear operator, rate of convergence, Voronovskaya theorem, modulus of continuity, modulus of smoothness, mixed modulus of smooth- ness, limit problem.

1. Introduction. In the paper [4] we studied the following operator L(f ; t, x) = f (0) e ⁻

^4tx

+

Z ∞

0 f (s)H(s, x, t)ds, x ­ 0, t > 0, where

H(s, x, t) = 1 4t e ⁻

^x+s4t

r x s I 1

 √ sx 2t

 , I ν is the modified Bessel function ([1]);

(1) I ν (z) =

X ∞ k=0

z 2

2k+ν

k! Γ(ν + k + 1) .

This operator is considered in the class of the measurable functions f on [0, +∞) with f(x) = O(e ^Kx ), where K ­ 0. The case t = 4n ¹ , n ∈ IN, was examined in [2]

by S. M. Mazhar and V. Totik.

In the present paper we will consider the bivariate version of the operator L. We define the operator

M (f ; t, x, y) = f (0, 0) e ⁻

^x+y4t

+ e ⁻

^4tx

Z ∞

0 f (0, s)H(s, y, t) ds + e ⁻

^4ty

Z ∞

0 f (z, 0)H(z, x, t) dz +

Z ∞ 0

Z ∞

0 f (z, s)H(z, x, t)H(s, y, t) dz ds, x ­ 0, y ­ 0, t > 0.

It is easy to remark that M is a positive linear operator. The purpose of this paper is to establish a convergence theorem for operator M. We also indicate the limit problem related to this operator.

2. Auxiliary results. In this section we shall give some properties of the above operator which we shall apply to the proofs of the main theorems.

Using (1) and the properties of the Gamma-Euler function we get the following lemma

Lemma 2.1 For 0 < t < _4K ¹ , x ­ 0 we have

(2) Z ∞ 0

z ⁿ e ^Kz H(z, x, t) dz = e ⁻

^4tx

X ∞ i=0

x ⁱ⁺¹ (4t) ^{n −i−1} (n + i)!

(1 − 4tK) ⁿ⁺ⁱ⁺¹ i! (i + 1)! , n ∈ IN.

In particular, from (2) we conclude that

(3) Z ∞

0

e ^Kz H(z, x, t) dz = e

^1−4tKxK

− e ⁻

^4tx

,

(4) Z ∞

0 z e ^Kz H(z, x, t) dz = x

(1 − 4tK) ² e

^1−4tKxK

,

(5) Z ∞

0 z ² e ^Kz H(z, x, t) dz =

 x ²

(1 − 4tK) ⁴ + 8tx (1 − 4tK) ³



e

^1−4tKxK

,

(6) Z ∞

0 z ³ e ^Kz H(z, x, t) dz =

 x ³

(1 − 4tK) ⁶ + 24x ² t

(1 − 4tK) ⁵ + 96xt ² (1 − 4tK) ⁴



e

^1−4tKxK

,

(7) Z ∞

0

z ⁴ e ^Kz H(z, x, t) dz

=

 x ⁴

(1 − 4tK) ⁸ + 48x ³ t

(1 − 4tK) ⁷ + 576x ² t ²

(1 − 4tK) ⁶ + 1536x ² t ³ (1 − 4tK) ⁵



× e

^1−4tKxK

.

Let Ω = [0, +∞) × [0, +∞). Using (3) - (7) it is easy to prove the following lemmas

Lemma 2.2 Let Q ⊂ Ω be a compact set. Then

(i) there exist numbers C > 0, 0 < α < _4K ¹ such that Z ∞

0 e ^Kz H(z, x, t) dz < C,

Z ∞

0 e ^Ks H(s, y, t) ds < C for (x, y) ∈ Q and 0 < t < α.

(ii)

t lim →0

Z ∞

0 e ^Kz (z − x) ² H(z, x, t) dz = 0, lim

t →0

Z ∞

0 e ^Ks (s − y) ² H(s, y, t) ds = 0 uniformly on Q.

Lemma 2.3 For x ­ 0, y ­ 0, t > 0 we get

M (f 00 ; t, x, y) = 1, M(f 10 ; t, x, y) = x, M(f 01 ; t, x, y) = y, M(f 11 ; t, x, y) = xy, M (f 20 ; t, x, y) = x ² + 8tx, M(f 02 ; t, x, y) = y ² + 8ty,

M (f 30 ; t, x, y) = x ³ + 24x ² t + 96xt ² , M (f 03 ; t, x, y) = y ³ + 24y ² t + 96yt ² , M (f 40 ; t, x, y) = x ⁴ + 48x ³ t + 576x ² t ² + 1536xt ³ ,

M (f 04 ; t, x, y) = y ⁴ + 48y ³ t + 576y ² t ² + 1536yt ³ ,

where f 00 (x, y) = 1, f 11 (x, y) = xy, f i0 (x, y) = x ⁱ , f 0i (x, y) = y ⁱ , i = 1, 2, 3, 4.

Lemma 2.4 For 0 < t < _4k ¹ , x ­ 0 we have

M (ϕ 10 ; t, x, y) = 0, M(ϕ 01 ; t, x, y) = 0, M(ϕ 11 ; t, x, y) = 0, M (ϕ 20 ; t, x, y) = 8tx, M(ϕ 02 ; t, x, y) = 8ty,

M (ϕ 22 ; t, x, y) = 8t(x + y), M(ϕ 44 ; t, x, y) = 192t ² (x ² + y ² ) + 1536t ³ (x + y), where

ϕ i0 (z, s; x, y) = (z − x) ⁱ , ϕ 0i (z, s; x, y) = (s − y) ⁱ , i = 1, 2, ϕ 11 (z, s; x, y) = (z − x)(s − y), ϕ ²² (z, s; x, y) = (z − x) ² + (s − y) ²

ϕ 44 (z, s : x, y) = (z − x) ⁴ + (s − y) ⁴ .

Now, we consider the class E of measurable functions f on Ω such that |f(x, y)| = O e ^K(x+y)

; K ­ 0.

We shall prove the following lemma.

Lemma 2.5 Let (x 0 , y 0 ) ∈ Ω. If f ∈ E and lim

(x,y)→(x

0

,y

0

) f (x, y) = 0, then

(t,x,y)→(0 lim

⁺

,x

0

,y

0

) M (f ; t, x, y) = 0.

Proof Let ε > 0. There exists a number δ > 0 such that

(8) |f(z, s)| < ε

8 for (z − x ⁰ ) ² + (s − y ⁰ ) ² < δ ² , (z, s) ∈ Ω.

Observe that

|M(f; t, x, y)| ¬ |f(0, 0)| e ⁻

^x+y4t

+ e ⁻

^4tx

Z ∞

0 |f(0, s)|H(s, y, t) ds + e ⁻

^4ty

Z ∞

0 |f(z, 0)|H(z, x, t) dz + Z ∞

0

Z ∞

0 |f(z, s)| H(z, x, t)H(s, y, t) dz ds = I ¹ + I 2 + I 3 + I 4 . Write

A ⁱ (x, y) = (

(z, s) : (z − x) ² + (s − y) ² ¬

 δ i

 2 )

, i = 1, 2.

It is easy to conclude that there exists a number 0 < δ ₁ < _4K ¹ such that I 1 = |f(0, 0)|e ⁻

^x+y4t

< ε

4 for 0 < t < δ ₁ and (x, y) ∈ A ¹ (x ₀ , y 0 ).

By Lemma 2.2 (i) we have I 2 = e ⁻

^4tx

Z ∞

0 |f(0, s)|H(s, y, t) ds ¬ M e ⁻

^4tx

Z ∞

0 e ^Ks H(s, y, t) ds < ε 4 for 0 < t < δ 2 and (x, y) ∈ A ¹ (x 0 , y 0 ) where δ 2 < _4K ¹ is a sufficiently small positive number.

Similarly, there exists a number 0 < δ ₃ < _4K ¹ such that I 3 < ε

4 for 0 < t < δ 3 and (x, y) ∈ A ¹ (x 0 , y 0 ).

Now we estimate I ₄ . We can write I 4 = Z ∞

0

Z ∞

0 |f(z, s)|H(z, x, t)H(s, y, t) dz ds

= Z Z

(z,s)∈A

¹

(x

0

,y

0

) |f(z, s)|H(z, x, t)H(s, y, t) dz ds + Z Z

(z,s)∈Ω\A

¹

(x

0

,y

0

) |f(z, s)|H(z, x, t)H(s, y, t) dz ds = I ⁴¹ + I 42 .

By (8) and (3) we have

I 41 ¬ ε 8

Z Z

(z,s)∈A

¹

(x

0

,y

0

) H(z, x, t)H(s, y, t) dz ds

¬ ε 8

Z ∞ 0

Z ∞ 0

H(z, x, t)H(s, y, t) dz ds ¬ ε 8 for 0 < t < α and (x, y) ∈ A ¹ (x 0 , y 0 ).

Now, we remark that if (x, y) ∈ A ² (x 0 , y 0 ) and (z, s) ∈ Ω \ A ¹ (x 0 , y 0 ), then (z, s) ∈ Ω \ A ² (x, y). Hence

I 42 = Z Z

(z,s)∈Ω\A

¹

(x

₀

,y

₀

) |f(z, s)|H(z, x, t)H(s, y, t) dz ds

¬ M Z Z

(z,s)∈Ω\A

²

(x,y) e ^K(z+s) H(z, x, t)H(s, y, t) dz ds

¬ 4M δ ²

Z ∞ 0

Z ∞

0 e ^K(z+s) (z − x) ² + (s − y) ²

H(z, x, t)H(s, y, t) dz ds

¬ 4M δ ²

Z _∞

0 e ^Kz (z − x) ² H(z, x, t) dz Z ∞

0 e ^Ks H(s, y, t) ds + Z ∞

0 e ^Kz H(z, x, t) dz Z ∞

0 e ^Ks (s − y) ² H(s, y, t) ds

 . From the above and Lemma 2.2 we obtain

I 42 < ε 8

for 0 < t < δ ₄ and (x, y) ∈ A ² (x ₀ , y 0 ), where δ ₄ < _4K ¹ is a sufficiently small positive number.

Finally

|M(f; t, x, y)| ¬ 3ε 4 + ε

8 + ε 8 = ε

for 0 < t < min (δ ₁ , δ 2 , δ 3 , α, δ 4 ) and (x, y) ∈ A ² (x ₀ , y 0 ). The proof of Lemma 2.4 is

completed. _■

3. Convergence theorems. Directly from linearity of the operator M, Lem- mas 2.2 and 2.4, we obtain the following theorems.

Theorem 3.1 If f ∈ E is continuous at a point (x ⁰ , y 0 ) ∈ Ω, then

(t,x,y)→(0 lim

⁺

,x

₀

,y

₀

) M (f ; x, y, t) = f (x 0 , y 0 ).

Theorem 3.2 Let U ⊂ Ω be a compact set. If f ∈ E is continuous on U, then

t →0 lim

⁺

M (f ; t, x, y) = f (x, y) uniformly on U.

Now, we shall prove the Voronovskaya type theorem for the operator M.

Theorem 3.3 If f ∈ E is of the class C ¹ in certain neighbourhood of a point (x ₀ , y 0 ), f ⁰⁰ (x ₀ , y 0 ) exists (in the Fr´echet sense), then

(9) lim

t →0

M (f ; t, x 0 , y 0 ) − f(x ⁰ , y 0 )

t = 4x 0 ∂ ² f

∂x ² (x 0 , y 0 ) + 4y 0 ∂ ² f

∂y ² (x 0 , y 0 ).

Proof We have

f (z, s) = f (x 0 , y 0 ) + (z − x ⁰ ) ∂f

∂x (x 0 , y 0 ) + (s − y ⁰ ) ∂f

∂y (x 0 , y 0 ) + 1

2



(z − x ⁰ ) ² ∂f

∂x ² (x 0 , y 0 ) + 2(z − x ⁰ )(s − y ⁰ ) ∂ ² f

∂x∂y (x 0 , y 0 ) +(s − y 0 ) ² ∂ ² f

∂y ² (x ₀ , y 0 )



+Ψ x

0

,y

0

(z, s) p

(z − x ⁰ ) ⁴ + (s − y ⁰ ) ⁴ , where Ψ is a function of class E and lim

(z,s)→(x

0

,y

0

) Ψ x

0

,y

0

(z, s) = 0.

By linearity of M, Lemmas 2.2 and 2.3 we obtain

M (f ; t, x 0 , y 0 ) − f(x ⁰ , y 0 )

= 4t

 x 0 ∂ ² t

∂x ² (x 0 , y 0 ) + y 0 ∂ ² t

∂y ² (x 0 , y 0 )



+ M (Ψ x

₀

,y

₀

√ ϕ 44 ; t, x 0 , y 0 ) . Using the H¨older inequality and Lemma 2.2 we get

|M (Ψ ^x

0

,y

₀

√ ϕ 44 ; t, x 0 , y 0 )|

¬ M Ψ ² x

₀

,y

₀

; t, x 0 , y 0

¹₂

(M (ϕ 44 ; t, x 0 , y 0 ))

¹²

¬ t M Ψ ² x

0

,y

0

; t, x ₀ , y 0

¹₂

q

192(x ² ₀ + y ² ₀ ) + 1536t(x ₀ + y ₀ ) The above and Lemma 2.3 imply

t→0 lim

M Ψ x

0

,y

0

√ ϕ 44 ; t, x ₀ , y 0

t = 0

which proves (9). _■

Corollary 3.4 If the function f satisfies the assumption of Theorem 3.3, then M (f ; t, x 0 , y 0 ) = f(x 0 , y 0 ) + O(t)

for t → 0.

4. Rate of convergence. In this part we shall give some estimates of the rate of convergence of the operator M. First we shall use the classical modulus of continuity defined by

ω(f, δ) = sup 

|f(z, s) − f(x, y)| : (z, s), (x, y) ∈ Ω, (z − x) ² + (s − y) ² < δ ² . By the above definition we have

(10) |f(z, s) − f(x, y)| ¬



1 + (z − x) ² + (s − y) ² δ ²

 ω(f, δ) for (z, s), (x, y) ∈ Ω and for every δ > 0.

Theorem 4.1 If f ∈ E is a continuous function on Ω, then (11) |M(f; t, x, y) − f(x, y)| ¬ (1 + 8(x + y))ω(f, √

t) and

(12) |M(f; t, x, y) − f(x, y)| ¬ 9ω  f, p

t(x + y)  for (x, y) ∈ Ω and t > 0.

Proof By (3), (10) and Lemma 2.3 we get

|M(f; t, x, y) − f(x, y)| ¬ M(|f − f(x, y)|; t, x, y)

¬ M

 1 + 1

δ ² ϕ 22 , t, x, y

 ω(f, δ)

¬



M (f 00 ; t, x, y) + 1

δ ² M (ϕ 22 ; t, x, y)



ω(f, δ) =



1 + 8t(x + y) δ ²

 ω(f, δ).

For δ = √

t we get (11).

If x + y > 0 then for δ = p

t(x + y) we obtain (12). If x + y = 0 i.e. x = 0 and y = 0 we have M (f ; t, 0, 0) = f (0, 0) and in this case (12) is valid too. _■

Theorem 4.2 If f ∈ E is a function continuously differentiable on Ω, then

(13)

|M(f; t, x, y) − f(x, y)|

¬ p

8(x + y) + p

8(x + y)  √ t

 ω

 ∂f

∂x , √ t

 + ω

 ∂f

∂y , √ t



.

and (14)

|M(f; t, x, y) − f(x, y)|

¬ (4 + 2 √ 2) p

t(x + y)

 ω

 ∂f

∂x , p

t(x + y)

 + ω

 ∂f

∂y , p

t(x + y)



. Proof Let (x, y) ∈ Ω. Observe that

f (z, s) − f(x, y) = (z − x) ∂f

∂x (x, y) + (s − y) ∂f

∂y (x, y) +

Z z x

 ∂f

∂u (u, s) − ∂f

∂x (x, y)

 du +

Z s y

 ∂f

∂v (x, v) − ∂f

∂y (x, y)

 dv

and    

Z z x

 ∂f

∂u (u, s) − ∂f

∂x (x, y)

 du

    ¬

   

Z z x

    ∂f

∂u (u, s) − ∂f

∂x (x, y)     du

   

¬ |z − x|ω

 ∂f

∂x , λ(z, s)



¬ λ(z, s)



1 + λ(z, s) δ

 ω

 ∂f

∂x , δ



for every δ > 0, where λ(z, s) = p

(z − x) ² + (s − y) ² . Similar reasoning gives

   

Z s y

 ∂f

∂v (x, v) − ∂f

∂y (x, y)

 dv

 

  ¬ λ(z, s)



1 + λ(z, s) δ

 ω

 ∂f

∂y , δ



for every δ > 0.

Hence, by Lemmas 2.2 and 2.3 we have

|M(f; t, x, y) − f(x, y)| ¬    M

Z z x

 ∂f

∂u (u, s) − ∂f

∂x (x, y)



du; t, x, y   

+    M

Z s y

 ∂f

∂v (x, v) − ∂f

∂y (x, y)



; t, x, y   

¬ M

 λ + λ ²

δ ; t, x, y

  ω

 ∂f

∂x , δ

 + ω

 ∂f

∂y , δ



¬



M (λ; t, x, y) + 1

δ M (λ ² ; t, x, y)

  ω

 ∂f

∂x , δ

 + ω

 ∂f

∂y , δ



. Applying the H¨older inequality, Lemmas 2.2 and 2.3 we get

M (λ; t, x, y) ¬ M λ ² ; t, x, y

¹₂

· (M(f ⁰ ; t, x, y))

¹²

= 2 p

2t(x + y).

Thus

|M(f; t, x, y)−f(x, y)| ¬ 2 p

2t(x + y) + 4t(x + y) δ

  ω

 ∂f

∂x , δ

 + ω

 ∂f

∂y , δ



. Setting δ = √

t we get (13). If x + y > 0, then for δ = p

2t(x + y) we obtain (14).

If x + y = 0 i.e. x = 0, y = 0, then M(f; t, 0, 0) = f(0, 0) and (14) holds. _■

We will now study the rate of convergence of the operator M by means of the Lipschitz class Lip _A α for 0 < α ¬ 1 and A > 0.

We recall that a function f ∈ E belongs to Lip A α if the inequality

(15) |f(z, s) − f(x, y)| ¬ A (z − x) ² + (s − y) ²

^α₂

holds for (x, y) ∈ Ω and (z, s) ∈ Ω.

Theorem 4.3 For any f ∈ Lip A α , t > 0 and for each (x, y) ∈ Ω we have

|M(f; t, x, y) − f(x, y)| ¬ M(8t(x + y))

^α2

.

Proof Let f ∈ Lip A α and (x, y) ∈ Ω, t > 0. M is a positive linear operator, then by (15) we get

|M(f; t, x, y) − f(x, y)| ¬ M(|f(·, ·) − f(x, y)|; t, x, y) ¬ AM 

(ϕ ₂₂ )

^α2

; t, x, y  . Applying the H¨older inequality with p = _α ² , q = _2−α ² we have

M 

(ϕ ₂₂ )

^α2

; t, x, y 

¬ (M (ϕ 22 ; t, x, y))

^α2

(M(f ₀ ; t, x, y)) ¹⁻

^α2

= (8t(x + y))

^α2

this ends the proof of Theorem 4.3. _■

Now we introduce the mixed modulus of smoothness and modulus of smoothness [3]. Let f : Ω −→ IR and δ ¹ > 0, δ > 0.

The function ω m (f) : Ω −→ IR defined by ω m (f)(δ 1 , δ 2 ) = ω m (f; δ 1 , δ 2 ) = sup

|x − z| ¬ δ1

|y − s| < δ2 (x, y) ∈ Ω (z, s) ∈ Ω

|f(z, s) − f(x, s) − f(z, y) + f(x, y)|

is called the mixed modulus of smoothness of order 1 of f.

The function ω 2 (f) : Ω −→ IR defined by

ω 2 (f)(δ 1 , δ 2 ) = ω 2 (f; δ 1 , δ 2 )

= sup

0 ¬ h ¬ δ1 0 ¬ k ¬ δ2 (x, y) ∈ Ω (x + 2h, y + 2k) ∈ Ω

|f(x + 2h, y + 2k) − 2f(x + h, y + k) + f(x, y)|

is called the modulus of smoothness of order 2 of the function f.

Moreover, we shall use the Stieklov function defined by

f hk (x, y) = 16 h ² k ²

Z

^k₂

0

Z

^k₂

0

Z

^h₂

0

Z

^h₂

0

2f(x + z 1 + z 2 , y + s 1 + s 2 )

−f(x + 2(z ¹ + z 2 ), y + 2(s 1 + s 2 ))dz 1 dz 2 ds 1 ds 2 . We observe that

|f ^hk (x, y) − f(x, y)| ¬ ω ² (f; h, k) and

f hk (x, y) = 32 h ² k ²

Z

^k₂

0

Z

^k₂

0

Z x+

^h₂

x

Z u+

^h₂

u

f (z, y + s 1 + s 2 )dz du ds 1 ds 2

− 4

h ² k ² Z

^k₂

0

Z

^k₂

0

Z x+h x

Z u+h u

f (z, y + 2(s 1 + s 2 ))dz du ds 1 ds 2

= 32 h ² k ²

Z y+

^k₂

y

Z v+

^k₂

v

Z

^h₂

0

Z

^h₂

0

f (x + z 1 + z 2 , w)dz 1 dz 2 dw dv

− 4

h ² k ² Z y+k

y

Z v+k v

Z

^h₂

0

Z

^h₂

0

f (x + 2z 1 + 2z 2 , w)dz 1 dz 2 dw dv

= 32 h ² k ²

Z

^k₂

0

Z y+

^k₂

y

Z

^h₂

0

Z x+

^h₂

x

f (u + z 2 , v + s 2 )du dz 2 dv ds 2

− 4

h ² k ² Z

^k₂

0

Z y+k y

Z

^h₂

0

Z x+h x

f (u + 2z 2 , v + 2s 2 )du dz ₂ dv ds 2 . Hence

∂ ²

∂x ² f hk (x, y)

= 32 h ² k ²

Z

^k₂

0

Z

^k₂

0 [f(x + h, y + s 1 + s 2 )

−2f(x + k

2 , y + s 1 + s 2 ) + f(x, y + s 1 + s 2 )

 ds 1 ds 2

− 4

h ² k ² Z

^k₂

0

Z

^k₂

0

[f(x + 2h, y + 2(s ₁ + s ₂ ))

−2f(x + h, y + 2(s ¹ + s 2 )) + f(x; y + 2(s 1 + s 2 ))] ds 1 ds 2

and (16)

    ∂ ²

∂x ² f hk (x, y)     ¬ 8

h ² ω 2

 h 2 , k

2

 + 1

h ² ω 2 (h, k) ¬ 9

h ² ω 2 (f; h, k) for (x, y) ∈ Ω.

Similarly (17)

    ∂ ²

∂y ² f hk (x, y)     ¬

9

k ² ω 2 (f; h, k), (x, y) ∈ Ω,

(18)

    ∂ ²

∂x∂y f hk (x, y)     ¬ 9

hk ω m (f; h, k), (x, y) ∈ Ω.

We shall prove

Theorem 4.4 If f ∈ E is a function continuous in Ω, then (19) |M(f; t, x, y)−f(x, y)| ¬ (2+36x+36y)ω 2 (f; √

t, √

t)+36(x+y)ω m (f; √ t, √

t), (x, y) ∈ Ω and t > 0.

Proof First we suppose that f is twice continuously differentiable on Ω. We get f (z, s) = f (x, y) + (z − x) ∂f

∂x (x, y) + (s − y) ∂f

∂y (x, y) + Z z

x (z − u) ∂ ² f

∂u ² (u, y) du + Z s

y (s − v) ∂ ² f

∂v ² (x, v) dv + Z z x

Z s y

∂ ² f

∂u∂v (u, v) du dv.

Hence, by Lemma 2.3 we have M (f ; t, x, y) = f (x, y) + M

Z z

x (z − u) ∂ ² f

∂u ² (u, y) du; t, x, y



+M

Z s

y (s − v) ∂ ² f

∂v ² (x, v) dv; t, x, y

 + M

Z z x

Z s y

∂ ² f

∂u∂v (u, v) du dv; t, x, y

 . Applying (16) - (18) we obtain

   M

Z z

x (z − u) ∂ ² f

∂u ² (u, y) du; t, x, y    ¬ M    Z z

x (z − u) ∂ ² f

∂u ² (u, y) du     ; t, x, y



¬ M

Z z x (z − u)

    ∂ ² f

∂u ² (u, y) du     ; t, x, y



¬ k ∂ ² f

∂u ² kM

Z z

x (z − u) du; t, x, y



=         ∂ ² f

∂u ²         M

 1

2 ϕ 20 (·, x, y); t, x, y



= 4k ∂ ² f

∂u ² ktx,

where

k ∂ ² f

∂u ² k = sup

(x,y)∈Ω

    ∂ ² f

∂u ² (u, v)     . Analogously

   M

Z s

y (s − v) ∂ ² f

∂v ² (x, v) dv; t, x, y    ¬ 4k ∂ ² f

∂v ² kty and    M

Z z x

Z s y

∂ ² f

∂u∂v (u, v) du dv; t, x, y    ¬ 4tk ∂ ² f

∂u∂v k(x + y), where

k ∂ ² f

∂v ² k = sup

(u,w)∈Ω

    ∂ ² f

∂v ² (u, v)  

 , k ∂ ² f

∂u∂v k = sup

(u,w)∈Ω

    ∂ ² f

∂u∂v (u, v)    .

Hence

|M(f; t, x, y) − f(x, y)| ¬ 4t

 x k ∂ ² f

∂u ² k + yk ∂ ² f

∂v ² k + (x + y)k ∂ ² f

∂u∂v k

 . Let f ∈ E be continuous on Ω. Thus f ^hk is twice continuously differentiable.

Using above for f hk we obtain

|M(f; t, x, y) − f(x, y)|

¬ M(|f − f ^hk |; t, x, y) + |M(f ^hk ; t, x, y) − f ^hk (x, y)| + |f ^hk (x, y) − f(x, y)|

¬ ω 2 (f; h, k)M(f ₀ ; t, x, y) + 4t



x k ∂ ² f hk

∂u ² k + yk ∂ ² f hk

∂v ² k + (x + y)k ∂ ² f

∂u∂v k



+ω 2 (f; h, k)

¬ 2ω ² (f; h, k) + 4t

 9x

h ² ω 2 (f; h, k) + 9y

k ² ω 2 (f; h, k) + 9(x + y)

hk ω m (f; h, k)



for x, y ∈ Ω, t > 0 and h > 0, k > 0. Putting h = k = √

t we obtain (19). _■ 5. Limit problem. In this section we indicate the limit problem related to the operator M. In the paper [4] we prove that L(f) is a solution of the problem

 



∂u(t, x)

∂t = 4x ∂ ² u(t, x)

∂t ² in D =



(t, x) : x > 0, 0 < t < 1 4K



u(t, x) | t=0 = f(x); x > 0.

Consequently by above and Theorem 3.1, for the operator M we obtain the following

theorem

Theorem 5.1 If f ∈ E, then M(f) is a solution of the problem (

_∂u(t,x,y)

∂t

= 4x

^∂2u(t,x,y)_∂x2

+ 4y

^∂2u(t,x,y)_∂y2

in ˜ D = 

(t, x, y) : 0 < t <

_4K¹

, x > 0, y > 0 u (t, x, y)|

t=0

= f(x, y); x > 0, y > 0.

References

[1] W. Magnus and F. Oberhettinger, Formelen und S¨atz f¨ur die speziellen Funktionen der ma- thematischen Physik, Springer-Verlag, Berlin 1948.

[2] S. M. Mazhar and V. Totik, Approximation by modified Sz´asz operators, Acta Sci. Math.

(Szeged) 49 (1985), 257–269.

[3] A. F. Timan, Theory of Approximation of Functions of a Real Variable, Moscow 1960 [in Russian].

[4] E. Wachnicki, On generalized Mazhar-Totik operators, Demonstratio Math. 38.1 (2005), 143–

151.

Eugeniusz Wachnicki

Institute of Mathematics, Pedagogical University ul. Podchora¸żych 2, 30-084 Kraków, Poland E-mail: euwachni@up.krakow.pl

(Received: 15.05.2010)