K a -Convergence For Double Sequences And Korovkin Type Approximation

(1)

K a -Convergence For Double Sequences And Korovkin Type Approximation

Sevda Y¬ld¬z

^y

Received 6 February 2020

Abstract

In this paper, we introduce the idea ofKa-convergence for double sequences. Then, we use this notion to prove a Korovkin type approximation theorem and present an application that satis…es our new main theorem but does not satisfy classical ones. Finally, we study the rate of convergence of positive linear operators.

1 Introduction and Preliminaries

The following, now a classical result, was proved by P. P. Korovkin [11]: Let I be a compact subset of the real numbers and(Ln)be a sequence of positive linear operators that mapsC(I);the space of all continuous real valued functions onI;into itself. Suppose that the sequence(Ln(f))converges tof uniformly onIfor the three special functionsfi:x!xⁱ;wherei= 0;1;2:Then this sequence converges tof uniformly onIfor everyf 2C(I):Because of its powerful applications, Korovkin’s result has been extended in many directions.

There is an extensive literature on Korovkin-type theorems (see for example [2,3, 4, 5,6,7,9, 10,15]). In this paper, we de…ne the concept ofK_a²-convergence that is a new convergence method and give an example in support of our de…nition. Then, we use this notion to prove a Korovkin type approximation theorem and present an application that satis…es our new main theorem but does not satisfy classical ones. Finally, we study the rate of convergence of positive linear operators.

Now, we recall well known and important convergence methods; statistical and almost convergence for double sequences.

A double sequence x = (x_ij) is said to be convergent in Pringsheim’s sense if, for every " > 0; there exists J = J(") 2 N; the set of all natural numbers, such that jx_ij Lj < " wheneveri; j > J; where L is called the P-limit of xand denoted by P lim

i;jxij =L (see [16]). We shall call such an x; brie‡y, “P- convergent”. A double sequence is called bounded if there exists a positive numberN such thatjxijj N for all(i; j)2N²=N N:Note that in contrast to the case for single sequences, a convergent double sequence need not to be bounded.

Statistical convergence of single sequences was introduced by Fast [8] and Steinhaus [17], independently and studied by many authors. This concept was extended to the double sequences by Moricz [13]. IfE N² is a two-dimensional subset of positive integers andjDjdenotes the cardinality ofD;then the double natural density ofEis given by

2(E) :=P lim

n;k

jfi n; j k: (i; j)2Egj

nk ;

if it exists. The number sequencex= (xij)is statistically convergent to L provided that for every " >0;

the set

E:=Enk(") :=fi n; j k: jxij Lj "g

Mathematics Sub ject Classi…cations: 40A35, 40B05, 41A25, 41A36.

ySinop University, Faculty of Arts and Science, Department of Mathematics, 57000, Sinop, TURKEY

62

(2)

has natural density zero; in that case we writest₂ lim

i;jx_ij=L: Clearly, aP-convergent double sequence is statistically convergent to the same value but its converse is not always true. Also, note that a statistically convergent double sequence may not be bounded.

The de…nition of almost convergence for double sequences was introduced by Moricz and Rhoades [14]

as follows.

A double sequence x= (x_ij)of real numbers is said to be almost convergent to a limitLif

P lim

n;k sup

p;q>0

1 nk

p+nX1 i=p

q+kX1 j=q

xij L = 0:

In this case L is called theF₂-limit ofxand denoted by F₂ lim

i;jx_ij =L: Note that a convergent double sequence need not be almost convergent. However every bounded convergent double sequence is almost convergent and every almost convergent double sequence is bounded.

Lazic and Jovovic de…ned theKa-convergence for single sequences in 1993 [12]. Now, we extend this idea to double sequences. This new convergence method is associated to the four dimensional matrix

A= 0 BB BB BB BB BB BB

@ 0

@ a₁₁ 0 0 :

0 0 0 :

: : : :

1 A

0

@ a₁₂ a₁₁ 0 :

0 0 0 :

: : : :

1 A

0

@ a₁₃ a₁₂ a₁₁ 0 :

0 0 0 0 :

: : : : :

1 A :::

0 BB

@

a12 0 0 : a11 0 0 :

0 0 0 :

: : : :

1 CC A

0 BB

@

a22 a21 0 : a12 a11 0 :

0 0 0 :

: : : :

1 CC A

0 BB

@

a23 a22 a21 0 : a13 a12 a11 0 :

0 0 0 0 :

: : : : :

1 CC A :::

: : : :::

1 CC CC CC CC CC CC A :

Leta= (a_nk)andx= (x_nk)are double sequences, set K_a²(x) =y;wherey= (y_nk)and

y_nk= Xn i=1

Xk j=1

a_{n i+1k j+1}x_ij(n; k= 1;2;3; :::):

Then it is said thaty= (ynk)is theK_a²-transformation of the double sequencex= (xnk):

De…nition 1 The double sequence x= (x_nk)of real numbers isK_a²-convergent to the numberL if, itsK_a²- transformation y = (y_nk) converges to the number L in Pringsheim’s sense, i.e. P lim

n;ky_nk =L; and we denote this limit byK_a² lim

n;kx_nk=L:

The proof of the following proposition can be easily established from the results concerning the general matrix transformation for double sequences. So, we omit it.

Proposition 1 Let a= (ank)be a double sequence and assume that

P lim

i;j

Xi n=1

Xj k=1

ja_nkj exists (1)

and

there exists a positive integer M such that X

(n;k)2N²

ja_nkj< M: (2)

(3)

(i) If x= (x_nk)isP-convergent, P lim

n;kx_nk=L and the conditions (1) and (2) are satis…ed, then K_a² lim

n;kx_nk=L X

(n;k)2N²

a_nk;

(ii) A convergence methodK_a² is regular if and only if the conditions (1), (2) and X

(n;k)2N²

a_nk= 1 (3)

are valid.

Now, the question arises in the theory of double sequences, which concerns the relationship, if any, between statistical convergence, almost convergence and Ka-convergence. Our answer is "these concepts overlap, but none is implied by the other" and it is important to say that, for these three convergence methods, if a double sequence is bounded convergent then it is statistical convergent, almost convergent and K_a-convergent, too. The following double sequence x= (x_nk)isK_a²-convergent, despite this, it is not P-convergent and also, it is not statistical and almost convergent.

Example 1 Let a= (a_nk)given by (a_nk) = 0 BB

@

1 0 0 :

0 1 0 :

0 0 0 :

: : : :

1 CC

A and letx= (x_nk) given by

(x_nk) = 0 BB BB

@

1 2

2 2

3 2

4 2

5 2

6 2 : 0 ¹₂ ²₂ ³₂ ⁴₂ ⁵₂ : 0 0 ¹₂ ²₂ ³₂ ⁴₂ :

0 0 0 ¹₂ ²₂ ³₂ :

: : : : : : :

1 CC CC A:

Then

(y_nk) = 0

@ Xn i=1

Xk j=1

a_{n i+1k j+1}x_ij 1 A=

0 BB

@

1 2

2 2

3 2

4 2 :

0 0 0 0 :

: : : : :

1 CC A:

Hence, we can write thatK_a² lim

n;kx_nk= 0:However, it can be easily seen that,x= (x_nk)is notP-convergent.

Also,xis neither statistical convergent nor almost convergent.

2 Korovkin Type Theorem via K

_a²

-convergence

In this section we study a Korovkin type approximation theorem via K_a²-convergence of positive linear operators.

Let I² = I I and I be a compact subset of the real numbers, C I² be the two-dimensional space of all continuous real valued functions on I² and kfkC(I²) denote the usual supremum norm off : Let L be a linear operator from C I² into itself. Then, we say that L is positive linear operator on condition that f 0 impliesL(f) 0: Also, we mean the value ofL(f)at a point (x; y)2I² byL(f(s; t);x; y) or L(f;x; y): Throughout the paper, we also use the following test functions

f₀(x; y) = 1; f₁(x; y) =x; f₂(x; y) =y; f₃(x; y) =x²+y²: Now, we begin with the following well-known Korovkin type theorems.

(4)

Theorem 2 ([18]) Suppose that (L_nk) is a double sequence of positive linear operators from C I² into itself, satisfying the following conditions:

P lim

n;kkLnk(fr) frkC(I²)= 0; (r= 0;1;2;3): Then, for all f 2C I² ;

P lim

n;kkLnk(f) fkC(I²)= 0:

Theorem 3 ([6]) Assume that (Lnk) is a double sequence of positive linear operators acting from C I² into itself, satisfying the following conditions:

st2 lim

Theorem 4 ([1]) Suppose that (Lnk) is a double sequence of positive linear operators from C I² into itself, satisfying the following conditions:

F2 lim

Now we give the following Korovkin type approximation theorem for methodK_a² that is our main result.

Theorem 5 Let a= (ank)be a double sequence and the conditions (1) and (2) are satis…ed. Suppose that (Lnk)is a double sequence of positive linear operators acting fromC I² into itself, satisfying the following conditions:

P lim

n;k

Xn i=1

Xk j=1

jan i+1k j+1jLij(fr) fr C(I²)

= 0; (r= 0;1;2;3): (4)

Then, for all f 2C I² ;we have

K_a² lim

n;kkL_nk(f) fkC(I²)= 0; i.e.,

P lim

n;k

Xn i=1

Xk j=1

an i+1k j+1Lij(f) f

C(I²)

= 0:

Proof. Letf 2C I² and (x; y)2I² be …xed. By the continuity off onI²;we can write

jf(x; y)j Mf: (5)

Therefore

jf(s; t) f(x; y)j 2M_f:

Also, since f is continuous on I²; we write that for every " > 0; there exists a number > 0 such that jf(s; t) f(x; y)j< "holds for all(s; t)2I² satisfyingjs xj< and jt yj< : Hence, we get

jf(s; t) f(x; y)j< "+2Mf 2

n

(s x)²+ (t y)² o

: (6)

(5)

This means

" 2M_f

2

n

(s x)²+ (t y)² o

< f(s; t) f(x; y)< "+2M_f

2

n

(s x)²+ (t y)² o

:

Using the linearity and the positivity of the operatorsLnk and the inequality (6), we get

Xn i=1

Xk j=1

a_{n i+1k j+1}L_ij(f;x; y) f(x; y) = Xn i=1

Xk j=1

a_{n i+1k j+1}(L_ij(f(s; t) ;x; y)

L_ij(f(x; y) ;x; y) +L_ij(f(x; y) ;x; y)) f(x; y)j Xn

i=1

Xk j=1

ja_{n i+1k j+1}jL_ij(jf(s; t) f(x; y)j;x; y)

+jf(x; y)j Xn i=1

Xk j=1

jan i+1k j+1jLij(f0;x; y) f0(x; y)

"+ "+Mf+2Mfkf3kC(I²) 2

! _n X

i=1

Xk j=1

+4Mfkf1kC(I²) 2

Xn i=1

Xk j=1

+4M_fkf₂kC(I²) 2

Xn i=1

Xk j=1

ja_{n i+1k j+1}jL_ij(f₂;x; y) f₂(x; y)

+2Mf 2

Xn i=1

Xk j=1

jan i+1k j+1jLij(f3;x; y) f3(x; y) :

Then taking supremum over(x; y)2I²;we have

Xn i=1

Xk j=1

an i+1k j+1Lij(f) f

C(I²)

"+K 8>

<

>: X3 r=0

Xn i=1

Xk j=1

jan i+1k j+1jLij(fr) fr C(I²)

9>

=

>;

where

K:= max (

"+Mf+2Mfkf3kC(I²)

2 ;4Mfkf1kC(I²)

2 ;4Mfkf1kC(I²) 2 ;2Mf

2

) :

Then using the hypothesis (4), we get

P lim

n;k

Xn i=1

Xk j=1

a_{n i+1k j+1}L_ij(f) f

C(I²)

= 0:

The proof is complete.

We now present an example of a sequence of positive linear operators that satis…es the conditions of Theorem5 but does not satisfy the conditions of Theorem2, Theorem3and Theorem 4.

(6)

Example 2 Let a= (ank)given by (ank) = 0 BB

@

1 0 0 :

0 1 0 :

0 0 0 :

: : : :

1 CC

A andx= (xnk) given by

(x_nk) = 0 BB BB BB

@

1 2

1

2 1 1 1 1 1 :

1 ¹₂ ¹₂ 0 0 0 0 : 1 0 ¹₂ ¹₂ 1 1 1 : 1 0 1 ¹₂ ¹₂ 0 0 : 1 0 1 0 ¹₂ ¹₂ 1 :

: : : : : : : :

1 CC CC CC A :

Observe now that, P

(n;k)2N²jankj= 2and P

(n;k)2N²

ank= 2:Then, we consider the double Bernstein operators:

Bnk(f;x; y) = Xn i=0

Xk j=0

f i n;j

k n

i k

j xⁱy^j(1 x)^{n i}(1 y)^{k j}

where (x; y)2I²= [0;1]² = [0;1] [0;1]; f 2C I² and n; k2N: Using these polynomials, we introduce the following positive linear operators on C I² :

T_nk(f;x; y) =x_nkB_nk(f;x; y); (x; y)2I²; f 2C I² : (7) We now claim that

P lim

n;k

Xn i=1

Xk j=1

jan i+1k j+1jTij(fr) fr C(I²)

= 0 ; (8)

for eachr= 0;1;2;3: Indeed, we …rst observe that

T_nk(f₀;x; y) = x_nkf₀(x; y); Tnk(f1;x; y) = xnkf1(x; y); T_nk(f₂;x; y) = x_nkf₂(x; y);

T_nk(f₃;x; y) = x_nk f₃(x; y) +x x²

n +y y²

k :

Hence,

Xn i=1

Xk j=1

ja_{n i+1k j+1}jT_ij(f₀;x; y) f₀(x; y) = Xn i=1

Xk j=1

ja_{n i+1k j+1}jx_nk 1 :

Then we get the following double sequence 0

@ Xn i=1

Xk j=1

jan i+1k j+1jxnk 1 1 A=

0 BB

@

1 2

1 2 0 : 0 0 0 : 0 0 0 :

: : : :

1 CC

A: (9)

We get

P lim

n;k

Xn i=1

Xk j=1

jan i+1k j+1jTij(f0) f0 C(I²)

= 0;

(7)

which guarantees that (8) holds true forr= 0: Also, since Xn

i=1

Xk j=1

jan i+1k j+1jTij(f1;x; y) f1(x; y) = Xn

i=1

Xk j=1

jan i+1k j+1jxxnk x

= jxj Xn i=1

Xk j=1

jan i+1k j+1jxnk 1 ;

then

Xn i=1

Xk j=1

= Xn i=1

Xk j=1

jan i+1k j+1jxnk 1 :

By (9), we have

P lim

n;k

Xn i=1

Xk j=1

= 0:

Similary, we have

P lim

n;k

Xn i=1

Xk j=1

ja_{n i+1k} _j+1jT_ij(f₂) f₂

C(I²)

= 0:

Hence (8) is valid for r= 1;2: Finally, we get Xn

i=1

Xk j=1

ja_{n i+1k} _j+1jT_ij(f₃) f₃

C(I²)

2 Xn i=1

Xk j=1

ja_{n i+1k} _j+1jx_nk 1

+1 4

8<

: Xn i=1

Xk j=1

jan i+1k j+1jxnk

n +

Xn i=1

Xk j=1

jan i+1k j+1jxnk

k 9=

;; (10)

xnk

n =

0 BB BB BB

@

1 2

1

2 1 1 1 1 1 :

1 2

1 4

1

4 0 0 0 0 :

1

3 0 ¹₆ ¹₆ ¹₃ ¹₃ ¹₃ :

1

4 0 ¹₄ ¹₈ ¹₈ 0 0 :

1

5 0 ¹₅ 0 ₁₀¹ ₁₀¹ ¹₅ :

: : : : : : : :

1 CC CC CC A

;

P lim

n;k xnk

n = 0 and similaryP lim

n;k xnk

k = 0: Then, from Proposition1, we obtain

K_a² lim

n;k

xnk

n = 0 andK_a² lim

n;k

xnk

k = 0: (11)

From the inequality (10) and using (9), (11), we have

P lim

n;k

Xn i=1

Xk j=1

= 0:

So, our claim (8) holds true for eachr= 0;1;2;3:Now, from (8), we can say that our sequence(Tnk)de…ned by (7) satis…es all assumptions of Theorem 5. Using these facts, we conclude that

P lim

n;k

Xn i=1

Xk j=1

an i+1k j+1Tij(f) f

C(I²)

= 0

(8)

holds for anyf 2C I² :However, sincekT_nk(f₀) f₀kC(I²)=jx_nk 1jand a double sequence kT_nk(f₀) f₀kC(I²)

does not converge in Pringsheim’s sense, Theorem 2 (the classical Korovkin theorem for double sequences) does not work for the sequence (Tnk): The double sequencex= (xnk)can be also given as follows:

xnk= 8>

>>

><

>>

:

1

2; ifk=n ork=n+ 1;

1; ifk > n+ 1 andnis odd , 0; ifk > n+ 1 andnis even, 1; ifk < n andkis odd, 0; ifk < n andkis even,

n; k= 1;2;3; :::: Hence, we get that the double sequence kT_nk(f₀) f₀kC(I²) is not statistically convergent and Theorem3(the statistical Korovkin theorem) does not work for the sequence(Tnk):Also, since

P lim

n;k

1 nk

p+nX1 i=p

q+kX1 j=q

Tij(f0) f0 C(I²)

6

= 0;(p; q2N);

Theorem4does not work for the sequence (Tnk); too.

3 Rate of Convergence

The main aim of this section is to study the rate ofK_a² convergence with the aid of the modulus of continuity that is de…ned by

!(f; ) =p sup

(s x)²+(t y)²

jf(s; t) f(x; y)j ( >0); f 2C(I²):

It is readily seen that, for any >0 and for allf 2C(I²)

!(f; ) (1 + [ ])!(f; )

where[ ]is de…ned to be the greatest integer less than or equal to :Then the result is stated as follows.

Theorem 6 Let a= (a_nk)be a double sequence and the conditions (1) and (2) are satis…ed. Assume that (L_nk)be a double sequence of positive linear operators acting fromC I² into itself, satisfying the following conditions:

(i) P lim

n;k

Pn i=1

Pk

j=1jan i+1k j+1jLij(f0) f0 C(I²)

= 0;

(ii) P lim

n;k!(f; _n;k) = 0;

where

n;k :=

vu uu t

Xn i=1

Xk j=1

jan i+1k j+1jLij (s :)²+ (t :)²

C(I²)

:

Then for allf 2C I² ;

K_a² lim

(9)

Proof. Let f 2 C I² and (x; y) 2 I² be …xed. Using (5), the properties of !; and the positivity and monotonicity ofLnk; we get that

Xn i=1

Xk j=1

an i+1k j+1Lij(f;x; y) f(x; y) Xn i=1

Xk j=1

jan i+1k j+1jLij(jf(s; t) f(x; y)j;x; y)

+M_f Xn i=1

Xk j=1

ja_{n i+1k} _j+1jL_ij(f₀;x; y) f₀(x; y)

Xn i=1

Xk j=1

jan i+1k j+1jLij

0

@!

0

@f; q

(s x)²+ (t y)² 1 A;x; y

1 A

+Mf

Xn i=1

Xk j=1

!(f; ) Xn i=1

Xk j=1

ja_{n i+1k} _j+1jL_ij 1 +(s x)²+ (t y)²

2 ;x; y

!

+Mf

Xn i=1

Xk j=1

!(f; ) Xn i=1

Xk j=1

jan i+1k j+1jLij(f0;x; y) f0(x; y) +!(f; )

+!(f; )

2

Xn i=1

Xk j=1

ja_{n i+1k j+1}jL_ij (s x)²+ (t y)²;x

+M_f Xn i=1

Xk j=1

ja_{n i+1k} _j+1jL_ij(f₀;x; y) f₀(x; y) :

Then taking supremum over(x; y)2I²;we have Xn

i=1

Xk j=1

an i+1k j+1Lij(f) f

C(I²)

!(f; ) Xn

i=1

Xk j=1

ja_{n i+1k j+1}jL_ij(f₀) f₀ + 2!(f; ) +M_f Xn

i=1

Xk j=1

ja_{n i+1k} _j+1jL_ij(f₀) f₀

where

:= n;k:=

vu uu t

Xn i=1

Xk j=1

jan i+1k j+1jLij (s :)²+ (t :)²

C(I²)

:

Then, from the hypotheses, we conclude that

P lim

n;k

Xn i=1

Xk j=1

an i+1k j+1Lij(f) f

C(I²)

= 0;

(10)

we obtain the assertion.

Acknowledgment. We would like to thank the referee(s) for reading carefully and making valuable suggestions.

References

[1] G. A. Anastassiou, M. Mursaleen and S. A. Mohiuddine, Some approximation theorems for functions of two variables through almost convergence of double sequences, J. Comput. Anal. Appl., 13(2011), 37–46.

[2] T. Acar and F. Dirik, Korovkin-type theorems in weighted Lp spaces via summation process, The Scienti…c World Journal, 2013(2013), 6pages.

[3] T. Acar and S. A. Mohiuddine, Statistical (C,1)(E,1) summability and Korovkin’s theorem, Filomat, 30(2016), 387–393.

[4] K. Demirci and F. Dirik, A Korovkin type approximation theorem for double sequences of positive linear operators of two variables inA-statistical sense, Bull. Korean Math. Soc., 47(2010), 825–837.

[5] F. Dirik and P. Okcu Sahin, Statistical relatively equal convergence and Korovkin-type approximation theorem, Results Math., 72(2017), 1613–1621.

[6] F. Dirik and K. Demirci, Korovkin type approximation theorem for functions of two variables in statistical sense, Turkish J. Math., 34(2010), 73–83.

[7] O. Duman, M.K. Khan and C. Orhan, A-statistical convergence of approximating operators, Math.

Inequal. Appl., 4(2003) 689–699.

[8] H. Fast, Sur la convergence statistique, Colloq. Math., 2(1951), 241–244.

[9] A. D. Gadjiev and C. Orhan, Some approximation theorems via statistical convergence, Rocky Mountain J. Math. 32(2002), 129–138.

[10] P. Garrancho, A general Korovkin result under generalized convergence, Constr. Math. Anal., 2(2019), 81–88.

[11] P. P. Korovkin, Linear Operators and Approximation Theory, Hindustan Publ. Co., Delhi, 1960.

[12] M. Lazic and V. Jovovic, Cauchy’s operators and convergence methods, Univ. Beograd. Publ. Elek- trothen Fak., 4(1993), 81–87.

[13] F. Moricz, Statistical convergence of multiple sequences. Arch. Math., 81(2004), 82–89.

[14] F. Moricz and B. E. Rhoades, Almost convergence of double sequences and strong regularity of summability matrices, Math. Proc. Camb. Phil. Soc., 104(1988), 283–294.

[15] S. Orhan, T. Acar and F. Dirik, Korovkin type theorems in weighted Lp spaces via statistical A- summability, An. Stiint. Univ. Al. I. Cuza Iasi. Mat., 42(2016), 537–546.

[16] A. Pringsheim, Zur theorie der zweifach unendlichen zahlenfolgen. Math. Ann., 53(1900), 289–321.

[17] H. Steinhaus, Sur la convergence ordinaire et la convergence asymtotique, Colloq. Math., 2(1951) 73–74.

[18] V. I. Volkov, On the convergence of sequences of linear positive operators in the space of two variables, Dokl. Akad. Nauk., 115(1957), 17–19.