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Rev. Anal. Num´er. Th´eor. Approx., vol. 41 (2012) no. 2, pp. 114–124 ictp.acad.ro/jnaat

SOME GENERAL KANTOROVICH TYPE OPERATORS

PETRU I. BRAICAand OVIDIU T. POP

Abstract. A general class of linear and positive operators of Kantorovich-type is constructed. The operators of this type which preserve exactly two test functions from the set{e0, e1, e2}are determined and their approximation properties and convergence theorems are studied.

MSC 2000. 41A10, 41A35, 41A36.

Keywords. Kantorovich operators, approximation and convergence theorem, Voronovskaja-type theorem.

1. INTRODUCTION

LetNbe the set of positive integers andN0 =N∪ {0}. In 1930, L. V. Kan- torovich in [3] introduced the linear and positive operators Km :L1([0,1])→ C([0,1]),m∈N0, defined for anyf ∈L1([0,1]),x∈[0,1] andm∈N0 by (1) (Kmf)(x) = (m+ 1)

m

X

k=0 m

k

xk(1−x)m−k Z k+1

m+1 k m+1

f(t)dt.

The operators from (1) are called Kantorovich operators. We remark that these operators preserve only the test functione0.

The aim of this paper is to construct and study a general class of lin- ear positive operators which preserve exactly two test functions from the set {e0, e1, e2}.

In [4], J. P. King introduced and studied a Bernstein type operator, which preserves only the test functionse0 and e2. Therefore, we say that the opera- tors constructed in this paper are operators of King’s type.

In Section 2, we recall some results from [6], which we use for obtaining the main results of this paper.

In Section 3, respectively 4, we determine the unique operators from Section 2, which preserve exactly the test functionse0ande1, respectivelye0ande2. In the Sections 4 and 5, we give approximation, convergence and Voronovskaja’s type theorems for the operators obtained.

Secondary School “Grigore Moisil”, 1 Mileniului 1 St. 440037 Satu Mare, Romania, e-mail: [email protected].

National College “Mihai Eminescu”, 5 Mihai Eminescu St., 440014 Satu Mare, Romania, e-mail: [email protected].

(2)

2. PRELIMINARIES

Let Nbe the set of positive integers and N0 =N∪ {0}. In this section we recall some results from [8], which we shall use in the present paper. LetI, J be real intervals with the property I ∩J 6=∅. For any m, k ∈N0,m 6= 0, we consider the functions ϕm,k :J →R, with the property thatϕm,k(x)≥0, for any x∈J and the linear positive functionals Am,k :E(I)→R.

For anym∈Nwe define the operatorLm:E(I)→F(J), by

(2) (Lmf)(x) =

X

k=0

ϕm,k(x)Am,k(f),

for any x ∈ J, where E(I) and F(J) are linear subspaces of real valued functions defined onI, resp. J, for which the sequence (Ln)n≥0 defined above is convergent (in the topology of F(J)).

Remark1. The operators (Lm)m∈Nare linear and positive onE(I∩J).

Form∈Nand i∈N0, we define (Tm,i) by (3) (Tm,iLm)(x) =mi(Lmψix)(x) =mi

X

k=0

ϕm,k(x)Am,kix) for any x∈I∩J,ψxi(t) = (t−x)i,t∈I.

In that followss∈N0 is even and we assume that the next two conditions:

• there exist the smallestαs, αs+2 ∈[0,+∞), so that

(4) lim

m→∞

(Tm,jLm)(x)

mαj =Bj(x)∈R for any x∈I∩J and j∈ {s, s+ 2}

(5) αs+2 < αs+ 2.

• I∩J is an interval.

Theorem 2. (see [8]) Let f ∈E(I) be a function. If x ∈I∩J and f is s times differentiable in a neighborhood of x, f(s) is continuous in x, then

(6) lim

m→∞ms−αs (Lmf)(x)−

s

X

i=0 f(i)(x)

mii! (Tm,jLm)(x)

!

= 0.

Assume thatf isstimes differentiable onI,f(s) is continuous onI and there exists a compact interval K ⊂ I ∩J, such that there exists m(s) ∈ N and constant kj ∈R depending on K, so for m ≥m(s) and x ∈K the following inequalities

(7) (Tm,jmLαjm)(x) ≤kj,

(3)

hold for j ∈ {s, s+ 2}.

Then the convergence expressed by (6) is uniform on K and ms−αs

(Lmf)(x)−

s

X

i=0 f(i)(x)

mii! (Tm,iLm)(x) (8)

s!1 (ks+ks+2

f(s);√ 1

m2+αsαs+2

,

for any x∈K, m≥m(s), where ω(f;·) denotes the modulus of continuity of the function f.

Corollary 3. Let f :I →R be a s times differentiable function onI ∩J with f(s) continuous on I∩J. Then

(9) lim

m→∞(Lmf)(x) =B0(x)f(x)

if s= 0 and α0= 0, where B0 is defined by (4). If s≥2, then (10) lim

m→∞ms−αs

"

(Lmf)(x)−

s−1

X

i=0 1

mii!(Tm,iLm)(x)f(i)(x)

#

=s!1 Bs(x)f(s)(x), where Bs are defined by (4).

If f is a s times differentiable function on I ∩J, with f(s) continuous and bounded on I ∩J and (7) takes place for an interval K ⊂ I ∩J, then the convergence in (9) and (10) are uniform on K.

3. THE CONSTRUCTION OF A GENERAL LINEAR AND POSITIVE OPERATORS

LetJ ⊂R be an interval, m0 ∈N0,m0 ≥2 given,N1 ={m∈N|m≥m0}, the function αm, βm : J → R, αm(x) ≥ 0, βm(x) ≥ 0 for any x ∈ J and m∈N1.

Definition 4. Form∈N1, we define the operator of the following form (11) (Kmf)(x) = (m+ 1)

m

X

k=0 m

k

αkm(x)βmm−k(x) Z m+1k+1

k m+1

f(t)dt for any f ∈L1([0,1]) and x∈J.

Lemma 5. The following identities

(12) (Kme0)(x) = (αm(x) +βm(x))m,

(13) (Kme1)(x) = m(x)+β2(m+1)m(x))m−1((2m+ 1)αm(x) +βm(x)), and

(Kme2)(x) = m(x)+β3(m+1)m(x))2 m−2 3m(m−1)αm2(x) (14)

+ 6mαm(x)(αm(x) +βm(x)) + (αm(x) +βm(x))2 , hold, for any x∈J and any m∈N1.

(4)

Proof. Form∈N1 andk∈ {0,1, . . . , m} we have Z k+1

m+1 k m+1

e0(t)dt= m+11 ,

Z k+1

m+1 k m+1

e1(t)dt= (m+1)2k+12

and

Z k+1

m+1 k m+1

e2(t)dt= 3k(m+1)2+3k+13 . Then

(Kme0)(x) = (m+ 1)

m

X

k=0 m

k

αkm(x)βmm−k(x) Z m+1k+1

k k+1

e0(t)dt

=

m

X

k=0 m

k

αkm(x)βmm−k(x), so (12) holds;

(Kme1)(x) = (m+ 1)

m

X

k=0 m

k

αkm(x)βmm−k(x) Z k+1

m+1 k m+1

e1(t)dt

= 2(m+1)1

2mαm(x)

m

X

k=1 m−1

k−1

αmk−1(x)βmm−k(x)

+

m

X

k=0 m

k

αkm(x)βmm−1(x)

,

so (13) holds and

(Kme2)(x) = (m+ 1)

m

X

k=0 m

k

αkm(x)βmm−k(x) Z k+1

m+1 k m+1

e2(t)dt

= 3(m+1)1 2

3m(m+ 1)αm2(x)

m

X

k=2 m−2

k−2

αk−2m (x)βmm−k(x)

+6mαm(x)

m

X

k=1 m−1

k−1

αmk−1(x)βmm−k(x) +

m

X

k=0 m

k

αkm(x)βmm−k(x)

,

from where (14) follows.

Remark 6. In the following, we will use Theorem 2, whereI = [0,1], (15) ϕm,k(x) = (m+ 1) mk

αmk(x)βmm−k(x) and

(16) Am,k(x) =

Z k+1

m+1 k m+1

f(t)dt

(5)

for any x∈J,f ∈L1([0,1]),m∈N1 and k∈ {0,1, . . . , m}.

4. KANTOROVICH-TYPE OPERATORS PRESERVING THE TEST FUNCTIONSe0 ANDe1

In this case, we impose the conditions (Kme0)(x) =e0(x) and (Kme1)(x) = e1(x), for anyx∈J andm∈N1. From the conditions above, taking (12) and (13) into account, we have

m(x) +β(x))m = 1 and

m(x)+βm(x))m−1

2(m+1) ((2m+ 1)αm(x) +βm(x)) =x, from where

(17) αm(x) = 2(m+1)x−12m ,

(18) βm(x) = 2m+1−2(m+1)x

2m ,

for any x∈[0,1] andm∈N1.

Fromαm(x)≥0 and βm(x)≥0, for anym∈N1, we have

(19) 2(m+1)1 ≤x≤ 2(m+1)2m+1 .

Lemma 7. The following (20)

h 1

2(m0+1);2(m2m0+1

0+1)

i

⊂h

1

2(m+1);2(m+1)2m+1 i

⊂[0,1]

hold for any m∈N1.

Proof. Because the function 2(m+1)1 is decreasing and the function

2m+1

2(m+1) is increasing, relation (20) follows.

Taking the remarks above, we construct the sequence of operators (K1,m )m≥m0. Definition 8. Ifm∈N1, we define the operator

(K1,m f)(x) (21)

= (2m)m+1m

m

X

k=0 m

k

(2(m+ 1)x−1)k(2m+ 1−2(m+ 1)x)m−k Z k+1

m+1 k m+1

f(t)dt for any f ∈L1([0,1]) and any x∈

1

2(m0+1);2(m2m0+1

0+1)

. Remark 9. In this case, we noteJ =

1

2(m0+1);2(m2m0+1

0+1)

=I(m(1)

0).

Lemma 10. We have

(22) (K1,m e0)(x) = 1,

(23) (K1,m e1)(x) =x

(6)

and

(24) (K1,m e2)(x) = m−1m x2+m1 x−12m(m+1)5m+3 2 for any x∈I(m(1)

0) and m∈N1.

Proof. Results immediately from the condition above and (14).

Lemma 11. The following identities

(25) (Tm,0K1,m )(x) = 1,

(26) (Tm,1K1,m )(x) = 0

and

(27) (Tm,2K1,m )(x) =mx(1−x)−12(m+1)m(5m+3)2 hold, for any x∈I(m(1)

0) and m∈N1

Proof. By using Lemma 10 and relation (3), we have (Tm,0K1,m )(x) = (K1,m e0)(x) = 1,

(Tm,1K1,m )(x) =m(K1,m ψx)(x) =m((K1,m e1)(x)−x(K1,m e0)(x)) = 0 and

(Tm,2K1,m (x) =m2(K1,m ψx2)(x) =m2 (K1,m e1)(x) +x2(K1,m e0)(x) ,

from where (27) follows.

Lemma 12. We have that

(28) lim

m→∞(Tm,0K1,m )(x) = 1,

(29) lim

m→∞

(Tm,2K1,m )(x)

m =x(1−x) for any x∈I(m(1)

0) and m(0)∈N exists such that

(30) (Tm,2K

1,m)(x) m54 for any x∈I(m(1)

0) and m∈N1, m≥m(0).

Proof. The relation (28) and (29) results taking (25) and (28) into account.

By using the definition of limit a function and because x(1−x) ≤ 14 for any x∈[0,1], from (29) the inequality (30) is obtained.

Theorem 13. Let f : [0,1]→Rbe a continuous function on [0,1]. Then

(31) lim

m→∞K1,m f =f

(7)

uniformly on I(m(1)

0) and there exists m(0)∈N1 such that

(32)

(K1,m f)(x)−f(x)

94ω

f;1m for any x∈I(m(1)

0) and m∈N1, m≥m0.

Proof. We apply Theorem 2 and Corollary 3 for s = 0, α0 = 0, α2 = 1,

k0= 1 and k2 = 94.

Theorem 14. If f ∈ C([0,1]), x ∈ I(m(1)

0), f is two times differentiable in neighborhood of x and f(2) is continuous onx, then

(33) lim

m→∞m (K1,m f)(x)−f(x)

= 12x(1−x)f(2)(x).

Proof. We use the results from Corollary 3 fors= 2.

5. KANTOROVICH-TYPE OPERATORS PRESERVING THE TEST FUNCTIONSe0 ANDe2

In this section, we impose the conditions (Kme0)(x) = e0(x) and (Kme2)(x) = e2(x), for any x ∈ J and m ∈ N1. Then, taking (12) and (14) into account, we have (αm(x) +βm(x))m = 1 and

m(x)+βm(x))m−2

3(m+1)2 3m(m−1)α2m(x) + 6mαm(x)(αm(x) +βm(x)) + (αm(x) +βm(x))2

=x2, from where

(34) αm(x) +βm(x) = 1

and

(35) 3m(m−1)α2m(x) + 6mαm(x) + 1−3(m+ 1)2x2 = 0.

The discriminant of the equation (35) is

m= 12m(2m+ 1 + 3(m−1)(m+ 1)2x2)≥0, for any x∈J and any m∈N1,m≥2 and we note

(36) δm(x) = 3m(2m+ 1 + 3(m−1)(m+ 1)2x2), x∈J,m∈N1,m≥2.

Ifm∈N1,m≥2, then for

(37) x≥ 1

(m+1) 3

the inequality 1−3(m+1)3m(m−1)2x2 ≤0 is true, so the equation from (35) has exactly one positive solution. This is

(38) αm(x) = −3m+

δm(x) 3m(m−1)

(8)

and then

(39) βm(x) = 3m

2

δm(x) 3m(m−1)

wherex∈J, and to satisfy (37),m∈N1,m≥2.

Lemma 15. Let m∈N1,m≥2. Then βm(x)≥0, x≥0 if and only if

(40) 0≤x≤

3m2+3m+1 (m+1)

3 . Proof. Fromβm(x) ≥0 we have 3m2 ≥p

δm(x), equivalent after calculus

tox23m3(m+1)2+3m+12 , from where (40) follows.

Lemma 16. Let m ∈ N1, m ≥ 2. If x ∈

1 (m+1)

3;

3m2+3m+1 (m+1)

3

, then αm(x)≥0 and βm(x)≥0.

Proof. Results immediately from (37) and (40).

Lemma 17. The following (41)

1 (m0+1)

3;

3m20+3m0+1 (m0+1)

3

⊂h

1 (m+1)

3;

3m2+3m+1 (m+1)

3

i

⊂[0,1]

hold, for any m∈N1.

Proof. By using that the functions 1

(m+1) 3 and

3m2+3m+1 (m+1)

3 are decreasing,

relations from (41) follows.

Definition 18. Ifm∈N1, we define the operator K2,m by (K2,m f)(x) = 3m(m−1))m+1 m

m

X

k=0 m

k

−3m+p δm(x)

k

(42)

×

3m2−p δm(x)

m−kZ k+1

m+1 k m+1

f(t)dt for any f ∈L1([0,1]) and any x∈

1 (m0+1)

3;

3m20+3m0+1 (m0+1)

3

. Remark 19. In this section, we note

J =

1 (m0+1)

3;

3m20+3m0+1 (m0+1)

3

=I(m(2)

0). Lemma 20. We have

(43) (K2,m e0)(x) = 1,

(44) (K2,m e1)(x) = 2

δm(x)−3m−3 6(m−1)(m+1)

(9)

and

(45) (K1,m e2)(x) =x2

for any x∈I(m(2)

0) and m∈N1.

Proof. It is inferred from the conditions above and (13).

Lemma 21. The following identities

(46) (Tm,0K1,m )(x) = 1,

(47) (Tm,1K2,m )(x) =m 2

δm(x)−3m−3 6(m−1)(m+1) −x

and

(48) (Tm,2K2,m )(x) = 2m2x

x−2

δm(x)−3m−3 6(m−1)(m+1)

.

hold, for any x∈I(m(2)

0) and m∈N1

Proof. By using Lemma 12 and (3), we have that (Tm,0K2,m )(x) = (K2,m e0)(x) = 1,

(Tm,1K2,m )(x) =m(K2,m ψx)(x) =m (K2,m e1)(x)−x(K2,m e0)(x) , so (47) holds and

(Tm,2K2,m )(x) =m2(K2,m ψx2)(x)

=m2 (K2,m e2)(x)−2x(K2,m e1)(x) +x2(K2,me0)(x) ,

from where (48) is obtained.

Lemma 22. The following identity

(49) lim

m→∞m 2

δm(x)−3m−3 6(m−1)(m+1) −x

= x−12 holds for any x∈I(m(2)

0). Proof. We have

m→∞lim

m2 (m−1)(m+1)·

δm(x)−3(m−1)(m+1)x

3m2(m−1)m

=−12 + lim

m→∞

δm(x)−3(m−1)(m+1)x 3m

=−12 + lim

m→∞

δm(x)−9(m−1)2(m+1)2x2 3m

δm(x)+3(m−1)(m+1)x

and after a few calculations, identity (49) follows.

(10)

Lemma 23. We have that

(50) lim

m→∞(Tm,0K2,m )(x) = 1,

(51) lim

m→∞

(Tm,2K2,m )(x)

m =x(1−x) for any x∈I(m(2)

0) and m(0)∈N exists such that

(52) (Tm,2K

2,m)(x) m54 for any x∈I(m(2)

0) and m∈N1, m≥m(0).

Proof. The relations (50) and (51) imply (46), (48) and (49). By using the definition of the limit of a function and becausex(1−x)≤ 14 for anyx∈[0,1],

from (51) the relation (52) is obtained.

Theorem 24. Let f : [0,1]→Rbe a continuous function on [0,1]. Then

(53) lim

m→∞K2,m f =f uniformly on x∈I(m(2)

0) and there existsm(0)∈N such that

(54)

(K2,m f)(x)−f(x) ≤ 94ω

f;1m

for any x∈I(m(2)

0) and any m∈N1, m≥m(0).

Proof. Theorem 24 is a results from Theorem 2 and Corollary 3 for s= 0,

α0= 0, α2 = 1,k0= 1 and k2= 54.

Theorem 25. If f ∈C([0,1]), x∈I(m(2)

0), f is two times differentiable in a neighborhood of x, f(2) is continuous in x, then

(55) lim

m→∞m (K2,m f)(x)−f(x)

= x−12 f(1)(x) + x(1−x)2 f(2)(x).

Proof. Taking Lemma 22 into account and applying Theorem 2 for s= 2,

we obtain the relation (55).

REFERENCES

[1] O. Agratini,An asymptotic formula for a class of approximation processes of King’s type, Studia Sci. Math. Hungar,47(2010) No. 4, pp. 435–444.

[2] P.I. Braica, O. T.Popand A. D. Indrea,About a King-type operator, Appl. Math.

Inf. Sci.,6(2012) No. 1, pp. 145–148.

[3] L.V. Kantorovich,Sur certain d´eveloppements suivant les polinˆomes de la forme de S. Bernstein, C. R. Acad. URSS,I,II(1930), pp. 563–568, pp. 595–600.

[4] J.P. King, Positive linear operators which preserve x2, Acta Math. Hungar.,99 (3) (2003), pp. 203–208.

[5] H. GonskaandP. Pitul,Remarks on an article of J. P. King, Comment. Math. Univ.

Carolina,46(2005) No. 4, pp. 645–665.

(11)

[6] O.T. Pop, About some linear and positive operators defined by infinite sum, Dem.

Math.,XXXIX(2006) No. 2, pp. 377–388.

[7] O.T. Pop, D. B˘arbosuandP.I. Braica,Some general linears and positive operators (to appear)

[8] O.T. Pop, The generalization of Voronovskaja’s theorem for a class of linear and positive operators, Rev. Anal. Num´er. Th´eor. Approx.,34(2005), pp. 79–91.

[9] E. Voronovskaja,etermination de la forme asymptotique d’approximation des fonc- tions par les polynˆomes de S. N. Bernstein, C. R. Acad. Sci. URSS, 1932, pp. 79–85.

Received by the editors: August 8, 2012.

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