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Rev. Anal. Num´er. Th´eor. Approx., vol. 34 (2005) no. 1, pp. 55–62 ictp.acad.ro/jnaat

ON THE Lp-SATURATION OF THE YE-ZHOU OPERATOR

ZOLT ´AN FINTA

Abstract. We solve the saturation problem for a class of Ye-Zhou operator Tn(f, x) =Pn(x)AnLn(f) with suitable sequence of matrices{An}n≥1.The solu- tion is based on the saturation theorem for the Kantorovich operator established by V. Maier and S. D. Riemenschneider.

MSC 2000. 41A36, 41A40.

Keywords. Ye-Zhou operator, Kantorovich operator, saturation theorem.

1. INTRODUCTION

ForfC[0,1] thenth Bernstein polynomial is defined by Bn(f, x) =

n

X

k=0

pn,k(x)fkn

n

X

k=0 n k

xk(1−x)n−kfnk. D. X. Zhou showed in [6] for the Kantorovich operator

(1) Kn(f, x) =

n

X

k=0

pn,k(x)(n+ 1)

Z (k+1)/(n+1) k/(n+1)

f(t) dt that if 0< α <1 then ω2(f, t) =O(tα) if and only if

|Kn(f, x)−f(x)| ≤M(x(1−x)/n+ 1/n2)α/2.

He also showed [6] that if 1 < α < 2 then there exist no functions {ϕn,α(x)}n≥1 such that

ω2(f, t) =O(tα)⇔ |Kn(f, x)−f(x)| ≤M ϕn,α(x).

Thus we cannot characterize the second orders of Lipschitz functions by the rate of convergence for the Kantorovich operators. To overcome this difficulty, M. D. Ye and D. X. Zhou [5] introduced a new method of linear approximation by means of matrices: let Pn(x) ≡ (pn,0(x), . . . , pn,n(x)) and

“Babe¸s-Bolyai” University, Faculty of Mathematics and Computer Science, 400084 Cluj- Napoca, Romania, e-mail: [email protected].

(2)

Ln(f) ≡ (Ln,0(f), . . . , Ln,n(f))T, where {pn,k(x)}nk=0 is a bases of the linear space span{1, x, . . . , xn}and

Ln,k(f) = (n+ 1)

Z (k+1)/(n+1) k/(n+1)

f(t) dt

are functionals onC[0,1],respectively. Then, for any (n+ 1)×(n+ 1) matrix A we get the so-called Ye-Zhou operator defined by

(2) Tn(f, x) =Pn(x)·A·Ln(f).

The aim of the present paper is to solve the saturation problem for the class of operator {Tn(f, x)}n≥1 for a suitable sequence of matrices {An}n≥1. By reason of (2), the saturation problem of the Ye-Zhou operator will be in connection with the saturation condition of the Kantorovich operator (see [2], [3], [4]).

2. THE CONSTRUCTION OF THE OPERATOR

LetAn= (ai,j)ni,j=0 be an (n+ 1)×(n+ 1) matrix with restriction ai,j = 0 for|i−j| ≥2.We denoteai,i=ai, ai,i+1=bi, ai,i−1 =ci and setc0 =bn= 0.

Now, we define the matrix An by

0≤ai, bi, ci≤1 for i= 0,1, . . . , n, (3)

ai+bi+ci = 1 for i= 0,1, . . . , n, (4)

ai+bi−1+ci+1 = 1 for i= 0,1, . . . , n (b−1 =cn+1 = 0), (5)

|ibi−(i+ 1)ci| ≤ 12 for i= 1,2, . . . , n−1, (6)

biMn2 and cinM2 for i= 0,1, . . . , n, (7)

with some absolute constant M >0.

The existence of a matrixAn with the properties (3)–(7) is guaranteed by the following numerical example:

a0 = 1−2n13, b0= 2n13,

ai = 1−bici, bi = i+12n3, ci = 2ni3, for i= 1,2, . . . , n−1, an= 1−2n12, cn= 2n12.

Then the explicit expression of the Ye-Zhou operator will be the following:

(8) Tn(f, x) =

n

X

i=0

pn,i(x)(ciLn,i−1(f) +aiLn,i(f) +biLn,i+1(f)), whereai, bi, ci (i= 0,1, . . . , n) satisfy the conditions (3)–(7).

Theorem 2.1. Let An and Tn be defined as above. Then we have (i) Tn is a positive operator on Lp[0,1];

(3)

(ii) Tn(f, x) =R01Ln(x, t)f(t) dt, where the kernel Ln(x, t) =

n

X

i=0

pn,i(x)(n+ 1) ciχIi−1(t) +aiχIi(t) +biχIi+1(t) satisfies

Z 1 0

Ln(x, t) dx= Z 1

0

Ln(x, t) dt= 1

withIi= (i/(n+ 1),(i+ 1)/(n+ 1))and χIi the characteristic function onIi, respectively.

Proof. (i) it is a direct consequence of (8) and (3);

(ii) a simple calculation shows that Z 1

0

pn,i(x) dx= n+11 for i= 0,1, . . . , n.

Then, by (5), Z 1

0

Ln(x, t) dx=

n

X

i=0

ciχIi−1(t) +aiχIi(t) +biχIi+1(t)

=ci+1+ai+bi−1 = 1 fortIi, i= 0,1, . . . , n, and

Z 1 0

Ln(x, t) dt=

n

X

i=0

pn,i(x)(n+ 1)n+1ci +n+1ai +n+1bi

=

n

X

i=0

pn,i(x)(ci+ai+bi)

= 1

in view of (4).

Theorem 2.2. For each fLp[0,1],1≤p≤ ∞, we have

n→∞lim kTn(f)−fkp = 0.

Proof. In view of (8), (4) and (1) we obtain Tn(f, x)−Kn(f, x) =

=

n

X

i=0

pn,i(x){ci[Ln,i−1(f)−Ln,i(f)] +bi[Ln,i+1(f)−Ln,i(f)]}.

Ifp=∞ then for 1≤inone has

|Ln,i(f)−Ln,i−1(f)| ≤

≤ (n+ 1)

Z (i+1)/(n+1) i/(n+1)

|f(t)|dt+ (n+ 1)

Z i/(n+1) (i−1)/(n+1)

|f(t)|dt

≤ 2kfk

(4)

and for 0≤in−1

|Ln,i+1(f)−Ln,i(f)| ≤

≤ (n+ 1)

Z (i+2)/(n+1) (i+1)/(n+1)

|f(t)|dt+ (n+ 1)

Z (i+1)/(n+1) i/(n+1)

|f(t)|dt

≤ 2 kfk.

Thus we have forx∈[0,1]

|Tn(f, x)−Kn(f, x)| ≤

n

X

i=0

pn,i(x)(ci+bi) 2 kfk4Mn2 kfk in view of (7). Hence

(9) kTn(f)−Kn(f)k4Mn2 kfk. Ifp= 1 then for 1≤inone has

|Ln,i(f)−Ln,i−1(f)| ≤

≤ (n+ 1)

Z (i+1)/(n+1) i/(n+1)

|f(t)|dt+ (n+ 1)

Z i/(n+1) (i−1)/(n+1)

|f(t)|dt

≤ (n+ 1)kfk1 and for 0≤in−1,

|Ln,i+1(f)−Ln,i(f)| ≤

≤ (n+ 1)

Z (i+2)/(n+1) (i+1)/(n+1)

|f(t)|dt+ (n+ 1)

Z (i+1)/(n+1) i/(n+1)

|f(t)|dt

≤ (n+ 1)kfk1.

Again, by (7), we have for x∈[0,1]

|Tn(f, x)−Kn(f, x)| ≤

n

X

i=0

pn,i(x)(ci+bi)(n+ 1)kfk14Mn kfk1. In conclusion

(10) kTn(f)−Kn(f)k14Mn kfk1.

By the interpolation theorem of Riesz-Thorin we get for 1≤p≤ ∞that kTn(f)−Kn(f)kp4Mn kfkp

in view of (9) and (10). Hence we obtain the assertion of the theorem, because

n→∞lim kKn(f)−fkp = 0, 1≤p≤ ∞.

(5)

3. THE SATURATION RESULT

The matrix An and the corresponding operatorsTn are determined in Sec- tion 2. We can now state our main result:

Theorem 3.1. Let {Tn}n≥1 be defined as above.

(i) ForfLp[0,1],1≤p≤ ∞,we have kTn(f)−fkp=O(n−1) if and only if

f(x) =k+ Z x

a h(u)

ϕ2(u) du a.e. x∈[0,1],

wherea∈(0,1), ϕ(x) =px(1x),k is a constant andh(0) =h(1) = 0. For 1 < p ≤ ∞, h is absolutely continuous with h0Lp[0,1]; for p= 1,h is of bounded variation on [0,1].

(ii) Moreover, ifkTn(f)−fkp =o(n−1) then f is a.e. constant.

Proof. The proof is based on the ideas of the saturation theorems for Kan- torovich polynomials established by Maier [2], [3] and Riemenschneider [4]

(see also [1, pp. 315–321]). Therefore we shall prove only the essential steps regarding the operators Tn.To be more precise, a Maier-type inequality (see [2, p. 225] or [1, p. 315, Lemma 6.1]) and a property in connection with the bilinear functional (16) will be established (see [1, p. 320, Lemma 6.6]).

Lemma 3.2. Let g1(x) = lnx, 0 < x ≤ 1. Then there exists an absolute constant C >0 such that

(11) |Tn(g1, x)g1(x)| ≤

X

j=n+1 (1−x)j

j +C

n

X

k=0

h 1

(k+1)2 + 1nipn,k(x).

Proof. LetS0 = 0 andSk =Pkj=11/j (k= 1,2, . . .).Then (12)

n

X

k=0

(SnSk) pn,k(x) =

n

X

k=1 1

k(1−x)k (see also [2, p. 225]). Because

g1(x) = lnx= ln(1−(1−x)) =

X

k=1 1

k(1−x)k, x∈(0,1],

(6)

and bn=c0 = 0,then, by (8), (4) and (12), we obtain Tn(g1, x)g1(x) =

=

n

X

k=0

pn,k(x) (ckLn,k−1(g1) +akLn,k(g1) +Ln,k+1(g1)) +

n

X

k=0

(SnSk)pn,k(x) +

X

j=n+1 (1−x)j

j

=

n

X

k=1

pn,k(x)ck{Ln,k−1(g1) + (SnSk)}

+

n

X

k=0

pn,k(x)ak{Ln,k(g1) + (SnSk)}

+

n−1

X

k=0

pn,k(x)bk{Ln,k+1(g1) + (SnSk)}+

X

j=n+1 (1−x)j

j

=

n

X

k=1

pn,k(x)ck{Ln,k−1(g1) + (SnSk−1)}+

n

X

k=1

pn,k(x)ck(Sk−1Sk) +

n

X

k=0

pn,k(x)ak{Ln,k(g1) + (SnSk)}

+

n−1

X

k=0

pn,k(x)bk{Ln,k+1(g1) + (SnSk+1)}

+

n−1

X

k=0

pn,k(x)bk(Sk+1Sk) +

X

j=n+1 (1−x)j

j . (13)

On the other hand, by [2, p. 225], one has

|Ln,k(g1) + (SnSk)|=

(n+ 1)

Z (k+1)/(n+1) k/(n+1)

lnt dt+ (SnSk)

6k52 +n+11 , (14)

wherek= 1,2, . . . , n−1 and, by (6) and (3),

n

X

k=1

pn,k(x)ck(Sk−1Sk) +

n−1

X

k=0

pn,k(x)bk(Sk+1Sk)

=

(7)

=

n

X

k=1

ckk pn,k(x) +

n−1

X

k=0 bk

k+1 pn,k(x)

=

cnn pn,n(x) +

n−1

X

k=1

ckk +k+1bk pn,k(x) +b0 pn,0(x)

pn,0(x) +

n−1

X

k=1

|kbk−(k+1)ck|

k(k+1) pn,k(x) + 1n pn,n(x)

pn,0(x) +

n−1

X

k=1 1

2k(k+1) pn,k(x) +n1 pn,n(x)

pn,0(x) +

n−1

X

k=1 1

(k+1)2 pn,k(x) +n1 pn,n(x)

n−1

X

k=0 1

(k+1)2 pn,k(x) + 1n pn,n(x).

(15)

Combining the relations (13), (14), (15) and (3) we arrive at (11).

Here we mention that we can deduce forTnsimilar statements to the lemmas established in [1, pp. 316–318, Lemmas 6.2, 6.4 and 6.5] for Kn from the application of Theorem 1.

Furthermore, to prove the necessity in Theorem 3.1 (i), we have to employ the bilinear functionals

(16) An(f, ψ) = 2n

Z 1 0

[Tn(f, x)−f(x)]ψ(x) dx.

Lemma 3.3. For each fixed ψC2[0,1] and 1 ≤ p ≤ ∞, the functionals (16) have bounded norm on Lp[0,1] :

(17) kAn(·, ψ)kpCψ.

Moreover, there exists the limit

(18) lim

n→∞ An(f, ψ) = Z 1

0

f(x)(ϕ2ψ0)0(x) dx.

Proof. We have An(f, ψ) = 2n

Z 1 0

[Kn(f, x)−f(x)]ψ(x) dx+2n Z 1

0

[Tn(f, x)−Kn(f, x)]ψ(x)dx.

(8)

Using [1, p. 320, Lemma 6.6] and (10) we get

|An(f, ψ)| ≤

2n Z 1

0

[Kn(f, x)−f(x)]ψ(x) dx

+ 2n

Z 1 0

[Tn(f, x)−Kn(f, x)]ψ(x) dx

C0 kfk1+ 8M kψk kfk1

C kfkp,

withC depending onψ.This inequality implies (17).

By [1, p. 320, (6.14)] we have

(19) lim

n→∞ 2n Z 1

0

[Kn(f, x)−f(x)]ψ(x) dx= Z 1

0

f(x)(ϕ2ψ0)0(x) dx,

wherefLp[0,1].On the other hand, by (4), (7) andc0 =bn= 0, we obtain

2n[Tn(f, x)−Kn(f, x)]≤2n n

X

i=1

pn,i(x)ci(|Ln,i−1(f)|+|Ln,i(f)|) +

n−1

X

i=0

pn,i(x)bi(|Ln,i+1(f)|+|Ln,i(f)|)

(9)

2Mn n

X

i=1

pn,i(x) (|Ln,i−1(f)|+|Ln,i(f)|) +

n−1

X

i=0

pn,i(x) (|Ln,i+1(f)|+|Ln,i(f)|)

. (20)

Furthermore,|Ln,i(f)| ≤ kfk forfC1[0,1]. Hence, by (20), one has

|2n[Tn(f, x)−Kn(f, x)] | ≤ 8Mn

n

X

i=0

pn,i(x)· kfk= 8Mn kfk. Thus

2n

Z 1 0

[Tn(f, x)−Kn(f, x)]ψ(x) dx

≤2n kTn(f)−Kn(f)k· kψk1

8Mn kfk· kψk1 which combined with (19) imply

n→∞lim 2n Z 1

0

[Tn(f, x)−f(x)]ψ(x) dx= Z 1

0

f(x)(ϕ2ψ0)0(x) dx=:A(f, ψ).

This A(f, ψ) is a linear functional on Lp[0,1]. By the Banach-Steinhaus theorem (see e.g. [1, p. 29]), limn→∞An(f, ψ) exists for allfLp[0,1] and is given by the integralR01f(x)(ϕ2ψ0)0(x) dx,that is (18).

REFERENCES

[1] DeVore, R. A. and Lorentz, G. G., Constructive Approximation, Springer-Verlag, Berlin Heidelberg New York, 1993.

[2] Maier, V.,TheL1-saturation class of the Kantorovich operator, J. Approx. Theory,22, pp. 223–232, 1978.

[3] Maier, V.,Lp-approximation by Kantorovich operators, Analysis Math.,4, pp. 289–295, 1978.

[4] Riemenschneider, S. D.,TheLp-saturation of the Bernstein-Kantorovich polynomials, J. Approx. Theory,23, pp. 158–162, 1978.

[5] Ye, M. D.andZhou, D. X.,A class of operators by means of three-diagonal matrices, J. Approx. Theory,78, pp. 239–259, 1994.

[6] Zhou, D. X.,On smoothness characterized by Bernstein type operators, J. Approx. The- ory,81, pp. 303–315, 1995.

Received by the editors: January 16, 2001.

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