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
Forf ∈C[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].
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)
bi ≤ Mn2 and ci≤ nM2 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−bi−ci, 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];
(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 fort∈Ii, 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 f ∈Lp[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≤i≤none 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∞
and for 0≤i≤n−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 kfk∞≤ 4Mn2 kfk∞ in view of (7). Hence
(9) kTn(f)−Kn(f)k∞≤ 4Mn2 kfk∞. Ifp= 1 then for 1≤i≤none 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≤i≤n−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)kfk1≤ 4Mn kfk1. In conclusion
(10) kTn(f)−Kn(f)k1≤ 4Mn kfk1.
By the interpolation theorem of Riesz-Thorin we get for 1≤p≤ ∞that kTn(f)−Kn(f)kp≤ 4Mn kfkp
in view of (9) and (10). Hence we obtain the assertion of the theorem, because
n→∞lim kKn(f)−fkp = 0, 1≤p≤ ∞.
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) Forf ∈Lp[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(1−x),k is a constant andh(0) =h(1) = 0. For 1 < p ≤ ∞, h is absolutely continuous with h0 ∈ Lp[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
(Sn−Sk) 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],
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
(Sn−Sk)pn,k(x) +
∞
X
j=n+1 (1−x)j
j
=
n
X
k=1
pn,k(x)ck{Ln,k−1(g1) + (Sn−Sk)}
+
n
X
k=0
pn,k(x)ak{Ln,k(g1) + (Sn−Sk)}
+
n−1
X
k=0
pn,k(x)bk{Ln,k+1(g1) + (Sn−Sk)}+
∞
X
j=n+1 (1−x)j
j
=
n
X
k=1
pn,k(x)ck{Ln,k−1(g1) + (Sn−Sk−1)}+
n
X
k=1
pn,k(x)ck(Sk−1−Sk) +
n
X
k=0
pn,k(x)ak{Ln,k(g1) + (Sn−Sk)}
+
n−1
X
k=0
pn,k(x)bk{Ln,k+1(g1) + (Sn−Sk+1)}
+
n−1
X
k=0
pn,k(x)bk(Sk+1−Sk) +
∞
X
j=n+1 (1−x)j
j . (13)
On the other hand, by [2, p. 225], one has
|Ln,k(g1) + (Sn−Sk)|=
(n+ 1)
Z (k+1)/(n+1) k/(n+1)
lnt dt+ (Sn−Sk)
≤ 6k52 +n+11 , (14)
wherek= 1,2, . . . , n−1 and, by (6) and (3),
n
X
k=1
pn,k(x)ck(Sk−1−Sk) +
n−1
X
k=0
pn,k(x)bk(Sk+1−Sk)
=
=
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 ψ ∈ C2[0,1] and 1 ≤ p ≤ ∞, the functionals (16) have bounded norm on Lp[0,1] :
(17) kAn(·, ψ)kp ≤ Cψ.
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.
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,
wheref ∈Lp[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)|)
≤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∞ forf ∈C1[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 allf ∈Lp[0,1] and is given by the integralR01f(x)(ϕ2ψ0)0(x) dx,that is (18).
REFERENCES
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[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.
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Received by the editors: January 16, 2001.