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数学物理学报, 2020, 40(3): 545-555 doi:

论文

移动紧圆盘上Bernstein-Durrmeyer型算子的逼近

庞兆鋆1, 虞旦盛,1, 周平2

On Approximation by Bernstein-Durrmeyer-Type Operators in Movable Compact Disks

Pang Zhaojun1, Yu Dansheng,1, Zhou Ping2

通讯作者: 虞旦盛,E-mail: dsyu@hznu.edu.cn

收稿日期: 2019-05-15  

Received: 2019-05-15  

摘要

为了逼近移动圆盘上的解析函数,构造了一种新的Bernstein-Durrmeyer型算子,并给出其在移动圆盘上同时逼近的一致逼近速度估计.

关键词: 复Bernstein-Durrmeyer型算子 ; 可移动紧圆盘 ; 逼近估计

Abstract

To approximate analytic functions in movable compact disks, we introduce a new kind of Bernstein-Durrmeyer-Type polynomials. The order of simultaneous approximation rate of the new polynomials in the movable compact disks is given.

Keywords: Complex Bernstein-Durrmeyer type operators ; Movable compact disks ; Approximation rates

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本文引用格式

庞兆鋆, 虞旦盛, 周平. 移动紧圆盘上Bernstein-Durrmeyer型算子的逼近. 数学物理学报[J], 2020, 40(3): 545-555 doi:

Pang Zhaojun, Yu Dansheng, Zhou Ping. On Approximation by Bernstein-Durrmeyer-Type Operators in Movable Compact Disks. Acta Mathematica Scientia[J], 2020, 40(3): 545-555 doi:

1 引言

定义在圆盘上的复Bernstein多项式的逼近性质是由Bernstein最早开始研究的.设f:GC是开集G上的解析函数,这里GC, ¯D1G (D1:={zC:|z|<1}). Bernstein[16]证明了经典复Bernstein多项式Bn(f,z):=nk=0(nk)zk(1z)nkf(kn); zC,在¯D1上一致收敛于f.有关Bernstein多项式逼近的精确量化估计以及Voronovaskaja型可参见文献[2-3].许多学者研究了诸如Bernstein-Stancu多项式, Kantorovich型Bernstein-Stancu多项式, Durrmeyer型Bernstein-Stancu多项式等Bernstein多项式的推广形式对圆盘上解析函数的逼近性质,建立了一些重要的结论(参见文献[1-21]).

最近,蒋和虞[17]引入了一类新的定义在移动圆盘上的复Bernstein-Stancu多项式.设α1,α2,β1,β2是任意给定的常数,且满足0α1β1, α2C, |α2|β2.蒋和虞[17]引入的新的复Bernstein-Stancu多项式Sn,α,β(f,z)定义如下:

Sn,α,β(f,z):=(n+β2n)nnk=0qn,k(z)f(k+α1n+β1),

其中

qn,k(z):=(nk)(zα2n+β2)k(n+α2n+β2z)nk,k=0,1,,n.

α2=β2=0时, Sn,α,β(f,z)即为经典复Bernstein-Stancu多项式.蒋和虞的主要结论如下:

定理1.1  设Rr是满足1r<R的两个常数, DR:={zC;|z|<R}.f:DRCDR上的解析函数,且对所有的zDRf(z)=k=0ckzk.如果nN充分大使得r+|α2|n+β2=:¯r<R1+β2n2,则对所有|zα2n+β2|r,有

|Sn,α,β(f,z)f(z)|M(α,β)r(f),

其中

M(α,β)r(f):=(1+β2n)nk=1|ck|[β2nk¯rk+2β1k+2k(k1)n+β1¯rk+|α2|n+β2k¯rk1]<.

同时,若1r<r1<R,nN充分大使得r1+|α2|n+β2=:¯r1<R1+β2n2,则对所有|zα2n+β2|r1, n,pN,有

|(Sn,α,β(f,z))(p)f(p)(z)|M(α,β)r1(f)p!r1(r1r)p+1.
(1.1)

应该注意的是:定理1.1中研究了Sn,α,β(f)在圆盘{zC:|zα2n+β2|r}上的逼近性质.这个圆盘是一个移动圆盘,即随着参数n,α2β2的改变而移动.

我们回顾由Mahmudov和Gupta给出的本征Durrmeyer-Stancu多项式(参见文献[15]):

U(α,β)n(f;z):=(n1)n1k=1pn,k(z)10pn2,k1(t)f(nt+αn+β)dt+f(αn+β)pn,0(z)+f(n+αn+β)pn,n(z),
(1.2)

其中α,β是满足0αβ的任意实数, zC, n=1,2,;pn,k(z)=(nk)zk(1z)nk.

α=β=0时,该多项式即通常的本征Bernstein-Durrmeyer算子Un(f;z)=U(0,0)n(f;z).这种情形更早在文献[1]中已有过研究.

Mahmudov和Gupta对U(α,β)n(f;z)在圆盘上对解析函数的逼近性质做了研究(参见文献[1]).以下引述其中一个关于U(α,β)n(f;z)逼近速度的结论:

定理1.2  设α,β是满足0αβ的两个任意给定的实数, Rr是满足1r<R的两个常数, DR:={zC:|z|<R}.f:DRCDR上的解析函数,且对所有的zDRf(z)=m=0cmzm,则对所有|z|r,有

|M(α,β)n(f;z)f(z)|Ar,β(f)n+β+αBr(f)n(n+β),

其中

Ar,β(f)=2(1+r)m=1|cm|(m(m1)2+βm)rm1,

Br(f)=α(1+r)m=2|cm|m(m1)rm2.

在本文中,我们将引入一种新的Durrmeyer-Stancu型多项式以进一步推广(1.2)所定义的多项式,并使之能够用以逼近移动圆盘上的解析函数.我们还将考查这一新多项式的逼近速度.

在下文中,记DR是以O(0,0)为圆心, R为半径的圆,这里R>1为常数,即DR:={zC:|z|<R}.α1,α2,β1,β2是任意给定的数且满足α2C,0α1β1,|α2|β2.

f:DRCDR上的解析函数,且对所有的zDRf(z)=m=0cmzm.定义新的Durrmeyer-Stancu如下:

¯Un,α,β(f;z)=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)f(nt+α1n+β1)dt+(n+β2n)n(zα2n+β2)nf(n2+nα1+nα2+α1β2(n+β1)(n+β2))+(n+β2n)n(n+α2n+β2z)nf(nα1+nα2+α1β2(n+β1)(n+β2)),

其中An:=[α2n+β2,n+α2n+β2],而

qn,k(z)=:(nk)(zα2n+β2)k(n+α2n+β2z)nk,k=0,1,,n,

qn2,k1(z)=:(n2k1)(zα2n+β2)k1(n+α2n+β2z)nk1,k=0,1,,n.

α2=β2=0时, ¯Un,α,β(f;z)即为(1.2)式中定义的多项式.

本文的主要结果为:

定理1.3  设Rr是满足1 rR的两个常数.设f:DRCDR上的解析函数,且对所有的zDRf(z)=m=0cmzm.如果nN充分大使得r(1+β2n)=:¯r<R,则对所有|zα2n+β2|r,有

|¯Un,α,β(f;z)f(z)|K(α,β)r,n(f),
(1.3)

其中

K(α,β)r,n(f):=m=2|cm|(2m(m1)(2+¯r)+2β1(m1)(1+¯r)2(n+β1)¯r+α1m(m1)2n(n+β1)¯r+m(m1)|α2|2(n+β1)(n+β2)¯r+m(m1)(α1+|α2|)2(n+β1)2+m(m1)α212n(n+β1)2+m(m1)α1|α2|2(n+β1)2(n+β2))¯rm2+|c1|(β1n+β1¯r+α1n+β1).

同时,若1r<r1<R,那么当nN充分大使得r(1+β2n):=¯r<R时,对所有|zα2n+β2|r1, n,pN,

|(¯Un,α,β(f;z))(p)f(z)(p)|K(α,β)r1,n(f)p!r1(r1r)p+1.
(1.4)

2 辅助引理

引理2.1  设m,nN0,r1,r(1+β2n):=¯r<R.那么对任意|zα2n+β2|r,|¯Un,α,β(em;z)|¯rm,其中em(z)=zm,m=0,1,2,.

  置

Un,α,β(f;z)=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)f(t)dt+(n+β2n)n(zα2n+β2)nf(n+α2n+β2)+(n+β2n)n(n+α2n+β2z)nf(α2n+β2).

直接计算可得

¯Un,α,β(em;z)=mp=0(mp)npαmp1(n+β1)mUn,α,β(ep;z).
(2.1)

注意到

Anqn2,k1(t)tpdt=An(n2k1)(tα2n+β2)k1(n+α2n+β2t)n1k(tα2n+β2+α2n+β2)pdt=An(n2k1)pj=0(pj)(α2n+β2)pj(tα2n+β2)k1+j(n+α2n+β2t)n1kdt,

u=n+β2ntα2n,则有

Anqn2,k1(t)tpdt=(n2k1)pj=0(pj)(pj)(α2n+β2)pj(nn+β2)n+j110uk1+j(1u)n1kdu=(n2k1)pj=0(pj)(α2n+β2)pj(nn+β2)n+j1(k1+j)!(nk1)!(n+j1)!=pj=0(pj)(α2n+β2)pj(nn+β2)n+j1(n2)!(n+j1)!Fj(k),

其中Fj(k)=j1s=0(k+s),k0,j1.因此

Un,α,β(ep;z)=pj=0(pj)(α2n+β2)pj(nn+β2)j(n1)!(n+j1)!(n+β2n)nn1k=1qn,k(z)Fj(k)+(n+β2n)n(zα2n+β2)n(n+α2n+β2)p+(n+β2n)n(n+α2n+β2z)n(α2n+β2)p.

˜z=n+β2nzα2n.(n+β2n)nqn,k(z)=pn,k(˜z),k=0,1,2,,从而

Un,α,β(ep;z)=pj=0(pj)(α2n+β2)pj(nn+β2)j×((n1)!(n+j1)!n1k=1pn,k(˜z)Fj(k)+pn,n(˜z)ej(1)+pn,0(˜z)ej(0))=pj=0(pj)(α2n+β2)pj(nn+β2)jU(0,0)n(ej;˜z).
(2.2)

1r<R,则对所有|z|r,[1]

|U(0,0)n(ej;z)|rj,j=0,1,2,,

结合下式

|˜z|=|n+β2nzα2n|=|n+β2n(zα2n+β2)|¯r,

|U(0,0)n(ej;˜z)|¯rj,j=0,1,2,.因此

|Un,α,β(ep;z)|pj=0(pj)(|α2|n+β2)pj(nn+β2)j¯rj¯rppj=0(pj)(|α2|n+β2)pj(nn+β2)j=¯rp(n+|α2|n+β2)p¯rp.

由(2.1)式,有

|¯Un,α,β(em;z)|mp=0(mp)npαmp1(n+β1)m¯rp¯rmmp=0(mp)npαmp1(n+β1)m=¯rm(n+α1n+β1)m¯rm.

引理2.1证毕.

引理2.2  对所有m,nN, zC,成立

¯Un,α,β(em+1;z)=n(zα2n+β2)(n+α2n+β2z)(n+β1)(n+m)(¯Un,α,β(em;z))+(α1(n+2m)+n2z(n+β1)(n+m)+mn(n+2α2)(n+β1)(n+β2)(n+m))¯Un,α,β(em;z)(mnα1(n+2α2)(n+β1)2(n+β2)(n+m)+mα21(n+β1)2(n+m)+mn2α2(n+α2)(n+β1)2(n+β2)2(n+m))¯Un,α,β(em1;z).
(2.3)

  令

Wn,α,β(em,z)=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)em(nt+α1n+β1),

Vn,α,β(em,z)=(n+β2n)n(zα2n+β2)nem(n2+nα1+nα2+α1β2(n+β1)(n+β2))+(n+β2n)n(n+α2n+β2z)nem(nα1+nα2+α1β2(n+β1)(n+β2)).

由计算可得

(zα2n+β2)(n+α2n+β2z)(Wn,α,β(em,z))=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)(nt+α1n+β1)m(kn+nα2n+β2nz)dt=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)(nt+α1n+β1)m×(kn+nα2n+β2nt+2tn+2α2n+β2)dt+(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)(nt+α1n+β1)m(nt2tnz)dt+(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)(nt+α1n+β1)m(n+2α2n+β2)dt=:J1+J2+J3.

对于J1,有

J1=(n+β2n)2n1(n1)n1k=1qn,k(z)An(qn2,k1(t))(tα2n+β2)(n+α2n+β2t)

(nt+α1n+β1)mdt=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)((2tn+2α2n+β2)(nt+α1n+β1)mmnn+β1(n+2α2n+β2tt2(n+α2)α2(n+β2)2)(nt+α1n+β1)m1)dt=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)(2tn+2α2n+β2)(nt+α1n+β1)mdt(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)mnn+β1×(n+2α2n+β2tt2(n+α2)α2(n+β2)2)(nt+α1n+β1)m1dt=:K1K2.

对于K1,有

K1=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)2t(nt+α1n+β1)mdt(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)(n+2α2n+β2)(nt+α1n+β1)mdt=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)2(nt+α1n+β1n+β1nα1n)(nt+α1n+β1)mdtn+2α2n+β2Wn,α,β(em,z)=2(n+β1)n(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)(nt+α1n+β1)m+1dt2α1n(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)(nt+α1n+β1)mdtn+2α2n+β2Wn,α,β(em,z)=2(n+β1)nWn,α,β(em+1,z)2α1nWn,α,β(em,z)n+2α2n+β2Wn,α,β(em,z).

对于K2,有

K2=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)mnn+β1×(n+2α2n+β2(nt+α1n+β1n+β1nα1n)(nt+α1n+β1n+β1nα1n)2α2(n+α2)(n+β2)2)×(nt+α1n+β1)m1dt=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)mnn+β1×(n+2α2n+β2(nt+α1n+β1n+β1nα1n))(nt+α1n+β1)m1dt(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)mnn+β1(nt+α1n+β1n+β1nα1n)2(nt+α1n+β1)m1dt(n+β2n)2n1(n1)n1k=1qn,k(z)×Anqn2,k1(t)mnn+β1α2(n+α2)(n+β2)2(nt+α1n+β1)m1dt=:L1L2L3.

容易观察到

L1=m(n+2α2)n+β2Wn,α,β(em,z)mα1(n+2α2)(n+β1)(n+β2)Wn,α,β(em1,z),L2=m(n+β1)nWn,α,β(em+1,z)2mα1nWn,α,β(em,z)+mα21n(n+β1)Wn,α,β(em1,z),L3=mnα2(n+α2)(n+β1)(n+β2)2Wn,α,β(em1,z).

对于J2J3,有

J2=(n+β2n)2n1(n1)n1k=1qn,k(z)Anqn2,k1(t)×((n2)(nt+α1n+β1n+β1nα1n)nz)(nt+α1n+β1)mdt=(n2)(n+β1)nWn,α,β(em+1,z)((n2)α1n+nz)Wn,α,β(em,z),

J3=n+2α2n+β2Wn,α,β(em,z).

因此

(zα2n+β2)(n+α2n+β2z)(Wn,α,β(em,z))=(n+β1)(m+n)nWn,α,β(em+1,z)(2mα1+nα1n+nz+m(n+2α2)n+β2)Wn,α,β(em,z)+(mα1(n+2α2)(n+β1)(n+β2)+mα21n(n+β1)+mnα2(n+α2)(n+β1)(n+β2)2)Wn,α,β(em1,z).

Wn,α,β(em,z)满足等式(2.3).

由直接计算可知Vn,α,β(em,z)也满足等式(2.3).引理2.2证毕.

3 定理1.3的证明

由引理2.2可得

¯Un,α,β(em;z)em(z)=n(zα2n+β2)(n+α2n+β2z)(n+β1)(n+m1)(¯Un,α,β(em1;z))+n(m1)+n2z+α1(n+m1)(n+β1)(n+m1)(¯Un,α,β(em1;z)em1)+α1(m1)(n+β1)(n+m1)¯Un,α,β(em1;z)+(m1)n(n+α2)(n+β1)(n+β2)(n+m1)¯Un,α,β(em1;z)+(m1)nα2(n+β1)(n+β2)(n+m1)¯Un,α,β(em1;z)n(m1)(n+β1)(n+m1)¯Un,α,β(em1;z)(m1)nα1(n+α2)(n+β1)2(n+β2)(n+m1)¯Un,α,β(em2;z)(m1)nα1α2(n+β1)2(n+β2)(n+m1)¯Un,α,β(em2;z)(m1)α21(n+β1)2(n+m1)¯Un,α,β(em2;z)(m1)n2α2(n+α2)(n+β1)2(n+β2)2(n+m1)¯Un,α,β(em2;z)+(m1)n(1z)+(α1zβ1)(n+m1)(n+β1)(n+m1)em1(z).

注意到

|z||zα2n+β2|+|α2n+β2|r+|α2|n+β2ˉr,

|n(m1)+n2z+α1(n+m1)(n+β1)(n+m1)|nn+β1m1+n|z|n+m1+α1n+β1n¯r+α1n+β1¯r.

同理可得

|(m1)n(1z)+(α1zβ1)(n+m1)(n+β1)(n+m1)|(m1)(1+¯r)n+β1+|α1zβ1n+β1|(m1)(1+¯r)n+β1+β1(1+¯r)n+β1=(m1+β1)(1+¯r)n+β1.

因此,对于m 2,有

|¯Un,α,β(em;z)em(z)|¯r|¯Un,α,β(em1;z)em1(z)|+(m1)(1+¯r)n+β1¯rm1+α1(m1)n(n+β1)¯rm1+m1n+β1¯rm1+(m1)|α2|(n+β1)(n+β2)¯rm1+m1n+β1¯rm1+(m1)α1(n+β1)2¯rm2+(m1)α1|α2|(n+β1)2(n+β2)¯rm2+(m1)α21n(n+β1)2¯rm2+(m1)|α2|(n+β1)2¯rm2+(m1+β1)(1+¯r)n+β1¯rm1.

因此,对于m=2,3,,可推得

|¯Un,α,β(em;z)em(z)|¯r(¯r|¯Un,α,β(em2;z)em2(z)|+(m2)(1+¯r)n+β1¯rm2+α1(m2)n(n+β1)¯rm2+m2n+β1¯rm2+(m2)|α2|(n+β1)(n+β2)¯rm2+m2n+β1¯rm2+(m2)α1(n+β1)2¯rm3+(m2)α1|α2|(n+β1)2(n+β2)¯rm3+(m2)α21n(n+β1)2¯rm3+(m2)|α2|(n+β1)2¯rm3+(m2+β1)(1+¯r)n+β1¯rm2)+(m1)(1+¯r)n+β1¯rm1+α1(m1)n(n+β1)¯rm1+m1n+β1¯rm1+(m1)|α2|(n+β1)(n+β2)¯rm1+m1n+β1¯rm1+(m1)α1(n+β1)2¯rm2+(m1)α1|α2|(n+β1)2(n+β2)¯rm2+(m1)α21n(n+β1)2¯rm2+(m1)|α2|(n+β1)2¯rm2+(m1+β1)(1+¯r)n+β1¯rm1m(m1)(1+¯r)2(n+β1)¯rm1+m(m1)2(n+β1)¯rm1+α1m(m1)2n(n+β1)¯rm1+m(m1)|α2|2(n+β1)(n+β2)¯rm1+m(m1)2(n+β1)¯rm1+m(m1)α12(n+β1)2¯rm2+m(m1)α1|α2|2(n+β1)2(n+β2)¯rm2+m(m1)α212n(n+β1)2¯rm2+m(m1)|α2|2(n+β1)2¯rm2+(m(m1)+2β1(m1))(1+¯r)2(n+β1)¯rm1.
(3.1)

由(3.1)式及¯Un,α,β(e0;z)=1;¯Un,α,β(e1;z)=nn+β1z+α1n+β1,得

|¯Un,α,β(f;z)f(z)|m=0|cm||¯Un,α,β(em;z)em(z)|=K(α,β)r,n(f),

(1.3)式得证.

Γ是以z0:=α2n+β2为圆心, r1为半径的圆周,且r1>r.因为对于任意的|zα2n+β2|r,vΓ,有|vz|r1r.由柯西公式,对于任意|zα2n+β2|r,n,pN,

|(¯Un,α,β(f;z))(p)f(p)(z)|=p!2π|Γ¯Un,α,β(f;z)f(z)(vz)p+1dv|K(α,β)r1,n(f)p!2π2πr1(r11)p+1=K(α,β)r1,n(f)p!r1(r1r)p+1.

这就证得了(1.4)式.

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