李霞
摘 要: 本文利用Salagean算子定義了一類在單位圓盤上解析的缺項負系數(shù)的函數(shù)族T(j)的兩個子族:T■(n,m,a)與L■(n,α,β,γ),研究了這兩個子族的鄰域的性質,在一定程度上推廣和優(yōu)化了之前學者的相關結論.
關鍵詞: 解析函數(shù) 鄰域 Salagean算子
1.引言
設A為在單位圓盤△={z:z∈C,|z|<1}內解析的解析函數(shù)族,記
T(j)={f∈A:f(z)=z-■a■z■;a■≥0,j∈N},
對于T(j)中的函數(shù)f(z),定義f(z)的(j,δ)鄰域:
N■(f)={g∈T(j):g(z)=z-■b■z■,{■k|a■-b■|■}■≤δ,p≥1,j∈N},
特別地,對于e(z)=z∈T(j),
N■(e)={g∈T(j):g(z)=z-■b■z■,{■k|b■|■}■≤δ,p≥1,j∈N},
對于f(z)∈T(j),Salagean[1]定義了Salagean算子:D■(z)= f(z),D1(z)=zf′(z),...,D■f(z)=D(D■(f(z))),(n∈N)由此可得:D■f(z)=z-■k■a■z■,(n∈N∪{0}).
S■(α)表示α(0≤α<1)階星型函數(shù)作成的類,f∈S■(α)當且僅當Re{■}>α;
K(α)表示a(0≤α<1)階凸函數(shù)作成的類,f∈K(α)當且僅當Re{1+■}>α.
S■■(α)={f:f∈S■(α)且f∈T(j)};K■(α)={f:f∈K(α)且f∈T(j)}.
本文將主要研究函數(shù)族T(j)的兩個子族:T■(n,m,α)與L■(n,α,β,γ)及其鄰域間的關系,在一定程度上推廣和優(yōu)化了之前學者的相關結論.
2.T■(n,m,α)的鄰域
定義T■(n,m,α)={f∈T(j):Re■>α,n,m∈N∪{0},0≤α<1,z∈△}
注意到T■(0,1,α)=S■■(α),T■(1,1,α)=K■(α);T■(0,1,α)=S■(α),T■(1,1,α)=K(α).
對于函數(shù)族T■(n,m,α),我們需要由Sekine給出下列引理:
引理2.1[2] f(z)∈T■(n,m,α)當且僅當■k■(k■-α)α■≤1-α,其中n,m∈N∪{0},0≤α<1,結果是精確的.
定理2.2 設j∈N,n,m∈N∪{0},0≤α<1,則T■(n,m,α)?奐N■(e),其中δ={■}■,(p≥1).
證明:若f(z)∈T■(n,m,α),且f(z)=z-■a■z■(a■≥0),
則由引理2.1得■k■a■≤1-α,
所以(j+1)■[(j+1)■-α]■a■≤1-α且(j+1)■[(j+1)■-α]■ka■≤1-α,
于是■a■≤■<1且■ka■≤■,
又因為p≥1,且注意到0≤a■<1,所以■ka■■≤■ka■≤■,
所以{■ka■■}■≤{■}■=δ,從而f(z)∈N■(e),
所以T■(n,m,α)?奐N■(e).
在定理2.2中取p=1,即得Aouf■研究所得結論;在定理2.2中取j=1,即得下面推論.
推論2.3 設n,m∈N∪{0},0≤α<1,則T■(n,m,α)?奐N■(e),其中
δ={■}■,(p≥1).
當p=1,n=0,m=1時,定理2.2與推論2.3即為Alt intas與Owa[4]研究所得結果.
當p=1,n=1,m=1時,定理2.2與推論2.3即為Alt intas與Owa[4]研究所得結果.
在定理中取n=0,m=1,即得下面推論2.4.
推論2.4 設0≤α<1,則S■■(α)?奐N■(e),其中δ={■}■,(p≥1).
在定理中取n=1,m=1,即得下面推論2.5.
推論2.5 設0≤α<1,則K■(α)?奐N■(e),其中δ={■}■,(p≥1).
3.L■(n,α,β,γ)的鄰域
定義L■(n,α,β,γ)
{f∈T(j)∶|■|<β,n∈N∪{0},z∈Δ,0≤α≤1,0<β≤1,0≤γ<1}.
定理3.1 設j∈N,n∈N∪{0},0≤α≤1,0<β≤1,0≤γ<1,則f(z)∈L■(n,α,β,γ)當且僅當(1+αβ)■k■a■≤β(1+α-γ).
證明:(必要性)若f(z)∈L■(n,α,β,γ),且f(z)=z-■a■z■(a■≥0),則
|■|=|■|<β,(z∈Δ)(3.1)
當z為實數(shù)時,■為實數(shù);當z從實軸上趨于1■時,得
■k■a■≤β(1+α-γ)-α■k■a■,即(1+αβ)■k■a■≤β(1+α-γ).
(充分性)若(1+αβ)■k■a■≤β(1+α-γ),則|■| =|■|≤■<■<β(|z|<1)
因此f(z)∈L■(n,α,β,γ).
取f(z)=z-■z■∈T(j)(k≥j+1),則在(3.1)式中等號成立,由此可知結果是精確的.
定理3.2 設j∈N,n∈N∪{0},0≤α≤1,0<β≤1,0≤γ<1,則
L■(n,α,β,γ)?奐N■(e),其中δ={■}■,(0≥1).
證明:若f(z)∈L■(n,α,β,γ),且f(z)=z-■a■z■(a■≥0),則由定理3.1得
(1+αβ)■k■a■≤β(1+α-γ),所以(1+αβ)(j+1)■■a■≤β(1+α-γ)且endprint
(1+αβ)(j+1)■■ka■≤β(1+α-γ),即
■a■≤■<1且■ka■≤■,又因為p≥1,0≤a■<1,所以
■ka■■≤■ka■≤■,故(■ka■)■≤(■)■=δ,從而f(z)∈N■(e),所以L■(n,α,β,γ)?奐N■(e).
在定理3.2中取j=1,即得下面推論.
推論3.3 設n∈N∪{0},0≤α≤1,0<β≤1,0≤γ<1,則
L■(n,α,β,γ)?奐N■(e),其中δ={■}■,(p≥1).
參考文獻:
[1]G.S., Subclass of univalent functions,Complexity and starlikeness for certain classes of analytic functions with negative coefficients.I,Atti della Accademia Nazionale dei Lincei 65(1978),no.1-2, 38-42(1979).
[2]T.Sekine, Generalization of certain subclasses of analytic functions, International Journel of Mathematics and Mathematical Sciences 10(1987), no.4, 725-732.
[3]M.K.Aouf,Neighbourhoods of certain classes of analytic functions with negative coefficients,International Journel of Mathematics and Mathematical Sciences 10(2006),1-6.
[4]O.Altintas and S.Owa,Neighborhoods of certain analytic functions with negative coefficients,International Journel of Mathematical Sciences 19 (1996),no.4,797-800.endprint
(1+αβ)(j+1)■■ka■≤β(1+α-γ),即
■a■≤■<1且■ka■≤■,又因為p≥1,0≤a■<1,所以
■ka■■≤■ka■≤■,故(■ka■)■≤(■)■=δ,從而f(z)∈N■(e),所以L■(n,α,β,γ)?奐N■(e).
在定理3.2中取j=1,即得下面推論.
推論3.3 設n∈N∪{0},0≤α≤1,0<β≤1,0≤γ<1,則
L■(n,α,β,γ)?奐N■(e),其中δ={■}■,(p≥1).
參考文獻:
[1]G.S., Subclass of univalent functions,Complexity and starlikeness for certain classes of analytic functions with negative coefficients.I,Atti della Accademia Nazionale dei Lincei 65(1978),no.1-2, 38-42(1979).
[2]T.Sekine, Generalization of certain subclasses of analytic functions, International Journel of Mathematics and Mathematical Sciences 10(1987), no.4, 725-732.
[3]M.K.Aouf,Neighbourhoods of certain classes of analytic functions with negative coefficients,International Journel of Mathematics and Mathematical Sciences 10(2006),1-6.
[4]O.Altintas and S.Owa,Neighborhoods of certain analytic functions with negative coefficients,International Journel of Mathematical Sciences 19 (1996),no.4,797-800.endprint
(1+αβ)(j+1)■■ka■≤β(1+α-γ),即
■a■≤■<1且■ka■≤■,又因為p≥1,0≤a■<1,所以
■ka■■≤■ka■≤■,故(■ka■)■≤(■)■=δ,從而f(z)∈N■(e),所以L■(n,α,β,γ)?奐N■(e).
在定理3.2中取j=1,即得下面推論.
推論3.3 設n∈N∪{0},0≤α≤1,0<β≤1,0≤γ<1,則
L■(n,α,β,γ)?奐N■(e),其中δ={■}■,(p≥1).
參考文獻:
[1]G.S., Subclass of univalent functions,Complexity and starlikeness for certain classes of analytic functions with negative coefficients.I,Atti della Accademia Nazionale dei Lincei 65(1978),no.1-2, 38-42(1979).
[2]T.Sekine, Generalization of certain subclasses of analytic functions, International Journel of Mathematics and Mathematical Sciences 10(1987), no.4, 725-732.
[3]M.K.Aouf,Neighbourhoods of certain classes of analytic functions with negative coefficients,International Journel of Mathematics and Mathematical Sciences 10(2006),1-6.
[4]O.Altintas and S.Owa,Neighborhoods of certain analytic functions with negative coefficients,International Journel of Mathematical Sciences 19 (1996),no.4,797-800.endprint