1. Introduction
Let U={z∈ℂ:|z|<1} be the unit disk of the complex plane and let 𝒜 be the class of functions f of the form
(1.1)f(z)=z+∑m=2∞amzm,
that are analytic in U and satisfy the usual normalization conditions f(0)=f′(0)-1=0. We denote by S the subclass of 𝒜 consisting of all univalent functions f in U and consider the class P of functions h that are analytic in U and that satisfy h(0)=1 and Reh(z)>0 for all z∈U.
In the present paper we obtain sufficient conditions for the following general integral operator to be in the class S (The univalent functions are of importance in geometric functions theory and may have some applications in fluid mechanics and physics):
(1.2)Hα1,…,αn,β,g1,…,gn,h1,…,hp(z)={β∫0zuβ-1∏i=1ngi′(u)αi∏j=1phj(u)du}1/β,
where αi,β are complex numbers, β≠0, the functions gi(u)∈𝒜 for all i=1,2,…,n and hj(u)∈P for all j=1,2,…,p, where n,p are positive integers.
For proving our main results we need the following theorems.
Theorem 1.1 (see [1]).
Let α be a complex number, let Reα>0; and let f∈𝒜. If
(1.3)1-|z|2ReαReα|zf′′(z)f′(z)|≤1,
for all z∈U, then for any complex number β with Reβ≥Reα, the function
(1.4)Fβ(z)={β∫0zuβ-1f′(u)du}1/β
is in the class S.
Theorem 1.2 (see [2]).
If the function g(z) is regular in U and |g(z)|<1 in U, then for all ξ∈U and z∈U, the following inequalities hold:
(1.5)|g(ξ)-g(z)1-g(z)¯⋅g(ξ)|≤|ξ-z1-z¯⋅ξ|,(1.6)|g′(z)|≤1-|g(z)|21-|z|2.
These are equalities if and only if g(z)=ɛ(z+u)/(1+¯uz), where |ɛ|=1 and |u|<1.
Remark 1.3 (see [2]).
For z=0, from inequality (1.5) we have
(1.7)|g(ξ)-g(0)1-g(0)¯⋅g(ξ)|≤|ξ|,
and hence,
(1.8)|g(ξ)|≤|ξ|+|g(0)|1+|g(0)|⋅|ξ|.
Writing g(0)=a and letting ξ=z, we get
(1.9)|g(z)|≤|z|+|a|1+|a|⋅|z|,
for all z∈U.
2. Main Results
Theorem 2.1.
Let δ>0. For i=1,2,…,n, let αi be a complex number, let Mi>0, and let gi∈𝒜. For j=1,2,…,p, let Nj>0 and hj∈P. If
(2.1)|zgi′′(z)gi′(z)|≤Mi, ∀i=1,2,…,n (z∈U),(2.2)|zhj′(z)hj(z)|≤Nj, ∀j=1,2,…,p (z∈U),(2.3)∑i=1n|αi|⋅Mi+∑j=1pNj≤δ,
then for every complex number β with Reβ≥δ>0, the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.
Proof.
Let us define the function:
(2.4)f(z)=∫0z∏i=1ngi′(u)αi∏j=1phj(u)du,
with gi∈𝒜, for all i=1,2,…,n and hj∈P, for all j=1,2,…,p, and, thus, we obtain
(2.5)f′(z)=∏i=1ngi′(z)αi∏j=1phj(z).
The function f is regular in U and f(0)=f′(0)-1=0. We have
(2.6)zf′′(z)f′(z)=∑i=1nαizgi′′(z)gi′(z)+∑j=1pzhj′(z)hj(z) (z∈U).
From (2.1), (2.2), and (2.6) we obtain
(2.7)1-|z|2δδ⋅|zf′′(z)f′(z)|≤1-|z|2δδ(∑i=1n|αi|⋅Mi+∑j=1pNj) (z∈U),
and by (2.3), we have
(2.8)1-|z|2δδ⋅|zf′′(z)f′(z)|≤1, ∀z∈U.
Using (2.8), by Theorem 1.1, it results that the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.
Letting p=1 in Theorem 2.1, we have the following.
Corollary 2.2.
Let δ>0, let N>0, and let h∈P. For i=1,2,…,n, let αi be a complex number, let Mi>0, and let gi∈𝒜. If
(2.9)|zgi′′(z)gi′(z)|≤Mi, ∀i=1,2,…,n (z∈U),|zh′(z)h(z)|≤N (z∈U)∑i=1n|αi|⋅Mi+N≤δ,
then for every complex number β with Reβ≥δ>0, the integral operator
(2.10)Fα1,…,αn,β,g1,…,gn,h(z)={β∫0zuβ-1∏i=1ngi′(u)αih(u)du}1/β
is in the class S.
Letting n=1 in Theorem 2.1, we have the following.
Corollary 2.3.
Let δ>0, let M>0, let g∈𝒜, and let α be a complex number. For j=1,2,…,p, let Nj>0 and let hj∈P. If
(2.11)|zg′′(z)g′(z)|≤M (z∈U),|zhj′(z)hj(z)|≤Nj, ∀j=1,2,…,p (z∈U),|α|⋅M+∑j=1pNj≤δ,
then for every complex number β with Reβ≥δ>0, the integral operator
(2.12)Gα,β,g,h1,…,hp(z)={β∫0zuβ-1g′(u)α∏j=1phj(u)du}1/β
is in the class S.
For M1=M2=⋯=Mn=M and N1=N2=⋯=Np=N in Theorem 2.1, we have the following.
Corollary 2.4.
Let δ>0, let M>0, and let N>0. For i=1,2,…,n, let αi be a complex number and let gi∈𝒜. For j=1,2,…,p, let hj∈P. If
(2.13)|zgi′′(z)gi′(z)|≤M, ∀i=1,2,…,n (z∈U),|zhj′(z)hj(z)|≤N, ∀j=1,2,…,p (z∈U),M∑i=1n|αi|+pN≤δ,
then for every complex number β with Reβ≥δ>0, the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.
Theorem 2.5.
Let δ>0. For i=1,2,…,n, let αi be a complex number, let Mi>0, and let gi∈𝒜,gi(z)=z+a2iz2+a3iz3+⋯. For j=1,2,…,p, let Nj>0 and hj∈P with hj′(0)=0, and c=∑i=1nαia2i/|∏i=1nαi|. If
(2.14)|gi′′(z)gi′(z)|<Mi, ∀i=1,2,…,n (z∈U),(2.15)|hj′(z)hj(z)|<Nj, ∀j=1,2,…,p (z∈U),(2.16)∑i=1n|αi|Mi+∑j=1pNj|∏i=1nαi|<1,(2.17)|∏i=1nαi|≤1max|z|<1[((1-|z|2δ)/δ)⋅|z|⋅((|z|+2|c|)/(1+2|c|⋅|z|))],
then for every complex number β with Reβ≥δ>0, the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.
Proof.
We define the function:
(2.18)f(z)=∫0z∏i=1ngi′(u)αi∏j=1phj(u)du (z∈U),
with gi∈𝒜, for all i=1,2,…,n and hj∈P, for all j=1,2,…,p.
We consider the function
(2.19)K(z)=1|∏i=1nαi|⋅f′′(z)f′(z), ∀z∈U.
We have
(2.20)f′′(z)f′(z)=∑i=1nαigi′′(z)gi′(z)+∑j=1phj′(z)hj(z) (z∈U).
From (2.14), (2.15), and (2.20) we obtain
(2.21)1|∏i=1nαi|⋅|f′′(z)f′(z)|≤∑i=1n|αi|Mi+∑j=1pNj|∏i=1nαi| (z∈U).
From (2.16), (2.19), and (2.21) we obtain |K(z)|<1 for all z∈U.
We have K(0)=2∑i=1nαia2i/|∏i=1nαi|=2c, and, using Remark 1.3 we get
(2.22)|K(z)|≤|z|+2|c|1+2|c|⋅|z| (z∈U).
From (2.19) and (2.22), we obtain
(2.23)1-|z|2δδ⋅|zf′′(z)f′(z)|≤|∏i=1nαi|⋅max|z|<1[1-|z|2δδ⋅|z|⋅|z|+2|c|1+2|c|⋅|z|],
for all z∈U.
From (2.17) and (2.23), we have
(2.24)1-|z|2δδ⋅|zf′′(z)f′(z)|≤1 (z∈U).
So, applying Theorem 1.1, we obtain that the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.
Letting p=1 in Theorem 2.5, we have the following.
Corollary 2.6.
Let δ>0, let N>0, and let h∈P with h′(0)=0. For i=1,2,…,n, let αi be a complex number, Mi>0 and gi∈𝒜,gi(z)=z+a2iz2+a3iz3+⋯, and c=∑i=1nαia2i/|∏i=1nαi|. If
(2.25)|gi′′(z)gi′(z)|<Mi, ∀i=1,2,…,n (z∈U),|h′(z)h(z)|<N (z∈U), ∑i=1n|αi|Mi+N|∏i=1nαi|<1,|∏i=1nαi|≤1max|z|<1[((1-|z|2δ)/δ)⋅|z|⋅((|z|+2|c|)/(1+2|c|⋅|z|))],
then for every complex number β with Reβ≥δ>0, the integral operator
(2.26)Fα1,…,αn,β,g1,…,gn,h(z)={β∫0zuβ-1∏i=1ngi′(u)αih(u)du}1/β
is in the class S.
Letting n=1 in Theorem 2.5, we have the following.
Corollary 2.7.
Let δ>0, M>0, g∈𝒜, g(z)=z+a2z2+a3z3+⋯ and let α be a complex number. For j=1,2,…,p, let Nj>0 and hj∈P with hj′(0)=0. If
(2.27)|g′′(z)g′(z)|<M (z∈U),|h′j(z)hj(z)|<Nj, ∀j=1,2,…,p (z∈U),|α|⋅M+∑j=1pNj<|α|≤1max|z|<1[((1-|z|2δ)/δ)⋅|z|⋅((|z|+2|a2|)/(1+2|a2|⋅|z|))],
then for every complex number β with Reβ≥δ>0, the integral operator
(2.28)Gα,β,g,h1,…,hp(z)={β∫0zuβ-1g′(u)α∏j=1phj(u)du}1/β
is in the class S.
For M1=M2=⋯=Mn=M and N1=N2=⋯=Np=N in Theorem 2.5, we have the following.
Corollary 2.8.
Let δ>0, M>0, N>0. For i=1,2,…,n, let αi be a complex number and gi∈𝒜, gi(z)=z+a2iz2+a3iz3+⋯. For j=1,2,…,p, let hj∈P with hj′(0)=0, and c=∑i=1nαia2i/|∏i=1nαi|.
If
(2.29)|gi′′(z)gi′(z)|<M ∀i=1,2,…,n (z∈U),|hj′(z)hj(z)|<N ∀j=1,2,…,p (z∈U),M∑i=1n|αi|+pN|∏i=1nαi|<1,|∏i=1nαi|≤1max|z|<1[((1-|z|2δ)/δ)⋅|z|⋅((|z|+2|c|)/(1+2|c|⋅|z|))],
then for every complex number β with Reβ≥δ>0, the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.