# On Certain Classes of Convex Functions

Authors: Young Jae Sim; Oh Sang Kwon
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

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## Abstract

For real numbersα and β such that 0≤α<1<β, we denote by 𝒦α,β the class of normalized analytic functions which satisfy the following two sided-inequality: α<ℜ1+zf′′z/f′z<βz∈𝕌, where 𝕌 denotes the open unit disk. We find some relationships involving functions in the class 𝒦(α,β). And we estimate the bounds of coefficients and solve the Fekete-Szegö problem for functions in this class. Furthermore, we investigate the bounds of initial coefficients of inverse functions or biunivalent functions.


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## Body

## 1. Introduction

Let𝒜 denote the class of analytic functions in the unit disc
(1)𝕌={z:z∈ℂ,|z|<1}
which is normalized by
(2)f(0)=0,f′(0)=1.
Also let 𝒮 denote the subclass of 𝒜 which is composed of functions which are univalent in 𝕌. And, as usual, we denote by 𝒦 the class of functions in 𝒜 which are convex in 𝕌.We say thatf is subordinate to F in 𝕌, written as f≺F(z∈𝕌), if and only if f(z)=F(w(z)) for some Schwarz function w(z) such that
(3)w(0)=0,|w(z)|<1(z∈𝕌).
If F is univalent in 𝕌, then the subordination f≺F is equivalent to
(4)f(0)=F(0),f(𝕌)⊂F(𝕌).Definition 1.
Letα and β be real numbers such that 0≤α<1<β. The function f∈𝒜 belongs to the class 𝒦(α,β) if f satisfies the following inequality:
(5)α<ℜ{1+zf′′(z)f′(z)}<β(z∈𝕌).It is clear that𝒦(α,β)⊂𝒦. And we remark that, for given real numbers α and β(0≤α<1<β), f∈𝒦(α,β) if and only if f satisfies each of the following two subordination relationships:
(6)1+zf′′(z)f′(z)≺1+(1-2α)z1-z(z∈𝕌),1+zf′′(z)f′(z)≺1+(1-2β)z1-z(z∈𝕌).Now, we define an analytic functionp:𝕌→ℂ by
(7)p(z)=1+β-απilog(1-e2πi((1-α)/(β-α))z1-z).
The above function p was introduced by Kuroki and Owa [1], and they proved that p maps 𝕌 onto a convex domain
(8)Λ={w:α<ℜ{w}<β}
conformally. Using this fact and the definition of subordination, we can obtain the following lemma, directly.Lemma 2.
Letf(z)∈𝒜 and 0≤α<1<β. Then f∈𝒦(α,β) if and only if
(9)1+zf′′(z)f′(z)≺1+β-απilog(1-e2πi((1-α)/(β-α))z1-z)(z∈𝕌).And we note that the functionp, defined by (7), has the form
(10)p(z)=1+∑n=1∞Bnzn,
where
(11)Bn=β-αnπi(1-e2nπi((1-α)/(β-α)))(n∈ℕ).For given real numbersα and β such that 0≤α<1<β, we denote by 𝒦σ(α,β) the class of biunivalent functions consisting of the functions in 𝒜 such that
(12)f∈𝒦(α,β),f-1∈𝒦(α,β),
where f-1 is the inverse function of f.In our present investigation, we first find some relationships for functions in bounded positive class𝒦(α,β). And we solve several coefficient problems including Fekete-Szegö problems for functions in the class. Furthermore, we estimate the bounds of initial coefficients of inverse functions and bi-univalent functions. For the coefficient bounds of functions in special subclasses of 𝒮, the readers may be referred to the works [2–4].


## 2. Relations Involving Bounds on the Real Parts

In this section, we will find some relations involving the functions in𝒦(α,β). And the following lemma will be needed in finding the relations.Lemma 3 (see Miller and Mocanu [5, Theorem 2.3.i]).
LetΞ be a set in the complex plane ℂ and let b be a complex number such that ℜ{b}>0. Suppose that a function ψ:ℂ2×𝕌→ℂ satisfies the condition
(13)ψ(iρ,σ;z)∉Ξ
for all real ρ,σ≤-|b-iρ|2/(2ℜ{b}) and all z∈𝕌. If the function p(z) defined by p(z)=b+b1z+b2z2+⋯ is analytic in 𝕌 and if
(14)ψ(p(z),zp′(z);z)∈Ξ,
then ℜ{p(z)}>0 in 𝕌.Theorem 4.
Letf∈𝒜, 0≤α<1 and
(15)ℜ{1+zf′′(z)f′(z)}>α(z∈𝕌).
Then
(16)ℜ{f′(z)}>12-α(z∈𝕌).Proof.
First of all, we putγ=1/(2-α) and note that 1/2≤γ<1 for 0≤α<1. Let
(17)p(z)=11-γ(f′(z)-γ).
Differentiating (17), we can obtain
(18)1+zf′′(z)f′(z)=1+2(1-γ)zp′(z)(1-γ)p(z)+γ=ψ(p(z),zp′(z)),
where
(19)ψ(r,s)=1+2(1-γ)s(1-γ)r+γ.
Using (15), we have
(20){ψ(p(z),zp′(z)):z∈𝕌}⊂{w∈ℂ:ℜ(w)>α}:=Ωα.
Now for all real ρ,σ∈ℝ with σ≤-(1+ρ2)/2,
(21)ℜ{ψ(iρ,σ)}=1+2γ(1-γ)σ(1-γ)2ρ2+γ2≤1-γ(1-γ)(1+ρ2)(1-γ)2ρ2+γ2.
Define a function g:ℝ→ℝ by
(22)g(ρ)=1+ρ2(1-γ)2ρ2+γ2.
Then g is a continuous even function and
(23)g′(ρ)=2(2γ-1)ρ((1-γ)2ρ2+γ2)2.
Hence g′(0)=0 and g is increasing on (0,∞), since 1/2≤γ<1. Hence g satisfies that
(24)1γ2≤g(ρ)<1(1-γ)2,
for all ρ∈ℝ. Therefore, by combining (21) and (24), we can get
(25)ℜ{ψ(iρ,σ)}≤1-γ(1-γ)g(ρ)≤2-1γ=α.
And this shows that ℜ{ψ(iρ,σ)}∉Ωα for all ρ,σ∈ℝ with σ≤-(1+ρ2)/2. By Lemma 3, we get ℜ{p(z)}>0 for all z∈𝕌, and this shows that the inequality (16) holds and the proof of Theorem 4 is completed.Theorem 5.
Letf∈𝒜, 1<β<2 and
(26)ℜ{1+zf′′(z)f′(z)}<β(z∈𝕌).
Then
(27)ℜ{f′(z)}<12-β(z∈𝕌).Proof.
We putδ=1/(2-β) and note that δ>1 for 1<β<2. And let
(28)p(z)=11-δ(f′(z)-δ),ψ(r,s)=1+2(1-δ)s(1-δ)r+δ.
As in the proof of Theorem 4, we can get
(29){ψ(p(z),zp′(z)):z∈𝕌}⊂{w∈ℂ:ℜ(w)<β}:=Ωβ,
by (26). And for all real ρ, σ with σ≤-(1+ρ2)/2,
(30)ℜ{ψ(iρ,σ)}=1+2δ(1-δ)σ(1-δ)2ρ2+δ2≥1-δ(1-δ)(1+ρ2)(1-δ)2ρ2+δ2=1-δ(1-δ)g(ρ),
where g(ρ) is given by
(31)g(ρ)=1+ρ2(1-δ)2ρ2+δ2.
Since δ>1,g satisfies the inequality
(32)1δ2≤g(ρ)<1(1-δ)2,
for all ρ∈ℝ. Therefore,
(33)ℜ{ψ(iρ,σ)}≥1-δ(1-δ)g(ρ)≥2-1δ=β.
And this shows that ℜ{ψ(iρ,σ)}∉Ωβ for all ρ,σ∈ℝ with σ≤-(1+ρ2)/2. By Lemma 3, we get ℜ{p(z)}>0 for all z∈𝕌, and this shows that the inequality (27) holds and the proof of Theorem 5 is completed.By combining Theorems4 and 5, we can obtain the following result.Theorem 6.
Letf∈𝒜, 0≤α<1<β<2 and
(34)α<ℜ{1+zf′′(z)f′(z)}<β(z∈𝕌).
Then
(35)12-α<ℜ{f′(z)}<12-β(z∈𝕌).


## 3. Coefficient Problems Involving Functions in𝒦(α,β)

In the present section, we will solve some coefficient problems involving functions in the class𝒦(α,β). And our first result on the coefficient estimates involves the function class 𝒦(α,β) and the following lemma will be needed.Lemma 7 (see Rogosinski [6, Theorem 10]).
Letp(z)=∑n=1∞‍Cnzn be analytic and univalent in 𝕌 and suppose that p(z) maps 𝕌 onto a convex domain. If q(z)=∑n=1∞‍Anzn is analytic in 𝕌 and satisfies the following subordination:
(36)q(z)≺p(z)(z∈𝕌),
then
(37)|An|≤|C1|(n=1,2,…).Theorem 8.
Letα and β be real numbers such that 0≤α<1<β. If the functions f(z)=z+∑n=2∞‍anzn∈𝒦(α,β), then
(38)|an|≤{12|B1|,n=2,|B1|n(n-1)∏k=1n-2(1+|B1|k),n=3,4,5,…,
where |B1| is given by
(39)|B1|=2(β-α)πsinπ(1-α)β-α.Proof.
Let us define(40)q(z)=1+zf′′(z)f′(z),(41)p(z)=1+β-απilog(1-e2πi((1-α)/(β-α))z1-z).
Then, the subordination (9) can be written as follows:
(42)q(z)≺p(z)(z∈𝕌).
Note that the function p(z) defined by (41) is convex in 𝕌 and has the form
(43)p(z)=1+∑n=1∞‍Bnzn,
where
(44)Bn=β-αnπi(1-e2nπi((1-α)/(β-α))).
If we let
(45)q(z)=1+∑n=1∞‍Anzn,
then by Lemma 7, we see that the subordination (42) implies that
(46)|An|≤|B1|(n=1,2,…),
where
(47)|B1|=2(β-α)πsinπ(1-α)β-α.
Now, equality (40) implies that
(48)f′(z)+zf′′(z)=q(z)f′(z)(z∈𝕌).
Then, the coefficients of zn-1 in both sides lead to
(49)an=1n(n-1)(An-1+2a2An-2+3a3An-3+⋯+(n-1)an-1A1).
A simple calculation combined with the inequality (46) yields that
(50)|an|=1n(n-1)×|An-1+2a2An-2+3a3An-3+⋯+(n-1)an-1A1|≤1n(n-1)×(|An-1|+2|a2||An-2|+3|a3||An-3|+⋯+(n-1)|an-1||A1|)≤|B1|n(n-1)∑k=2n‍(k-1)|ak-1|,
where |B1| is given by (47) and |a1|=1. Hence, we have |a2|≤|B1|/2. To prove the assertion of the theorem, we need to show that
(51)|an|≤|B1|n(n-1)∑k=2n‍(k-1)|ak-1|≤|B1|n(n-1)∏k=1n-2‍(1+|B1|k)(n=3,4,5,…).
We now use the mathematical induction for the proof of the theorem. For the case n=3, it is clear. We assume that the inequality (51) holds for n=m. Then, some calculation gives us that
(52)|am+1|≤|B1|(m+1)m∑k=2m+1‍(k-1)|ak-1|≤|B1|(m+1)m∏k=1m-2‍(1+|B1|k)+|B1|2m(m+1)(m-1)∏k=1m-2‍(1+|B1|k)=|B1|(m+1)m∏k=1m-1‍(1+|B1|k),
which implies that the inequality (51) is true for n=m+1. Hence, by the mathematical induction, we prove that
(53)|an|≤|B1|n(n-1)∏k=1n-2‍(1+|B1|k)(n=3,4,5,…),
where |B1| is given by (47). This completes the proof of Theorem 8.And now, we will solve the Fekete-Szegö problem forf∈𝒦(α,β), and we will need the following lemma.Lemma 9 (see Keogh and Merkes [7]).
Letp(z)=1+c1z+c2z2+⋯ be a function with positive real part in 𝕌. Then, for any complex number ν,
(54)|c2-νc12|≤2max{1,|1-2ν|}.Now, the following result holds for the coefficient off∈𝒦(α,β).Theorem 10.
Let0≤α<1<β and let the function f(z) given by f(z)=z+∑n=2∞‍anzn be in the class 𝒦(α,β). Then, for a complex number μ,
(55)|a3-μa22|≤β-α3πsin(1-αβ-απ)·max{1;|12+(1-32μ)β-απi+(12-(1-32μ)β-απi)×e2πi((1-α)/(β-α))12+(1-32μ)β-απi|}.
The result is sharp.Proof.
Let us consider a functionq(z) given by q(z)=1+zf′′(z)/f′(z). Then, since f∈𝒦(α,β), we have q(z)≺p(z)(z∈𝕌), where
(56)p(z)=1+β-απilog(1-e2πi((1-α)/(β-α))z1-z)=1+∑n=1∞‍Bnzn,
where Bn is given by (11). Let
(57)h(z)=1+p-1(q(z))1-p-1(q(z))=1+h1z+h2z2+⋯(z∈𝕌).
Then h is analytic and has positive real part in the open unit disk 𝕌. We also have
(58)q(z)=p(h(z)-1h(z)+1)(z∈𝕌).
We find from (58) that
(59)a2=14B1h1,a3=112B1h2-124B1h12+124B2h12+124B12h12,
which imply that
(60)a3-μa22=112B1(h2-νh12),
where
(61)ν=12(1-B2B1-B1+32μB1).
Applying Lemma 9, we can obtain
(62)|a3-μa22|=112|B1||h2-νh12|≤16|B1|·max{1;|1-2ν|}.
And substituting
(63)B1=β-απi(1-e2πi((1-α)/(β-α))),(64)B2=β-α2πi(1-e4πi((1-α)/(β-α)))
in (62), we can obtain the result as asserted. The estimate is sharp for the function f:𝕌→ℂ defined by
(65)f(z)=∫0z‍{exp(∫0ζ‍p(ξ)-1ξdξ)}dζ,
where the function p is given by (7). Hence the proof of Theorem 10 is completed.Using Theorem10, we can get the following result.Corollary 11.
Let0≤α<1<β and let the function f, given by f(z)=z+∑n=2∞‍anzn, be in the class 𝒦(α,β). Also let the function f-1, defined by
(66)f-1(f(z))=z=f(f-1(z)),
be the inverse of f. If
(67)f-1(w)=w+∑n=2∞‍bnwn(|w|<r0;r0>14),
then
(68)|b2|≤β-απsin(1-αβ-απ),|b3|≤β-α3πsin(1-αβ-απ)·max{1;|{(12+2(β-α)πi)}12-2(β-α)πi+(12+2(β-α)πi)e2πi((1-α)/(β-α))|}.Proof.
The relations (66) and (67) give
(69)b2=-a2,b3=2a22-a3.
Thus, we can get the estimate for |b2| by
(70)|b2|=|a2|≤12|B1|=β-απsin(1-αβ-απ),
immediately. Furthermore, an application of Theorem 10 (with μ=2) gives the estimates for |b3|; hence, the proof of Corollary 11 is completed.Finally, we will estimate some initial coefficients for the bi-univalent functionsf∈𝒦σ(α,β).Theorem 12.
For givenα and β such that 0≤α<1<β, let f given by f(z)=z+∑n=2∞‍anzn, be in the class 𝒦σ(α,β). Then
(71)|a2|≤|B1||B1|2|B12-2B1+2B2|,(72)|a3|≤12(|B1|+|B2-B1|),
where B1 and B2 are given by (63) and (64).Proof.
Iff∈𝒦σ(α,β), then f∈𝒦(α,β) and g∈𝒦(α,β), where g=f-1. Hence
(73)Q(z):=1+zf′′(z)f′(z)≺p(z)(z∈𝕌),L(z):=1+zg′′(z)g′(z)≺p(z)(z∈𝕌),
where p(z) is given by (7). Let
(74)h(z)=1+p-1(Q(z))1-p-1(Q(z))=1+h1z+h2z2+⋯(z∈𝕌),k(z)=1+p-1(L(z))1-p-1(L(z))=1+k1z+k2z2+⋯(z∈𝕌).
Then h and k are analytic and have positive real part in 𝕌. Also, we have
(75)Q(z)=p(h(z)-1h(z)+1)(z∈𝕌),L(z)=p(k(z)-1k(z)+1)(z∈𝕌).
By suitably comparing coefficients, we get
(76)a2=14B1h1,(77)3a3-2a22=14B1h2-18B1h12+18B2h12,(78)b2=14B1k1,(79)3b3-2b22=14B1k2-18B1k12+18B2k12,
where B1 and B2 are given by (63) and (64), respectively. Now, considering (76) and (78), we get
(80)h1=-k1.
Also, from (77), (78), (79), and (80), we find that
(81)a22=B13(h2+k2)8(B12-2B1+2B2).
Therefore, we have
(82)|a22|≤|B1|38|B12-2B1+2B2|(|h2|+|k2|)≤|B1|32|B12-2B1+2B2|.
This gives the bound on |a2| as asserted in (71). Now, further computations from (77), (79), (80), and (81) lead to
(83)a3=112B1(2h2+k2)+18h12(B2-B1).
This equation, together with the well-known estimates
(84)|h1|≤2,|h2|≤2,|k2|≤2,
leads us to the inequality (72). Therefore, the proof of Theorem 12 is completed.



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