# The Distortion Theorems for Harmonic Mappings with Analytic Parts Convex or Starlike Functions of Orderβ

Authors: Mengkun Zhu; Xinzhong Huang
Journal: Journal of Mathematics
 (2015)

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## Abstract

Some sharp estimates of coefficients, distortion, and growth for harmonic mappings with analytic parts convex or starlike functions of orderβ are obtained. We also give area estimates and covering theorems. Our main results generalise those of Klimek and Michalski.


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## Body

## 1. Introduction

LetS denote the class of functions of the form f(z)=z+∑n=2∞anzn that are analytic and univalent in the unit disk D≔z:z<1. An analytic function f(z)=z+∑n=2∞anzn is said to be convex of order β if(1)Re⁡1+zf′′zf′z≥β>0,z∈Dfor which we write f(z)∈Cβ⊂S. An analytic function f(z)=z+∑n=2∞anzn is said to be starlike of order β if(2)Re⁡zf′zfz≥β>0,z∈Dfor which we write f(z)∈Sβ∗⊂S, where β∈(0,1]. Moreover, when the positive constant β vanishes in (1), (2), the function f turns to be starlike or convex function. That is to say, when an analytic function f only satisfies Re{zf′(z)/f(z)}>0 or Re{1+zf′′(z)/f′(z)}>0, we call f belongs to starlike or convex function, for which we write f∈S0∗ or f∈C0 for convenience.A complex-valued harmonic functionf in the open unit disk D⊂C has a canonical decomposition(3)fz=hz+gz¯,where h and g are analytic in D. Generally, we call h(z) the analytic part and g(z) the coanalytic part of f(z)=h(z)+g(z)¯. An elegant and complete account of the theory of planar harmonic mappings is given in Duren’s monograph [1].Lewy [2] proved in 1936 that a necessary and sufficient condition for f(z)=h(z)+g(z)¯ to be locally univalent and sense-preserving in D is Jf(z)>0, where(4)Jfz=h′z2-g′z2,z∈D.To such a function f, h′ does not vanish in D; let(5)ωz=g′zh′z,z∈D,and then the second complex dilatation ω(z) is analytic with ωz<1.Clunie and Sheil-Small introduced the class of all sense-preserving univalent harmonic mappings ofD with h(0)=g(0)=h′(0)-1=0 that is denoted by SH [3]. In [4], Hotta and Michalski denoted the class LH of all locally univalent and sense-preserving harmonic functions in the unit disk with h(0)=g(0)=h′0-1=0. Obviously, SH⊂LH. Every function f∈LH is uniquely determined by coefficients of the following power series expansions:(6)hz=z+∑n=2∞anzn,gz=∑n=1∞bnzn,z∈D,where an,bn∈C,  n=2,3,4,….In [5], the classes of starlike and convex functions of order β were first introduced by Robertson. Then, such functions have been studied and used in [6–9], and so forth. In [10, 11], Klimek and Michalski studied the cases when the analytic part h is the identity mapping (convex of order 1) and convex mapping (convex of order 0), respectively. In [4], Hotta and Michalski considered the case when the analytic part h is a starlike analytic mapping (starlike of order 0). The main idea of this paper is to characterize the subclasses of SH when h∈Cβ and the subclasses of LH when h∈Sβ∗, where β∈[0,1].In order to establish our main results, we need the following lemma.Lemma 1 (see [12]).
Iff(z)=a0+a1z+⋯+anzn+⋯ is analytic and fz≤1 on D, then(7)an≤1-a02,n=1,2,….


## 2. Main Results and Their Proofs

In what follows, the harmonic mappings that we consider are all normalized locally univalent and sense-preserving.Definition 2.
Forα∈[0,1), let(8)SHαCβ≔fz=hz+gz¯:hz∈Cβ,b1=α,0≤β≤1⊂SH.By [3, Theorem  5.7], if h(z)∈Cβ with ωz=g′z/h′z<1, then f(z)=h(z)+g(z)¯∈SH; hence, the class SHα(Cβ) is well-defined.Definition 3.
Forα∈[0,1), let(9)LHαSβ∗≔fz=hz+gz¯:hz∈Sβ∗,b1=α,0≤β≤1⊂LH.In particular, we establish a smaller subclass of SH,(10)SHαSβ∗≔fz=hz+gz¯∈SH:hz∈Sβ∗,b1=α,0≤β≤1.Lemma 4.
Iff(z)=h(z)+g(z)¯∈LHα(Sβ∗), then F(z)=H(z)+G(z)¯∈SHα(Cβ), where h(z), g(z), and H(z), G(z) are related by zH′z=h(z), zG′z=g(z), z∈D.Proof.
By the definition ofLHα(Sβ∗), h(z)∈Sβ∗. Using classical Alexander’s theorem [13, page 43], the function H(z)∈Cβ. Also, H(0)=0, H′0=limz→0h(z)/z=h′0=1, and G′0=|limz→0g(z)/z|=g′0=α. Let Γ:=[0,h(z)]⊂h(D), z∈D∖{0}; then(11)gz=∫Γdg∘h-1w≤∫Γdg∘h-1wdwdw<∫Γdw=hz.
Hence,(12)G′z=limt→z⁡gtt<limt→z⁡htt=H′z,z∈D∖0,which implies that F(z) is a sense-preserving and locally univalent harmonic mapping in D. By [11, Corollary  2.3], we obtain that F∈SHα(Cβ).Applying Lemma1, we can prove the following theorem.Theorem 5.
Iff(z)=h(z)+g(z)¯=z+∑n=2∞anzn+∑n=1∞bnzn¯∈SHα(Cβ), then(13)an≤∏k=2nk-2βn!,n=2,3,4,…,(14)bn≤α∏k=2nk-2βn!+1-α2n1+∑k=2n-1∏t=2kt-2βk-1!,n=3,4,5,….Specially,(15)b2≤1+2α1-β-α22,whereb1=α,0≤β≤1.The estimate for b2 is sharp; the extremal functions are(16)
Ω
z
≔
H
0
z
+
G
0
z
¯
=
1
-
1
-
z
2
β
-
1
2
β
-
1
+
∫
0
z
ξ
+
α
1
+
α
ξ
1
-
ξ
2
-
2
β
d
ξ
¯
,
β
≠
1
2
,
log
⁡
1
1
-
z
+
∫
0
z
ξ
+
α
1
-
1
-
α
ξ
-
α
ξ
2
d
ξ
¯
,
β
=
1
2
.Proof.
Assumingf(z)=h(z)+g(z)¯=z+∑n=2∞anzn+∑n=1∞bnzn¯∈SHα(Cβ), z∈D, then by [7] we have (13). Let g′z=ωzh′z, where ω(z) is the dilatation of f. Since ω(z) is analytic in D, it has a power series expansion(17)ωz=∑n=0∞cnzn,z∈D,where cn∈C, n=0,1,2,…, and c0=ω0=g′0=b1=α. Recall that ωz<1 for all z∈D; then, by Lemma 1, we have(18)cn≤1-c02,n=1,2,3,….
Together with formulas (5), (6), and (17) we give(19)∑n=1∞nbnzn-1=∑n=0∞cnzn∑n=1∞nanzn-1,z∈D,which leads to(20)∑n=0∞n+1bn+1zn=∑n=0∞∑k=0nk+1ak+1cn-kzn,z∈D.Comparing coefficients, we obtain(21)n+1bn+1=∑k=0nk+1ak+1cn-k,n=0,1,2,….Applying formulas (13) and (18) and by simple calculation, we have(22)nbn≤cn-1+2a2cn-2+⋯+n-1an-1c1+nanc0≤1+∑k=2n-1kak1-c02+nanc0≤1+∑k=2n-1∏t=2kt-2βk-1!1-α2+∏k=2nk-2βn-1!α,n=3,4,5,….In particular,(23)2b2≤a1c1+2a2c0≤1-c02+2a2c0≤1+21-βα-α2.
Next, we will prove the estimate is sharp. Forα∈[0,1)⊂D, consider a function Ω(z)≔H0(z)+G0(z)¯, such that(24)
C
β
∋
H
0
z
≔
1
-
1
-
z
2
β
-
1
2
β
-
1
,
β
≠
1
2
,
log
⁡
1
1
-
z
,
β
=
1
2
,
z
∈
D
.and suppose that the dilatation of Ω(z) satisfies(25)ω0z≔z+α1+αz,z∈D.Applying formula (5), we obtain(26)
G
0
′
z
=
z
+
α
1
+
α
z
1
-
z
2
-
2
β
=
α
+
1
+
2
α
1
-
β
-
α
2
z
+
⋯
,
β
≠
1
2
,
z
+
α
1
+
α
z
1
-
z
=
α
+
1
+
α
-
α
2
z
+
⋯
,
β
=
1
2
.which implies the estimate of (15) is sharp. Obviously, ω0z<1, z∈D, which means Ω(z)≔H0(z)+G0(z)¯∈SHα(Cβ). Hence, the proof is completed.Corollary 6.
Iff(z)=h(z)+g(z)¯=z+∑n=2∞anzn+∑n=1∞bnzn¯∈LHα(Sβ∗), then |an|≤∏k=2n(k-2β)/(n-1)!, n=2,3,4,…,(27)b2≤21-βα+1-α22,bn≤α∏k=2nk-2βn-1!+1-α21+∑k=2n-1∏t=2kt-2βk-1!,n=3,4,5,….Proof.
Iff(z)∈LHα(Sβ∗), then, by Lemma 4, the function F(z)≔H(z)+G(z)¯∈SHα(Cβ), where zH′z=h(z), zG′z=g(z), z∈D. Let G(z) be expanded in the power series(28)Gz=∑n=1∞Bnzn,z∈D,Bn∈C.Together with expansion (6) of g(z) and formula g(z)=zG′z, we have bn=nBn; then by Theorem 5 we can easily obtain the coefficient estimates of f(z)∈LHα(Sβ∗).
Specially, by comparing coefficients, we have2b2=2a2c0+c1, which easily leads to the estimate b2≤2(1-β)α+1-α2/2 by the condition of Corollary 6.Since the analytic parth of f∈SHα(Cβ) belongs to Cβ, then, by [7], we have the following distortion estimate of h:(29)11+r21-β≤h′z≤11-r21-β,z=reiθ∈D.Our next aim is to give the distortion estimate of the coanalytic partg of f∈SHα(Cβ).Theorem 7.
Iff(z)=h(z)+g(z)¯∈SHα(Cβ), then(30)g′z≥α-r1-αr1+r21-β,z=reiθ∈D,(31)g′z≤α+r1+αr1-r21-β,z=reiθ∈D.These inequalities are sharp. The equalities hold for the harmonic function Ω(z) which is defined in (16).Proof.
Letg′0=αeiμ. Consider the function(32)f0z≔e-iμωz-α1-αe-iμωz,z=reiθ∈D,which satisfies assumptions of the Schwarz lemma; then we have(33)e-iμωz-α≤r1-αe-iμωz,z=reiθ∈D.It is equivalent to(34)e-iμωz-α1-r21-α2r2≤r1-α21-α2r2,z=reiθ∈D.
Hence, applying the triangle inequalities to formula (34) we have(35)α-r1-αr≤ωz≤α+r1+αr,z=reiθ∈D.Finally, applying formula (29) together with (35) to the identity g′=ωh′, we obtain (30) and (31). The function Ω(z) defined in (16) shows that inequalities (30) and (31) are sharp. The proof is completed.Corollary 8.
Iff(z)=h(z)+g(z)¯∈LH(Sβ∗), then(36)g′z≥α-r1-r+2βr1+r3-2β,z=reiθ∈D,g′z≤α+r1+r-2βr1-r3-2β,z=reiθ∈D.Proof.
In [7], we know that if f(z)=h(z)+g(z)¯∈LH(Sβ∗), then(37)1-1-2βr1+r3-2β≤h′z≤1+1-2βr1-r3-2β,z=reiθ∈D.Using inequality (35) to identity g′z=ωzh′z, then the corollary can be obtained immediately.By [7], we have the following growth estimate of h∈Cβ, where f(z)=h(z)+g(z)¯∈SHα(Cβ).In the caseβ≠1/2,(38)1+r2β-1-12β-1≤hz≤1-1-r2β-12β-1,z=reiθ∈D.In the case β=1/2,(39)log⁡1+r≤hz≤-log⁡1-r,z=reiθ∈D.In the next results, we give the growth estimate of coanalytic part g of f∈SHα(Cβ).Theorem 9.
Iff(z)=h(z)+g(z)¯∈SHα(Cβ), then(40)gz≤∫0rα+ρ1+αρ1-ρ21-βdρ,z=reiθ∈D.The inequality is sharp. The equality holds for the harmonic function Ω(z) which is defined in (16).Proof.
Letγ≔[0,z]; applying estimate (31) we have(41)gz=∫γg′ξdξ≤∫γg′ξdξ≤∫0rα+ρ1+αρ1-ρ21-βdρ,where z=reiθ∈D. The function Ω(z) defined (16) shows that inequality (40) is sharp.Forf(z)=h(z)+g(z)¯∈LH(Sβ∗), by [7], we have(42)r1+r21-β≤hz≤r1-r21-β,z=reiθ∈D.Now, we give the growth estimates of coanalytic part g(z) of f(z)∈LH(Sβ∗).Corollary 10.
Iff(z)=h(z)+g(z)¯∈LH(Sβ∗), then(43)gz≤∫0rα+ρ1+ρ-2βρ1-ρ3-2βdρ,z=reiθ∈D.Using the distortion estimates in (29) and (35), we can easily deduce the following area estimates of f(z)∈SHα(Cβ).Theorem 11.
Letβ∈(1/2,1] and A≔∬DJf(z)dxdy; if f(z)=h(z)+g(z)¯∈SHα(Cβ), then(44)2π∫01r1-r21-α21+r41-β1+αr2dr≤A≤2π∫01r1-r21-α21-r41-β1-αr2dr,where z=reiθ∈D.Proof.
Observe that iff(z)∈SHα(Cβ), then h′z does not vanish in D. We can give the Jacobian of f(z)=h(z)+g(z)¯ in the form(45)Jfz=h′z21-ωz2,z∈D,where ω(z) is the dilatation of f(z). Applying (29) and (35) to (45) we obtain (46)A≔∬DJfzdxdy=∫02πdθ∫01Jfreiθrdr=2π∫01rJfreiθdr=2π∫01rh′reiθ21-ωreiθ2dr≥2π∫01r11+r21-β21-α+r1+αr2dr=2π∫01r1-α21-r21+r41-β1+αr2dr,A≔2π∫01rh′reiθ21-ωreiθ2dr≤2π∫01r11-r21-β21-α-r1-αr2dr=2π∫01r1-α21-r21-r41-β1-αr2dr,where z=reiθ∈D; this completes the proof.Remark 12.
To avoid the maximum ofA having no sense, we give the limiting condition β∈(1/2,1] in Theorem 11.Corollary 13.
Letβ∈[0,1) and A≔∬DJf(z)dxdy; if f(z)=h(z)+g(z)¯∈LH(Sβ∗), then(47)A≥2π∫011-α2r1-r21-r+2βr1+r6-4β1+αr2dr,A≤2π∫011-α2r1-r21+r-2βr21-r6-4β1-αr2dr.Theorem 14.
Iff(z)∈SHα(Cβ), then(48)
f
z
≥
∫
0
r
1
-
α
1
-
ρ
1
+
α
ρ
1
+
ρ
2
1
-
β
d
ρ
,
z
=
r
e
i
θ
∈
D
,
(49)
f
z
≤
1
-
1
-
r
2
β
-
1
2
β
-
1
+
∫
0
r
α
+
ρ
1
+
α
ρ
1
-
ρ
2
1
-
β
d
ρ
,
β
≠
1
2
,
log
⁡
1
+
α
r
1
-
r
,
β
=
1
2
,
z
=
r
e
i
θ
∈
D
.Proof.
For any pointz=reiθ∈D, let Dr≔D(0,r)={z∈D:z<r} and denote(50)d≔minz∈Dr⁡fDrand then D(0,d)⊆f(Dr)⊆f(D). Hence, there exists zr∈∂Dr such that d=fzr. Let L(t)≔tf(zr), t∈[0,1]; then l(t)≔f-1(L(t)) and t∈[0,1] is a well-defined Jordan arc. Since f=h+g¯, then we can obtain(51)d=fzr=∫Ldw=∫ldf=∫lh′ξdξ+g′ξ¯dξ¯≥∫lh′ξ-g′ξdξ.By ω=g′/h′  with formulas (29) and (35), we have(52)h′ξ-g′ξ=h′ξ1-ωξ≥11+ξ21-β1-α+ξ1+αξ=1-α1-ξ1+αξ1+ξ21-β.Hence, we obtain(53)d≥∫l1-α1-ξ1+αξ1+ξ21-βdξ=∫011-α1-lt1+αlt1+lt21-βdt≥∫0r1-α1-ρ1+αρ1+ρ21-βdρ.
To prove (49) we simply use the inequality(54)fz=hz+gz¯≤hz+gz.By formulas (38), (39), and (40) with simple calculation we have (49); this completes the proof.Corollary 15.
Iff(z)∈SH(Sβ∗), then(55)fz≥∫0r1-α1-ρ1-ρ+2βρ1+αρ1+ρ3-2βdρ,z=reiθ∈D,(56)fz≤r1-r21-β+∫0rα+ρ1+ρ-2βρ1-ρ3-2βdρ,z=reiθ∈D.Finally, the growth estimate off∈SHα(Cβ) yields the following covering estimate.Theorem 16.
Iff(z)∈SHα(Cβ), then(57)D0,R⊂fD,where(58)R≔∫011-α1-ρ1+αρ1+ρ21-βdρ.Proof.
Letr tend to 1 in estimate (48); then Theorem 16 follows immediately from the argument principle for harmonic mappings.Corollary 17.
Iff(z)∈SH(Sβ∗), then(59)D0,R⊂fD,where(60)R≔∫011-α1-ρ1-ρ+2βρ1+αρ1+ρ3-2βdρ.Proof.
Letr tend to 1 in estimate (55); then Corollary 17 follows immediately from the argument principle for harmonic mappings.Remark 18.
The univalence problem of a locally univalent harmonic mapping with starlike analytic parts is an open problem. ThoughSβ∗ have stronger properties than S0∗, we cannot obtain the sharp value of β such that f(z)=h(z)+g(z)¯∈LH(Sβ∗) is univalent. It has important sense to study. Moreover, case of β=0,1 was given a systematic study in [4, 10, 11].



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