# Properties of a Class ofp-Harmonic Functions

Authors: Elif Yaşar; Sibel Yalçın
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Ap times continuously differentiable complex-valued function
F=u+iv in a domain D⊆ℂ is p-harmonic if F satisfies the p-harmonic equation Δ⋯ΔF=0, where p is a positive integer. By using the generalized Salagean differential operator, we introduce a
class of p-harmonic functions and investigate necessary and sufficient
coefficient conditions, distortion bounds, extreme points, and convex
combination of the class.


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## Body

## 1. Introduction

A continuous complex-valued functionf=u+iv in a domain D⊆ℂ is harmonic if both u and v are real harmonic in D; that is, Δu=0 and Δv=0. Here Δ represents the complex Laplacian operator
(1)Δ=4∂2∂z∂z¯:=∂2∂x2+∂2∂y2.In any simply connected domainD we can write f=h+g¯, where h and g are analytic in D. We call h the analytic part and g the coanalytic part of f. A necessary and  sufficient condition for f to be locally univalent and sense preserving in D is that Jf=|fz|2-|fz¯|2>0 in D. See [1, 2].Denote bySH the class of functions f=h+g¯ that are harmonic, univalent, and sense preserving in the unit disk U={z:|z|<1} for which f(0)=fz(0)-1=0. Then for f=h+g¯∈SH we may express the analytic functions h and g as
(2)h(z)=z+∑‍j=2∞ajzj,g(z)=∑‍j=1∞bjzj,|b1|<1.The properties of the classSH and its geometric subclasses have been investigated by many authors; see ([1–6]). Note that SH reduces to the class S of normalized analytic univalent functions in U if the coanalytic part of f is identically zero.Ap times continuously differentiable complex-valued function F=u+iv in a domain D⊆ℂ is p-harmonic if F satisfies the p-harmonic equation Δ⋯ΔF=0, where p is a positive integer.A functionF is p-harmonic in a simply connected domain D if and only if F has the following representation:
(3)F(z)=∑‍pk=1|z|2(k-1)fp-k+1(z),
where Δfp-k+1(z)=0 in D for each k∈{1,…,p}. fp-k+1 has the form
(4)fp-k+1=hp-k+1+g¯p-k+1,
where
(5)hp(z)=z+∑‍j=2∞aj,pzj,hp-k+1(z)=∑‍j=1∞aj,p-k+1zj,(k≥2),gp-k+1(z)=∑‍j=1∞bj,p-k+1zj,(k≥1).Denote bySHp the class of functions F of the form (3) that are harmonic, univalent, and sense-preserving in the unit disk. Apparently, if p=1 and p=2,  F is harmonic and biharmonic, respectively.Biharmonic functions have been studied by several authors, such as, [7–9]. Also, biharmonic functions arise in many physical situations, particularly, in fluid dynamics and elasticity problems. They have many important applications in engineering, biology, and medicine, such as in [10, 11].For a functionf in S, differential operator Dn(n∈ℕ0) was introduced by Sălăgean [12]. Al-Oboudi [13] generalized Dn as follows:
(6)Dλ0f(z)=f(z),Dλ1f(z)=(1-λ)f(z)+λzf′(z),λ≥0,Dλnf(z)=Dλ1(Dλn-1f(z)).
When λ=1, we get the Salagean differential operator.Forf(z)=h(z)+g(z)¯ given by (2), Li and Liu [14] defined the following generalized Salagean operator Dλn in SH:
(7)Dλnf(z)=Dλnh(z)+Dλng(z)¯,λ≥0,
where
(8)Dλnh(z)=z+∑‍j=2∞[1+(k-1)λ]najzj,Dλng(z)=∑‍j=1∞[1+(k-1)λ]nbjzj.For ap-harmonic function F given by (3), we define the following operator:
(9)Dλ0F(z)=F(z),(10)Dλ1F(z)=(1-λ)Dλ0F(z)+λ[z(Dλ0F(z))z+z¯(Dλ0F(z))z¯],λ≥0,DλnF(z)=Dλ1(Dλn-1F(z)),(n∈ℕ).
If F is given by (3), then from (10) we see that
(11)DλnF(z)=∑‍pk=1|z|2(k-1)∑‍j=1∞[1+(j-1)λ+2(k-1)λ]n×aj,p-k+1zj+∑‍pk=1|z|2(k-1)∑‍j=1∞[1+(j-1)λ+2(k-1)λ]n×b¯j,p-k+1z¯j,(a1,p=1,|b1,p|<1).
When p=1, we get the generalized Salagean operator for harmonic univalent functions defined by Li and Liu [14].Denote bySHp(n,λ,α) the class of functions F of the form (3) which satisfy the condition
(12)Re{Dλn+1F(z)DλnF(z)}≥α,0≤α<1,
where DλnF(z) is defined by (11).We let the subclassSH¯p of SHp consist of functions F of the form (3) which include fp-k+1=hp-k+1+g¯p-k+1, where
(13)hp(z)=z-∑‍j=2∞|aj,p|zj,hp-k+1(z)=-∑‍j=1∞|aj,p-k+1|zj,(k≥2),(14)gp-k+1(z)=-∑‍j=1∞|bj,p-k+1|zj,(k≥1).
Define SH¯p(n,λ,α):=SHp(n,λ,α)∩SH¯p.The main object of the paper is to introduce a class ofp-harmonic functions by using the generalized Salagean operator which was defined by Li and Liu [14]. We investigate necessary and sufficient coefficient conditions, extreme points, distortion bounds, and convex combination of the class.


## 2. Main Results

Theorem 1.
LetF be a p-harmonic function given by (3). Furthermore, let
(15)∑‍pk=1∑‍∞j=1[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n[|aj,p-k+1|+|bj,p-k+1|]≤2(1-α),
where λ≥1,n∈ℕ,0≤α<1,  and  a1,p=1. Then F is sense preserving, p-harmonic, and univalent inU and F∈SHp(n,λ,α).Proof.
Supposez1,z2∈U and z1≠z2, so that |z1|≤|z2|<1:
(16)|F(z1)-F(z2)|≥|fp(z1)-fp(z2)|-|∑‍pk=2[|z1|2(k-1)fp-k+1(z1)-|z2|2(k-1)fp-k+1(z2)]|≥|z1-z2|[1-∑‍∞j=2|z1j-z2j||z1-z2||aj,p|-∑‍∞j=1|z1j-z2j||z1-z2||bj,p|-∑‍pk=2∑‍∞j=1|z2|2(k-1)|z1j-z2j||z1-z2|×[|aj,p-k+1|+|bj,p-k+1|]∑‍∞j=2|z1j-z2j||z1-z2|]>|z1-z2|[1-∑‍∞j=2j|aj,p|-∑‍∞j=1j|bj,p|-∑‍pk=2∑‍∞j=1j[|aj,p-k+1|+|bj,p-k+1|]]≥|z1-z2|[2-∑‍pk=1∑‍∞j=1([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)∑‍pk=1∑‍∞j=1×(1-α)-1[|aj,p-k+1|+|bj,p-k+1|]]≥0,
which proves univalence.
In order to prove thatF is sense preserving, we need to show that |Fz(z)|-|Fz¯(z)|>0:
(17)|Fz(z)|-|Fz¯(z)|=|1+∑‍∞j=2jaj,pzj-1+∑‍pk=2|z|2(k-1)∑‍∞j=1(j+k-1)aj,p-k+1zj-1+∑‍pk=1|z|2(k-1)z∑‍∞j=1(k-1)b¯j,p-k+1z¯j|-|∑‍pk=1|z|2(k-1)z¯∑‍∞j=1(k-1)aj,p-k+1zj+∑‍pk=1|z|2(k-1)∑‍∞j=1(j+k-1)b¯j,p-k+1z¯j-1|>2-∑‍pk=1∑‍∞j=1[j+2(k-1)][|aj,p-k+1|+|bj,p-k+1|]≥2-∑‍pk=1∑‍∞j=1([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)×(1-α)-1[|aj,p-k+1|+|bj,p-k+1|]≥0,
for all z∈U.
Using the fact thatRew≥α if and only if |1-α+w|≥|1+α-w|, it suffices to show that
(18)|(1-α)DλnF(z)+Dλn+1F(z)|-|(1+α)DλnF(z)-Dλn+1F(z)|≥0.
Substituting for DλnF in (18), we obtain
(19)|(1-α)DλnF(z)+Dλn+1F(z)|-|(1+α)DλnF(z)-Dλn+1F(z)|≥2(1-α)|z|-2∑‍∞j=2[1+(j-1)λ-α][1+(j-1)λ]n|aj,p||z|j-2∑‍∞j=1[1+(j-1)λ-α][1+(j-1)λ]n|bj,p||z|j-2∑‍pk=2|z|2(k-1)∑‍∞j=1[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n×[|aj,p-k+1|+|bj,p-k+1|]|z|j>2(1-α)|z|[2-∑‍pk=1∑‍∞j=1([1+(j-1)λ+2(k-1)λ]n][1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)×(1-α)-1[|aj,p-k+1|+|bj,p-k+1|]∑‍pk=1∑‍∞j=1].
This last expression is nonnegative by (15), and so the proof is complete.Theorem 2.
LetF be given by (13) and (14). Then F∈SH¯p(n,λ,α) if and only if
(20)∑‍pk=1∑‍∞j=1[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n×[|aj,p-k+1|+|bj,p-k+1|]≤2(1-α),
where λ≥1,  n∈ℕ,0≤α<1,  and a1,p=1.Proof.
The “if" part follows from Theorem1 upon noting that SH¯p(n,λ,α)⊂SHp(n,λ,α). For the “only if" part, we show that F∉SH¯p(n,λ,α) if the condition (20) does not hold.
Note that a necessary and sufficient condition forF given by (13) and (14) to be in SH¯p(n,λ,α) is that the condition (12) should be satisfied.
This is equivalent toRe{A(z)/B(z)}≥0, where
(21)A(z)=(1-α)z-∑‍∞j=2(1+(j-1)λ-α)[1+(j-1)λ]n|aj,p|zj-∑‍∞j=1(1+(j-1)λ-α)[1+(j-1)λ]n|bj,p|z¯j-∑‍pk=2|z|2(k-1)∑‍∞j=1[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n×[|aj,p-k+1|zj+|bj,p-k+1|z¯j],B(z)=z-∑‍∞j=2[1+(j-1)λ]n|aj,p|zj-∑‍∞j=1[1+(j-1)λ]n|bj,p|z¯j-∑‍pk=2|z|2(k-1)∑‍∞j=1[1+(j-1)λ+2(k-1)λ]n×[|aj,p-k+1|zj+|bj,p-k+1|z¯j].
The above condition must hold for all values of z, |z|=r<1. Upon choosing the values of z on the positive real axis, where 0≤z=r<1 we must have(22)((1-α)-∑‍∞j=2(1+(j-1)λ-α)[1+(j-1)λ]n|aj,p|rj-1-∑‍∞j=1(1+(j-1)λ-α)[1+(j-1)λ]n|bj,p|rj-1)×(1-∑‍∞j=2[1+(j-1)λ]n|aj,p|-∑‍∞j=1[1+(j-1)λ]n|bj,p|-∑‍pk=2∑‍∞j=1[1+(j-1)λ+2(k-1)λ]n×[|aj,p-k+1|+|bj,p-k+1|]∑‍∞)-1-(∑‍pk=2∑‍∞j=1[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n×[|aj,p-k+1|+|bj,p-k+1|]rj+2k-3∑‍pk=2∑‍∞j=1)×(1-∑‍∞j=2[1+(j-1)λ]n|aj,p|-∑‍∞j=1[1+(j-1)λ]n|bj,p|-∑‍pk=2∑‍∞j=1[1+(j-1)λ+2(k-1)λ]n×[|aj,p-k+1|+|bj,p-k+1|]∑‍∞j=2)-1≥0.
If the condition (20) does not hold, then the numerator in (22) is negative for r is sufficiently close to 1. Hence there exist z0=r0 in (0,1) for which the quotient in (22) is negative. This contradicts the required condition for F∈SH¯p(n,λ,α) and so the proof is complete.Theorem 3.
LetF be given by (13) and (14). Then F∈SH¯p(n,λ,α) if and only if
(23)F(z)=∑‍k=1p∑‍j=1∞(Xj,p-k+1hj,p-k+1(z)+Yj,p-k+1gj,p-k+1(z)),
where
(24)h1,p(z)=z,hj,p(z)=z-2(1-α)(1+(j-1)λ-α)[1+(j-1)λ]nzj,(j≥2),gj,p(z)=z-2(1-α)(1+(j-1)λ-α)[1+(j-1)λ]nz¯j,(j≥1),hj,p-k+1(z)=z-|z|2(k-1)(2(1-α))×([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)-1zj,(j≥1,2≤k≤p),gj,p-k+1(z)=z-|z|2(k-1)(2(1-α))×([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)-1z¯j,(j≥1,2≤k≤p),
and ∑k=1p∑j=1∞(Xj,p-k+1+Yj,p-k+1)=1,   Xj,p-k+1≥0,  Yj,p-k+1≥0.
In particular, the extreme points ofSH¯p(n,λ,α) are {hj,p-k+1(z)} and {gj,p-k+1(z)}, where j≥1 and 1≤k≤p.Proof.
For functionsF of the form (13) and (14) we have
(25)F(z)=∑‍k=1p∑‍j=1∞(Xj,p-k+1hj,p-k+1(z)+Yj,p-k+1gj,p-k+1(z))=z-∑‍j=2∞(2(1-α))((1+(j-1)λ-α)[1+(j-1)λ]n)-1×Xj,pzj-∑‍j=1∞(2(1-α))((1+(j-1)λ-α)[1+(j-1)λ]n)-1×Yj,pz¯j-∑‍k=2p|z|2(k-1)×∑‍j=1∞(2(1-α))(×[1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)-1×[Xj,p-k+1zj+Yj,p-k+1z¯j].
Then
(26)∑‍j=2∞(1+(j-1)λ-α)[1+(j-1)λ]n2(1-α)×(2(1-α)(1+(j-1)λ-α)[1+(j-1)λ]nXj,p)+∑‍j=1∞(1+(j-1)λ-α)[1+(j-1)λ]n2(1-α)×(2(1-α)(1+(j-1)λ-α)[1+(j-1)λ]nYj,p)+∑‍k=2p∑‍j=1∞([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)(2(1-α))-1×([Xj,p-k+1+Yj,p-k+1](2(1-α))([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α](2(1-α))×[1+(j-1)λ+2(k-1)λ]n)-1×[Xj,p-k+1+Yj,p-k+1])=∑‍j=2∞Xj,p+∑‍j=1∞Yj,p+∑‍k=2p∑‍j=1∞[Xj,p-k+1+Yj,p-k+1]=1-X1,p≤1,
and so F∈SH¯p(n,λ,α). Conversely, if F∈SH¯p(n,λ,α), then
(27)|aj,p|≤2(1-α)(1+(j-1)λ-α)[1+(j-1)λ]n,(j≥2),|aj,p-k+1|≤(2(1-α))([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α](2(1-α))×[1+(j-1)λ+2(k-1)λ]n)-1,(j≥1,2≤k≤p),|bj,p|≤2(1-α)(1+(j-1)λ-α)[1+(j-1)λ]n,(j≥2),|bj,p-k+1|≤(2(1-α))([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α](2(1-α))×[1+(j-1)λ+2(k-1)λ]n)-1,[1+(j-1)λ+2(k-1)λ]n)-1(j≥1,2≤k≤p).
Set
(28)Xj,p=([1+(j-1)λ]n(1+(j-1)λ-α)×[1+(j-1)λ]n)(2(1-α))-1|aj,p|,[1+(j-1)λ]n)(2(1-α))-1|aj,p|(j≥2),Yj,p=((1+(j-1)λ-α)[1+(j-1)λ]n)×(2(1-α))-1|bj,p|,(j≥1),Xj,p-k+1=([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)×(2(1-α))-1|aj,p-k+1|,(j≥1,2≤k≤p),Yj,p-k+1=([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)×(2(1-α))-1|bj,p-k+1|,(j≥1,2≤k≤p),X1,p=1-∑‍j=2∞Xj,p-∑‍j=1∞Yj,p-∑‍k=2p∑‍j=1∞[Xj,p-k+1+Yj,p-k+1],
where X1,p≥0. Then, as required, we obtain
(29)F(z)=∑‍k=1p∑‍j=1∞(Xj,p-k+1hj,p-k+1(z)+Yj,p-k+1gj,p-k+1(z)).Theorem 4.
LetF∈SH¯p(n,λ,α). Then for |z|=r<1 we have
(30)|F(z)|≤(1+|b1,p|+∑‍pk=2(|a1,p-k+1|+|b1,p-k+1|))r+(-∑‍pk=1[1+2(k-1)λ-α][1+2(k-1)λ]n(1+λ-α)[1+λ]n2(1-α)(1+λ-α)[1+λ]n-∑‍pk=1[1+2(k-1)λ-α][1+2(k-1)λ]n(1+λ-α)[1+λ]n-∑‍pk=1[1+2(k-1)λ-α][1+2(k-1)λ]n(1+λ-α)[1+λ]n×[|a1,p-k+1|+|b1,p-k+1|])r2,|F(z)|≥(1-|b1,p|-∑‍pk=2(|a1,p-k+1|+|b1,p-k+1|))r-(2(1-α)(1+λ-α)[1+λ]n-∑‍pk=1[1+2(k-1)λ-α][1+2(k-1)λ]n(1+λ-α)[1+λ]n×[|a1,p-k+1|+|b1,p-k+1|]2(1-α)(1+λ-α)[1+λ]n)r2.Proof.
We only prove the right-hand inequality. The proof for the left-hand inequality is similar and will be omitted. LetF∈SH¯p(n,λ,α). Taking the absolute value of F we have
(31)|F(z)|≤(∑‍pk=1(|a1,p-k+1|+|b1,p-k+1|))r+(∑‍pk=1∑‍j=2∞(|aj,p-k+1|+|bj,p-k+1|))r2≤(∑‍pk=1(|a1,p-k+1|+|b1,p-k+1|))r+2(1-α)r2(1+λ-α)[1+λ]n×(∑‍pk=1∑‍j=2∞([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)×(2(1-α))-1[|aj,p-k+1|+|bj,p-k+1|]∑‍pk=1∑‍j=2∞)≤(1+|b1,p|+∑‍pk=2(|a1,p-k+1|+|b1,p-k+1|))r+2(1-α)r2(1+λ-α)[1+λ]n×(1-∑‍pk=1[1+2(k-1)λ-α][1+2(k-1)λ]n2(1-α)×[|a1,p-k+1|+|b1,p-k+1|]∑‍pk=1).The following covering result follows from the left-hand inequality in Theorem4.Corollary 5.
LetF of the form (13) and (14) be so that F∈SH¯p(n,λ,α). Then
(32){w:|w|<(1+λ-α)[1+λ]n-(1-α)(1+λ-α)[1+λ]n+(1-α)-(1+λ-α)[1+λ]n(1+λ-α)[1+λ]n|b1,p|+∑‍pk=2((1+λ-α)[1+λ]n[1+2(k-1)λ-α][1+2(k-1)λ]n-(1+λ-α)[1+λ]n)((1+λ-α)[1+λ]n)-1×(|a1,p-k+1|+|b1,p-k+1|)(1+λ-α)[1+λ]n-(1-α)(1+λ-α)[1+λ]n}⊂F(U).Theorem 6.
The classSH¯p(n,λ,α) is closed under convex combinations.Proof.
LetFi∈SH¯p(n,λ,α) for i=1,2,…, where Fi is given by
(33)Fi(z)=z-∑‍j=2∞|aij,p|zj-∑‍j=1∞|bij,p|z¯j-∑‍pk=2|z|2(k-1)∑‍j=1∞[|aij,p-k+1|zj+|bij,p-k+1|z¯j].
Then by (20),
(34)∑‍pk=1∑‍∞j=1(×[1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)(2(1-α))-1×[|aij,p-k+1|+|bij,p-k+1|]≤1.
For ∑i=1∞ti=1, 0≤ti≤1, the convex combination of Fi may be written as
(35)∑‍∞i=1tiFi(z)=z-∑‍j=2∞(∑‍∞i=1ti[|aij,p|zj+|bij,p|z¯j])-∑‍pk=2|z|2(k-1)∑‍j=1∞(∑‍∞i=1ti[|aij,p-k+1|zj+|bij,p-k+1|z¯j]).
Then by (34),
(36)∑‍pk=1∑‍∞j=1([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)(2(1-α))-1×(∑‍∞i=1ti[|aij,p-k+1|+|bij,p-k+1|])=∑‍∞i=1ti[∑‍pk=1∑‍∞j=1([1+(j-1)λ+2(k-1)λ]n[1+(j-1)λ+2(k-1)λ-α]×[1+(j-1)λ+2(k-1)λ]n)×(2(1-α))-1[|aij,p-k+1|+|bij,p-k+1|]∑‍pk=1∑‍∞j=1]≤∑‍∞i=1ti=1.
This is the condition required by (20) and so ∑i=1∞tiFi(z)∈SH¯p(n,λ,α).



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*Source: 968627-2013-06-11.xml*