# Normality of the p-Harmonic and Log-p-Harmonic Mappings

Authors: Qinghua Jiang; Fanning Meng; Guoqiang Dang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, the concepts of p-harmonic mappings and log-p-harmonic mappings in the unit disk have been introduced and studied by many researchers. We proved the normality of the p-harmonic mappings and log-p-harmonic mappings, which extend the related results of harmonic mappings of earlier authors.


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## Body

## 1. Introduction and Preliminaries

For real-valued harmonic functions defined inD, Lappan [1] established that φ is normal if
(1)supz∈D1−z2∣gradφz∣1+φ2z<∞,where gradφ is the gradient vector of φ. In [2], the authors also proved geometric properties of real-valued harmonic normal functions. Namely, a real-valued harmonic function φ with the property
(2)∬D∣gradφz∣1+φ2z2dΩ<∞,is normal.Recently, many authors considered the properties of the complex-valued harmonic mappings and harmonic quasiconformal mappings in [3–13]. We are motivated to establish the topic of normality for complex-valued p-harmonic mappings and log-p-harmonic mappings defined in the unit disk. An important concept related with normal harmonic functions is the Bloch function, which was studied by Colonna in [14]. It is a classical result of Lewy [15] that a harmonic mapping is locally univalent in a domain Ω if and only if its Jacobian does not vanish. In terms of the canonical decomposition, the Jacobian of harmonic mappings f=h+g¯ is given by Jf=h′2−g′2, and thus, a locally univalent harmonic mapping in a simply connected domain Ω will be sense-preserving if ∣h′∣>∣g′∣.Following the above ideas, particularly the definition of Bloch harmonic function given by Colonna [14], we will prove that the polyharmonic mapping F and log-p-harmonic mapping f defined in the unit disk D are normal if they satisfy a Lipschitz type condition. Further, for the complex-valued polyharmonic mappings and log-p-harmonic mappings, we give out some additional conditions for which are normal. These conditions cannot be omitted. A 2p-times continuously differentiable complex-valued function Fz=uz+ivz in a domain D⊆ℂ is polyharmonic mapping or p-harmonic if Fz satisfies the p-harmonic equation
(3)ΔpF=ΔΔp−1F=0,where the Laplacian operator
(4)ΔF=4Fzz¯=∂2F∂x2+∂2F∂y2.As we see in Proposition 1 in [16], we know that a mapping F is polyharmonic in a simply connected domain D⊆ℂ if and only if F has the following representation
(5)Fz=∑k=1pz2k−1Gp−k+1z,where each Gp−k+1 is harmonic for k∈1,⋯,p. When p=1, the mapping F is called harmonic. When p=2, the mapping F is called biharmonic. f is called log-p-harmonic mapping if logf is p-harmonic mapping. When p=1, the mapping f is called log-harmonic. When p=2, the mapping f is called log-biharmonic, which can be regarded as generalizations of holomorphic functions. So we say that f is called log-p-harmonic mapping in a simply connected domain D⊆ℂ if and only if f has the form
(6)fz=∏k=1pgp−k+1zz2k−1,where each gp−k+1 is log-harmonic for k∈1,⋯,p.For a continuously differentiable mappingf in D, we define
(7)Λfz=max0≤θ≤2πfzz+exp−2iθfz¯z=fzz+fz¯z,λfz=min0≤θ≤2π∣fzz+exp−2iθfz¯z∣=fzz−fz¯z.Recently, many authors considered Landau-type theorems for harmonic mappings, biharmonic mappings, and p-harmonic mappings [16–23]. Li and Wang [24] introduced the log-p-harmonic mappings and derived two versions of Landau-type theorems. However, in virtue of being inspired by these results, we establish the normality of polyharmonic mappings and log-p-harmonic mappings.


## 2. Necessary Lemmas

In order to derive our main results, we need the following lemmas.Lemma 1. [14].
Suppose thatfz=hz+gz¯ is a harmonic mapping of the unit disk D with hz=∑n=1∞anzn and gz=∑n=1∞bnzn are analytic on D. If∣fz∣<M for all z∈D, then
(8)Λfz≤4Mπ1−z2.Lemma 2. [22].
Suppose thatfz=hz+gz¯ is a harmonic mapping of the unit disk D with hz=∑n=1∞anznandgz=∑n=1∞bnzn are analytic on D. If∣fz∣<M for all z∈D, then for ∣z∣=r<1, we have
(9)∣fz∣≤4Mrπ1−r.Lemma 3. [25].
Suppose thatfz=hz+gz¯ is a harmonic mapping of the unit disk D with hz=∑n=1∞anzn and gz=∑n=1∞bnzn are analytic on D and λf0=1. IfΛfz≤Λ for all z∈D, then
(10)∣an∣+∣bn∣≤Λ2−1nΛ,n=2,3,⋯.
WhenΛ>1, the above estimates are sharp for all n=2,3,⋯, with the extremal functions fnz and fnz¯ as follows
(11)fnz=Λ2z−Λ3−Λ∫0z1Λ+zn−1dz.
WhenΛ=1, then fz=a1z+b1¯z¯ with a1−b1=1.Lemma 4. [19].
Suppose thatfz=hz+gz¯ is a harmonic mapping of the unit disk D with hz=∑n=1∞anzn and gz=∑n=1∞bnzn are analytic on D. IfΛfz≤Λ for all z∈D, then for each z∈D,
(12)∣fz∣≤Λz.
We recall that the chordal distance on the generalized complex planeℂ^, which is defined by
(13)χz1,z2=∣z1−z2∣1+z121+z22;z1,z2∈ℂ,χz,∞=11+z2.
IfPz1,Pz2 are the two points on the Riemann sphere, under stereographic projection, corresponding to z1 and z2, respectively, we have
(14)∣Pz1−Pz2∣=χz1,z2.
Therefore,(15)LΓ≥ρz1,z2≥χz1,z2,
whereϱz1,z2 is the spherical distance of z1 and z2, Γ is any rectifiable curve in ℂ with endpoints z1,z2, and
(16)LΓ=∫Γ∣du∣1+u2
is the spherical length ofΓ. On the basis of the paper, given z1,z2∈D, ρz1,z2 denotes the hyperbolic distance between z1,z2. Therefore, if τ denotes the hyperbolic geodesic joining z1 to z2, then
(17)ρz1,z2=∫τ∣dv∣1−v2.
More explicitly,(18)ρz1,z2=12log1+λ1−λ,where
(19)λ=z1−z21−z¯1z2.
With these notations, a polyharmonic mapping or log-p-harmonic mappingf:D→ℂ is called a normal polyharmonic mapping or normal log-p-harmonic mapping, if
(20)supz1≠z2χfz1,fz2ρz1,z2<∞.The following lemma provides an alternative method for deciding when a polyharmonic mapping or log-p-harmonic mapping is normal.Lemma 5.
Letfz be a polyharmonic mapping or log-p-harmonic mapping in the unit disk D, then f is normal if
(21)f≔supz∈D1−z2∣fz∣+∣fz¯∣1+f2<∞.Proof.
Suppose that∥f∥<∞ and let z1,z2∈D. If τ:0,1→D is the hyperbolic geodesic with endpoints z1 and z2,
(22)χfz1,fz2≤∫f∘τ∣du∣1+u2=∫01∣dfτtτ′t∣1+fτt2dt,where df stands for the differential of f. From here and (21), we have
(23)χfz1,fz2≤∫01∣dfτtτ′t∣1+fτt2dt≤∫01supz∈D1−z21−τt2∣dfτtτ′t1+fτt2dt=∫01supz∈D1−z2∣dfτt∣1+fτt2∣τ′t∣1−τt2≤∥f∥·∫01∣τ′t∣1−τt2dt=∥f∥·ρz1,z2.
Hence, we obtain(24)supz1≠z2χfz1,fz2ρz1,z2<∞.
So it implies thatf is normal.


## 3. Main Results and Their Proofs

In this section, we prove the normality of the polyharmonic mappings and log-p-harmonic mappings as follows.Theorem 6.
LetFz=∑k=1pz2k−1Gp−k+1z be a polyharmonic mapping in the unit disk D satisfying F0=λF0−1=0. Suppose that for each k∈1,⋯,p, we have
(25)Gp−k+1zisharmonicinD,andGp−k+10=0;Gp−k+1z≤Mp−k+1,andΛGpz≤Λp,whereMp−k+1≥0,Λp≥1.
Then,F is normal polyharmonic mapping of the unit disk D.Proof.
We may represent the harmonic functionsGp−k+1z in series form as
(26)Gp−k+1z=∑j=1∞aj,p−k+1zj+∑j=1∞bj,p−k+1¯z¯j,k∈1,2,⋯,p;a1,p−b1,p=λF0=Gpz0−Gpz¯0=λGp0=1,∣a1,p∣+∣b1,p∣=∣Gpz0∣+∣Gpz¯0∣=ΛGp0≤Λp.
Firstly, we calculate the boundedness of the derivative ofF.
(27)∣Fzz+Fz¯z∣≤∣Fzz∣+∣Fz¯z∣=∣Gpzz+∑k=1p−1z2kGp−kzz+kGp−kzz¯kzk−1∣+∣Gpz¯z+∑k=1p−1z2kGp−kz¯z+kGp−kzz¯k−1zk∣≤A1+A2+A3+A4,where
(28)A1=Gpz0+Gpz¯0,A2=∑k=1p−1z2kGp−kzz+Gp−kz¯z,A3=∑k=1p−1kGp−kzz¯kzk−1+z¯k−1zk,A4=Gpzz−Gpz0+Gpz¯z−Gpz¯0.
By a simple calculation, we have(29)A1≤ΛGp0≤Λp.
Using Lemma1, we have
(30)A2≤∑k=1p−1z2k∣Gp−kzz∣+∣Gp−kz¯z∣≤∑k=1p−1r2kΛGp−kz≤∑k=1p−1r2k4Mp−kπ1−r2.
By Lemma2, we have
(31)A3≤∑k=1p−1∣kGp−kz∣∣z¯kzk−1∣+∣z¯k−1zk∣≤8π1−r∑k=1p−1kMp−kr2k.
Using Lemma3, we have
(32)A4≤∣Gpzz−Gpz0∣+∣Gpz¯z−Gpz¯0∣≤∑n=2∞∣an,p∣+∣bn,p∣nrn−1≤∑n=2∞Λp2−1rn−1Λp=Λp2−1Λpr1−r.
By the above estimates, we obtain the following result(33)∣Fzz+Fz¯z∣≤S1r,where
(34)S1r=4π1−r2∑k=1p−1Mp−kr2k+8π1−r∑k=1p−1kMp−kr2k+Λp2−1rΛp1−r+Λp.
Now, differentiatingS1r, we have
(35)S′1r=∑k=1p−14Mp−k2kr2k−11−r2+2r2k+1π1−r22+∑k=1p−18kMp−k2kr2k+11−r+r2kπ1−r2+Λp2−1Λp1−r2.
In view ofΛp≥1 and r∈0,1, after a simple calculation, it shows that S′1r>0. It is simple to verify that S1r is strictly increasing in 0,1.
(36)limr→0S1r=Λp,limr→1S1r=+∞.
Obviously,S1r has only one pole r=1∣z∣=r. In other words, S1r is bounded in the interval 0,1.
Finally, we consider the boundedness ofF for any ∣z∣=r1, then we have
(37)∣Fz∣=∣∑n=1∞an,pzn+bn,p¯z¯n+∑k=1p−1z2kGp−kz∣≤∣a1,pz+b1,p¯z¯∣+∣∑n=2∞an,pzn+bn,p¯z¯n∣+∣∑k=1p−1z2kGp−kz∣≤Λpr1+Λp2−1Λp∑n=2∞r1nn+∑k=1p−1r12k4Mp−kr1π1−r1=Λpr1−Λp2−1Λpr1+ln1−r1+4r1π1−r1∑k=1p−1Mp−kr12k.
So∣Fz∣≤S2r1, where
(38)S2r1=Λpr1−Λp2−1Λpr1+ln1−r1+4r1π1−r1∑k=1p−1Mp−kr12k.
By the similar approach for differentiatingS2r1, we have the following one
(39)S′2r1=∑k=1p−14Mp−k2k+1r12k1−r1+r12k+1π1−r12+Λp2−1Λp21−r1+1Λp
By elementary calculations, we getS2′r1>0. It implies that S2r1 is increasing for Λp≥1 and r1∈0,1. It is simple to verify that S2r1 is bounded in 0,1. Combined with Lemma 5 and Estimation (33), we conclude ultimately that F is normal polyharmonic mapping in the unit disk D. The proof of this theorem is complete.Theorem 7.
LetFz=∑k=1pz2k−1Gp−k+1z, be a polyharmonic mapping of D satisfying F0=λF0−1=0. Suppose that for k∈1,2,⋯,p, we have
(40)Gp−k+1zisharmonicinD,andGp−k+10=0;ΛGp−k+1z≤Λp−k+1forallz∈D,whereΛp−k+1≥0,k=2,3,⋯,p,andΛp≥1.
Then,F is normal polyharmonic mapping in the unit disk D.Proof.
We represent the harmonic functionsGp−k+1z in series form as
(41)Gp−k+1z=∑j=1∞aj,p−k+1zj+∑j=1∞bj,p−k+1¯z¯j,
for eachk∈1,2,⋯,p, and
(42)∣a1,p∣+∣b1,p∣=ΛF0=∣Gpz0∣+∣Gpz¯0∣=ΛGp0≤Λp,a1,p−b1,p=λF0=Gpz0−Gpz¯0=λGp0=1
To prove the normality ofF, we first determine that the derivative of F is bounded in D. Then, as in the proof of Theorem 6, we have
(43)∣Fzz+Fz¯z∣≤B1+B2+B3+B4,where
(44)B1≤∣Gpz0∣+∣Gpz¯0∣=ΛGp0≤Λp,B2≤∑k=1p−1z2k∣Gp−kzz∣+∣Gp−kz¯z∣≤∑k=1p−1r2kΛGp−kz≤∑k=1p−1Λp−kr2k.
By the condition (2) of Theorem 7 and Lemma 4, we have
(45)B3≤∑k=1p−1∣kGp−kz∣∣z¯kzk−1∣+∣z¯k−1zk∣≤∑k=1p−12kΛp−kr2k.
Applying Lemma3, we have
(46)B4≤∣Gpzz−Gpz0∣+∣Gpz¯z−Gpz¯0∣≤∑n=2∞n∣an,p∣+∣bn,p∣rn−1≤Λp2−1Λp∑n=2∞rn−1=Λp2−1Λpr1−r.
By above estimates, we obtain that(47)∣Fzz+Fz¯z∣≤∑k=1p−1Λp−kr2k+∑k=1p−12kΛp−kr2k+Λp2−1Λpr1−r+Λp=∑k=1p−12k+1Λp−kr2k+Λp2−1Λpr1−r+Λp.
Set(48)S3r=∑k=1p−12k+1Λp−kr2k+Λp2−1Λpr1−r+Λp.
For differentiatingS3r, we have
(49)S3′r=∑k=1p−12k2k+1Λp−kr2k−1+Λp2−1Λp1−r2.
By a simple calculation, we obtainS3′r>0. It shows that S3r is strictly increasing in r∈0,1, and
(50)limr→0S3r=Λp,limr→1S3r=+∞.
This shows thatS3r has only one pole r=1, and it says that S3r is bounded in 0,1. So ∣Fzz+Fz¯z∣ is a finite value in the unit disk D.
Now, we consider anyz with ∣z∣=r2. Then, we have
(51)∣Fz∣=∣∑n=1∞an,pzn+bn,p¯z¯n+∑k=1p−1z2kGp−kz∣≤∣a1,pz+b1,p¯z¯∣+∣∑n=2∞an,pzn+bn,p¯z¯n∣+∣∑k=1p−1z2kGp−kz∣≤Λpr2+Λp2−1Λp∑n=2∞r2nn+∑k=1p−1Λp−kr22k+1=∑k=1p−1Λp−kr22k+1−Λp2−1Λpr2+ln1−r2+Λpr2.≔S4r2
DifferentiatingS4r2, we have
(52)S4′r2=∑k=1p−12k+1Λp−kr2k+Λp2−1Λp1−r2+1Λp
After elementary calculations, we have thatS4′r2>0. It implies that S4r2 is strictly increasing. It is simple to verify that S4r2 is finite for r2∈0,1. It says that F is bounded in D. Using these above estimates and Condition (21), we conclude that F is normal polyharmonic mapping in D. This proof of Theorem 7 is complete.Finally, we establish the normality of log-p-harmonic mappings as follows.Theorem 8.
Letfz=∏k=1pgp−k+1zz2k−1 be a log-p-harmonic mapping in the unit disk D satisfying f0=gp0=λf0=1. Suppose that for each k∈1,⋯,p we have
(53)gp−k+1zislog−harmonicinD,gp−k+1z≤Mp−k+1,letGp=loggp,ΛGp≤Λp,whereMp−k+1≥1,Λp≥1.Then,f is a normal log-p-harmonic mapping in the unit disk D.Proof.
LetFz=∑k=1pz2k−1Gp−k+1z, for each k∈1,⋯,p. We may represent the harmonic functions Gp−k+1z=loggp−k+1 in series form as
(54)Gp−k+1z=∑n=1∞aj,p−k+1zj+∑n=1∞bj,p−k+1¯z¯j.
Then,F=logf is a polyharmonic mapping in D. We know that
(55)λf0=fz0−fz¯0=f0Fz0−Fz¯0,and f0=1, so it follows from gp0=λf0=1, we have Gp0=λF0−1=0.
Obviously,(56)∣Gp−k+1∣=∣loggp−k+1∣=∣log∣gp−k+1∣+iarggp−k+1∣≤∣log∣gp−k+1+π,so we have
(57)∣Gp−k+1∣≤logMp−k+1+π=M~p−k+1.
In order to prove the normality off, it follows from Theorem 6 that we have
(58)Fzz+Fz¯z≤S5r,where
(59)S5r=4π1−r2∑k=1p−1r2kM~p−k+8π1−r∑k=1p−1kM~p−kr2k+Λp2−1Λpr1−r+Λp.
So we obtain(60)∣fzzfz+fz¯zfz∣≤S5r.
DifferentiatingS5r, we have
(61)S5′r=∑k=1p−14Mp−k2kr2k−11−r2+2r2k+1π1−r22+∑k=1p−18kM~p−k2kr2k−11−r+r2kπ1−r2+Λp2−1Λp1−r2.
By elementary calculations, we getS5′r>0. It follows that S5r is strictly increasing in 0,1, and
(62)limr→0S5r=Λp,limr→1S5r=+∞.
It implies thatS5r is bounded in 0,1. Hence, there exists a finite value m1 such that
(63)∣fzzfz+fz¯zfz∣≤m1.
Finally, we consider anyz with ∣z∣=r3; then, we have
(64)∣logfz∣=∣Fz∣=∣∑n=1∞an,pzn+bn,p¯z¯n+∑k=1p−1z2kGp−kz∣≤∣a1,pz+b1,p¯z¯∣+∣∑n=2∞an,pzn+bn,p¯z¯n∣+∣∑k=1p−1z2kGp−kz∣≤Λpr3+Λp2−1Λp∑n=2∞r3nn+∑k=1p−14M~p−kr32k+1π1−r3=∑k=1p−14M~p−kr32k+1π1−r3−Λp2−1Λpr3+ln1−r3+Λpr3=S6r3.
DifferentiatingS6r3, have the following result
(65)S6′r3=∑k=1p−14M~p−k2k+11−r3r32k+r32k+1π1−r32+Λp2−1Λp1−r3+1Λp
It is not difficult to verify thatS6′r3>0, which means that S6r3 is strictly increasing. It is also easily seen that S6r3 is bounded in 0,1. Hence, there is a finite value m2 such that
(66)∣logfz∣≤m2.
Applying Lemma5 and (63), (66), we obtain f is normal in D. The proof of Theorem 8 is complete.



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