A Note on Schwarz-Pick Lemma for Bounded Complex-Valued Harmonic Functions in the Unit Ball of Rn∗

Shaoyu DAI Yifei PAN

1 Introduction

This paper is a note about Chen’s paper(see[1]).Using the same method as in[1],we obtain Theorem 1.1,which extends the Schwarz-Pick lemma(see[1])for planar harmonic mappings to bounded complex-valued harmonic functions in the unit ball of Rn.In addition,motivated by[1]and this paper,we consider a Schwarz lemma for harmonic mappings between real unit balls in another paper.Now we introduce some denotations and the background.

Letnbe a positive integer greater than 1.Rnis the real space of dimensionn.ForLetbe the unit ball of Rn.The unit sphere,i.e.,the boundary of Bnis denoted byS;the normalized surfacearea measure onSis denoted byσ(so thatσ(S)=1).LetS+denote the northern hemisphereandS−denote the southern hemispheredenotes the north pole ofis the open ball centered at origin of radiusr;its closure is the closed ballA twice continuously differentiable,complex-valued functionFdefined on Bnis harmonic on Bnif and only ifwheredenotes the second partial derivative with respect to thej-th coordinate variablexj.By Ωn,we denote the class of all complex-valued harmonic functionsF(x)on Bnwithforx∈Bn.

Let D be the unit disk in the complex plane C.Denote the diskbyDr;its closure is the closed disk

For a holomorphic functionffrom D into D,the classical Schwarz lemma says that iff(0)=0,then

holds forz∈D.For 0

So the classical Schwarz lemma can be regarded as concerning the region ofIf the conditionf(0)=0 is relaxed,then what the region ofis?The answer can be found in the classical Schwarz-Pick lemma.By Schwarz-Pick lemma(see[2]),it is known that

holds forz1,z2∈D.Using the notations

for the pseudo-distance betweenwe know that

forz1,z2∈D by(1.3).Denotefor the closed pseudo-disk with center atzand pseudo-radiusr.Then(1.4)may be written in the following form:

forz∈D and 0

Whenf(0)=0,(1.5)becomes(1.2).

For a complex-valued harmonic functionFon D such thatF(D)⊂D andF(0)=0,it is known(see[3])that

holds forz∈D.For 0

If the conditionF(0)=0 is relaxed,then what the region ofF(Dr)is?Unfortunately,the compositionf◦Fof a harmonic functionFand a holomorphic functionfdo not need to be harmonic,so it is a serious problem to seek the estimate corresponding to(1.5)for a harmonic functionFwithout the assumptionF(0)=0.Fortunately,Chen resolved this problem in[1].In[1],for any 0

which is sharp.(1.8)is the estimate for complex-valued harmonic functions corresponding to(1.5).Note that a complex-valued harmonic functionFon D such thatF(D)⊂D can be seen asF∈Ω2.So it is natural to consider the same problem as in Ωn.

ForF∈Ωn,the harmonic Schwarz lemma(see[4])says that ifF(0)=0,then

holds forx∈Bn,whereUis the Poisson integral of the function that equals 1 onS+and−1 onS−.For 0

If the conditionF(0)=0 is relaxed,then what the region ofis?This problem will be solved in this paper.

In this paper,by the same method as in[1],we obtain the following theorem about the region ofThe result is sharp.Whenn=2,our result is coincident with(1.8).And whenF(0)=0,our result is coincident with(1.10).Note that in the following theorem,Er,ρis defined as(3.1).

Theorem 1.1Let0≤ρ<1,α∈Rand0

The theorem above will be proved in three steps as follows:

Step 1 Find the extremal line ofin the normal direction of e0i,which is related to the value ofF(0).

Step 2 Find the extremal line ofin the normal direction of a given direction.For a given direction of eiβwithconstruct a new harmonic functionFβ=e−iβFthrough rotatingby an anti-clockwise rotation of angleβ.Using the result of Step 1,we will have the the extremal line ofin the normal direction of e0i,which is denoted by.Note thatcan be obtained fromby a clockwise rotation of angleβ.Then the extremal line ofin the normal direction of eiβ,which is denoted bylβ,can be obtained fromby a clockwise rotation of angleβ.

Step 3 Using the result of Step 2,we will obtain all the extremal lines ofin every normal direction,with which we can wrapand obtain the region of

Step 1 will be solved in Section 2.Step 2 and Step 3 will be solved in Section 3.

2 Some Lemmas

In this section,we will introduce some lemmas,which are important for the proof of Theorem 3.1.Lemma 2.1 will be used in Lemma 2.2.Lemma 2.2 will be used in Lemma 2.3.Lemmas 2.3–2.4 will be used in Theorem 3.1.

Now we give Lemma 2.1 first.Lemma 2.1 constructs a bijection(R,I)from R×R+onto the upper half disk{(a,b):a∈R,b∈R,a2+b2<1,b>0},which will be used to constructua,b,rin Lemma 2.2 for the caseb>0.

For 00 and a real numberλ,define

and

The idea of the conformation ofAr,λ,μ(ω),R(r,λ,μ)andI(r,λ,μ)originates from the needs of(2.16)and(2.21).

Lemma 2.1Let00,defined on the upper half disk{(a,b):a2+b2<1,b>0}and analytic in the real sense,such that R(r,λ(r,a,b),μ(r,a,b))=a and I(r,λ(r,a,b),μ(r,a,b))=b for any point(a,b)in the half disk.

ProofA simple calculation gives

It is easy to see that(i)by(2.3),for anyλandμ>0,R(r,λ,μ)is strictly decreasing as a function ofλfor a fixedμ;

(ii)by(2.2),for a fixedμ,or 1 according to(iii)by(2.3)–(2.6)and the convexity of the square function,

for anyλandμ>0;

(iiii)by(2.2),for anyλandμ>0.

By(i)and(ii),we know that for fixedμ,R(r,λ,μ)is strictly decreasing from 1 to−1 asλincreases from−∞to+∞.Then for any−1

Further,using the implicit function theorem,we have that the functionλ=λ(μ,a)defined on{(μ,a):μ>0,−1

Next,we consider the functionI(r,λ(μ,a),μ)forμ>0.

By(i)and(iii),we havewhich shows thatI(r,λ(μ,a),μ)is strictly increasing as a function ofμon(0,+∞)for a fixeda.Note(iiii).Thus,for a fixedhas a respectively finite limit asμ→0 and

For a fixeda,we claim thatas,andas

Asμ→0,there exists a subsequencesuch thatλ(μk,a)has a finite limittor tends to∞.We only need to prove thatSincewe only need to prove thatalmost everywhere onS.Note that

and

Ifthenis bounded andalmost everywhere onS.Thusalmost everywhere onS.Ifasthen it is obvious thatThe first claim is proved.

Asuniformly forω∈S.If there exists a subsequencesuch thatthenuniformly for,anda contradiction.This shows thatis bounded asThus there exists a subsequencesuch thattends to a finite limitt.That is

By(2.1),(2.8)andwe obtain

uniformly forω∈S,and

uniformly forω∈S.By the Lebesgue’s dominated convergence theorem,(2.2)and(2.9)–(2.10),we have

and

Note thatby(2.7),andThen by(2.11)we obtain thatConsequently by(2.12),

The second claim is proved.

Further,using the implicit function theorem,we have that the functionμ(a,b)defined onis a continuous function.

Denoteλ(μ(a,b),a)byλ(r,a,b).Denoteμ(a,b)byμ(r,a,b).We have proved that there exists a unique pair of functionsλ=λ(r,a,b)andμ=μ(r,a,b)such that

on the upper half disk.The real analyticity ofandis asserted by the implicit function theorem.The lemma is proved.

Letaandbbe two numbers such that 0≤b<1,−1

Every functiondefines a harmonic function

Let 0

Obviously,Ua,bis a closed set,andLris a continuous functional onUa,b.Then there exists an extremal function such thatLrattains its maximum onUa,bat the extremal function.We will claim in the following lemma that the extremal function is unique.In the proof of the following lemma,we will construct a functionu0first and then prove thatu0is the unique extremal function,which will be denoted byua,b,r.

Lemma 2.2For any a,b and r satisfying the above conditions,there exists a unique extremal function such that Lrattains its maximum onat

ProofLeta,bandrbe fixed.First assume thatb>0.From Lemma 2.1,we haveλ=λ(r,a,b)andμ=μ(r,a,b)>0 such thatR(r,λ,μ)=aandI(r,λ,μ)=b.For the need of(2.21),let

whereis defined as in(2.1).Thenand by(2.2),and we know

This means that

Let.By(2.14)and(2.17),we have

whereξis a real number betweenu0(ω)andu(ω).By(2.1)and(2.16),we have

Then by(2.15)and(2.18)–(2.21),we obtain that

Thuswith equality if and only ifThereforeLr(u0)≥with equality if and only ifu(ω)=u0(ω)almost everywhere.This shows thatu0(ω)is the unique extremal function,which will be denoted by

Next we consider the caseb=0.For a real numberd,let

For a fixed real numberasuch that−1

We want to prove thatu0is just the unique extremal function,which will be denoted byua,0,r(ω).

It is obvious thatu0∈Ua,0.Letu∈Ua,0.By(2.14)and(2.25),we have

Let

Note that

Then by(2.15)and(2.26)–(2.31),we obtain that

ThusLr(u0)≥Lr(u)with equality if and only ifu(ω)=u0(ω)almost everywhere.The lemma is proved.

Letaandbbe two real numbers witha2+b2<1 and 0

and

Forb<0,let

Then for anya∈R,b∈R anda2+b2<1,let

The harmonic functionsatisfiesand⊂D,since we will show that.By the convexity of the square function,

with equality if and only ifua,b,r(ω)andva,b,r(ω)are constants almost everywhere onS.Howeverua,b,r(ω)is not possible to be a constant almost everywhere onS.Thus

The functionsFa,b,rare the extremal functions in the following lemma.

Lemma 2.3Let F(x)=U(x)+iV(x)be a harmonic function such that F(Bn)⊂D,F(0)=a+bi.Then,for0

with equality at some point rω if and only ifwhere A is an orthogonal matrix such that rωA=rN,Ua,b,ris defined as in(2.33)and(2.35),and is defined as in(2.36).Further,for|x|

ProofStep 1 First,the caserω=rNwill be proved.Letbe fixed.Construct a function

G(x)is harmonic onandLet.Then

So by(2.14)we know thatand by Lemma 2.2,we havewith equality if and only ifalmost everywhere onS.Forby(2.17)and(2.25),we have

Ifalmost everywhere onS,then by(2.33)and(2.35),we have

and by(2.32),we have

Note that by(2.37)–(2.39),we have

Then

So

Forit is proved thatwith equality if and only if.Now letNote that

Then by the result forG(x),we havewith equality if and only ifF(x)=

Step 2 Now we prove the caseConstruct a function

whereAis an orthogonal matrix such thatandA−1is the inverse matrix ofA.By[4],we know thatis also a harmonic function.LetNote thatThen by the result of Step 1,we havewith equality if and only ifNote thatandThuswith equality if and only ifIt is just thatwith equality if and only if

Step 3 We will show thatforBy the result of Step 2 and the maximum principle,we havefor|x|≤r.If the equality holds for somex0with|x0|

Lemma 2.4For fixed00,a2+b2<1}and is continuous to the real diameter.

ProofLet 00,a2+b2<1}.We only need to prove thatis continuous at the points of the real diameter.Note(2.36).Then we only need to prove thatandare continuous at the points of the real diameter.

Let−1

Step 1 For the casewithb=0,by(2.33)–(2.34),we only need to provealmost everywhere onSas.Recall that

whereandare defined as in(2.22)–(2.24).This shows thatalmost everywhere onSas

Step 2 For the casewithb>0,by(2.33)–(2.34),we only need to provefor anyaswithb>0.

First we want to prove thataswithb>0,whereis defined asμ(a,b)in(2.13).Assume thataswithThen there exists a sequencewithsuch thathas a positive lower bound sinceμ(r,a,b)>0.Then by(2.2)and(2.13),we have

whereis defined asin(2.13).ThusAssume thatThen by(2.1)and(2.16)–(2.17),we obtain

uniformly for,and,a contradiction.

Now,we want to prove that

aswithb>0,whereis defined as in(2.29).On the contrary,assume thataswithb>0.Then there is a sequencewithbk>0 such thatIfthen,as above,a contradiction.In the case thatλ?is finite,by(2.1)and(2.16)–(2.17),we have

This contradicts

It is proved thatandaswithb>0.Thus,

Step 3 For the case thatwithb<0,by the result of Step 2,we know thatandaswith−b>0.Note thatandThen we haveandaswithb<0.

It is proved thatandare continuous at(a0,0).The lemma is proved.

3 Main Results

Fora real numberδ,denote the straight linel(β,δ)and the closed half planeP(β,δ)by

and

respectively.

Theorem 3.1LetandDenote

and define

where Uρcosβ,−ρsinβ,ris defined as in(2.33)and(2.35),andis defined as in(2.36).Then

(1)for any harmonic function F such that andwe have

(2)Er,ρis a closed convex domain and is symmetrical with respect to the real axis,and ρis an interior point of Er,ρ;

(3) Γr,ρis a convex Jordan closed curve and∂Er,ρ= Γr,ρ;

(4)for any w?∈Er,ρ,there is a harmonic function F such that F(Bn)⊂D,F(0)=ρ and F(rN)=w?.

Proof(1)Denote

Pβandlβare obtained fromandby an anti-clockwise rotation of angleβ,respectively.

LetFbe a harmonic function such thatF(Bn)⊂D andF(0)=ρ.ForletThen,andUsing Lemma 2.3 to the harmonic functionFβ,we haveand consequently,This shows(1).

(2)It is obvious thatEr,ρis a closed convex set and is symmetrical with respect to the real axis.We only need to prove thatρis an interior point ofEr,ρ.

First we want to prove thatforsinceLetThe harmonic functionGsatisfies the conditionsandG(0)=ρ.By(1),Note thatandwhich was proved above.Then we have

Thenρis an interior point ofsinceEr,ρis a convex set.

(3)First we want to prove that Γr,ρis a Jordan closed curve. Γr,ρis close and continuous by Lemma 2.4.Assume that there existsuch thatThenandw0is the vertex of the angular domainFurther,it is easy to see thatforsinceandis analytic on(0,π)in the real sense by Lemma 2.4.Then we haveforand by the continuity,This is a contraction,sinceby(3.2)–(3.3).This shows thatis a Jordan curve.For the same reason,is also a Jordan curve.Then Γr,ρis a Jordan closed curve.

For−π≤β≤π,it is proved in(2)thatThen.Note that∂Er,ρmust be a convex Jordan closed curve.Thus

(4)For,draw a straight linelpassing throughw?and intersect∂Er,ρatw1andw2.Letwithandk1+k2=1.There are two real numbersβ1andβ2such thatfr,ρ(β1)=w1andfr,ρ(β2)=w2.Then the harmonic functionF=satisfiesandThe theorem is proved.

Whenρ=0,we have a corollary as follows,which is coincident with(1.10).

Corollary 3.1Let0

where U is the Poisson integral of the function that equals1on S+and−1on S−.

ProofBy Theorem 3.1,we only need to prove thatFurther,by the definition ofEr,ρin Theorem 3.1,we only need to prove thatNote that by(2.25),

Then by(2.33)we know thatThe corollary is proved.

From Theorem 3.1,we obtain Theorem 1.1,which is the general version of the above Theorem 3.1.

[1]Chen,H.H.,The Schwarz-Pick lemma for planar harmonic mappings,Science China Mathematics,54(6),2011,1101–1118.

[2]Ahlfors,L.V.,Conformal Invariants:Topics in Geometric Function Theory,McGraw-Hill,New York,1973,1–3.

[3]Heinz,E.,On one-to-one harmonic mappings,Paci fi c J.Math.,9,1959,101–105.

[4]Axler,S.,Bourdon,P.,Wade,R.,Harmonic Function Theory,Second Edition,Springer-Verlag,New York,2001.