# Characterizations of Bloch-Type Spaces of Harmonic Mappings

Authors: Munirah Aljuaid; Flavia Colonna
Journal: Journal of Function Spaces
 (2019)

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## Abstract

We study the Banach spaceBHα (α>0) of the harmonic mappings h on the open unit disk D satisfying the condition supz∈D⁡(1-z2)α(hzz+hz¯z)<∞, where hz and hz¯ denote the first complex partial derivatives of h. We show that several properties that are valid for the space of analytic functions known as the α-Bloch space extend to BHα. In particular, we prove that for α>0 the mappings in BHα can be characterized in terms of a Lipschitz condition relative to the metric defined by dH,α(z,w)=sup⁡{hz-hw:h∈BHα,hBHα≤1}. When α>1, the harmonic α-Bloch space can be viewed as the harmonic growth space of order α-1, while for 0<α<1, BHα is the space of harmonic mappings that are Lipschitz of order 1-α.


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## Body

## 1. Introduction

Given a regionΩ in the complex plane C, aharmonic mapping with domain Ω is a complex-valued function h defined on Ω satisfying the Laplace equation (1)Δh≔4hzz¯≡0onΩ,having denoted by hzz¯ the mixed complex second partial derivatives of h.It is well known that a harmonic mappingh admits a representation of the form f+g¯, where f and g are analytic functions. This representation is unique if, fixing a base point z0, the function g is chosen so that g(z0)=0.In the last several decades, much research has been carried out on the study of Banach spaces of analytic functions on the open unit diskD in the complex plane. Since analytic functions are clearly harmonic, a natural question is whether such spaces X are subspaces of some Banach space XH of harmonic mappings on D in such a way that the norm on the larger space agrees with the norm of X when restricting to the elements of X.A space that has been thoroughly studied in complex function theory is the classical Bloch spaceB defined as the set of analytic functions f on D such that (2)βf≔supz∈D⁡1-z2f′z<∞.In [1], Theorem 10, the second author observed that the functions f in the Bloch space are precisely the analytic Lipschitz maps when regarded as functions between the metric spaces (D,ρ) (where ρ denotes the hyperbolic distance) and (C,·) (see also Theorem 5.5 of [2]). Moreover, the correspondence f↦βf is a seminorm on B and is equal to the Lipschitz number, namely, for f∈B, (3)βf=supz≠w⁡fz-fwρz,w,where we recall (4)ρz,w=12log⁡1+ϕzw1-ϕzw,withϕzw=z-w1-z¯w.This result motivated the following notion of Bloch harmonic mapping in [3].A harmonic mappingh on D is calledBloch if there exists a constant c>0 such that (5)hz-hw≤cρz,wfor  allz,w∈D,where for h=f+g¯, with f,g analytic on D, the Lipschitz number (6)Lh=supz≠w⁡hz-hwρz,wwas shown in [3] to be equal to the quantity(7)βh≔supz∈D⁡1-z2f′z+g′z=supz∈D⁡1-z2hzz+hz¯z,thereby extending to harmonic mappings the corresponding result valid for analytic functions.In this work, we expand the research done in [3] by focusing on the study of the harmonic extensions of the α-Bloch spaces (for α>0) introduced by Zhu in [4].Research on the study of harmonic mappings has been conducted extensively in the last two decades by mostly analyzing the function theoretic aspects. The study of the extensions of classical Banach spaces of analytic functions is still relatively limited. Besides [3], references in the unit disk setting include [5] for the study of harmonic Bloch and Besov spaces, [6] for the study of harmonic ν-Bloch mappings, [7] on planar harmonic Lipschitz and Hardy classes, and [8] for the study of harmonic Lipschitz-type spaces. In the setting of the unit ball in Cn, see [9] for the study of the harmonic Bloch spaces, [10, 11] for the study of the harmonic Bergman spaces and [12] for extensions of the main results in [5]. For a general reference on harmonic mappings in the plane we refer the interested reader to [13].After giving in Section2 some preliminaries and background on the (analytic) α-Bloch spaces, we introduce in Section 3 the harmonic α-Bloch spaces and study their properties. In particular, we show that as done by the second author in [3] for the harmonic Bloch space (case α=1), the α-Bloch space can be characterized by Lipschitz-type conditions similar to the conditions obtained by Zhu in [4] for other positive values of α.In Section4, we show that for α≠1, as for the analytic counterparts described by Zhu in [4], such spaces can be divided into two classes: the space of Lipschitz harmonic mappings of order 1-α for 0<α<1, and the harmonic weighted Banach space of harmonic mappings with weight given by the Bergman weight z↦(1-z2)α-1 for α>1.Finally, in Section5, we give some properties of the harmonic growth spaces that are useful to extend a characterization of functions in the Zygmund space to the harmonic space counterpart. This topic is treated in [14].


## 2. Preliminaries and Background

LetH(D) denote the class of analytic functions on D, and for R>0, let D(0,R) be the open disk centered at 0 of radius R.In [4], for α>0, Zhu introduced the α-Bloch space Bα as the collection of functions f∈H(D) such that (8)βfα≔supz∈D⁡1-z2αf′z<∞.The correspondence f↦βfα is a seminorm and Bα is a Banach space under the norm (9)fBα=f0+βfα.For α=1, Bα is with the classical Bloch space B. Thus the α-Bloch space Bα can be considered as the space of functions f∈H(D) such that f′ is in the growth space A-α, defined as the collection of functions g∈H(D) satisfying the growth condition (10)gz=O1-z-α.The subspace B0α of Bα consisting of the functions f satisfying the condition (11)limz→1⁡1-z2αf′z=0,known as the little α-Bloch space, is the closure in Bα of the polynomials and hence separable. Again, this subspace can be viewed as the collection of functions in H(D) whose derivative is in the little growth space A0-α, whose members f satisfy the “little oh" version of (10) as z→1.Zhu proved that for a positive numberα≠1, the α-Bloch space Bα can be identified with one of two families of spaces, depending on whether 0<α<1 or α>1. Specifically, (see Proposition 9 in [4]) if 0<α<1, then an analytic function f on D belongs to Bα if and only if (12)supz≠w⁡fz-fwz-w1-α<∞.In particular, the space Bα is contained in the disk algebra.In Proposition 7 of [4] it was shown that for α>1, the space Bα (respectively, B0α) is the growth space A1-α (respectively, the little growth space A01-α) and the corresponding norms are equivalent, where for γ>0 the norm of a function f∈A-γ is defined as (13)fA-γ≔supz∈D⁡1-z2γfz.For more information on the growth spaces we refer the interested reader to [15].The following result proved by Zhu in [4] shows that, for any integer n≥2, the elements of the α-Bloch space can be characterized in terms of their nth derivative and a certain Bergman weight dependent of n and α.Theorem 1 ([4], Proposition 8).
Letα>0, n≥2 be an integer, and f∈H(D). Then
(i)f∈Bα if and only if (14)βf,nα≔supz∈D⁡1-z2α+n-1fnz<∞.
(ii)f∈B0α if and only if (15)limz→1⁡1-z2α+n-1fnz=0.GivenR>0, we recall the Poisson integral representation of bounded real-valued harmonic functions on the disk D(0,R)≔{z∈C:|z|<R} with continuous boundary values (see, e.g., [16], (2.13), p. 260), extended in the obvious way to harmonic mappings.Theorem 2.
ForR>0, a complex-valued continuous function h on D(0,R)¯ and harmonic on D(0,R), admits the Poisson integral representation: (16)hz=12π∫-ππR2-z2Reit-z2hReitdtfor z<R.This representation yields the following integral formulas for the complex partial derivatives of a harmonic mapping onD(0,R) with continuous extension to D(0,R)¯.Theorem 3.
Leth be a complex-valued continuous function on D(0,R)¯ and harmonic on D(0,R). Then for z∈D(0,R),(17)hzz=12π∫-ππReitReit-z2hReitdt,(18)hz¯z=12π∫-ππRe-itRe-it-z¯2hReitdt.Proof.
Let us evaluate the partial derivatives with respect toz and z¯ of the Poisson kernel: (19)∂∂zR2-z2Reit-z2=ReitRe-it-z¯Re-it-z¯Reit-z2=ReitReit-z2,(20)∂∂z¯R2-z2Reit-z2=Re-itReit-zReit-zRe-it-z¯2=Re-itRe-it-z¯2.The results follow by applying (19) and (20) after differentiating (16) with respect to z and z¯ under the integral sign.Remark 4.
In the special case when the functionh is constant on the unit circle, the harmonic extension to D is constant as well, so its complex partial derivatives are identically 0.
With the goal of characterizing functions inBα by a Lipschitz type condition, Zhu proved the following result.Proposition 5 ([4], Proposition 16).
Forα>0, and z,w∈D, let (21)dαz,w=sup⁡fz-fw:βfα≤1.Then dα defines a distance on D.In [4], Zhu observed that for α=1, the metric dα is precisely the hyperbolic metric ρ. To the best of our knowledge, an explicit formula of dα for the case α≠1 has not been determined.The following result shows that the ratio of the distancedα between two points in D to their Euclidean distance yields in the limit the reciprocal of the Bergman weight of order α.Theorem 6 ([4], Theorem 17).
For anyα>0 and z∈D, (22)limw→z⁡dαz,wz-w=1-z2-α.This leads to the following Lipschitz-type characterization of analytic functions inBα.Theorem 7 ([4], Theorem 18).
Forα>0 and f∈H(D) the following statements are equivalent: (1)
f∈Bα.(2)
There exists a constantc>0 such that for all z,w∈D(23)fz-fw≤cdαz,w.Moreover, for each f∈Bα, (24)βfα=supz≠w⁡fz-fwdαz,w.In the next section, after giving the notions of harmonic growth space and of harmonicα-Bloch space, we follow a similar framework that will then allow us to extend Theorem 7 to harmonic mappings.


## 3. Harmonic Growth Spaces andα-Bloch Spaces

Forα>0, we define theharmonic growth spaceAH-α as the collection of all harmonic mappings h on D such that (25)hAH-α≔supz∈D⁡1-z2αhz<∞.Theorem 8.
The mappingh↦hAH-α defines a Banach space structure on AH-α that extends the corresponding structure on A-α.Proof.
It is straightforward to verify that·AH-α is a norm. The space A-α is clearly a subspace of AH-α and the respective norms coincide. To prove completeness, assume (hn) is a Cauchy sequence in AH-α. Fix z∈D and ε>0. Choose N∈N such that (26)hn-hmAH-α<ε1-z2-α+1for all n,m≥N. Then for all n,m≥N, (27)hnz-hmz≤1-z2-αhn-hmAH-α<ε.Thus, (hn(z)) is a Cauchy sequence in C, which is complete, and we may define h(z)≔limn→∞⁡hn(z). Then h is continuous on D and since harmonic mappings satisfy the mean value property, for each z∈D and r∈(0,1-z) and n∈N, (28)hnz=12π∫02πhnz+reitdt.Passing to the limit as n→∞, it follows that h satisfies the mean value property. By Theorem 2.11 in [16] extended to complex-valued functions, h is harmonic.
Since Cauchy sequences are bounded,L≔supn∈N⁡hnAH-α is finite. Fixing z∈D, we have (29)1-z2αhz=limn→∞⁡1-z2αhnz≤L.Therefore, h∈AH-α and hAH-α≤L.
To prove thathn converges to h in norm, note that for z∈D, and n,m≥N, (30)1-z2αhnz-hmz≤hn-hmAH-α<ε.Fixing n≥N and letting m→∞, we get (31)1-z2αhnz-hz≤ε.Taking the supremum over all z∈D, we obtain hn-hAH-α→0, as desired.Define thelittle harmonic growth spaceAH,0-α as the subspace of AH-α whose elements h satisfy (32)limz→1⁡1-z2αhz=0.It is evident from the definition that iff,g∈A-α (respectively, A0-α), then f+g¯∈AH-α (respectively, AH,0-α). We shall prove that the converse holds as well.Forα>0, we now define the harmonic α-Bloch space BHα and in Section 4, we shall prove that in analogy to the analytic case, for α>1, as sets, BHα and the harmonic growth space AH-(α-1) are equal, whereas for 0<α<1, the space BHα is the Lipschitz space of harmonic mappings of order 1-α.Definition 9.
Forα>0, we define theharmonicα-Bloch spaceBHα as the collection of all harmonic mappings h on D such that (33)βhα≔supz∈D⁡1-z2αhzz+hz¯z<∞.Clearly, when α=1 this definition agrees with expression (7).
Theharmonic littleα-Bloch space is defined as the subspace BH,0α of BHα consisting of the mappings h such that (34)limz→1⁡1-z2αhzz+hz¯z=0.Observe that ifh=f+g¯ with f,g∈H(D), then hz=f′ and hz¯=g′¯. Thus (35)βhα=supz∈D⁡1-z2αf′z+g′z.This implies that(36)12βfα+βgα≤max⁡βfα,βgα≤βhα≤βfα+βgα.We deduce the following result.Proposition 10.
Letα>0 and let h be a harmonic mapping, h=f+g¯ with f,g∈H(D). Then h∈BHα (respectively, h∈BH,0α) if and only if f and g are in Bα (respectively, f and g are in B0α).In particular, since for0<α<1, the space Bα is contained in the disk algebra, the space BHα is contained in the space of complex-valued harmonic functions in D which are continuous on D¯.It is straightforward to verify that the mapping(37)h↦hBHα≔h0+βhαdefines a Banach space structure on BHα, which extends to harmonic mappings the norm on Bα.A natural question that arises is whether Theorems6 and 7 extend to harmonic mappings. We shall show that this is indeed the case. We make use of the following result whose proof is elementary.Proposition 11.
Forα>0, the function dH,α defined by (38)dH,αz,w=sup⁡hz-hw:h∈BHα,βhα≤1,forz,w∈D,is a distance on D.Theorem 12.
For anyα>0 and z∈D, (39)limw→z⁡dH,αz,wz-w=1-z2-α.Proof.
By Theorem 17 of [4], for z∈D, (40)limw→z⁡dαz,wz-w=1-z2-α.From the definitions of dH,α and dα, it is immediate to see that dH,α(z,w)≥dα(z,w), for all z,w∈D. Hence, fixing z∈D, we have (41)liminfw→z⁡dH,αz,wz-w≥limw→z⁡dαz,wz-w=1-z2-α.Therefore, to prove the result, it suffices to show that for each z∈D, (42)limsupw→z⁡dH,αz,wz-w≤1-z2-α.Fix z∈D and let h∈BHα with βhα≤1. Replacing h with h-h(0), we may assume h(0)=0, so that hBHα=βhα. Represent h as f+g¯ with f,g analytic and f(0)=g(0)=0. Proceeding as in the proof of Theorem 17 of [4], fix z∈D and let r=1-z/2. Then for w-z<r, (43)hw=hz+f′zw-z+g′z¯w¯-z¯+f2zw-z2+g2z¯w¯-z¯2,where(44)f2w=12πi∫ζ-z=rfζζ-z2ζ-wdζandg2w=12πi∫ζ-z=rgζζ-z2ζ-wdζ.By Proposition 5 of [4], there exists some positive constant C only dependent on z such that (45)f2w≤CfBαandg2w≤CgBαfor w-z<r. Therefore, since the assumption βhα≤1 implies that (46)f′z+g′z≤1-z2-α,and noting that by (36), fBα+gBα≤2hBHα≤2, using (46) we obtain (47)hz-hw≤z-wf′z+g′z+Cz-w2fBα+gBα≤z-w1-z2-α+2Cz-w2.Hence, (48)hz-hwz-w≤1-z2-α+2Cz-w.Taking the supremum over all such mappings h, we obtain (49)dH,αz,wz-w≤1-z2-α+2Cz-wfor w-z<r. Letting w→z, we obtain (50)limsupw→z⁡dH,αz,wz-w≤1-z2-α,as desired.We can now prove one of our main results in this section.Theorem 13.
Leth be a harmonic mapping on D and α>0. Then h∈BHα if and only if there exists a constant C>0 such that for all z,w∈D(51)hz-hw≤CdH,αz,w.Moreover, for each h∈BHα, (52)βhα=supz≠w⁡hz-hwdH,αz,w.Proof.
Assumeh∈BHα and let Mhα denote the right-hand side of (52). If h is constant, then Mhα=0=βhα, and we are done. So assume h is nonconstant so that βhα>0. Then, the mapping defined by (53)h~=1βhαhis in BHα and has seminorm 1, so, fixing distinct points z,w∈D, by the definition of dH,α, we have (54)hz-hw=βhαh~z-h~w≤βhαdH,αz,w.Dividing by dH,α(z,w) and taking the supremum over all distinct points z and w, we obtain Mhα≤βhα.
Next, note that fixingz∈D, (55)Mhα≥limsupw→z⁡hz-hwdH,αz,w=limsupw→z⁡hz-hwz-wlimw→z⁡z-wdH,αz,w.For w=z+reiθ with 0<r<1-z and θ∈R fixed, (56)hw-hzw-z=fw-fzw-z+gw¯-gz¯w¯-z¯e-2iθ→f′z+g′z¯e-2iθas r→0. Taking the supremum over θ, we obtain (57)limsupw→z⁡hz-hwz-w=f′z+g′z.Hence, by Theorem 12, from (55) we obtain (58)Mhα≥f′z+g′z1-z2α.Taking the supremum over all z∈D, we conclude that Mhα≥βhα, completing the proof.In [17], the norm of the point-evaluation functional on the Bloch space was calculated precisely. Specifically, it was shown that for z∈D, the quantity (59)supf∈B,fB≤1⁡fz=max⁡1,ρz,0,where we recall that ρ is the distance dα for α=1.We now provide a point-evaluation estimate for the harmonicα-Bloch space in terms of the metric dH,α valid for all α>0. For the case α=1 we obtain an extension of (59) to the harmonic Bloch space BH.Theorem 14.
Forh∈BHα and z∈D, (60)hz≤max⁡1,dH,αz,0hBHα.In the case when α=1, for all z∈D, (61)suph∈BH,hBH≤1⁡hz=max⁡1,ρz,0.Proof.
Leth∈BHα and fix z∈D. The inequality is clear for z=0. So assume z≠0. Then, by (52), we have (62)hz≤h0+hz-h0=h0+hz-h0dH,αz,0dH,αz,0≤h0+βhαdH,αz,0=h0+hBHα-h0dH,αz,0=1-dH,αz,0h0+dH,αz,0hBHα=max⁡1,dH,αz,0hBHαproving the estimate.
Whenα=1, by (59), we have (63)max⁡1,ρz,0=supf∈B,fBα≤1⁡fz≤suph∈BH,hBHα≤1⁡hz,which, combined with (62), yields the conclusion.We now prove that for eachα>0, the map h→βhα is lower semicontinuous on the space BHα.Theorem 15.
Let(hn) be a sequence in BHα converging uniformly on compact subsets of D to some function h. If the sequence (βhnα) is bounded, then h∈BHα and (64)βhα≤liminfn→∞⁡βhnα.Proof.
Let(hn) be as in the statement of the theorem. Applying the mean value property of harmonic mappings to each hn and then passing to the limit as n→∞, we see that h also satisfies the mean value property and hence it is likewise harmonic. Set (65)β=liminfn→∞⁡βhnα.Then, there exists a subsequence (βhnkα) converging to β. We wish to show that for all z,w∈D, (66)hz-hw≤βdH,αz,w.Fix z,w∈D. Since the above inequality is trivial if z=w, assume z≠w and fix ε>0. Since βhnkα→β, there exists ν∈N such that for all k≥ν(67)βhnkα<β+ε2dH,αz,w,hnkz-hz<ε4,andhnkw-hw<ε4.Therefore, for k≥ν, and having shown in Theorem 13 that the seminorm in BHα equals the Lipschitz number with respect to the distance dH,α and the Euclidean distance in C, we have (68)hz-hw≤hz-hnkz+hnkz-hnkw+hnkw-hw<ε2+βhnkαdH,αz,w<ε2+β+ε2dH,αz,wdH,αz,w=ε+βdH,αz,w.Since ε is arbitrary, we conclude that hz-hw≤βdH,α(z,w) for all z,w∈D.
This implies that the Lipschitz number ofh is no greater than β. In particular, h∈BHα and βhα≤β, completing the proof.


## 4. Characterizations of theα-Bloch Harmonic Mappings (α≠1)

We now introduce the space of harmonic Lipschitz mappings and prove that it is a complex Banach space.Forα>0, let LipHα denote the collection of harmonic mappings h on D satisfying the condition (69)hα≔supz≠w⁡hz-hwz-wα<∞.Define hLipHα≔h0+hα.Theorem 16.
Forα>0, under the above norm, LipHα is a Banach space.Proof.
It is immediate to verify thatLipHα is a normed linear space. To prove completeness, suppose (hn) is a Cauchy sequence in LipHα. Fix z∈D and ε>0. Choose N∈N such that (70)hn-hmLipHα<εzα+1for all n,m≥N. Then for all n,m≥N, (71)hnz-hmz≤hnz-hmz-hn0-hm0+hn0-hm0≤hn-hmαzα+hn0-hm0≤zα+1hn-hmLipHα<ε.Thus, (hn(z)) is a Cauchy sequence in C, which is complete, and so h(z)≔limn→∞⁡hn(z) exists. Arguing as in the proof of Theorem 8, we see that h is harmonic.
Since Cauchy sequences are bounded,(72)L≔supk∈N⁡hkα≤supk∈N⁡hkLipHα<∞.Then for z≠w fixed in D, we have (73)hz-hwz-wα=limn→∞⁡hnz-hnwz-wα≤L.Taking the supremum over all pairs of distinct points z,w∈D, we obtain h∈LipHα and hLipHα≤L.
Lastly, to show thathn-hLipHα→0, note that with z∈D fixed, for w∈D, with w≠z, and n,m≥N, (74)hn0-hm0+hnz-hmz-hnw-hmwz-wα<ε.Letting m→∞, we get (75)hn0-h0+hnz-hz-hnw-hwz-wα≤ε,whence hn-hLipHα→0.Theorem 17.
For0<α<1, as sets, BHα=LipH1-α, and the respective norms are equivalent.Proof.
First, assumeh∈BHα and let f,g∈H(D) such that h=f+g¯ with g(0)=0. Then f,g∈Bα, so by Proposition 9 in [4], f and g belong to the Lipschitz space of analytic functions of order 1-α. Thus, for z≠w, (76)hz-hwz-w1-α≤fz-fwz-w1-α+gz-gwz-w1-α≤f1-α+g1-α.Taking the supremum over all pairs of distinct points z,w∈D, we obtain (77)h1-α≤f1-α+g1-α.Hence h∈LipH1-α. Since the space Bα and the space Lip1-α of analytic functions on D which are Lipschitz of order 1-α have equivalent norms, where for f∈Lip1-α, (78)fLip1-α=f0+f1-α,using the equalities h(0)=f(0) and g(0)=0 and (36), we deduce that (79)hLipH1-α=h0+h1-α≤f0+f1-α+g1-α=fLip1-α+gLip1-α≤CfBα+gBα≤2Cf0+max⁡βfα,βgα≤2Ch0+βhα=2ChBHαfor some positive constant C.
Conversely, assumeh1-α<∞. Fix z=zeiτ∈D. By Theorem 3, making a change of variable and using Remark 4, and since eit+τ-z2=eit-z2=e-it+τ-z¯2, we have (80)hzz+hz¯z=12π∫02πeiteit-z2heitdt+12π∫02πe-ite-it-z¯2heitdt=12π∫02πeiteit-z2heit-heiτdt+12π∫02πe-ite-it-z¯2heit-heiτdt≤12π∫-ππheit+τ-heiτeit+τ-z2dt+12π∫-ππheit+τ-heiτe-it+τ-z¯2dt.=1π∫-ππheit+τ-heiτeit-z2dt.By the definition of h1-α, and expanding eit as a power series in t about 0, for t≤π, we have (81)heit+τ-heiτ≤eit+τ-eiτ1-αh1-α=eit-11-αh1-α≤t1-αeπ-11-απ1-αh1-α.From (80) and (81), noting that for 0≤t≤π, (82)eit-z2=1-z2+4zsin2⁡t2≥1-z2+4zt2π2,we deduce(83)hzz+hz¯z≤eπ-11-αh1-απ2-α∫-ππt1-αeit-z2dt≤2eπ-11-αh1-απ2-α∫0πt1-α1-z2+4zt2/π2dt.If z≤1/2, then 1-z2≥1/4, so (84)∫0πt1-α1-z2+4zt2/π2dt≤∫0πt1-α1-z2dt≤4∫0πt1-αdt=4π2-α2-α,so from (83), we obtain (85)1-z2αhzz+hz¯z≤8eπ-11-α2-αh1-α.Next assume 1/2≤z<1. From (83), making the substitution s=t/1-|z|, and noting that 1>4z/π2, we have (86)hzz+hz¯z≤2eπ-11-αh1-απ2-α1-z-α∫0∞s1-α1+4zs2/π2ds≤2eπ-11-αh1-απ2-α1-z-α∫0∞s1-α4z/π21+s2ds=παeπ-11-αh1-α2z1-z-α∫0∞s1-α1+s2ds=C1-z2-αh1-α∫01s1-α1+s2ds+∫1∞s1-α1+s2ds≤C1-z2-αh1-απ4+∫1∞s1-α1+s2ds,where C=πα(eπ-1)1-α. Noting that (87)∫1∞s1-α1+s2ds≤∫1∞1s1+αds=1α,it follows from (86) that for 1/2≤z<1, (88)1-z2αhzz+hz¯z≤Ch1-α.Therefore, combining the results in the two cases z≤1/2 and 1/2≤z<1, we obtain (89)supz∈D⁡1-z2αhzz+hz¯z≤const.h1-α.This proves that h∈BHα and (90)hBHα≤const.h0+h1-α=const.hLipH1-α.Combining (79) and (90), we obtain the equivalence of the norms ·BHα and ·LipH1-α.We next show that, in analogy to the analytic case, forα>1 the space BHα can be identified with the harmonic growth space AH1-α.Theorem 18.
Forα>1, as sets, BHα=AH1-α, and the respective norms are equivalent. The little subspaces BH,0α and AH,01-α are equal as well.Proof.
First, assumeh∈BHα and let f,g∈H(D) such that h=f+g¯ and g(0)=0. Then f,g∈Bα, so by Proposition 7 in [4], f and g belong to the growth space A1-α. Thus, (91)supz∈D⁡1-z2α-1hz≤supz∈D⁡1-z2α-1fz+supz∈D⁡1-z2α-1gz,which is finite. Therefore, h∈AH1-α. The inclusion BH,0α⊆AH,01-α follows as well.
Moreover, from (91), since the norms in Bα and A1-α are equivalent, h(0)=f(0) and g(0)=0, for some positive constant C, we have (92)hAH1-α≤fA1-α+gA1-α≤CfBα+gBα=Cf0+βfα+βgα≤2Ch0+max⁡βfα,βgα≤2Ch0+βhα=2ChBHα.
Conversely, supposeh∈AH1-α. Fix z∈D and set R=1+z2/2. So 1/2≤R<1. Using (17) and (18), we have (93)hzz+hz¯z≤Rπ∫-ππhReitReit-z2dt.Since h∈AH1-α, we have (94)hReit≤hAH1-α1-R2α-1.Since by our choice of R, R2=1+z2/2, we have (95)1-R2=R2-z2=1-z22.By these facts and multiplying (93) by 1-z2α, it follows that (96)1-z2αhzz+hz¯z≤Rπ∫-ππhAH1-α1-R2Reit-z21-z21-R2αdt.≤2αhAH1-α1π∫-ππR2-z2Reit-z2dt.=2α+1hAH1-α,where we used the identity (97)12π∫-ππR2-z2Reit-z2dt=1,since the mean of the Poisson kernel for the disk of radius R over the circle of radius R is 1. Therefore, h∈BHα.
Finally, sinceh0≤hAH1-α, from (96) it follows that (98)hBHα≤1+2α+1hAH1-α,which, combined with (92), proves that the norms of BHα and AH1-α are equivalent.
Lastly, assumeh∈AH,01-α. Fix ε>0 and let r∈(0,1) be chosen so that (99)hz<ε1-z2α-1whenever r<z<1. Choosing z with r<z<1 and defining R in terms of z as done previously, from (93) arguing as above, we see that (100)1-z2αhzz+hz¯z<Rπ∫-ππε1-R2Reit-z21-z21-R2αdt≤2α+1ε.Hence, h∈BH,0α.Givenα>0, n∈N, and h harmonic on D, let us define (101)βh,nα≔supz∈D⁡1-z2α+n-1fnz+gnz.The following result shows that, in analogy to the analytic case, for any integern≥2, the elements of the harmonic α-Bloch space and its subspace BH,0α can be characterized in terms of the nth derivatives of f and g and the Bergman weight z↦1-z2γ, whose exponent γ is a certain function of n and α.Theorem 19.
Supposeα>0, n≥2 is an integer, and h is a harmonic mapping on D. Then (i)
h∈BHα if and only if βh,nα<∞.(ii)
h∈BH,0α if and only if (102)limz→1⁡1-z2α+n-1fnz+gnz=0.Proof.
To prove (i), assumeh∈BHα. By Theorem 1, there exists C>0 such that for each f analytic on D, βf,nα≤Cβfα.
Leth∈BHα such that h=f+g¯ for some f,g∈Bα. Then by Theorem 1 and the inequality (36) (103)βh,nα≤βf,nα+βg,nα≤Cβfα+βgα≤2Cβhα.In particular, if h∈BHα, then βh,nα is finite.
Conversely, assumeβh,nα<∞. Then βf,nα≤βh,nα and βg,nα≤βh,nα. Thus, βf,nα and βg,nα are finite. Therefore, again by Theorem 1, f,g∈Bα, and thus by Proposition 10, h∈BHα.
To prove (ii), assume first thath∈BH,0α. Then f,g∈B0α, so by Theorem 1, (104)limz→1⁡1-z2α+n-1fnz+gnz≤limz→1⁡1-z2α+n-1fnz+limz→1⁡1-z2α+n-1gnz=0,proving that (102) holds.
Conversely, suppose (102) holds. Then (105)limz→1⁡1-z2α+n-1fnz≤limz→1⁡1-z2α+n-1fnz+gnz=0and the same property holds for g(n). Therefore, f,g∈B0α, hence h∈BH,0α.


## 5. Remarks on the Harmonic Growth Spaces

An immediate consequence of the identification of the harmonicα-Bloch space with the harmonic growth space of order α-1 is the following result.Corollary 20.
Givenα>0 and f,g∈H(D), f+g¯∈AH-α (respectively, f+g¯∈AH,0-α) if and only if f,g∈A-α (respectively, f,g∈A0-α).Proof.
We only need to prove the necessity. Assumeh:=f+g¯∈AH-α. Then by Theorem 18, h∈BHα+1. So by Proposition 10, f,g∈Bα+1. Therefore, by Proposition 7 of [4], f,g∈A-α. The proof for the little spaces case is similar.We now apply this result to prove a property of the harmonic growth spaces and their “little” subspaces.Proposition 21.
Letα>0 and ϕ,f,g∈H(D).
(a) Ifϕ is bounded and f+g¯∈AH-α (respectively, AH,0-α), then ϕf+ϕg¯∈AH-α (respectively, AH,0-α).
(b) Ifϕ is bounded away from 0 near the unit circle and ϕf+ϕg¯∈AH-α (respectively, AH,0-α), then f+g¯∈AH-α (respectively, AH,0-α).Proof.
Assumeϕ is bounded and f+g¯∈AH-α. Then by Corollary 20, f,g∈A-α, so by the triangle inequality, (106)supz∈D⁡1-z2αϕzfz+ϕz¯gz¯≤ϕ∞supz∈D⁡1-z2αfz+supz∈D⁡1-z2αgz,which is finite.
Next, assumef+g¯∈AH,0-α. Then f,g∈A0-α, so by the triangle inequality and the boundedness of ϕ, (107)limz→1⁡1-z2αϕzfz+ϕz¯gz¯≤ϕ∞limz→1⁡1-z2αfz+limz→1⁡1-z2αgz=0.Therefore, ϕf+ϕg¯∈AH,0-α, proving (a).
Next, assumeϕf+ϕg¯∈AH-α, where ϕ is bounded away from zero. Then, again by Corollary 20, the functions ϕf and ϕg are in A-α, and there exist constants r∈(0,1) and δ>0 such that ϕz>δ whenever r<z<1.
We now show thatf,g∈A-α. Since f is analytic, it is bounded on each compact subset of D, and thus (108)supz≤r⁡1-z2αfz≤maxz=r⁡fz<∞.On the other hand,(109)supr<z<1⁡1-z2αfz=supr<z<1⁡1ϕz1-z2αϕzfz≤1δsupr<z<1⁡1-z2αϕzfz<∞.Therefore, combining (108) and (109), we have (110)supz∈D⁡1-z2αfz≤supz≤r⁡1-z2αfz+supr<z<1⁡1-z2αfz<∞,proving that f∈A-α. The same argument shows that g∈A-α. Therefore, by Corollary 20, f+g¯∈AH-α.
Next, assumeϕf+ϕg¯∈AH,0-α. Then, by Corollary 20, the functions ϕf and ϕg are in A0-α. Fix ε>0 and choose r∈(0,1) and δ>0 such that ϕz>δ and 1-z2αϕzfz<ε whenever r<z<1. Then, for r<z<1, (111)1-z2αfz≤1δ1-z2αϕzfz<εδ,proving that f∈A0-α. Arguing similarly, we see that g∈A0-α. Therefore, again by Corollary 20, we have f+g¯∈AH,0-α, completing the proof of (b).The following result is an immediate consequence of Proposition21.Corollary 22.
Letα>0, f,g∈H(D), and p a polynomial in z with no zeros on the unit circle. Then, f+g¯∈AH-α (respectively, AH,0-α) if and only if pf+pg¯∈AH-α (respectively, AH,0-α).The results of Proposition21 and its corollary do not hold if ϕ is unbounded or ϕ(z) is not bounded away from 0 as z approaches the unit circle. For example, if ϕ(z)=f(z)=g(z)=(1-z)-α, then(112)fz+gz¯=2Re⁡1-z-α,which is in AH-α, but ϕf+ϕg¯∉AH-α. On the other hand, if we choose ϕ(z)=(1-z)α and f(z)=g(z)=(1-z)-2α, then ϕf+ϕg¯∈AH-α while, f+g¯∉AH-α.



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