# A Class of Unbounded Fourier Multipliers on the Unit Complex Ball

Authors: Pengtao Li; Jianhao Lv; Tao Qian
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We introduce a class of Fourier multiplier operatorsMb on n-complex unit sphere, where the symbol b∈Hs(Sω). We obtained the Sobolev boundedness of Mb. Our result implies that the operators Mb take a role of fractional differential operators on ∂𝔹.


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## Body

## 1. Introduction

In this paper, we introduce a class of unbounded holomorphic Fourier multipliersMb on n-complex unit sphere. We further study the boundedness of Mb on Sobolev spaces. Our results generalize the theory of Fourier multipliers on Lipschitz curves in ℂ to n-complex unit sphere 𝔹n. We refer the reader to Gaudry et al. [1], McIntosh and Qian [2], and Qian [3, 4] for further information on multipliers on Lipschitz curves.Our motivation originates from the following example on the unit sphere inℂn. The explicit formula of the Cauchy-Szegö kernel is as follows:
(1)H(z,ξ¯)=1ω2n-11(1-zξ¯′)n.
Let {pkv} denote the orthonormal system in the space of holomorphic functions in 𝔹n. The following result is well-known:
(2)H(z,ξ¯)=∑k=0∞∑v=1Nkpvk(z)pvk(ξ)¯,z∈𝔹n,ξ∈∂𝔹n.
See Theorem 1 and (16) below for details. Formally, (2) can be seen as the special case of (4) below. Let Sω be the sector defined as
(3)Sω={z∈ℂ:z≠0,|argz|<ω}.
Assume that (1)
bis holomorphic on Sω;(2)
b is bounded near the origin;(3)
|b(z)|≤C|z|s for |z|>1.We consider the function:(4)Hb(z,ξ¯)=∑k=1∞b(k)∑v=1Nkpvk(z)pvk(ξ)¯.
If b(z)≡1, then (4) becomes (2). For s=0, Cowling and Qian [5] introduced a class of bounded holomorphic multipliers on L2(∂𝔹n). In this paper, we consider the case s≠0. For this case, b is unbounded on {z:|z|>1}. We prove that if b∈Hs(Sω), then
(5)|Hb(z,ξ¯)|=Cμ′δ(ν,μ′)|1-zξ′¯|n+s.
See Theorem 5.In Section4, we introduce a class of Fourier multipliers Mb with b∈Hs(Sω),s≠0. Unlike the ones of Cowling and Qian [5], our multipliers b are unbounded on Sω. Take b(k)=ks. Plancherel's theorem implies that Mb is not bounded on L2(∂𝔹n). Hence for such Mb, we need to consider their boundedness on some function spaces with higher regularity. Let r,s∈[0,∞). We prove that if b∈Hs(Sω), Mb is bounded from Sobolev space Wp,r+s(∂𝔹n) to Sobolev space Wp,r(∂𝔹n), 1<p<∞. Our result implies that the operators Mb take a role of fractional differential operators on ∂𝔹n. See Theorem 11.The rest of this paper is organized as follows. In Section2, we state some basic preliminaries and notations which will be used in the sequel. In Section 3, we estimate the kernels generated by holomorphic multipliers b∈Hs(Sω). The Sobolev boundedness of the operators Mb is given in Section 4.Notations. U≈V represents that there is a constant c>0 such that c-1V≤U≤cV whose right inequality is also written as U≲V. Similarly, one writes V≳U for V≥cU.


## 2. Preliminaries and Notations

In this section, we state some preliminaries and notations and refer the reader to Gong [6], Hua [7], and Rudin [8] for further information. We use z as a general element of ℂn; that is, z=(z1,…,zn), zi∈ℂ, i=1,2,…,n, n≥2. Denote z¯=[z1¯,…,zn¯]. The notation z is considered to be a row vector. Denote by 𝔹n the open unit ball {z∈ℂn:|z|<1}, where |z|=(∑i=1n|zi|2)1/2. The unit sphere in ℂn is denoted by
(6)∂𝔹n=𝕊2n-1={z∈ℂn:|z|=1}.
The open ball centered at z with radius r will be denoted by B(z,r). A general element on ∂𝔹n is usually denoted by ξ. The constant ω2n-1 involved in the Cauchy-Szegö kernel is the surface area of ∂𝔹n and is equal to 2πn/Γ(n). For z,w∈ℂn, we use the notation zw′=∑k=1nzkwk. The theory developed in this paper is relevant to the radial Dirac operator
(7)D=∑k=1nzk∂∂zk.Now we state some basis knowledge of basic functions in the space of holomorphic function in𝔹n and some relevant function spaces on ∂𝔹n. We refer to Hua [7] for details. Let k be a nonnegative integer. We consider the column vector z[k] with components
(8)k!k1!⋯kn!z1k1⋯znkn,k1+⋯kn=k.
The dimension of z[k] is
(9)Nk=1k!n(n+1)·(n+k-1)=Cn+k-1k.Letdz and dσ(ξ) be the Lebesgue volume element of ℂn and the Lebesgue area element of ∂𝔹n, respectively. Define
(10)H1k=∫𝔹nz[k]′¯·z[k]dz,H2k=∫∂𝔹nξ[k]′¯·ξ[k]dσ(ξ).
It is easy to prove that H1k and H2k are positive definite Hermitian matrices of order Nk. There exists a matrix Γ such that
(11)Γ′¯·H1k·Γ=Λ,Γ′¯·H2k·Γ=I,
where Λ=[β1k,…,βnk] is a diagonal matrix and I is the identity matrix. Set
(12)z[k]=z[k]·Γ;ξ[k]=ξ[k]·Γ.
Denote by pvk(z) the components of the vectors z[k]. From (11), we can see that
(13)∫𝔹npvk(z)pμk(z)¯dz=δvμ·δkl·βvk,∫∂𝔹npvk(ξ)pμk(ξ)¯dσ(ξ)=δvμ·δkl.
The following theorem is well known.Theorem 1.
The system of functions(14){(βνk)-12pνk,k=0,1,2,…,ν=1,2,…,Nk}
is a complete orthonormal system in the space of holomorphic functions in 𝔹n. The system {pνk} is orthonormal, but it is not complete in the space of continuous functions on ∂𝔹n.The explicit formula of the Cauchy-Szegö kernel(15)H(z,ξ¯)=1ω2n-11(1-zξ¯′)n
on ∂𝔹n was first deduced in Hua [7] by using the system {pvk} and the relation
(16)H(z,ξ¯)=∑k=0∞∑v=1Nkpvk(z)pvk(ξ)¯,z∈𝔹n,ξ∈𝔹n.
For z,ω∈𝔹n∪∂𝔹n, the nonisotropic distance d(z,ω) is defined as
(17)d(z,ω)=|1-zω¯′|1/2.
It can be easily shown that d(·,·) is a metric on ∂𝔹n. For ξ∈∂𝔹n and ɛ>0, we define the ball corresponding to d(·,·) as
(18)S(ξ,ɛ)={η∈∂𝔹n,d(ξ,η)≤ɛ}.
The complement set of S(ξ,ɛ) in ∂𝔹n is denoted by Sc(ξ,ɛ).Set(19)𝒜={f:fisholomorphicinB(0,1+δ)forsomeδ>0}.
If f∈𝒜, then
(20)f(z)=∑k=0∞∑v=0Nkckvpvk(z),
where ckv are the Fourier coefficients of f:
(21)ckv=∫∂𝔹npvk(ξ)¯f(ξ)dσ(ξ),
and for any positive integer l, the series
(22)∑k=0∞kl∑v=0Nkckvpvk(z)
is uniformaly and absolutely convergent in any compact ball contained in B(0,1+δ) in which f is defined.Denote by𝒰 the unitary group of ℂn consisting of all unitary operators on the Hilbert space ℂn under the complex inner product 〈z,w〉=zw¯′. These are the linear operators U that preserve inner products:
(23)〈Uz,Uw〉=〈z,w〉.
Clearly, 𝒰 is a compact subset of O(2n). It is easy to verify that 𝒜 is invariant under U∈𝒰. If f∈𝒜, then f is defined by its values on ∂𝔹n. In Section 3, we treat f|∂𝔹n as identical to f∈𝒜.


## 3. The Kernel Generated by Holomorphic Multipliers

Set(24)Sω={z∈ℂz≠0and|argz|<ω},Sω(π)={z∈ℂz≠0,|Re(z)|≤π,|arg(±z)|<ω},Wω(π)={z∈ℂ∣z≠0,|Re(z)|≤π,Im⁡(z)>0}⋃Sω(π),Hω={z∈ℂz=eiω,ω∈Wω(π)}.The following function space is relevant.Definition 2.
Let-1<s<∞. Hs(Sω) is defined as the set of all holomorphic functions in Sω such that (a)
b is bounded for |z|≤1;(b)
|b(z)|≤Cμ|z|s, z∈Sμ, 0<μ<ω.Remark 3.
The classesHs(Sω) are generalizations of H∞(Sω) which is introduced by McIntosh and his collaborators. We refer to Li et al. [9], McIntosh [10], McIntosh and Qian [2], Qian [11], and the reference therein for further information on H∞(Sω).Let(25)φb(z)=∑k=1∞b(k)zk.Lemma 4.
Letb∈Hs(Sω), -1<s<∞. Then φb can be holomorphically extended to Hω. Moreover, for 0<μ<μ′<ω and l=0,1,2,…,
(26)|(zddz)lφb(z)|≲Cμ′l!δl(μ,μ′)|1-z|l+1+s,z∈Hμ,
where δ(μ,μ′)=min⁡{1/2,tan(μ,μ′}; Cμ′ are the constants in Definition 2.Proof.
Let(27)Vω={z∈ℂ:Im⁡(z)>0}⋃‍Sω⋃‍(-Sω),Wω=Vω⋂{z∈ℂ:-π≤Rez≤π}
and ρθ is the ray rexp⁡(iθ), 0<r<∞, where θ is chosen so that ρθ⊆Sω. Define
(28)Ψb(z)=12π∫ρ(θ)exp⁡(iξz)b(ξ)dξ,z∈Vω,
where exp⁡(izξ) is exponentially decaying as ξ→∞ along ρθ. Then we get
(29)||z|1+sΨb(z)|=|12π∫ρ(θ)exp⁡(iξz)|z|1+sb(ξ)dz|≲Cμ′2π∫0∞exp⁡(-r|z|sin(θ+argz))(r|z|)sd(r|z|)s≲Cμ′,
which implies |Ψb(z)|≲1/|z|1+s. Define
(30)ψb(z)=2π∑n=-∞∞‍Ψb(z+2nπ),z∈⋃n=-∞∞(2nπ+Wω).
It is easy to see that ψb is holomorphically and 2π-periodically defined in the described region, and |ψb(z)|≲1/|z|1+s. Let
(31)φb(z)=ψb(log⁡zi).
For z∈exp⁡(iSω), we write z=eiu, where u∈Sω. Then sin⁡(|u|/2)≲|u|/2. This implies that 2-2cos⁡|u|≲|u|2 and |1-ei|u||≲|u|. Therefore, (29) gives
(32)|φb(z)|≲Cμ′|log⁡z|1+s≲Cμ′|log⁡|z||1+s≲Cμ′|1-z|1+s.
Take the ball(33)B(z,r)={ξ:|z-ξ|<δ(μ,μ′)|1-z|}.
Applying Cauchy's integral formula, we obtain
(34)φb(l)(z)=l!2πi∫∂B(z,r)φ(η)(η-z)1+ldη.
For any η∈∂B(z,r), we have |η-z|≥(1-δ(μ,μ′))|1-z|. Then we have
(35)|φb(l)(z)|≲l!∥b∥Hs(Sωc)δl(μ,μ′)|1-z|l|∫∂B(z,r)1|1-η|1+sdη|≲l!δl(μ,μ′)|1-z|l+1+s.Theorem 5.
Letb∈Hs(Sω) and
(36)Hb(z,ξ¯)=∑k=1∞b(k)∑v=1Nkpvk(z)pvk(ξ)¯,z∈𝔹n,ξ∈∂𝔹n.
Then
(37)Hb(z,ξ¯)=1(n-1)!ω2n-1(rn-1φb(r))(n-1)|r=zξ¯′
is holomorphically defined for z∈𝔹n, ξ∈∂𝔹n such that zξ¯′∈Hω, where φb is the function defined in Lemma 4, Moreover, for 0<μ<μ′<ω and l=0,1,2,…,
(38)|DzlHb(z,ξ¯)|≲Cμ′l!δl(μ,μ′)|1-zξ¯′|n+l+s,zξ¯′∈Hμ,
where δ(μ,μ′)=min⁡{1/2,tan(μ′-μ)}, Cμ′ are the constant in the definition of the function space Hs(Sω).Proof.
Recall that(39)φb(z)=∑k=1∞b(k)zk;rn-1φb(r)=∑k=1∞b(k)rn+k-1.
Then we have
(40)1(n-1)!(rn-1φb(r))(n-1)=1(n-1)!∑k=1∞b(k)(n+k-1)(n+k-2)⋯(k+1)rk=∑k=1∞b(k)rk(n+k-1)!(n-1)!k!=∑k=1∞(n+k-1)(n+k-2)(n+1)nk!b(k)rk.
Therefore,
(41)1(n-1)!(rn-1φb(r))(n-1)|r=zξ¯′=∑k=1∞b(k)(n+k-1)(n+k-2)(n+1)nk!(zξ¯′)k=ω2n-1∑k=1∞b(k)∑v=1Nkpvk(z)pvk(ξ)¯=ω2n-1Hb(z,ξ¯).By [4, Theorem 3], we could obtain the following result.Theorem 6.
Lets be a negative integer. If b∈Hs(Sω,±),
(42)Hb(z,ξ)=∑k=1∞b(k)∑v=1Nkpvk(z)pμl(ξ),z∈𝔹,ξ∈∂𝔹n,
then
(43)|DzlHb(z,ξ¯)|≲Cμl![|ln⁡|1-zξ¯′||+1]δl(μ,μ′)|1-zξ¯′|n+l+s.Proof.
The proof is similar to Theorem5. we omit it.


## 4. Sobolev Spaces and Unbounded Fourier Multipliers

### 4.1. Integral Representation of Multipliers

Givenb∈Hs(Sω), we define the Fourier multiplier operator Mb:𝒜→𝒜 by
(44)Mb(f)(ξ)=∑k=1∞b(k)∑v=0Nkckvpvk(ξ),ξ∈∂𝔹n,
where {ckv} are the Fourier coefficients of the test function f∈𝒜.For the above operatorMb, the Plemelj type formula holds.Theorem 7.
Letb∈Hs(Sω),s>0. Take b1(z)=z-s1b(z), where s1=[s]+1. Operator Mb has a singular integral expression. For f∈𝒜,
(45)Mb(f)(ξ)=lim⁡ɛ→0[∫Sc(ξ,ɛ)Hb1(ξ,η¯)Dηs1f(η)dσ(η)+(Dzs1f)(ξ)∫Sc(ξ,ɛ)Hb1(ξ,η¯)dσ(η)],
where ∫S(ξ,ɛ)‍Hb1(ξ,η¯)dσ(η) is a bounded function of ξ∈∂𝔹n and ɛ.Proof.
Let(46)Mb(f)(ρξ)=∑k=1∞b(k)∑v=1Nkckvpvk(ρξ),ξ∈∂𝔹n,
where
(47)ckv=∫∂Bpvk(η)¯f(η)dσ(η).
We can see that
(48)Dzz[l]=l!l1!l2!⋯ln!∑k=1nzk∂∂zk(z1l1z2l2⋯znln)=l!l1!l2!⋯ln!∑k=1nzklkz1l1z2l2⋯zk-1lk-1zklk-1zk+1lk+1⋯znln=l!l1!l2!⋯ln!(∑k=1nlk)z1l1z2l2⋯znln=lz[l],
which implies that Dzpvk=kpvk. Then, we have
(49)Mb(f)(ρξ)=∑k=1∞b(k)∑v=1Nk∫∂Bpvk(ρξ)pvk(η)¯f(η)dσ(η)=∑k=1∞b(k)1ks1∑v=1Nk∫∂Bpvk(ρξ)ks1pvk(η)¯f(η)dσ(η)=∑k=1∞b(k)1ks1∑v=1Nk∫∂Bpvk(ρξ)Dηs1pvk(η)¯f(η)dσ(η).
By integration by parts,
(50)Mb(f)(ρξ)=∑k=1∞b(k)1ks1∑v=1Nk∫∂Bpvk(ρξ)pvk(η)¯(Dηs1f)(η)dσ(η)=∑k=1∞b1(k)∑v=1Nk∫∂Bpvk(ρξ)pvk(η)¯(Dηs1f)(η)dσ(η).
For any ɛ>0, we have
(51)Mb(f)(ρξ)=∫Sc(ξ,ɛ)Hb1(ρξ,η¯)Dηs1f(η)dσ(η)+∫S(ξ,ɛ)Hb1(ρξ,η¯)(-Dξs1f(ξ)+Dηs1f(η))dσ(η)+Dξs1f(ξ)∫S(ξ,ɛ)Hb1(ρξ,η¯)dσ(η)=:I1(ρ,ɛ)+I2(ρ,ɛ)+Dξs1f(ξ)I3(ρ,ɛ),
where
(52)I1(ρ,ɛ)=∫Sc(ξ,ɛ)Hb1(ρξ,η¯)Dηs1f(η)dσ(η),I2(ρ,ɛ)=∫S(ξ,ɛ)Hb1(ρξ,η¯)(-Dξs1f(ξ)+Dηs1f(η))dσ(η),I3(ρ,ɛ)=∫S(ξ,ɛ)Hb1(ρξ,η¯)dσ(η).
Forρ→1-0, we have
(53)lim⁡ρ→1-0I1(ρ,ɛ)=lim⁡ρ→1-0∫Sc(ξ,ɛ)Hb1(ρξ,η¯)Dηs1f(η)dσ(η)=∫Sc(ξ,ɛ)Hb1(ξ,η¯)Dηs1f(η)dσ(η).
Now we consider I2(ρ,ɛ). Let ξ=[1,0,…,0]. For η∈∂𝔹n, write
(54)η1=reiθ,η2=v2,η3=v3,…,ηn=vn;v=[v2,v3,…,vn].
For such η∈∂𝔹n, vv¯′=1-r2. Without loss of generality, assume ξ=1. We get
(55)|1-ξη¯′|1/2=|1-reiθ|1/2=[(1-rcos⁡θ)2+(rsinθ)2]1/4≤ɛ,
which implies that
(56)cos⁡θ≥1+r2-ɛ42r.
The above estimate implies
(57)S(ξ,ɛ)={η∣vv¯′=1-r2,cos⁡θ≥1+r2-ɛ42r}.
Since
(58)1+r2-ɛ42r≤cos⁡θ≤1,
we obtain 1-r≤ɛ2 and then
(59)vv¯′=1-r2≤1-(1-ɛ2)2=2ɛ2-ɛ4.
Denote
(60)a=a(r,ɛ)=arccos(1+r2-ɛ42r).
Since (1-r)2≤ɛ4 and 1-y=O(arccos2y), we get a=O(ɛ2). It is easy to see
(61)|ξ-η|2=|1-reiθ|2+∑k=2n|vk|2=(1+r2-2rcos⁡θ)+(1-r2)=2-2rcos⁡θ,d4(ξ,η)=1+r2-2rcos⁡θ=(2-2rcos⁡θ)-(1-r2)=|ξ-η|2-(1+r)(1-r),
that is, d2(ξ,η)≤|ξ-η|. Because
(62)d2(ξ,η)=[1+r2-2rcos⁡θ]1/2≥1-r,
then we have 1-r≤d2(ξ,η), so
(63)|ξ-η|2≤d4(ξ,η)+(1+r)d2(ξ,η).
Since d2(ξ,η)≤2, then
(64)|ξ-η|2≤2d2(ξ,η)+2d2(ξ,η)=4d2(ξ,η),
that is
(65)|ξ-η|≤2d(ξ,η).
Since f∈𝒜, we have
(66)|f(ξ)-f(η)|≲|ξ-η|≲d(ξ,η).
For ρ∈(0,1),
(67)|I2(ρ,ɛ)|≲∫S(ξ,ɛ)|Hb1(ρξ,η¯)||f(ξ)-f(η)|dσ(η)≲∫S(ξ,ɛ)d(ξ,η)|1-ξη¯′|ndσ(η)≲∫vv¯′≤2ɛ2-ɛ4∫-aa1|1-reiθ|n-1/2dθdv.
For n=2,
(68)12a∫-aa1|1-reiθ|2-1/2dθ≤(12a∫-aa1|1-reiθ|2dθ)3/4≤(12a∫-ππ1|1-reiθ|2dθ)3/4≤(12a)3/41(1-r2)3/4.
Then we get
(69)|I2(ρ,ɛ)|≲∫vv¯′≤2ɛ2-ɛ4a1/41(1-r2)3/4dv≲ɛ1/2∫vv¯′≤2ɛ2-ɛ41(vv¯′)3/4dv=ɛ1/2∫02ɛ2-ɛ4tt3/2dt≲ɛ⟶0.
For n>2, we have
(70)∫-aa1|1-reiθ|n-1/2dθ≲∫-aa|1-r2|n-1/2-2|1-reiθ|n-1/21|1-r2|n-1/2-2dθ≲1|1-r2|n-1/2-1∫-ππ1|1-reiθ|2dθ≲1|1-r2|n-1/2-1.
Then, we get
(71)|I2(ρ,ɛ)|≲∫02ɛ2-ɛ4t2n-31t2n-3dt≲2ɛ2⟶0.
Now we prove if ρ→1-0, I3(ρ,ɛ) has a limit uniformly bounded for ɛ near 0. Integrating as before, we have
(72)I3(ρ,ɛ)=∫S(ξ,ɛ)Hb1(ρξ,η¯)dσ(η)=∫vv¯′≤2ɛ2-ɛ4∫-aa(tn-1φb1(t))(n-1)|t=ρreiθdθdv.
Let s=ρreiθ. Then ds=isdθ. We get
(73)I3(ρ,ɛ)=-i∫vv¯′≤2ɛ2-ɛ4∫ρre-iaρreia(sn-1φb1(s))(n-1)dsdv.
By integration by parts, the inside integral with respect to the variable t becomes
(74)∫-aa(tn-1φb1(t))(n-1)|t=ρreiθdθ=[∑k=1n-1(k-1)!(tn-1φb1(t))(n-k-1)tk]|ρre-iaρreia+(n-1)!∫ρre-iaρreiaφb1(t)tdt=∑k=1n-1[Jk(t)]ρre-iaρreia+L(r,a).
We first estimate Jk,
(75)∫vv¯′≤2ɛ2-ɛ4Jk(ρre±ia)dv≲∫vv¯′≤2ɛ2-ɛ4(k-1)!(ρre±ia)k(ρre±ia)k1|1-ρre±ia|n-kdv≲∫vv¯′≤2ɛ2-ɛ41|1-ρre±ia|n-kdv.
Because|1-ρre±ia|2=1+ρ2r2-2ρrcos⁡a,
(76)|1-ρre±ia|2-|1-re±ia|2=ρ2r2-2ρrcos⁡a-(r2-2rcos⁡a)=r2(ρ2-1)+2rcos⁡a(1-ρ).
Since cos⁡a=(1+r2-ɛ4)/2r, we have
(77)|1-ρre±ia|2-|1-re±ia|2=r2(ρ2-1)+(1+r2-ɛ4)(1-ρ)=(1-ρ)[1+r2-ɛ4-(1+ρ)r2]=(1-ρ)(1-ρr2-ɛ4)>0.
So
(78)|1-ρre±ia|≥|1-re±ia|=ɛ2.
For k, when ɛ→0, we obtain
(79)∫vv¯′≤2ɛ2-ɛ4Jk(ρre±ia)dv≲1ɛ2n-2k∫vv¯′≤2ɛ2-ɛ4dv≲1ɛ2n-2k∫02ɛ2-ɛ4t2n-3dt≲ɛ2n-2ɛ2n-2k≲1.
On the other hand,
(80)(n-1)!∫ρre-iaρreiaφb1(t)tdt=i(n-1)!∫-aaφb1(t)|t=ρreiθdθ≲1,(whenρ⟶0)
that implies
(81)∫vv¯′≤2ɛ2-ɛ4L(ρr,a)dv.


### 4.2. Sobolev Spaces on∂𝔹n via Fourier Multipliers

Sobolev spaces on then-complex unit sphere ∂𝔹n are defined as follows. We define the fractional integral operator ℐs on ∂𝔹n as follows. Let
(82)f(z)=∑k=0∞∑v=0Nkckvpvk(z).
For -∞<s<∞, the operator ℐs is defined by
(83)ℐsf(z)=∑k=0∞∑v=0Nkksckvpvk(z).Fors∈ℤ+, we can see that the operators ℐs become the ordinary differential operators with higher orders.Theorem 8.
Lets∈ℤ+. Dzs=ℐs on L2(∂𝔹n).Proof.
Without loss of generality, we assume thatf∈𝒜. Then
(84)f(z)=∑k=0∞∑v=0Nkckvpvk(z),
where ckv are the Fourier coefficients of f:
(85)ckv=∫∂𝔹npvk(ξ)¯f(ξ)dσ(ξ).
So
(86)Dzsf(z)=∑k=0∞∑v=0Nk∫∂𝔹npvk(ξ)¯f(ξ)dσ(ξ)Dzs(pvk)(z)=∑k=0∞ks∑v=0Nk∫∂𝔹npvk(ξ)¯f(ξ)dσ(ξ)pvk(z).Definition 9.
Lets∈[0,+∞). The Sobolev norm ∥·∥W2,s(∂𝔹n) on ∂𝔹n is defined as
(87)∥f∥W2,s(∂𝔹n)=:∥ℐsf∥2<∞.
The Sobolev spaces on ∂𝔹n are defined as the closure of 𝒜 under the norm ∥·∥W2,s(∂𝔹n), that is, W2,s(∂𝔹n)=𝒜¯∥·∥W2,s(∂𝔹n).Remark 10.
By the Plancherel theorem,f∈W2,s(∂𝔹n) if and only if
(88)(∑k=1∞k2s∑v=0Nk|ckv|2)1/2<∞.Now we consider the Sobolev boundedness ofMb.Theorem 11.
Givenr,s∈[0,+∞) and b∈Hs(Sω). The Fourier multiplier operator Mb is bounded from W2,r+s(∂𝔹n) to W2,r(∂𝔹n).Proof.
Write(89)ℐsf(z)=∑k=0∞∑v=0Nkckvspvk(z).
By the orthogonality of {pvk}, we can see that ckvs=ksckv. Let b(z)=z-sb(z). Because b∈Hs(Sω), we can see that b1∈H∞(Sω). This implies that
(90)ℐr(Mb(f))(ξ)=∑k=1∞b(k)kr∑v=0Nkckvpvk(ξ)=∑k=1∞b1(k)kr+s∑v=0Nkckvpvk(ξ)=Mb1(ℐr+sf)(ξ).
Finally, by [5, Theorem 3], we can see that
(91)∥Mb(f)∥W2,r=∥ℐr(Mb(f))∥2=∥Mb1(ℐr+sf)∥2≤C∥ℐr+sf∥2.
This completes the proof of Theorem 11.


## 4.1. Integral Representation of Multipliers

Givenb∈Hs(Sω), we define the Fourier multiplier operator Mb:𝒜→𝒜 by
(44)Mb(f)(ξ)=∑k=1∞b(k)∑v=0Nkckvpvk(ξ),ξ∈∂𝔹n,
where {ckv} are the Fourier coefficients of the test function f∈𝒜.For the above operatorMb, the Plemelj type formula holds.Theorem 7.
Letb∈Hs(Sω),s>0. Take b1(z)=z-s1b(z), where s1=[s]+1. Operator Mb has a singular integral expression. For f∈𝒜,
(45)Mb(f)(ξ)=lim⁡ɛ→0[∫Sc(ξ,ɛ)Hb1(ξ,η¯)Dηs1f(η)dσ(η)+(Dzs1f)(ξ)∫Sc(ξ,ɛ)Hb1(ξ,η¯)dσ(η)],
where ∫S(ξ,ɛ)‍Hb1(ξ,η¯)dσ(η) is a bounded function of ξ∈∂𝔹n and ɛ.Proof.
Let(46)Mb(f)(ρξ)=∑k=1∞b(k)∑v=1Nkckvpvk(ρξ),ξ∈∂𝔹n,
where
(47)ckv=∫∂Bpvk(η)¯f(η)dσ(η).
We can see that
(48)Dzz[l]=l!l1!l2!⋯ln!∑k=1nzk∂∂zk(z1l1z2l2⋯znln)=l!l1!l2!⋯ln!∑k=1nzklkz1l1z2l2⋯zk-1lk-1zklk-1zk+1lk+1⋯znln=l!l1!l2!⋯ln!(∑k=1nlk)z1l1z2l2⋯znln=lz[l],
which implies that Dzpvk=kpvk. Then, we have
(49)Mb(f)(ρξ)=∑k=1∞b(k)∑v=1Nk∫∂Bpvk(ρξ)pvk(η)¯f(η)dσ(η)=∑k=1∞b(k)1ks1∑v=1Nk∫∂Bpvk(ρξ)ks1pvk(η)¯f(η)dσ(η)=∑k=1∞b(k)1ks1∑v=1Nk∫∂Bpvk(ρξ)Dηs1pvk(η)¯f(η)dσ(η).
By integration by parts,
(50)Mb(f)(ρξ)=∑k=1∞b(k)1ks1∑v=1Nk∫∂Bpvk(ρξ)pvk(η)¯(Dηs1f)(η)dσ(η)=∑k=1∞b1(k)∑v=1Nk∫∂Bpvk(ρξ)pvk(η)¯(Dηs1f)(η)dσ(η).
For any ɛ>0, we have
(51)Mb(f)(ρξ)=∫Sc(ξ,ɛ)Hb1(ρξ,η¯)Dηs1f(η)dσ(η)+∫S(ξ,ɛ)Hb1(ρξ,η¯)(-Dξs1f(ξ)+Dηs1f(η))dσ(η)+Dξs1f(ξ)∫S(ξ,ɛ)Hb1(ρξ,η¯)dσ(η)=:I1(ρ,ɛ)+I2(ρ,ɛ)+Dξs1f(ξ)I3(ρ,ɛ),
where
(52)I1(ρ,ɛ)=∫Sc(ξ,ɛ)Hb1(ρξ,η¯)Dηs1f(η)dσ(η),I2(ρ,ɛ)=∫S(ξ,ɛ)Hb1(ρξ,η¯)(-Dξs1f(ξ)+Dηs1f(η))dσ(η),I3(ρ,ɛ)=∫S(ξ,ɛ)Hb1(ρξ,η¯)dσ(η).
Forρ→1-0, we have
(53)lim⁡ρ→1-0I1(ρ,ɛ)=lim⁡ρ→1-0∫Sc(ξ,ɛ)Hb1(ρξ,η¯)Dηs1f(η)dσ(η)=∫Sc(ξ,ɛ)Hb1(ξ,η¯)Dηs1f(η)dσ(η).
Now we consider I2(ρ,ɛ). Let ξ=[1,0,…,0]. For η∈∂𝔹n, write
(54)η1=reiθ,η2=v2,η3=v3,…,ηn=vn;v=[v2,v3,…,vn].
For such η∈∂𝔹n, vv¯′=1-r2. Without loss of generality, assume ξ=1. We get
(55)|1-ξη¯′|1/2=|1-reiθ|1/2=[(1-rcos⁡θ)2+(rsinθ)2]1/4≤ɛ,
which implies that
(56)cos⁡θ≥1+r2-ɛ42r.
The above estimate implies
(57)S(ξ,ɛ)={η∣vv¯′=1-r2,cos⁡θ≥1+r2-ɛ42r}.
Since
(58)1+r2-ɛ42r≤cos⁡θ≤1,
we obtain 1-r≤ɛ2 and then
(59)vv¯′=1-r2≤1-(1-ɛ2)2=2ɛ2-ɛ4.
Denote
(60)a=a(r,ɛ)=arccos(1+r2-ɛ42r).
Since (1-r)2≤ɛ4 and 1-y=O(arccos2y), we get a=O(ɛ2). It is easy to see
(61)|ξ-η|2=|1-reiθ|2+∑k=2n|vk|2=(1+r2-2rcos⁡θ)+(1-r2)=2-2rcos⁡θ,d4(ξ,η)=1+r2-2rcos⁡θ=(2-2rcos⁡θ)-(1-r2)=|ξ-η|2-(1+r)(1-r),
that is, d2(ξ,η)≤|ξ-η|. Because
(62)d2(ξ,η)=[1+r2-2rcos⁡θ]1/2≥1-r,
then we have 1-r≤d2(ξ,η), so
(63)|ξ-η|2≤d4(ξ,η)+(1+r)d2(ξ,η).
Since d2(ξ,η)≤2, then
(64)|ξ-η|2≤2d2(ξ,η)+2d2(ξ,η)=4d2(ξ,η),
that is
(65)|ξ-η|≤2d(ξ,η).
Since f∈𝒜, we have
(66)|f(ξ)-f(η)|≲|ξ-η|≲d(ξ,η).
For ρ∈(0,1),
(67)|I2(ρ,ɛ)|≲∫S(ξ,ɛ)|Hb1(ρξ,η¯)||f(ξ)-f(η)|dσ(η)≲∫S(ξ,ɛ)d(ξ,η)|1-ξη¯′|ndσ(η)≲∫vv¯′≤2ɛ2-ɛ4∫-aa1|1-reiθ|n-1/2dθdv.
For n=2,
(68)12a∫-aa1|1-reiθ|2-1/2dθ≤(12a∫-aa1|1-reiθ|2dθ)3/4≤(12a∫-ππ1|1-reiθ|2dθ)3/4≤(12a)3/41(1-r2)3/4.
Then we get
(69)|I2(ρ,ɛ)|≲∫vv¯′≤2ɛ2-ɛ4a1/41(1-r2)3/4dv≲ɛ1/2∫vv¯′≤2ɛ2-ɛ41(vv¯′)3/4dv=ɛ1/2∫02ɛ2-ɛ4tt3/2dt≲ɛ⟶0.
For n>2, we have
(70)∫-aa1|1-reiθ|n-1/2dθ≲∫-aa|1-r2|n-1/2-2|1-reiθ|n-1/21|1-r2|n-1/2-2dθ≲1|1-r2|n-1/2-1∫-ππ1|1-reiθ|2dθ≲1|1-r2|n-1/2-1.
Then, we get
(71)|I2(ρ,ɛ)|≲∫02ɛ2-ɛ4t2n-31t2n-3dt≲2ɛ2⟶0.
Now we prove if ρ→1-0, I3(ρ,ɛ) has a limit uniformly bounded for ɛ near 0. Integrating as before, we have
(72)I3(ρ,ɛ)=∫S(ξ,ɛ)Hb1(ρξ,η¯)dσ(η)=∫vv¯′≤2ɛ2-ɛ4∫-aa(tn-1φb1(t))(n-1)|t=ρreiθdθdv.
Let s=ρreiθ. Then ds=isdθ. We get
(73)I3(ρ,ɛ)=-i∫vv¯′≤2ɛ2-ɛ4∫ρre-iaρreia(sn-1φb1(s))(n-1)dsdv.
By integration by parts, the inside integral with respect to the variable t becomes
(74)∫-aa(tn-1φb1(t))(n-1)|t=ρreiθdθ=[∑k=1n-1(k-1)!(tn-1φb1(t))(n-k-1)tk]|ρre-iaρreia+(n-1)!∫ρre-iaρreiaφb1(t)tdt=∑k=1n-1[Jk(t)]ρre-iaρreia+L(r,a).
We first estimate Jk,
(75)∫vv¯′≤2ɛ2-ɛ4Jk(ρre±ia)dv≲∫vv¯′≤2ɛ2-ɛ4(k-1)!(ρre±ia)k(ρre±ia)k1|1-ρre±ia|n-kdv≲∫vv¯′≤2ɛ2-ɛ41|1-ρre±ia|n-kdv.
Because|1-ρre±ia|2=1+ρ2r2-2ρrcos⁡a,
(76)|1-ρre±ia|2-|1-re±ia|2=ρ2r2-2ρrcos⁡a-(r2-2rcos⁡a)=r2(ρ2-1)+2rcos⁡a(1-ρ).
Since cos⁡a=(1+r2-ɛ4)/2r, we have
(77)|1-ρre±ia|2-|1-re±ia|2=r2(ρ2-1)+(1+r2-ɛ4)(1-ρ)=(1-ρ)[1+r2-ɛ4-(1+ρ)r2]=(1-ρ)(1-ρr2-ɛ4)>0.
So
(78)|1-ρre±ia|≥|1-re±ia|=ɛ2.
For k, when ɛ→0, we obtain
(79)∫vv¯′≤2ɛ2-ɛ4Jk(ρre±ia)dv≲1ɛ2n-2k∫vv¯′≤2ɛ2-ɛ4dv≲1ɛ2n-2k∫02ɛ2-ɛ4t2n-3dt≲ɛ2n-2ɛ2n-2k≲1.
On the other hand,
(80)(n-1)!∫ρre-iaρreiaφb1(t)tdt=i(n-1)!∫-aaφb1(t)|t=ρreiθdθ≲1,(whenρ⟶0)
that implies
(81)∫vv¯′≤2ɛ2-ɛ4L(ρr,a)dv.


## 4.2. Sobolev Spaces on∂𝔹n via Fourier Multipliers

Sobolev spaces on then-complex unit sphere ∂𝔹n are defined as follows. We define the fractional integral operator ℐs on ∂𝔹n as follows. Let
(82)f(z)=∑k=0∞∑v=0Nkckvpvk(z).
For -∞<s<∞, the operator ℐs is defined by
(83)ℐsf(z)=∑k=0∞∑v=0Nkksckvpvk(z).Fors∈ℤ+, we can see that the operators ℐs become the ordinary differential operators with higher orders.Theorem 8.
Lets∈ℤ+. Dzs=ℐs on L2(∂𝔹n).Proof.
Without loss of generality, we assume thatf∈𝒜. Then
(84)f(z)=∑k=0∞∑v=0Nkckvpvk(z),
where ckv are the Fourier coefficients of f:
(85)ckv=∫∂𝔹npvk(ξ)¯f(ξ)dσ(ξ).
So
(86)Dzsf(z)=∑k=0∞∑v=0Nk∫∂𝔹npvk(ξ)¯f(ξ)dσ(ξ)Dzs(pvk)(z)=∑k=0∞ks∑v=0Nk∫∂𝔹npvk(ξ)¯f(ξ)dσ(ξ)pvk(z).Definition 9.
Lets∈[0,+∞). The Sobolev norm ∥·∥W2,s(∂𝔹n) on ∂𝔹n is defined as
(87)∥f∥W2,s(∂𝔹n)=:∥ℐsf∥2<∞.
The Sobolev spaces on ∂𝔹n are defined as the closure of 𝒜 under the norm ∥·∥W2,s(∂𝔹n), that is, W2,s(∂𝔹n)=𝒜¯∥·∥W2,s(∂𝔹n).Remark 10.
By the Plancherel theorem,f∈W2,s(∂𝔹n) if and only if
(88)(∑k=1∞k2s∑v=0Nk|ckv|2)1/2<∞.Now we consider the Sobolev boundedness ofMb.Theorem 11.
Givenr,s∈[0,+∞) and b∈Hs(Sω). The Fourier multiplier operator Mb is bounded from W2,r+s(∂𝔹n) to W2,r(∂𝔹n).Proof.
Write(89)ℐsf(z)=∑k=0∞∑v=0Nkckvspvk(z).
By the orthogonality of {pvk}, we can see that ckvs=ksckv. Let b(z)=z-sb(z). Because b∈Hs(Sω), we can see that b1∈H∞(Sω). This implies that
(90)ℐr(Mb(f))(ξ)=∑k=1∞b(k)kr∑v=0Nkckvpvk(ξ)=∑k=1∞b1(k)kr+s∑v=0Nkckvpvk(ξ)=Mb1(ℐr+sf)(ξ).
Finally, by [5, Theorem 3], we can see that
(91)∥Mb(f)∥W2,r=∥ℐr(Mb(f))∥2=∥Mb1(ℐr+sf)∥2≤C∥ℐr+sf∥2.
This completes the proof of Theorem 11.



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