# Weighted Iterated Radial Composition Operators between Some Spaces of Holomorphic Functions on the Unit Ball

Authors: Stevo Stević
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

The boundedness and compactness of weighted iterated radial
composition operators from the mixed-norm space to the weighted-type space
and the little weighted-type space on the unit ball are characterized here. We also calculate the Hilbert-Schmidt norm of the operator on the weighted Bergman-Hilbert space as well as on the HardyH2 space.


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## Body

## 1. Introduction

Letz=(z1,…,zn) and w=(w1,…,wn) be points in ℂn, 〈z,w〉=∑k=1nzkw¯k, and |z|=〈z,z〉. Let 𝔹={z∈ℂn:|z|<1} be the open unit ball in ℂn, ∂𝔹 its boundary, and H(𝔹) the class of all holomorphic functions on 𝔹.For anf∈H(𝔹) with the Taylor expansion f(z)=∑|β|≥0aβzβ, let (1.1)Rf(z)=∑|β|≥0|β|aβzβ
be the radial derivative of f, where β=(β1,β2,…,βn) is a multi-index, |β|=β1+⋯+βn and zβ=z1β1···znβn [1]. It is easy to see that (1.2)Rf(z)=〈∇f(z),z¯〉,
where ∇f is the complex gradient of function f.The iterated radial derivative operatorℜmf is defined inductively by (1.3)Rmf=R(Rm-1f),m∈N∖{1}.A positive continuous functionν on the interval [0,1) is called normal [2] if there are δ∈[0,1) and τ and t, 0<τ<t such that  (1.4)ν(r)(1-r)τisdecreasingon[δ,1),lim⁡r→1ν(r)(1-r)τ=0,ν(r)(1-r)tisincreasingon[δ,1),lim⁡r→1ν(r)(1-r)t=∞.
If we say that a function ν:𝔹→[0,∞) is normal, we also assume that it is radial, that is, ν(z)=ν(|z|), z∈𝔹.Strictly positive continuous functions on𝔹 are called weights.The weighted-type spaceHμ∞(𝔹)=Hμ∞ consists of all f∈H(𝔹) such that (1.5)‖f‖Hμ∞:=sup⁡z∈Bμ(z)|f(z)|<∞,
where μ is a weight (see, e.g., [3, 4] as well as [5] for a related class of spaces).The little weighted-type spaceHμ,0∞(𝔹)=Hμ,0∞ is a subspace of Hμ∞ consisting of all f∈H(𝔹) such that (1.6)lim⁡|z|→1μ(z)|f(z)|=0.For0<p, q<∞, and ϕ normal, the mixed-norm space H(p,q,ϕ)(𝔹)=H(p,q,ϕ) consists of all functions f∈H(𝔹) such that(1.7)‖f‖H(p,q,ϕ)=(∫01Mqp(f,r)ϕp(r)1-rdr)1/p<∞,
where (1.8)Mq(f,r)=(∫∂B|f(rζ)|qdσ(ζ))1/q,
and dσ is the normalized surface measure on ∂𝔹. For p=q, ϕ(r)=(1-r2)(α+1)/p, and α>-1, the space is equivalent with the weighted Bergman space Aαp(𝔹)=Aαp, which is defined as the class of all f∈H(𝔹) such that (1.9)‖f‖Aαpp:=∫B|f(z)|p(1-|z|2)αdV(z)<∞,
where dV(z) is the Lebesgue volume measure on 𝔹. Some facts on mixed-norm spaces in various domains in ℂn can be found, for example, in [6–8] (see also the references therein).For0<p<∞ the Hardy space Hp(𝔹)=Hp consists of all f∈H(𝔹) such that (1.10)‖f‖Hp:=sup⁡0<r<1(∫∂B|f(rζ)|pdσ(ζ))1/p<∞.
For p=2 the Hardy and the weighted Bergman space are Hilbert.Letφ be a holomorphic self-map of 𝔹, u∈H(𝔹), and m∈ℕ0. For f∈H(𝔹), the weighted iterated radial composition operator is defined by (1.11)Ru,φm(f)(z)=u(z)Rmf(φ(z)),z∈B.
Note that the operator is the composition of the multiplication, composition and the iterated radial operator, that is (1.12)Ru,φm=Mu∘Cφ∘Rm.
This is one of the product operators suggested by this author to be investigated at numerous talks (e.g., in [9]). Note that for m=0 the operator ℜu,φm becomes the weighted composition operator (see, e.g., [4, 8, 10]). It is of interest to provide function-theoretic characterizations for when φ and u induce bounded or compact weighted iterated radial composition operators on spaces of holomorphic functions. Studying products of some concrete linear operators on spaces of analytic functions attracted recently some attention see, for example, [11–32] as well as the related references therein. Some operators on mixed-norm spaces have been studied, for example, in [8, 10, 11, 16, 25, 26, 29, 33] (see also the references therein).Here we study the boundedness and compactness of weighted iterated radial composition operators from mixed-norm spaces to weighted-type spaces on the unit ball for the casem∈ℕ. We also calculate the Hilbert-Schmidt norm of the operator on the weighted Bergman-Hilbert space Aα2(𝔹) as well as on the Hardy H2(𝔹) space.In this paper, constants are denoted byC, they are positive and may differ from one occurrence to the other. The notation a⪯b means that there is a positive constant C such that a≤Cb. If both a⪯b and b⪯a hold, then one says that a≍b.


## 2. Auxiliary Results

In this section we quote several lemmas which are used in the proofs of the main results.The next characterization of compactness is proved in a standard way, hence we omit its proof (see, e.g., [34]).Lemma 2.1.
Assumep,q>0, φ is a holomorphic self-map of 𝔹, u∈H(𝔹), ϕ is normal and μ is a weight. Then the operator ℜu,φm:H(p,q,ϕ)→Hμ∞ is compact if and only if for every bounded sequence (fk)k∈ℕ⊂H(p,q,ϕ) converging to 0 uniformly on compacts of 𝔹 as k→∞, one has
(2.1)lim⁡k→∞‖Ru,φmfk‖Hμ∞=0.The following lemma is a slight modification of Lemma 2.5 in [8] and is proved similar to Lemma 1 in [35].Lemma 2.2.
Assumeμ is a normal weight. Then a closed set K in Hμ,0∞ is compact if and only if it is bounded and
(2.2)lim⁡|z|→1sup⁡f∈Kμ(z)|f(z)|=0.The following lemma is folklore and in the next form it can be found in [36].Lemma 2.3.
Assume that0<p, q<∞, ϕ is normal, and m∈ℕ. Then for every f∈H(𝔹) the following asymptotic relationship holds:
(2.3)∫01Mqp(f,r)ϕp(r)1-rdr≍|f(0)|p+∫01Mqp(Rmf,r)(1-r)mpϕp(r)1-rdr.Lemma 2.4.
Assume thatm∈ℕ, 0<p, q<∞, ϕ is normal and f∈H(p,q,ϕ). Then, there is a positive constant C independent of f such that
(2.4)|Rmf(z)|≤C‖f‖H(p,q,ϕ)|z|ϕ(|z|)(1-|z|2)n/q+m.Proof.
Letg=ℜm-1f and z∈𝔹. By the definition of the radial derivative, the Cauchy-Schwarz inequality and the Chauchy inequality, we have that
(2.5)|Rg(z)|≤|z||∇g(z)|≤C|z|sup⁡B(z,1-|z|)/4)⁡|g(w)|1-|z|.
From (2.3) with m→m-1 we easily obtain the following inequality (see, e.g., [8, Lemma 2.1]):(2.6)|g(z)|≤C‖f‖H(p,q,ϕ)ϕ(|z|)(1-|z|2)n/q+m-1.
From (2.5) and (2.6) and the asymptotic relations
(2.7)1-|w|≍1-|z|,ϕ(|z|)≍ϕ(|w|),forw∈B(z,1-|z|2),
inequality (2.4) follows.Lemma 2.5.
Let(2.8)fa,s(z)=1(1-〈z,a〉)s,z∈B.
Then,
(2.9)Rmfa,s(z)=sPm(〈z,a〉)(1-〈z,a〉)s+m,
where
(2.10)Pm(w)=sm-1wm+pm-1(m)(s)wm-1+⋯+p2(m)(s)w2+w,
and where pj(m)(s), j=2,…,m-1 are nonnegative polynomials for s>0.Proof.
We prove the lemma by induction. Form=1,
(2.11)Rfa,s(z)=s〈z,a〉(1-〈z,a〉)s+1,z∈B,
which is formula (2.9) with P1(w)=w.
Assume (2.9) is true for every m∈{1,…,l}. Taking the radial derivative operator on equality (2.9) with m=l, we obtain (2.12)Rl+1fa,s(z)=sRsl-1〈z,a〉l+pl-1(l)(s)〈z,a〉l-1+⋯+p2(l)(s)〈z,a〉2+〈z,a〉(1-〈z,a〉)s+l=s(s+l)(sl-1〈z,a〉l+1+pl-1(l)(s)〈z,a〉l+⋯+p2(l)(s)〈z,a〉3+〈z,a〉2)(1-〈z,a〉)s+l+1+s(lsl-1〈z,a〉l+(l-1)pl-1(l)(s)〈z,a〉l-1+⋯+2p2(l)(s)〈z,a〉2+〈z,a〉)(1-〈z,a〉)(1-〈z,a〉)s+l+1,=ssl〈z,a〉l+1+⋯+[(s+l-2)p2(l)(s)+3p3(l)(s)]〈z,a〉3(1-〈z,a〉)s+l+1+[(s+l-1)+2p2(l)(s)]〈z,a〉2+〈z,a〉(1-〈z,a〉)s+l+1
from which the inductive proof easily follows.


## 3. Boundedness and Compactness ofℜu,φm:H(p,q,ϕ)→Hμ∞(or Hμ,0∞)

This section characterizes the boundedness and compactness ofℜu,φm:H(p,q,ϕ)→Hμ∞(orHμ,0∞).Theorem 3.1.
Assumem∈ℕ, 0<p, q<∞, ϕ is normal, μ is a weight, φ is a holomorphic self-map of 𝔹, and u∈H(𝔹). Then ℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded if and only if
(3.1)M1:=sup⁡z∈Bμ(z)|u(z)||φ(z)|ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m<∞.
Moreover, if ℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded, then the following asymptotic relationship holds
(3.2)‖Ru,φm‖H(p,q,ϕ)→Hμ∞≍M1.Proof.
Assume (3.1) holds. Then by Lemma 2.4 for each f∈H(p,q,ϕ), we have that
(3.3)μ(z)|Ru,φmf(z)|≤C‖f‖H(p,q,ϕ)μ(z)|u(z)||φ(z)|ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m.
Taking the supremum over the unit ball in (3.3) and using (3.1) the boundedness of operator ℜu,φm:H(p,q,ϕ)→Hμ∞ follows and
(3.4)‖Ru,φm‖H(p,q,ϕ)→Hμ∞≤CM1.
Now assume that operatorℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded. By using the test functions(3.5)fj(z)=zj∈H(p,q,ϕ),j=1,…,n,
we obtain
(3.6)Ru,φmfj∈Hμ∞,j=1,…,n,
that is, for each j=1,…,n, holds
(3.7)‖Ru,φmfj‖Hμ∞=sup⁡z∈Bμ(z)|u(z)||φj(z)|≤‖zj‖H(p,q,ϕ)‖Ru,φm‖H(p,q,ϕ)→Hμ∞<∞,
which implies that
(3.8)sup⁡z∈Bμ(z)|u(z)||φ(z)|≤∑j=1nsup⁡z∈Bμ(z)|u(z)||φj(z)|≤‖Ru,φm‖H(p,q,ϕ)→Hμ∞∑j=1n‖zj‖H(p,q,ϕ)<∞.
Let(3.9)fw(z)=(1-|w|2)t+1ϕ(w)(1-〈z,w〉)n/q+t+1.
It is known that L1:=supw∈𝔹∥fw∥H(p,q,ϕ)<∞ (see [8, Theorem 3.3]). From this, using the boundedness of ℜu,φm:H(p,q,ϕ)→Hμ∞ and by Lemma 2.5, we have that for each a∈𝔹(3.10)L1‖Ru,φm‖H(p,q,ϕ)→Hμ∞≥‖Ru,φm(fφ(a))‖Hμ∞=sup⁡z∈Bμ(z)|u(z)||Rmfφ(a)(φ(z))|≥(nq+t+1)μ(a)|u(a)|Pm(|φ(a)|2)ϕ(|φ(a)|)(1-|φ(a)|2)n/q+m≥(nq+t+1)μ(a)|u(a)||φ(a)|2ϕ(|φ(a)|)(1-|φ(a)|2)n/q+m.
From (3.10), we have that  (3.11)∞>sup⁡|φ(z)|>1/2μ(z)|u(z)||φ(z)|2ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m≥sup⁡|φ(z)|>1/2μ(z)|u(z)||φ(z)|2ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m.
On the other hand, from (3.8) and since ϕ is normal, we obtain  (3.12)sup⁡|φ(z)|≤1/2μ(φ(z))|u(z)||φ(z)|2ϕ(φ(z))(1-|φ(z)|2)n/q+m≤Csup⁡z∈Bμ(φ(z))|u(z)||φ(z)|<∞.
From (3.8), (3.10), (3.11), and (3.12) condition (3.1) follows, and moreover
(3.13)M1≤C‖Ru,φm‖H(p,q,ϕ)→Hμ∞.
From (3.4) and (3.13) asymptotic relationship (3.2) follows, finishing the proof of the theorem.Theorem 3.2.
Assumem∈ℕ, 0<p,  q<∞, ϕ is normal, μ is a weight, φ is a holomorphic self-map of 𝔹 and u∈H(𝔹). Then ℜu,φm:H(p,q,ϕ)→Hμ∞ is compact if and only if ℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded and
(3.14)lim⁡|φ(z)|→1μ(z)|u(z)||φ(z)|ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m=0.Proof.
Suppose thatℜu,φm:H(p,q,ϕ)→Hμ∞ is compact. Then it is clear that ℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded. If ∥φ∥∞<1, then (3.14) is vacuously satisfied. Hence assume that ∥φ∥∞=1. Let (zk)k∈ℕ be a sequence in 𝔹 such that |φ(zk)|→1 as k→∞, and fk(z)=fφ(zk)(z),k∈ℕ, where fw is defined in (3.9). Then supk∈ℕ∥fk∥H(p,q,ϕ)<∞, fk→0 uniformly on compacts of 𝔹 as k→∞ since
(3.15)lim⁡k→∞(1-|φ(zk)|2)t+1ϕ(|φ(zk)|)=0,
so that
(3.16)lim⁡k→∞‖Ru,φmfk‖Hμ∞=0.
On the other hand, by (3.10), we have
(3.17)μ(zk)|u(zk)||φ(zk)|2ϕ(|φ(zk)|)(1-|φ(zk)|2)n/q+m≤C‖Ru,φmfk‖Hμ∞.
From (3.16) and (3.17), equality (3.14) easily follows.
Conversely, assume thatℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded and (3.14) holds. From the proof of Theorem 3.1 we know that (3.1) holds. On the other hand, from (3.14), we have that, for every ɛ>0, there is a δ∈(0,1), such that(3.18)μ(z)|u(z)||φ(z)|ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m<ɛ
whenever δ<|φ(z)|<1.
Assume(fk)k∈ℕ is a sequence in H(p,q,ϕ) such that supk∈ℕ∥fk∥H(p,q,ϕ)≤L and fk converges to 0 uniformly on compact subsets of 𝔹 as k→∞. Let K={z∈𝔹:|φ(z)|≤δ}. Then from (3.18), and by Lemma 2.4, it follows that(3.19)‖Ru,φmfk‖Hμ∞=sup⁡z∈Bμ(z)|u(z)||Rmfk(φ(z))|≤sup⁡z∈Kμ(z)|u(z)||Rmfk(φ(z))|+sup⁡z∈B∖Kμ(z)|u(z)||Rmfk(φ(z))|≤sup⁡z∈Kμ(z)|u(z)||φ(z)||∇Rm-1fk(φ(z))|+C‖fk‖H(p,q,ϕ)sup⁡z∈B∖Kμ(z)|u(z)||φ(z)|ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m≤K2Csup⁡|ζ|≤δ|∇Rm-1fk(ζ)|+Cε‖fk‖H(p,q,ϕ),
where K2:=supz∈𝔹μ(z)|u(z)||φ(z)| (see (3.8)). Therefore
(3.20)‖Ru,φmfk‖Hμ∞≤K2sup⁡|ζ|≤δ|∇Rm-1fk(ζ)|+CLɛ.
Since (fk)k∈ℕ converges to zero on compact subsets of 𝔹 as k→∞, by Cauchy's estimates it follows that the sequence (|∇ℜm-1fk|)k∈ℕ also converges to zero on compact subsets of 𝔹 as k→∞, in particular
(3.21)lim⁡k→∞sup⁡|ζ|≤δ|∇Rm-1fk(ζ)|=0.
Using these facts and letting k→∞ in (3.20), we obtain that
(3.22)limsup⁡k→∞‖Ru,φmfk‖Hμ∞≤CLɛ.
Since ɛ is an arbitrary positive number it follows that the last limit is equal to zero. Applying Lemma 2.1, the implication follows.Theorem 3.3.
Assumem∈ℕ, 0<p, q<∞, ϕ, μ are normal, φ is a holomorphic self-map of 𝔹 and u∈H(𝔹). Then ℜu,φm:H(p,q,ϕ)→Hμ,0∞ is bounded if and only if ℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded and
(3.23)lim⁡|z|→1μ(z)|u(z)||φ(z)|=0.Proof.
First assume thatℜu,φm:H(p,q,ϕ)→Hμ,0∞ is bounded. Then, it is clear that ℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded, and as in the proof of Theorem 3.1, by taking the test functions fj(z)=zj, j=1,…,n, we obtain (3.23).
Conversely, assume that the operatorℜu,φm:H(p,q,ϕ)→Hμ∞ is bounded and condition (3.23) holds. Then, for each polynomial p, we have(3.24)μ(z)|(Ru,φmp)(z)|≤μ(z)|u(z)||(Rmp)(φ(z))|≤μ(z)|u(z)||φ(z)|‖∇Rm-1p‖∞,
from which along with condition (3.23) it follows that ℜu,φmp∈Hμ,0∞. Since the set of all polynomials is dense in H(p,q,ϕ), we see that for every f∈H(p,q,ϕ) there is a sequence of polynomials (pk)k∈ℕ such that
(3.25)lim⁡k→∞‖f-pk‖H(p,q,ϕ)=0.
From this and by the boundedness of the operator ℜu,φm:H(p,q,ϕ)→Hμ∞, we have that
(3.26)‖Ru,φmf-Ru,φmpk‖Hμ∞≤‖Ru,φm‖H(p,q,ϕ)→Hμ∞‖f-pk‖H(p,q,ϕ)⟶0
as k→∞. Hence ℜu,φm(H(p,q,ϕ))⊆Hμ,0∞, and consequently ℜu,φm:H(p,q,ϕ)→Hμ,0∞ is bounded.Theorem 3.4.
Assumem∈ℕ, 0<p, q<∞, ϕ, μ are normal, φ is a holomorphic self-map of 𝔹 and u∈H(𝔹). Then ℜu,φm:H(p,q,ϕ)→Hμ,0∞ is compact if and only if
(3.27)lim⁡|z|→1μ(z)|u(z)||φ(z)|ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m=0.Proof.
From (3.27), we see that (3.1) hold. This fact along with (3.3) implies that the set ℜu,φm({f|∥f∥H(p,q,ϕ)≤1}) is bounded in Hμ∞, moreover in Hμ,0∞. By taking the supremum in (3.3) over the unit ball in H(p,q,ϕ), using (3.27) and applying Lemma 2.2 we obtain that the operator ℜu,φm:H(p,q,ϕ)→Hμ,0∞ is compact.
Ifℜu,φm:H(p,q,ϕ)→Hμ,0∞ is compact, then by Theorem 3.2, we have that condition (3.14) holds, which implies that for every ɛ>0 there is an r∈(0,1) such that  (3.28)μ(z)|u(z)||φ(z)|ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m<ɛ,
for r<|φ(z)|<1.
As in Theorem3.3, we have that (3.23) holds. Thus there is a σ∈(0,1) such that  (3.29)μ(z)|u(z)||φ(z)|<ɛ(1-r2)n/q+minf⁡|φ(z)|≤rϕ(|φ(z)|),
for σ<|z|<1.
If|φ(z)|≤r and σ<|z|<1, then from (3.29), we obtain(3.30)μ(z)|u(z)||φ(z)|(1-|φ(z)|2)n/q+mϕ(|φ(z)|)≤μ(z)|u(z)||φ(z)|(1-r2)n/q+minf⁡|φ(z)|≤r⁡ϕ(|φ(z)|)<ɛ.
Using (3.30) and the fact that from (3.28), we have(3.31)μ(z)|u(z)||φ(z)|ϕ(|φ(z)|)(1-|φ(z)|2)n/q+m<ɛ,
for σ<|z|<1 and r<|φ(z)|<1, we get (3.27).


## 4. Hilbert-Schmidt Norm ofℜu,φm:Aα2→Aα2 and ℜu,φm:H2→H2

In this section we calculate Hilbert-Schmidt norm of the operatorsℜu,φm:Aα2→Aα2 and ℜu,φm:H2→H2. For some related results see [37, 38].Ifℋ is a separable Hilbert space, then the Hilbert-Schmidt norm ∥T∥HS of an operator T:ℋ→ℋ is defined by  (4.1)‖T‖HS=(∑n=1∞‖Ten‖2)1/2,
where {en}n∈ℕ is an orthonormal basis on ℋ. The right-hand side in (4.1) does not depend on the choice of basis. Hence, it is larger than the operator norm ∥T∥op of T.Let〈·,·〉α, α≥-1, be the usual scalar product on Aα2, where we regard that   A-12=H2.  Since for each multi-index β=(β1,…,βn)∈ℕ0n(4.2)∫B|zβ|2dvα(z)=β!Γ(n+α+1)Γ(n+|β|+α+1),
where β!=β1!⋯βn!, and (4.3)∫Bzβzγdvα(z)=0,β≠γ
(see, e.g., [1]), we have that the vectors (4.4)eβ(z)=Γ(n+|β|+α+1)β!Γ(n+α+1)zβ,β∈N0n,
form an orthonormal basis in Aα2.Theorem 4.1.
Letm∈ℕ0. Then Hilbert-Schmidt norm of the operator ℜu,φm on Aα2, α>-1 is
(4.5)‖Ru,φm‖HS=(∫B|u(z)|2(Rm1(1-∑j=1nwj)n+α+1)|wj=|φj(z)|2dvα(z))1/2.Proof.
By using the definition of the Hilbert-Schmidt norm and the monotone convergence theorem, we have(4.6)‖Ru,φm‖HS2=∑β‖Ru,φm(eβ)‖2,α2=∑βΓ(n+|β|+α+1)β!Γ(n+α+1)‖Ru,φm(zβ)‖2,α2=∑βΓ(n+|β|+α+1)β!Γ(n+α+1)|β|m∫B|u(z)|2∏j=1n|φj(z)|2βjdvα(z)=∫B|u(z)|2∑β|β|mΓ(n+|β|+α+1)β!Γ(n+α+1)∏j=1n|φj(z)|2βjdvα(z).
We also have that
(4.7)(1-∑j=1nwj)-(n+α+1)=∑k=0∞(∑j=1nwj)kΓ(n+α+k+1)k!Γ(n+α+1)=∑k=0∞∑|l|=kk!l!∏j=1nwjljΓ(n+α+k+1)k!Γ(n+α+1)=∑lΓ(n+α+|l|+1)l!Γ(n+α+1)∏j=1nwjlj,
from which by taking the radial derivatives it follows that
(4.8)Rm1(1-∑j=1nwj)n+α+1=∑l|l|mΓ(n+α+|l|+1)l!Γ(n+α+1)∏j=1nwjlj.
From (4.6) and (4.8) the result easily follows.Similar to Theorem4.1 the following result regarding the case of the Hardy space is proved. We omit the proof.Theorem 4.2.
Letm∈ℕ0. Then, Hilbert-Schmidt norm of the operator ℜu,φm on H2, is
(4.9)‖Ru,φm‖HS=sup⁡0<r<1(∫S|u(rζ)|2(Rm1(1-∑j=1nwj)n)|wj=|φj(rζ)|2dσ(ζ))1/2.



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*Source: 801264-2010-12-08.xml*