# Weighted Differentiation Composition Operators from the Mixed-Norm Space to thenth Weigthed-Type Space on the Unit Disk

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

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## Abstract

The boundedness and compactness of the weighted differentiation composition operator from the mixed-norm space to thenth weighted-type space on the unit disk are characterized.


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## Body

## 1. Introduction

Throughout this paper𝔻 will denote the open unit disk in the complex plane ℂ, H(𝔻) the class of all holomorphic functions on 𝔻, and H∞=H∞(𝔻) the space of all bounded holomorphic functions on 𝔻 with the norm ∥f∥∞=supz∈𝔻|f(z)|.The mixed norm spaceHp,q,γ=Hp,q,γ(𝔻),  0<p,q<∞,  -1<γ<∞, consists of all f∈H(𝔻) such that (1.1)∥f∥Hp,q,γq=∫01Mpq(f,r)(1-r)γdr<∞,
where (1.2)Mp(f,r)=(12π∫02π|f(reiθ)|pdθ)1/p.A positive continuous function on𝔻 is called weight. Let μ(z) be a weight and n∈ℕ0. The  nthweighted-type space on 𝔻, denoted by 𝒲μ(n)(𝔻), consists of all f∈H(𝔻) such that (1.3)b𝒲μ(n)(𝔻)(f):=supz∈𝔻μ(z)|f(n)(z)|<∞.
The space was recently introduced by this author in [1] as an extension of several weighted-type spaces which attracted a lot of attention in last few decades. For instance, when n=0, the space becomes the weighted-type space Hμ∞(𝔻) (see, e.g., [2–4]), when n=1, the Bloch-type space ℬμ(𝔻) (see, e.g., [5–7]), and for n=2, the Zygmund-type space 𝒵μ(𝔻). Some information on Zygmund-type spaces on 𝔻 and some operators on them can be found, for example, in [8–10] and on the unit ball, for example, in [11, 12].The quantityb𝒲μ(n)(𝔻)(f) is a seminorm on the nth weighted-type space 𝒲μ(n)(𝔻) and a norm on 𝒲μ(n)(𝔻)/ℙn-1, where ℙn-1 is the set of all polynomials whose degrees are less than or equal to n-1. A natural norm on the nth weighted-type space is introduced as follows: (1.4)∥f∥𝒲μ(n)(𝔻)=∑j=0n-1|f(j)(0)|+b𝒲μ(n)(𝔻)(f).
With this norm the nth weighted-type space becomes a Banach space.The littlenth weighted-type space, denoted by 𝒲μ,0(n)(𝔻), is a closed subspace of 𝒲μ(n)(𝔻) consisting of those f for which (1.5)lim|z|→1μ(z)|f(n)(z)|=0.An analytic self-mapφ:𝔻→𝔻 induces the composition operator Cφ on H(𝔻), defined by Cφ(f)(z)=f(φ(z)) for f∈H(𝔻) (see, e.g., [8, 13–16]).Letφ be an analytic self-map of 𝔻, u∈H(𝔻), and m∈ℕ. Then the weighted differentiation composition operator, denoted by Dφ,um, is defined on H(𝔻) by (1.6)Dφ,umf(z)=u(z)f(m)(φ(z)),f∈H(𝔻).Recently there has been some interest in studying some particular cases of operatorDφ,um (see, e.g., [17–25]). For some other products of linear operators on spaces of holomorphic functions see also recent papers [11, 26–32].Here we study the boundedness and compactness of the operatorDφ,um from Hp,q,γ to nth weighted-type spaces, where n∈ℕ.Throughout 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 B/C≤A≤CB.


## 2. Auxiliary Results

Here we quote some auxiliary results which will be used in the proofs of the main results. The first lemma can be proved in a standard way (see, e.g., in [13, Proposition 3.11] or in [15, Lemma 3]).Lemma 2.1.
Assume thatm∈ℕ0, n∈ℕ, p,q>0, γ>-1, φ is an analytic self-map of 𝔻 and u∈H(𝔻). Then the operator Dφ,um:Hp,q,γ→𝒲μ(n) is compact if and only if Dφ,um:Hp,q,γ→𝒲μ(n) is bounded and for any bounded sequence (fk)k∈ℕ in Hp,q,γ which converges to zero uniformly on compact subsets of 𝔻, Dφ,umfk→0 in 𝒲μ(n) as k→∞.The next lemma is known, but we give a proof of it for the benefit of the reader.Lemma 2.2.
Assume thatn∈ℕ0,  0<p,q<∞,  -1<γ<∞ and  f∈Hp,q,γ. Then there is a positive constant  C independent of  f such that
(2.1)|f(n)(z)|≤C∥f∥Hp,q,γ(1-|z|2)(γ+1)/q+1/p+n.Proof.
By the monotonicity of the integral means, using the well-known asymptotic formula(2.2)∫01Mpq(f,r)(1-r)γdr≍|f(0)|q+∫01Mpq(f(n),r)(1-r)γ+nqdr,
and Theorem 7.2.5 in [33], we have that
(2.3)∥f∥Hp,q,γq≥∫(1+|z|)/21Mpq(f(n),r)(1-r)γ+nqdr≥CMpq(f(n),1+|z|2)(1-|z|2)γ+1+nq≥C(1-|z|2)γ+1+nq+q/p|f(n)(z)|q,
from which the result follows.The following lemma can be found in [34].Lemma 2.3.
Forβ>-1 and m>1+β one has
(2.4)∫01(1-r)β(1-ρr)mdr≤C(1-ρ)1+β-m,0<ρ<1.A proof of the next lemma can be found in [35, Lemma 2.3].Lemma 2.4.
Assumea>0 and
(2.5)Dn(a)=|11⋯1aa+1⋯a+n-1a(a+1)(a+1)(a+2)⋯(a+n-1)(a+n)⋯∏j=0n-2(a+j)∏j=0n-2(a+j+1)⋯∏j=0n-2(a+j+n-1)|.
Then Dn(a)=∏j=1n-1j!.The following formula(2.6)(f∘φ)(n)(z)=∑k=1nf(k)(φ(z))∑k1,…,knn!k1!⋯kn!∏j=1n(φ(j)(z)j!)kj,
where the second sum is over all nonnegative integers k1,k2,…,kn satisfying k=k1+k2+⋯+kn and k1+2k2+⋯+nkn=n, is attributed to Faà di Bruno [36]. By using Bell polynomials Bn,k(x1,…,xn-k+1) it can be written as follows: (2.7)(f∘φ)(n)(z)=∑k=0nf(k)(φ(z))Bn,k(φ′(z),φ′′(z),…,φ(n-k+1)(z)).
For n∈ℕ the last sum can go from k=1 since Bn,0(φ′(z),φ′′(z),…,φ(n+1)(z))=0; however we will keep the summation since for n=0 the only existing term B0,0 is equal to 1 and we will use it.The Leibnitz formula along with (2.6) yields (2.8)(u(z)g(φ(z)))(n)=∑l=0n‍Clnu(n-l)(z)∑k=0lg(k)(φ(z))Bl,k(φ′(z),…,φ(l-k+1)(z)).
Hence we have the next result.Lemma 2.5.
Assume thatg,u∈H(𝔻) and φ is an analytic self-map of 𝔻. Then
(2.9)(u(z)g(φ(z)))(n)=∑k=0n‍g(k)(φ(z))∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z)).


## 3. The Boundedness and Compactness ofDφ,um:Hp,q,γ→𝒲μ(n)

This section characterizes the boundedness and compactness of the operatorDφ,um:Hp,q,γ→𝒲μ(n).Theorem 3.1.
Suppose thatm,n∈ℕ, 0<p,q<∞, -1<γ<∞, φ is an analytic self-map of the unit disk, u∈H(𝔻), and μ is a weight. Then the operator Dφ,um:Hp,q,γ→𝒲μ(n) is bounded if and only if for each k∈{0,1,…,n}(3.1)Ik:=supz∈𝔻μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+m+k<∞.
Moreover if Dφ,um:Hp,q,γ→𝒲μ(n) is bounded, then the following asymptotic relation holds
(3.2)∥Dφ,um∥Hp,q,γ→𝒲μ(n)/ℙn-1≍∑k=0n‍Ik.Proof.
First assume thatDφ,um:Hp,q,γ→𝒲μ(n) is bounded; then there exists a constant C such that
(3.3)∥Dφ,umf∥𝒲μ(n)≤C∥f∥Hp,q,γ
for all f∈Hp,q,γ.
For a fixedw∈𝔻, t≥(γ+1)/q, and constants c1,…,cn+1, set (3.4)gw(z)=∑j=1n+1cj∏l=0m-1(j+t+1/p+l)ĝw,j(z),
where
(3.5)ĝw,j(z)=(1-|w|2)j+t-(γ+1)/q(1-w¯z)1/p+j+t,j=1,…,n+1.
By [33, Theorem 1.4.10], we get
(3.6)Mp(ĝw,j,r)≤C(1-|w|2)j+t-(γ+1)/q(1-r|w|)j+t,j=1,…,n+1.
Applying Lemma 2.3, we have that
(3.7)∥ĝw,j∥Hp,q,γq=∫01Mpq(ĝw,j,r)(1-r)γdr≤C∫01(1-|w|2)q(j+t)-(γ+1)(1-r|w|)q(j+t)(1-r)γdr≤C.
Therefore gw∈Hp,q,γ, and moreover supw∈𝔻∥gw∥Hp,q,γ<∞.
Now we show that for eachs∈{m,m+1,…,m+n}, there are constants c1,c2,…,cn+1, such that(3.8)gw(s)(w)=w¯s(1-|w|2)s+(γ+1)/q+1/p,gw(t)(w)=0,t∈{m,…,m+n}∖{s}.
By differentiating function gw, for each s∈{m,…,m+n}, (3.8) becomes
(3.9)c1+c2+⋯+cn+1=0,(t+p-1+m+1)c1+(t+p-1+m+2)c2+⋯+(t+p-1+m+n+1)cn+1=0,⋮∏j=1s-m(t+p-1+m+j)c1+⋯+∏j=1s-m(t+p-1+m+n+j)cn+1=1,⋮∏j=1n(t+p-1+m+j)c1+⋯+∏j=1n(t+p-1+m+n+j)cn+1=0.
Applying Lemma2.4 with a=t+1/p+m+1>0 and where n→n+1, we see that the determinant of system (3.9) is different from zero, as claimed.
Bygw,k, k∈{0,1,…,n}, denote the corresponding family of functions which satisfy (3.8) with s=m+k. Then, for each fixed k∈{0,1,…,n}, inequality (3.3) along with (2.9) and (3.8) implies that for each φ(w)≠0(3.10)μ(w)|φ(w)|k+m|∑l=knClnu(n-l)(w)Bl,k(φ′(w),…,φ(l-k+1)(w))|(1-|φ(w)|2)(γ+1)/q+1/p+k+m≤Csupw∈𝔻∥Dφ,um(gφ(w),k)∥𝒲μ(n)≤C∥Dφ,um∥Hp,q,γ→𝒲μ(n).
From (3.10) it follows that for each k∈{0,1,…,n},(3.11)sup|φ(z)|>1/2μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m≤C∥Dφ,um∥Hp,q,γ→𝒲μ(n).
Let(3.12)hk(z)=zk,k=m,…,n+m.
Then clearly
(3.13)∥hk∥Hp,q,γ≤1,foreachk∈ℕ.
By formula (2.9) applied to the function f(z)=hm(z) we get(3.14)(Dφ,umhm)(n)(z)=hm(m)(φ(z))∑l=0nClnu(n-l)(z)Bl,0(φ′(z),…,φ(l+1)(z))=m!∑l=0nClnu(n-l)(z)Bl,0(φ′(z),…,φ(l+1)(z)),
which along with the boundedness of the operator Dφ,um:Hp,q,γ→𝒲μ(n) and (3.13) implies that
(3.15)m!supz∈𝔻μ(z)|∑l=0nClnu(n-l)(z)Bl,0(φ′(z),…,φ(l+1)(z))|≤∥Dφ,um(zm)∥𝒲μ(n)≤∥Dφ,um∥Hp,q,γ→𝒲μ(n).
Now assume that we have proved that forj∈{0,1,…,k-1} and a k≤n(3.16)supz∈𝔻μ(z)|∑l=jnClnu(n-l)(z)Bl,j(φ′(z),…,φ(l-j+1)(z))|≤C∥Dφ,um∥Hp,q,γ→𝒲μ(n).
Applying (2.9) to the function f(z)=hm+k(z),  k∈{0,1,…,n}, and noticing that hm+k(s)(z)≡0 for s>m+k, we get(3.17)(Dφ,umhm+k)(n)(z)=∑j=0khm+k(m+j)(φ(z))∑l=jnClnu(n-l)(z)Bl,j(φ′(z),…,φ(l-j+1)(z))=∑j=0k(m+k)⋯(k-j+1)(φ(z))k-j∑l=jnClnu(n-l)(z)Bl,j(φ′(z),…,φ(l-j+1)(z)).
From (3.17), the boundedness of the operator Dφ,um:Hp,q,γ→𝒲μ(n), the fact that ∥φ∥∞≤1, the triangle inequality, noticing that (m+k)! is the coefficient at ∑l=knClnu(n-l)(z)Bl,k(φ'(z),…,φ(l-k+1)(z)), and finally using hypothesis (3.16) we get
(3.18)supz∈𝔹μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|≤C∥Dφ,um∥Hp,q,γ→𝒲μ(n).
Hence by induction, (3.18) holds for each k∈{0,1,…,n}.
From (3.18), for each fixed k∈{0,1,…,n}(3.19)sup|φ(z)|≤1/2μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m≤Csupz∈𝔹μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|≤C∥Dφ,um∥Hp,q,γ→𝒲μ(n).
Inequalities (3.11) and (3.19) imply
(3.20)∑k=0nIk≤C∥Dφ,um∥Hp,q,γ→𝒲μ(n).
Now assume that (3.1) holds. Then for any f∈Hp,q,γ, by (2.9) and Lemma 2.2 we have
(3.21)μ(z)|(Dφ,umf)(n)(z)|=μ(z)|∑k=0nf(m+k)(φ(z))∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|≤μ(z)∑k=0n|f(m+k)(φ(z))||∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(3.22)≤C∥f∥Hp,q,γ∑k=0nμ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m(3.23)≤C∥f∥Hp,q,γ∑k=0nsupz∈𝔻μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m
We also have that for each s∈{1,…,n-1}(3.24)|(Dφ,umf)(s)(0)|=|∑k=0sf(m+k)(φ(0))∑l=ksClsu(s-l)(0)Bl,k(φ′(0),…,φ(l-k+1)(0))|≤C∥f∥Hp,q,γ∑k=0s|∑l=ksClsu(s-l)(0)Bl,k(φ′(0),…,φ(l-k+1)(0))|(1-|φ(0)|2)(γ+1)/q+1/p+m+k,|(Dφ,umf)(0)|=|u(0)||f(m)(φ(0))|≤C|u(0)|∥f∥Hp,q,γ(1-|φ(0)|2)(γ+1)/q+1/p+m.
Using (3.23), (3.24), and (3.1) it follows that the operator Dφ,um:Hp,q,γ→𝒲μ(n) is bounded.
From (3.23) and (3.20) the asymptotic relation (3.2) follows.Theorem 3.2.
Suppose thatm,n∈ℕ, 0<p,q<∞, -1<γ<∞, φ is an analytic self-map of the unit disk, u∈H(𝔻), and μ is a weight. Then the operator Dφ,um:Hp,q,γ→𝒲μ,0(n) is bounded if and only if Dφ,um:Hp,q,γ→𝒲μ(n) is bounded and for each k∈{0,1,…,n}(3.25)lim|z|→1μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|=0.Proof.
The boundedness ofDφ,um:Hp,q,γ→𝒲μ,0(n) clearly implies that Dφ,um:Hp,q,γ→𝒲μ(n) is bounded. Applying (2.9) to the function f(z)=hm(z) and using the assumption Dφ,um(hm)∈𝒲μ,0(n) it follows that
(3.26)μ(z)|(Dφ,umhm)(n)(z)|=m!μ(z)|∑l=0nClnu(n-l)(z)Bl,0(φ′(z),…,φ(l+1)(z))|→0,
as |z|→1, which is (3.25) for k=0.
Assume that we have proved the following inequalities:(3.27)lim|z|→1μ(z)|∑l=jnClnu(n-l)(z)Bl,j(φ′(z),…,φ(l-j+1)(z))|=0,
for j∈{0,1,…,k-1} and a k≤n.
Applying formula (2.9) to the function f(z)=hm+k(z),k∈{0,1,…,n}, we get (3.17). From (3.17), by using the boundedness of function φ, the triangle inequality, noticing that the coefficient at ∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z)) is independent of z, and finally using hypothesis (3.27), we easily obtain(3.28)lim|z|→1μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|=0.
Hence by induction we get that (3.25) holds for each k∈{0,1,…,n}.
Now assume thatDφ,um:Hp,q,γ→𝒲μ(n) is bounded and (3.25) holds for each k∈{0,1,…,n}. For each polynomial p we have(3.29)μ(z)|(Dφ,ump)(n)(z)|=μ(z)|∑k=0np(k)(φ(z))∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|≤∑k=0n∥p(k)∥∞μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|→0,
as |z|→1.
From (3.29) we have that, for each polynomial p, Dφ,ump∈𝒲μ,0(n). The set of all polynomials is dense in Hp,q,γ, so we have that for each f∈Hp,q,γ, there is a sequence of polynomials (pk)k∈ℕ such that ∥f-pk∥Hp,q,γ→0 as k→∞. Thus the boundedness of Dφ,um:Hp,q,γ→𝒲μ(n) implies (3.30)∥Dφ,umf-Dφ,umpk∥𝒲μ(n)≤∥Dφ,um∥Hp,q,γ→𝒲μ(n)∥f-pk∥Hp,q,γ→0,ask→∞.
Hence Dφ,um(Hp,q,γ)⊆𝒲μ,0(n), from which the boundedness of Dφ,um:Hp,q,γ→𝒲μ,0(n) follows, completing the proof of the theorem.Theorem 3.3.
Suppose thatm,n∈ℕ, 0<p,q<∞, -1<γ<∞, φ is an analytic self-map of the unit disk, u∈H(𝔻), and μ is a weight. Then the operator Dφ,um:Hp,q,γ→𝒲μ(n) is compact if and only if Dφ,um:Hp,q,γ→𝒲μ(n) is bounded and for each k∈{0,1,…,n}(3.31)lim|φ(z)|→1μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m=0.Proof.
First assume thatDφ,um:Hp,q,γ→𝒲μ(n) is bounded and (3.31) holds. By Theorem 3.1 we have that for each k∈{0,1,…,n}, (3.1) holds.
Let(fi)i∈ℕ be a sequence in Hp,q,γ such that supi∈ℕ∥fi∥Hp,q,γ≤L and fi converges to 0 uniformly on compact subsets of 𝔻 as i→∞. By the assumption, for any ɛ>0, there is a δ∈(0,1), such that for each k∈{0,1,…,n} and δ<|φ(z)|<1(3.32)μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m<ɛ.
We have
(3.33)∥Dφ,umfi∥𝒲μ(n)=supz∈𝔻μ(z)|(Dφ,umfi)(n)(z)|+∑j=0n-1|(Dφ,umfi)(j)(0)|=supz∈𝔻μ(z)|∑k=0nfi(m+k)(φ(z))∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|+∑j=0n-1|∑k=0jfi(m+k)(φ(0))∑l=kjClju(j-l)(0)Bl,k(φ′(0),…,φ(l-k+1)(0))|≤(sup|φ(z)|≤δ+sup|φ(z)|>δ)μ(z)∑k=0n|fi(m+k)(φ(z))||∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|+∑j=0n-1|∑k=0jfi(m+k)(φ(0))∑l=kjClju(j-l)(0)Bl,k(φ′(0),…,φ(l-k+1)(0))|=J1+J2+J3.
Now we estimateJ1, J2, and J3:(3.34)J1=sup|φ(z)|≤δμ(z)∑k=0n|fi(m+k)(φ(z))||∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|≤∑k=0n‍sup|w|≤δ|fi(m+k)(w)|sup|φ(z)|≤δμ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|≤∑k=0nsup|w|≤δ|fi(m+k)(w)|supz∈𝔻μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+m+k=∑k=0n‍sup|w|≤δ|fi(m+k)(w)|Ik→0,asi→∞,
where in (3.34) we have used the fact that from fi→0 uniformly on compact subsets of 𝔻 as i→∞ it follows that for each  s∈ℕ, fi(s)→0 uniformly on compact subsets of 𝔻 as i→∞.
The fact that(3.35)J3=∑j=0n-1|∑k=0jfi(m+k)(φ(0))∑l=kj‍Clju(j-l)(0)Bl,k(φ′(0),…,φ(l-k+1)(0))|→0,
as i→∞, is proved similarly; so we omit it.
By Lemma2.2 and (3.32) we have that(3.36)J2≤C∥fi∥Hp,q,γ∑k=0n‍sup|φ(z)|>δμ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m<Cɛ(n+1)L.
From (3.34), (3.35), and (3.36) we obtain(3.37)limi→∞∥Dφ,umfi∥𝒲μ(n)=0.
From this and applying Lemma 2.1 the implication follows.
Now assume thatDφ,um:Hp,q,γ→𝒲μ(n) is compact; then clearly Dφ,um:Hp,q,γ→𝒲μ(n) is bounded. Let (zi)i∈ℕ be a sequence in 𝔻 such that |φ(zi)|→1 as i→∞. If such a sequence does not exist, then the conditions in (3.31) automatically hold.
Letgw,k, k∈{0,1,…,n} be as in Theorem 3.1. Then the sequences (gφ(zi),k)i∈ℕ are bounded and gφ(zi),k→0 uniformly on compact subsets of 𝔻 as i→∞. Since Dφ,um:Hp,q,γ→𝒲μ(n) is compact, we have that for each k∈{0,1,…,n}(3.38)limi→∞∥Dφ,umgφ(zi),k∥𝒲μ(n)=0.
On the other hand, from (3.10) we obtain
(3.39)∥Dφ,umgφ(zi),k∥𝒲μ(n)≥Cμ(zi)|φ(zi)|k+m|∑l=knClnu(n-l)(zi)Bl,k(φ′(zi),…,φ(l-k+1)(zi))|(1-|φ(zi)|2)(γ+1)/q+1/p+k+m,
which along with |φ(zi)|→1 as i→∞ and (3.38) implies that
(3.40)limi→∞μ(zi)|∑l=knClnu(n-l)(zi)Bl,k(φ′(zi),…,φ(l-k+1)(zi))|(1-|φ(zi)|2)(γ+1)/q+1/p+k+m,
for each k∈{0,1,…,n}, from which (3.31) holds in this case.


## 4. The Compactness of the OperatorDφ,um:Hp,q,γ→𝒲μ,0(n)

The compactness ofDφ,um:Hp,q,γ→𝒲μ,0(n) is characterized here. The proof of the next lemma is similar to the proof of the corresponding result in [14].Lemma 4.1.
Suppose thatn∈ℕ0 and μ is a radial weight such that lim|z|→1μ(z)=0. A closed set K in 𝒲μ,0(n) is compact if and only if it is bounded and satisfies
(4.1)lim|z|→1supf∈Kμ(z)|f(n)(z)|=0.Theorem 4.2.
Suppose thatm,n∈ℕ, 0<p,q<∞, -1<γ<∞, φ is an analytic self-map of the unit disk, u∈H(𝔻) and μ is a radial weight such that lim|z|→1μ(z)=0. Then the operator Dφ,um:Hp,q,γ→𝒲μ,0(n) is compact if and only if for each k∈{0,1,…,n}(4.2)lim|z|→1μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m=0.Proof.
First assume thatDφ,um:Hp,q,γ→𝒲μ,0(n) is compact. Then it is bounded and since the test functions in (3.12) belong to Hp,q,γ(𝔻), we have that (3.25) holds. Beside this the operator Dφ,um:Hp,q,γ→𝒲μ(n) is compact too, so that (3.31) holds. Hence, if ∥φ∥∞<1, from (3.25) for each k∈{0,1,…,n} we get
(4.3)μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m≤μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-∥φ∥∞2)(γ+1)/q+1/p+k+m→0,
as |z|→1, hence we obtain (4.2) in this case.
Now assume∥φ∥∞=1. Let (φ(zi))i∈ℕ be a sequence such that |φ(zi)|→1 as i→∞. Then from (3.31) we have that for every ɛ>0, there is an r∈(0,1) such that for each k∈{0,1,…,n}(4.4)μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m<ɛ
when r<|φ(z)|<1, and from (3.25) there exists a σ∈(0,1) such that for σ<|z|<1(4.5)μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|<ɛ(1-r2)(γ+1)/q+1/p+k+m.
Therefore, whenσ<|z|<1 and r<|φ(z)|<1, we have that(4.6)μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m<ɛ.
On the other hand, if |φ(z)|≤r and σ<|z|<1, from (4.5) we obtain
(4.7)μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-|φ(z)|2)(γ+1)/q+1/p+k+m<μ(z)|∑l=knClnu(n-l)(z)Bl,k(φ′(z),…,φ(l-k+1)(z))|(1-r2)(γ+1)/q+1/p+k+m<ɛ.
Combining the last two inequalities we obtain (4.2), as desired.
Now assume that (4.2) holds. Taking the supremum in (3.22) over f in the unit ball of Hp,q,γ, then letting |z|→1 is such obtained inequality and using (4.2) we get (4.8)lim|z|→1sup∥f∥Hp,q,γ≤1μ(z)|(Dφ,umf)(n)(z)|=0.
Hence by Lemma 4.1 the compactness of the operator Dφ,um:Hp,q,γ→𝒲μ,0(n) follows.



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*Source: 246287-2010-07-27.xml*