# On the Generalized Hardy Spaces

Authors: M. Fatehi
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We introduce new spaces that are extensions of the Hardy spaces and
we investigate the continuity of the point evaluations as well as the boundedness
and the compactness of the composition operators on these spaces.


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## Body

## 1. Introduction

LetU be the open unit disk in the complex plane ℂ, ∂U its boundary, and H(U) the space of all analytic functions on the unit disk.For an analytic functionf on the unit disk and 0<r<1, we define the delay function fr by fr(eiθ)=f(reiθ). It is easy to see that the functions fr are continuous for each r, hence they are in Lp(∂U,dθ/2π), where dθ/2π is the normalized arc length on the unit circle.For0<p<∞, the Hardy space Hp(U)=Hp is the set of all f∈H(U) such that(1.1)∥f∥pp=sup0<r<1∫02π|fr(eiθ)|pdθ2π<∞.
Also we recall that H∞(U)=H∞ is the space of all bounded analytic functions defined on U, with supremum norm ∥f∥∞=supz∈U|f(z)|. We know that for p≥1, Hp is a Banach space (see, e.g., [1, page 37]).By the Littlewood Subordination Theorem (see [2, Corollary 2.23]), we see that the supremum in the above definition of Hp is actually a limit, that is,(1.2)∥f∥pp=limr→1∫02π|fr(eiθ)|pdθ2π<∞.
Another important result is Fatou's Radial Limit Theorem (see [1, Theorems 2.2 and 2.6]), which says, if f is in Hp for some p>0, then the radial limit(1.3)f*(eiθ)=limr→1f(reiθ)
exists for almost all θ and the mapping f→f* is an isometry of Hp to a closed subspace of Lp(∂U,dθ/2π). Therefore,(1.4)∥f∥pp=∫02π|f*(eiθ)|pdθ2π<∞.
We will also write f(eiθ) for f*(eiθ). If p=2 and f̂(n) are the nth coefficients of f in its Maclaurin series, then we have another representation for the norm of f on H2 as follows:(1.5)∥f∥22=∑n=0∞|f̂(n)|2<∞.
The formula above defines a norm that turns H2 into a Hilbert space whose inner product is given by(1.6)〈f,g〉H2=∑n=0∞f̂(n)ĝ(n)¯=∫02πf(eiθ)g(eiθ)¯dθ2π
for each f,g∈H2 (see, e.g., [2]).Letew be the point evaluation at w, that is, ew(f)=f(w). It is well known that point evaluations at the points of U are all continuous on Hp (see, e.g., [1, page 36]).Letw∈U and H be a Hilbert space of analytic functions on U. If ew is a bounded linear functional on H, then the Riesz Representation Theorem implies that there is a function (which is usually called Kw) in H that induces this linear functional, that is, ew(f)=〈f,Kw〉.Letφ be an analytic self-map of the unit disk. The linear composition operator Cφ is defined by Cφ(f)=f∘φ for f∈H(U). It is well known (see, e.g., [1, page 29] or [3, Theorem 1]) that the composition operators are bounded on each of the Hardy spaces Hp (0<p<∞). One of the first papers in this research area is [3], while Schwartz in [4] begun the research on compact composition operators on Hp. Shapiro and Taylor in [5] have studied the role of angular derivative for compactness of Cφ in Hp. For some other classical results, see [2, 6].The boundedness and compactness of composition operators, as well as weighted composition operators and other natural extensions of them, on various spaces of analytic functions have been investigated by many authors; see, books [2, 6], and, for example, the following recent papers [7–28] and the references therein.Throughout this paper,P denotes the set of all analytic polynomials and for a function F, RF denotes the range of F.


## 2. Generalized Hardy Spaces

In this section, we define new spaces and investigate some basic properties of these spaces.Definition 2.1.
LetF:H(U)→H(U) be a linear operator such that F(f)=0 if and only if f=0, that is, F is 1-1. For 1≤p<∞, the generalized Hardy space HF,p(U)=HF,p is defined to be the collection of all analytic functions f on U for which
(2.1)sup0<r<1∫02π|(F(f))r(eiθ)|pdθ2π<∞.
Denote the pth root of this supremum by ∥f∥HF,p. Since |F(f)|p is a subharmonic function, so by [2, Corollary 2.23], we have
(2.2)∥f∥HF,pp=limr→1∫02π|(F(f))r(eiθ)|pdθ2π.
Therefore, f∈HF,p if and only if F(f)∈Hp and
(2.3)∥f∥HF,pp=∥F(f)∥pp=∫02π|F(f)(eiθ)|pdθ2π.
It is easy to see that HF,p is a normed space with the norm ∥·∥HF,p.From now on, unless otherwise stated, we assume thatF satisfies the conditions of Definition 2.1.In this section, we first set some conditions onF such that HF,p becomes a Banach space. In the following theorem, we obtain a necessary and sufficient condition for HF,p to be a Banach space.Theorem 2.2.
Let1≤p<∞ and P⊆RF. Then Hp is a subspace of RF if and only if HF,p is a Banach space.Proof.
Suppose thatHp⊆RF. Since HF,p is a normed space, it suffices to show that it is complete. Let {fn} be a Cauchy sequence in HF,p and set F(fn)=gn. Then {gn} is a Cauchy sequence in Hp. Since Hp is complete, there is a g∈Hp such that
(2.4)∥gn-g∥p→0,asn→∞.
Since Hp⊆RF, there is an f∈H(U) such that F(f)=g. Now we show that this f is the HF,p-limit of {fn}. We have
(2.5)∥fn-f∥HF,p=∥gn-g∥p→0,asn→∞.
Hence fn-f∈HF,p for sufficiently large positive integer n, which implies that f∈HF,p. So fn→f in HF,p as n→∞.
Conversely, suppose thatHF,p is a Banach space. If Hp⊈RF, then there is a g∈Hp such that g is not in RF. Since the polynomials are dense in Hp, there is a sequence {pn} in P such that ∥pn-g∥p→0 as n→∞. Let qn=F-1(pn). Then {qn} is a Cauchy sequence in HF,p and so there is a q∈HF,p such that ∥qn-q∥HF,p→0 as n→∞. Hence ∥F(qn)-F(q)∥p→0 as n→∞. On the other hand, ∥F(qn)-g∥p→0 as n→∞. This shows that g=F(q) which is a contradiction.Example 2.3.
LetF(∑j=0∞ajzj)=∑j=0∞(aj/2j)zj, where ∑j=0∞ajzj∈H(U). It is not hard to see that g(z)=∑n=0∞2-n/2zn∈H2 and g is not in RF. So by Theorem 2.2, HF,2 is not a Banach space.Proposition 2.4.
IfH2⊆RF, then HF,2 is a Hilbert space.Proof.
We define a scalar product onHF,2 by
(2.6)〈f,g〉HF,2=〈F(f),F(g)〉H2.
It is easy to show that this scalar product defines an inner product on HF,2.There is a Banach spaceHF,p such that it does not satisfy the condition of Theorem 2.2. For example, let 1≤p<∞, F(f)=zf for each f∈H(U). Then 1∉RF. By the following proposition, we see that although Hp⊈RF, HF,p is a Banach space.Proposition 2.5.
Suppose1≤p<∞,  h∈H(U),h≢0, and F(f)=fh for every f∈H(U). Then HF,p is a Banach space.Proof.
IfHp⊆RF, then by Theorem 2.2, the proposition holds. Otherwise, let {fn} be a Cauchy sequence in HF,p. Seting F(fn)=gn, so {gn} is a Cauchy sequence in Hp. Therefore, there is a g∈Hp such that ∥gn-g∥p→0 as n→∞. If g∈RF, then the proof is similar to the proof of Theorem 2.2.
Now suppose thatg is not in RF. Then there are z0∈U, m1≥0, and m2>m1 such that
(2.7)g(z)=(z-z0)m1g0(z),h(z)=(z-z0)m2h0(z),
where h0,g0∈H(U), g0(z0)≠0, and h0(z0)≠0. Therefore, we have
(2.8)∥gn-g∥p=∥hfn-g∥p=∫02π|(eiθ-z0)m2h0(eiθ)fn(eiθ)-(eiθ-z0)m1g0(eiθ)|pdθ2π=∫02π|(eiθ-z0)|m1|(eiθ-z0)m2-m1h0(eiθ)fn(eiθ)-g0(eiθ)|pdθ2π≥(1-|z0|)m1∫02π|(eiθ-z0)m2-m1h0(eiθ)fn(eiθ)-g0(eiθ)|pdθ2π.
Since ∥gn-g∥p→0 as n→∞, we have
(2.9)limn→∞∫02π|(eiθ-z0)m2-m1h0(eiθ)fn(eiθ)-g0(eiθ)|pdθ2π=0.
Hence ∥(z-z0)m2-m1h0fn-g0∥p→0 as n→∞. Since the point evaluation at z0 is a bounded linear functional on Hp, we have
(2.10)(z0-z0)m2-m1h0(z0)fn(z0)-g0(z0)→0,asn→∞.
So g0(z0)=0, which is a contradiction.The set of all analytic polynomials is dense inHp, but this is not the case for each space HF,p. Also it is possible that P⊈HF,p. For example, let p=2, g(z)=1/(1-z), and F(f)=fg. Then 1 is not in HF,2, for if 1∈HF,2, then F(1)=g∈H2, which is a contradiction.In the following proposition, we will find a dense subset inHF,p, whenever P⊆RF.Proposition 2.6.
Suppose1≤p<∞ and P⊆RF. Then {F-1(p):p∈P}¯=HF,p.Proof.
It is clear that{F-1(p):p∈P}⊆HF,p. Suppose that f∈HF,p. Then there is a sequence {hn} in P such that ∥hn-F(f)∥p→0 as n→∞. Setting fn=F-1(hn), we have
(2.11)∥fn-f∥HF,p=∥hn-F(f)∥p,
so the result follows.Corollary 2.7.
Suppose1≤p<∞,  P⊆RF, and F-1(p)∈P for each p∈P. Then P∩HF,p¯=HF,p.


## 3. Point Evaluations

In this section, we investigate the continuity of the point evaluations onHF,p. The idea behind Theorem 3.1 is similar to the one found in [29, page 51].Theorem 3.1.
Suppose thatw∈U and H2⊆RF. Then we have the following.(a)
Ifp≥2 and ∑j=0∞F-1(zj)(w)¯zj∈H2, then ew is continuous on HF,p.(b)
Let1≤p<2 and ∑j=0∞F-1(zj)(w)¯zj∈H∞. If for each 0<r<1 and f∈HF,1, (F(f))r=F(fr), then ew is continuous on HF,p.Proof.
(a) By Proposition 2.4, HF,2 is a Hilbert space. Since
(3.1)〈F-1(zj),F-1(zk)〉HF,2=〈zj,zk〉H2=0,∥F-1(zj)∥HF,2=∥zj∥H2=1
for each j,k∈ℕ∪{0}, where j≠k, {F-1(zj):j∈ℕ∪{0}} is an orthonormal set in HF,2. Also if g∈HF,2 and for each j∈ℕ∪{0}, 〈g,F-1(zj)〉HF,2=0, then for each j∈ℕ∪{0}, 〈F(g),zj〉H2=0. Since {zj:j∈ℕ∪{0}} is a basis for H2, g≡0. Therefore, {F-1(zj):j∈ℕ∪{0}} is a basis for HF,2. Since ∑j=0∞F-1(zj)(w)¯zj∈H2, there is an h∈HF,2 such that F(h)=∑j=0∞F-1(zj)(w)¯zj. For each j∈ℕ∪{0}, we have
(3.2)〈F-1(zj),h〉HF,2=〈zj,F(h)〉H2=〈zj,∑k=0∞F-1(zk)(w)¯zk〉H2=F-1(zj)(w).
Hence h=Kw∈HF,2 and ew is continuous on HF,2.
Letp≥2. If f∈HF,p, then
(3.3)|f(w)|≤∥Kw∥HF,2∥f∥HF,2≤∥Kw∥HF,2∥f∥HF,p,
so ew is continuous on HF,p.
(b) Let f∈HF,1. Then for each 0<r<1, fr∈HF,2 and so
(3.4)fr(w)=〈fr,Kw〉HF,2=〈F(fr),F(Kw)〉H2=∫02πF(fr)(eiθ)F(Kw)¯(eiθ)dθ2π.
Also by [1, Theorem 2.6], ∥(F(f))r-F(f)∥1→0 as r→1. Therefore, by Holder's inequality and the fact that F(Kw)=∑j=0∞F-1(zj)(w)¯zj, we have
(3.5)|∫02π(F(f))r(eiθ)F(Kw)¯(eiθ)dθ2π-∫02πF(f)(eiθ)F(Kw)¯(eiθ)dθ2π|≤∥F(Kw)∥∞∫02π|(F(fr)-F(f))(eiθ)|dθ2π≤∥F(Kw)∥∞∥(F(f))r-F(f)∥1→0,
as r→1. So we obtain
(3.6)f(w)=limr→1fr(w)=limr→1∫02πF(fr)(eiθ)F(Kw)¯(eiθ)dθ2π=∫02πF(f)(eiθ)F(Kw)¯(eiθ)dθ2π.
Hence
(3.7)|f(w)|=|∫02πF(f)(eiθ)F(Kw)¯(eiθ)dθ2π|≤∥F(Kw)∥∞∫02π|F(f)(eiθ)|dθ2π=∥F(Kw)∥∞∥f∥HF,1
for each f∈HF,1. Now let 1≤p<2. If f∈HF,p, then
(3.8)|f(w)|≤∥F(Kw)∥∞∥f∥HF,1≤∥F(Kw)∥∞∥f∥HF,p,
so the result follows.Suppose thatw and F satisfy the hypotheses in the first line of Theorem 3.1. It is easy to see that if ew is continuous on HF,2, then according to the proof of the previous theorem, F(Kw)=∑j=0∞F-1(zj)(w)¯zj.Now we give two examples for the preceding theorem.Example 3.2.
(a) Let w∈U and F(∑j=0∞ajzj)=a0+∑j=1∞jajzj, where ∑j=0∞ajzj∈H(U). It is easy to see that ∑j=0∞F-1(zj)(w)¯zj=1+∑j=1∞((w¯)j/j)zj∈H∞ and for each f∈HF,1, F(fr)=(F(f))r. Therefore, by Theorem 3.1, ew is continuous on HF,p for each 1≤p<∞.
(b) Let p≥2 and F(∑j=0∞ajzj)=∑j=0∞bjzj such that (b0,b1,…,bn) is a permutation of (a0,a1,…,an) for some n∈ℕ and bk=ak for k>n. Then by Theorem 3.1, for each point w in the open unit disk, ew is continuous on HF,p. Also if p=2 and S is the above permutation, then
(3.9)F(Kw)=∑j=0∞F-1(zj)(w)¯zj=∑j=0∞(w̅)S-1(j)zj.
Therefore, Kw(z)=∑j=0∞(w̅)S-1(j)zS-1(j).There is a Banach spaceHF,p such that it does not satisfy the conditions of Theorem 3.1, but for each w∈U, ew is continuous on HF,p. For example, let 1≤p<∞, g(z)=1/2-z, and F(f)=fg for each f∈H(U). Then 1 is not in RF and F((1)r)≢(F(1))r. By the following theorem, we see that although the hypotheses of Theorem 3.1 do not hold, ew is continuous on HF,p for each w∈U.Theorem 3.3.
Let1≤p<∞,  w∈U,  h∈H(U),  h≢0, and for each f∈H(U),  F(f)=fh. Then ew is continuous on HF,p.Proof.
We break the proof into two parts.
(1) Let h(w)≠0. If |w|<r<1 and Γr is the circle of radius r with center at the origin, then the Cauchy formula shows that for any f in HF,p,
(3.10)f(w)h(w)=12πi∫Γrf(ζ)h(ζ)ζ-wdζ=12πi∫02πf(reiθ)h(reiθ)reiθ-wrieiθdθ=∫02πf(reiθ)h(reiθ)rr-we-iθdθ2π.
It follows that
(3.11)|f(w)||h(w)|≤∥(fh)r∥p∥rr-we-iθ∥q,
where 1/p+1/q=1. Now if r tends to 1, |r/(r-we-iθ)| converges uniformly to the bounded function |1-weiθ|-1 and ∥(fh)r∥p≤∥fh∥p. Hence there is an M<∞ such that
(3.12)|f(w)|≤M|h(w)|∥f∥HF,p,
and the result follows.
(2) Let h(w)=0. Then h(z)=(z-w)mh0(z), where m∈ℕ, h0∈H(U), and h0(w)≠0. Let F1(f)=fh0 for each f∈H(U), it is easy to see that HF,p⊆HF1,p. Then by the preceding part, there is a constant 0<C<∞ such that
(3.13)|f(w)|p≤C∥fh0∥pp=C∫02π|f(eiθ)|p|h0(eiθ)|p|eiθ-w|mp|eiθ-w|mpdθ2π≤C(1-|w|)mp∫02π|f(eiθ)h(eiθ)|pdθ2π=C(1-|w|)mp∥f∥HF,pp
for each f∈HF,p. So ew is continuous on HF,p.There is an example ofHF,p such that HF,2 is not a Hilbert space and ew is continuous for some w∈U.Example 3.4.
Letp=2 and F(∑j=0∞ajzj)=∑j=0∞(aj/2j)zj, where ∑j=0∞ajzj∈H(U). We can show that for each w∈(1/2)U, ew is continuous on HF,2. Suppose f(z)=∑j=0∞ajzj∈HF,2, we have
(3.14)|f(w)|=|∑j=0∞ajwj|=|∑j=0∞aj2j(2w)j|≤(∑j=0∞|aj2j|2)1/2(∑j=0∞|(2w)j|2)1/2=∥f∥HF,2(∑j=0∞|2w|2j)1/2,
as desired. Also it is easy to see that ew is continuous on HF,p for each p>2 and w∈(1/2)U.


## 4. Continuity of the Composition Operators onHF,p

In the most important classical spaces, all analytic maps of the unit disk into itself induce bounded composition operators, but there are analytic self-maps of the unit disk which do not give bounded composition operators on some generalized Hardy spaces.Example 4.1.
(a) Suppose that φ(z)=z2 and for each f∈H(U), F(f)=fg, where g(z)=1-z. Let fj(z)=1+z+z2+⋯+zj for each j∈ℕ∪{0}. Then fj∈HF,2. We see that
(4.1)∥Cz2(fj)∥HF,22=∥fj(z2)g(z)∥22=∥(1+z2+⋯+z2j)(1-z)∥22=2j+2,∥fj∥HF,22=∥fjg∥22=∥1-zj+1∥22=2.
So ∥Cz2∥2≥j+1 and Cz2 is not bounded on HF,2.
(b) Let F(∑j=0∞ajzj)=∑j=0∞ajβ(j)zj, where ∑j=0∞ajzj∈H(U) and {β(j)} is given in [2, Example 3.4]. If φ(z)=z2, then in [2, Example 3.4] shows that Cφ is not bounded on HF,2.In this section, we investigate the continuity of the composition operator on some spacesHF,p in terms of a Carleson measure criterion. This criterion has been used to characterize the boundedness and the compactness of the composition operators in different papers (see, e.g., [16, 30]).Definition 4.2.
A positive measureμ on U is called a Carleson measure (in U) if there is a constant K<∞ such that μ(S(b,h))<Kh for all b∈∂U and 0<h<1, where S(b,h)={z∈U:|z-b|<h}.Proposition 4.3.
Let1≤p<∞ and P⊆RF. If μ is a finite, positive Borel measure on U, then the following conditions are equivalent.(a)
μ is a Carleson measure in U.(b)
There is a constantC<∞ such that
(4.2)∫U|F(f)|pdμ≤C∥f∥HF,pp
for all f in HF,p.Proof.
By [2, Theorem 2.33], (a)⇒(b) is clear. Now we prove that (b) implies (a). For each p0∈P, there is a qp0∈HF,p such that F(qp0)=p0 and
(4.3)∫U|p0|pdμ≤C∥p0∥pp.
If g∈Hp, then there is a sequence {pn} in P such that ∥pn-g∥p→0 as n→∞. So
(4.4)∥pn∥pp→∥g∥pp,asn→∞.
By (4.3), we obtain
(4.5)∫U|pn|pdμ≤C∥pn∥pp
for each n∈ℕ. The right-hand side of (4.5) is bounded, hence there is a subsequence {pnk} such that ∫U|pnk|pdμ converges. Since point evaluations at the points of U are all continuous on Hp, the Principle of Uniform Boundedness implies that {pnk} converges to g uniformly on compact subsets of U. In particular, for each 0<r<1, the convergence is uniform on rU. Therefore, for each 0<r<1, ∫rU|pnk|pdμ→∫rU|g|pdμ as k→∞. Since {∫U|pnk|pdμ} is a bounded sequence, there is an M>0 such that ∫rU|g|pdμ≤M for each 0<r<1. Hence ∫U|g|pdμ≤M and g∈Lp(μ). Now let {rm} be an increasing sequence which converges to 1 and let ϵ>0. By [31, page 32], there is an integer m0 such that ∫A|g|pdμ<ϵ, where A=U-rm0U. Let B=rm0U. Then by [31, page 73], we have lim supk→∞∫A|pnk|pdμ≤ϵ. So there is a K0 such that for each k>K0, ∫A|pnk|pdμ≤ϵ. Again apply [31, page 32] for |pn1|p,|pn2|p,…,|pnK0|p. Therefore, there are m1,m2,…,mK0 such that
(4.6)∫U-rmkU|pnk|pdμ≤ϵ
for each 1≤k≤K0. Let m′=max{m0,m1,…,mK0}. Therefore, for each m>m′ and k∈ℕ, ∫U-rmU|pnk|pdμ≤ϵ. So by [32, Theorem 19.10], we have
(4.7)limk→∞∫U|pnk|pdμ=limk→∞limm→∞∫rmU|pnk|pdμ=limm→∞limk→∞∫rmU|pnk|pdμ=limm→∞∫rmU|g|pdμ=∫U|g|pdμ.
So by the preceding relation and (4.4) and (4.5), we have
(4.8)∫U|g|pdμ≤C∥g∥pp
for every g∈Hp. Thus the result follows from [2, Theorem 2.33].Definition 4.4.
Forb on the unit circle and 0<h<1, let
(4.9)𝒮(b,h)={z∈U̅:|z-b|<h}.
From now on, unless otherwise stated, we assume thatHF,p is a Banach space.Proposition 4.5.
Let1≤p<∞ and P⊆RF. If μ is a finite, positive Borel measure on U̅, then the following conditions are equivalent.(a)
There is a constantK<∞ such that μ(𝒮(b,h))<Kh for all b∈∂U and 0<h<1.(b)
There is a constantC<∞ such that
(4.10)∫U¯|F(f)|pdμ≤C∥f∥HF,pp
for all f∈HF,p.Proof.
The implication (a)⇒(b) follows by Proposition 4.3 and is exactly the same as the proof of [2, Theorem 2.35(1)⇒(2)]. For the other direction, by Theorem 2.2, Hp⊆RF. Hence for each g∈Hp, we have
(4.11)∫U¯|g|pdμ≤C∥g∥pp.
Then the result follows from [2, Theorem 2.35(2)⇒(1)].In the following theorem, we use the techniques used in [2, Theorem 3.12 part (1)].Theorem 4.6.
Suppose1≤p<∞,  φ and φ0 are analytic self-maps of U. Let P⊆RF and for each f∈HF,p,  F(f∘φ)=F(f)∘φ0. Then Cφ is a bounded operator on HF,p if and only if μ(𝒮(ζ,h))=◯(h) for all ζ in ∂U and 0<h<1, where μ(E)=σ((φ0*)-1(E)),  σ is normalized Lebesgue measure on the unit circle, and E is a subset of the closed disk U¯.Proof.
Letf∈HF,p. By [2, Theorem 2.25] and [33, page 163], we obtain
(4.12)∫∂U|(F(f∘φ))*|pdσ=∫∂U|(F(f)∘φ0)*|pdσ=∫∂U|(F(f))*∘φ0*|pdσ=∫U̅|(F(f))*|pdμ.
So the result follows from Proposition 4.5.Remark 4.7.
Ifp, φ, φ0 and F satisfy the hypotheses in Theorem 4.6, then by [2, Theorem 3.12, part (1)], Cφ is a bounded operator on HF,p if and only if Cφ0 is a bounded operator on Hp. Since we know that the composition operators are always bounded on each of the Hardy spaces Hp, Cφ is a bounded operator on HF,p.


## 5. Compactness of the Composition Operators onHF,p

In this section, we investigate the compactness of the composition operators onHF,p.The idea of the proof of the following theorem is similar to the proof of Proposition3.11 in [2]. A detailed proof of another similar result can be found, for example, in [18].Theorem 5.1.
Suppose1≤p<∞ and for each w∈U,  ew is continuous. Also if fn→f as n→∞ uniformly on compact subsets of U, then F(fn)→F(f) in H(U) as n→∞. Then the following conditions are equivalent. (a)
Cφ is compact on HF,p.(b)
If{fn} is a bounded sequence in HF,p and fn→0 as n→∞ uniformly on compact subsets of U, then ∥fn∘φ∥HF,p→0 as n→∞.Proof.
The implication (a)⇒(b) follows exactly as the proof of [2, Theorem 3.11].
Now we show that (b)⇒(a). Let {fn} be a bounded sequence in HF,p. Since ew is continuous for each w∈U, {fn} is a normal family. So there is a function f∈H(U) and a subsequence {fnk} such that fnk→f as k→∞ uniformly on compact subsets of U. Hence F(fnk)→F(f) in H(U) as k→∞. It is easy to see that f∈HF,p. Therefore, {fnk-f} is a bounded sequence in HF,p such that it converges uniformly to 0 on compact subsets of U. By the hypotheses, we conclude that fnk∘φ→f∘φ in HF,p as k→∞. Thus Cφ is a compact operator.Corollary 5.2.
Let1≤p<∞,  g∈Hp,  g≢0, and F(f)=fg for each f∈H(U). Let φ be an analytic self-map of U and φ(U)¯⊆U. Then Cφ is a compact operator on HF,p.In the rest of this section, we investigate the relation between compactness ofCφ on HF,p and Hp.Theorem 5.3.
Letp,  φ0,  φ,  F, and μ satisfy the hypotheses in Theorem 4.6. Then Cφ is compact on HF,p if and only if μ(𝒮(ζ,h))=∘(h) as  h→0 uniformly in ζ in ∂U.Proof.
Letμ(𝒮(ζ,h))=∘(h) as h→0 uniformly in ζ in ∂U and {fn} be a bounded sequence in HF,p. Then by [2, Theorem 3.12, part (2)], there is a subsequence {F(fnk)} such that {Cφ0(F(fnk))} converges in Hp. Since F(fnk∘φ)=F(fnk)∘φ0, {fnk∘φ} converges in HF,p. Therefore, Cφ is a compact operator on HF,p.
Conversely, let{fn} be a bounded sequence in Hp. By Theorem 2.2, Hp⊆RF, so there is a sequence {gn} in HF,p such that F(gn)=fn. Since Cφ is compact on HF,p, we may extract a subsequence {gnk} such that {Cφ(gnk)} converges in HF,p. So Cφ0 is a compact operator on Hp and [2, Theorem 3.12, part (2)] implies the result.Remark 5.4.
Ifp, φ, φ0, and F satisfy the hypotheses in Theorem 5.3, then by [2, Theorem 3.12, part (2)], Cφ is a compact operator on HF,p if and only if Cφ0 is a compact operator on Hp.Now we present an example for Remarks4.7 and 5.4.Example 5.5.
Letp=2, φ0(z)=φ(z)=z2, and F(Σj=0∞ajzj)=∑j=0∞ajcjzj, where Σj=0∞ajzj∈H(U) and for every odd positive integer j and every n∈ℕ, c2nj=cj=1/j and c0=1. It is easy to see that by Remark 4.7, Cφ is a bounded operator on HF,2 and by Remark 5.4 and [2, Corollary 3.14] Cφ is not a compact operator on HF,2.



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*Source: 803230-2010-02-10.xml*