1.Introduction ShanliYeandZhengyuanZhuo WeightedCompositionOperatorsfromHardytoZygmundTypeSpaces ResearchArticle

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Volume 2013, Article ID 365286,10pages http://dx.doi.org/10.1155/2013/365286

Research Article

Weighted Composition Operators from Hardy to Zygmund Type Spaces

Shanli Ye and Zhengyuan Zhuo

Department of Mathematics, Fujian Normal University, Fuzhou 350007, China

Correspondence should be addressed to Shanli Ye; ye [email protected] Received 20 January 2013; Accepted 24 March 2013

Academic Editor: Yansheng Liu

Copyright © 2013 S. Ye and Z. Zhuo. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper aims at studying the boundedness and compactness of weighted composition operator between spaces of analytic functions. We characterize boundedness and compactness of the weighted composition operator𝑢𝐶_𝜙from the Hardy spaces𝐻^𝑝 to the Zygmund type spacesZ_𝛼= {𝑓 ∈ 𝐻(𝐷) :sup_𝑧∈𝐷(1 − |𝑧|²)^𝛼|𝑓^󸀠󸀠(𝑧)| < ∞}and the little Zygmund type spacesZ_𝛼,0in terms of function theoretic properties of the symbols𝑢and𝜙.

1. Introduction

Let𝐷 = {𝑧 : |𝑧| < 1}be the open unit disk in the complex planeC and𝑇 = {𝑧 : |𝑧| = 1}its boundary, and𝐻(𝐷)denotes the set of all analytic functions on𝐷. An analytic self-map 𝜑 : 𝐷 → 𝐷induces the composition operator𝐶_𝜑on𝐻(𝐷), defined by𝐶_𝜑(𝑓) = 𝑓(𝜑(𝑧))for𝑓analytic on𝐷. It is a well- known consequence of Littlewood’s subordination principle that the composition operator𝐶_𝜑is bounded on the classical Hardy𝐻^𝑝 (0 < 𝑝 ≤ ∞)spaces, Bergman𝐴^𝑝 (0 < 𝑝 ≤ ∞) spaces, and Bloch spaces (see, e.g., [1–4]).

Let 𝑢 be a fixed analytic function on the open unit disk. Define a linear operator𝑢𝐶_𝜑 on the space of analytic functions on𝐷, called a weighted composition operator, by 𝑢𝐶_𝜑𝑓 = 𝑢 ⋅ (𝑓 ∘ 𝜑), where𝑓is an analytic function on𝐷. We can regard this operator as a generalization of a multiplication operator and a composition operator. In recent years the weighted composition operator has received much attention and appears in various settings in the literature. For example, it is known that isometries of many analytic function spaces are weighted composition operators (see [5], for instance).

Their boundedness and compactness have been studied on various Banach spaces of analytic functions, such as Hardy, Bergman, BMOA, Bloch-type, and Zygmund spaces; see, for example, [6–11]. Also, it has been studied from one Banach space of analytic functions to another; one may see [12–23].

The purpose of this paper is to consider the weighted composition operators from the Hardy space𝐻^𝑝 (0 < 𝑝 <

∞) to the Zygmund type spacesZ_𝛼. Our main goal is to characterize boundedness and compactness of the operators 𝑢𝐶_𝜑from𝐻^𝑝toZ_𝛼in terms of function theoretic properties of the symbols𝑢and𝜑.

Now we give a detailed definition of these spaces. For0 ≤ 𝑟 < 1,𝑓(𝑧) ∈ 𝐻(𝐷), we set

𝑀_𝑝(𝑟, 𝑓) = ( 1 2𝜋∫^2𝜋

0 󵄨󵄨󵄨󵄨󵄨𝑓 (𝑟𝑒^𝑖𝜃)󵄨󵄨󵄨󵄨󵄨^𝑝𝑑𝜃)

1/𝑝

, 0 < 𝑝 < ∞, 𝑀_∞(𝑟, 𝑓) = max

0≤𝜃≤2𝜋󵄨󵄨󵄨󵄨󵄨𝑓 (𝑟𝑒^𝑖𝜃)󵄨󵄨󵄨󵄨󵄨.

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For0 < 𝑝 ≤ ∞, the Hardy space 𝐻^𝑝 consists of those functions𝑓 ∈ 𝐻(𝐷), for which

󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝= sup

0≤𝑟<1𝑀_𝑝(𝑟, 𝑓) < ∞. (2) It is well known that with norm (2) the𝐻^𝑝space is a Banach space if1 ≤ 𝑝 ≤ ∞, for0 < 𝑝 < 1,𝐻^𝑝space is a nonlocally convex topological vector space, and𝑑(𝑓, 𝑔) = ‖𝑓 − 𝑔‖^𝑝_𝑝is a complete metric for it. For more information about the𝐻^𝑝 space, one may see these books, for example, [24,25].

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For𝛼 > 0the𝛼-Bloch space𝛽_𝛼 consists of all analytic functions𝑓defined on𝐷such that

󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝛽𝛼 =sup{(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑓^󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 : 𝑧 ∈ 𝐷} , 0 < 𝛼 < +∞.

(3) The spaceZ_𝛼consists of all analytic functions𝑓defined on 𝐷such that

󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩Z𝛼 =sup{(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑓^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 : 𝑧 ∈ 𝐷} , 0 < 𝛼 < +∞.

(4) When𝛼 = 1, it is called the Zygmund space. From a theorem by Zygmund (see [26, vol. I, p. 263] or [24, Theorem 5.3]), we see that𝑓 ∈Z₁if and only if𝑓is continuous in the close unit disk𝐷 = {𝑧 : |𝑧| ≤ 1}and the boundary function𝑓(𝑒^𝑖𝜃)such that

󵄨󵄨󵄨󵄨󵄨𝑓 (𝑒^{𝑖(𝜃+ℎ)}) + 𝑓 (𝑒^{𝑖(𝜃−ℎ)}) − 2𝑓 (𝑒^𝑖𝜃)󵄨󵄨󵄨󵄨󵄨

ℎ < ∞. (5)

When𝛼 > 1, from Proposition 8 of [27], we know thatZ_𝛼= 𝛽_𝛼−1. Then the spaceZ_𝛼 is called a Zygmund type space if 0 < 𝛼 ≤ 1. However, all results in this paper are valid for all Z_𝛼spaces (𝛼 > 0). An analytic function𝑓 ∈ 𝐻(𝐷)is said to belong to the little Zymund type spaceZ_𝛼,0which consists of all𝑓 ∈Z_𝛼satisfying lim_{|𝑧| → 1}−(1 − |𝑧|²)^𝛼|𝑓^󸀠󸀠(𝑧)| = 0. It can be easily proved thatZ_𝛼is a Banach space under the norm

󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩∗ = 󵄨󵄨󵄨󵄨𝑓 (0)󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨󵄨𝑓^󸀠(0)󵄨󵄨󵄨󵄨󵄨 +󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩Z𝛼. (6) And the polynomials are norm-dense in closed subspace Z_𝛼,0. For some other information on this space and some operators on it, see, for example, [28–31].

Throughout this paper, constants are denoted by𝐶,𝐶(𝑝), they are positive, and𝐶(𝑝)are only depending on𝑝and may differ from one occurrence to the another.

2. Auxiliary Results

In order to prove the main results of this paper. We need some auxiliary results. The first lemma is well known.

Lemma 1 (see [24, p. 65]). For𝑝 > 1, there exists a constant 𝐶(𝑝)such that

∫^2𝜋

0

𝑑𝜃

|1 − 𝑧|^𝑝 ≤ 𝐶 (𝑝)

(1 − |𝑧|²)^𝑝−1 for every 𝑧 ∈ 𝐷. (7) Lemma 2. Suppose that0 < 𝑝 < ∞,𝑓 ∈ 𝐻^𝑝; then

󵄨󵄨󵄨󵄨󵄨𝑓^(𝑛)(𝑧)󵄨󵄨󵄨󵄨󵄨 ≤ 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

(1 − |𝑧|²)^1/𝑝+𝑛 (8)

for every𝑧 ∈ 𝐷and all nonnegative integer𝑛 = 0, 1, 2, . . ..

Proof. We use induction on𝑛. The case𝑛 = 0holds because it is Exercise 5 in [25, p. 85]. Assume the case𝑛 = 𝑘holds. Fix

0 < 𝑟 < 1and let𝑔(𝑧) = 𝑓^(𝑘)(𝑟𝑧). Then𝑔(𝑧)is in𝐻^∞ ⊂ 𝛽₁, and‖𝑔‖_𝛽₁ ≤ ‖𝑔‖_∞. It follows that

(1 − |𝑧|²) 󵄨󵄨󵄨󵄨󵄨𝑔^󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 = (1 − |𝑧|²) 󵄨󵄨󵄨󵄨󵄨𝑟𝑓^(𝑘+1)(𝑟𝑧)󵄨󵄨󵄨󵄨󵄨

≤ 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

(1 − |𝑟𝑧|²)^1/𝑝+𝑘

≤ 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

(1 − |𝑧|²)^1/𝑝+𝑘.

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Let𝑟 → 1⁻; we have

󵄨󵄨󵄨󵄨󵄨𝑓^(𝑘+1)(𝑧)󵄨󵄨󵄨󵄨󵄨 ≤ 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

(1 − |𝑧|²)^{1/𝑝+𝑘+1}. (10) Then the case𝑛 = 𝑘 + 1holds. Hence (8) holds.

Lemma 3. For0 < 𝑝 < ∞, suppose𝑢𝐶_𝜑 : 𝐻^𝑝 → Z_𝛼,0 is a bounded operator. Then𝑢𝐶_𝜑 : 𝐻^𝑝 → Z_𝛼is a bounded operator.

This is obvious.

3. Boundedness of 𝑢𝐶

_𝜑

from 𝐻

^𝑝

(0 < 𝑝 < ∞) to Z

_𝛼

and Z

_𝛼,0

In this section we characterize bounded weighted composition operators from the Hardy space𝐻^𝑝 (0 < 𝑝 < ∞)to the Zygmund spacesZ_𝛼.

Theorem 4. Let𝛼 > 0,0 < 𝑝 < ∞, and𝑢be an analytic function on the unit disc𝐷and𝜑an analytic self-map of𝐷.

Then 𝑢𝐶_𝜑 is a bounded operator from𝐻^𝑝 to the Zygmund spacesZ_𝛼if and only if the following are satisfied:

sup𝑧∈𝐷

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 < ∞, (11) sup𝑧∈𝐷

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 < ∞, (12) sup𝑧∈𝐷

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 < ∞. (13) Proof . Suppose𝑢𝐶_𝜑 is bounded from𝐻^𝑝 to the Zygmund spacesZ_𝛼. Then we can easily obtain the following results by taking𝑓(𝑧) = 1and𝑓(𝑧) = 𝑧in𝐻^𝑝, respectively:

𝑢 ∈Z_𝛼,

sup𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧) + 𝜑 (𝑧) 𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨

< +∞.

(14)

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By (14) and the boundedness of the function𝜑(𝑧), we get 𝐾₁=sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨 < +∞.

(15) Let𝑓(𝑧) = 𝑧²in𝐻^𝑝again; in the same way we have

sup𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨󵄨 4𝜑(𝑧)𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑²(𝑧) 𝑢^󸀠󸀠(𝑧) +2𝑢 (𝑧) (𝜑 (𝑧) 𝜑^󸀠󸀠(𝑧) + (𝜑^󸀠(𝑧))²)󵄨󵄨󵄨󵄨󵄨󵄨󵄨 < ∞.

(16) Using these facts and the boundedness of the function𝜑(𝑧) again, we get

𝐾₂=sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨 < +∞. (17) Fix𝑎 ∈ 𝐷; we take the test functions

𝑓_𝑎(𝑧) = − 1 − |𝑎|²

(1 − 𝑎𝑧)^1+1/𝑝 + 2 (1 − |𝑎|²)²

(1 − 𝑎𝑧)^2+1/𝑝 − (1 − |𝑎|²)³ (1 − 𝑎𝑧)^3+1/𝑝

(18) for 𝑧 ∈ 𝐷. From Lemma 1we obtain that 𝑓_𝑎 ∈ 𝐻^𝑝 and sup_𝑎‖𝑓_𝑎‖_𝑝 ≤ 𝐶(𝑝) < ∞ with a direct calculation. Since 𝑓_𝑎(𝑎) = 0,𝑓_𝑎^󸀠(𝑎) = 0, and𝑓_𝑎^󸀠󸀠(𝑎) = −2𝑎²/(1 − |𝑎|²)^2+1/𝑝, it follows that, for all𝜆 ∈ 𝐷with|𝜑(𝜆)| > 1/2, we have

𝐶󵄩󵄩󵄩󵄩𝑓^𝑎󵄩󵄩󵄩󵄩∗ ≥ 󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑𝑓_𝑎󵄩󵄩󵄩󵄩󵄩∗≥sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝑢𝐶^𝜑𝑓_𝑎)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

=sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑓_𝑎^󸀠(𝜑 (𝑧)) + 𝑓_𝑎^󸀠󸀠(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)

+ 𝑢^󸀠󸀠(𝑧) 𝑓_𝑎(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨󵄨󵄨.

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Let𝑎 = 𝜑(𝜆); it follows that 𝐶󵄩󵄩󵄩󵄩𝑓^𝑎󵄩󵄩󵄩󵄩∗≥ (1 − |𝜆|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝜆) 𝑢^󸀠(𝜆) + 𝜑^󸀠󸀠(𝜆) 𝑢 (𝜆)) 𝑓_𝜑(𝜆)^󸀠 (𝜑 (𝜆)) + 𝑓_𝜑(𝜆)^󸀠󸀠 (𝜑 (𝜆)) (𝜑^󸀠(𝜆))²𝑢 (𝜆)

+ 𝑢^󸀠󸀠(𝜆) 𝑓_𝜑(𝜆)(𝜑 (𝜆))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

= (1−|𝜆|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨

(𝜑^󸀠(𝜆))²𝑢 (𝜆) 2(𝜑 (𝜆))² (1−󵄨󵄨󵄨󵄨𝜑 (𝜆)󵄨󵄨󵄨󵄨²)^2+1/𝑝

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨

≥ 1 2

(1 − |𝜆|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝜆) (𝜑^󸀠(𝜆))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝜆)󵄨󵄨󵄨󵄨²)^2+1/𝑝 .

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For all𝜆 ∈ 𝐷with|𝜑(𝜆)| ≤ 1/2, by (17), we have

sup𝜆∈𝐷

(1 − |𝜆|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝜆) (𝜑^󸀠(𝜆))²󵄨󵄨󵄨󵄨󵄨󵄨

1 − 󵄨󵄨󵄨󵄨𝜑 (𝜆)󵄨󵄨󵄨󵄨^1/𝑝+2

≤ (4

3)^1/𝑝+2sup

𝜆∈𝐷(1 − |𝜆|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝜆) (𝜑^󸀠(𝜆))²󵄨󵄨󵄨󵄨󵄨󵄨

< +∞.

(21)

Hence (12) holds.

Next, fix𝑎 ∈ 𝐷; we take another test functions

𝑔_𝑎(𝑧)= −3 1 − |𝑎|²

(1−𝑎𝑧)^1/𝑝+1+5 (1 − |𝑎|²)²

(1 − 𝑎𝑧)^1/𝑝+2−2 (1−|𝑎|²)³ (1−𝑎𝑧)^1/𝑝+3

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for 𝑧 ∈ 𝐷. From Lemma1 we obtain that 𝑔_𝑎 ∈ 𝐻^𝑝 and sup_𝑎‖𝑔_𝑎‖_𝑝 ≤ 𝐶(𝑝) < ∞with a direct calculation. Since 𝑔_𝑎(𝑎) = 0,𝑔^󸀠󸀠_𝑎(𝑎) = 0, and𝑔^󸀠_𝑎(𝑎) = 𝑎/(1 − |𝑎|²)^1/𝑝+1, it follows that, for all𝜆 ∈ 𝐷with|𝜑(𝜆)| > 1/2, we obtain that

𝐶󵄩󵄩󵄩󵄩𝑔^𝑎󵄩󵄩󵄩󵄩∗≥ 󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑𝑔_𝑎(𝑧)󵄩󵄩󵄩󵄩󵄩∗

≥sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝑢𝐶^𝜑𝑔_𝑎)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

=sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑔_𝑎^󸀠(𝜑 (𝑧)) + 𝑔^󸀠󸀠_𝑎 (𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)

+ 𝑢^󸀠󸀠(𝑧) 𝑔_𝑎(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

≥ (1 − |𝜆|²)^𝛼

(4)

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝜆) 𝑢^󸀠(𝜆) + 𝜑^󸀠󸀠(𝜆) 𝑢 (𝜆)) 𝑔_𝜑(𝜆)^󸀠 (𝜑 (𝜆)) + 𝑔^󸀠󸀠_𝜑(𝜆)(𝜑 (𝜆)) (𝜑^󸀠(𝜆))²𝑢 (𝜆)

+ 𝑢^󸀠󸀠(𝜆) 𝑔_𝜑(𝜆)(𝜑 (𝜆))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

= (1 − |𝜆|²)^𝛼

×󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

󵄨

(2𝜑^󸀠(𝜆) 𝑢^󸀠(𝜆) + 𝜑^󸀠󸀠(𝜆) 𝑢 (𝜆))

× 𝜑 (𝜆)

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝜆)󵄨󵄨󵄨󵄨²)^1/𝑝+1

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨

≥ 1 2

(1 − |𝜆|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝜆) 𝑢^󸀠(𝜆) + 𝜑^󸀠󸀠(𝜆) 𝑢 (𝜆)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝜆)󵄨󵄨󵄨󵄨²)^1/𝑝+1 .

(23)

For all𝜆 ∈ 𝐷with|𝜑(𝜆)| ≤ 1/2, by (15), we have

|^𝜑(𝜆)sup|^≤1/2

(1 − |𝜆|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝜆) 𝑢^󸀠(𝜆) + 𝜑^󸀠󸀠(𝜆) 𝑢 (𝜆)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝜆)󵄨󵄨󵄨󵄨²)^1/𝑝+1

≤ (4

3)^1/𝑝+1 sup

|^𝜑(𝜆)|^≤1/2(1 − |𝜆|²)^𝛼

× 󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝜆) 𝑢^󸀠(𝜆) + 𝜑^󸀠󸀠(𝜆) 𝑢 (𝜆)󵄨󵄨󵄨󵄨󵄨

< ∞.

(24)

Hence (13) holds.

Finally, fix𝑎 ∈ 𝐷, and, for all𝑧 ∈ 𝐷, let

ℎ_𝑎(𝑧) = −𝑝 + 3 𝑝 + 1

1 − |𝑎|² (1 − 𝑎𝑧)^1/𝑝+1 +2 (𝑝 + 3)

𝑝 + 2

(1 − |𝑎|²)²

(1 − 𝑎𝑧)^1/𝑝+2 − (1 − |𝑎|²)³ (1 − 𝑎𝑧)^1/𝑝+3.

(25)

From Lemma1 we obtain thatℎ_𝑎 ∈ 𝐻^𝑝 and sup_𝑎‖ℎ_𝑎‖_𝑝 ≤ 𝐶(𝑝) < ∞ with a direct calculation. Since ℎ^󸀠_𝑎(𝑎) = 0,

ℎ^󸀠󸀠_𝑎(𝑎) = 0, andℎ_𝑎(𝑎) = −2/(𝑝 + 1)(𝑝 + 2)(1 − |𝑎|²)^1/𝑝, it follows that, for all𝜆 ∈ 𝐷, we obtain that

𝐶󵄩󵄩󵄩󵄩ℎ^𝑎󵄩󵄩󵄩󵄩∗≥ 󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑ℎ_𝑎(𝑧)󵄩󵄩󵄩󵄩󵄩∗

≥sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝑢𝐶^𝜑ℎ_𝑎)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

=sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) ℎ^󸀠_𝑎(𝜑 (𝑧)) + ℎ^󸀠󸀠_𝑎(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)

+ 𝑢^󸀠󸀠(𝑧) ℎ_𝑎(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

≥ (1 − |𝜆|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝜆) 𝑢^󸀠(𝜆) + 𝜑^󸀠󸀠(𝜆) 𝑢 (𝜆)) ℎ^󸀠_𝜑(𝜆)(𝜑 (𝜆)) + ℎ^󸀠󸀠_𝜑(𝜆)(𝜑 (𝜆)) (𝜑^󸀠(𝜆))²𝑢 (𝜆)

+ 𝑢^󸀠󸀠(𝜆) ℎ_𝜑(𝜆)(𝜑 (𝜆))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

= (1 − |𝜆|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝜆) ℎ_𝜑(𝜆)(𝜑 (𝜆))󵄨󵄨󵄨󵄨󵄨

= 2

(𝑝 + 1) (𝑝 + 2)

(1 − |𝜆|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝜆)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝜆)󵄨󵄨󵄨󵄨²)^1/𝑝 .

(26) Then (11) holds.

Conversely, suppose that (11), (12), and (13) hold. For𝑓 ∈ 𝐻^𝑝, by Lemma2, we have the following inequality:

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝑢𝐶^𝜑𝑓)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

= (1 − |𝑧|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑓^󸀠(𝜑 (𝑧)) + 𝑓^󸀠󸀠(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)

+ 𝑢^󸀠󸀠(𝑧) 𝑓 (𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

≤ (1 − |𝑧|²)^𝛼

× 󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑓^󸀠(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨 + (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑓^󸀠󸀠(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

+ (1 − |𝑧|²) 󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧) 𝑓 (𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨

(5)

≤ (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

+(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

+(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

≤ 𝐶󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝,

󵄨󵄨󵄨󵄨𝑢(0)𝑓(𝜑(0))󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨󵄨𝑢^󸀠(0) 𝑓 (𝜑 (0))󵄨󵄨󵄨󵄨󵄨 +󵄨󵄨󵄨󵄨󵄨𝑢 (0) 𝑓^󸀠(𝜑 (0))󵄨󵄨󵄨󵄨󵄨

≤ ( |𝑢 (0)|

(1 − 󵄨󵄨󵄨󵄨𝜑 (0)󵄨󵄨󵄨󵄨²)^1/𝑝 + 󵄨󵄨󵄨󵄨󵄨𝑢^󸀠(0)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (0)󵄨󵄨󵄨󵄨²)^1/𝑝

+ |𝑢 (0)|

(1 − 󵄨󵄨󵄨󵄨𝜑 (0)󵄨󵄨󵄨󵄨²)^1/𝑝+1) 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝.

(27)

This shows that𝑢𝐶_𝜑is bounded. This completes the proof of Theorem4.

Theorem 5. Let𝛼 > 0,0 < 𝑝 < ∞, and𝑢be an analytic function on the unit disc𝐷and𝜑an analytic self-map of𝐷.

Then𝑢𝐶_𝜑 : 𝐻^𝑝 → Z_𝛼,0is a bounded operator provided that the following are satisfied:

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 = 0, (28)

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 = 0, (29)

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 = 0. (30) Conversely, if 𝑢𝐶_𝜑 : 𝐻^𝑝 → Z_𝛼,0 is a bounded operator, then𝑢 ∈ Z_𝛼,0,(11),(12), and(13)hold, and the following are satisfied:

|𝑧| → 1lim⁻(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨 = 0, (31)

|𝑧| → 1lim⁻(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨 = 0. (32)

Proof. Assume that (28), (29), and (30) hold. Then for any𝜖 >

0, there is a constant𝛿,0 < 𝛿 < 1, such that𝛿 < |𝑧| < 1 implies

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 < 𝜖, (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 < 𝜖, (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 < 𝜖.

(33)

Then, for any𝑓 ∈ 𝐻^𝑝, from Lemma2we obtain that (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝑢𝐶^𝜑𝑓)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

= (1 − |𝑧|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑓^󸀠(𝜑 (𝑧)) + 𝑓^󸀠󸀠(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)

+ 𝑢^󸀠󸀠(𝑧) 𝑓 (𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

≤ (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨

× 1

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

+ (1−|𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨 1

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

+(1 − |𝑧|²) 󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝 ≤ 3󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝𝜖.

(34) Hence𝑢𝐶_𝜑𝑓 ∈ Z_𝛼,0for all𝑓 ∈ Z_𝛼,0. On the other hand, (25), (28), and (29) imply that (11), (12), and (13) hold; then 𝑢𝐶_𝜑: 𝐻^𝑝 → Z_𝛼is bounded by Theorem4. So𝑢𝐶_𝜑: 𝐻^𝑝 → Z_𝛼,0is bounded.

Conversely, assume that𝑢𝐶_𝜑is bounded from𝐻^𝑝to the little Zygmund type spaceZ_𝛼,0. Then𝑢 = 𝑢𝐶_𝜑1 ∈Z_𝛼,0. Also 𝑢𝜑 = 𝑢𝐶_𝜑𝑧 ∈Z_𝛼,0; thus

(1−|𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧)+𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧) + 𝜑 (𝑧) 𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 󳨀→ 0 (|𝑧| 󳨀→ 1⁻) .

(35) Since |𝜑| ≤ 1 and 𝑢 ∈ Z_𝛼,0, we have lim_{|𝑧| → 1}⁻(1 −

|𝑧|²)^𝛼|2𝜑^󸀠(𝑧)𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧)𝑢(𝑧)| = 0. Hence (32) holds.

(6)

Similarly,𝑢𝐶_𝜑𝑧²∈Z_𝛼,0; then

(1 − |𝑧|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨4𝜑 (𝑧) 𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑²(𝑧) 𝑢^󸀠󸀠(𝑧)

+2𝑢 (𝑧) (𝜑 (𝑧) 𝜑^󸀠󸀠(𝑧) + (𝜑^󸀠(𝑧))²)󵄨󵄨󵄨󵄨󵄨󵄨󵄨 󳨀→ 0 (|𝑧| 󳨀→ 1⁻) .

(36)

By (32),|𝜑| ≤ 1, and𝑢 ∈ Z_𝛼,0, we get that lim_{|𝑧| → 1}⁻(1 −

|𝑧|²)^𝛼|𝑢(𝑧)(𝜑^󸀠(𝑧))²| = 0; that is, (31) holds.

On the other hand, from Lemma3and Theorem4, we obtain that (11), (12), and (13) hold.

4. Compactness of 𝑢𝐶

_𝜑

In order to prove the compactness of𝑢𝐶_𝜑 from𝐻^𝑝 to the Zygmund spacesZ_𝛼, we require the following lemmas.

Lemma 6. Let 𝛼 > 0,0 < 𝑝 < ∞, and𝑢be an analytic function on the unit disc𝐷and𝜑an analytic self-map of𝐷.

Suppose that𝑢𝐶_𝜑is a bounded operator from𝐻^𝑝toZ_𝛼. Then 𝑢𝐶_𝜑is compact if and only if, for any bounded sequence{𝑓_𝑛} in𝐻^𝑝which converges to0uniformly on compact subsets of𝐷, one has‖𝑢𝐶_𝜑(𝑓_𝑛)‖_∗ → 0as𝑛 → ∞.

The proof is similar to that of Proposition 3.11 in [32]. The details are omitted.

Theorem 7. Let𝛼 > 0, 0 < 𝑝 < ∞,𝑢be an analytic function on the unit disc𝐷and𝜑an analytic self-map of𝐷. Then𝑢𝐶_𝜑 is a compact operator from𝐻^𝑝toZ_𝛼if and only if𝑢𝐶_𝜑is a bounded operator and the following are satisfied:

(i) lim

|^𝜑(𝑧)|^{→ 1}⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 = 0, (ii) lim

|^𝜑(𝑧)|^{→ 1}⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 = 0, (iii) lim

|𝜑(𝑧)| → 1⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 = 0.

(37)

Proof. Suppose that𝑢𝐶_𝜑is compact from𝐻^𝑝to the Zygmund type space Z_𝛼. Let {𝑧_𝑛} be a sequence in 𝐷 such that

|𝜑(𝑧_𝑛)| → 1 as 𝑛 → ∞. If such a sequence does not exist, then (37) are automatically satisfied. Without loss of

generality we may suppose that|𝜑(𝑧_𝑛)| > 1/2for all𝑛. We take the test functions

𝑓_𝑛(𝑧) = − 1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²

(1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+1 + 2 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)² (1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+2

− (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)³ (1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+3.

(38)

By a direct calculation, we may easily prove that {𝑓_𝑛} converges to 0 uniformly on compact subsets of 𝐷 and sup_𝑛‖𝑓_𝑛‖_𝑝 ≤ 𝐶(𝑝) < ∞. Then{𝑓_𝑛}is a bounded sequence in𝐻^𝑝which converges to0uniformly on compact subsets of 𝐷. Then lim_{𝑛 → ∞}‖𝑢𝐶_𝜑(𝑓_𝑛)‖_∗= 0by Lemma6. Note that

𝑓_𝑛(𝜑 (𝑧_𝑛)) = 0, 𝑓_𝑛^󸀠(𝜑 (𝑧_𝑛)) = 0, 𝑓_𝑛^󸀠󸀠(𝜑 (𝑧_𝑛)) = − 2𝜑 (𝑧_𝑛)²

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)^1/𝑝+2. (39) It follows that

󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑𝑓_𝑛󵄩󵄩󵄩󵄩󵄩∗≥ 󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑𝑓_𝑛󵄩󵄩󵄩󵄩󵄩Z𝛼

≥ (1 − 󵄨󵄨󵄨󵄨𝑧^𝑛󵄨󵄨󵄨󵄨²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝑢^󸀠(𝑧_𝑛) 𝜑^󸀠(𝑧_𝑛) + 𝜑^󸀠󸀠(𝑧_𝑛) 𝑢 (𝑧_𝑛))

× 𝑓_𝑛^󸀠(𝜑 (𝑧_𝑛)) + 𝑢 (𝑧_𝑛) 𝑓_𝑛^󸀠󸀠(𝜑 (𝑧_𝑛)) (𝜑^󸀠(𝑧_𝑛))² + 𝑢^󸀠󸀠(𝑧_𝑛) 𝑓_𝑛(𝜑 (𝑧_𝑛))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

= (1 − 󵄨󵄨󵄨󵄨𝑧^𝑛󵄨󵄨󵄨󵄨²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨

(𝜑^󸀠(𝑧_𝑛))²𝑢 (𝑧_𝑛) 2𝜑 (𝑧_𝑛)² (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)^1/𝑝+2

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨 .

(40) Then

𝑛 → ∞lim

(1 − 󵄨󵄨󵄨󵄨𝑧^𝑛󵄨󵄨󵄨󵄨²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧^𝑛) (𝜑^󸀠(𝑧_𝑛))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)^1/𝑝+2 = 0. (41) Next, let

𝑔_𝑛(𝑧) = − 3 1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²

(1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+1 + 5 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)² (1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+2

− 2 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)³ (1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+3.

(42)

(7)

By a direct calculation we obtain that𝑔_𝑛 󴁂󴀱 0 (𝑛 → ∞) on compact subsets of𝐷 and sup_𝑛‖𝑔_𝑛‖_𝑝 ≤ 𝐶(𝑝) < +∞.

Consequently, {𝑔_𝑛} is a bounded sequence in 𝐻^𝑝 which converges to 0 uniformly on compact subsets of 𝐷. Then lim_{𝑛 → ∞}‖𝑢𝐶_𝜑(𝑔_𝑛)‖_∗ = 0by Lemma6. Note that𝑔_𝑛(𝜑(𝑧_𝑛)) ≡ 0,𝑔^󸀠󸀠_𝑛(𝜑(𝑧_𝑛)) ≡ 0and𝑔_𝑛^󸀠(𝜑(𝑧_𝑛)) = 𝜑(𝑧_𝑛)/(1 − |𝜑(𝑧_𝑛)|²)^1/𝑝+1; it follows that

󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑𝑔_𝑛󵄩󵄩󵄩󵄩󵄩∗ ≥ 󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑𝑔_𝑛󵄩󵄩󵄩󵄩󵄩Z_𝛼

≥ (1 − 󵄨󵄨󵄨󵄨𝑧^𝑛󵄨󵄨󵄨󵄨²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝑢^󸀠(𝑧_𝑛) 𝜑^󸀠(𝑧_𝑛) + 𝜑^󸀠󸀠(𝑧_𝑛) 𝑢 (𝑧_𝑛))

× 𝑔^󸀠_𝑛(𝜑 (𝑧_𝑛)) + 𝑢 (𝑧_𝑛) 𝑔^󸀠󸀠_𝑛 (𝜑 (𝑧_𝑛)) (𝜑^󸀠(𝑧_𝑛))² + 𝑢^󸀠󸀠(𝑧_𝑛) 𝑔_𝑛(𝜑 (𝑧_𝑛))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

= (1 − 󵄨󵄨󵄨󵄨𝑧^𝑛󵄨󵄨󵄨󵄨²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝑢^󸀠(𝑧_𝑛) 𝜑^󸀠(𝑧_𝑛) + 𝜑^󸀠󸀠(𝑧_𝑛) 𝑢 (𝑧_𝑛))

× 𝜑 (𝑧_𝑛)

(1 −󵄨󵄨󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨󵄨󵄨²)^1/𝑝+1

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨.

(43)

Then lim_{𝑛 → ∞}(1 − |𝑧_𝑛|²)^𝛼(2𝑢^󸀠(𝑧_𝑛)𝜑^󸀠(𝑧_𝑛) + 𝜑^󸀠󸀠(𝑧_𝑛)𝑢(𝑧_𝑛))/(1 −

|𝜑(𝑧_𝑛)|²)^1/𝑝+1= 0.

Finally, let

ℎ_𝑛(𝑧) = −𝑝 + 3

𝑝 + 1 1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨² (1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+1

+2 (𝑝 + 3)

𝑝 + 2 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)² (1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+2

− (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)³ (1 − 𝜑 (𝑧_𝑛)𝑧)^1/𝑝+3.

(44)

By a direct calculation we obtain thatℎ_𝑛 󴁂󴀱 0 (𝑛 → ∞)on compact subsets of𝐷 and sup_𝑛‖ℎ_𝑛‖_𝑝 < ∞. Consequently, {ℎ_𝑛} is a bounded sequence in 𝐻^𝑝 which converges to 0 uniformly on compact subsets of 𝐷. Then lim_{𝑛 → ∞}‖𝑢𝐶_𝜑(ℎ_𝑛)‖_∗ = 0 by Lemma 6. Note that

ℎ_𝑛(𝜑(𝑧_𝑛)) = 2/(𝑝 + 1)(𝑝 + 2)(1 − |𝜑(𝑧_𝑛)|²)^1/𝑝,ℎ^󸀠_𝑛(𝜑(𝑧_𝑛)) ≡ 0, andℎ^󸀠󸀠_𝑛(𝜑(𝑧_𝑛)) ≡ 0; it follows that

󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑ℎ_𝑛󵄩󵄩󵄩󵄩󵄩∗ ≥ 󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑ℎ_𝑛󵄩󵄩󵄩󵄩󵄩Z𝛼

≥ (1 − 󵄨󵄨󵄨󵄨𝑧^𝑛󵄨󵄨󵄨󵄨²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝑢^󸀠(𝑧_𝑛) 𝜑^󸀠(𝑧_𝑛) + 𝜑^󸀠󸀠(𝑧_𝑛) 𝑢 (𝑧_𝑛))

× ℎ^󸀠_𝑛(𝜑 (𝑧_𝑛))+𝑢 (𝑧_𝑛) ℎ^󸀠󸀠_𝑛 (𝜑 (𝑧_𝑛))(𝜑^󸀠(𝑧_𝑛))² + 𝑢^󸀠󸀠(𝑧_𝑛) ℎ_𝑛(𝜑 (𝑧_𝑛))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

= 2(1 − 󵄨󵄨󵄨󵄨𝑧^𝑛󵄨󵄨󵄨󵄨²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧_𝑛)󵄨󵄨󵄨󵄨󵄨

(𝑝 + 1) (𝑝 + 2) (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧^𝑛)󵄨󵄨󵄨󵄨²)^1/𝑝.

(45)

Then lim_{𝑛 → ∞}((1 − |𝑧_𝑛|²)^𝛼|𝑢^󸀠󸀠(𝑧_𝑛)|/(1 − |𝜑(𝑧_𝑛)|²)^1/𝑝) = 0. The proof of the necessary is completed.

Conversely, Suppose that (37) hold. Since 𝑢𝐶_𝜑 is a bounded operator, from Theorem4, we have

𝑀₁≜sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨 < ∞, 𝑀₂≜sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 < ∞, 𝑀₃≜sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨 < ∞.

(46)

Let{𝑓_𝑛}be a bounded sequence in𝐻^𝑝with‖𝑓_𝑛‖_𝑝 ≤ 1and 𝑓_𝑛 → 0uniformly on compact subsets of𝐷. We only prove lim_{𝑛 → ∞}‖𝑢𝐶_𝜑(𝑓_𝑛)‖_∗ = 0by Lemma6. By the assumption, for any𝜖 > 0, there is a constant𝛿,0 < 𝛿 < 1, such that 𝛿 < |𝜑(𝑧)| < 1implies

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 < 𝜖, (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 < 𝜖, (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 < 𝜖.

(47)

(8)

Let𝐾 = {𝑤 ∈ 𝐷 : |𝑤| ≤ 𝛿}. Note that𝐾is a compact subset of𝐷. Then from Lemma2it follows that

󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑𝑓_𝑛󵄩󵄩󵄩󵄩󵄩Z_𝛼 =sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝑢𝐶^𝜑𝑓_𝑛)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

= (1 − |𝑧|²)^𝛼

×󵄨󵄨󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑓_𝑛^󸀠(𝜑 (𝑧)) + 𝑓_𝑛^󸀠󸀠(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)

+ 𝑢^󸀠󸀠(𝑧) 𝑓_𝑛(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨󵄨󵄨

≤sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼

× 󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑓_𝑛^󸀠(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨 +sup

𝑧∈𝐷(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑓^𝑛^󸀠󸀠(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

+sup

𝑧∈𝐷(1 − |𝑧|²) 󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧) 𝑓_𝑛(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨

≤ 3𝜖 + sup

|𝜑(𝑧)|≤𝛿(1 − |𝑧|²)^𝛼

× 󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑓_𝑛^󸀠(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨

+ sup

|𝜑(𝑧)|≤𝛿(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑓^𝑛^󸀠󸀠(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

+ sup

|𝜑(𝑧)|≤𝛿(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧) 𝑓_𝑛(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨

≤ 3𝜖 + 𝑀₃sup

𝑤∈𝐾󵄨󵄨󵄨󵄨󵄨𝑓^𝑛^󸀠(𝑤)󵄨󵄨󵄨󵄨󵄨 + 𝑀¹sup

𝑤∈𝐾󵄨󵄨󵄨󵄨󵄨𝑓^𝑛^󸀠󸀠(𝑤)󵄨󵄨󵄨󵄨󵄨 + 𝑀₂sup

𝑤∈𝐾󵄨󵄨󵄨󵄨𝑓^𝑛(𝑤)󵄨󵄨󵄨󵄨 .

(48)

As𝑛 → ∞,

󵄩󵄩󵄩󵄩󵄩𝑢𝐶^𝜑𝑓_𝑛󵄩󵄩󵄩󵄩󵄩Z_𝛼󳨀→ 0. (49) Hence𝑢𝐶_𝜑 is compact. This completes the proof of Theo- rem7.

In order to prove the compactness of 𝑢𝐶_𝜑 on the little Zygmund spacesZ_𝛼,0, we require the following lemma.

Lemma 8. Let𝑈 ⊂Z_𝛼,0. Then𝑈is compact if and only if it is closed, bounded and satisfies

|𝑧| → 1lim⁻sup

𝑓∈𝑈(1 − |𝑧|²) 󵄨󵄨󵄨󵄨󵄨𝑓^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 = 0. (50) The proof is similar to that of Lemma 1 in [1], but we omit it.

Theorem 9. Let𝛼 > 0,0 < 𝑝 < ∞,𝑢be an analytic function on the unit disc𝐷and𝜑an analytic self-map of𝐷. Then𝑢𝐶_𝜑

is compact from𝐻^𝑝to the little Zygmund type spacesZ_𝛼,0if and only if (28),(29), and(30)hold.

Proof. Assume that (28), (29), and (30) hold. By Theorem5, we know that𝑢𝐶_𝜑is bounded from𝐻^𝑝to the little Zygmund type spacesZ_𝛼,0. Suppose that𝑓 ∈ 𝐻^𝑝with‖𝑓‖_𝑝 ≤ 1. From Lemmas1and2we obtain that

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝑢𝐶^𝜑𝑓)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

≤ (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨(2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)) 𝑓^󸀠(𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨 + (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑓^󸀠󸀠(𝜑 (𝑧)) (𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

+ (1 − |𝑧|²) 󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧) 𝑓 (𝜑 (𝑧))󵄨󵄨󵄨󵄨󵄨

≤ (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨

× 1

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

+(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝

+(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩^𝑝,

(51) thus

sup{󵄨󵄨󵄨󵄨󵄨󵄨(1 − |𝑧|²)^𝛼(𝑢𝐶_𝜑𝑓)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨 : 𝑓 ∈ 𝐻^𝑝, 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝≤ 1}

≤ (1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨 1 1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨² +(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝜑^󸀠(𝑧))²𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 +(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 ,

(52) and it follows that

|𝑧| → 1lim⁻sup{󵄨󵄨󵄨󵄨󵄨󵄨(1 − |𝑧|²)^𝛼(𝑢𝐶_𝜑𝑓)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨 : 𝑓 ∈ 𝐻^𝑝, 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝≤ 1}

= 0.

(53) Hence𝑢𝐶_𝜑: 𝐻^𝑝 → Z_𝛼,0is compact by Lemma8.

Conversely, suppose that𝑢𝐶_𝜑: 𝐻^𝑝 → Z_𝛼,0is compact.

Firstly, it is obvious𝑢𝐶_𝜑is bounded, from Theorem5we have𝑢 ∈ Z_𝛼,0, and (31), (32) hold. On the other hand, we have

(9)

|𝑧| → 1lim⁻sup{󵄨󵄨󵄨󵄨󵄨󵄨(1 − |𝑧|²)^𝛼(𝑢𝐶_𝜑𝑓)^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨󵄨 : 𝑓 ∈ 𝐻^𝑝, 󵄩󵄩󵄩󵄩𝑓󵄩󵄩󵄩󵄩𝑝≤ 𝑀}

= 0,

(54) for some𝑀 > 0by Lemma6.

Next, note that the proof of Theorem4and the fact that the functions given in (18) are in𝐻^𝑝and have norms bounded independently of𝑎; we obtain that

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 = 0 (55) for|𝜑(𝑧)| > 1/2. However, if|𝜑(𝑧)| ≤ 1/2, by (31), we easily have

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2

≤ lim

|𝑧| → 1⁻(4

3)^1/𝑝+2(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨𝑢 (𝑧) (𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

= 0.

(56)

Similarly, note that the functions given in (22) and (25) are in𝐻^𝑝 and have norms bounded independently of𝑎, we obtain that

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 = 0,

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝 = 0

(57)

for|𝜑(𝑧)| > 1/2. However, if|𝜑(𝑧)| ≤ 1/2, from𝑢 ∈Z_𝛼,0and (32), we easily have

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1

≤ (4

3)^1/𝑝+1 lim

|𝑧| → 1⁻(1 − |𝑧|²)^𝛼

× 󵄨󵄨󵄨󵄨󵄨2𝜑^󸀠(𝑧) 𝑢^󸀠(𝑧) + 𝜑^󸀠󸀠(𝑧) 𝑢 (𝑧)󵄨󵄨󵄨󵄨󵄨

= 0,

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝

≤ lim

|𝑧| → 1⁻(4

3)^1/𝑝(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝑢^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨

= 0.

(58)

This completes the proof of Theorem9.

Remark 10. From Theorems5and9, we conjecture that𝑢𝐶_𝜑: 𝐻^𝑝 → Z_𝛼,0is compact if and only if𝑢𝐶_𝜑 : 𝐻^𝑝 → Z_𝛼,0is bounded.

Taking 𝑢(𝑧) = 1 from Theorems 4, 7, and 9, we obtain the following results about the characterization of the boundedness and compactness of the composition operator 𝐶_𝜑: 𝐻^𝑝 → Z_𝛼(orZ_𝛼,0).

Corollary 11. Let𝛼 > 0,0 < 𝑝 < ∞, and𝜑be an analytic self-map of𝐷. Then𝐶_𝜑 : 𝐻^𝑝 → Z_𝛼is a bounded operator if and only if the following are satisfied:

sup𝑧∈𝐷

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 < ∞, sup𝑧∈𝐷

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝜑^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 < ∞.

(59)

Corollary 12. Let𝛼 > 0,0 < 𝑝 < ∞, and𝜑be an analytic self-map of𝐷. Then𝐶_𝜑 : 𝐻^𝑝 → Z_𝛼is a compact operator if and only if𝐶_𝜑is bounded and the following are satisfied:

|𝜑(𝑧)| → 1lim ⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 = 0,

|𝜑(𝑧)| → 1lim ⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝜑^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 = 0.

(60)

Corollary 13. Let𝛼 > 0,0 < 𝑝 < ∞, and𝜑be an analytic self-map of𝐷. Then𝐶_𝜑 : 𝐻^𝑝 → Z_𝛼,0is a compact operator if and only if

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨󵄨(𝜑^󸀠(𝑧))²󵄨󵄨󵄨󵄨󵄨󵄨

(1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+2 = 0,

|𝑧| → 1lim⁻

(1 − |𝑧|²)^𝛼󵄨󵄨󵄨󵄨󵄨𝜑^󸀠󸀠(𝑧)󵄨󵄨󵄨󵄨󵄨 (1 − 󵄨󵄨󵄨󵄨𝜑 (𝑧)󵄨󵄨󵄨󵄨²)^1/𝑝+1 = 0.

(61)

In the formulation of corollary, we use the notation𝑀_𝑢on 𝐻(𝐷)defined by𝑀_𝑢𝑓 = 𝑢𝑓for𝑓 ∈ 𝐻(𝐷). Taking𝜑(𝑧) = 𝑧 from Theorems4,5,7, and9, we obtain the following results about the characterization of the boundedness and compactness of pointwise multiplier𝑀_𝑢: 𝐻^𝑝 → Z_𝛼(𝑜𝑟 Z_𝛼,0).

Corollary 14. Let𝛼 > 0,0 < 𝑝 < ∞. Then the pointwise multiplier𝑀_𝑢: 𝐻^𝑝 → Z_𝛼is a bounded operator if and only if

(i)𝑢 = 0if𝛼 < 2 + 1/𝑝;

(ii)𝑢 ∈ 𝐻^∞if𝛼 = 2 + 1/𝑝;

(iii) sup_𝑧∈𝐷(1 − |𝑧|²)^{𝛼−2−1/𝑝}|𝑢(𝑧)| < ∞if𝛼 > 2 + 1/𝑝.