Norm of a Volterra Integral Operator on Some Analytic Function Spaces

Abstract

Let 𝑓 be an analytic function in the unit disc 𝔻. The Volterra integral operator 𝐼_𝑓 is defined as follows: 𝐼_𝑓(ℎ)(𝑧)=∫^𝑧₀𝑓(𝑤)ℎ^′(𝑤)𝑑𝑤,ℎ∈𝐻(𝔻),𝑧∈𝔻. In this paper, we compute the norm of 𝐼_𝑓 on some analytic function spaces.

1. Introduction

Let 𝔻={𝑧∶|𝑧|<1} be the unit disk of complex plane ℂ and 𝐻(𝔻) the class of functions analytic in 𝔻. Denote by 𝑑𝜎 the normalized Lebesgue area measure in 𝔻 and 𝑔(𝑎,𝑧) the Green function with logarithmic singularity at 𝑎; that is, 𝑔(𝑎,𝑧)=−log|𝜑_𝑎(𝑧)|, where 𝜑_𝑎(𝑧)=(𝑎−𝑧)/(1−𝑎𝑧) is the Möbius transformation of 𝔻.

Let 0<𝑝<∞. The 𝑄_𝑝 is the space of all functions 𝑓∈𝐻(𝔻) such that 𝑓²_𝑄𝑝=𝑓(0)²+sup_{𝑎∈𝔻𝔻}𝑓^′(𝑧)²1−𝜑_𝑎(𝑧)^2𝑝𝑑𝜎(𝑧)<∞. (1) From [1, 2], we see that 𝑄₁= BMOA, the space of all analytic functions of bounded mean oscillation. When 𝑝>1, the space 𝑄_𝑝 is the same and equal to the Bloch space 𝔅, which consists of all 𝑓∈𝐻(𝔻) for which 𝑓_𝔅=𝑓(0)+sup_𝑧∈𝔻𝑓^′(𝑧)1−|𝑧|²<∞. (2) See [3, 4] for the theory of Bloch functions.

For 𝛼>0, the 𝛼-Bloch space, denoted by 𝔅^𝛼, is the space of all 𝑓∈𝐻(𝔻) such that 𝑓_𝔅^𝛼=𝑓(0)+sup_𝑧∈𝔻𝑓^′(𝑧)1−|𝑧|^2𝛼<∞. (3) It is clear that 𝔅^𝛼1⫋𝔅⫋𝔅^𝛼2 for 0<𝛼₁<1<𝛼₂<∞.

Let 1⩽𝑞⩽∞ and let 0⩽𝛼⩽1. The mean Lipschitz space Λ(𝑞,𝛼) consists of those functions 𝑓∈𝐻(𝔻) for which 𝑓_{Λ(𝑞,𝛼)}=𝑓(0)+sup_0⩽𝑟<11−𝑟^21−𝛼×12𝜋₀^2𝜋𝑓^′𝑟𝑒^𝑖𝜑𝑞𝑑𝜑^1/𝑞<∞. (4) It is obvious that Λ(∞,0) is just the Bloch space 𝔅, which is contained in Λ(𝑞,0) for every 1<𝑞<∞. Note that Λ(𝑞,1/𝑞) increases with 𝑞∈(1,∞). We refer to [5] for more information of mean Lipschitz spaces.

For 0⩽𝑠<∞, we say that an 𝑓∈𝐻(𝔻) belongs to the growth space 𝐻^∞_𝑠 if 𝑓_𝐻∞𝑠=sup_0⩽𝑟<1𝑓𝑟𝑒^𝑖𝜃1−𝑟^2𝑠<∞. (5) It is easy to see that 𝐻^∞₀=𝐻^∞.

For −1<𝛼<∞, an 𝑓∈𝐻(𝔻) is said to belong to the 𝒟^𝛼 space if 𝑓²_𝒟𝛼=𝑓(0)²+_𝔻𝑓^′(𝑧)²1−|𝑧|^2𝛼𝑑𝜎(𝑧)<∞. (6)

For 1<𝑝<∞, the Besov space ℬ_𝑝 is defined to be the space of all analytic functions 𝑓 in 𝔻 such that 𝑓^𝑝_ℬ𝑝=𝑓(0)^𝑝+_𝔻𝑓^′(𝑧)^𝑝1−|𝑧|^2𝑝−2𝑑𝜎(𝑧)<∞. (7)

Let 𝑓∈𝐻(𝔻). The Volterra integral operators 𝐼_𝑓 and 𝐽_𝑓 are defined as follows: 𝐼_𝑓(ℎ)(𝑧)=^𝑧₀ℎ^′(𝑤)𝑓(𝑤)𝐽𝑑𝑤,_𝑓(ℎ)(𝑧)=^𝑧₀ℎ(𝑤)𝑓^′(𝑤)().𝑑𝑤,𝑧∈𝔻 (8) It is easy to see that 𝐼_𝑓+𝐽_𝑓ℎ+𝑓(0)ℎ(0)=𝑀_𝑓(ℎ), (9) where 𝑀_𝑓 denotes the multiplication operator; that is, 𝑀_𝑓(ℎ)=𝑓ℎ. If 𝑓 is a constant, then all results about 𝐼_𝑓, 𝐽_𝑓, or 𝑀_𝑓 are trivial. In this paper, we assume that 𝑓 is a nonconstant. Both operators have been studied by many authors. See [6–23] and the references therein.

Norms of some special operators, such as composition operator, weighted composition operator, and some integral operators, have been studied by many authors. The interested readers can refer [13, 24–32], for example. Recently, Liu and Xiong studied the norm of integral operators 𝐼_𝑓 and 𝐽_𝑓 on the Bloch space, Dirichlet space, BMOA space, and so on in [13]. In this paper, we study the norm of integral operator 𝐼_𝑓 on some function spaces in the unit disk.

2. Main Results

In this section, we state and prove our main results. In order to formulate our main results, we need some auxiliary results which are incorporated in the following lemmas.

Lemma 1 (see [5, page 144]). If 𝑓∈𝐻^𝑝  (0<𝑝⩽∞), then |𝑓(𝑧)|⩽(1−|𝑧|²)^−1/𝑝‖𝑓‖_𝑝, |𝑧|<1, and the inequality is sharp for each fixed 𝑧.

Lemma 2. Let −1<𝛼<∞ and 0<𝑝<∞. For any 𝑓∈𝐻(𝔻), the following one has: 𝑓(𝑎)^𝑝⩽(𝛼+1)_𝔻𝑓(𝑧)^𝑝1−|𝑎|^22+𝛼|1−𝑎𝑧|^4+2𝛼1−|𝑧|^2𝛼𝑑𝜎(𝑧), (10) where 𝑎 is any point in 𝔻.

Proof. For any 𝑓∈𝐻(𝔻), taking 𝑧=𝑟𝑒^𝑖𝜃 and the subharmonicity of |𝑓(𝑧)|^𝑝, we get 𝑓(0)^𝑝⩽12𝜋₀^2𝜋𝑓𝑟𝑒^𝑖𝜃𝑝𝑑𝜃, (11) and so 𝑓(0)^𝑝⩽()1𝛼+1𝜋¹₀₀^2𝜋𝑓𝑟𝑒^𝑖𝜃𝑝×1−𝑟^2𝛼=()𝑑𝜃𝑟𝑑𝑟𝛼+1_𝔻𝑓(𝑧)^𝑝1−|𝑧|^2𝛼𝑑𝜎(𝑧). (12) For any 𝑎∈𝔻, let 𝜑_𝑎(𝑧)=(𝑎−𝑧)/(1−𝑎𝑧). Replacing 𝑓 by 𝑓∘𝜑_𝑎(𝑧) and applying the change of variable formula give the following: 𝑓(𝑎)^𝑝⩽(𝛼+1)_𝔻𝑓∘𝜑_𝑎(𝑧)^𝑝1−|𝑧|^2𝛼(𝑧)=()𝑑𝜎𝛼+1_𝔻𝑓(𝑧)^𝑝1−𝜑_𝑎(𝑧)^2𝛼1−|𝑎|²²|1−𝑎𝑧|⁴(𝑧)=()𝑑𝜎𝛼+1_𝔻𝑓(𝑧)^𝑝1−|𝑎|^22+𝛼1−|𝑧|^2𝛼|1−𝑎𝑧|^2𝛼+4𝑑𝜎(𝑧). (13) The proof is complete.

Theorem 3. Let 𝑓∈𝐻(𝔻). The integral operator 𝐼_𝑓 is bounded on Λ(1,1) if and only if 𝑓∈𝐻^∞. Moreover, one has 𝐼_𝑓=𝑓_𝐻∞. (14)

Proof. If 𝑓∈𝐻^∞, by (4), we have 𝐼_𝑓ℎ_Λ(1,1)=sup_0⩽𝑟<112𝜋₀^2𝜋𝑓𝑟𝑒^𝑖𝜑ℎ^′𝑟𝑒^𝑖𝜑⩽𝑓𝑑𝜑_𝐻∞sup_0⩽𝑟<112𝜋₀^2𝜋ℎ^′𝑟𝑒^𝑖𝜑⩽𝑓𝑑𝜑_𝐻∞‖ℎ‖_Λ(1,1). (15) Thus ‖𝐼_𝑓‖⩽‖𝑓‖_𝐻^∞.
On the other hand, denote 𝑐=sup_𝑧∈𝔻|𝑓(𝑧)|. Given any 𝜖>0, there exists 𝑧₁∈𝔻 such that |𝑓(𝑧₁)|>𝑐−𝜖. Let ℎ(𝑧)=𝑧₁−𝑧1−𝑧₁𝑧−𝑧₁. (16) Then we have ‖ℎ‖_Λ(1,1)=1. In fact, taking 𝑧₁=𝑟₁𝑒^𝑖𝜑1 and 𝑧=𝑟𝑒^𝑖𝜑 and using Poisson integral, we get ‖ℎ‖_Λ(1,1)=sup_0⩽𝑟<112𝜋₀^2𝜋ℎ^′𝑟𝑒^𝑖𝜑𝑑𝜑=sup_0⩽𝑟<112𝜋₀^2𝜋1−𝑟²₁1−𝑟𝑟₁𝑒^{𝑖(𝜑−𝜑1)2}𝑑𝜑=sup_0⩽𝑟<11−𝑟²₁1−𝑟²𝑟²₁=1. (17) Taking 𝑧₁=𝑟₁𝑒^𝑖𝜑1, we obtain ∞>𝐼_𝑓⩾𝐼_𝑓ℎ_Λ(1,1)=sup_0⩽𝑟<112𝜋₀^2𝜋𝑓𝑟𝑒^𝑖𝜑ℎ^′𝑟𝑒^𝑖𝜑𝑑𝜑=sup_0⩽𝑟<112𝜋₀^2𝜋𝑓𝑟𝑒^𝑖𝜑×1−𝑟²₁1−𝑟₁𝑟𝑒^{𝑖(𝜑−𝜑1)2}𝑑𝜑. (18) So 𝑓ℎ′∈𝐻¹ and 𝑓∈𝐻¹. Thus Theorem 2.6 in [5] yields 𝐼_𝑓⩾12𝜋₀^2𝜋𝑓𝑒^𝑖𝜑1−𝑟²₁1−𝑟₁𝑒^{𝑖(𝜑−𝜑1)2}𝑑𝜑. (19) By Theorem 2.12 in [5], we have 𝐼_𝑓⩾𝑓𝑧₁>𝑐−𝜖, (20) so the arbitrariness of 𝜖 gives ‖𝐼_𝑓‖⩾‖𝑓‖_𝐻∞ and the proof is complete.

Lemma 3 in [13] gives the norm of 𝐼_𝑓 on Dirichlet space. Here, we consider the norm of 𝐼_𝑓 on 𝛼-Dirichlet space 𝒟^𝛼.

Theorem 4. Let 𝑓∈𝐻(𝔻) and −1<𝛼<∞. Then 𝐼_𝑓 is bounded on 𝒟^𝛼 if and only if 𝑓∈𝐻^∞. Moreover, one has 𝐼_𝑓=𝑓_𝐻∞. (21)

Proof. First, we assume that 𝑓∈𝐻^∞. Let ℎ∈𝒟^𝛼. Then (6) gives 𝐼_𝑓ℎ²_𝒟𝛼=_𝔻𝑓(𝑧)ℎ^′(𝑧)²1−|𝑧|^2𝛼(𝑧)⩽𝑓𝑑𝜎_2𝐻∞𝔻ℎ^′(𝑧)²1−|𝑧|^2𝛼(𝑧)⩽𝑓𝑑𝜎_2𝐻∞‖ℎ‖²_𝒟^𝛼, (22) and so we have ‖𝐼_𝑓‖⩽‖𝑓‖_𝐻∞.
Now we need only to show the reverse inequality. Denote 𝑐=sup_𝑧∈𝔻|𝑓(𝑧)|. Given any 𝜖>0, there exists 𝑧₁∈𝔻 such that |𝑓(𝑧₁)|>𝑐−𝜖. Let ℎ₁(𝑧)=_Γ(𝑧)1−𝑧₁^21+(𝛼/2)1−𝑧₁𝜁^2+𝛼𝑑𝜁, (23) where Γ(𝑧) is any path in 𝔻 from 0 to 𝑧. By Theorem 13.11 in [33, page 274], we know ℎ₁ is an analytic function in 𝔻 and ℎ₁′(𝑧)=(1−|𝑧₁|²)^1+(𝛼/2)/(1−𝑧₁𝑧)^2+𝛼. Also it is easy to check that ‖ℎ₁‖²_𝒟^𝛼=1/(𝛼+1). Indeed, by using the method of the proof of Lemma 4.2.2 in [4], we have ℎ₁²_𝒟𝛼=_𝔻1−𝑧₁^22+𝛼1−𝑧₁𝑧^4+2𝛼1−|𝑧|^2𝛼(𝑧)=𝑧𝑑𝜎1−₁^22+𝛼_𝔻1−|𝑧|^2𝛼1−𝑧₁𝑧^4+2𝛼(𝑧)=𝑧𝑑𝜎1−₁^22+𝛼Γ(𝛼+1)Γ²(𝛼+2)^∞_𝑛=0Γ²(𝑛+2+𝛼)()𝑧𝑛!Γ𝑛+2+𝛼^12𝑛=1−𝑧₁^22+𝛼𝛼+1^∞_𝑛=0Γ(𝑛+2+𝛼)()𝑧𝑛!Γ𝛼+2^12𝑛=1−𝑧₁^22+𝛼1𝛼+11−𝑧₁^22+𝛼=1.𝛼+1 (24) Let ℎ(𝑧)=ℎ₁(𝑧)/‖ℎ₁‖_𝒟^𝛼, and so ‖ℎ‖_𝒟^𝛼=1. Thus by Lemma 2 we have 𝐼_𝑓²≥𝐼_𝑓ℎ(𝑧)²_𝒟𝛼=_𝔻𝑓(𝑧)ℎ^′(𝑧)²1−|𝑧|^2𝛼(𝑧)=()𝑑𝜎𝛼+1_𝔻𝑓(𝑧)²1−𝑧₁^22+𝛼1−𝑧₁𝑧^4+2𝛼1−|𝑧|^2𝛼(𝑧)⩾𝑓𝑧𝑑𝜎₁₂>(𝑐−𝜖)². (25) Since 𝜖 is arbitrary, we get 𝐼_𝑓⩾𝑓_𝐻∞, (26) which implies the desired result.

Theorem 5. Let 𝑓∈𝐻(𝔻) and let 1<𝑝<∞. The integral operator 𝐼_𝑓 is bounded on ℬ_𝑝 if and only if 𝑓∈𝐻^∞. Moreover, one has 𝐼_𝑓=𝑓_𝐻∞. (27)

Proof. If 𝑓∈𝐻^∞, then by (7), we have 𝐼_𝑓ℎ^𝑝_ℬ𝑝=_𝔻𝑓(𝑧)ℎ^′(𝑧)^𝑝1−|𝑧|^2𝑝−2(𝑧)⩽𝑓𝑑𝜎_{𝑝𝐻∞𝔻}ℎ^′(𝑧)^𝑝1−|𝑧|^2𝑝−2(𝑧)⩽𝑓𝑑𝜎_𝑝𝐻∞‖ℎ‖^𝑝_ℬ𝑝, (28) and so ‖𝐼_𝑓‖⩽‖𝑓‖_𝐻∞.
Now we need only to show the reverse inequality. Denote 𝑐=sup_𝑧∈𝔻|𝑓(𝑧)|. Given any 𝜖>0, there exists 𝑧₁∈𝔻 such that |𝑓(𝑧₁)|>𝑐−𝜖. Let ℎ₁(𝑧)=𝑧₁−𝑧1−𝑧₁𝑧−𝑧₁,𝑧∈𝔻. (29) We see that ‖ℎ₁‖^𝑝_ℬ𝑝=1/(𝑝−1). Indeed, ℎ₁^𝑝_ℬ𝑝=_𝔻1−𝑧₁^2𝑝1−𝑧₁𝑧^2𝑝1−|𝑧|^2𝑝−2(𝑧)=𝑧𝑑𝜎1−₁^2𝑝_𝔻1−|𝑧|^2𝑝−21−𝑧₁𝑧^2𝑝(𝑧)=𝑧𝑑𝜎1−₁^2𝑝Γ𝑝−1Γ²𝑝^∞_𝑛=0Γ𝑛+𝑝²𝑧𝑛!Γ𝑛+𝑝^12𝑛=1−𝑧₁^2𝑝𝑝−1^∞_𝑛=0Γ𝑛+𝑝𝑝𝑧𝑛!Γ^12𝑛=1−𝑧₁^2𝑝1𝑝−11−𝑧₁^2𝑝=1.𝑝−1 (30) Let ℎ(𝑧)=ℎ₁(𝑧)/‖ℎ₁‖_ℬ𝑝. Then ‖ℎ‖_ℬ𝑝=1. Thus by Lemma 2, we have 𝐼_𝑓^𝑝≥𝐼_𝑓ℎ(𝑧)^𝑝_ℬ𝑝=_𝔻𝑓(𝑧)ℎ^′(𝑧)^𝑝1−|𝑧|^2𝑝−2(𝑧)=𝑑𝜎𝑝−1_𝔻𝑓(𝑧)^𝑝1−𝑧₁^2𝑝1−𝑧₁𝑧^2𝑝×1−|𝑧|^2𝑝−2(𝑧)⩾𝑓𝑧𝑑𝜎_1𝑝>(𝑐−𝜖)^𝑝. (31) Since 𝜖 is arbitrary, we get ‖𝐼_𝑓‖⩾‖𝑓‖_𝐻∞. The proof is complete.

Theorem 6. Let 𝑓∈𝐻(𝔻) and let 0<𝛼⩽𝛽<∞. The integral operator 𝐼_𝑓 is bounded from 𝔅^𝛼 to 𝔅^𝛽 if and only if 𝑓∈𝐻^∞_𝛽−𝛼. Moreover, one has 𝐼_𝑓=sup_𝑧∈𝔻𝑓(𝑧)1−|𝑧|^2𝛽−𝛼. (32)

Proof. If 𝑓∈𝐻^∞_𝛽−𝛼, then by (3), we have 𝐼_𝑓ℎ_𝔅𝛽=sup_𝑧∈𝔻𝑓(𝑧)ℎ^′(𝑧)1−|𝑧|^2𝛽⩽𝑓_{𝐻∞𝛽−𝛼}sup_𝑧∈𝔻ℎ^′(𝑧)1−|𝑧|^2𝛼⩽𝑓_{𝐻∞𝛽−𝛼}‖ℎ‖_𝔅^𝛼. (33) Hence ‖𝐼_𝑓‖⩽‖𝑓‖_𝐻^∞_𝛽−𝛼.
For the converse, denote 𝑐=sup_𝑧∈𝔻|𝑓(𝑧)|(1−|𝑧|²)^𝛽−𝛼. Given any 𝜖>0, there exists 𝑧₁∈𝔻 such that |𝑓(𝑧₁)|(1−|𝑧₁|²)^𝛽−𝛼>𝑐−𝜖. Set ℎ(𝑧)=_Γ(𝑧)1−𝑧₁^2𝛼1−𝑧₁𝜁^2𝛼𝑑𝜁, (34) where Γ(𝑧) is any path in 𝔻 from 0 to 𝑧. By Theorem 13.11 in [33, page 274], we know that ℎ is an analytic function in 𝔻 and ℎ′(𝑧)=(1−|𝑧₁|²)^𝛼/(1−𝑧₁𝑧)^2𝛼, and it is easy to check that ‖ℎ‖_𝔅^𝛼=1. Thus 𝐼_𝑓⩾𝐼_𝑓ℎ_𝔅𝛽=sup_𝑧∈𝔻𝑓(𝑧)ℎ^′(𝑧)1−|𝑧|^2𝛽⩾𝑓𝑧₁ℎ^′𝑧₁1−𝑧₁^2𝛽⩾𝑓𝑧₁1−𝑧₁^2𝛼1−𝑧₁^22𝛼1−𝑧₁^2𝛽=𝑓𝑧₁1−𝑧₁^2𝛽−𝛼>𝑐−𝜖. (35) Since 𝜖 is arbitrary, we obtain the desired result. The proof is complete.

Theorem 7. Let 𝑓∈𝐻(𝔻). The integral operator 𝐼_𝑓 is bounded from Λ(1,1) to 𝔅 if and only if 𝑓∈𝐻^∞. Moreover, one has 𝐼_𝑓=𝑓_𝐻∞. (36)

Proof. If 𝑓∈𝐻^∞, then by Lemma 1, we have 𝐼_𝑓ℎ_𝔅=sup_𝑧∈𝔻𝑓(𝑧)ℎ^′(𝑧)1−|𝑧|²⩽‖ℎ‖_Λ(1,1)sup_𝑧∈𝔻1−|𝑧|²⁻¹1−|𝑧|²𝑓(𝑧=‖ℎ‖_Λ(1,1))sup_𝑧∈𝔻𝑓(𝑧), (37) and hence ‖𝐼_𝑓‖⩽‖𝑓‖_𝐻∞.
For the converse, denote 𝑐=sup_𝑧∈𝔻|𝑓(𝑧)|. Given any 𝜖>0, there exists 𝑧₁∈𝔻 such that |𝑓(𝑧₁)|>𝑐−𝜖. Let ℎ(𝑧)=𝑧₁−𝑧1−𝑧₁𝑧−𝑧₁,𝑧∈𝔻. (38) Then by the proof of Theorem 3, we see that ‖ℎ‖_Λ(1,1)=1. In the meantime, we know that |ℎ′(𝑧₁)|(1−|𝑧₁|²)=1, which gives 𝐼_𝑓⩾𝐼_𝑓ℎ_𝔅=sup_𝑧∈𝔻𝑓(𝑧)ℎ^′(𝑧)1−|𝑧|²⩾𝑓𝑧₁ℎ^′𝑧₁1−𝑧₁²=𝑓𝑧₁>𝑐−𝜖. (39) Since 𝜖 is arbitrary, we get the desired result. The proof is complete.

Finally, we consider the norm of 𝐼_𝑓 from Λ(∞,1) to some Banach spaces.

Theorem 8. If 𝑓∈𝐻(𝔻), then the following assertions hold.(1)Let 0<𝑝<∞. The integral operator 𝐼_𝑓 is bounded from Λ(∞,1) space to 𝑄_𝑝 space if and only if 𝑓 satisfies sup_{𝑎∈𝔻𝔻}𝑓(𝑧)²1−𝜑_𝑎(𝑧)^2𝑝𝑑𝜎(𝑧)<∞. (40) Moreover, one has 𝐼_𝑓=sup_{𝑎∈𝔻𝔻}𝑓(𝑧)²1−𝜑_𝑎(𝑧)^2𝑝𝑑𝜎(𝑧). (41)(2)Let 0⩽𝛼⩽1 and let 0⩽𝑞⩽∞. The integral operator 𝐼_𝑓 is bounded from Λ(∞,1) space to Λ(𝑞,𝛼) space if and only if 𝑓 satisfies sup_0⩽𝑟<11−𝑟^21−𝛼12𝜋₀^2𝜋𝑓𝑟𝑒^𝑖𝜑𝑞𝑑𝜑^1/𝑞<∞. (42) Moreover, one has 𝐼_𝑓=sup_0⩽𝑟<11−𝑟^21−𝛼12𝜋₀^2𝜋𝑓𝑟𝑒^𝑖𝜑𝑞𝑑𝜑^1/𝑞. (43)(3)Let 0<𝛼<∞. The integral operator 𝐼_𝑓 is bounded from Λ(∞,1) space to 𝔅^𝛼 space if and only if 𝑓 satisfies sup_𝑧∈𝔻|𝑓(𝑧)|(1−|𝑧|²)^𝛼<∞. Moreover, one has 𝐼_𝑓=sup_𝑧∈𝔻𝑓(𝑧)1−|𝑧|^2𝛼. (44)(4)Let −1<𝛼<∞. The integral operator 𝐼_𝑓 is bounded from Λ(∞,1) space to 𝒟^𝛼 space if and only if 𝑓 satisfies ∫_𝔻|𝑓(𝑧)|²(1−|𝑧|²)^𝛼𝑑𝜎(𝑧)<∞. Moreover, one has 𝐼_𝑓=_𝔻𝑓(𝑧)²1−|𝑧|^2𝛼𝑑𝜎(𝑧). (45)

Proof. The assertion (1) will be proved only here, and the conclusions of (2), (3), and (4) follow by using the similar arguments to that used in proving (1), and so the proofs are omitted.
If ℎ∈Λ(∞,1), then by (1), we have 𝐼_𝑓ℎ²_𝑄𝑝=sup_{𝑎∈𝔻𝔻}𝑓(𝑧)ℎ^′(𝑧)²1−𝜑_𝑎(𝑧)^2𝑝(𝑧)⩽‖ℎ‖𝑑𝜎_2Λ()∞,1sup_{𝑎∈𝔻𝔻}𝑓(𝑧)²1−𝜑_𝑎(𝑧)^2𝑝𝑑𝜎(𝑧), (46) and so 𝐼_𝑓⩽sup_{𝑎∈𝔻𝔻}𝑓(𝑧)²1−𝜑_𝑎(𝑧)^2𝑝𝑑𝜎(𝑧). (47)
For the converse, let ℎ(𝑧)=𝑧. It is easy to see that ‖ℎ‖_Λ(∞,1)=1. Thus 𝐼_𝑓⩾𝐼_𝑓ℎ²_𝑄𝑝=sup_{𝑎∈𝔻𝔻}𝑓(𝑧)ℎ^′(𝑧)²1−𝜑_𝑎(𝑧)^2𝑝𝑑𝜎(𝑧)=sup_{𝑎∈𝔻𝔻}𝑓(𝑧)²1−𝜑_𝑎(𝑧)^2𝑝𝑑𝜎(𝑧). (48) The desired result follows by (47) and (48). The proof is complete.

Acknowledgments

Hao Li is supported by the National Natural Science Foundation of China (no. 11126284). Songxiao Li is supported by the project of Department of Education of Guangdong Province (no. 2012KJCX0096).