Integral-Type Operators from Bloch-Type Spaces to QK Spaces

and Applied Analysis 3 a There is a constant C > 0 such that for all t > 0 K 2t ≤ CK t . 1.10 b The auxiliary function λK satisfies the following condition: ∫1 0 λK s s ds < ∞. 1.11 Let Ω 0,∞ denote the class of all nondecreasing continuous functions on 0,∞ satisfying conditions 1.8 , 1.10 , and 1.11 . A positive Borel measure μ on D is called a K-Carleson measure 3 if


Introduction
Let D be the open unit disk in the complex plane, ∂D be its boundary, D w, r be disk centered at w of radius r, and let H D be the class of all holomorphic functions on D. Let be the involutive M öbius transformation which interchanges points 0 and a.If X is a Banach space, then by B X we will denote the closed unit ball in X.
The α-Bloch space, B α D B α , α > 0, consists of all f ∈ H D such that sup Abstract and Applied Analysis The little α-Bloch space B α 0 D B α 0 consists of all functions f holomorphic on D such that lim |z| → 1 1 − |z| 2 α |f z | 0. The norm on B α is defined by With this norm, B α is a Banach space, and the little α-Bloch space B α 0 is a closed subspace of the α-Bloch space.Note that B 1 B is the usual Bloch space.
Given a nonnegative Lebesgue measurable function K on 0, 1 the space Q K consists of those f ∈ H D for which where dm z 1/π dx dy 1/π r dr dθ is the normalized area measure on where Aut D is the group of all automorphisms of the unit disk D. It is a Banach space with the norm defined by The space Q K,0 consists of all f ∈ H D such that It is known that Q K,0 is a closed subspace of Q K .For classical Q spaces, see 2 .
It is clear that each Q K contains all constant functions.If Q K consists of just constant functions, we say that it is trivial.Q K is nontrivial if and only if sup t∈ 0, 1 Throughout this paper, we assume that condition 1.8 is satisfied, so that the space Q K is nontrivial.An important tool in the study of Q K spaces is the auxiliary function λ K defined by where the domain of K is extended to 0, ∞ by setting K t K 1 when t > 1.The next two conditions play important role in the study of Q K spaces.

Abstract and Applied Analysis
3 a There is a constant C > 0 such that for all t > 0 K 2t ≤ CK t .
1.10 b The auxiliary function λ K satisfies the following condition: Let Ω 0, ∞ denote the class of all nondecreasing continuous functions on 0, ∞ satisfying conditions 1.8 , 1.10 , and 1.11 .
A positive Borel measure μ on D is called a K-Carleson measure 3 if sup where the supermum is taken over all subarcs I ⊂ ∂D, |I| is the length of I, and S I is the Carleson box defined by We also need the following results of Wulan and Zhu in 3 , in which Q K spaces are characterized in terms of K-Carleson measures.
Also, μ is a vanishing K-Carleson measure if and only if From Theorem 1.1 and the definition of the spaces Q K and Q K,0 , we see that when Let ϕ ∈ S D be the family of all holomorphic self-maps of D, g ∈ H D , and n ∈ N 0 .We define an integral-type operator as follows: 1.17 Operator

Auxiliary Results
Here, we quote several lemmas which will be used in the proofs of the main results in this paper.The following lemma is folklore see, e.g., 30 .
Lemma 2.1.For any f ∈ H D and z ∈ D, the following inequalities hold

2.2
The next lemma is obtained in 31, 32 .

2.3
Also, if α / 1, then there are two functions f 3 , f 4 ∈ B α and C > 0, such that The next Schwartz-type lemma 33 is proved in a standard way, so we omit the proof.
has the Schur property, it follows that it is equivalent to Proof.We follow the lines of Lemma 2.3 in 24 .From the elementary inequality and since K is nondecreasing and satisfies 1.10 , we easily get From 2.7 and since Φ ϕ,g,K a is finite, it follows that Φ ϕ,g,K is finite at each point a 1 ∈ D. Let a ∈ D be fixed, and let a l l∈N ⊂ D be a sequence converging to a.We have From 2.6 , we have that for l such that 1 and consequently for l ≥ l 0 , it holds

2.10
From this and since Φ ϕ,g,K is finite at a, by the Lebesgue dominated convergence theorem, we get that the integral in 2.8 converges to zero as l → ∞ which implies that Φ ϕ,g,K a l → Φ ϕ,g,K a as l → ∞, from which the lemma follows.

Boundedness and Compactness of
In this section, we characterize the boundedness and compactness of the operators I n ϕ,g : , and n ∈ N, or n 0 and α > 1.
Then the following statements are equivalent.
from which the boundedness of I n ϕ,g : B α → Q K follows, and moreover 3 holds, and if n 0 and α > 1 such that 2.4 holds.Let It is known see 30 that for each f ∈ H D and n ∈ N, we have From this, Lemma 2.2, and since z , for |z| > δ.

3.7
Now note that for any f ∈ B α , the functions f r z f rz , r ∈ 0, 1 belong to B α , and moreover, sup 0<r<1 f r B α ≤ f B α .Applying 3.7 , using an elementary inequality, the boundedness of I n ϕ,g : B α 0 → Q K , and the last inequality, we obtain

3.8
Letting r → 1 in 3.8 and using the monotone convergence theorem, we get On the other hand, for

3.10
From 3.9 and 3.10 , c follows.Moreover we get Proof.By Lemma 2.4, we have that b is equivalent to c .d ⇒ a .Let f l l∈N be a bounded sequence in B α , say by L, converging to zero uniformly on compacts of D. Then f n l also converges to zero uniformly on compacts of D. From 3.11 we have that for every ε > 0 there is an r 1 ∈ 0, 1 such that for r ∈ r 1 , 1 sup Therefore, by Lemma 2.1 and 3.12 , we have that for r ∈ r 1 , 1

3.13
Letting l → ∞ in 3.13 , using the first condition in d and sup |w|≤r |f we have that the first condition in d holds.Let f l z z l /l, l ∈ N. It is easy to see that f l l∈N is a bounded sequence in B α 0 converging to zero uniformly on compacts of D. Hence, by Lemma 2.3, it follows that I n ϕ,g f l Q K → 0 as l → ∞.Thus, for every ε > 0, there is an l 0 ∈ N, l 0 > n such that for l ≥ l 0 From 3.14 we have that for each r ∈ 0, 1 and l ≥ l 0 3.17 Hence, for every ε > 0, there is a t ∈ 0, 1 such that From this and 3.16 , we have that for such t and each r ∈

3.19
From 3.19 we conclude that for every f ∈ B B α 0 , there is a δ 0 ∈ 0, 1 and δ 0 δ 0 f, ε such that for r ∈ δ 0 , 1 On the other hand, from 3.20 , it follows that if δ : max 1≤j≤k δ j f j , ε , then for r ∈ δ, 1 and all j ∈ {1, . . ., k}, we have sup If we apply 3.23 to the delays of the functions in 3.5 which are normalized so that they belong to B B α and then use the monotone convergence theorem, we easily get that for r > max{δ, δ} where δ is chosen as in 3.7 from which 3.11 follows, as desired.
Then the next statements are equivalent.
Proof.By Theorem 1.1, e and f are equivalent; by Lemma 2.4, c is equivalent to d , while, by Lemma 2.5, a is equivalent to c .Also b obviously implies a .a ⇒ e Let h 1 and h 2 be as in 3.5 .Then from 3.7 and an elementary inequality, we get

3.25
For From this and since I n ϕ,g h j ∈ Q K,0 , j 1, 2, by letting |a| → 1, we get that e holds.e ⇒ b .We have that for every ε > 0 there is a δ ∈ 0, 1 so that for |a| > δ Φ ϕ,g,K a < ε.

3.26
On the other hand, by Lemma 2.6, Φ ϕ,g,K is continuous on |a| ≤ δ, so uniformly bounded therein, which along with 3.26 gives the boundedness of Φ ϕ,g,K on D. Hence, by Theorem 3.1, Now assume that f l l∈N is a bounded sequence in B α , say by L, converging to zero uniformly on compacta of D as l → ∞.To show that the operator I n ϕ,g : B α → Q K,0 is compact, it is enough to prove that there is a subsequence f l k k∈N of f l l∈N such that I n ϕ,g f l k converges in Q K,0 as k → ∞.By Lemma 2.1 and Montel's theorem, it follows that there is a subsequence, which we may denote again by f l l∈N converging uniformly on compacta of 3.28

3.29
For a ∈ D and t ∈ 0, 1 , let Lemma 2.6 essentially shows that Ψ t is continuous on D. Hence, for each a ∈ D, there is a t a ∈ r, 1 such that Ψ t a a < ε/2.Moreover, there is a neighborhood O a of a such that, for every b ∈ O a , Ψ t a b < ε.From this and since the set |a| ≤ δ is compact, it follows that there is a t 0 ∈ 0, 1 such that Ψ t 0 a < ε when |a| ≤ δ.This along with Lemma 2.1 implies that sup

3.31
Abstract and Applied Analysis 13 By the Weierstrass theorem f n l → f n uniformly on compacta as l → ∞, from which along with 2.2 and since ϕ t 0 D is compact, for r sup w∈ϕ t 0 D |w|, it follows that sup Then the following statements are equivalent.
Proof.Suppose b holds and f ∈ B α 0 .Then by Theorem 3.1, Since f ∈ B α 0 , we have that, for every ε > 0, there is an r ∈ 0, 1 such that see, e.g., the idea in 35, Lemma 2.4

3.35
We also have

3.36
Combining 3.35 and 3.36 , we get e I ϕ,g : B α → Q K is compact.
Moreover, if I ϕ,g : B α → Q K is bounded, then the next asymptotic relations hold

3.37
Proof.The proof of the equivalence of statements a -d of this theorem is similar to the proof of Theorem 3.1; moreover, the implication b ⇒ c is much simpler since it follows by using the test function f 0 z ≡ 1.That c is equivalent to e -g is proved similarly as in Theorem 3.2, by using the well-known fact that if a bounded sequence f l l∈N in B α , α ∈ 0, 1 converges to zero uniformly on compacts of D, then it converges to zero uniformly on the whole D. The details are omitted.
The proof of the next theorem is similar to the proofs of Theorems 3.3 and 3.4 and will be omitted.
and only if it is weakly compact, which, by Gantmacher's theorem 34 , is equivalent to I n ϕ,g * * B α 0 * * ⊆ Q K,0 .Since B α on B α , this can be written as I n ϕ,g B α ⊆ Q K,0 , which by the closed graph theorem is equivalent to

Theorem 3 . 2 .
relations in 3.2 follow, finishing the proof of the theorem.Let α > 0, K ∈ Ω 0, ∞ , ϕ ∈ S D , g ∈ H D , and n ∈ N, or n 0 and α > 1.Let I n ϕ,g : B α → Q K be bounded.Then the following statements are equivalent.

From 3 .
21 and 3.22 , we have that for r ∈ δ, 1 and every f ∈ B B α 0 sup a∈D |ϕ z |>r

Theorem 3 . 5 .
0 is bounded, then I n ϕ,g : B α 0 → Q K is bounded too.Thus, by Theorem 3.1, we get the first condition in b .For f 0 z z n /n! ∈ B α 0 , we get I n ϕ,g f 0 ∈ Q K,0 , which is equivalent to 3.33 , finishing the proof of the theorem.If n 0, we simply denote the operator I 0 ϕ,g by I ϕ,g .Let α ∈ 0, 1 , K ∈ Ω 0, ∞ , ϕ ∈ S D , and g ∈ H D .Then the following statements are equivalent.aI ϕ,g : B α → Q K is bounded.b I ϕ,g : B α 0 → Q K is bounded.c M 1 : sup a∈D D K 1 − |η a z | 2 |g z | 2 dm z < ∞. d dμ 1 z |g z | 2 dm z is a K-Carleson measure.
Proof.By Theorem 1.1, it is clear that c and d are equivalent.c ⇒ a .Let f ∈ B B α .First note that I ∈ H B and n ∈ N 0 .From this and by Lemma 2.1, we have 2. 3.2