QK type spaces of analytic functions

For a nondecreasing function K : [0,∞) → [0,∞) and 0 < p < ∞ , −2 < q < ∞ , we introduce QK(p, q) , a QK type space of functions analytic in the unit disk and study the characterizations of QK(p, q) . Necessary and sufficient conditions on K such that QK(p, q) become some known spaces are given.

Through this paper, we assume that K : [0, ∞) → [0, ∞) is a rightcontinuous and nondecreasing function.For 0 < p < ∞, −2 < q < ∞, we say that a function f analytic in Δ belongs to the space Q K (p, q) if (1.1) ||f || p K,p,q = sup a∈Δ Δ |f (z)| p (1 − |z| 2 ) q K(g(z, a))dA(z) < ∞, where dA(z) is the Euclidean area element on Δ.It is clear that Q K (p, q) is a Banach space with the norm ||f || = |f (0)| + ||f || K,p,q when p ≥ 1.If q + 2 = p, Q K (p, q) is Möbius-invariant, i.e., ||f • ϕ a || K,p,q = ||f || K,p,q for all a ∈ Δ.Now we consider some special cases.If p = 2, q = 0, we have that Q K (p, q) = Q K (cf.[2] and [4]).If K(t) = t s , then Q K (p, q) = F (p, q, s) (cf.[7]).We know from [7] that In this paper we show some relations between Q K (p, q) and B α for a nondecreasing K , and give a general way to construct different spaces Q K1 (p, q) and Q K2 (p, q) by using some functions K 1 and K 2 .We assume throughout the paper that 3) does not hold and assume that f is a non-constant function.We show that f can not belong to Q K (p, q).By assumption there exists at least one point a ∈ Δ such that f (a) = 0.By if and only if Proof.(i) For a fixed r ∈ (0, 1), let We know that E(a, r) ⊂ Δ(a, r) and for any z ∈ E(a, r), we have , then by estimate above we have sup a∈Δ The proof of (i) is complete.

Now we assume B q+2 p
= Q K (p, q) and we show that (2.1) holds.In fact, by [5] there exist two functions Thus (2.1) holds.We complete the proof.
Moreover, we say that f ∈ Q K (p, q) belongs to the space As a subspace of B α , the little α -Bloch space B α 0 consists of analytic functions f on Δ such that if and only if (2.1) holds.

.11)
Since f 0 is defined by a gap series with Hadamard condition, we have where It means that f 0 ∈ B q+2 p 0 \Q K0 (p, q), which is a contradiction.Hence (2.1) holds.

More results on Q K (p, q)-spaces
The following result means that the kernel functions K can be chosen as bounded.
To prove Theorem 3.3 we need the following lemma, which is also an characterization of α -Bloch function.
if and only if there exists ρ ∈ (0, 1) such that if and only if there exists ρ ∈ (0, 1) such that For any ρ ∈ (0, 1) and a ∈ Δ, we have Conversely, suppose that (3.2) holds for some ρ, 0 < ρ < 1.By the proof of Theorem 2.1 (i) we have which shows that f ∈ B q+2 p .(ii) We omit the proof since its proof is similar to that of (i) above.
Proof of Theorem 3.3.It is clear that Q K2 (p, q) ⊂ Q K1 (p, q).Suppose that Q K2 (p, q) = Q K1 (p, q).By the Open Mapping Theorem we know that the identity map from one of these spaces into the other one is continuous.Thus there exists a constant C such that || • || K2,p,q ≤ C|| • || K1,p,q .Since K1(r) K2(r) → 0 as r → 0 , then there exists r 0 ∈ (0, 1) such that By Lemma 3.4 there exists a constant C such that for f ∈ Q K2 (p, q), We have proved that h ∈ Q K2 (p, q).It means that Q K2 (p, q) = B q+2 p .It follows from Theorem 2.1 that the integral (2.1) with K = K 2 must be convergent, a contradiction.We obtain that Q K2 (p, q) Q K1 (p, q).We finish the proof of Theorem 3.3.Corollary 3.5.Let 0 < p < ∞ and −2 < q < ∞.Then F (p, q, 0) ⊆ Q K (p, q) and the equality holds if and only if K(0) > 0 .
Proof.By Theorem 3.1 the kernel function K can be viewed as bounded, which means that F (p, q, 0) ⊆ Q K (p, q).If K(0) > 0, F (p, q, 0) = Q K (p, q) holds immediately by the following estimates.