Generalized Composition Operators from Area Nevanlinna Spaces to Zygmund-type Spaces

The boundedness and compactness of generalized composition operators from area Nevanlinna spaces to Zygmund-type spaces and little Zygmund-type spaces are characterized.


Introduction
Let C be the complex plane of the open unit disk, H(D) the space of all analytic function on D. An function f ∈ H(D) is said to belong to the Zygmund-type space Z µ , if sup where µ is the normal function(see [6]).It is a Banach space with norm The little Zygmund-type space Z µ,0 consists of those functions f in Z µ satisfying lim |z|→1 − µ(|z|)|f (z)| = 0 and it is easy to see that Z µ,0 is a closed subspace of Z µ .For α > −1, let dA α (z) = (α + 1)(1 − |z| 2 ) α dA(z) be the weighted Lebesgue measure on D. Let 1 ≤ p < ∞ and α > −1, an function f ∈ H(D) is said to belong to the area Nevanlinna space N p α = N p α (D), if Let ϕ be an analytic self-map of D. Then composition operator C ϕ is defined by (C ϕ f ) = f (ϕ(z)), f ∈ H(D).The composition operator has been studied by many researchers on various space.(see for example, [1,2,3]).In [4], the authors defined and studied the generalized composition operator and the boundedness and compactness of C g ϕ on Zygmund spaces and Bloch spaces were investigated in it.The purpose of this paper is to characterize the boundedness and compactness of the generalized composition operator C g ϕ from area Nevanlinna spaces to Zygmund-type spaces and little Zygmund-type spaces.In what follows, we use the letter C to denote a positive constant whose value may change its value at each occurrence.

Main results and proofs
Lemma 1. Suppose that ϕ is an analytic self-map of D, g ∈ H(D), 1 ≤ p < ∞, α > −1.Then C g ϕ : N p α → Z µ is compact if and only if for each sequence {f k } k∈N which is bounded in N p α and converges to zero uniformly on compact subsets of D as k →∞, we have C g ϕ f k Zµ → 0 as k →∞.Since the proof of Lemma 1 is similar to Proposition 3.11 in [1], it is omitted.
The proof of it is similar to that of [5].The next result can be found in [7,8].Lemma 3. Let n be a nonnegative integer, 1 ≤ p < ∞ and α > −1.Then there exists some C such that for each f ∈ N p α and z ∈ D, Theorem 4. Let g ∈ H(D) and ϕ be an analytic self-map of D , 1 ≤ p < ∞ and α > −1.Then C g ϕ : N p α → Z µ is bounded if and only if and Proof.Suppose that C g ϕ : N p α →Z µ is bounded.Now taking f (z) = z and f (z) = z 2 , and obviously each of them belongs to N p α , and using the boundedness of the function ϕ(z), we get For ω ∈ D, set For any fixed r ∈ (0, 1) , Therefore, ( 8) and ( 9) yield (3). Next, set From ( 7) and (10) , we get Combining ( 11) with ( 5), similar to the former proof, we get (4).
For the converse, suppose that (3) and ( 4) hold.By Lemma 3, we have and Proof.Suppose that C g ϕ : Then f k ∈ N p α by Theorem 4, and converges to 0 uniformly on compact subsets of D as k → ∞.From the compactness of C g ϕ , we have lim k→∞ C g ϕ f k Zµ = 0. On the other hand, similar to the proof of Theorem 4, we have From which we get (12).Next, set Then h k ∈ N p α and converges to 0 uniformly on compact subsets of D as k → ∞.From the compactness of C g ϕ , we have lim k→∞ C g ϕ h k Zµ = 0. On the other hand, From which we get (13).
Conversely, suppose that C g ϕ : N p α → Z µ is bounded and ( 12) and (13) hold.Assume {f k } k∈N is a bounded sequence in N p α such that f k converges to 0 uniformly on compact subsets of D as k → ∞.By the assumption, for any ε > 0, there exists a δ ∈ (0, 1) such that Then by Lemma 3 , (15), ( 16), ( 17) and (18), we have that Then we get Since f k converges to 0 uniformly on compact subsets of D as k → ∞, Cauchy's estimation gives that f k and f k also do as k → ∞ , and both {z ∈ D : |z| < δ} and {0} are compact subsets of D. Hence, letting k → ∞ in (19), we get lim k→∞ C g ϕ f k Zµ = 0.By Lemma 1, we see that C g ϕ : N p α → Z µ is compact.