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Weighted differentiation composition operators from weighted bergman space to n th weighted space on the unit disk
Journal of Inequalities and Applications volume 2011, Article number: 65 (2011)
Abstract
This paper characterizes the boundedness and compactness of the weighted differentiation composition operator from weighted Bergman space to n th weighted space on the unit disk of ≤.
2000 Mathematics Subject Classification: Primary: 47B38; Secondary: 32A37, 32H02, 47G10, 47B33.
1. Introduction
Let be the open unit disk in the complex space ≤, dA the Lebesegue measure on normalized so that . Let be the space of all analytic functions on .
Let α > -1, p > 0. f is said to belong to the weighted Bergman space, denoted by , if and
When 0 < p < 1, it is complete metric space; when p ≥1, it is a Banach space.
Let μ(z)(weight) be a positive continuous function on and n ∈ ℕ0. The n th weighted space on the unit disk, denoted by , consists of all such that
For n = 0, the space becomes the weighted-type space ; for n = 1, the Bloch-type space ; and for n = 2, the Zygmud-type . For more details about these spaces, we recommend the readers to ([1, 2]).
The expression defines a semi-norm on the n th weighted space , while the natural norm is given by
With this norm, becomes a Banach space. The little n th weighted space, denoted by , is a closed subspace of , consisting of those f for which
Let φ be a non-constant analytic self-map of , , and m ∈ ℕ. The weighted differentiation composition operator is defined by
for , . If m = 1, u (z) = φ' (z), then ; if let m = 1, u (z) = 1, then .
Recently, there have been some interests in studying some particular cases of operators, such as DC φ , C φ D and , between different function spaces. From those studies, they gave some sufficient and necessary conditions for these operators to be bounded and compact. Concerning these results, we also recommend the interested readers to ([3–9]).
In this paper, we characterize the boundedness and compactness of the operator from to n th weighted space. For the case of , we have the following results:
Theorem 1. Assume that p > 0, α > -1, n, m ∈ ℕ, μ is a weight on, φ is a non-constant analytic self-map of, and . Then,
(1a) is bounded if and only if for each k ∈ {0, 1,..., n}
(1b) is compact if and only ifis bounded and for each k ∈ {0, 1,..., n}
For the case of , our main results are the following:
Theorem 2. Assume that p > 0, α > -1, n, m ∈ ℕ, μ is a weight on, φ is a non-constant analytic self-map of, and. Then,
(2a) is bounded if and only ifis bounded and for each k ∈ {0, 1,..., n},
(2b) is compact if and only ifis bounded and for each k ∈ {0, 1,..., n},
The organization of this paper is as follows: we give some lemmas in Section 2, and then prove Theorem 1 in Section 3 and Theorem 2 in Section 4, respectively.
Throughout this paper, we will use the symbol C to denote a finite positive number, and it may differ from one occurrence to the other.
2. Some Lemmas
Lemma 1 is a direct consequence of the well-known estimate in ([10], Proposition 1.4.10). Hence, we omit its proof.
Lemma 1. Assume that p > 0, α > - 1, n ∈ ℕ, n > 0, and. Then the function
belongs to. Moreover, .
The next lemma comes from ([11]).
Lemma 2. Assume that p > 0, α > -1, n ∈ ℕ, and. Then, there is a positive constant C independent of f such that
Lemma 3. Let p > 0, α > -1, m ∈ ℕ, and
Then, .
Proof. With and replacing n by n + 1 in ([12], Lemma 2.3), the lemma easily follows. □
The next lemma can be found in ([7], Lemma 4).
Lemma 4. Assume n ∈ ℕ, g, and φ is an analytic self-map of. Then,
where
and the sum in (5) is overall non-negative integers k1,..., k l satisfying k1+ k2 + ⋯ + k1 = k and k1 + 2k2 + ⋯ + lk l = l.
By a proof in a standard way ([1], Proposition 3.11), we can get the next lemma.
Lemma 5. Suppose, p > 0, α > - 1, n, m ∈ ℕ and φ is a non-constant analytic self-map of. Then the operatoris compact if and only ifis bounded and for any bounded sequence {f k }k∈ℕinwhich converges to zero uniformly on compact subsets ofas k → ∞, we haveinas k → ∞.
Lemma 6. Suppose n ∈ ℕ and μ is a radial weight such that lim|z|→1μ (z) = 0. A closed set K inis compact if and only if it is bounded and satisfies
Proof. The proof of this Lemma is followed by standard arguments similar to those outlined in ([13]). We omit the details. □
3. The Proof of Theorem 1
(1a) Boundedness of .
We will prove the sufficiency first. Suppose that the conditions in (1) hold. Then, for any , from Lemma 2 and Lemma 4, we obtain
For z = 0 and every d ∈ {0, 1,..., n - 1},
From (1), (6), and (7), we know that is bounded.
Conversely, suppose that is bounded. Then, there exists a constant C such that , for all .
For a fixed , and constants c1, c2,..., cn + 1, set
Applying Lemma 1 and triangle inequality, it is easy to get that for every . Moreover, we have that
Now we show that for each s ∈ {m, m + 1,..., m + n}, there are constants c1, c2,..., cn+1, such that,
Indeed, by differentiating function g w for each s ∈ {m, m + 1,..., m + n}, the system in (10) becomes
By Lemma 3, the determinant of system (11) is different from zero, which implies the statement. For each k ∈ {0, 1, 2,..., n}, we choose the corresponding family of functions that satisfy (10) with s = m + k and denote it by g w,k . For each fixed k ∈ {0, 1,..., n}, the boundedness of the operator , along with Lemma 4 and (8) implies that for each φ(w) ≠ 0,
From (12), it follows that for each k ∈ {0, 1,..., n},
Now we use the test functions
For each k ∈ ℕ, it is easy to get that
By applying Lemma 4 to the h0 (z) = zm , we get
which along with boundedness of the operator implies that
Assume now that we have proved the following inequalities
for j ∈ {0, 1,..., k - 1}, k ≤ n.
Apply Lemma 4 to the h k (z) = zm+k, and knowing that z(s)≡ 0 for s > m + k and the boundedness of the operator , we get
Using hypothesis (16), we can know that
for each k ∈ {0, 1,..., n}. Then, for each k ∈ {0, 1,..., n},
From (13) and (19), we know that (1) holds.
(1b) Compactness of .
Suppose is bounded and (2) holds. Then, by (1a), (1) holds. Let be a sequence in , such that, and f i converges to 0 uniformly on compact subsets of as i → ∞. By the assumption, for any ε > 0, there is a δ ∈ (0, 1), such that, for each k ∈ {0, 1,..., n} and δ < |φ(z) | < 1,
From Lemma 2 and (20), we have
If |φ(z) | ≤ r, then by Cauchy's estimate and (19), we have
For j = 0, 1,..., n - 1, we have
Applying (21), (22), and (23), we know that . From Lemma 5, is compact.
Conversely, suppose that is compact, then is bounded. Let (z i )i∈ℕbe a sequence in such that |φ(z i )| → 1, i → ∞. If such a sequence does not exist, then the condition in (2) is easily satisfied. Now, assume that when ||φ||∞ = 1 and (2) does not hold, then there is k ∈ {0, 1,..., n} and δ > 0 such that
Let , k ∈ 0, 1,..., n be as in Theorem 1. Then, and g i → 0 uniformly on compact subsets of as i → ∞. By the assumption and Lemma 5, we have that for k ∈ {0, 1,..., n}
On the other hand, from (12), we obtain
for large enough i. From (24) and (25), this is a contradiction. So, (2) holds.
Now the proof of Theorem 1 is completed.
4. The Proof of Theorem 2
(2a) Boundedness of .
First, suppose that is bounded and (3) holds. For each polynomial p(z), we obtain |p(m+k)(z)| ≤ C p , z ∈ D, C p is s constant depending on p.
And
From (26), we have that for each polynomial p(z), . Since the set of polynomials is dense in , we have that for each , there is a sequence of polynomials , such that as k → ∞. From the boundness of , we have that
Then, , from which the boundedness of follows.
Conversely, suppose that is bounded. It is clear that is bounded. Then, taking the test functions h k (z) = zm+k for each k ∈ {0, 1,..., n}, we obtain . By the proof of Theorem 1, for each k ∈ {0, 1,..., n},
(2a) is completed.
(2b) Compactness of .
First, assume that is compact, so it is bounded and (3) holds. Hence, if ||φ||∞ < 1,
If ||φ||∞ = 1, since is compact too and (2) holds, then for all ε > 0, there is an r ∈ (0, 1), such that when r < |φ (z) | < 1, for k ∈ {0, 1,..., n},
By (3), we know there is a δ ∈ (0, 1), such that δ < |z| < 1, for k ∈ {0, 1,..., n},
Then, when δ < |z| < 1 and r < |φ (z) | < 1 for k ∈ {0, 1,..., n}, we get
In addition, when |φ (z) | ≤ r and δ < |z| < 1, we have
Combining (30) and (31), we know (4) holds.
Conversely, assume is bounded and (4) holds. Taking the supremum in (6) for all f in the unit ball of , and using the condition (4), we have , from which by Lemma 6, the compactness of follows.
Now the proof of Theorem 2 is finished.
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Acknowledgements
We are grateful to the referee(s) for many helpful comments on the manuscript. Hong-Gang Zeng is supported in part by the National Natural Science Foundation of China (Grand Nos. 10971153, 10671141).
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LZ found the question and drafted the manuscript. HGZ joined in the discussion about the question and revised the paper. All authors read and approved the final manuscript
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Zhang, L., Zeng, HG. Weighted differentiation composition operators from weighted bergman space to n th weighted space on the unit disk. J Inequal Appl 2011, 65 (2011). https://doi.org/10.1186/1029-242X-2011-65
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DOI: https://doi.org/10.1186/1029-242X-2011-65
Keywords
- weighted differentiation composition operators
- weighted Bergman space
- n th weighted space
- boundedness
- compactness