We characterized the boundedness and compactness of weighted differentiation composition operators from BMOA and the Bloch space to Bloch-type spaces. Moreover, we obtain new characterizations of boundedness and compactness of weighted differentiation composition operators.

1. Introduction

Let 𝔻 be the open unit disk in the complex plane ℂ, H(𝔻) the space of all functions holomorphic on 𝔻, dA(z) = (1/π)dxdy the normalized area measure on 𝔻, and H^∞ the space of all bounded holomorphic functions with the norm ∥f∥_∞ = sup _z∈𝔻 | f(z)|.

Let α > 0. The α-Bloch space ℬ^α on 𝔻 is the space of all holomorphic functions f on 𝔻 such that

()

The little α-Bloch space

consists of all f ∈ ℬ^α such that

()

Both spaces ℬ^α and

are Banach spaces with the norm

()

and

is a closed subspace of ℬ^α. If α = 1, they become the classical Bloch space ℬ and little Bloch space ℬ₀, respectively. For any α > 0, the space

consists of functions f ∈ H(𝔻) such that

()

For information of such spaces, see, for example, [1–4].

For a ∈ 𝔻, let

be the automorphism of 𝔻 that interchanges 0 and a. Let the Green function in 𝔻 with logarithmic singularity at a be given by

()

The space BMOA consists of all f in the Hardy space H² such that

()

BMOA is a Banach space under following norm (see, e.g., [5]):

()

Let φ and ψ be holomorphic maps on the open unit disk 𝔻 such that φ(𝔻) ⊂ 𝔻. For a nonnegative integer n, we define a linear operator

as follows:

()

We call it weighted differentiation composition operators, which was defined in [6, 7]. If n = 0 and ψ ≡ 1,

becomes C_φ induced by φ, defined as C_φf = f ∘ φ, f ∈ H(𝔻). If ψ = 1 and φ(z) = z, then

is the differentiation operator defined as Dⁿf = f⁽ⁿ⁾. If n = 0, then we get the weighted composition operator ψC_φ defined as ψC_φf = ψ · (f∘φ). If n = 1 and ψ(z) = φ^′(z), then

reduces to DC_φ. When ψ ≡ 1, then

reduces to differentiation composition operator C_φDⁿ (also named as product of differentiation and composition operator). If we put φ(z) = z, then

, the product of multiplication and differentiation operator.

The boundedness and compactness of differentiation composition operator between spaces of holomorphic functions have been studied extensively. For example, Hibschweiler; Portnoy and Ohno studied differentiation composition operator C_φD on Hardy and Bergman spaces in [8, 9]; Li; Stević and Ohno studied C_φD on Bloch type spaces in [10–12]; Wu and Wulan gave a new compactness criterion of C_φD^m on the Bloch space in [13]. Recently, the weighted differentiation composition operator between different function spaces has also been investigated by several authors (see, for example, [14–21]).

Boundedness, compactness, and essential norm of weighted composition operator ψC_φ between Bloch-type spaces have been studied in [22–24]. Recently, Manhas and Zhao [25] and Hyvärinen and Lindström [26] gave a new characterization of boundedness and compactness of ψC_φ in terms of the norm of φⁿ (for the compactness of composition operator, see [27, 28]).

Motivated by [13, 25, 26], we study the operator from BMOA and Bloch space to Bloch-type spaces.

Throughout this paper, constants are denoted by C; they are positive and not necessarily the same at each occurrence. The notation A≲B means that there is a positive constant C such that A ≤ CB. When A≲B and B≲A, we write A ≈ B.

2. Some Lemmas

It is well known that H^∞ ⊂ BMOA ⊂ ℬ. From the definition of the norm, we know

()

Indeed, Girela proved that

()

in Corollary 5.2 of [5]. The following lemma is from Lemma 5 in [29] (see also Lemma 4.12 of [4]).

Lemma 1. If f ∈ H(𝔻), then

()

The following lemma may be known, but we fail to find its reference; so we give a proof for the completeness of the paper.

Lemma 2. Let f ∈ H(𝔻). Then,

()

Proof. Applying Littlewood-Paley identity

()

and Lemma 1, we have

()

It follows from the definitions of Bloch space and BMOA space that

()

By Theorem 6.2 of [5] and the proof of Theorem 1 of [30], we have the following lemma.

Lemma 3. Let n be a fixed positive integer and f ∈ ℬ with f(0) = f^′(0) = ··· = f⁽ⁿ⁻¹⁾(0) = 0. If

()

then ∥f∥_BMOA≲1.

Lemma 4. Suppose that n is a fixed positive integer. Let k ∈ ℕ⁺, 0 ≤ x ≤ 1, and

()

If k ≥ n, then there are two positive constants c_n and C_n, depending only on n, such that

()

Proof. The proof is similar to that of Lemma 2.2 of [13] and is so omitted.

3. Boundedness of

In this section, we characterize the boundedness of from BMOA and the Bloch space to Bloch-type spaces.

Theorem 5. Let α > 0, ψ ∈ H(𝔻), n ∈ ℕ⁺, and φ a holomorphic self-map of 𝔻. Then, the following statements are equivalent:

(a)
is bounded.
(b)
and are bounded.
(c)
is bounded.
(d)
and are bounded.
(e)
is bounded.
(f)
and are bounded.
(g)
sup _z∈𝔻((1−|z|²) ^α/(1−|φ(z)|²) ⁿ)|ψ^′(z)| < ∞ and sup _z∈𝔻((1−|z|²) ^α/(1−|φ(z)|²) ⁿ⁺¹) | ψ(z)φ^′(z)| < ∞.
(h)
and .

Proof. It is obvious that (f)⇒(b), (f)⇒(d), (e)⇒(c), and (e)⇒(a). Thus, we will prove the theorem according to the following steps. (I): (a)⇒(g), (c)⇒(g). (II): (b)⇒(g), (d)⇒(g). (III): (g)⇒(e), (g)⇒(f). (IV): (f)⇔(h).

(I): (a)⇒(g), (c)⇒(g). Suppose that (a) or (c) holds. We choose the test function g₁(z) = zⁿ. By Lemma 2, we get

()

Taking g₂(z) = zⁿ⁺¹ and using the fact that |φ(z)| < 1, we have

()

We now consider the function

()

It is easy to check that f_λ ∈ ℬ₀∩BMOA and

. Moreover,

()

Thus,

and

()

We obtain

()

Thus, for any r₀ ∈ (0,1), we have

()

Using (21) yields

()

Combining (26) with (27), we get

()

We next consider the function

()

Similarly, we get g_λ ∈ ℬ₀∩BMOA and

()

Moreover,

()

and

. We have, as above,

()

Thus, for any s₀ ∈ (0,1),

()

Applying (20), we get

()

Combining (34) with (35) yields

()

(II): (b)⇒(g) and (d)⇒(g). Suppose that

is bounded or

is bounded. Set

()

If λ = ∞, then for any positive integer N, we can find b ∈ 𝔻 such that

()

If φ(b) = 0, then choose the test function g(z) = zⁿ. It is clear that g ∈ ℬ₀. From Lemma 2, we have

()

If φ(b) ≠ 0, consider the function

()

where a = φ(b). Let

. Then, F(0) = F^′(0) = ··· = F⁽ⁿ⁻¹⁾(0) = 0 and

()

It is easy to see that

()

So, by Theorems 5.4 and 5.13 of [4], we have F ∈ ℬ₀ and ∥F∥_ℬ≲1. By Lemma 1 of [31] and Lemma 3, we get ∥F∥_BMOA≲1. We have

()

Since N is arbitrary, we get

. This contradicts the boundedness of

and that of

Now, suppose that is bounded or is bounded. Set

()

If η = ∞, then for any positive integer M, exists u ∈ 𝔻 such that

()

If φ(u) = 0, then set g(z) = zⁿ⁺¹. The process as above gives

()

If φ(u) ≠ 0, consider the function

()

where a = φ(u). Let

. Then, F(0) = F^′(0) = ··· = F⁽ⁿ⁾(0) = 0 and

()

Applying Theorems 5.4 and 5.13 of [4] again yields F ∈ ℬ₀ and ∥F∥_ℬ≲1. We get ∥F∥_BMOA≲1 and

()

Since M is arbitrary, we have

. This contradicts the boundedness of

(III): (g)⇒(e), (g)⇒(f). Note that

()

The desired results follow.

(IV): (f)⇔(h). Suppose that (f) is true. It follows from Proposition 5.1 of [4] that . So,

()

Conversely, assume that (h) is true. It is easy to see that

()

If ∥φ∥_∞ < 1, then

()

Hence, (g) is true. From (g)⇒(f), we obtain that (f) is also true.

From now on, we assume that ∥φ∥_∞ = 1. For any integer k ≥ n, let

()

Let m with m ≥ n be the smallest positive integer such that

. Since

is not empty for every integer k ≥ m and

. By Lemma 4, for f ∈ ℬ,

()

So,

is bounded. Similar argument implies

()

Thus,

is bounded. Theorem 5 is proved.

4. Compactness of

The following criterion for the compactness is a useful tool and it follows from standard arguments, for example, Proposition 3.11 of [32] or Lemma 2.10 of [33].

Lemma 6. Let α > 0, n ∈ ℕ⁺, and X = ℬ₀, ℬ, or BMOA. Suppose that ψ and φ are in H(𝔻) such that φ(𝔻) ⊂ 𝔻. Then, is compact if and only if for any sequence {f_m} in X with , which converges to zero locally uniformly on 𝔻; we have .

We now give the compactness of from BMOA and the Bloch space to Bloch-type spaces.

Theorem 7. Let α > 0, ψ ∈ H(𝔻), n ∈ ℕ⁺, and φ a holomorphic self-map of 𝔻. Then, the following statements are equivalent:

(a)
is compact.
(b)
is compact and is compact.
(c)
is compact.
(d)
is compact and is compact.
(e)
is compact.
(f)
is compact and is compact.
(g)
ψ ∈ ℬ^α, ,
()
(h)
and .

Proof. The proof is a modification of that of Theorem 5; so we give a sketch of the proof. We will prove the theorem according to the following steps. (I): (a)⇒(g), (c)⇒(g). (II): (b)⇒(g), (d)⇒(g). (III): (g)⇒(e), (g)⇒(f). (IV): (f)⇔(h).

(I): (a)⇒(g), (c)⇒(g). Suppose that (a) or (c) holds. Then by Theorem 5, we have

()

That is, ψ ∈ ℬ^α,

Let {z_j} be a sequence in 𝔻 such that |φ(z_j)| → 1 as j → ∞. Now, we consider the function

()

Simple computation shows that f_j ∈ ℬ₀∩BMOA and

()

It is also easy to check that f_j → 0 uniformly on compact subsets of 𝔻 as j → ∞. Moreover,

()

We have

()

By Lemma 6, we get

()

We next consider the function

()

Similarly, we get g_j ∈ ℬ₀∩BMOA and

()

It is easy to see that g_j converges to zero uniformly on compact subsets of 𝔻 as j → ∞ and

()

Thus,

()

Applying Lemma 6 again, we have

()

Since z_j ∈ 𝔻 is arbitrary, we proved that (g) is true.

(II) (b)⇒(g), (d)⇒(g). Suppose that (b) or (d) holds. A similar argument to (I) shows that ψ ∈ ℬ^α, . Now, suppose that the equations in (g) are not true. Then, there exists a sequence {z_j} in 𝔻 and δ > 0 such that |φ(z_j)| → 1 as j → ∞ and

()

Choose a subsequence of {z_j} if necessary and suppose that inf _j | φ(z_j)| > 1/2. Let

()

Then, it is easy to check that f_j ∈ ℬ₀∩BMOA, f_j → 0, uniformly on compact subsets of 𝔻 and

()

Thus,

()

Those contradict the compactness of

and

(III) (g)⇒(e), (g)⇒(f). Let {f_m} be a norm bounded sequence in ℬ that converges to zero uniformly on compact subsets of 𝔻. Let . For ɛ > 0, then there exists r₀ ∈ (0,1) such that for |φ(z)| > r₀, we have

()

Thus, for z ∈ 𝔻, we have

()

where K₁ = sup _z∈𝔻(1 − |z|²) ^α|ψ^′(z)| and K₂ = sup _z∈𝔻(1 − |z|²) ^α|ψ(z)φ^′(z)|. Since

uniformly on compact subsets of 𝔻 as m → ∞, we have

as m → ∞. It follows from Lemma 6 that

is compact.

Similar as above, we know

()

From

uniformly on compact subsets of 𝔻, we have

and

as m → ∞. So,

are compact.

(IV): (f)⇔(h). Suppose that (f) is true. Note that and z^k → 0 uniformly on compact subsets of 𝔻 as k → ∞; by Lemma 6, we have

()

Conversely, assume that (h) is true. It is easy to see that

()

If ∥φ∥_∞ < 1, from (g)⇒(f), we get that (f) is true. If ∥φ∥_∞ = 1, as in the proof of Theorem 5, let

()

And let m with m ≥ n be the smallest positive integer such that

. For given ɛ > 0, there exists a large enough integer M₁ with M₁ > m such that

()

whenever k > M₁. Let {f_j} be a norm bounded sequence in ℬ that converges to zero uniformly on compact subsets of 𝔻 as j → ∞. Denote

. We get

()

Then,

()

where

()

Since

uniformly on compact subsets of 𝔻, then

as j → ∞. Thus, by Lemma 6,

is compact. Similar as above, we can prove that

is compact. The proof is complete.

Acknowledgments

This work was supported by NNSF of China (Grant no. 11171203) and NSF of Guangdong Province (Grant nos. 10151503101000025 and S2011010004511).

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Weighted Differentiation Composition Operators to Bloch-Type Spaces

Abstract

1. Introduction

2. Some Lemmas

3. Boundedness of

4. Compactness of

Acknowledgments

References

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Weighted Differentiation Composition Operators to Bloch-Type Spaces

Abstract

1. Introduction

2. Some Lemmas

3. Boundedness of

4. Compactness of

Acknowledgments

References

Citing Literature

References

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