Differences of Composition Operators Followed by Differentiation between Weighted Banach Spaces of Holomorphic Functions

Lemma 1. Let v be a radial weight satisfying condition (L1). There is a constant C > 0 (depending only on the weight v) such that for all $$,

for every z ∈ 𝔻.

Lemma 2. Let v be a radial weight satisfying condition (L1). There is a constant C > 0 such that for all $$,

for all z, ω ∈ 𝔻.

Proof. For $$, let u(z) = v(z)(1−|z|²), by Lemma 1, we obtain $$, so by Lemma 3.2 in [31] and Lemma 1, there is a constant C > 0 such that

for each z, ω ∈ 𝔻. This completes the proof.

Remark 3. From Lemma 2, it is not hard to see that for any z, ω ∈ r𝔻 = {z ∈ 𝔻:|z|² < r < 1}, then

for any $$, where $$.

Lemma 4. Assume that v is a normal weight. Then for every f ∈ H(𝔻) the following asymptotic relationship holds:

Lemma 5. Suppose that u, v ∈ H(𝔻) and φ, ψ ∈ S(𝔻). Then the operator $$ is compact if and only if whenever {f_n} is a bounded sequence in $$ with f_n → 0 uniformly on compact subsets of 𝔻, and then ∥(DC_φ − DC_ψ)f_n∥ → 0, as n → ∞.

Theorem 6. Suppose that v is an arbitrary weight and that u is a normal and radial weight. Then the following statements are equivalent.

• (i)

  $$ is bounded.

• (ii)

  The conditions (12) and (14) hold.

• (iii)

  The conditions (13) and (14) hold.

Proof. First, we prove the implication (i) ⇒ (ii). Assume that $$ is bounded. Fixing w ∈ 𝔻, we consider the function f_w defined by

Next prove that $$. In fact,

By Lemma 4 we have thus $$, and $$. Note that $$, and $$. So by the boundedness of $$, it then follows that for any w ∈ 𝔻. Since w ∈ 𝔻 is an arbitrary element, then from (18) and (4), we can obtain (12).

Next we prove (14). For given w ∈ 𝔻, we consider the function

Like for f_w above, we can show that $$ with $$. One sees that $$. Then where By Lemma 2 and (12), we conclude that |I(w)| < ∞, which combines with (20), and we obtain that for all w ∈ 𝔻; thus (14) holds.

(ii) ⇒ (iii). Assume that (12) and (14) hold, we need only to show that (13) holds. In fact,

from which, using (12) and (14), the desired condition (13) holds.

(iii) ⇒ (i). Assume that (13) and (14) hold. By Lemmas 1 and 2, for any $$, we have

from which it follows that $$ is bounded. The whole proof is complete.

Corollary 7. Suppose that v is an arbitrary weight and that u is a normal and radial weight satisfying condition (L1). Then the following statements are equivalent.

• (i)

  $$ is bounded.

• (ii)

  The conditions (12) and (14) hold.

• (iii)

  The conditions (13) and (14) hold.

Theorem 8. Suppose that v is an arbitrary weight and that u is a normal and radial weight. Then $$ is compact if and only if $$ is bounded and the conditions (25)–(27) hold.

Proof. First we suppose that $$ is bounded and the conditions (25)–(27) hold. Then the conditions (12)–(14) hold by Theorem 6. From (25)–(27), it follows that for any ε > 0, there exists 0 < r < 1 such that

Now, let {f_n} be a sequence in $$ such that $$ (constant) and {f_n} → 0 uniformly on compact subsets of 𝔻. By Lemma 5 we need only to show that $$ as n → ∞. A direct calculation shows that

where

We divide the argument into a few cases.

Case 1 (|φ(z)| ≤ r and |ψ(z)| ≤ r). By the assumption, note that {f_n} converges to zero uniformly on E = {w:|w | ≤ r} as n → ∞; using (14) and Cauchy′s integral formula, it is easy to check that J_n(z) → 0,   n → ∞ uniformly for all z with |ψ(z)| ≤ r.

On the other hand, it follows from Remark 3 after Lemma 2 and (12) that

Case 2 (|φ(z)| > r and |ψ(z)| ≤ r). As in the proof of Case 1, J_n(z) → 0 uniformly as n → ∞. On the other hand, using Lemma 2 and (28) we obtain |I_n(z)| ≤ CLε.

Case 3 (|φ(z)| > r and |ψ(z)| > r). For n sufficiently large, by Lemma 2 and (28) we obtain that |I_n(z)| ≤ CLε. Meanwhile, |J_n(z)| ≤ CLε by Lemma 1 and (30).

Case 4 (|φ(z)| ≤ r and |ψ(z)| > r). We rewrite

where

The desired result follows by an argument analogous to that in the proof of Case 2. Thus, together with the above cases, we conclude that

for sufficiently large n. Employing Lemma 5 we obtain the compactness of $$.

For the converse direction, we suppose that $$ is compact. From which we can easily obtain the boundedness of $$. Next we only need to show that (25)–(27) hold.

Let {z_n} be a sequence of points in 𝔻 such that |φ(z_n)| → 1 as n → ∞. Define the functions

Clearly, {f_n} converges to 0 uniformly on compact subsets of 𝔻 as n → ∞ and $$ with $$ for all n. Moreover,

By the compactness of $$ and Lemma 5, it follows that $$. On the other hand, using (38) we have

Letting n → ∞ in (39), it follows that (25) holds. The condition (26) holds for the similar arguments.

Now we need only to show the condition (27) holds. Assume that {z_n} is a sequence in 𝔻 such that |φ(z_n)| → 1 and |ψ(z_n)| → 1 as n → ∞. Define the function

It is easy to check that {h_n} converges to 0 uniformly on compact subsets of 𝔻 as n → ∞ and $$ with $$ for all n ∈ N. Note that $$, then $$, and it follows from Lemma 5 that $$, n → ∞. On the other hand we obtain that where

By Lemma 2 and the condition (25) that has been proved, we get I(z_n) → 0, n → ∞. This combines with (41), and we obtain J(z_n) → 0, n → ∞. This shows that (27) is true. The whole proof is complete.

Corollary 9. Suppose that v is an arbitrary weight and that u is a normal and radial weight satisfying condition (L1). Then $$ is compact if and only if $$ is bounded and the conditions (25)–(27) hold.

Example 1. In this example we will show that there exist weight u (normal, radial) and v, analytic self-maps on the unit disk φ, ψ such that the conditions (25)–(27) in Theorem 8 are satisfied while $$ is not compact.

Let

and ψ(z) = −φ(z), where $$.

Since for |z | < 1, we have |φ(z)| < M/(M + 1) so φ belongs to S(𝔻), as well as ψ. Moreover, |ψ(z)| and |ψ(z)| can never tend to 1 for any z ∈ 𝔻, which means that conditions (25)–(27) hold trivially.

Now we will show that $$ is not bounded, and then not compact. Let z_k = 1 − 1/k, and then it is easy to check that φ(z_k) → M/(M + 1) and ψ(z_k)→−M/(M + 1) as k → ∞. So

However, since $$, then |φ^′(z_k)| → ∞ as k → ∞. Thus choose u(z) = 1 − |z|² and v(z) = u(φ(z)), and we can obtain Hence DC_φ − DC_ψ does not map $$ boundedly into $$ by Theorem 6.