A new product of weighted differentiation and superposition operators between Hardy and Zygmund Spaces

1.   Introduction

Let $\mathbb{D} = \{z\in\mathbb{C}: \vert z\vert < 1\}$ be the open unit disk in the complex plane $\mathbb{C},$ $H(\mathbb{D})$ denote the class of all analytic functions in $\mathbb{D}.$ Let $\mathcal\phi$ be a complex-valued function in the plane $\mathbb{C}$. The superposition operator $S_{\phi}$ defined as follow (see^[2]):

Definition 1. Let $X$ and $Y$ be two metric spaces of ${H}\mathbb{(D)}$ and $\phi$ denote a complex-valued function in the plane $\mathbb{C}$ such that $\phi \circ f \in Y$ whenever $f \in X,$ we say that $\phi$ acts by superposition from $X$ into $Y$ and the superposition operator $S_{\phi}$ on $X$ is defined by

Observe that if, X contains the linear functions and $S_\phi$ maps X into Y, then $S_\phi$ must be an entire function.

The problem of boundedness and compactness of $S_{\phi}$ has been studied in many Banach spaces of analytic functions and the study of such operators has recently attracted the most attention (see ^{[1,6,7,8,15,16]} and others).

Now, we will introduce a class of nonlinear operators as follows:

Definition 2. Let n be a nonnegative integer, ${u\in H(\mathbb{D}) }$ and $\phi$ is a non constant analytic self-map of $\mathbb{D}$. The weighted differentiation superposition operators $({D^n_u} S_{\phi }f)(z)$ defined as

The Bloch type space defined as follows (see ^[13,16]).

Definition 3. An analytic function f is said to belong to the Bloch space $\mathcal{B}$ if

while the little Bloch spaces $\mathcal{B}_{0 } \subset \mathcal{B}$ consisting of all functions analytic in $\mathbb{D}$ for which

Now we present the needed spaces and some facts. The Hardly space can be defined as follows (see^[17]).

Definition 4. The space $H^\infty$ denotes the space of all bounded analytic functions f on the unit disk $\mathbb{D}$ such that

Definition 5. Let $\mathcal{Z}$ denote the space of all f $\in H(\mathbb{D})\cap {C}(\mathbb{\bar{D}})$ such that

where the supremum is taken over all $e^{i\theta }\in \partial \mathbb{D}$ which denote the boundary of $\mathbb{D}$ and $h > 0$. By the Zygmund theorem and the closed-graph theorem (see ^[5], Theorem 5.3), we see that f $\in \mathcal{Z}$ if and only if

Moreover, the following asymptotic relation holds:

Therefore, $\mathcal{Z}$ is called the Zygmund class. Since the quantities in $(1.4)$ are semi norms, it is natural to add them the quantity $\vert f(0)\vert+\vert f'(0)\vert$ to obtain two equivalent norms on the Zygmund class.The Zygmund class with such defined norm will be called the Zygmund space. Some information on Zygmund type spaces on the unit disk and some operators on them can be found in (see ^[3,10,11,12]). This norm will be again denoted by $\parallel.\parallel_\mathcal{Z}$. The little Zygmund space $\mathcal{Z}_0$ was introduced by Li and Stević (see ^[9]) in the following natural way:

It is easy to see that $\mathcal{Z}_0$ is a closed subspace of $\mathcal{Z}$ and the set of all polynomials is dense in $\mathcal{Z}_0$.

Now, we will introduce the definition of boundedness and compactness of the operator ${D^n_u} S_{\phi }: {H}^{\infty} \rightarrow \mathcal{Z}$.

Definition 6. The operators ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is said to be bounded, if there is a positive constant $C$ such that $\vert|{D^n_u} S_{\phi }\; f\vert|_{\mathcal{Z}}\leq C||{f}||_{\infty}$ for all $f\in {H}^{\infty}.$

Definition 7. The operators ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is said to be compact, if it maps any function in unit disk in ${H}^{\infty}$ onto a pre-compact set in $\mathcal{Z}$.

The notation a $\preceq b$ means that there is a positive constant C such that a $\preceq C b$. Also, the notation a $\asymp$ b means that a $\preceq$ b and b $\preceq$ a hold.

In this paper, we study a concerned class of weighted differentiation superposition operators ${D^n_u} S_{\phi}$. Furthermore, It has made the discussions on the boundedness and compactness property of the new class of operators from $H^\infty$ to Zygmund spaces. Finally, it has also provided the conditions which grant the product operators ${D^n_u} S_{\phi}$ be bounded and compact.

2.   The Boundedness of ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$

Now we characterize the boundedness of the operators ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$.

First we enumerate several useful lemmas. The first one below is well-known.

Lemma 1. (see(^[13])) Assume that $f\in{H}^{\infty}$. Then for each $n\in\mathbb{N}$, there is a positive constant $C$ independent of f such that

The following lemma is introduced in (see^[18]).

Lemma 2. Assume that $f \in \mathcal{B}$. Then for each $n \in \mathbb{N}$.

Theorem 1. Suppose $\phi$ be an entire function and $u\in H(\mathbb{D})$. Then ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ bounded if and only if the following conditions are satisfied,

and

Proof. First direction, we assume that conditions (2.1)-(2.3) hold. So, for every $z \in\mathbb{D}$ and $f \in H^\infty,$ by using Lemma 1, we have

Since, if we take $f(z) = z^{n},$ we have

On the other hand, we obtain

and

From the fact $\mid f(0)\mid < 1$ and by applying the conditions (2.1)-(2.3), it follows that the operators ${D^n_u} S_{\phi } : {H}^{\infty} \; or \; \mathcal{B} \rightarrow \mathcal{Z}$ is bounded.

Now, we will prove the second direction, assume that ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is bounded, this means that there exists a constant $C$ such that

For all $f \in {H}^{\infty}.$ From the above inequality and by taking the function $\phi (z) = z^{n}$ we have

By taking the function $\phi (z) = z^{n+1}.$ From the fact that $\parallel \phi \parallel_\infty \leq 1$ and using (2.5), it follows that

Similarly, by taking the function $\phi (z) = z^{n+2}.$ From the fact that $\parallel \phi \parallel_\infty \leq 1$ and by using (2.5), (2.6), it follows that

For a fixed $w \in \mathbb D,$ we consider the following test functions

By the triangle inequality, we can see that

Hence

and we have

Which follows that

If we take $f(w) = w^{n}$, we get

For a fixed $\delta \in(0, 1)$ and by using (2.1), (2.5), we obtain

It follows that the condition(2.1) holds as desired.

Next, we prove the condition (2.3). To see this, for a fixed $w \in \mathbb{D}$, put

It is easy to prove that

and we have

Which follows that

If we take $f(w) = w^{n}$, we get

For a fixed $\delta \in(0, 1)$ and by using (2.7), (2.17), we obtain

It follows that the condition (2.3) holds as desired.

Now, we will prove the condition (2.2), for a fixed $w \in \mathbb{D}$, put

It is easy to see that

and we have

Which follows that

If we take $f(w) = w^{n}$, we get

from (2.6) and (2.23) simliar to (2.12) we obtain (2.2), finishing the proof of the theorem.

3.   The compactness of ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$

Now we characterize the compactness of the operators ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$. The next Lemma is often used in dealing the compactness of operators on analytic function spaces. Since the proof standard (see Proposition 3.11 in ^[4]).

Lemma 3. Suppose $\phi$ be an entire function and $u \in H(\mathbb{D}).$ Then ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is compact if and only if ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is bounded and for any bounded sequence $\{f_k\}$ in $H^{\infty}$ which converges to zero uniformly on compact subsets of $\mathbb{D}$ as $k \rightarrow \infty,$ we have $\parallel {D^n_u}S_{\phi} f_n\parallel_\mathcal{Z} \rightarrow 0$ as $n\rightarrow \infty.$

The second following lemma was introduced and proved in ^[9] which is similar to the corresponding lemma in ^[14].

Lemma 4. A closed set $K$in $\mathcal{Z}_0$ is compact if and only if $K$ is bounded and satisfies

Now, we begin with the sufficient and necessary condition for the compactness of ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$

Theorem 2. Suppose that $\phi$ be an entire function and $u \in H(\mathbb{D}).$ Thus ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is compact if and only if ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is bounded and the following conditions are satisfied,

and

Proof. Suppose that ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is bounded and that conditions (3.1)-(3.3) hold. For any bounded sequence $\{f_k\}$ in $H^\infty$ with ${f_k}\rightarrow 0$ uniformly on compact subsets of $\mathbb{D}.$ To establish the assertion, it suffices, in view of Lemma 3, to show that

We assume that $||f_k||_\infty \leq 1$. From (3.1)-(3.3), we have given $\epsilon > 0,$ there exists a $\delta \in (0, 1),$ when $\delta < |f(z)| < 1$, we have

From the boundedness of ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ by Theorem 1 we see that (2.5)-(2.7) hold. Since $f_k \rightarrow 0$ uniformly on compact subsets of $\mathbb{D},$ Cauchy's estimate gives that $f^{'}_k, f^{''}_k \; and \; f^{'''}_k$ converges to 0 uniformly on compact subsets of $\mathbb{D}.$ Hence, there exists a $K_0 \in \mathbb{N}$ such that for for $k > K_0.$

From (3.4) and (3.5), we have

If we take $f(z) = z^n,$ we get

when $k > K_0.$ It follows that the operators ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is compact.

Conversely, suppos that ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is compact. Therefore it is clear that ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ is bounded. Let $\{z_k\}$ be a sequence in $\mathbb{D}$ such that $|f(z_k)|\rightarrow 1$ as $k\rightarrow \infty.$ If such a sequence does not exist, thus (3.1)-(3.3) are automatically holding. Now, we consider the test functions

From (2.9) and (2.10), we have

And

For $|z| = r < 1,$ we get

that is, $\phi_{f_k(z_k)}^{*(n)}$ converges to 0 uniformly on compact subsets of $\mathbb{D},$ using (2.10) and the compactness of ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ we get

If we take $f_k(z_k) = z_k^n,$ we get

From this, and $|z_k^n|\rightarrow 1,$ it follows that

We get (3.1).

In order to prove (3.2), consider

It follows from (2.14) and (2.16) that

and

That is, $\phi_{f_k(z_k)}^{**(n)}$ converges to 0 uniformly on compact subsets of $\mathbb{D},$ using (2.11) and the compactness of ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ tends to

From (2.16), we obtain

If we take $f_k(z_k) = z_k^n$ we have

Thus,

Eq (3.3) satisfied. Next, consider

From (2.20) and (2.22), we get

and

and ${\phi_{f_k(z_k)}^{***}}(f_k(z_k))$ converges to 0 uniformly on compact subsets of $\mathbb{D},$ the compactness of ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}$ implies that

From this and (2.23), we obtain (3.2), the proof of the theorem is complete.

Theorem 3. Suppose that $\phi$ be an entire function and $u \in H(\mathbb{D}).$ Thus, ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}_{0}$ is compact if and only if the following conditions are holding.

and

Proof. Suppose that conditions (3.9)-(3.11) are satisfied. Consider the supremum in inequality (2.4) over all $f \in H ^\infty$ such that $||f||_\infty \leq 1$ and letting $|z| \rightarrow 1$ yield

Therfore, by Lemma 3, we see that the operators ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}_{0}$is compact. Now suppose that ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}_{0}$ is compact. Then ${D^n_u} S_{\phi } : {H}^{\infty} \rightarrow \mathcal{Z}_{0}$ is bounded, and by considering the function $\phi (z) = z^{n}$ we have

By considering the function $\phi (z) = z^{n+1}.$ We have

From (3.12), (3.13) and the fact that $||f||_\infty\leq1$, we get

By taking the function $\phi (z) = z^{n+2}.$ From (3.12), (3.14) and the fact that $||f||_\infty\leq1$, we get

By (2.11), (2.18), (2.23), and observing that ${D^n_u} S_{\phi^{*} } \phi^{*}_{f(w)}$, ${D^n_u} S_{\phi^{**} } \phi^{**}_{f(w)}$ and ${D^n_u} S_{\phi^{***} } \phi^{***}_{f(w)}$ we know that

and

We prove that (3.12) and(3.16) imply (3.9). The proof of (3.10) and (3.11) by the same way. Then, it will be held.

From (3.16), it follows that for every $\epsilon > 0,$ there exists $\delta \in (0, 1)$ such that

when $\delta < |z| < 1.$ Using (3.12), we see that there exists $\tau \in (0, 1)$ such that

when $\tau < \mid z\mid < 1$.

Thus, when $\tau < \mid z\mid < 1$ and $\delta < |z| < 1,$ by (3.19) we have

On the other hand, when $\tau < \mid z\mid < 1$ and $|z| \leq \delta,$ by (3.20) we obtain

From (3.21) and (3.22), we obtain (3.9) as desired. This is the end of the proof.

4.   Conclusions

The present study dealt with a radical study of a concerned class of weighted differentiation superposition operators ${D^n_u} S_{\phi}$. Furthermore, It has made the discussions on the boundedness and compactness property of the new class of operators from $H^\infty$ to Zygmund spaces. Finally, it has also provided the conditions which grant the product operators ${D^n_u} S_{\phi}$ be bounded and compact.

Acknowledgments

Researchers would like to thank the Deanship of Scientific Research, Qassim University for funding publication of this project. The authors would like to thank the anonymous concerned reviewers for their valuable remarks on this study.

Conflict of interest

There is no any conflict.