Besov Spaces Induced by Doubling Weights

Abstract

Let $1\leqslant p<\infty$, $0<q<\infty$, and $\nu$ be a two-sided doubling weight satisfying

$$\begin{aligned} \sup _{0\leqslant r<1}\frac{(1-r)^q}{\int _r^1\nu (t)\,dt}\int _0^r\frac{\nu (s)}{(1-s)^q}\,ds<\infty . \end{aligned}$$

The weighted Besov space $\mathcal {B}_{\nu }^{p,q}$ consists of those $f\in H^p$ such that

$$\begin{aligned} \int _0^1 \left( \int _{0}^{2\pi } |f'(re^{i\theta })|^p\,d\theta \right) ^{q/p}\nu (r)\,dr<\infty . \end{aligned}$$

Our main result gives a characterization for $f\in \mathcal {B}_{\nu }^{p,q}$ depending only on |f|, p, q, and $\nu$. As a consequence of the main result and inner-outer factorization, we obtain several interesting by-products. For instance, we show the following modification of a classical factorization by F. and R. Nevanlinna: If $f\in \mathcal {B}_{\nu }^{p,q}$, then there exist $f_1,f_2\in \mathcal {B}_{\nu }^{p,q} \cap H^\infty$ such that $f=f_1/f_2$. Moreover, we give a sufficient and necessary condition guaranteeing that the product of $f\in H^p$ and an inner function belongs to $\mathcal {B}_{\nu }^{p,q}$. Applying this result, we make some observations on zero sets of $\mathcal {B}_{\nu }^{p,p}$.

1 Introduction and Characterizations

Let $\mathbb {D}$ be the open unit disc of the complex plane $\mathbb {C}$ and $\mathbb {T}$ be the boundary of $\mathbb {D}$. The set of all analytic functions in $\mathbb {D}$ is denoted by $\mathcal {H}(\mathbb {D})$. For $0<p<\infty$, the Hardy space $H^p$ consists of those $f\in \mathcal {H}(\mathbb {D})$ such that

$$\begin{aligned} \Vert f\Vert _{H^p} =\sup _{0\leqslant r<1} M_p(r,f)<\infty , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} M_p(r,f)=\left( \frac{1}{2\pi }\int _{0}^{2\pi } |f(re^{i\theta })|^p\,d\theta \right) ^{1/p}. \end{aligned} \end{aligned}$$

The Hardy space $H^\infty$ is the set of all bounded functions in $\mathcal {H}(\mathbb {D})$. Moreover, we recall that a measurable function f on $\mathbb {T}$ belongs to $L^p(\mathbb {T})$ for some $p\in (0,\infty )$ if

$$\begin{aligned} \Vert f\Vert _{L^p}^p = \frac{1}{2\pi }\int _0^{2\pi } |f(e^{i\theta })|^p\,d\theta <\infty . \end{aligned}$$

Alternatively, the Hardy space $H^p$ for $0<p<\infty$ can be characterized as follows: $f\in H^p$ if and only if $f\in \mathcal {H}(\mathbb {D})$, the nontangential limit $f(e^{i\theta })$ exists almost everywhere on $\mathbb {T}$, and $f(e^{i\theta })\in L^p(\mathbb {T})$. In particular, $\Vert f\Vert _{H^p}=\Vert f\Vert _{L^p}$ for $0<p<\infty$ and $f\in H^p$. This is due to Hardy’s convexity and the mean convergence theorems. These results and much more can be found in the classic book [8] by P. Duren.

A function $\nu : \mathbb {D}\rightarrow [0,\infty )$ is called a (radial) weight if it is integrable over $\mathbb {D}$ and $\nu (z)=\nu (|z|)$ for all $z\in \mathbb {D}$. For $0<p,q<\infty$, and a weight $\nu$, the weighted mixed norm space $A^{p,q}_\nu$ consists of those $f\in \mathcal {H}(\mathbb {D})$ such that

$$\begin{aligned} \Vert f\Vert _{A^{p,q}_\nu }^q=\int _0^1 M_p^q(r,f)\, \nu (r)\,dr<\infty . \end{aligned}$$

If $\nu (z)=(1-|z|)^\alpha$ for $-1<\alpha <\infty$, then the notation $A_\alpha ^{p,q}$ is used for $A^{p,q}_\nu$. In this paper, we study class $\mathcal {D}$ of so-called two-sided doubling weights, which originates from the work of J. A.  Peláez and J. Rättyä [19, 20]. For the definition of $\mathcal {D}$ we have to define two wider classes. For a weight $\nu$, set

$$\begin{aligned} \widehat{\nu }(z)=\widehat{\nu }{(|z|)}=\int _{|z|}^1\nu (s)\,ds,\quad z\in \mathbb {D}. \end{aligned}$$

If a weight $\nu$ satisfies the condition $\widehat{\nu }(r)\leqslant C\widehat{\nu }(\frac{1+r}{2})$ for all $0\leqslant r<1$ and some $C=C(\nu )>0$, then we write $\nu \in \widehat{\mathcal {D}}$. Correspondingly, $\nu \in \check{\mathcal {D}}$ if there exist $K=K(\nu )>1$ and $C=C(\nu )>1$ such that

$$\begin{aligned} \begin{aligned} \widehat{\nu }(r)\geqslant C \widehat{\nu }\left( 1-\frac{1-r}{K}\right) , \quad 0\leqslant r<1. \end{aligned} \end{aligned}$$

Class $\mathcal {D}$ is the intersection of $\widehat{\mathcal {D}}$ and $\check{\mathcal {D}}$. In addition, we define the following subclass of $\widehat{\mathcal {D}}$: $\nu \in \widehat{\mathcal {D}}_p$ for some $p\in (0,\infty )$ if the condition

$$\begin{aligned} \widehat{\mathcal {D}}_p(\nu )=\sup _{0\leqslant r<1}\frac{(1-r)^p}{\widehat{\nu }(r)}\int _0^r\frac{\nu (s)}{(1-s)^p}\,ds<\infty \end{aligned}$$

(1.1)

is satisfied. As a concrete example, we mention that $\nu _1(z)=(1-|z|)^\alpha$ and $\nu _2(z)=(1-|z|)^\alpha \left( \log \frac{e}{1-|z|}\right) ^\beta$ for any $\beta \in \mathbb {R}$ belonging to $\mathcal {D}\cap \widehat{\mathcal {D}}_p$ if and only if $-1<\alpha <p-1$. Additional information about weights can be found in [18,19,20]. Some basic properties are recalled also in Sect. 2.

Define the weighted Besov space $\mathcal {B}_{\nu }^{p,q}$ by $\mathcal {B}_{\nu }^{p,q}=\{f:f'\in A_\nu ^{p,q}\}\cap H^p$. For $-1<\alpha <\infty$ and $\nu (z)=(1-|z|)^\alpha$, the notation $\mathcal {B}_{\alpha }^{p,q}$ is used for $\mathcal {B}_{\nu }^{p,q}$. The space $\mathcal {B}_{\nu }^{p,q}$ is the main research objective of this paper. Hence it is worth pointing out that the definition is rational, which means that $H^p$ is not a subset of $\{f:f'\in A_\nu ^{p,q}\}$ in general, or conversely. The family of Blaschke products offers examples for the case where $f\in H^\infty$ and $f'\notin A_\nu ^{p,q}$; see for instance [24]. Moreover, it would be natural that certain lacunary series g lie out of $H^p$, while $g'\in A_\nu ^{p,q}$. Arguments for these kinds of examples can be found in M. Pavlović’s book [16], which contains numerous important observations on the topic of this paper. The existence of both examples, of course, depends on p, q, and $\nu$. In other words, under certain hypotheses for p, q, and $\nu$, an inclusion relation between $\{f:f'\in A_\nu ^{p,q}\}$ and $H^p$ might be valid. However, this is not the case in general.

For $0{<}p{<}\infty$ and $f{\in } L^p(\mathbb {T})$, the $L^p$ modulus of continuity $\omega _p(t,f)$ is defined by

$$\begin{aligned} \omega _p(t,f)=\sup _{0<h<t} \left( \int _0^{2\pi } |f(e^{i(\theta +h)})-f(e^{i\theta })|^p \,d\theta \right) ^{1/p}, \quad 0<t\leqslant 2\pi . \end{aligned}$$

We interpret $\omega _p(t, f)=\omega _p(2\pi , f)$ for $t>2\pi$. It is a well-known fact that, for $0<p,q<\infty$, $-1<\alpha <q-1$, and $f\in H^p$, the derivative of f belongs to $A_\alpha ^{p,q}$ if and only if

$$\begin{aligned} \int _0^\infty \frac{\omega _p(t,f)^q}{t^{q-\alpha }}\,dt<\infty . \end{aligned}$$

This result originates from E. M. Stein’s book [26, Chapter V, Section 5], and the complete version is a consequence of [17, Theorems 2.1 and 5.1] or [12, Theorem 1.2] by M. Pavlović and M. Jevtić. Our first theorem is a partial generalization of the result. Its proof uses some ideas from [8, 21, 22].

Theorem 1

Let $1\leqslant p<\infty$, $0<q<\infty$, and $\nu \in \mathcal {D}$. Then $\nu \in \widehat{\mathcal {D}}_q$ if and only if there exists a constant $C=C(p,q,\nu )>0$ such that

$$\begin{aligned} \int _{1/2}^1 \omega _p(1-r, f)^q \frac{\nu (r)}{(1-r)^q}\,dr \leqslant C \Vert f'\Vert _{A_\nu ^{p,q}}^q \end{aligned}$$

(1.2)

for all $f\in H^p$.

Note that (1.2) is valid also if $0<p,q<\infty$, $\nu$ is a weight, and there exists $\beta =\beta (q,\nu )<q-1$ such that $\nu (r)/(1-r)^{\beta }$ is increasing for $0\leqslant r<1$. This is due to [17, Theorem 5.1] and its proof. Even though the result is valid also for $0<p<1$, Theorem 1 is more useful for our purposes. In particular, it is worth underlining that the hypothesis $1\leqslant p<\infty$ in Theorems 2 and 3 below is natural.

Theorem 2 gives a practical estimate for $\Vert f'\Vert _{A_\nu ^{p,q}}^q+\Vert f\Vert _{H^p}^q$ when $f\in H^p$. As a by-product of its argument, we deduce that also the converse inequality of (1.2) holds if the norm $\Vert f\Vert _{H^p}^q$ is added into the right-hand side. Setting

$$\begin{aligned} d\mu _{z}(\theta )=\frac{1-|z|^2}{|e^{i\theta }-z|^2}\frac{d\theta }{2\pi }, \quad z\in \mathbb {D}, \quad 0\leqslant \theta <2\pi , \end{aligned}$$

Theorem 2 reads as follows.

Theorem 2

Let $1\leqslant p<\infty$, $0<q<\infty$, and $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_q$. Then there exist positive constants $C_1$ and $C_2$ depending only on p, q, and $\nu$ such that

$$\begin{aligned} \begin{aligned} \Vert f'\Vert _{A_\nu ^{p,q}}^q&\leqslant \int _0^1 \left( \int _0^{2\pi } \left( \int _0^{2\pi } |f(e^{i\theta })-f(re^{it})| d\mu _{re^{it}}(\theta ) \right) ^p dt\right) ^{q/p} \frac{\nu (r)}{(1-r)^q}\,dr \\&\leqslant C_1 \left( \int _0^1 \omega _p(1-s, f)^q \frac{\nu (s)}{(1-s)^{q}}\,ds + \Vert f\Vert _{H^p}^q \right) \\&\leqslant C_2 \left( \Vert f'\Vert _{A_\nu ^{p,q}}^q+\Vert f\Vert _{H^p}^q \right) \end{aligned} \end{aligned}$$

(1.3)

for all $f\in H^p$.

By studying the classical weight $\nu (z)=(1-|z|)^\alpha$, where $-1<\alpha <q-1$, we obtain K. M. Dyakonov’s [9, Proposition 2.2(a)] as a direct consequence of Theorem 2. Hence it does not come as a surprise that the proofs of [9, Theorem 2.1] and Theorem 2 have some similarities. Nonetheless, it is worth mentioning that the presence of general weights complicates the argument, and consequently, our proof is quite technical. Note also that Theorem 1 plays an essential role in the proof.

Our main result below gives a characterization for functions f in $\mathcal {B}_{\nu }^{p,q}$ depending only |f|, p, q, and $\nu$. This result improves B. Bøe’s [6, Theorem 1.1], which concentrates only on the case where $1\leqslant p,q<\infty$, $-1<\alpha <q-1$, and $\nu (z)=(1-|z|)^\alpha$. It also generalizes the essential contents of [2, Proposition 2.4] and [9, Proposition 2.2(b)] made by A. Aleman and K. M. Dyakonov, respectively.

Theorem 3

Let $1\leqslant p<\infty$, $0<q<\infty$, and $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_q$. Then there exist positive constants $C_1$ and $C_2$ depending only on p, q, and $\nu$ such that

$$\begin{aligned} \begin{aligned}&\Vert f'\Vert _{A_\nu ^{p,q}}^q \leqslant C_1\left( F_1(f)+F_2(f)\right) \leqslant C_2\left( \Vert f'\Vert _{A_\nu ^{p,q}}^q + \Vert f\Vert _{H^p}^q\right) , \quad f\in H^p, \end{aligned} \end{aligned}$$

(1.4)

where and

Before we talk about the argument of Theorem 3, recall the inner-outer factorization. An inner function is a member of $H^\infty$ having unimodular radial limits almost everywhere on $\mathbb {T}$. For $0<p\leqslant \infty$, an outer function for $H^p$ takes the form

$$\begin{aligned} O_\phi (z)=\exp \left( \frac{1}{2\pi }\int _0^{2\pi } \frac{e^{i\theta }+z}{e^{i\theta }-z}\log \phi (e^{i\theta })\,d\theta \right) , \quad z\in \mathbb {D}, \end{aligned}$$

where $\phi$ is a non-negative function in $L^p(\mathbb {T})$ and $\log \phi \in L^1(\mathbb {T})$. The inner-outer factorization asserts that $f\in H^p$ can be represented as the product of an inner and outer function; see for instance [8, Theorem 2.8]. It is worth noting that the factorization is unique, and

$$\begin{aligned} \begin{aligned} |f(\xi )|=|O_\phi (\xi )|=\phi (\xi ) \end{aligned} \end{aligned}$$

(1.5)

for almost every $\xi \in \mathbb {T}$ if $O_\phi$ is the outer function from the factorization of f. Equation (1.5) is due to the definition of inner functions, the Poisson integral formula, the harmonicity of $\log |O_{\phi }(z)|$ and the fact that

$$\begin{aligned} |O_\phi (z)|=\exp \left( \int _0^{2\pi } \log \phi (e^{i\theta })\,d\mu _z(\theta )\right) , \quad z\in \mathbb {D}. \end{aligned}$$

The last inequality in (1.4) can be proved by applying Theorem 2. In the argument of the first inequality, the inner-outer factorization, the Schwarz–Pick lemma, and an upper estimate for $|O_\phi '|$ from [6] are the main tools. It is worth underlining that Bøe’s idea to make an upper estimate for $|f'|$ by using the factorization seems to be quite effective. Another way to prove results like Theorem 3 is to use a modification of Theorem 2 together with the well-known equation

$$\begin{aligned} \begin{aligned} \int _0^{2\pi } |f(e^{i\theta })-f(z)|^2 d\mu _z(\theta )=\int _0^{2\pi } |f(e^{i\theta })|^2 d\mu _z(\theta )-|f(z)|^2, \quad z\in \mathbb {D}; \end{aligned} \end{aligned}$$

but this method of Dyakonov has the obvious defect that it works only when $f\in H^2$. The advantage of this method in the case where $2\leqslant p<\infty$, $0<q<\infty$, $q/2-1<\alpha <q-1$, and $\nu (z)=(1-|z|)^\alpha$ is that $F_1(f)+F_2(f)$ in Theorem 3 can be replaced by

$$\begin{aligned} \begin{aligned} \int _0^1 \left( \int _0^{2\pi } \left( \int _0^{2\pi } |f(e^{i\theta })|^2 d\mu _{re^{it}}(\theta )-|f(re^{it})|^2\right) ^{p/2} dt\right) ^{q/p} \frac{\nu (r)}{(1-r)^q}\,dr; \end{aligned} \end{aligned}$$

see [9, Proposition 2.2(b)]. It is an open problem to prove a corresponding estimate for general weights.

Next we give an example that shows that the hypothesis $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_q$ in Theorem 3 for $p\geqslant 2$ is sharp in a certain sense. Note that the example is a modification of [22, Example 8]. Before the statement we fix some notation. Write $f\lesssim g$ if there exists a constant $C>0$ such that $f\leqslant Cg$, while $f > rsim g$ is understood analogously. If $f\lesssim g$ and $f > rsim g$, then we write $f\asymp g$.

Example 4

Let $2\leqslant p<\infty$, $q=p$, $\nu (z)=(1-|z|)^{p-1}$, and dA(z) be the two-dimensional Lebesgue measure dxdy. Let f be an inner function such that

$$\begin{aligned} \int _{\{z\in \mathbb {D}:|f(z)|<\varepsilon \}}\frac{dA(z)}{1-|z|}=\infty \end{aligned}$$

for some $\varepsilon \in (0,1)$. The existence of such f is guaranteed by [7, Theorem 5]. Then

$$\begin{aligned} \begin{aligned} F_1(f)+F_2(f)=F_1(f)\asymp \int _{\mathbb {D}} (1-|f(z)|)^p (1-|z|)^{-1}\,dA(z)=\infty , \end{aligned} \end{aligned}$$

while

$$\begin{aligned} \begin{aligned} \Vert f'\Vert _{A_\nu ^{p,q}}^q + \Vert f\Vert _{H^p}^q=\Vert f'\Vert _{A_{p-1}^{p,p}}^p+1<\infty \end{aligned} \end{aligned}$$

by the well-known inclusion

$$\begin{aligned} H^p \subset \{g:g'\in A_{p-1}^{p,p}\}, \quad 2\leqslant p<\infty , \end{aligned}$$

which originates from [13].

We close the section by explaining how the remainder of this paper is organized. Auxiliary results on weights are recalled in the next section. The utility of Theorem 3 is demonstrated in Sects. 3 and 4. More precisely, in Sect. 3, we prove the factorization that states that, for any $f\in \mathcal {B}_{\nu }^{p,q}$, there exist $f_1,f_2\in \mathcal {B}_{\nu }^{p,q} \cap H^\infty$ such that $f=f_1/f_2$. Section 4 begins with a result giving a sufficient and necessary condition guaranteeing that the product of $f\in H^p$ and an inner function belongs to $\mathcal {B}_{\nu }^{p,q}$. As a consequence of this theorem, we obtain some results on zero sets of $\mathcal {B}_{\nu }^{p,p}$. Sections 5, 6, and 7 consist of the proofs of Theorems 1, 2, and 3, respectively.

2 Auxiliary Results on Weights

In this section, we recall some basic properties of weights in $\widehat{\mathcal {D}}$ and $\check{\mathcal {D}}$. These properties are needed in subsequent sections. Another reason for these results is to help the reader to understand the nature of weights in $\mathcal {D}$. We begin with a result that is essentially [19, Lemma 3]; see also [18, Lemma 2.1].

Lemma A

Let $\nu$ be a weight. Then the following statements are equivalent:

• (i) $\nu \in \widehat{\mathcal {D}}$.

• (ii) There exist $C=C(\nu )>0$ and $\beta =\beta (\nu )>0$ such that

  $$\begin{aligned} \widehat{\nu }(r)\leqslant C\left( \frac{1-r}{1-s}\right) ^\beta \widehat{\nu }(s), \quad 0\leqslant r\leqslant s<1. \end{aligned}$$

• (iii) There exist $C=C(\nu )>0$ and $\gamma =\gamma (\nu )>0$ such that

  $$\begin{aligned} \int _0^r \left( \frac{1-r}{1-s}\right) ^\gamma \nu (s)\,ds\leqslant C\widehat{\nu }(r), \quad 0\leqslant r<1. \end{aligned}$$

• (iv) The estimate

  $$\begin{aligned}\int _0^1 s^x\nu (s)\,ds \asymp \widehat{\nu }\left( 1-\frac{1}{x}\right) , \quad 1\leqslant x<\infty ,\end{aligned}$$

  is satisfied.

From the point view of our main results, Lemma A(iii) is interesting because it states that $\nu \in \widehat{\mathcal {D}}$ if and only if $\nu \in \widehat{\mathcal {D}}_{p}$ for some $p>0$. This means that $\widehat{\mathcal {D}}=\bigcup _{p>0} \widehat{\mathcal {D}}_p$. Nevertheless, Lemma A(ii) gives maybe the most interesting description for $\widehat{\mathcal {D}}$. Together with its $\check{\mathcal {D}}$ counterpart below it offers a very practical characterization for weights in $\mathcal {D}$. Essentially this characterization says that $\widehat{\nu }$ is normal in the sense of A. L. Shields and D. L. Williams [25].

Lemma B

Let $\nu$ be a weight. Then $\nu \in \check{\mathcal {D}}$ if and only if there exist $C=C(\nu )>0$ and $\alpha =\alpha (\nu )>0$ such that

$$\begin{aligned} \widehat{\nu }(s)\leqslant C\left( \frac{1-s}{1-r}\right) ^\alpha \widehat{\nu }(r), \quad 0\leqslant r\leqslant s<1. \end{aligned}$$

Lemma B originates from [20], and it can be proved in a corresponding manner as Lemma A(ii). See in particular the proof of [18, Lemma 2.1].

By the definition of class $\widehat{\mathcal {D}}_p$, it is clear that $\widehat{\mathcal {D}}_p \subset \widehat{\mathcal {D}}_{p+\varepsilon }$ for any $\varepsilon >0$. Next we state [21, Lemma 3], which shows that also the converse inclusion is true for sufficiently small $\varepsilon =\varepsilon (\nu , p)>0$. The proof of this result is based on integration by parts. Note that $\widehat{\mathcal {D}}_p(\nu )$ in the statement is defined by (1.1).

Lemma C

If $0<p<\infty$ and $\nu \in \widehat{\mathcal {D}}_p$, then $\nu \in \widehat{\mathcal {D}}_{p-\varepsilon }$ for any $\varepsilon \in \left( 0,\frac{p}{\widehat{\mathcal {D}}_p(\nu )+1}\right)$.

The last result of this section is [22, Lemma 5], which shows that $\nu \in \mathcal {D}$ in the norm $\Vert f\Vert _{A_\nu ^{p,q}}$ can be replaced by $\widehat{\nu }(z)/(1-|z|)$ without losing any essential information.

Lemma D

Let $0<p,q<\infty$ and $\nu$ be a weight.

• (i) If $\nu \in \check{\mathcal {D}}$, then there exists $C=C(\nu )>0$ such that

  $$\begin{aligned} \Vert f\Vert _{A_\nu ^{p,q}}^q\geqslant C \int _0^1 M_p^q(r,f)\,\frac{\widehat{\nu }(r)}{1-r}\,dr, \quad f\in \mathcal {H}(\mathbb {D}). \end{aligned}$$

• (ii) If $\nu \in \widehat{\mathcal {D}}$, then there exists $C=C(\nu )>0$ such that

  $$\begin{aligned} \Vert f\Vert _{A_\nu ^{p,q}}^q\leqslant C \int _0^1 M_p^q(r,f)\,\frac{\widehat{\nu }(r)}{1-r}\,dr, \quad f\in \mathcal {H}(\mathbb {D}). \end{aligned}$$

For $\nu \in \mathcal {D}$, Lemmas A(ii) and B yield

$$\begin{aligned} \widehat{\nu }(r)\asymp \int _r^1\frac{\widehat{\nu }(s)}{1-s}\,ds, \quad 0\leqslant r<1. \end{aligned}$$

(2.1)

In [22], Lemma D is proved by applying this fact together with partial integrations. An alternative way to prove results like Lemma D is to split the integral with respect to dr into infinitely many parts by using a dyadic partition, and then apply (2.1) together with the monotonicity of $M_p^q(r,f)$. An advantage of the last method is that $M_p^q(r,f)$ can be easily replaced by a certain monotonic function g(r). This observation will be utilized several times in the argument of Theorem 2.

3 Quotient Factorization

Recall that if $f\in H^p$ for some $p>0$, then there exist $f_1,f_2\in H^\infty$ such that $f=f_1/f_2$. This is an important consequence of classical factorization [8, Theorem 2.1] by F. and R. Nevanlinna. The main purpose of this section is to give the following $\mathcal {B}_{\nu }^{p,q}$ counterpart to the above-mentioned result.

Theorem 5

Let $1\leqslant p<\infty$, $0<q<\infty$, and $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_q$. If $f\in \mathcal {B}_{\nu }^{p,q}$, then there exist $f_1,f_2\in \mathcal {B}_{\nu }^{p,q} \cap H^\infty$ such that $f=f_1/f_2$ and $f_2$ is an outer function.

It is worth mentioning that [16, Theorem 9.19] is a similar type of result as Theorem 5 with a different hypothesis for $\nu$. Moreover, we note that Theorem 5 generalizes [2, Corollary 2.7], [6, Theorem 3.4] and [9, Corollary 3.4]. For its argument we need an extension of [6, Theorem 3.3]. Note that a part of what follows is really inspired by [6].

Proposition 6

Let $1\leqslant p<\infty$, $0<q<\infty$, $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_q$ and $f\in H^p$ be the product of an inner function I and an outer function $O_\phi$. Then there exists a constant $C=C(p,q,\nu )>0$ such that

$$\begin{aligned} \begin{aligned}&\Vert O_{\max \{\phi ,1\}}'\Vert _{A_\nu ^{p,q}}^q + \Vert (IO_{\min \{\phi ,1\}})'\Vert _{A_\nu ^{p,q}}^q + \Vert O_{\max \{\phi ,1\}}\Vert _{H^p}^q\\&\quad \leqslant C \left( \Vert f'\Vert _{A_\nu ^{p,q}}^q + \Vert f\Vert _{H^p}^q + 1\right) . \end{aligned} \end{aligned}$$

Before the proof of Proposition 6, we note that the quantities $F_1(f)$ and $F_2(f)$ in Theorem 3 are used repeatedly hereafter.

Proof

Let us begin by noting that $|O_{\phi }(e^{i\theta })|=\phi (e^{i\theta })$, $|O_{\max \{\phi ,1\}}(e^{i\theta })|=\max \{\phi (e^{i\theta }),1\}$, and

$$\begin{aligned} \begin{aligned} \max \{\phi (e^{i\theta }),1\}-\phi (e^{i\theta })&=\frac{\max \{\phi (e^{i\theta }),1\}-\phi (e^{i\theta })}{\max \{\phi (e^{i\theta }),1\}} \\&\leqslant |O_{\max \{\phi ,1\}}(z)|\left( 1-\frac{\phi (e^{i\theta })}{\max \{\phi (e^{i\theta }),1\}}\right) \end{aligned} \end{aligned}$$

for all $z\in \mathbb {D}$ and almost every $\theta \in [0,2\pi )$. Using these facts together with Jensen’s inequality [10, Chapter I, Lemma 6.1] and the definition of outer functions, we obtain

$$\begin{aligned} \begin{aligned}&\int _0^{2\pi } |O_{\max \{\phi ,1\}}(e^{i\theta })|\,d\mu _z(\theta ) - \int _0^{2\pi }|O_\phi (e^{i\theta })|\,d\mu _z(\theta ) \\&\quad \leqslant |O_{\max \{\phi ,1\}}(z)|\left( 1-\int _0^{2\pi } \frac{\phi (e^{i\theta })}{\max \{\phi (e^{i\theta }),1\}}\,d\mu _z(\theta )\right) \\&\quad = |O_{\max \{\phi ,1\}}(z)|\left( 1-\int _0^{2\pi } \exp \left( \log \phi (e^{i\theta })-\log \max \{\phi (e^{i\theta }),1\}\right) \,d\mu _z(\theta )\right) \\&\quad \leqslant |O_{\max \{\phi ,1\}}(z)|\left( 1-\frac{|O_{\phi }(z)|}{|O_{\max \{\phi ,1\}}(z)|}\right) , \quad z\in \mathbb {D}. \end{aligned} \end{aligned}$$

Consequently, the obvious inequality $|f(z)|\leqslant |O_\phi (z)|$ yields

$$\begin{aligned} \begin{aligned}&\int _0^{2\pi } |O_{\max \{\phi ,1\}}(e^{i\theta })|\,d\mu _z(\theta ) - |O_{\max \{\phi ,1\}}(z)|\\&\quad \leqslant \int _0^{2\pi } |f(e^{i\theta })|d\mu _z(\theta )-|f(z)|, \quad z\in \mathbb {D}. \end{aligned} \end{aligned}$$

(3.1)

Write $z=re^{it}$. Raising both sides of (3.1) to power p, integrating from 0 to $2\pi$ with respect to dt, then raising both sides to power q/p, and finally integrating from 0 to 1 with respect to $\nu (r) dr/(1-r)^q$, we obtain $F_1(O_{\max \{\phi ,1\}})\leqslant F_1(f)$.

Next we show $F_2(O_{\max \{\phi ,1\}})\leqslant F_2(f)$. Set

$$\begin{aligned} \Gamma _1=\Gamma _1(z,\phi )=\left\{ \theta \in [0,2\pi ):\int _0^{2\pi } \max \{\phi (e^{is}),1\} d\mu _z(s)\leqslant \phi (e^{i\theta })\right\} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Gamma _2=\Gamma _2(z,\phi )=\left\{ \theta \in [0,2\pi ):\int _0^{2\pi } \phi (e^{is}) d\mu _z(s)\leqslant \phi (e^{i\theta })\right\} , \quad z\in \mathbb {D}. \end{aligned} \end{aligned}$$

Then elementary calculations yield

$$\begin{aligned} \begin{aligned}&\int _0^{2\pi } \left| \max \{\phi (e^{i\theta }),1\} -\int _0^{2\pi } \max \{\phi (e^{is}),1\} d\mu _z(s)\right| d\mu _z(\theta ) \\&\quad =2\int _{\Gamma _1}\left( \max \{\phi (e^{i\theta }),1\} -\int _0^{2\pi } \max \{\phi (e^{is}),1\} d\mu _z(s)\right) \,d\mu _z(\theta )\\&\quad \quad + \int _0^{2\pi }\left( \int _0^{2\pi } \max \{\phi (e^{is}),1\} d\mu _z(s)-\max \{\phi (e^{i\theta }),1\} \right) \,d\mu _z(\theta ) \\&\quad = 2\int _{\Gamma _1}\left( \phi (e^{i\theta })-\int _0^{2\pi } \max \{\phi (e^{is}),1\} d\mu _z(s)\right) \,d\mu _z(\theta ) \\&\quad \leqslant 2\int _{\Gamma _2}\left( \phi (e^{i\theta })-\int _0^{2\pi } \phi (e^{is}) d\mu _z(s)\right) \,d\mu _z(\theta ) \\&\quad =\int _0^{2\pi } \left| \phi (e^{i\theta }) -\int _0^{2\pi } \phi (e^{is}) d\mu _z(s)\right| d\mu _z(\theta ), \quad z\in \mathbb {D}. \end{aligned} \end{aligned}$$

(3.2)

Consequently, we obtain $F_2(O_{\max \{\phi ,1\}})\leqslant F_2(O_\phi )=F_2(f)$ by doing a similar integral procedure as above. Now Theorem 3 together with the inequalities for $F_1(f)$ and $F_2(f)$ gives

$$\begin{aligned} \begin{aligned} \Vert O_{\max \{\phi ,1\}}'\Vert _{A_\nu ^{p,q}}^q + \Vert O_{\max \{\phi ,1\}}\Vert _{H^p}^q \lesssim \Vert f'\Vert _{A_\nu ^{p,q}}^q + \Vert f\Vert _{H^p}^q + 1. \end{aligned} \end{aligned}$$

(3.3)

By (3.3) it suffices to show

$$\begin{aligned} \begin{aligned} \Vert (IO_{\min \{\phi ,1\}})'\Vert _{A_\nu ^{p,q}}^q \lesssim \Vert f'\Vert _{A_\nu ^{p,q}}^q + \Vert f\Vert _{H^p}^q. \end{aligned} \end{aligned}$$

(3.4)

Since

$$\begin{aligned} \begin{aligned} \phi (e^{i\theta })-\min \{\phi (e^{i\theta }),1\}\geqslant |O_{\min \{\phi ,1\}}(z)|\left( \frac{\phi (e^{i\theta })}{\min \{\phi (e^{i\theta }),1\}}-1\right) , \end{aligned} \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned}&\int _0^{2\pi } |O_{\phi }(e^{i\theta })|d\mu _z(\theta )-|O_{\phi }(z)| \\&\quad \geqslant \int _0^{2\pi } |O_{\min \{\phi ,1\}}(e^{i\theta })|\,d\mu _z(\theta ) - |O_{\min \{\phi ,1\}}(z)|, \quad z\in \mathbb {D}, \end{aligned} \end{aligned}$$

by arguing as above using Jensen’s inequality. It follows that

$$\begin{aligned} \begin{aligned}&\int _0^{2\pi } |f(e^{i\theta })|d\mu _z(\theta )-|f(z)|\\&\quad =\left( \int _0^{2\pi } |O_{\phi }(e^{i\theta })|d\mu _z(\theta )-|O_{\phi }(z)|\right) +|O_{\phi }(z)|(1-|I(z)|) \\&\quad \geqslant \left( \int _0^{2\pi } |O_{\min \{\phi ,1\}}(e^{i\theta })|d\mu _z(\theta )-|O_{\min \{\phi ,1\}}(z)|\right) +|O_{\min \{\phi ,1\}}(z)|(1-|I(z)|) \\&\quad = \int _0^{2\pi } |IO_{\min \{\phi ,1\}}(e^{i\theta })|d\mu _z(\theta )-|IO_{\min \{\phi ,1\}}(z)|, \quad z\in \mathbb {D}. \end{aligned} \end{aligned}$$

Hence it is easy to deduce $F_1(IO_{\min \{\phi ,1\}})\leqslant F_1(f)$. Since

$$\begin{aligned} F_2(IO_{\min \{\phi ,1\}})=F_2(O_{\min \{\phi ,1\}})\leqslant F_2(O_\phi ) =F_2(f) \end{aligned}$$

can be shown by using a modification of (3.2), the desired estimate (3.4) follows from Theorem 3. This completes the proof. $\square$

Now we can easily prove Theorem 5 by using Proposition 6.

Proof of Theorem 5

By the inner-outer factorization, there exist an inner function I and an outer function $O_{\phi }$ such that $f=IO_{\phi }$. Since $O_{\phi }=O_{\min \{\phi ,1\}} O_{\max \{\phi ,1\}}$, we have $f=f_1/f_2$, where $f_1=IO_{\min \{\phi ,1\}}$ and $f_2=1/O_{\max \{\phi ,1\}}$. Applying Proposition 6 together with the inequalities

$$\begin{aligned} |O_{\min \{\phi ,1\}}(z)|\leqslant 1 \leqslant |O_{\max \{\phi ,1\}}(z)| \end{aligned}$$

and

$$\begin{aligned} |f_2'(z)| \leqslant |O_{\max \{\phi ,1\}}(z)|^2|f_2'(z)| =|O_{\max \{\phi ,1\}}'(z)|, \quad z\in \mathbb {D}, \end{aligned}$$

we can check that $f_1$ and $f_2$ belong to $\mathcal {B}_{\nu }^{p,q}\cap H^\infty$. Moreover, it is obvious that $f_2$ is an outer function. Hence the proof is complete. $\square$

4 Product of $f\in H^p$ and an Inner Function in $\mathcal {B}_{\nu }^{p,q}$

Theorem 7 below gives a sufficient and necessary condition guaranteeing that the product of $f\in H^p$ and an inner function belongs to $\mathcal {B}_{\nu }^{p,q}$. This result generalizes [6, Corollary 3.2], the essential contents of [6, Corollary 3.1] and [9, Theorem 3.2].

Theorem 7

Let $1\leqslant p<\infty$, $0<q<\infty$, $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_q$, $f\in H^p$, and I be an inner function. Then $fI\in \mathcal {B}_{\nu }^{p,q}$ if and only if $f\in \mathcal {B}_{\nu }^{p,q}$ and

$$\begin{aligned} \begin{aligned} \int _0^1 \left( \int _0^{2\pi } \left( \frac{|f(re^{it})|(1-|I(re^{it})|)}{1-r}\right) ^p\,dt \right) ^{q/p}\,\nu (r)\,dr<\infty . \end{aligned} \end{aligned}$$

Proof

We have

$$\begin{aligned} \int _0^{2\pi } |fI(e^{i\theta })|d\mu _z(\theta )-|fI(z)|= & {} \left( \int _0^{2\pi } |f(e^{i\theta })|d\mu _z(\theta )-|f(z)|\right) \nonumber \\&+|f(z)|(1-|I(z)|) \end{aligned}$$

(4.1)

for all $z\in \mathbb {D}$. Write $z=re^{it}$. Raising both sides of (4.1) to power p, integrating from 0 to $2\pi$ with respect to dt, then raising both sides to power q/p, integrating from 0 to 1 with respect to $\nu (r) dr/(1-r)^q$ and finally splitting the right-hand side into two parts by using well-known inequalities, we obtain

$$\begin{aligned} \begin{aligned} F_1(fI)\asymp F_1(f)+\int _0^1 \left( \int _0^{2\pi } \left( \frac{|f(re^{it})|(1-|I(re^{it})|)}{1-r}\right) ^p\,dt\right) ^{q/p}\,\nu (r)\,dr. \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} F_2(fI)+\Vert fI\Vert _{H^p}^q=F_2(f)+\Vert f\Vert _{H^p}^q, \end{aligned}$$

the assertion follows from Theorem 3. $\square$

Recall that a subspace X of $H^p$ satisfies the F-property if the hypothesis $fI\in X$, where $f\in H^p$ and I is an inner function, implies $f\in X$. The F-property for $\mathcal {B}_{\nu }^{p,q}$ is a direct consequence of Theorem 7. However, it is worth mentioning that if one just aims to prove the F-property for $\mathcal {B}_{\nu }^{p,q}$, our argument is maybe not the simplest one, taking into account the length of proofs of Theorem 3 and its auxiliary results. Ideas for an alternative proof can be found, for instance, in [16, Section 5.8.3].

A sequence $\{z_n\}\subset \mathbb {D}$ is said to be a zero set of $\mathcal {B}_{\nu }^{p,q}$ if there exists $f\in \mathcal {B}_{\nu }^{p,q}$ such that $\{z:f(z)=0\}=\{z_n\}$. Here each zero $z_n$ is repeated according to its multiplicity and function f is not identically zero. Applying Theorem 7, we make some observations on zero sets of $\mathcal {B}_{\nu }^{p,p}$. More precisely, we concentrate on the case where $\{z_n\}$ is separated, which means that there exists $\delta =\delta (\{z_n\})>0$ such that $d(z_n,z_k)>\delta$ for all $n\ne k$, where

$$\begin{aligned} d(z,w)=\left| \frac{z-w}{1-\overline{z}w}\right| ,\quad z,w\in \mathbb {D}, \end{aligned}$$

is the pseudo-hyperbolic distance between points z and w. Before these results some basic properties of Hardy spaces are recalled.

For $\{z_n\}\subset \mathbb {D}$ satisfying the Blaschke condition $\sum _n (1-|z_n|)<\infty$ and a point $\theta \in [0,2\pi )$, the Blaschke product with zeros $\{z_n\}$ is defined by

$$\begin{aligned} B(z)=e^{i\theta } \prod _{n}\frac{|z_n|}{z_n}\frac{z_n-z}{1-\overline{z}_nz}, \quad z\in \mathbb {D}. \end{aligned}$$

For $z_n=0$, the interpretation $|z_n|/z_n=-1$ is used. By factorization [8, Theorem 2.5] made by F. Riesz, we know that any $f\in H^p$ for some fixed $p\in (0,\infty ]$ can be represented in the form $f=Bg$, where B is a Blaschke product and $g\in H^p$ does not vanish in $\mathbb {D}$. More precisely, Beurling factorization [8, Theorem 2.8] asserts that g is the product of an outer function and a singular inner function

$$\begin{aligned} S(z)=\exp \left( \int _{\mathbb {T}} \frac{z+\xi }{z-\xi }\, d\sigma (\xi )+i\theta \right) ,\quad z\in \mathbb {D}, \end{aligned}$$

where $\theta \in [0,2\pi )$ is a constant and $\sigma$ is a positive measure on $\mathbb {T}$, singular with respect to the Lebesgue measure. Consequently, every zero set of $\mathcal {B}_{\nu }^{p,q}$ satisfies the Blaschke condition. With these preparations we are ready to state and prove the following result.

Corollary 8

Let $1\leqslant p<\infty$, $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_p$, and assume that $\{z_n\}$ is a finite union of separated sequences and zero set of $\mathcal {B}_{\nu }^{p,p}$. Then there exists an outer function $O_{\phi } \in \mathcal {B}_{\nu }^{p,p}$ such that

$$\begin{aligned} \begin{aligned} \sum _n |O_{\phi }(z_n)|^p \frac{\widehat{\nu }(z_n)}{(1-|z_n|)^{p-1}}<\infty . \end{aligned} \end{aligned}$$

Proof

Let $\{z_n\}=\bigcup _{j=1}^M \{z_n^j\}$, where $M\in \mathbb {N}$ and each $\{z_n^j\}$ is separated. Let B be the Blaschke product with zeros $\{z_n\}$, S a singular inner function and $O_{\phi }$ an outer function such that $BSO_{\phi }\in \mathcal {B}_{\nu }^{p,q}$. By Theorem 7, we know that $O_{\phi }$ and $BO_{\phi }$ belong to $\mathcal {B}_{\nu }^{p,p}$. For $w\in \mathbb {D}$ and $0<r<1$, set

$$\begin{aligned} \Delta (w,r)=\{z:d(z,w)<r\} \quad \text {and} \quad \Lambda (w,r)=\{z:|w-z|<r(1-|w|)\}. \end{aligned}$$

Since each $\{z_n^j\}$ is separated, we find $R_j,\delta _j\in (0,1)$ such that, for a fixed j, discs $\Lambda (z_n^j, R_j)$ are pairwise disjoint and the inclusion $\Delta (z_n^j, \delta _j) \subset \Lambda (z_n^j, R_j)$ is valid for every n. Hence $\widehat{\nu }$ is essentially constant in each disc $\Delta (z_n^j, \delta _j)$ by Lemma A(ii). Moreover,

$$\begin{aligned} |B(z)|\leqslant \left| \frac{z_n^j-z}{1-\overline{z}_n^j z}\right| \leqslant \delta _j, \quad z\in \Delta (z_n^j, \delta _j). \end{aligned}$$

Using these facts together with the subharmonicity of $|O_{\phi }|^p$, we obtain

$$\begin{aligned}&\sum _n |O_{\phi }(z_n)|^p \frac{\widehat{\nu }(z_n)}{(1-|z_n|)^{p-1}}=\sum _{j=1}^{M}\sum _n |O_{\phi }(z_n^j)|^p \frac{\widehat{\nu }(z_n^j)}{(1-|z_n^j|)^{p-1}} \nonumber \\&\quad \lesssim \sum _{j=1}^{M} \sum _n \int _{\Delta (z_n^j, \delta _j)} |O_{\phi }(z)|^p\,dA(z) \frac{\widehat{\nu }(z_n^j)}{(1-|z_n^j|)^{p+1}} \nonumber \\&\quad \asymp \sum _{j=1}^{M} \sum _n \int _{\Delta (z_n^j, \delta _j)} |O_{\phi }(z)|^p \frac{\widehat{\nu }(z)}{(1-|z|)^{p+1}} \,dA(z) \nonumber \\&\quad \leqslant \sum _{j=1}^{M} (1-\delta _j)^{-p}\sum _n \int _{\Delta (z_n^j, \delta _j)}\left( \frac{|O_{\phi }(z)|(1-|B(z)|)}{1-|z|}\right) ^p \frac{\widehat{\nu }(z)}{1-|z|} \,dA(z) \nonumber \\&\quad \lesssim \int _\mathbb {D}\left( \frac{|O_{\phi }(z)|(1-|B(z)|)}{1-|z|}\right) ^p \frac{\widehat{\nu }(z)}{1-|z|} \,dA(z), \end{aligned}$$

(4.2)

where dA(z) is the two-dimensional Lebesgue measure. Now it suffices to show that the last integral in (4.2) is finite.

Set $\psi (z)=\widehat{\nu }(z)/(1-|z|)$ for $z\in \mathbb {D}$. Note that $\widehat{\nu }(r)\asymp \widehat{\psi }(r)$ for $0\leqslant r<1$ by Lemmas A(ii) and B. Moreover, integrating by parts, one can show that $\nu \in \widehat{\mathcal {D}}_p$ if and only if

$$\begin{aligned} \begin{aligned} \frac{(1-r)^p}{\widehat{\nu }(r)}\int _0^r \frac{\widehat{\nu }(s)}{(1-s)^{p+1}}\,ds\asymp 1, \quad r \rightarrow 1^-. \end{aligned} \end{aligned}$$

(4.3)

In particular, $\psi \in \mathcal {D}\cap \widehat{\mathcal {D}}_p$ by the hypotheses of $\nu$. Since Lemma D implies $BO_{\phi }$ in $\mathcal {B}_{\psi }^{p,p}$, Theorem 7 gives

$$\begin{aligned} \int _\mathbb {D}\left( \frac{|O_{\phi }(z)|(1-|B(z)|)}{1-|z|}\right) ^p \frac{\widehat{\nu }(z)}{1-|z|} \,dA(z)<\infty . \end{aligned}$$

This completes the proof. $\square$

Recall that a sequence $\{z_n\}\subset \mathbb {D}$ is said to be uniformly separated if

$$\begin{aligned} \inf _{n\in \mathbb {N}}\prod _{k\ne n}\left| \frac{z_k-z_n}{1-\overline{z}_kz_n}\right| >0; \end{aligned}$$

and a finite union of uniformly separated sequences is called a Carleson–Newman sequence. It is worth mentioning that any Carleson–Newman sequence is a finite union of separated sequences satisfying the Blaschke condition, but the converse statement is not true. For $1<p<\infty$, $p-2<\alpha <p-1$, and a Carleson–Newman sequence $\{z_n\}$, we can give a sufficient and necessary condition for $\{z_n\}$ to be a zero set of $\mathcal {B}_{\alpha }^{p,p}$. This is a straightforward consequence of Theorem 7, Corollary 8, and the reasoning made in paper [4] by N. Arcozzi, D. Blasi and J. Pau.

Corollary 9

Let $1<p<\infty$, $p-2<\alpha <p-1$, and $\{z_n\}$ be a Carleson–Newman sequence. Then $\{z_n\}$ is a zero set of $\mathcal {B}_{\alpha }^{p,p}$ if and only if there exists an outer function $O_{\phi }\in \mathcal {B}_{\alpha }^{p,p}$ such that

$$\begin{aligned} \begin{aligned} \sum _n |O_{\phi }(z_n)|^p (1-|z_n|)^{\alpha +2-p}<\infty . \end{aligned} \end{aligned}$$

(4.4)

Proof

Let B be the Blaschke product with zeros $\{z_n\}$ and $O_{\phi }\in \mathcal {B}_{\alpha }^{p,p}$ an outer function satisfying (4.4). Then [14, Theorem 3.5] together with some elementary calculations gives

$$\begin{aligned} \begin{aligned}&\int _\mathbb {D}\left( \frac{|O_{\phi }(z)|(1-|B(z)|)}{1-|z|}\right) ^p (1-|z|)^{\alpha }\,dA(z) \\&\quad \leqslant 2\int _\mathbb {D}|O_{\phi }(z)|^p \sum _n \frac{1-|z_n|^2}{|1-\overline{z}_nz|^2} (1-|z|)^{\alpha +1-p}\,dA(z) \\&\quad \lesssim \sum _n \int _\mathbb {D}|O_{\phi }(z_n)|^p \frac{1-|z_n|^2}{|1-\overline{z}_nz|^2} (1-|z|)^{\alpha +1-p}\,dA(z) \\&\quad \quad + \sum _n \int _\mathbb {D}|O_{\phi }(z)-O_{\phi }(z_n)|^p\frac{1-|z_n|^2}{|1-\overline{z}_nz|^2} (1-|z|)^{\alpha +1-p}\,dA(z) \\&\quad =:\mathcal {I}_1+\mathcal {I}_2. \end{aligned} \end{aligned}$$

(4.5)

Following the reasoning in the proof of [4, Proposition 3.2], it is easy to check that $\mathcal {I}_1$ and $\mathcal {I}_2$ are finite. More precisely, estimating in a natural manner, one can show

$$\begin{aligned} \mathcal {I}_1 \lesssim \sum _n |O_{\phi }(z_n)|^p (1-|z_n|)^{\alpha +2-p}<\infty . \end{aligned}$$

In the argument of $\mathcal {I}_2\lesssim \Vert O_{\phi }'\Vert _{A_\alpha ^{p,p}}^p<\infty$, [5, Lemma 2.1] and the hypothesis that $\{z_n\}$ is a Carleson–Newman sequence play key roles.

Since $O_{\phi }\in \mathcal {B}_{\alpha }^{p,p}$ and the first integral in (4.5) is finite, $BO_{\phi }$ belongs to $\mathcal {B}_{\alpha }^{p,p}$ by Theorem 7. Consequently, the implication $\Leftarrow$ is valid. The converse implication is a direct consequence of Corollary 8. Hence the proof is complete. $\square$

It is an open problem to prove a $\mathcal {B}_{\nu }^{p,p}$ counterpart of Corollary 9. One could try to prove such a result, for instance, assuming $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_p$ and

$$\begin{aligned} \sup _{0\leqslant r<1}\frac{(1-r)^{p-1}}{\widehat{\nu }(r)}\int _r^1\frac{\nu (s)}{(1-s)^{p-1}}\,ds<\infty . \end{aligned}$$

In this case, the implication $\Leftarrow$ is the problematic part. An idea to approach this problem is to follow the argument of [4, Proposition 3.2] and aim to apply therein [3, Theorem 3.1] instead of [5, Lemma 2.1]. The downside of this method is that it leads to laborious computations of Bekollé–Bonami weights.

Corollaries 8 and 9 are related to some main results in [15] by J. Pau and J. A.  Peláez. In particular, the equivalence $\mathrm (i) \Leftrightarrow (ii)$ in [15, Theorem 1] follows from Corollary 9 by setting $p=2$. Moreover, Corollary 8 shows that the implication $\mathrm (i) \Rightarrow (ii)$ in [15, Theorem 1] is valid also if $\{z_n\}$ in the statement is a finite union of separated sequences. Applying the last observation, we can also replace a Carleson–Newman sequence in [15, Corollary 1] by a finite union of separated sequences: If $0<\alpha <1$, $\{z_n\}$ is a finite union of separated sequences and zero set of $\mathcal {B}_{\alpha }^{2,2}$, then

$$\begin{aligned} \int _0^{2\pi } \log \left( \sum _n \frac{(1-|z_n|)^{\alpha +1}}{|e^{i\theta }-z_n|^2}\right) \,d\theta <\infty . \end{aligned}$$

This result offers a practical way to construct Blaschke sequences that are not zero sets of $\mathcal {B}_{\alpha }^{2,2}$; see [15, Theorem 2] and its proof.

Note that (4.2) and (4.5) together with the estimates for $\mathcal {I}_1$ and $\mathcal {I}_2$ are valid also if outer function $O_\phi$ is replaced by an arbitrary $f\in H^p$. Using this observation and Theorem 7, we can rewrite Corollary 9 in the following form.

Corollary 10

Let $1<p<\infty$, $p-2<\alpha <p-1$, $f\in H^p$, and B be a Blaschke product associated with a Carleson–Newman sequence $\{z_n\}$. Then $fB\in \mathcal {B}_{\alpha }^{p,p}$ if and only if $f\in \mathcal {B}_{\alpha }^{p,p}$ and

$$\begin{aligned} \begin{aligned} \sum _n |f(z_n)|^p (1-|z_n|)^{\alpha +2-p}<\infty . \end{aligned} \end{aligned}$$

Corollary 10 is a partial improvement of the main result in M. Jevtić’s paper [11]. More precisely, this paper contains an extended counterpart of Corollary 10 (in the sense of p and q) with the defect $f \equiv 1$. It is also worth mentioning that Corollary 10 is not valid if the Carleson–Newman sequence $\{z_n\}$ is replaced by an arbitrary Blaschke sequence. This can be shown by studying the case where $f\equiv 1$ and B is a Blaschke product with zeros on the positive real axis. More precisely, the counter example follows from [23, Theorem 1], which asserts that all such Blaschke products belong to $\mathcal {B}_{\alpha }^{p,p}$ for $1/2<p<\infty$ and $p-3/2<\alpha <\infty$.

Theorem 7 for $f\equiv 1$ (or Theorem 3 for inner functions) also has the extended counterpart [22, Theorem 1].

Theorem E

Let $0<p,q<\infty$, and $\nu \in \mathcal {D}$. Then $\nu \in \widehat{\mathcal {D}}_q$ if and only if

$$\begin{aligned} \Vert I'\Vert _{A^{p,q}_\nu }^q \asymp \int _0^1 \left( \int _0^{2\pi } \left( \frac{1-|I(re^{i\theta })|}{1-r}\right) ^p d\theta \right) ^{q/p} \nu (r)\, dr \end{aligned}$$

for all inner functions I. Here the comparison constants may depend only on p, q, and $\nu$.

Theorem E confirms that the hypothesis $\nu \in \widehat{\mathcal {D}}_q$ in Theorems 3 and 7 is sharp in a certain sense. Studying the argument of this result in [22], we can also deduce that the proof of Theorem 3 is more straightforward when f is an inner function, and the statement is valid for all $0<p<\infty$. It is also worth mentioning that results like Theorem E have turned out to be useful in the theory of inner functions. Several by-products of Theorem E can be found in [22, 24].

5 Proof of Theorem 1

Before the proof of Theorem 1 we recall [22, Lemma 6], which is a modification of [1, Lemma 5].

Lemma F

If $0<p\leqslant 1$ and $g:[0,1) \rightarrow [0,\infty )$ is measurable, then

$$\begin{aligned} \left( \int _r^1 g(s)ds\right) ^p\leqslant 2\int _r^1 \sup _{0\leqslant x\leqslant s} g(x)^p(1-s)^{p-1}ds \end{aligned}$$

for $0\leqslant r<1$.

Proof of Theorem 1

Let $\frac{4}{5}\leqslant s<1$ and choose $n=n(s)\in \mathbb {N}\setminus \{1,2,3,4\}$ such that $1-\frac{1}{n}\leqslant s<1-\frac{1}{n+1}$. Set $f_n(z)=z^n$ for $z\in \mathbb {D}$. Since

$$\begin{aligned} \begin{aligned} \left| e^{in(\theta +h)}-e^{in\theta }\right| ^2&= \left| 1-e^{inh}\right| ^2= 2(1-\cos (nh))=2n^2h^2\sum _{k=1}^\infty (-1)^{k-1} \frac{(nh)^{2(k-1)}}{(2k)!} \\&\geqslant 2n^2h^2 \left( \frac{1}{2}-\frac{n^2h^2}{24}\right) \geqslant \frac{2}{3}n^2h^2, \quad 0<h<\frac{2}{n}, \end{aligned} \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&\int _{1/2}^1 \omega _p(1-r, f_n)^q \frac{\nu (r)}{(1-r)^q}\,dr\\&\quad \asymp \left( \int _{1/2}^{1-2/n} + \int _{1-2/n}^1\right) \sup _{0<h<1-r} |1-e^{inh}|^q \frac{\nu (r)}{(1-r)^q}\,dr \\&\quad \geqslant \int _{1/2}^{1-2/n} |1-e^{i2}|^q \frac{\nu (r)}{(1-r)^q}\,dr + \int _{1-2/n}^s |1-e^{in(1-r)}|^q \frac{\nu (r)}{(1-r)^q}\,dr \\&\quad > rsim \int _{1/2}^{1-2/n} \frac{\nu (r)}{(1-r)^q}\,dr + \int _{1-2/n}^s (n+1)^q \nu (r)\,dr \\&\quad \geqslant \int _{1/2}^{1-2/n} \frac{\nu (r)}{(1-r)^q}\,dr+(1-s)^{-q}\int _{1-2/n}^s \nu (r)\,dr \\&\quad \geqslant \int _{1/2}^s \frac{\nu (r)}{(1-r)^q}\,dr \asymp \int _0^s \frac{\nu (r)}{(1-r)^q}\,dr. \end{aligned} \end{aligned}$$

Using the hypothesis $\nu \in \widehat{\mathcal {D}}$ together with Lemma A(iv)(ii) in a similar manner as in the proof of [21, Theorem 1], we obtain

$$\begin{aligned} \begin{aligned} \Vert f_n'\Vert _{A_\nu ^{p,q}}^q&\asymp n^q \int _0^1 r^{q(n-1)+1}\nu (r)\,dr \asymp n^q \int _{1-\frac{1}{q(n-1)+1}}^1 \nu (r)\,dr \\&\asymp n^q \int _{1-\frac{1}{n+1}}^1 \nu (r)\,dr \leqslant \frac{\widehat{\nu }(s)}{(1-s)^q}. \end{aligned} \end{aligned}$$

Finally combining the estimates above and using the inequality

$$\begin{aligned} \int _0^t \frac{\nu (r)}{(1-r)^q}\,dr \lesssim 1 \asymp \frac{\widehat{\nu }(t)}{(1-t)^q}, \quad 0<t<\frac{4}{5}, \end{aligned}$$

we deduce that if $\nu \in \widehat{\mathcal {D}}$ and (1.2) is satisfied for all $f\in H^p$, then $\nu \in \widehat{\mathcal {D}}_p$. Hence it suffices to prove the converse statement.

Let $f\in H^p$, $0\leqslant \theta <2\pi$, $\frac{1}{2}<r<1$, and $0<h<\frac{1}{2}$. Set $\rho =r-h$ and let $\Gamma$ be the contour that goes first rapidly from $re^{i\theta }$ to $\rho e^{i\theta }$, then along the circle $\{z:|z|=\rho \}$ to $\rho e^{i(\theta +h)}$, and finally rapidly to $re^{i(\theta +h)}$. Since

$$\begin{aligned} f(re^{i(\theta +h)})-f(re^{i\theta })=\int _{\Gamma } f'(z)\,dz, \end{aligned}$$

we have

$$\begin{aligned} |f(re^{i(\theta +h)})-f(re^{i\theta })|\leqslant & {} \int _\rho ^r |f'(se^{i\theta })|\,ds + \int _\theta ^{\theta +h} |f'(\rho e^{it})|\,dt\\&+ \int _\rho ^r |f'(se^{i(\theta +h)})|\,ds. \end{aligned}$$

Consequently, the discrete and continuous forms of Minkowski’s inequality, a change of variable, and Hardy’s convexity theorem yield

$$\begin{aligned} \begin{aligned}&\left( \int _0^{2\pi } |f(re^{i(\theta +h)})-f(re^{i\theta })|^p\,d\theta \right) ^{1/p}\leqslant \left( \int _0^{2\pi } \left( \int _\rho ^r |f'(se^{i\theta })|\,ds\right) ^p d\theta \right) ^{1/p} \\&\qquad + \left( \int _0^{2\pi } \left( \int _0^{h} |f'(\rho e^{i(x+\theta )})|\,dx \right) ^p d\theta \right) ^{1/p}\\&\qquad +\left( \int _0^{2\pi } \left( \int _\rho ^r |f'(se^{i(\theta +h)})|\,ds \right) ^p d\theta \right) ^{1/p} \\&\quad \leqslant 2 \int _\rho ^r M_p(s,f')\,ds + h M_p(\rho ,f') \leqslant 3\int _{r-h}^r M_p(s,f')\,ds. \end{aligned} \end{aligned}$$

(5.1)

Note that the deduction above can be found, for instance, in the proof of [8, Theorem 5.4].

By raising both sides of (5.1) to power q, adding $\sup _{0<h<1-t}$, and then integrating from 1/2 to r with respect to $\nu (t)\,dt/(1-t)^{q}$, we obtain

$$\begin{aligned} \begin{aligned}&\int _{1/2}^r \sup _{0<h<1-t} \left( \int _0^{2\pi } |f(re^{i(\theta +h)})-f(re^{i\theta })|^p \,d\theta \right) ^{q/p} \frac{\nu (t)}{(1-t)^q}\,dt \\&\quad \lesssim \int _{1/2}^r \left( \int _{r-(1-t)}^r M_p(s,f')\,ds\right) ^q \frac{\nu (t)}{(1-t)^q}\,dt. \end{aligned} \end{aligned}$$

Letting $r \rightarrow 1^-$ and using the monotone and mean convergence theorems together with the hypothesis $f\in H^p$, we deduce

$$\begin{aligned} \begin{aligned} \int _{1/2}^1 \omega _p(1-t, f)^q \frac{\nu (t)}{(1-t)^q}\,dt \lesssim \int _{1/2}^1 \left( \int _{t}^1 M_p(s,f')\,ds\right) ^q \frac{\nu (t)}{(1-t)^q}\,dt=:\mathcal {I}. \end{aligned} \end{aligned}$$

Hence it suffices to show $\mathcal {I}\lesssim \Vert f'\Vert _{A_\nu ^{p,q}}^q$. Note that the argument of this estimate uses ideas from [22].

If $q\leqslant 1$, then Lemma F with the choice $g(s)=M_p(s,f')$, Hardy’s convexity theorem, Fubini’s theorem, the hypothesis $\nu \in \widehat{\mathcal {D}}_q$, and Lemma D give

$$\begin{aligned} \begin{aligned} \mathcal {I}&\lesssim \int _0^1 \frac{\nu (t)}{(1-t)^q} \int _t^1 \sup _{0\leqslant x\leqslant s}M_p^q(x,f')(1-s)^{q-1}\,ds \,dt \\&=\int _0^1 \frac{\nu (t)}{(1-t)^q} \int _t^1 M_p^q(s,f')(1-s)^{q-1}\,ds \,dt \\&=\int _0^1 M_p^q(s,f')(1-s)^{q-1} \int _0^s \frac{\nu (t)}{(1-t)^q}\,dt\,ds \\&\lesssim \int _0^1 M_p^q(s,f')\frac{\widehat{\nu }(s)}{1-s}\,ds \asymp \Vert f'\Vert _{A_\nu ^{p,q}}^q \end{aligned} \end{aligned}$$

for all $f \in \mathcal {H}(\mathbb {D})$. Hence the assertion for $q\leqslant 1$ is proved. If $q>1$, $0<\varepsilon <q/(\widehat{\mathcal {D}}_q(\nu )+1)$, and $h(s)=(1-s)^{\frac{q-1-\varepsilon }{q}}$, then Hölder’s inequality and Fubini’s theorem yield

$$\begin{aligned} \begin{aligned} \mathcal {I}&\lesssim \int _0^1 \int _t^1 M_p^q(s,f')h(s)^q\,ds \left( \int _t^1 h(r)^{-\frac{q}{q-1}}dr\right) ^{q-1}\frac{\nu (t)}{(1-t)^q}\,dt \\&\asymp \int _0^1 \frac{\nu (t)}{(1-t)^{q-\varepsilon }} \int _t^1 M_p^q(s,f')(1-s)^{q-1-\varepsilon }\,ds\,dt \\&= \int _0^1 M_p^q(s,f')(1-s)^{q-1-\varepsilon } \int _0^s \frac{\nu (t)}{(1-t)^{q-\varepsilon }}\,dt \,ds \end{aligned} \end{aligned}$$

for all $f \in \mathcal {H}(\mathbb {D})$. Since $\nu \in \widehat{\mathcal {D}}_{q-\varepsilon }$ by Lemma C, the assertion for $q>1$ follows from Lemma D. This completes the proof. $\square$

Since

$$\begin{aligned} \begin{aligned} \int _0^{1/2} \omega _p(1-t, f)^q \frac{\nu (t)}{(1-t)^q}\,dt \leqslant 2^{2q} \Vert f\Vert _{H^p}^q \int _0^{1/2} \nu (t)\,dt \end{aligned} \end{aligned}$$

by Minkowski’s inequality, Theorem 1 has the following consequence.

Corollary 11

Let $1\leqslant p<\infty$, $0<q<\infty$, and $\nu \in \mathcal {D}\cap \widehat{\mathcal {D}}_q$. Then there exists a constant $C=C(p,q,\nu )>0$ such that

$$\begin{aligned} \int _{0}^1 \omega _p(1-r, f)^q \frac{\nu (r)}{(1-r)^q}\,dr \leqslant C \left( \Vert f'\Vert _{A_\nu ^{p,q}}^q+\Vert f\Vert _{H^p}^q \right) \end{aligned}$$

for all $f\in H^p$.

Note that Corollary 11 is a part of Theorem 2. We state it here as an independent result because it is needed for the proof of Theorem 2.

6 Proof of Theorem 2

We go directly to the proof of Theorem 2.

Proof of Theorem 2

Let $0\leqslant r<1$ and $0\leqslant t<2\pi$. Since

$$\begin{aligned} \int _0^{2\pi } \frac{e^{i\theta }d\theta }{(e^{i\theta }-re^{it})^2}=0, \end{aligned}$$

Cauchy’s integral formula gives

$$\begin{aligned} \begin{aligned} |f'(re^{it})|&=\frac{1}{2\pi (1-r^2)} \left| \int _0^{2\pi }\left( f(e^{i\theta })-f(re^{it})\right) \frac{e^{i\theta }(1-r^2)}{(e^{i\theta }-re^{it})^2}\,d\theta \right| \\&\leqslant \frac{1}{1-r} \int _0^{2\pi } \left| f(e^{i\theta })-f(re^{it})\right| d\mu _{re^{it}}(\theta ), \quad f\in H^1. \end{aligned} \end{aligned}$$

Raising both sides to power p, integrating from 0 to $2\pi$ with respect to dt, then raising both sides to power q/p, and finally integrating from 0 to 1 with respect to $\nu (r)\,dr$, we obtain

$$\begin{aligned} \Vert f'\Vert _{A_\nu ^{p,q}}^q \leqslant \int _0^1 \left( \int _0^{2\pi } \left( \int _0^{2\pi } |f(e^{i\theta })-f(re^{it})| d\mu _{re^{it}}(\theta ) \right) ^p dt\right) ^{q/p} \frac{\nu (r)}{(1-r)^q}\,dr\nonumber \\ \end{aligned}$$

(6.1)

for all $f\in H^p$.

Let $f\in H^p$, $0<q\leqslant 1$, and set

$$\begin{aligned} \mathcal {I}(r)=\left( \int _0^{2\pi } \left( \int _0^{2\pi } |f(e^{i\theta })-f(re^{it})| d\mu _{re^{it}}(\theta ) \right) ^p dt\right) ^{q/p}. \end{aligned}$$

By the proof of [9, Theorem 2.1], we know that

$$\begin{aligned} \begin{aligned} \mathcal {I}(r)\lesssim \left( \sum _{k=0}^\infty 2^{-k} \omega _p(2^k(1-r), f) \right) ^q. \end{aligned} \end{aligned}$$

(6.2)

Hence the sub-additivity of $g(x)=x^q$ for $x\geqslant 0$ and Fubini’s theorem give

$$\begin{aligned} \begin{aligned} \int _0^1 \frac{\mathcal {I}(r)\nu (r)}{(1-r)^{q}}\,dr&\lesssim \sum _{k=0}^\infty 2^{-qk} \int _0^1 \omega _p(2^{k}(1-r), f)^q \frac{\nu (r)}{(1-r)^{q}}\,dr. \end{aligned} \end{aligned}$$

(6.3)

Next we show that the weight $\nu (r)$ in the right-hand side can be replaced by $\frac{\widehat{\nu }(r)}{1-r}$ without losing any essential information.

Set $\psi (z)=\widehat{\nu }(z)/(1-|z|)$ for $z\in \mathbb {D}$, and recall that $\widehat{\nu }(r)\asymp \widehat{\psi }(r)$ for $0\leqslant r<1$ by Lemmas A(ii) and B. In particular, $\psi$ belongs to class $\mathcal {D}$, and thus, there exist $K=K(\psi )>1$ and $C=C(\psi )>1$ such that

$$\begin{aligned} \begin{aligned} \widehat{\psi }(r)\geqslant C \widehat{\psi }\left( 1-\frac{1-r}{K}\right) , \quad 0\leqslant r<1. \end{aligned} \end{aligned}$$

(6.4)

Let $k\in \mathbb {N}\cup \{0\}$ and $r_n=1-K^{-n}$ for $n\in \mathbb {N}\cup \{0\}$. Using (6.4) together with Lemma A(ii), we obtain

$$\begin{aligned} \begin{aligned} (C-1)\widehat{\psi }(r_{n+1})&=C\widehat{\psi }\left( 1-\frac{1-r_n}{K}\right) -\widehat{\psi }(r_{n+1})\leqslant \widehat{\psi }(r_{n})-\widehat{\psi }(r_{n+1})\\&=\int _{r_n}^{r_{n+1}}\psi (r)\,dr\leqslant \widehat{\psi }(r_{n}) \asymp \widehat{\psi }(r_{n+1}). \end{aligned} \end{aligned}$$

(6.5)

Now Minkowski’s inequality, the monotonicity of $\omega _p(s, f)$ with s and (6.5) yield

$$\begin{aligned} \begin{aligned}&\int _{0}^1 \omega _p(2^k(1-r), f)^q \frac{\nu (r)}{(1-r)^{q}}\,dr\\&\quad \lesssim \sum _{n=1}^\infty \int _{r_n}^{r_{n+1}} \omega _p(2^k(1-r), f)^q \frac{\nu (r)}{(1-r)^{q}}\,dr + \Vert f\Vert _{H^p}^q \\&\quad \leqslant \sum _{n=1}^\infty \omega _p(2^k(1-r_n), f)^q \frac{\widehat{\nu }(r_n)}{(1-r_{n+1})^{q}} + \Vert f\Vert _{H^p}^q \\&\quad \asymp \sum _{n=1}^\infty \omega _p(2^k(1-r_n), f)^q \frac{\widehat{\psi }(r_n)}{(1-r_n)^{q}} + \Vert f\Vert _{H^p}^q \\&\quad \asymp \sum _{n=0}^\infty \omega _p(2^k(1-r_{n+1}), f)^q \frac{\int _{r_n}^{r_{n+1}}\psi (r)\,dr}{(1-r_{n})^{q}} + \Vert f\Vert _{H^p}^q \\&\quad \leqslant \sum _{n=0}^\infty \int _{r_n}^{r_{n+1}} \omega _p(2^k(1-r), f)^q \frac{\psi (r)}{(1-r)^{q}}\,dr + \Vert f\Vert _{H^p}^q \\&\quad = \int _0^1 \omega _p(2^k(1-r), f)^q \frac{\widehat{\nu }(r)}{(1-r)^{q+1}}\,dr + \Vert f\Vert _{H^p}^q. \end{aligned} \end{aligned}$$

(6.6)

It is worth noting that a similar deduction works also in the opposite direction.

Using (6.3) and (6.6), we obtain

$$\begin{aligned} \begin{aligned} \int _0^1 \frac{\mathcal {I}(r)\nu (r)}{(1-r)^{q}}\,dr&\lesssim \left[ \sum _{k=0}^\infty 2^{-qk} \int _0^{1-2^{-k}} \omega _p(2^{k}(1-r), f)^q \frac{\widehat{\nu }(r)}{(1-r)^{q+1}}\,dr + \Vert f\Vert _{H^p}^q \right] \\&\quad + \sum _{k=0}^\infty 2^{-qk} \int _{1-2^{-k}}^1 \omega _p(2^{k}(1-r), f)^q \frac{\widehat{\nu }(r)}{(1-r)^{q+1}}\,dr \\&=:\mathcal {I}_1+\mathcal {I}_2. \end{aligned} \end{aligned}$$

Minkowski’s inequality, (4.3) with p being replaced by q, and Lemma B yield

$$\begin{aligned} \begin{aligned} \mathcal {I}_1&\lesssim \Vert f\Vert _{H^p}^q\sum _{k=0}^\infty 2^{-qk}\int _0^{1-2^{-k}} \frac{\widehat{\nu }(r)}{(1-r)^{q+1}}\,dr \\&\lesssim \Vert f\Vert _{H^p}^q\sum _{n=0}^\infty \widehat{\nu }(1-2^{-k}) \\&\lesssim \widehat{\nu }(0)\Vert f\Vert _{H^p}^q\sum _{n=0}^\infty 2^{-\alpha k}\asymp \Vert f\Vert _{H^p}^q \end{aligned} \end{aligned}$$

for some $\alpha =\alpha (\nu )>0$. The continuity of $\widehat{\nu }$, changes of variables, Fubini’s theorem, and the hypothesis $\nu \in \mathcal {D}$ give

$$\begin{aligned} \begin{aligned} \mathcal {I}_2&\asymp \int _0^1 \frac{\omega _p(1-s, f)^q}{(1-s)^{q+1}}\int _0^\infty \widehat{\nu }\left( 1-2^{-k}(1-s)\right) \,dk \,ds \\&=\frac{1}{\log 2}\int _0^1 \frac{\omega _p(1-s, f)^q}{(1-s)^{q+1}}\int _s^1 \frac{\widehat{\nu }(x)}{1-x}\,dx\,ds \\&\asymp \int _0^1 \omega _p(1-s, f)^q \frac{\widehat{\nu }(s)}{(1-s)^{q+1}}\,ds. \end{aligned} \end{aligned}$$

Summarizing, we have shown

$$\begin{aligned} \begin{aligned} \int _0^1 \frac{\mathcal {I}(r)\nu (r)}{(1-r)^{q}}\,dr \lesssim \int _0^1 \omega _p(1-s, f)^q \frac{\widehat{\nu }(s)}{(1-s)^{q+1}}\,ds + \Vert f\Vert _{H^p}^q. \end{aligned} \end{aligned}$$

(6.7)

Applying a similar argument as in (6.6), we can replace $\widehat{\nu }(s)$ in the right-hand side of (6.7) by $\nu (s)(1-s)$. Consequently, (6.1) and Corollary 11 imply (1.3) for all $f\in H^p$. Hence the assertion for $q\leqslant 1$ is proved.

Let $1<q<\infty$. Then (6.2), the continuous form of Minkowski’s inequality, (6.6), and well-known inequalities give

$$\begin{aligned}&\int _0^1 \frac{\mathcal {I}(r)\nu (r)}{(1-r)^{q}}\,dr \lesssim \int _0^1 \left( \sum _{k=0}^\infty 2^{-k} \omega _p(2^{k}(1-r), f) \right) ^q\frac{\nu (r)}{(1-r)^{q}}\,dr \\&\quad \leqslant \left( \sum _{k=0}^\infty 2^{-k}\left( \int _{0}^1 \omega _p(2^{k}(1-r), f)^q\frac{\nu (r)}{(1-r)^{q}} \,dr\right) ^{1/q}\right) ^{q} \\&\quad \lesssim \left( \sum _{k=0}^\infty 2^{-k}\left( \int _{0}^1 \omega _p(2^{k}(1-r), f)^q\frac{\widehat{\nu }(r)}{(1-r)^{q+1}} \,dr\right) ^{1/q}\right) ^{q} + \Vert f\Vert _{H^p}^q \\&\quad \asymp \left[ \left( \sum _{k=0}^\infty 2^{-k}\left( \int _0^{1-2^{-k}} \omega _p(2^{k}(1-r), f)^q\frac{\widehat{\nu }(r)}{(1-r)^{q+1}} \,dr\right) ^{1/q}\right) ^{q} + \Vert f\Vert _{H^p}^q \right] \\&\qquad +\left( \sum _{k=0}^\infty 2^{-k}\left( \int _{1-2^{-k}}^1 \omega _p(2^{k}(1-r), f)^q\frac{\widehat{\nu }(r)}{(1-r)^{q+1}} \,dr\right) ^{1/q}\right) ^{q} \\&\quad =:\mathcal {I}_3 + \mathcal {I}_4. \end{aligned}$$

Minkowski’s inequality, (4.3) with p being replaced by q, and Lemma B yield

$$\begin{aligned} \begin{aligned} \mathcal {I}_3&\lesssim \Vert f\Vert _{H^p}^q\left( \sum _{k=0}^\infty 2^{-k}\left( \int _0^{1-2^{-k}} \frac{\widehat{\nu }(r)}{(1-r)^{q+1}} \,dr\right) ^{1/q}\right) ^{q} \\&\lesssim \Vert f\Vert _{H^p}^q\left( \sum _{k=0}^\infty \widehat{\nu }(1-2^{-k})^{1/q}\right) ^{q} \asymp \Vert f\Vert _{H^p}^q. \end{aligned} \end{aligned}$$

By Lemma B, there exists a constant $\alpha =\alpha (\nu )>0$ such that

$$\begin{aligned} \widehat{\nu }\left( 1-2^{-k}(1-s)\right) \lesssim 2^{-\alpha k}\widehat{\nu }(s), \quad 0\leqslant s\leqslant 1-2^{-k}(1-s)<1. \end{aligned}$$

Using this together with a change of variable and modification of (6.6), we get

$$\begin{aligned} \begin{aligned} \mathcal {I}_4&= \left( \sum _{k=0}^\infty \left( \int _0^1 \omega _p(2^{k}(1-s), f)^q\frac{\widehat{\nu }\left( 1-2^{-k}(1-s)\right) }{(1-s)^{q+1}} \,ds\right) ^{1/q}\right) ^{q} \\&\lesssim \left( \sum _{k=0}^\infty 2^{-\alpha k/q}\left( \int _0^1 \omega _p(1-s, f)^q\frac{\widehat{\nu }(s)}{(1-s)^{q+1}} \,ds\right) ^{1/q}\right) ^{q} \\&\asymp \int _0^1\omega _p(1-s, f)^q \frac{\widehat{\nu }(s)}{(1-s)^{q+1}}\,ds \\&\lesssim \int _0^1\omega _p(1-s, f)^q \frac{\nu (s)}{(1-s)^{q}}\,ds + \Vert f\Vert _{H^p}^q. \end{aligned} \end{aligned}$$

(6.8)

Finally (6.1), (6.8), and Corollary 11 imply (1.3) for all $f\in H^p$. This completes the proof. $\square$

7 Proof of Theorem 3

Before the proof of Theorem 3 we recall the following result, which is a part of the argument of [6, Theorem 1.1].

Lemma G

If $O_{\phi }$ is an outer function, then

$$\begin{aligned} \begin{aligned}&|O_{\phi }'(z)|\leqslant \frac{4}{1-|z|} \left( \int _0^{2\pi } \left| \phi (e^{i\theta })-\int _0^{2\pi } \phi (e^{is}) d\mu _z(s)\right| d\mu _z(\theta )\right. \\&\quad \left. +\int _0^{2\pi } \phi (e^{ih}) d\mu _z(h) - |O_{\phi }(z)| \right) \end{aligned} \end{aligned}$$

for all $z\in \mathbb {D}$.

Proof of Theorem 3

Let $f \in H^p$. Then there exist an inner function I and an outer function $O_{\phi }$ such that $f=IO_{\phi }$. Hence the Schwarz–Pick lemma, Lemma G, and the fact that $\phi (\xi )=|f(\xi )|$ for almost every $\xi \in \mathbb {T}$ yield

$$\begin{aligned} \begin{aligned}&|f'(z)|(1-|z|)\leqslant \left( |I(z)O_{\phi }'(z)| + |I'(z)O_{\phi }(z)|\right) (1-|z|) \\&\quad \leqslant |O_{\phi }'(z)|(1-|z|) + 2|O_{\phi }(z)|(1-|I(z)|)\\&\quad \leqslant 4\int _0^{2\pi } \left| |f(e^{i\theta })|-\int _0^{2\pi } |f(e^{is})| d\mu _z(s)\right| d\mu _z(\theta )\\&\qquad +4\left( \int _0^{2\pi } |f(e^{ih})| d\mu _z(h) - |f(z)|\right) \end{aligned} \end{aligned}$$

(7.1)

for all $z\in \mathbb {D}$. Write $z=re^{it}$. Raising both sides of (7.1) to power p, integrating from 0 to $2\pi$ with respect to dt, then raising both sides to power q/p, integrating from 0 to 1 with respect to $\nu (r) dr/(1-r)^q$, and finally splitting the right-hand side into two parts, we obtain

$$\begin{aligned} \Vert f'\Vert _{A_\nu ^{p,q}}^q \lesssim F_1(f)+F_2(f), \end{aligned}$$

which is the first inequality in (1.4).

Set

$$\begin{aligned} \Gamma =\Gamma (z,f)=\left\{ \theta \in [0,2\pi ):\int _0^{2\pi } |f(e^{is})| d\mu _z(s)\leqslant |f(e^{i\theta })|\right\} , \quad z\in \mathbb {D}. \end{aligned}$$

Then elementary calculations together with the subharmonicity of |f| yield

$$\begin{aligned} \begin{aligned}&\int _0^{2\pi } \left| |f(e^{i\theta })|-\int _0^{2\pi } |f(e^{is})| d\mu _z(s)\right| d\mu _z(\theta ) \\&\quad = 2\int _{\Gamma }\left( |f(e^{i\theta })|-\int _0^{2\pi } |f(e^{is})| d\mu _z(s)\right) \,d\mu _z(\theta ) \\&\quad \leqslant 2 \int _{\Gamma } \left( |f(e^{i\theta })|-|f(z)|\right) d\mu _{z}(\theta ), \quad z\in \mathbb {D}. \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \begin{aligned}&\int _0^{2\pi } \left| |f(e^{i\theta })|-\int _0^{2\pi } |f(e^{is})| d\mu _z(s)\right| d\mu _z(\theta ) +\left( \int _0^{2\pi } |f(e^{ih})| d\mu _z(h) - |f(z)|\right) \\&\quad \leqslant 2 \left( \int _{\Gamma }+\int _0^{2\pi }\right) \left| f(e^{i\theta })-f(z)\right| d\mu _{z}(\theta ) \\&\quad \leqslant 4 \int _0^{2\pi } \left| f(e^{i\theta })-f(z)\right| d\mu _{z}(\theta ), \quad z\in \mathbb {D}. \end{aligned} \end{aligned}$$

Doing a corresponding integration procedure for this estimate as above and applying Theorem 2, we obtain

$$\begin{aligned} F_1(f)+F_2(f)\lesssim \Vert f'\Vert _{A_\nu ^{p,q}}^q + \Vert f\Vert _{H^p}^q, \end{aligned}$$

which is the last inequality in (1.4). This completes the proof. $\square$