Hilbert-type operator induced by radial weight on Hardy spaces

Abstract

We consider the Hilbert-type operator defined by

$$\begin{aligned} H_{\omega }(f)(z)=\int _0^1 f(t)\left( \frac{1}{z}\int _0^z B^{\omega }_t(u)\,du\right) \,\omega (t)dt, \end{aligned}$$

where $\{B^{\omega }_\zeta \}_{\zeta \in \mathbb {D}}$ are the reproducing kernels of the Bergman space $A^2_\omega$ induced by a radial weight $\omega$ in the unit disc $\mathbb {D}$. We prove that $H_{\omega }$ is bounded on the Hardy space $H^p$, $1<p<\infty$, if and only if

$$\begin{aligned} \sup _{0\le r<1} \frac{\widehat{\omega }(r)}{\widehat{\omega }\left( \frac{1+r}{2}\right) }<\infty , (\dag )\end{aligned}$$

and

$$\begin{aligned} \sup \limits _{0<r<1}\left( \int _0^r \frac{1}{\widehat{\omega }(t)^p} dt\right) ^{\frac{1}{p}} \left( \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt\right) ^{\frac{1}{p'}} <\infty , \end{aligned}$$

where $\widehat{\omega }(r)=\int _r^1 \omega (s)\,ds$. We also prove that $H_\omega : H^1\rightarrow H^1$ is bounded if and only if ($\dag$) holds and

$$\begin{aligned} \sup \limits _{r \in [0,1)} \frac{\widehat{\omega }(r)}{1-r} \left( \int _0^r \frac{ds}{\widehat{\omega }(s)}\right) <\infty . \end{aligned}$$

As for the case $p=\infty$, $H_\omega$ is bounded from $H^\infty$ to $\mathord \textrm{BMOA}$, or to the Bloch space, if and only if ($\dag$) holds. In addition, we prove that there does not exist radial weights $\omega$ such that $H_{\omega }: H^p \rightarrow H^p$, $1\le p<\infty$, is compact and we consider the action of $H_{\omega }$ on some spaces of analytic functions closely related to Hardy spaces.

1 Introduction

For $0<p<\infty$, let $L^p_{ [0,1)}$ be the Lebesgue space of measurable functions such that

$$\begin{aligned} \Vert f\Vert ^p_{L^p_{[0,1)}}=\int _0^1 |f(t)|^p\,dt<\infty , \end{aligned}$$

and let $\mathcal {H}(\mathbb {D})$ denote the space of analytic functions in the unit disc $\mathbb {D}=\{z\in \mathbb {C}:|z|<1\}$. The Hardy space $H^p$ consists of $f\in \mathcal {H}(\mathbb {D})$ for which

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

where

$$\begin{aligned} M_p(r,f)=\left( \frac{1}{2\pi }\int _0^{2\pi } |f(re^{i\theta })|^p\,d\theta \right) ^{\frac{1}{p}},\quad 0<p<\infty , \end{aligned}$$

and

$$\begin{aligned} M_\infty (r,f)=\max _{0\le \theta \le 2\pi }|f(re^{i\theta })|. \end{aligned}$$

For a nonnegative function $\omega \in L^1_{[0,1)}$, the extension to $\mathbb {D}$, defined by $\omega (z)=\omega (|z|)$ for all $z\in \mathbb {D}$, is called a radial weight. Let $A^2_{\omega }$ denote the weighted Bergman space of $f\in \mathcal {H}(\mathbb {D})$ such that $\Vert f\Vert _{A^2_\omega }^2=\int _\mathbb {D}|f(z)|^2\omega (z)\,dA(z)<\infty$, where $dA(z)=\frac{dx\,dy}{\pi }$ is the normalized area measure on $\mathbb {D}$. Throughout this paper we assume $\widehat{\omega }(z)=\int _{|z|}^1\omega (s)\,ds>0$ for all $z\in \mathbb {D}$, for otherwise $A^2_\omega =\mathcal {H}(\mathbb {D})$.

The Hilbert matrix is the infinite matrix whose entries are $h_{n,k}=(n+k+1)^{-1},$  $k,n\in \mathbb {N}\cup \{0\}$. It can be viewed as an operator on spaces of analytic functions, by its action on the Taylor coefficients

$$\begin{aligned} \widehat{f}(n)\mapsto \sum _{k=0}^{\infty } \frac{\widehat{f}(k)}{n+k+1}, \quad n\in \mathbb {N}\cup \{0\}, \end{aligned}$$

called the Hilbert operator. That is, if $f(z)=\sum _{k=0}^\infty \widehat{f}(k)z^k\in \mathcal {H}(\mathbb {D})$

$$\begin{aligned} H(f)(z)= \sum _{n=0}^{\infty }\left( \sum _{k=0}^{\infty } \frac{\widehat{f}(k)}{n+k+1}\right) z^n, \end{aligned}$$

(1.1)

whenever the right hand side makes sense and defines an analytic function in $\mathbb {D}$.

The Hilbert operator H is bounded on Hardy spaces $H^p$ if and only if $1<p<\infty$ [4]. A proof of this result can be obtained using the following integral representation, valid for any $f\in H^1$,

$$\begin{aligned} H(f)(z)=\int _0^1f(t)\frac{1}{1-tz}\,dt. \end{aligned}$$

(1.2)

Going further, the formula (1.2) has been employed to solve a good number of questions in operator theory related to the boundedness, the operator norm and the spectrum of the Hilbert operator on classical spaces of analytic functions [1, 3, 5, 24]. During the last decades several generalizations of the Hilbert operator have attracted a considerable amount of attention [9, 11, 24, 26]. We will focus on the following, introduced in [26]. For a radial weight $\omega$, we consider the Hilbert-type operator

$$\begin{aligned} H_{\omega }(f)(z)=\int _0^1 f(t)\left( \frac{1}{z}\int _0^z B^{\omega }_t(\zeta )d\zeta \right) \,\omega (t)dt, \end{aligned}$$

where $\{B^\omega _z\}_{z\in \mathbb {D}}\subset A^2_\omega$ are the Bergman reproducing kernels of $A^2_\omega$. The choice $\omega =1$ gives the integral representation (1.2) of the classical Hilbert operator, therefore it is natural to think of the features of a radial weight $\omega$ so that $H_\omega$ has some of the nice properties of the (classical) Hilbert operator. In this paper, among other results, we describe the radial weights $\omega$ such that the Hilbert-type operator $H_\omega$ is bounded on $H^p$, $1\le p<\infty$.

In order to state our results some more notation is needed. For $0<p<\infty$, the Dirichlet-type space $D^p_{p-1}$ is the space of $f\in \mathcal {H}(\mathbb {D})$ such that

$$\begin{aligned} \Vert f\Vert ^p_{D^p_{p-1}}=|f(0)|^p+\int _\mathbb {D}|f'(z)|^p(1-|z|)^{p-1}\,dA(z)<\infty , \end{aligned}$$

and the Hardy–Littlewood space HL(p) consists of the $f(z)=\sum \nolimits _{n=0}^{\infty } \widehat{f}(n) z^n\in \mathcal {H}(\mathbb {D})$ such that

$$\begin{aligned} \Vert f\Vert ^p_{HL(p)}=\sum \limits _{n=0}^{\infty } |\widehat{f}(n)|^p (n+1)^{p-2}<\infty . \end{aligned}$$

We will also consider the space $H(\infty ,p)=\{f\in \mathcal {H}(\mathbb {D}): \Vert f\Vert ^p_{H(\infty ,p)}=\int _0^1\,M^p_\infty (r,f)\,dr<\infty \}.$ These spaces satisfy the well-known inclusions

$$\begin{aligned}{} & {} D^p_{p-1}\subset H^p\subset HL(p),\quad 0<p\le 2, \end{aligned}$$

(1.3)

$$\begin{aligned}{} & {} HL(p)\subset H^p\subset D^p_{p-1},\quad 2\le p<\infty , \end{aligned}$$

(1.4)

and

$$\begin{aligned} H^p\subset H(\infty ,p), \quad D^p_{p-1}\subset H(\infty ,p),\quad 0<p<\infty . \end{aligned}$$

(1.5)

See [6, 7, 14] for proofs of (1.3) and (1.4), and [27, p. 127] and [8, Lemma 4] for a proof of (1.5).

The Bergman reproducing kernels, induced by a radial weight $\omega$, can be written as $B^\omega _z(\zeta )=\sum \overline{e_n(z)}e_n(\zeta )$ for each orthonormal basis $\{e_n\}$ of $A^2_\omega$, and therefore using the basis induced by the normalized monomials,

$$\begin{aligned} B^\omega _z(\zeta )=\sum _{n=0}^\infty \frac{\left( \overline{z}\zeta \right) ^n}{2\omega _{2n+1}}, \quad z,\zeta \in \mathbb {D}. \end{aligned}$$

(1.6)

Here $\omega _{2n+1}$ are the odd moments of $\omega$, and in general from now on we write $\omega _x=\int _0^1r^x\omega (r)\,dr$ for all $x\ge 0$. A radial weight $\omega$ belongs to the class $\widehat{\mathcal {D}}$ if $\widehat{\omega }(r)\le C\widehat{\omega }(\frac{1+r}{2})$ for some constant $C=C(\omega )>1$ and all $0\le r <1$. If there exist $K=K(\omega )>1$ and $C=C(\omega )>1$ such that $\widehat{\omega }(r)\ge C\widehat{\omega }\left( 1-\frac{1-r}{K}\right)$ for all $0\le r<1$, then $\omega \in \check{\mathcal {D}}$. Further, we write $\mathcal {D}=\widehat{\mathcal {D}}\cap \check{\mathcal {D}}$ for short. Recall that $\omega \in \mathcal {M}$ if there exist constants $C=C(\omega )>1$ and $K=K(\omega )>1$ such that $\omega _{x}\ge C\omega _{Kx}$ for all $x\ge 1$. It is known that $\check{\mathcal {D}}\subset \mathcal {M}$ [23, Proof of Theorem 3] but $\check{\mathcal {D}}\subsetneq \mathcal {M}$ [23, Proposition 14]. However, [23, Theorem 3] ensures that $\mathcal {D}=\widehat{\mathcal {D}}\cap \check{\mathcal {D}}=\widehat{\mathcal {D}}\cap \mathcal {M}$. These classes of weights arise in meaningful questions concerning radial weights and classical operators, such as the differentiation operator $f^{(n)}$ or the Bergman projection $P_\omega (f)(z)=\int _{\mathbb {D}}f(\zeta ) \overline{B^\omega _{z}(\zeta )}\,\omega (\zeta )dA(\zeta )$ [23]. We will also deal with the sublinear Hilbert-type operator

$$\begin{aligned} \widetilde{H_{\omega }}(f)(z)=\int _0^1 |f(t)|\left( \frac{1}{z}\int _0^z B^{\omega }_t(\zeta )d\zeta \right) \,\omega (t)\,dt. \end{aligned}$$

If $X,Y \subset \mathcal {H}(\mathbb {D})$ are normed vector spaces, and T is a sublinear operator, we denote $\Vert T\Vert _{X\rightarrow Y}=\sup _{\Vert f\Vert _X\le 1}\Vert T(f)\Vert _{Y}$.

Theorem 1

Let $\omega$ be a radial weight and $1<p<\infty$. Let $X_p,Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$ and $T\in \{ H_{\omega }, \widetilde{H_{\omega }}\}$. Then the following statements are equivalent:

• (i) $T:\,X_p \rightarrow Y_p$ is bounded;

• (ii) $\omega \in \mathcal {D}$ and $M_p(\omega )=\sup \nolimits _{N\in \mathbb {N}} \left( \sum \nolimits _{n=0}^N \frac{1}{(n+1)^2 \omega _{2n+1}^p}\right) ^{\frac{1}{p}}\left( \sum \nolimits _{n=N}^{\infty } \omega _{2n+1}^{p'}(n+1)^{p'-2}\right) ^{\frac{1}{p'}}<\infty ;$

• (iii) $\omega \in \widehat{\mathcal {D}}$ and $M_p(\omega )=\sup \nolimits _{N\in \mathbb {N}} \left( \sum \nolimits _{n=0}^N \frac{1}{(n+1)^2 \omega _{2n+1}^p}\right) ^{\frac{1}{p}}\left( \sum \nolimits _{n=N}^{\infty } \omega _{2n+1}^{p'}(n+1)^{p'-2}\right) ^{\frac{1}{p'}}<\infty ;$

• (iv) $\omega \in \widehat{\mathcal {D}}$ and $M_{p,c}(\omega )= \sup \nolimits _{0<r<1} \left( \int _0^r \frac{1}{\widehat{\omega }(t)^p} dt\right) ^{\frac{1}{p}} \left( \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt\right) ^{\frac{1}{p'}}<\infty$.

The proof of (i)$\Rightarrow$(iii) of Theorem 1 has two steps. Firstly, we prove that $\omega \in \widehat{\mathcal {D}}$, and later on the condition $M_p(\omega )<\infty$ is obtained by using polynomials of the form $f_{N, M}(z)=\sum \nolimits _{k=N}^M \omega _{2k}^{\alpha } (k+1)^{\beta }z^k,\, N,M\in \mathbb {N},\; \alpha , \beta \in \mathbb {R}$ as test functions. Then, we see that any radial weight $\omega$ satisfying the condition $M_p(\omega )<\infty$, belongs to $\mathcal {M}$. This proves (ii)$\Leftrightarrow$(iii). The proof of (iii)$\Leftrightarrow$(iv) is a calculation based on known descriptions of the class $\widehat{\mathcal {D}}$ [21, Lemma 2.1]. Finally, we prove (iv)$\Rightarrow$(i) which is the most involved implication in the proof of Theorem 1. In order to obtain it, we merge techniques coming from complex and harmonic analysis, such as a very convenient description of the class $\mathcal {D}$, see Lemma 14 below, precise estimates of the integral means of order p of the derivative of the kernels $K^\omega _u(z)=\frac{1}{z}\int _0^z B^\omega _u(z)\,du$, decomposition norm theorems and classical weighted inequalities for Hardy operators.

Observe that both, the discrete condition $M_p(\omega )<\infty$ and its continuous version $M_{p,c}(\omega )<\infty$, are used in the proof of Theorem 1. The first one follows from (i), and the condition $M_{p,c}(\omega )<\infty$ is employed to prove that $T:\,X_p \rightarrow Y_p$ is bounded.

As for the case $p=1$ we obtain the following result.

Theorem 2

Let $\omega$ be a radial weight, $X_1,Y_1\in \{H(\infty ,1),H^1, D^1_{0}, HL(1)\}$ and $T\in \{ H_{\omega }, \widetilde{H_{\omega }}\}$. Then the following statements are equivalent:

• (i) $T: X_1\rightarrow Y_1$ is bounded;

• (ii) $\omega \in \widehat{\mathcal {D}}$ and the measure $\mu _\omega$ defined as $d\mu _\omega (z)= \omega (z)\left( \int _0^{|z|} \frac{ds}{\widehat{\omega }(s)}\right) \,\chi _{[0,1)}(z)\, dA(z)$ is a 1-Carleson measure for $X_1$;

• (iii) $\omega \in \widehat{\mathcal {D}}$ and satisfies the condition

  $$\begin{aligned} M_{1,c}(\omega )= \sup \limits _{a \in [0,1)} \frac{1}{1-a}\int _a^1 \omega (t)\left( \int _0^t \frac{ds}{\widehat{\omega }(s)}\right) \,dt<\infty \textit{;} \end{aligned}$$

• (iv) $\omega \in \mathcal {D}$ and satisfies the condition $M_{1,c}(\omega )<\infty$;

• (v) $\omega \in \widehat{\mathcal {D}}$ and satisfies the condition $M_{1,d}(\omega )= \sup \nolimits _{a \in [0,1)} \frac{\widehat{\omega }(a)}{1-a} \left( \int _0^a \frac{ds}{\widehat{\omega }(s)}\right) <\infty$;

• (vi) $\omega \in \widehat{\mathcal {D}}$ and satisfies the condition

  $$\begin{aligned} M_1(\omega )= \sup \limits _{N \in \mathbb {N}} (N+1)\omega _{2N}\sum \limits _{k=0}^{N}\frac{1}{(k+1)^2 \omega _{2k}}<\infty . \end{aligned}$$

We recall that given a Banach space (or a complete metric space) X of analytic functions on $\mathbb {D}$, a positive Borel measure $\mu$ on $\mathbb {D}$ is called a q-Carleson measure for X if the identity operator $I_d:\, X\rightarrow L^q(\mu )$ is bounded. Carleson provided a geometric description of p-Carleson measures for Hardy spaces $H^p$, $0<p<\infty$, [6, Chapter 9]. These measures are called classical Carleson measures. The proof of Theorem 2 uses characterizations of Carleson measures for $X_1$-spaces, universal Cesàro basis of polynomials and some of the main ingredients of the proofs of Theorem 1 and [26, Theorem 2].

Concerning the classes of radial weights $\widehat{\mathcal {D}}$ and $M_{p,c}=\{ \omega : M_{p,c}(\omega )<\infty \}$, $1\le p<\infty$, a standard weight, $\omega (z)=(1-|z|)^{\beta }$, $\beta >-1$, satisfies the condition $M_{p,c}(\omega )<\infty$ if and only if $\beta >\frac{1}{p}-1$, so $H_\omega : H^p\rightarrow H^p$ is bounded if and only if $\beta >\frac{1}{p}-1$. Moreover, a calculation shows that the exponential type weight $\omega (r)=\exp \left( -\frac{1}{1-r}\right) \in M_{p,c}$ for any $p\in [1,\infty )$, but $\omega \notin \widehat{\mathcal {D}}$, see [28, Example 3.2] for further details. So, $\widehat{\mathcal {D}}$ and $M_{p,c}$ are not included in each other.

The study of the radial weights $\omega$ such that $H_\omega : H^p\rightarrow H^p$ is bounded, has been previously considered in [26]. Indeed, Theorem 2 improves [26, Theorem 2], by removing the initial hypothesis $\omega \in \widehat{\mathcal {D}}$. On the other hand, [26, Theorem 3] describes the weights $\omega \in \widehat{\mathcal {D}}$ such that $H_\omega : L^p_{[0,1)}\rightarrow H^p$ is bounded, and consequently gives a sufficient condition for the boundedness of $H_\omega : H^p\rightarrow H^p$, $1<p<\infty$. The following improvement of [26, Theorem 3] is a byproduct of Theorem 1.

Corollary 3

Let $\omega$ be a radial weight and $1<p<\infty$. Let $Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$ and $T\in \{ H_{\omega }, \widetilde{H_{\omega }}\}$. Then the following statements are equivalent:

• (i) $T:L^p_{[0,1)} \rightarrow Y_p$ is bounded;

• (ii) $\omega \in \mathcal {D}$ and satisfies the condition

  $$\begin{aligned} m_p(\omega )=\sup \limits _{0<r<1}\left( 1+\int _0^r \frac{1}{\widehat{\omega }(t)^p} dt\right) ^{\frac{1}{p}} \left( \int _r^1 \omega (t)^{p'}\,dt\right) ^{\frac{1}{p'}} <\infty ; \end{aligned}$$

• (iii) $\omega \in \widehat{\mathcal {D}}$ and satisfies the condition $m_p(\omega ) <\infty$.

In relation to an analogous result to Corollary 3 for $p=1$, Theorem 26 below shows that the radial weights such that $T:L^1_{[0,1)} \rightarrow Y_1$ is bounded, where $Y_1\in \{H(\infty ,1),H^1, D^1_{0}, HL(1)\}$ and $T\in \{ H_{\omega }, \widetilde{H_{\omega }}\}$, are the weights $\omega \in \mathcal {D}$ such that $m_1(\omega )= \mathop {\mathrm {ess\,sup}}\limits _{t \in [0,1)} \omega (t)\left( 1+\int _0^t \frac{ds}{\widehat{\omega }(s)}\right) <\infty .$

In view of the above findings, we compare the conditions $M_{p,c}(\omega )<\infty$, $M_{p,d}(\omega )<\infty$ and $m_p(\omega )<\infty$ in order to put the boundedness of $T:\,X_p \rightarrow Y_p$ alongside the boundedness of $T:L^p_{[0,1)} \rightarrow Y_p$, where $X_p, Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$ and $T\in \{ H_{\omega }, \widetilde{H_{\omega }}\}$ for $1\le p<\infty$. Bearing in mind (1.5), it is clear that the condition $m_p(\omega )<\infty$ implies that $M_{p,c}(\omega )<\infty$, for any weight $\omega \in \widehat{\mathcal {D}}$. Moreover, observe that $M_{p,c}(\omega )<\infty$ if and only if

$$\begin{aligned} \sup \limits _{0<r<1} \left( 1+\int _0^r \frac{1}{\widehat{\omega }(t)^p} dt\right) ^{\frac{1}{p}} \left( \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt\right) ^{\frac{1}{p'}}<\infty ,\quad \text {when}\,1<p<\infty \end{aligned}$$

and $\sup \nolimits _{a \in [0,1)} \frac{\widehat{\omega }(a)}{1-a} \left( 1+\int _0^a \frac{ds}{\widehat{\omega }(s)}\right) <\infty$ if and only if $M_{1,d}(\omega )<\infty$. So, the conditions $M_{p,c}(\omega )<\infty$ and $m_p(\omega )<\infty$, are equivalent for any $1\le p < \infty$ whenever $\omega$ satisfies the pointwise inequality

$$\begin{aligned} \omega (t)\lesssim \frac{\widehat{\omega }(t)}{1-t},\quad t \in [0,1), \end{aligned}$$

(1.7)

and $\omega \in \widehat{\mathcal {D}}$. The condition (1.7) implies restrictions on the decay and on the regularity of the weight, in fact if $\omega$ fulfills (1.7) then $\omega$ cannot decrease rapidly and cannot oscillate strongly. For instance, the exponential type weight $\omega (r)=\exp \left( -\frac{1}{1-r}\right)$, which is a prototype of rapidly decreasing weight (see [18]), has the property

$$\begin{aligned} \widehat{\omega }(r)\asymp \omega (r) (1-r)^2,\quad 0\le r<1, \end{aligned}$$

so it does not satisfy (1.7). On the other hand, any regular or rapidly increasing weight satisfies (1.7). Regular and rapidly increasing weights are large subclasses of $\widehat{\mathcal {D}}$, see [25, Section 1.2] for the definitions and examples of these classes of radial weights. However, we construct in Corollaries 19 and 28 weights $\omega \in \mathcal {D}$ with a strong oscillatory behaviour so that $M_{p,c}(\omega )<\infty$ and $m_p(\omega )=\infty$, and consequently they do not satisfy (1.7).

With the aim of discussing some results concerning the case $p=\infty$, we recall that the space $\mathord \textrm{BMOA}$ consists of those functions in the Hardy space $H^1$ that have bounded mean oscillation on the boundary of $\mathbb {D}$ [10], and the Bloch space $\mathcal {B}$ is the space of all analytic functions on $\mathbb {D}$ such that

$$\begin{aligned} \Vert f\Vert _{\mathcal {B}}=|f(0)|+\sup _{z\in \mathbb {D}}(1-|z|^2)\,|f'(z)|<\infty . \end{aligned}$$

We also consider the space $HL(\infty )$ of the $f(z)=\sum _{n=0}^\infty \widehat{f}(n) z^n \in \mathcal {H}(\mathbb {D})$ such that

$$\begin{aligned} \Vert f\Vert _{HL(\infty )}=\sup _{n\in \mathbb {N}\cup \{0\}}(n+1)\left| \widehat{f}(n)\right| <\infty . \end{aligned}$$

The following chain of inclusions hold [10]

$$\begin{aligned} HL(\infty )\subsetneq \mathord \textrm{BMOA}\subsetneq \mathcal {B}. \end{aligned}$$

(1.8)

If $\omega$ is a radial weight

$$\begin{aligned} H_{\omega }(1)(x) =\sum \limits _{n=0}^{\infty }\frac{\omega _n}{2\omega _{2n+1}(n+1)}x^n&\ge \frac{1}{2x}\sum \limits _{n=0}^{\infty }\frac{x^{n+1}}{n+1}=\frac{1}{2x}\log \left( \frac{1}{1-x}\right) , \quad x\in (0,1), \end{aligned}$$

so $H_{\omega }$ is not bounded on $H^\infty$. As for the classical Hilbert matrix H, it is bounded from $H^\infty$ to BMOA [13, Theorem 1.2]. So, it is natural wondering about the radial weights such that $H_{\omega }:H^{\infty }\rightarrow \mathord \textrm{BMOA}$ is bounded. The next result answers this question.

Theorem 4

Let $\omega$ be a radial weight and let $T\in \{H_{\omega },\widetilde{H_{\omega }}\}$. Then, the following statements are equivalent:

• (i) $T: H^{\infty }\rightarrow HL(\infty )$ is bounded;

• (ii) $T:H^{\infty }\rightarrow \mathord \textrm{BMOA}$ is bounded;

• (iii) $T: H^{\infty }\rightarrow \mathcal {B}$ is bounded;

• (iv) $\omega \in \widehat{\mathcal {D}}$.

The equivalence (iii)$\Leftrightarrow$(iv) was proved in [26, Theorem 1], so our contribution in Theorem 4 consists on proving the rest of equivalences.

Bearing in mind Theorems 1, 2 and 4, we deduce that $T\in \{H_{\omega },\widetilde{H_{\omega }}\}$ is bounded from $H^\infty$ to $HL(\infty )$ if $T: X_p\rightarrow Y_p$ is bounded, where $X_p,Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$, $1\le p<\infty$. We prove that this is a general phenomenon for Hilbert-type operators and parameters $1\le q<p$.

Theorem 5

Let $\omega$ be radial weight, $T\in \{H_\omega ,\widetilde{H_\omega }\}$ and $1\le q<p< \infty$. Further, let $X_q,Y_q\in \{H^q, D^q_{q-1}, HL(q), H(\infty ,q)\}$ and $X_p,Y_p\in \{H^p, D^p_{p-1}, HL(p), H(\infty ,p)\}$. If $T: X_q\rightarrow Y_q$ is bounded, then $T: X_p\rightarrow Y_p$ is bounded.

We also prove that that there does not exist radial weights $\omega$ such that $H_\omega : X_p\rightarrow Y_p$ is compact, where $X_p,Y_p\in \{H^p, D^p_{p-1}, HL(p), H(\infty ,p)\}$ and $1\le p< \infty$, neither radial weights such that $H_{\omega }:H^{\infty } \rightarrow \mathcal {B}$ is compact, see Theorems 22, 31, 34 below.

The letter $C=C(\cdot )$ will denote an absolute constant whose value depends on the parameters indicated in the parenthesis, and may change from one occurrence to another. We will use the notation $a\lesssim b$ if there exists a constant $C=C(\cdot )>0$ such that $a\le Cb$, and $a\gtrsim b$ is understood in an analogous manner. In particular, if $a\lesssim b$ and $a\gtrsim b$, then we write $a\asymp b$ and say that a and b are comparable. We remark that if a or b are quantities which depends on a radial weight $\omega$, the constant C such that $a\lesssim b$ or $a\gtrsim b$ may depend on $\omega$ but it does not depend on a neither on b.

The rest of the paper is organized as follows. Section 2 is devoted to prove some auxiliary results. We prove Theorem 1 and Corollary 3 in Sect. 3, and Theorem 2 is proved in Sect. 4. Section 5 contains a proof of Theorem 4 and Theorem 5 is proved in Sect. 6 together with some reformulations of the condition $M_{p,c}(\omega )<\infty$.

2 Preliminary results

In this section, we will prove some convenient preliminary results which will be repeatedly used throughout the rest of the paper. The first auxiliary lemma contains several characterizations of upper doubling radial weights. For a proof, see [21, Lemma 2.1].

Lemma 6

Let $\omega$ be a radial weight on $\mathbb {D}$. Then, the following statements are equivalent:

• (i) $\omega \in \widehat{\mathcal {D}}$;

• (ii) There exist $C=C(\omega )\ge 1$ and $\beta _0=\beta _0(\omega )>0$ such that

  $$\begin{aligned} \widehat{\omega }(r)\le C \left( \frac{1-r}{1-t}\right) ^{\beta }\widehat{\omega }(t), \quad 0\le r\le t<1; \end{aligned}$$

  for all $\beta \ge \beta _0$.

• (iii)
  $$\begin{aligned} \int _0^1 s^x \omega (s) ds\asymp \widehat{\omega }\left( 1-\frac{1}{x}\right) ,\quad x \in [1,\infty ); \end{aligned}$$

• (iv) There exists $C=C(\omega )>0$ and $\beta =\beta (\omega )>0$ such that

  $$\begin{aligned} \omega _x\le C \left( \frac{y}{x}\right) ^{\beta }\omega _y,\quad 0<x\le y<\infty ; \end{aligned}$$

• (v) $\widehat{\mathcal {D}}(\omega )=\sup \nolimits _{n\in \mathbb {N}}\frac{\omega _n}{\omega _{2n}}<\infty .$

We will also use the following characterizations of the class $\check{\mathcal {D}}$, see [23, (2.27)].

Lemma 7

Let $\omega$ be a radial weight. The following statements are equivalent:

• (i) $\omega \in \check{\mathcal {D}}$;

• (ii) There exist $C=C(\omega )>0$ and $\alpha _0=\alpha _0(\omega )>0$ such that

  $$\begin{aligned} \widehat{\omega }(s)\le C \left( \frac{1-s}{1-t}\right) ^{\alpha }\widehat{\omega }(t), \quad 0\le t\le s<1 \end{aligned}$$

  for all $0<\alpha \le \alpha _0$;

• (iii) There exist $K=K(\omega )>1$ and $C=C(\omega )>0$ such that

  $$\begin{aligned} \int _r^{1-\frac{1-r}{K}}\omega (s)ds \ge C \widehat{\omega }(r), \quad 0\le r<1. \end{aligned}$$

  (2.1)

Embedding relations among spaces $X_p,Y_p\in \{H^p, D^p_{p-1}, HL(p), H(\infty ,p)\}$ are quite useful in the study of operators acting on them. In particular, we recall that

$$\begin{aligned} \Vert f\Vert _{H(\infty ,p)}\le C_p \Vert f\Vert _{X_p},\quad 0< p<\infty , \end{aligned}$$

(2.2)

for $X_p\in \{ H^p,D^p_{p-1}\}$, see [27, p. 127] and [8, Lemma 4].

This inequality is no longer true for $X_p=HL(p)$ if $0<p<1$. In fact, take $f(z)=\sum _{n=0}^\infty 2^{\frac{n}{p}}z^{2^n}$. A calculation shows that $f\in HL(p)$, if $0<p<1$. However, using [15, Theorem 1],

$$\begin{aligned} \Vert f\Vert ^p_{H(\infty ,p)}= \int _0^1 \left( \sum _{n=0}^\infty 2^{\frac{n}{p}}s^{2^n}\right) ^p\,ds \asymp \sum _{n=0}^\infty 1=\infty . \end{aligned}$$

Our following result extends the inequality (2.2) to $X_p=HL(p)$ and $1\le p<\infty$.

Lemma 8

Let $1\le p<\infty$. Then, there is $C_p>0$ such that

$$\begin{aligned} \Vert f\Vert _{H(\infty ,p)}\le C_p \Vert f\Vert _{X_p},\quad f\in \mathcal {H}(\mathbb {D}), \end{aligned}$$

where $X_p\in \{ H^p,D^p_{p-1}, HL(p)\}$.

Proof

By (2.2) it is enough to prove the inequality for $X_p=HL(p)$. By [15, Theorem 1] and Hölder’s inequality

$$\begin{aligned} \int _0^1 M^p_\infty (t,f)\,dt&\le \int _{0}^1 \left( \sum _{n=0}^\infty |\widehat{f}(n)|t^n\right) ^p\,dt\\&\lesssim |\widehat{f}(0)|^p+ \sum _{n=0}^\infty 2^{-n}\left( \sum _{k=2^n}^{2^{n+1}-1} |\widehat{f}(k)|\right) ^p\\&\le |\widehat{f}(0)|^p+\sum _{n=0}^\infty 2^{n(p-2)}\sum _{k=2^n}^{2^{n+1}-1} |\widehat{f}(k)|^p\\&\lesssim |\widehat{f}(0)|^p+\sum _{n=0}^\infty \sum _{k=2^n}^{2^{n+1}-1} (k+1)^{p-2}|\widehat{f}(k)|^p = \Vert f\Vert _{HL(p)}^p. \end{aligned}$$

This finishes the proof.$\square$

For $0<p<\infty$ and $\omega$ a radial weight, let $L^p_{\omega , [0,1)}$ be the Lebesgue space of measurable functions such that

$$\begin{aligned} \Vert f\Vert ^p_{L^p_{\omega , [0,1)}}=\int _0^1 |f(t)|^p\omega (t)\,dt<\infty . \end{aligned}$$

Next, we will prove that the sublinear operator $\widetilde{H_{\omega }}$ does not distinguish the norm of the spaces $H(\infty ,p), HL(p), D^p_{p-1}, H^p,$ when $1<p<\infty$ and $\omega \in \widehat{\mathcal {D}}$.

Lemma 9

Let $\omega \in \widehat{\mathcal {D}}$, $1<p<\infty$ and $X_p, Y_p\in \{H(\infty ,p), HL(p), D^p_{p-1}, H^p\}$. Then,

$$\begin{aligned} \Vert \widetilde{H_{\omega }} (f)\Vert _{X_p}\asymp \Vert \widetilde{H_{\omega }} (f)\Vert _{Y_p},\, \quad f \in L^1_{\omega ,[0,1)}. \end{aligned}$$

Proof

Here and on the following, let us denote $I(n)=\{k\in \mathbb {N}: 2^n\le k<2^{n+1}\}$,  $n\in \mathbb {N}\cup \{0\}$. By Lemma 6

$$\begin{aligned} \omega _{2^{n+2}}\asymp \omega _{2k+2}\asymp \omega _{2k}\asymp \omega _{2^{n}},\quad \text {for any} n\in \mathbb {N}\cup \{0\} \text {and} k\in I(n). \end{aligned}$$

(2.3)

The above equivalences and [15, Theorem 1], yield

$$\begin{aligned} \begin{aligned} \Vert \widetilde{H_{\omega }} (f)\Vert _{H(\infty ,p)}^p&\asymp \sum _{n=0}^\infty 2^{-n}\left( \sum _{k\in I(n)} \frac{\int _0^1 |f(t)|t^k\omega (t)\,dt}{(k+1)\omega _{2k+1}} \right) ^p+ \left( \int _0^1 |f(t)|\omega (t)\,dt\right) ^p\\&\asymp \sum _{n=0}^\infty 2^{-n} \left( \frac{\int _0^1 |f(t)|t^{2^n}\omega (t)\,dt}{\omega _{2^{n+1}}} \right) ^p+ \left( \int _0^1 |f(t)|\omega (t)\,dt\right) ^p\\&\asymp \Vert \widetilde{H_{\omega }} (f)\Vert _{HL(p)}^p, \quad f \in L^1_{\omega ,[0,1)}. \end{aligned} \end{aligned}$$

This, together with [26, Lemma 8], finishes the proof.$\square$

3 Hilbert-type operators acting on $X_p$-spaces, $1<p<\infty$

3.1 Necessity part of Theorem 1

We begin this section with the construction of appropriate families of test functions to be used in the proof of Theorem 1. To do this, some notation and previous results are needed. Let $g(z)=\sum \nolimits _{k=0}^{\infty } \widehat{g}(k) z^k \in \mathcal {H}(\mathbb {D})$, and denote ${\Delta }_n g(z) =\sum \nolimits _{k \in I(n)} \widehat{g}(k)z^k$. In the particular case $g(z)=\frac{1}{1-z}$, we simply write ${\Delta }_n(z)={\Delta }_n(g)(z) =\sum \nolimits _{k \in I(n)}z^k$. We recall that

$$\begin{aligned} \Vert {\Delta }_n\Vert _{H^p}\asymp 2^{n(1-1/p)}, \quad n\in \mathbb {N}\cup \{0\},\quad 1<p<\infty , \end{aligned}$$

(3.1)

see [2, Lemma 2.7].

For any $n_1, \,n_2 \in \mathbb {N}\cup \left\{ 0\right\}$, $n_1<n_2$, write $S_{n_1,n_2} g(z)=\sum \nolimits _{k=n_1}^{n_2-1} \widehat{g}(k) z^k$. The next known result can be proved mimicking the proof of [13, Lemma 3.4] (see also [24, Lemma E]), that is, by summing by parts and using the M. Riesz projection theorem.

Lemma 10

Let $1<p<\infty$ and $\lambda =\left\{ \lambda _k\right\} _{k=0}^{\infty }$ be a positive and monotone sequence. Let $g(z)=\sum \nolimits _{k=0}^{\infty }b_k z^k$ and $(\lambda g)(z)=\sum \nolimits _{k=0}^{\infty }\lambda _k b_k z^k$.

• (a) If $\left\{ \lambda _k\right\} _{k=0}^{\infty }$ is nondecreasing, there exists a constant $C>0$ such that

  $$\begin{aligned} C^{-1}\lambda _{n_1}\Vert S_{n_1,n_2}g\Vert _{H^p}\le \Vert S_{n_1,n_2}(\lambda g)\Vert _{H^p}\le C \lambda _{n_2}\Vert S_{n_1,n_2}g\Vert _{H^p}. \end{aligned}$$

• (b) If $\left\{ \lambda _k\right\} _{k=0}^{\infty }$ is nonincreasing, there exists a constant $C>0$ such that

  $$\begin{aligned} C^{-1}\lambda _{n_2}\Vert S_{n_1,n_2}g\Vert _{H^p}\le \Vert S_{n_1,n_2}(\lambda g)\Vert _{H^p}\le C \lambda _{n_1}\Vert S_{n_1,n_2}g\Vert _{H^p}.\end{aligned}$$

Lemma 11

Let $\omega \in \widehat{\mathcal {D}}$, $1<p<\infty$, $\alpha ,\beta \in \mathbb {R}$ and $M, N \in \mathbb {N}\cup \{0\}$ such that $0\le N<4N+1\le M$. Let us consider the function

$$\begin{aligned} f_{N,M}(z)=\sum \limits _{k=N}^M \omega _{2k}^{\alpha } (k+1)^{\beta }z^k. \end{aligned}$$

Then,

$$\begin{aligned} \Vert f_{ N, M}\Vert _{HL(p)}\asymp \Vert f_{ N, M}\Vert _{H^p}\asymp \Vert f_{ N, M}\Vert _{D^p_{p-1}}, \end{aligned}$$

(3.2)

where the constants involved do not depend on M or N. In particular, if $\alpha =0$ then (3.2) holds for any radial weight.

Proof

Firstly, let us show that for all $N, M \in \mathbb {N}$, $M>N$,

$$\begin{aligned} \Vert f_{ 2^N +1, 2^M }\Vert _{D^p_{p-1}}\asymp \Vert f_{2^N+1, 2^M}\Vert _{HL(p)}. \end{aligned}$$

(3.3)

[16, Theorem 2.1(b)] (see also [20, 7.5.8]), Lemma 10, (2.3) and (3.1) implies

$$\begin{aligned} \Vert f_{2^N+1, 2^M}\Vert _{D^p_{p-1}}^p&\asymp \sum \limits _{n=N}^{M-1} 2^{-np} \left\| \sum \limits _{k\in I(n)} \omega _{2k+2}^{\alpha } (k+2)^{\beta } (k+1) z^k\right\| _{H^p}^p\\&\asymp \sum \limits _{n=N}^{M-1} 2^{np\beta } \omega _{2^{n+1}}^{p\alpha }\Vert \Delta _n\Vert ^p_{H^p}\\&\asymp \sum \limits _{n=N}^{M-1} 2^{n(p\beta +p-1)} \omega _{2^{n+1}}^{p\alpha }\\&\asymp \sum \limits _{k=2^N+1}^{2^M}(k+1)^{p\beta +p-2}\omega _{2k}^{p\alpha } = \Vert f_{2^N+1,2^M}\Vert _{HL(p)}^p, \end{aligned}$$

A similar calculation shows that

$$\begin{aligned} \Vert f_{2^{N+1}+1, 2^M}\Vert _{D^p_{p-1}} \asymp \Vert f_{2^{N}+1, 2^{M+1}}\Vert _{D^p_{p-1}},\quad M>N+1. \end{aligned}$$

(3.4)

Next, if $N> 2$, there is $N^\star , M^\star \in \mathbb {N}$ such that $2^{N^\star }\le N-1<2^{N^\star +1}$ and $2^{M^\star }\le M-1<2^{M^\star +1}$, so $N^\star +1<M^\star$. Then, by [16, Theorem 2.1(b)] and the boundedness of the Riesz projection, (3.3) and (3.4)

$$\begin{aligned} \Vert f_{N, M}\Vert _{D^p_{p-1}}^p&\asymp 2^{-pN^\star } \left\| \sum \limits _{k=N-1}^{2^{N^\star +1}-1} (k+1)\widehat{f}_{N,M}(k+1) z^k\right\| _{H^p}^p + \Vert f_{2^{{N^\star }+1}+1, 2^{M^\star }}\Vert ^p_{D^p_{p-1}}\\&\quad + 2^{-pM^\star } \left\| \sum \limits _{k=2^{M^\star }}^{M-1} (k+1)\widehat{f}_{N,M}(k+1) z^k\right\| _{H^p}^p \\&\lesssim \Vert f_{2^{N^\star }+1, 2^{M^\star +1}}\Vert _{D^p_{p-1}}^p\\ {}&\asymp \Vert f_{2^{N^\star +1}+1, 2^{M^\star }}\Vert _{D^p_{p-1}}^p \asymp \Vert f_{2^{N^\star +1}+1, 2^{M^\star } }\Vert _{HL(p)}^p\lesssim \Vert f_{N,M}\Vert _{HL(p)}^p. \end{aligned}$$

On the other hand,

$$\begin{aligned} \Vert f_{N, M}\Vert _{D^p_{p-1}}^p\gtrsim & {} \Vert f_{2^{N^\star +1}+1, 2^{M^\star }}\Vert _{D^p_{p-1}}^p \asymp \Vert f_{2^{N^\star }+1, 2^{M^\star +1}}\Vert _{D^p_{p-1}}^p\\ {}&\asymp&\Vert f_{2^{N^\star }+1, 2^{M^\star +1}}\Vert _{HL(p)}^p \ge \Vert f_{N,M}\Vert _{HL(p)}^p. \end{aligned}$$

Then, bearing in mind (1.3) and (1.4), we obtain $\Vert f_{N, M}\Vert _{HL(p)}\asymp \Vert f_{N, M}\Vert _{H^p}\asymp \Vert f_{ N, M}\Vert _{D^p_{p-1}}$ for each $N> 2$.

If $N\in \{0,1,2\}$, the previous argument together with minor modifications implies (3.2). This finishes the proof. $\square$

Now we are ready to prove the necessity part of Theorem 1.

Proposition 12

Let $\omega$ be a radial weight and $1<p<\infty$. If $X_p, Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$, $T\in \{H_{\omega }, \widetilde{H_{\omega }}\}$, and $T: X_p\rightarrow Y_p$ is a bounded operator. Then, $\omega \in \mathcal {D}$ and

$$\begin{aligned} M_p(\omega )=\sup \limits _{N\in \mathbb {N}} \left( \sum \limits _{n=0}^N \frac{1}{(n+1)^2 \omega _{2n+1}^p}\right) ^{\frac{1}{p}}\left( \sum \limits _{n=N}^{\infty } \omega _{2n+1}^{p'}(n +1)^{p'-2}\right) ^{\frac{1}{p'}}<\infty . \end{aligned}$$

(3.5)

Proof

In order to obtain both conditions, $\omega \in \mathcal {D}$ and $M_p(\omega )<\infty$, we are going to work with families of test functions constructed in Lemma 11. Since they have non-negative Maclaurin coefficients, it is enough to prove the result for $T= H_{\omega }$. Take $f \in \mathcal {H}(\mathbb {D})$ such that $\widehat{f}(n) \ge 0$ for all $n \in \mathbb {N}$.

First Step. We will prove that $\omega \in \widehat{\mathcal {D}}$. By Lemma 8, it is enough to deal with the case $Y_p=H(\infty ,p)$.

Observe that $M_\infty (r,H_\omega (f))=\sum _{n=0}^\infty \frac{1}{2(n+1)\omega _{2n+1}}\left( \sum _{k=0}^\infty \widehat{f}(k)\omega _{n+k}\right) r^n$. Now, consider the test functions $f_N(z)=\sum \nolimits _{n=0}^N \frac{1}{(n+1)^{1-\frac{1}{p-1}}} z^n, N\in \mathbb {N}$. Given that $\sum \nolimits _{k=0}^N \frac{1}{(k+1)^{1-\frac{1}{p-1}}}\asymp (N+1)^{\frac{1}{p-1}}$,

$$\begin{aligned}\begin{aligned} M_\infty (r,H_\omega (f_N))&\ge \sum _{n=6N}^{7N}\frac{1}{2(n+1)\omega _{2n+1}}\left( \sum _{k=0}^N \widehat{f_N}(k)\omega _{n+k}\right) r^n \\ {}&\gtrsim \sum _{n=6N}^{7N}\frac{\omega _{n+N}}{(n+1)\omega _{2n+1}}\left( \sum \limits _{k=0}^N \frac{1}{(k+1)^{1-\frac{1}{p-1}}}\right) r^n \\ {}&\gtrsim (N+1)^{\frac{1}{p-1}} \frac{\omega _{8N}}{\omega _{12N}}r^{7N}, \quad N\in \mathbb {N},\quad 0\le r<1. \end{aligned} \end{aligned}$$

So,

$$\begin{aligned} \Vert H_\omega (f_N)\Vert ^p_{H(\infty ,p)}\gtrsim (N+1)^{\frac{1}{p-1}}\left( \frac{\omega _{8N}}{\omega _{12N}}\right) ^p,\quad N\in \mathbb {N}. \end{aligned}$$

By Lemmas 11 and 8,

$$\begin{aligned} \Vert f_N\Vert ^p_{X_p}\lesssim \Vert f_N\Vert ^p_{HL(p)}= \sum \limits _{n=0}^N \frac{1}{(n+1)^{1-\frac{1}{p-1}}}\asymp (N+1)^{\frac{1}{p-1}}. \end{aligned}$$

Consequently,

$$\begin{aligned} \begin{aligned} (N+1)^{\frac{1}{p-1}}\left( \frac{\omega _{8N}}{\omega _{12N}}\right) ^p&\lesssim \Vert H_\omega (f_N)\Vert ^p_{H(\infty ,p)} \lesssim \Vert f_N\Vert ^p_{X_p}\lesssim (N+1)^{\frac{1}{p-1}},\quad N\in \mathbb {N}. \end{aligned} \end{aligned}$$

Therefore, there is $C=C(\omega ,p)$ such that $\omega _{8N}\le C \omega _{12N},\quad N \in \mathbb {N}$. From now on, for each $x\in \mathbb {R}$, $\lfloor x\rfloor$ denotes the biggest integer $\le x$. For any $x\ge 120$, take $N\in \mathbb {N}$ such that $8N\le x<8N+8$, and then

$$\begin{aligned} \omega _x\le \omega _{8N}\le C \omega _{12N}\le C \omega _{8 \lfloor \frac{3N}{2} \rfloor } \le C^2 \omega _{12 \lfloor \frac{3N}{2} \rfloor } \le C^2 \omega _{18N-12}\le C^2 \omega _{16N+16}\le C^2 \omega _{2x}. \end{aligned}$$

So, $\omega \in \widehat{\mathcal {D}}$ by Lemma 6.

Second Step. We will prove that $M_p(\omega ) < \infty$.

Case $\varvec{Y_p=HL (p)}$. Set an arbitrary $N \in \mathbb {N}$. Then, bearing in mind that $\{\omega _k\}_{k=0}^\infty$ is decreasing,

$$\begin{aligned} \begin{aligned} \left( \sum \limits _{n=0}^{N}\frac{1}{(n+1)^2\omega _{2n+1}^p}\right) \left( \sum \limits _{k=N}^{\infty }\widehat{f}(k)\omega _{2k+1}\right) ^p&\le \sum \limits _{n=0}^{\infty }\frac{1}{(n+1)^2\omega _{2n+1}^p} \left( \sum \limits _{k=0}^{\infty }\widehat{f}(k)\omega _{n+k}\right) ^p \\ {}&\lesssim \Vert H_{\omega }\Vert _{X_p\rightarrow HL(p) }^p \Vert f\Vert ^p_{X_p}. \end{aligned}\nonumber \\ \end{aligned}$$

(3.6)

Take $M, N \in \mathbb {N}$, $M>4N+1$, and consider the family of test polynomials

$$\begin{aligned} f_{N,M}(z)=\sum \limits _{k=N}^{M}\omega _{2k+1}^{p'-1}(k+1)^{p'-2}z^k,\, z \in \mathbb {D}. \end{aligned}$$

(3.7)

Then, Lemmas 8 and 11 yield

$$\begin{aligned} \sum \limits _{k=N}^{M}\omega _{2k+1}^{p'}(k+1)^{p'-2}= \Vert f_{ N, M}\Vert ^p_{HL(p)}\gtrsim \Vert f_{N,M}\Vert ^p_{X_p} \end{aligned}$$

where the constants do not depend on M or N.

So, testing this family of functions in (3.6), there exists $C=C(p, \omega )>0$ such that

$$\begin{aligned} \left( \sum \limits _{n=0}^{N}\frac{1}{(n+1)^2\omega _{2n+1}^p}\right) \left( \sum \limits _{k=N}^{M}\omega _{2k+1}^{p'}(k+1)^{p'-2}\right) ^{p-1} \le C, \quad \text {for any} M, N \in \mathbb {N}, M>4N+1. \end{aligned}$$

By letting $M\rightarrow \infty$, and taking the supremum in $N\in \mathbb {N}$, (3.5) holds.

Case $\varvec{Y_p\in \{H(\infty ,p),H^p,D^p_{p-1}\}}$. Let $f_{ N,M}$ be the functions defined in (3.7), then $H_{\omega }(f_{N,M})=\widetilde{H_{\omega }}(f_{N,M})$. This together with the fact that $\omega \in \widehat{\mathcal {D}}$ and Lemma 9, yields

$$\begin{aligned} \Vert H_{\omega }(f_{N,M})\Vert _{Y_p}\asymp \Vert H_{\omega }(f_{N,M})\Vert _{HL(p)} \end{aligned}$$

where the constants in the inequalities do not depend on M or N.

Therefore, using Lemmas 8,  9 and 11, there exists $C=C(p, \omega )>0$ such that

$$\begin{aligned} \Vert H_{\omega }(f_{N,M})\Vert _{HL(p)} \le C \Vert f_{ N,M}\Vert _{HL(p)}. \end{aligned}$$

So, arguing as in the case $Y_p=HL(p)$, we obtain $M_p(\omega )<\infty .$

Third Step. We will prove that the condition $M_p(\omega )<\infty$ implies that $\omega \in \mathcal {M}$. Indeed, set $K,M \in \mathbb {N}$, $K,M >1$ and $N \in \mathbb {N}$. By (3.5),

$$\begin{aligned} \infty&> M_p(\omega ) \ge \left( \sum \limits _{j=N}^{KN} \frac{1}{(j+1)^2 \omega _{2j+1}^p}\right) ^{\frac{1}{p}}\left( \sum \limits _{j=KN}^{(K+M)N-1} \omega _{2j+1}^{p'}(j+1)^{p'-2}\right) ^{\frac{1}{p'}} \\ {}&\ge \frac{\omega _{2(K+M)N}}{\omega _{2N}} \left( \sum \limits _{j=N}^{KN} \frac{1}{(j+1)^2 }\right) ^{\frac{1}{p}}\left( \sum \limits _{j=KN}^{(K+M)N-1}(j+1)^{p'-2}\right) ^{\frac{1}{p'}}, \end{aligned}$$

So, there is $C=C(p)>0$ such that

$$\begin{aligned} \omega _{2N}\ge \omega _{2(K+M)N}\frac{1}{M_p(\omega )}C\left( (K+M)^{p'-1}-K^{p'-1}\right) ^{\frac{1}{p'}}, \text {for all} N\in \mathbb {N}. \end{aligned}$$

(3.8)

Now, fix $K>1$ and take $M\in \mathbb {N}$ large enough such that

$$\begin{aligned} \frac{1}{M_p(\omega )}C\left( (K+M)^{p'-1}-K^{p'-1}\right) ^{\frac{1}{p'}}=C(K,M,p, \omega )>1. \end{aligned}$$

Let $x\ge 1$ and take $N \in \mathbb {N}$ such that $2N-2\le x <2N$. Then, by (3.8)

$$\begin{aligned} \omega _x\ge & {} \omega _{2N}\ge C(K,M,p, \omega ) \omega _{2(K+M)N}\ge C(K,M,p, \omega ) \omega _{(K+M)x+2(K+M)}\\ {}\ge & {} C(K,M,p, \omega ) \omega _{3(K+M)x}, \end{aligned}$$

so $\omega \in \mathcal {M}$. Since $\omega \in \widehat{\mathcal {D}}$, [23, Theorem 3] yields $\omega \in \mathcal {D}$. The proof is finished. $\square$

3.2 Sufficiency part of Theorem 1

For the purpose of proving Theorem 1 we need some additional preparations. In particular, we aim for reformulating the necessary discrete condition on the moments of the radial weight $\omega$, $M_p(\omega )<\infty$, as a continuous inequality in terms of $\widehat{\omega }(r)$. Observe that a radial weight $\omega$ satisfies the condition

$$\begin{aligned} K_{p,c}(\omega )= \sup \limits _{0<r<1} \left( 1+\int _0^r \frac{1}{\widehat{\omega }(t)^p} dt\right) ^{\frac{1}{p}} \left( \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt\right) ^{\frac{1}{p'}}<\infty \end{aligned}$$

if and only if $M_{p,c}(\omega )<\infty .$ This fact will be used repeatedly throughout the paper.

Lemma 13

Let $1<p<\infty$ and $\omega \in \widehat{\mathcal {D}}$. Set

$$\begin{aligned} K_{p,c}(\omega )= \sup \limits _{0<r<1} \left( 1+\int _0^r \frac{1}{\widehat{\omega }(t)^p} dt\right) ^{\frac{1}{p}} \left( \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt\right) ^{\frac{1}{p'}}. \end{aligned}$$

Then,

$$\begin{aligned} \int _0^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt \asymp \sum \limits _{k=0}^{\infty } \omega _{2k+1}^{p'}(k+1)^{p'-2} \end{aligned}$$

and

$$\begin{aligned} M_p(\omega )\asymp K_{p,c}(\omega ). \end{aligned}$$

Proof

Let $0<r<1$ and set $N \in \mathbb {N}$ such that $1-\frac{1}{N}\le r < 1-\frac{1}{N+1}$. Then, by using Lemma 6,

$$\begin{aligned} \sum \limits _{k=0}^N \frac{1}{(k+1)^2 \omega _{2k+1}^p}&\asymp \sum \limits _{k=0}^N \frac{1}{(k+1)^2 \widehat{\omega }\left( 1-\frac{1}{k+1}\right) ^p} \gtrsim 1+ \sum \limits _{k=1}^N \int _k^{k+1} \frac{1}{x^2 \widehat{\omega }\left( 1-\frac{1}{x}\right) ^p} dx \\ {}&= 1+ \int _0^{1-\frac{1}{N+1}} \frac{1}{\widehat{\omega }(s)^p} ds \ge 1+ \int _0^{r} \frac{1}{\widehat{\omega }(s)^p} ds. \end{aligned}$$

In addition, by Lemma 6 again,

$$\begin{aligned} \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt&\le \int _{1-\frac{1}{N}}^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt= \sum \limits _{k=N}^{\infty } \int _{1-\frac{1}{k}}^{1-\frac{1}{k+1}} \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt \nonumber \\ {}&\le \sum \limits _{k=N}^{\infty } \widehat{\omega }\left( 1-\frac{1}{k}\right) ^{p'}\int _{1-\frac{1}{k}}^{1-\frac{1}{k+1}} \frac{1}{(1-t)^{p'}}\,dt \lesssim \sum \limits _{k=N}^{\infty } \omega _{2k+1}^{p'}(k+1)^{p'-2}. \end{aligned}$$

(3.9)

Therefore, $K_{p,c}(\omega )\lesssim M_p(\omega )$.

Conversely, in order to obtain the reverse inequality, a similar argument to (3.9) yields

$$\begin{aligned} \sum \limits _{k=0}^N \frac{1}{(k+1)^2 \omega _{2k+1}^p}&\asymp 1+ \sum \limits _{k=1}^N \frac{1}{(k+1)^2 \widehat{\omega }\left( 1-\frac{1}{k}\right) ^p} \lesssim 1+ \int _0^{1-\frac{1}{N+1}} \frac{1}{\widehat{\omega }(s)^p} ds. \end{aligned}$$

Now, on the one hand, if $r\le \frac{1}{2}$ then $\sum \nolimits _{k=0}^N \frac{1}{(k+1)^2 \omega _{2k+1}^p} \lesssim 1\le 1+ \int _0^{r} \frac{1}{\widehat{\omega }(s)^p} ds.$ On the other hand, if $\frac{1}{2}\le r <1$,

$$\begin{aligned} \sum \limits _{k=0}^N \frac{1}{(k+1)^2 \omega _{2k+1}^p}\lesssim & {} 1+ \int _0^{r} \frac{1}{\widehat{\omega }(s)^p} ds + \int _{1-\frac{1}{N}}^{1-\frac{1}{N+1}} \frac{1}{\widehat{\omega }(s)^p}ds\\ {}\lesssim & {} 1+ \int _0^{r} \frac{1}{\widehat{\omega }(s)^p} ds +\frac{1}{N \widehat{\omega }\left( 1-\frac{1}{N+1}\right) ^{p}} \end{aligned}$$

So, Lemma 6 yields

$$\begin{aligned} \sum \limits _{k=0}^N \frac{1}{(k+1)^2 \omega _{2k+1}^p}&\lesssim 1+ \int _0^{r} \frac{1}{\widehat{\omega }(s)^p} ds +\frac{1-r}{ \widehat{\omega }\left( 2r-1 \right) ^{p}}\\&\asymp 1+ \int _0^{r} \frac{1}{\widehat{\omega }(s)^p} ds + \int _{2r-1}^{r} \frac{1}{\widehat{\omega }(s)^p} ds\\&\lesssim 1+ \int _0^{r} \frac{1}{\widehat{\omega }(s)^p} ds,\quad \frac{1}{2}\le r <1. \end{aligned}$$

Next,

$$\begin{aligned} \sum \limits _{k=N}^{\infty } \omega _{2k+1}^{p'}(k+1)^{p'-2}&\asymp \sum \limits _{k=N}^{\infty } \widehat{\omega }\left( 1-\frac{1}{k+2}\right) ^{p'}(k+1)^{p'} \int _{1-\frac{1}{k+1}}^{1-\frac{1}{k+2}}dt \nonumber \\&\lesssim \int _{1-\frac{1}{N+1}}^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt \le \int _{r}^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt, \end{aligned}$$

(3.10)

and consequently, $M_p(\omega )\lesssim K_{p,c}(\omega )$. Finally, (3.9) and (3.10) imply

$$\begin{aligned} \int _0^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt \asymp \sum \limits _{k=0}^{\infty } \omega _{2k+1}^{p'}(k+1)^{p'-2}. \end{aligned}$$

This finishes the proof. $\square$

We will also need the following description of the class $\mathcal {D}$.

Lemma 14

Let $\omega$ be a radial weight. Then the following conditions are equivalent:

• (i) $\omega \in \mathcal {D}$;

• (ii) The function defined as $\widetilde{\omega }(r)=\frac{\widehat{\omega }(r)}{1-r}$, $0\le r<1$, is a radial weight and satisfies

  $$\begin{aligned} \widehat{\omega }(r)\asymp \widehat{\widetilde{\omega }}(r),\quad 0\le r<1. \end{aligned}$$

Proof

(i)$\Rightarrow$(ii). By Lemma 7, there is $\alpha >0$ such that

$$\begin{aligned} \int _{r}^1 \widetilde{\omega }(s)\,ds\lesssim \frac{\widehat{\omega }(r)}{(1-r)^\alpha }\int _r^1 (1-s)^{\alpha -1}\,ds\lesssim \widehat{\omega }(r),\quad 0\le r<1, \end{aligned}$$

which, in particular, implies that $\widetilde{\omega }$ is a radial weight. On the other hand, by Lemma 6, there is $\beta >0$ such that

$$\begin{aligned} \int _{r}^1 \widetilde{\omega }(s)\,ds \gtrsim \frac{\widehat{\omega }(r)}{(1-r)^\beta } \int _{r}^1 (1-s)^{\beta -1}\,ds\gtrsim \widehat{\omega }(r),\quad 0\le r<1. \end{aligned}$$

Reciprocally, if (ii) holds, there are $C_1,C_2>0$ such that

$$\begin{aligned} C_1\widehat{\omega }(r)\le \widehat{\widetilde{\omega }}(r) \le C_2\widehat{\omega }(r),\quad 0\le r<1. \end{aligned}$$

So, for any $K>1$,

$$\begin{aligned} \widehat{\omega }(r)\ge \frac{1}{C_2} \int _{r}^{1-\frac{1-r}{K}}\widetilde{\omega }(s)\,ds\ge \frac{\log K}{C_2} \widehat{\omega }\left( 1-\frac{1-r}{K}\right) ,\quad 0\le r<1. \end{aligned}$$

Therefore, taking K such that $\frac{\log K}{C_2}>1$, $\omega \in \check{\mathcal {D}}$.

Moreover, for any $K>1$

$$\begin{aligned} \begin{aligned} \widehat{\omega }(r) \le \frac{1}{C_1}\widehat{\widetilde{\omega }}(r) \le \frac{\log K}{C_1}\widehat{\omega }(r)+\frac{1}{C_1}\int _{1-\frac{1-r}{K}}^1\widetilde{\omega }(s)\,ds\quad 0\le r<1. \end{aligned} \end{aligned}$$

If $\frac{\log K}{C_1}<1$, then

$$\begin{aligned} \widehat{\omega }(r)\le \frac{C_2}{1-\frac{\log K}{C_1}}\widehat{\omega }\left( 1-\frac{1-r}{K}\right) ,\quad 0\le r<1. \end{aligned}$$

So, $\omega \in \widehat{\mathcal {D}}$. This finishes the proof. $\square$

The previous lemma may be used to prove that a differentiable non-decreasing function $h:[0,1)\rightarrow [0,\infty )$ belongs to $L^p_{\omega , [0,1)}$ if and only if it belongs to $L^p_{\widetilde{\omega }, [0,1)}$. This result is essential for our purposes. In particular, bearing in mind Lemma 14 and two integration by parts,

$$\begin{aligned} \int _0^1 h(t)\omega (t)\,dt\lesssim h(0)\widehat{\omega }(0) +\int _0^1 h(t)\widetilde{\omega }(t)\,dt, \end{aligned}$$

(3.11)

for any differentiable non-decreasing function $h:[0,1)\rightarrow [0,\infty )$.

Bearing in mind Lemma 13, our next result ensures that the Hilbert-type operators $H_\omega$ and $\widetilde{H_\omega }$ are well defined on $X_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$, $1<p<\infty$ when $\omega \in \mathcal {D}$ and $M_p(\omega )<\infty$.

Lemma 15

Let $\omega \in \mathcal {D}$ and $1<p<\infty$ such that $\int _0^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt<\infty$.

Then

$$\begin{aligned} \int _0^1 M_\infty (t,f)\omega (t)\,dt\lesssim \Vert f\Vert _{H(\infty ,p)}\left( \int _0^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt \right) ^{1/p'},\quad f\in \mathcal {H}(\mathbb {D}). \end{aligned}$$

In particular, $T(f)\in \mathcal {H}(\mathbb {D})$ for any $f\in X_p$, where $X_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$ and $T\in \{H_\omega , \widetilde{H_\omega }\}$.

Proof

By (3.11)

$$\begin{aligned} \int _0^1 M_\infty (t,f)\omega (t)\,dt \le |f(0)|\widehat{\omega }(0)+\int _0^1 M_\infty (t,f)\widetilde{\omega }(t)\,dt. \end{aligned}$$

(3.12)

Then, by Hölder’s inequality

$$\begin{aligned} \int _0^1 M_\infty (t,f)\omega (t)\,dt&\le |f(0)|\widehat{\omega }(0)+ \left( \int _0^1 M^p_\infty (t,f)\,dt\right) ^{1/p}\left( \int _0^1 \widetilde{\omega }(t)^{p'}\,dt \right) ^{1/p'}\\&\lesssim \Vert f\Vert _{H(\infty ,p)}\left( \int _0^1 \widetilde{\omega }(t)^{p'}\,dt \right) ^{1/p'}<\infty , \quad f\in H(\infty ,p). \end{aligned}$$

Joining the above chain of inequalities with Lemma 8, the proof is finished. $\square$

Next, for $p,q>0$ and $\alpha >-1$, let $H^1(p,q, \alpha )$ denote the space of $f \in \mathcal {H}(\mathbb {D})$ such that

$$\begin{aligned} \Vert f\Vert _{H^1(p,q, \alpha )}=\left( |f(0)|^p +\int _0^1 M_q^p(r,f')(1-r)^{\alpha }dr\right) ^{\frac{1}{p}}<\infty . \end{aligned}$$

It is worth mentioning that $H^1\left( p,p, p-1\right) =D^p_{p-1}$.

The following inequality will be used in the proof of Theorem 1. It was proved in [16, Corollary 3.1].

Lemma 16

Let $1<q<p<\infty$. Then,

$$\begin{aligned} \Vert f\Vert _{H^p}\lesssim \Vert f\Vert _{H^1\left( p,q, p\left( 1-\frac{1}{q}\right) \right) }, \quad f\in \mathcal {H}(\mathbb {D}). \end{aligned}$$

Now, we are ready to prove the main result of this section.

Proof of Theorem 1

The implication (i)$\Rightarrow$(ii) was proved in Proposition 12. The implication (ii)$\Rightarrow$(iii) is clear, and (iii)$\Rightarrow$(ii) follows from the third step in the proof of Proposition 12. On the other hand, bearing in mind that $M_{p,c}(\omega )<\infty$ if and only if $K_{p,c}(\omega )<\infty$, (iii)$\Leftrightarrow$(iv) follows from Lemma 13. Then, it is enough to prove (ii)$\Rightarrow$(i).

(ii)$\mathbf {\Rightarrow }$(i).

First Step. We will prove the inequality

$$\begin{aligned} \Vert T(f)\Vert _{Y_p}\lesssim \Vert f\Vert _{X_p}+ \Vert \widetilde{H_\omega }(f)\Vert _{Y_p},\quad f\in X_p. \end{aligned}$$

(3.13)

By Lemmas 8 and  16, it is enough to prove

$$\begin{aligned} \Vert H_\omega (f)\Vert _{H^1\left( p,q, p\left( 1-\frac{1}{q}\right) \right) }\lesssim \Vert f\Vert _{X_p}+ \Vert \widetilde{H_\omega }(f)\Vert _{Y_p},\quad 1<q,p<\infty ,\quad f\in X_p.\nonumber \\ \end{aligned}$$

(3.14)

Let $f \in X_p$. Then, Lemmas 13 and  15 ensure that $H_\omega (f)\in \mathcal {H}(\mathbb {D})$. By [16, Theorem 2.1]

$$\begin{aligned} \Vert H_{\omega } (f)\Vert _{H^1\left( p,q, p\left( 1-\frac{1}{q}\right) \right) }^p&\asymp |H_{\omega }(f)(0)|^p+ |H_{\omega }(f)'(0)|^p\nonumber \\ {}&\quad + \sum \limits _{n=0}^{\infty } 2^{-n\left( p\left( 1-\frac{1}{q}\right) +1\right) }\Vert {\Delta }_n (H_{\omega }(f))'\Vert _{H^q}^p. \end{aligned}$$

(3.15)

Due to

$$\begin{aligned} (H_{\omega } (f))'(z)=\sum \limits _{n=0}^{\infty }\frac{n+1}{2(n+2)\omega _{2n+3}}\left( \int _0^1 f(t) t^{n+1}\omega (t)dt\right) z^n, \end{aligned}$$

and using the proof of [8, Lemma 7], Lemma 10 and (3.1),

$$\begin{aligned} \Vert {\Delta }_n (H_{\omega }(f))'\Vert _{H^q}^p\lesssim \frac{\left( \int _0^1 t^{2^{n-2}+1} |f(t)| \omega (t)dt\right) ^p}{\omega _{2^{n+2}+3}^p}2^{np\left( 1-\frac{1}{q}\right) }, \quad n \ge 3. \end{aligned}$$

Hence, by using Lemma 6,

$$\begin{aligned} \sum \limits _{n=3}^{\infty } 2^{-n\left( p\left( 1-\frac{1}{q}\right) +1\right) }\Vert {\Delta }_n (H_{\omega }(f))'\Vert _{H^q}^p&\lesssim \sum \limits _{n=3}^{\infty } \frac{2^{-2n}}{\omega _{2^{n+2}}^p} \sum \limits _{k=2^{n-3}}^{2^{n-2}} \left( \int _0^1 t^{k} |f(t)| \omega (t)dt\right) ^p \nonumber \\&\lesssim \sum \limits _{k=1}^{\infty } \frac{\left( \int _0^1 t^{k} |f(t)| \omega (t)dt\right) ^p}{\omega _{2k+1}^p (k+1)^2}\lesssim \Vert \widetilde{H_{\omega }} (f)\Vert _{HL(p)}^p. \end{aligned}$$

(3.16)

In addition, by Lemmas 8 and  15

$$\begin{aligned} |H_{\omega }(f)(0)|^p +|H_{\omega }(f)'(0)|^p +\sum \limits _{n=0}^2 2^{-np}\Vert {\Delta }_n (H_{\omega }(f))'\Vert _{H^p}^p \lesssim \Vert f\Vert _{X_p}^p. \end{aligned}$$

(3.17)

Therefore, by putting together (3.15), (3.16) and (3.17)

$$\begin{aligned} \Vert H_{\omega } (f)\Vert _{H^1\left( p,q, p\left( 1-\frac{1}{q}\right) \right) }^p \lesssim \Vert f\Vert _{X_p}^p + \Vert \widetilde{H_{\omega }} (f)\Vert _{HL(p)}^p. \end{aligned}$$

The above inequality, together with Lemma 9, yields (3.14).

Second Step. We will prove the inequality

$$\begin{aligned} \Vert \widetilde{H_\omega }(f)\Vert _{D^p_{p-1}}\lesssim \Vert f\Vert _{H(\infty ,p)},\quad f\in H(\infty ,p). \end{aligned}$$

(3.18)

We denote by

$$\begin{aligned} G^\omega _t(z)=\frac{d}{dz}\left( \frac{1}{z}\int _0^z B^{\omega }_t(\zeta )d\zeta \right) . \end{aligned}$$

(3.19)

By [26, Lemma B]

$$\begin{aligned} M_p (r, G_t^{\omega })\lesssim \left( 1+\int _0^{rt}\frac{ds}{\widehat{\omega }(s)(1-s)^p} \right) ^{1/p} \lesssim \frac{1}{\widehat{\omega }(rt) (1-rt)^{1-\frac{1}{p}}},\quad 0\le r,t<1, \end{aligned}$$

which together with Minkowski’s inequality yields

$$\begin{aligned} \Vert \widetilde{H_{\omega }}(f)\Vert _{D^p_{p-1}}^p&\lesssim |\widetilde{H_{\omega }}(f)(0)|^p +\int _0^1 \left( \int _0^1 \vert f(t)\vert \omega (t) M_p (r, G_t^{\omega } ) dt\right) ^p (1-r)^{p-1}\,dr\\&\lesssim |\widetilde{H_{\omega }}(f)(0)|^p +\int _0^1 \left( \int _0^1 \frac{\vert f(t)\vert \omega (t)}{\widehat{\omega }(rt) (1-rt)^{1-\frac{1}{p}}} dt\right) ^p (1-r)^{p-1}\,dr\\&\le |\widetilde{H_{\omega }}(f)(0)|^p +\int _0^1 \left( \int _0^1 \frac{ M_{\infty }(t,f) \omega (t)}{\widehat{\omega }(rt) (1-rt)^{1-\frac{1}{p}}} dt\right) ^p (1-r)^{p-1}\,dr. \end{aligned}$$

Now, by (3.11)

$$\begin{aligned} \Vert \widetilde{H_{\omega }}(f)\Vert _{D^p_{p-1}}^p{} & {} \lesssim |\widetilde{H_{\omega }}(f)(0)|^p + |f(0)|^p \nonumber \\{} & {} \quad +\int _0^1 \left( \int _0^1 \frac{ M_{\infty }(t,f) }{\widehat{\omega }(rt) (1-rt)^{1-\frac{1}{p}}} \frac{\widehat{\omega }(t)}{1-t} dt\right) ^p (1-r)^{p-1} \,dr. \end{aligned}$$

(3.20)

Next, by Lemma 13, $M_{p,c}(\omega )<\infty$ holds, so [17, Theorem 2] yields

$$\begin{aligned}{} & {} \int _0^1 \left( \int _r^1 \frac{ M_{\infty }(t,f) }{\widehat{\omega }(rt) (1-rt)^{1-\frac{1}{p}}} \frac{\widehat{\omega }(t)}{1-t} dt\right) ^p (1-r)^{p-1}\, dr\nonumber \\{} & {} \asymp \int _0^1 \left( \int _r^1 M_{\infty }(t,f) \frac{\widehat{\omega }(t)}{1-t} dt\right) ^p \frac{1}{\widehat{\omega }(r)^{p}} \,dr\nonumber \\{} & {} \lesssim \Vert f\Vert ^p_{H(\infty ,p)} \end{aligned}$$

(3.21)

On the other hand, by [17, Theorem 1],

$$\begin{aligned}{} & {} \int _0^1 \left( \int _0^r \frac{ M_{\infty }(t,f) }{\widehat{\omega }(rt) (1-rt)^{1-\frac{1}{p}}} \frac{\widehat{\omega }(t)}{1-t} dt\right) ^p (1-r)^{p-1} \,dr\nonumber \\{} & {} \asymp \int _0^1 \left( \int _0^r \frac{ M_{\infty }(t,f) }{ (1-t)^{2-\frac{1}{p}}} dt\right) ^p (1-r)^{p-1} \,dr \nonumber \\{} & {} \lesssim \Vert f\Vert ^p_{H(\infty ,p)}, \end{aligned}$$

(3.22)

where in the last inequality we have used that $\sup \nolimits _{0<r<1} \left( \int _r^1(1-t)^{p-1}\right) ^{\frac{1}{p}} \left( \int _0^r(1-t)^{-1-p'}\right) ^{\frac{1}{p'}}<\infty$. So, joining Lemma 15, (3.20), (3.21) and (3.22), we get (3.18).

Third Step. Since $\omega \in \widehat{\mathcal {D}}$, by Lemma 9

$$\begin{aligned} \Vert \widetilde{H_\omega }(f)\Vert _{D^p_{p-1}}\asymp \Vert \widetilde{H_\omega }(f)\Vert _{Y_p}, \quad f\in X_p, \end{aligned}$$

for $Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$. This, together with (3.13), (3.18) and Lemma 8 implies

$$\begin{aligned} \begin{aligned} \Vert T(f)\Vert _{Y_p}&\lesssim \Vert f\Vert _{X_p}+ \Vert \widetilde{H_\omega }(f)\Vert _{Y_p}\\&\asymp \Vert f\Vert _{X_p}+ \Vert \widetilde{H_\omega }(f)\Vert _{D^p_{p-1}}\\&\lesssim \Vert f\Vert _{X_p}+\Vert f\Vert _{H(\infty ,p)}\\&\lesssim \Vert f\Vert _{X_p}, \quad f\in X_p. \end{aligned} \end{aligned}$$

This finishes the proof. $\square$

3.3 $\varvec{H_\omega : X_p\rightarrow Y_p}$ versus $\varvec{H_\omega : L^p_{[0,1)}\rightarrow Y_p,\, 1<p<\infty }$

An additional byproduct of Theorem 1 is the following improvement of [26, Theorem 3].

Proof of Corollary 3

(i)$\Rightarrow$(ii). By Lemma 8, $T:Y_p \rightarrow Y_p$ is bounded, and so by Theorem 1, $\omega \in \mathcal {D}$. Next, by the proofs of [26, Theorems 3 and 4] we obtain $m_p(\omega )<\infty$.

(ii)$\Rightarrow$(iii) is clear. Finally, (iii)$\Rightarrow$(i) follows from Lemma 8 and [26, Theorem 3].

Putting together Lemma 8, Theorem 1 and Corollary 3, we deduce the following result.

Corollary 17

Let $\omega$ be a radial weight and $1<p<\infty$. Let $X_p,Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$ and let $T \in \{H_{\omega }, \widetilde{H_{\omega }}\}$. If there exists $C>0$ such that

$$\begin{aligned} \omega (t)\le C \frac{\widehat{\omega }(t)}{1-t}\text {for all} 0\le t <1. \end{aligned}$$

(3.23)

Then, the following statements are equivalent:

• (i) $T:L^p_{[0,1)}\rightarrow Y_p$ is bounded;

• (ii) $T: X_p \rightarrow Y_p$ is bounded;

• (iii) $\omega \in \mathcal {D}$ and $M_{p,c}(\omega )<\infty$;

• (iv) $\omega \in \widehat{\mathcal {D}}$ and $M_{p,c}(\omega )<\infty$;

• (v) $\omega \in \mathcal {D}$ and $m_p(\omega )<\infty$;

• (vi) $\omega \in \widehat{\mathcal {D}}$ and $m_p(\omega )<\infty$.

Proof

The implication (i)$\Rightarrow$(ii) follows from Lemma 8, and (ii)$\Leftrightarrow$(iii)$\Leftrightarrow$(iv) follows from Theorem 1. Next, (iii)$\Rightarrow$(v) is a byproduct of the hypothesis (3.23). The equivalence (v)$\Leftrightarrow$(vi) and the implication (vi)$\Rightarrow$(i) have been proved in Corollary 3. This finishes the proof. $\square$

Next, we will prove that there are weights $\omega \in \mathcal {D}$, such that $M_{p,c}(\omega )<\infty$ and $m_p(\omega )=\infty$, so in particular they do not satisfy (3.23). Consequently, the boundedness of the operator $H_\omega : L^p_{[0,1)}\rightarrow Y_p$ is not equivalent to the boundedness of the the operator $H_\omega : X_p\rightarrow Y_p$, where $X_p,Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$. With this aim we prove the next result, which shows that despite its innocent looking condition, the class $\mathcal {D}$ has in a sense a complex nature.

Lemma 18

Let $1<p<\infty$ and $\nu \in \mathcal {D}$. Then, there exists $\omega \in \mathcal {D}$ such that

$$\begin{aligned} \widehat{\omega }(t)\asymp \widehat{\nu }(t), \quad t\in [0,1), \end{aligned}$$

$\omega \in L^{p'}_{[0,r_0]}$ for any $r_0\in (0,1)$ and $\omega \notin L^{p'}_{[0,1)}$.

Proof

By Lemma 14, $\widetilde{\nu }\in \mathcal {D}$. So, we can choose $K>1$ so that $\widetilde{\nu }$ satisfies (2.1). Next, consider the sequences $r_n= 1-\frac{1}{K^n},\,t_n=r_n+a_n$, with

$$\begin{aligned} 0<a_n<\min \left( r_{n+1}-r_n, \frac{\left( \widehat{\widetilde{\nu }}(r_n)\right) ^p}{(n+1)^{p-1}}\right) , \quad n\in \mathbb {N}\cup \{0\}. \end{aligned}$$

Let

$$\begin{aligned} \omega (t)=\sum _{n=0}^\infty h_n\chi _{[r_n,t_n]}(t)\widetilde{\nu }(t),\, t\in [0,1),\quad \text {where}\quad h_n=\frac{\widehat{\widetilde{\nu }}(r_n)-\widehat{\widetilde{\nu }}(r_{n+1})}{\widehat{\widetilde{\nu }}(r_n)-\widehat{\widetilde{\nu }}(t_{n})}, \,n\in \mathbb {N}\cup \{0\}. \end{aligned}$$

Observe that the sequence $\{h_n\}_{n=0}^\infty$ is well-defined because

$$\begin{aligned} \widehat{\widetilde{\nu }}(r_n)-\widehat{\widetilde{\nu }}(t_{n})=\int _{r_n}^{t_n} \frac{\widehat{\nu }(s)}{1-s}\,ds \ge \widehat{\nu }(t_n)\log \left( 1+\frac{a_n}{1-t_n} \right) >0,\,n\in \mathbb {N}\cup \{0\}. \end{aligned}$$

Moreover, $\omega$ is non-negative and

$$\begin{aligned} \int _0^1 \omega (t)\,dt= & {} \sum _{n=0}^\infty h_n\left( \widehat{\widetilde{\nu }}(r_n)-\widehat{\widetilde{\nu }}(t_{n})\right) \\ {}= & {} \sum _{n=0}^\infty \left( \widehat{\widetilde{\nu }}(r_n)-\widehat{\widetilde{\nu }}(r_{n+1})\right) =\widehat{\widetilde{\nu }}(0)=\int _0^1 \widetilde{\nu }(t)\, dt\asymp \int _0^1 \nu (t)\,dt<\infty , \end{aligned}$$

where in the last equivalence we have used Lemma 14.

Next, take $t\in [0,1)$ and $N\in \mathbb {N}\cup \{ 0\}$ such that $r_N\le t<r_{N+1}$. By Lemmas 14 and  6,

$$\begin{aligned} \widehat{\omega }(t)&\le \widehat{\omega }(r_N)= \sum _{n=N}^\infty \left( \widehat{\widetilde{\nu }}(r_n)-\widehat{\widetilde{\nu }}(r_{n+1})\right) =\widehat{\widetilde{\nu }}(r_{N})\asymp \widehat{\nu }(r_N)\lesssim \widehat{\nu }(t) \quad \text {and}\\ \widehat{\omega }(t)&\ge \widehat{\omega }(r_{N+1})=\sum _{n=N+1}^\infty \left( \widehat{\widetilde{\nu }}(r_n)-\widehat{\widetilde{\nu }}(r_{n+1})\right) =\widehat{\widetilde{\nu }}(r_{N+1})\asymp \widehat{\nu }(r_{N+1})\gtrsim \widehat{\nu }(t), \end{aligned}$$

so $\widehat{\omega }(t)\asymp \widehat{\nu }(t)$ and hence $\omega \in \mathcal {D}$.

It is clear that $\omega \in L^{p'}_{[0,r_0]}$ for any $r_0\in (0,1)$, so it only remains to prove that $\omega \notin L^{p^\prime }_{[0,1)}$. Bearing mind (2.1), we get that

$$\begin{aligned} h_n\asymp \frac{\widehat{\widetilde{\nu }}(r_n)}{\widehat{\widetilde{\nu }}(r_n)-\widehat{\widetilde{\nu }}(t_{n})}, \quad N\in \mathbb {N}\cup \{ 0\}. \end{aligned}$$

This, together with Lemma 6 and Hölder’s inequality, implies

$$\begin{aligned} \int _0^1 \omega (t)^{p^\prime }\,dt&=\sum _{n=0}^\infty h_n^{p^\prime } \int _{r_n}^{t_n}\left( \frac{\widehat{\nu }(t)}{1-t}\right) ^{p^\prime }\, dt \asymp \sum _{n=0}^\infty \left( \widehat{\widetilde{\nu }}(r_n)\right) ^{p^\prime } \frac{\int _{r_n}^{t_n}\left( \frac{\widehat{\nu }(t)}{1-t}\right) ^{p^\prime }\, dt}{\left( \int _{r_n}^{t_n} \frac{\widehat{\nu }(t)}{1-t}\, dt\right) ^{p^\prime }}\\&\ge \sum _{n=0}^\infty \left( \frac{\widehat{\widetilde{\nu }}(r_n)}{(t_n-r_n)^{1/p}}\right) ^{p^\prime } =\sum _{n=0}^\infty \left( \frac{\widehat{\widetilde{\nu }}(r_n)}{a_n^{1/p}}\right) ^{p^\prime }\ge \sum _{n=0}^\infty (n+1)=\infty . \end{aligned}$$

$\square$

Corollary 19

Let $1<p<\infty$ and $X_p,Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$. For each radial weight $\nu$ such that $Q: L^p_{[0,1)}\rightarrow Y_p$ is bounded, where $Q\in \{ H_\nu ,\widetilde{H_\nu }\}$, there is a radial weight $\omega$ such that

$$\begin{aligned} \widehat{\omega }(t)\asymp \widehat{\nu }(t),\quad t\in [0,1), \end{aligned}$$

$\omega \in L^{p'}_{[0,r_0]}$ for any $r_0\in (0,1)$, $T: X_p\rightarrow Y_p$ is bounded and $T: L^p_{[0,1)}\rightarrow Y_p$ is not bounded. Here $T\in \{ H_\omega ,\widetilde{H_\omega }\}$.

Proof

Since $Q: L^p_{[0,1)}\rightarrow Y_p$ is bounded, by Theorem 1, $\nu \in \mathcal {D}$ and $M_{p,c}(\nu )<\infty$. Now, by Lemma 18 there is a radial weight $\omega$ such that $\widehat{\omega }(t)\asymp \widehat{\nu }(t)$, $\omega \in L^{p'}_{[0,r_0]}$ for any $r_0\in (0,1)$ and $\omega \not \in L^{p'}_{[0,1)}$. So, $m_p(\omega )=\infty$ and by Corollary 3, $T: L^p_{[0,1)}\rightarrow Y_p$ is not bounded. Moreover, $\omega \in \mathcal {D}$ and $M_{p,c}(\omega )<\infty$ because $\nu$ satisfies both properties, so Theorem 1 yields $T: X_p\rightarrow Y_p$ is bounded. $\square$

3.4 Compactness of Hilbert-type operators on $X_p$-spaces. Case  $\varvec{1< p<\infty }$

For X, Y two Banach spaces, a sublinear operator $L: X\rightarrow Y$ is said to be compact provided L(A) has compact closure for any bounded set $A\subset X$. Once it has been understood the radial weights $\omega$ such that $H_\omega : X_p\rightarrow Y_p$ is bounded, $X_p,Y_p\in \{ H(\infty ,p),D^p_{p-1}, H^p, HL(p)\}$, $1<p<\infty$, it is natural to consider the analogous problem, replacing boundedness by compactness. Theorem 22 in this section answers this question, but firstly we need some previous results.

Lemma 20

Let $1< p< \infty$ and $\omega \in \mathcal {D}$ such that $\Vert \widetilde{\omega }\Vert _{L^{p'}_{[0,1)}}<\infty$. Let $\{f_k\}_{k=0}^\infty \subset X_p \in \{ H(\infty ,p),D^p_{p-1}, H^p, HL(p)\}$ such that $\sup \nolimits _{k\in \mathbb {N}} \Vert f_k\Vert _{X_p}<\infty$ and $f_k\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$. Then the following statements hold:

• (i) $\int _0^1 |f_k(t)|\omega (t)dt\rightarrow 0$ when $k\rightarrow \infty$.

• (ii) If $T\in \{H_\omega , \widetilde{H_\omega }\}$, then $T(f_k)\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$.

Proof

(i). Let $\varepsilon >0$. By hypothesis $\int _0^1 \widetilde{\omega }(t)^{p'} dt < \infty$, so there exists $0<\rho _0<1$ such that $\int _{\rho _0}^1 \widetilde{\omega }(t)^{p'} dt < \varepsilon$. Moreover, there exists $k_0$ such that for every $k\ge k_0$ and $z\in M=\overline{D(0,\rho _0)}$, $|f_k(z)|<\varepsilon$. Then, by Lemma 14, (3.11), and Hölder inequality

$$\begin{aligned} \int _0^1|f_k(t)|\omega (t)dt&\le |f_k(0)|\widehat{\omega }(0) + \int _0^1 M_{\infty }(t,f_k)\widetilde{\omega }(t)dt\\&\lesssim \int _0^{\rho _0} M_{\infty }(t,f_k)\widetilde{\omega }(t)dt + \int _{\rho _0}^1 M_{\infty }(t,f_k)\widetilde{\omega }(t)dt\\&<\varepsilon \int _0^{\rho _0}\widetilde{\omega }(t)dt + \sup _{k\in \mathbb {N}} \Vert f_k\Vert _{H(\infty ,p)} \int _{\rho _0}^1 \widetilde{\omega }(t)^{p'} dt\\&<\varepsilon \left( \int _0^{1}\widetilde{\omega }(t)dt + \sup _{k\in \mathbb {N}} \Vert f_k\Vert _{H(\infty ,p)}\right) =C\varepsilon , \end{aligned}$$

where in the last step we have used Lemma 8.

(ii). Let be $M\subset \mathbb {D}$ a compact set and $K_t^\omega (z)=\frac{1}{z}\int _0^z B_t^\omega (u)\,du$. If $z\in M$

$$\begin{aligned} |T(f_k)(z)| \le \int _0^1 |f_k(t)|\,|K_t^\omega (z)|\omega (t)dt \le \sup _{\begin{array}{c} z\in M\\ t\in [0,1) \end{array}} |K_t^\omega (z)|\int _0^1 |f_k(t)|\omega (t)\,dt. \end{aligned}$$

Since, $M\subset \overline{D(0,\rho _0)}$, for some $\rho _0\in (0,1)$, then

$$\begin{aligned} \sup _{\begin{array}{c} z\in M\\ t\in [0,1) \end{array}} |K_t^\omega (z)|&=\sup _{\begin{array}{c} z\in M\\ t\in [0,1) \end{array}} \left| \sum _{k=0}^\infty \frac{t^kz^k}{2(k+1)\omega _{2k+1}} \right| \le \sum _{k=0}^\infty \frac{{\rho _0}^k}{2(k+1)\omega _{2k+1}}= C(\omega ,\rho _0)<\infty , \end{aligned}$$

so, by (i), $T(f_k)\rightarrow 0$ uniformly on M. This finishes the proof. $\square$

Theorem 21

Let $\omega$ be a radial weight, $1<p<\infty$, $X_p, Y_p \in \{ H(\infty ,p),D^p_{p-1}, H^p, HL(p)\}$ and let $T\in \{H_\omega , \widetilde{H_\omega }\}$. Then, the following assertions are equivalent:

• (i) $T: X_p \rightarrow Y_p$ is compact;

• (ii) For every sequence $\{f_k\}_{k=0}^\infty \subset X_p$ such that $\sup \nolimits _{k\in \mathbb {N}} \Vert f_k\Vert _{X_p}<\infty$ and $f_k\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$, $\lim \nolimits _{k\rightarrow \infty } \Vert T(f_k)\Vert _{Y_p} =0$.

Proof

(i)$\Rightarrow$(ii). Let $\{f_n\}_{n=0}^\infty \subset X_p$ such that $\sup \nolimits _{n\in \mathbb {N}} \Vert f_n\Vert _{X_p}<\infty$ and $f_n\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$. Assume there exist $\varepsilon >0$ and a subsequence $\{n_k\}_k \subset \mathbb {N}$ such that

$$\begin{aligned} \Vert T(f_{n_k})\Vert _{Y_p} >\varepsilon , \quad \text {for any} k. \end{aligned}$$

(3.24)

Since T is compact, there exists a subsequence $\{n_{k_j}\}_j\subset \mathbb {N}$ and $g \in Y_p$ such that $\lim \nolimits _{j\rightarrow \infty } \Vert T(f_{n_{k_j}})-g\Vert _{Y_p}=0$. Moreover, Theorem 1 ensures that $\omega \in \mathcal {D}$ and $M_{p,c}(\omega )<\infty$, so $\Vert \widetilde{\omega }\Vert _{L^{p'}_{[0,1)}}<\infty$. Therefore Lemma 20, implies that $T(f_{n_{k_j}})\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$, so $\lim \nolimits _{j\rightarrow \infty } \Vert T(f_{n_{k_j}})\Vert _{Y_p}=0$ which yields a contradiction with (3.24).

(ii)$\Rightarrow$(i). Let $\{f_n\}\subset X_p$ such that $\sup \nolimits _{n\in \mathbb {N}}\Vert f_n\Vert _{X_p}<\infty .$ Then, $\{f_n\}$ is uniformly bounded on compact subsets of $\mathbb {D}$. Then, by Montel’s Theorem there exists $\{f_{n_k}\}_k$ and $f\in \mathcal {H}(\mathbb {D})$ such that $f_{n_k}\rightarrow f$ uniformly on compact subsets of $\mathbb {D}$. Let $g_{n_k}=f_{n_k}-f$, then $g_{n_k}\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$ and $\sup \nolimits _{k\in \mathbb {N}}\Vert g_{n_k}\Vert _{X_p}<\infty$. Therefore, by hypothesis $\lim \nolimits _{k\rightarrow \infty } \Vert T(g_{n_k})\Vert _{Y_p} =0$, that is, T is compact. $\square$

Theorem 22

Let $\omega$ be a radial weight, $1< p<\infty$, $X_p,Y_p\in \{ H(\infty ,p),D^p_{p-1}, H^p, HL(p)\}$, and let $T\in \{H_{\omega }, \widetilde{H_{\omega }}\}$. Then, $T:X_p\rightarrow Y_p$ is not compact.

Proof

Assume that $T:X_p\rightarrow Y_p$ is compact. For each $0<a<1$, set

$$\begin{aligned} f_a(z)&=\left( \frac{1-a^2}{(1-az)^2}\right) ^{1/p}=\sum _{n=0}^\infty \widehat{f_a}(n)z^n,\quad z\in \mathbb D, \end{aligned}$$

where $\widehat{f_a}(n)=(1-a^2)^{1/p}\frac{\Gamma (n+2/p)}{n!\Gamma (2/p)} a^n \ge 0$. So, by Stirling’s formula

$$\begin{aligned} \widehat{f_a}(n)\asymp (1-a^2)^{1/p}(n+1)^{2/p -1}a^n,\quad n\in \mathbb {N}\cup \{0\}. \end{aligned}$$

(3.25)

Consequently, $\Vert f_a\Vert _{HL(p)}\asymp 1,\, a\in (0,1)$. Moreover, $\Vert f_a\Vert _{H(\infty ,p)}\asymp \Vert f_a\Vert _{D^p_{p-1}}\asymp \Vert f_a\Vert _{H^p}=1.$ Furthermore, it is clear that $f_a\rightarrow 0$, as $a\rightarrow 1$ uniformly on compact subsets of $\mathbb {D}$, and $H_\omega (f_a)=\widetilde{H_\omega }(f_a)$. Since $T:X_p\rightarrow Y_p$ is compact, $\omega \in \widehat{\mathcal {D}}$ by Theorem 1. So, Lemma 9 implies that

$$\begin{aligned} \Vert T(f_a)\Vert _{Y_p}\asymp \Vert T(f_a)\Vert _{HL(p)}, \quad a\in (0,1). \end{aligned}$$

Therefore, by using (3.25) we have

$$\begin{aligned} \Vert H_{\omega }(f_a)\Vert ^p_{Y_p}\gtrsim & {} \Vert H_{\omega }(f_a)\Vert ^p_{HL(p)}=\sum _{n=0}^\infty (n+1)^{p-2}\left( \frac{\sum \nolimits _{k=0}^\infty \widehat{f_a}(k)\omega _{k+n}}{2(n+1)\omega _{2n+1}} \right) ^p\nonumber \\\gtrsim & {} (1-a^2)\sum _{n=0}^\infty \frac{1}{(n+1)^{2}}\left( \sum \limits _{k=0}^\infty (k+1)^{2/p-1}a^k\,\frac{\omega _{k+n}}{\omega _{2n}}\right) ^p\nonumber \\\ge & {} (1-a)\sum _{n=0}^\infty \frac{1}{(n+1)^{2}}\left( \sum \limits _{k=0}^n (k+1)^{2/p-1}a^k\,\frac{\omega _{k+n}}{\omega _{2n}}\right) ^p\nonumber \\\ge & {} (1-a)\sum _{n=0}^\infty \frac{1}{(n+1)^{2}}\left( \sum \limits _{k=0}^n (k+1)^{2/p-1}a^k\right) ^p\nonumber \\\ge & {} (1-a)\sum _{n=0}^\infty \frac{a^{pn}}{(n+1)^{2}}\left( \sum \limits _{k=0}^n (k+1)^{2/p-1}\right) ^p\nonumber \\\gtrsim & {} (1-a)\sum _{n=0}^\infty a^{pn}=\frac{1-a}{1-a^p}\gtrsim 1, \end{aligned}$$

(3.26)

so using Theorem 21 we deduce that $T: X_p\rightarrow Y_p$ is not a compact operator. $\square$

4 Hilbert type operators acting on $X_1$-spaces

The first result of this section gives the equivalence of conditions (iii)–(vi) of Theorem 2.

Lemma 23

Let $\omega \in \widehat{\mathcal {D}}$. Then, the following conditions are equivalent:

• (i) $K_{1,c}(\omega )=\sup \nolimits _{a \in [0,1)} \frac{1}{1-a}\int _a^1 \omega (t)\left( 1+\int _0^t \frac{ds}{\widehat{\omega }(s)}\right) \,dt<\infty ;$

• (ii) $K_{1,d}(\omega )= \sup \nolimits _{a \in [0,1)} \frac{\widehat{\omega }(a)}{1-a} \left( 1+\int _0^a \frac{ds}{\widehat{\omega }(s)}\right) <\infty ;$

• (iii) $M_1(\omega )=\sup \nolimits _{N \in \mathbb {N}} (N+1)\omega _{2N}\sum \nolimits _{k=0}^{N}\frac{1}{(k+1)^2 \omega _{2k}}<\infty$.

Moreover,

$$\begin{aligned} K_{1,c}(\omega )\asymp K_{1,d}(\omega )\asymp M_1(\omega ), \end{aligned}$$

(4.1)

and $\omega \in \check{\mathcal {D}}$ when $\omega$ satisfies any of the three previous conditions.

Observe that for any radial weight, $K_{1,c}(\omega )<\infty$ holds if and only if $M_{1,c}(\omega )<\infty$, and analogously $K_{1,d}(\omega )<\infty$ if and only if $M_{1,d}(\omega )<\infty$. This fact will be used repeatedly throughout the paper.

Proof

On the one hand,

$$\begin{aligned} \frac{\widehat{\omega }(a)}{1-a} \left( 1+\int _0^a \frac{ds}{\widehat{\omega }(s)}\right)&\le \frac{1}{1-a} \int _a^1 \omega (t)\left( 1+\int _0^t \frac{ds}{\widehat{\omega }(s)}\right) \,dt \\&\le K_{1,c}(\omega ), \quad a\in [0,1), \end{aligned}$$

so (i)$\Rightarrow$(ii) and $K_{1,d}(\omega )\lesssim K_{1,c}(\omega )$.

On the other hand assume that (ii) holds. Since $\omega \in \widehat{\mathcal {D}}$, [22, Lemma 3(ii)] (for $\nu (t)=1$) yields

$$\begin{aligned} \frac{1}{1-a}\int _a^1 \omega (t)\left( 1+\int _0^t \frac{ds}{\widehat{\omega }(s)}\right) \,dt&\le K_{1,d}(\omega ) \left( \frac{1}{1-a}\int _a^1\frac{ \omega (t) (1-t)}{\widehat{\omega }(t)} dt\right) \\&\lesssim K_{1,d}(\omega ),\quad 0<a<1, \end{aligned}$$

that is (i) holds and $K_{1,c}(\omega )\lesssim K_{1,d}(\omega )$. Finally, by mimicking the proof of Lemma 13,

$$\begin{aligned} K_{1,d}(\omega )\asymp M_1(\omega ), \end{aligned}$$

so (ii)$\Leftrightarrow$(iii) and (4.1) holds.

Next, for any $K>1$ and $r\in (0,1)$

$$\begin{aligned} \begin{aligned} M_{1,d}(\omega )&\ge \frac{K\widehat{\omega }\left( 1-\frac{1-r}{K}\right) }{1-r} \int _r^{1-\frac{1-r}{K}} \frac{ds}{\widehat{\omega }(s)}\\&\ge (K-1)\frac{\widehat{\omega }\left( 1-\frac{1-r}{K}\right) }{\widehat{\omega }\left( r\right) }, \end{aligned} \end{aligned}$$

that is

$$\begin{aligned} \widehat{\omega }\left( r\right) \ge \frac{K-1}{M_{1,d}(\omega )}\widehat{\omega }\left( 1-\frac{1-r}{K}\right) ,\quad 0<r<1, \end{aligned}$$

so taking $K>M_{1,d}(\omega )+1$, we get $\omega \in \check{\mathcal {D}}$. This finishes the proof.$\square$

The following result will be used to prove the equivalence (ii)$\Leftrightarrow$(iii) of Theorem 2.

Proposition 24

Let $\mu$ be a finite positive Borel measure on [0, 1) and $X_1\in \{H(\infty ,1), HL(1)\}$. Then $\mu$ is a 1-Carleson measure for $X_1$ if and only if $\mu$ is a classical Carleson measure. Moreover,

$$\begin{aligned} \Vert I_d\Vert _{X_1\rightarrow L^1(\mu )}\asymp \sup _{a\in [0,1)}\frac{\mu ([a,1))}{1-a}. \end{aligned}$$

Proof

If $\mu$ is a 1-Carleson measure for $X_1$, then by (2.2) and (1.3), $\mu$ is a 1-Carleson measure for $H^1$. So, by [6, Theorem 9.3] and its proof, $\mu$ is a classical Carleson measure and

$$\begin{aligned} \sup _{a\in [0,1)}\frac{\mu ([a,1))}{1-a}\lesssim \Vert I_d\Vert _{X_1\rightarrow L^1(\mu )}. \end{aligned}$$

Conversely, if $\mu$ is a classical Carleson measure, two integration by parts yield

$$\begin{aligned} \int _{\mathbb {D}} |f(z)|d \mu (z)=\int _0^1 |f(t)|d \mu (t) \le \int _0^1 M_\infty (t,f)\,d \mu (t) \lesssim \Vert f\Vert _{H(\infty ,1)} \sup _{a\in [0,1)}\frac{\mu ([a,1))}{1-a}. \end{aligned}$$

This inequality, together with Lemma 8, finishes the proof. $\square$

We introduce some more notation to prove Theorem 2. For any $C^\infty$-function $\Phi :\mathbb {R}\rightarrow \mathbb {C}$ with compact support, define the polynomials

$$\begin{aligned} W_n^\Phi (z)=\sum _{k\in \mathbb Z}\Phi \left( \frac{k}{n}\right) z^{k},\quad n\in \mathbb {N}. \end{aligned}$$

A particular case of the previous construction is useful for our purposes. Some properties of these polynomials have been gathered in the next lemma, see [12, Section 2] or [19, p. 143–144] for a proof.

Lemma 25

Let $\Psi :\mathbb {R}\rightarrow \mathbb {R}$ be a $C^\infty$-function such that $\Psi \equiv 1$ on $(-\infty ,1]$, $\Psi \equiv 0$ on $[2,\infty )$ and $\Psi$ is decreasing and positive on (1, 2). Set $\psi (t)=\Psi \left( \frac{t}{2}\right) -\Psi (t)$ for all $t\in \mathbb {R}$. Let $V_{0}(z)=1+z$ and

$$\begin{aligned} V_{n}(z)=W^\psi _{2^{n-1}}(z)=\sum _{j=0}^\infty \psi \left( \frac{j}{2^{n-1}}\right) z^j=\sum _{j=2^{n-1}}^{2^{n+1}-1} \psi \left( \frac{j}{2^{n-1}}\right) z^j,\quad n\in \mathbb {N}. \end{aligned}$$

Then,

$$\begin{aligned} f(z)=\sum _{n=0}^\infty (V_{n}*f)(z),\quad z\in \mathbb {D},\quad f\in \mathcal {H}(\mathbb {D}), \end{aligned}$$

(4.2)

and for each $0<p<\infty$ there exists a constant $C=C(p,\Psi )>0$ such that

$$\begin{aligned} \Vert V_{n}*f\Vert _{H^p}\le C\Vert f\Vert _{H^p},\quad f\in H^p, \quad n \in \mathbb {N}. \end{aligned}$$

(4.3)

In addition

$$\begin{aligned} \Vert V_n\Vert _{H^p}\asymp 2^{n(1-1/p)}, \quad 0< p<\infty . \end{aligned}$$

(4.4)

Let us denote $f_r(z)=f(rz)$, $z\in \mathbb {D}$, $r\in (0,1)$. Now we are ready to prove the main theorem of this section.

Proof of Theorem 2

First of all, recall that $M_{1,c}(\omega )<\infty$ if and only if $K_{1,c}(\omega )<\infty$ and analogously $M_{1,d}(\omega )<\infty$ if and only if $K_{1,d}(\omega )< \infty$, so that the equivalences (iii)$\Leftrightarrow$(iv)$\Leftrightarrow$(v)$\Leftrightarrow$(vi) follow from Lemma 23. The equivalence between (ii) and (iii) is a consequence of [6, Theorem 9.3] when $X_1=H^1$, [29, Theorem 2.1] when $X_1=D^1_0$ and Proposition 24 when $X_1\in \{H(\infty ,1),HL(1)\}$.

(i)$\Rightarrow$(iii). In order to obtain both conditions, $\omega \in \widehat{\mathcal {D}}$ and $M_{1,c}(\omega )<\infty$, we are going to deal with functions $f \in \mathcal {H}(\mathbb {D})$ such that $\widehat{f}(n) \ge 0$ for all $n \in \mathbb {N}\cup \{0\}$, so it is enough to prove the result for $T= H_{\omega }$.

First Step. Let us prove $\omega \in \widehat{\mathcal {D}}$. Bearing in mind Lemma 8 and (1.4)

$$\begin{aligned} \sum \limits _{n=0}^{\infty }\frac{\omega _{n+N}}{(n+1)^2 \omega _{2n+1}} \left( \sum \limits _{k=0}^{N}\widehat{f}(k)\right){} & {} \le \Vert H_{\omega }(f)\Vert _{H(\infty ,1)} \lesssim \Vert H_{\omega }(f)\Vert _{Y_1} \lesssim \Vert f\Vert _{X_1}\nonumber \\{} & {} \lesssim \Vert f \Vert _{D_0^1}, \quad N\in \mathbb {N}, \end{aligned}$$

(4.5)

for any $f\in \mathcal {H}(\mathbb {D})$ such that $\widehat{f}(n)\ge 0$, $n\in \mathbb {N}\cup \{0\}$. Next, for each $N \in \mathbb {N}$, consider the test functions $f_{\alpha , N}(z)=\sum \nolimits _{k=0}^N (k+1)^{\alpha }z^k,\, \alpha >0$. Set $M \in \mathbb {N}$ such that $2^M <N \le 2^{M+1}$. Then, bearing in mind (4.2),

$$\begin{aligned} (f_{\alpha , N}')_s(z)=\sum \limits _{n=0}^\infty (V_{n}*(f_{\alpha ,N}')_s)(z)=\sum \limits _{n=0}^M (V_{n}*(f_{\alpha ,N}')_s)(z), \end{aligned}$$

which together with [16, Lemma 3.1], [19, Lemma 5.4] and (4.4) gives

$$\begin{aligned} \Vert f_{\alpha , N}\Vert _{D_0^1}&\le \int _0^1 M_1(s, f_{\alpha ,N}')ds \le \sum \limits _{n=0}^M \int _0^1 \Vert V_{n}*(f_{\alpha ,N}')_s \Vert _{H^1}ds\\&\lesssim \sum \limits _{n=0}^M \int _0^1 s^{2^{n-1}} 2 ^{n(\alpha +1)} \Vert V_n\Vert _{H^1} ds \asymp \sum \limits _{n=0}^M 2^{n\alpha }\asymp 2^{M\alpha } \asymp (N+1)^{\alpha }. \end{aligned}$$

Testing the functions $f_{\alpha , N}$ in (4.5), $\sup \nolimits _{N\in \mathbb {N}}(N+1)\sum \nolimits _{n=0}^{\infty }\frac{\omega _{n+N}}{(n+1)^2 \omega _{2n+1}} <\infty .$ Therefore, there exists $C=C(\omega )>0$

$$\begin{aligned} \frac{\omega _{8N}}{\omega _{12N}} \lesssim \frac{\omega _{8N}}{\omega _{12N}} (N+1)\sum \limits _{n=6N}^{7N}\frac{1}{(n+1)^2} \le (N+1) \sum \limits _{n=6N}^{7N}\frac{\omega _{n+N}}{(n+1)^2 \omega _{2n+1}} \le C. \end{aligned}$$

So, arguing as in the first step proof of Proposition 12, $\omega \in \widehat{\mathcal {D}}.$

Second Step. We will prove $M_{1,c}(\omega )<\infty$. Let us consider the test functions $f_a(z)=\frac{1-a^2}{(1-az)^2}$, $a\in (0,1)$. A calculation shows that $\Vert f_a\Vert _{D^1_0}\asymp 1$, $a\in (0,1)$. Then, by Lemma 8 and (1.3),

$$\begin{aligned} \Vert H_\omega \Vert _{X_1\rightarrow Y_1}\gtrsim \sup _{a\in (0,1)} \Vert H_{\omega }(f_a)\Vert _{Y_1} \gtrsim \sup _{a\in (0,1)} \Vert H_{\omega }(f_a)\Vert _{L^1_{[0,1)}}. \end{aligned}$$

Consequently, using that $\omega \in \widehat{\mathcal {D}}$ and mimicking the proof (4.2) of [26, Theorem 2], we get $M_{1,c}(\omega )<\infty .$

Now let us prove (iv) $\Rightarrow$ (i). Firstly, observe that the condition $M_{1,c}(\omega )<\infty$ implies $K_{1,c}(\omega )<\infty$ so that $\widetilde{\omega }(t)=\frac{\widehat{\omega }(t)}{1-t}$ is bounded on [0, 1). So, using Lemma  8 and (3.11),

$$\begin{aligned} \int _0^1 M_\infty (t,f)\omega (t)\,dt\lesssim \int _0^1 M_\infty (t,f)\widetilde{\omega }(t)\,dt\lesssim \Vert f\Vert _{H(\infty ,1)}\lesssim \Vert f\Vert _{X_1}, \end{aligned}$$

that is $H_\omega (f)\in \mathcal {H}(\mathbb {D})$ for any $f\in X_1$. Secondly, by (1.4) and Lemma 8, it is enough to prove the inequality

$$\begin{aligned} \Vert H_{\omega }(f) \Vert _{D^1_0}\lesssim \Vert f\Vert _{H(\infty ,1)},\quad f\in H(\infty ,1), \end{aligned}$$

to end the proof. Indeed,

$$\begin{aligned} \Vert H_{\omega }(f) \Vert _{D^1_0}&\le \int _0^1 M_1(s, H_{\omega }(f)') ds \le \int _0^1 \left( \int _0^1 |f(t)| \, \omega (t) M_1(s, G^\omega _t ) dt \right) ds\\&= \int _0^1 |f(t)| \omega (t) \left( \int _0^1 M_1(s, G^\omega _t ) ds \right) dt . \end{aligned}$$

Then by (3.19) and [26, Lemma B]

$$\begin{aligned} M_1(s,G^\omega _t) \asymp 1+\int _0^{st}\frac{dx}{\widehat{\omega }(x)(1-x)}, \quad 0\le s,t<1. \end{aligned}$$

Bearing in mind that $M_{1,c}(\omega )<\infty$ implies $K_{1,c}(\omega )<\infty$ and applying Proposition 24, the measure $\mu _\omega$ defined as $d\mu _{\omega }(z)= \omega (z)\left( 1+\int _0^{|z|} \frac{ds}{\widehat{\omega }(s)}\right) \,\chi _{[0,1)}(z)\, dA(z)$ is a 1-Carleson measure for $H(\infty , 1)$, so by Tonelli’s theorem,

$$\begin{aligned} \Vert H_{\omega }(f) \Vert _{D^1_0}&\lesssim \int _0^1 |f(t)| \, \omega (t) \left( 1+ \int _0^1 \left( \int _0^{st} \frac{dx}{\widehat{\omega }(x)(1-x)} \right) ds \right) dt \nonumber \\&= \int _0^1 |f(t)| \, \omega (t) \left( 1+ \int _0^t \frac{(1-\frac{x}{t})}{\widehat{\omega }(x)(1-x)} dx \right) dt \nonumber \\&\le \int _0^1 |f(t)| \, \omega (t) \left( 1+ \int _0^t \frac{dx}{\widehat{\omega }(x)} \right) dt \lesssim \Vert f\Vert _{H(\infty ,1)}. \end{aligned}$$

(4.6)

This finishes the proof. $\square$

4.1 $\varvec{H_\omega : X_1\rightarrow Y_1}$ versus $\varvec{H_\omega : L^1_{[0,1)}\rightarrow Y_1}$

Firstly, we will study the boundedness of $T: L^1_{[0,1)}\rightarrow Y_1$, $T\in \{H_\omega ,\widetilde{H_{\omega }}\}$, $Y_1\in \{H(\infty ,1),H^1, D^1_{0}, HL(1)\}$.

Theorem 26

Let $\omega$ be a radial weight, let $Y_1\in \{H(\infty ,1),H^1, D^1_{0}, HL(1)\}$ and let $T \in \{ H_{\omega }, \widetilde{H_{\omega }}\}$. Then the following statements are equivalent:

• (i) $T: L^1_{[0,1)}\rightarrow Y_1$ is bounded;

• (ii) $\omega \in \mathcal {D}$ and $m_1(\omega )= \mathop {\mathrm {ess\,sup}}\limits \nolimits _{t \in [0,1)} \omega (t)\left( 1+\int _0^t \frac{ds}{\widehat{\omega }(s)}\right) <\infty .$

• (iii) $\omega \in \widehat{\mathcal {D}}$ and $m_1(\omega )= \mathop {\mathrm {ess\,sup}}\limits \nolimits _{t \in [0,1)} \omega (t)\left( 1+\int _0^t \frac{ds}{\widehat{\omega }(s)}\right) <\infty .$

Proof

(i)$\Rightarrow$(ii). If (i) holds, then $\omega \in \mathcal {D}$ by Lemma 8 and Theorem 2. Next, using Lemma 8 again and making minor modifications in the proof of [26, Theorem 2] we get

$$\begin{aligned} \Vert T (f)\Vert _{Y_1}\gtrsim \Vert H_\omega (f)\Vert _{L^1_{[0,1)}}\gtrsim \int _{0}^1 f(t)\omega (t)\left( 1+\int _0^t \frac{ds}{\widehat{\omega }(s)}\right) \,dt, \end{aligned}$$

for any $f\ge 0$, $f\in L^1_{[0,1)}$. So,

$$\begin{aligned} \int _{0}^1 f(t)\omega (t)\left( 1+\int _0^t \frac{ds}{\widehat{\omega }(s)}\right) \,dt\lesssim \Vert f\Vert _{L^1_{[0,1)}},\quad f\ge 0 \end{aligned}$$

which implies that $m_1(\omega )<\infty$.

(ii)$\Rightarrow$(iii) is clear.

(iii)$\Rightarrow$(i). If (iii) holds, then $H_\omega (f)\in \mathcal {H}(\mathbb {D})$ for any $f\in L^1_{[0,1)}$ and arguing as in (4.6)

$$\begin{aligned} \Vert H_{\omega }(f) \Vert _{D^1_0}&\lesssim \int _0^1 |f(t)| \, \omega (t) \left( 1+ \int _0^t \frac{dx}{\widehat{\omega }(x)} \right) dt \lesssim \Vert f\Vert _{L^1_{[0,1)}}. \end{aligned}$$

This together with (1.3) and Lemma 8 gives that $H_{\omega }: L^1_{[0,1)}\rightarrow Y_1$ is bounded. This finishes the proof. $\square$

Joining Theorems 2,  26 and Lemma 8 we deduce the following.

Corollary 27

Let $\omega$ be a radial weight, $X_1,Y_1\in \{H(\infty ,1),H^1, D^1_{0}, HL(1)\}$ and let $T\in \{H_{\omega }, \widetilde{H_{\omega }}\}$. If $\omega$ satisfies the condition (3.23), then the following statements are equivalent:

• (i) $T:L^1_{[0,1)}\rightarrow Y_1$ is bounded;

• (ii) $T: X_1 \rightarrow Y_1$ is bounded;

• (iii) $\omega \in \widehat{\mathcal {D}}$ and $M_{1,c}(\omega )<\infty$;

• (iv) $\omega \in \mathcal {D}$ and $M_{1,c}(\omega )<\infty$;

• (v) $\omega \in \widehat{\mathcal {D}}$ and $m_1(\omega )<\infty$;

• (vi) $\omega \in \mathcal {D}$ and $m_1(\omega )<\infty$.

Proof

(i)$\Rightarrow$(ii) follows from Lemma 8, and (ii)$\Leftrightarrow$(iii)$\Leftrightarrow$(iv) were proved in Theorem 2. Next, since $M_{1,c}(\omega )<\infty$ implies $K_{1,c}(\omega )<\infty$, (iii)$\Rightarrow$(v) is a byproduct of the hypothesis (3.23). Finally, the equivalences (v)$\Leftrightarrow$(vi)$\Leftrightarrow$(i) follow from Theorem 26. This finishes the proof. $\square$

A similar comparison between the conditions $M_{1,c}(\omega )<\infty$ and $m_1(\omega )<\infty$, to that made for the conditions $M_{p,c}(\omega )<\infty$ and $m_p(\omega )<\infty$, $1<p<\infty$, can also be considered. The following result shows that they are not equivalent.

Corollary 28

Let $X_1,Y_1\in \{H(\infty ,1),H^1, D^1_{0}, HL(1)\}$. For each radial weigth $\nu$ such that $Q: L^1_{[0,1)}\rightarrow Y_1$ is bounded, where $Q\in \{ H_\nu ,\widetilde{H_\nu }\}$, there is a radial weight $\omega$ such that

$$\begin{aligned} \widehat{\omega }(t)\asymp \widehat{\nu }(t),\quad t\in [0,1), \end{aligned}$$

$\omega \in L^\infty _{[0,r_0]}$ for any $r_0\in (0,1)$, $T: X_1\rightarrow Y_1$ is bounded and $T: L^1_{[0,1)}\rightarrow Y_1$ is not bounded. Here $T\in \{ H_\omega ,\widetilde{H_\omega }\}$.

Proof

Since $Q: L^1_{[0,1)}\rightarrow Y_1$ is bounded, by Theorem 2, $\nu \in \mathcal {D}$ and $M_{1,c}(\nu )<\infty$. Now, by Lemma 18 and its proof, there is a radial weight $\omega$ such that $\widehat{\omega }(t)\asymp \widehat{\nu }(t)$, $\omega \in L^\infty _{[0,r_0]}$ for any $r_0\in (0,1)$ and $\omega \not \in L^{\infty }_{[0,1)}$. So, $m_1(\omega )=\infty$ and by Theorem 26, $T: L^1_{[0,1)}\rightarrow Y_1$ is not bounded. Moreover, $\omega \in \mathcal {D}$ and $M_{1,c}(\omega )<\infty$ because $\nu$ satisfies both properties, consequently $T: X_1\rightarrow Y_1$ is bounded. $\square$

4.2 Compactness of Hilbert-type operators on $X_1$-spaces

Lemma 29

Let $\omega \in \mathcal {D}$ such that $M_{1,d}(\omega )<\infty$. Let $\{f_k\}_{k=0}^\infty \subset X_1 \in \{ H(\infty ,1),D^1_{0}, H^1, HL(1)\}$ such that $\sup \nolimits _{k\in \mathbb {N}} \Vert f_k\Vert _{X_1}<\infty$ and $f_k\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$. Then the following statements hold:

• (i) $\int _0^1 |f_k(t)|\omega (t)dt\rightarrow 0$ when $k\rightarrow \infty$.

• (ii) If $T\in \{H_\omega , \widetilde{H_\omega }\}$, then $T(f_k)\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$.

Proof

Firstly, let us prove that

$$\begin{aligned} \lim _{a\rightarrow 1^-} \frac{\widehat{\omega }(a)}{1-a}=0. \end{aligned}$$

(4.7)

Since $M_{1,d}(\omega )<\infty$, then

$$\begin{aligned}\begin{aligned} \int _0^{\frac{1+a}{2}} \frac{ds}{\widehat{\omega }(s)}&\ge \left( M_{1,d}(\omega )\right) ^{-1}\int _0^{\frac{1+a}{2}} \left( \int _{0}^s\frac{dt}{\widehat{\omega }(t)}\right) \frac{ds}{1-s}\\&\ge \left( M_{1,d}(\omega )\right) ^{-1}\int _{\frac{1}{2}}^{\frac{1+a}{2}} \left( \int _{0}^s\frac{dt}{\widehat{\omega }(t)}\right) \frac{ds}{1-s}\\&\ge \left( M_{1,d}(\omega )\right) ^{-1} \left( \int _{0}^{\frac{1}{2}}\frac{dt}{\widehat{\omega }(t)}\right) \int _{\frac{1}{2}}^{\frac{1+a}{2}} \frac{ds}{1-s}\\&= \left( M_{1,d}(\omega )\right) ^{-1}\left( \int _{0}^{\frac{1}{2}}\frac{dt}{\widehat{\omega }(t)}\right) \log \frac{1}{1-a}, \quad 0<a<1. \end{aligned} \end{aligned}$$

So $\lim _{a\rightarrow 1^-} \int _0^{a} \frac{ds}{\widehat{\omega }(s)}=\infty$, and then using again the condition $M_{1,d}(\omega )<\infty$, (4.7) holds.

From now on, the proof follow the lines of Lemma 20. Let $\varepsilon >0$. By (4.7) there exists $0<\rho _0<1$ such that $\widetilde{\omega }(t) < \varepsilon$ for any $t\in [\rho _0,1)$. Moreover, there exists $k_0$ such that for every $k\ge k_0$ and $z\in M=\overline{D(0,\rho _0)}$, $|f_k(z)|<\varepsilon$. Then, by Lemma 14 and (3.11)

$$\begin{aligned} \int _0^1|f_k(t)|\omega (t)dt&\le |f_k(0)|\widehat{\omega }(0) + \int _0^1 M_{\infty }(t,f_k)\widetilde{\omega }(t)dt\\&\lesssim \int _0^{\rho _0} M_{\infty }(t,f_k)\widetilde{\omega }(t)dt + \int _{\rho _0}^1 M_{\infty }(t,f_k)\widetilde{\omega }(t)dt\\&\le \varepsilon \left( \int _0^{1}\widetilde{\omega }(t)dt + \sup _{k\in \mathbb {N}} \Vert f_k\Vert _{X_1}\right) =C\varepsilon , \end{aligned}$$

where in the second to last step we have used Lemma 8.

The proof of (ii) is analogous to that of Lemma 20 so we omit its proof. $\square$

Using the previous lemma and Theorem 2 we obtain the following by mimicking the proof of Theorem 21.

Theorem 30

Let $\omega$ be a radial weight and $X_1, Y_1 \in \{ H(\infty ,1),D^1_{0}, H^1, HL(1)\}$ and let $T\in \{H_\omega , \widetilde{H_\omega }\}$. Then, the following assertions are equivalent:

• (i) $T: X_1 \rightarrow Y_1$ is compact;

• (ii) For every sequence $\{f_k\}_{k=0}^\infty \subset X_1$ such that $\sup \nolimits _{k\in \mathbb {N}} \Vert f_k\Vert _{X_1}<\infty$ and $f_k\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$, $\lim \nolimits _{k\rightarrow \infty } \Vert T(f_k)\Vert _{Y_1} =0$.

Theorem 31

Let $\omega$ be a radial weight and $X_1,Y_1\in \{ H(\infty ,1),D^1_{0}, H^1, HL(1)\}$, and let $T\in \{H_{\omega }, \widetilde{H_{\omega }}\}$. Then, $T:X_1\rightarrow Y_1$ is not compact.

Proof

The proof is analogous to that of Theorem 22, so we provide a sketch of the proof. Assume that $T:X_1\rightarrow Y_1$ is compact. For each $0<a<1$, set $f_a(z)=\frac{1-a^2}{(1-az)^2},\, z\in \mathbb {D}.$ A calculation shows that $\sup _{a\in (0,1)}\Vert f_a\Vert _{X_1}\asymp 1$ and $f_a\rightarrow 0$, as $a\rightarrow 1$ uniformly on compact subsets of $\mathbb {D}$. Moreover, since $T(f_a)$ has non-negative Taylor coefficients,

$$\begin{aligned} \Vert T(f_a)\Vert _{Y_1}\gtrsim \Vert H_{\omega }(f_a)\Vert _{H(\infty ,1)}\asymp \Vert H_{\omega }(f_a)\Vert _{HL(1)}\gtrsim 1,\quad a\in (0,1), \end{aligned}$$

where the last inequality follows taking $p=1$ in (3.26). So, using Theorem 30 we deduce that $T: X_1\rightarrow Y_1$ is not a compact operator. $\square$

5 Hilbert-type operators acting on $H^{\infty }$

We will prove a result which includes Theorem 4. With this aim we need some more notation. The space $Q_p$,  $0\le p<\infty$, consists of those $f\in H(\mathbb {D})$ such that

$$\begin{aligned} \Vert f\Vert ^2_{Q_p}=|f(0)|^2+\sup _{a\in \mathbb {D}} \int _\mathbb {D}|f'(z)|^2 (1-|\varphi _a(z)|^2)^p\,dA(z)<\infty , \end{aligned}$$

where $\varphi _a(z)= \frac{a-z}{1-\overline{a} z}, \; z, a \in \mathbb {D}.$ If $p>1$, $Q_p$ coincides with the Bloch space $\mathcal {B}$. The space $Q_1$ coincides with BMOA (see, e. g., [10, Theorem 5.2]). However, if $0<p<1$, $Q_p$ is a proper subspace of BMOA [30]. The space $Q_0$ reduces to the classical Dirichlet space $\mathcal {D}$.

We recall that

$$\begin{aligned} \begin{aligned}&Q_p\subsetneq \mathord \textrm{BMOA}\subsetneq \mathcal {B},\quad 0<p<1. \\ {}&H^\infty \subsetneq \mathord \textrm{BMOA}\subsetneq \mathcal {B}, \end{aligned} \end{aligned}$$

(5.1)

however if $0<p<1$, $H^\infty \not \subset Q_p$, and $Q_p \not \subset H^\infty$, see [30].

Moreover, $HL(\infty )\subsetneq Q_p$. This embedding might have been proved in some previous paper, however we include a short direct proof for the sake of completeness.

Lemma 32

Let $0<p\le \infty$, then $HL(\infty )\subsetneq Q_p$ and

$$\begin{aligned} \Vert f\Vert _{Q_p}\lesssim \Vert f\Vert _{HL(\infty )},\quad f\in \mathcal {H}(\mathbb {D}). \end{aligned}$$

Proof

Let $f\in HL(\infty )$, then

$$\begin{aligned} M_2^2(\rho ,f^\prime )= \sum _{n=1}^\infty n^2|\widehat{f}(n)|^2 \rho ^{2n-2}\le \Vert f\Vert _{HL(\infty )}^2\sum _{n=1}^\infty \rho ^{2n-2}=\frac{\Vert f\Vert _{HL(\infty )}^2}{1-\rho ^2}, \end{aligned}$$

So for any $0<p<\infty$,

$$\begin{aligned} \sup _{a\in \mathbb {D}}\int _\mathbb {D}(1-|\varphi _a(z)|^2)^p|f^\prime (z)|^2\,dA(z)&\lesssim \sup _{a\in \mathbb {D}} \int _0^1 \left( \frac{(1-|a|)(1-s^2)}{(1-|a|s)^2}\right) ^p M_2^2(s,f^\prime )\,ds\\&\le \Vert f\Vert _{HL(\infty )}^2 \sup _{a\in \mathbb {D}}\int _0^1 \frac{(1-|a|)^p(1-s^2)^{p-1}}{(1-|a|s)^{2p}}\, ds\\&\lesssim \Vert f\Vert _{HL(\infty )}^2, \end{aligned}$$

so $\Vert f\Vert _{Q_p}\lesssim \Vert f\Vert _{HL(\infty )}$. The lacunary series $f(z)=\sum _{k=0}^\infty 2^{-k}\log {(k+2)}z^{2^k}\in \bigcap \nolimits _{0<p} Q_p{\setminus } HL(\infty )$. This finishes the proof.$\square$

Now we will prove the main result of this section, which is an extension of Theorem 4.

Theorem 33

Let $\omega$ be a radial weight and let $T \in \{H_\omega , \widetilde{H_\omega }\}$. Then, the following statements are equivalent:

• (i) $T: H^{\infty }\rightarrow HL(\infty )$ is bounded;

• (ii) $T: H^{\infty }\rightarrow Q_p$ is bounded for $0<p<1$;

• (iii) $T: H^{\infty }\rightarrow \mathord \textrm{BMOA}$ is bounded;

• (iv) $T: H^{\infty }\rightarrow \mathcal {B}$ is bounded;

• (v) $\omega \in \widehat{\mathcal {D}}$.

Proof of Theorem 33

By Lemma 6,

$$\begin{aligned} \begin{aligned} \Vert H_{\omega }(f)\Vert _{HL(\infty )}&= \sup \limits _{k\in \mathbb {N}\cup \{0\}} (k+1)\left| \frac{\int _0^1 f(t)t^k \omega (t)dt}{2(k+1)\omega _{2k+1}}\right| \\&\le \sup \limits _{k\in \mathbb {N}\cup \{0\}} (k+1) \frac{\int _0^1 |f(t)|t^k \omega (t)dt}{2(k+1)\omega _{2k+1}}\\&=\Vert \widetilde{H_{\omega }}(f)\Vert _{HL(\infty )} \lesssim \Vert f\Vert _{H^{\infty }}, \end{aligned} \end{aligned}$$

so (v)$\Rightarrow$ (i). The implications (i)$\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(iv) follow from (5.1) and Lemma 32.

The implication (iv)$\Rightarrow$(v) was proved in [26, Theorem 1], and this finishes the proof. $\square$

It is worth mentioning that for $f(z)=\log \frac{1}{1-z}\in HL(\infty )$ and $\omega$ a radial weight,

$$\begin{aligned} \begin{aligned} H_\omega (f)'(x)&=\sum _{n=1}^\infty \frac{n}{2(n+1)\omega _{2n+1}}\left( \sum _{k=1}^\infty \frac{\omega _{n+k}}{k} \right) x^{n-1}\\&\ge \sum _{n=1}^\infty \frac{n}{2(n+1)}\left( \sum _{k=1}^n\frac{1}{k} \right) x^{n-1}\\&\gtrsim \sum _{n=1}^\infty \log (n+1) x^{n-1}, \end{aligned} \end{aligned}$$

so $H_\omega (f)\notin \mathcal {B}$. So, the space $H^\infty$ cannot be replaced by $HL(\infty )$ and by any $Q_p$ space, $0<p<\infty$, in the statement of Theorem 33. That is, the remaining cases for $p=\infty$, analogous to those of Theorems 1 and 2, which do not appear in Theorem 33, simply do not hold for any radial weight.

Finally, we will prove the analogous result to Theorem 22 for $p=\infty$.

Theorem 34

Let $\omega$ be a radial weight and let $T \in \{H_\omega , \widetilde{H_\omega }\}$. Then $T:H^{\infty }\rightarrow Y_\infty$ is not a compact operator, where $Y_\infty \in \{Q_p,\mathcal {B},\mathord \textrm{BMOA}, HL(\infty )\}$, $0<p<1$.

We need the following result, whose proof can be obtained by mimicking the proof Theorem 21.

Theorem 35

Let $\omega$ be a radial weight and $T\in \{H_\omega , \widetilde{H_\omega }\}$. Then, the following assertions are equivalent:

• (i) $T: H^{\infty }\rightarrow \mathcal {B}$ is compact;

• (ii) For every sequence $\{f_k\}_{k=0}^\infty \subset H^\infty$ such that $\sup \nolimits _{k\in \mathbb {N}} \Vert f_k\Vert _{H^\infty }<\infty$ and $f_k\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$, $\lim \nolimits _{k\rightarrow \infty } \Vert T(f_k)\Vert _\mathcal {B}=0$.

Proof of Theorem 34

By (5.1) and Lemma 32, it is enough to prove that $T: H^\infty \rightarrow \mathcal {B}$ is not compact. Let consider for every $k\in \mathbb {N}$ the function $f_k(z)=z^k$, $z\in \mathbb {D}$. It is clear that $\Vert f_k\Vert _{H^\infty }=1$ for every $k\in \mathbb {N}$ and $f_k\rightarrow 0$ uniformly on compact subsets of $\mathbb {D}$. Since

$$\begin{aligned} T(f_k)(z)=\sum _{n=0}^\infty \frac{\int _0^1 f_k(t)t^n\omega (t)\,dt}{2(n+1)\omega _{2n+1}}z^n=\sum _{n=0}^\infty \frac{\omega _{n+k}}{2(n+1)\omega _{2n+1}}z^n, \end{aligned}$$

for any $k\ge 2$

$$\begin{aligned} \Vert T(f_k)\Vert _{\mathcal {B}}&\ge \sup _{x\in (0,1)}(1-x)\sum _{n=1}^\infty \frac{n\omega _{n+k}}{2(n+1)\omega _{2n+1}}x^{n-1} \ge \frac{1}{4}\sup _{x\in (0,1)}(1-x)\sum _{n=1}^\infty \frac{\omega _{n+k}}{\omega _{2n}}x^{n-1}\\&\ge \frac{1}{4}\sup _{x\in (0,1)}(1-x)\sum _{n=k}^{2k} \frac{\omega _{n+k}}{\omega _{2n}}x^{n-1} \ge \frac{1}{4}\sup _{x\in (0,1)}(1-x)\sum _{n=k}^{2k}x^{n-1}\\&\ge \frac{1}{4}\sup _{x\in (0,1)}(1-x)kx^{2k-1} \ge \frac{1}{8}\left( 1-\frac{1}{2k}\right) ^{2k-1}\ge \frac{1}{8}\inf _{m\ge 2}\left( 1-\frac{1}{m}\right) ^{m}\ge C>0 \end{aligned}$$

so $\lim \nolimits _{k\rightarrow \infty }\Vert T(f_k)\Vert _\mathcal {B}\ne 0$ and hence, by Theorem 35, $T: H^\infty \rightarrow \mathcal {B}$ is not compact.$\square$

Before ending this section, we briefly compare the action of the Hilbert-type operator $H_\omega$ and the Bergman projection

$$\begin{aligned} P_\omega (f)(z)=\int _{\mathbb {D}}f(\zeta ) \overline{B^\omega _{z}(\zeta )}\,\omega (\zeta )dA(\zeta ), \end{aligned}$$

induced by a radial weight $\omega$. As a consequence of Theorem 33 and [23, Theorem 1], the condition $\omega \in \widehat{\mathcal {D}}$ characterizes the boundedness of the operators $H_{\omega }:H^{\infty }\rightarrow \mathcal {B}$ and $P_{\omega }:L^{\infty }\rightarrow \mathcal {B}$. Moreover, $P_{\omega }:L^{\infty }\rightarrow \mathcal {B}$ is bounded and onto if and only if $\omega \in \mathcal {D}$ [23, Theorem 3]. So, it is natural to think about the radial weights such that the operator $H_{\omega }:H^{\infty }\rightarrow \mathcal {B}$ is bounded and onto. A straightforward argument proves there is no radial weights such that $H_{\omega }:H^{\infty }\rightarrow \mathcal {B}$ satisfies both properties: If $H_{\omega }:H^{\infty }\rightarrow \mathcal {B}$ is bounded, Theorem 33 yields that $H_{\omega }:H^{\infty }\rightarrow \mathord \textrm{BMOA}$ is also bounded, so if $g\in \mathcal {B}\setminus \mathord \textrm{BMOA}$, e.g. $g(z)=\sum \nolimits _{k=0}^\infty z^{2^k}$, there does not exist $f\in H^\infty$ such that $H_{\omega }(f)=g$. Consequently, $H_{\omega }:H^{\infty }\rightarrow \mathcal {B}$ is not surjective.

6 Comparisons and reformulations of the $M_{p,c}$-conditions

In order to prove Theorem 5 we will study the relationship between some of the conditions which describe the boundedness of the Hilbert-type operators $H_\omega$ and $\widetilde{H_\omega }$ from $X_p$ to $Y_p$, and $X_q$ to $Y_q$.

Theorem 5

Firstly, assume $1<q<p<\infty$. Since $T:X_q\rightarrow Y_q$ is bounded, Theorem 1 yields $\omega \in \widehat{\mathcal {D}}$ and $M_{q,c}(\omega )<\infty$, and as a consequence, $K_{q,c}(\omega )<\infty$. By using Lemma 6,

$$\begin{aligned} 1+\int _0^r \frac{ds}{\widehat{\omega }(s)^q}\gtrsim \frac{1-r}{\widehat{\omega }(r)^q},\quad 0\le r<1. \end{aligned}$$

(6.1)

On the other hand, Hölder’s inequality with exponents $x=\frac{q'}{p'}>1$ and $x'=\frac{x}{x-1}$, implies

$$\begin{aligned} \left( \int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{p'}\,ds \right) ^{\frac{1}{p'}} \le \left( \int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{q'}\,ds \right) ^{\frac{1}{q'}}(1-r)^{\frac{1}{p'}-\frac{1}{q'}}, \quad 0\le r<1.\nonumber \\ \end{aligned}$$

(6.2)

Moreover,

$$\begin{aligned} \left( 1+\int _0^r \frac{1}{\widehat{\omega }(s)^p}\,ds \right) ^{\frac{1}{p}}\lesssim \widehat{\omega }(r)^{\frac{q}{p}-1} \left( 1+\int _0^r \frac{1}{\widehat{\omega }(s)^q}\,ds \right) ^{\frac{1}{p}}, \quad 0\le r<1. \end{aligned}$$

(6.3)

So, the identity $\frac{1}{p'}-\frac{1}{q'}=\frac{1}{q}-\frac{1}{p}$, together with (6.1), (6.2) and (6.3) yield

$$\begin{aligned}&\left( 1+\int _0^r \frac{1}{\widehat{\omega }(t)^p} dt\right) ^{\frac{1}{p}} \left( \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt\right) ^{\frac{1}{p'}}\\&\quad \lesssim \left( \frac{1-r}{\widehat{\omega }(r)^q}\right) ^{\frac{1}{q}-\frac{1}{p}} \left( 1+\int _0^r \frac{1}{\widehat{\omega }(s)^q}\,ds \right) ^{\frac{1}{p}}\left( \int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{q'}\,ds \right) ^{\frac{1}{q'}}\\&\quad \lesssim \left( 1+\int _0^r \frac{1}{\widehat{\omega }(s)^q}\,ds \right) ^{\frac{1}{q}}\left( \int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{q'}\,ds \right) ^{\frac{1}{q'}}. \end{aligned}$$

Consequently $K_{p,c}(\omega )<\infty$, and by Theorem 1, $T:X_p\rightarrow Y_p$ is bounded.

Assume that $q=1$, that is, $T:X_1\rightarrow Y_1$ is bounded. By Theorem 2, $\omega \in \widehat{\mathcal {D}}$ and $M_{1,d}(\omega )<\infty$, so $K_{1,d}(\omega )<\infty$. Then,

$$\begin{aligned} \begin{aligned} \int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{p'}\,ds&\le K^{p'}_{1,d}(\omega ) \int _r^1 \left( 1+\int _0^s \frac{1}{\widehat{\omega }(t)}\,dt \right) ^{-p'} \,ds \\ {}&\le K^{p'}_{1,d}(\omega )(1-r)\left( 1+\int _0^r \frac{1}{\widehat{\omega }(t)}\,dt \right) ^{-p'} \\ {}&\lesssim K^{p'}_{1,d}(\omega ) \frac{\widehat{\omega }(r)^{p'}}{(1-r)^{p'-1}}, \quad 0\le r<1, \end{aligned} \end{aligned}$$

where in the last inequality we have used (6.1) with $q=1$. Moreover,

$$\begin{aligned} \left( 1+\int _0^r \frac{1}{\widehat{\omega }(s)^p}\,ds \right) ^{\frac{1}{p}}\lesssim & {} \frac{1}{\widehat{\omega }(r)^{1-\frac{1}{p}}}\left( 1+\int _0^r \frac{1}{\widehat{\omega }(s)}\,ds \right) ^{\frac{1}{p}}\\ {}\le & {} \frac{(1-r)^{\frac{1}{p}}}{\widehat{\omega }(r)} K_{1,d}(\omega )^{1/p}, \quad 0\le r<1. \end{aligned}$$

So, $M_{p,c}(\omega )<\infty$, and by Theorem 1, $T:X_p\rightarrow Y_p$ is bounded. This finishes the proof.

Finally, we present two more conditions which characterize the radial weights $\omega$ such that $T: X_p\rightarrow Y_p$, $1<p<\infty$, is bounded, where $X_p,Y_p\in \{H(\infty ,p),H^p, D^p_{p-1}, HL(p)\}$ and $T\in \{ H_{\omega }, \widetilde{H_{\omega }}\}$.

Proposition 36

Let $\omega$ be a radial weight and $1<p<\infty$. Then, the following conditions are equivalent:

• (i) $\omega \in \widehat{\mathcal {D}}$ and $K_{p,d}(\omega )=\sup \nolimits _{0< r<1}\frac{\widehat{\omega }(r)}{(1-r)^{\frac{1}{p}}}\left( 1+\int _0^r \frac{1}{\widehat{\omega }(s)^p}\,ds \right) ^{\frac{1}{p}}<\infty$;

• (ii) $\omega \in \widehat{\mathcal {D}}$ and $K_{p,e}(\omega )=\sup \nolimits _{0< r<1}\frac{(1-r)^{\frac{1}{p}}}{\widehat{\omega }(r)} \left( \int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{p'}\,ds \right) ^{\frac{1}{p'}}<\infty$;

• (iii) $\omega \in \widehat{\mathcal {D}}$ and $K_{p,c}(\omega )= \sup \nolimits _{0<r<1} \left( 1+\int _0^r \frac{1}{\widehat{\omega }(t)^p} dt\right) ^{\frac{1}{p}} \left( \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt\right) ^{\frac{1}{p'}}<\infty .$

Proof

Assume that (i) holds. A calculation shows that $F(r)=(1-r)^{\kappa }\left( 1+\int _0^r \frac{ds}{\widehat{\omega }(s)^p}\, \right)$, with $\kappa =\frac{1}{K^p_{p,d}(\omega )}$, is non-decreasing in [0, 1). So,

$$\begin{aligned} \begin{aligned} \int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{p'}\,ds&\le K^{p'}_{p,d}(\omega ) \int _r^1 \frac{1}{1-s}\left( 1+\int _0^s \frac{1}{\widehat{\omega }(t)^p}\,dt \right) ^{-\frac{p'}{p}} \,ds \\ {}&\le K^{p'}_{p,d}(\omega )F(r)^{-\frac{p'}{p}} \int _r^1 (1-s)^{\frac{\kappa p'}{p}-1} ds \\ {}&\lesssim K^{p'+p}_{p,d}(\omega )\left( 1+\int _0^r \frac{1}{\widehat{\omega }(t)^p}\,dt \right) ^{-\frac{p'}{p}} \\ {}&\lesssim K^{p'+p}_{p,d}(\omega ) \frac{\widehat{\omega }(r)^{p'}}{(1-r)^{p'-1}}, \quad 0\le r<1, \end{aligned} \end{aligned}$$

where in the last inequality we have used (6.1). That is (ii) holds.

Now, assume that (ii) holds. Since $\omega \in \widehat{\mathcal {D}}$

$$\begin{aligned} \int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{p'}\,ds \gtrsim \frac{\widehat{\omega }(r)^{p'}}{(1-r)^{p'-1}},\quad 0\le r<1. \end{aligned}$$

Moreover, $H(r)=(1-r)^{-\eta }\int _r^1 \left( \frac{\widehat{\omega }(s)}{1-s}\right) ^{p'}\,ds$, with $\eta =\frac{1}{K^{p'}_{p,e}(\omega )}$, is non-increasing in [0, 1). So,

$$\begin{aligned}\begin{aligned} 1+\int _0^r \frac{ds}{\widehat{\omega }(s)^p}\,&\le 1+ K^{p}_{p,e}(\omega ) \int _0^r \frac{1}{1-s}\left( \int _s^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt \right) ^{-\frac{p}{p'}}\,ds \\ {}&\le 1+ K^{p}_{p,e}(\omega ) H(r)^{-\frac{p}{p'}} \int _0^r \frac{1}{(1-s)^{1+\eta \frac{p}{p'}}} ds \\ {}&\lesssim K^{p+p'}_{p,e}(\omega )\left( \int _r^1 \left( \frac{\widehat{\omega }(t)}{1-t}\right) ^{p'}\,dt \right) ^{-\frac{p}{p'}} \\ {}&\lesssim K^{p+p'}_{p,e}(\omega ) \frac{1-r}{\widehat{\omega }(r)^p},\quad 0\le r<1, \end{aligned} \end{aligned}$$

therefore (i) holds.

Next, if (i) holds, then (ii) holds and so it is clear that (iii) holds. Finally, (iii) together with (6.1) implies (ii). This finishes the proof. $\square$