Mean Lipschitz spaces and a generalized Hilbert operator

Abstract

If $\mu$ is a positive Borel measure on the interval [0, 1) we let $\mathcal {H}_\mu$ be the Hankel matrix $\mathcal {H}_\mu =(\mu _{n, k})_{n,k\ge 0}$ with entries $\mu _{n, k}=\mu _{n+k}$, where, for $n\,=\,0, 1, 2, \dots$, $\mu _n$ denotes the moment of order n of $\mu$. This matrix induces formally the operator

$$\begin{aligned}\mathcal {H}_\mu (f)(z)= \sum _{n=0}^{\infty }\left( \sum _{k=0}^{\infty } \mu _{n,k}{a_k}\right) z^n\end{aligned}$$

on the space of all analytic functions $f(z)=\sum _{k=0}^\infty a_kz^k$, in the unit disc ${\mathbb {D}}$. This is a natural generalization of the classical Hilbert operator. In this paper we study the action of the operators $\mathcal {H}_\mu$ on mean Lipschitz spaces of analytic functions.

1 Introduction and main results

Let $\mathbb {D}$ be the unit disc in the complex plane $\mathbb {C}$, and let ${\mathcal {H}ol}(\mathbb {D} )$ denote the space of all analytic functions in $\mathbb {D}$. For $0<r<1$ and $f\in {\mathcal {H}ol}(\mathbb {D} )$, we set

$$\begin{aligned} M_ p(r, f)= & {} \left( \frac{1}{2\pi }\int _ {-\pi }^\pi \left| f(re^{i\theta }) \right| ^pd\theta \right) ^{1/p}, \quad 0<p<\infty ,\\ M_ \infty (r, f)= & {} \max _ {\vert z\vert =r}\vert f(z)\vert .\end{aligned}$$

For $0<p\le \infty$ the Hardy space $H^p$ consists of those functions f, analytic in $\mathbb {D}$, for which

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

We refer to [9] for the theory of Hardy spaces.

The space BMOA consists of those functions $f\in H^1$ whose boundary values have bounded mean oscillation on $\partial \mathbb {D}$. The Bloch space $\mathcal {B}$ consists of all analytic functions f in ${\mathbb {D}}$ with bounded invariant derivative:

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

We mention [1, 13, 23] as excellent references for these spaces. Let us recall that $BMOA \subsetneq \mathcal {B}$.

If $\mu$ is a finite positive Borel measure on [0, 1) and $n\, = 0, 1, 2, \dots$, we let $\mu _n$ denote the moment of order n of $\mu$, that is, $\mu _n=\int _{[0,1)}t^n\,d\mu (t),$ and we let $\mathcal {H}_\mu$ be the Hankel matrix $(\mu _{n,k})_{n,k\ge 0}$ with entries $\mu _{n,k}=\mu _{n+k}$. The matrix $\mathcal {H}_\mu$ induces formally an operator, also denoted $\mathcal {H}_\mu$, on spaces of analytic functions in the following way: if  $f(z)=\sum _{k=0}^\infty a_kz^k\in {\mathcal {H}ol}({\mathbb {D}})$ we define

$$\begin{aligned} \mathcal {H}_\mu (f)(z)= \sum _{n=0}^{\infty }\left( \sum _{k=0}^{\infty } \mu _{n,k}{a_k}\right) z^n, \end{aligned}$$

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

If $\mu$ is the Lebesgue measure on [0, 1) the matrix $\mathcal {H}_\mu$ reduces to the classical Hilbert matrix   $\mathcal {H}= \left( {(n+k+1)^{-1}}\right) _{n,k\ge 0}$, which induces the classical Hilbert operator $\mathcal {H}$. The Hilbert operator is known to be well defined on $H^1$ and bounded from $H^p$ into itself, if $1<p<\infty$, but not if $p=1$ or $p=\infty$ [8].

The question of describing the measures $\mu$ for which the operator $\mathcal {H}_\mu$ is well defined and bounded on distinct spaces of analytic functions has been studied in a good number of papers (see [2, 7, 10, 15, 16, 19,20,21]). The measures in question are Carleson-type measures.

If $I\subset \partial {\mathbb {D}}$ is an interval, $\vert I\vert$ will denote the length of I. The Carleson squareS(I) is defined as $S(I)=\{re^{it}:\,e^{it}\in I,\quad 1-\frac{|I|}{2\pi }\le r <1\}$.

If $\, s>0$ and $\mu$ is a positive Borel measure on ${\mathbb {D}}$, we shall say that $\mu$ is an s-Carleson measure if there exists a positive constant C such that

$$\begin{aligned} \mu \left( S(I)\right) \le C{|I|^s}, \quad \hbox {for any interval } I\subset \partial {\mathbb {D}}. \end{aligned}$$

A 1-Carleson measure will be simply called a Carleson measure.

If $\mu$ is a positive Borel measure on ${\mathbb {D}}$, $0\le \alpha <\infty$, and $0<s<\infty$ we say that $\mu$ is an $\alpha$-logarithmic s-Carleson measure [22] if there exists a positive constant C such that

$$\begin{aligned}\frac{ \mu \left( S(I)\right) \left( \log \frac{2\pi }{\vert I\vert }\right) ^\alpha }{|I|^s}\le C, \quad \hbox {for any interval } I\subset \partial {\mathbb {D}}. \end{aligned}$$

A positive Borel measure $\mu$ on [0, 1) can be seen as a Borel measure on $\mathbb {D}$ by identifying it with the measure $\tilde{\mu }$ defined by

$$\begin{aligned} \tilde{\mu } (A)\,=\,\mu \left( A\cap [0,1)\right) ,\quad \text {for any Borel subset }A\hbox { of }\mathbb {D}. \end{aligned}$$

In this way a positive Borel measure $\mu$ on [0, 1) is an s-Carleson measure if and only if there exists a positive constant C such that

$$\begin{aligned} \mu \left( [t,1)\right) \le C(1-t)^s, \quad 0\le t<1. \end{aligned}$$

We have a similar statement for $\alpha$-logarithmic s-Carleson measures.

Widom [21, Theorem 3. 1] (see also [20, Theorem 3] and [19, p. 42, Theorem 7. 2]) proved that $\mathcal {H}_\mu$ is a bounded operator from $H^2$ into itself if and only $\mu$ is a Carleson measure. Galanopoulos and Peláez [10] studied the operators $\mathcal {H}_\mu$ acting on $H^1$. The action of $\mathcal {H}_\mu$ on the Hardy spaces $H^p$, $0<p\le \infty$, has been studied in [7, 15, 16]. The papers [15] and [16] study also the operators $\mathcal {H}_\mu$ acting on distinct subspaces of the Bloch space, including BMOA, Besov spaces, and the $Q_s$-spaces.

In this paper we shall study the operators $\mathcal {H}_\mu$ acting on mean Lipschitz spaces of analytic functions.

If $f\in {\mathcal {H}ol}(\mathbb {D} )$ has a non-tangential limit $f(e^{i\theta })$ at almost every $e^{i\theta }\in \partial \mathbb {D}$ and $\delta >0$, we define

$$\begin{aligned}\omega _ p(\delta , f)=&\sup _{0<\vert t\vert \le \delta }\left( \frac{1}{2\pi }\int _{-\pi }^\pi \left| f(e^{i(\theta +t)})-f(e^{i\theta })\right| ^p\, d\theta \right) ^{1/p}, \quad \text {if }1\le p<\infty ,\\ \omega _ \infty (\delta , f)=&\sup _ {0<\vert t\vert \le \delta }\left( \mathop {\hbox {ess.sup}}\limits _ {\theta \in [-\pi , \pi ]} \vert f(e^{i(\theta +t)})-f(e^{i\theta })\vert \right) .\end{aligned}$$

Then $\omega _ p(\cdot , f)$ is the integral modulus of continuity of order p of the boundary values $f(e^{i\theta })$ of f.

Given $1\le p\le \infty$ and $0<\alpha \le 1$, the mean Lipschitz space $\varLambda _ \alpha ^p$ consists of those functions $f\in {\mathcal {H}ol}(\mathbb {D})$ having a non-tangential limit almost everywhere for which $\omega _ p(\delta , f)=O(\delta ^\alpha )$, as $\delta \rightarrow 0$. If $p=\infty$ we write $\varLambda _ \alpha$ instead of $\varLambda _ \alpha ^\infty$. This is the usual Lipschitz space of order $\alpha$.

A classical result of Hardy and Littlewood [17] (see also [9, Chapter 5]) asserts that for $1\le p\le \infty$ and $0<\alpha \le 1$, we have that $\varLambda _ \alpha ^p\subset H^p$ and

$$\begin{aligned} \varLambda _ \alpha ^p=\left\{ f\in {\mathcal {H}ol}(\mathbb {D}) : M_ p(r, f^\prime )={{\mathrm{O}}}\left( \frac{1}{(1-r)^{1-\alpha }}\right) \right\} . \end{aligned}$$

(1)

It is known that if $1<p<\infty$ and $\alpha >\frac{1}{p}$ then each $f\in \varLambda ^p_\alpha$ is bounded and has a continuous extension to the closed unit disc ([6], p. 88). This is not true for $\alpha =\frac{1}{p}$, because the function $f(z)=\log (1-z)$ belongs to $\varLambda ^p_{1/p}$ for all $p\in (1,\infty )$. By a theorem of Hardy and Littlewood [9, Theorem 5.9] and of [6, Theorem 2.5] we have

$$\begin{aligned} \varLambda ^p_{1/p}\subset \varLambda ^q_{1/q}\subset BMOA\quad 1\le p<q<\infty . \end{aligned}$$

The inclusion $\varLambda ^p_{1/p}\subset BMOA$, $1\le p<\infty$ was proved to be sharp in a very strong sense in [3, 11, 12] using the following generalization of the spaces $\varLambda _ \alpha ^p$ which occurs frequently in the literature. Let $\omega :[0, \pi ]\rightarrow [0, \infty )$ be a continuous and increasing function with $\omega (0)=0$ and $\omega (t)>0$ if $t>0$. Then, for $1\le p\le \infty$, the mean Lipschitz space $\varLambda (p, \omega )$ consists of those functions $f\in H^p$ such that

$$\begin{aligned} \omega _ p(\delta , f)=O(\omega (\delta )),\quad \hbox {as }\delta \rightarrow 0. \end{aligned}$$

With this notation we have $\varLambda ^p_ \alpha =\varLambda (p, \delta ^\alpha )$.

The question of finding conditions on $\omega$ so that it is possible to obtain results on the spaces $\varLambda (p, \omega )$ analogous to those proved by Hardy and Littlewood for the spaces $\varLambda ^p_ \alpha$ has been studied by several authors (see [4,5,6]). We shall say that $\omega$ satisfies the Dini condition or that $\omega$ is a Dini-weight if there exists a positive constant C such that

$$\begin{aligned} \int _ 0^\delta \frac{\omega (t)}{t}\, dt\le C\omega (\delta ), \quad 0<\delta <1. \end{aligned}$$

We shall say that $\omega$ satisfies the condition $b_1$ or that $\omega \in b_1$ if there exists a positive constant C such that

$$\begin{aligned} \int _ \delta ^\pi \frac{\omega (t)}{t^{2}}\, dt\le C\frac{\omega (\delta )}{\delta }, \quad 0<\delta <1. \end{aligned}$$

In order to simplify our notation, let $\mathcal A\mathcal W$ denote the family of all functions $\omega :[0, \pi ]\rightarrow [0, \infty )$ which satisfy the following conditions:

• (i) $\omega$ is continuous and increasing in $[0, \pi ]$.

• (ii) $\omega (0)=0$ and $\omega (t)>0$ if $t>0$.

• (iii) $\omega$ is a Dini-weight.

• (iv) $\omega$ satisfies the condition $b_1$.

The elements of $\mathcal A\mathcal W$ will be called admissible weights. Characterizations and examples of admissible weights can be found in [4, 5].

Blasco and de Souza extended the above mentioned result of Hardy and Littlewood showing in [4, Th. 2.1] that if $\omega \in \mathcal A\mathcal W$ then,

$$\begin{aligned} \varLambda (p, \omega )=\left\{ f\text { analytic in }{\mathbb {D}}: M_ p(r, f^\prime )={{\mathrm{O}}}\left( \frac{\omega (1-r)}{1-r}\right) ,\text { as }r\rightarrow 1\right\} . \end{aligned}$$

In [3, 11, 12] it is proved that if $1\le p<\infty$ and $\omega$ is an admissible weight such that

$$\begin{aligned} \frac{\omega (\delta )}{\delta ^{1/p}}\rightarrow \infty ,\text { as }\delta \rightarrow 0, \end{aligned}$$

then there exists a function $f\in \varLambda (p,\omega )$ which is a not a normal function (see [1] for the definition). Since any Bloch function is normal, if follows that for such admissible weights $\omega$ one has that $\varLambda (p, \omega )\not \subset \mathcal {B}$.

One of the main results in [16] is the following one.

Theorem A

[16] Let $\mu$ be a positive Borel measure on [0, 1) and let X be a Banach space of analytic functions in $\mathbb {D}$ with $\varLambda ^2_{1/2}\subset X\subset \mathcal {B}$. Then the following conditions are equivalent.

• (i) The operator $\mathcal {H}_\mu$ is well defined in X and, furthermore, it is a bounded operator from X into the Bloch space $\mathcal {B}$.

• (ii) The operator $\mathcal {H}_\mu$ is well defined in X and, furthermore, it is a bounded operator from X into $\varLambda ^2_{1/2}$.

• (iii) The measure $\mu$ is a 1-logarithmic 1-Carleson measure.

• (iv) $\int _{[0,1)}t^n\log \frac{1}{1-t}d\mu (t)\,=\,{{\mathrm{O}}}\left( \frac{1}{n}\right)$.

A key ingredient in the proof of Theorem A is the fact that for any space X with $\varLambda ^2_{1/2}\subset X\subset \mathcal {B}$ the functions $f\in X$ of the form $f(z)=\sum _{n=0}^\infty a_nz^n$ whose sequence of Taylor coefficients $\{ a_n\}$ is a decreasing sequence of non-negative numbers are the same. Indeed, for such a function f and such a space X we have that $f\in X\,\,\Leftrightarrow \,\, a_n={{\mathrm{O}}}\left( \frac{1}{n}\right) .$ This result remains true if we substitute $\varLambda ^2_{1/2}$ by $\varLambda ^p_{1/p}$ for any $p>1$. That is, the following result holds:

Lemma 1

Suppose that $1<p<\infty$ and let $f\in {\mathcal {H}ol}({\mathbb {D}})$ be of the form $f(z)=\sum _{n=0}^\infty a_nz^n$ with $\{ a_n\}_{n=0}^\infty$ being a decreasing sequence of nonnegative numbers. If X is a subspace of ${\mathcal {H}ol}({\mathbb {D}})$ with $\varLambda ^p_{1/p}\subset X\subset \mathcal {B}$, then

$$\begin{aligned} f\in X\quad \Leftrightarrow \quad a_n={{\mathrm{O}}}\left( \frac{1}{n}\right) . \end{aligned}$$

Lemma 1 is a consequence of the following one which will be proved in Sect. 2.

Lemma 2

Let $1<p<\infty$, $\omega \in \mathcal {AW}$ and let $f(z)=\sum _{n=0}^\infty a_nz^n$ with $\{ a_n\}_{n=0}^\infty$ being a decreasing sequence of nonnegative numbers. Then

$$\begin{aligned} f\in \varLambda (p,\omega )\quad \Leftrightarrow \quad a_n\,=\,{{\mathrm{O}}}\left( \frac{\omega (1/n)}{n^{1-1/p}}\right) . \end{aligned}$$

(2)

Using Lemma 1 and following the proof of Theorem A in [16], we obtain

Theorem 1

Suppose that $1<p<\infty$. Let $\mu$ be a positive Borel measure on [0, 1) and let X be a Banach space of analytic functions in $\mathbb {D}$ with $\varLambda ^p_{1/p}\subset X\subset \mathcal {B}$. Then the following conditions are equivalent.

• (i) The operator $\mathcal {H}_\mu$ is well defined in X and, furthermore, it is a bounded operator from X into the Bloch space $\mathcal {B}$.

• (ii) The operator $\mathcal {H}_\mu$ is well defined in X and, furthermore, it is a bounded operator from X into $\varLambda ^p_{1/p}$.

• (iii) The measure $\mu$ is a 1-logarithmic 1-Carleson measure.

• (iv) $\int _{[0,1)}t^n\log \frac{1}{1-t}d\mu (t)\,=\,{{\mathrm{O}}}\left( \frac{1}{n}\right)$.

As an immediate consequence of Theorem 1 we obtain the following result.

Corollary 1

Let $\mu$ be a positive Borel measure on [0, 1) and $1<p<\infty$. Then the operator $\mathcal {H}_\mu$ is well defined in $\varLambda ^p_{1/p}$ and, furthermore, it is a bounded operator from $\varLambda ^p_{1/p}$ into itself if and only if $\mu$ is a 1-logarithmic 1-Carleson measure.

Let us turn our attention now to the spaces $\varLambda (p,\omega )$ with $\frac{\omega (\delta )}{\delta ^{1/p}}\nearrow \infty$, $\delta \searrow 0$ which, as noted before, are not included in the Bloch space. We have the following result which shows that the situation is different from the one covered in Theorem 1.

Theorem 2

Let $1<p<\infty$, $\omega \in \mathcal {AW}$ with $\frac{\omega (\delta )}{\delta ^{1/p}}\nearrow \infty$ when $\delta \searrow 0$. The following conditions are equivalent:

• (i) The operator $\mathcal {H}_\mu$ is well defined in $\varLambda (p,\omega )$ and, furthermore, it is a bounded operator from $\varLambda (p,\omega )$ into itself.

• (ii) The measure $\mu$ is a Carleson measure.

The proofs of Lemma 2 and Theorem 2 will be presented in Sect. 2. We close this section noticing that, as usual, we shall be using the convention that $C=C(p, \alpha ,q,\beta , \dots )$ will denote a positive constant which depends only upon the displayed parameters $p, \alpha , q, \beta \dots$ (which sometimes will be omitted) but not necessarily the same at different occurrences. Moreover, for two real-valued functions $E_1, E_2$ we write $E_1\lesssim E_2$, or $E_1\gtrsim E_2$, if there exists a positive constant C independent of the arguments such that $E_1\le C E_2$, respectively $E_1\ge C E_2$. If we have $E_1\lesssim E_2$ and $E_1\gtrsim E_2$ simultaneously then we say that $E_1$ and $E_2$ are equivalent and we write $E_1\asymp E_2$.

2 Proofs of the main results

We start recalling that for a function $f(z)=\sum _{n=0}^\infty a_nz^n$ analytic in ${\mathbb {D}},$ the polynomials $\varDelta _jf$ are defined as follows:

$$\begin{aligned} \varDelta _jf(z)= & {} \sum _{k=2^j}^{2^{j+1}-1} a_kz^k, \quad \text {for }j\ge 1,\\ \varDelta _0f(z)= & {} a_0+a_1z. \end{aligned}$$

The proof of Lemma 2 is based in the following result of Girela and González [14, Theorem 2].

Theorem B

Let $1<p<\infty$ and let $\omega$ be an admissible weight. If $f\in {\mathcal {H}ol}({\mathbb {D}})$ with $f(z)=\sum _{n=0}^\infty a_nz^n$ then

$$\begin{aligned} f\in \varLambda (p,\omega )\,\,\Leftrightarrow \,\,\Vert \varDelta _N f\Vert _{H^p}=O\left( \omega \left( \frac{1}{2^N}\right) \right) . \end{aligned}$$

Proof of Lemma 2

By Lemma A of [18], since $a_n\searrow 0$, we have

$$\begin{aligned} \Vert \varDelta _N f\Vert _{H^p}\asymp a_{2^N} 2^{N(1-1/p)},\quad N\ge 1. \end{aligned}$$

So by Theorem B we have that

$$\begin{aligned} f\in \varLambda (p,\omega )\Leftrightarrow a_{2^N} \lesssim \frac{\omega \left( 1/2^N\right) }{2^{N(1-1/p)}},\quad N\ge 1. \end{aligned}$$

This easily implies (2). $\square$

Lemma 3

Suppose that $1<p<\infty$. Let $\nu$ be a positive Borel measure on [0, 1), and let $\omega \in \mathcal {AW}$ satisfying that $x^{-1/p}\omega (x)\nearrow \infty$, as $x\searrow 0$. Then following conditions are equivalent:

• (i) $\nu _n\lesssim \frac{\omega (1/n)}{n^{{1-1/p}}},\, n\ge 2$.

• (ii) $\nu ([b,1))\lesssim (1-b)^{1-1/p} \omega (1-b),\, b\in [0,1).$

Proof

Suppose (i). Then we have that

$$\begin{aligned} 1&\gtrsim \frac{n^{1-1/p}\,\nu _n}{\omega (1/n)}=\frac{n^{1-1/p}}{\omega (1/n)}\int _{[0,1)} t^n\,d\nu (t)\ge \frac{n^{1-1/p}}{\omega (1/n)}\int _{[1-1/n,1)} t^n\,d\nu (t) \\&\ge \frac{n^{1-1/p}}{\omega (1/n)}\nu ([1-1/n,1))\,\left( 1-\frac{1}{n}\right) ^n \\&\ge \frac{n^{1-1/p}}{\omega (1/n)}\nu ([1-1/n,1))\,\inf \limits _{m\ge 2}\left( 1-\frac{1}{m}\right) ^m \\&\gtrsim \frac{n^{1-1/p}}{\omega (1/n)}\nu ([1-1/n,1)). \end{aligned}$$

So $\nu ([1-1/n,1))\lesssim \frac{\omega (1/n)}{n^{1-1/p}}$ for $n\ge 2$.

Let now $b\in [1/2,1)$. There exists $n\ge 2$ such that $1-\frac{1}{n}\le b<1-\frac{1}{n+1}$ so using the above we have that

$$\begin{aligned} \nu ([b,1))\le \nu ([1-1/n,1))\lesssim \frac{\omega (1/n)}{n^{1-1/p}}. \end{aligned}$$

This, and the facts that $\omega (1/n)n^{1/p}\le \omega (1/(n+1))(n+1)^{1/p}$ and that the weight $\omega$ increases give (ii).

Suppose now (ii). Then

$$\begin{aligned} \nu _n&=\int _{[0,1)} t^n\,d\nu (t)=n\int _0^1 \nu ([t,1))t^{n-1}\,dt \\&\lesssim n\int _0^1 (1-t)^{1-1/p} \omega (1-t)t^{n-1}\,dt \\&= n \int _0^{1-\frac{1}{n}}\,+\,\int _{1-\frac{1}{n}}^1\left( (1-t)^{1-1/p} \omega (1-t)t^{n-1}\, \right) dt. \end{aligned}$$

The first integral can be estimated bearing in mind that $(1-t)^{-1/p}\omega (1-t)\nearrow \infty$ when $t\nearrow 1$ as follows

$$\begin{aligned}&n \int _0^{1-\frac{1}{n}} (1-t)^{1-1/p} \omega (1-t)t^{n-1}\,dt\\&\quad \le n^{1+1/p}\omega (1/n)\int _0^{1-\frac{1}{n}} (1-t)t^{n-1}\,dt\\&\quad = n^{1+1/p}\omega (1/n)\left( 1-\frac{1}{n}\right) ^n\left( \frac{1}{n} -\frac{n-1}{n(n+1)}\right) \\&\quad \lesssim \frac{\omega (1/n)}{n^{1-1/p}}. \end{aligned}$$

To estimate of the second integral we use that $(1-t)^{1-1/p}\omega (1-t)\searrow 0$ when $t\nearrow 1$ to obtain

$$\begin{aligned}&n \int _{1-\frac{1}{n}}^1 (1-t)^{1-1/p} \omega (1-t)t^{n-1}\,dt\\&\quad \le n^{1/p}\omega (1/n)\int _{1-\frac{1}{n}}^1 t^{n-1}\,dt\\&\quad = \frac{\omega (1/n)}{n^{1-1/p}}\left( 1-\left( 1-\frac{1}{n}\right) ^n\right) \\&\quad \lesssim \frac{\omega (1/n)}{n^{1-1/p}}. \end{aligned}$$

Then (i) follows. $\square$

Proof of Theorem 2

(i) $\Rightarrow$ (ii) Suppose that $\mathcal {H}_\mu :\varLambda (p,\omega )\rightarrow \varLambda (p,\omega )$ is bounded. By Lemma 2 we have that the function f defined by $f(z)=\sum _{n=1}^\infty \frac{\omega (1/n)}{n^{1-1/p}}z^n$ belongs to the space $\varLambda (p,\omega )$ so, by the hypothesis, $\mathcal {H}_\mu (f)$ belongs also to $\varLambda (p,\omega )$. Now

$$\begin{aligned} \mathcal {H}_\mu (f)(z)=\sum _{n=0}^\infty \left( \sum _{k=1}^\infty \frac{\omega (1/k)}{k^{1-1/p}}\,\mu _{n+k} \right) z^n. \end{aligned}$$

Notice that $\sum _{k=1}^\infty \frac{\omega (1/k)}{k^{1-1/p}}\,\mu _{n+k}\searrow 0,\, n\rightarrow \infty$, so using again Lemma 2 it holds that

$$\begin{aligned} \sum _{k=1}^\infty \frac{\omega (1/k)}{k^{1-1/p}}\,\mu _{n+k}=\int _{[0,1)} t^n\sum _{k=1}^\infty \frac{\omega (1/k)}{k^{1-1/p}}\,t^k\,d\mu (t) \lesssim \frac{\omega (1/n)}{n^{1-1/p}}, \end{aligned}$$

that is, the moments of the measure $\nu$ defined by

$$\begin{aligned} d\nu (t)=\sum _{k=1}^\infty \frac{\omega (1/k)}{k^{1-1/p}}\,t^k\,d\mu (t) \end{aligned}$$

satisfy that

$$\begin{aligned} \nu _n\lesssim \frac{\omega (1/n)}{n^{1-1/p}}, \end{aligned}$$

so by Lemma 3 we have that $\nu ([b,1))\lesssim (1-b)^{1-1/p} \omega (1-b)$, $b\in [0,1)$.

According to the definition of the measure

$$\begin{aligned} (1-b)^{1-1/p} \omega (1-b)&\gtrsim \nu ([b,1))=\int _{[b,1)}d\nu (t) \\&=\int _{[b,1)}\sum _{k=1}^\infty \frac{\omega (1/k)}{k^{1-1/p}}\,t^k\,d\mu (t) \\&\ge \mu \left( [b,1)\right) \sum _{k=1}^\infty \frac{\omega (1/k)}{k^{1-1/p}}\,b^k \end{aligned}$$

and the sum can be estimated as follows

$$\begin{aligned} \sum _{k=1}^\infty \frac{\omega (1/k)}{k^{1-1/p}}\,b^k&\asymp \int _1^\infty \frac{\omega (1/x)}{x^{1-1/p}}\,b^x\,dx \\&\ge \int _1^{\frac{1}{1-b}} \frac{\omega (1/x)}{x^{1-1/p}}\,b^x\,dx \\&\ge (1-b)^{1-1/p}\omega (1-b)b^{\frac{1}{1-b}}\left( \frac{1}{1-b}-1\right) \\&\gtrsim \frac{\omega (1-b)}{(1-b)^{1/p}}. \end{aligned}$$

Finally, putting all together we have that

$$\begin{aligned} \mu ([b,1))\lesssim 1-b \end{aligned}$$

so $\mu$ is a Carleson measure.

(ii) $\Rightarrow$ (i) To prove this implication we need to use the integral operator $I_\mu$ considered in [7, 10, 15, 16] which is closely related to the operator $\mathcal {H}_\mu$.

If $\mu$ is a positive Borel measure on [0, 1) and $f\in {\mathcal {H}ol}({\mathbb {D}})$, we shall write throughout the paper

$$\begin{aligned} I_\mu (f)(z)=\int _{[0,1)}\frac{f(t)}{1-tz}\,d\mu (t), \end{aligned}$$

whenever the right hand side makes sense and defines an analytic function in ${\mathbb {D}}$. It turns out that the operators $\mathcal {H}_\mu$ and $I_\mu$ are closely related. Indeed, as shown in the just mentioned papers, it turns out that if f is good enough $\mathcal {H}_\mu$ and $I_\mu (f)$ are well defined and coincide.

Suppose that $\mu$ is a Carleson measure supported on [0, 1) and let $f\in \varLambda (p,\omega )$. We claim that

$$\begin{aligned} \int _{[0,1)}\frac{|f(t)|}{|1-tz|}\,d\mu (t)<\infty . \end{aligned}$$

(3)

Indeed, using Lemma 3 of [14] we have that

$$\begin{aligned} f\in \varLambda (p,\omega )\Rightarrow |f(z)|\lesssim \frac{\omega (1-|z|)}{(1-|z|)^{1/p}},\quad z\in {\mathbb {D}}. \end{aligned}$$

(4)

Then we obtain

$$\begin{aligned} \int _{[0,1)}\frac{|f(t)|}{|1-tz|}\,d\mu (t)&\le \frac{1}{1-|z|}\int _{[0,1)}|f(t)|\,d\mu (t) \\&\lesssim \frac{1}{1-|z|}\int _{[0,1)} \frac{\omega (1-t)}{(1-t)^{1/p}}\,d\mu (t). \end{aligned}$$

If we choose $r\in [0,1)$ we can split the integral in the intervals [0, r) and [r, 1). In the first one, as $\omega$ is an increasing weight we have

$$\begin{aligned} \int _{[0,r)} \frac{\omega (1-t)}{(1-t)^{1/p}}\,d\mu (t)&\le \omega (1)\int _{[0,r)} \frac{d\mu (t)}{(1-t)^{1/p}} \\&\le \omega (1)\int _{[0,1)} \frac{d\mu (t)}{(1-t)^{1/p}} \\&\lesssim 1, \end{aligned}$$

because $\mu$ is a Carleson measure. Using this and the condition $\frac{\omega (\delta )}{\delta ^{1/p}}\nearrow \infty$, as $\delta \searrow 0$ we can estimate the other integral as follows

$$\begin{aligned} \int _{[r,1)} \frac{\omega (1-t)}{(1-t)^{1/p}}\,d\mu (t)&\le \frac{\omega (1-r)}{(1-r)^{1/p}} \int _{[r,1)} d\mu (t) \\&\lesssim \omega (1-r)(1-r)^{1-1/p} \\&\lesssim 1. \end{aligned}$$

So we have that for $f\in \varLambda (p,\omega )$ and $z\in {\mathbb {D}}$, (3) holds. This implies that $I_\mu (f)$ is well defined, and, using Fubini’s theorem and standard arguments it follows easily that $\mathcal {H}_\mu (f)$ is also well defined and that, furthermore,

$$\begin{aligned} \mathcal {H}_\mu (f)(z)=I_\mu (f)(z),\quad z\in \mathbb {D}. \end{aligned}$$

Now we have,

$$\begin{aligned} I_\mu (f)^\prime (z)=\int _{[0,1)}\frac{tf(t)}{(1-tz)^2}\,d\mu (t),\quad z\in {\mathbb {D}}, \end{aligned}$$

so the mean of order p of $I_\mu (f)^\prime$ has the form

$$\begin{aligned} M_p\left( r,I_\mu (f)^\prime \right) =\left( \frac{1}{2\pi }\int _{-\pi }^\pi \left| \int _{[0,1)}\frac{tf(t)}{(1-tre^{i\theta })^2}\,d\mu (t)\right| ^p\,d\theta \right) ^{1/p}. \end{aligned}$$

Using again (4), the Minkowski inequality and a classical estimation of integrals we obtain that

$$\begin{aligned} M_p\left( r,I_\mu (f)^\prime \right)&\lesssim \int _{[0,1)} |f(t)|\left( \int _{-\pi }^\pi \frac{d\theta }{|1-tre^{i\theta }|^{2p}}\right) ^{1/p}\,d\mu (t) \\&\lesssim \int _{[0,1)}\frac{ |f(t)|}{(1-tr)^{2-1/p}}\,d\mu (t) \\&\lesssim \int _{[0,1)}\frac{ \omega (1-t)}{(1-t)^{1/p}(1-tr)^{2-1/p}}\,d\mu (t). \end{aligned}$$

At this point we split the integrals on the sets [0, r) and [r, 1).

In the first integral we use that $x^{-1/p}\omega (x)\nearrow \infty$, as $x\searrow 0$, and the fact that if $\mu$ is a Carleson measure (so that $\mu _n=\int _{[0,1)} t^n\,d\mu (t)\lesssim \frac{1}{n}$) to obtain

$$\begin{aligned} \int _{[0,r)}\frac{ \omega (1-t)}{(1-t)^{1/p}(1-tr)^{2-1/p}}\,d\mu (t)&\le \frac{ \omega (1-r)}{(1-r)^{1/p}}\int _{[0,r)}\frac{d\mu (t)}{(1-tr)^{2-1/p}}\\&\le \frac{ \omega (1-r)}{(1-r)^{1/p}}\int _{[0,1)}\frac{d\mu (t)}{(1-tr)^{2-1/p}} \\&\lesssim \frac{ \omega (1-r)}{(1-r)^{1/p}}\sum \limits _{n=1}^\infty n^{1-1/p}r^n\int _{[0,1)} t^n\,d\mu (t) \\&\lesssim \frac{ \omega (1-r)}{(1-r)^{1/p}}\sum \limits _{n=1}^\infty \frac{r^n}{n^{1/p}} \\&\lesssim \frac{ \omega (1-r)}{(1-r)} . \end{aligned}$$

In the second integral we use that $\omega$ is an increasing weight and the fact that the measure $\mu$ being a Carleson measure is equivalent to saying that the measure $\nu$ defined by $d\nu (t)=\frac{d\mu (t)}{(1-t)^{1/p}}$ is a $1-\frac{1}{p}$-Carleson measure so that the moments $\nu _n$ of $\nu$ satisfy $\nu _n\lesssim \frac{1}{n^{1-\frac{1}{p}}}$. Then we obtain

$$\begin{aligned} \int _{[r,1)}\frac{ \omega (1-t)}{(1-t)^{1/p}(1-tr)^{2-1/p}}\,d\mu (t)&\le \omega (1-r)\int _{[r,1)}\frac{d\nu (t)}{(1-tr)^{2-1/p}} \\&\le \omega (1-r)\int _{[0,1)}\frac{d\nu (t)}{(1-tr)^{2-1/p}} \\&\lesssim \omega (1-r)\sum \limits _{n=1}^\infty n^{1-1/p}r^n\int _{[0,1)}t^n\,d\nu (t) \\&\lesssim \omega (1-r)\sum \limits _{n=1}^\infty r^n \\&=\frac{ \omega (1-r)}{(1-r)} . \end{aligned}$$

Therefore $I_\mu (f)\in \varLambda (p,\omega )$ and then the operator $I_\mu$ (and hence the operator $\mathcal {H}_\mu$) is bounded from $\varLambda (p,\omega )$ into itself. $\square$