Essential norm of generalized Hilbert matrix from Bloch type spaces to BMOA and Bloch space

1.   Introduction

Denote by $H({\mathbb{D}})$ the space of all analytic functions on the unit disk ${\mathbb{D}} = \{z:|z| < 1\}$ in the complex plane. For $0 < p \leq \infty$, we let $H^p$ denote the classical Hardy space. If $f\in H({\mathbb{D}})$ and

we say that $f\in BMOA$. Here $\varphi_a(z) = \frac{a-z}{1-\bar{a}z}, \; a\in {\mathbb{D}},$ is a Möbius transformation of ${\mathbb{D}}$. Fefferman's duality theorem says that $BMOA = (H^1)^*$. We refer to ^[10] about the theory of $BMOA$.

Let $0 < \alpha < \infty$. An $f\in H({\mathbb{D}})$ is said to belong to the Bloch type space (or called the $\alpha-$Bloch space), denoted by ${\mathcal{B}}^\alpha$, if

The classical Bloch space ${\mathcal{B}}$ is just ${\mathcal{B}}^1$. It is clear that ${\mathcal{B}}^\alpha$ is a Banach space with the norm $\|f\| = |f(0)|+\|f\|_{{\mathcal{B}}^\alpha}$. See ^[21] for the theory of Bloch type spaces.

For a subarc $I\subset \partial\mathbb{D}$, let $S(I)$ be the Carleson box based on $I$ with

Here $|I| = (2\pi)^{-1}\int_I|d\xi|$ is the normalized length of the arc $I$. If $I = {\partial {\mathbb{D}}}$, let $S(I) = {\mathbb{D}}$. For $0 < s < \infty$, we say that a positive Borel measure $\mu$ is an $s-$Carleson measure on $\mathbb{D}$ if (see ^[7])

We say that a positive Borel measure $\mu$ is a vanishing $s-$Carleson measure on ${\mathbb{D}}$ if

Here and henceforth $\sup_{I\subset\partial\mathbb{D}}$ indicates the supremum taken over all subarcs $I$ of $\partial\mathbb{D}$. When $s = 1$, $\mu$ is called a Carleson measure on $\mathbb{D}$. It is well known that, for any $f\in H^p(0 < p < \infty)$,

if and only if $\mu$ is a Carleson measure. See, for example, ^[8].

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

for any Borel subset $E$ of ${\mathbb{D}}$. Then a positive Borel measure $\mu$ on [0, 1) is an $s$-Carleson measure if there exists a constant $C > 0$ such that (see ^[11])

A vanishing $s-$Carleson measure on $[0, 1)$ can be defined similarly.

Let $\mu$ be a finite positive measure on $[0, 1)$ and $n = 0, 1, 2, \cdots$. Denote $\mu_n$ the moment of order $n$ of $\mu$, that is, $\mu_n = \int_{[0, 1)} t^nd\mu(t).$ Let ${\mathcal H}_{\mu}$ be the Hankel matrix $(\mu_{n, k})_{n, k\geq 0}$ with entries $\mu_{n, k} = \mu_{n+k}$. The matrix ${\mathcal H}_{\mu}$ induces an operator, denoted also by ${\mathcal H}_{\mu}$, on $H({\mathbb{D}})$ by its action on the Taylor coefficient:

More precisely, if $f(z) = \sum^\infty_{k = 0}a_kz^k\in H({\mathbb{D}})$, then

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

As in ^[9], to obtain an integral representation of ${\mathcal H}_{\mu}$, we write

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

If $\mu$ is the Lebesgue measure on $[0, 1)$, then the matrix ${\mathcal H}_\mu$ is just the classical Hilbert matrix $H = \big(\frac{1}{n+k+1} \big)_{n, k\geq 0}$, which induces the classical Hilbert operator $H$. The Hilbert operator $H$ was studied in ^{[1,2,4,5,6,14]}. A generalized Hilbert operator was studied in ^{[11,12,14,15]}.

The operator ${\mathcal H}_\mu$ acting on analytic functions spaces has been studied by many authors. Galanopoulos and Peláez ^[9] obtained a characterization that ${\mathcal H}_\mu$ is bounded or compact on $H^1$. Chatzifountas, Girela and Peláez ^[3] described the measure $\mu$ for which ${\mathcal H}_\mu$ is bounded (compact) operator from $H^p$ into $H^q, 0 < p, q < \infty$. See ^[13] about the Hankel matrix acting on the Dirichlet space.

Let $X$ and $Y$ be two Banach spaces. The essential norm of a continuous linear operator $T$ between normed linear spaces $X$ and $Y$ is the distance to the set of compact operators $K$, that is, $\|T\|^{X\rightarrow Y}_e = \inf\{\|T-K\|: K$ is compact$\}$, where $\|\cdot\|$ is the operator norm. It is easy to see that $\|T\|^{X\rightarrow Y}_e = 0$ if and only if $T$ is compact. See ^[16,19] for the study of essential norm of some operators.

In ^[11,12], Girela and Merchán studied the operator ${\mathcal H}_\mu$ acting on spaces of analytic functions on ${\mathbb{D}}$ such as the Bloch space, BMOA, the Besov space and Hardy spaces. The paper generalizes some results of ^[11]. Moreover we also characterize the essential norm of ${\mathcal H}_\mu$ from ${\mathcal{B}}^\alpha$ to ${\mathcal{B}}$ and BMOA. We first acknowledge that the proof of part result are suggested by the technique of ^[11].

In this paper, $C$ denotes a constant which may be different in each case.

2.   The operator ${\mathcal H}_\mu:{\mathcal{B}}^\alpha\rightarrow BMOA$ (${\mathcal{B}}$), $0 < \alpha < 1$

In this section, we characterize the boundedness of ${\mathcal H}_\mu$ from ${\mathcal{B}}^\alpha$ into the $BMOA$ and the Bloch space when $0 < \alpha < 1$. For this purpose, we need some auxiliary results.

Lemma 2.1. ^[21] If $0 < \alpha < 1$, then $f\in {\mathcal{B}}^\alpha$ are bounded. If $\alpha > 1$, then $f\in{\mathcal{B}}^\alpha$ if and only if there exists some constant $C$ such that

The following lemma can be found in ^[18] (see Corollary 3.3.1 in ^[18]).

Lemma 2.2. If $a_n\downarrow 0$, then $f(z) = \sum^\infty_{n = 0}a_nz^n\in{\mathcal{B}}$ if and only if $\sup_nna_n < \infty.$

Theorem 2.3. Let $\mu$ be a positive measure on $[0, 1)$ and $0 < \alpha < 1$. Then the following statements are equivalent.

(1) The operator ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ into ${\mathcal{B}}$.

(2) The operator ${\mathcal H}_\mu$ is compact from ${\mathcal{B}}^\alpha$ into ${\mathcal{B}}$.

(3) The operator ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ into $BMOA$.

(4) The operator ${\mathcal H}_\mu$ is compact from ${\mathcal{B}}^\alpha$ into $BMOA$.

(5) The measure $\mu$ is a Carleson measure.

Proof. (1)$\Rightarrow$ (5). Assume that the operator ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ into ${\mathcal{B}}$. Let $f(z) = 1\in{\mathcal{B}}^\alpha$. Then

Note that $\mu_{n, 0}$ is positive and decreasing. For any $0 < \lambda < 1$, we choose $n$ such that $1-\frac{1}{n}\leq\lambda < 1-\frac{1}{n+1}$. Lemma 2.2 gives that

The above estimate gives that $\mu$ is a Carleson measure.

(5)$\Rightarrow$(3). Assume that $\mu$ is a Carleson measure. Lemma 2.1 implies that ${\mathcal{B}}^\alpha$ is a subspace of $H^1$ for $0 < \alpha < 1$. Then ${\mathcal H}_\mu(f)$ is an analytic function for any $f\in{\mathcal{B}}^\alpha$ by Proposition 1 in ^[9]. Moreover, ${\mathcal H}_\mu(f) = I_\mu(f)$ for any $f\in{\mathcal{B}}^\alpha$.

For any given $f\in{\mathcal{B}}^\alpha$,

Then we have

for any $f\in{\mathcal{B}}^\alpha, g\in H^1$ and $0 < r < 1$. It is easy to obtain that

whenever $f\in{\mathcal{B}}^\alpha$ and $g\in H^1$. The reader can refer to the proof of Theorem 2.2 in ^[11]. Using (2.1), we have

We obtain ${\mathcal H}_\mu(f) = I_\mu(f)\in BMOA$ for any $f\in{\mathcal{B}}^\alpha$ by Fefferman's duality Theorem.

(5)$\Rightarrow$(4). Assume that $\mu$ is a Carleson measure. Then ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ to $BMOA$ and ${\mathcal H}_\mu(f) = I_\mu(f)$ for any $f\in{\mathcal{B}}^\alpha, 0 < \alpha < 1$. Let $\{f_n\}$ be any sequence with $\sup_n\|f_n\|_{{\mathcal{B}}^\alpha}\leq 1$ and $\lim_{n\rightarrow\infty}f_n(z) = 0$ on any compact subset of ${\mathbb{D}}$. Then we have $\sup_{z\in{\mathbb{D}}}|f_n(z)|\rightarrow 0$ as $n\rightarrow\infty$ by Lemma 3.2 in ^[20]. Applying (2.1) again, we have

Then

This prove that $\lim_{n\rightarrow\infty}{\mathcal H}_\mu(f_n) = \lim_{n\rightarrow\infty}I_\mu(f_n) = 0$. So ${\mathcal H}_\mu$ is compact.

The other cases are trivial. The proof is complete.

Corollary 2.4. Let $\mu$ be a positive Borel measure on $[0, 1)$. If ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ to ${\mathcal{B}}$ for any $0 < \alpha < 1,$ then

3.   The operator ${\mathcal H}_\mu:{\mathcal{B}}^\alpha\rightarrow BMOA$ (${\mathcal{B}}$), $\alpha > 1$

In this section, we will give the essential norm of the operator ${\mathcal H}_\mu$ from ${\mathcal{B}}^\alpha$ to $BMOA$ and ${\mathcal{B}}$ when $\alpha > 1$. The following lemma will be needed in the proof of the main results.

Lemma 3.1. Let $\mu$ be a positive Borel measure on $[0, 1)$ and $\alpha > 1$. Then the following conditions are equivalent.

(1) $\int_{[0, 1)}(1-t)^{1-\alpha}d\mu(t) < \infty$.

(2) For any given $f\in{\mathcal{B}}^\alpha$, the integral in (1.1) converges for all $z\in{\mathbb{D}}$ and the resulting function $I_\mu(f)$ is analytic on ${\mathbb{D}}$.

Proof. (1)$\Rightarrow$(2). We assume that (1) holds. Lemma 2.1 gives

This implies that

for any $f\in {\mathcal{B}}^\alpha$ and $z\in{\mathbb{D}}$. By (3.1) we have

(3.2) and Fubini's Theorem give that the integral $\int_{[0, 1)}\frac{f(t)}{1-tz}d\mu(t)$ converges absolutely for any fixed $z\in{\mathbb{D}}$. Then we have

Hence $I_{\mu}(f)$ is a well defined analytic function in ${\mathbb{D}}$ and

(2)$\Rightarrow$(1). Let $f(z) = (1-z)^{1-\alpha}$. Then $f$ belongs to ${\mathcal{B}}^\alpha$. So $I_{\mu}(f)$ is well defined for every $z\in{\mathbb{D}}$. In particular,

is a complex number. Since $\mu$ is a positive Borel measure on $[0, 1)$, we get the desired result. The proof is complete.

Lemma 3.2. Let $\mu$ be a positive measure on $[0, 1)$ and $\alpha > 1$. Let $v$ be the positive measure on $[0, 1)$ defined by

Then the following conditions are equivalent.

(1) $\mu$ is an $\alpha$-Carleson measure.

(2) $v$ is a Carleson measure.

Proof. (2)$\Rightarrow$(1) Note that $v([t, 1)\lesssim (1-t)$ and $d\mu(t) = (1-t)^{\alpha-1}dv(t)$. We have

(1)$\Rightarrow$(2) Note that $\mu([t, 1))\lesssim (1-t)^\alpha$. Integrating by parts, we obtain

The proof is complete.

Lemma 3.3. Let $f(z) = \sum^\infty_{n = 0}a_nz^n \in {\mathcal{B}}^\alpha$ for any $\alpha > 0$. Then

Proof. For any $0 < r < 1$ and $f(z) = \sum^\infty_{k = 0}a_k z^k\in{\mathcal{B}}^\alpha$, we have

This gives that

Choosing $r = 1-2^{-n}$ for any fixed $n$, we obtain

Then (3.3) follows by (3.4).

A complex sequence $\{\lambda_n\}^\infty_{n = 0}$ is a multiplier from $l(2, \infty)$ to $l^1$ if and only if there exists a positive constant $C$ such that whenever $\{a_n\}^\infty_{n = 0}\in l(2, \infty)$, we have $\sum^\infty_{n = 0}|\lambda_na_n|\leq C\|\{a_n\}\|_{l(2, \infty)}$. $l(2, \infty)$ consists all the sequences $\{b_k\}^\infty_{k = 0}$ for which

The following result can be found in ^[17].

Lemma 3.4. A complex sequence $\{\lambda_n\}^\infty_{n = 0}$ is a multiplier from $l(2, \infty)$ to $l^1$ if and only if

Theorem 3.5. Let $\mu$ be a positive measure on $[0, 1)$ and $\alpha > 1$. Then the following statements are equivalent.

(1) The measure $\mu$ is an $\alpha$-Carleson measure.

(2) The operator ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ into ${\mathcal{B}}$.

(3) The operator ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ into BMOA.

Proof. (3)$\Rightarrow$ (2). It is trivial.

(2)$\Rightarrow$(1). We suppose that ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ into ${\mathcal{B}}$ for $\alpha > 1$. For any $0 < \lambda < 1$, let

where $a_{k, \lambda} = O((1-r^2)k^{\alpha-1}\lambda^k)$. It is easy to see that $f_\lambda\in{\mathcal{B}}^\alpha.$ Then

Lemma 2.2 gives that

We choose $n$ such that $1-\frac{1}{n}\leq\lambda < 1-\frac{1}{n+1}.$ We have

So $\mu$ is an $\alpha-$Carleson measure.

(1)$\Rightarrow$(3). Assume that the condition (1) holds. Lemma 3.1 shows that $I_\mu(f)$ is analytic on ${\mathbb{D}}$. Let $f(z) = \sum^\infty_{n = 0}a_nz^n\in{\mathcal{B}}^\alpha.$ By Lemma 3.3 we have that the sequence $\{a_k/k^{\alpha-1}\}\in l(2, \infty)$. Since $\mu$ is an $\alpha$-Carleson measure, we have $\mu_{k}\leq\frac{C}{k^\alpha}$ by Lemma 2.7 in ^[11]. There exists a constant $C$ such that

This shows that the sequence $\{\mu_k k^{\alpha-1}\}$ is a multiplier from $l(2, \infty)$ to $l^1$ by Lemma 3.4. Note that $\{\mu_n\}^\infty_{n = 1}$ is a decreasing sequence of positive numbers. Given any $f(z) = \sum^\infty_{n = 0}a_nz^n\in{\mathcal{B}}^\alpha$ for $\alpha > 1$, we have

This implies that ${\mathcal H}_\mu(f)(z)$ is well defined for all $z\in{\mathbb{D}}$ and ${\mathcal H}_\mu(f)$ is an analytic function in ${\mathbb{D}}$. Applying Fubini's Theorem, we get

Note that $|f(t)|\lesssim(1-t)^{1-\alpha}$ by Lemma 2.1. Applying (2.1) and Lemma 3.2, we have

We obtain ${\mathcal H}_\mu(f) = I_\mu(f)\in BMOA$ by Fefferman's duality Theorem for any $f\in{\mathcal{B}}^\alpha$. The proof is complete.

Theorem 3.6. Let $\mu$ be a positive measure on $[0, 1)$ and $\alpha > 1$. Then the following statements are equivalent.

(1) The measure $\mu$ is a vanishing $\alpha$-Carleson measure.

(2) The operator ${\mathcal H}_\mu$ is compact from ${\mathcal{B}}^\alpha$ spaces into ${\mathcal{B}}$.

(3) The operator ${\mathcal H}_\mu$ is compact from ${\mathcal{B}}^\alpha$ spaces into $BMOA$.

Proof. (3)$\Rightarrow$(2). It is trivial.

(2)$\Rightarrow$(1). Suppose that ${\mathcal H}_\mu:{\mathcal{B}}^\alpha\rightarrow {\mathcal{B}}$ is compact. Let $f_{\lambda}$ be defined by (3.5). Then $\{f_{\lambda}\}$ is a bounded sequence in ${\mathcal{B}}^\alpha$ and $\lim_{r\rightarrow 1}f_{\lambda}(z) = 0$ on any compact subset of ${\mathbb{D}}$. Then we have

The proof of Theorem 3.5 gives that

Consequently, $\mu$ is a vanishing $\alpha$-Carleson measure.

(1)$\Rightarrow$(3). Assume that $\mu$ is a vanishing $\alpha$-Carleson measure. The proof of the sufficiency for the boundedness gives that ${\mathcal H}_\mu(f) = I_\mu(f)$ and

for all $f\in{\mathcal{B}}^\alpha$ and $g\in H^1$. Let $\{f_n\}$ be any sequence with $\sup_n\|f_n\|_{{\mathcal{B}}^\alpha}\leq 1$ and $\lim_{n\rightarrow\infty}f_n(z) = 0$ on any compact subset of ${\mathbb{D}}$. Then we have

Since $v$ is a vanishing Carleson measure, where $v$ is defined by $dv(t) = (1-t)^{1-\alpha}d\mu(t)$. We obtain

where $dv_r(t) = \chi_{0 < t < r}dv(t)$. It is well known that $v$ is a vanishing Carleson measure if and only if

See p. 283 of ^[22]. Combining (3.7) and (3.8), then

This prove that $\lim_{n\rightarrow\infty}{\mathcal H}_\mu(f_n) = 0$. So ${\mathcal H}_\mu$ is compact. The proof is complete.

Theorem 3.7. Let $\mu$ be a positive measure on $[0, 1)$. If ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ to ${\mathcal{B}}$ for any $\alpha > 1$, then

Proof. For any $f\in{\mathcal{B}}^\alpha$, we have

This gives that

We now give the upper estimate of ${\mathcal H}_\mu$ from ${\mathcal{B}}^\alpha$ to BMOA. Since ${\mathcal H}_\mu$ is bounded from ${\mathcal{B}}^\alpha$ to ${\mathcal{B}}$, then the operator ${\mathcal H}_\mu$ from ${\mathcal{B}}^\alpha$ to BMOA is bounded and $\mu$ is an $\alpha-$Carleson measure by Theorem 3.5. For any $0 < r < 1$, the positive measure $\mu_r$ is defined by

It is easy to check that $\mu_r$ is a vanishing $\alpha$-Carleson measure. We have that ${\mathcal H}_{\mu_r}$ is compact from ${\mathcal{B}}^\alpha$ to BMOA by Theorem 3.6. Then

By (2.1) we have

for any $g\in H^1$, where $dv(t) = (1-t)^{1-\alpha}d\mu(t)$ and $dv_r(t) = (1-t)^{1-\alpha}d\mu_r(t)$. The above estimate gives

We now give the lower estimate of ${\mathcal H}_\mu$ from ${\mathcal{B}}^\alpha$ to ${\mathcal{B}}$. For any $0 < \lambda < 1$, let $f_\lambda$ be defined by (3.5). Then $f_\lambda\in{\mathcal{B}}^\alpha.$ Since $f_\lambda\rightarrow 0$ weakly in ${\mathcal{B}}^\alpha$, we have that $\|Kf_\lambda\|\rightarrow 0$ as $\lambda\rightarrow 1$ for any compact operator $K$ on ${\mathcal{B}}^\alpha$. Moreover

By the proof of Theorem 3.5 we have

Let $r = \lambda$ and we choose $n$ such that $1-\frac{1}{n+1}\leq\lambda < 1-\frac{1}{n}$. We have

Then

The proof is complete.

4.   Essential norm of ${\mathcal H}_\mu$ on ${\mathcal{B}}$

The reader can refer to ^[11,12] for the results of ${\mathcal H}_\mu:{\mathcal{B}}\rightarrow BMOA$ and ${\mathcal H}_\mu:{\mathcal{B}}\rightarrow{\mathcal{B}}$. In this section, we characterize the essential of norm of ${\mathcal H}_\mu$ on ${\mathcal{B}}$. The following results will be needed in the proof of the main result.

Lemma 4.1. ^[11] Let $\mu$ be a positive Borel measure on $[0, 1)$. Let $v$ be the Borel measure on $[0, 1)$ defined by

Then the following statements are equivalent.

(1) $v$ is a Carleson measure.

(2) $\mu$ is a $1-$logarithmic $1-$Carleson measure.

Lemma 4.2. ^[11] Let $\mu$ be a positive Borel measure on $[0, 1)$. Then the following statements are equivalent.

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

(2) The operator ${\mathcal H}_\mu$ is compact on ${\mathcal{B}}$.

(3) The operator ${\mathcal H}_\mu$ is compact from ${\mathcal{B}}$ to BMOA.

Theorem 4.3. Let $\mu$ be an $1-$logarithmic $1-$Carleson measure on $[0, 1)$. Then

Proof. For any $f\in{\mathcal{B}}$, we have

This gives that

We now give the upper estimate of ${\mathcal H}_\mu$ from ${\mathcal{B}}$ to BMOA. Since $\mu$ is an $1-$logarithmic $1-$Carleson measure on $[0, 1)$, the operator ${\mathcal H}_\mu$ from ${\mathcal{B}}$ to BMOA is bounded by Theorem 2.8 of ^[11]. For any $0 < r < 1$, let the positive measure $\mu_r$ defined by (3.10). It is easy to check that $\mu_r$ is a vanishing $1-$logarithmic $1-$Carleson measure. We have that ${\mathcal H}_{\mu_r}$ is compact from ${\mathcal{B}}$ to BMOA by Lemma 4.2. Then

By (2.1) we have

where $dv(t) = \log\frac{e}{1-t}d\mu(t)$ and $dv_r(t) = \log\frac{e}{1-t}d\mu_r(t)$. The positive measure $v-v_r$ is a Carleson measure by Lemma 4.1. The above estimate gives

We will give the lower estimate for ${\mathcal H}_\mu$. Let $0 < \lambda < 1$ and

where $\beta_\lambda = \log^{-1}\frac{e}{1-\lambda^2}$. Then $\{f_\lambda\}$ is a bounded sequence in ${\mathcal{B}}$ and $\lim_{\lambda\rightarrow 1^{-}}f_\lambda(z) = 0$ on any compact subset of ${\mathbb{D}}$. Since $f_\lambda\rightarrow 0$ weakly in ${\mathcal{B}}$, we have that $\|Kf_\lambda\|\rightarrow 0$ as $\lambda\rightarrow 1$ for any compact operator $K$ on ${\mathcal{B}}$. Moreover

Note that ${\mathcal H}_\mu(f_\lambda) = I_\mu(f_\lambda)$. We have

The above estimate shows that

The proof is complete.

Acknowledgments

The authors thank the referee for useful remarks and comments that led to the improvement of this paper. The first author was partially supported by NNSF of China (No. 11720101003).

Conflict of interest

The authors declare that they have no conflict of interest.