Generalized Hilbert operator on analytic function spaces

Sunanda Naik can be contacted at: snaik@gauhati.ac.in

Competing interests: The author declares that there is no conflict of interest regarding the publication of this article.

Purpose

The main objective of this work is to study the boundedness property of a generalized Hilbert operator $H_{β}$ for β ≥ 0 defined by, if $f (z) = \sum_{n = 0}^{\infty} a_{n} z^{n}$ be any analytic function on the unit disk D then $H_{β} f (z) = \sum_{n = 0}^{\infty} (\sum_{k = 0}^{\infty} \frac{Γ (n + β + 1) Γ (n + k + 1)}{Γ (n + 1) Γ (n + k + β + 2)} a_{k}) z^{n}$ on few analytic function spaces such as Hardy space H¹, weighted Hardy spaces $H_{α}^{P}$ and the Bloch space.

Design/methodology/approach

The authors shall use a few well-known results obtained by using Hardy’s inequality and the Hardy–Littlewood theorem to find the coefficient conditions of an analytic function f on the unit disk D so that f is in Hardy spaces. Also, the authors use many properties of the Gaussian hypergeometric function to prove the main results.

Findings

We find conditions on the parameter β and establish criteria for the boundedness of $H_{β}$ on certain function spaces such as Hardy space H¹, weighted Hardy spaces $H_{α}^{P}$ and the Bloch space.

Originality/value

In particular, β = 0 gives the classical Hilbert operator $H$ ⁠. So our results extend some results of Lanucha et al. [10]. The authors use of the Gaussian hypergeometric function in the proof is a different approach.

1. Introduction

We denote D as the unit disk in the complex plane C and H(D) as the class of all analytic functions on D. For 0 < p ≤ ∞, Hardy space H^p is the space of all analytic functions f ∈ H(D) for which

‖ f ‖_{p} = \sup_{0 \leq r < 1} M_{p} (r, f) < \infty,

where

\begin{matrix} M_{p} (r, f) & = & {(\frac{1}{2 π} \int_{0}^{2 π} | f (r e^{i θ}) |^{p} d θ)}^{\frac{1}{p}}, 0 < p < \infty; \\ M_{\infty} (r, f) & = & \sup_{0 \leq θ \leq 2 π} | f (r e^{i θ}) | . \end{matrix}

More properties of Hardy space can be found in [1]. For α > 0 and 0 < p ≤ ∞, we define the weighted Hardy spaces $H_{α}^{P}$ as follows:

H_{α}^{P} = {f \in H (D) : M_{p} (r, f) = O {(1 - r)}^{- α}} .

The norm in these spaces is defined by

‖ f ‖_{p, α} = \sup_{0 < r < 1} {(1 - r)}^{α} M_{p} (r, f) .

The Bloch space is all f ∈ H(D) such that

\sup_{z \in D} (1 - | z |^{2}) | f^{'} (z) | < \infty .

More about the Bloch space can be found in [2].

The classical/Gaussian hypergeometric series is defined by the power series expansion

F (a, b; c; z) = \sum_{n = 0}^{\infty} \frac{(a, n) (b, n) z^{n}}{(c, n) (1, n) n!}, (| z | < 1) .

Here a, b and c are complex numbers such that c ≠ − m, m = 0,1,2,3…and (a, n) is the Pochhammer’s symbol, which is defined as

(a, n) ≔ a (a + 1) \dots (a + n - 1) = \frac{Γ (a + n)}{Γ (a)}, n \in N

and (a, 0) = 1 for a ≠ 0. For more details see [3].

We consider the generalized Hilbert’s inequality, as stated in [4] for β ≥ 0, if 1 < p < ∞ and ${(a_{k})}_{k \in N_{0}} \in l^{p}$ then the following inequality holds:

{(\sum_{n = 0}^{\infty} {|\sum_{k = 0}^{\infty} \frac{Γ (n + β + 1) Γ (n + k + 1)}{Γ (n + 1) Γ (n + k + β + 2)} a_{k}|}^{p})}^{\frac{1}{p}} \leq \frac{Γ (\frac{1}{p}) Γ (\frac{1}{q} + β)}{Γ (1 + β)} {(\sum_{k = 0}^{\infty} | a_{k} |^{p})}^{\frac{1}{p}} .

We call this the generalized Hilbert inequality, since when β = 0, it is the well-known Hilbert inequality.

{(\sum_{n = 0}^{\infty} {|\sum_{k = 0}^{\infty} \frac{a_{k}}{n + k + 1}|}^{p})}^{\frac{1}{p}} \leq \frac{π}{\sin (π / p)} {(\sum_{k = 0}^{\infty} | a_{k} |^{p})}^{\frac{1}{p}} .

The constant $\frac{π}{\sin (π / p)}$ is best possible (see [5]). Thus the so-called generalized Hilbert matrix for β ≥ 0, $H_{β} = (\frac{Γ (n + β + 1) Γ (n + k + 1)}{Γ (n + 1) Γ (n + k + β + 2)})$ ⁠, n, k = 0, 1, 2… can be viewed as an operator on spaces of analytic functions by its action on the Taylor coefficients. If f ∈ H(D) with $f (z) = \sum_{n = 0}^{\infty} a_{n} z^{n}$ ⁠, then for β ≥ 0, so-called generalized Hilbert operator denoted by $H_{β}$ is defined by

H_{β} f (z) = \sum_{n = 0}^{\infty} (\sum_{k = 0}^{\infty} \frac{Γ (n + β + 1) Γ (n + k + 1)}{Γ (n + 1) Γ (n + k + β + 2)} a_{k}) z^{n} .

(1)

The operator $H_{β}$ was introduce in [6]. In particular β = 0, gives the classical Hilbert operator $H$ ⁠. A simple computation using the definition of the Gamma function shows $H_{β}$ has integral representation.

H_{β} f (z) = \int_{0}^{1} \frac{f (t) {(1 - t)}^{β}}{{(1 - t z)}^{β + 1}} d t .

(2)

This is an integral-type operator. Recently there has been a huge interest in studying integral-type operators on spaces of analytic functions. For some results in this area, see [7–15] and more reference therein.

It was proved by Diamantopoulos and Siskakis ([10]) that the operator $H$ is bounded on H^p, 1 < p < ∞ and not bounded on H¹. In [16] the authors proved that if f ∈ H¹, then $H$ f extends to a continuous function on $\bar{D} \ {1}$ and gives a sufficient condition for $H f \in H^{1}$ and they find that the Hilbert operator is bounded on weighted Hardy spaces $H_{α}^{p}$ ⁠, 0 < p ≤ ∞ and α > 0 if and only if α + 1/p < 1. Finally, the authors investigate the action of the operator $H$ on the Bloch space and Besov spaces. It was proved in [4] that the operator $H_{β}$ is bounded on H^p, 1 < p < ∞ and not bounded on H¹. Therefore, motivated by the results of [16] we want to extend and generalize a few results by finding the action of the operator $H_{β}$ on certain analytic function spaces.

Throughout the text, the letter C will denote the constant depending only on related parameters such as p, α, β and so on; C may differ at different occurrences.

2. Main results

In this section, we show that if f ∈ H¹, then $H_{β} f$ extends to a continuous function on $\bar{D} \ {1}$ and give a sufficient condition for $H_{β} f \in H^{1}$ ⁠. We also find that the generalized Hilbert operator is bounded on weighted Hardy spaces $H_{α}^{p}$ ⁠, 0 < p ≤ ∞ and α > 0. Finally, we obtain the action of the operator $H_{β} f$ on Bloch space. In order to prove our result, we will make use of the following lemma.

Lemma 1.

If f(z) = ∑a_nzⁿ ∈ H¹, then

\sum_{n = 0}^{\infty} \frac{| a_{n} |}{n + 1} \leq π ‖ f ‖_{1} .

(ii)
If $f (z) = \sum_{n = 0}^{\infty} a_{n} z^{n} \in H^{p}$ , 0 < p ≤ 2 then

\sum n^{p - 2} | a_{n} |^{p} < \infty .

and

{\{\sum_{n = 0}^{\infty} {(n + 1)}^{p - 2} | a_{n} |^{p}\}}^{\frac{1}{p}} \leq C_{p} ‖ f ‖_{p},

where C_p depends only on p.

(iii)
If $f (z) = \sum_{n = 0}^{\infty} a_{n} z^{n} \in H^{p}$ , 0 < p ≤ 1, then

a_{n} = o (n^{\frac{1}{p} - 1})

and

| a_{n} | \leq C n^{\frac{1}{p} - 1} ‖ f ‖_{p} .

Lemma 1(i) and (ii) are well-known results, respectively, referred to as Hardy’s inequality and the Hardy–Littlewood Theorem. We refer to p. 48 and p. 95 of [1] for parts (i) and (ii) of Lemma 1, whereas for Lemma 1 (iii), see ([1] p. 98).

Lemma 2.

If γ > 1,

\int_{0}^{2 π} \frac{d t}{| z - e^{i t} |^{γ}} = O (\frac{1}{{(1 - r)}^{γ - 1}}),

where |z| = r.

Lemma 2 is due to Hardy and Littlewood, and we refer to ([1]). The first main result of this section we shall prove here is the following.

Proposition 3.

If f ∈ H¹ then $H_{β}$ f (β ≥ 0) extends to a continuous function on $\bar{D} \ {1}$ ⁠.

Proof: From (1), we have $H_{β} f (z) = \frac{1}{1 - z} F_{f} (z)$ ⁠, where

F_{f} (z) = (1 - z) \sum_{n = 0}^{\infty} (\sum_{k = 0}^{\infty} \frac{Γ (n + β + 1) Γ (n + k + 1)}{Γ (n + 1) Γ (n + k + β + 2)} a_{k}) z^{n} .

Let $A_{n, k}^{β} = \frac{Γ (n + β) Γ (n + k)}{Γ (n) Γ (n + k + β + 1)}$ ⁠. For z ∈ D, we have

\begin{matrix} F_{f} (z) & = & \sum_{n = 0}^{\infty} (\sum_{k = 0}^{\infty} A_{n + 1, k}^{β} a_{k}) z^{n} - \sum_{n = 0}^{\infty} (\sum_{k = 0}^{\infty} A_{n + 1, k}^{β} a_{k}) z^{n + 1}, \\ = & \sum_{k = 0}^{\infty} \frac{Γ (β + 1) Γ (k + 1)}{Γ (k + β + 2)} a_{k} + \sum_{n = 1}^{\infty} (\sum_{k = 0}^{\infty} A_{n + 1, k}^{β} a_{k}) z^{n} - \sum_{n = 1}^{\infty} (\sum_{k = 0}^{\infty} A_{n, k}^{β} a_{k}) z^{n}, \\ = & Γ (β + 1) \sum_{k = 0}^{\infty} \frac{Γ (k + 1)}{Γ (k + β + 2)} a_{k} + \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty} A_{n, k}^{β} (\frac{β k - n}{n (n + k + β + 1)}) a_{k} z^{n} . \end{matrix}

For the case β = 0, we get

F_{f} (z) = \sum_{k = 0}^{\infty} \frac{1}{k + 1} a_{k} - \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty} (\frac{a_{k}}{(n + k) (n + k + 1)}) z^{n} .

The proof is similar to Lemma 2.1 of [16].

Now for β > 0, double series $S = \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty} A_{n, k}^{β} (\frac{β k - n}{n (n + k + β + 1)}) a_{k} z^{n}$ ⁠, we have

\begin{matrix} S & = & \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty} \frac{Γ (β + n) Γ (k + n)}{Γ (n) Γ (n + k + β + 1)} (\frac{β k - n}{n (n + k + β + 1)}) a_{k} z^{n}, \\ = & \sum_{k = 0}^{\infty} (\sum_{n = 1}^{\infty} \frac{Γ (β + n)}{Γ (β + 2) Γ (n + 1)} (\int_{0}^{1} t^{β + 2 - 1} {(1 - t)}^{n + k - 1} d t) (β k - n)) a_{k} z^{n}, \\ = & \frac{1}{Γ (β + 2)} \sum_{k = 0}^{\infty} \int_{0}^{1} t^{β + 1} {(1 - t)}^{k - 1} \{\sum_{n = 1}^{\infty} \frac{Γ (β + n)}{Γ (n + 1)} (β k - n) {(1 - t)}^{n}\} a_{k} z^{n} d t, \\ = & \frac{1}{Γ (β + 2)} \sum_{k = 0}^{\infty} \int_{0}^{1} t^{β + 1} {(1 - t)}^{k - 1} \sum_{n = 0}^{\infty} {(1 - t)}^{n + 1} (\frac{Γ (β + n + 1)}{Γ (n + 2)} β k - \frac{Γ (β + n + 1)}{Γ (n + 1)}) a_{k} z^{n} d t . \end{matrix}

From the definition of hypergeometric functions, a simple calculation of the above equality gives

\begin{matrix} S & = & \frac{1}{(β + 1)} \int_{0}^{1} t^{β + 1} \sum_{k = 0}^{\infty} a_{k} {(1 - t)}^{k} \sum_{n = 0}^{\infty} \{β k \frac{{(β + 1)}_{n} {(1)}_{n}}{{(2)}_{n} {(1)}_{n}} - \frac{{(β + 1)}_{n} {(1)}_{n}}{{(1)}_{n} {(1)}_{n}}\} {(1 - t)}^{n} z^{n} d t, \\ = & \frac{1}{(β + 1)} \int_{0}^{1} t^{β + 1} \sum_{k = 0}^{\infty} a_{k} {(1 - t)}^{k} {β k F (β + 1,1; 2; (1 - t) z) - F (β + 1,1; 1; (1 - t) z)} d t, \\ = & \frac{β}{β + 1} \int_{0}^{1} t^{β + 1} F (β + 1,1; 2; (1 - t) z) \sum_{k = 0}^{\infty} k a_{k} {(1 - t)}^{k} d t \\ - \frac{1}{β + 1} \int_{0}^{1} t^{β + 1} F (β + 1,1; 1; (1 - t) z) \sum_{k = 0}^{\infty} a_{k} {(1 - t)}^{k} d t . \end{matrix}

So |S| ≤ I₁ + I₂, where

I_{1} = \frac{β}{β + 1} \int_{0}^{1} t^{β + 1} F (β + 1,1; 2; (1 - t) z) \sum_{k = 0}^{\infty} k a_{k} {(1 - t)}^{k} d t .

and

I_{2} = \frac{1}{β + 1} \int_{0}^{1} t^{β + 1} F (β + 1,1; 1; (1 - t) z) \sum_{k = 0}^{\infty} a_{k} {(1 - t)}^{k} d t .

Using Theorem 2.1.3 of [3], there exists M > 0 such that |F(a, b; c; z)| < M|1 − z|^c−a−b, for c < a + b and since t < 1 − (1 − t)|z|. For β > 0 we have

\begin{matrix} I_{1} \leq C \frac{β}{β + 1} \int_{0}^{1} t^{β + 1} | 1 - (1 - t) z |^{- β} \sum_{k = 0}^{\infty} k | a_{k} | {(1 - t)}^{k} d t, \\ \leq C \frac{β}{β + 1} \sum_{k = 0}^{\infty} k | a_{k} | \int_{0}^{1} t^{β + 1} \frac{1}{t^{β}} {(1 - t)}^{k} d t, \\ \leq C \frac{β}{β + 1} \sum_{k = 0}^{\infty} k | a_{k} | \int_{0}^{1} t {(1 - t)}^{k} d t, \\ = C \frac{β}{β + 1} \sum_{k = 1}^{\infty} k | a_{k} | β (2, k + 1), \\ = C \frac{β}{β + 1} \sum_{k = 1}^{\infty} \frac{k Γ (k + 1) | a_{k} |}{Γ (k + 3)} . \end{matrix}

Using Stirling’s formula, we have

k \frac{Γ (k + 1)}{Γ (k + 3)} \leq k k^{1 - 3} = C_{1} k^{- 1} .

From Lemma 1(ii), together with the above estimate gives the integral I₁ is finite.

Similarly for

\begin{matrix} I_{2} \leq C \frac{1}{β + 1} \int_{0}^{1} t^{β + 1} | 1 - (1 - t) z |^{- (β + 1)} \sum_{k = 0}^{\infty} | a_{k} | {(1 - t)}^{k} d t, \\ \leq C \frac{1}{β + 1} \int_{0}^{1} t^{β + 1} \frac{1}{t^{β + 1}} \sum_{k = 0}^{\infty} | a_{k} | {(1 - t)}^{k} d t, \\ \leq \frac{1}{β + 1} \int_{0}^{1} {(1 - t)}^{k} \sum_{k = 0}^{\infty} a_{k} | d t, \\ = C \frac{1}{β + 1} \sum_{k = 0}^{\infty} | a_{k} | β (1, k + 1), \\ = C \frac{1}{β + 1} \sum_{k = 0}^{\infty} \frac{| a_{k} |}{k + 1} . \end{matrix}

Lemma 1(i) we have I₂ < ∞. This implies that the last double series S converges absolutely and uniformly. Now for the first summation from Lemma 1(iii), we can write

| a_{k} | < c ‖ f ‖_{1} .

and using Stirling’s formula

\frac{Γ (k + 1)}{Γ (k + β + 2)} \leq C_{1} k^{1 - β - 2} = C_{1} k^{- β - 1},

we have

\frac{Γ (β + 1) Γ (k + 1)}{Γ (k + β + 2)} | a_{k} | < C_{2} \frac{‖ f ‖_{1}}{k^{β + 1}} .

Since β > 0 the series converges. Hence the function F_f can be continuously extended to $\bar{D}$ ⁠. This completes the proof.

Since H¹ ⊂ H^p for 0 < p < 1, as a consequence of the above proposition, we get the following.

Corollary 4.

The operator $H_{β}$ acts as a bounded operator from H¹ into H^p for 0 < p < 1.

The next result gives a sufficient condition for the generalized Hilbert operator $H_{β}$ to be in H¹ space.

Theorem 5.

Let β ≥ 0. If f ∈ H¹ and

\int_{- π}^{π} | f (e^{i θ}) | \log \frac{π}{| θ |} d θ < \infty,

then $H_{β} f \in H^{1}$ ⁠.

Proof: Using (2),

H_{β} f (z) = \int_{0}^{1} \frac{f (t) {(1 - t)}^{β}}{{(1 - t z)}^{β + 1}} d t, z \in D,

Since f ∈ H¹, the convergence of the above integral is guaranteed by the Fejér–Riesz inequality. Since β ≥ 0, (1 − t)^β ≤ |1 − te^iθ|^β and using the Fubini theorem, we obtain

\begin{array}{l} \frac{1}{2 π} \int_{0}^{2 π} | H_{β} f (e^{i θ}) | d θ & \leq & \frac{1}{2 π} \int_{0}^{2 π} \int_{0}^{1} \frac{| f (t) | {(1 - t)}^{β} d t}{{| 1 - t e^{i θ} |}^{β + 1}} d θ, \\ \leq & \frac{1}{2 π} \int_{0}^{2 π} \int_{0}^{1} \frac{| f (t) d t}{| 1 - t e^{i θ} |} d θ, \\ = & \int_{0}^{1} | f (t) | \frac{1}{2 π} \int_{0}^{2 π} \frac{d θ}{| 1 - t e^{i θ} |} d t, \\ \leq & C \int_{0}^{1} | f (t) | \log \frac{2}{1 - t} d t . \end{array}

The remaining part of the proof follows if we proceed similar to Theorem 2.3 of [16].

Many authors have studied the boundedness properties of various generalized Hilbert operators on the Hardy spaces, on the weighted Bergman spaces and on the Dirichlet spaces; see [17–20] and references therein. The following result gives the action of the generalized Hilbert operator on the weighted Hardy spaces.

Theorem 6.

For α > 0, β ≥ 0. Suppose $α + \frac{1}{p} < 1$ and $α + \frac{1}{p} - 1 < β \leq α + \frac{1}{p}$ then $H_{β}$ $m a p s H_{α}^{p}$ into $H_{α}^{p}$ ⁠.

Proof: Let $f \in H_{α}^{p}$ ⁠, and h = $H_{β}$ f, Then from (2), we have

h^{'} (z) = \int_{0}^{1} \frac{f (t) {(1 - t)}^{β} (β + 1) t}{{(1 - t z)}^{β + 2}} d t .

From the above formula, we obtain

\begin{matrix} M_{p} (r, h^{'}) = {(\frac{1}{2 π} \int_{0}^{2 π} {|\int_{0}^{1} \frac{f (t) {(1 - t)}^{β} (β + 1) t d t}{{(1 - t z)}^{β + 2}}|}^{P} d θ)}^{\frac{1}{p}}, \\ \leq {(\frac{1}{2 π} \int_{0}^{2 π} {(\int_{0}^{1} \frac{| f (t) | {(1 - t)}^{β} (β + 1) t d t}{| 1 - t r e^{i θ} |^{(β + 2)}})}^{p} d θ)}^{\frac{1}{p}} . \end{matrix}

Using Lemma 2, Minkowski’s inequality and inequality, for $f \in H_{α}^{p}$

| f (r) | \leq C {(1 - r)}^{- α - \frac{1}{p}},

we get

\begin{matrix} M_{p} (r, h^{'}) \leq \int_{0}^{1} | f (t) | {(1 - t)}^{β} {(\frac{{(β + 1)}^{p}}{{(2 π)}^{p}} \int_{0}^{2 π} | 1 - t r e^{i θ} |^{{- (β + 2))}^{p}} d θ)}^{\frac{1}{p}} d t, \\ \leq (β + 1) \int_{0}^{1} | f (t) | {(1 - t)}^{β} {(1 - t r)}^{\frac{1 - (β + 2) p}{p}} d t, \\ \leq C \int_{0}^{1} {(1 - t)}^{- α - \frac{1}{p} + β} {(1 - t r)}^{\frac{1}{p} - (β + 2)} d t . \end{matrix}

For the integral of the last inequality, we have

\int_{0}^{1} {(1 - t)}^{- α - \frac{1}{p} + β} {(1 - t r)}^{\frac{1}{p} - (β + 2)} d t = I_{3} + I_{4},

where

\begin{matrix} I_{3} & = & \int_{0}^{r} {(1 - t)}^{- α + β - \frac{1}{p}} {(1 - t r)}^{\frac{1}{p} - (β + 2)} d t, \\ \leq & C \int_{0}^{r} {(1 - r)}^{- α + β - \frac{1}{p}} {(1 - t)}^{\frac{1}{p} - (β + 2)} d t, \\ \leq & C {(1 - r)}^{- α + β - \frac{1}{p}} \int_{0}^{r} {(1 - t)}^{\frac{1}{p} - (β + 2)} d t, \\ \leq & C {(1 - r)}^{- α + β - \frac{1}{p}} [{(1 - r)}^{\frac{1}{p} - (β + 2) + 1} - 1], \\ \leq & C {(1 - r)}^{- α - 1} . \end{matrix}

and

\begin{matrix} I_{4} & = & \int_{r}^{1} {(1 - t)}^{- α + β - \frac{1}{p}} {(1 - t r)}^{\frac{1}{p} - (β + 2)} d t, \\ \leq & C \int_{r}^{1} {(1 - t)}^{- α + β - \frac{1}{p}} {(1 - r)}^{\frac{1}{p} - (β + 2)} d t, \\ \leq & C {(1 - r)}^{\frac{1}{p} - (β + 2)} \int_{r}^{1} {(1 - t)}^{- α + β - \frac{1}{p}} d t, \\ \leq & C {(1 - r)}^{\frac{1}{p} - (β + 2)} {(1 - r)}^{- α + β - \frac{1}{p} + 1,} \\ \leq & C {(1 - r)}^{- α - 1} . \end{matrix}

Therefore

M_{p} (r, h^{'}) \leq C {(1 - r)}^{- (α + 1)} .

Hence by using Theorem 5.5 [1], we have

M_{p} (r, h) < C {(1 - r)}^{- α} .

This completes the proof

Let f(z) = − log(1 − z). It is easy to see $f \in B$ ⁠. A simple calculation using Gamma function in (1) gives

H_{β} f (z) = \frac{1}{Γ (β + 1)} \int_{0}^{1} {(1 - t)}^{β} \log \frac{1}{1 - t} \sum_{n = 0}^{\infty} \frac{Γ (n + β + 1)}{Γ (n + 1)} t^{n} z^{n} d t .

Since −log(1 − t) ≥ t for 0 ≤ t < 1, the above equation gives

H_{β}^{'} f (r) \geq \frac{1}{Γ (β + 1)} \int_{0}^{1} {(1 - t)}^{β} \sum_{n = 1}^{\infty} \frac{Γ (n + β + 1)}{Γ (n + 1)} n t^{n + 1} r^{n - 1} d t .

A few steps of simple calculations give

H_{β}^{'} f (r) \geq \frac{2 F (β + 2,3; β + 4; r)}{(β + 2) (β + 3)} .

Using Theorem 2.1.3 of [3], we have $(1 - r) H_{β}^{'} f (r) \geq C$ as r → 1. This gives $H_{β} \notin B$ ⁠. This implies the Bloch space is not mapped into itself. Therefore, in our next result, we describe a space of analytic functions in D that are mapped by $H_{β}$ into the Bloch space.

Theorem 7.

If f ∈ H(D) satisfies the condition

(3.1) \sup_{z \in D} | f^{'} (z) | (1 - | z |) {(\log \frac{2}{1 - | z |})}^{1 + ϵ} < \infty,

for an ɛ > 0, then $H_{β} f \in B$ for β ≥ 0.

Proof: Assume that f ∈ H(D) satisfies (3.1) and set F(z) = f(z) − f(0). It is enough to show that $H_{β} F \in B$ ⁠. Clearly, F also satisfies (3.1). Then by Lemma 4.2.8 in [2], we can write

F (z) = \int_{D} \frac{F^{'} (ω) (1 - | ω |^{2})}{\bar{ω} {(1 - \bar{ω} z)}^{2}} d A (ω) .

Consequently,

\begin{matrix} | {(H_{β} F)}^{'} (z) | = |\int_{0}^{1} \int_{D} \frac{F^{'} (ω) (1 - {| ω |}^{2}) t {(1 - t)}^{β} (β + 1) d A (ω) d t}{\bar{ω} {(1 - \bar{ω} t)}^{2} {(1 - t z)}^{β + 2}}|, \\ \leq \int_{0}^{1} \int_{D} \frac{| f^{'} (ω) | (1 - | ω |^{2}) t {(1 - t)}^{β} (β + 1) | d A (ω) | d t}{| \bar{ω} {(1 - \bar{ω} t)}^{2} {(1 - t z)}^{β + 2} |} . \end{matrix}

Using (3.1)

\begin{matrix} | {(H_{β} F)}^{'} (z) | \leq C \int_{0}^{1} \int_{D} \frac{(1 - {| ω |}^{2}) {(1 - t)}^{β} | d A (ω) | d t}{| \bar{ω} {(1 - \bar{ω} t)}^{2} {(1 - t z)}^{β + 2} | (1 - | ω |) {(\log \frac{2}{1 - | ω |})}^{1 + ϵ}}, \\ \leq C \int_{0}^{1} \int_{D} \frac{{(1 - t)}^{β} | d A (ω) | d t}{| \bar{ω} {(1 - \bar{ω} t)}^{2} {(1 - t z)}^{β + 2} | {(\log \frac{2}{1 - | ω |})}^{1 + ϵ}}, \\ \leq C \int_{0}^{1} \int_{0}^{1} \int_{0}^{2 π} \frac{\frac{1}{2 π} {(1 - t)}^{β} s d s d θ d t}{s \log {(\frac{2}{1 - | s e^{i θ} |})}^{1 + ϵ} | {(1 - s e^{- i θ} t)}^{2} {(1 - t z)}^{β + 2} |} . \end{matrix}

Since (1 − t)^β ≤ |1 − tz|^β and Lemma 2, we have

\begin{array}{l} | {(H_{β} F)}^{'} (z) | \leq C \int_{0}^{1} \int_{0}^{1} \frac{d s d t}{\log {(\frac{2}{1 - s})}^{1 + ϵ} (1 - s t) {(1 - t | z |)}^{2}}, \\ \leq C \int_{0}^{1} {(\log \frac{2}{1 - s})}^{- 1 - ϵ} \frac{d s}{(1 - s)} \int_{0}^{1} \frac{d t}{{(1 - t | z |)}^{2}}, \\ \leq \frac{C}{1 - | z |} \int_{0}^{1} {(\log \frac{2}{1 - s})}^{- 1 - ϵ} \frac{d s}{(1 - s)} . \end{array}

Since the last integral is finite, our claim is proved.

References

Duren

Theory of H^p spaces

New York and London

Academic Press

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1970

Generalized Hilbert operator on analytic function spaces

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