New Upper Bound for the Hilbert Matrix Norm on Negatively Indexed Weighted Bergman Spaces

Abstract

In this note, we obtain a new upper bound for the norm of the Hilbert matrix H on the weighted Bergman spaces ${A}^p_\alpha$ when $-1<\alpha <0$, which is better than the known one in the case $-1/2<\alpha <0$.

1 Introduction

In recent years, the Hilbert matrix has been intensively studied on various spaces of holomorphic functions on the unit disk of the complex plane. In particular, questions related to the boundedness and exact norm of the Hilbert matrix on weighted Bergman spaces have also been very topical in recent years. See [1,2,3,4,5,6,7,8,9] and references therein. For information related to the spectrum of Hilbert matrix operator, see [10]. It is known [5], that the Hilbert matrix is bounded on weighted Bergman spaces ${A}^p_\alpha$ if and only if $1<\alpha +2<p$. It was obtained in [6] that

$$\begin{aligned} \Vert {H}\Vert _{{{A}^p_\alpha }\rightarrow {A}^p_\alpha }\ge \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}, \end{aligned}$$

for all $1<\alpha +2<p$, and it was conjectured that this lower bound is the exact norm of the Hilbert matrix on weighted Bergman space ${A}^p_\alpha$. This immediately implies, that in order to prove the conjecture, it is necessary to show that the following inequality is valid

$$\begin{aligned} \Vert {H}\Vert _{{{A}^p_\alpha }\rightarrow {A}^p_\alpha }\le \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}, \end{aligned}$$

for $1<\alpha +2<p$. The conjecture is completely resolved in the case $\alpha =0$, that is, in the case of unweighted Bergman spaces. See [1, 2, 4, 8]. Also, the conjecture is resolved in the case when $\alpha >0$ and $2(\alpha +2)\le p$ in [6], which reduces the conjecture in the case $\alpha >0$, to the following interval $\alpha +2<p<2(\alpha +2)$. Lindström, Miihkinen and Wikman in [9] resolved the conjecture in the case $\alpha >0$ when

$$\begin{aligned} \alpha +2+\sqrt{(\alpha +2)^2-\frac{1}{2}(\alpha +2)}\le p<2(\alpha +2). \end{aligned}$$

Moreover, it was very recently shown in [7] that the conjecture is correct in the case $\alpha >0$ when

$$\begin{aligned} \alpha +2+\sqrt{(\alpha +2)^2-\left( \sqrt{2}-\frac{1}{2}\right) (\alpha +2)}\le p<2(\alpha +2). \end{aligned}$$

In the case $-1<\alpha <0$ an explicit upper bound for the norm of the Hilbert matrix on weighted Bergman spaces ${A}^p_\alpha$ is obtained for the first time in [7, Theorem 1.3]. In this note, we provide a new upper bound for the norm of the Hilbert matrix H on the weighted Bergman spaces ${A}^p_\alpha$ when $-1<\alpha <0$. Namely, we have the following main result of this note.

Theorem 1.1

Let $-1<\alpha <0$ and $\alpha +2<p$.

• (i) If $2(\alpha +2)\le p$ then

  $$\begin{aligned} \left\| {H}\right\| _{{A}^p_\alpha \rightarrow {A}^p_\alpha }\le 2^\frac{1-\alpha }{p}\frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}. \end{aligned}$$

• (ii) If $\alpha +2<p<2(\alpha +2)$ then

  $$\begin{aligned} \left\| {H}\right\| _{{A}^p_\alpha \rightarrow {A}^p_\alpha }\le 2^\frac{1-\alpha }{p}\left( 1+2^{\frac{2(\alpha +2)}{p}-1}\right) \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}. \end{aligned}$$

We note that in [7, Theorem 1.3], the constant $2^\frac{1-\alpha }{p}$ from Theorem 1.1, in both of its parts, is replaced by the constant $2^\frac{\alpha +2}{p}$. Since $2^\frac{1-\alpha }{p}<2^\frac{\alpha +2}{p}$, when $-1/2<\alpha <0$, Theorem 1.1 is a slight improvement of [7, Theorem 1.3] in that case.

1.1 Weighted Bergman Spaces

Let $\mathrm {Hol}({\mathbb {D}})$ be the space of all holomorphic functions on the unit disk ${\mathbb {D}}=\left\{ z\in {\mathbb {C}}:|z|<1 \right\}$ of the complex plane ${\mathbb {C}}$. For $0<r<1$, we write ${\mathbb {D}}_r=r{\mathbb {D}}$. The weighted Bergman space ${A}^p_\alpha$, where $0<p<\infty$ and $-1<\alpha <\infty$, is defined as follows

$$\begin{aligned} {A}^p_\alpha =\left\{ f\in \mathrm {Hol}({\mathbb {D}}):\Vert f\Vert _{{A}^p_\alpha }=\left( \frac{\alpha +1}{\pi }\int _{\mathbb {D}}|f(z)|^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z)\right) ^{\frac{1}{p}}<\infty \right\} , \end{aligned}$$

where $\mathrm {dm}$ is the Euclidean area measure in the complex plane, that is,

$$\begin{aligned} \mathrm {dm}(z)=\mathrm {d}x\,\mathrm {d}y=r\,\mathrm {d}r\,\mathrm {d}\theta , \,\,\, \text {for} \,\,\, z=x+\mathrm {i}y=r\mathrm {e}^{\mathrm {i}\theta }\in {\mathbb {C}}. \end{aligned}$$

For further information related to weighted Bergman spaces, see monograph [11].

1.2 The Hilbert Matrix

The infinite matrix

$$\begin{aligned} {H}=\left[ \frac{1}{n+k+1}\right] _{n,k=0}^\infty , \end{aligned}$$

is called the Hilbert matrix. We note that the Hilbert matrix can be viewed as an operator on spaces of holomorphic functions on the unit disk of the complex plane, by its action on their Taylor coefficients. Namely, if

$$\begin{aligned} f(z)=\sum _{n=0}^\infty a_nz^n, \,\,\, z\in {\mathbb {D}}, \end{aligned}$$

is a holomorphic function on the unit disk ${\mathbb {D}}$, then

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

For other valuable information related to the Hilbert matrix see [3, 4, 7, 9].

2 Preliminaries

In this section, we first prove one auxiliary assertion, which will be useful later. Namely, we have the following result.

Lemma 2.1

Let $0<c<1$, $-1<\alpha <0$ and

$$\begin{aligned} {F}(s)={F}_{c,\alpha }(s)=\frac{\int _{\max \lbrace s,c \rbrace }^1(r+c)\left( \left( 1-c\right) ^2-\left( r-c\right) ^2\right) ^\alpha \mathrm {d}r}{\int _s^1r\left( 1-r^2\right) ^\alpha \mathrm {d}r}, \,\,\, s\in [0,1). \end{aligned}$$

Then

$$\begin{aligned} \underset{s\in [0,1)}{\max }{F}(s)={F}(c)\le \frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}. \end{aligned}$$

Proof

Let

$$\begin{aligned} \Phi (r)=(r+c)\left( \left( 1-c\right) ^2-\left( r-c\right) ^2\right) ^\alpha , \,\,\, \Psi (r)=r\left( 1-r^2\right) ^\alpha , \,\,\, r\in (0,1). \end{aligned}$$

Then

$$\begin{aligned} {F}(s)=\frac{\int _{\max \lbrace s,c \rbrace }^1\Phi }{\int _s^1\Psi }, \,\,\, s\in [0,1). \end{aligned}$$

If $s\in [0,c]$, then we have

$$\begin{aligned} {F}(s)=\frac{\int _c^1\Phi }{\int _s^1\Psi }=\frac{\int _{c}^1(r+c)\left( \left( 1-c\right) ^2-\left( r-c\right) ^2\right) ^\alpha \mathrm {d}r}{\int _s^1r\left( 1-r^2\right) ^\alpha \mathrm {d}r}, \end{aligned}$$

which immediately implies that F is an increasing function on the interval [0, c], because $s\mapsto \int _s^1\Psi$ is a decreasing function. On the other hand, when $s\in [c,1)$, we get

$$\begin{aligned} {F}(s)=\frac{\int _s^1\Phi }{\int _s^1\Psi }=\frac{\int _{s}^1(r+c)\left( \left( 1-c\right) ^2-\left( r-c\right) ^2\right) ^\alpha \mathrm {d}r}{\int _s^1r\left( 1-r^2\right) ^\alpha \mathrm {d}r}. \end{aligned}$$

Let

$$\begin{aligned} h(r)=\frac{\Phi (r)}{\Psi (r)}=\frac{(r+c)(1+r-2c)^\alpha }{r(1+r)^\alpha }, \,\,\, r\in [c,1). \end{aligned}$$

It is easy to check that

$$\begin{aligned} h'(r)=\frac{c(1+r-2c)^{\alpha -1}\xi (r)}{r^2(1+r)^{\alpha +1}}, \end{aligned}$$

where $\xi (r)=2c-1+(2\alpha c+2c-2)r+(2\alpha -1)r^2$ on the interval [c, 1). Since $0<c<1$ and $-1<\alpha <0$, we have $\xi ''(r)=2(2\alpha -1)<0$, which implies $\xi '(r)\le \xi '(c)=6\alpha c-2<0$ on [c, 1). Therefore, $\xi (r)\le \xi (c)=(4\alpha +1)c^2-1<0$ on [c, 1). This implies $h'\le 0$ on [c, 1). Hence, h is a decreasing function on the interval [c, 1). Let $c\le t\le s<1$. Then, we obtain

$$\begin{aligned} \int _s^1\Phi \int _t^s\Psi =\int _s^1h\Psi \int _t^s\Psi \le \int _s^1\Psi h(s)\int _t^s\Psi \le \int _s^1\Psi \int _t^sh\Psi =\int _s^1\Psi \int _t^s\Phi , \end{aligned}$$

or equivalently

$$\begin{aligned} \int _s^1\Phi \left( \int _t^1\Psi -\int _s^1\Psi \right) \le \int _s^1\Psi \left( \int _t^1\Phi -\int _s^1\Phi \right) , \end{aligned}$$

which implies

$$\begin{aligned} {F}(s)=\frac{\int _s^1\Phi }{\int _s^1\Psi }\le \frac{\int _t^1\Phi }{\int _t^1\Psi }={F}(t). \end{aligned}$$

This means that F is decreasing function on [c, 1). We have already proved that F is increasing on [0, c]. Therefore,

$$\begin{aligned} \underset{s\in [0,1)}{\max }{F}(s)={F}(c)=\frac{\int _c^1\Phi }{\int _c^1\Psi }. \end{aligned}$$

Note that

$$\begin{aligned} \int _c^1\Psi =\int _c^1r\left( 1-r^2\right) ^\alpha \mathrm {d}r=\frac{\left( 1-c^2\right) ^{\alpha +1}}{2(\alpha +1)}, \end{aligned}$$

and

$$\begin{aligned} \int _c^1\Phi= & {} \int _{c}^1(r+c)\left( \left( 1-c\right) ^2-\left( r-c\right) ^2\right) ^\alpha \mathrm {d}r = \int _0^{1-c}(t+2c)\left( \left( 1-c\right) ^2-t^2\right) ^\alpha \mathrm {d}t \\= & {} \left( 1-c\right) ^{2\alpha +1}\int _0^1 \left( (1-c)x+2c\right) \left( 1-x^2\right) ^\alpha \mathrm {d}x\\= & {} \left( 1-c\right) ^{2\alpha +1}\left( \frac{1-c}{2(\alpha +1)}+2c\int _0^1\left( 1-x^2\right) ^\alpha \mathrm {d}x\right) \\\le & {} \left( 1-c\right) ^{2\alpha +1}\left( \frac{1-c}{2(\alpha +1)}+2c\int _0^1\left( 1-x\right) ^\alpha \mathrm {d}x\right) =\frac{\left( 1-c\right) ^{2\alpha +1}(1+3c)}{2(\alpha +1)}. \end{aligned}$$

Hence

$$\begin{aligned} {F}(c)\le \frac{\left( 1-c\right) ^{2\alpha +1}(1+3c)}{2(\alpha +1)}\bigg /\frac{\left( 1-c^2\right) ^{\alpha +1}}{2(\alpha +1)}=\frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}. \end{aligned}$$

This completes the proof. $\square$

Let $\eta :{\mathbb {D}}\rightarrow {\mathbb {D}}$ be a holomorphic function, $0<p<\infty$, $-1<\alpha <0$ and $f\in \mathrm {Hol}({\mathbb {D}})$. Then we have the following well-known inequality

$$\begin{aligned} \int _{\mathbb {D}}\left| (f\circ \eta )(z)\right| ^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z)\!\le \!\left( \frac{1+|\eta (0)|}{1-|\eta (0)|}\right) ^{\alpha +2}\int _{\mathbb {D}}\left| f(z)\right| ^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z), \end{aligned}$$

which can be viewed as a consequence of Littlewood Subordination Principle. See [11, Theorem 11.6]. In the special case, when

$$\begin{aligned} \eta (z)=\rho z+c, \,\,\, \text {for} \,\,\, z\in {\mathbb {D}}, \end{aligned}$$

where $0<c<1$ and $\rho =1-c$, we obtain

$$\begin{aligned} \rho ^{\alpha +2}\int _{\mathbb {D}}\left| (f\circ \eta )(z)\right| ^p\left( 1-|z|^2\right) ^\alpha \!\!\mathrm {dm}(z)\!\le \!\left( 1+c\right) ^{\alpha +2}\int _{\mathbb {D}}\!\!\left| f(z)\right| ^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z). \end{aligned}$$

We note that this inequality, but only for $-1<\alpha <0$, was used in the proof of [7, Theorem 1.3]. Under the above conditions, we are now able to show a slightly different inequality, that we will use in the proof of our Theorem 1.1. Actually, we obtain the following result.

Lemma 2.2

Let $0<p<\infty$, $-1<\alpha <0$, $0<c<1$ and $\eta (z)=\rho z+c$ for $z\in {\mathbb {D}}$, where $\rho =1-c$. Then

$$\begin{aligned} \rho ^{\alpha +2}\int _{\mathbb {D}}\!\!|(f\circ \eta )(z)|^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z)\!\le \!\frac{1+3c}{(1+c)^{\alpha +1}}\int _{\mathbb {D}}|f(z)|^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z), \end{aligned}$$

where $f\in \mathrm {Hol}({\mathbb {D}})$.

Proof

Let $0<r<1$. We define ${R}={R}(r)=\rho r+c$. Note that $c<{R}<1$. If $z\in \overline{{\mathbb {D}}}_r$, then $|\eta (z)|\le \rho |z|+c\le \rho r+c={R}$, that is, $\eta (z)\in \overline{{\mathbb {D}}}_{{R}}$. Let u be the harmonic function in ${\mathbb {D}}_{{R}}$ with boundary values $|f|^p$ on $\partial {\mathbb {D}}_{{R}}$. Since $|f|^p$ is a subharmonic function in the unit disk ${\mathbb {D}}$, we have $|f|^p\le u$ on $\overline{{\mathbb {D}}}_{{R}}$. Therefore, $|f\circ \eta |^p\le u\circ \eta$ on $\overline{{\mathbb {D}}}_r$. From this and by using the fact that $u\circ \eta$ is also a harmonic function, we obtain

$$\begin{aligned} \int _0^{2\pi }\left| \left( f\circ \eta \right) \left( r\mathrm {e}^{\mathrm {i}\theta }\right) \right| ^p\mathrm {d}\theta \le \int _0^{2\pi }\left( u\circ \eta \right) \left( r\mathrm {e}^{\mathrm {i}\theta }\right) \mathrm {d}\theta =2\pi \left( u\circ \eta \right) (0)=2\pi u(c).\nonumber \\ \end{aligned}$$

(2.1)

By using Harnack inequality we get

$$\begin{aligned} u(c)\le \frac{{R}+c}{{R}-c}\,u(0)=\frac{1}{2\pi }\frac{{R}+c}{{R}-c}\int _0^{2\pi }u\left( {R}\mathrm {e}^{\mathrm {i}\theta }\right) \mathrm {d}\theta . \end{aligned}$$

(2.2)

From (2.1) and (2.2), we derive

$$\begin{aligned} \int _0^{2\pi }\left| \left( f\circ \eta \right) \left( r\mathrm {e}^{\mathrm {i}\theta }\right) \right| ^p\mathrm {d}\theta \le \frac{{R}+c}{{R}-c}\int _0^{2\pi }u\left( {R}\mathrm {e}^{\mathrm {i}\theta }\right) \mathrm {d}\theta \end{aligned}$$

and since $u=|f|^p$ on $\partial {\mathbb {D}}_{\mathrm {R}}$, we find

$$\begin{aligned} \int _0^{2\pi }\left| \left( f\circ \eta \right) \left( r\mathrm {e}^{\mathrm {i}\theta }\right) \right| ^p\mathrm {d}\theta \le \frac{{R}+c}{{R}-c}\int _0^{2\pi }\left| f\left( {R}\mathrm {e}^{\mathrm {i}\theta }\right) \right| ^p\mathrm {d}\theta , \end{aligned}$$

(2.3)

where $0<r<1$ and ${R}={R}(r)=\rho r+c$. On the other hand, we denote

$$\begin{aligned} \varphi (r)=\int _0^{2\pi }\left| f\left( r\mathrm {e}^{\mathrm {i}\theta }\right) \right| ^p\mathrm {d}\theta \,\,\, \text {and} \,\,\, \chi (r)=\varphi (r)-\varphi (0), \end{aligned}$$

for $0<r<1$. Then $\varphi$ is an increasing and differentiable function on (0, 1), which implies that $\chi$ is also increasing and differentiable on (0, 1). Hence

$$\begin{aligned} \chi '\ge 0 \,\,\, \text {on} \,\,\, (0,1) \,\,\, \text {and} \,\,\, \chi (r)=\int _0^r\chi '\left( s\right) \mathrm {d}s, \end{aligned}$$

for $0<r<1$. It follows from (2.3) that

$$\begin{aligned} \int _{\mathbb {D}}|(f\circ \eta )(z)|^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z)= & {} \int _0^1r\left( 1-r^2\right) ^\alpha \int _0^{2\pi }\left| \left( f\circ \eta \right) \left( r\mathrm {e}^{\mathrm {i}\theta }\right) \right| ^p\mathrm {d}\theta \,\mathrm {d}r \\&{\le }&\int _0^1r\left( 1-r^2\right) ^\alpha \frac{{R}+c}{{R}-c}\int _0^{2\pi }\left| f\left( {R}\mathrm {e}^{\mathrm {i}\theta }\right) \right| ^p\mathrm {d}\theta \,\mathrm {d}r\\= & {} \int _0^1r\left( 1-r^2\right) ^\alpha \frac{{R}+c}{{R}-c}\,\varphi \left( {R}\right) \mathrm {d}r, \end{aligned}$$

and by using change of variable ${R}={R}(r)=\rho r+c$, we get

$$\begin{aligned} \int _0^1r\left( 1-r^2\right) ^\alpha \frac{{R}+c}{{R}-c}\,\varphi \left( {R}\right) \mathrm {d}r\!=\!\int _c^1\frac{{R}-c}{\rho }\left( 1-\left( \frac{{R}-c}{\rho }\right) ^2\right) ^\alpha \frac{{R}+c}{{R}-c}\,\varphi \left( {R}\right) \frac{\mathrm {d}{R}}{\rho }, \end{aligned}$$

which implies

$$\begin{aligned} \int _{\mathbb {D}}|(f\circ \eta )(z)|^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z)\le \frac{1}{\rho ^{2\alpha +2}}\int _c^1\left( {R}+c\right) \left( \rho ^2-\left( {R}-c\right) ^2\right) ^\alpha \varphi \left( {R}\right) \mathrm {d}{R}, \end{aligned}$$

or equivalently

$$\begin{aligned} \int _{\mathbb {D}}|(f\circ \eta )(z)|^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z)\le \frac{\int _c^1\left( r+c\right) \left( (1-c)^2-\left( r-c\right) ^2\right) ^\alpha \varphi \left( r\right) \mathrm {d}r}{(1-c)^{2\alpha +2}}. \end{aligned}$$

That is, we have

$$\begin{aligned} \rho ^{\alpha +2}\int _{\mathbb {D}}|(f\circ \eta )(z)|^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z)\le \frac{1}{(1-c)^\alpha }\int _c^1\Phi \varphi , \end{aligned}$$

where we used the notation from Lemma 2.1. Also

$$\begin{aligned} \frac{1+3c}{(1+c)^{\alpha +1}}\int _{\mathbb {D}}|f(z)|^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(z)=\frac{1+3c}{(1+c)^{\alpha +1}}\int _0^1\Psi \varphi . \end{aligned}$$

Based on the previous formulas, it is enough to prove

$$\begin{aligned} \int _c^1\Phi \varphi \le \frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}\int _0^1 \Psi \varphi , \end{aligned}$$

or equivalently

$$\begin{aligned} \int _c^1\Phi \chi +\varphi (0)\int _c^1\Phi \le \frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}\left( \int _0^1\Psi \chi +\varphi (0)\int _0^1\Psi \right) . \end{aligned}$$

Based on Lemma 2.1, we have

$$\begin{aligned} \varphi (0)\int _c^1\Phi ={F}\left( 0\right) \varphi \left( 0\right) \int _0^1\Psi \le \frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}\,\varphi \left( 0\right) \int _0^1\Psi . \end{aligned}$$

So, it is enough to prove that

$$\begin{aligned} \int _c^1\Phi \chi \le \frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}\int _0^1\Psi \chi . \end{aligned}$$

But, by using Fubini theorem and Lemma 2.1, we obtain

$$\begin{aligned} \int _c^1\Phi \chi= & {} \int _c^1\Phi \left( r\right) \chi \left( r\right) \mathrm {d}r=\int _c^1\int _0^r\Phi \left( r\right) \chi '\left( s\right) \mathrm {d}s\,\mathrm {d}r \\= & {} \int _0^1\chi '\left( s\right) \int _{\max \lbrace s,c \rbrace }^1\Phi \left( r\right) \mathrm {d}r\,\mathrm {d}s = \int _0^1\chi '\left( s\right) {F}\left( s\right) \int _{s}^1\Psi \left( r\right) \mathrm {d}r\,\mathrm {d}s \\\le & {} \frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}\int _0^1\chi '\left( s\right) \int _{s}^1\Psi \left( r\right) \mathrm {d}r\,\mathrm {d}s \\= & {} \frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}\int _0^1\int _0^r\Psi \left( r\right) \chi '\left( s\right) \mathrm {d}s\,\mathrm {d}r\\= & {} \frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}\int _0^1\Psi \left( r\right) \chi \left( r\right) \mathrm {d}r=\frac{(1-c)^\alpha (1+3c)}{(1+c)^{\alpha +1}}\int _0^1\Psi \chi . \end{aligned}$$

This finishes the proof. $\square$

3 Proof of the Main Result

As stated before, the Hilbert matrix H is bounded on weighted Bergman space ${A}^p_\alpha$ if and only if $1<\alpha +2<p$. It is a well-known fact [2, 3, 6], that if $f\in {A}^p_\alpha$, then

$$\begin{aligned} {H}f(z)=\int _0^1{T}_tf(z)\,\mathrm {d}t, \,\,\, z\in {\mathbb {D}}, \end{aligned}$$

where

$$\begin{aligned} {T}_tf(z)=\frac{1}{1-(1-t)z}f\left( \frac{t}{1-(1-t)z}\right) , \,\,\, z\in {\mathbb {D}}, \,\,\, 0<t<1. \end{aligned}$$

By following [6], of the Minkowski inequality we have the estimate

$$\begin{aligned} \left\| {H}f\right\| _{{A}^p_\alpha }\le \int _0^1\left\| {T}_tf\right\| _{{A}^p_\alpha }\mathrm {d}t. \end{aligned}$$

It can be shown [7] that

$$\begin{aligned} \left\| {T}_tf\right\| _{{A}^p_\alpha }=\psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\int _{\eta _t({\mathbb {D}})}|w|^{p-2(\alpha +2)}|f(w)|^p\left( \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\right) ^\alpha \mathrm {dm}(w)\right) ^{\frac{1}{p}}, \end{aligned}$$

where

$$\begin{aligned} \psi _{p,\alpha }(t)=t^{\frac{\alpha +2}{p}-1}(1-t)^{-\frac{\alpha +2}{p}}, \,\, c_t=\frac{1}{2-t}, \,\, \rho _t=1-c_t=\frac{1-t}{2-t}, \,\, \text {for} \,\, 0<t<1, \end{aligned}$$

and

$$\begin{aligned} \eta _t(z)=\rho _tz+c_t, \,\,\, \text {for} \,\,\, z\in {\mathbb {D}} \,\,\, \text {and} \,\,\, 0<t<1. \end{aligned}$$

Before the proof of Theorem 1.1, we will need the following elementary results.

Lemma 3.1

Let $-1<\alpha <0$ and $\kappa (x)=\frac{1+3x}{(1+x)^{\alpha +1}}$ for $x\in [0,1]$. Then

$$\begin{aligned} \underset{x\in [0,1]}{\max }\kappa (x)=\kappa (1)=2^{1-\alpha }. \end{aligned}$$

Proof

Note that $\kappa '(x)=\frac{2-\alpha -3\alpha x}{(1+x)^{\alpha +2}}>0$ for $x\in [0,1]$, which implies that $\kappa$ is an increasing function on the interval [0, 1]. This completes the proof. $\square$

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

Proof of Theorem 1.1

We note that it suffices to proceed similarly to the proof of [7, Theorem 1.3] and apply Lemma 2.2 and Lemma 3.1, instead of [11, Theorem 11.6]. For the sake of completeness, we provide details below. Let $f\in {A}^p_\alpha$, where $-1<\alpha <0$ and $\alpha +2<p$. Then we have the following two cases.

(i) ${2(\alpha +2)\le p}$. By using Lemma 2.2, Lemma 3.1 and the fact that in this case we have $|w|^{p-2(\alpha +2)}\le 1$ for $w\in \eta _t({\mathbb {D}})\subset {\mathbb {D}}$, we obtain

$$\begin{aligned} \left\| {T}_tf\right\| _{{A}^p_\alpha }\le & {} \psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\right) ^\frac{1}{p}\left( \int _{\eta _t({\mathbb {D}})}|f(w)|^p\left( \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\right) ^\alpha \mathrm {dm}(w)\right) ^{\frac{1}{p}} \\= & {} \psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\right) ^\frac{1}{p}\left( \rho ^{\alpha +2}_t\int _{{\mathbb {D}}}\left| \left( f\circ \eta _t\right) (z)\right| ^p\left( 1-|z|^2\right) ^\alpha \mathrm {dm}(w)\right) ^{\frac{1}{p}} \\\le & {} \psi _{p,\alpha }\left( t\right) \kappa \left( c_t\right) ^\frac{1}{p}\Vert f\Vert _{{A}^p_\alpha }\le 2^\frac{1-\alpha }{p}\psi _{p,\alpha }\left( t\right) \Vert f\Vert _{{A}^p_\alpha }, \end{aligned}$$

which immediately implies

$$\begin{aligned} \left\| {H}f\right\| _{{A}^p_\alpha }\le \int _0^1\left\| {T}_tf\right\| _{{A}^p_\alpha }\mathrm {d}t\le 2^\frac{1-\alpha }{p}\frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}\Vert f\Vert _{{A}^p_\alpha }. \end{aligned}$$

(ii) ${\alpha +2<p<2(\alpha +2)}$. If $w\in \eta _t({\mathbb {D}})$, then $w=\eta _t(z)$ for some $z\in {\mathbb {D}}$, whence it follows $|w|=\left| \rho _tz+c_t\right| \ge c_t-\rho _t|z|>c_t-\rho _t=t/(2-t)$, that is,

$$\begin{aligned} |w|^{p-2(\alpha +2)}\le \left( \frac{2-t}{t}\right) ^{2(\alpha +2)-p}. \end{aligned}$$

Similar to the case (i) we obtain the following inequality

$$\begin{aligned} \left\| {T}_tf\right\| _{{A}^p_\alpha }\le 2^\frac{1-\alpha }{p}\left( \frac{2-t}{t}\right) ^{\frac{2(\alpha +2)}{p}-1}\psi _{p,\alpha }\left( t\right) \Vert f\Vert _{{A}^p_\alpha }. \end{aligned}$$

Since $\frac{2(\alpha +2)}{p}-1\in (0,1)$, we find

$$\begin{aligned} (2-t)^{\frac{2(\alpha +2)}{p}-1}=\left( t+2(1-t)\right) ^{\frac{2(\alpha +2)}{p}-1}\le t^{\frac{2(\alpha +2)}{p}-1}+2^{\frac{2(\alpha +2)}{p}-1}(1-t)^{\frac{2(\alpha +2)}{p}-1}, \end{aligned}$$

which implies

$$\begin{aligned} \left( \frac{2-t}{t}\right) ^{\frac{2(\alpha +2)}{p}-1}\psi _{p,\alpha }(t)\le \psi _{p,\alpha }(t)+2^{\frac{2(\alpha +2)}{p}-1}\psi _{p,\alpha }(1-t). \end{aligned}$$

Then

$$\begin{aligned} \left\| {T}_tf\right\| _{{A}^p_\alpha }\le 2^\frac{1-\alpha }{p}\left( \psi _{p,\alpha }(t)+2^{\frac{2(\alpha +2)}{p}-1}\psi _{p,\alpha }(1-t)\right) \Vert f\Vert _{{A}^p_\alpha }, \end{aligned}$$

which leads to

$$\begin{aligned} \left\| {H}f\right\| _{{A}^p_\alpha }\le \int _0^1\left\| {T}_tf\right\| _{{A}^p_\alpha }\mathrm {d}t\le 2^\frac{1-\alpha }{p}\left( 1+2^{\frac{2(\alpha +2)}{p}-1}\right) \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}} \Vert f\Vert _{{A}^p_\alpha }. \end{aligned}$$

This finishes the proof. $\square$