Hilbert Matrix and Its Norm on Weighted Bergman Spaces

Abstract

It is well known that the Hilbert matrix ${\mathrm {H}}$ is bounded on weighted Bergman spaces $A^p_\alpha$ if and only if $1<\alpha +2<p$ with the conjectured norm $\pi /\sin \frac{(\alpha +2)\pi }{p}$. The conjecture was confirmed in the case when $\alpha =0$ and also in the case when $\alpha >0$ and $p\ge 2(\alpha +2)$, which reduces the conjecture in the case when $\alpha >0$ to the interval $\alpha +2<p<2(\alpha +2)$. In the remaining case when $-1<\alpha <0$ and $p>\alpha +2$ there has been no progress so far in proving the conjecture, moreover, there is no even an explicit upper bound for the norm of the Hilbert matrix ${\mathrm {H}}$ on weighted Bergman spaces $A^p_\alpha$. In this paper we obtain results which are better than known related to the validity of the mentioned conjecture in the case when $\alpha >0$ and $\alpha +2<p<2(\alpha +2)$. On the other hand, we also provide for the first time an explicit upper bound for the norm of the Hilbert matrix ${\mathrm {H}}$ on weighted Bergman spaces $A^p_\alpha$ in the case when $-1<\alpha <0$ and $p>\alpha +2$.

1 Introduction

The Hilbert matrix ${\mathrm {H}}$ and its action on the space $\ell ^2$ consisting of square summable sequences was first studied in [11], where Magnus described the spectrum of the Hilbert matrix. Thereafter Diamantopoulos and Siskakis in [3, 4] begin to study the action of the Hilbert matrix on Hardy and Bergman spaces, which can be seen as the beginning of studying of the Hilbert matrix as an operator on spaces of holomorphic functions. They obtained some partial results concerning the questions of boundedness and exact norm of the Hilbert matrix on Hardy and Bergman spaces, which have been improved in [5] by Dostanić, Jevtić and Vukotić. We note also that Aleman, Montes-Rodríguez and Sarafoleanu provide a closed formula for the eigenvalues of the Hilbert matrix in a more general context (see [1]). Following the above results, it was known that Hilbert matrix ${\mathrm {H}}$ is bounded on Bergman space $A^p$ if and only if $2<p<\infty$ and

$$\begin{aligned} \left\| {\mathrm {H}}\right\| _{A^p\rightarrow A^p}=\frac{\pi }{\sin \frac{2\pi }{p}}, \end{aligned}$$

when $4\le p<\infty$. It was also conjectured that previous equality remains valid in the remaining case when $2<p<4$. This conjecture was actually proven in [2], where the new method based on the new way to use monotonicity of the integral means was introduced (see also [9]).

The starting point for studying the boundedness of the Hilbert matrix ${\mathrm {H}}$ on weighted Bergman spaces $A^p_\alpha$ was paper [6] by Galanopoulos, Girela, Peláez and Siskakis, where the corresponding partial results were obtained. A complete characterization of the boundedness of the Hilbert matrix ${\mathrm {H}}$ on the spaces $A^p_\alpha$ is given in [7], where it is proved

$$\begin{aligned} {\mathrm {H}} \,\,\,\text {is bounded on}\,\,\, A^p_\alpha \,\,\,\text {if and only if}\,\,\, 1<\alpha +2<p. \end{aligned}$$

On the other hand, the preceding result opened the way to the question of the exact norm of the Hilbert matrix acting on the weighted Bergman spaces. In [8] it was proved that

$$\begin{aligned} \Vert {\mathrm {H}}\Vert _{{A^p_\alpha }\rightarrow A^p_\alpha }\ge \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}} \,\,\, \text {for} \,\,\, 1<\alpha +2<p, \end{aligned}$$

(1.1)

and it was conjectured that this lower bound is the exact norm of the Hilbert matrix. This implies that it is necessary to have the following upper bound

$$\begin{aligned} \Vert {\mathrm {H}}\Vert _{{A^p_\alpha }\rightarrow A^p_\alpha }\le \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}} \,\,\, \text {for} \,\,\, 1<\alpha +2<p, \end{aligned}$$

to prove mentioned conjecture. The conjecture was confirmed [8] in the case when $\alpha \ge 0$ and $p\ge 2(\alpha +2)$, which reduces the conjecture in the case $\alpha \ge 0$ to the interval $\alpha +2<p<2(\alpha +2)$. When $\alpha =0$ this was completely solved in [2] (see also [9]). Very recently, Lindström, Miihkinen and Wikman in [10] confirmed the conjecture in the case $\alpha >0$ when

$$\begin{aligned} \alpha +2+\sqrt{\alpha ^2+\frac{7}{2}\alpha +3}=\alpha +2+\sqrt{(\alpha +2)^2-\frac{1}{2}(\alpha +2)}\le p<2(\alpha +2). \end{aligned}$$

Among other things in this paper, we improved the previous result by confirming the conjecture in the case $\alpha >0$ when

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

On the other hand, we note that in the case $-1<\alpha <0$ there has been no progress so far in proving the conjecture, moreover, there is no even an explicit upper bound for the norm of the Hilbert matrix ${\mathrm {H}}$ on weighted Bergman spaces $A^p_\alpha$. Finally, in this paper we also provide an explicit upper bound for the norm of the Hilbert matrix ${\mathrm {H}}$ on weighted Bergman spaces $A^p_\alpha$ in the case $-1<\alpha <0$ when $p>\alpha +2$.

1.1 Basic Notation

Let ${\mathrm {D}}\left( z_0,r\right) =\left\{ z\in \mathbb {C}:\left| z-z_0\right| <r \right\}$ be the open Euclidean disc of radius $r>0$ centered at the point $z_0$ in the complex plane $\mathbb {C}$. Let also $\mathcal {H}(\mathbb {D})$ be the space of all holomorphic functions in the open unit disc $\mathbb {D}={\mathrm {D}}(0,1)$. An annulus centered at the point $z_0$ in the complex plane is defined as follows ${\mathrm {A}}(z_0,r,R)=\left\{ z\in \mathbb {C}: r<\left| z-z_0\right| <R\right\}$ where $r<R$. The Euclidean area measure on the complex plane will be denoted by ${\mathrm {dm}}$, that is

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

Given a function f holomorphic in the unit disc $\mathbb {D}$, then for $0<p<\infty$ and $0<r<1$, we consider its integral means of order p defined in the following way

$$\begin{aligned} {\mathrm {M}}_p(r,f)=\left( \frac{1}{2\pi }\int _0^{2\pi }\left| f\left( r\mathrm{{e}}^{{\mathrm {i}}\theta }\right) \right| ^p{\mathrm {d}}\theta \right) ^\frac{1}{p}. \end{aligned}$$

It is well known that $r\mapsto {\mathrm {M}}_p(r,f)$ is an nondecreasing function. This is a simple consequence of the subharmonicity of $|f|^p$. The Beta function is defined by

$$\begin{aligned} {\mathrm {B}}(a,b)=\int _0^1 t^{a-1}(1-t)^{b-1}\,{\mathrm {d}}t, \end{aligned}$$

where a and b are real numbers such that $a>0$ and $b>0$. If $0<a<1$ then we will use the following well known formula

$$\begin{aligned} {\mathrm {B}}(a,1-a)=\frac{\pi }{\sin \pi a}. \end{aligned}$$

1.2 Hilbert Matrix and Weighted Bergman Spaces

The Hilbert matrix is an infinite matrix

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

If $f(z)=\sum _{n=0}^\infty a_nz^n$ is a holomorphic function in the unit disc $\mathbb {D}$, that is $f\in \mathcal {H}(\mathbb {D})$, then the Hilbert matrix can be viewed as an operator on spaces of holomorphic functions in the following way

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

For $0<p<\infty$ and $\alpha >-1$ the weighted Bergman space is defined as follows

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

We note that if $\alpha =0$ then $A^p=A^p_0$ are standard unweighted Bergman spaces. It is well known (see [3, 4, 8]) that if a function f belongs to weighted Bergman space $A^p_\alpha$ then we have

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

where

Recall that the Hilbert matrix ${\mathrm {H}}$ is bounded on weighted Bergman space $A^p_\alpha$ if and only if $1<\alpha +2<p$. In that case, by following [8], from the continuous version of Minkowski inequality we have estimate

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

and

$$\begin{aligned} \left\| {\mathrm {T}}_tf\right\| _{A^p_\alpha }=t^{\frac{\alpha +2}{p}-1}(1-t)^{-\frac{\alpha +2}{p}}\left( \frac{\alpha +1}{\pi }\int _{{\mathrm {D}}_t}|w|^{p-4}|f(w)|^pg_t\left( w\right) ^\alpha {\mathrm {dm}}(w)\right) ^{\frac{1}{p}}, \end{aligned}$$

(1.2)

where

$$\begin{aligned} g_t(w)=\frac{2{\text {Re}}w-t-(2-t)|w|^2}{(1-t)|w|^2}, \,\,\ {\mathrm {D}}_t={\mathrm {D}}\left( c_t,\rho _t\right) , \,\,\, c_t=\frac{1}{2-t} \,\,\, \text {and} \,\,\ \rho _t=\frac{1-t}{2-t}. \end{aligned}$$

In the rest of the paper we will use the following function

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

where $0<t<1$. It is easy to check that

$$\begin{aligned} g_t(w)=\frac{1}{|w|^2}\cdot \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t} \,\,\, \text {for} \,\,\, w\in {\mathrm {D}}\left( c_t,\rho _t\right) . \end{aligned}$$

Therefore we conclude

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

The previous formula is valid for all $1<\alpha +2<p$ and it will be used in the last section of this paper. On the other hand, following [8] in the special case when $\alpha >0$ we obtain

$$\begin{aligned} g_t(w)^\alpha \le \left( \frac{1+|w|^2-t-(2-t)|w|^2}{(1-t)|w|^2}\right) ^\alpha =\left( \frac{1-|w|^2}{|w|^2}\right) ^\alpha . \end{aligned}$$

(1.3)

By combining (1.2) and (1.3) we get

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

1.3 The Functions $\Psi _\alpha$ and $\Phi _\alpha$

Let $\alpha >0$. Then we define the functions $\Psi _\alpha$ and $\Phi _\alpha$ as follows

$$\begin{aligned} \Psi _\alpha (x)=2x^2-\left( 4(\alpha +2)+1\right) x -2\sqrt{\alpha +2}\sqrt{x}+\alpha +2, \end{aligned}$$

and

$$\begin{aligned} \Phi _\alpha (x)=2x^2-\left( 4(\alpha +2)+1\right) x +2\sqrt{\alpha +2}\sqrt{x}+\alpha +2, \end{aligned}$$

where $x\in \left( \alpha +2, 2(\alpha +2)\right)$. We note that these functions will play a crucial role in our paper. Next we obtain

$$\begin{aligned} \Psi _\alpha '(x)=4x-4(\alpha +2)-1-\frac{\sqrt{\alpha +2}}{\sqrt{x}} \,\,\, \text {and} \,\,\, \Psi _\alpha ''(x)=4+\frac{\sqrt{\alpha +2}}{2x\sqrt{x}}, \end{aligned}$$

for $x\in \left( \alpha +2, 2(\alpha +2)\right)$. Therefore

$$\begin{aligned} \Psi _\alpha ''(x)>0, \end{aligned}$$

for every $x\in \left( \alpha +2, 2(\alpha +2)\right)$. This leads that function $\Psi _\alpha '$ is increasing on interval $\left( \alpha +2, 2(\alpha +2)\right)$. On the other hand, we find

$$\begin{aligned} \Psi _\alpha '(\alpha +2)=-2<0 \,\,\, \text {and} \,\,\, \Psi _\alpha '(2(\alpha +2))=4(\alpha +2)-1-\frac{1}{\sqrt{2}}>0. \end{aligned}$$

Based on the above considerations we can conclude that it is valid

$$\begin{aligned} \Psi _\alpha (x)\le \max \left\{ \Psi _\alpha (\alpha +2), \Psi _\alpha (2(\alpha +2)) \right\} , \end{aligned}$$

for every $x\in \left( \alpha +2, 2(\alpha +2)\right)$. Since

$$\begin{aligned} \Psi _\alpha (\alpha +2)=-2(\alpha +2)(\alpha +3)<0 \,\,\, \text {and} \,\,\, \Psi _\alpha (2(\alpha +2))=-\left( 2\sqrt{2}+1\right) (\alpha +2)<0, \end{aligned}$$

we have

$$\begin{aligned} \Psi _\alpha (x)<0, \end{aligned}$$

(1.4)

for every $x\in \left( \alpha +2, 2(\alpha +2)\right)$. By straightforward calculations we also derive

$$\begin{aligned} \Phi _\alpha '(x)=4x-4(\alpha +2)-1+\frac{\sqrt{\alpha +2}}{\sqrt{x}} \,\,\, \text {and} \,\,\, \Phi _\alpha ''(x)=4-\frac{\sqrt{\alpha +2}}{2x\sqrt{x}}, \end{aligned}$$

for $x\in \left( \alpha +2, 2(\alpha +2)\right)$. Function $\Phi _\alpha ''$ is increasing on interval $\left( \alpha +2, 2(\alpha +2)\right)$ which implies

$$\begin{aligned} \Phi _\alpha ''(x)>\Phi _\alpha ''(\alpha +2)=4-\frac{1}{2(\alpha +2)}>0, \end{aligned}$$

for every $x\in \left( \alpha +2, 2(\alpha +2)\right)$. Hence function $\Phi _\alpha '$ is also increasing on interval $\left( \alpha +2, 2(\alpha +2)\right)$. This leads to

$$\begin{aligned} \Phi _\alpha '(x)>\Phi _\alpha '(\alpha +2)=4(\alpha +2)-4(\alpha +2)-1+1=0, \end{aligned}$$

where $x\in \left( \alpha +2, 2(\alpha +2)\right)$, whence it follows that $\Phi _\alpha$ is an increasing function on interval $\left( \alpha +2, 2(\alpha +2)\right)$. Then

$$\begin{aligned} \Phi _\alpha (\alpha +2)=-2(\alpha +1)(\alpha +2)<0 \,\,\, \text {and} \,\,\, \Phi _\alpha (2(\alpha +2))=\left( 2\sqrt{2}-1\right) (\alpha +2)>0. \end{aligned}$$

This means that function $\Phi _\alpha$ has a unique zero $\alpha _0$ on the interval $\left( \alpha +2, 2(\alpha +2)\right)$. Moreover, we get $\Phi _\alpha <0$ on $\left( \alpha +2, \alpha _0\right)$ and $\Phi _\alpha >0$ on $\left( \alpha _0, 2(\alpha +2)\right)$. The previous notation will be used in the rest of the paper.

1.4 The Main Results

Let $\alpha >0$ and let $\alpha _0$ be a unique zero of the function $\Phi _\alpha$ on the interval $\left( \alpha +2, 2(\alpha +2)\right)$. We are now ready to state the main results of the paper.

Theorem 1.1

Let $\alpha >0$ and $\alpha _0\le p<2(\alpha +2)$. Then $\displaystyle \left\| {\mathrm {H}}\right\| _{A^p_\alpha \rightarrow A^p_\alpha }=\frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}$.

An immediate consequence we obtain the following result.

Corollary 1.1

Let $\alpha >0$ and

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

Then

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

Proof

It is enough to prove that

$$\begin{aligned} \Phi _\alpha \left( \alpha +2+\sqrt{(\alpha +2)^2-\left( \sqrt{2}-\frac{1}{2}\right) (\alpha +2)}\right) >0. \end{aligned}$$

Namely, the previous inequality implies

$$\begin{aligned} \alpha _0<\alpha +2+\sqrt{(\alpha +2)^2-\left( \sqrt{2}-\frac{1}{2}\right) (\alpha +2)}, \end{aligned}$$

whence by Theorem 1.1 it follows the required conclusion. We can split the function $\Phi _\alpha$ into two parts

$$\begin{aligned} \Phi _\alpha (x)=\Upsilon _\alpha (x)+\Lambda _\alpha (x), \end{aligned}$$

where we denoted

$$\begin{aligned} \Upsilon _\alpha (x)=2\left[ \left( x-(\alpha +2)\right) ^2-\left( (\alpha +2)^2-\left( \sqrt{2}-\frac{1}{2}\right) (\alpha +2)\right) \right] , \end{aligned}$$

and

$$\begin{aligned} \Lambda _\alpha (x)=2\sqrt{\alpha +2}\sqrt{x}-x-2\left( \sqrt{2}-1\right) (\alpha +2). \end{aligned}$$

Note that

$$\begin{aligned} \Upsilon _\alpha \left( \alpha +2+\sqrt{(\alpha +2)^2-\left( \sqrt{2}-\frac{1}{2}\right) (\alpha +2)}\right) =0. \end{aligned}$$

So it is enough to prove

$$\begin{aligned} \Lambda _\alpha \left( \alpha +2+\sqrt{(\alpha +2)^2-\left( \sqrt{2}-\frac{1}{2}\right) (\alpha +2)}\right) >0, \end{aligned}$$

or equivalently

$$\begin{aligned} \Lambda _\alpha \left( a+\sqrt{a^2-\left( \sqrt{2}-\frac{1}{2}\right) a}\right) >0, \end{aligned}$$

where $a=\alpha +2$. This leads to

$$\begin{aligned} 2\sqrt{a}\sqrt{a+\sqrt{a^2-\left( \sqrt{2}-\frac{1}{2}\right) a}}>\left( 2\sqrt{2}-1\right) a+\sqrt{a^2-\left( \sqrt{2}-\frac{1}{2}\right) a}, \end{aligned}$$

whence after squaring we get the following equivalent form

$$\begin{aligned} \left( 3-2\sqrt{2}\right) \sqrt{a^2-\left( \sqrt{2} -\frac{1}{2}\right) a}>\left( 3-2\sqrt{2}\right) a-\left( \frac{\sqrt{2}}{2}-\frac{1}{4}\right) . \end{aligned}$$

Since

$$\begin{aligned} \left( 3-2\sqrt{2}\right) \sqrt{a^2-\left( \sqrt{2}-\frac{1}{2}\right) a}>0, \end{aligned}$$

it is enough to prove

$$\begin{aligned} \left( \left( 3-2\sqrt{2}\right) \sqrt{a^2-\left( \sqrt{2} -\frac{1}{2}\right) a}\right) ^2>\left( \left( 3-2\sqrt{2}\right) a-\left( \frac{\sqrt{2}}{2}-\frac{1}{4}\right) \right) ^2, \end{aligned}$$

which after some calculations reduces to

$$\begin{aligned} a>\frac{\sqrt{2}-\frac{1}{2}}{8\left( 5\sqrt{2}-7\right) } =\frac{13+9\sqrt{2}}{16}. \end{aligned}$$

The last inequality is true, since

$$\begin{aligned} a=\alpha +2>2>\frac{13+9\sqrt{2}}{16}. \end{aligned}$$

This completes the proof. $\square$

Remark 1.1

Note that Corollary 1.1 improves the last best known result recently obtained by Lindström, Miihkinen and Wikman in [10], where they get the same conclusion under the assumptions $\alpha >0$ and

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

which was discussed earlier. $\square$

On the other hand, let

$$\begin{aligned} \beta =\alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}. \end{aligned}$$

Then by straightforward calculations we obtain

$$\begin{aligned} \Phi _\alpha (\beta )=2\underbrace{\left( \beta ^2-2(\alpha +2) \beta +\alpha +2\right) }_{=0}-\left( \sqrt{\beta }-\sqrt{\alpha +2}\right) ^2<0, \end{aligned}$$

which implies that $\beta <\alpha _0$. In the case when $\alpha >0$ and $\alpha +2<p\le \beta$ we obtain the following partial result.

Theorem 1.2

Let $\alpha >0$, $\alpha +2<p\le \alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}$ and suppose that the following condition holds

$$\begin{aligned} \int _0^1\psi _{p,\alpha }(t)\xi _{p,\alpha }(t)\,{\mathrm {d}}t\le \frac{1}{\alpha +1}\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t=\frac{1}{\alpha +1}{\mathrm {B}}\left( \frac{\alpha +2}{p}, 1-\frac{\alpha +2}{p}\right) , \end{aligned}$$

(1.5)

where we denoted

$$\begin{aligned} \psi _{p,\alpha }(t)=t^{\frac{\alpha +2}{p}-1}(1-t)^{-\frac{\alpha +2}{p}} \,\,\, \text {and} \,\,\, \xi _{p,\alpha }(t)=\int _{\left( \frac{t}{2-t}\right) ^2}^1\rho ^{\frac{p}{2}-(\alpha +2)}(1-\rho )^{\alpha }\,{\mathrm {d}}\rho , \end{aligned}$$

for $0<t<1$. Then

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

Remark 1.2

We note that condition (1.5) is not always satisfied under the given conditions even when $p/2-(\alpha +2)+1>0$, that is $p>2\alpha +2$, which actually allows the convergence of the integral $\xi _{p,\alpha }(0)$. Namely a calculation involving Mathematica shows that when $\alpha =1$ then $\beta \approx 5.449$ and $\alpha _0\approx 5.487$ and also for

$$\begin{aligned} \alpha +2<2\alpha +2<p=4.4<\beta <\alpha _0, \end{aligned}$$

we have

$$\begin{aligned} \int _0^1\psi _{p,\alpha }(t)\xi _{p,\alpha }(t)\,{\mathrm {d}}t-\frac{1}{\alpha +1}\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t\approx 0.962>0. \end{aligned}$$

On the other hand, if $\alpha =1$ and $\alpha +2<2\alpha +2<p=5.2<\beta <\alpha _0$ then

$$\begin{aligned} \int _0^1\psi _{p,\alpha }(t)\xi _{p,\alpha }(t)\,{\mathrm {d}}t-\frac{1}{\alpha +1}\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t\approx -0.103<0, \end{aligned}$$

which allows the application of Theorem 1.2 in some cases where it is not possible to apply Theorem 1.1. We also have

$$\begin{aligned} \beta =\alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}<\alpha _0<\alpha +2+\sqrt{(\alpha +2)^2-\left( \sqrt{2}-\frac{1}{2}\right) (\alpha +2)}, \end{aligned}$$

and since $\sqrt{2}-1/2\approx 0.914$ (actually $\sqrt{2}-1/2> 0.914$) we can write

$$\begin{aligned} \alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}<\alpha _0 <\alpha +2+\sqrt{(\alpha +2)^2-0.914(\alpha +2)}. \end{aligned}$$

(1.6)

From (1.6) we can conclude that there remains a small gap between $\beta$ and $\alpha _0$ to which we cannot apply the above Theorem 1.1 and Theorem 1.2. $\diamond$

Finally, in the case when

$$\begin{aligned} -1<\alpha <0 \,\,\, \text {and} \,\,\, p>\alpha +2, \end{aligned}$$

for the first time we obtain an explicit upper bound for the norm of the Hilbert matrix ${\mathrm {H}}$ on weighted Bergman spaces $A^p_\alpha$. Namely, we have the following result.

Theorem 1.3

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

• (i) If $p\ge 2(\alpha +2)$ then

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

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

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

In the rest of the paper, we present the proofs of Theorem 1.1, Theorem 1.2, and Theorem 1.3.

2 Preliminaries

Let $\alpha >0$ and $\alpha +2<p<2(\alpha +2)$. We consider functions

$$\begin{aligned} \psi _{p,\alpha }(t)=t^{\frac{\alpha +2}{p}-1}(1-t)^{-\frac{\alpha +2}{p}} \,\,\, \text {and} \,\,\, \xi _{p,\alpha }(t)=\int _{\left( \frac{t}{2-t}\right) ^2}^1\rho ^{\frac{p}{2}-(\alpha +2)}(1-\rho )^{\alpha }\,{\mathrm {d}}\rho , \end{aligned}$$

where $0<t<1$ as in the statement of Theorem 1.2. For $s\in [0,1]$ we denote

$$\begin{aligned} F_{p,\alpha }(s)= & {} \xi _{p,\alpha }(s)\int _0^s\psi _{p,\alpha }(t)\,{\mathrm {d}}t+\int _s^1 \psi _{p,\alpha }(t)\xi _{p,\alpha }(t)\,{\mathrm {d}}t\\&-\frac{1}{\alpha +1}\left( 1-\left( \frac{s}{2-s}\right) ^2\right) ^{\alpha +1} \int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t. \end{aligned}$$

As we will see later it is of interest to examine under what conditions the function $F_{p,\alpha }$ is nonpositive on the segment [0, 1]. Let also

$$\begin{aligned} B_{p,\alpha }={\mathrm {B}}\left( \frac{\alpha +2}{p}, 1-\frac{\alpha +2}{p}\right) =\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t. \end{aligned}$$

Then

$$\begin{aligned} F_{p,\alpha }'(s)= & {} \xi _{p,\alpha }'(s)\int _0^s\psi _{p,\alpha }(t)\,{\mathrm {d}}t+\xi _{p,\alpha }(s)\psi _{p,\alpha }(s)-\psi _{p,\alpha }(s)\xi _{p,\alpha }(s)\\&- \frac{1}{\alpha +1}\cdot (\alpha +1)\cdot \left( 1-\left( \frac{s}{2-s}\right) ^2\right) ^{\alpha }\cdot (-1)\cdot \frac{4s}{(2-s)^3}\cdot \int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t\\= & {} -\left( \left( \frac{s}{2-s}\right) ^2\right) ^{\frac{p}{2}-(\alpha +2)}\left( 1-\left( \frac{s}{2-s}\right) ^2\right) ^{\alpha }\frac{4s}{(2-s)^3}\int _0^s\psi _{p,\alpha }(t)\,{\mathrm {d}}t\\&+ \frac{4s}{(2-s)^3}\left( 1-\left( \frac{s}{2-s}\right) ^2\right) ^{\alpha }\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t\\= & {} \frac{4s\left( 1-\left( \frac{s}{2-s}\right) ^2\right) ^{\alpha }}{(2-s)^3}\left( B_{p,\alpha }-\left( \frac{s}{2-s}\right) ^{p-2(\alpha +2)}\int _0^s\psi _{p,\alpha }(t)\,{\mathrm {d}}t\right) . \end{aligned}$$

Therefore

$$\begin{aligned} F_{p,\alpha }'(s)=\frac{4s}{(2-s)^3}\left( 1-\left( \frac{s}{2-s}\right) ^2\right) ^{\alpha }\left( \frac{s}{2-s}\right) ^{p-2(\alpha +2)}G_{p,\alpha }(s), \end{aligned}$$

(2.1)

where we denote

$$\begin{aligned} G_{p,\alpha }(s)=\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p}B_{p,\alpha }-\int _0^s\psi _{p,\alpha }(t)\,{\mathrm {d}}t. \end{aligned}$$

(2.2)

We find

$$\begin{aligned} G_{p,\alpha }'(s)= & {} \left( 2(\alpha +2)-p\right) \left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p-1}\frac{2}{(2-s)^2}B_{p,\alpha }-\psi _{p,\alpha }(s)\\= & {} \left( 4(\alpha +2)-2p\right) B_{p,\alpha }\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p-1}\frac{1}{(2-s)^2}-\psi _{p,\alpha }(s). \end{aligned}$$

Hence

$$\begin{aligned} G_{p,\alpha }'(s)=\left( 4(\alpha +2)-2p\right) B_{p,\alpha }\psi _{p,\alpha }(s)E_{p,\alpha }(s), \end{aligned}$$

(2.3)

where

$$\begin{aligned} E_{p,\alpha }(s)=\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p-1}\frac{s^{-\frac{\alpha +2}{p}+1}(1-s)^{\frac{\alpha +2}{p}}}{(2-s)^2}-\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}, \end{aligned}$$

or equivalently

$$\begin{aligned} E_{p,\alpha }(s)=\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p+1}s^{-\frac{\alpha +2}{p}-1}(1-s)^{\frac{\alpha +2}{p}}-\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}. \end{aligned}$$

(2.4)

By differentiation we obtain

$$\begin{aligned} E_{p,\alpha }'(s)= & {} \left( 2(\alpha +2)-p+1\right) \left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p}\frac{2}{(2-s)^2}s^{-\frac{\alpha +2}{p}-1}(1-s)^{\frac{\alpha +2}{p}} \\&+ \left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p+1}\left( -\frac{\alpha +2}{p}-1\right) s^{-\frac{\alpha +2}{p}-2}(1-s)^{\frac{\alpha +2}{p}}\\&+ \left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p+1}s^{-\frac{\alpha +2}{p}-1}\frac{\alpha +2}{p}(1-s)^{\frac{\alpha +2}{p}-1}(-1)\\= & {} \left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p+1}s^{-\frac{\alpha +2}{p}-2}(1-s)^{\frac{\alpha +2}{p}}K_{p,\alpha }(s), \end{aligned}$$

where

$$\begin{aligned} K_{p,\alpha }(s)=\left( 4(\alpha +2)-2p+2\right) \frac{1}{2-s}-\frac{\alpha +2}{p}-1-\frac{\alpha +2}{p}\frac{s}{1-s}, \end{aligned}$$

or

$$\begin{aligned} K_{p,\alpha }(s)=\left( 4(\alpha +2)-2p+2\right) \frac{1}{2-s}-1-\frac{\alpha +2}{p}\frac{1}{1-s}. \end{aligned}$$

Next we can write

$$\begin{aligned} K_{p,\alpha }(s)=\frac{1}{(2-s)(1-s)}L_{p,\alpha }(s), \end{aligned}$$

where we denote

$$\begin{aligned} L_{p,\alpha }(s)=\left( 4(\alpha +2)-2p+2\right) (1-s)-(2-s)(1-s) -\frac{\alpha +2}{p}(2-s). \end{aligned}$$

Finally we obtain

$$\begin{aligned} L_{p,\alpha }(s)=-s^2-\left( 4(\alpha +2)-2p-1 -\frac{\alpha +2}{p}\right) s+4(\alpha +2)-2p-\frac{2(\alpha +2)}{p}, \end{aligned}$$

and

$$\begin{aligned} E_{p,\alpha }'(s)=\left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p+2}s^{-\frac{\alpha +2}{p}-3}(1-s)^{\frac{\alpha +2}{p}-1}L_{p,\alpha }(s). \end{aligned}$$

(2.5)

Let

$$\begin{aligned} A_{p,\alpha }=4(\alpha +2)-2p-\frac{2(\alpha +2)}{p}. \end{aligned}$$

Note that

$$\begin{aligned} L_{p,\alpha }(s)=-s^2-\left( A_{p,\alpha }+\frac{\alpha +2}{p}-1\right) s+A_{p,\alpha }. \end{aligned}$$

(2.6)

We denote the discriminant of the quadratic equation $L_{p,\alpha }(s)=0$ by $D_{p,\alpha }$. Then

$$\begin{aligned} D_{p,\alpha }=\left( A_{p,\alpha }+\frac{\alpha +2}{p}-1\right) ^2+4A_{p,\alpha }, \end{aligned}$$

or

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

Equivalently, we get

$$\begin{aligned} D_{p,\alpha }=\left( A_{p,\alpha }+\left( 1+\sqrt{\frac{\alpha +2}{p}}\right) ^2\right) \left( A_{p,\alpha }+\left( 1-\sqrt{\frac{\alpha +2}{p}}\right) ^2\right) . \end{aligned}$$

In view of Sect. 1.3 we have

$$\begin{aligned} A_{p,\alpha }+\left( 1+\sqrt{\frac{\alpha +2}{p}}\right) ^2=-\frac{1}{p}\cdot \Psi _\alpha (p)>0, \end{aligned}$$

where we used the fact that $\alpha +2<p<2(\alpha +2)$ and inequality (1.4). This means that discriminant $D_{p,\alpha }$ has the same sign as the factor

$$\begin{aligned} A_{p,\alpha }+\left( 1-\sqrt{\frac{\alpha +2}{p}}\right) ^2=-\frac{1}{p}\cdot \Phi _\alpha (p). \end{aligned}$$

We actually have

$$\begin{aligned} D_{p,\alpha }=\frac{1}{p^2}\cdot \Psi _\alpha (p)\cdot \Phi _\alpha (p). \end{aligned}$$

Now we are ready to state our first preliminary result.

Lemma 2.1

Let $\alpha >0$ and $\alpha _0\le p<2(\alpha +2)$. Then $F_{p,\alpha }(s)\le 0$ for all $s\in [0,1]$.

Proof

Since $\alpha _0\le p<2(\alpha +2)$ we have (see Sect. 1.3) that

$$\begin{aligned} \Psi _\alpha (p)<0 \,\,\, \text {and} \,\,\, \Phi _\alpha (p)\ge 0, \end{aligned}$$

which implies

$$\begin{aligned} D_{p,\alpha }=\frac{1}{p^2}\cdot \Psi _\alpha (p)\cdot \Phi _\alpha (p)\le 0. \end{aligned}$$

On the other hand, since $D_{p,\alpha }$ is a discriminant of a quadratic function $L_{p,\alpha }$ given by (2.6) we can conclude that

$$\begin{aligned} L_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, [0,1]. \end{aligned}$$

By using Eq. (2.5) we get

$$\begin{aligned} E_{p,\alpha }'\le 0 \,\,\, \text {on} \,\,\, [0,1], \end{aligned}$$

which in turn implies that $E_{p,\alpha }$ is nonincreasing function on [0, 1]. By Eq. (2.4) we obtain

$$\begin{aligned} E_{p,\alpha }(s)= & {} \left( \frac{s}{2-s}\right) ^{2(\alpha +2)-p+1}s^{-\frac{\alpha +2}{p}-1} (1-s)^{\frac{\alpha +2}{p}} -\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}\\= & {} \left( \frac{1}{2-s}\right) ^{2(\alpha +2)-p+1}s^{2(\alpha +2) -p-\frac{\alpha +2}{p}}(1-s)^{\frac{\alpha +2}{p}} -\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}. \end{aligned}$$

Hence

$$\begin{aligned} E_{p,\alpha }(s)=\left( \frac{1}{2-s}\right) ^{2(\alpha +2)-p+1}s^{\frac{A_{p,\alpha }}{2}}(1-s)^{\frac{\alpha +2}{p}}-\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}. \end{aligned}$$

(2.7)

Also

$$\begin{aligned} A_{p,\alpha }+\underbrace{\left( 1-\sqrt{\frac{\alpha +2}{p}}\right) ^2}_{>0}=-\frac{1}{p}\cdot \Phi _\alpha (p)\le 0, \end{aligned}$$

which leads to

$$\begin{aligned} A_{p,\alpha }<0. \end{aligned}$$

(2.8)

Combining (2.7) and (2.8) we obtain

$$\begin{aligned} \lim _{s\rightarrow 0^+}E_{p,\alpha }(s)=+\infty \,\,\, \text {and} \,\,\, E_{p,\alpha }(1)=-\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}<0. \end{aligned}$$

As we have already concluded that the function $E_{p,\alpha }(s)$ is nonincreasing on [0, 1] we derive that there exists $s_0\in (0,1)$ such that

$$\begin{aligned} E_{p,\alpha }\ge 0 \,\,\, \text {on} \,\,\, \left[ 0,s_0\right] \,\,\, \text {and} \,\,\, E_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ s_0,1\right] . \end{aligned}$$

From (2.3) we find that

$$\begin{aligned} G_{p,\alpha }'\ge 0 \,\,\, \text {on} \,\,\, \left[ 0,s_0\right] \,\,\, \text {and} \,\,\, G_{p,\alpha }'\le 0 \,\,\, \text {on} \,\,\, \left[ s_0,1\right] . \end{aligned}$$

Therefore the function $G_{p,\alpha }$ is nondecreasing on $\left[ 0,s_0\right]$ and nonincreasing on $\left[ s_0,1\right]$. So we have

$$\begin{aligned} G_{p,\alpha }(s)\ge \min \left\{ G_{p,\alpha }(0), G_{p,\alpha }(1)\right\} , \end{aligned}$$

for all $s\in [0,1]$. Since (see (2.2))

$$\begin{aligned} G_{p,\alpha }(0)=0 \,\,\, \text {and} \,\,\, G_{p,\alpha }(1)=B_{p,\alpha }-\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t=0, \end{aligned}$$

we get $G_{p,\alpha }\ge 0$ on [0, 1]. From (2.1) we find $F_{p,\alpha }'\ge 0$ on [0, 1] which implies that $F_{p,\alpha }$ is nondecreasing function on [0, 1]. Finally

$$\begin{aligned} F_{p,\alpha }(s)\le F_{p,\alpha }(1), \end{aligned}$$

for all $s\in [0,1]$. Since

$$\begin{aligned} F_{p,\alpha }(1)=\underbrace{\xi _{p,\alpha }(1)}_{=0}\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t=0\cdot B_{p,\alpha }=0, \end{aligned}$$

we conclude that

$$\begin{aligned} F_{p,\alpha }(s)\le 0, \end{aligned}$$

for all $s\in [0,1]$. This finishes the proof. $\square$

We need also the following preliminary result which will be used later.

Lemma 2.2

Let $\alpha >0$, $\alpha +2<p\le \beta =\alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}$ and suppose that the following condition holds

$$\begin{aligned} \int _0^1\psi _{p,\alpha }(t)\xi _{p,\alpha }(t)\, {\mathrm {d}}t\le \frac{1}{\alpha +1}\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t. \end{aligned}$$

Then $F_{p,\alpha }(s)\le 0$ for all $s\in [0,1]$.

Proof

Note that given condition

$$\begin{aligned} \int _0^1\psi _{p,\alpha }(t)\xi _{p,\alpha }(t)\, {\mathrm {d}}t\le \frac{1}{\alpha +1}\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t, \end{aligned}$$

is actually equivalent to

$$\begin{aligned} F_{p,\alpha }(0)\le 0. \end{aligned}$$

(2.9)

In view of Sect. 1.3 we have

$$\begin{aligned} \Psi _\alpha (p)<0 \,\,\, \text {and} \,\,\, \Phi _\alpha (p)<0, \end{aligned}$$

which implies

$$\begin{aligned} D_{p,\alpha }=\frac{1}{p^2}\cdot \Psi _\alpha (p)\cdot \Phi _\alpha (p)> 0. \end{aligned}$$

We also notice that

$$\begin{aligned} L_{p,\alpha }(s)=-s^2-C_{p,\alpha }s+A_{p,\alpha } \,\,\, \text {for} \,\,\, s\in [0,1], \end{aligned}$$

where we denote

$$\begin{aligned} C_{p,\alpha }=A_{p,\alpha }+\frac{\alpha +2}{p}-1. \end{aligned}$$

The roots of the quadratic equation $L_{p,\alpha }(s)=0$ are given by

$$\begin{aligned} x_{p,\alpha }=\frac{-C_{p,\alpha }-\sqrt{D_{p,\alpha }}}{2} \,\,\, \text {and} \,\,\ X_{p,\alpha }=\frac{-C_{p,\alpha }+\sqrt{D_{p,\alpha }}}{2}. \end{aligned}$$

Then

$$\begin{aligned} \left( C_{p,\alpha }+2\right) ^2=C_{p,\alpha }^2 +4\left( C_{p,\alpha }+1\right) >C_{p,\alpha }^2+4A_{p,\alpha }=D_{p,\alpha }. \end{aligned}$$

(2.10)

We also have

$$\begin{aligned} C_{p,\alpha }+2= & {} 4(\alpha +2)-2p-\frac{2(\alpha +2)}{p}+\frac{\alpha +2}{p}-1+2 \\= & {} 4(\alpha +2)-2p-\frac{\alpha +2}{p}+1 \\= & {} -\frac{1}{p}\cdot \left( 2p^2-\left( 4(\alpha +2)+1\right) p +2\sqrt{\alpha +2}\sqrt{p}+\alpha +2-2\sqrt{\alpha +2}\sqrt{p}\right) , \end{aligned}$$

that is

$$\begin{aligned} C_{p,\alpha }+2=-\frac{1}{p}\cdot \left( \underbrace{\Phi _\alpha (p)}_{<0} -2\sqrt{\alpha +2}\sqrt{p}\right) >0. \end{aligned}$$

(2.11)

From (2.10) and (2.11) we find

$$\begin{aligned} C_{p,\alpha }+2>\sqrt{D_{p,\alpha }}, \end{aligned}$$

which leads to

$$\begin{aligned} X_{p,\alpha }=\frac{-C_{p,\alpha }+\sqrt{D_{p,\alpha }}}{2}<1. \end{aligned}$$

Therefore

$$\begin{aligned} x_{p,\alpha }<X_{p,\alpha }<1, \end{aligned}$$

(2.12)

and

$$\begin{aligned} L_{p,\alpha }(s)=-\left( s-x_{p,\alpha }\right) \left( s-X_{p,\alpha }\right) , \end{aligned}$$

(2.13)

for $s\in [0,1]$. It is also easy to see that

$$\begin{aligned} A_{p,\alpha }=4(\alpha +2)-2p-\frac{2(\alpha +2)}{p} =-\frac{2}{p}\left( p^2-2(\alpha +2)p+\alpha +2\right) , \end{aligned}$$

and

$$\begin{aligned} A_{p,\alpha }=-\frac{2}{p} \underbrace{\left( p-\left( \alpha +2-\sqrt{(\alpha +2)^2 -(\alpha +2)}\right) \right) }_{>0}(p-\beta ), \end{aligned}$$

(2.14)

where we used the fact that $p>\alpha +2>\alpha +2-\sqrt{(\alpha +2)^2-(\alpha +2)}$. Then we can consider the following two cases.

Case $\varvec{\alpha +2<p<\beta }$ In this case from (2.14) we obtain that $A_{p,\alpha }>0$. Hence

$$\begin{aligned} D_{p,\alpha }=C_{p,\alpha }^2+4A_{p,\alpha }>C_{p,\alpha }^2, \end{aligned}$$

which implies $\sqrt{D_{p,\alpha }}>-C_{p,\alpha }$ and $\sqrt{D_{p,\alpha }}>C_{p,\alpha }$. Therefore

$$\begin{aligned} x_{p,\alpha }=\frac{-C_{p,\alpha }-\sqrt{D_{p,\alpha }}}{2}<0 \,\,\, \text {and} \,\,\, X_{p,\alpha }=\frac{-C_{p,\alpha }+\sqrt{D_{p,\alpha }}}{2}>0, \end{aligned}$$

and by using (2.12) we have

$$\begin{aligned} x_{p,\alpha }<0<X_{p,\alpha }<1. \end{aligned}$$

Recall that $L_{p,\alpha }(s)=-\left( s-x_{p,\alpha }\right) \left( s-X_{p,\alpha }\right)$ for $s\in [0,1]$ (see (2.13)). So we get

$$\begin{aligned} L_{p,\alpha }\ge 0 \,\,\, \text {on} \,\,\, \left[ 0,X_{p,\alpha }\right] \,\,\, \text {and} \,\,\, L_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ X_{p,\alpha },1\right] . \end{aligned}$$

By using (2.5) we have

$$\begin{aligned} E_{p,\alpha }'\ge 0 \,\,\, \text {on} \,\,\, \left[ 0,X_{p,\alpha }\right] \,\,\, \text {and} \,\,\, E_{p,\alpha }'\le 0 \,\,\, \text {on} \,\,\, \left[ X_{p,\alpha },1\right] . \end{aligned}$$

We can conclude that function $E_{p,\alpha }$ is nondecreasing on $\left[ 0,X_{p,\alpha }\right]$ and nonincreasing on $\left[ X_{p,\alpha },1\right]$. Therefore

$$\begin{aligned} E_{p,\alpha }\left( X_{p,\alpha }\right) =\underset{s\in [0,1]}{\max }\,E_{p,\alpha }(s). \end{aligned}$$

We recall that (see (2.7))

$$\begin{aligned} E_{p,\alpha }(s)=\left( \frac{1}{2-s}\right) ^{2(\alpha +2)-p+1} s^{\frac{A_{p,\alpha }}{2}}(1-s)^{\frac{\alpha +2}{p}} -\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}, \end{aligned}$$

which leads to

$$\begin{aligned} E_{p,\alpha }(0)=E_{p,\alpha }(1)=-\frac{1}{\left( 4(\alpha +2) -2p\right) B_{p,\alpha }}<0, \end{aligned}$$

where we used the fact that $A_{p,\alpha }>0$ in this case. Next we claim $E_{p,\alpha }\left( X_{p,\alpha }\right) >0$. Assume to the contrary that $E_{p,\alpha }\left( X_{p,\alpha }\right) \le 0$. This implies that $E_{p,\alpha }(s)\le 0$ for all $s\in [0,1]$. From (2.3) we get $G_{p,\alpha }'\le 0$ on [0, 1]. Hence function $G_{p,\alpha }$ is nonincreasing on [0, 1]. Since

$$\begin{aligned} G_{p,\alpha }(0)=G_{p,\alpha }(1)=0, \end{aligned}$$

we find that $G_{p,\alpha }\equiv 0$ on [0, 1] which in turn implies $G_{p,\alpha }'\equiv 0$ on [0, 1]. This leads to $E_{p,\alpha }\equiv 0$ on [0, 1] (see again (2.3)). This is in a contradiction with (2.7), because formula (2.7) implies that function $E_{p,\alpha }$ cannot be identically equal to zero on [0, 1]. In this way we have proved that

$$\begin{aligned} E_{p,\alpha }\left( X_{p,\alpha }\right) >0. \end{aligned}$$

We have already proved that $E_{p,\alpha }(0)=E_{p,\alpha }(1)<0$ and that $E_{p,\alpha }$ is nondecreasing on $\left[ 0,X_{p,\alpha }\right]$ and nonincreasing on $\left[ X_{p,\alpha },1\right]$. Thus there exists $s_1\in \left( 0,X_{p,\alpha }\right)$ such that

$$\begin{aligned} E_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ 0,s_1\right] \,\,\, \text {and} \,\,\, E_{p,\alpha }\ge 0 \,\,\, \text {on} \,\,\, \left[ s_1, X_{p,\alpha }\right] , \end{aligned}$$

and there exists $s_2\in \left( X_{p,\alpha },1\right)$ such that

$$\begin{aligned} E_{p,\alpha }\ge 0 \,\,\, \text {on} \,\,\, \left[ X_{p,\alpha },s_2\right] \,\,\, \text {and} \,\,\, E_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ s_2, 1\right] . \end{aligned}$$

So we obtain

$$\begin{aligned} E_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ 0,s_1\right] , \,\,\, E_{p,\alpha }\ge 0 \,\,\, \text {on} \,\,\, \left[ s_1,s_2\right] \,\,\, \text {and} \,\,\, E_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ s_2, 1\right] . \end{aligned}$$

By using (2.3) we find

$$\begin{aligned} G_{p,\alpha }'\le 0 \,\,\, \text {on} \,\,\, \left[ 0,s_1\right] , \,\,\, G_{p,\alpha }'\ge 0 \,\,\, \text {on} \,\,\, \left[ s_1,s_2\right] \,\,\, \text {and} \,\,\, G_{p,\alpha }'\le 0 \,\,\, \text {on} \,\,\, \left[ s_2, 1\right] . \end{aligned}$$

Therefore function $G_{p,\alpha }$ is nonincreasing on $\left[ 0,s_1\right]$, nondecreasing on $\left[ s_1,s_2\right]$ and nonincreasing on $\left[ s_2,1\right]$. Since $G_{p,\alpha }(0)=G_{p,\alpha }(1)=0$ there exists $s_3\in \left[ s_1,s_2\right]$ such that

$$\begin{aligned} G_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ 0,s_3\right] \,\,\, \text {and} \,\,\, G_{p,\alpha }\ge 0 \,\,\, \text {on} \,\,\, \left[ s_3,1\right] . \end{aligned}$$

From (2.1) we can conclude that

$$\begin{aligned} F_{p,\alpha }'\le 0 \,\,\, \text {on} \,\,\, \left[ 0,s_3\right] \,\,\, \text {and} \,\,\, F_{p,\alpha }'\ge 0 \,\,\, \text {on} \,\,\, \left[ s_3,1\right] . \end{aligned}$$

Hence function $F_{p,\alpha }$ is nonincreasing on $\left[ 0,s_3\right]$ and nondecreasing on $\left[ s_3,1\right]$. This implies that

$$\begin{aligned} F_{p,\alpha }(s)\le \max \left\{ F_{p,\alpha }(0), F_{p,\alpha }(1) \right\} , \end{aligned}$$

for all $s\in [0,1]$. By (2.9) we have $F_{p,\alpha }(0)\le 0$ and since $F_{p,\alpha }(1)=0$ we finally conclude that $F_{p,\alpha }(s)\le 0$ for all $s\in [0,1]$.

Case $\varvec{p=\beta }$ It remains for us to consider what is happening in this case. From (2.14) we find that $A_{p,\alpha }=0$. Hence we have $D_{p,\alpha }=C_{p,\alpha }^2+4A_{p,\alpha }=C_{p,\alpha }^2$ and

$$\begin{aligned} C_{p,\alpha }=A_{p,\alpha }+\frac{\alpha +2}{p}-1=\frac{\alpha +2}{p}-1<0, \end{aligned}$$

which implies $\sqrt{D_{p,\alpha }}=-C_{p,\alpha }$. Therefore

$$\begin{aligned} x_{p,\alpha }=\frac{-C_{p,\alpha }-\sqrt{D_{p,\alpha }}}{2}=0 \,\,\, \text {and} \,\,\, X_{p,\alpha }=\frac{-C_{p,\alpha }+\sqrt{D_{p,\alpha }}}{2}=-C_{p,\alpha }>0, \end{aligned}$$

and by using (2.12) again we get

$$\begin{aligned} x_{p,\alpha }=0<X_{p,\alpha }<1. \end{aligned}$$

Since $L_{p,\alpha }(s)=-\left( s-x_{p,\alpha }\right) \left( s-X_{p,\alpha }\right)$ for $s\in [0,1]$ (see (2.13)) we find that

$$\begin{aligned} L_{p,\alpha }\ge 0 \,\,\, \text {on} \,\,\, \left[ 0,X_{p,\alpha }\right] \,\,\, \text {and} \,\,\, L_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ X_{p,\alpha },1\right] , \end{aligned}$$

and by using (2.5) we have

$$\begin{aligned} E_{p,\alpha }'\ge 0 \,\,\, \text {on} \,\,\, \left[ 0,X_{p,\alpha }\right] \,\,\, \text {and} \,\,\, E_{p,\alpha }'\le 0 \,\,\, \text {on} \,\,\, \left[ X_{p,\alpha },1\right] . \end{aligned}$$

Therefore we can conclude that function $E_{p,\alpha }$ is nondecreasing on $\left[ 0,X_{p,\alpha }\right]$ and nonincreasing on $\left[ X_{p,\alpha },1\right]$. Recall that (see (2.7))

$$\begin{aligned} E_{p,\alpha }(s)=\left( \frac{1}{2-s}\right) ^{2(\alpha +2) -p+1}s^{\frac{A_{p,\alpha }}{2}}(1-s)^{\frac{\alpha +2}{p}} -\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}, \end{aligned}$$

and since

$$\begin{aligned} A_{p,\alpha }=4(\alpha +2)-2p-\frac{2(\alpha +2)}{p}=0, \end{aligned}$$

we obtain

$$\begin{aligned} E_{p,\alpha }(s)=\left( \frac{1}{2-s}\right) ^{\frac{\alpha +2}{p}+1}(1-s)^{\frac{\alpha +2}{p}}-\frac{1}{\frac{2(\alpha +2)}{p}B_{p,\alpha }}. \end{aligned}$$

Let

$$\begin{aligned} c=\frac{\alpha +2}{p}=\frac{\alpha +2}{\beta }=\frac{\alpha +2}{\alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}}. \end{aligned}$$

It is easy to check that

$$\begin{aligned} \frac{1}{2}<c=\frac{\alpha +2}{\alpha +2+\sqrt{(\alpha +2)^2-(\alpha +2)}}<2-\sqrt{2}, \end{aligned}$$

because of $\alpha +2>2$. Then

$$\begin{aligned} E_{p,\alpha }(s)=\left( \frac{1}{2-s}\right) ^{c+1}(1-s)^{c}-\frac{\sin {\pi }c}{2\pi c}, \end{aligned}$$

which implies that

$$\begin{aligned} E_{p,\alpha }(0)=\frac{\pi c-2^c\sin {\pi }c}{2^{c+1}\pi c} \,\,\, \text {and} \,\,\, E_{p,\alpha }(1)=-\frac{\sin {\pi }c}{2\pi c}<0. \end{aligned}$$

Since $1/2<c<2-\sqrt{2}$ we have

$$\begin{aligned} 2^c\sin {\pi }c\le 2^c<2^{2-\sqrt{2}}<\frac{\pi }{2}<\pi c. \end{aligned}$$

Therefore

$$\begin{aligned} E_{p,\alpha }(0)=\frac{\pi c-2^c\sin {\pi }c}{2^{c+1}\pi c}>0 \,\,\, \text {and} \,\,\, E_{p,\alpha }(1)=-\frac{\sin {\pi }c}{2\pi c}<0. \end{aligned}$$

We have also already proved that function $E_{p,\alpha }$ is nondecreasing on $\left[ 0,X_{p,\alpha }\right]$ and nonincreasing on $\left[ X_{p,\alpha },1\right]$. Thus there exists $S\in \left( X_{p,\alpha },1\right)$ such that

$$\begin{aligned} E_{p,\alpha }\ge 0 \,\,\, \text {on} \,\,\, \left[ 0,S\right] \,\,\, \text {and} \,\,\, E_{p,\alpha }\le 0 \,\,\, \text {on} \,\,\, \left[ S,1\right] . \end{aligned}$$

Then from (2.3) we obtain that

$$\begin{aligned} G_{p,\alpha }'\ge 0 \,\,\, \text {on} \,\,\, \left[ 0,S\right] \,\,\, \text {and} \,\,\, G_{p,\alpha }'\le 0 \,\,\, \text {on} \,\,\, \left[ S,1\right] . \end{aligned}$$

Hence the function $G_{p,\alpha }$ is nondecreasing on $\left[ 0,S\right]$ and nonincreasing on $\left[ S,1\right]$. So we conclude that

$$\begin{aligned} G_{p,\alpha }(s)\ge \min \left\{ G_{p,\alpha }(0), G_{p,\alpha }(1)\right\} , \end{aligned}$$

for all $s\in [0,1]$. On the other hand, since (see (2.2))

$$\begin{aligned} G_{p,\alpha }(0)=0 \,\,\, \text {and} \,\,\, G_{p,\alpha }(1)=B_{p,\alpha }-\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t=0, \end{aligned}$$

we obtain $G_{p,\alpha }\ge 0$ on [0, 1]. Then from (2.1) we find $F_{p,\alpha }'\ge 0$ on [0, 1] which implies that $F_{p,\alpha }$ is nondecreasing function on [0, 1]. Therefore $F_{p,\alpha }(s)\le F_{p,\alpha }(1)$ for all $s\in [0,1]$. Since

$$\begin{aligned} F_{p,\alpha }(1)={\xi _{p,\alpha }(1)}\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t=0\cdot B_{p,\alpha }=0, \end{aligned}$$

we have that $F_{p,\alpha }(s)\le 0$ for all $s\in [0,1]$. On the other hand, note that in this case it was not necessary to further assume at the beginning that inequality (2.9) is valid. This completes the proof. $\square$

Remark 2.1

It will be interesting to see what happens in the case when $\beta<p<\alpha _0$. Then we can apply the same procedure as in the proof of Lemma 2.2. Namely, in this case from (2.14) we conclude that $A_{p,\alpha }<0$. Therefore from (2.7) we obtain

$$\begin{aligned} \lim _{s\rightarrow 0^+}E_{p,\alpha }(s)=+\infty \,\,\, \text {and} \,\,\, E_{p,\alpha }(1)=-\frac{1}{\left( 4(\alpha +2)-2p\right) B_{p,\alpha }}<0. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} C_{p,\alpha }=\underbrace{A_{p,\alpha }}_{<0} -\underbrace{\left( 1-\frac{\alpha +2}{p}\right) }_{>0}<0, \end{aligned}$$

and $D_{p,\alpha }=C_{p,\alpha }^2+4A_{p,\alpha }<C_{p,\alpha }^2$ which implies $\sqrt{D_{p,\alpha }}<-C_{p,\alpha }$. Hence

$$\begin{aligned} x_{p,\alpha }=\frac{-C_{p,\alpha }-\sqrt{D_{p,\alpha }}}{2}>0, \end{aligned}$$

and by using (2.12) we get

$$\begin{aligned} 0<x_{p,\alpha }<X_{p,\alpha }<1. \end{aligned}$$

This already complicates the determination of the sign of the quadratic function $L_{p,\alpha }(s)$ on the interval [0, 1] and proceeding further similarly as in the proof of Lemma 2.2, it turns out that it is not easy to conclude under which conditions the function $F_{p,\alpha }(s)$ is nonpositive on the interval [0, 1] in all possible cases. $\diamond$

3 Proofs of Theorem 1.1 and Theorem 1.2

Let us first consider the case when $\alpha >0$ and $\alpha +2<p<2(\alpha +2)$. Later we will focus on the special cases when $\alpha _0\le p<2(\alpha +2)$ or $\alpha +2<p\le \beta$ as in the statements of Theorem 1.1 or Theorem 1.2, respectively. Let $f\in A^p_\alpha$. In view of (1.1) we have the corresponding lower bound, so we need to prove

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

(3.1)

We will use the technique developed in the paper [2]. Denote

$$\begin{aligned} \varphi (r)=2{\mathrm {M}}_p^p(r,f) \,\,\, \text {and} \,\,\, \chi (r)=\varphi (r)-\varphi (0), \end{aligned}$$

for $0\le r<1$. Of course, functions $\varphi$ and $\chi$ depend on the choice of the initial function f which we assume to be fixed in the considerations that follow. Then $\varphi$ is nondecreasing and differentiable function on the interval (0, 1), which implies that function $\chi$ is also nondecreasing and differentiable on (0, 1). Therefore

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

(3.2)

for $0\le r<1$. As shown in Sect. 1.2 we know that

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

(3.3)

and

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

where

$$\begin{aligned} {\mathrm {D}}_t={\mathrm {D}}\left( c_t,\rho _t\right) , \,\,\, c_t=\frac{1}{2-t} \,\,\, \text {and} \,\,\ \rho _t=\frac{1-t}{2-t}. \end{aligned}$$

Note that it is valid

$$\begin{aligned} {\mathrm {D}}_t\subset {\mathrm {A}}\left( 0,c_t-\rho _t, c_t+\rho _t\right) ={\mathrm {A}}\left( 0, \frac{t}{2-t}, 1\right) . \end{aligned}$$

We denote

$$\begin{aligned} {\mathrm {A}}_t={\mathrm {A}}\left( 0, \frac{t}{2-t}, 1\right) . \end{aligned}$$

Consequently

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

or equivalently

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

(3.4)

On the other hand

$$\begin{aligned} \frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}\Vert f\Vert _{{A^p_\alpha }}=\int _0^1\psi _{p,\alpha }(t)\left( (\alpha +1)\int _0^1r\left( 1-r^2\right) ^\alpha \varphi (r)\,{\mathrm {d}}r\right) ^\frac{1}{p}{\mathrm {d}}t. \end{aligned}$$

(3.5)

Because of (3.3), (3.4) and (3.5), we can conclude that (3.1) holds if the following inequality is true

$$\begin{aligned} \int _0^1 \psi _{p,\alpha }(t)\left( I_{p,\alpha }(t)^{1/p}-J_\alpha ^{1/p}\right) {\mathrm {d}}t\le 0, \end{aligned}$$

(3.6)

where

$$\begin{aligned} I_{p,\alpha }(t)=\int _{\frac{t}{2-t}}^1 r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha \varphi (r)\,{\mathrm {d}}r, \end{aligned}$$

and

$$\begin{aligned} J_\alpha =\int _0^1r\left( 1-r^2\right) ^\alpha \varphi (r)\,{\mathrm {d}}r. \end{aligned}$$

Note that $I_{p,\alpha }(t)$ and $J_\alpha$ also depend on the function $\varphi$, that is on the function f that was initially selected. We obtain

$$\begin{aligned} I_{p,\alpha }(t)^{1/p}-J_\alpha ^{1/p}\le \frac{1}{p}J_\alpha ^{\frac{1}{p}-1}\left( I_{p,\alpha }(t)-J_\alpha \right) , \end{aligned}$$

where we used the well known fact that $x^\gamma -y^\gamma \le \gamma y^{\gamma -1}(x-y)$ for $x\ge 0$, $y\ge 0$ and $\gamma \in (0,1)$ as well as the fact that it is valid $1/p\in (0,1)$ because of $p>\alpha +2>2$. Thus we have that (3.6) holds if the following inequality is true

$$\begin{aligned} \int _0^1 \psi _{p,\alpha }(t)\left( I_{p,\alpha }(t)-J_\alpha \right) {\mathrm {d}}t\le 0, \end{aligned}$$

or

$$\begin{aligned} \int _0^1 \psi _{p,\alpha }(t)\left( \int _{\frac{t}{2-t}}^1 r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha \varphi (r)\,{\mathrm {d}}r-\int _0^1r\left( 1-r^2\right) ^\alpha \varphi (r)\,{\mathrm {d}}r\right) {\mathrm {d}}t\le 0, \end{aligned}$$

or equivalently

$$\begin{aligned} V_{p,\alpha }+\varphi (0)W_{p,\alpha }\le U_{p,\alpha }, \end{aligned}$$

(3.7)

where we denoted

$$\begin{aligned} V_{p,\alpha }=\int _0^1 \psi _{p,\alpha }(t)\int _{\frac{t}{2-t}}^1 r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha \chi (r)\,{\mathrm {d}}r{\mathrm {d}}t, \end{aligned}$$

and

$$\begin{aligned} W_{p,\alpha }= & {} \int _0^1 \psi _{p,\alpha }(t)\int _{\frac{t}{2-t}}^1 r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha {\mathrm {d}}r{\mathrm {d}}t\\&- \int _0^1 \psi _{p,\alpha }(t)\int _0^1r\left( 1-r^2\right) ^\alpha {\mathrm {d}}r{\mathrm {d}}t, \end{aligned}$$

and

$$\begin{aligned} U_{p,\alpha }=\int _0^1 \psi _{p,\alpha }(t)\int _0^1r\left( 1-r^2\right) ^\alpha \chi (r)\,{\mathrm {d}}r{\mathrm {d}}t. \end{aligned}$$

Then by using a change of variable $\rho =r^2$ we get

$$\begin{aligned} W_{p,\alpha }= & {} \frac{1}{2}\int _0^1 \psi _{p,\alpha }(t)\int _{\frac{t}{2-t}}^1 \left( r^2\right) ^{\frac{p}{2}-(\alpha +2)}\left( 1-r^2\right) ^\alpha {\mathrm {d}}\left( r^2\right) {\mathrm {d}}t\\&- \frac{1}{2}\int _0^1\left( 1-r^2\right) ^\alpha {\mathrm {d}}\left( r^2\right) \int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t \\= & {} \frac{1}{2}\int _0^1 \psi _{p,\alpha }(t)\int _{\left( \frac{t}{2-t}\right) ^2}^1 \rho ^{\frac{p}{2}-(\alpha +2)}\left( 1-\rho \right) ^\alpha {\mathrm {d}}\rho {\mathrm {d}}t\\&-\frac{1}{2}\int _0^1\left( 1-\rho \right) ^\alpha {\mathrm {d}}\rho \int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t \\= & {} \frac{1}{2}\left( \int _0^1 \psi _{p,\alpha }(t)\xi _{p,\alpha }(t)\,{\mathrm {d}}t-\frac{1}{\alpha +1}\int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t\right) \\= & {} \frac{1}{2}F_{p,\alpha }(0). \end{aligned}$$

On the other hand by using Fubini theorem we obtain

$$\begin{aligned} V_{p,\alpha }= & {} \int _0^1 \psi _{p,\alpha }(t)\left( \int _{\frac{t}{2-t}}^1 r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha \chi (r)\,{\mathrm {d}}r\right) {\mathrm {d}}t\\= & {} \int _0^1 \psi _{p,\alpha }(t)\left( \int _{\frac{t}{2-t}}^1 r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha \int _0^r\chi '(s)\,{\mathrm {d}}s{\mathrm {d}}r\right) {\mathrm {d}}t\\= & {} \int _0^1 \psi _{p,\alpha }(t)\left( \int _{\frac{t}{2-t}}^1\int _0^r r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha \chi '(s)\,{\mathrm {d}}s{\mathrm {d}}r\right) {\mathrm {d}}t\\= & {} \int _0^1 \psi _{p,\alpha }(t)\left( \int _{0}^1\int _{\max \left\{ s,\frac{t}{2-t}\right\} }^1 r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha \chi '(s)\,{\mathrm {d}}r{\mathrm {d}}s\right) {\mathrm {d}}t\\= & {} \int _0^1 \psi _{p,\alpha }(t)\left( \int _{0}^1\chi '(s)\int _{\max \left\{ s,\frac{t}{2-t}\right\} }^1 r^{p-2(\alpha +2)+1}\left( 1-r^2\right) ^\alpha \, {\mathrm {d}}r{\mathrm {d}}s\right) {\mathrm {d}}t\\= & {} \int _0^1 \psi _{p,\alpha }(t)\left( \int _{0}^1\frac{\chi '(s)}{2}\int _{\max ^2\left\{ s,\frac{t}{2-t}\right\} }^1 \rho ^{\frac{p}{2}-(\alpha +2)}\left( 1-\rho \right) ^\alpha \,{\mathrm {d}}\rho {\mathrm {d}}s\right) {\mathrm {d}}t, \end{aligned}$$

and by using change of variable $s=u/(2-u)$ we get

$$\begin{aligned} V_{p,\alpha }=\int _0^1 \psi _{p,\alpha }(t)\left( \int _{0}^1\frac{\chi '\left( \frac{u}{2-u}\right) }{(2-u)^2}\int _{\max ^2\left\{ \frac{u}{2-u},\frac{t}{2-t}\right\} }^1 \rho ^{\frac{p}{2}-(\alpha +2)}\left( 1-\rho \right) ^\alpha \,{\mathrm {d}}\rho {\mathrm {d}}u\right) {\mathrm {d}}t, \end{aligned}$$

or equivalently

$$\begin{aligned} V_{p,\alpha }=\int _0^1 \psi _{p,\alpha }(t)\left( \int _{0}^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) \xi _{p,\alpha }\left( \max \lbrace u,t\rbrace \right) {\mathrm {d}}u\right) {\mathrm {d}}t. \end{aligned}$$

This implies that

$$\begin{aligned} V_{p,\alpha }=\int _{0}^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) \left( \int _0^1 \psi _{p,\alpha }(t)\xi _{p,\alpha }\left( \max \lbrace u,t\rbrace \right) {\mathrm {d}}t\right) {\mathrm {d}}u, \end{aligned}$$

or

$$\begin{aligned} V_{p,\alpha }=\int _{0}^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) \left( \xi _{p,\alpha }(u)\int _0^u \psi _{p,\alpha }(t){\mathrm {d}}t + \int _u^1 \psi _{p,\alpha }(t)\xi _{p,\alpha }\left( t\right) {\mathrm {d}}t \right) {\mathrm {d}}u. \end{aligned}$$

Similarly we have

$$\begin{aligned} U_{p,\alpha }= & {} \int _0^1r\left( 1-r^2\right) ^\alpha \chi (r)\,{\mathrm {d}}r\int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t\\= & {} \int _0^1r\left( 1-r^2\right) ^\alpha \int _0^r\chi '(s)\,{\mathrm {d}}s{\mathrm {d}}r\int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t\\= & {} \int _0^1\chi '(s)\int _s^1r\left( 1-r^2\right) ^\alpha \,{\mathrm {d}}r{\mathrm {d}}s\int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t\\= & {} \int _0^1\frac{\chi '(s)}{2}\frac{1}{\alpha +1}\left( 1-s^2\right) ^{\alpha +1}{\mathrm {d}}s\int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t, \end{aligned}$$

which leads to

$$\begin{aligned} U_{p,\alpha }=\int _0^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) \frac{1}{\alpha +1}\left( 1-\left( \frac{u}{2-u}\right) ^2\right) ^{\alpha +1}{\mathrm {d}}u\int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t, \end{aligned}$$

or equivalently

$$\begin{aligned} U_{p,\alpha }=\int _0^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) \left( \frac{1}{\alpha +1}\left( 1-\left( \frac{u}{2-u}\right) ^2\right) ^{\alpha +1}\int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t\right) {\mathrm {d}}u. \end{aligned}$$

Based on the foregoing considerations we can conclude

$$\begin{aligned} V_{p,\alpha }-U_{p,\alpha }=\int _0^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) F_{p,\alpha }(u)\,{\mathrm {d}}u. \end{aligned}$$

Finally to obtain (3.7) it is enough to prove that

$$\begin{aligned} \int _0^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) F_{p,\alpha }(u)\,{\mathrm {d}}u+\frac{\varphi (0)}{2}F_{p,\alpha }(0)\le 0. \end{aligned}$$

Note also that $\varphi (0)=2|f(0)|^p$. So we can conclude the following. In the case when $\alpha >0$ and $\alpha +2<p<2(\alpha +2)$ if

$$\begin{aligned} \int _0^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) F_{p,\alpha }(u)\,{\mathrm {d}}u+|f(0)|^pF_{p,\alpha }(0)\le 0, \end{aligned}$$

(3.8)

then it is valid inequality (3.1). Now we are ready to prove Theorem 1.1.

Proof of Theorem 1.1

Let $f\in A^p_\alpha$ where $\alpha >0$ and $\alpha _0\le p<2(\alpha +2)$. In view of (1.1) it is enough to prove that inequality (3.1) is valid. By using Lemma 2.1 we have that

$$\begin{aligned} F_{p,\alpha }(u)\le 0, \end{aligned}$$

(3.9)

for all $u\in [0,1]$. From (3.2) we conclude that

$$\begin{aligned} \frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) \ge 0. \end{aligned}$$

(3.10)

Combining (3.9) and (3.10) we obtain

$$\begin{aligned} \int _0^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) F_{p,\alpha }(u)\,{\mathrm {d}}u+|f(0)|^pF_{p,\alpha }(0)\le 0, \end{aligned}$$

which leads that inequality (3.8) is valid. This implies that inequality (3.1) is also valid, which completes the proof. $\square$

Next we will use the previously presented results as well as Lemma 2.2 from Sect. 2. Thus we are ready to prove Theorem 1.2.

Proof of Theorem 1.2

Let $f\in A^p_\alpha$ where $\alpha >0$ and $\alpha +2<p\le \beta$. We can use Lemma 2.2, because we know that under the assumptions of Theorem 1.2 the following condition is satisfied

$$\begin{aligned} \int _0^1\psi _{p,\alpha }(t)\xi _{p,\alpha }(t)\,{\mathrm {d}}t\le \frac{1}{\alpha +1}\int _0^1\psi _{p,\alpha }(t)\,{\mathrm {d}}t=\frac{1}{\alpha +1}{\mathrm {B}}\left( \frac{\alpha +2}{p}, 1-\frac{\alpha +2}{p}\right) . \end{aligned}$$

Therefore

$$\begin{aligned} F_{p,\alpha }(u)\le 0, \end{aligned}$$

for all $u\in [0,1]$. Since $\chi '\ge 0$ we derive that

$$\begin{aligned} \int _0^1\frac{1}{(2-u)^2}\chi '\left( \frac{u}{2-u}\right) F_{p,\alpha }(u)\,{\mathrm {d}}u+|f(0)|^pF_{p,\alpha }(0)\le 0, \end{aligned}$$

which implies validity of the inequality (3.1). This finishes the proof. $\square$

4 Proof of Theorem 1.3

In this section we consider the case when $-1<\alpha <0$ and $p>\alpha +2$. Let also $f\in A^p_\alpha$. As we showed in Sect. 1.2 we have

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

and

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

where

$$\begin{aligned} {\mathrm {D}}_t={\mathrm {D}}\left( c_t,\rho _t\right) , \,\,\, c_t=\frac{1}{2-t} \,\,\, \text {and} \,\,\ \rho _t=\frac{1-t}{2-t}. \end{aligned}$$

Let

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

Note that $\eta _t(\mathbb {D})={\mathrm {D}}_t\subset \mathbb {D}$. We will also use the following well known result, which is a consequence of Littlewood Subordination Principle (see Chapter 11 in [12]).

Lemma 4.1

(see Theorem 11.6 in [12]) If $\eta :\mathbb {D}\rightarrow \mathbb {D}$ is holomorphic function, $p>0$ and $\alpha >-1$ then

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

for all holomorphic functions f on $\mathbb {D}$.

After the preliminary results mentioned above, we are now ready to prove Theorem 1.3 from the Sect. 1.4.

Proof of Theorem 1.3

Let $f\in A^p_\alpha$. Then we consider the following two cases as in the statement of Theorem 1.3.

Case (i) $\varvec{p\ge 2(\alpha +2)}$ In this case we actually have that $|w|^{p-2(\alpha +2)}\le 1$ for all $w\in {\mathrm {D_t}}=\eta _t(\mathbb {D})$. Therefore

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

After change of variable $w=\eta _t(z)$ we obtain

$$\begin{aligned} \left\| {\mathrm {T}}_tf\right\| _{A^p_\alpha }\le & {} \psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\right) ^\frac{1}{p}\left( \rho _t^\alpha \int _\mathbb {D}\left| \left( f\circ \eta _t\right) (z)\right| ^p\left( 1-|z|^2\right) ^\alpha \left| \eta _t'(z)\right| ^2{\mathrm {dm}}(z)\right) ^{\frac{1}{p}} \\= & {} \psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\right) ^\frac{1}{p}\left( \rho _t^{\alpha +2} \int _\mathbb {D}\left| \left( f\circ \eta _t\right) (z)\right| ^p\left( 1-|z|^2\right) ^\alpha {\mathrm {dm}}(z)\right) ^{\frac{1}{p}}. \end{aligned}$$

Since

$$\begin{aligned} \rho _t^{\alpha +2}\left( \frac{1+\left| \eta _t(0)\right| }{1-\left| \eta _t(0)\right| }\right) ^{\alpha +2}=\left( \rho _t\cdot \frac{1+c_t}{1-c_t}\right) ^{\alpha +2}=\left( \frac{1-t}{2-t}\cdot \frac{3-t}{1-t}\right) ^{\alpha +2}=\left( \frac{3-t}{2-t}\right) ^{\alpha +2}, \end{aligned}$$

by using Lemma 4.1 we obtain

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

Hence we conclude

$$\begin{aligned} \left\| {\mathrm {T}}_tf\right\| _{A^p_\alpha }\le \left( \frac{3-t}{2-t}\right) ^{\frac{\alpha +2}{p}}\psi _{p,\alpha }(t)\left( \frac{\alpha +1}{\pi }\int _\mathbb {D}\left| f(z)\right| ^p\left( 1-|z|^2\right) ^\alpha {\mathrm {dm}}(z)\right) ^\frac{1}{p}, \end{aligned}$$

or

$$\begin{aligned} \left\| {\mathrm {T}}_tf\right\| _{A^p_\alpha }\le \left( \frac{3-t}{2-t}\right) ^{\frac{\alpha +2}{p}}\psi _{p,\alpha }(t)\Vert f\Vert _{A^p_\alpha }. \end{aligned}$$

Note that

$$\begin{aligned} \left( \frac{3-t}{2-t}\right) ^{\frac{\alpha +2}{p}}\le 2^\frac{\alpha +2}{p}, \end{aligned}$$

for all $0< t< 1$. This leads to

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

Finally we obtain

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

Therefore

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

Case (ii) $\varvec{\alpha +2<p<2(\alpha +2)}$ In this case we have

$$\begin{aligned} |w|\ge c_t-\left| w-c_t\right| >c_t-\rho _t=\frac{t}{2-t} \,\,\, \text {for all} \,\,\, w\in {\mathrm {D}}_t. \end{aligned}$$

Therefore

$$\begin{aligned} |w|^{p-2(\alpha +2)}\le \left( \frac{t}{2-t}\right) ^{p-2(\alpha +2)}=\left( \frac{2-t}{t}\right) ^{2(\alpha +2)-p} \,\,\, \text {for all} \,\,\, w\in {\mathrm {D}}_t. \end{aligned}$$

This implies that $\left\| {\mathrm {T}}_tf\right\| _{A^p_\alpha }$ is not greater than

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

Similar to the Case (i) we get that it is valid

$$\begin{aligned} \left( \frac{\alpha +1}{\pi }\int _{\eta _t(\mathbb {D})}|f(w)|^p\left( \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\right) ^\alpha {\mathrm {dm}}(w)\right) ^{\frac{1}{p}}\le \left( \frac{3-t}{2-t}\right) ^{\frac{\alpha +2}{p}}\Vert f\Vert _{A^p_\alpha }, \end{aligned}$$

which leads to

$$\begin{aligned} \left( \frac{\alpha +1}{\pi }\int _{\eta _t(\mathbb {D})}|f(w)|^p\left( \frac{\rho _t^2-\left| w-c_t\right| ^2}{\rho _t}\right) ^\alpha {\mathrm {dm}}(w)\right) ^{\frac{1}{p}}\le 2^\frac{\alpha +2}{p}\Vert f\Vert _{A^p_\alpha }. \end{aligned}$$

Hence we obtain

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

(4.1)

On the other hand, we have

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

where we actually used the known fact which states that the following inequality is valid

$$\begin{aligned} (x+y)^\gamma \le x^\gamma +y^\gamma , \end{aligned}$$

for all $x\ge 0$, $y\ge 0$ and $\gamma \in (0,1)$ as well as the fact that

$$\begin{aligned} \frac{2(\alpha +2)}{p}-1\in (0,1), \end{aligned}$$

because of $\alpha +2<p<2(\alpha +2)$. So we get

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

or

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

(4.2)

From (4.1) and (4.2) we obtain

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

Since

$$\begin{aligned} \int _0^1 \psi _{p,\alpha }(1-t)\,{\mathrm {d}}t=\int _0^1 \psi _{p,\alpha }(t)\,{\mathrm {d}}t={\mathrm {B}}\left( \frac{\alpha +2}{p}, 1-\frac{\alpha +2}{p}\right) =\frac{\pi }{\sin \frac{(\alpha +2)\pi }{p}}, \end{aligned}$$

we find

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

Finally we obtain

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

This completes the proof. $\square$