1 Introduction and Preliminaries

In recent years, the study for Hilbert matrix operator \(\mathcal {H}\)’s boundedness and norm on different analytic function spaces has been under active investigation (see [1,2,3,4,5,6,7,8]). Diamantopoulus and Siskakis [7] studied \(\mathcal {H}\) is bounded on Hardy space \(H^p(1<p<\infty )\) and also obtained an upper bound estimate for its norm. In [6], Diamantopoulus began consider the boundedness of \(\mathcal {H}\) on the Bergman spaces \(A^p(2<p<\infty )\), and obtained the upper bound estimate for the norm of \(\mathcal {H}\). Then Dostanić, Jevtić, Vukotić [8] established the precise value of the norm of \(\mathcal {H}\) in the Hardy space \(H^p(1<p<\infty )\) and also gave exact value of the norm of \(\mathcal {H}\) in the Bergman space \(A^p(4<p<\infty )\). In 2017, Boz̆in and Karapetrović [3] solved the question of exact value of the norm of \(\mathcal {H}\) in the Bergman space \(A^p\), for \(2<p<4\). The norm of Hilbert matrix operator \(\mathcal {H}\) has also been studied in other analytic function spaces like Korenblum spaces \(H^\infty _\alpha \) [5, 19].

The study of boundedness of \(\mathcal {H}\) on \(A^p_\alpha \) was initiated in [10] and some partial results were obtained. The boundedness of the Hilbert matrix on \(A^p_\alpha \) for \(1< 2 + \alpha <p\) was also studied by Jevtić M and Karapetrović B in [12]. In [13], Karapetrović obtained the exact norm of \(\mathcal {H}\) on \(A^p_\alpha \) when \(4 \le 2(2+\alpha ) \le p < \infty \). Additionally, he showed that the same lower bound holds for all \(p> 2+\alpha > 1\). He also conjectured that the upper bound for the norm of \(\mathcal {H}\) is the same as above also for the case \(1< 2+\alpha<p < 2(2+\alpha )\) in [13]. In [18] Lindström, Miihkinen and Wikman confirmed the conjecture in the positive for \(2 + \alpha + \sqrt{ \alpha ^2 + \frac{7}{2}\alpha + 3} \le p < 2(2 + \alpha )\). Recently Karapetrović generalized the work of [18] by showing that the conjecture holds for \(2 + \alpha + \sqrt{ (2 + \alpha )^2 - (\sqrt{2} - \frac{1}{2} )(2 + \alpha )} \le p < 2(2 + \alpha )\). In [2], Bralović and Karapetrović provide a new upper bound for the norm of the Hilbert matrix \(\mathcal {H}\) on the weighted Bergman spaces \(A^p_\alpha \) when \(-1<\alpha < 0\), which represents an improvement.

We also realized the Hilbert matrix operator is unbounded on \(A^2\) in [6]. And the situation is actually even worse: the series defining \(\mathcal {H}f (0)\) is divergent in [8]. Then in [16] Łanucha, Nowak and Pavlović considered \(\mathcal {H}\) acts as a bounded operator from \(A^2_{\log ^\alpha }\) to \(A^2\) for \(\alpha >3\) and this was improved in [11, Theorem 4.5], where it is proved that \(\mathcal {H}\) maps \(A^2_{\log ^\alpha }\) into \(A^2\) for \(\alpha >2\). The last result is also improved in [15, Theorem 3.2] by Karapetrović, where it is proved that \(\mathcal {H}\) maps \(A^2_{\log ^\alpha }\) into \(A^2_{\log ^{\alpha -2-\varepsilon }}\) for \(\alpha >2\) and \(0<\varepsilon \le \alpha -2\). In this paper, we obtained the upper bound for the norm from logarithmically weighted Bergman spaces \(A^2_{\log ^\alpha }\) into Bergman spaces \(A^2\) for \(\alpha >2\). We also find \(\mathcal {H}\) acts as a bounded operator from \(A^p_{\log ^\alpha }\) into \(A^p\) for \(2<p<\infty \) and \(\alpha >0\), and also calculate the upper bound for the norm of \(\mathcal {H}\).

Let \(\mathbb {D}\) denote the open unit disk of the complex plane \(\mathbb {C}\), and let \(H(\mathbb {D})\) denote the set of all analytic functions in \(\mathbb {D}\).

For \(0 < p \le \infty \), the Hardy space \(H^p\) is the space of all functions \(f \in H(\mathbb {D} )\) for which

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

where

$$\begin{aligned}{} & {} M_p(r,f)=\left( \frac{1}{2\pi }\int _0^{2\pi } |f(re^{it})|^p dt\right) ^{\frac{1}{p}}, \quad 0<p<\infty ; \\{} & {} M_\infty (r,f)=\sup _{0\le t<2\pi }|f(re^{it})|. \end{aligned}$$

For \(0< p < \infty \) the Bergman space \(A^p\) consists of those \(f \in H(\mathbb {D})\) such that

$$\begin{aligned} \Vert f\Vert _{A^p}\overset{def}{=}\ \left( \int _\mathbb {D}|f(z)|^p dA(z) \right) ^{1/p}< \infty . \end{aligned}$$

Here, dA stands for the area measure on \(\mathbb {D}\), normalized so that the total area of \(\mathbb {D}\) is 1. Thus \(dA(z) = \frac{1}{\pi }dxdy = \frac{1}{\pi }rdrd\theta \).

Then for \(0< p < \infty \) and \(\alpha >0\) the logarithmically weighted Bergman space \(A^p_{\log ^\alpha }\) consists of those \(f \in H(\mathbb {D})\) such that

$$\begin{aligned} \Vert f\Vert _{A_{\log ^\alpha }^p}\overset{def}{=}\ \left( \int _\mathbb {D}|f(z)|^p \left( \log \frac{2}{1-|z|^2} \right) ^\alpha dA(z) \right) ^{1/p}< \infty . \end{aligned}$$

The relation between these spaces we introduced above is well known that \(A^p_{\log ^\alpha } \subset A^p\).

The Hilbert matrix is an infinite matrix \(\mathcal {H}\) whose entries are \(a_{n,k} = \frac{1}{n+k+1} \), \(n,k \ge 0\). The Hilbert matrix \(\mathcal {H}\) can be also viewed as an operator on spaces of analytic functions by its action on their Taylor coefficients. Hence for those \(f \in H(\mathbb {D})\), \(f(z) = \sum _{k=0}^\infty a_kz^k\), then we define a transformation \(\mathcal {H}\) by

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

As usual, throughout this paper, C denotes a positive constant which depends only on the displayed parameters but not necessarily the same from one occurrence to the next.

2 Norm Estimates of the Hilbert Matrix \(\Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2}\)

In this section, we drive norm estimates for Hilbert matrix operator acting from \(A_{\log ^{\alpha }}^2\) into \(A^2\) for \(\alpha >2\).

According to [16, Lemma 4.2], we obtain that there exists a constant \(C > 0\) such that

$$\begin{aligned} \sum _{k=0}^\infty \frac{|a_k|}{k+1}\le C \Vert f\Vert _{A^2_{\log ^\alpha }} \end{aligned}$$

for every \(f(z)= \sum _{k=0}^\infty a_k z^k\) that belongs to \(A_{\log ^\alpha }^2\), \(\alpha >2\). Then we obtain a well-defined analytic function \(\mathcal {H}f(z)\) on \(\mathbb {D}\). Hence, we have that

$$\begin{aligned} \mathcal {H}f(z)&= \sum _{n=0}^\infty \left( \sum _{k=0}^\infty \frac{a_k}{n+k+1} \right) z^n \nonumber \\&= \sum _{n=0}^\infty \left( \sum _{k=0}^\infty a_k \int _0^1 t^{n+k}dt \right) z^n \nonumber \\&= \int _0^1 \sum _{k=0}^\infty a_k t^k \sum _{n=0}^\infty t^n z^n dt \nonumber \\&= \int _0^1 \frac{f(t)}{1-tz} dt. \end{aligned}$$
(2.1)

2.1 Upper Bound for the Norm \(\Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2}\)

We know that the Hilbert matrix operator \(\mathcal {H}\) has an integral representation in terms of weighted composition operators \(T_t\) (see [6]):

$$\begin{aligned} \mathcal {H} f(z)=\int _0^1 T_t f(z) dt, \end{aligned}$$
(2.2)

where

$$\begin{aligned} T_tf(z)=w_t(z)f(\phi _t(z)), \quad w_t(z)=\frac{1}{1-(1-t)z}, \quad \phi _t(z)=\frac{t}{1-(1-t)z}. \end{aligned}$$

Theorem 2.1

Let \(\alpha >2\). Then the norm of the Hilbert matrix operator acting from \(A_{\log ^{\alpha }}^2\) into \(A^2\) satisfies the upper estimate

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2} \le \int _0^1 \frac{2}{ \left( \log \frac{2}{1-x^2}\right) ^{\frac{\alpha }{2}}} \left( \frac{1}{1-x}+\frac{1}{(x(1-x))^{\frac{1}{2}}}\right) dx. \end{aligned}$$

Proof

By Minkowski’s inequality, we have

$$\begin{aligned} \Vert \mathcal {H}f\Vert _{A^2}&=\left( \int _{\mathbb {D}} |\mathcal {H}f(z)|^2 dA(z)\right) ^{\frac{1}{2}} \nonumber \\&=\left( \int _{\mathbb {D}} \left| \int _0^1 T_t(z)dt\right| ^2 dA(z)\right) ^{\frac{1}{2}} \nonumber \\&\le \int _0^1 \left( \int _{\mathbb {D}} |T_tf(z)|^2 dA(z)\right) ^{\frac{1}{2}} dt \nonumber \\&= \int _0^1 \Vert T_t f\Vert _{A^2} dt. \end{aligned}$$
(2.3)

Using linear fractional change of variable \(w = \phi _t(z), z \in \mathbb {D}\), we obtain that

$$\begin{aligned} \Vert T_t f\Vert ^2_{A^2}&= \int _{\mathbb {D}} |w_t(z)|^2|f(\phi _t(z))|^2 dA(z) \\&= \int _{\phi _t(\mathbb {D})} |w_t(\phi ^{-1}_t(w))|^2 \frac{|f(w)|^2}{|\phi _t'(\phi _t^{-1}(w))|^2} dA(w) \\&= \frac{1}{(1-t)^2} \int _{\phi _t(\mathbb {D})}|w|^{-2} |f(w)|^2 dA(w). \end{aligned}$$

Therefore

$$\begin{aligned} \Vert T_t f\Vert _{A^2}= \frac{1}{1-t}\left( \int _{D_t} |w|^{-2} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}}, \end{aligned}$$

here \(D_t = \phi _t(\mathbb {D})\). It is easy to find that \(D_t = D\left( \frac{1}{2-t}, \frac{1-t}{2-t} \right) \), i.e. \(D_t\) is the Euclidean disc with center on \(\frac{1}{2-t}\) and of radius \(\frac{1-t}{2-t}\). It is easy to see that \(|w| \ge \frac{t}{ 2-t}\), for \(w \in D_t\), and \(D_t \subset E_t \), where \(E_t = \{w \in \mathbb {C}: \frac{t}{2-t}< |w| < 1\}\). Hence, we obtain

$$\begin{aligned} \Vert T_t f\Vert _{A^2}&\le \frac{1}{1-t} \left( \int _{E_t} |w|^{-2} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}}. \end{aligned}$$
(2.4)

On the other hand, we also have

$$\begin{aligned} \left( \int _{E_t} |w|^{-2} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}}=\left( 2\int _{\frac{t}{2-t}}^1\frac{1}{r^2}\cdot rM_2^2(r,f)dr \right) ^{\frac{1}{2}}. \end{aligned}$$

It is easy to find function \(r \rightarrow \frac{1}{r^2}\) is decreasing and function \(r \rightarrow rM^2_2(r,f )\) is increasing, by using Chebyshev’s inequality, we get

$$\begin{aligned} \left( \int _{E_t} |w|^{-2} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}}&\le \left( \frac{2}{1-\frac{t}{2-t}} \int _{\frac{t}{2-t}}^1 \frac{1}{r^2}dr \int _{\frac{t}{2-t}}^1 rM^2_2(r,f) dr\right) ^{\frac{1}{2}} \nonumber \\&= \left( \frac{2-t}{t} \cdot 2 \int _{\frac{t}{2-t}}^1 rM^2_2(r,f) dr \right) ^{\frac{1}{2}}\nonumber \\&= \left( \frac{2-t}{t} \int _{E_t} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}}. \end{aligned}$$
(2.5)

And by using (2.5), we have that

$$\begin{aligned} \int _0^1 \Vert T_t f\Vert _{A^2} dt&\le \int _0^1 \left( \frac{1}{1-t}\cdot \frac{(2-t)^{\frac{1}{2}}}{t^{\frac{1}{2}}}\right) \left( \int _{E_t} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}} dt \nonumber \\&= \int _0^1 \left( \frac{1}{1-t}\cdot \left( 1+\frac{2(1-t)}{t}\right) ^{\frac{1}{2}}\right) \left( \int _{E_t} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}} dt \nonumber \\&\le \int _0^1 \left( \frac{1}{1-t}\cdot \left( 1+\left( \frac{2(1-t)}{t}\right) ^{\frac{1}{2}}\right) \right) \left( \int _{E_t} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}} dt \nonumber \\&= \int _0^1 \left( \frac{1}{1-t}+ \frac{\sqrt{2}}{(t(1-t))^{\frac{1}{2}}}\right) \left( \int _{E_t} |f(w)|^2 dA(w)\right) ^{\frac{1}{2}} dt \nonumber \\&=\int _0^1 \left( \frac{1}{1-t}+ \frac{\sqrt{2}}{(t(1-t))^{\frac{1}{2}}}\right) \left( \int _{E_t} \left( \log \frac{2}{1-|w|^2}\right) ^{-\alpha } |f(w)|^2\right. \nonumber \\&\times \left. \left( \log \frac{2}{1-|w|^2}\right) ^\alpha dA(w)\right) ^{\frac{1}{2}} dt. \end{aligned}$$
(2.6)

Since function \(w\rightarrow \left( \log \frac{2}{1-|w|^2}\right) ^{-\alpha }\) is decreasing, by simple calculation we found that,

$$\begin{aligned} \int _0^1 \Vert T_t f\Vert _{A^2} dt&\le \int _0^1 \frac{\left( \frac{1}{1-t}+ \frac{\sqrt{2}}{(t(1-t))^{\frac{1}{2}}}\right) }{\left( \log \frac{2}{1-(\frac{t}{2-t})^2}\right) ^{\frac{\alpha }{2}}} \left( \int _{E_t} |f(w)|^2 \left( \log \frac{2}{1-|w|^2}\right) ^\alpha dA(w)\right) ^{\frac{1}{2}} dt \nonumber \\&\le \int _0^1 \left( \frac{1}{(1-t)\left( \log \frac{2}{1-(\frac{t}{2-t})^2}\right) ^{\frac{\alpha }{2}}}+ \frac{\sqrt{2}}{(t(1-t))^{\frac{1}{2}}\left( \log \frac{2}{1-(\frac{t}{2-t})^2}\right) ^{\frac{\alpha }{2}} } \right) dt\Vert f\Vert _{A_{\log ^{\alpha }}^2}, \end{aligned}$$
(2.7)

Making the change of variable in (2.7), we obtain that

$$\begin{aligned} \int _0^1 \Vert T_t f\Vert _{A^2} dt&\le \int _0^1 \frac{2}{(1-t^2)\left( \log \frac{2}{1-t^2}\right) ^{\frac{\alpha }{2}}}dt \Vert f\Vert _{A_{\log ^{\alpha }}^2}\nonumber \\&+ \int _0^1 \frac{2}{(t+1)(t(1-t))^{\frac{1}{2}}\left( \log \frac{2}{1-t^2}\right) ^{\frac{\alpha }{2}}}dt \Vert f\Vert _{A_{\log ^{\alpha }}^2} \nonumber \\&= \int _0^1 \frac{2}{ (1+t)\left( \log \frac{2}{1-t^2}\right) ^{\frac{\alpha }{2}}} \left( \frac{1}{1-t}+\frac{1}{(t(1-t))^{\frac{1}{2}}}\right) dt \Vert f\Vert _{A_{\log ^{\alpha }}^2}. \end{aligned}$$
(2.8)

From (2.3) and (2.8), we have that

$$\begin{aligned} \Vert \mathcal {H}f\Vert _{A^2}&\le \int _0^1 \frac{2}{ (1+t)\left( \log \frac{2}{1-t^2}\right) ^{\frac{\alpha }{2}}} \left( \frac{1}{1-t}+\frac{1}{(t(1-t))^{\frac{1}{2}}}\right) dt\Vert f\Vert _{A_{\log ^{\alpha }}^2} \\&\le \int _0^1 \frac{2}{ \left( \log \frac{2}{1-t^2}\right) ^{\frac{\alpha }{2}}} \left( \frac{1}{1-t}+\frac{1}{(t(1-t))^{\frac{1}{2}}}\right) dt\Vert f\Vert _{A_{\log ^{\alpha }}^2}. \end{aligned}$$

It means that

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2} \le \int _0^1 \frac{2}{ \left( \log \frac{2}{1-x^2}\right) ^{\frac{\alpha }{2}}} \left( \frac{1}{1-x}+\frac{1}{(x(1-x))^{\frac{1}{2}}}\right) dx, \end{aligned}$$

and the last integral converges for \(\alpha >2\).

This finishes the proof of the theorem. \(\square \)

This result improves Theorem 4.3 in [16], and we also give a new proof method of Theorem 4.5 in [11].

2.2 Lower Bound for the Norm \(\Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2}\)

Before we get lower bound for the norm \(\Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2}\), we need find a special function in \(A^2_{\log ^\alpha }\).

Lemma 2.1

Let \(\alpha >2\), \(b\ge 1\) and \(1<\gamma <2\). Then the function

$$\begin{aligned} f (z) = \left( \frac{1}{z}\log \frac{b}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{2}}, \end{aligned}$$

belongs to \(A_{\log ^{\alpha }}^2\).

Proof

First we recall a well known result of Littlewood [17, pp.93–96]: Shows the function has an an integral mean with growth [9, p49]

$$\begin{aligned} M_2(r,f)^2&=\frac{1}{2\pi }\int _0^{2\pi } \left| \left( \frac{1}{re^{i\theta }}\log \frac{b}{1-re^{i\theta }} \right) ^{-\alpha }(1 - re^{i\theta })^{-\gamma }\right| d\theta \\&\le \frac{1}{2\pi }\int _0^{2\pi }\left| \left( \frac{1}{re^{i\theta }}\log \frac{1}{1-re^{i\theta }} \right) ^{-\alpha }(1 - re^{i\theta })^{-\gamma }\right| d\theta \\&\sim C\frac{1}{(1-r)^{\gamma -1}} \left( \log {\frac{1}{1-r}}\right) ^{-\alpha }, \quad r \rightarrow 1. \end{aligned}$$

Since

$$\begin{aligned} (a+b)^p\le 2^p(a^p + b^p), \quad a\ge 0, \quad b\ge 0, \end{aligned}$$

we have

$$\begin{aligned} M_2(r,f)^2\left( \log {\frac{2}{1-r^2}}\right) ^{\alpha }&\le C\frac{1}{(1-r)^{\gamma -1}} \left( \log {\frac{1}{1-r}}\right) ^{-\alpha } 2^\alpha \left( \log 2^\alpha +\left( \log \frac{1}{1-r}\right) ^\alpha \right) \\&=C\frac{2^\alpha \log 2^\alpha }{(1-r)^{\gamma -1}} \left( \log {\frac{1}{1-r}}\right) ^{-\alpha } +C\frac{2^\alpha }{(1-r)^{\gamma -1}}, \quad r \rightarrow 1. \end{aligned}$$

Thus we can find the integral \(\int _0^1 M_2(r,f)^2\left( \log {\frac{2}{1-r^2}}\right) ^{\alpha } dr \) converges while \(\alpha >2\), \(b\ge 1\) and \(1<\gamma <2\), this shows that \(f (z) \in A_{\log ^{\alpha }}^2\). It is also easy to see that

$$\begin{aligned} \lim _{\gamma \rightarrow 2}\Vert f\Vert _{A_{\log ^{\alpha }}^2}=\infty . \end{aligned}$$

\(\square \)

Corollary 2.1

Let \(\alpha >1\), \(b\ge 1\) and \(1<\gamma <2\). Then the function

$$\begin{aligned} f (z) = \left( \frac{1}{z}\log \frac{b}{1-z} \right) ^{-\frac{\alpha }{p}}(1 - z)^{-\frac{\gamma }{p}}, \end{aligned}$$

belongs to \(A_{\log ^{\alpha }}^p(p>2)\).

Theorem 2.2

Let \(\alpha >2\). Then the norm of the Hilbert matrix operator acting from \(A_{\log ^{\alpha }}^2\) into \(A^2\) satisfies the lower estimate

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2}\ge C_{\alpha }\int _0^1 \frac{x^{\frac{\alpha }{2}}}{(1-x) \left( \log \frac{1}{1-x}\right) ^{\frac{\alpha }{2}}}dx. \end{aligned}$$

where

$$\begin{aligned} C_{\alpha }=\limsup _{\gamma \rightarrow 2}\frac{\left\| (1-z)^{-\frac{\gamma }{2}}\right\| _{A^2}}{\left\| \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{2}}\right\| _{A_{\log ^{\alpha }}^2}}. \end{aligned}$$

Proof

Let \(\alpha >2\), we begin by selecting a family of test functions. Choose an arbitrary \(\gamma \) such that \(1<\gamma <2\). It is a standard exercise to check that the function

$$\begin{aligned} f_\gamma (z) = \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{2}}, \end{aligned}$$

and Lemma 2.1 shows that \(f_\gamma (z) \in A_{\log ^{\alpha }}^2\). It is also easy to see that

$$\begin{aligned} \lim _{\gamma \rightarrow 2}\Vert f_\gamma \Vert _{A_{\log ^{\alpha }}^2}=\infty . \end{aligned}$$

And we let

$$\begin{aligned} F_\gamma (z)=(1 - z)^{-\frac{\gamma }{2}}, \end{aligned}$$

belong to \(A^2\), and

$$\begin{aligned} \lim _{\gamma \rightarrow 2}\Vert F_\gamma \Vert _{A^2}=\infty , \end{aligned}$$

we obtain a relationship between \(f_\gamma (z)\) and \(F_\gamma (z)\) by the proof of Lemma 2.1, that

$$\begin{aligned} \limsup _{\gamma \rightarrow 2}\frac{\Vert F_\gamma \Vert _{A^2}}{\Vert f_\gamma \Vert _{A_{\log ^{\alpha }}^2}}&=\limsup _{\gamma \rightarrow 2}\frac{\left\| (1-z)^{-\frac{\gamma }{2}}\right\| _{A^2}}{\left\| \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{2}}\right\| _{A_{\log ^{\alpha }}^2}} \nonumber \\&\le C\limsup _{\gamma \rightarrow 2}\frac{\int _0^1\frac{1}{(1-r)^{\gamma -1}}dr}{\int _0^1 M_2(r,f)^2\left( \log {\frac{1}{1-r}}\right) ^{\alpha } dr} \nonumber \\&\le C\limsup _{\gamma \rightarrow 2}\frac{\int _0^1\frac{1}{(1-r)^{\gamma -1}}dr}{ \int _0^1\frac{1}{(1-r)^{\gamma -1}}dr}=C<\infty . \end{aligned}$$
(2.9)

Thus, we let \(C_{\alpha }=\limsup _{\gamma \rightarrow 2}\frac{\left\| (1-z)^{-\frac{\gamma }{2}}\right\| _{A^2}}{\left\| \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{2}}\right\| _{A_{\log ^{\alpha }}^2}}\), and \(C_\alpha \) is a constant, depending only on \(\alpha \).

Using (2.1), we find that

$$\begin{aligned} \mathcal {H}f_\gamma (z)&= \int _0^1 \frac{f_\gamma (t)}{1-tz}dt \nonumber \\&= \int _0^1 \frac{t^{\frac{\alpha }{2}}dt}{\left( \log \frac{1}{1-t} \right) ^{\frac{\alpha }{2}}(1-t)^{\frac{\gamma }{2}}(1-tz)}. \end{aligned}$$
(2.10)

Then making the change of variable \(w=(1-tz)/(1-t)\), we calculate that

$$\begin{aligned} \mathcal {H}f_\gamma (z)&= (1 - z)^{-\frac{\gamma }{2}} \int _1^\infty \frac{(w-1)^{\frac{\alpha }{2}}dw}{\left( \log \frac{1}{1-(\frac{w-1}{w-z})} \right) ^{\frac{\alpha }{2}}w(w-z)^{\frac{\alpha }{2}+1-\frac{\gamma }{2}}}, \end{aligned}$$
(2.11)

we define

$$\begin{aligned} \phi _\gamma (z)=\int _1^\infty \frac{(w-1)^{\frac{\alpha }{2}}dw}{\left( \log \frac{1}{1-(\frac{w-1}{w-z})^2} \right) ^{\frac{\alpha }{2}}w(w-z)^{\frac{\alpha }{2}+1-\frac{\gamma }{2}}}. \end{aligned}$$

for every z in \(\mathbb {D}\), which shows \(\mathcal {H}f_\gamma (z)=F_\gamma (z)\phi _\gamma (z).\)

Knowing that in the definition (2.10) of the function \(\phi _\gamma \) is similar to the function \(\phi _\gamma \) defined in the [8, Theorm 4], we can let w to be a real number \(s \ge 1\). Thus we obtained that \(\phi _\gamma \) belongs to the disk algebra whenever \(\gamma \le 2\), (the case \(\gamma = 2\) will also be useful to us although \(f_2 \notin A_{\log ^{\alpha }}^2\)), we can view \(\phi _\gamma \) is an analytic function of z that

$$\begin{aligned} \phi _\gamma (z)=\int _1^\infty \frac{(s-1)^{\frac{\alpha }{2}}ds}{\left( \log \frac{1}{1-(\frac{s-1}{s-z})^2} \right) ^{\frac{\alpha }{2}}s(s-z)^{\frac{\alpha }{2}+1-\frac{\gamma }{2}}}. \end{aligned}$$
(2.12)

We will use the test function \(g_\gamma (z)=\frac{f_\gamma (z)}{\Vert F_\gamma \Vert _{A^2}}\) and \(G_\gamma (z)=\frac{F_\gamma (z)}{\Vert F_\gamma \Vert _{A^2}}\), we obtain that

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2} \ge \frac{\Vert \mathcal {H}(g_\gamma )\Vert _{A^2}}{\Vert g_\gamma \Vert _{A_{\log ^{\alpha }}^2}} = \frac{\Vert F_\gamma \Vert _{A^2}}{\Vert f_\gamma \Vert _{A_{\log ^\alpha }^2}} \frac{\Vert \mathcal {H}(f_\gamma )\Vert _{A^2}}{\Vert F_\gamma \Vert _{A^2}} = \frac{\Vert F_\gamma \Vert _{A^2}}{\Vert f_\gamma \Vert _{A_{\log ^\alpha }^2}} \frac{\Vert F_\gamma (z)\phi _\gamma (z)\Vert _{A^2}}{\Vert F_\gamma \Vert _{A^2}}. \end{aligned}$$

Letting \(\gamma \rightarrow 2\), and by [8, Theorem 4] we get,

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2}&\ge \limsup _{\gamma \rightarrow 2 } \frac{\Vert F_\gamma \Vert _{A^2}}{\Vert f_\gamma \Vert _{A_{\log ^\alpha }^2}} \frac{\Vert F_\gamma (z)\phi _\gamma (z)\Vert _{A^2}}{\Vert F_\gamma \Vert _{A^2}}= C_{\alpha } \lim _{\gamma \rightarrow 2}\Vert G_\gamma \phi _\gamma \Vert _{A^2} =C_{\alpha } \Vert \phi _2\Vert _\infty \\&= C_{\alpha } \sup _{z\in \mathbb {D}}\int _1^\infty \frac{(s-1)^{\frac{\alpha }{2}}}{\left( \log \frac{1}{1-(\frac{s-1}{s-z})^2} \right) ^{\frac{\alpha }{2}}s(s-z)^{\frac{\alpha }{2}}}ds\\&\ge C_{\alpha } \sup _{0\le r \le 1}\int _1^\infty \frac{(s-1)^{\frac{\alpha }{2}}}{\left( \log \frac{1}{1-(\frac{s-1}{s-z})^2} \right) ^{\frac{\alpha }{2}}s(s-z)^{\frac{\alpha }{2}}}ds\\&\ge C_{\alpha } \int _1^\infty \frac{(s-1)^{\frac{\alpha }{2}}}{s^{\frac{\alpha }{2}+1}\left( \log \frac{1}{1-(\frac{s-1}{s})^2} \right) ^{\frac{\alpha }{2}}} ds. \end{aligned}$$

Then making the change of variable \(x=(s-1)/s\), we calculate that

$$\begin{aligned} \int _1^\infty \frac{(s-1)^{\frac{\alpha }{2}}}{s^{\frac{\alpha }{2}+1}\left( \log \frac{1}{1-(\frac{s-1}{s})^2} \right) ^{\frac{\alpha }{2}}} ds=\int _0^1 \frac{x^{\frac{\alpha }{2}}}{(1-x) \left( \log \frac{1}{1-x}\right) ^{\frac{\alpha }{2}}}dx. \end{aligned}$$

Thus we obtained that,

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A_{\log ^\alpha }^2\rightarrow A^2} \ge C_{\alpha }\int _0^1 \frac{x^{\frac{\alpha }{2}}}{(1-x) \left( \log \frac{1}{1-x}\right) ^{\frac{\alpha }{2}}}dx, \end{aligned}$$

where

$$\begin{aligned} C_{\alpha }=\limsup _{\gamma \rightarrow 2}\frac{\left\| (1-z)^{-\frac{\gamma }{2}}\right\| _{A^2}}{\left\| \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{2}}\right\| _{A_{\log ^{\alpha }}^2}}. \end{aligned}$$

and the last integral converges for \(\alpha >2\).

This concludes the proof. \(\square \)

Corollary 2.2

Let \(\alpha >2\). Then the norm of the Hilbert matrix operator acting from \(A_{\log ^{\alpha }}^2\) into \(A^2\) satisfies

$$\begin{aligned} C_{\alpha }\int _0^1 \frac{x^{\frac{\alpha }{2}}}{(1-x) \left( \log \frac{1}{1-x}\right) ^{\frac{\alpha }{2}}}dx{} & {} \le \Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^2 \rightarrow A^2} \\{} & {} \le \int _0^1 \frac{2}{ \left( \log \frac{2}{1-x^2}\right) ^{\frac{\alpha }{2}}} \left( \frac{1}{1-x}+\frac{1}{(x(1-x))^{\frac{1}{2}}}\right) dx. \end{aligned}$$

where

$$\begin{aligned} C_{\alpha }=\limsup _{\gamma \rightarrow 2}\frac{\left\| (1-z)^{-\frac{\gamma }{2}}\right\| _{A^2}}{\left\| \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{2}}\right\| _{A_{\log ^{\alpha }}^2}}. \end{aligned}$$

3 Norm Estimates of the Hilbert Matrix \(\Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^p \rightarrow A^p}\)

Then we consider the boundedness of Hilbert matrix from into \(A_{\log ^\alpha }^p\) into \(A^p\), for \(p>2\) and \(\alpha > 0\). That can easy obtain Lemma 3.1.

Lemma 3.1

If \(p>2\) and \(\alpha > 0\), then \(\mathcal {H}\) acts as a bounded operator from \(A_{\log ^\alpha }^p\) into \(A^p\).

Since \(A_{\log ^\alpha }^p \subset A^p\), that the lemma is obviously established.

Lemma 3.2

[3, 6] Let \(2<p<\infty \). Then the norm of the Hilbert matrix operator acting on \(A^p\) satisfies

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A^p \rightarrow A^p}=\frac{\pi }{\sin \frac{2\pi }{p}}. \end{aligned}$$

Then we give the upper bound for the norm estimates of the Hilbert matrix \(\Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^p \rightarrow A^p}\).

Theorem 3.1

Let \(2<p<\infty \) and \(\alpha >1\). Then the norm of the Hilbert matrix operator acting from \(A_{\log ^{\alpha }}^p\) into \(A^p\) satisfies

$$\begin{aligned} C_{\alpha ,p} \int _0^1 \frac{x^{\frac{\alpha }{p}}}{(1-x)^{\frac{2}{p}} \left( \log \frac{1}{1-x}\right) ^{\frac{\alpha }{p}}}dx \le \Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^p \rightarrow A^p} \le \int _0^1 \frac{2^{\frac{2}{p}}x^{\frac{2}{p}-1}}{(1-x)^{\frac{2}{p}} \left( \log {\frac{2}{1-x^2}}\right) ^{\frac{\alpha }{p}}} dt, \end{aligned}$$

where

$$\begin{aligned} C_{\alpha ,p}=\limsup _{\gamma \rightarrow 2}\frac{\left\| (1-z)^{-\frac{\gamma }{p}}\right\| _{A^p}}{\left\| \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{p}}\right\| _{A_{\log ^{\alpha }}^2}}. \end{aligned}$$

Proof

First, we establish the lower bound for the norm \(\Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^p\rightarrow A^p}\). We also construct a family of test functions like Theorem 2.2. Choose an arbitrary \(\gamma \) such that \(1<\gamma <2\). It is a standard exercise to check that the function

$$\begin{aligned} f^1_\gamma (z) = \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{p}}(1 - z)^{-\frac{\gamma }{p}}, \end{aligned}$$

After some elementary calculations, we can also establish that \(f^1_\gamma (z)\) belongs to \(A_{\log ^{\alpha }}^p\). It is also easy to observe that

$$\begin{aligned} \lim _{\gamma \rightarrow 2}\Vert f^1_\gamma \Vert _{A_{\log ^{\alpha }}^p}=\infty . \end{aligned}$$

And we let

$$\begin{aligned} F^1_\gamma (z)=(1 - z)^{-\frac{\gamma }{p}}, \end{aligned}$$

belong to \(A^p\), have that

$$\begin{aligned} \lim _{\gamma \rightarrow 2}\Vert F^1_\gamma \Vert _{A^p}=\infty . \end{aligned}$$

We also obtain a relationship between \(f^1_\gamma (z)\) and \(F^1_\gamma (z)\) by the Theorem 2.2 and Corollary 2.1, that

$$\begin{aligned} \limsup _{\gamma \rightarrow 2} \frac{\Vert F^1_\gamma \Vert _{A^p}}{\Vert f^1_\gamma \Vert _{A_{\log ^{\alpha }}^p}}\le C<\infty . \end{aligned}$$

Thus, we let \(C_{\alpha ,p}=\limsup _{\gamma \rightarrow 2}\frac{\left\| (1-z)^{-\frac{\gamma }{p}}\right\| _{A^p}}{\left\| \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{p}}\right\| _{A_{\log ^{\alpha }}^2}}\) and \(C_{\alpha ,p}\) is a constant, depending only on \(\alpha \) and p.

It can be seen from the proof of Theorem 2.2, we define

$$\begin{aligned} \phi ^1_\gamma (z)=\int _1^\infty \frac{(w-1)^{\frac{\alpha }{p}}}{\left( \log \frac{1}{1-(\frac{w-1}{w-z})} \right) ^{\frac{\alpha }{p}}w(w-z)^{\frac{\alpha }{p}+1-\frac{\gamma }{p}}}dw \end{aligned}$$

for every z in \(\mathbb {D}\), which shows \(\mathcal {H}f^1_\gamma (z)=F^1_\gamma (z)\phi ^1_\gamma (z).\)

According the proof of Theorem 2.2, we can let w to be a real number \(s \ge 1\). And we view \(\phi ^1_\gamma \) is an analytic function of z that

$$\begin{aligned} \phi ^1_\gamma (z)=\int _1^\infty \frac{(s-1)^{\frac{\alpha }{p}}}{\left( \log \frac{1}{1-(\frac{s-1}{s-z})^2} \right) ^{\frac{\alpha }{p}}s(s-z)^{\frac{\alpha }{p}+1-\frac{\gamma }{p}}}ds. \end{aligned}$$

We will use the test function \(g^1_\gamma (z)=\frac{f^1_\gamma (z)}{\Vert F_\gamma \Vert _{A^p}}\) and \(G^1_\gamma (z)=\frac{F^1_\gamma (z)}{\Vert F^1_\gamma \Vert _{A^p}}\), we have that

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A^p_{\log ^{\alpha }} \rightarrow A^p} \ge \frac{\Vert \mathcal {H}(g^1_\gamma )\Vert _{A^p}}{\Vert g^1_\gamma \Vert _{A_{\log ^{\alpha }}^p}} = \frac{\Vert F^1_\gamma \Vert _{A^p}}{\Vert f^1_\gamma \Vert _{A_{\log ^\alpha }^p}} \frac{\Vert \mathcal {H}(f^1_\gamma )\Vert _{A^p}}{\Vert F^1_\gamma \Vert _{A^p}} = \frac{\Vert F^1_\gamma \Vert _{A^p}}{\Vert f^1_\gamma \Vert _{A_{\log ^\alpha }^p}} \frac{\Vert F^1_\gamma (z)\phi ^1_\gamma (z)\Vert _{A^p}}{\Vert F^1_\gamma \Vert _{A^p}}. \end{aligned}$$

Letting \(\gamma \rightarrow 2\), and by [8, Theorem 4] we get,

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A^p_{\log ^{\alpha }} \rightarrow A^p}&\ge \limsup _{\gamma \rightarrow 2} \frac{\Vert F^1_\gamma \Vert _{A^p}}{\Vert f^1_\gamma \Vert _{A_{\log ^\alpha }^p}} \frac{\Vert F^1_\gamma (z)\phi _\gamma (z)\Vert _{A^p}}{\Vert F^1_\gamma \Vert _{A^p}}\\&= C_{\alpha ,p}\lim _{\gamma \rightarrow 2}\Vert G^1_\gamma \phi ^1_\gamma \Vert _{A^p} = C_{\alpha ,p}\Vert \phi ^1_2\Vert _\infty \\&= C_{\alpha ,p}\sup _{z\in \mathbb {D}}\int _1^\infty \frac{(s-1)^{\frac{\alpha }{p}}}{\left( \log \frac{1}{1-(\frac{s-1}{s-z})} \right) ^{\frac{\alpha }{p}}s(s-z)^{\frac{\alpha -2}{p}+1}}ds \\&\ge C_{\alpha ,p}\sup _{0\le r \le 1}\int _1^\infty \frac{(s-1)^{\frac{\alpha }{p}}}{\left( \log \frac{1}{1-(\frac{s-1}{s-r})} \right) ^{\frac{\alpha }{p}}s(s-r)^{\frac{\alpha -2}{p}+1}} ds \\&\ge C_{\alpha ,p} \int _1^\infty \frac{(s-1)^{\frac{\alpha }{p}}}{s^{\frac{\alpha -2}{p}+2}\left( \log \frac{1}{1-(\frac{s-1}{s})} \right) ^{\frac{\alpha }{p}}} ds. \end{aligned}$$

Then making the change of variable \(x=(s-1)/s\), we calculate that

$$\begin{aligned} \int _1^\infty \frac{(s-1)^{\frac{\alpha }{p}}}{s^{\frac{\alpha -2}{p}+2}\left( \log \frac{1}{1-(\frac{s-1}{s})} \right) ^{\frac{\alpha }{p}}} ds=\int _0^1 \frac{x^{\frac{\alpha }{p}}}{(1-x)^{\frac{2}{p}} \left( \log \frac{1}{1-x}\right) ^{\frac{\alpha }{p}}}dx. \end{aligned}$$

Thus we obtained that,

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A_{\log ^\alpha }^p\rightarrow A^p} \ge C_{\alpha ,p} \int _0^1 \frac{x^{\frac{\alpha }{p}}}{(1-x)^{\frac{2}{p}} \left( \log \frac{1}{1-x}\right) ^{\frac{\alpha }{p}}}dx. \end{aligned}$$

where

$$\begin{aligned} C_{\alpha ,p}=\limsup _{\gamma \rightarrow 2}\frac{\left\| (1-z)^{-\frac{\gamma }{p}}\right\| _{A^p}}{\left\| \left( \frac{1}{z}\log \frac{1}{1-z} \right) ^{-\frac{\alpha }{2}}(1 - z)^{-\frac{\gamma }{p}}\right\| _{A_{\log ^{\alpha }}^2}}. \end{aligned}$$

On the other hand, we give the upper bound for the norm \(\Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^p\rightarrow A^p}\). We using the method in the proof of Theorem 2.1 and Lemma 3.2, by simple calculation we found that

$$\begin{aligned} \Vert \mathcal {H}f\Vert _{A^p}&\le \int _0^1 \Vert T_t f\Vert _{A^p} dt =\int _0^1 \frac{t^{\frac{2}{p}-1}}{(1-t)^{\frac{2}{p}}}\left( \int _{D_t} |w|^{p-4} |f(w)|^p dA(w)\right) ^{\frac{1}{p}} dt \\&\le \int _0^1 \frac{t^{\frac{2}{p}-1}}{(1-t)^{\frac{2}{p}} \left( \log {\frac{2}{1-(\frac{t}{2-t})^2}}\right) ^{\frac{\alpha }{p}}} dt \left( \int _{D_t} |w|^{p-4} |f(w)|^p \left( \log {\frac{2}{1-|w|^2}} \right) ^{\alpha } dA(w)\right) ^{\frac{1}{p}}. \end{aligned}$$

By Theorem 3.2 in [3] and Lemma 2 in [6] we can find when \(2<p<4\) and \(4\le p < \infty \), we have that

$$\begin{aligned} \Vert \mathcal {H}f\Vert _{A^p}&\le \int _0^1 \frac{t^{\frac{2}{p}-1}}{(1-t)^{\frac{2}{p}} \left( \log {\frac{2}{1-(\frac{t}{2-t})^2}}\right) ^{\frac{\alpha }{p}}} dt \left( \int _{\mathbb {D}} |f(w)|^p \left( \log {\frac{2}{1-|w|^2}} \right) ^{\alpha } dA(w)\right) ^{\frac{1}{p}} \\&= \int _0^1 \frac{2^{\frac{2}{p}}t^{\frac{2}{p}-1}}{(1+t)(1-t)^{\frac{2}{p}} \left( \log {\frac{2}{1-t^2}}\right) ^{\frac{\alpha }{p}}} dt \Vert f\Vert _{A_{\log ^{\alpha }}^p}. \\&\le \int _0^1 \frac{2^{\frac{2}{p}}t^{\frac{2}{p}-1}}{(1-t)^{\frac{2}{p}} \left( \log {\frac{2}{1-t^2}}\right) ^{\frac{\alpha }{p}}} dt \Vert f\Vert _{A_{\log ^{\alpha }}^p}. \end{aligned}$$

Hence, in this case, we conclude that

$$\begin{aligned} \Vert \mathcal {H}\Vert _{A_{\log ^{\alpha }}^p\rightarrow A^p} \le \int _0^1 \frac{2^{\frac{2}{p}}x^{\frac{2}{p}-1}}{(1-x)^{\frac{2}{p}} \left( \log {\frac{2}{1-x^2}}\right) ^{\frac{\alpha }{p}}} dt. \end{aligned}$$

and this concludes the proof. \(\square \)