1 Introduction
Let \({\mathbb {D}}=\{z\in {\mathbb {C}}:\vert z\vert <1\}\) be the open unit disk of the complex plane \({\mathbb {C}}\), \(r{\mathbb {D}}=\{z\in {\mathbb {D}}:\vert z\vert < r\}\), \(H({\mathbb {D}})\) be the space of all analytic functions on \({\mathbb {D}}\), and \(dA(z)=\frac{1}{\pi }r\,dr\,d\theta \) be the normalized area measure on \({\mathbb {D}}\) (i.e., \(A({\mathbb {D}})=1\)).
A positive continuous function on \({\mathbb {D}}\) is called weight. Let \(\mu (z)\) be a weight function on \({\mathbb {D}}\). The weighted-type space \(H_{\mu }^{\infty }({\mathbb {D}})=H_{\mu }^{\infty }\) consists of \(f\in H({\mathbb {D}})\) such that
The little weighted-type space on \({\mathbb {D}}\), \(H_{\mu ,0}^{\infty }({\mathbb {D}})=H _{\mu ,0}^{\infty }\) consists of all \(f\in H({\mathbb {D}})\) such that
(see, e.g., [1] and the related references therein). For \(\mu (z)=(1-\vert z\vert )^{\alpha }\), \(\alpha >0\), the classical weighted-type space \(H_{\alpha }^{\infty }({\mathbb {D}})=H_{\alpha }^{\infty }\) and the classical little weighted-type space \(H_{\alpha ,0}^{\infty }({\mathbb {D}})=H _{\alpha ,0}^{\infty }\) are obtained whose special cases frequently appear in the literature, which is also the case in this paper.
$$ \Vert f \Vert _{H_{\mu }^{\infty }}:=\sup_{z\in {\mathbb {D}}} \mu (z) \bigl\vert f(z) \bigr\vert < \infty . $$
$$ \lim_{ \vert z \vert \to 1-0} \mu (z) \bigl\vert f(z) \bigr\vert =0 $$
Anzeige
Let \(-1<\gamma <\infty \) and \(\delta \le 0\). The logarithmic Hilbert–Bergman space, denoted by \(A^{2}_{\gamma ,\delta }({\mathbb {D}})=A ^{2}_{\gamma ,\delta }\), consists of all \(f\in H({\mathbb {D}})\) such that
where
$$ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}:= \biggl( \int _{{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z) \biggr)^{1/2}< \infty , $$
$$ \omega _{\gamma ,\delta }(z)= \biggl(\ln \frac{1}{ \vert z \vert } \biggr)^{\gamma } \biggl[\ln \biggl(1-\frac{1}{\ln \vert z \vert } \biggr) \biggr]^{\delta }. $$
Under the inner product
\(A^{2}_{\gamma ,\delta }\) is a Hilbert space. This is why it is called the logarithmic Hilbert–Bergman space. The space \(A^{2}_{\gamma , \delta }\) has been recently introduced in [2] as a special case of the logarithmic Bergman space \(A^{p}_{\gamma ,\delta }\), \(p\in (0,\infty )\).
$$ \langle f,g\rangle = \int _{{\mathbb {D}}}f(z)\overline{g(z)} \omega _{\gamma ,\delta }(z)\,dA(z), $$
For a given function \(g\in H({\mathbb {D}})\), the integral-type operator \(I_{g}\) is defined by
for \(f\in H({\mathbb {D}})\).
$$\begin{aligned} I_{g}(f) (z)= \int _{0}^{z}f(\zeta )g(\zeta )\,d\zeta \end{aligned}$$
(1)
As usual, throughout the paper, we will write frequently \(I_{g} f\) instead of \(I_{g}(f)\).
Anzeige
The operator (1) is clearly a natural generalization of the integral operator (the one obtained for \(g(z)\equiv 1\)). The operator can be regarded as a classical/folklore one. A variant of the operator can be found in [3], which could be one of the first papers studying such an operator on concrete spaces of analytic functions on \({\mathbb {D}}\) (see [3, Lemma 1]). The topic of studying integral-type operators on spaces of analytic functions has attracted some considerable recent attention. Much information on the topic, including a large list of references up to the end of 2006, can be found in [4]. Some product-type generalizations of the operator on the unit disc were later introduced and studied, for example, in [5], while the corresponding operator for the case of the unit ball was introduced in [6] and later studied in many papers to mention, for example, [7] (for the case of the polydisc, see, e.g., [8]). For some further generalizations, related operators, and related results, see also [9‐11] and the references therein. We would like to point out that a great majority of these papers are devoted to characterizing some function-theoretic properties of these operators in terms of the involved symbols.
Let X, Y be two Banach spaces and \(T:X\to Y\) be a bounded linear operator (an operator which maps bounded sets in X to bounded sets in Y). Recall that the essential norm of the bounded linear operator \(T:X\to Y\), denoted by \(\Vert T\Vert _{e}\), is defined as
where \(\Vert \cdot \Vert \) denotes the usual operator norm. For some results on the essential norms of concrete operators (such as the composition, multiplication, weighted composition, differentiation, integral, and their various products and relatives) see, for example, [7, 12‐18] and the related references therein. From the definition of the essential norm and since the set of all compact operators is a closed subset of the space of bounded linear operators, it follows that the operator \(T:X\to Y\) is compact if and only if \(\Vert T\Vert _{e}=0\).
$$ \Vert T \Vert _{e}:=\inf \bigl\{ \Vert T-K \Vert :K \text{ is compact from $X$ to $Y$} \bigr\} , $$
In this paper, first, we characterize the boundedness and compactness of \(I_{g}\) on \(A^{2}_{\gamma ,\delta }\), when \(\delta <0\). As it is shown, when \(\delta =0\), the space is equivalent to the classical weighted Hilbert–Bergman space for which the corresponding results are known even in much more general settings [19, 20]. We also estimate essential norm of the operator, which is practically the main result in the paper. This paper, among others, can be regarded as a continuation of our investigations of integral-type operators (see [4‐7] and the references therein), essential norms of concrete operators on spaces of analytic functions (see [7, 15, 16] and the references therein), as well as the investigation of concrete operators on \(A^{2}_{\gamma ,\delta }\). Before this work, composition operators on, or between, \(A^{2}_{\gamma ,\delta }\) were studied in [2, 18], and product-type operators from \(A^{2}_{\gamma ,\delta }\) to Zygmund–Orlicz spaces were studied in [21]. We also present some basic results on the space \(A^{2}_{\gamma ,\delta }\). For example, we give a completely analytic proof why the space \(A^{2}_{\gamma ,\delta }\) is the same as the space consisting of all \(f\in H({\mathbb {D}})\) such that
It is well known that \(f\in A^{2}_{\gamma }({\mathbb {D}})\) if and only if \(f\in {\mathcal{D}}_{\gamma +2}^{2}({\mathbb {D}})\) (the classical weighted Dirichlet space); moreover,
Here, we prove a similar result for the space \(A^{2}_{\gamma ,\delta }\). We obtain pointwise estimates for functions in \(A^{2}_{\gamma , \delta }\), as well as for their derivatives. We also give a complete orthonormal system in \(A^{2}_{\gamma ,\delta }\). These basic results on the space \(A^{2}_{\gamma ,\delta }\) seem new (we could not find them in the literature, so far).
$$\begin{aligned} \int _{{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{\gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr) ^{\delta }\,dA(z)< +\infty . \end{aligned}$$
(2)
$$ \Vert f \Vert _{A_{\gamma }^{2}}^{2}\asymp \bigl\vert f(0) \bigr\vert ^{2}+ \int _{{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2}\bigl(1- \vert z \vert ^{2}\bigr)^{ \gamma +2} \,dA(z). $$
In this paper, the letter C denotes a positive constant which may differ from one occurrence to the other. The notation \(a\lesssim b\) means that there exists a positive constant C independent of the essential variables in the quantities a and b such that \(a\leq Cb\). If \(a\lesssim b\) and \(b\lesssim a\), then we write \(a\asymp b\).
2 Some basic results on \(A^{2}_{\gamma,\delta}\) and auxiliary ones
This section presents several auxiliary results which are employed in the proofs of the main ones, as well as several basic results on the space \(A^{2}_{\gamma ,\delta }\). First, we present a completely analytic proof of the fact that the space is the same as the space consisting of all \(f\in H({\mathbb {D}})\) satisfying (2), connecting it to a more familiar weight.
If we use the well-known asymptotic relation \(\ln (1+x)=x+o(x)\), as \(x\to 0\), we easily obtain
as \(z\to 0\), and
as \(\vert z\vert \to 1-0\).
$$\begin{aligned} \omega _{\gamma ,\delta }(z) = \biggl(\ln \frac{1}{ \vert z \vert } \biggr)^{\gamma } \biggl[\ln \biggl(1+\frac{1}{\ln \frac{1}{ \vert z \vert }} \biggr) \biggr]^{ \delta }\sim \biggl( \ln \frac{1}{ \vert z \vert } \biggr)^{\gamma -\delta } \end{aligned}$$
(3)
$$\begin{aligned} \omega _{\gamma ,\delta }(z) &= \biggl(\ln \biggl(1+\frac{1- \vert z \vert }{ \vert z \vert } \biggr) \biggr) ^{\gamma } \biggl[\ln \biggl(1+ \frac{1}{\ln (1+\frac{1- \vert z \vert }{ \vert z \vert } )} \biggr) \biggr]^{ \delta } \\ &= \biggl(\frac{1- \vert z \vert }{ \vert z \vert }\bigl(1+O\bigl(1- \vert z \vert \bigr)\bigr) \biggr)^{\gamma } \biggl[\ln \biggl(1+\frac{1}{\frac{1- \vert z \vert }{ \vert z \vert }(1+O(1- \vert z \vert ))} \biggr) \biggr]^{ \delta } \\ &\sim \bigl(1- \vert z \vert \bigr)^{\gamma } \biggl(\ln \biggl( \frac{1}{1- \vert z \vert }+O(1) \biggr) \biggr) ^{\delta } \end{aligned}$$
(4)
From (3), (4) and the continuity of the functions therein, it is easy to see that
for every \(\vert z\vert \le r_{1}<1\) and each fixed \(r_{1}\in (0,1)\), while
for every \(r_{2}\le \vert z\vert <1\) and each fixed \(r_{2}\in (0,1)\), where in the proof of relation (6) the fact that, for each \(0< m\le M\), there are \(\alpha _{1}\in (0,1]\) and \(\alpha _{2}\ge 1\) such that
for z sufficiently close to the unit circle is also used.
$$\begin{aligned} \omega _{\gamma ,\delta }(z)\asymp \biggl(\ln \frac{1}{ \vert z \vert } \biggr) ^{\gamma -\delta } \end{aligned}$$
(5)
$$\begin{aligned} \omega _{\gamma ,\delta }(z)\asymp \bigl(1- \vert z \vert \bigr)^{\gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }, \end{aligned}$$
(6)
$$\begin{aligned} \ln \biggl(\frac{e}{1- \vert z \vert } \biggr)^{\alpha _{1}}\le \ln \frac{m}{1- \vert z \vert }\le \ln \frac{M}{1- \vert z \vert }\le \ln \biggl(\frac{e}{1- \vert z \vert } \biggr) ^{\alpha _{2}} \end{aligned}$$
(7)
Proposition 1
Let
\(-1<\gamma <\infty \)
and
\(\delta \le 0\). Then the asymptotic relation
holds for every
\(f\in A^{2}_{\gamma ,\delta }\).
$$\begin{aligned} \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\asymp \int _{{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{ \gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }\,dA(z) \end{aligned}$$
(8)
Proof
From (6), we have
for each \(r_{0}\in (0,1)\).
$$\begin{aligned} \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2}\omega _{\gamma , \delta }(z)\,dA(z)\asymp \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{ \gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }\,dA(z) \end{aligned}$$
(9)
It is easy to see that, for each fixed \(r_{0}\in (0,1)\), we have
From (5), we have
$$\begin{aligned} \int _{r_{0}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{\gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }\,dA(z)\asymp \int _{r_{0}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \,dA(z). \end{aligned}$$
(10)
$$\begin{aligned} \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2}\omega _{\gamma ,\delta }(z)\,dA(z) \asymp \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr) ^{\gamma -\delta }\,dA(z). \end{aligned}$$
(11)
If \(\gamma -\delta \ge 0\), then
From (9) and (10) with \(r_{0}=1/e\), (11) and (12), it follows that
$$\begin{aligned} \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2}\,dA(z) \le \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr)^{\gamma -\delta }\,dA(z). \end{aligned}$$
(12)
$$\begin{aligned} \int _{{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{\gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr) ^{\delta }\,dA(z)\lesssim \Vert f \Vert _{A^{2}_{\gamma ,\delta }}. \end{aligned}$$
(13)
Further, let
It is easy to see that \(C_{1}\in (0,+\infty )\).
$$ C_{1}:=\sup_{0< r< 1/e} \biggl(\ln \frac{1}{r} \biggr)^{\gamma -\delta } \sqrt{r}. $$
We have
where we have used the polar coordinates \(z=re^{i\theta }\) and the monotonicity of the integral means
$$\begin{aligned}& \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr) ^{\gamma -\delta }\,dA(z) \\ & \quad =2 \int _{0}^{1/e}M_{2}^{2}(f,r) \biggl(\ln \frac{1}{r} \biggr)^{\gamma -\delta }r\,dr \\ & \quad \le 2C_{1} \int _{0}^{1/e}M_{2}^{2}(f,r) \sqrt{r}\,dr \\ & \quad =4C_{1} \int _{0}^{1/\sqrt{e}}M_{2}^{2}\bigl(f, \rho ^{2}\bigr)\rho ^{2} \,d\rho \\ & \quad \le 4C_{1} \int _{0}^{1/\sqrt{e}}M_{2}^{2}(f,\rho ) \rho \,d\rho \\ & \quad \le 2C_{1} \int _{\frac{1}{\sqrt{e}} {\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2}\,dA(z) \\ & \quad \le 2C_{1} \biggl( \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \,dA(z)+ \int _{\frac{1}{\sqrt{e}} {\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2}\,dA(z) \biggr) \\ & \quad \lesssim \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \,dA(z)+ \int _{\frac{1}{\sqrt{e}} {\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{ \gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }\,dA(z) \\ & \quad \lesssim \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \,dA(z)+ \int _{{\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{\gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }\,dA(z), \end{aligned}$$
(14)
$$ M_{2}^{2}(f,r)=\frac{1}{2\pi } \int _{0}^{2\pi } \bigl\vert f\bigl(re^{i\theta } \bigr) \bigr\vert ^{2}\,d \theta ,\quad r\in [0,1). $$
From (9) and (10) with \(r_{0}=1/e\), (11), and (14), it follows that
From (13) and (15), we obtain that the asymptotic relation (8) holds in this case.
$$\begin{aligned} \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\lesssim \int _{{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{ \gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }\,dA(z). \end{aligned}$$
(15)
If \(\gamma -\delta <0\), then clearly
Hence, from (9) and (10) with \(r_{0}=1/e\), (11), and (16), it follows that
$$\begin{aligned} \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr) ^{\gamma -\delta }\,dA(z)\le \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \,dA(z). \end{aligned}$$
(16)
$$\begin{aligned} \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\lesssim \int _{{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{ \gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }\,dA(z). \end{aligned}$$
(17)
On the other hand, we have
from which, along with the monotonicity of the integral means and use of the polar coordinates, it follows that
$$ C_{2}:=\sup_{0< r< 1/e^{2/3}}r \biggl(\ln \frac{1}{r} \biggr)^{\delta - \gamma }\in (0,+\infty ), $$
$$\begin{aligned} \int _{\frac{1}{e^{2/3}}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr) ^{\gamma -\delta }\,dA(z) &=2 \int _{0}^{1/e^{2/3}}M_{2}^{2}(f,r) \biggl(\ln \frac{1}{r} \biggr)^{\gamma -\delta }r\,dr \\ &\ge \frac{2}{C_{2}} \int _{0}^{1/e^{2/3}}M_{2}^{2}(f,r)r^{2} \,dr \\ &\ge \frac{4}{3C_{2}} \int _{0}^{1/e}M_{2}^{2}\bigl(f, \rho ^{2/3}\bigr)\rho \,d \rho \\ &\ge \frac{4}{3C_{2}} \int _{0}^{1/e}M_{2}^{2}(f,\rho ) \rho \,d\rho \\ &\ge \frac{2}{3C_{2}} \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \,dA(z). \end{aligned}$$
(18)
From (18), we have
$$\begin{aligned} \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2}\,dA(z) &\lesssim \int _{\frac{1}{e^{2/3}}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr) ^{\gamma -\delta }\,dA(z) \\ &\lesssim \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr)^{\gamma -\delta }\,dA(z) + \int _{\frac{1}{e^{2/3}}{\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr)^{\gamma -\delta }\,dA(z) \\ &\lesssim \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr)^{\gamma -\delta }\,dA(z) + \int _{\frac{1}{e^{2/3}}{\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2}\,dA(z) \\ &\lesssim \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr)^{\gamma -\delta }\,dA(z) + \int _{\frac{1}{e^{2/3}}{\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z) \\ &\lesssim \int _{\frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \biggl(\ln \frac{1}{ \vert z \vert } \biggr)^{\gamma -\delta }\,dA(z) + \int _{{\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z). \end{aligned}$$
(19)
From (9) and (10) with \(r_{0}=1/e\), (11), and (19), it follows that
$$\begin{aligned} \int _{{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1- \vert z \vert \bigr)^{\gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr) ^{\delta }\,dA(z)\lesssim \Vert f \Vert _{A^{2}_{\gamma ,\delta }}. \end{aligned}$$
(20)
Remark 1
In [18] it is said that the space \(A^{2}_{\gamma ,\delta }\) is the same as the space consisting of all \(f\in H({\mathbb {D}})\) satisfying condition (2) not giving a proof of the claim. Due to the estimates in (7), Proposition 1 gives a pure analytic proof of the equivalence of these two spaces. Note also that from Proposition 1 with \(\delta =0\) it is obtained that the space \(A^{2}_{\gamma ,0}\) is equivalent to the weighted Bergman space \(A^{2}_{\gamma }\).
Remark 2
Note that, by using (5), (6), and the polar coordinates, we have
since \(\gamma >-1\).
$$\begin{aligned} \Vert 1 \Vert _{A^{2}_{\gamma ,\delta }} &= \int _{{\mathbb {D}}}\omega _{\gamma , \delta }(z)\,dA(z) = \int _{\frac{1}{e}{\mathbb {D}}}\omega _{\gamma ,\delta }(z)\,dA(z)+ \int _{{\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}}\omega _{\gamma ,\delta }(z)\,dA(z) \\ &\lesssim \int _{\frac{1}{e}{\mathbb {D}}} \biggl(\ln \frac{1}{ \vert z \vert } \biggr) ^{\gamma -\delta } \,dA(z)+ \int _{{\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}}\bigl(1- \vert z \vert \bigr)^{ \gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta }\,dA(z) \\ &\le \frac{1}{\pi } \int _{0}^{1/e} \int _{0}^{2\pi } \biggl(\ln \frac{1}{ \rho } \biggr)^{\gamma -\delta } \,d\theta \rho \,d\rho + \biggl(\ln \frac{e ^{2}}{e-1} \biggr)^{\delta } \int _{{\mathbb {D}}\setminus \frac{1}{e}{\mathbb {D}}}\bigl(1- \vert z \vert \bigr)^{ \gamma } \,dA(z) \\ &=2 \int _{1}^{\infty }\frac{s^{\gamma -\delta }}{e^{2s}}\,ds+2 \biggl(\ln \frac{e ^{2}}{e-1} \biggr)^{\delta } \int _{1/e}^{1}(1-\rho )^{\gamma }\rho \,d \rho < \infty \end{aligned}$$
(21)
Proposition 2
Let
g
be a nondecreasing integrable function on
\([0,1)\)
and
h
be a positive function on
\((0,1)\)
which is continuous on
\([0,1)\). Then, for each fixed
\(r_{0}\in [0,1)\), we have
$$\begin{aligned} \int _{0}^{1} g(r)h(r)\,dr\asymp \int _{r_{0}}^{1}g(r)h(r)\,dr. \end{aligned}$$
(22)
Proof
If \(r_{0}=0\), then the result is obvious. Hence, assume that \(r_{0}\in (0,1)\). We have
Now note that there is unique \(n_{0}\in {\mathbb {N}}_{0}\) such that
Note that by the choice of \(n_{0}\) we have
$$\begin{aligned} \int _{0}^{1} g(r)h(r)\,dr= \int _{0}^{r_{0}} g(r)h(r)\,dr+ \int _{r_{0}}^{1} g(r)h(r)\,dr. \end{aligned}$$
(23)
$$ n_{0}\frac{1-r_{0}}{2}\le r_{0}< (n_{0}+1) \frac{1-r_{0}}{2}. $$
$$ \frac{(n_{0}+1)(1-r_{0})}{2}< \frac{1+r_{0}}{2}< 1. $$
Since g is a nondecreasing function and h is positive on \((0,1)\) and continuous on \([0,1)\), we have
$$\begin{aligned} \int _{0}^{r_{0}} g(r)h(r)\,dr &\le \max _{0\le r\le r_{0}}h(r) \int _{0} ^{r_{0}}g(r)\,dr \\ &\le \max_{0\le r\le r_{0}}h(r) \int _{0}^{(1-r_{0})(n_{0}+1)/2}g(r)\,dr \\ &=\max_{0\le r\le r_{0}}h(r)\sum_{j=0}^{n_{0}} \int _{(1-r_{0})j/2} ^{(1-r_{0})(j+1)/2} g(r)\,dr \\ &\le (n_{0}+1)\max_{0\le r\le r_{0}}h(r) \int _{r_{0}}^{(1+r_{0})/2} g(r)\,dr \\ &\le (n_{0}+1)\frac{\max_{0\le r\le r_{0}}h(r)}{ \min_{r_{0}\le r\le (1+r_{0})/2}h(r)} \int _{r_{0}}^{(1+r_{0})/2} g(r)h(r)\,dr \\ &\le C(r_{0},h) \int _{r_{0}}^{1} g(r)h(r)\,dr. \end{aligned}$$
(24)
Corollary 1
Let
g
be an arbitrary nondecreasing integrable function on
\([0,1)\)
and
h
be a fixed positive function on
\((0,1)\)
which is continuous on
\([0,1)\). Then, for each fixed
\(r_{0}\in [0,1)\), we have
for some positive constant
\(C(r_{0},h)\)
depending on
\(r_{0}\)
and h.
$$ \int _{r_{0}}^{1}g(r)h(r)\,dr\le \int _{0}^{1} g(r)h(r)\,dr\le C(r_{0},h) \int _{r_{0}}^{1}g(r)h(r)\,dr $$
Lemma 1
Let
\(-1<\gamma <\infty \)
and
\(\delta \le 0\). Then
for every
\(f\in A^{2}_{\gamma ,\delta }\).
$$\begin{aligned} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }}\asymp \bigl\vert f(0) \bigr\vert ^{2}+ \int _{{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \end{aligned}$$
(25)
Proof
Since
for \(a,b\in {\mathbb {C}}\), we have
$$\begin{aligned} \sqrt{ \vert a \vert ^{2}+ \vert b \vert ^{2}}\le \vert a \vert + \vert b \vert \le \sqrt{2\bigl( \vert a \vert ^{2}+ \vert b \vert ^{2}\bigr)} \end{aligned}$$
(26)
$$\begin{aligned} \begin{aligned}[b] & \biggl( \bigl\vert f(0) \bigr\vert ^{2}+ \int _{{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(z)\,dA(z) \biggr)^{1/2} \\ &\quad \asymp \bigl\vert f(0) \bigr\vert + \biggl( \int _{{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \biggr)^{1/2}.\end{aligned} \end{aligned}$$
(27)
Let
It is clear that \(\Vert f\Vert _{1}\ge 0\) for every \(f\in A^{2}_{\gamma , \delta }\), that \(\Vert 0\Vert _{1}=0\) and \(\Vert \lambda f\Vert _{1}=\vert \lambda \vert \Vert f\Vert _{1}\) for every \(f\in A^{2}_{\gamma ,\delta }\) and \(\lambda \in {\mathbb {C}}\). If \(\Vert f\Vert _{1}=0\), then obviously \(f(0)=0\), and since \(\omega _{\gamma +2,\delta }(z)>0\), \(z\in {\mathbb {D}}\setminus \{0 \}\), it follows that \(f'(z)=0\), \(z\in {\mathbb {D}}\setminus \{0\}\), and since \(f\in H({\mathbb {D}})\), it must be \(f'(z)=0\), \(z\in {\mathbb {D}}\), from which it follows that \(f(z)\equiv f(0)=0\). By using the triangle inequality and the Cauchy–Schwarz inequality it easily follows that \(\Vert f+g\Vert _{1}\le \Vert f\Vert _{1}+\Vert g\Vert _{1}\) for every \(f,g\in A^{2}_{\gamma ,\delta }\). Hence, \(\Vert \cdot \Vert _{1}\) is a norm on \(A^{2}_{\gamma , \delta }\).
$$ \Vert f \Vert _{1}:= \bigl\vert f(0) \bigr\vert + \biggl( \int _{{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \biggr)^{1/2}. $$
By using the polar coordinates, we have
Employing Proposition 2 with \(g(r)=M_{2}^{2}(f',r)\) (here the well-known fact that the integral means of holomorphic functions are nondecreasing functions is used, which in this case is a simple statement due the easily checked equality \(M_{2}^{2}(\hat{f},r)=\sum_{j=0}^{\infty }\vert a _{k}\vert ^{2}r^{2k}\), if \(\hat{f}(z)=\sum_{j=0}^{\infty }a_{k}z^{k}\)) and \(h(r)=r\omega _{\gamma +2,\delta }(r)\), which is positive and continuous on \((0,1)\) and if it is naturally defined by
it becomes continuous on \([0,1)\), we have
independently on f.
$$\begin{aligned} \int _{{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(z)\,dA(z)=2 \int _{0}^{1}M_{2}^{2} \bigl(f',r\bigr)\omega _{\gamma +2,\delta }(r)r\,dr. \end{aligned}$$
(28)
$$ h(0):=\lim_{r\to +0}h(r)=0 $$
$$\begin{aligned} \int _{0}^{1}M_{2}^{2} \bigl(f',r\bigr)\omega _{\gamma +2,\delta }(r)r\,dr\asymp \int _{r_{0}}^{1}M_{2}^{2} \bigl(f',r\bigr)\omega _{\gamma +2,\delta }(r)r\,dr, \end{aligned}$$
(29)
From (27)–(29) and by using the polar coordinates, it follows that
for each \(r_{0}\in [0,1)\).
$$\begin{aligned} & \biggl( \bigl\vert f(0) \bigr\vert ^{2}+ \int _{{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(z)\,dA(z) \biggr)^{1/2} \\ &\quad \asymp \bigl\vert f(0) \bigr\vert + \biggl( \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \biggr)^{1/2} \end{aligned}$$
(30)
Since
similar as above we obtain
for each \(r_{0}\in [0,1)\).
$$\begin{aligned} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }}=2 \int _{0}^{1}M_{2}^{2}(f,r) \omega _{\gamma ,\delta }(r)r\,dr, \end{aligned}$$
(31)
$$\begin{aligned} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }}\asymp \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2}\omega _{\gamma , \delta }(z)\,dA(z) \end{aligned}$$
(32)
By using the following known formula
which can be easily checked by direct calculation, in (32) and employing polar coordinates and the Fubini theorem, we get
$$\begin{aligned} M_{2}^{2}(f,r)= \bigl\vert f(0) \bigr\vert ^{2}+2 \int _{r{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2}\ln \frac{r}{ \vert z \vert }\,dA(z), \end{aligned}$$
(33)
$$\begin{aligned} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }} &\asymp 2 \int _{r_{0}}^{1}M_{2}^{2}(f,r) \omega _{\gamma ,\delta }(r)r\,dr \\ &\asymp 2 \bigl\vert f(0) \bigr\vert ^{2} \int _{r_{0}}^{1}\omega _{\gamma ,\delta }(r)r\,dr+4 \int _{r_{0}}^{1}\omega _{\gamma ,\delta }(r)r \int _{r{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2} \ln \frac{r}{ \vert z \vert }\,dA(z)\,dr \\ &= 2 \bigl\vert f(0) \bigr\vert ^{2} \int _{r_{0}}^{1}\omega _{\gamma ,\delta }(r)r\,dr+8 \int _{r_{0}}^{1}\omega _{\gamma ,\delta }(r)r \int _{0}^{r}M_{2}^{2} \bigl(f', \rho \bigr)\ln \frac{r}{\rho } \rho \,d\rho \,dr \\ &= 2 \bigl\vert f(0) \bigr\vert ^{2} \int _{r_{0}}^{1}\omega _{\gamma ,\delta }(r)r\,dr+8 \int _{0}^{r_{0}} \int _{r_{0}}^{1}\omega _{\gamma ,\delta }(r)r\ln \frac{r}{ \rho } \,dr M_{2}^{2}\bigl(f',\rho \bigr)\rho \,d\rho \\ &\quad{} +8 \int _{r_{0}}^{1} \int _{\rho }^{1}\omega _{\gamma ,\delta }(r)r \ln \frac{r}{\rho } \,dr M_{2}^{2}\bigl(f',\rho \bigr)\rho \,d\rho . \end{aligned}$$
(34)
By a slight modification of the proof of Lemma 3.1 in [2], we have that
for \(0< r_{0}\le \vert z\vert <1\).
$$\begin{aligned} \int _{ \vert z \vert }^{1}\omega _{\gamma ,\delta }(r)r\ln \frac{r}{ \vert z \vert } \,dr \asymp \omega _{\gamma +2,\delta }(z) \end{aligned}$$
(35)
From (34) and (35), and by using (26), it follows that
$$\begin{aligned} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }} &\gtrsim \bigl\vert f(0) \bigr\vert ^{2}+ \int _{r_{0}} ^{1}M_{2}^{2} \bigl(f',\rho \bigr)\omega _{\gamma +2,\delta }(\rho ) \rho \,d\rho \\ &\gtrsim \bigl\vert f(0) \bigr\vert ^{2}+ \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \\ &\gtrsim \biggl( \bigl\vert f(0) \bigr\vert + \biggl( \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \biggr)^{1/2} \biggr)^{2}. \end{aligned}$$
(36)
From (30) and (36), we have
for every \(f\in A^{2}_{\gamma ,\delta }\).
$$\begin{aligned} \Vert f \Vert _{1}\lesssim \Vert f \Vert _{A^{2}_{\gamma ,\delta }} \end{aligned}$$
(37)
Applying the open mapping theorem to the identity map
we have that
from which along with (37) the asymptotic relation (25) follows. □
$$ I:\bigl(A^{2}_{\gamma ,\delta }, \Vert \cdot \Vert _{A^{2}_{\gamma ,\delta }} \bigr) \to \bigl(A^{2}_{\gamma ,\delta }, \Vert \cdot \Vert _{1}\bigr), $$
$$ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\lesssim \Vert f \Vert _{1}, $$
By repeating use of Lemma 1, it follows that the following corollary holds.
Corollary 2
Let
\(-1<\gamma <\infty \), \(\delta \le 0\), and
\(m\in {\mathbb {N}}\). Then
for every
\(f\in A^{2}_{\gamma ,\delta }\).
$$\begin{aligned} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }}\asymp \sum _{j=0}^{m-1} \bigl\vert f^{(j)}(0) \bigr\vert ^{2}+ \int _{{\mathbb {D}}} \bigl\vert f^{(m)}(z) \bigr\vert ^{2}\omega _{\gamma +2m,\delta }(z)\,dA(z) \end{aligned}$$
From Corollary 2, we obtain the following corollary.
Corollary 3
Let
\(-1<\gamma <\infty \), \(\delta \le 0\), \(m\in {\mathbb {N}}\), and
\((f_{n})_{n\in {\mathbb {N}}}\subset A^{2}_{\gamma , \delta }\). Then
if and only if
and
for
\(j=\overline{0,m-1}\).
$$ \lim_{n\to \infty } \Vert f_{n}-f \Vert _{A^{2}_{\gamma ,\delta }}=0 $$
$$ \lim_{n\to \infty } \int _{{\mathbb {D}}} \bigl\vert f^{(m)}_{n}(z)-f^{(m)}(z) \bigr\vert ^{2} \omega _{\gamma +2m,\delta }(z)\,dA(z)=0 $$
$$\begin{aligned} \lim_{n\to \infty }f^{(j)}_{n}(0)=f^{(j)}(0) \end{aligned}$$
Note that from the proof of Lemma 1 we see that the following result was also proved.
Corollary 4
Let
\(-1<\gamma <\infty \)
and
\(\delta \le 0\). Then, for fixed
\(r_{0}\in [0,1)\), we have
$$\begin{aligned} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }}\asymp \bigl\vert f(0) \bigr\vert ^{2}+ \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert f'(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z). \end{aligned}$$
We need also the following estimate for the functions in \(A^{2}_{ \gamma ,\delta }\).
Lemma 2
Let
\(-1<\gamma <\infty \), \(\delta \le 0\), \(r_{0}\in (0,1)\), and
\(m\in {\mathbb {N}}_{0}\). Then, if
\(\vert z\vert \geq r_{0}\), it follows that
for
\(f\in A^{2}_{\gamma ,\delta }\).
$$\begin{aligned} \bigl\vert f^{(m)}(z) \bigr\vert \lesssim \bigl[\omega _{\gamma +2m+2,\delta }(z) \bigr] ^{-\frac{1}{2}} \Vert f \Vert _{A^{2}_{\gamma ,\delta }} \end{aligned}$$
Proof
From Corollary 2, since
for \(\vert \zeta -z\vert <\frac{1-\vert z\vert }{2}\), and by the subharmonicity of the function \(\vert f^{(m)}(z)\vert ^{2}\), we have
from which the corollary follows. □
$$ \omega _{\gamma +2m,\delta }(\zeta )\asymp \omega _{\gamma +2m,\delta }(z) $$
$$\begin{aligned} \Vert f \Vert _{A^{2}_{\gamma ,\delta }}^{2} &\gtrsim \int _{{\mathbb {D}}} \bigl\vert f^{(m)}( \zeta ) \bigr\vert ^{2}\omega _{\gamma +2m,\delta }(\zeta )\,dA(\zeta ) \\ &\gtrsim \int _{ \vert \zeta -z \vert < \frac{1- \vert z \vert }{2}} \bigl\vert f^{(m)}(\zeta ) \bigr\vert ^{2} \omega _{\gamma +2m,\delta }(\zeta )\,dA(\zeta ) \\ &\gtrsim \omega _{\gamma +2m,\delta }(z) \int _{ \vert \zeta -z \vert < \frac{1- \vert z \vert }{2}} \bigl\vert f^{(m)}(\zeta ) \bigr\vert ^{2}\,dA(\zeta ) \\ &\gtrsim \omega _{\gamma +2m,\delta }(z) \bigl(1- \vert z \vert \bigr)^{2} \bigl\vert f^{(m)}(z) \bigr\vert ^{2} \\ &\gtrsim \omega _{\gamma +2m+2,\delta }(z) \bigl\vert f^{(m)}(z) \bigr\vert ^{2}, \end{aligned}$$
Remark 3
Lemma 2 with \(m=0\) shows that the point evaluations \(\varLambda _{z}\) at the point \(z\in {\mathbb {D}}\setminus \{0\}\) are bounded linear functions on \(A^{2}_{\gamma ,\delta }\). Then there are reproducing kernels for these z, denoted by \(K_{\gamma ,\delta }(z, \cdot )\), such that
and
$$ \varLambda _{z}f=\bigl\langle f,K_{\gamma ,\delta }(z,\cdot )\bigr\rangle = \int _{{\mathbb {D}}}f(w)\overline{K_{\gamma ,\delta }(z,w)} \omega _{\gamma ,\delta }(w)\,dA(w) $$
$$ \Vert \varLambda _{z} \Vert = \bigl\Vert K_{\gamma ,\delta }(z, \cdot ) \bigr\Vert _{A^{2}_{\gamma , \delta }}. $$
Therefore, from Lemma 2 with \(m=0\), we have
for \(z\in {\mathbb {D}}\setminus \{0\}\).
$$\begin{aligned} \sqrt{K_{\gamma ,\delta }(z,z)}= \bigl\Vert K_{\gamma ,\delta }(z,\cdot ) \bigr\Vert _{A^{2}_{\gamma ,\delta }} \lesssim \bigl[\omega _{\gamma +2,\delta }(z) \bigr] ^{-\frac{1}{2}} \end{aligned}$$
(38)
Remark 4
From the proof of Lemma 2 we see that if \(-1<\gamma <\infty \), \(\delta \le 0\), and \(m\in {\mathbb {N}}_{0}\), then for each \(r\in (0,1)\), there is a constant \(c_{m}(r)\) such that
for \(\vert z\vert \ge r\) and every \(f\in A^{2}_{\gamma ,\delta }\).
$$\begin{aligned} \bigl\vert f^{(m)}(z) \bigr\vert \le c_{m}(r) \bigl[ \omega _{\gamma +2m+2,\delta }(z) \bigr] ^{-\frac{1}{2}} \Vert f \Vert _{A^{2}_{\gamma ,\delta }} \end{aligned}$$
Lemma 3
Let
\(-1<\gamma <\infty \), \(\delta \le 0\), and
\(g\in H({\mathbb {D}})\). Then a bounded operator
K
is compact on
\(A^{2}_{\gamma ,\delta }\)
if and only if, for every bounded sequence
\((f_{n})_{n\in {\mathbb {N}}}\)
in
\(A^{2}_{\gamma ,\delta }\)
such that
\(f_{n}\to 0\)
uniformly on every compact subset of
\({\mathbb {D}}\)
as
\(n\to \infty \), it follows that
\(\lim_{n\to \infty }\Vert Kf_{n}\Vert _{A^{2} _{\gamma ,\delta }}=0\).
Proof
We will apply Lemma 3 in [8]. Two cases are considered separately, when \(\gamma \ge \delta \) and \(\gamma <\delta\).
Case
\(\gamma <\delta \). We show that \(\Vert f_{n}-f\Vert _{A^{2} _{\gamma ,\delta }}\to 0\) as \(n\to \infty \) implies \(f_{n}\to f\) uniformly on compacts of \({\mathbb {D}}\).
Since \(\gamma <\delta \), then there exists a positive integer m such that \(\gamma +2m\geq \delta \). Therefore, by Corollary 2, we have that
for \(j=\overline{0,m-1}\), for some nonnegative constants \(c_{j}\), \(j=\overline{0,m-1}\).
$$\begin{aligned} \bigl\vert f^{(j)}(0) \bigr\vert \le c_{j} \Vert f \Vert _{A^{2}_{\gamma ,\delta }} \end{aligned}$$
(39)
From Lemma 2, we see that for each fixed \(r\in [1/2,1)\) it follows that
where
which is a finite constant due to the continuity of the function under the sign of maximum on a compact set.
$$\begin{aligned} \max_{1/2\le \vert z \vert \le r} \bigl\vert f^{(m)}(z) \bigr\vert \lesssim \max_{1/2\le \vert z \vert \le r} \bigl[\omega _{\gamma +2m+2,\delta }(z) \bigr]^{-\frac{1}{2}} \Vert f \Vert _{A^{2}_{\gamma ,\delta }} =\widehat{c}_{m}(r) \Vert f \Vert _{A^{2}_{\gamma , \delta }}, \end{aligned}$$
(40)
$$ \widehat{c}_{m}(r)=\max_{1/2\le t\le r} \bigl[\omega _{\gamma +2m+2,\delta }(t) \bigr] ^{-\frac{1}{2}}, $$
On the other hand, by the subharmonicity of the function \(\vert f^{(m)}(z)\vert ^{2}\), we have that
from which it follows that
for \(m\in {\mathbb {N}}_{0}\).
$$ \bigl\vert f^{(m)}(z) \bigr\vert ^{2}\le 16 \int _{ \vert \zeta -z \vert \le 1/4} \bigl\vert f^{(m)}(\zeta ) \bigr\vert ^{2}\,dA( \zeta ), $$
$$\begin{aligned} \max_{ \vert z \vert \le 1/2} \bigl\vert f^{(m)}(z) \bigr\vert ^{2}\le 16 \int _{ \vert \zeta \vert \le 3/4} \bigl\vert f^{(m)}( \zeta ) \bigr\vert ^{2}\,dA(\zeta ) \end{aligned}$$
(41)
Since \(\gamma +2m\ge \delta \), then there is a positive constant \(\widehat{c}_{1}\) such that
Hence, from (41) and (42), we have that
From (40) and (43) we have that, for each fixed \(r\in [0,1)\), there is a constant \(\widetilde{c}_{m}(r)\) such that
from which it further follows that
and consequently \(\Vert f_{n}-f\Vert _{A^{2}_{\gamma ,\delta }}\to 0\) as \(n\to \infty \) implies \(f^{(m)}_{n}\to f^{(m)}\) uniformly on compacts of \({\mathbb {D}}\). Moreover, (44) means that the point evaluation functionals \(f \mapsto f^{(m)}(z)\) are continuous.
$$\begin{aligned} 0< \widehat{c}_{1}\le \inf_{\vert z\vert \le 3/4}\omega _{\gamma +2m,\delta }(z). \end{aligned}$$
(42)
$$\begin{aligned} \max_{ \vert z \vert \le 1/2} \bigl\vert f^{(m)}(z) \bigr\vert ^{2}\le \frac{16}{\widehat{c}_{1}} \int _{ \vert \zeta \vert \le 3/4} \bigl\vert f^{(m)}(\zeta ) \bigr\vert ^{2}\omega _{\gamma +2m,\delta }(\zeta )\,dA(\zeta )\le \frac{16}{\widehat{c}_{1}} \Vert f \Vert _{A^{2}_{\gamma ,\delta }}^{2}. \end{aligned}$$
(43)
$$\begin{aligned} \max_{ \vert z \vert \le r} \bigl\vert f^{(m)}(z) \bigr\vert \le \widetilde{c}_{m}(r) \Vert f \Vert _{A^{2}_{ \gamma ,\delta }}, \end{aligned}$$
(44)
$$ \max_{ \vert z \vert \le r} \bigl\vert \bigl(f^{(m)}-f^{(m)}_{n} \bigr) (z) \bigr\vert \le \widetilde{c}_{m}(r) \Vert f-f _{n} \Vert _{A^{2}_{\gamma ,\delta }}, $$
Since
for each \(k\in \{1,\ldots, m\}\), by using (39), (44), and some simple estimates, we obtain
for every \(z\in r{\mathbb {D}}\).
$$ f^{(m-k)}(z)=\sum_{j=0}^{k-1} \frac{f^{(m-k+j)}(0)}{j!}z^{j}+ \int _{0} ^{z} \int _{0}^{\zeta _{k}}\cdots \int _{0}^{\zeta _{2}}f^{(m)}(\zeta _{1}) \,d\zeta _{1}\cdots d\zeta _{k-1}\,d\zeta _{k} $$
$$\begin{aligned} \bigl\vert f^{(m-k)}(z) \bigr\vert &\le \sum _{j=0}^{k-1}\frac{ \vert f^{(m-k+j)}(0) \vert }{j!} \vert z \vert ^{j}+ \sup_{z\in r{\mathbb {D}}} \biggl\vert \int _{0}^{z} \int _{0}^{\zeta _{k}}\cdots \int _{0}^{\zeta _{2}}f^{(m)}(\zeta _{1})\,d\zeta _{1}\cdots d\zeta _{k-1}\,d\zeta _{k} \biggr\vert \\ &\le \sum_{j=0}^{k-1}\frac{ \vert f^{(m-k+j)}(0) \vert }{j!}r^{j}+ \sup_{z\in r{\mathbb {D}}} \bigl\vert f^{(m)}(z) \bigr\vert \biggl\vert \frac{z^{k}}{k!} \biggr\vert \\ &\le \Biggl(\sum_{j=0}^{k-1} \frac{c_{m-k+j}}{j!}r^{j}+\widetilde{c} _{m}(r) \frac{r^{k}}{k!} \Biggr) \Vert f \Vert _{A^{2}_{\gamma ,\delta }} \end{aligned}$$
(45)
If we take \(k=m\) in (45), it follows that
from which it follows that
and consequently \(\Vert f_{n}-f\Vert _{A^{2}_{\gamma ,\delta }}\to 0\) as \(n\to \infty \) implies \(f_{n}\to f\) uniformly on compacts of \({\mathbb {D}}\). Moreover, from (46) we see that \(\vert f(z)\vert \lesssim \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\), which means that the point evaluation functionals are continuous. Hence, Lemma 3 in [8] can be applied in the case.
$$\begin{aligned} \max_{ \vert z \vert \leq r} \bigl\vert f(z) \bigr\vert \leq \Biggl(\sum _{j=0}^{m-1}\frac{c_{j}}{j!}r ^{j}+\widetilde{c}_{m}(r)\frac{r^{m}}{m!} \Biggr) \Vert f \Vert _{A^{2}_{\gamma ,\delta }}, \end{aligned}$$
(46)
$$ \max_{ \vert z \vert \le r} \bigl\vert (f-f_{n}) (z) \bigr\vert \le \Biggl(\sum_{j=0}^{m-1} \frac{c_{j}}{j!}r^{j}+\widetilde{c}_{m}(r) \frac{r^{m}}{m!} \Biggr) \Vert f-f _{n} \Vert _{A^{2}_{\gamma ,\delta }}, $$
Case
\(\gamma \ge \delta \). From Lemma 2 we have that, for each fixed \(r\in [1/2,1)\), it follows that
where
which is a finite constant due to the continuity of the function under the sign of maximum on a compact set.
$$\begin{aligned} \max_{1/2\le \vert z \vert \le r} \bigl\vert f(z) \bigr\vert \lesssim \max _{1/2\le \vert z \vert \le r} \bigl[\omega _{\gamma +2,\delta }(z) \bigr]^{-\frac{1}{2}} \Vert f \Vert _{A^{2}_{\gamma , \delta }} =\widehat{c}_{0}(r) \Vert f \Vert _{A^{2}_{\gamma ,\delta }}, \end{aligned}$$
(47)
$$ \widehat{c}_{0}(r):=\max_{1/2\le t\le r} \bigl[\omega _{\gamma +2, \delta }(t) \bigr]^{-\frac{1}{2}}, $$
Now note that (41) holds for \(m=0\), that is, we have
Since \(\gamma \ge \delta \), by using (3), it is not difficult to see that there is a positive constant \(c_{2}\) such that
Hence, from (48) and (49), we have
From (47) and (50) we have that, for each fixed \(r\in [0,1)\), there is a constant \(\widetilde{c}_{0}(r)\) such that
from which it follows that
and consequently \(\Vert f_{n}-f\Vert _{A^{2}_{\gamma ,\delta }}\to 0\) as \(n\to \infty \) implies \(f_{n}\to f\) uniformly on compacts of \({\mathbb {D}}\). Moreover, (51) means that the point evaluation functionals are continuous, which is the other necessary condition for applying Lemma 3 in [8], finishing the proof of the lemma. □
$$\begin{aligned} \max_{ \vert z \vert \le 1/2} \bigl\vert f(z) \bigr\vert ^{2}\le 16 \int _{ \vert \zeta \vert \le 3/4} \bigl\vert f(\zeta ) \bigr\vert ^{2} \,dA( \zeta ). \end{aligned}$$
(48)
$$\begin{aligned} 0< c_{2}\le \inf_{\vert z\vert \le 3/4}\omega _{\gamma ,\delta }(z). \end{aligned}$$
(49)
$$\begin{aligned} \max_{ \vert z \vert \le 1/2} \bigl\vert f(z) \bigr\vert ^{2}\le \frac{16}{c_{2}} \int _{ \vert \zeta \vert \le 3/4} \bigl\vert f( \zeta ) \bigr\vert ^{2} \omega _{\gamma ,\delta }(\zeta )\,dA(\zeta )\le \frac{16}{c _{2}} \Vert f \Vert _{A^{2}_{\gamma ,\delta }}^{2}. \end{aligned}$$
(50)
$$\begin{aligned} \max_{ \vert z \vert \le r} \bigl\vert f(z) \bigr\vert \le \widetilde{c}_{0}(r) \Vert f \Vert _{A^{2}_{\gamma , \delta }}, \end{aligned}$$
(51)
$$ \max_{ \vert z \vert \le r} \bigl\vert (f-f_{n}) (z) \bigr\vert \le \widetilde{c}_{0}(r) \Vert f-f_{n} \Vert _{A ^{2}_{\gamma ,\delta }}, $$
Remark 5
Note that the proof of Lemma 3 is considerably more complex than those for the case of some other spaces of analytic functions. For example, the corresponding result for the weighted Bloch space can be found in [8], for the Zygmund-type space in [12], while for the Besov and BMOA spaces in [22]. The first result of the type seems the one in [22], but it has some inaccuracies.
Lemma 4
Let
\(-1<\gamma <\infty \), \(\delta \le 0\), \(r_{0}\in (0,1)\), and
\(g\in H({\mathbb {D}})\). If
\(I_{g}\)
is bounded on
\(A^{2}_{\gamma ,\delta }\), then there exists a positive constant
C
independent of
\(f\in A^{2}_{\gamma ,\delta }\)
and
\(a\in {\mathbb {D}}\setminus r_{0}{\mathbb {D}}\)
such that
$$\begin{aligned} \bigl\vert f(a) \bigr\vert \bigl\vert g(a) \bigr\vert \leq C \Vert I_{g}f \Vert _{A^{2}_{\gamma ,\delta }} \bigl\Vert K_{\gamma +2, \delta }(a,\cdot ) \bigr\Vert _{A^{2}_{\gamma +2,\delta }}. \end{aligned}$$
(52)
Proof
Since \(I_{g}\) is bounded on \(A^{2}_{\gamma ,\delta }\), by Lemma 1 we have that \(f(z)g(z)=(I_{g}f)'(z)\in A^{2}_{\gamma +2, \delta }\) for every \(f\in A^{2}_{\gamma ,\delta }\). Then, by the reproducing kernel of \(A^{2}_{\gamma +2,\delta }\), for each \(a\in {\mathbb {D}}\setminus r_{0}{\mathbb {D}}\) it follows that
From (53), the Cauchy–Schwarz inequality, and Lemma 1, it follows that
This finishes the proof of the lemma. □
$$\begin{aligned} f(a)g(a)= \int _{{\mathbb {D}}}f(z)g(z)\overline{K_{\gamma +2,\delta }(a,z)} \omega _{\gamma +2,\delta }(z)\,dA(z). \end{aligned}$$
(53)
$$\begin{aligned} \bigl\vert f(a)g(a) \bigr\vert &\leq \int _{{\mathbb {D}}} \bigl\vert f(z)g(z)K_{\gamma +2,\delta }(a,z) \bigr\vert \omega _{\gamma +2,\delta }(z)\,dA(z) \\ &\leq \biggl( \int _{{\mathbb {D}}} \bigl\vert f(z)g(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \biggr)^{\frac{1}{2}} \\ & \quad {}\times \biggl( \int _{{\mathbb {D}}} \bigl\vert K_{\gamma +2,\delta }(a,z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \biggr)^{\frac{1}{2}} \\ &= \bigl\Vert (I_{g}f)' \bigr\Vert _{A^{2}_{\gamma +2,\delta }} \bigl\Vert K_{\gamma +2,\delta }(a, \cdot ) \bigr\Vert _{A^{2}_{\gamma +2,\delta }} \\ &\leq C \Vert I_{g}f \Vert _{A^{2}_{\gamma ,\delta }} \bigl\Vert K_{\gamma +2,\delta }(a, \cdot ) \bigr\Vert _{A^{2}_{\gamma +2,\delta }}. \end{aligned}$$
For fixed \(a\in {\mathbb {D}}\), let
It is easy to see that
It was essentially proved in [2] that for each \(r\in (0,1)\)
When \(\delta <0\), it is easily seen that \(k_{a}\to 0\) uniformly on compacts of \({\mathbb {D}}\) as \(\vert a\vert \to 1\). Note that this is not true if \(\delta =0\), a claim which is stated in [2] as a minor oversight.
$$\begin{aligned} k_{a}(z)=\frac{(1- \vert a \vert )^{-\delta }}{(1-\overline{a}z)^{ \frac{\gamma }{2}+1-\delta }} \biggl[\ln \biggl(1-\frac{1}{\ln \vert a \vert } \biggr) \biggr] ^{-\frac{\delta }{2}},\quad z\in {\mathbb {D}}. \end{aligned}$$
(54)
$$\begin{aligned} k_{a}(a)\asymp \frac{1}{(1- \vert a \vert )^{\frac{\gamma }{2}+1}} \biggl[\ln \biggl(1- \frac{1}{\ln \vert a \vert } \biggr) \biggr]^{-\frac{\delta }{2}}. \end{aligned}$$
(55)
$$\begin{aligned} \sup_{ \vert a \vert \ge r} \Vert k_{a} \Vert _{A^{2}_{\gamma ,\delta }} \lesssim 1. \end{aligned}$$
(56)
Let
and \(D(w,\delta )=\{z:\vert \varphi _{w}(z)\vert <\delta \}\).
$$ \varphi _{w}(z)=\frac{w-z}{1-\overline{w}z} $$
The following simple lemma is proved by a slight modification of the proof of Lemma 3.2 in [18], so the proof is omitted.
Lemma 5
Let
\(r_{0}\in (0,1)\), \(a\in {\mathbb {D}}\setminus r_{0}{\mathbb {D}}\). Then
for
\(w\in D(a,1/2)\).
$$\begin{aligned} \ln \biggl(1-\frac{1}{\ln \vert w \vert } \biggr)\asymp \ln \biggl(1-\frac{1}{ \ln \vert a \vert } \biggr) \end{aligned}$$
Now we present a lower bound estimate for the quantity \(\Vert k_{a}\Vert _{A ^{2}_{\gamma ,\delta }}\).
Lemma 6
Let
\(-1<\gamma <\infty \), \(\delta <0\), \(r_{0}\in (0,1)\), and
\(a\in {\mathbb {D}}\setminus r_{0}{\mathbb {D}}\). Then
\(\Vert k_{a}\Vert _{A^{2}_{\gamma ,\delta }}\gtrsim 1\).
Proof
By Lemma 1 and the expression of \(k_{a}\), we have
where
On the other hand,
Then, from Lemma 5 and since \(1-\vert w\vert \asymp 1-\vert a\vert \) and \(A(D(a,1/2)) \asymp (1-\vert a\vert ^{2})^{2}\), when \(w\in D(a,1/2)\), it follows that
Hence, from (57)–(59), we have that
for \(\vert a\vert \ge r_{0}\), from which the desired result follows. □
$$\begin{aligned} \Vert k_{a} \Vert ^{2}_{A^{2}_{\gamma ,\delta }} &\geq C_{1} \int _{{\mathbb {D}}} \bigl\vert k _{a}'(w) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(w)\,dA(w) \\ &=c(\gamma ,\delta ,a) \int _{{\mathbb {D}}}\frac{1}{ \vert 1-\overline{a}w \vert ^{ \gamma -2\delta +4}}\omega _{\gamma +2,\delta }(w)\,dA(w), \end{aligned}$$
(57)
$$ c(\gamma ,\delta ,a)= \biggl(\frac{\gamma }{2}+1-\delta \biggr)^{2} \vert a \vert ^{2}\bigl(1- \vert a \vert ^{2} \bigr)^{-2 \delta } \biggl[\ln \biggl(1-\frac{1}{\ln \vert a \vert } \biggr) \biggr]^{-\delta }C_{1}. $$
$$\begin{aligned} \int _{{\mathbb {D}}}\frac{1}{ \vert 1-\overline{a}u \vert ^{\gamma -2\delta +4}} \omega _{\gamma +2,\delta }(u)\,dA(u) \gtrsim \frac{1}{(1- \vert a \vert ^{2})^{ \gamma -2\delta +4}} \int _{D(a,1/2)}\omega _{\gamma +2,\delta }(u)\,dA(u). \end{aligned}$$
(58)
$$\begin{aligned} \int _{D(a,1/2)}\omega _{\gamma +2,\delta }(w)\,dA(w) \geq C_{2}\bigl(1- \vert a \vert ^{2}\bigr)^{ \gamma +4} \biggl[\ln \biggl(1-\frac{1}{\ln \vert a \vert } \biggr) \biggr]^{\delta }. \end{aligned}$$
(59)
$$\begin{aligned} \Vert k_{a} \Vert ^{2}_{A^{2}_{\gamma ,\delta }}\ge r_{0}^{2}C_{1}C_{2} \biggl( \frac{ \gamma }{2}+1-\delta \biggr)^{2} \end{aligned}$$
Remark 6
With the help of Lemma 6, we see that when \(\delta <0\) the functions
converge uniformly to zero on the compact subsets of \({\mathbb {D}}\) as \(\vert a\vert \to 1\). The functions \(f_{a}\) will be used in the characterization of the compactness of \(I_{g}\) on \(A^{2}_{\gamma ,\delta }\).
$$ f_{a}(z)=\frac{k_{a}(z)}{ \Vert k_{a} \Vert _{A^{2}_{\gamma ,\delta }}},\quad z \in {\mathbb {D}}, $$
3 Main results and proofs
First, we characterize the boundedness of the operator \(I_{g}\) on \(A^{2}_{\gamma ,\delta }\).
Theorem 1
Let
\(-1<\gamma <\infty \), \(\delta <0\), and
\(g\in H({\mathbb {D}})\). Then the operator
\(I_{g}\)
is bounded on
\(A^{2}_{\gamma ,\delta }\)
if and only if
\(g\in H_{1}^{\infty }\).
Proof
Suppose that \(g\in H_{1}^{\infty }\). First, note that
Using (60), Corollary 4, and the fact that
for \(x\in (r,1)\), where r is a fixed number in the interval \((0,1)\), it follows that
From (62), it follows that \(I_{g}\) is bounded on \(A^{2}_{ \gamma ,\delta }\), and moreover that \(\Vert I_{g}\Vert \lesssim \Vert g\Vert _{H_{1} ^{\infty }}\).
$$\begin{aligned} (I_{g}f)'(z)=f(z)g(z)\quad \text{and}\quad I_{g}f(0)=0. \end{aligned}$$
(60)
$$\begin{aligned} \ln \frac{1}{x}\asymp 1-x \end{aligned}$$
(61)
$$\begin{aligned} \Vert I_{g}f \Vert ^{2}_{A^{2}_{\gamma ,\delta }} &\asymp \int _{{\mathbb {D}}\setminus r{\mathbb {D}}} \bigl\vert (I_{g}f)'(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \\ &= \int _{{\mathbb {D}}\setminus r{\mathbb {D}}} \bigl\vert f(z) \bigr\vert ^{2} \bigl\vert g(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \\ &\leq \sup_{\{z\in {\mathbb {D}}: \vert z \vert \geq r\}} \biggl(\ln \frac{1}{ \vert z \vert } \biggr) ^{2} \bigl\vert g(z) \bigr\vert ^{2} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }} \\ &\lesssim \sup_{\{z\in {\mathbb {D}}: \vert z \vert \geq r\}}\bigl(1- \vert z \vert \bigr)^{2} \bigl\vert g(z) \bigr\vert ^{2} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }} \\ &\le \Vert g \Vert _{H_{1}^{\infty }}^{2} \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }}. \end{aligned}$$
(62)
Conversely, suppose that \(I_{g}\) is bounded on \(A^{2}_{\gamma ,\delta }\) and that \(r\in (0,1)\). By Lemma 4, we obtain
for \(f\in A^{2}_{\gamma ,\delta }\) and \(a\in {\mathbb {D}}\setminus r{\mathbb {D}}\). Specially, for every \(f\in A^{2} _{\gamma ,\delta }\) with \(f(a)\neq 0\), by (63) we have
Applying (64) to the function \(k_{a}\), from (38), (55), and (56), it follows that
from which it follows that
$$\begin{aligned} \bigl\vert f(a) \bigr\vert \bigl\vert g(a) \bigr\vert \leq C \Vert I_{g}f \Vert _{A^{2}_{\gamma ,\delta }} \bigl\Vert K_{\gamma +2, \delta }(a,\cdot ) \bigr\Vert _{A^{2}_{\gamma +2,\delta }} \end{aligned}$$
(63)
$$\begin{aligned} \bigl(1- \vert a \vert \bigr) \bigl\vert g(a) \bigr\vert \leq C \frac{(1- \vert a \vert )}{ \vert f(a) \vert } \Vert I_{g}f \Vert _{A^{2}_{\gamma , \delta }} \bigl\Vert K_{\gamma +2,\delta }(a,\cdot ) \bigr\Vert _{A^{2}_{\gamma +2,\delta }}. \end{aligned}$$
(64)
$$\begin{aligned} \bigl(1- \vert a \vert \bigr) \bigl\vert g(a) \bigr\vert &\leq C \frac{(1- \vert a \vert )}{ \vert k_{a}(a) \vert } \Vert I_{g} \Vert \Vert k_{a} \Vert _{A ^{2}_{\gamma ,\delta }} \bigl\Vert K_{\gamma +2,\delta }(a,\cdot ) \bigr\Vert _{A^{2}_{ \gamma +2,\delta }}\lesssim \Vert I_{g} \Vert , \end{aligned}$$
$$\begin{aligned} \sup_{ \vert a \vert \ge r}\bigl(1- \vert a \vert \bigr) \bigl\vert g(a) \bigr\vert < +\infty . \end{aligned}$$
(65)
From (65) along with the obvious fact
we obtain that \(g\in H_{1}^{\infty }\). □
$$ \sup_{ \vert a \vert \le r}\bigl(1- \vert a \vert \bigr) \bigl\vert g(a) \bigr\vert < +\infty , $$
Next we characterize the compactness of the operator \(I_{g}\) on \(A^{2}_{\gamma ,\delta }\). We will present a direct proof of it for the presentational reasons and an obvious connectedness with Theorem 1, although it will be also a consequence of our next theorem.
Theorem 2
Let
\(-1<\gamma <\infty \), \(\delta <0\), and
\(g\in H({\mathbb {D}})\). Then the operator
\(I_{g}\)
is compact on
\(A^{2}_{\gamma ,\delta }\)
if and only if
\(g\in H_{1,0}^{\infty }({\mathbb {D}})\).
Proof
First, assume that \(g\in H_{1,0}^{\infty }({\mathbb {D}})\). Then, for sufficiently small \(\varepsilon >0\), there exists \(\delta _{0}\in (0,1)\) such that
for \(z\in {\mathbb {D}}\setminus \delta _{0}{\mathbb {D}}\).
$$\begin{aligned} \bigl(1- \vert z \vert \bigr) \bigl\vert g(z) \bigr\vert < \varepsilon \end{aligned}$$
(66)
Now note that by (3) and since \(\gamma -\delta >-1\), we have
$$\begin{aligned} \int _{\delta _{0}{\mathbb {D}}}\omega _{\gamma +2,\delta }(z)\,dA(z)\asymp \int _{0}^{\delta _{0}} \int _{0}^{2\pi } \biggl(\ln \frac{1}{\rho } \biggr) ^{\gamma -\delta +2}\rho \,d\theta \,d\rho =2\pi \int _{\ln \frac{1}{\delta _{0}}}^{\infty }\frac{s^{\gamma -\delta +2}}{e ^{2s}}\,ds< \infty . \end{aligned}$$
(67)
Let \((f_{n})_{n\in {\mathbb {N}}}\) be a sequence in \(A^{2}_{\gamma ,\delta }\) such that \(\Vert f_{n}\Vert _{A^{2}_{\gamma ,\delta }}\leq M\) and \(f_{n}\to 0\) uniformly on compact subsets of \({\mathbb {D}}\).
For above chosen ε, choose a positive integer N such that
for all \(z\in \delta _{0}{\mathbb {D}}\) and \(n\geq N\), which is possible since due to (67) we have
$$\begin{aligned} \bigl\vert f_{n}(z) \bigr\vert < \frac{\varepsilon }{\int _{\delta _{0}{\mathbb {D}}} \vert g(z) \vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z)} \end{aligned}$$
(68)
$$ \int _{\delta _{0}{\mathbb {D}}} \bigl\vert g(z) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(z)\,dA(z) \le \max_{ \vert z \vert \le \delta _{0}} \bigl\vert g(z) \bigr\vert ^{2} \int _{\delta _{0}{\mathbb {D}}} \omega _{\gamma +2,\delta }(z)\,dA(z)< \infty . $$
By using (61), (66), and (68), we have
Thus, the condition \(g\in H_{1,0}^{\infty }({\mathbb {D}})\) implies the compactness of \(I_{g}\) on \(A^{2}_{\gamma ,\delta }\).
$$\begin{aligned} \Vert I_{g}f_{n} \Vert ^{2}_{A^{2}_{\gamma ,\delta }} &= \int _{{\mathbb {D}}} \bigl\vert I_{g}f _{n}(z) \bigr\vert ^{2}\omega _{\gamma ,\delta }(z)\,dA(z) \\ &\leq C \int _{{\mathbb {D}}} \bigl\vert f_{n}(z)g(z) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(z)\,dA(z) \\ &\lesssim \int _{\delta _{0}{\mathbb {D}}}+ \int _{{\mathbb {D}}\setminus \delta _{0}{\mathbb {D}}} \bigl\vert f_{n}(z)g(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \\ &\lesssim \varepsilon ^{2}+\varepsilon ^{2} \Vert f_{n} \Vert ^{2}_{A^{2}_{ \gamma ,\delta }}. \end{aligned}$$
Now suppose that \(I_{g}\) is compact on \(A^{2}_{\gamma ,\delta }\). Put
As we have seen in Remark 6, \(f_{a}\to 0\) uniformly on compacts of \({\mathbb {D}}\) as \(\vert a\vert \to 1\). Since \(I_{g}\) is compact on \(A^{2}_{\gamma ,\delta }\), by Lemma 3 we have
$$ f_{a}(z)=\frac{k_{a}(z)}{ \Vert k_{a} \Vert _{A^{2}_{\gamma ,\delta }}}. $$
$$ \lim_{ \vert a \vert \to 1^{-}} \Vert I_{g}f_{a} \Vert _{A^{2}_{\gamma ,\delta }}=0. $$
From (64) we have
as \(\vert a\vert \to 1\), from which it follows that \(g\in H_{1,0}^{\infty }({\mathbb {D}})\). □
$$\begin{aligned} \bigl(1- \vert a \vert \bigr) \bigl\vert g(a) \bigr\vert &\lesssim \frac{1- \vert a \vert }{ \vert f_{a}(a) \vert } \Vert I_{g} f_{a} \Vert _{A^{2} _{\gamma ,\delta }} \bigl\Vert K_{\gamma +2,\delta }(a,\cdot ) \bigr\Vert _{A^{2}_{\gamma +2,\delta }} \\ &\lesssim \Vert I_{g} f_{a} \Vert _{A^{2}_{\gamma ,\delta }}\to 0 \end{aligned}$$
(69)
We now consider the essential norm of the operator \(I_{g}\) on \(A^{2}_{\gamma ,\delta }\). Before this we formulate and prove an auxiliary result on a complete orthonormal system in \(A^{2}_{\gamma , \delta }\). For some basics on the topic see, for example, [23] and [24].
Lemma 7
Let
Then the sequence
\((\varphi _{n}(z))_{n\in {\mathbb {N}}_{0}}\)
is a complete orthonormal system in
\(A^{2}_{\gamma ,\delta }\).
$$\begin{aligned} \varphi _{n}(z)=\frac{z^{n}}{ \Vert z^{n} \Vert _{A^{2}_{\gamma ,\delta }}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(70)
Proof
By using the polar coordinates, we have
from which it easily follows that
where \(\delta _{n,m}\) is the δ Kronecker symbol. Hence, the system is orthonormal.
$$\begin{aligned} \langle \varphi _{n}, \varphi _{m}\rangle &= \int _{{\mathbb {D}}}\varphi _{n}(z) \overline{\varphi _{m}(z)}\omega _{\gamma ,\delta }(z)\,dA(z) \\ &=\frac{1}{\pi \Vert z^{n} \Vert _{A^{2}_{\gamma ,\delta }} \Vert z^{m} \Vert _{A^{2} _{\gamma ,\delta }}} \int _{0}^{1} \int _{0}^{2\pi } r^{n} e^{in\theta } r^{m}e^{-im\theta }\omega _{\gamma ,\delta }(r)r\,dr\,d\theta \\ &=\frac{1}{\pi \Vert z^{n} \Vert _{A^{2}_{\gamma ,\delta }} \Vert z^{m} \Vert _{A^{2} _{\gamma ,\delta }}} \int _{0}^{1} r^{n+m+1}\omega _{\gamma ,\delta }(r)\,dr \int _{0}^{2\pi }e^{i(n-m)\theta }\,d\theta , \end{aligned}$$
(71)
$$ \bigl\langle z^{n}, z^{m}\bigr\rangle =\delta _{n,m}, $$
To prove that the system is complete, we prove that the system span \(A^{2}_{\gamma ,\delta }\), for which it is enough to prove that the polynomials are dense in \(A^{2}_{\gamma ,\delta }\).
Let \(f\in A^{2}_{\gamma ,\delta }\). Then, for every \(\varepsilon >0\), there is \(r_{0}\in (0,1)\) such that
which is equivalent to
$$ \int _{{\mathbb {D}}\setminus \overline{r_{0}{\mathbb {D}}}} \bigl\vert f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z)< \varepsilon , $$
$$\begin{aligned} 2 \int _{r_{0}}^{1}M_{2}^{2}(f,r) \omega _{\gamma ,\delta }(r)\,dr< \varepsilon . \end{aligned}$$
(72)
Since \(M_{2}(f,r)\) is a nondecreasing function, we have also that
for every \(\rho \in [0,1)\), where \(f_{\rho }(z)=f(\rho z)\).
$$\begin{aligned} 2 \int _{r_{0}}^{1}M_{2}^{2}(f_{\rho },r) \omega _{\gamma ,\delta }(r)\,dr< \varepsilon \end{aligned}$$
(73)
Due to the uniform continuity of f on the compact \(\overline{r_{0}{\mathbb {D}}}\), we have that, for every \(\varepsilon >0\), there is \(\delta _{0}\in (0,1)\) such that
for every \(\rho \in (1-\delta _{0},1)\) and \(z\in \overline{r_{0}{\mathbb {D}}}\).
$$\begin{aligned} \bigl\vert f_{\rho }(z)-f(z) \bigr\vert < \sqrt{\varepsilon } \end{aligned}$$
(74)
By using (72)–(74), we have
$$\begin{aligned} \Vert f_{\rho }-f \Vert ^{2}_{A^{2}_{\gamma ,\delta }} &= \int _{{\mathbb {D}}} \bigl\vert f( \rho z)-f(z) \bigr\vert ^{2}\omega _{\gamma ,\delta }(z)\,dA(z) \\ &= \int _{\overline{r_{0}{\mathbb {D}}}} \bigl\vert f(\rho z)-f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z)+ \int _{{\mathbb {D}}\setminus \overline{r_{0}{\mathbb {D}}}} \bigl\vert f(\rho z)-f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z) \\ &< \varepsilon \int _{\overline{r_{0}{\mathbb {D}}}}\omega _{\gamma ,\delta }(z)\,dA(z)+2 \int _{{\mathbb {D}}\setminus \overline{r_{0}{\mathbb {D}}}}\bigl( \bigl\vert f( \rho z) \bigr\vert ^{2}+ \bigl\vert f(z) \bigr\vert ^{2}\bigr)\omega _{\gamma ,\delta }(z)\,dA(z) \\ &< \varepsilon \biggl( \int _{{\mathbb {D}}}\omega _{\gamma ,\delta }(z)\,dA(z)+2 \biggr). \end{aligned}$$
(75)
Since \(f\in H({\mathbb {D}})\), the Taylor polynomials \(s_{n}(f_{\rho })= \sum_{j=0}^{n-1}a_{j}(f_{\rho })z^{j}\) of \(f_{\rho }\) converge uniformly to the function on \({\mathbb {D}}\). Hence, for every \(\varepsilon >0\), there is \(n_{0}\in {\mathbb {N}}_{0}\) such that
for \(n\ge n_{0}\) and \(z\in {\mathbb {D}}\).
$$\begin{aligned} \bigl\vert f_{\rho }(z)-s_{n}(f_{\rho }) (z) \bigr\vert < \sqrt{\varepsilon } \end{aligned}$$
(76)
From (76), we have
$$\begin{aligned} \bigl\Vert f_{\rho }-s_{n}(f_{\rho }) \bigr\Vert ^{2}_{A^{2}_{\gamma ,\delta }} &= \int _{{\mathbb {D}}} \bigl\vert f(\rho z)-s_{n}(f_{\rho }) (z) \bigr\vert ^{2}\omega _{\gamma , \delta }(z)\,dA(z) \\ &< \varepsilon \int _{{\mathbb {D}}}\omega _{\gamma ,\delta }(z)\,dA(z). \end{aligned}$$
(77)
From (75) and (77), it follows that
for \(\rho \in (1-\delta _{0},1)\), from which the result follows. □
$$\begin{aligned} \bigl\Vert f-s_{n}(f_{\rho }) \bigr\Vert _{A^{2}_{\gamma ,\delta }} < \sqrt{\varepsilon } \biggl( \biggl( \int _{{\mathbb {D}}}\omega _{\gamma ,\delta }(z)\,dA(z) \biggr) ^{1/2}+ \biggl( \int _{{\mathbb {D}}}\omega _{\gamma ,\delta }(z)\,dA(z)+2 \biggr) ^{1/2} \biggr) \end{aligned}$$
(78)
Theorem 3
Let
\(-1<\gamma <\infty \), \(\delta <0\), and
\(g\in H({\mathbb {D}})\). If the operator
\(I_{g}\)
is bounded on
\(A^{2}_{ \gamma ,\delta }\), then
$$ \Vert I_{g} \Vert _{e}\asymp A:=\limsup _{ \vert z \vert \to 1}\bigl(1- \vert z \vert \bigr) \bigl\vert g(z) \bigr\vert . $$
Proof
Let \(K:A^{2}_{\gamma ,\delta }\to A^{2}_{\gamma ,\delta }\) be compact and \(k_{a}\) be the functions defined in (54). Then by Lemma 3 we have that \(\Vert Kk_{a}\Vert _{A^{2}_{\gamma ,\delta }}\to 0\) as \(\vert a\vert \to 1-0\), from which it follows that
By (55) and Remark 3, we have that
Hence, from Lemma 4 it follows that
for \(\vert a\vert \ge r>0\), from which we obtain
From (79) and (81), we have
which shows that
$$\begin{aligned} \Vert I_{g}-K \Vert &\geq \limsup_{ \vert a \vert \to 1} \Vert I_{g}k_{a}-Kk_{a} \Vert _{A^{2}_{ \gamma ,\delta }} \\ &\geq \limsup_{ \vert a \vert \to 1} \Vert I_{g}k_{a} \Vert _{A^{2}_{\gamma ,\delta }}- \limsup_{ \vert a \vert \to 1} \Vert Kk_{a} \Vert _{A^{2}_{\gamma ,\delta }} \\ &=\limsup_{ \vert a \vert \to 1} \Vert I_{g}k_{a} \Vert _{A^{2}_{\gamma ,\delta }}. \end{aligned}$$
(79)
$$\begin{aligned} \bigl\Vert K_{\gamma +2,\delta }(a,\cdot ) \bigr\Vert _{A^{2}_{\gamma +2,\delta }} \lesssim \bigl(1- \vert a \vert \bigr)^{-1}k_{a}(a). \end{aligned}$$
(80)
$$\begin{aligned} \bigl\vert k_{a}(a)g(a) \bigr\vert &\leq C \Vert I_{g}k_{a} \Vert _{A^{2}_{\gamma ,\delta }} \bigl\Vert K_{ \gamma +2,\delta }(a,\cdot ) \bigr\Vert _{A^{2}_{\gamma +2,\delta }}\lesssim \Vert I _{g}k_{a} \Vert _{A^{2}_{\gamma ,\delta }}\bigl(1- \vert a \vert \bigr)^{-1}k_{a}(a) \end{aligned}$$
$$\begin{aligned} \bigl(1- \vert a \vert \bigr) \bigl\vert g(a) \bigr\vert \lesssim \Vert I_{g}k_{a} \Vert _{A^{2}_{\gamma ,\delta }}. \end{aligned}$$
(81)
$$\begin{aligned} \Vert I_{g}-K \Vert \gtrsim \limsup_{ \vert a \vert \to 1} \bigl(1- \vert a \vert \bigr) \bigl\vert g(a) \bigr\vert , \end{aligned}$$
$$\begin{aligned} \Vert I_{g} \Vert _{e}\gtrsim \limsup _{ \vert a \vert \to 1}\bigl(1- \vert a \vert \bigr) \bigl\vert g(a) \bigr\vert . \end{aligned}$$
(82)
For a holomorphic function \(f(z)=\sum_{m=0}^{\infty }a_{m}z ^{m}\) on \({\mathbb {D}}\), let
and
As a finite-rank operator, \(P_{j}\) is compact on \(A^{2}_{\gamma , \delta }\), and
for each \(j\in {\mathbb {N}}\).
$$ P_{j}f(z)=\sum_{m=0}^{j}a_{m}z^{m} $$
$$ R_{j}f(z)=\sum_{m=j+1}^{\infty }a_{m}z^{m}. $$
$$\begin{aligned} \Vert I_{g} \Vert _{e}= \bigl\Vert I_{g}(P_{j}+R_{j}) \bigr\Vert _{e} \leq \Vert I_{g}P_{j} \Vert _{e}+ \Vert I _{g}R_{j} \Vert _{e}= \Vert I_{g}R_{j} \Vert _{e}\leq \Vert I_{g}R_{j} \Vert \end{aligned}$$
(83)
Thus
Hence, we have
$$ \Vert I_{g} \Vert _{e}\leq \liminf _{j\to \infty } \Vert I_{g}R_{j} \Vert . $$
$$\begin{aligned} \Vert I_{g} \Vert _{e}^{2} &\leq \liminf _{j\to \infty } \Vert I_{g}R_{j} \Vert ^{2}= \liminf_{j\to \infty }\sup_{ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\leq 1} \Vert I _{g}R_{j}f \Vert ^{2}_{A^{2}_{\gamma ,\delta }} \\ &\asymp \liminf_{j\to \infty } \sup_{ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\leq 1} \int _{{\mathbb {D}}} \bigl\vert (I_{g}R _{j}f)'(z) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(z)\,dA(z) \\ &=\liminf_{j\to \infty }\sup_{ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\leq 1} \int _{{\mathbb {D}}} \bigl\vert R_{j}f(z)g(z) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(z)\,dA(z). \end{aligned}$$
(84)
For every \(\varepsilon >0\), there is \(r_{0}\in (0,1)\) such that
for \(r_{0}\le \vert z\vert <1\).
$$\begin{aligned} \bigl(1- \vert z \vert \bigr) \bigl\vert g(z) \bigr\vert < A+\varepsilon \end{aligned}$$
(85)
From (84) and (85), it follows that
$$\begin{aligned} \Vert I_{g} \Vert _{e}^{2} &\leq \liminf _{j\to \infty } \biggl( \sup_{ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\leq 1} \int _{r_{0}{\mathbb {D}}} \bigl\vert R _{j}f(z)g(z) \bigr\vert ^{2}\omega _{\gamma +2,\delta }(z)\,dA(z) \\ & \quad{} +\sup_{ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\leq 1} \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert R_{j}f(z)g(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \biggr) \\ &\lesssim \liminf_{j\to \infty } \biggl( \int _{r_{0}{\mathbb {D}}} \bigl\vert g(z) \bigr\vert ^{2} \omega _{\gamma +2,\delta }(z)\,dA(z) \sup_{ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\leq 1}M^{2}_{\infty }(R_{j}f,r _{0}) \\ & \quad{} +(A+\varepsilon )^{2}\sup_{ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\leq 1} \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert R_{j}f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z) \biggr). \end{aligned}$$
(86)
By Lemma 7 we know that the system \(\varphi _{n}(z)=z^{n}/\Vert z^{n}\Vert _{A^{2}_{\gamma ,\delta }}\), \(n\in {\mathbb {N}}_{0}\), is a complete orthonormal system in the space \(A^{2}_{\gamma ,\delta }\). Then, by a known theory, we have that the reproducing kernel of \(A^{2}_{\gamma , \delta }\) is
Now we estimate the quantity of \(\Vert z^{n}\Vert ^{2}_{A^{2}_{\gamma ,\delta }}\). From Proposition 1, we have
$$ K_{\gamma ,\delta }(z,w)=\sum_{n=0}^{\infty } \varphi _{n}(z)\overline{ \varphi _{n}(w)} =\sum _{n=0}^{\infty }\frac{z^{n}\overline{w}^{n}}{ \Vert z ^{n} \Vert ^{2}_{A^{2}_{\gamma ,\delta }}}. $$
$$\begin{aligned} \bigl\Vert z^{n} \bigr\Vert ^{2}_{A^{2}_{\gamma ,\delta }} &\asymp \int _{{\mathbb {D}}} \vert z \vert ^{2n}\bigl(1- \vert z \vert \bigr)^{ \gamma } \biggl(\ln \frac{e}{1- \vert z \vert } \biggr)^{\delta } \,dA(z) \\ &= \int ^{1}_{0} \int _{0}^{2\pi }r^{2n+1}(1-r)^{\gamma } \biggl(\ln \frac{e}{1-r} \biggr)^{\delta }\frac{dr\,d\theta }{\pi } \\ &\ge 2 \int ^{(1+r_{0})/2}_{0}r^{2n+1}(1-r)^{\gamma } \biggl(\ln \frac{e}{1-r} \biggr)^{\delta }\,dr \\ &\gtrsim \int ^{(1+r_{0})/2}_{0}r^{2n+1}\,dr \\ &=\frac{1}{2n+2} \biggl(\frac{1+r_{0}}{2} \biggr)^{2n+2}. \end{aligned}$$
(87)
Since \(K_{\gamma ,\delta }(z,w)\) is the kernel of \(A^{2}_{\gamma , \delta }\) and \(R_{j}f\in A^{2}_{\gamma ,\delta }\) for \(f\in A^{2}_{ \gamma ,\delta }\) (see (95) below), we have
The orthogonality of \(w^{n}\) with respect to \(\omega _{\gamma ,\delta }(w)\,dA(w)\) shows that
(for a related idea, see [25]).
$$\begin{aligned} R_{j}f(z)= \int _{{\mathbb {D}}}R_{j}f(w)\overline{K_{\gamma ,\delta }(z,w)} \omega _{\gamma ,\delta }(w)\,dA(w). \end{aligned}$$
$$\begin{aligned} R_{j}f(z)= \int _{{\mathbb {D}}}f(w)R_{j}\overline{K_{\gamma ,\delta }(z,w)} \omega _{\gamma ,\delta }(w)\,dA(w) \end{aligned}$$
(88)
From the expression of \(K_{\gamma ,\delta }(z,w)\), the Cauchy–Schwarz inequality, and by using (21), we have
Therefore, from (87) and (89) it follows that
Now, from (90) we have
Since for \(r_{0}\in (0,1)\) we have
the series
converges, which shows that
$$\begin{aligned} \bigl\vert R_{j}f(z) \bigr\vert &\leq \int _{{\mathbb {D}}} \bigl\vert f(w) \bigr\vert \bigl\vert R_{j}\overline{K_{\gamma , \delta }(z,w)} \bigr\vert \omega _{\gamma ,\delta }(w)\,dA(w) \\ &\leq \Vert f \Vert _{A^{2}_{\gamma ,\delta }} \Biggl( \int _{{\mathbb {D}}} \Biggl(\sum_{n=j+1}^{\infty } \frac{ \vert z^{n} \vert \vert w^{n} \vert }{ \Vert z^{n} \Vert ^{2}_{A_{\gamma , \delta }}} \Biggr)^{2}\omega _{\gamma ,\delta }(w)\,dA(w) \Biggr)^{ \frac{1}{2}} \\ &\leq \Vert f \Vert _{A^{2}_{\gamma ,\delta }} \biggl( \int _{{\mathbb {D}}} \omega _{\gamma ,\delta }(w)\,dA(w) \biggr)^{1/2}\sum_{n=j+1}^{\infty } \frac{ \vert z \vert ^{n}}{ \Vert z^{n} \Vert ^{2}_{A_{\gamma ,\delta }}} \\ &\lesssim \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\sum_{n=j+1}^{\infty } \frac{ \vert z \vert ^{n}}{ \Vert z^{n} \Vert ^{2}_{A_{\gamma ,\delta }}}. \end{aligned}$$
(89)
$$\begin{aligned} \bigl\vert R_{j}f(z) \bigr\vert \lesssim \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\sum_{n=j+1}^{ \infty }(2n+2) \biggl(\frac{2}{1+r_{0}} \biggr)^{2n} \vert z \vert ^{n}. \end{aligned}$$
(90)
$$\begin{aligned} M_{\infty }(R_{j}f,r_{0})=\max_{ \vert z \vert =r_{0}} \bigl\vert R_{j}f(z) \bigr\vert \lesssim \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\sum_{n=j+1}^{\infty }(2n+2) \biggl( \frac{4r _{0}}{(1+r_{0})^{2}} \biggr)^{n}. \end{aligned}$$
(91)
$$ \frac{4r_{0}}{(1+r_{0})^{2}}\in (0,1), $$
$$ \sum_{n=0}^{\infty }(2n+2) \biggl( \frac{4r_{0}}{(1+r_{0})^{2}} \biggr) ^{n} $$
$$\begin{aligned} \lim_{j\to \infty }\sum_{n=j+1}^{\infty }(2n+2) \biggl(\frac{4r_{0}}{(1+r _{0})^{2}} \biggr)^{n}=0. \end{aligned}$$
(92)
Using (92) in (91), it follows that
Employing (93) in (86), we obtain
On the other hand, since
we have
From (94) and (95), we obtain
From this and since ε is an arbitrary positive number, it follows that \(\Vert I_{g}\Vert _{e}\le 2A\), finishing the proof of the theorem. □
$$\begin{aligned} \lim_{j\to \infty }M_{\infty }(R_{j}f,r_{0})=0. \end{aligned}$$
(93)
$$\begin{aligned} \Vert I_{g} \Vert _{e}^{2}\le (A+\varepsilon )^{2}\liminf_{j\to \infty } \sup_{ \Vert f \Vert _{A^{2}_{\gamma ,\delta }}\leq 1} \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert R_{j}f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z). \end{aligned}$$
(94)
$$\begin{aligned} \int _{{\mathbb {D}}\setminus r_{0}{\mathbb {D}}} \bigl\vert P_{j}f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z) &\leq \int _{{\mathbb {D}}} \bigl\vert P_{j}f(z) \bigr\vert ^{2} \omega _{\gamma ,\delta }(z)\,dA(z) \\ &= \int ^{1}_{0} \int _{0}^{2\pi } \Biggl\vert \sum _{n=0}^{j}a_{n}r^{n}e^{in \theta } \Biggr\vert ^{2}\omega _{\gamma ,\delta }(r)r\frac{dr\,d\theta }{ \pi } \\ &=2\sum_{n=0}^{j} \vert a_{n} \vert ^{2} \int ^{1}_{0}r^{2n+1}\omega _{\gamma , \delta }(r)\,dr \\ &\leq 2\sum_{n=0}^{\infty } \vert a_{n} \vert ^{2} \int ^{1}_{0}r^{2n+1} \omega _{\gamma ,\delta }(r)\,dr \\ &= \Vert f \Vert ^{2}_{A^{2}_{\gamma ,\delta }}, \end{aligned}$$
$$\begin{aligned} \Vert R_{j}f \Vert _{A^{2}_{\gamma ,\delta }}= \Vert f-P_{j}f \Vert _{A^{2}_{\gamma , \delta }}\leq 2 \Vert f \Vert _{A^{2}_{\gamma ,\delta }}. \end{aligned}$$
(95)
$$ \Vert I_{g} \Vert ^{2}_{e}\le 4(A+ \varepsilon )^{2}. $$
Remark 7
If \(\delta =0\), then by using the following test functions
for \(a\in {\mathbb {D}}\) and some fixed \(c>0\), it can be proved in the same fashion that Theorems 1–3 also hold in the case.
$$ k_{a}(z)= \frac{(1- \vert a \vert )^{c}}{(1-\overline{a}z)^{\frac{\gamma }{2}+1+c}},\quad z \in {\mathbb {D}}, $$
Acknowledgements
The study in the paper is a part of the investigation under the projects III 41025 and III 44006 by the Serbian Ministry of Education and Science, Sichuan Science and Technology Program (No. 2018JY0200), and the project of Education Department of Sichuan Province (No. 18ZA0339).
Availability of data and materials
Not applicable.
Competing interests
The authors declare that they have no competing interests.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.